[ { "id": 1, "subject": "Chemistry", "question": "During dehydration of alcohols to alkenes by heating with conc. H2SO4 the initiation step is", "options": [ { "text": "formation of carbocation" }, { "text": "elimination of water" }, { "text": "formation of an ester" }, { "text": "protonation of alcohol molecule" } ], "answer": "protonation of alcohol molecule", "solution": "**Answer:** protonation of alcohol molecule\n\nThe dehydration of alcohol to form alkene occurs in following three step. Step $$(1)$$ is initiation step. \n

\"AIEEE ", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 2, "subject": "Chemistry", "question": "Acetyl bromide reacts with excess of CH3MgI followed by treatment with a saturated solution\nof NH4Cl given ", "options": [ { "text": "acetone" }, { "text": "acetyl iodide " }, { "text": "2- methyl -2- propanol " }, { "text": "acetamide " } ], "answer": "2- methyl -2- propanol ", "solution": "**Answer:** 2- methyl -2- propanol \n\n\"AIEEE ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 3, "subject": "Chemistry", "question": "Acid catalyzed hydration of alkenes except ethene leads to the formation of ", "options": [ { "text": "primary alcohol " }, { "text": "secondary or tertiary alcohol " }, { "text": "mixture of primary and secondary alcohols " }, { "text": "mixture of secondary and tertiary alcohols " } ], "answer": "mixture of secondary and tertiary alcohols ", "solution": "**Answer:** mixture of secondary and tertiary alcohols \n\nWater adds directly to the more reactive alkene in presence of a strongly acidic catalyst forming alcohols. Addition occurs according to Markonikov's rule.\n

\"AIEEE ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 4, "subject": "Chemistry", "question": "The best reagent to convert pent -3- en-2-ol into pent -3-en-2-one is", "options": [ { "text": "Acidic permanganate" }, { "text": "Acidic dichromate" }, { "text": "Chromic anhydride in glacial acetic acid " }, { "text": "Pyridinium chloro – chromate" } ], "answer": "Pyridinium chloro – chromate", "solution": "**Answer:** Pyridinium chloro – chromate\n\n\"AIEEE\n

Pyridiminum chloro-chromate $$(PCC)$$ is specific for the conversion. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 5, "subject": "Chemistry", "question": "In the following sequence of reactions,
\nCH3CH2OH $$\\buildrel {P + {I_2}} \\over\n \\longrightarrow $$ A $$\\mathrel{\\mathop{\\kern0pt\\longrightarrow}\n\\limits_{ether}^{Mg}} $$ B $$\\buildrel {HCHO} \\over\n \\longrightarrow $$ C $$\\buildrel {{H_2}O} \\over\n \\longrightarrow $$ D
\nthe compound ‘D’ is ", "options": [ { "text": "butanal" }, { "text": "n-butyl alcohol " }, { "text": "n-propyl alcohol " }, { "text": "propanal " } ], "answer": "n-propyl alcohol ", "solution": "**Answer:** n-propyl alcohol \n\n\"AIEEE ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 6, "subject": "Chemistry", "question": "From amongst the following alcohols the one that would react fastest with conc. HCl and anhydrous\nZnCl2, is", "options": [ { "text": "2–Butanol" }, { "text": "2–Methylpropan–2–ol" }, { "text": "2–Methylpropanol" }, { "text": "1–Butanol" } ], "answer": "2–Methylpropan–2–ol", "solution": "**Answer:** 2–Methylpropan–2–ol\n\nTertiary alcohols react fastest with conc. $$HCl$$ and anhydrous $$ZnC{l_2}$$ (lucas reagent) as its mechanism proceeds through the formation of stable tertiary carbocation.\n

Mechanism \n

\"AIEEE ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 7, "subject": "Chemistry", "question": "An unknown alcohol is treated with the “Lucas reagent” to determine whether the alcohol is primary,\nsecondary or tertiary. Which alcohol reacts fastest and by what mechanism: ", "options": [ { "text": "tertiary alcohol by SN1" }, { "text": "secondary alcohol by SN2" }, { "text": "tertiary alcohol by SN2" }, { "text": "secondary alcohol by SN1 " } ], "answer": "tertiary alcohol by SN1", "solution": "**Answer:** tertiary alcohol by SN1\n\nTertiary alcohols reacts fastest with lucas reagnet as the rate of reaction is directly proportional to the stability of carbocation formed in the reaction. Since most stable $${3^ \\circ }$$ carbocation is formed in the reaction hence it will react fastest further tetriary alcohols appears to react by $${S_N}1$$ mechanism.\n

\"JEE ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 8, "subject": "Chemistry", "question": "The most suitable reagent for the conversion of R - CH2 - OH $$\\to$$ R - CHO is:", "options": [ { "text": "CrO3" }, { "text": "PCC (Pyridinium Chlorochromate)" }, { "text": "KMnO4" }, { "text": "K2Cr2O7" } ], "answer": "PCC (Pyridinium Chlorochromate)", "solution": "**Answer:** PCC (Pyridinium Chlorochromate)\n\nAn excellent reagent for oxidation of $${1^ \\circ }$$ alcohols to aldehydes is $$PCC.$$ \n

\"JEE ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 9, "subject": "Chemistry", "question": "The gas evolved on heating CH3MgBr in methanol is :", "options": [ { "text": "HBr" }, { "text": "Methane" }, { "text": "Ethane" }, { "text": "Propane" } ], "answer": "Methane", "solution": "**Answer:** Methane\n\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 10, "subject": "Chemistry", "question": "Bouveault-Blanc reduction reaction involves :", "options": [ { "text": "Reduction of an acyl halide with H2/Pd." }, { "text": "Reduction of an ester with Na/C2H5OH." }, { "text": "Reduction of a carbonyl compound with Na/Hg and HCl." }, { "text": "Reduction of an anhydride with LiAlH4." } ], "answer": "Reduction of an ester with Na/C2H5OH.", "solution": "**Answer:** Reduction of an ester with Na/C2H5OH.\n\n

Bourveault–Blanc reduction involves the reduction of esters to primary alcohols in the presence of sodium and alcohol.

\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 11, "subject": "Chemistry", "question": "The major product of the following reaction is :
\n$${\\rm{C}}{{\\rm{H}}_3}{\\rm{CH = CHC}}{{\\rm{O}}_2}{\\rm{CH_3 }}\\buildrel {LiAl{H_4}} \\over\n \\longrightarrow $$", "options": [ { "text": "CH3CH2CH2CHO" }, { "text": "CH3CH2CH2CO2CH3" }, { "text": "CH3CH = CHCH2OH" }, { "text": "CH3CH2CH2CH2OH" } ], "answer": "CH3CH = CHCH2OH", "solution": "**Answer:** CH3CH = CHCH2OH\n\nLiAlH4 reduces esters to alcohols but does not\nreduce C = C.\n

$${\\rm{C}}{{\\rm{H}}_3}{\\rm{CH = CHC}}{{\\rm{O}}_2}{\\rm{CH_3 }}\\buildrel {LiAl{H_4}} \\over\n \\longrightarrow $$ CH3CH = CHCH2OH + CH3OH", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 12, "subject": "Chemistry", "question": "Ceric ammonium nitrate and CHCl3/alc. KOH are used for the identification of functional groups present in __________ and _______ respectively.", "options": [ { "text": "amine, alcohol" }, { "text": "alcohol, phenol" }, { "text": "alcohol, amine" }, { "text": "amine, phenol" } ], "answer": "alcohol, amine", "solution": "**Answer:** alcohol, amine\n\nAlcohol give positive test with ceric ammonium nitrate and primary amines gives carbyl amine\ntest with CHCl3, KOH.", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 13, "subject": "Chemistry", "question": "In the given reaction \n

3-Bromo-2, 2-dimethyl butane $$\\buildrel {{C_2}{H_5}OH} \\over\n \\longrightarrow \\mathop {'A'}\\limits_{(Major\\,\\Pr oduct)} $$

Product A is :", "options": [ { "text": "2-Ethoxy-3, 3-dimethyl butane" }, { "text": "1-Ethoxy-3, 3-dimethyl butane" }, { "text": "2-Ethoxy-2, 3-dimethyl butane" }, { "text": "2-Hydroxy-3, 3-dimethyl butane" } ], "answer": "2-Ethoxy-2, 3-dimethyl butane", "solution": "**Answer:** 2-Ethoxy-2, 3-dimethyl butane\n\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 14, "subject": "Chemistry", "question": "To synthesize 1.0 mole of 2-methylpropan-2-ol from Ethylethanoate ___________ equivalents of CH3MgBr Reagent will be required. (Integer value)", "options": [], "answer": "2", "solution": "**Answer:** 2\n\n

2 moles of CH3MgBr\nreagent will be used. Chemical reaction is as\nfollows

\n\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 15, "subject": "Chemistry", "question": "Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).

Assertion (A) : Treatment of bromine water with propene yields 1-bromopropan-2-ol.

Reason (R) : Attack of water on bromonium ion follows Markovnikov rule and results in 1-bromopropan-2-ol.

In the light of the above statements, choose the most appropriate answer from the options given below :", "options": [ { "text": "Both (A) and (R) are true but (R) is not the correct explanation of (A)" }, { "text": "(A) is false but (R) is true" }, { "text": "Both (A) and (R) are true and (R) is the correct explanation of (A)" }, { "text": "(A) is true but (R) is false" } ], "answer": "Both (A) and (R) are true and (R) is the correct explanation of (A)", "solution": "**Answer:** Both (A) and (R) are true and (R) is the correct explanation of (A)\n\n\"JEE

Its IUPAC name 1-bromopropan-2-ol

A and R are true and (R) is the correct explanation of (A).", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 16, "subject": "Chemistry", "question": "

The number of chiral alcohol(s) with molecular formula C4H10O is ________.

", "options": [], "answer": "2", "solution": "**Answer:** 2\n\n\"JEE
Out of which only two are chiral ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 17, "subject": "Chemistry", "question": "

Hex-4-ene-2-ol on treatment with PCC gives 'A'. 'A' on reaction with sodium hypoiodite gives 'B', which on further heating with soda lime gives 'C'. The compound 'C' is :

", "options": [ { "text": "2-pentene" }, { "text": "proponaldehyde" }, { "text": "2-butene" }, { "text": "4-methylpent-2-ene" } ], "answer": "2-butene", "solution": "**Answer:** 2-butene\n\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 18, "subject": "Chemistry", "question": "

Given below are two statements :

\n

Statement I : On heating with $$\\mathrm{KHSO}_{4}$$, glycerol is dehydrated and acrolein is formed.

\n

Statement II : Acrolein has fruity odour and can be used to test glycerol's presence.

\n

Choose the correct option.

", "options": [ { "text": "Both Statement I and Statement II are correct." }, { "text": "Both Statement I and Statement II are incorrect." }, { "text": "Statement I is correct but Statement II is incorrect." }, { "text": "Statement I is incorrect but Statement II is correct." } ], "answer": "Statement I is correct but Statement II is incorrect.", "solution": "**Answer:** Statement I is correct but Statement II is incorrect.\n\n

Glycerol, on heating with KHSO4, undergoes\ndehydration to give unsaturated aldehyde called\nacrolein. So, the statement I is correct.

\n

\"JEE

\n

Acrolein has a piercing unpleasant smell. So,\nstatement II is incorrect.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 19, "subject": "Chemistry", "question": "

When ethanol is heated with conc. $$\\mathrm{H}_{2} \\mathrm{SO}_{4}$$, a gas is produced. The compound formed, when this gas is treated with cold dilute aqueous solution of Baeyer's reagent, is

", "options": [ { "text": "formaldehyde" }, { "text": "formic acid" }, { "text": "glycol" }, { "text": "ethanoic acid" } ], "answer": "glycol", "solution": "**Answer:** glycol\n\n$$$\\mathrm{CH}_{3}-\\mathrm{CH}_{2}-\\mathrm{OH} \\frac{\\text { Conc. } \\mathrm{H}_{2} \\mathrm{SO}_{4}}{\\Delta} \\mathrm{CH}_{2}=\\mathrm{CH}_{2} \\frac{\\text { Cold dil solution of }}{\\text { Bayers reagent }} \\mathrm{OH}-\\mathrm{CH}_{2}-\\mathrm{CH}_{2}-\\mathrm{OH}\\,(\\text {Glycol})$$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 20, "subject": "Chemistry", "question": "

A solution of $$\\mathrm{Cr O_5}$$ in amyl alcohol has a __________ colour.

", "options": [ { "text": "Yellow" }, { "text": "Orange-Red" }, { "text": "Green" }, { "text": "Blue" } ], "answer": "Blue", "solution": "**Answer:** Blue\n\n

CrO$$_5$$ is blue in colour in amyl alcohol.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 21, "subject": "Chemistry", "question": "

The denticity of the ligand present in the Fehling's reagent is ___________.

", "options": [], "answer": "4", "solution": "**Answer:** 4\n\nThe ligand present in Fehling's reagent is tartrate, specifically potassium sodium tartrate (KNaC4H4O6•4H2O), which acts as a chelating agent. The denticity of tartrate is four, meaning that it can bond to a metal ion through four different atoms, in this case, two oxygen atoms from each of the two carboxylate groups and two oxygen atoms from the two hydroxyl groups. This chelation allows the copper ions in Fehling's A to remain in solution, preventing them from precipitating out as insoluble copper hydroxides. The chelation also helps to stabilize the intermediate copper complexes formed during the redox reaction with reducing sugars, allowing for the formation of the brick-red precipitate of copper(I) oxide (Cu2O).", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 22, "subject": "Chemistry", "question": "

Incorrect method of preparation for alcohols from the following is:

", "options": [ { "text": "Hydroboration-oxidation of alkene." }, { "text": "Reaction of Ketone with $$\\mathrm{RMgBr}$$ followed by hydrolysis." }, { "text": "Ozonolysis of alkene." }, { "text": "Reaction of alkyl halide with aqueous $$\\mathrm{NaOH}$$." } ], "answer": "Ozonolysis of alkene.", "solution": "**Answer:** Ozonolysis of alkene.\n\nReductive ozonolysis of alkenes will lead to\nformation of aldehyde or ketones, oxidative\nozonolysis of alkenes will lead to formation of\ncarboxylic acids or ketones.

\nSo, alcohol is not formed by ozonolysis of alkenes.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 23, "subject": "Chemistry", "question": "

Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R.

\n

Assertion A: Butan -1- ol has higher boiling point than ethoxyethane.

\n

Reason R: Extensive hydrogen bonding leads to stronger association of molecules.

\n

In the light of the above statements, choose the correct answer from the options given below:

", "options": [ { "text": "A is false but R is true" }, { "text": "Both A and R are true but R is not the correct explanation of A" }, { "text": "Both A and R are true and R is the correct explanation of A" }, { "text": "A is true but R is false" } ], "answer": "Both A and R are true and R is the correct explanation of A", "solution": "**Answer:** Both A and R are true and R is the correct explanation of A\n\n

The boiling point of a substance is dependent on the strength of the intermolecular forces between its molecules. Stronger intermolecular forces require more energy to overcome, thus resulting in a higher boiling point.

\n

In the case of butan-1-ol and ethoxyethane (diethyl ether), they are both composed of similar numbers of similar atoms, but the type of intermolecular force between the molecules differs due to their functional groups.

\n

Butan-1-ol is an alcohol and has a -OH group, which allows for hydrogen bonding, a strong type of dipole-dipole interaction. This happens when the hydrogen atom in a polar bond (especially -OH, -NH, -HF) is attracted to some electronegative atom (like O, N, or F) in another molecule. Because of this, butan-1-ol molecules "stick together" more strongly, requiring more heat energy to separate them and thus exhibiting a higher boiling point.

\n

On the other hand, ethoxyethane, being an ether, does not have the ability to form hydrogen bonds since it lacks the -OH group. The primary intermolecular forces in ethers are relatively weaker van der Waals forces. Therefore, it requires less heat energy to break these forces, resulting in a lower boiling point compared to butan-1-ol.

\n

So, both A and R are true and R is indeed the correct explanation for A.

\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 24, "subject": "Chemistry", "question": "

The water gas on reacting with cobalt as a catalyst forms

", "options": [ { "text": "Methanoic acid" }, { "text": "Methanol" }, { "text": "Methanal" }, { "text": "Ethanol" } ], "answer": "Methanol", "solution": "**Answer:** Methanol\n\n

Water gas, also known as synthesis gas or "syngas", is a mixture of carbon monoxide (CO) and hydrogen (H2). When water gas reacts with a catalyst, typically consisting of a combination of cobalt and molybdenum, it can undergo a reaction to form methanol (CH3OH).

\n

This is known as the methanol synthesis reaction and can be written as:

\n

$$\\mathrm{CO(g)} + 2\\mathrm{H}_{2}(g) \\rightarrow \\mathrm{CH}_{3}\\mathrm{OH(g)}$$

\n

This reaction is widely used in industry for the production of methanol, which can be used as a solvent, an antifreeze, a fuel, and as a feedstock for the production of chemicals.

\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 25, "subject": "Chemistry", "question": "

Among the following, the number of compounds which will give positive iodoform reaction is _________

\n

(a) 1-Phenylbutan-2-one

\n

(b) 2-Methylbutan-2-ol

\n

(c) 3-Methylbutan-2-ol

\n

(d) 1-Phenylethanol

\n

(e) 3,3-dimethylbutan-2-one

\n

(f) 1-Phenylpropan-2-ol

", "options": [], "answer": "4", "solution": "**Answer:** 4\n\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 26, "subject": "Chemistry", "question": "

Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R:

\n

Assertion A: Alcohols react both as nucleophiles and electrophiles.

\n

Reason R: Alcohols react with active metals such as sodium, potassium and aluminum to yield corresponding alkoxides and liberate hydrogen.

\n

In the light of the above statements, choose the correct answer from the options given below:

", "options": [ { "text": "$$A$$ is false but $$R$$ is true.\n" }, { "text": "Both $$A$$ and $$R$$ are true but $$R$$ is NOT the correct explanation of $$A$$.\n" }, { "text": "Both $$A$$ and $$R$$ are true and $$R$$ is the correct explanation of $$A$$.\n" }, { "text": "$$A$$ is true but $$R$$ is false." } ], "answer": "Both $$A$$ and $$R$$ are true but $$R$$ is NOT the correct explanation of $$A$$.\n", "solution": "**Answer:** Both $$A$$ and $$R$$ are true but $$R$$ is NOT the correct explanation of $$A$$.\n\n\n

Assertion A is that alcohols can act as both nucleophiles and electrophiles. This is indeed true.

\n

Alcohols can act as nucleophiles because they have a polar O-H bond; the oxygen atom has a pair of nonbonding electrons that can be donated to electrophiles. For example, in the presence of a strong acid such as hydrochloric acid, the lone pair on the oxygen can attack the positively charged hydrogen atom to form an oxonium ion ($$R-OH_2^+$$), which then acts as a good leaving group in substitution reactions.

\n

Alcohols can also behave as electrophiles. An example of an alcohol acting as an electrophile is when it reacts with a strong base, such as sodium hydride (NaH), to undergo deprotonation forming an alkoxide ion ($$R-O^-$$). The alkoxide ion is a strong nucleophile, but the original alcohol molecule in the presence of a base acts as an electrophile due to the acidity of the hydroxyl hydrogen.

\n

Reason R posits that alcohols react with active metals such as sodium, potassium and aluminum to yield corresponding alkoxides and liberate hydrogen. This is also true. When alcohols react with active metals, the alcohol acts as an acid, donating a hydrogen ion ($$H^+$$) to form hydrogen gas ($$H_2$$), while the remaining alkoxide ion ($$R-O^-$$) bonds with the metal cation ($$M^+$$, where M is an active metal like sodium) to form the alkoxide salt ($$MOR$$).

\n

Now, considering the above explanations, the relationship between Assertion A and Reason R needs to be evaluated:

\n

While both Assertion A and Reason R are indeed true, Reason R is not the correct explanation of Assertion A. Reason R describes a reaction (alcohol with active metal) where the alcohol functions only as an acid (not as a nucleophile or an electrophile). Hence, Reason R does not explain how alcohols can act as both nucleophiles and electrophiles, which is what Assertion A states.

\n

Therefore, the correct answer is:

\nOption B :\nBoth $$A$$ and $$R$$ are true but $$R$$ is NOT the correct explanation of $$A$$.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 27, "subject": "Chemistry", "question": "An ethar is more volatile than an alcohol having the same molecular formula. This is due to", "options": [ { "text": "alcohols having resonance structures" }, { "text": "inter-molecular hydrogen bonding in ethers" }, { "text": "inter-molecular hydrogen bonding in alcohols" }, { "text": "dipole characters of ethers" } ], "answer": "inter-molecular hydrogen bonding in alcohols", "solution": "**Answer:** inter-molecular hydrogen bonding in alcohols\n\nAlcohol and ether are isomer with each other. So, with same molecular formula we can make ether as well as alcohol. \n

For ex, \n

With molecular formula C2H6O \n

(1) $$\\,\\,\\,$$ alcohol will be CH3CH2 OH\n

(2) $$\\,\\,\\,$$ ether will be CH3 $$-$$ O $$-$$ CH3 \n

In Alcohol there is hydrogen bond and in Ether there is Van der walls force of attraction. \n

We know that H bond is stronger bond than van der walls force of attraction as the atoms of alcohol are strongly attached with each other by hydrogen bonding so tendency of vaporization of alcohol is less compared to ether.\n

In alcohol inter-molecular hydrogen bonding look like this -\n

\"AIEEE ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 28, "subject": "Chemistry", "question": "HBr reacts with CH2 = CH – OCH3 under anhydrous conditions at room temperature to give ", "options": [ { "text": "CH3CHO and CH3Br " }, { "text": "BrCH2CHO and CH3OH" }, { "text": "BrCH2 – CH2 – OCH3 " }, { "text": "H3C – CHBr – OCH3" } ], "answer": "H3C – CHBr – OCH3", "solution": "**Answer:** H3C – CHBr – OCH3\n\nMethyl vinyl ether under anhydrous condition at room temperature undergoes addition reaction .\n

\"AIEEE ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 29, "subject": "Chemistry", "question": "Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).

Assertion (A) : Synthesis of ethyl phenyl ether may be achieved by Williamson synthesis.

Reason (R) : Reaction of bromobenzene with sodium ethoxide yields ethyl phenyl ether.

In the light of the above statements, choose the most appropriate answer from the options given below :", "options": [ { "text": "Both (A) and (R) are correct and (R) is the correct explanation of (A)" }, { "text": "(A) is correct but (R) is not correct" }, { "text": "(A) is not correct but (R) is correct" }, { "text": "Both (A) and (R) are correct but (R) is not the correct explanation of (A)." } ], "answer": "(A) is correct but (R) is not correct", "solution": "**Answer:** (A) is correct but (R) is not correct\n\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 30, "subject": "Chemistry", "question": "

2-Methyl propyl bromide reacts with $$\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{O}^{-}$$ and gives 'A' whereas on reaction with $$\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}$$ it gives 'B'. The mechanism followed in these reactions and the products 'A' and 'B' respectively are :

", "options": [ { "text": "$$\\mathrm{S}_{N} 1, A=$$ tert-butyl ethyl ether; $$\\mathrm{S}_{N} 2, B=$$ iso-butyl ethyl ether" }, { "text": "$$\\mathrm{S}_{\\mathrm{N}} 1, \\mathrm{~A}=$$ tert-butyl ethyl ether; $$\\mathrm{S}_{\\mathrm{N}} 1, \\mathrm{~B}=$$ 2-butyl ethyl ether" }, { "text": "$$\\mathrm{S}_{\\mathrm{N}} 2, \\mathrm{~A}=$$ iso-butyl ethyl ether; $$\\mathrm{S}_{\\mathrm{N}} 1, \\mathrm{~B}=$$ tert-butyl ethyl ether" }, { "text": "$$\\mathrm{S}_{\\mathrm{N}} 2, \\mathrm{~A}=$$ 2-butyl ethyl ether; $$\\mathrm{S}_{\\mathrm{N}} 2, \\mathrm{~B}=$$ iso-butyl ethyl ether" } ], "answer": "$$\\mathrm{S}_{\\mathrm{N}} 2, \\mathrm{~A}=$$ iso-butyl ethyl ether; $$\\mathrm{S}_{\\mathrm{N}} 1, \\mathrm{~B}=$$ tert-butyl ethyl ether", "solution": "**Answer:** $$\\mathrm{S}_{\\mathrm{N}} 2, \\mathrm{~A}=$$ iso-butyl ethyl ether; $$\\mathrm{S}_{\\mathrm{N}} 1, \\mathrm{~B}=$$ tert-butyl ethyl ether\n\n2-Methyl propyl bromide (also known as isobutyl bromide) has the formula (CH₃)₂CHCH₂Br.\n

\nWhen it reacts with C₂H₅O⁻ (ethoxide ion), it undergoes an SN2 reaction because ethoxide ion is a strong nucleophile. The reaction proceeds with a direct exchange of the leaving group (Br⁻) and the nucleophile (C₂H₅O⁻). The product 'A' would be iso-butyl ethyl ether (CH₃CH₂OCH(CH₃)CH₃).\n

\nWhen it reacts with C₂H₅OH (ethanol), the reaction proceeds via an SN1 mechanism because ethanol is a weak nucleophile. In this case, the bromide ion leaves first, forming a carbocation intermediate, which is then attacked by the nucleophile (C₂H₅OH). The product 'B' would be tert-butyl ethyl ether ((CH₃)₃COCH₂CH₃).\n

\nSo, the correct option is:\n

\nSN2, A = iso-butyl ethyl ether; SN1, B = tert-butyl ethyl ether", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 31, "subject": "Chemistry", "question": "

Suitable reaction condition for preparation of Methyl phenyl ether is

", "options": [ { "text": "Benzene, MeBr" }, { "text": "$$\\mathrm{Ph{O^\\Theta }N{a^ \\oplus }}$$, MeBr" }, { "text": "Ph - Br, $$\\mathrm{Me{O^\\Theta }N{a^ \\oplus }}$$" }, { "text": "$$\\mathrm{Ph{O^\\Theta }N{a^ \\oplus }}$$, MeOH" } ], "answer": "$$\\mathrm{Ph{O^\\Theta }N{a^ \\oplus }}$$, MeBr", "solution": "**Answer:** $$\\mathrm{Ph{O^\\Theta }N{a^ \\oplus }}$$, MeBr\n\n$\\mathrm{PhONa}+\\mathrm{MeBr} \\rightarrow \\mathrm{PhOMe}+\\mathrm{NaBr}$

Williamson's Synthesis

\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 32, "subject": "Chemistry", "question": "Phenyl magnesium bromide reacts with methanol to give ", "options": [ { "text": "a mixture of anisole and Mg(OH)Br" }, { "text": "a mixture of benzene and Mg(OMe)Br " }, { "text": "a mixture of toluene and Mg(OH)Br" }, { "text": "a mixture of phenol and Mg(Me)Br" } ], "answer": "a mixture of benzene and Mg(OMe)Br ", "solution": "**Answer:** a mixture of benzene and Mg(OMe)Br \n\n$$C{H_3}OH + {C_6}{H_5}MgBr\\buildrel \\, \\over\n \\longrightarrow C{H_3}O.MgBr + {C_6}{H_6}$$ v", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 33, "subject": "Chemistry", "question": "Phenol, when it first reacts with concentrated sulphuric acid and then with concentrated nitric acid,\ngives", "options": [ { "text": "2,4,6-trinitrobenzene" }, { "text": "o-nitrophenol " }, { "text": "p-nitrophenol " }, { "text": "nitrobenzene " } ], "answer": "o-nitrophenol ", "solution": "**Answer:** o-nitrophenol \n\nPhenol on reaction with conc. $${H_2}S{O_4}$$ gives a mixture of $$o$$- and $$p$$- products (i.e., $$ - S{O_3}H$$ group, occupies $$o$$-, $$p$$- position). At room temperature $$o$$-product is more stable, which on treatment with conc. $$HN{O_3}$$ will yield $$o$$-nitrophenol. \n

\"AIEEE ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 34, "subject": "Chemistry", "question": "The major product obtained on interaction of phenol with sodium hydroxide and carbon dioxide is : ", "options": [ { "text": "benzoic acid " }, { "text": "salicylaldehyde " }, { "text": "salicylic acid" }, { "text": "phthalic acid " } ], "answer": "salicylic acid", "solution": "**Answer:** salicylic acid\n\n\"AIEEE ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 35, "subject": "Chemistry", "question": "Phenol is heated with a solution of mixture of KBr and KBrO3. The major product obtained in the above\nreaction is ", "options": [ { "text": "3-Bromophenol" }, { "text": "4-Bromophenol" }, { "text": "2, 4, 6- Tribromophenol " }, { "text": "2-Bromophenol " } ], "answer": "2, 4, 6- Tribromophenol ", "solution": "**Answer:** 2, 4, 6- Tribromophenol \n\n\"AIEEE ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 36, "subject": "Chemistry", "question": "Ortho–Nitrophenol is less soluble in water than p– and m– Nitrophenols because :", "options": [ { "text": "o–Nitrophenol is more volatile in steam than those of m – and p–isomers" }, { "text": "o–Nitrophenol shows Intramolecular H–bonding" }, { "text": "o–Nitrophenol shows Intermolecular H–bonding" }, { "text": "Melting point of o–Nitrophenol is lower than those of m–and p–isomers." } ], "answer": "o–Nitrophenol shows Intramolecular H–bonding", "solution": "**Answer:** o–Nitrophenol shows Intramolecular H–bonding\n\nAs in ortho-Nitrophenol H of -OH is already in Intra-moleculer H Bond and become stable so it will not perticipate in H bonding with H2O by breaking this H bonding. That is why solublity of Ortho–Nitrophenol is less in water.\n\"AIEEE\n

para– and meta– Nitrophenols both shows inter-moleculer H bonding with H2O. That is why both are more soluble in water.\n\"AIEEE\n\"AIEEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 37, "subject": "Chemistry", "question": "Arrange the following compounds in increasing order of C–OH bond length :
methanol, phenol,\np-ethoxyphenol", "options": [ { "text": "phenol < methanol < p-ethoxyphenol" }, { "text": "methanol < p-ethoxyphenol < phenol" }, { "text": "methanol < phenol < p-ethoxyphenol" }, { "text": "phenol < p-ethoxyphenol < methanol" } ], "answer": "phenol < p-ethoxyphenol < methanol", "solution": "**Answer:** phenol < p-ethoxyphenol < methanol\n\nC-OH bond length in Among methanol, phenol and\np-ethoxyphenol, C-OH bond length is least in phenol due to\nresonance and maximum in methanol due to\nlack of resonance whereas it will have some\nintermediate value in p-ethoxyphenol.\n

phenol < p-ethoxyphenol < methanol", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 38, "subject": "Chemistry", "question": "A solution of phenol in chloroform when treated\nwith aqueous NaOH gives compound P as a\nmajor product. The mass percentage of carbon\nin P is ______. (to the nearest integer)\n
(Atomic mass : C = 12; H = 1; O = 16)", "options": [], "answer": "69", "solution": "**Answer:** 69\n\n\"JEE\n

mass of Salicylaldehyde = 12 × 7 + 6 × 1 + 16 × 2 = 122\n

mass of carbon = 12 × 7 = 84\n

$$ \\therefore $$ mass % of C in P = $${{84} \\over {122}}$$ $$ \\times $$ 100 \n

= 68.85 % $$ \\simeq $$ 69 %", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 39, "subject": "Chemistry", "question": "Given below are two statements :

Statement I : o-Nitrophenol is steam volatile due to intramolecular hydrogen bonding.

Statement II : o-Nitrophenol has high melting due to hydrogen bonding.

In the light of the above statements, choose the most appropriate answer from the options given below :", "options": [ { "text": "Both Statement I and Statement II are true" }, { "text": "Statement I is true but Statement II is false" }, { "text": "Statement I is false but Statement II is true" }, { "text": "Both Statement I and Statement II are false" } ], "answer": "Statement I is true but Statement II is false", "solution": "**Answer:** Statement I is true but Statement II is false\n\n\"JEE\n

So it is more volatile due to intramolecular\nH-bonding.\n

Melting point depends on packing efficiency\nnot on H-bonding thus statement II is false.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 40, "subject": "Chemistry", "question": "

Given below are two statements :

\n

Statement I : Phenols are weakly acidic.

\n

Statement II : Therefore they are freely soluble in NaOH solution and are weaker acids than alcohols and water.

\n

Choose the most appropriate option :

", "options": [ { "text": "Both Statement I and Statement II are correct." }, { "text": "Both Statement I and Statement II are incorrect." }, { "text": "Statement I is correct but Statement II is incorrect." }, { "text": "Statement I is incorrect but Statement II is correct." } ], "answer": "Statement I is correct but Statement II is incorrect.", "solution": "**Answer:** Statement I is correct but Statement II is incorrect.\n\nPhenol are weakly acidic. Phenol is more acidic\nthan alcohol & H2O statement (I) is correct. (II) is\nincorrect.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 41, "subject": "Chemistry", "question": "

Compound 'P' on nitration with dil. HNO3 yields two isomers (A) and (B). These isomers can be separated by steam distillation. Isomers (A) and (B) show the intramolecular and intermolecular hydrogen bonding respectively. Compound (P) on reaction with conc. HNO3 yields a yellow compound 'C', a strong acid. The number of oxygen atoms is present in compound 'C' _____________.

", "options": [], "answer": "7", "solution": "**Answer:** 7\n\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 42, "subject": "Chemistry", "question": "

The difference in the reaction of phenol with bromine in chloroform and bromine in water medium is due to :

\n", "options": [ { "text": "Hyperconjugation in substrate" }, { "text": "Polarity of solvent" }, { "text": "Free radical formation" }, { "text": "Electromeric effect the substrate" } ], "answer": "Polarity of solvent", "solution": "**Answer:** Polarity of solvent\n\nPhenol gives different products with bromine in\nchloroform and water medium due to the polarity\ndifference between chloroform and water acting as\nsolvent", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 43, "subject": "Chemistry", "question": "

Hydrolysis of which compound will give carbolic acid?

", "options": [ { "text": "Cumene" }, { "text": "Benzenediazonium chloride" }, { "text": "Benzal chloride" }, { "text": "Ethylene glycol ketal" } ], "answer": "Benzenediazonium chloride", "solution": "**Answer:** Benzenediazonium chloride\n\n

Phenol, is known as Carbolic acid.

\n

Diazonium salt are hydrolysed to phonols.

\n

\"JEE

\n

Benzal chloride on hydrolysis gives benzaldehyde

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 44, "subject": "Chemistry", "question": "

The increasing order of $$\\mathrm{pK_a}$$ for the following phenols is

\n

(A) 2, 4 - Dinitrophenol

\n

(B) 4 - Nitrophenol

\n

(C) 2, 4, 5 - Trimethylphenol

\n

(D) Phenol

\n

(E) 3-Chlorophenol

\n

Choose the correct answer from the option given below :

", "options": [ { "text": "(A), (E), (B), (D), (C)" }, { "text": "(C), (E), (D), (B), (A)" }, { "text": "(C), (D), (E), (B), (A)" }, { "text": "(A), (B), (E), (D), (C)" } ], "answer": "(A), (B), (E), (D), (C)", "solution": "**Answer:** (A), (B), (E), (D), (C)\n\n\"JEE\n

Order of acidity for following phenol is A > B > E > D > C.

\n- M and – I increases acidity.

\n+ M and + I decreases acidity.

\n$$ \\therefore $$ Their pKa value is A < B < E < D < C.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 45, "subject": "Chemistry", "question": "

Given below are two statements, one is labelled as Assertion A and the other is labelled as Reason R

\n

Assertion A : Butylated hydroxy anisole when added to butter increases its shelf life.

\n

Reason R : Butylated hydroxy anisole is more reactive towards oxygen than food.

\n

In the light of the above statements, choose the most appropriate answer from the options given below

", "options": [ { "text": "A is correct but R is not correct" }, { "text": "Both A and R are correct but R is NOT the correct explanation of A" }, { "text": "Both A and R are correct and R is the correct explanation of A" }, { "text": "A is not correct but R is correct" } ], "answer": "Both A and R are correct and R is the correct explanation of A", "solution": "**Answer:** Both A and R are correct and R is the correct explanation of A\n\nButylated hydroxy anisole is added to butter to\nincrease its shelf life from months to years as it is\nmore reactive towards oxygen than food.\nTherefore, both Assertion and Reason are correct\nand Reason is the correct explanation of Assertion.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 46, "subject": "Chemistry", "question": "

Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R :

\n

Assertion A : Acetal / Ketal is stable in basic medium.

\n

Reason R : The high leaving tendency of alkoxide ion gives the stability to acetal/ketal in basic medium.

\n

In the light of the above statements, choose the correct answer from the options given below :

", "options": [ { "text": "A is false but R is true." }, { "text": "Both A and R are true but R is NOT the correct explanation of A" }, { "text": "Both A and R are true and R is the correct explanation of A" }, { "text": "A is true but R is false." } ], "answer": "A is true but R is false.", "solution": "**Answer:** A is true but R is false.\n\nFor Assertion: Acetal and ketals are basically\nethers hence they must be stable in basic medium\nbut should break down in acidic medium.\nHence assertion is correct.

\nFor reason: Alkoxide ion (RO–) is not considered a good leaving group hence reason must be false.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 47, "subject": "Chemistry", "question": "Match List - I with List - II.\n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
List I (Reactants)List II (Product)
(A) Phenol, Zn/Δ(I) Salicylaldehyde
(B) Phenol, CHCl3, NaOH, HCl(II) Salicylic acid
(C) Phenol, CO2, NaOH, HCl(III) Benzene
(D) Phenol, Conc. HNO3(IV) Picric acid
\n
\nChoose the correct answer from the options given below :", "options": [ { "text": "(A)-(IV), (B)-(I), (C)-(II), (D)-(III)" }, { "text": "(A)-(III), (B)-(I), (C)-(II), (D)-(IV)" }, { "text": "(A)-(IV), (B)-(II), (C)-(I), (D)-(III)" }, { "text": "(A)-(III), (B)-(IV), (C)-(I), (D)-(II)" } ], "answer": "(A)-(III), (B)-(I), (C)-(II), (D)-(IV)", "solution": "**Answer:** (A)-(III), (B)-(I), (C)-(II), (D)-(IV)\n\nLet's analyze each reaction given in List I to identify the correct product from List II.\n\n

(A) Phenol, Zn/Δ: When phenol is reacted with zinc dust (Zn) upon heating (Δ), it undergoes a reduction reaction known as the Clemmensen reduction, which results in the removal of the oxygen from the hydroxyl group (-OH) attached to the benzene ring, thus forming benzene. Hence, (A) corresponds to (III) Benzene.

\n\n

(B) Phenol, CHCl3, NaOH, HCl: This reaction is known as the Reimer-Tiemann reaction. Phenol, when reacted with chloroform (CHCl3) in the presence of an aqueous base (NaOH), followed by acidification with HCl, forms salicylaldehyde. Therefore, (B) corresponds to (I) Salicylaldehyde.

\n\n

(C) Phenol, CO2, NaOH, HCl: This is the Kolbe-Schmitt reaction. In this reaction, phenol reacts with carbon dioxide (CO2) under pressure at high temperatures in the presence of sodium hydroxide (NaOH), followed by acidification with HCl, to yield salicylic acid. Thus, (C) corresponds to (II) Salicylic acid.

\n\n

(D) Phenol, Conc. HNO3: The nitration of phenol with concentrated nitric acid (HNO3) produces picric acid (2,4,6-trinitrophenol). So, (D) corresponds to (IV) Picric acid.

\n\nBased on the above reactions, the correct matching would be:\n\n

(A)-(III) Benzene
\n(B)-(I) Salicylaldehyde
\n(C)-(II) Salicylic acid
\n(D)-(IV) Picric acid

\n\nTherefore, the correct answer is Option B: \n(A)-(III), (B)-(I), (C)-(II), (D)-(IV).", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 48, "subject": "Chemistry", "question": "

Given below are two statements :

\n

Statement (I) : p-nitrophenol is more acidic than m-nitrophenol and o-nitrophenol.

\n

Statement (II) : Ethanol will give immediate turbidity with Lucas reagent.

\n

In the light of the above statements, choose the correct answer from the options given below :

", "options": [ { "text": "Statement I is true but Statement II is false\n" }, { "text": "Both Statement I and Statement II are true\n" }, { "text": "Both Statement I and Statement II are false\n" }, { "text": "Statement I is false but Statement II is true" } ], "answer": "Statement I is true but Statement II is false\n", "solution": "**Answer:** Statement I is true but Statement II is false\n\n\n

Acidic strength

\n

\"JEE

\n

Ethanol give lucas test after long time

\n

Statement (I) $$\\rightarrow$$ correct

\n

Statement (II) $$\\rightarrow$$ incorrect

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 49, "subject": "Chemistry", "question": "

Phenolic group can be identified by a positive:

", "options": [ { "text": "Tollen's test\n" }, { "text": "Phthalein dye test\n" }, { "text": "Carbylamine test\n" }, { "text": "Lucas test" } ], "answer": "Phthalein dye test\n", "solution": "**Answer:** Phthalein dye test\n\n\n

The correct answer for identifying the phenolic group is Option B, the Phthalein dye test.

\n\n

Option A: Tollen's test - This test is used to identify aldehydes. Aldehydes are oxidized to carboxylic acids by the Tollen's reagent (ammoniacal silver nitrate), and this reaction results in a characteristic silver mirror or a black precipitate of silver metal, which is not relevant to phenolic compounds.

\n\n

Option B: Phthalein dye test - The phenolic group can be identified using the phthalein dye test. In this test, phenols react with phthalic anhydride in the presence of a dehydrating agent, typically concentrated sulfuric acid, to form phthalein dyes. A common example of a phthalein dye that can be formed in this manner is phenolphthalein, which is synthesized using phenol and phthalic anhydride. These dyes exhibit color changes in different pH environments, which is a characteristic helpful in identifying phenolic compounds.

\n\n

Option C: Carbylamine test - The carbylamine test, also known as the isocyanide test, is specific to primary amines. In this reaction, primary amines are heated with chloroform and an alkali base, typically potassium hydroxide, to produce a foul-smelling isocyanide or carbylamine. Since phenolic compounds do not have an amine group, they will not give a positive carbylamine test.

\n\n

Option D: Lucas test - The Lucas test is used to classify alcohols of low molecular weight based on their reactivity with Lucas reagent, a solution of zinc chloride in concentrated hydrochloric acid. The test differentiates between primary, secondary, and tertiary alcohols. Phenols, which have an -OH group bonded to an aromatic ring, do not react in the same way as aliphatic alcohols in the Lucas test, and therefore, this test is not used to identify phenolic groups.

\n\n

Thus, the test that is specifically used to identify a phenolic group is the Phthalein dye test.

", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 50, "subject": "Chemistry", "question": "

Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R:

\n

Assertion A: $$\\mathrm{pK}_{\\mathrm{a}}$$ value of phenol is 10.0 while that of ethanol is 15.9 .

\n

Reason R: Ethanol is stronger acid than phenol.

\n

In the light of the above statements, choose the correct answer from the options given below:

", "options": [ { "text": "A is false but R is true.\n" }, { "text": "Both $$A$$ and $$R$$ are true but $$R$$ is NOT the correct explanation of $$A$$.\n" }, { "text": "Both $$\\mathrm{A}$$ and $$\\mathrm{R}$$ are true and $$\\mathrm{R}$$ is the correct explanation of $$\\mathrm{A}$$.\n" }, { "text": "$$A$$ is true but $$R$$ is false." } ], "answer": "$$A$$ is true but $$R$$ is false.", "solution": "**Answer:** $$A$$ is true but $$R$$ is false.\n\n

Phenol is more acidic than ethanol because conjugate base of phenoxide is more stable than ethoxide.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 51, "subject": "Chemistry", "question": "

Phenol treated with chloroform in presence of sodium hydroxide, which further hydrolyzed in presence of an acid results

", "options": [ { "text": "Salicylic acid\n" }, { "text": "Benzene-1,2-diol\n" }, { "text": "2-Hydroxybenzaldehyde\n" }, { "text": "Benzene-1,3-diol" } ], "answer": "2-Hydroxybenzaldehyde\n", "solution": "**Answer:** 2-Hydroxybenzaldehyde\n\n\n

\"JEE

\n

It is Reimer Tiemann Reaction

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 52, "subject": "Chemistry", "question": "

Match List I with List II

\n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
List - I
(Compound)		List - II
($$\\mathrm{pK_a}$$ value)
(A)Ethanol(I)10.0
(B)Phenol(II)15.9
(C)m-Nitrophenol(III)7.1
(D)p-Nitrophenol(IV)8.3

\n

Choose the correct answer from the options given below:

", "options": [ { "text": "A-I, B-II, C-III, D-IV\n" }, { "text": "A-IV, B-I, C-II, D-III\n" }, { "text": "A-II, B-I, C-IV, D-III\n" }, { "text": "A-III, B-IV, C-I, D-II" } ], "answer": "A-II, B-I, C-IV, D-III\n", "solution": "**Answer:** A-II, B-I, C-IV, D-III\n\n\n

$$\\begin{aligned}\n& \\text { Ethanol } \\rightarrow 15.9 \\\\\n& \\text { Phenol } \\rightarrow 10 \\\\\n& \\text { M-Nitrophenol } \\rightarrow 8.3 \\\\\n& \\text { P-Nitrophenol } \\rightarrow 7.1\n\\end{aligned}$$

", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 53, "subject": "Chemistry", "question": "

Common name of Benzene - 1,2 - diol is -

", "options": [ { "text": "catechol\n" }, { "text": "quinol\n" }, { "text": "o-cresol\n" }, { "text": "resorcinol" } ], "answer": "catechol\n", "solution": "**Answer:** catechol\n\n\n

Benzene-1, 2-diol is also called catechol

\n

\"JEE

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 54, "subject": "Chemistry", "question": "

Which one the following compounds will readily react with dilute $$\\mathrm{NaOH}$$ ?

", "options": [ { "text": "$$\\mathrm{C}_6 \\mathrm{H}_5 \\mathrm{CH}_2 \\mathrm{OH}$$\n" }, { "text": "$$\\mathrm{C}_2 \\mathrm{H}_5 \\mathrm{OH}$$\n" }, { "text": "$$\\mathrm{C}_6 \\mathrm{H}_5 \\mathrm{OH}$$\n" }, { "text": "$$\\left(\\mathrm{CH}_3\\right)_3 \\mathrm{COH}$$" } ], "answer": "$$\\mathrm{C}_6 \\mathrm{H}_5 \\mathrm{OH}$$\n", "solution": "**Answer:** $$\\mathrm{C}_6 \\mathrm{H}_5 \\mathrm{OH}$$\n\n\n

To determine which of the given compounds will readily react with dilute $$\\mathrm{NaOH}$$, we need to understand the chemical reactivity of these compounds towards bases like sodium hydroxide ($$\\mathrm{NaOH}$$).

\n\n

Here's a brief overview of each compound's reactivity towards $$\\mathrm{NaOH}$$:

\n\n\n\n

Given the choices, the correct answer is Option C: $$\\mathrm{C}_6 \\mathrm{H}_5 \\mathrm{OH}$$ (Phenol), as it will readily react with dilute $$\\mathrm{NaOH}$$ due to the increased acidity of its hydroxyl group caused by the resonance stability of the phenoxide ion formed in the reaction.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 55, "subject": "Chemistry", "question": "

Given below are two statement :

\n

Statements I : Bromination of phenol in solvent with low polarity such as $$\\mathrm{CHCl}_3$$ or $$\\mathrm{CS}_2$$ requires Lewis acid catalyst.

\n

Statements II : The Lewis acid catalyst polarises the bromine to generate $$\\mathrm{Br}^{+}$$.

\n

In the light of the above statements, choose the correct answer from the options given below :

", "options": [ { "text": "Statement I is true but Statement II is false\n" }, { "text": "Both Statement I and Statement II are true\n" }, { "text": "Statement I is false but Statement II is true\n" }, { "text": "Both Statement I and Statement II are false" } ], "answer": "Statement I is false but Statement II is true\n", "solution": "**Answer:** Statement I is false but Statement II is true\n\n\n

\"JEE

\n

Presence of Lewis acid promotes the reaction due \nto ease in formation of electrophile.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 56, "subject": "Chemistry", "question": "Which one of the following reduced with zinc and hydrochloric acid to give the corresponding\nhydrocarbon? ", "options": [ { "text": "Ethyl acetate " }, { "text": "Butan -2-one " }, { "text": "Acetamide" }, { "text": "Acetic acid" } ], "answer": "Butan -2-one ", "solution": "**Answer:** Butan -2-one \n\nIt is Clemmensen's reduction\n

\"AIEEE ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 57, "subject": "Chemistry", "question": "Trichloroacetaldehyde was subjected to Cannizzaro’s reaction by using NaOH. The mixture of the\nproducts contains sodium trichloroacetate and another compound. The other compound is :", "options": [ { "text": "Trichloromethanol" }, { "text": "2, 2, 2-Trichloropropanol" }, { "text": "Chloroform" }, { "text": "2, 2, 2-Trichloroethanol " } ], "answer": "2, 2, 2-Trichloroethanol ", "solution": "**Answer:** 2, 2, 2-Trichloroethanol \n\n\"AIEEE \n
In Cannizzaro's reaction the compounds which do not contain $$\\alpha $$-hydrogen atoms undergo oxidation and reduction simultaneously i.e. undergo disproportionation and form one molecule of sodium salt of carboxylic acid as oxidation product and one molecule of alcohol as reduction product. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 58, "subject": "Chemistry", "question": "Which of the following compounds will most readily be dehydrated to give alkene under acidic condition ? ", "options": [ { "text": "1- Pentanol " }, { "text": "4-Hydroxypentan - 2-one" }, { "text": "3-Hydroxypentan - 2-one" }, { "text": "2-Hydroxycyclopentanone" } ], "answer": "4-Hydroxypentan - 2-one", "solution": "**Answer:** 4-Hydroxypentan - 2-one\n\nAmong those compounds which have most acidic hydrogen will undergo dehydration most readily. That is why 4 $$-$$ Hydroxypentan $$-$$ 2 $$-$$ one will most readily dehydrated to give unsaturated ketone. \n

\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 59, "subject": "Chemistry", "question": "In the following reaction : \n

Aldehyde + Alcohol   $$\\buildrel {HCl} \\over\n \\longrightarrow $$  Acetal \n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
Aldehyde                 Alcohol
HCHOtBuOH
CH3CHOMeOH
\n
The best combination is ", "options": [ { "text": "CH3CHO and MeOH" }, { "text": "HCHO and MeOH" }, { "text": "CH3CHO and tBuOH" }, { "text": "HCHO and tBuOH " } ], "answer": "HCHO and MeOH", "solution": "**Answer:** HCHO and MeOH\n\nIn Formaldehyde (HCHO) both side of the C atom H presents but in Acetaldehyde (CH3CHO) one side of C atom -CH3 group and other side H atom presents. So Formaldehyde is less crowded then it is more reactive.\n

CH3OH is 1o alcohol and tBuOH is 3o alcohol. As we know on 3o alcohol steric hindrance is very high compare to 1o alcohol and due to steric hindrance reactivity of 3o alcohol is less compare to 1o alcohol.", "topic": "Discrete Mathematics", "subtopic": "Combinatorics" }, { "id": 60, "subject": "Chemistry", "question": "The increasing order of the reactivity of the\nfollowing compound in nucleophilic addition\nreaction is :\n

Propanal, Benzaldehyde, Propanone, Butanone", "options": [ { "text": "Butanone < Propanone < Benzaldehyde <\nPropanal" }, { "text": "Benzaldehyde < Butanone < Propanone <\nPropanal" }, { "text": "Propanal < Propanone < Butanone <\nBenzaldehyde" }, { "text": "Benzaldehyde < Propanal < Propanone <\nButanone" } ], "answer": "Butanone < Propanone < Benzaldehyde <\nPropanal", "solution": "**Answer:** Butanone < Propanone < Benzaldehyde <\nPropanal\n\nRate of Nucleophilic addition :\n

1. Aldehyde > Ketone\n

2. Aliphatic aldehyde > Aromatic aldehyde\n

$$ \\therefore $$ Correct order of nucleophilic addition reaction is :\n
\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 61, "subject": "Chemistry", "question": "Assertion A : Enol form of acetone [CH3COCH3] exists in < 0.1% quantity. However, the enol form of acetyl acetone [CH3COCH2OCCH3] exists in approximately 15% quantity.

Reason R : Enol form of acetyl acetone is stabilized by intramolecular hydrogen bonding, which is not possible in enol form of acetone.

Choose the correct statement :", "options": [ { "text": "Both A and R are true but R is not the correct explanation of A" }, { "text": "A is true but R is false" }, { "text": "A is false but R is true" }, { "text": "Both A and R are true and R is the correct explanation of A" } ], "answer": "Both A and R are true and R is the correct explanation of A", "solution": "**Answer:** Both A and R are true and R is the correct explanation of A\n\n\"JEE
\nAcetyl acetone in enol form have intramolecular\nH-bonding, which is absent in acetone.\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 62, "subject": "Chemistry", "question": "Experimentally reducing a functional group cannot be done by which one of the following reagents?", "options": [ { "text": "Pt-C/H2" }, { "text": "Na/H2" }, { "text": "Pd-C/H2" }, { "text": "Zn/H2O" } ], "answer": "Na/H2", "solution": "**Answer:** Na/H2\n\nNa in presence of H2, will not release electron\nwhich are required for reduction.\n

H2 gas also not get adsorbed on Na. Hence\nNa/H2 cannot be used as a reducing agent.", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 63, "subject": "Chemistry", "question": "Given below are two statements :

Statement I : The nucleophilic addition of sodium hydrogen sulphite to an aldehyde or a ketone involves proton transfer to form a stable ion.

Statement II : The nucleophilic addition of hydrogen cyanide to an aldehyde or a ketone yields amine as final product.

In the light of the above statements, choose the most appropriate answer from the options given below :", "options": [ { "text": "Both Statement I and Statement are true." }, { "text": "Statement I is true but Statement II is false." }, { "text": "Statement I is false but Statement II is true." }, { "text": "Both Statement I and Statement II are false." } ], "answer": "Statement I is true but Statement II is false.", "solution": "**Answer:** Statement I is true but Statement II is false.\n\nStatement I : Correct

\"JEE
Statement II : Wrong

\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 64, "subject": "Chemistry", "question": "

A hydrocarbon 'X' is found to have molar mass of 80. A 10.0 mg of compound 'X' on hydrogenation consumed 8.40 mL of H2 gas (measured at STP). Ozonolysis of compound 'X' yields only formaldehyde and dialdehyde. The total number of fragments/molecules produced from the ozonolysis of compound 'X' is _____________.

", "options": [], "answer": "4", "solution": "**Answer:** 4\n\nMoles of $\\mathrm{X}=\\frac{10 \\mathrm{mg}}{80}=0.125 \\mathrm{~m} \\mathrm{~mol}$.

moles consumed of $\\mathrm{H}_{2}=\\frac{8.4}{22.4} 0.375 \\mathrm{~m} \\mathrm{~mol}$.

$\\frac{\\mathrm{n}_{\\mathrm{H}_{2}}}{\\mathrm{n}_{\\mathrm{X}}}=\\frac{0.375}{0.125}=3$\n

\nSo, the compound $\\mathrm{X}$ have 3 double bond.

Ozonolysis of the compound yield formaldehyde and\ndialdehyde.\n

\nThe compound is\n

\n$$\n\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CH}-\\mathrm{CH}=\\mathrm{CH}-\\mathrm{CH}=\\mathrm{CH}_{2}\n$$\n

\nMolecular mass $=(12 \\times 6)+1 \\times 8=72+8=80 \\,\\mathrm{amu}$\n

\nOzonolysis form:\n

\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 65, "subject": "Chemistry", "question": "

The number of stereoisomers formed in a reaction of $$(±)\\mathrm{Ph}(\\mathrm{C}=\\mathrm{O}) \\mathrm{C}(\\mathrm{OH})(\\mathrm{CN}) \\mathrm{Ph}$$ with $$\\mathrm{HCN}$$ is ___________.

\n

$$\\left[\\right.$$where $$\\mathrm{Ph}$$ is $$-\\mathrm{C}_{6} \\mathrm{H}_{5}$$]

", "options": [], "answer": "3", "solution": "**Answer:** 3\n\n\"JEE\n

Number of stereoisomers = 3", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 66, "subject": "Chemistry", "question": "

Given below are two statements :

\n

Statements I : Acidity of $$\\alpha$$-hydrogens of aldehydes and ketones is responsible for Aldol reaction.

\n

Statement II : Reaction between benzaldehyde and ethanal will NOT give Cross - Aldol product.

\n

In the light of the above statements, choose the most appropriate answer from the options given below :

", "options": [ { "text": "Both Statement I and Statement II are incorrect\n" }, { "text": "Statement I is correct but Statement II is incorrect\n" }, { "text": "Both Statement I and Statement II are correct\n" }, { "text": "Statement I is incorrect but Statement II is correct" } ], "answer": "Statement I is correct but Statement II is incorrect\n", "solution": "**Answer:** Statement I is correct but Statement II is incorrect\n\n\n

Let's analyze the given statements in detail:

\n\n

Statement I: Acidity of $$\\alpha$$-hydrogens of aldehydes and ketones is responsible for Aldol reaction.

\n\n

This statement is correct. The presence of acidic $$\\alpha$$-hydrogens in aldehydes and ketones makes it possible for these compounds to undergo deprotonation, forming enolates. These enolates can then attack the carbonyl carbon of another aldehyde or ketone, leading to the formation of the Aldol product. Therefore, the acidity of the $$\\alpha$$-hydrogens is crucial for the Aldol reaction.

\n\n

Statement II: Reaction between benzaldehyde and ethanal will NOT give Cross-Aldol product.

\n\n

This statement is incorrect. Benzaldehyde (which lacks an $$\\alpha$$-hydrogen) can indeed react with ethanal (which has an $$\\alpha$$-hydrogen) in a Cross-Aldol reaction. In a Cross-Aldol reaction, one molecule (ethanal) forms an enolate and reacts with another molecule (benzaldehyde), resulting in the formation of a Cross-Aldol product.

\n\n

Based on these explanations, the most appropriate answer is:

\n\n

Option B

\n\n

Statement I is correct but Statement II is incorrect.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 67, "subject": "Chemistry", "question": "

Vanillin compound obtained from vanilla beans, has total sum of oxygen atoms and $$\\pi$$ electrons is __________.

", "options": [], "answer": "11", "solution": "**Answer:** 11\n\n

To determine the total sum of oxygen atoms and $$\\pi$$ electrons in vanillin, let's first consider the structure of vanillin. Vanillin is an organic compound with the chemical formula $$C_8H_8O_3$$. It consists of a benzene ring attached to a methoxy group (-OCH3), a hydroxyl group (-OH), and an aldehyde group (-CHO).

\n\n

Firstly, let's count the oxygen atoms:

\n\n\n\n

Thus, there are a total of 3 oxygen atoms in vanillin.

\n\n

Next, let's calculate the total number of $$\\pi$$ electrons:

\n\n\n\n

Therefore, vanillin has a total of 8 $$\\pi$$ electrons from the benzene ring and the aldehyde group.

\n\n

Adding the number of oxygen atoms (3) and the $$\\pi$$ electrons (8) gives us a total sum of 11.

\n\n

Thus, the total sum of oxygen atoms and $$\\pi$$ electrons in vanillin is 11.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 68, "subject": "Chemistry", "question": "

Two moles of benzaldehyde and one mole of acetone under alkaline conditions using aqueous $$\\mathrm{NaOH}$$ after heating gives $$x$$ as the major product. The number of $$\\pi$$ bonds in the product $$x$$ is ______.

", "options": [], "answer": "9", "solution": "**Answer:** 9\n\n

\"JEE

\n

Total number of $$\\pi$$ bonds in $$X=9$$

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 69, "subject": "Chemistry", "question": "Reaction of cyclohexanone with dimethylamine in the presence of catalytic amount of\nan acid forms a compound if water during the reaction is continuously removed. The\ncompound formed is generally known as", "options": [ { "text": "a Schiff’s base" }, { "text": "an enamine " }, { "text": "an imine" }, { "text": "an amine" } ], "answer": "an enamine ", "solution": "**Answer:** an enamine \n\n\"AIEEE ", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 70, "subject": "Chemistry", "question": "The correct match between Item-I (starting\nmaterial) and Item-II (reagent) for the\npreparation of benzaldehyde is :\n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
Item-IItem-II
(I) Benzene(P) HCl and SnCl2,\nH3O+
(II) Benzonitrile(Q) H2, Pd-BaSO4, S\nand quinoline
(III) Benzoyl Chloride(R) CO, HCl and AlCl3
", "options": [ { "text": "(I) - (R), (II) - (P) and (III) - (Q)" }, { "text": "(I) - (P), (II) - (Q) and (III) - (R)" }, { "text": "(I) - (Q), (II) - (R) and (III) - (P)" }, { "text": "(I) - (R), (II) - (Q) and (III) - (P)" } ], "answer": "(I) - (R), (II) - (P) and (III) - (Q)", "solution": "**Answer:** (I) - (R), (II) - (P) and (III) - (Q)\n\n\"JEE", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 71, "subject": "Chemistry", "question": "Which one of the following reactions will not form acetaldehyde?", "options": [ { "text": "$$C{H_3}C{H_2}OH\\mathrel{\\mathop{\\kern0pt\\longrightarrow}\n\\limits_{573K}^{Cu}} $$" }, { "text": "$$C{H_3}C{H_2}OH\\buildrel {Cr{O_3} - {H_2}S{O_4}} \\over\n \\longrightarrow $$" }, { "text": "$$C{H_2} = C{H_2} + {O_2}\\mathrel{\\mathop{\\kern0pt\\longrightarrow}\n\\limits_{{H_2}O}^{Pd(II)/Cu(II)}} $$" }, { "text": "$$C{H_3}CN\\mathrel{\\mathop{\\kern0pt\\longrightarrow}\n\\limits_{ii)\\,{H_2}O}^{i)\\,DIBAL - H}} $$" } ], "answer": "$$C{H_3}C{H_2}OH\\buildrel {Cr{O_3} - {H_2}S{O_4}} \\over\n \\longrightarrow $$", "solution": "**Answer:** $$C{H_3}C{H_2}OH\\buildrel {Cr{O_3} - {H_2}S{O_4}} \\over\n \\longrightarrow $$\n\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 72, "subject": "Chemistry", "question": "One mole of a symmetrical alkene on ozonolysis gives two moles of an aldehyde having a molecular\nmass of 44 u. The alkene is", "options": [ { "text": "propene " }, { "text": "1–butene" }, { "text": "2–butene" }, { "text": "ethene" } ], "answer": "2–butene", "solution": "**Answer:** 2–butene\n\nThe given molecular formula suggests that the aldehyde formed will be acetaldehyde hence the alkene will be \n

\"AIEEE ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 73, "subject": "Chemistry", "question": "The predominant intermolecular forces present in ethyl acetate, a liquid, are:", "options": [ { "text": "Dipole-dipole and hydrogen bonding" }, { "text": "hydrogen bonding and London dispersion" }, { "text": "London dispersion, dipole-dipole and hydrogen bonding" }, { "text": "London dispersion and dipole-dipole" } ], "answer": "London dispersion and dipole-dipole", "solution": "**Answer:** London dispersion and dipole-dipole\n\n

Ethyl acetate (CH3COOCH2CH3) is an organic compound with a molecular structure that does not include any hydrogen atoms bonded directly to highly electronegative atoms like nitrogen, oxygen, or fluorine. Because of this, it cannot form hydrogen bonds. However, ethyl acetate does possess a polar carbonyl group (C=O), which contributes to dipole-dipole interactions, and like all molecules, it can experience London dispersion forces. Therefore, the predominant intermolecular forces present in ethyl acetate are dipole-dipole interactions and London dispersion forces.

\n\n

Hence, the correct option is :

\n\n

Option D: London dispersion and dipole-dipole

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 74, "subject": "Chemistry", "question": "Which of the following reagent is used for the following reaction ?

\n$$\n\\mathrm{CH}_3 \\mathrm{CH}_2 \\mathrm{CH}_3 \\stackrel{?}{\\longrightarrow} \\mathrm{CH}_3 \\mathrm{CH}_2 \\mathrm{CHO}\n$$", "options": [ { "text": "Molybdenum oxide" }, { "text": "Manganese acetate" }, { "text": "Copper at high temperature and pressure" }, { "text": "Potassium permanganate" } ], "answer": "Molybdenum oxide", "solution": "**Answer:** Molybdenum oxide\n\nMolybdenum oxide (Mo2O3 ) is used for oxidising alkanes to\naldehyde. It used to manufacture molybdenum metal, which serves\nas an additive to steel and corrosive resistant alloys.\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 75, "subject": "Chemistry", "question": "Mesityl oxide is a common name of :", "options": [ { "text": "3-Methyl cyclohexane carbaldehyde" }, { "text": "4-Methyl pent-3-en-2-one" }, { "text": "2,4-Dimethyl pentan-3-one" }, { "text": "2-Methyl cyclohexanone" } ], "answer": "4-Methyl pent-3-en-2-one", "solution": "**Answer:** 4-Methyl pent-3-en-2-one\n\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 76, "subject": "Chemistry", "question": "The total number of C-C sigma bond/s in mesityl oxide (C6H10O) is __________. (Round off to the Nearest Integer).", "options": [], "answer": "5", "solution": "**Answer:** 5\n\nStructure of mesityl oxide is\n

\"JEE\n

Number of C–C sigma bonds = 5", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 77, "subject": "Chemistry", "question": "

Oxidation of toluene to benzaldehyde can be easily carried out with which of the following reagents?

", "options": [ { "text": "CrO3/acetic acid, H3O+" }, { "text": "CrO3/acetic anhydride, H3O+" }, { "text": "KMnO4/HCl, H3O+" }, { "text": "CO/HCl, anhydrous AlCl3" } ], "answer": "CrO3/acetic anhydride, H3O+", "solution": "**Answer:** CrO3/acetic anhydride, H3O+\n\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 78, "subject": "Chemistry", "question": "

A trisubstituted compound '$$\\mathrm{A}$$', $$\\mathrm{C}_{10} \\mathrm{H}_{12} \\mathrm{O}_{2}$$ gives neutral $$\\mathrm{FeCl}_{3}$$ test positive. Treatment of compound 'A' with $$\\mathrm{NaOH}$$ and $$\\mathrm{CH}_{3} \\mathrm{Br}$$ gives $$\\mathrm{C}_{11} \\mathrm{H}_{14} \\mathrm{O}_{2}$$, with hydroiodic acid gives methyl iodide and with hot conc. $$\\mathrm{NaOH}$$ gives a compound $$\\mathrm{B}, \\mathrm{C}_{10} \\mathrm{H}_{12} \\mathrm{O}_{2}$$. Compound 'A' also decolorises alkaline $$\\mathrm{KMnO}_{4}$$. The number of $$\\pi$$ bond/s present in the compound '$$\\mathrm{A}$$' is _____________.

", "options": [], "answer": "4", "solution": "**Answer:** 4\n\n

$$\\mathrm{A:C_{10}H_{12}O_2}$$

\n

DU of A $$=\\frac{22-12}{2}=5$$

\n

1 DU is due to Ring (Benzene ring)

\n

4 $$\\pi$$-bonds will be there

\n

(3 $$\\pi$$-bonds in ring and 1 $$\\pi$$-bond outside ring) as it decolorises alkaline KMnO$$_4$$.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 79, "subject": "Chemistry", "question": "The compound formed as a result of oxidation of ethyl benzene by KMnO4 is :", "options": [ { "text": "benzophenone" }, { "text": "acetophenone" }, { "text": "benzoic acid" }, { "text": "benzyl alcohol " } ], "answer": "benzoic acid", "solution": "**Answer:** benzoic acid\n\nWhen alkyl benzene are oxidised with alkaline $$KMn{O_4},$$ (strong oxidising agent) the entire alkyl group is oxidised to $$-COOH$$ group regardless of length of side chain. \n

\"AIEEE ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 80, "subject": "Chemistry", "question": "A liquid was mixed with ethanol and a drop of concentrated H2SO4 was added. A compound with a\nfruity smell was formed. The liquid was : ", "options": [ { "text": "CH3OH " }, { "text": "HCHO" }, { "text": "CH3CO CH3" }, { "text": "CH3COOH" } ], "answer": "CH3COOH", "solution": "**Answer:** CH3COOH\n\nFruity smell is due to ester formation which is formed between ethanol and acid. \n

\"AIEEE ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 81, "subject": "Chemistry", "question": "The number of compound/s given below which contain/s -COOH group is __________. (Integer answer)

(A) Sulphanilic acid

(B) Picric acid

(C) Aspirin

(D) Ascorbic acid", "options": [], "answer": "1", "solution": "**Answer:** 1\n\nThe structures of the given compounds are

\n(A) Sulphanilic acid

\n\"JEE
\n(B) Picric acid

\n\"JEE
\n(C) Aspirin

\n\"JEE
\n(D) Ascorbic Acid

\n\"JEE
$$ \\therefore $$ Only 1 compound has –COOH group\n", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 82, "subject": "Chemistry", "question": "The correct sequential addition of reagents in the preparation of 3-nitrobenzoic acid from benzene is :", "options": [ { "text": "Br2/AlBr3, HNO3/H2SO4, Mg/ether, CO2, H3O+" }, { "text": "Br2/AlBr3, NaCN, H3O+, HNO3/H2SO4" }, { "text": "Br2/AlBr3, HNO3/H2SO4, NaCN, H3O+" }, { "text": "HNO3/H2SO4, Br2/AlBr3, Mg/ether, CO2, H3O+" } ], "answer": "HNO3/H2SO4, Br2/AlBr3, Mg/ether, CO2, H3O+", "solution": "**Answer:** HNO3/H2SO4, Br2/AlBr3, Mg/ether, CO2, H3O+\n\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 83, "subject": "Chemistry", "question": "Given below are two statements :

Statement I : Ethyl pent-4-yn-oate on reaction with CH3MgBr gives a 3$$^\\circ$$-alcohol.

Statement II : In this reaction one mole of ethyl pent-4-yn-oate utilizes two moles of CH3MgBr.

In the light of the above statements, choose the most appropriate answer from the options given below :", "options": [ { "text": "Both Statement I and Statement II are false." }, { "text": "Statement I is false but Statement II is true." }, { "text": "Statement I is true but Statement II is false." }, { "text": "Both Statement I and Statement II are true." } ], "answer": "Statement I is true but Statement II is false.", "solution": "**Answer:** Statement I is true but Statement II is false.\n\nStatement I is true

But it consume 3 moles of CH3MgBr.

So Statement II is false.

\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 84, "subject": "Chemistry", "question": "Which one of the following reactions will not yield propionic acid?", "options": [ { "text": "CH3CH2COCH3 + OI$$-$$/H3O+" }, { "text": "CH3CH2CH3 + KMnO4(Heat), OH$$-$$/H3O+" }, { "text": "CH3CH2CCl3 + OH$$-$$/H3O+" }, { "text": "CH3CH2CH2Br + Mg, CO2 dry ether/H3O+" } ], "answer": "CH3CH2CH2Br + Mg, CO2 dry ether/H3O+", "solution": "**Answer:** CH3CH2CH2Br + Mg, CO2 dry ether/H3O+\n\nAll gives propanoic acid as product but option (d) gives butanoic as product

\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 85, "subject": "Chemistry", "question": "

Given below are two statements :

\n

Statement I : The esterification of carboxylic acid with an alcohol is a nucleophilic acyl substitution.

\n

Statement II : Electron withdrawing groups in the carboxylic acid will increase the rate of esterification reaction.

\n

Choose the most appropriate option :

", "options": [ { "text": "Both Statement I and Statement II are correct." }, { "text": "Both Statement I and Statement II are incorrect." }, { "text": "Statement I is correct but Statement II is incorrect." }, { "text": "Statement I is incorrect but Statement II is correct." } ], "answer": "Both Statement I and Statement II are correct.", "solution": "**Answer:** Both Statement I and Statement II are correct.\n\nEsterification of carboxylic acid with an alcohol is\nnucleophilic acyl substitution and presence of\nelectron withdrawing group in the carboxylic acid\nincreases the rate of esterification reaction.", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 86, "subject": "Chemistry", "question": "

Two statements are given below :

\n

Statement I : The melting point of monocarboxylic acid with even number of carbon atoms is higher than that of with odd number of carbon atoms acid immediately below and above it in the series.

\n

Statement II : The solubility of monocarboxylic acids in water decreases with increase in molar mass.

\n

Choose the most appropriate option :

", "options": [ { "text": "Both Statement I and Statement II are correct." }, { "text": "Both Statement I and Statement II are incorrect." }, { "text": "Statement I is correct but Statement II is incorrect." }, { "text": "Statement I is incorrect but Statement II is correct." } ], "answer": "Both Statement I and Statement II are correct.", "solution": "**Answer:** Both Statement I and Statement II are correct.\n\nStatement (I) is correct as monocarboxylic acids\nwith even number of carbon atoms show better\npacking efficiency in the solid state, statement (II) is\nalso correct as the solubility of carboxylic acids\ndecreases with an increase in molar mass due to\nincrease in the hydrophobic portion with an increase in\nthe number of carbon atoms.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 87, "subject": "Chemistry", "question": "

Given below are two statements, one is labelled as Assertion A and the other is labelled as Reason R.

\n

Assertion A : A solution of the product obtained by heating a mole of glycine with a mole of chlorine in presence of red phosphorous generates chiral carbon atom.

\n

Reason R : A molecule with 2 chiral carbons is always optically active.

\n

In the light of the above statements, choose the correct answer from the options given below:

", "options": [ { "text": "Both A and R are true and R is the correct explanation of A" }, { "text": "Both A and R are true but R is NOT the correct explanation of A" }, { "text": "A is true but R is false" }, { "text": "A is false but R is true" } ], "answer": "A is true but R is false", "solution": "**Answer:** A is true but R is false\n\n\"JEE\n

As for Reason R, it is false. A molecule with 2 chiral carbons is not necessarily optically active. It may be if the molecule is not superimposable on its mirror image (i.e., it's chiral). But if the molecule is superimposable on its mirror image (i.e., it's achiral), then it will not be optically active. This can occur if the molecule has a plane of symmetry, in which case it is a meso compound.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 88, "subject": "Chemistry", "question": "Which among the followng has highest boiling point?", "options": [ { "text": "$\\mathrm{CH}_3 \\mathrm{CH}_2 \\mathrm{CH}_2 \\mathrm{CH}_2-\\mathrm{OH}$" }, { "text": "$\\mathrm{CH}_3 \\mathrm{CH}_2 \\mathrm{CH}_2 \\mathrm{CH}_3$" }, { "text": "$\\mathrm{CH}_3 \\mathrm{CH}_2 \\mathrm{CH}_2 \\mathrm{CHO}$" }, { "text": "$\\mathrm{H}_5 \\mathrm{C}_2-\\mathrm{O}-\\mathrm{C}_2 \\mathrm{H}_5$" } ], "answer": "$\\mathrm{CH}_3 \\mathrm{CH}_2 \\mathrm{CH}_2 \\mathrm{CH}_2-\\mathrm{OH}$", "solution": "**Answer:** $\\mathrm{CH}_3 \\mathrm{CH}_2 \\mathrm{CH}_2 \\mathrm{CH}_2-\\mathrm{OH}$\n\n

To predict the boiling point of these compounds, we need to consider the types of intermolecular forces present in each molecule. These intermolecular forces include van der Waals forces, dipole-dipole interactions, and hydrogen bonding.

\n\n

The compounds given are:

\n\n\n

Now let's analyze each one:

\n\n

Option A: Butanol is capable of hydrogen bonding because it has an $\\mathrm{-OH}$ group. Hydrogen bonding is the strongest intermolecular force among the forces affecting these molecules. Therefore, butanol will have the highest boiling point among the compounds without considering the others.

\n\n

Option B: Butane only has van der Waals forces because it is a non-polar molecule. Van der Waals forces are the weakest intermolecular forces, resulting in a lower boiling point compared to molecules that can hydrogen bond or have permanent dipoles.

\n\n

Option C: Butyraldehyde can form dipole-dipole interactions because of its polar carbonyl ($\\mathrm{C=O}$) group, but it cannot form hydrogen bonds with itself since there is no $\\mathrm{OH}$ group. Dipole-dipole interactions are stronger than van der Waals forces but weaker than hydrogen bonds. As a result, butyraldehyde will have a higher boiling point than butane but a lower boiling point than butanol.

\n\n

Option D: Diethyl ether has an oxygen atom, which makes it polar and enables dipole-dipole interactions. However, ethers cannot form hydrogen bonds with themselves, so diethyl ether will have a higher boiling point than butane but lower than butanol.

\n\n

Considering the intermolecular forces, the compound with the highest boiling point is the one that can form hydrogen bonds, which in this list is the primary alcohol, butanol.

\n\n

Therefore, the compound with the highest boiling point is:

\n

Option A : Butanol ($\\mathrm{CH}_3 \\mathrm{CH}_2 \\mathrm{CH}_2 \\mathrm{CH}_2-\\mathrm{OH}$).

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 89, "subject": "Chemistry", "question": "

The molecular formula of second homologue in the homologous series of mono carboxylic acids is

", "options": [ { "text": "$$\\mathrm{C}_2 \\mathrm{H}_4 \\mathrm{O}_2$$\n" }, { "text": "$$\\mathrm{C}_2 \\mathrm{H}_2 \\mathrm{O}_2$$\n" }, { "text": "$$\\mathrm{CH}_2 \\mathrm{O}$$\n" }, { "text": "$$\\mathrm{C}_3 \\mathrm{H}_6 \\mathrm{O}_2$$" } ], "answer": "$$\\mathrm{C}_2 \\mathrm{H}_4 \\mathrm{O}_2$$\n", "solution": "**Answer:** $$\\mathrm{C}_2 \\mathrm{H}_4 \\mathrm{O}_2$$\n\n\n

\"JEE

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 90, "subject": "Chemistry", "question": "

The total number of 'Sigma' and 'Pi' bonds in 2-formylhex-4-enoic acid is _________.

", "options": [], "answer": "22", "solution": "**Answer:** 22\n\n

\"JEE

\n

Sigma bonds : 19\n

pi bonds : 3

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 91, "subject": "Chemistry", "question": "The correct statement about the synthesis of erythritol (C(CH2OH)4) used in the preparation of PETN is : ", "options": [ { "text": "The synthesis requires four aldol condensations between methanol\nand ethanol." }, { "text": "The synthesis requires two aldol condensations and two Cannizzaro\nreactions." }, { "text": "The synthesis requires three aldol condensations and one Cannizzaro\nreaction." }, { "text": "Alpha hydrogens of ethanol and methanol are involved in this\nreaction." } ], "answer": "The synthesis requires three aldol condensations and one Cannizzaro\nreaction.", "solution": "**Answer:** The synthesis requires three aldol condensations and one Cannizzaro\nreaction.\n\n

The synthesis of erythritol from aldehyde requires three aldol condensation and one Cannizzaro reaction.

\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 92, "subject": "Chemistry", "question": "

In the Claisen-Schmidt reaction to prepare $$351 \\mathrm{~g}$$ of dibenzalacetone using $$87 \\mathrm{~g}$$ of acetone, the amount of benzaldehyde required is _________ g. (Nearest integer)

", "options": [], "answer": "318", "solution": "**Answer:** 318\n\n

\"JEE

\n

$$\\begin{aligned}\n& \\text { mol of benzaldehyde required }=1.5 \\times 2 \\\\\n&=3 \\mathrm{~mol} \\\\\n& \\text { mass }=318 \\mathrm{~g}\n\\end{aligned}$$

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 93, "subject": "Chemistry", "question": "The correct match between Item I and Item II is :\n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
Item - IItem - II
(A)Benzaldehyde(P)Mobile phase
(B)Alumina(Q)Adsorbent
(C)Acetonitrile(R)Adsorbate
", "options": [ { "text": "(A) $$ \\to $$ (Q); (B) $$ \\to $$ (P); (C) $$ \\to $$ (R)" }, { "text": "(A) $$ \\to $$ (R); (B) $$ \\to $$ (Q); (C) $$ \\to $$ (P)" }, { "text": "(A) $$ \\to $$ (Q); (B) $$ \\to $$ (R); (C) $$ \\to $$ (P)" }, { "text": "(A) $$ \\to $$ (P); (B) $$ \\to $$ (R); (C) $$ \\to $$ (Q)" } ], "answer": "(A) $$ \\to $$ (R); (B) $$ \\to $$ (Q); (C) $$ \\to $$ (P)", "solution": "**Answer:** (A) $$ \\to $$ (R); (B) $$ \\to $$ (Q); (C) $$ \\to $$ (P)\n\n

To determine the correct match between Item I and Item II, let's analyze each item and its corresponding match :

\n\n

1. Benzaldehyde: Benzaldehyde is an aromatic aldehyde, which is typically considered as an adsorbate in chromatography. An adsorbate is a substance that is adsorbed on a surface.

\n\n

2. Alumina: Alumina (Aluminum oxide) is well known for its use as an adsorbent in chromatography processes. An adsorbent is a material that adsorbs another substance.

\n\n

3. Acetonitrile: Acetonitrile is an organic solvent often used as a mobile phase in chromatography. The mobile phase is the phase that moves in chromatography, helping to carry the sample through the stationary phase.

\n\n

Now, let's match the items :

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
Item - IItem - II
(A)Benzaldehyde(R)Adsorbate
(B)Alumina(Q)Adsorbent
(C)Acetonitrile(P)Mobile phase
\n\n

Therefore, the correct match is :

\n\n

(A) $$\\to$$ (R); (B) $$\\to$$ (Q); (C) $$\\to$$ (P)

\n\n

This corresponds to Option B.

", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 94, "subject": "Chemistry", "question": "2, 4-DNP test can be used to identify :", "options": [ { "text": "ether" }, { "text": "amine" }, { "text": "aldehyde" }, { "text": "halogens" } ], "answer": "aldehyde", "solution": "**Answer:** aldehyde\n\n2,4 DNP test is used to identify\n– (CO) – group. It\ngives addition reaction with carbonyl\ncompounds. So, it can be used to identify\naldehyde in the given option. It gives yellow/\norange ppt with carbonyl containing\ncompounds.\n
\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 95, "subject": "Chemistry", "question": "In Tollen's test for aldehyde, the overall number of electron(s) transferred to the Tollen's reagent formula [Ag(NH3)2]+ per aldehyde group to form silver mirror is ___________. (Round off to the Nearest Integer).", "options": [], "answer": "2", "solution": "**Answer:** 2\n\n\"JEE\n

Total 2e-\ntransfer to Tollen's reagent.", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 96, "subject": "Chemistry", "question": "

The spin only magnetic moment of the complex present in Fehling's reagent is __________ B.M. (Nearest integer).

", "options": [], "answer": "2", "solution": "**Answer:** 2\n\n In the complex present in Fehling's reagent, $\\mathrm{Cu}^{+2}$ ion is present.\n

\nSo, spin only magnetic moment\n

\n$$\n=\\sqrt{1(1+2)}\n$$\n

\n$$\n\\begin{aligned}\n& =\\sqrt{3} \\simeq 2 \\mathrm{~B} . \\mathrm{M}\n\\end{aligned}\n$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 97, "subject": "Chemistry", "question": "

Number of isomeric compounds with molecular formula $$\\mathrm{C}_{9} \\mathrm{H}_{10} \\mathrm{O}$$ which (i) do not dissolve in $$\\mathrm{NaOH}$$ (ii) do not dissolve in $$\\mathrm{HCl}$$. (iii) do not give orange precipitate with 2,4-DNP (iv) on hydrogenation give identical compound with molecular formula $$\\mathrm{C}_{9} \\mathrm{H}_{12} \\mathrm{O}$$ is ____________.

", "options": [], "answer": "2", "solution": "**Answer:** 2\n\n

2 possibilities

\n

\"JEE

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 98, "subject": "Chemistry", "question": "

The mass of NH$$_3$$ produced when 131.8 kg of cyclohexanecarbaldehyde undergoes Tollen's test is ________ kg. (Nearest Integer)

\n

Molar Mass of

\n

C = 12g/mol

\n

N = 14g/mol

\n

O = 16g/mol

", "options": [], "answer": "60", "solution": "**Answer:** 60\n\n$$\n\\begin{aligned}\n& \\mathrm{RCHO}+2\\left[\\mathrm{Ag}\\left(\\mathrm{NH}_3\\right)_2\\right] \\mathrm{OH} \\rightarrow \\mathrm{RCOONH}_4+2 \\mathrm{Ag} +3 \\mathrm{NH}_3+\\mathrm{H}_2 \\mathrm{O} \\\\\\\\\n& \\text {moles }=\\frac{131.8 \\times 10^3}{112}=1.176 \\times 10^3 \\\\\\\\\n& 1 \\text { mole of aldehyde produces } 3 \\text { moles of } \\mathrm{NH}_3 \\\\\\\\\n& 1.176 \\times 10^3 \\mathrm{~mol} \\text { aldehyde will produce } \\\\\\\\\n& =3 \\times 1.176 \\times 10^3 \\text { moles of } \\mathrm{NH}_3 \\\\\\\\\n& \\text {Mass of } \\mathrm{NH}_3 \\text { produced }=3 \\times 1.176 \\times 10^3 \\times 17 \\\\\\\\\n& =59.97 \\times 10^3 \\mathrm{~g} \\\\\\\\\n& \\approx 60~ \\mathrm{kg}\n\\end{aligned}\n$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 99, "subject": "Chemistry", "question": "

From the compounds given below, number of compounds which give positive Fehling's test is _________.

\n

Benzaldehyde, Acetaldehyde, Acetone, Acetophenone, Methanal, 4nitrobenzaldehyde, cyclohexane carbaldehyde.

", "options": [], "answer": "3", "solution": "**Answer:** 3\n\n

Acetaldehyde $$(\\mathrm{CH}_3 \\mathrm{CHO})$$, Methanal $$(\\mathrm{HCHO})$$, and cyclohexane carbaldehyde \"JEE

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 100, "subject": "Chemistry", "question": "

Which of the following compounds will give silver mirror with ammoniacal silver nitrate?

\n

A. Formic acid

\n

B. Formaldehyde

\n

C. Benzaldehyde

\n

D. Acetone

\n

Choose the correct answer from the options given below :

", "options": [ { "text": "B and C only" }, { "text": "C and D only" }, { "text": "A, B and C only" }, { "text": "A only" } ], "answer": "A, B and C only", "solution": "**Answer:** A, B and C only\n\n

The reaction involved here is the Tollen's test, which detects aldehydes and alpha-hydroxy ketones, as they are able to reduce the ammoniacal silver nitrate solution to metallic silver, producing a mirror-like coating on the glassware. This test specifically distinguishes these reducing sugars from non-reducing sugars and is often used to identify the presence of an aldehyde functional group.

\n\n

Let's analyze each option:

\n\n

A. Formic acid - Formic acid ($HCOOH$) is a carboxylic acid but importantly, it can also reduce silver ions to metallic silver because it can act as a reducing agent just like aldehydes. Thus, it will give a positive Tollen's test.

\n\n

B. Formaldehyde ($HCHO$) is the simplest aldehyde, and aldehydes are known to give positive results with the Tollen's test since they can be oxidized to carboxylic acids, reducing the silver ions in the process to metallic silver.

\n\n

C. Benzaldehyde ($C_6H_5CHO$) is an aromatic aldehyde and, similar to formaldehyde, it can be oxidized to an acid, thereby reducing the silver ions to silver in the Tollen's test, giving a positive result.

\n\n

D. Acetone ($CH_3COCH_3$) is a ketone. Generally, ketones do not undergo oxidation easily and do not reduce ammoniacal silver nitrate, and thus, do not give a silver mirror with the Tollen's reagent.

\n\n

Therefore, the compounds that will give a silver mirror with ammoniacal silver nitrate (positive Tollen's test) are Formic acid, Formaldehyde, and Benzaldehyde, but not Acetone.

\n\n

The correct answer from the options given is therefore:

\n\n

Option C: A, B, and C only.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 101, "subject": "Chemistry", "question": "Consider the acidity of the carboxylic acids:
\n(a) PhCOOH
\n(b) o – NO2C6H4COOH
\n(c) p – NO2C6H4COOH
\n(d) m – NO2C6H4COOH
\nWhich of the following order is correct?", "options": [ { "text": "(b) > (d) > (a) > (c)" }, { "text": "(b) > (d) > (c) > (a)" }, { "text": "(a) > (b) > (c) > (d)" }, { "text": "(b) > (c) > (d) > (a)" } ], "answer": "(b) > (c) > (d) > (a)", "solution": "**Answer:** (b) > (c) > (d) > (a)\n\nIn aromatic acids presence of electron withdrawing substituent e.g. $$ - N{O_2}$$ disperses the negative charge of the anion and stablises it and hence increases the acidity of the parent benzoic acid.\n

Further $$O$$- isomer will have higher acidity than corresponding $$m$$ and $$p$$ isomers. Since nitro group at $$p$$-position have more pronounced electron withdrawing than $$ - N{O_2}$$ group at $$m$$-position hence the correct order is the one given above. \n

\"AIEEE ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 102, "subject": "Chemistry", "question": "CH3Br + Nu- $$\\to$$ CH3 - Nu + Br- The decreasing order of the rate of the above reaction with nucleophiles (Nu–) A to D is
\n[Nu– = (A) PhO–, (B) AcO–, (C) HO–, (D) CH3O–]", "options": [ { "text": "A > B > C > D" }, { "text": "B > D > C > A" }, { "text": "D > C > A > B" }, { "text": "D > C > B > A" } ], "answer": "D > C > A > B", "solution": "**Answer:** D > C > A > B\n\nTIPS/Formulae : \n

The stronger the acid, the weaker the conjugate base formed. \n

The acid character follows the order : \n

$$C{H_3}COOH > {C_6}{H_5}OH > {H_2}O > C{H_3}OH$$\n

The basic character will follow the order \n

$$C{H_3}CO{O^ - } < {C_6}{H_5}{O^ - } < {O^ - }H < C{H_3}{O^ - }$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 103, "subject": "Chemistry", "question": "The correct order of increasing basicity of the given conjugate bases (R = CH3) is", "options": [ { "text": "$$RCO\\overline O$$ < HC = $$\\overline C$$ < $$\\overline R$$ < $$\\overline NH_2$$" }, { "text": "$$\\overline R$$ < HC = $$\\overline C$$ < $$RCO\\overline O$$ < $$\\overline NH_2$$" }, { "text": "$$RCO\\overline O$$ < $$\\overline NH_2$$ < HC = $$\\overline C$$ < $$\\overline R$$" }, { "text": "$$RCO\\overline O$$ < HC = $$\\overline C$$ < $$\\overline NH_2$$ < $$\\overline R$$" } ], "answer": "$$RCO\\overline O$$ < HC = $$\\overline C$$ < $$\\overline NH_2$$ < $$\\overline R$$", "solution": "**Answer:** $$RCO\\overline O$$ < HC = $$\\overline C$$ < $$\\overline NH_2$$ < $$\\overline R$$\n\nThe correct order of basicity is \n

\"AIEEE ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 104, "subject": "Chemistry", "question": "Which amongst the following is the strongest acid ? ", "options": [ { "text": "CHBr3" }, { "text": "CHI3 " }, { "text": "CH(CN)3" }, { "text": "CHCl3" } ], "answer": "CH(CN)3", "solution": "**Answer:** CH(CN)3\n\nAmong those $$-$$CN has highest $$-$$M and $$-$$I effect. So after losing proton (H+) anion become more stable.\n

So,   CH(CN)3 has the highest tendency to donate proton that is why it is most acidic. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 105, "subject": "Chemistry", "question": "The correct decreasing order for acid strength is : ", "options": [ { "text": "NO2CH2COOH > FCH2COOH > CNCH2COOH > ClCH2COOH" }, { "text": "FCH2COOH > NCCH2COOH > NO2CH2COOH > ClCH2COOH" }, { "text": "CNCH2COOH > O2NCH2COOH > FCH2COOH > ClCH2COOH" }, { "text": "NO2CH2COOH > NCCH2COOH > FCH2COOH > ClCH2COOH" } ], "answer": "NO2CH2COOH > NCCH2COOH > FCH2COOH > ClCH2COOH", "solution": "**Answer:** NO2CH2COOH > NCCH2COOH > FCH2COOH > ClCH2COOH\n\nAfter releasing proton (H+) the more stable anion is more acidic nature. The 4 anions are $$ \\to $$\n

\"JEE\n

Here $$-$$NO2, $$-$$ CN, $$-$$Cl, $$-$$F all of them have $$-$$ I effective and all have same distance from the anion ($$-$$ O$$-$$).\n

So depending on which one have more power $$-$$ I effect will be more stable. \n

Acidic nature $$ \\propto $$ $$-$$ I power.\n

$$-$$ I power order : \n

$$-$$ NO2 > $$-$$ C $$ \\equiv $$ N > $$-$$ F > $$-$$ Cl\n

So, acidic strength order $$ \\to $$\n

NO2 $$-$$ CH2 $$-$$ COOH > NC $$-$$ CH2 $$-$$ COOH\n
> F $$-$$ CH2 $$-$$ COOH > Cl $$-$$ CH2 $$-$$ COOH", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 106, "subject": "Chemistry", "question": "In the following compounds, the decreasing order of basic strength will be :", "options": [ { "text": "(C2H5)2NH > NH3 > C2H5NH2" }, { "text": "NH3 > C2H5NH2 > (C2H5)2NH" }, { "text": "(C2H5)2NH > C2H5NH2 > NH3" }, { "text": "C2H5NH2 > NH3 > (C2H5)2NH" } ], "answer": "(C2H5)2NH > C2H5NH2 > NH3", "solution": "**Answer:** (C2H5)2NH > C2H5NH2 > NH3\n\nBecause of +I effect of -C2H5 group, C2H5NH2 and (C2H5)2NH are more basic than NH3.\n

When with amine ethyl group (-C2H5) present then basic strength order :\n

2o amine > 3o amine > 1o amine\n

Note :\n

When with amine methyl group (-CH3) present then basic strength order :\n

2o amine > 1o amine > 3o amine", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 107, "subject": "Chemistry", "question": "Which of the following compounds will show\nthe maximum 'enol' content?", "options": [ { "text": "CH3COCH3" }, { "text": "CH3COCH2COCH3" }, { "text": "CH3COCH2COOC2H5" }, { "text": "CH3COCH2CONH2" } ], "answer": "CH3COCH2COCH3", "solution": "**Answer:** CH3COCH2COCH3\n\nHydrogen removed from the $$\\alpha $$ carbon should be acidic in nature because ehen hydrogen is more acidic then enol formation is easy.\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 108, "subject": "Chemistry", "question": "The increasing order of nucleophilicity of the following nucleophiles is :\n
(a) CH3CO2$$-$$           (b) H2O\n
(c) CH3SO3$$-$$           (d) $$\\mathop O\\limits^ - H$$", "options": [ { "text": "(b) < (c) < (a) < (d)" }, { "text": "(b) < (c) < (d) < (a)" }, { "text": "(a) < (d) < (c) < (b)" }, { "text": "(d) < (a) < (c) < (b)" } ], "answer": "(b) < (c) < (a) < (d)", "solution": "**Answer:** (b) < (c) < (a) < (d)\n\nWhen stability of nucleophile increases then nucleophilicity decreases.\n

Stability order :\n
H2O > CH3SO3$$-$$ > CH3CO2$$-$$ > $$\\mathop O\\limits^ - H$$\n

So Nucleophilicity order :\n
H2O < CH3SO3$$-$$ < CH3CO2$$-$$ < $$\\mathop O\\limits^ - H$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 109, "subject": "Chemistry", "question": "

The correct order in aqueous medium of basic strength in case of methyl substituted amines is :

", "options": [ { "text": "$$\\mathrm{NH_3 > Me_3N > MeNH_2 > Me_2NH}$$" }, { "text": "$$\\mathrm{Me_3N > Me_2NH > MeNH_2 > NH_3}$$" }, { "text": "$$\\mathrm{Me_2NH > MeNH_2 > Me_3N > NH_3}$$" }, { "text": "$$\\mathrm{Me_2NH > Me_3N > MeNH_2 > NH_3}$$" } ], "answer": "$$\\mathrm{Me_2NH > MeNH_2 > Me_3N > NH_3}$$", "solution": "**Answer:** $$\\mathrm{Me_2NH > MeNH_2 > Me_3N > NH_3}$$\n\nIn aqueous medium basic strength is dependent on electron density on nitrogen as well as solvation of cation formed after accepting $\\mathrm{H}^{+}$. After considering all these factors overall basic strength order is

\n$$\n\\mathrm{Me}_2 \\mathrm{NH}>\\mathrm{MeNH}_2>\\mathrm{Me}_3 \\mathrm{N}>\\mathrm{NH}_3\n$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 110, "subject": "Chemistry", "question": "

The descending order of acidity for the following carboxylic acid is-

\n

A. $$\\mathrm{CH}_{3} \\mathrm{COOH}$$

\n

B. $$\\mathrm{F}_{3} \\mathrm{C}-\\mathrm{COOH}$$

\n

C. $$\\mathrm{ClCH}_{2}-\\mathrm{COOH}$$

\n

D. $$\\mathrm{FCH}_{2}-\\mathrm{COOH}$$

\n

E. $$\\mathrm{BrCH}_{2}-\\mathrm{COOH}$$

\n

Choose the correct answer from the options given below:

", "options": [ { "text": "B > C > D > E > A" }, { "text": "D > B > A > E > C" }, { "text": "B > D > C > E > A" }, { "text": "E > D > B > A > C" } ], "answer": "B > D > C > E > A", "solution": "**Answer:** B > D > C > E > A\n\n

The acidity of carboxylic acids is influenced by the electron-withdrawing ability of the groups attached to the carboxylic acid. The stronger the electron-withdrawing ability of the group, the more it stabilizes the carboxylate anion (formed after losing the acidic proton), and hence the stronger the acid.

\n

Let's look at the groups attached to the carboxylic acid in each option:

\n

A. $$\\mathrm{CH}_{3}\\mathrm{-}$$: Methyl group. It's electron-donating due to the hyperconjugation effect.

\n

B. $$\\mathrm{F}_{3}\\mathrm{C}-$$: Trifluoromethyl group. Strong electron-withdrawing group due to the electronegativity of fluorine atoms and inductive effect.

\n

C. $$\\mathrm{ClCH}_{2}-$$: Chloromethyl group. It has some electron-withdrawing ability due to the electronegativity of the chlorine atom, but it's less than that of fluorine.

\n

D. $$\\mathrm{FCH}_{2}-$$: Fluoromethyl group. It has electron-withdrawing ability due to the electronegativity of the fluorine atom, but less than that of the trifluoromethyl group.

\n

E. $$\\mathrm{BrCH}_{2}-$$: Bromomethyl group. It has some electron-withdrawing ability due to the electronegativity of the bromine atom, but it's less than that of chlorine and fluorine.

\n

So the order of acidity will be: B > D > C > E > A

\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 111, "subject": "Chemistry", "question": "

What will be the decreasing order of basic strength of the following conjugate bases? $${ }^{-} \\mathrm{OH}, \\mathrm{R} \\overline{\\mathrm{O}}, \\mathrm{CH}_3 \\mathrm{CO} \\overline{\\mathrm{O}}, \\mathrm{Cl}$$\n\n

", "options": [ { "text": "$$\\mathrm{C} \\overline{\\mathrm{I}}>\\mathrm{R} \\overline{\\mathrm{O}}>{ }^{-} \\mathrm{OH}>\\mathrm{CH}_3 \\mathrm{CO} \\overline{\\mathrm{O}}$$\n" }, { "text": "$${ }^{-} \\mathrm{OH}>\\mathrm{R} \\overline{\\mathrm{O}}>\\mathrm{CH}_3 \\mathrm{CO} \\overline{\\mathrm{O}}>\\mathrm{C} \\overline{1}$$\n\n" }, { "text": "$$\\mathrm{C} \\overline{\\mathrm{I}}>{ }^{-} \\mathrm{OH}>\\mathrm{R} \\overline{\\mathrm{O}}>\\mathrm{CH}_3 \\mathrm{CO} \\bar{O}$$\n" }, { "text": "$$\\mathrm{R} \\overline{\\mathrm{O}}>{ }^{-} \\mathrm{OH}>\\mathrm{CH}_3 \\mathrm{CO} \\overline{\\mathrm{O}}>\\mathrm{C} \\overline{1}$$" } ], "answer": "$$\\mathrm{R} \\overline{\\mathrm{O}}>{ }^{-} \\mathrm{OH}>\\mathrm{CH}_3 \\mathrm{CO} \\overline{\\mathrm{O}}>\\mathrm{C} \\overline{1}$$", "solution": "**Answer:** $$\\mathrm{R} \\overline{\\mathrm{O}}>{ }^{-} \\mathrm{OH}>\\mathrm{CH}_3 \\mathrm{CO} \\overline{\\mathrm{O}}>\\mathrm{C} \\overline{1}$$\n\n

To determine the decreasing order of basic strength of the given conjugate bases, we need to understand the concept of basicity. A base is defined by its ability to accept protons (H+). The stronger the conjugate base, the more readily it accepts protons. The basicity of conjugate bases is also influenced by the stability of the anion (conjugate base); the more stable the anion, the weaker the base, and vice versa.

\n\n

Let's analyze each conjugate base:

\n\n

1. $${ }^{-} \\mathrm{OH}$$ (Hydroxide ion): This is a strong base because it is less stabilized by resonance or inductive effects.

\n\n

2. $$\\mathrm{R} \\overline{\\mathrm{O}}$$ (Alkoxide ion): This also is a strong base. However, alkyl groups (R) are electron donating, which can increase the electron density on oxygen, making it a stronger base compared to hydroxide ion.

\n\n

3. $$\\mathrm{CH}_3 \\mathrm{CO} \\overline{\\mathrm{O}}$$ (Acetate ion): This is a much weaker base because the negative charge is delocalized over the two oxygen atoms through resonance, which stabilizes the ion.

\n\n

4. $$\\mathrm{Cl}$$ (Chloride ion): Chloride ion is very weakly basic since it is highly stable due to its high electronegativity and full octet.

\n\n

Based on these observations, the decreasing order of basic strength should be:

\n\n

\n\n

$$\\mathrm{R} \\overline{\\mathrm{O}}>^{-} \\mathrm{OH}>\\mathrm{CH}_3 \\mathrm{CO} \\overline{\\mathrm{O}}>\\mathrm{Cl}$$

\n\n

\n\n

Therefore, the correct choice is:

\n\n

Option D

\n\n

\n\n

$$\\mathrm{R} \\overline{\\mathrm{O}}>{ }^{-} \\mathrm{OH}>\\mathrm{CH}_3 \\mathrm{CO} \\overline{\\mathrm{O}}>\\mathrm{Cl}$$

\n\n

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 112, "subject": "Chemistry", "question": "

The correct sequence of acidic strength of the following aliphatic acids in their decreasing order is:

\n

$$\\mathrm{CH}_3 \\mathrm{CH}_2 \\mathrm{COOH}, \\mathrm{CH}_3 \\mathrm{COOH}, \\mathrm{CH}_3 \\mathrm{CH}_2 \\mathrm{CH}_2 \\mathrm{COOH}, \\mathrm{HCOOH}$$

", "options": [ { "text": "$$\\mathrm{HCOOH}>\\mathrm{CH}_3 \\mathrm{CH}_2 \\mathrm{CH}_2 \\mathrm{COOH}>\\mathrm{CH}_3 \\mathrm{CH}_2 \\mathrm{COOH}>\\mathrm{CH}_3 \\mathrm{COOH}$$" }, { "text": "$$\\mathrm{HCOOH}>\\mathrm{CH}_3 \\mathrm{COOH}>\\mathrm{CH}_3 \\mathrm{CH}_2 \\mathrm{COOH}>\\mathrm{CH}_3 \\mathrm{CH}_2 \\mathrm{CH}_2 \\mathrm{COOH}$$" }, { "text": "$$\\mathrm{CH}_3 \\mathrm{CH}_2 \\mathrm{CH}_2 \\mathrm{COOH}>\\mathrm{CH}_3 \\mathrm{CH}_2 \\mathrm{COOH}>\\mathrm{CH}_3 \\mathrm{COOH}>\\mathrm{HCOOH}$$" }, { "text": "$$\\mathrm{CH}_3 \\mathrm{COOH}>\\mathrm{CH}_3 \\mathrm{CH}_2 \\mathrm{COOH}>\\mathrm{CH}_3 \\mathrm{CH}_2 \\mathrm{CH}_2 \\mathrm{COOH}>\\mathrm{HCOOH}$$" } ], "answer": "$$\\mathrm{HCOOH}>\\mathrm{CH}_3 \\mathrm{COOH}>\\mathrm{CH}_3 \\mathrm{CH}_2 \\mathrm{COOH}>\\mathrm{CH}_3 \\mathrm{CH}_2 \\mathrm{CH}_2 \\mathrm{COOH}$$", "solution": "**Answer:** $$\\mathrm{HCOOH}>\\mathrm{CH}_3 \\mathrm{COOH}>\\mathrm{CH}_3 \\mathrm{CH}_2 \\mathrm{COOH}>\\mathrm{CH}_3 \\mathrm{CH}_2 \\mathrm{CH}_2 \\mathrm{COOH}$$\n\n

$$\\begin{aligned}\n& \\mathrm{HCOOH}>\\mathrm{CH}_3 \\mathrm{COOH}>\\mathrm{CH}_3 \\mathrm{CH}_2 \\mathrm{COOH}> \\\\\n& \\mathrm{CH}_3 \\mathrm{CH}_2 \\mathrm{CH}_2 \\mathrm{COOH}\n\\end{aligned}$$

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 113, "subject": "Chemistry", "question": "

Given below are two statements :

\n

Statement I : Tropolone is an aromatic compound and has $$8 \\pi$$ electrons.

\n

Statement II : $$\\pi$$ electrons of $$ > \\mathrm{C}=\\mathrm{O}$$ group in tropolone is involved in aromaticity.

\n

In the light of the above statements, choose the correct answer from the options given below:

", "options": [ { "text": "Statement I is false but Statement II is true" }, { "text": "Both Statement I and Statement II are false" }, { "text": "Statement I is true but Statement II is false" }, { "text": "Both Statement I and Statement II are true" } ], "answer": "Both Statement I and Statement II are false", "solution": "**Answer:** Both Statement I and Statement II are false\n\n

Let's analyze the structure of tropolone to evaluate the given statements.

\n\n

1. Statement I: \"Tropolone is an aromatic compound and has $$8 \\pi$$ electrons.\"

\n\n

Tropolone is a seven-membered ring compound with an oxygen atom attached to it through a double bond, creating a carbonyl group (C=O). To determine if it is aromatic, we should check if it meets Hückel's rule for aromaticity, which states that a molecule is aromatic if it is cyclic, planar, and has a $$4n + 2$$ $\\pi$ electrons (where n is a non-negative integer).

\n\n

Upon examining the structure of tropolone:

\n\n\n\n\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n
ElectronsCount
Electrons from double bonds in the ring6
Electrons from the carbonyl $$\\pi$$ bond2
Total $$\\pi$$ electrons8
\n\n

According to the table, tropolone has a total of 8 $$\\pi$$ electrons. Since 8 does not satisfy the $$4n + 2$$ rule (where n would be 1 or 2, leading to 6 or 10 $$\\pi$$ electrons), tropolone does not qualify as aromatic. Therefore, Statement I is false.

\n\n

2. Statement II: \"$$\\pi$$ electrons of $$ > \\mathrm{C}=\\mathrm{O}$$ group in tropolone is involved in aromaticity.\"

\n\n

As established, tropolone does not meet the criteria for aromaticity. Additionally, the $$\\pi$$ electrons of the carbonyl group indeed contribute to the total count of $$\\pi$$ electrons in the ring system, but since tropolone isn't aromatic, Statement II is false.

\n\n

Given the analysis, the correct answer is:

\n\n

Option B: Both Statement I and Statement II are false

", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 114, "subject": "Chemistry", "question": "Arrangement of (CH3)3C-, (CH3)2CH-, CH3-CH2- when attached to benzyl or an unsaturated group in increasing order of inductive effect is", "options": [ { "text": "(CH3)3C- < (CH3)2CH- < CH3-CH2" }, { "text": "CH3-CH2- < (CH3)2CH- < (CH3)3C-" }, { "text": "(CH3)2CH- < (CH3)3C- < CH3-CH2" }, { "text": "(CH3)3C- < CH3-CH2- < CH3)2CH-" } ], "answer": "CH3-CH2- < (CH3)2CH- < (CH3)3C-", "solution": "**Answer:** CH3-CH2- < (CH3)2CH- < (CH3)3C-\n\n$$ - C{H_3}$$ group has $$+I$$ effect, as number of $$ - C{H_3}$$ group increases, the inductive effect increases. \n

Therefore the correct order is \n

$$C{H_3} - C{H_2} - < {\\left( {C{H_3}} \\right)_2}CH - < {\\left( {C{H_3}} \\right)_3}C - $$ \n", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 115, "subject": "Chemistry", "question": "In the anion HCOO$$-$$\n the two carbon-oxygen bonds are found to be of equal length. What is the reason for it?", "options": [ { "text": "The C = O bond is weaker than the C-O bond" }, { "text": "The anion HCOO-\n has two resonating structures" }, { "text": "The anion is obtained by removal of a proton from the acid molecule" }, { "text": "Electronic orbitals of carbon atom are hybridised" } ], "answer": "The anion HCOO-\n has two resonating structures", "solution": "**Answer:** The anion HCOO-\n has two resonating structures\n\nThe anion is \n

\"AIEEE\n

$$HCO{O^ - }$$ exists in following resonating structures \n

\"AIEEE \n

And actual structure will be, \n

\"AIEEE\n

Bond order of this anion = $${3 \\over 2}$$ = 1.5.\n

As 1.5 is exactly in the middle of 1 and 2 so, the bond length will be equal between all carbon and oxygen atom. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 116, "subject": "Chemistry", "question": "Presence of a nitro group in a benzene ring", "options": [ { "text": "activates the ring towards electrophilic substitution " }, { "text": "renders the ring basic " }, { "text": "deactivates the ring towards nucleophilic substitution " }, { "text": "deactivates the ring towards electrophilic substitution " } ], "answer": "deactivates the ring towards electrophilic substitution ", "solution": "**Answer:** deactivates the ring towards electrophilic substitution \n\nNitro group is electron withdrawing group, so it deactivities the ring towards electrophilic substitution.", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 117, "subject": "Chemistry", "question": "Which of the following has the shortest C-Cl\nbond?", "options": [ { "text": "Cl–CH=CH–CH3" }, { "text": "Cl–CH=CH–NO2" }, { "text": "Cl–CH=CH–OCH3" }, { "text": "Cl–CH=CH2" } ], "answer": "Cl–CH=CH–NO2", "solution": "**Answer:** Cl–CH=CH–NO2\n\n\"JEE\ndue to strong – R effect of – NO2 group and M effect of Cl, partial\ndouble bond character between C and Cl\nincreases. So\nshortest bond length.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 118, "subject": "Chemistry", "question": "Given below are two statements :

Statement I : Hyperconjugation is a permanent effect.

Statement II : Hyperconjugation in ethyl cation $$\\left( {C{H_3} - \\mathop C\\limits^ + {H_2}} \\right)$$ involves the overlapping of $${C_{s{p^2}}} - {H_{1s}}$$ bond with empty 2p orbital of other carbon.

Choose the correct option :", "options": [ { "text": "Both statement I and statement II are false" }, { "text": "Statement I is incorrect but statement II is true" }, { "text": "Statement I is correct but statement II is false" }, { "text": "Both statement I and statement II are true." } ], "answer": "Statement I is correct but statement II is false", "solution": "**Answer:** Statement I is correct but statement II is false\n\nStatement I : It is correct statement

Statement II : $$C{H_3} - \\mathop {C{H_2}}\\limits^ \\oplus $$ involve $${C_{s{p^3}}} - {H_{1s}}$$ bond with empty 2p orbital hence given statement is false. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 119, "subject": "Chemistry", "question": "

Arrange the following in increasing order of reactivity towards nitration

\n

A. p-xylene

\n

B. bromobenzene

\n

C. mesitylene

\n

D. nitrobenzene

\n

E. benzene

\n

Choose the correct answer from the options given below

", "options": [ { "text": "C < D < E < A < B" }, { "text": "D < B < E < A < C" }, { "text": "D < C < E < A < B" }, { "text": "C < D < E < B < A" } ], "answer": "D < B < E < A < C", "solution": "**Answer:** D < B < E < A < C\n\nThe correct order of reactivity towards nitration is

\n\"JEE\n
as electron releasing groups on benzene ring facilitate the nitration at benzene ring.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 120, "subject": "Chemistry", "question": "

Given below are two statements, one is labelled as Assertion A and the other is labelled as Reason R

\n

Assertion A : Benzene is more stable than hypothetical cyclohexatriene

\n

Reason R : The delocalised $$\\pi$$ electron cloud is attracted more strongly by nuclei of carbon atoms.

\n

In the light of the above statements, choose the correct answer from the options given below:

", "options": [ { "text": "A is true but R is false" }, { "text": "Both A and R are correct and R is the correct explanation of A" }, { "text": "Both A and R are correct but R is NOT the correct explanation of A" }, { "text": "A is false but R is true" } ], "answer": "Both A and R are correct and R is the correct explanation of A", "solution": "**Answer:** Both A and R are correct and R is the correct explanation of A\n\n

Benzene is more stable than hypothetical cyclohexatriene due to the delocalization of electrons in the π electron cloud of the aromatic ring. This delocalization of electrons leads to a more stable electronic configuration in benzene, as the electrons are shared more evenly between all six carbon atoms in the ring. This is in agreement with Assertion A.\n

\n

Reason R is also correct, as the delocalized π electron cloud in the aromatic ring is attracted more strongly by the nuclei of the carbon atoms. This is due to the fact that the π electrons in the ring are more spread out and are not localized to a single atom, which makes them more attracted to the positively charged nuclei of the carbon atoms. This explanation supports Assertion A.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 121, "subject": "Chemistry", "question": "The set of meta directing functional groups from the following sets is :", "options": [ { "text": "$-\\mathrm{CN},-\\mathrm{NH}_2,-\\mathrm{NHR},-\\mathrm{OCH}_3$" }, { "text": "$-\\mathrm{CN},-\\mathrm{CHO},-\\mathrm{NHCOCH}_3,-\\mathrm{COOR}$" }, { "text": "$-\\mathrm{NO}_2,-\\mathrm{NH}_2,-\\mathrm{COOH},-\\mathrm{COOR} $" }, { "text": "$-\\mathrm{NO}_2,-\\mathrm{CHO},-\\mathrm{SO}_3 \\mathrm{H},-\\mathrm{COR}$" } ], "answer": "$-\\mathrm{NO}_2,-\\mathrm{CHO},-\\mathrm{SO}_3 \\mathrm{H},-\\mathrm{COR}$", "solution": "**Answer:** $-\\mathrm{NO}_2,-\\mathrm{CHO},-\\mathrm{SO}_3 \\mathrm{H},-\\mathrm{COR}$\n\nThe term meta-directing refers to substituents on a benzene ring that direct incoming electrophiles to the meta position (relative to themselves) during an electrophilic aromatic substitution reaction.\n\n

Meta directors are typically deactivating groups with respect to the aromatic ring and they often posses an atom with a partial positive charge or a full formal positive charge directly attached to the ring, which tends to withdraw electron density from the ring through the inductive effect or resonance. This decreased electron density in the ring makes the ortho and para positions less reactive, thereby directing new substituents to the meta position. Furthermore, the presence of these withdrawing groups can stabilize the intermediate cation (whichever is formed) better when the substituent is at the meta position relative to the ortho or para positions.

\n\nLet's examine the options:\n\n

Option A (Incorrect):\n

\n

\n\n

Option B (Possibly correct):\n

\n

\n\n

Option C (Incorrect):\n

\n

\n\n

Option D (Correct):\n

\n\n

Based on the description above, the set of functional groups that are meta directing from the given options is Option D: $-\\mathrm{NO}_2$, $-\\mathrm{CHO}$, $-\\mathrm{SO}_3 \\mathrm{H}$, $-\\mathrm{COR}$. These groups are all electron-withdrawing and would direct incoming substituents to the meta position during electrophilic aromatic substitution.

", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 122, "subject": "Chemistry", "question": "The functional group that shows negative resonance effect is :", "options": [ { "text": "$-\\mathrm{OH}$" }, { "text": "$-\\mathrm{OR}$" }, { "text": "$-\\mathrm{COOH}$" }, { "text": "$-\\mathrm{NH}_2$" } ], "answer": "$-\\mathrm{COOH}$", "solution": "**Answer:** $-\\mathrm{COOH}$\n\n

The negative resonance effect, also called the -I effect, refers to the ability of a functional group to withdraw electron density through the p-orbitals via resonance. It is often the characteristic of groups that have atoms with high electronegativity or multiple bonds that allow for electron delocalization away from the conjugated system.

\n\n

To determine which functional group shows a negative resonance effect from the options provided, let's review them one by one:

\n\n\n\n

Given the choices, Option C - $-\\mathrm{COOH}$ is the functional group that shows a negative resonance effect because of its ability to withdraw electron density through resonance with its carbonyl group which is highly electronegative and capable of delocalizing electrons.

", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 123, "subject": "Chemistry", "question": "Ionic reactions with organic compounds proceed through :

\n(A) homolytic bond cleavage

\n(B) heterolytic bond cleavage

\n(C) free radical formation

\n(D) primary free radical

\n(E) secondary free radical\n

\nChoose the correct answer from the options given below :\n", "options": [ { "text": "(A) only" }, { "text": "(B) only" }, { "text": "(C) only" }, { "text": "(D) and (E) only" } ], "answer": "(B) only", "solution": "**Answer:** (B) only\n\n

The correct answer is Option B: (B) only.

\n\n

In organic chemistry, ionic reactions typically involve heterolytic bond cleavage. This is when a bond between two atoms breaks and both electrons of the shared pair go to one of the atoms, creating ions. One atom becomes a positively charged ion (cation), as it loses an electron, and the other becomes a negatively charged ion (anion), as it gains an electron.

\n\n

In contrast, homolytic bond cleavage occurs when each atom gets one electron from the shared pair, leading to the formation of two free radicals, which are highly reactive species with unpaired electrons. However, this is not typically how ionic reactions proceed in organic compounds.

\n\n

The options (C), (D), and (E) mention free radicals. Since homolytic cleavage leads to free radicals, and we've established that ionic reactions in organic chemistry generally proceed through heterolytic cleavage, free radicals are not typically involved in these reactions. Therefore, these options are not correct in the context of ionic reactions with organic compounds.

\n\n

Thus, the process of heterolytic bond cleavage (B) is the mechanism through which ionic reactions usually occur in organic chemistry.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 124, "subject": "Chemistry", "question": "

Among the following, total number of meta directing functional groups is (Integer based)

\n

$$-\\mathrm{OCH}_3,-\\mathrm{NO}_2,-\\mathrm{CN},-\\mathrm{CH}_3-\\mathrm{NHCOCH}_3, -\\mathrm{COR},-\\mathrm{OH},-\\mathrm{COOH},-\\mathrm{Cl}$$

", "options": [], "answer": "4", "solution": "**Answer:** 4\n\n

Here's how to identify the meta-directing functional groups and determine the total :

\n\nMeta-Directing Groups:\n\n

\n
Identifying Meta-Directing Groups in the List:\n\n

\n

Total Count:\n\n

There are four meta-directing groups in the list: -NO₂, -CN, -COR, -COOH

\n\nAnswer:\n\n

The total number of meta-directing functional groups is 4.

\n\n

", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 125, "subject": "Chemistry", "question": "

The interaction between $$\\pi$$ bond and lone pair of electrons present on an adjacent atom is responsible for

", "options": [ { "text": "Resonance effect\n" }, { "text": "Electromeric effect\n" }, { "text": "Hyperconjugation" }, { "text": "Inductive effect" } ], "answer": "Resonance effect\n", "solution": "**Answer:** Resonance effect\n\n\n

It is a type of conjugation responsible for resonance.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 126, "subject": "Chemistry", "question": "

Which among the following is incorrect statement?

", "options": [ { "text": "The electromeric effect is, temporary effect\n" }, { "text": "Hydrogen ion $$(\\mathrm{H}^{+})$$ shows negative electromeric effect\n" }, { "text": "The organic compound shows electromeric effect in the presence of the reagent only.\n" }, { "text": "Electromeric effect dominates over inductive effect" } ], "answer": "Hydrogen ion $$(\\mathrm{H}^{+})$$ shows negative electromeric effect\n", "solution": "**Answer:** Hydrogen ion $$(\\mathrm{H}^{+})$$ shows negative electromeric effect\n\n\n

Let's analyze each statement to determine which is incorrect:

\n\n

Option A: The electromeric effect is a temporary effect.

\n\n

This statement is correct. The electromeric effect is indeed a temporary effect that occurs in the presence of a reagent and disappears when the reagent is removed.

\n\n

Option B: Hydrogen ion $$(\\mathrm{H}^{+})$$ shows negative electromeric effect.

\n\n

To understand this, let's define the electromeric effect. The electromeric effect involves the transfer of electron pairs within a molecule in the presence of an attacking reagent. There are two types of electromeric effects: the positive electromeric effect ($$E^+$$) and the negative electromeric effect ($$E^-$$). A positive electromeric effect occurs when electrons are transferred towards the reagent, while a negative electromeric effect occurs when electrons are transferred away from the reagent.

\n\n

Since a hydrogen ion $$(\\mathrm{H}^{+})$$ is an electron-deficient species (it lacks electrons), it would generally induce the transfer of electrons towards itself, resulting in a positive electromeric effect. Hence, this statement is incorrect.

\n\n

Option C: The organic compound shows electromeric effect in the presence of the reagent only.

\n\n

This statement is correct. The electromeric effect is observed only in the presence of an attacking reagent. Once the reagent is removed, the effect ceases to exist.

\n\n

Option D: Electromeric effect dominates over inductive effect.

\n\n

This statement is correct. The electromeric effect generally has a stronger influence on the electronic distribution within a molecule compared to the inductive effect, especially when both effects are present simultaneously.

\n\n

Therefore, the incorrect statement is:

\n\n

Option B: Hydrogen ion $$(\\mathrm{H}^{+})$$ shows negative electromeric effect.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 127, "subject": "Chemistry", "question": "A similarity between optical and geometrical isomerism is that", "options": [ { "text": "each forms equal number of isomers for a given compound" }, { "text": "if in a compound one is present then so is the other" }, { "text": "both are included in stereoisomerism" }, { "text": "they have no similarity" } ], "answer": "both are included in stereoisomerism", "solution": "**Answer:** both are included in stereoisomerism\n\nStereoisomerism involve those isomers which contain same ligands in their co-ordination spheres but differ in the arrangement of those ligands in space. Stereo-isomerism is of two type geometrical isomerism ligands occupy different positions around the central metal atom or ion. \n

NOTE : In optical isomerism isomers have same formula but differ in their ability to rotate directions of the plane of polarized light. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 128, "subject": "Chemistry", "question": "Racemic mixture is formed by mixing two", "options": [ { "text": "isomeric compounds" }, { "text": "enantiomers with chiral compounds" }, { "text": "meso compounds" }, { "text": "optical isomers" } ], "answer": "enantiomers with chiral compounds", "solution": "**Answer:** enantiomers with chiral compounds\n\nA mixture of equal amount of two enantiomers is called a racemic mixture. A racemic mixture does not rotate plane polarized light. They are optically inactive because for every molecule in a racemic mixture that rotate plane of polarized light in one direction, there is a mirror image molecule that rotates the plane in opposite direction. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 129, "subject": "Chemistry", "question": "Which of the following does not show geometrical isomerism?", "options": [ { "text": "1,2-dichloro-1-pentene" }, { "text": "1,3-dichloro-2-pentene" }, { "text": "1,1-dichloro-1-pentene" }, { "text": "1,4-dichloro-2-pentene" } ], "answer": "1,1-dichloro-1-pentene", "solution": "**Answer:** 1,1-dichloro-1-pentene\n\n\"AIEEE \n

does not show geometrical isomerism due to presence of two similar $$Cl$$ atoms on the same $$C$$-atom. Geometrical isomerism is shown by compounds in which the groups/atoms attached to $$C=C$$ are different. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 130, "subject": "Chemistry", "question": "The general formula CnH2nO2 could be for open chain", "options": [ { "text": "carboxylic acids" }, { "text": "diols" }, { "text": "dialdehydes" }, { "text": "deketones" } ], "answer": "carboxylic acids", "solution": "**Answer:** carboxylic acids\n\n$${C_n}{H_{2n}}{O_2}$$ is general formula for carboxylic acid ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 131, "subject": "Chemistry", "question": "Which of the following compound is not chiral? ", "options": [ { "text": "1- chloropentane" }, { "text": "3-chloro-2- methyl pentane " }, { "text": "1-chloro -2- methyl pentane " }, { "text": "2- chloropentane " } ], "answer": "1- chloropentane", "solution": "**Answer:** 1- chloropentane\n\n$$1$$ - chloropentane is not chiral while others are chiral in nature\n

\"AIEEE ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 132, "subject": "Chemistry", "question": "Which of the following will have meso-isomer also? ", "options": [ { "text": "2- chlorobutane" }, { "text": "2- hydroxyopanoic acid" }, { "text": "2,3 – dichloropentane" }, { "text": "2-3- dichlorobutane" } ], "answer": "2-3- dichlorobutane", "solution": "**Answer:** 2-3- dichlorobutane\n\nNOTE : The compounds containing two similar assymmetric $$C$$-atoms have plane of symmetry and exist in Meso form. \n

\"AIEEE ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 133, "subject": "Chemistry", "question": "Increasing order of stability among the three main conformations (i.e. Eclipse, Anti, Gauche) of\n2-fluoroethanol is", "options": [ { "text": "Eclipse, Gauche, Anti" }, { "text": "Gauche, Eclipse, Anti " }, { "text": "Eclipse, Anti, Gauche " }, { "text": "Anti, Gauche, Eclipse " } ], "answer": "Eclipse, Anti, Gauche ", "solution": "**Answer:** Eclipse, Anti, Gauche \n\n\"AIEEE \n

Due to hydrogen bonding between $$H$$ & $$F$$ gauche conformation is most stable hence the correct order is Eclipse, Anti, Gauche\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 134, "subject": "Chemistry", "question": "Which one of the following conformation of cyclohexane is chiral? ", "options": [ { "text": "Twist boat " }, { "text": "Rigid " }, { "text": "Chair " }, { "text": "Boat" } ], "answer": "Twist boat ", "solution": "**Answer:** Twist boat \n\nChiral conformation will not have plane of symmetry. Since twist boat does not have plane of symmetry it is chiral. \n

\"AIEEE ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 135, "subject": "Chemistry", "question": "The number of stereoisomers possible for a compound of the molecular formula\nCH3 - CH = CH - CH(OH) - Me is :", "options": [ { "text": "2" }, { "text": "4" }, { "text": "6" }, { "text": "3" } ], "answer": "4", "solution": "**Answer:** 4\n\n\"AIEEE \n

exhibits both geometrical as well as optical isomerism. \n

$$cis\\,\\,$$ - $$R$$ $$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,cis\\,\\,$$ $$S$$\n

$$trans\\,\\,$$ - $$R$$ $$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,trans\\,\\,$$ - $$S$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 136, "subject": "Chemistry", "question": "The alkene that exhibits geometrical isomerism is :", "options": [ { "text": "propene" }, { "text": "2-methyl propene " }, { "text": "2-butene" }, { "text": "2- methyl -2- butene " } ], "answer": "2-butene", "solution": "**Answer:** 2-butene\n\n\"AIEEE ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 137, "subject": "Chemistry", "question": "Out of the following, the alkene that exhibits optical isomerism is", "options": [ { "text": "3–methyl–2–pentene " }, { "text": "4–methyl–1–pentene" }, { "text": "3–methyl–1–pentene" }, { "text": "2–methyl–2–pentene" } ], "answer": "3–methyl–1–pentene", "solution": "**Answer:** 3–methyl–1–pentene\n\nFor a compound to show optical isomerism, presence of chiral carbon atom is a necessary condition.\n

\"AIEEE ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 138, "subject": "Chemistry", "question": "Identify the compound that exhibits tautomerism.", "options": [ { "text": "Lactic acid" }, { "text": "2-Pentanone" }, { "text": "Pheno" }, { "text": "2- Butene " } ], "answer": "2-Pentanone", "solution": "**Answer:** 2-Pentanone\n\n\"AIEEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 139, "subject": "Chemistry", "question": "Which of the following compounds will exhibit geometrical isomerism?", "options": [ { "text": "3-Phenyl-1-butene" }, { "text": "2-Phenyl-1-butene" }, { "text": "1,1-Diphenyl-1-propane" }, { "text": "1-Phenyl-2-butene" } ], "answer": "1-Phenyl-2-butene", "solution": "**Answer:** 1-Phenyl-2-butene\n\n\"JEE \n

$$1$$- Phenyl- $$2$$ -butene the two groups around each of the doubly bonded carbon\n

Because, all are different. This compound can show $$cis$$-and $$trans$$- isomerism. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 140, "subject": "Chemistry", "question": "Which of these factors does not govern the stability of a conformation in acyclic compounds?", "options": [ { "text": "Angle strain" }, { "text": "Torsional strain" }, { "text": "Electrostatic forces of interaction" }, { "text": "Steric interactions" } ], "answer": "Angle strain", "solution": "**Answer:** Angle strain\n\nAngle strain is not present in acyclic\ncompounds.", "topic": "Number Theory", "subtopic": "Divisibility / Factorization" }, { "id": 141, "subject": "Chemistry", "question": "Which of the following compounds shows\ngeometrical isomerism?", "options": [ { "text": "2-methylpent-1-ene" }, { "text": "4-methylpent-1-ene" }, { "text": "2-methylpent-2-ene" }, { "text": "4-methylpent-2-ene" } ], "answer": "4-methylpent-2-ene", "solution": "**Answer:** 4-methylpent-2-ene\n\n

Only 4-methylpent-2-ene shows geometrical isomerism.

\n\n\"JEE\n\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 142, "subject": "Chemistry", "question": "Compound with molecular formula C3H6O can show :", "options": [ { "text": "Positional isomerism" }, { "text": "Both positional isomerism and metamerism" }, { "text": "Metamerism" }, { "text": "Functional group isomerism" } ], "answer": "Functional group isomerism", "solution": "**Answer:** Functional group isomerism\n\nC3H6O (degree of unsaturation = 1)

\n\"JEE
Functional group isomerism.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 143, "subject": "Chemistry", "question": "Which of the following molecules does not show stereo isomerism?", "options": [ { "text": "3, 4-Dimethylhex-3-ene" }, { "text": "3-Methylhex-1-ene" }, { "text": "3-Ethylhex-3-ene" }, { "text": "4-Methylhex-1-ene" } ], "answer": "3-Ethylhex-3-ene", "solution": "**Answer:** 3-Ethylhex-3-ene\n\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 144, "subject": "Chemistry", "question": "The number of acyclic structural isomers (including geometrical isomers) for pentene are ___________", "options": [], "answer": "6", "solution": "**Answer:** 6\n\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 145, "subject": "Chemistry", "question": "Staggered and eclipsed conformers of ethane are :", "options": [ { "text": "Polymers" }, { "text": "Rotamers" }, { "text": "Enantiomers" }, { "text": "Mirror images" } ], "answer": "Rotamers", "solution": "**Answer:** Rotamers\n\nStaggered and eclipsed conformers of ethane also known as rotamers.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 146, "subject": "Chemistry", "question": "The dihedral angle in staggered form of Newman projection of 1, 1, 1-Trichloro ethane is ___________ degree. (Round off to the nearest integer)", "options": [], "answer": "60", "solution": "**Answer:** 60\n\n1, 1, 1-Trichloro ethane [CCl3-CH3]

\"JEE", "topic": "Geometry", "subtopic": "Plane Geometry" }, { "id": 147, "subject": "Chemistry", "question": "The number of stereoisomers possible for 1, 2-dimethyl cyclopropane is :", "options": [ { "text": "One" }, { "text": "Four" }, { "text": "Two" }, { "text": "Three" } ], "answer": "Three", "solution": "**Answer:** Three\n\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 148, "subject": "Chemistry", "question": "

Total number of possible stereoisomers of dimethyl cyclopentane is ____________.

", "options": [], "answer": "6", "solution": "**Answer:** 6\n\nDimethyl cyclopentane

\"JEE\n\"JEE
\nwill show stereo isomerism, Its stereo isomers are

\n\"JEE
\n\"JEE
will show stereo isomerism, Its stereo isomers are

\"JEE\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 149, "subject": "Chemistry", "question": "

The total number of monobromo derivatives formed by the alkanes with molecular formula C5H12 is (excluding stereo isomers) __________.

", "options": [], "answer": "8", "solution": "**Answer:** 8\n\n

Total monobromo derivatives = 8

\"JEE

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 150, "subject": "Chemistry", "question": "

Total number of isomers (including stereoisomers) obtained on monochlorination of methylcyclohexane is ___________.

", "options": [], "answer": "12", "solution": "**Answer:** 12\n\n

Compounds formed on mono-chlorination of\nmethylcyclohexane are :

\"JEE

\n

$\\therefore$ Total mono-chlorinated products formed = 12

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 151, "subject": "Chemistry", "question": "

Optical activity of an enantiomeric mixture is $$+12.6^{\\circ}$$ and the specific rotation of $$(+)$$ isomer is $$+30^{\\circ}$$. The optical purity is __________$$\\%$$.

", "options": [], "answer": "42", "solution": "**Answer:** 42\n\nOptical purity $=\\frac{\\text { Total rotation }}{\\text { Specific rotation }} \\times 100$\n

\n$$\n\\begin{aligned}\n&=\\frac{12 \\cdot 6}{30} \\times 100 \\\\\\\\\n&=42 \\%\n\\end{aligned}\n$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 152, "subject": "Chemistry", "question": "Total number of isomeric compounds (including stereoisomers) formed by monochlorination of 2-methylbutane is _______ .", "options": [], "answer": "6", "solution": "**Answer:** 6\n\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 153, "subject": "Chemistry", "question": "Number of optical isomers possible for 2-chlorobutane ________.", "options": [], "answer": "2", "solution": "**Answer:** 2\n\n\"JEE\n
There is one chiral centre present in given compound.\n\n

So, Total optical isomers $=2$\n

\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 154, "subject": "Chemistry", "question": "

3-Methylhex-2-ene on reaction with $$\\mathrm{HBr}$$ in presence of peroxide forms an addition product (A). The number of possible stereoisomers for '$$\\mathrm{A}$$' is ________.

", "options": [], "answer": "4", "solution": "**Answer:** 4\n\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 155, "subject": "Chemistry", "question": "

The incorrect statement regarding conformations of ethane is :

", "options": [ { "text": "The dihedral angle in staggered conformation is $$60^{\\circ}$$.\n" }, { "text": "Ethane has infinite number of conformations.\n" }, { "text": "Eclipsed conformation is the most stable conformation.\n" }, { "text": "The conformations of ethane are inter-convertible to one-another." } ], "answer": "Eclipsed conformation is the most stable conformation.\n", "solution": "**Answer:** Eclipsed conformation is the most stable conformation.\n\n\n

The statement regarding conformations of ethane that is incorrect is:

\n

Option C - Eclipsed conformation is the most stable conformation.

\n\n

To explain why Option C is incorrect, let's analyze all the options:

\n\n

Option A: \"The dihedral angle in staggered conformation is $$60^{\\circ}$$.\"\nThis statement is incorrect. In staggered conformation, the dihedral angle, or the angle between the planes containing the hydrogen atoms, is actually $$60^{\\circ}$$. However, the staggered form is not unique; it refers to the form where any given pair of hydrogens on adjacent carbons has this dihedral angle, but rotating the molecule can yield a variety of staggered conformations with different specific hydrogen atoms at the $$60^{\\circ}$$ dihedral angle. The full rotation of one ethyl group relative to the other through $$360^{\\circ}$$ creates a cycle where the staggered conformation occurs every $$120^{\\circ}$$.

\n\n

Option B: \"Ethane has infinite number of conformations.\"\nThis statement is correct. The rotation around the carbon-carbon single bond (sigma bond) is relatively free, so ethane can theoretically adopt an infinite number of conformations between the two extremes of staggered and eclipsed. These conformations can have very slight differences in angles, resulting in finely varied overlap of electron clouds.

\n\n

Option C: \"Eclipsed conformation is the most stable conformation.\"\nThis statement is incorrect, which makes it the answer to the question. The eclipsed conformation is actually the least stable conformation of ethane due to the increased torsional strain from electron repulsion. The electron clouds of the bonding pairs of hydrogens on adjacent carbons are aligned in such a way as to be as close to each other as possible, increasing repulsion. The most stable conformation is the staggered conformation, where the groups are as far apart as possible, reducing repulsion.

\n\n

Option D: \"The conformations of ethane are inter-convertible to one-another.\"\nThis statement is correct. As you rotate around the carbon-carbon single bond, the conformations continuously change from one to another. This interconversion can be quite rapid and occurs without the need for a high energy input under normal conditions.

\n\n

Thus, the incorrect statement is Option C.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 156, "subject": "Chemistry", "question": "

Number of isomeric products formed by monochlorination of 2-methylbutane in presence of sunlight is ________.

", "options": [], "answer": "6", "solution": "**Answer:** 6\n\n

\"JEE

\n

$$\\therefore$$ Number of isomeric products = 6

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 157, "subject": "Chemistry", "question": "

The number of different chain isomers for C$$_7$$H$$_{16}$$ is __________.

", "options": [], "answer": "9", "solution": "**Answer:** 9\n\n

(1) heptane

\n

(2) 2-methylhexane

\n

(3) 3-methylhexane

\n

(4) 2,2-dimethylpentane

\n

(5) 2,3-dimethylpentane

\n

(6) 2,4-dimethylpentane

\n

(7) 3,3-dimethylpentane

\n

(8) 3-ethylpentane

\n

(9) 2,2,3-trimethylbutane

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 158, "subject": "Chemistry", "question": "

Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).

\n

Assertion (A) : Cis form of alkene is found to be more polar than the trans form.

\n

Reason (R) : Dipole moment of trans isomer of 2-butene is zero.

\n

In the light of the above statements, choose the correct answer from the options given below :

", "options": [ { "text": "(A) is true but (R) is false" }, { "text": "(A) is false but (R) is true" }, { "text": "Both $$(\\mathbf{A})$$ and $$(\\mathbf{R})$$ are true and $$(\\mathbf{R})$$ is the correct explanation of $$(\\mathbf{A})$$" }, { "text": "Both $$(\\mathbf{A})$$ and $$(\\mathbf{R})$$ are true but $$(\\mathbf{R})$$ is NOT the correct explanation of $$(\\mathbf{A})$$" } ], "answer": "Both $$(\\mathbf{A})$$ and $$(\\mathbf{R})$$ are true and $$(\\mathbf{R})$$ is the correct explanation of $$(\\mathbf{A})$$", "solution": "**Answer:** Both $$(\\mathbf{A})$$ and $$(\\mathbf{R})$$ are true and $$(\\mathbf{R})$$ is the correct explanation of $$(\\mathbf{A})$$\n\n

\"JEE

\n

$$\\begin{equation}\n\\text { Dipole moment of cis }>\\text { trans and } \\mu_{\\text {trans }}=0\n\\end{equation}$$

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 159, "subject": "Chemistry", "question": "

Match List - I with List - II.

\n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
LIST I
(Pair of Compounds)		LIST II
(Isomerism)
A.n-propanol and IsopropanolI.Metamerism
B.Methoxypropane and ethoxyethaneII.Chain Isomerism
C.Propanone and propanalIII.Position Isomerism
D.Neopentane and IsopentaneIV.Functional Isomerism

\n

Choose the correct answer from the options given below :

", "options": [ { "text": "(A)-(II), (B)-(I), (C)-(IV), (D)-(III)\n" }, { "text": "(A)-(III), (B)-(I), (C)-(II), (D)-(IV)\n" }, { "text": "(A)-(III), (B)-(I), (C)-(IV), (D)-(II)\n" }, { "text": "(A)-(I), (B)-(III), (C)-(IV), (D)-(II)" } ], "answer": "(A)-(III), (B)-(I), (C)-(IV), (D)-(II)\n", "solution": "**Answer:** (A)-(III), (B)-(I), (C)-(IV), (D)-(II)\n\n\n

\"JEE

", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 160, "subject": "Chemistry", "question": "

The incorrect statement regarding the geometrical isomers of 2-butene is :

", "options": [ { "text": "cis-2-butene and trans-2-butene are not interconvertible at room temperature.\n" }, { "text": "trans-2-butene is more stable than cis-2-butene.\n" }, { "text": "cis-2-butene has less dipole moment than trans-2-butene.\n" }, { "text": "cis-2-butene and trans-2-butene are stereoisomers." } ], "answer": "cis-2-butene has less dipole moment than trans-2-butene.\n", "solution": "**Answer:** cis-2-butene has less dipole moment than trans-2-butene.\n\n\n

Let's analyze each statement to determine which one is incorrect regarding the geometrical isomers of 2-butene.

\n\n

Option A: \"Cis-2-butene and trans-2-butene are not interconvertible at room temperature.\"

\n\n

This statement is correct. Cis and trans isomers of 2-butene are not readily interconvertible at room temperature due to the energy barrier associated with the rotation around the double bond. The double bond ensures rigidity, making interconversion between these isomers difficult without sufficient energy input.

\n\n

Option B: \"Trans-2-butene is more stable than cis-2-butene.\"

\n\n

This statement is correct. The trans isomer is more stable than the cis isomer due to reduced steric hindrance. In the cis isomer, the bulkier substituents (methyl groups) are on the same side, leading to increased steric repulsion.

\n\n

Option C: \"Cis-2-butene has less dipole moment than trans-2-butene.\"

\n\n

This statement is incorrect. The cis isomer typically has a higher dipole moment than the trans isomer because the substituents on the same side of the double bond create a net dipole moment. In the trans isomer, the dipole moments of the substituents cancel each other out, resulting in a lower overall dipole moment.

\n\n

Option D: \"Cis-2-butene and trans-2-butene are stereoisomers.\"

\n\n

This statement is correct. Cis-2-butene and trans-2-butene are indeed stereoisomers because they have the same molecular formula and connectivity of atoms but differ in the spatial arrangement of the substituents around the double bond.

\n\n

Therefore, the incorrect statement is:

\n\n

Option C: Cis-2-butene has less dipole moment than trans-2-butene.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 161, "subject": "Chemistry", "question": "The IUPAC name of CH3COCH(CH3)2 is", "options": [ { "text": "2-methyl-3-butaone" }, { "text": "4-methylisopropyl 1 ketone" }, { "text": "3-methyl-2-butaone" }, { "text": "Isopropylmethyl ketone" } ], "answer": "3-methyl-2-butaone", "solution": "**Answer:** 3-methyl-2-butaone\n\n\"AIEEE ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 162, "subject": "Chemistry", "question": "Which one the following does not have sp2 hybridized carbon?", "options": [ { "text": "Acetone" }, { "text": "Acetamide" }, { "text": "Acetonitrile" }, { "text": "Acetic acid" } ], "answer": "Acetonitrile", "solution": "**Answer:** Acetonitrile\n\n\"AIEEE ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 163, "subject": "Chemistry", "question": "The correct decreasing order of priority for the functional groups of organic compounds in the IUPAC\nsystem of nomenclature is", "options": [ { "text": "−COOH, −SO3H, −CONH2, −CHO " }, { "text": "−SO3H, −COOH, −CONH2, −CHO " }, { "text": "−CHO, −COOH, −SO3H, −CONH2 " }, { "text": "−CONH2, −CHO, −SO3H, −COOH " } ], "answer": "−COOH, −SO3H, −CONH2, −CHO ", "solution": "**Answer:** −COOH, −SO3H, −CONH2, −CHO \n\nThe correct order of priority for the given functional group is \n

\"AIEEE\n\n
\n\n\n \n \n \n \n \n \n\n\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\n
Functional group (priority order)suffix(s)prefix
(1) \"AIEEE Oic acid(when carbon(C) of this functional group present in the main chain)
or
Carboxylic acid (when carbon(C) of this functional group is not present in the main chain) 		Carboxy
(2) \"AIEEE sulphonic acidsulpho
(3) \"AIEEE Oic anhydrid (when carbon(C) of this functional group present in the main chain)
or
Carboxylic anhydride (when carbon(C) of this functional group is not present in the main chain) 		$$ - $$
(4) \"AIEEEOate (when carbon(C) of this functional group present in the main chain)
or
Carboxylate (when carbon(C) of this functional group is not present in the main chain) 		$$ - $$
(5) \"AIEEE Oyl halide (when carbon(C) of this functional group present in the main chain)
or
Carboxyl halide (when carbon(C) of this functional group is not present in the main chain) 		haloflormyl
(6) \"AIEEEamide (when carbon(C) of this functional group present in the main chain)
or
carbamide (when carbon(C) of this functional group is not present in the main chain) 		carbamoyl
(7)cynide ($$ - $$ C $$ \\equiv $$ N)nitrile (when carbon(C) of this functional group present in the main chain)
or
carbonitrile (when carbon(C) of this functional group is not present in the main chain) 		cyano
(8) \"AIEEECarbylamineisocyano
(9) \"AIEEEal (when carbon(C) of this functional group present in the main chain)
or
Carbaldehyde (when carbon(C) of this functional group is not present in the main chain) 		formyl (oxo)
(10) \"AIEEEoneoxo
(11)$$ - $$ OHolhydroxy
(12)$$ - $$ SHthiolmercapto
(13)$$ - $$ NH$$_2$$amineamino
\n

From this table you can see -COOH has highest priority and -NH2 has lowest priority among all functional groups.

", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 164, "subject": "Chemistry", "question": "The IUPAC name of neopentane is ", "options": [ { "text": "2-methylbutane" }, { "text": "2, 2-dimethylpropane" }, { "text": "2-methylpropane " }, { "text": "2,2-dimethylbutane" } ], "answer": "2, 2-dimethylpropane", "solution": "**Answer:** 2, 2-dimethylpropane\n\n\"AIEEE ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 165, "subject": "Chemistry", "question": "The hydrocarbon with seven carbon atoms containing a neopentyl and a vinyl group\nis :", "options": [ { "text": "2, 2-dimethyl-4-pentene " }, { "text": "Isopropyl-2-butene" }, { "text": "4, 4-dimethylpentene" }, { "text": "2, 2-dimethyl-3-pentene" } ], "answer": "4, 4-dimethylpentene", "solution": "**Answer:** 4, 4-dimethylpentene\n\n\"JEE", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 166, "subject": "Chemistry", "question": "

The IUPAC name of ethylidene chloride is :

", "options": [ { "text": "1-Chloroethene" }, { "text": "1-Chloroethyne" }, { "text": "1,2-Dichloroethane" }, { "text": "1,1-Dichloroethane" } ], "answer": "1,1-Dichloroethane", "solution": "**Answer:** 1,1-Dichloroethane\n\n

Ethylidene chloride is CH3– CHCl2, its IUPAC name is 1,1-Dichloromethane.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 167, "subject": "Chemistry", "question": "

Given below are two statements:

\n

Statement I: IUPAC name of $$\\mathrm{HO}-\\mathrm{CH}_2-\\left(\\mathrm{CH}_2\\right)_3-\\mathrm{CH}_2-\\mathrm{COCH}_3$$ is 7-hydroxyheptan-2-one.

\n

Statement II: 2-oxoheptan-7-ol is the correct IUPAC name for above compound. In the light of the above statements, choose the most appropriate answer from the options given below:

", "options": [ { "text": "Statement I is incorrect but Statement II is correct.\n" }, { "text": "Statement I is correct but Statement II is incorrect.\n" }, { "text": "Both Statement I and Statement II are correct.\n" }, { "text": "Both Statement I and Statement II are incorrect." } ], "answer": "Statement I is correct but Statement II is incorrect.\n", "solution": "**Answer:** Statement I is correct but Statement II is incorrect.\n\n\n

7-Hydroxyheptan-2-one is correct IUPAC name

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 168, "subject": "Chemistry", "question": "

The total number of 'sigma' and 'pi' bonds in 2-oxohex-4-ynoic acid is ______.

", "options": [], "answer": "18", "solution": "**Answer:** 18\n\n

\"JEE

\n

Total number of $$\\sigma$$ & $$\\pi$$ bonds = 18

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 169, "subject": "Chemistry", "question": "For the estimation of nitrogen, 1.4 g of organic compound was digested by Kjeldahl method and the\nevolved ammonia was absorbed in 60 mL of M/10 sulphuric acid. The unreacted acid required 20 ml of M/10\nsodium hydroxide for complete neutralization. The percentage of nitrogen in the compound is:", "options": [ { "text": "3%" }, { "text": "5%" }, { "text": "6%" }, { "text": "10%" } ], "answer": "10%", "solution": "**Answer:** 10%\n\n$$\\% \\,\\,$$ of $$\\,\\,N = {{1.4\\,\\, \\times \\,\\,meq.of\\,\\,acid} \\over {mass\\,\\,of\\,\\,organic\\,\\,compound}}$$ \n

meq. of $$\\,\\,{H_2}S{O_4} = 60 \\times {M \\over {10}} \\times 2 = 12$$ \n

meq. of $$\\,\\,NaOH = 20 \\times {M \\over {10}} = 2$$ \n

$$\\therefore$$ meq. of acid consumed $$ = 12 - 2 = 10$$ \n

$$\\therefore$$ $$\\,\\,\\,\\% \\,\\,\\,$$ $$N = {{1.4 \\times 10} \\over {1.4}} = 10\\% $$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 170, "subject": "Chemistry", "question": "In Carius method of estimation of halogens, 250 mg of an organic compound gave 141 mg of AgBr. The\npercentage of bromine in the compound is: (at. Mass Ag = 108; Br = 80)", "options": [ { "text": "48" }, { "text": "60" }, { "text": "36" }, { "text": "24" } ], "answer": "24", "solution": "**Answer:** 24\n\nMass of substance $$=250$$ $$mg$$ $$=0.250$$ $$g$$ \n

Mass of $$AgBr$$ $$=141$$ $$mg=0.141$$ $$g$$ \n

$$1$$ mole of $$AgBr$$ $$=1$$ $$g$$ atom of $$Br$$\n

$$188$$ $$g$$ of $$AgBr$$ $$=80$$ $$g$$ of $$Br$$\n

$$188$$ $$g$$ of $$AgBr$$ contain bromine $$=80$$ $$g$$ \n

$$0.141$$ $$g$$ of $$AgBr$$ contain bromine $$ = {{80} \\over {188}} \\times 0.141$$ \n

This much amount of bromine present in $$0.250$$ $$g$$ of organic compound \n

$$\\therefore$$ $$\\% $$ of bromine $$ = {{80} \\over {188}} \\times {{0.414} \\over {0.250}} \\times 100 = 24\\% $$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 171, "subject": "Chemistry", "question": "The distillation technique most suited for separating glycerol from spent-lye in the soap industry is :", "options": [ { "text": "Simple distillation" }, { "text": "Fractional distillation" }, { "text": "Steam distillation" }, { "text": "Distillation under reduced pressure" } ], "answer": "Distillation under reduced pressure", "solution": "**Answer:** Distillation under reduced pressure\n\n

The best technique for separating glycerol from spent-lye in the soap industry is Distillation under Reduced Pressure, also known as Vacuum Distillation (Option D).

\n

The reason for this is that glycerol has a high boiling point (~$${290^ \\circ }$$$$C$$). If we were to use simple or fractional distillation, which involves boiling at standard atmospheric pressure, the high temperature could degrade or decompose the glycerol.

\n

In contrast, vacuum distillation allows the separation to occur at a much lower temperature ( around $${180^ \\circ }C$$) by reducing the pressure. This helps to prevent degradation of the glycerol, making it a better method for this particular separation. Steam distillation isn't suitable here, as it's typically used for temperature-sensitive, volatile substances, which isn't the case for glycerol.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 172, "subject": "Chemistry", "question": "Which of the following statements is not true about partition chromatography?", "options": [ { "text": "Mobile phase can be a gas" }, { "text": "Stationary phase is a finely divided solid adsorbent" }, { "text": "Separation depends upon equilibration of solute between a mobile and a stationary phase\n" }, { "text": "Paper chromatography is an example of partition chromatography" } ], "answer": "Stationary phase is a finely divided solid adsorbent", "solution": "**Answer:** Stationary phase is a finely divided solid adsorbent\n\n

In partition chromatography, which includes paper chromatography, the stationary phase is typically water held by cellulose fibers in the paper. This paper acts as an inert support. When a mixture is applied to the paper, its components partition between this stationary phase and the mobile phase (which can be a liquid or a gas). The varying affinities of the mixture's components for the stationary and mobile phases allow for their separation.

\n

Now, let's address the options :

\n

Option A : Mobile phase can be a gas - This is true. In partition chromatography, the mobile phase can be either a liquid or a gas.

\n

Option B : Stationary phase is a finely divided solid adsorbent - This is typically false for partition chromatography. Here, the stationary phase is a liquid (like water in paper chromatography) that's held by an inert solid support (like cellulose paper).

\n

Option C : Separation depends upon equilibration of solute between a mobile and a stationary phase - This is true. In partition chromatography, the mixture's components equilibrate between the mobile and stationary phases, leading to their separation.

\n

Option D : Paper chromatography is an example of partition chromatography - This is true. As explained above, in paper chromatography, the stationary phase (water) is held by an inert support (cellulose paper), fitting the definition of partition chromatography.

\n", "topic": "Discrete Mathematics", "subtopic": "Other" }, { "id": 173, "subject": "Chemistry", "question": "Two compounds I and II are eluted by column chromatography (adsorption of I > II). Which one of following is a correct statement ? ", "options": [ { "text": "I moves faster and has higher Rf value than II" }, { "text": "II movesfaster and has higher  Rf  value than I" }, { "text": "I moves slower and has higher  Rf  value than II" }, { "text": "II moves slower and has higher  Rf  value than I" } ], "answer": "II movesfaster and has higher  Rf  value than I", "solution": "**Answer:** II movesfaster and has higher  Rf  value than I\n\nAccording to question, \n

Absorption of I > II, it means I is firmly attached to column. Hence it will move slowly and move little distance. As II is loosely attached to column.\n

So, it will move faster and will move larger distance.", "topic": "Discrete Mathematics", "subtopic": "Other" }, { "id": 174, "subject": "Chemistry", "question": "The correct match between items of List-I and List-II is :\n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
List - IList - II
(A)Coloured impurity(P)Steam distilation
(B)Mixture of o-nitrophenol
and p-nitrophenol		(Q)Fractional distilation
(C)Crude Naphtha (R)Charcoal treatment
(d)Mixture of glycerol
and sugars		(S)Distillation under
reduced pressure
", "options": [ { "text": "(A)-(R), (B)-(S), (C)-(P), (D)-(Q)" }, { "text": "(A)-(R), (B)-(P), (C)-(S), (D)-(Q)" }, { "text": "(A)-(R), (B)-(P), (C)-(Q), (D)-(S)" }, { "text": "(A)-(P), (B)-(S), (C)-(R), (D)-(Q)" } ], "answer": "(A)-(R), (B)-(P), (C)-(Q), (D)-(S)", "solution": "**Answer:** (A)-(R), (B)-(P), (C)-(Q), (D)-(S)\n\n(a) Charcoal treatment removes coloured impurity through adsorption.\n

(b) Steam distillation separates the mixture of o-nitrophenol and p-nitrophenol. The o-nitrophenol is\nsteam volatile (due to intramolecular hydrogen bonding), and the para isomer is not volatile.\n

(c) Fractional distillation separates crude naphtha. Naphtha is a flammable liquid hydrocarbon\nmixture.\n

(d) Distillation under reduced pressure separates mixture of glycerol and sugars. Vacuum distillation\nlowers the boiling point and prevents decomposition.\n", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 175, "subject": "Chemistry", "question": "If dichloromethane (DCM) and water (H2O) are used for differential extraction, which one of the following statements is correct ?", "options": [ { "text": "DCM and H2O would stay as upper and lower layer respectively in the separating funnel (S.F.)" }, { "text": "DCM and H2O will make turbid/colloidal mixture" }, { "text": "DCM and H2O would stay as lower and upper layer respectively in the separating funnel(S.F)." }, { "text": "DCM and H2O will be miscible clearly " } ], "answer": "DCM and H2O would stay as lower and upper layer respectively in the separating funnel(S.F).", "solution": "**Answer:** DCM and H2O would stay as lower and upper layer respectively in the separating funnel(S.F).\n\nDensity of DCM = 1.39 and density of H2O = 1.0.\n

Dichloromethane (DCM) is a denser solvent than water. Therefore, when a mixture of DCM and water is allowed to settle in a separating funnel, the DCM forms the lower layer due to its higher density, while the water forms the upper layer. The two solvents are immiscible, meaning they do not mix and form two distinct layers.

\n

So, the correct answer is :

\n

Option C : DCM and H2O would stay as lower and upper layer respectively in the separating funnel(S.F).

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 176, "subject": "Chemistry", "question": "The correct match between items I and II is : \n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
Item-I (Mixture)Item-II (Seperation method)
(A)H2O : Sugar(P)Sublimation
(B)H2O : Aniline(Q)Recrystallization
(C)H2O : Toluene
(R)Steam distillation
(S)Differential extraction
", "options": [ { "text": "A$$ \\to $$(R); (B)$$ \\to $$(P); (C)$$ \\to $$(S)" }, { "text": "A$$ \\to $$(Q); (B)$$ \\to $$(R); (C)$$ \\to $$(S)" }, { "text": "A$$ \\to $$(Q); (B)$$ \\to $$(P); (C)$$ \\to $$(R)" }, { "text": "A$$ \\to $$(S); (B)$$ \\to $$(R); (C)$$ \\to $$(P)" } ], "answer": "A$$ \\to $$(Q); (B)$$ \\to $$(R); (C)$$ \\to $$(S)", "solution": "**Answer:** A$$ \\to $$(Q); (B)$$ \\to $$(R); (C)$$ \\to $$(S)\n\n

To find the correct match, let's understand what each separation method does :

\n\n

So, the correct match would be :

\n

A -> Q : Water and sugar can be separated by recrystallization.

\n

B -> R : Water and aniline can be separated by steam distillation.

\n

C -> S : Water and toluene can be separated by differential extraction.

\n

So, the correct option is Option C : A$$ \\to $$(Q); (B)$$ \\to $$(R); (C)$$ \\to $$(S)

\n", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 177, "subject": "Chemistry", "question": "The principle of column chromatography is :", "options": [ { "text": "Gravitational force." }, { "text": "Capillary action." }, { "text": "Differential adsorption of the substances on the solid phase." }, { "text": "Differential absorption of the substances on the solid phase. " } ], "answer": "Differential adsorption of the substances on the solid phase.", "solution": "**Answer:** Differential adsorption of the substances on the solid phase.\n\n

The principle of column chromatography is based on the differential adsorption of substances on the solid phase. This means that different substances in a mixture will interact with the solid stationary phase to different extents. Some substances will adsorb more strongly to the solid phase and will therefore move more slowly through the column, while others will adsorb less strongly and will therefore move more quickly. This differential adsorption allows for the separation of the substances in the mixture.

\n

So, the correct answer is :

\n

Option C : Differential adsorption of the substances on the solid phase.

", "topic": "Discrete Mathematics", "subtopic": "Other" }, { "id": 178, "subject": "Chemistry", "question": "In chromatography, which of the following statements is incorrect for Rf ?", "options": [ { "text": "Rf value depends on the type of chromatography." }, { "text": "Higher Rf\n value means higher adsorption." }, { "text": "Rf\n value is dependent on the mobile phase." }, { "text": "The value of Rf\n can not be more than one. " } ], "answer": "Higher Rf\n value means higher adsorption.", "solution": "**Answer:** Higher Rf\n value means higher adsorption.\n\nRf = $${x \\over y}$$\n

$$x$$ = Distance moved by the substance from baseline\n

$$y$$ = Distance moved by the solvent from baseline\n

Option A : Correct. The Rf value can depend on the type of chromatography being used because different methods can use different stationary and mobile phases, which can influence how far a substance travels.\n\n

Option B : Incorrect. The Rf value is actually inversely related to adsorption. A higher Rf value indicates that the substance is less strongly adsorbed to the stationary phase, and therefore moves further along with the mobile phase.\n\n

Option C : Correct. The nature of the mobile phase can greatly influence the Rf value. The more compatible the substance is with the mobile phase, the further it will travel, leading to a higher Rf value.\n\n

Option D : Correct. By definition, the Rf value cannot exceed one, as a substance cannot travel further than the solvent.\n\n

So, Option B is incorrect for the Rf value in chromatography.", "topic": "Discrete Mathematics", "subtopic": "Other" }, { "id": 179, "subject": "Chemistry", "question": "Kjeldahl's method cannot be used to estimate\nnitrogen for which of the following\ncompounds?", "options": [ { "text": "C6H5NH2" }, { "text": "$${\\rm{N}}{{\\rm{H}}_2} - \\mathop C\\limits^{\\mathop \\parallel \\limits^O } - {\\rm{N}}{{\\rm{H}}_2}$$" }, { "text": "C6H5NO2" }, { "text": "CH3CH2-C $$ \\equiv $$ N" } ], "answer": "C6H5NO2", "solution": "**Answer:** C6H5NO2\n\nKjeldahl’s method is not applicable to\ncompounds containing nitrogen in nitro, azo\ngroups and nitrogen present in ring (pyridine).", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 180, "subject": "Chemistry", "question": "Glycerol is separated in soap industries by :", "options": [ { "text": "Fractional distillation" }, { "text": "Distillation under reduced pressure" }, { "text": "Differential extraction" }, { "text": "Steam distillation" } ], "answer": "Distillation under reduced pressure", "solution": "**Answer:** Distillation under reduced pressure\n\nGlycerol is separated in soap industries by distillation under reduced pressure.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 181, "subject": "Chemistry", "question": "A flask contains a mixture of isohexane and 3-methylpentane. One of the liquids boils at 63oC while the\nother boils at 60oC What is the best way to separate the two liquids and which one will be distilled out first ?", "options": [ { "text": "fractional distillation, 3-methylpentane" }, { "text": "simple distillation, 3-methylpentane" }, { "text": "fractional distillation, isohexane" }, { "text": "simple distillation, isohexane" } ], "answer": "fractional distillation, isohexane", "solution": "**Answer:** fractional distillation, isohexane\n\nIsohexane and 3 Methylpentane Having same molecular formula.\n

Isohexane boil at 60oC and 3-Methyl pentane boil at 63oC. Both Having low Boiling point difference so\nFractional distillation is useful for separation and Isohexane Having low Boiling point So comes out first.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 182, "subject": "Chemistry", "question": "A chromatography column, packed with silica gel as stationary phase, was used to separate a\nmixture of compounds consisting of (A) benzanilide (B) aniline and (C) acetophenone. When the\ncolumn is eluted with a mixture of solvents, hexane : ethylacelate (20:80), the sequence of\nobtained compounds is :", "options": [ { "text": "(A), (B) and (C)" }, { "text": "(B), (A) and (C)" }, { "text": "(C), (A) and (B)\n" }, { "text": "(B), (C) and (A) " } ], "answer": "(C), (A) and (B)\n", "solution": "**Answer:** (C), (A) and (B)\n\n\n

In column chromatography, we separate different parts of a mixture by washing it with solvents. The part of the mixture that gets most strongly attached to the moving solvent (or mobile phase) moves down the column quicker and comes out first. This is marked by a high Rf value. Other parts of the mixture follow in the order of decreasing Rf values.

\n

In this particular separation, we're using a very polar solvent because most of it is ethyl acetate (80%). This means that the most polar part of the mixture (the one with the highest dipole moment) will have the highest Rf value and will be the first to come out. We measure the dipole moments of the mixture's components to determine this order.

\n

\"JEE\n
$\\therefore$ The sequence of obtained compounds is\n

$$\n(C)>(A)>(B)\n$$", "topic": "Discrete Mathematics", "subtopic": "Other" }, { "id": 183, "subject": "Chemistry", "question": "Given below are two statements :

Statement I : A mixture of chloroform and aniline can be separated by simple distillation.

Statement II : When separating aniline from a mixture of aniline and water by steam distillation aniline boils below its boiling point.

In the light of the above statements, choose the most appropriate answer from the options given below :", "options": [ { "text": "Both Statement I and Statement II are true" }, { "text": "Both Statement I and Statement II are false" }, { "text": "Statement I is true but Statement II is false" }, { "text": "Statement I is false but Statement II is true" } ], "answer": "Both Statement I and Statement II are true", "solution": "**Answer:** Both Statement I and Statement II are true\n\nMixture of chloroform and aniline can be\nseparated by simple distillation as these\ntwo liquids have sufficient difference in\nboiling point.\nChloroform (b.p. 334 K), aniline (b.p. 457 K)\n

In steam distillation, if one of the\nsubstances is water and the other, a water\ninsoluble substance (like aniline) then the\nmixture will boil close to but below 373 K.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 184, "subject": "Chemistry", "question": "In chromotography technique, the purification of compound is independent of :", "options": [ { "text": "Solubility of the compound" }, { "text": "Mobility or flow of solvent system" }, { "text": "Physical state of the pure compound" }, { "text": "Length of the column or TLC plate" } ], "answer": "Physical state of the pure compound", "solution": "**Answer:** Physical state of the pure compound\n\n

The purification of a compound using chromatography is dependent on several factors, including the solubility of the compound, the mobility or flow of the solvent system, and the length of the column or TLC plate. However, it is not typically dependent on the physical state of the pure compound. Here's why :

\n

Option A : The solubility of the compound in the mobile phase and its interaction with the stationary phase are key factors in determining how the compound will migrate in a chromatographic system. Hence, the solubility of the compound does affect its purification.

\n

Option B : The mobility or flow of the solvent system, often referred to as the mobile phase in chromatography, also plays a significant role in the separation of components in the mixture. The rate at which the mobile phase moves can influence how far and how quickly the components of the mixture will travel.

\n

Option C : The physical state (solid, liquid, gas) of the pure compound is usually not a factor in the purification process in chromatography. Whether a compound is solid, liquid, or gas does not typically affect its ability to be separated by chromatography, as it is the interactions between the compound and the stationary/mobile phases that are key.

\n

Option D : The length of the column or TLC plate can impact the degree of separation achieved in chromatography. A longer column or TLC plate can provide more opportunities for the interactions that drive the separation process, often leading to better separation of the components.

\n

So, Option C is correct : The physical state of the pure compound is usually not a factor in its purification by chromatography.

", "topic": "Discrete Mathematics", "subtopic": "Other" }, { "id": 185, "subject": "Chemistry", "question": "Given below are two statements :

Statement I : Retardation factor (Rf) can be measured in meter/centimeter.

Statement II : Rf value of a compound remains constant in all solvents.

Choose the most appropriate answer from the options given below :", "options": [ { "text": "Statement I is true but statement II is false" }, { "text": "Statement I is false but statement II is true" }, { "text": "Both statement I and statement II are true" }, { "text": "Both statement I and statement II are false" } ], "answer": "Both statement I and statement II are false", "solution": "**Answer:** Both statement I and statement II are false\n\nRetardation factor (Rf) is the ratio of the distance\nmoved by the substance from the baseline to the\ndistance moved by the solvent from the baseline.\nSo, it is dimensionless. The distance moved by the\nsubstance is due to the adsorption of the substance on\nthe stationary phase. It does not depend on the\nnature of the solvent. But the distance moved by the\nsolvent will change with the nature of the solvent.\nTherefore, Rf will vary with the change in the solvent.

\nSo, both the statements are false.", "topic": "Number Theory", "subtopic": "Divisibility / Factorization" }, { "id": 186, "subject": "Chemistry", "question": "When 0.15 g of an organic compound was analyzed using Carius method for estimation of bromine, 0.2397 g of AgBr was obtained. The percentage of bromine in the organic compound is ______________. (Nearest integer)

[Atomic mass : Silver = 108, Bromine = 80]", "options": [], "answer": "68", "solution": "**Answer:** 68\n\nMoles of Br = Moles of AgBr obtained

$$\\Rightarrow$$ Mass of Br = $${{0.2397} \\over {188}}$$ $$\\times$$ 80 g

therefore % Br in the organic compound

$$ = {{{W_{Br}}} \\over {{W_T}}}$$ $$\\times$$ 100

$${{0.2397 \\times 80} \\over {188 \\times 0.15}}$$ $$\\times$$ 100 = 0.85 $$\\times$$ 80 = 68

$$\\Rightarrow$$ Nearest integer is '68'.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 187, "subject": "Chemistry", "question": "Which purification technique is used for high boiling organic liquid compound (decomposes near its boiling point)?", "options": [ { "text": "Simple distillation" }, { "text": "Steam ditillation" }, { "text": "Fractional distillation" }, { "text": "Reduced pressure distillation" } ], "answer": "Reduced pressure distillation", "solution": "**Answer:** Reduced pressure distillation\n\nReduced pressure distillation or vacuum distillation is used for the purification of high boiling organic liquids which decomposes at or below their boiling point.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 188, "subject": "Chemistry", "question": "Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R) :

Assertion (A) : A simple distillation can be used to separate a mixture of propanol and propanone.

Reason (R) : Two liquids with a difference of more than 20$$^\\circ$$ in their boiling points can be separated by simple distillations.

In the light of the above statements, choose the most appropriate answer from the options given below :", "options": [ { "text": "(A) is false but (R) is true." }, { "text": "Both (A) and (R) are correct but (R) is not the correct explanation of (A)" }, { "text": "(A) is true but (R) is false" }, { "text": "Both (A) and (R) are correct and (R) is the correct explanation of (A)." } ], "answer": "Both (A) and (R) are correct and (R) is the correct explanation of (A).", "solution": "**Answer:** Both (A) and (R) are correct and (R) is the correct explanation of (A).\n\nSimple distillation can be used to separate a\nmixture of propanol and propanone.\n

• Two liquid with difference in B.P. around 20°C\ncan be separated by simple distillation.\n

• B.P. of propanol is 370 K.\n
B.P. of acetone is 329 K.\n\n

Both assertion & reason are correct & (R) is the correct explanation of (A).", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 189, "subject": "Chemistry", "question": "

Which one of the following techniques is not used to spot components of a mixture separated on thin layer chromatographic plate?

", "options": [ { "text": "I2 (Solid)" }, { "text": "U.V. Light" }, { "text": "Visualisation agent as a component of mobile phase" }, { "text": "Spraying of an appropriate reagent" } ], "answer": "Visualisation agent as a component of mobile phase", "solution": "**Answer:** Visualisation agent as a component of mobile phase\n\nTLC is a technique used to separate mixture of compounds based on differences in polarity. In TLC a glass\nplate coated with a stationary phase is spotted with the mixture to be separated.", "topic": "Discrete Mathematics", "subtopic": "Other" }, { "id": 190, "subject": "Chemistry", "question": "

Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R.

\n

Assertion A : A mixture contains benzoic acid and napthalene. The pure benzoic acid can be separated out by the use of benzene.

\n

Reason R : Benzoic acid is soluble in hot water.

\n

In the light of the above statements, choose the most appropriate answer from the options given below.

", "options": [ { "text": "Both A and R are true and R is the correct explanation of A." }, { "text": "Both A and R are true but R is NOT the correct explanation of A." }, { "text": "A is true but R is false." }, { "text": "A is false but R is true." } ], "answer": "A is false but R is true.", "solution": "**Answer:** A is false but R is true.\n\nSince, both benzoic acid and naphthalene will\ndissolve in benzene. Hence assertion is wrong.

\nBenzoic acid is almost insoluble in cold water but\nsoluble in hot water. Hence Reason is true.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 191, "subject": "Chemistry", "question": "

Phenol on reaction with dilute nitric acid, gives two products. Which method will be most efficient for large scale separation?

", "options": [ { "text": "Chromatographic separation" }, { "text": "Fractional Crystallisation" }, { "text": "Steam distillation" }, { "text": "Sublimation" } ], "answer": "Steam distillation", "solution": "**Answer:** Steam distillation\n\n\"JEE
\no-Nitrophenol and p-Nitrophenol can be easily separated by steam distillation.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 192, "subject": "Chemistry", "question": "

The separation of two coloured substances was done by paper chromatography. The distances travelled by solvent front, substance A and substance B from the base line are 3.25 cm, 2.08 cm and 1.05 cm, respectively. The ratio of Rf values of A to B is _____________.

", "options": [], "answer": "2", "solution": "**Answer:** 2\n\n$$\\mathrm{R}_{\\mathrm{f}}=\\frac{\\text { Distance travelled by the substance }}{\\text { Distance travelled by the solvent front }}$$\n

\n$$\\left(R_{f}\\right)_{A}=\\frac{2.08}{3.25}$$\n

\n$$\\left(R_{f}\\right)_{B}=\\frac{1.05}{3.25}$$\n

\n$$\\frac{\\left(R_{f}\\right)_{A}}{\\left(R_{f}\\right)_{B}} \\simeq 2$$", "topic": "Discrete Mathematics", "subtopic": "Other" }, { "id": 193, "subject": "Chemistry", "question": "

Which technique among the following, is most appropriate in separation of a mixture of $$100 \\,\\mathrm{mg}$$ of $$p$$-nitrophenol and picric acid ?

", "options": [ { "text": "Steam distillation" }, { "text": "2-5 ft long column of silica gel" }, { "text": "Sublimation" }, { "text": "Preparative TLC (Thin Layer Chromatography)" } ], "answer": "Preparative TLC (Thin Layer Chromatography)", "solution": "**Answer:** Preparative TLC (Thin Layer Chromatography)\n\nThin layer chromatography is a technique used to\nisolate non-volatile mixtures.

\n Hence, mixture of p-nitrophenol and Picric acid is\nseparated by TLC.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 194, "subject": "Chemistry", "question": "

Match List - I with List - II.

\n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
List - I
(Mixture)		List - II
(Purification Process)
(A)Chloroform & Aniline(I)Steam distillation
(B)Benzoic acid & Napthalene(II)Sublimation
(C)Water & Aniline(III)Distillation
(D)Napthalene & Sodium chloride(IV)Crystallisation

\n

Choose the correct answer from the options given below :

", "options": [ { "text": "(A) - (IV), (B) - (III), (C) - (I), (D) - (II)" }, { "text": "(A) - (III), (B) - (I), (C) - (IV), (D) - (II)" }, { "text": "(A) - (III), (B) - (IV), (C) - (II), (D) - (I)" }, { "text": "(A) - (III), (B) - (IV), (C) - (I), (D) - (II)" } ], "answer": "(A) - (III), (B) - (IV), (C) - (I), (D) - (II)", "solution": "**Answer:** (A) - (III), (B) - (IV), (C) - (I), (D) - (II)\n\n(A) Chloroform + Aniline $\\rightarrow$ (III) Distillation

\n(B) Benzoic acid + Napthalene $\\rightarrow$ (IV) Crystallisation

\n(C) Water + Aniline $\\rightarrow$ (I) Steam distillation

\n(D) Napthalene + Sodium chloride $\\rightarrow$ (II) Sublimation", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 195, "subject": "Chemistry", "question": "

Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R

\n

Assertion A : Thin layer chromatography is an adsorption chromatography.

\n

Reason R : A thin layer of silica gel is spread over a glass plate of suitable size in thin layer chromatography which acts as an adsorbent.

\n

In the light of the above statements, choose the correct answer from the options given below

", "options": [ { "text": "Both A and R are true and R is the correct explanation of A" }, { "text": "Both A and R are true but R is NOT the correct explanation of A" }, { "text": "A is true but R is false" }, { "text": "A is false but R is true" } ], "answer": "Both A and R are true and R is the correct explanation of A", "solution": "**Answer:** Both A and R are true and R is the correct explanation of A\n\nThin layer chromoatography (TLC) is another type\nof adsorption chromatography, which involve\nsepration of substance of a mixture ovel a thin\nlayer of an adsorbent coated on glass plate.

\n A thin layer (about 0.2 mm thick) of an adsorbent\n(silica gel) or (Alumina) in spread overa glass plate\nof suitable size. Hence Assertion (A) is correct and\nReason (R) is correct explanation of (A) ", "topic": "Discrete Mathematics", "subtopic": "Combinatorics" }, { "id": 196, "subject": "Chemistry", "question": "

Match List I with List II:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
List I (Mixture) List II (Separation Technique)
A.$\\mathrm{CHCl}_3+\\mathrm{C}_6 \\mathrm{H}_5 \\mathrm{NH}_2$I.Steam distillation
B.$\\mathrm{C}_6 \\mathrm{H}_{14}+\\mathrm{C}_5 \\mathrm{H}_{12}$II.Differential extraction
C.$\\mathrm{C}_6 \\mathrm{H}_5 \\mathrm{NH}_2+\\mathrm{H}_2 \\mathrm{O}$III.Distillation
D.$\\text { Organic compound in } \\mathrm{H}_2 \\mathrm{O}$IV.Fractional distillation
", "options": [ { "text": "A-III, B-I, C-IV, D-II" }, { "text": "A-II, B-I, C-III, D-IV" }, { "text": "A-IV, B-I, C-III, D-II" }, { "text": "A-III, B-IV, C-I, D-II" } ], "answer": "A-III, B-IV, C-I, D-II", "solution": "**Answer:** A-III, B-IV, C-I, D-II\n\n

\n\n\n\n\n\n\n\n\n \n \n \n \n \n \n\n\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\n
MixtureSeparation technique
(A)$$\\mathrm{CHCl_3+C_6H_5NH_2}$$$$\\to$$Distillation
(B)$$\\mathrm{C_6H_4+C_5H_{12}}$$$$\\to$$Fractional distillation
(C)$${C_6H_NH_2+H_2O}$$$$\\to$$Steam distillation
(D)Organic compound in $$\\mathrm{H_2O}$$$$\\to$$Differential extraction

", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 197, "subject": "Chemistry", "question": "Which of the following statement is correct for paper chromatography?", "options": [ { "text": "Water present in the mobile phase gets absorbed by the paper which then forms the stationary phase." }, { "text": "Water present in the pores of the paper forms the stationary phase." }, { "text": " Paper sheet forms the stationary phase." }, { "text": "Paper and water present in its pores together form the stationary phase." } ], "answer": "Water present in the pores of the paper forms the stationary phase.", "solution": "**Answer:** Water present in the pores of the paper forms the stationary phase.\n\nThe correct statement for paper chromatography is:\n

\n(B) Water present in the pores of the paper forms the stationary phase.\n

\nIn paper chromatography, the stationary phase is the water that is absorbed and trapped in the pores of the paper, while the mobile phase is a liquid solvent that moves through the paper carrying the sample with it. The different components of the sample have different affinities for the stationary phase and the mobile phase, causing them to move at different rates and separate from each other on the paper.", "topic": "Discrete Mathematics", "subtopic": "Other" }, { "id": 198, "subject": "Chemistry", "question": "

Using column chromatography, mixture of two compounds 'A' and 'B' was separated. 'A' eluted first, this indicates 'B' has

", "options": [ { "text": "high Rf, weaker adsorption" }, { "text": "high Rf, stronger adsorption" }, { "text": "low Rf, weaker adsorption" }, { "text": "low Rf, stronger adsorption" } ], "answer": "low Rf, stronger adsorption", "solution": "**Answer:** low Rf, stronger adsorption\n\n

In column chromatography, the compound that elutes first is the one that is less strongly adsorbed to the stationary phase. Hence, 'A' has weaker adsorption than 'B'. Therefore, 'B' has stronger adsorption.

\n

The retention factor (Rf) is a measure of how much a compound moves up the column; a higher Rf means the compound travels further and is less strongly adsorbed to the stationary phase. Since 'A' elutes first, 'A' has a higher Rf than 'B'. Hence, 'B' has a lower Rf.

\n

So, the correct answer is 'Option D: low Rf, stronger adsorption'.

\n", "topic": "Discrete Mathematics", "subtopic": "Other" }, { "id": 199, "subject": "Chemistry", "question": "Lassaigne's test is used for detection of :", "options": [ { "text": "Phosphorous and halogens only" }, { "text": "Nitrogen, Sulphur and Phosphorous only" }, { "text": "Nitrogen, Sulphur, phosphorous and halogens" }, { "text": "Nitrogen and Sulphur only" } ], "answer": "Nitrogen, Sulphur, phosphorous and halogens", "solution": "**Answer:** Nitrogen, Sulphur, phosphorous and halogens\n\n

Lassaigne's test is a qualitative test in analytical chemistry used to detect the presence of certain elements, namely nitrogen, sulfur, and halogens in an organic compound. The test involves heating the organic compound with sodium metal to convert these elements to sodium salts, which can then react with specific reagents to yield visibly identifiable compounds.

\n

Let's go through the options one by one:

\n

Option A suggests that Lassaigne's test is used for the detection of phosphorous and halogens only. This is incorrect because the test is also used to detect nitrogen and sulfur.

\n

Option B suggests that the test is for nitrogen, sulfur, and phosphorous only. However, Lassaigne's test does not typically involve a specific test for phosphorous.

\n

Option C, which is the correct answer, indicates that this test is used to detect nitrogen, sulfur, phosphorous, and halogens. Even though phosphorous detection isn't as common as the detection of nitrogen, sulfur, and halogens in the typical Lassaigne's test, the option stating all four elements should be considered the most complete answer among those provided, given its inclusion of all elements that could potentially be detected with adaptations of the test.

\n

Option D states that the test is for nitrogen and sulfur only, which is not entirely accurate because the test can also be used to detect halogens, and potentially, albeit less commonly, phosphorous.

\n

Therefore, the most accurate answer is Option C: Nitrogen, Sulphur, phosphorous, and halogens.

\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 200, "subject": "Chemistry", "question": "

Which among the following purification methods is based on the principle of \"Solubility\" in two different solvents?

", "options": [ { "text": "Distillation\n" }, { "text": "Sublimation\n" }, { "text": "Column Chromatography\n" }, { "text": "Differential Extraction" } ], "answer": "Differential Extraction", "solution": "**Answer:** Differential Extraction\n\n

The purification method among the ones listed that is based on the principle of \"Solubility\" in two different solvents is Option D, Differential Extraction.

\n\n

Differential extraction, also known as liquid-liquid extraction or solvent extraction, is a technique to separate compounds based on their relative solubilities in two different immiscible liquids, usually water and an organic solvent. It exploits the fact that the compound of interest has different solubility properties in two solvents which are not miscible with each other. A solution containing the desired compound and impurities is mixed with a second solvent in which the desired compound is more soluble. The two solvents form layers due to their immiscibility, and the compound partitions into the layer where it is more soluble. By separating the layers and repeating the process if necessary, a higher degree of purification can be achieved for the target compound.

\n\n

Let's quickly review why the other options do not primarily use solubility in two different solvents:

\n\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 201, "subject": "Chemistry", "question": "

Methods used for purification of organic compounds are based on :

", "options": [ { "text": "neither on nature of compound nor on the impurity present.\n" }, { "text": "presence of impurity only.\n" }, { "text": "nature of compound and presence of impurity.\n" }, { "text": "nature of compound only." } ], "answer": "nature of compound and presence of impurity.\n", "solution": "**Answer:** nature of compound and presence of impurity.\n\n\n

The methods used for the purification of organic compounds are fundamentally based on both the nature of the compound and the presence of the impurity. This is because the selection of a suitable purification technique requires understanding not only what the organic compound is (i.e., its physical and chemical properties) but also the type of impurities that need to be removed.

\n\n

For example, if an organic compound is a solid with a high melting point and the impurity is a low melting solid or a liquid, techniques like sublimation or crystallization could be used, which rely on the differences in physical properties between the compound and the impurity. On the other hand, if the compound needs to be purified from substances with similar boiling points, then a more refined method such as fractional distillation might be employed, relying on slight differences in boiling points. In cases where the compound is mixed with substances that have similar physical properties but different chemical reactivities, chemical methods such as extraction or chromatography could be used, which exploit these chemical differences.

\n\n

Therefore, the correct answer is:

\n

Option C: nature of compound and presence of impurity.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 202, "subject": "Chemistry", "question": "Due to the presence of an unpaired electron, free radicals are: ", "options": [ { "text": "Chemically reactive " }, { "text": "Chemically inactive " }, { "text": "Anions " }, { "text": "Cations" } ], "answer": "Chemically reactive ", "solution": "**Answer:** Chemically reactive \n\nFree radicals are electrically neutral, unstable and very reactive on account of the presence of odd electrons. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 203, "subject": "Chemistry", "question": "The increasing order of stability of the following free radicals is", "options": [ { "text": "$${({C_6}{H_5})_2}\\mathop C\\limits^ \\bullet H$$ < $${({C_6}{H_5})_3}\\mathop C\\limits^ \\bullet$$ < $${(C{H_3})_3}\\mathop C\\limits^ \\bullet $$ < $${(C{H_3})_2}\\mathop C\\limits^ \\bullet H$$" }, { "text": "$${(C{H_3})_2}\\mathop C\\limits^ \\bullet H$$ < $${(C{H_3})_3}\\mathop C\\limits^ \\bullet $$ < $${({C_6}{H_5})_2}\\mathop C\\limits^ \\bullet H$$ < $${({C_6}{H_5})_3}\\mathop C\\limits^ \\bullet$$ " }, { "text": "$${(C{H_3})_2}\\mathop C\\limits^ \\bullet H$$ < $${(C{H_3})_3}\\mathop C\\limits^ \\bullet $$ < $${({C_6}{H_5})_3}\\mathop C\\limits^ \\bullet$$ < $${({C_6}{H_5})_2}\\mathop C\\limits^ \\bullet H$$" }, { "text": "$${({C_6}{H_5})_3}\\mathop C\\limits^ \\bullet$$ < $${({C_6}{H_5})_2}\\mathop C\\limits^ \\bullet H$$ < $${(C{H_3})_3}\\mathop C\\limits^ \\bullet $$ < $${(C{H_3})_2}\\mathop C\\limits^ \\bullet H$$" } ], "answer": "$${(C{H_3})_2}\\mathop C\\limits^ \\bullet H$$ < $${(C{H_3})_3}\\mathop C\\limits^ \\bullet $$ < $${({C_6}{H_5})_2}\\mathop C\\limits^ \\bullet H$$ < $${({C_6}{H_5})_3}\\mathop C\\limits^ \\bullet$$ ", "solution": "**Answer:** $${(C{H_3})_2}\\mathop C\\limits^ \\bullet H$$ < $${(C{H_3})_3}\\mathop C\\limits^ \\bullet $$ < $${({C_6}{H_5})_2}\\mathop C\\limits^ \\bullet H$$ < $${({C_6}{H_5})_3}\\mathop C\\limits^ \\bullet$$ \n\nThe order of stability of free radicals \n

$${\\left( {{C_6}{H_5}} \\right)_3}\\,\\mathop C\\limits^ \\bullet > {\\left( {{C_6}{H_5}} \\right)_2}\\mathop C\\limits^ \\bullet H > {\\left( {C{H_3}} \\right)_3}\\mathop C\\limits^ \\bullet > {\\left( {C{H_3}} \\right)_2}\\mathop C\\limits^ \\bullet H$$ \n

The stabilisation of first two is due to resonance and last two is due to inductive effect. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 204, "subject": "Chemistry", "question": "Arrange the carbanions, $${(C{H_3})_3}\\overline C $$, $$\\overline C C{l_3}$$ , $${(C{H_3})_2}\\overline C H$$, $${C_6}{H_5}\\overline C H{}_2$$ in order of their decreasing stability : ", "options": [ { "text": "$${(C{H_3})_2}\\overline C H$$ > $$\\overline C C{l_3}$$ > $${C_6}{H_5}\\overline C H{}_2$$ > $${(C{H_3})_3}\\overline C $$" }, { "text": "$$\\overline C C{l_3}$$ > $${C_6}{H_5}\\overline C H{}_2$$ > $${(C{H_3})_2}\\overline C H$$ > $${(C{H_3})_3}\\overline C $$" }, { "text": "$${(C{H_3})_3}\\overline C $$ > $${(C{H_3})_2}\\overline C H$$ > $${C_6}{H_5}\\overline C H{}_2$$ > $$\\overline C C{l_3}$$" }, { "text": "$${C_6}{H_5}\\overline C H{}_2$$ > $$\\overline C C{l_3}$$ > $${(C{H_3})_3}\\overline C $$ > $${(C{H_3})_2}\\overline C H$$" } ], "answer": "$$\\overline C C{l_3}$$ > $${C_6}{H_5}\\overline C H{}_2$$ > $${(C{H_3})_2}\\overline C H$$ > $${(C{H_3})_3}\\overline C $$", "solution": "**Answer:** $$\\overline C C{l_3}$$ > $${C_6}{H_5}\\overline C H{}_2$$ > $${(C{H_3})_2}\\overline C H$$ > $${(C{H_3})_3}\\overline C $$\n\n\"AIEEE ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 205, "subject": "Chemistry", "question": "

A species having carbon with sextet of electrons and can act as electrophile is called

", "options": [ { "text": "pentavalent carbon\n" }, { "text": "carbon free radical\n" }, { "text": "carbanion\n" }, { "text": "carbocation" } ], "answer": "carbocation", "solution": "**Answer:** carbocation\n\n

The species described as having a carbon atom with a sextet of electrons—that is, six electrons in its valence shell—capable of acting as an electrophile (an electron-pair acceptor) is called a carbocation. Therefore, the correct answer is:

\n\n\nOption D\ncarbocation\n\"JEE\n\n

Here's a brief explanation of each option:

\n\n

Option A: Pentavalent carbon - This term is not commonly used in organic chemistry, and it refers incorrectly to a carbon with five bonds, which contradicts the tetravalency of carbon.

\n\n

Option B: Carbon free radical - This species has a carbon atom with an unpaired electron (a septet, or seven valence electrons). It is highly reactive but is not typically classified as an electrophile because it seeks to pair its single unpaired electron rather than accepting a pair of electrons.

\n\n

Option C: Carbanion - This anion has a carbon atom with three bonds and a pair of electrons, which in total gives eight valence electrons. It is nucleophilic (electron-pair donor) rather than electrophilic.

\n\n

Option D: Carbocation - A carbocation is indeed a carbon species with only six valence electrons and thus has an empty p-orbital that can accept a pair of electrons, making it electrophilic. Carbocations are usually trigonal planar in structure and carry a positive charge.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 206, "subject": "Chemistry", "question": "Complete hydrolysis of cellulose gives", "options": [ { "text": "D-ribose" }, { "text": "D-glucose" }, { "text": "L-glucose" }, { "text": "D-fructose" } ], "answer": "D-glucose", "solution": "**Answer:** D-glucose\n\nCellulose is a linear polymer of $$\\beta $$ $$ - D - $$ glucose in which $${C_1}$$ of one glucose unit is connected to $${C_4}$$ of the other through $$\\beta - D$$ glucosidic linkage. It does not undergo hydrolysis easily. However on heating with dilute $${H_2}S{O_4}$$ under pressure. It does undergo hydrolysis to give only $$D-$$ glucose. \n

$$$\\left( {{C_6}{H_{10}}{O_5}} \\right)n + n{H_2}O\\buildrel {\\,H + } \\over\n \\longrightarrow \\mathop {n{C_6}{H_{12}}{O_6}}\\limits_{D - Glu\\cos e} $$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 207, "subject": "Chemistry", "question": "The term anomers of glucose refers to ", "options": [ { "text": "isomers of glucose that differ in configurations at carbons one and four (C-1 and C-4) " }, { "text": "a mixture of (D)-glucose and (L)-glucose" }, { "text": "enantiomers of glucose" }, { "text": "isomers of glucose that differ in configuration at carbon one (C-1) " } ], "answer": "isomers of glucose that differ in configuration at carbon one (C-1) ", "solution": "**Answer:** isomers of glucose that differ in configuration at carbon one (C-1) \n\nCyclization of the open chain structure of $$D$$-$$(+)$$- glucose has created a new stereocenter at $${C_1}$$ which explains the existance of two cyclic forms of $$D$$-$$(+)$$- glucose, namely $$\\alpha - $$ and $$\\beta - $$. These two cyclic forms are diasteromers, such diasteromers which differ only in the configuration of chiral carbon developed on hemiacetal formation (it is $${C_1}$$ in glucose and $${C_2}$$ in fructose) are called anomers and the hemiacetal carbon ($${C_1}$$ or $${C_2}$$) is called the anomeric carbon.\n

\"AIEEE ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 208, "subject": "Chemistry", "question": "$$\\alpha$$-D-(+)-glucose and $$\\beta$$-D-(+)-glucose are ", "options": [ { "text": "conformers " }, { "text": "epimers" }, { "text": "anomers" }, { "text": "enantiomers " } ], "answer": "anomers", "solution": "**Answer:** anomers\n\nSince $$\\alpha - D - \\left( + \\right) - \\,\\,$$ glucose and $$\\beta - D - \\left( + \\right)\\,\\,$$ glucose differ in configuration at $$C-1$$ atom so they are anomers. \n

NOTE : Anomers are those diastereomers that differ in configuration at $$C-1$$ atom. i.e., $$(c)$$ in the correct answer. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 209, "subject": "Chemistry", "question": "The two functional groups present in a typical carbohydrate are ", "options": [ { "text": "-OH and -COOH " }, { "text": "-CHO and -COOH " }, { "text": "> C = O and - OH " }, { "text": "- OH and -CHO" } ], "answer": "- OH and -CHO", "solution": "**Answer:** - OH and -CHO\n\nNOTE : Glucose is considered as a typical carbohydrate which contains $$-CHO$$ and $$-OH$$ group. ", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 210, "subject": "Chemistry", "question": "Which of the following compounds can be detected by Molisch’s test ?", "options": [ { "text": "Nitro compounds" }, { "text": "Sugars" }, { "text": "Amines " }, { "text": "Primary alcohols" } ], "answer": "Sugars", "solution": "**Answer:** Sugars\n\nMolisch's Test : This is a general test for carbohydrates. One or two drops of alcoholic solution of $$\\alpha $$-naphthol is added to $$2$$ ml glucose solution. $$1$$ ml of conc. $${H_2}S{O_4}$$ solution is added carefully along the sides of the test-tube. The formation of a violet ring at the junction of two liquids confirms the presence of a carbohydrare or sugar.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 211, "subject": "Chemistry", "question": "The incorrect statement among the following is :", "options": [ { "text": "$$\\alpha $$-D-glucose and $$\\beta $$-D-glucose are anomers." }, { "text": "$$\\alpha $$-D-glucose and $$\\beta $$-D-glucose are enantiomers.\n" }, { "text": "Cellulose is a straight chain polysaccharide made up of only $$\\beta $$-D-glucose units." }, { "text": "The penta acetate of glucose does not react with hydroxyl amine" } ], "answer": "$$\\alpha $$-D-glucose and $$\\beta $$-D-glucose are enantiomers.\n", "solution": "**Answer:** $$\\alpha $$-D-glucose and $$\\beta $$-D-glucose are enantiomers.\n\n\n$$\\alpha $$-D-glucose and $$\\beta $$-D-glucose are anomers not enantiomers.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 212, "subject": "Chemistry", "question": "Glucose on prolonged heating with HI gives", "options": [ { "text": "6-iodohexanal" }, { "text": "n-Hexane" }, { "text": "1-Hexene" }, { "text": "Hexanoic acid" } ], "answer": "n-Hexane", "solution": "**Answer:** n-Hexane\n\nAs HI is a strong reducing agent, it reduces all the aldehydes and ketones to acohol and alcohols are further reduce to alkanes. \n

\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 213, "subject": "Chemistry", "question": "Among the following, the incorrect statement is : ", "options": [ { "text": "Maltose and lactose has 1, 4 - glycosidic linkage. " }, { "text": "Sucrose and amylose has 1, 2 - glycosidic linkage" }, { "text": "Cellulose and amylose has 1, 4 - glycosidic linkage. " }, { "text": "Lactose contains $$\\beta $$-D-galactose and $$\\beta $$-D-glucose. " } ], "answer": "Sucrose and amylose has 1, 2 - glycosidic linkage", "solution": "**Answer:** Sucrose and amylose has 1, 2 - glycosidic linkage\n\n(a) Both maltose and lactose have 1,4-glycosidic linkage. \n

(b) In sucrose, C1-$$\\alpha $$ of glucose is connected to C2-$$\\beta $$ of fructose. In amylose, C1 of one glucose unit is attached to C4 of other glucose through $$\\alpha $$-glycosidic linkage.\n

(c) Cellulose has 1,4-$$\\beta $$-D-glycosidic linkage, but amylose has 1,4-$$\\alpha $$-D-glycosidic linkage. \n

(d) In lactose, C1-$$\\beta $$ of galactose is linked to C4-$$\\beta $$ of glucose.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 214, "subject": "Chemistry", "question": "Which of the following statements is not true\nabout sucrose?", "options": [ { "text": "The glycosidic linkage is present between\nC1 of $$\\alpha $$-glucose and C1 of $$\\beta $$-fructose" }, { "text": "It is also named as invert sugar" }, { "text": "It is a non reducing sugar" }, { "text": "On hydrolysis, it produces glucose and\nfructose" } ], "answer": "The glycosidic linkage is present between\nC1 of $$\\alpha $$-glucose and C1 of $$\\beta $$-fructose", "solution": "**Answer:** The glycosidic linkage is present between\nC1 of $$\\alpha $$-glucose and C1 of $$\\beta $$-fructose\n\nThe glycosidic linkage is present between\nC1 of $$\\alpha $$-D-glucose and C2 of $$\\beta $$-D-fructose.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 215, "subject": "Chemistry", "question": "Which of the given statements is INCORRECT about glycogen? ", "options": [ { "text": "It is a straight chain polymer similar to amylose" }, { "text": "It is present in animal cells" }, { "text": "It is present in some yeast and fungi" }, { "text": "Only $$\\alpha $$-linkages are present in the molecule" } ], "answer": "It is a straight chain polymer similar to amylose", "solution": "**Answer:** It is a straight chain polymer similar to amylose\n\nGlycogen is branch chain polymer.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 216, "subject": "Chemistry", "question": "Glucose and Galactose are having identical configuration in all the positions except position", "options": [ { "text": "C–5" }, { "text": "C–3" }, { "text": "C–2" }, { "text": "C–4" } ], "answer": "C–4", "solution": "**Answer:** C–4\n\nGlucose and galactose differs in configuration on C4.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 217, "subject": "Chemistry", "question": "Number of stereo centers present in linear and cyclic structures of glucose are respectively :", "options": [ { "text": "4 & 4\n" }, { "text": "4 & 5" }, { "text": "5 & 4" }, { "text": "5 & 5" } ], "answer": "4 & 5", "solution": "**Answer:** 4 & 5\n\n\"JEE\n
In this linear structure 4 chiral carbons present.\n\"JEE\n
In this cyclic structure 5 chiral carbons present.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 218, "subject": "Chemistry", "question": "Amylopectin is compound of :", "options": [ { "text": "$$\\alpha $$-D-glucose, C1 – C4 and C1 – C6 linkages" }, { "text": "$$\\beta $$-D-glucose, C1 – C4 and C2 – C6 linkages" }, { "text": "$$\\beta $$-D-glucose, C1 – C4 and C1 – C6 linkages" }, { "text": "$$\\alpha $$-D-glucose, C1 – C4 and C2 – C6 linkages" } ], "answer": "$$\\alpha $$-D-glucose, C1 – C4 and C1 – C6 linkages", "solution": "**Answer:** $$\\alpha $$-D-glucose, C1 – C4 and C1 – C6 linkages\n\nAmylopectin is a component of starch and starch is a polymer of $$\\alpha $$-D-glucose.\n

Amylopectin has $$\\alpha $$-1, 6-glycosidic linkage in\naddition to $$\\alpha $$-1,4-glycosidic linkage and is a\ncross-linked polymer.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 219, "subject": "Chemistry", "question": "Fructose and glucose can be distinguished by :", "options": [ { "text": "Barfoed's test" }, { "text": "Fehling's test" }, { "text": "Seliwanoff's test" }, { "text": "Benedict's test" } ], "answer": "Seliwanoff's test", "solution": "**Answer:** Seliwanoff's test\n\nSeliwanoff's test is used to distinguish aldose\nand ketose.\n

When Seliwanoff's reagent added to a solution containing ketoses, red colour is formed rapidly indicating a positive test. \n
While in case of aldose, a slower forming light pink colour observed.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 220, "subject": "Chemistry", "question": "Maltose on treatment with dilute HCI gives :", "options": [ { "text": "D-Galactose" }, { "text": "D-Glucose and D-Fructose" }, { "text": "D-Glucose" }, { "text": "D-Fructose" } ], "answer": "D-Glucose", "solution": "**Answer:** D-Glucose\n\nMaltose on treatment with dil. HCl gives D-glucose. Hydrolysis of\nmaltose yields two moles of $$\\alpha $$- D-glucose. Thus, it is composed of two $$\\alpha $$-D-glucose units in\nwhich C-1 of one glucose unit (I) is linked to C-4 of another\nglucose unit (II). The free aldehyde group can be produced at C-1\nof second glucose in solution and it shows reducing properties.\n
So, it is a reducing sugar.

\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 221, "subject": "Chemistry", "question": "Which of the following statement is not true for glucose?", "options": [ { "text": "Glucose gives Schiff's test for aldehyde" }, { "text": "Glucosse reacts with hydroxylamine to form oxime" }, { "text": "The pentaacetate of glucose does not react with hydroxylamine to give oxime" }, { "text": "Glucose exists in two crystalline forms $$\\alpha $$ and $$\\beta $$" } ], "answer": "Glucose gives Schiff's test for aldehyde", "solution": "**Answer:** Glucose gives Schiff's test for aldehyde\n\nGlucose gives negative test with Schiff reagent.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 222, "subject": "Chemistry", "question": "Which one of the following statements is not\ntrue?", "options": [ { "text": "Lactose contains $$\\alpha $$-glycosidic linkage\nbetween C1 of galactose and C4 of glucose" }, { "text": "Lactose is a reducing sugar and it gives\nFehling’s test" }, { "text": "On acid hydrolysis, lactose gives one\nmolecule of D(+)-glucose and one molecule\nof D(+)-galactose" }, { "text": "Lactose (C11H22O11) is a disaccharide and it\ncontains 8 hydroxyl groups" } ], "answer": "Lactose contains $$\\alpha $$-glycosidic linkage\nbetween C1 of galactose and C4 of glucose", "solution": "**Answer:** Lactose contains $$\\alpha $$-glycosidic linkage\nbetween C1 of galactose and C4 of glucose\n\n\"JEE\nLactose contains $$\\beta $$-glycosidic linkage between C1\n of galactose and C4\n of glucose.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 223, "subject": "Chemistry", "question": "The number of chiral carbons present in\nsucrose is _____.", "options": [], "answer": "9", "solution": "**Answer:** 9\n\n\"JEE\n

Total number of chiral carbon in sucrose = 9", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 224, "subject": "Chemistry", "question": "What are the functional groups present in the\nstructure of maltose?", "options": [ { "text": "One acetal and one ketal" }, { "text": "One ketal and one hemiketal" }, { "text": "Two acetals" }, { "text": "One acetal and one hemiacetal" } ], "answer": "One acetal and one hemiacetal", "solution": "**Answer:** One acetal and one hemiacetal\n\n\"JEE\n

Maltose contains one acetal and one\nhemiacetal.", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 225, "subject": "Chemistry", "question": "Two monomers in maltose are :", "options": [ { "text": "$$\\alpha $$-D-glucose and $$\\alpha $$-D-galactose" }, { "text": "$$\\alpha $$-D-glucose and $$\\alpha $$-D-Fructose" }, { "text": "$$\\alpha $$-D-glucose and $$\\alpha $$-D-glucose" }, { "text": "$$\\alpha $$-D-glucose and $$\\beta $$-D-glucose" } ], "answer": "$$\\alpha $$-D-glucose and $$\\alpha $$-D-glucose", "solution": "**Answer:** $$\\alpha $$-D-glucose and $$\\alpha $$-D-glucose\n\nTwo monomers in maltose are $$\\alpha $$-D-glucose &\n$$\\alpha $$-D-glucose.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 226, "subject": "Chemistry", "question": "Which of the following statements is correct ?\n", "options": [ { "text": "Gluconic acid can from cyclic (acteal/hemiacetal) structure" }, { "text": "Gluconic acid is a partial oxidation product of glucose" }, { "text": "Gluconic acid is obtained by oxidation of glucose with (HNO3)" }, { "text": "Gluconic acid is a dicarboxylic acid." } ], "answer": "Gluconic acid is a partial oxidation product of glucose", "solution": "**Answer:** Gluconic acid is a partial oxidation product of glucose\n\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 227, "subject": "Chemistry", "question": "Which of the glycosidic linkage between galactose and glucose is present in lactose?", "options": [ { "text": "C-1 of galactose and C-6 of glucose" }, { "text": "C-1 of glucose and C-6 of galactose" }, { "text": "C-1 of galactose and C-4 of glucose" }, { "text": "C-1 of glucose and C-4 of galactose" } ], "answer": "C-1 of galactose and C-4 of glucose", "solution": "**Answer:** C-1 of galactose and C-4 of glucose\n\n

Glycosidic linkage is a type of covalent bond that joins\ncarbohydrate molecules to another group.

\n

\"JEE

\n

In lactose, the glycosidic linkage is formed between C-1 of galactose\nand C-4, glycosidic of glucose.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 228, "subject": "Chemistry", "question": "Match List - I with List - II.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
List - I List - II
(a)Sucrose(i)$$\\beta $$-D-Galactose and $$\\beta $$-D-Glucose
(b)Lactose(ii)$$\\alpha $$-D-Glucose and $$\\beta $$-D-Fructose
(c)Maltose(iii)$$\\alpha $$-D-Glucose and $$\\alpha $$-D-Glucose

Choose the correct answer from the options given below :", "options": [ { "text": "(a) $$ \\to $$ (iii), (b) $$ \\to $$ (ii), (c) $$ \\to $$ (i)" }, { "text": "(a) $$ \\to $$ (i), (b) $$ \\to $$ (iii), (c) $$ \\to $$ (ii)" }, { "text": "(a) $$ \\to $$ (ii), (b) $$ \\to $$ (i), (c) $$ \\to $$ (iii)" }, { "text": "(a) $$ \\to $$ (iii), (b) $$ \\to $$ (i), (c) $$ \\to $$ (ii)" } ], "answer": "(a) $$ \\to $$ (ii), (b) $$ \\to $$ (i), (c) $$ \\to $$ (iii)", "solution": "**Answer:** (a) $$ \\to $$ (ii), (b) $$ \\to $$ (i), (c) $$ \\to $$ (iii)\n\nSucrose $$ \\to \\alpha $$-D- Glucose and $$\\beta $$-D- Fructose

\nLactose $$ \\to \\beta $$-D- Galactose and $$\\beta $$-D- Glucose

\nMaltose $$ \\to \\alpha $$-D- Glucose and $$\\alpha $$-D- Glucose", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 229, "subject": "Chemistry", "question": "$${C_{12}}{H_{22}}{O_{11}} + {H_2}O\\buildrel {Enzyme\\,A} \\over\n \\longrightarrow \\mathop {{C_6}{H_{12}}{O_6}}\\limits_{Glu\\cos e} + \\mathop {{C_6}{H_{12}}{O_6}}\\limits_{Fructose} $$

$$\\mathop {{C_6}{H_{12}}{O_6}}\\limits_{Glu\\cos e} \\buildrel {Enzyme\\,B} \\over\n \\longrightarrow 2{C_2}{H_5}OH + 2C{O_2}$$

In the above reactions, the enzyme A and enzyme B respectively are :", "options": [ { "text": "Zymase and Invertase" }, { "text": "Invertase and Amylase" }, { "text": "Amylase and Invertase" }, { "text": "Invertase and Zymase" } ], "answer": "Invertase and Zymase", "solution": "**Answer:** Invertase and Zymase\n\n\"JEE

$$ \\therefore $$ Enzyme A is Invertase
\nEnzyme B is Zymase", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 230, "subject": "Chemistry", "question": "Fructose is an example of :", "options": [ { "text": "Heptose" }, { "text": "Pyranose" }, { "text": "Ketohexose" }, { "text": "Aldohexose" } ], "answer": "Ketohexose", "solution": "**Answer:** Ketohexose\n\nFructose contains ketone group so it is ketose.

\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 231, "subject": "Chemistry", "question": "A non-reducing sugar \"A\" hydrolyses to give two reducing mono saccharides. Sugar A is ", "options": [ { "text": "Fructose" }, { "text": "Galactose" }, { "text": "Glucose" }, { "text": "Sucrose" } ], "answer": "Sucrose", "solution": "**Answer:** Sucrose\n\nSucrose $$\\buildrel {Hydrolyses} \\over\n \\longrightarrow $$ glucose + fructose

\nGlucose and fructose both are monosaccharides.\nSucrose is non-reducing sugar.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 232, "subject": "Chemistry", "question": "Identify the incorrect statement from the following", "options": [ { "text": "Amylose is a branched chain polymer of glucose" }, { "text": "Starch is a polymer of $$\\alpha$$-D glucose" }, { "text": "$$\\beta$$-Glycosidic linkage makes cellulose polymer" }, { "text": "Glycogen is called as animal starch" } ], "answer": "Amylose is a branched chain polymer of glucose", "solution": "**Answer:** Amylose is a branched chain polymer of glucose\n\nAmylose is a linear chain polymer of $$\\alpha $$-D-glucose\nwhile amylopectine is branched chain polymer of\n$$\\alpha $$-D-glucose.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 233, "subject": "Chemistry", "question": "Which one among the following chemical tests is used to distinguish monosaccharide from disaccharide?", "options": [ { "text": "Seliwanoff's test" }, { "text": "Iodine test" }, { "text": "Barfoed test" }, { "text": "Tollen's test" } ], "answer": "Barfoed test", "solution": "**Answer:** Barfoed test\n\nBarford test is used for distinguish monosaccharide from disaccharide.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 234, "subject": "Chemistry", "question": "Compound A gives D-Galactose and D-Glucose on hydrolysis. The compound A is :", "options": [ { "text": "Amylose" }, { "text": "Sucrose" }, { "text": "Maltose" }, { "text": "Lactose" } ], "answer": "Lactose", "solution": "**Answer:** Lactose\n\nLactose : It is a disaccharide of $$\\beta$$-D-Galactose and $$\\beta$$-D-Glucose with C1 of galactose and C4 of glucose link.

Lactose : $$\\beta$$-D-Galactose + $$\\beta$$-D-Glucose", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 235, "subject": "Chemistry", "question": "Given below are two statements : one is labelled as Assertion (A) and other is labelled as Reason (R).

Assertion (A) : Sucrose is a disaccharide and a non-reducing sugar.

Reason (R) : Sucrose involves glycosidic linkage between C1 of $$\\beta$$-glucose and C2 of $$\\alpha$$-fructose.

Choose the most appropriate answer from the options given below :", "options": [ { "text": "Both (A) and (R) are true but (R) is not the true explanation of (A)." }, { "text": "(A) is false but (R) is true." }, { "text": "(A) is true but (R) is false." }, { "text": "Both (A) and (R) are true and (R) is the true explanation of (A)" } ], "answer": "(A) is true but (R) is false.", "solution": "**Answer:** (A) is true but (R) is false.\n\nSucrose is example of disaccharide & non reducing sugar

Assertion : Correct

Sucrose involves glycosidic linkage between C1 of $$\\alpha$$-D-glucose C2 of $$\\beta$$-D-fructose

Reason : Incorrect", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 236, "subject": "Chemistry", "question": "Hydrolysis of sucrose gives :", "options": [ { "text": "$$\\alpha$$-D-($$-$$)-Glucose and $$\\beta$$-D-($$-$$)-Fructose" }, { "text": "$$\\alpha$$-D-(+)-Glucose and $$\\alpha$$-D-($$-$$)-Fructose" }, { "text": "$$\\alpha$$-D-($$-$$)-Glucose and $$\\alpha$$-D-(+)-Fructose" }, { "text": "$$\\alpha$$-D-(+)-Glucose and $$\\beta$$-D-($$-$$)-Fructose" } ], "answer": "$$\\alpha$$-D-(+)-Glucose and $$\\beta$$-D-($$-$$)-Fructose", "solution": "**Answer:** $$\\alpha$$-D-(+)-Glucose and $$\\beta$$-D-($$-$$)-Fructose\n\nSucrose is formed by $$\\alpha$$-D(+). Glucose + $$\\beta$$-D($$-$$) Fructose.

We obtain these monomers on hydrolysis.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 237, "subject": "Chemistry", "question": "Which one of the following compounds contains $$\\beta$$-C1-C4 glycosidic linkage?", "options": [ { "text": "Lactose" }, { "text": "Sucrose" }, { "text": "Maltose" }, { "text": "Amylose" } ], "answer": "Lactose", "solution": "**Answer:** Lactose\n\nIn Lactose it is $$\\beta$$ C1 - C4 glycosidic linkage.

In Maltose, Amylose $$\\alpha$$ C1 - C4 glycosidic linkage is present.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 238, "subject": "Chemistry", "question": "

C6H12O6 $$\\buildrel \\text{Zymase} \\over\n \\longrightarrow $$ A $$\\mathrel{\\mathop{\\kern0pt\\longrightarrow}\n\\limits_\\Delta ^\\text{NaOI}} $$ B + CHI3

\n

The number of carbon atoms present in the product B is _______________.

", "options": [], "answer": "1", "solution": "**Answer:** 1\n\n$$\\mathrm{C}_6 \\mathrm{H}_{12} \\mathrm{O}_6 \\stackrel{\\text { Zymase }}{\\longrightarrow} \\mathrm{C}_2 \\mathrm{H}_5 \\mathrm{OH} \\stackrel{\\mathrm{NaOl}}{\\longrightarrow} \\underset{\\text { (B) }}{\\mathrm{HCOONa}}+\\mathrm{CHI}_3\n$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 239, "subject": "Chemistry", "question": "

When sugar 'X' is boiled with dilute H2SO4 in alcoholic solution, two isomers 'A' and 'B' are formed. 'A' on oxidation with HNO3 yields saccharic acid where as 'B' is laevorotatory. The compound 'X' is :

", "options": [ { "text": "Maltose" }, { "text": "Sucrose" }, { "text": "Lactose" }, { "text": "Starch" } ], "answer": "Sucrose", "solution": "**Answer:** Sucrose\n\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 240, "subject": "Chemistry", "question": "

Given below are two statements.

\n

Statement I : Maltose has two $$\\alpha$$-D-glucose units linked at C1 and C4 and is a reducing sugar.

\n

Statement II : Maltose has two monosaccharides : $$\\alpha$$-D-glucose and $$\\beta$$-D-glucose linked at C1 and C6 and it is a non-reducing sugar.

\n

In the light of the above statements, choose the correct answer from the options given below :

", "options": [ { "text": "Both Statement I and Statement II are true." }, { "text": "Both Statement I and Statement II are false." }, { "text": "Statement I is true but Statement II is false." }, { "text": "Statement I is false but Statement II is true." } ], "answer": "Statement I is true but Statement II is false.", "solution": "**Answer:** Statement I is true but Statement II is false.\n\nMaltose is composed of two $\\alpha$-D-glucose units in which $C_{1}$ of one glucose unit and $C_{4}$ of second glucose unit are linked.

\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 241, "subject": "Chemistry", "question": "

A polysaccharide 'X' on boiling with dil H2SO4 at 393 K under 2-3 atm pressure yields 'Y'. 'Y' on treatment with bromine water gives gluconic acid. 'X' contains $$\\beta$$-glycosidic linkages only. Compound 'X' is :

", "options": [ { "text": "starch" }, { "text": "cellulose" }, { "text": "amylose" }, { "text": "amylopectin" } ], "answer": "cellulose", "solution": "**Answer:** cellulose\n\nCellulose contains $\\beta$-glycosidic linkages only. Structure of cellulose

\n\"JEE
\nOn boiling with dil. $\\mathrm{H}_{2} \\mathrm{SO}_{4}$ at $393 \\mathrm{~K}$ under 2-3 atm, ' $X$ ' forms glucose, which given gluconic acid on treatment with bromine water.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 242, "subject": "Chemistry", "question": "

Glycosidic linkage between C1 of $$\\alpha$$-glucose and C2 of $$\\beta$$-fructose is found in

", "options": [ { "text": "maltose" }, { "text": "sucrose" }, { "text": "lactose" }, { "text": "amylose" } ], "answer": "sucrose", "solution": "**Answer:** sucrose\n\n

\"JEE

\n

Hence in sucrose glycosidic linkage between $\\mathrm{C}_{1}$ of $\\alpha$-glucose and $C_{2}$ of $\\beta$-D-fructose is found\n

\nMaltose $\\Rightarrow$ Glycosidic linkage between $\\mathrm{C}_{1}$ and $\\mathrm{C}_{4}$\n

\nLactose $\\Rightarrow$ Glycosidic linkage between $\\mathrm{C}_{1}$ and $\\mathrm{C}_{4}$\n

\nAmylose $\\Rightarrow$ Glycosidic linkage between $\\mathrm{C}_{1}$ and $\\mathrm{C}_{4}$\n

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 243, "subject": "Chemistry", "question": "

Animal starch is the other name of

", "options": [ { "text": "amylose." }, { "text": "maltose." }, { "text": "glycogen." }, { "text": "amylopectin." } ], "answer": "glycogen.", "solution": "**Answer:** glycogen.\n\nAnimal starch is the other name of glycogen\nbecause its structure is similar to amylopectin.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 244, "subject": "Chemistry", "question": "

A sugar 'X' dehydrates very slowly under acidic condition to give furfural which on further reaction with resorcinol gives the coloured product after sometime. Sugar 'X' is

", "options": [ { "text": "Aldopentose" }, { "text": "Aldotetrose" }, { "text": "Oxalic acid" }, { "text": "Ketotetrose" } ], "answer": "Aldopentose", "solution": "**Answer:** Aldopentose\n\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 245, "subject": "Chemistry", "question": "

Match List - I with Match List - II.

\n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
List - IList - II
(A)Glucose + HI(I)Gluconic acid
(B)Glucose + Br$$_2$$ water(II)Glucose pentacetate
(C)Glucose + acetic anhydride(III)Saccharic acid
(D)Glucose + HNO$$_3$$(IV)Hexane

\n

Choose the correct answer from the options given below:

", "options": [ { "text": "(A) - (IV), (B) - (I), (C) -(II), (D) - (III)" }, { "text": "(A) - (IV), (B) - (III), (C) -(II), (D) - (I)" }, { "text": "(A) - (III), (B) - (I), (C) -(IV), (D) - (II)" }, { "text": "(A) - (I), (B) - (III), (C) -(IV), (D) - (II)" } ], "answer": "(A) - (IV), (B) - (I), (C) -(II), (D) - (III)", "solution": "**Answer:** (A) - (IV), (B) - (I), (C) -(II), (D) - (III)\n\n\"JEE", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 246, "subject": "Chemistry", "question": "

Given below are two statements. One is labelled as Assertion A and the other is labelled as Reason R.

\n

Assertion A: Amylose is insoluble in water.

\n

Reason R: Amylose is a long linear molecule with more than 200 glucose units.

\n

In the light of the above statements, choose the correct answer from the options given below.

", "options": [ { "text": "Both $$\\mathbf{A}$$ and $$\\mathbf{R}$$ are correct and $$\\mathbf{R}$$ is the correct explanation of $$\\mathbf{A}$$." }, { "text": "Both $$\\mathbf{A}$$ and $$\\mathbf{R}$$ are correct but $$\\mathbf{R}$$ is NOT the correct explanation of $$\\mathbf{A}$$." }, { "text": "$$\\mathbf{A}$$ is correct but $$\\mathbf{R}$$ is not correct" }, { "text": "$$\\mathbf{A}$$ is not correct but $$\\mathbf{R}$$ is correct." } ], "answer": "$$\\mathbf{A}$$ is not correct but $$\\mathbf{R}$$ is correct.", "solution": "**Answer:** $$\\mathbf{A}$$ is not correct but $$\\mathbf{R}$$ is correct.\n\n

Let's analyze the given statements:

\n

Assertion A: "Amylose is insoluble in water." This statement is incorrect. Amylose is actually soluble in water, although its solubility is lower compared to other carbohydrates like amylopectin.

\n

Reason R: "Amylose is a long linear molecule with more than 200 glucose units." This statement is correct, as amylose is indeed a linear polysaccharide made up of glucose units.

\n

Given this information, the correct option is:

\n

Option D: $\\mathbf{A}$ is not correct but $\\mathbf{R}$ is correct.

\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 247, "subject": "Chemistry", "question": "

Match List I with List II

\n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
List IList II
TestFunctional group / Class of Compound
A.Molisch's TestI.Peptide
B.Biuret TestII.Carbohydrate
C.Carbylamine TestIII.Primary amine
D.Schiff's TestIV.Aldehyde

\n

Choose the correct answer from the options given below :

", "options": [ { "text": "(A) - III, (B) - IV, (C) - II, (D) - I" }, { "text": "(A) - II, (B) - I, (C) - III, (D) - IV" }, { "text": "(A) - III, (B) - IV, (C) - I, (D) - II" }, { "text": "(A) - I, (B) - II, (C) - III, (D) - IV" } ], "answer": "(A) - II, (B) - I, (C) - III, (D) - IV", "solution": "**Answer:** (A) - II, (B) - I, (C) - III, (D) - IV\n\n(A) Molisch test is for carbohydrates\n

(B) Biuret test is for proteins/peptide\n

(C) Carbylamine test is for primary amine\n

(D) Schiff’s test is for aldehyde", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 248, "subject": "Chemistry", "question": "

Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).

\n

Assertion (A) : Ketoses give Seliwanoff's test faster than Aldoses.

\n

Reason (R) : Ketoses undergo $$\\beta$$-elimination followed by formation of furfural.

\n

In the light of the above statements, choose the correct answer from the options given below :

", "options": [ { "text": "(A) is false but (R) is true" }, { "text": "Both (A) and (R) are true but (R) is not the correct explanation of (A)" }, { "text": "(A) is true but (R) is false" }, { "text": "Both (A) and (R) are true and (R) is the correct explanation of (A)" } ], "answer": "(A) is true but (R) is false", "solution": "**Answer:** (A) is true but (R) is false\n\nSeliwanoff 's test is a differentiating test for Ketose\nand aldose. This test relies on the principle that the\nketo hexose are more rapidly dehydrated to form\n5-hydroxy methyl furfural when heated in acidic\nmedium which on condensation with resorcinol,\nCherry red or brown red coloured complex is\nformed rapidly indicating a positive test.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 249, "subject": "Chemistry", "question": "

Number of compounds from the following which will not produce orange red precipitate with Benedict solution is ___________.

\n

Glucose, maltose, sucrose, ribose, 2-deoxyribose, amylose, lactose

", "options": [], "answer": "2", "solution": "**Answer:** 2\n\nBenedict's solution is a reagent that tests for reducing sugars. A reducing sugar is one that has a free aldehyde or ketone group, which can reduce the copper(II) ions in Benedict's reagent to copper(I) oxide, which forms a brick-red or orange-red precipitate.\n

Given the compounds :\n\n

1. Glucose - is a reducing sugar, as it has a free aldehyde group.\n

2. Maltose - is a reducing sugar. It has a free anomeric carbon that can form an aldehyde group.\n

3. Sucrose - is not a reducing sugar. Its glycosidic bond locks the anomeric carbons, preventing them from forming an aldehyde or ketone.\n

4. Ribose - is a reducing sugar, as it has an aldehyde group.\n

5. 2-deoxyribose - is a reducing sugar. It has a free anomeric carbon that can form an aldehyde group.\n

6. Amylose - is a polysaccharide and does not typically behave as a reducing sugar in Benedict's test. Even though one end of the amylose chain can technically open to reveal an aldehyde group (the so-called reducing end), the reaction would likely be much less dramatic than with true reducing sugars, and might not occur at all under typical test conditions.\n

7. Lactose - is a reducing sugar, as it has a free anomeric carbon that can form an aldehyde group.\n\n

Hence, out of the seven compounds listed, sucrose and amylose will not likely produce an orange-red precipitate with Benedict's solution. Thus, the number of compounds that will not produce an orange-red precipitate with Benedict's solution is 2.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 250, "subject": "Chemistry", "question": "

Match List I with List II

\n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
List - IList - II
(A)Glucose/$$\\mathrm{NaHCO_3/\\Delta}$$(I)Gluconic acid
(B)$$\\text { Glucose } / \\mathrm{HNO}_3$$(II)No reaction
(C)$$\\text { Glucose } / \\mathrm{HI} / \\Delta$$(III)n-hexane
(D)Glucose/Bromine water(IV)Saccharic acid

\n

Choose the correct answer from the options given below:

", "options": [ { "text": "A-II, B-IV, C-III, D-I" }, { "text": "A-III, B-II, C-I, D-IV\n" }, { "text": "A-IV, B-I, C-III, D-II\n" }, { "text": "A-I, B-IV, C-III, D-II" } ], "answer": "A-II, B-IV, C-III, D-I", "solution": "**Answer:** A-II, B-IV, C-III, D-I\n\n

Glucose $$\\xrightarrow[\\Delta]{\\mathrm{NaHCO}_3}$$ no reaction

\n

Glucose $$\\xrightarrow[\\Delta]{\\mathrm{HNO}_3}$$ saccharic acid

\n

Glucose $$\\xrightarrow[\\Delta]{\\mathrm{HI}}$$ n-hexane

\n

Glucose $$\\xrightarrow[\\Delta]{\\mathrm{Br}_2}$$ Gluconic acid

", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 251, "subject": "Chemistry", "question": "

Match List I with List II

\n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
List - I
(Bio Polymer)		List - II
(Monomer)
(A)Starch(I)nucleotide
(B)Cellulose(II)$$\\alpha$$-glucose
(C)Nucleic acid(III)$$\\beta$$-glucose
(D)Protein(IV)$$\\alpha$$-amino acid

\n

Choose the correct answer from the options given below:

", "options": [ { "text": "A-IV, B-II, C-I, D-III\n" }, { "text": "A-I, B-III, C-IV, D-II\n" }, { "text": "A-II, B-I, C-III, D-IV\n" }, { "text": "A-II, B-III, C-I, D-IV" } ], "answer": "A-II, B-III, C-I, D-IV", "solution": "**Answer:** A-II, B-III, C-I, D-IV\n\n

A-II, B-III, C-I, D-IV

\n

Fact based.

", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 252, "subject": "Chemistry", "question": "

Sugar which does not give reddish brown precipitate with Fehling's reagent, is :

", "options": [ { "text": "Glucose\n" }, { "text": "Sucrose\n" }, { "text": "Lactose\n" }, { "text": "Maltose" } ], "answer": "Sucrose\n", "solution": "**Answer:** Sucrose\n\n\n

Sucrose do not contain hemiacetal group.

\n

Hence it does not give test with Fehling solution.

\n

While all other give positive test with Fehling solution

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 253, "subject": "Chemistry", "question": "

The incorrect statement about Glucose is :

", "options": [ { "text": "Glucose remains in multiple isomeric form in its aqueous solution\n" }, { "text": "Glucose is soluble in water because of having aldehyde functional group\n" }, { "text": "Glucose is an aldohexose\n" }, { "text": "Glucose is one of the monomer unit in sucrose" } ], "answer": "Glucose is soluble in water because of having aldehyde functional group\n", "solution": "**Answer:** Glucose is soluble in water because of having aldehyde functional group\n\n\n

Let's analyze each statement to identify the incorrect one about glucose.

\n\n

Option A: Glucose remains in multiple isomeric forms in its aqueous solution. This statement is correct. Glucose exists in equilibrium with various isomeric forms in solution, including its linear form and two cyclic forms (alpha and beta anomers) that result from the formation of a hemiacetal bond within the molecule itself. The interconversion between these forms is known as mutarotation.

\n\n

Option B: Glucose is soluble in water because of having an aldehyde functional group. This statement is incorrect though it might seem partially true at first glance. While the aldehyde group can participate in hydrogen bonding, the primary reason glucose is highly soluble in water is due to its multiple hydroxyl (-OH) groups that can form hydrogen bonds with water molecules. The ability to form numerous hydrogen bonds makes glucose highly water-soluble, not just the presence of an aldehyde group.

\n\n

Option C: Glucose is an aldohexose. This statement is correct. Glucose is classified as an aldohexose because it contains an aldehyde group (-CHO) on its first carbon and has a total of six carbon atoms, making it a hexose sugar.

\n\n

Option D: Glucose is one of the monomer units in sucrose. This statement is correct. Sucrose is a disaccharide composed of two monomers: glucose and fructose. Thus, glucose is indeed one of the monomer units in sucrose.

\n\n

In summary, the incorrect statement about glucose is Option B: Glucose is soluble in water because of having an aldehyde functional group. The primary reason for its solubility in water is the hydroxyl groups' ability to form extensive hydrogen bonding, not solely the presence of an aldehyde functional group.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 254, "subject": "Chemistry", "question": "

Match List I with List II

\n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
LIST ILIST II
A.$$\\alpha$$ - Glucose and $$\\alpha$$ - GalactoseI.Functional isomers
B.\t$$
\\alpha \\text { - Glucose and } \\beta \\text { - Glucose }
$$		II.Homologous
C.$$
\\alpha \\text { - Glucose and } \\alpha \\text { - Fructose }
$$		III.Anomers
D.$$
\\alpha \\text { - Glucose and } \\alpha \\text { - Ribose }
$$		IV.Epimers

\n

Choose the correct answer from the options given below:

", "options": [ { "text": "A-III, B-IV, C-II, D-I\n" }, { "text": "A-IV, B-III, C-I, D-II\n" }, { "text": "A-III, B-IV, C-I, D-II\n" }, { "text": "A-IV, B-III, C-II, D-I" } ], "answer": "A-IV, B-III, C-I, D-II\n", "solution": "**Answer:** A-IV, B-III, C-I, D-II\n\n\n

$$\\alpha$$-Glucose and $$\\alpha$$-Galactose differ in configuration of one asymmetric carbon. Thus they are epimers.

\n

$$\\alpha$$ and $$\\beta$$-glucose differ at anomeric carbon. Thus they are anomers.

\n

$$\\alpha$$-Glucose and $$\\alpha$$-Fructose have different functional groups. Thus they are functional isomers.

", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 255, "subject": "Chemistry", "question": "The functional group, which is found in amino acid is", "options": [ { "text": "-COOH group" }, { "text": "-NH2 group" }, { "text": "-CH3 group" }, { "text": "both (a) and (b)" } ], "answer": "both (a) and (b)", "solution": "**Answer:** both (a) and (b)\n\nAmino acids contain $$-$$ $$N{H_2}$$ and $$-$$ $$COOH$$ groups $$e.g$$ \n

\"AIEEE ", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 256, "subject": "Chemistry", "question": "Identify the correct statements regarding enzymes", "options": [ { "text": "Enzymes are specific biological catalysts that can normally function at very high\ntemperature (T ~ 1000 K) " }, { "text": "Enzymes are specific biological catalysts that the posses well – defined active sites " }, { "text": "Enzymes are specific biological catalysts that can not be poisoned " }, { "text": "Enzymes are normally heterogeneous catalysts that are very specific in their action " } ], "answer": "Enzymes are specific biological catalysts that the posses well – defined active sites ", "solution": "**Answer:** Enzymes are specific biological catalysts that the posses well – defined active sites \n\nEnzymes are shape selective specific biological catalysts which normally functions effectively at body temperature.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 257, "subject": "Chemistry", "question": "Insulin production and its action in human body are responsible for the level of diabetes. This\ncompound belongs to which of the following categories? ", "options": [ { "text": "A co- enzyme" }, { "text": "An antibiotic " }, { "text": "An enzyme" }, { "text": "A hormone " } ], "answer": "A hormone ", "solution": "**Answer:** A hormone \n\nInsulin is a proteinaceous hormone secreted by $$\\beta $$-cells by islet of Langerhans of pancreas in our body.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 258, "subject": "Chemistry", "question": "The secondary structure of a protein refers to ", "options": [ { "text": "$$\\alpha$$-helical backbone" }, { "text": "hydrophobic interactions" }, { "text": "sequence of $$\\alpha$$-amino acids" }, { "text": "fixed configuration of the polypeptide backbone" } ], "answer": "$$\\alpha$$-helical backbone", "solution": "**Answer:** $$\\alpha$$-helical backbone\n\nThe secondary structure of a protein refers to the shape in which a long peptide chain can exist. There are two different conformations of the peptide linkage present in protein, these are $$\\alpha $$-helix and $$\\beta $$-conformation. The $$\\alpha $$-helix always has a right handed arrangement. In $$\\beta $$-conformation all peptide chains are streched out to nearly maximum extension and then laid side by side and held together by intermolecular hydrogen bonds. The structure resembles the pleated folds of drapery and therefore is known as $$\\beta $$-pleated sheet.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 259, "subject": "Chemistry", "question": "Biuret test is not given by", "options": [ { "text": "carbohydrates" }, { "text": "polypeptides" }, { "text": "urea" }, { "text": "proteins" } ], "answer": "carbohydrates", "solution": "**Answer:** carbohydrates\n\nBiuret test produces violet colour on addition of dilute $$CaS{O_4}$$ to alkaline solution of a compound containing perptide linkage.\n

Polypeptides, proteins and urea have -\n

\"AIEEE \n

(peptide) linkage while carbohydrates have glycosidic linkages. So, test of carbohydrates should be different from that of other three. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 260, "subject": "Chemistry", "question": "Which one of the following statements is correct ?", "options": [ { "text": "All amino acids except lysine are optically active" }, { "text": "All amino acids are optically active\n" }, { "text": "All amino acids except glycine are optically active" }, { "text": "All amino acids except glutamic acid are optically active\n" } ], "answer": "All amino acids except glycine are optically active", "solution": "**Answer:** All amino acids except glycine are optically active\n\nWith the exception of glycine all the $$19$$ other common amino acids have a uniquely different functional group on the central tetrahedral alpha carbon. \n

\"AIEEE ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 261, "subject": "Chemistry", "question": "Synthesis of each molecule of glucose in photosynthesis involves: ", "options": [ { "text": "10 molecules of ATP" }, { "text": "8 molecules of ATP " }, { "text": "6 molecules of ATP " }, { "text": "18 molecules of ATP" } ], "answer": "18 molecules of ATP", "solution": "**Answer:** 18 molecules of ATP\n\n$$6C{O_2} + 12NADPH + 18ATP$$ + 6RuBP\n

$$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, \\to {C_6}{H_{12}}{O_6} + 12NADP + 18ADP$$ + 6RuBP + 18P\n

One molecule of glucose is formed from 6CO2 by utilising 18ATP and 12NADPH.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 262, "subject": "Chemistry", "question": "Observation of “Rhumann’s purple” is a confirmatory test for the presence of :\n", "options": [ { "text": "Reducing sugar" }, { "text": "Cupric ion" }, { "text": "Protein" }, { "text": "Starch" } ], "answer": "Protein", "solution": "**Answer:** Protein\n\nRuhemann’s purple is ninhydrin.\n
\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 263, "subject": "Chemistry", "question": "Thiol group is present in :", "options": [ { "text": "Cytosine" }, { "text": "Cystine" }, { "text": "Cysteine " }, { "text": "Methionine" } ], "answer": "Cysteine ", "solution": "**Answer:** Cysteine \n\nAmong $$20$$ naturally occuring amino acids ''Cysteine'' has' $$-$$ $$SH'$$ or thiol functional group.\n

\"JEE ", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 264, "subject": "Chemistry", "question": "Among the following, the essential amino acid is :", "options": [ { "text": "Alanine" }, { "text": "Valine" }, { "text": "Aspartic acid " }, { "text": "Serine" } ], "answer": "Valine", "solution": "**Answer:** Valine\n\n

Valine is considered as an essential amino acid since the human body is not capable of synthesising it. Alanine, aspartic acid and serine can be synthesised by all living organisms, plants and animals. Therefore, they are known as non-essential amino acids.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 265, "subject": "Chemistry", "question": "The correct match between Item I and Item II is :\n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
Item IItem II
(A)Ester test(P)Tyr
(B)Carbylamine test(Q)Asp
(C)Phthalein dye test
(R)Ser
(S)Lys
", "options": [ { "text": "(A)$$ \\to $$(Q); (B)$$ \\to $$(S); (C)$$ \\to $$(P)" }, { "text": "(A)$$ \\to $$(R); (B)$$ \\to $$(S); (C)$$ \\to $$(Q)" }, { "text": "(A)$$ \\to $$(R); (B)$$ \\to $$(Q); (C)$$ \\to $$(P)" }, { "text": "(A)$$ \\to $$(Q); (B)$$ \\to $$(S); (C)$$ \\to $$(R)" } ], "answer": "(A)$$ \\to $$(Q); (B)$$ \\to $$(S); (C)$$ \\to $$(P)", "solution": "**Answer:** (A)$$ \\to $$(Q); (B)$$ \\to $$(S); (C)$$ \\to $$(P)\n\n\"JEE", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 266, "subject": "Chemistry", "question": "The peptide that gives positive ceric\nammonium nitrate and carbylamine tests is :", "options": [ { "text": "Ser-Lys" }, { "text": "Asp-Gln" }, { "text": "Gln-Asp" }, { "text": "Lys-Asp" } ], "answer": "Ser-Lys", "solution": "**Answer:** Ser-Lys\n\nPositive Ceric ammonium nitrate test is given by\nalcohol (- OH).\n

Positive Carbylamine test is given by 1o amine (-NH2).\n\"JEE\n

Among those peptide you can see only in (A) Serine(Ser)Lysine(Lys) has alcohol (- OH) and 1o amine (-NH2).", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 267, "subject": "Chemistry", "question": "The correct statement (s) among I to III with respect to potassium ions that are abundant within the cell fluid\nis /are \n
I. They activate many enzymes \n
II. They participate in the oxidation of glucose to produce ATP \n
III. Along with sodium ions they are responsible for the transmission of nerve signals", "options": [ { "text": "III only " }, { "text": "I and II only " }, { "text": "I and III only " }, { "text": "I, II and III only " } ], "answer": "I, II and III only ", "solution": "**Answer:** I, II and III only \n\n

All the statements are correct. K+ being metallic unipositive ions work as enzyme activators. These also participate in many reactions of glycolysis and Kreb’s cycle to produce ATP from glucose.

\n

Being unipositive these are also equally responsible for nerve signal transmission along with Na+.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 268, "subject": "Chemistry", "question": "Among the following compounds most basic amino acid is - ", "options": [ { "text": "Lysine" }, { "text": "Asparagine " }, { "text": "Histidine " }, { "text": "Serine" } ], "answer": "Lysine", "solution": "**Answer:** Lysine\n\nCompound      Pl value\n
Histidine           7.6\n
Serine               5.7\n
Lysine               9.8\n
Asparagine       5.4", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 269, "subject": "Chemistry", "question": "Which of the following tests cannot be used for identifying amino acids? ", "options": [ { "text": "Ninhydrin test " }, { "text": "Barfoed test \n" }, { "text": "Xanthoproteic test" }, { "text": "Biuret test \n" } ], "answer": "Barfoed test \n", "solution": "**Answer:** Barfoed test \n\n\nBarfoed test is used to detect the presence of monosaccharide (reducing sugars) in solutions not amino acids.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 270, "subject": "Chemistry", "question": "The increasing order of pka of the following amino acids in aqueous solution is :

\nGly, Asp, Lys, Arg
", "options": [ { "text": "Asp < Gly < Arg < Lys" }, { "text": "Gly < Asp < Arg < Lys" }, { "text": "Asp < Gly < Lys < Arg" }, { "text": "Arg < Lys < Gly < Asp" } ], "answer": "Asp < Gly < Lys < Arg", "solution": "**Answer:** Asp < Gly < Lys < Arg\n\n\"JEE\n
\"JEE\n
\"JEE\n
\"JEE\n
pKa = $$-$$ logKa\n

$$ \\therefore $$   Acidic strength (Ka) $$ \\propto $$ $${1 \\over {pKa}}$$\n

$$ \\therefore $$   Order of acidic strength : \n

Asp > Gly > LYs > Arg\n

$$ \\therefore $$   Order of pKa is : \n

Asp < Gly < Lys < Arg.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 271, "subject": "Chemistry", "question": "The number of chiral centres present in\nthreonine is ________.", "options": [], "answer": "2", "solution": "**Answer:** 2\n\nThe structure of threonine is\n\"JEE\n
No. of chiral centres present in it = 2", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 272, "subject": "Chemistry", "question": "Which of the following is not an essential\namino acid :", "options": [ { "text": "Valine" }, { "text": "Leucine" }, { "text": "Lysine" }, { "text": "Tyrosine" } ], "answer": "Tyrosine", "solution": "**Answer:** Tyrosine\n\nTyrosine is not an essential amino acid.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 273, "subject": "Chemistry", "question": "The number of chiral carbon(s) present in\n
peptide, Ile-Arg-Pro, is _____.", "options": [], "answer": "4", "solution": "**Answer:** 4\n\nThe amino acids present in the given tripeptide\nI$$l$$e–Arg–Pro are isoleucine, arginine and proline.\n
\"JEE\n\"JEE\n\"JEE\n
Number of chiral carbons present in the given\ntripeptide is 4.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 274, "subject": "Chemistry", "question": "Out of the following, which type of interaction is reponsible for the stabilisation of $$\\alpha $$-helix structure of proteins ?", "options": [ { "text": "Covalent bonding" }, { "text": "vander Waals forces" }, { "text": "Ionic bonding" }, { "text": "Hydrogen bonding" } ], "answer": "Hydrogen bonding", "solution": "**Answer:** Hydrogen bonding\n\n$$\\alpha $$-helix structure of proteins is stabilised by H-bonds between\nhydrogen of —NH and oxygen of - C = O group is amino acids.\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 275, "subject": "Chemistry", "question": "The secondary structure of protein is stabilised by:", "options": [ { "text": "Hydrogen bonding" }, { "text": "glycosidic bond" }, { "text": "Peptide bond" }, { "text": "van der Waals forces" } ], "answer": "Hydrogen bonding", "solution": "**Answer:** Hydrogen bonding\n\nThe secondary structure of a protein is stabilised by\nhydrogen bonding.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 276, "subject": "Chemistry", "question": "Which one of the following statements is not true about enzymes?", "options": [ { "text": "Enzymes are non-specific for a reaction and substrate." }, { "text": "Almost all enzymes are proteins." }, { "text": "Enzymes work as catalysts by lowering the activation energy of a biochemical reaction." }, { "text": "The action of enzymes is temperature and pH specific." } ], "answer": "Enzymes are non-specific for a reaction and substrate.", "solution": "**Answer:** Enzymes are non-specific for a reaction and substrate.\n\nEnzymes are mostly proteins. They function as catalysts in biochemical reactions by lowering the energy of activation. They are highly specific w.r.t. temperature and pH in their action.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 277, "subject": "Chemistry", "question": "The water soluble protein is :", "options": [ { "text": "Fibrin" }, { "text": "Albumin" }, { "text": "Myosin" }, { "text": "Collagen" } ], "answer": "Albumin", "solution": "**Answer:** Albumin\n\nAlbumin is water soluble.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 278, "subject": "Chemistry", "question": "The total number of negative charge in the tetrapeptide, Gly-Glu-Asp-Tyr, at pH 12.5 will be ______________. (Integer answer)", "options": [], "answer": "4", "solution": "**Answer:** 4\n\n\"JEE

Total negative charge produced = 4.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 279, "subject": "Chemistry", "question": "Which of the following is not an example of fibrous protein?", "options": [ { "text": "Keratin" }, { "text": "Albumin" }, { "text": "Collagen" }, { "text": "Myosin" } ], "answer": "Albumin", "solution": "**Answer:** Albumin\n\n

Fibrous proteins are characterized by their elongated, insoluble structure and often serve structural or mechanical roles. Among the options provided, keratin, collagen, and myosin are examples of fibrous proteins.

\n

Albumin, on the other hand, is a globular protein that is soluble in water. It has a more spherical shape compared to the elongated shape of fibrous proteins. Albumin is found in blood plasma and has various functions including maintaining osmotic pressure.

\n

So, the correct answer is:

\n

Option B: Albumin.

\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 280, "subject": "Chemistry", "question": "A peptide synthesized by the reactions of one molecule each of Glycine, Leucine, Aspartic acid and Histidine will have _________ peptide linkages.", "options": [], "answer": "3", "solution": "**Answer:** 3\n\nCombination of n amino acids gives a polypeptide\nwith (n – 1) peptide linkages.\n

Similarly combination of four amino acids gives a\ntetrapeptide with three peptide linkages.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 281, "subject": "Chemistry", "question": "

The structure of protein that is unaffected by heating is

", "options": [ { "text": "secondary structure" }, { "text": "tertiary structure" }, { "text": "primary structure" }, { "text": "quaternary structure" } ], "answer": "primary structure", "solution": "**Answer:** primary structure\n\n

The primary structure of protein is unaffected by physical 'or' chemical changes.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 282, "subject": "Chemistry", "question": "

Stability of $$\\alpha$$-Helix structure of proteins depends upon

", "options": [ { "text": "dipolar interaction" }, { "text": "H-bonding interaction" }, { "text": "van der Waals forces" }, { "text": "$$\\pi$$-stacking interaction" } ], "answer": "H-bonding interaction", "solution": "**Answer:** H-bonding interaction\n\nMostly $\\mathrm{H}$-bonding is responsible for the stability of $\\alpha$-helix form.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 283, "subject": "Chemistry", "question": "

Match List-I with List-II

\n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
List - I
Enzyne		List - II
Conversion of
A.InvertaseI.Starch into maltose
B.ZymaseII.Maltose into glucose
C.DiastaseIII.Glucose into ethanol
D.MaltaseIV.Cane sugar into glucose

\n

Choose the most appropriate answer from the options given below :

", "options": [ { "text": "A-III, B-IV, C-II, D-I" }, { "text": "A-III, B-II, C-I, D-IV" }, { "text": "A-IV, B-III, C-I, D-II" }, { "text": "A-IV, B-II, C-III, D-I" } ], "answer": "A-IV, B-III, C-I, D-II", "solution": "**Answer:** A-IV, B-III, C-I, D-II\n\n(A) Invertase → Cane sugar into glucose

\n(B) Zymase → Glucose into ethanol

\n(C) Diastase → Starch into maltose

\n(D) Maltase → Maltose into glucose", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 284, "subject": "Chemistry", "question": "

How many of the given compounds will give a positive Biuret test ____________ ?

\n

Glycine, Glycylalanine, Tripeptide, Biuret

", "options": [], "answer": "2", "solution": "**Answer:** 2\n\nSince dipeptides and free amino acids do not give\nbiuret test. Hence glycine and glycylalanine do not\ngive this test.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 285, "subject": "Chemistry", "question": "

In alanylglycyl leucyl alanyl valine, the number of peptide linkages is ___________.

", "options": [], "answer": "4", "solution": "**Answer:** 4\n\nThe given pentapeptide is\n

\n$$\\text { ALA - GLY - LEU - ALA - VAL }$$\n

\nIt has 4 peptide linkages.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 286, "subject": "Chemistry", "question": "

During the denaturation of proteins, which of these structures will remain intact?

", "options": [ { "text": "Primary" }, { "text": "Secondary" }, { "text": "Tertiary" }, { "text": "Quaternary" } ], "answer": "Primary", "solution": "**Answer:** Primary\n\nDuring the denaturation of proteins hydrogen bonds\nare disturbed. As a result, the secondary and\ntertiary structures are destroyed but the primary\nstructures remain intact. ", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 287, "subject": "Chemistry", "question": "

In a linear tetrapeptide (Constituted with different amino acids), (number of amino acids) $$-$$ (number of peptide bonds) is ________.

", "options": [], "answer": "1", "solution": "**Answer:** 1\n\nIn tetrapeptide

\n No. of amino acids = 4

\n No. of peptide bonds = 3

\n Hence, (1)", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 288, "subject": "Chemistry", "question": "

Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).

\n

Assertion (A) : $$\\alpha$$-halocarboxylic acid on reaction with dil. $$\\mathrm{NH_3}$$ gives good yield of $$\\alpha$$-amino carboxylic acid whereas the yield of amines is very low when prepared from alkyl halides.

\n

Reason (R) : Amino acids exist in zwitter ion form in aqueous medium.

\n

In the light of the above statements, choose the correct answer from the options given below :

", "options": [ { "text": "(A) is not correct but (R) is correct" }, { "text": "(A) is correct but (R) is not correct" }, { "text": "Both (A) and (R) are correct but (R) is not the correct explanation of (A)" }, { "text": "Both (A) and (R) are correct and (R) is the correct explanation of (A)" } ], "answer": "Both (A) and (R) are correct and (R) is the correct explanation of (A)", "solution": "**Answer:** Both (A) and (R) are correct and (R) is the correct explanation of (A)\n\n\"JEE\n

The reaction of $\\alpha$-halocarboxylic acid with dilute ammonia is known as the haloform reaction. In this reaction, the halogen atom of $\\alpha$-halocarboxylic acid is replaced by an amino group to form $\\alpha$-amino carboxylic acid. The yield of $\\alpha$-amino carboxylic acid is good because the amino group is a better nucleophile than the halogen atom, and it attacks the carbonyl carbon of the carboxylic acid to form an intermediate, which is then hydrolyzed to form the product. On the other hand, when alkyl halides are reacted with ammonia, the yield of amines is very low because alkyl halides are less reactive than $\\alpha$-halocarboxylic acids.

\n\n

The reason (R) is also true. Amino acids exist in the zwitterion form in an aqueous medium. In this form, the amino group has a positive charge, and the carboxyl group has a negative charge. The zwitter ion form is stabilized by the interaction between the opposite charges, which is known as an electrostatic interaction. This interaction makes the zwitterion form more stable than the neutral form.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 289, "subject": "Chemistry", "question": "A short peptide on complete hydrolysis produces 3 moles of glycine (G), two moles of leucine (L) and two moles of valine (V) per mole of peptide. The number of peptide linkages in it are ________.", "options": [], "answer": "6", "solution": "**Answer:** 6\n\n

The peptide has seven amino acid units therefore it has six peptide bonds.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 290, "subject": "Chemistry", "question": "

Number of cyclic tripeptides formed with 2 amino acids A and B is :

", "options": [ { "text": "2" }, { "text": "3" }, { "text": "5" }, { "text": "4" } ], "answer": "4", "solution": "**Answer:** 4\n\nThe cyclic tripeptides possible with amino acids $A$ and $B$ will be\n

\nAAA, BBB, ABA, BAB\n

\nThere are are 4 possible cyclic\nstructures.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 291, "subject": "Chemistry", "question": "

Total number of tripeptides possible by mixing of valine and proline is ___________

", "options": [], "answer": "8", "solution": "**Answer:** 8\n\n

To calculate the total number of tripeptides possible by mixing valine and proline, we need to use the fundamental principle of counting.

\n\n

Since a tripeptide is made up of three amino acids, and we have two choices for each position (either valine or proline), we can use the multiplication principle to find the total number of possible tripeptides.

\n\n

That is, the total number of tripeptides possible by mixing valine and proline is:

\n\n

2 x 2 x 2 = 8

\n\n

Therefore, there are a total of 8 possible tripeptides that can be formed by mixing valine and proline.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 292, "subject": "Chemistry", "question": "

In an oligopeptide named Alanylglycylphenyl alanyl isoleucine, the number of $$\\mathrm{sp}^{2}$$ hybridised carbons is __________.

", "options": [], "answer": "10", "solution": "**Answer:** 10\n\n

The given Oligopeptide has the following\nstructure

\"JEE
\nIt has 10 sp2 hybridised C atoms given with star in the above structure.\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 293, "subject": "Chemistry", "question": "

The one that does not stabilize 2$$^\\circ$$ and 3$$^\\circ$$ structures of proteins is

", "options": [ { "text": "$$-$$O $$-$$O $$-$$linkage" }, { "text": "$$-$$S $$-$$S $$-$$linkage" }, { "text": "van der Waals forces" }, { "text": "H-bonding" } ], "answer": "$$-$$O $$-$$O $$-$$linkage", "solution": "**Answer:** $$-$$O $$-$$O $$-$$linkage\n\n

The stabilization of secondary (2°) and tertiary (3°) structures of proteins mainly involves interactions such as disulfide bridges (S-S linkages), van der Waals forces, hydrogen bonding, and hydrophobic interactions.

\n

Option A: O-O linkage is not usually involved in the stabilization of protein structures. Oxygen-oxygen single bonds are generally unstable and rarely observed in nature.

\n

Option B: S-S linkage (disulfide bridges) between cysteine residues in a protein can help stabilize the protein's structure, particularly its tertiary structure.

\n

Option C: Van der Waals forces can contribute to the stabilization of protein structures, particularly in the interior of the protein where they can promote hydrophobic interactions.

\n

Option D: Hydrogen bonding can play a critical role in the formation and stabilization of secondary structures such as alpha-helices and beta-sheets.

\n

Therefore, the correct answer is Option A (-O-O- linkage).

\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 294, "subject": "Chemistry", "question": "

Sulphur (S) containing amino acids from the following are:

\n

(a) isoleucine (b) cysteine (c) lysine (d) methionine (e) glutamic acid

", "options": [ { "text": "b, d" }, { "text": "a, d" }, { "text": "b, c, e" }, { "text": "a, b, c" } ], "answer": "b, d", "solution": "**Answer:** b, d\n\nIsoleucine is an $\\alpha$-amino acid that is used in the biosynthesis of proteins. It contains an $\\alpha$ amino group, an $\\alpha$-carboxylic acid group, and a hydrocarbon side chain with a branch. It is classified as a non-polar, uncharged, branchedchain, aliphatic amino acid.

\nLysine is an $\\alpha$-amino acid that is a precursor to many proteins. It contains an $\\alpha$-amino group, an alpha-carboxylic acid group, and a side chain lysyl, classifying it as a basic, charged, aliphatic amino acid. It is encoded by the codons AAA and AAG. Glutamic acid is an amino acid used to form proteins.

\nCysteine and methionine are sulphur containing amino acids. Amino acids get linked to one another by peptide bond formation and form a polypeptide chain of proteins. Hence cysteine and methionine are found in several proteins.

\n(a) isoleucine

\n\"JEE
\n(b) cysteine

\n\"JEE
\n(c) lysine

\n\"JEE
\n(d) methionine

\n\"JEE
\n(e) glutamic acid

\n\"JEE\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 295, "subject": "Chemistry", "question": "

Match List I with List II

\n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
LIST I
Enzymatic reaction		LIST II
Enzyme
A.Sucrose $$\\to$$ Glocuse and FructoseI.Zymase
B.Glucose $$\\to$$ ethyl alcohol and CO$$_2$$II.Pepsin
C.Starch $$\\to$$ MaltoseIII.Invertase
D.Proteins $$\\to$$ Amino acidsIV.Diastase

\n

Choose the correct answer from the options given below:

", "options": [ { "text": "A-III, B-I, C-II, D-IV" }, { "text": "A-III, B-I, C-IV, D-II" }, { "text": "A-I, B-IV, C-III, D-II" }, { "text": "A-I, B-II, C-IV, D-III" } ], "answer": "A-III, B-I, C-IV, D-II", "solution": "**Answer:** A-III, B-I, C-IV, D-II\n\n

Let's match the enzymes with the corresponding reactions:

\n\n

So, the correct option is:

\n

Option B\nA-III, B-I, C-IV, D-II

\n", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 296, "subject": "Chemistry", "question": "The number of tripeptides formed by three different amino acids using each amino acid once is ______.", "options": [], "answer": "6", "solution": "**Answer:** 6\n\n

To understand the number of tripeptides formed by three different amino acids using each amino acid once, consider the amino acids as distinct objects we need to arrange. In this case, we have 3 amino acids (let’s name them A, B, and C for simplicity). The question asks how many unique sequences (tripeptides) we can form if we use each amino acid exactly once.

\n\n

Arranging 3 distinct items in a sequence is a fundamental combinatorial problem. The number of ways to arrange n distinct objects is given by the factorial of n, denoted as n!. The factorial function (n!) means multiplying all whole numbers from n down to 1. Therefore, for 3 amino acids, the number of unique tripeptides we can form is calculated by 3! (3 factorial).

\n\n

3! = 3 × 2 × 1 = 6

\n\n

This means there are 6 possible unique tripeptides that can be formed using these three different amino acids exactly once, which are: ABC, ACB, BAC, BCA, CAB, CBA.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 297, "subject": "Chemistry", "question": "

Which structure of protein remains intact after coagulation of egg white on boiling?

", "options": [ { "text": "Quaternary\n" }, { "text": "Primary" }, { "text": "Secondary\n" }, { "text": "Tertiary" } ], "answer": "Primary", "solution": "**Answer:** Primary\n\n

The structure of protein that remains intact after the coagulation of egg white on boiling is the Primary structure. The correct answer is Option B.

\n\n

The primary structure of a protein is the sequence of amino acids that are linked by peptide bonds. This sequence dictates the protein's higher levels of structure including secondary, tertiary, and quaternary structures. Regardless of the change in the protein's conformation due to denaturation (such as what happens during boiling of egg white), the sequence of amino acids remains the same since the peptide bonds do not get broken under such conditions.

\n\n

When egg white is boiled, the proteins denature, which means the structure of the protein becomes altered due to the breakdown of non-covalent interactions that hold the protein's higher-order structures together. This includes:

\n\n\n\n

During coagulation caused by heat, the weak hydrogen and ionic bonds of secondary and tertiary structures are broken and the proteins can aggregate or form gel-like structures, changing the consistency of the egg white. However, the covalent peptide bonds of the primary structure stay intact throughout this process.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 298, "subject": "Chemistry", "question": "

Type of amino acids obtained by hydrolysis of proteins is :

", "options": [ { "text": "$$\\gamma$$\n" }, { "text": "$$\\beta$$\n" }, { "text": "$$\\alpha$$\n" }, { "text": "$$\\delta$$" } ], "answer": "$$\\alpha$$\n", "solution": "**Answer:** $$\\alpha$$\n\n\n

Proteins are natural polymers composed of $$\\alpha$$-amino acids which are connected by peptide linkages.

\n

Hence proteins upon acidic hydrolysis produce $$\\alpha$$-amino acids.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 299, "subject": "Chemistry", "question": "

Number of compounds among the following which contain sulphur as heteroatom is ___________.

\n

Furan, Thiophene, Pyridine, Pyrrole, Cysteine, Tyrosine

", "options": [], "answer": "2", "solution": "**Answer:** 2\n\n

\"JEE

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 300, "subject": "Chemistry", "question": "

Total number of essential amino acid among the given list of amino acids is ________.

\n

Arginine, Phenylalanine, Aspartic acid, Cysteine, Histidine, Valine, Proline

", "options": [], "answer": "4", "solution": "**Answer:** 4\n\n

Arginine, Phenylalanine, Histidine, Valine are essential amino acids.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 301, "subject": "Chemistry", "question": "

The total number of carbon atoms present in tyrosine, an amino acid, is ________.

", "options": [], "answer": "9", "solution": "**Answer:** 9\n\n

The structure of tyrosine is :

\n

\"JEE

\n

Number of C-atoms = 9

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 302, "subject": "Chemistry", "question": "

Which of the following gives a positive test with ninhydrin ?

", "options": [ { "text": "Polyvinyl chloride\n" }, { "text": "Cellulose\n" }, { "text": "Egg albumin\n" }, { "text": "Starch" } ], "answer": "Egg albumin\n", "solution": "**Answer:** Egg albumin\n\n\n

The ninhydrin test is a chemical test which is used to detect the presence of amino acids or proteins. When amino acids or proteins react with ninhydrin, a deep blue or purple color is produced, which is indicative of a positive test. This reaction primarily occurs due to the reaction of ninhydrin with the free amino groups (-NH2) present in amino acids or proteins.

\n\n

Let's examine each option:

\n\n

Option A: Polyvinyl chloride - Polyvinyl chloride, commonly known as PVC, is a synthetic polymer. It does not contain amino acids or free amino groups. Therefore, it does not give a positive test with ninhydrin.

\n\n

Option B: Cellulose - Cellulose is a carbohydrate polymer consisting of glucose units. It is the main constituent of the cell wall in plants. Like PVC, cellulose does not contain amino acids or free amino groups, so it also does not give a positive test with ninhydrin.

\n\n

Option C: Egg albumin - Egg albumin is a protein found in egg whites. Proteins are made up of amino acids linked by peptide bonds, and free amino groups are present in the structure of proteins. Therefore, egg albumin reacts with ninhydrin and gives a positive test by producing a deep blue or purple color.

\n\n

Option D: Starch - Starch is a polysaccharide composed of glucose units and is a carbohydrate like cellulose. It does not possess amino acids or free amino groups. Consequently, starch does not give a positive test with ninhydrin.

\n\n

Based on the explanation, the correct answer is Option C: Egg albumin, as it is the only option among the ones listed that contains proteins, which have free amino groups capable of reacting with ninhydrin to give a positive test.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 303, "subject": "Chemistry", "question": "

Coagulation of egg, on heating is because of :

", "options": [ { "text": "Denaturation of protein occurs\n" }, { "text": "Biological property of protein remains unchanged\n" }, { "text": "Breaking of the peptide linkage in the primary structure of protein occurs\n" }, { "text": "The secondary structure of protein remains unchanged" } ], "answer": "Denaturation of protein occurs\n", "solution": "**Answer:** Denaturation of protein occurs\n\n\n

The coagulation of an egg on heating is because of Option A: Denaturation of protein occurs. When an egg is heated, the proteins in the egg (primarily albumins) undergo denaturation. This process involves the structural alteration of the protein molecules without breaking the peptide bonds that hold the amino acid sequences together. Denaturation causes the proteins to unfold and then aggregate into a new, solid structure, which is what you observe as the egg white and yolk changing from liquid to solid. This change is principally caused by the disruption of non-covalent bonds in the protein, such as hydrogen bonds, ionic bonds, and hydrophobic interactions.

\n\n

Option B is incorrect because the biological property of the protein does indeed change during the process. The protein in its denatured (coagulated) form does not function in the same way as it did in its native form.

\n\n

Option C is also incorrect because the primary structure of the protein, defined by its amino acid sequence linked by peptide bonds, remains unaltered during denaturation. The process does not involve the breaking of these peptide linkages.

\n\n

Lastly, Option D is incorrect because the secondary structure of the protein, which includes α-helices and β-pleated sheets formed by hydrogen bonding, typically undergoes significant alteration during denaturation, contributing to the observable change in the egg's texture.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 304, "subject": "Chemistry", "question": "

The incorrect statements regarding enzymes are :

\n

(A) Enzymes are biocatalysts.

\n

(B) Enzymes are non-specific and can catalyse different kinds of reactions.

\n

(C) Most Enzymes are globular proteins.

\n

(D) Enzyme - oxidase catalyses the hydrolysis of maltose into glucose.

\n

Choose the correct answer from the option given below :

", "options": [ { "text": "(B) and (D)\n" }, { "text": "(A), (B) and (C)\n" }, { "text": "(B) and (C)\n" }, { "text": "(B), (C) and (D)" } ], "answer": "(B) and (D)\n", "solution": "**Answer:** (B) and (D)\n\n\n

The correct answer is Option A: (B) and (D).

\n

Let’s analyze each statement:

\n

(A) Enzymes are biocatalysts. - This statement is correct. Enzymes are indeed biocatalysts, meaning they are natural catalytic substances produced by living organisms to speed up chemical reactions in cells without being consumed in the process. They are crucial for a myriad of biological processes.

\n

(B) Enzymes are non-specific and can catalyse different kinds of reactions. - This statement is incorrect. Enzymes are highly specific in nature; they typically catalyze only one kind of reaction or a very limited range of reactions. This specificity is a result of the unique 3D structure of the enzyme molecule, particularly the active site where substrate molecules bind.

\n

(C) Most enzymes are globular proteins. - This statement is correct. The vast majority of enzymes are indeed globular proteins. This means they are folded into a compact shape which is crucial for their function. Their globular structure allows for the creation of specific active sites where the chemical reactions take place.

\n

(D) Enzyme - oxidase catalyses the hydrolysis of maltose into glucose. - This statement is incorrect. The enzyme that catalyses the hydrolysis of maltose into glucose is maltase, not oxidase. Oxidases are a class of enzymes that catalyze the transfer of electrons to oxygen, and are involved in oxidation reactions, not in the hydrolysis of sugars.

\n

Therefore, the incorrect statements regarding enzymes are (B) Enzymes are non-specific and can catalyse different kinds of reactions, and (D) Enzyme - oxidase catalyses the hydrolysis of maltose into glucose, making Option A the correct choice.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 305, "subject": "Chemistry", "question": "RNA is different from DNA because RNA contains", "options": [ { "text": "ribose sugar and thymine" }, { "text": "ribose sugar and uracil" }, { "text": "deoxyribose sugar and thymine" }, { "text": "deoxyribose sugar and uracil" } ], "answer": "ribose sugar and uracil", "solution": "**Answer:** ribose sugar and uracil\n\nIn $$RNA,$$ the sugar is $$D$$-ribose and base is uracil where as in $$DNA$$, the sugar is $$D$$-$$2$$ deoxyribose and the nitrogeneous base is thymine. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 306, "subject": "Chemistry", "question": "The reason for double helical structure of $$DNA$$ is operation of ", "options": [ { "text": "dipole-dipole interaction" }, { "text": "hydrogen bonding" }, { "text": "electrostatic attractions " }, { "text": "van der Waals' forces" } ], "answer": "hydrogen bonding", "solution": "**Answer:** hydrogen bonding\n\n$$DNA$$ consists of two polynucleotide chains, each chain forms a right handed spiral with ten bases in one turn of the spiral. The two chains coil to double helix and run in opposite direction held together by hydrogen bonding. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 307, "subject": "Chemistry", "question": "Which base is present in RNA but not in DNA? ", "options": [ { "text": "Uracil " }, { "text": "Thymine" }, { "text": "Guanine" }, { "text": "Cytosine" } ], "answer": "Uracil ", "solution": "**Answer:** Uracil \n\n$$RNA$$ contains cytosine and uracil as pyrimidine bases while $$DNA$$ has cytosine and thymine.\n

Both have the same purine bases i.e., Guanine and adenine.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 308, "subject": "Chemistry", "question": "In both DNA and RNA, heterocyclic base and phosphate ester linkages are at- ", "options": [ { "text": "$$C_5^{'}$$ and $$C_2^{'}$$ respectively of the sugar molecule" }, { "text": "$$C_2^{'}$$ and $$C_5^{'}$$ respectively of the sugar molecule" }, { "text": "$$C_1^{'}$$ and $$C_5^{'}$$ respectively of the sugar molecule" }, { "text": "$$C_5^{'}$$ and $$C_1^{'}$$ respectively of the sugar molecule " } ], "answer": "$$C_1^{'}$$ and $$C_5^{'}$$ respectively of the sugar molecule", "solution": "**Answer:** $$C_1^{'}$$ and $$C_5^{'}$$ respectively of the sugar molecule\n\nIn $$DNA$$ and $$RNA$$ heterocyclic base and phosphate ester are at $${C_1}'$$ and $${C_5}'$$ respectively of the sugar molecule. \n

\"AIEEE ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 309, "subject": "Chemistry", "question": "The pyrimidine bases present in DNA are ", "options": [ { "text": "cytosine and adenine" }, { "text": "cytosine and guanine" }, { "text": "cytosine and thymine " }, { "text": "cytosine and uracil " } ], "answer": "cytosine and thymine ", "solution": "**Answer:** cytosine and thymine \n\nDNA contains cytosine and thymine as pyrimidine bases and guanine and adenine as purine bases.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 310, "subject": "Chemistry", "question": "The presence or absence of hydroxyl group on which carbon atom of sugar differentiates RNA and DNA ? ", "options": [ { "text": "2nd" }, { "text": "3rd " }, { "text": "4th" }, { "text": "1st " } ], "answer": "2nd", "solution": "**Answer:** 2nd\n\n$$RNA$$ has $$D(-)$$- Ribose and the $$DNA$$ has $$2$$-Dexoy $$D(-)$$ $$-$$ribose as the carbohydrate unit.\n

\"AIEEE \n

From the structures it is clear that $${2^{nd}}$$ carbon in $$DNA$$ do not have $$OH$$ group.", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 311, "subject": "Chemistry", "question": "Which one of the following bases is not present in DNA?", "options": [ { "text": "Cytosine" }, { "text": "Thymine" }, { "text": "Quinoline" }, { "text": "Adenine" } ], "answer": "Quinoline", "solution": "**Answer:** Quinoline\n\n$$DNA$$ contains adenine (A), thymine (T), guanine (G) and cytosine (C) bases.\n

So quinoline is not present in $$DNA.$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 312, "subject": "Chemistry", "question": "Which of the vitamins given below is water soluble?", "options": [ { "text": "Vitamin D" }, { "text": "Vitamin E" }, { "text": "Vitamin K" }, { "text": "Vitamin C" } ], "answer": "Vitamin C", "solution": "**Answer:** Vitamin C\n\nWater-soluble vitamins dissolve in water and are not stored by the body. The water soluble vitamins include the vitamin $$B$$-complex group and vitamin $$C.$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 313, "subject": "Chemistry", "question": "Which of the following statements is not true about RNA?", "options": [ { "text": "It controls the synthesis of protein" }, { "text": "It is present in the nucleus of the cell" }, { "text": "It usually does not replicate" }, { "text": "It has always double stranded $$\\alpha $$-helix structure" } ], "answer": "It has always double stranded $$\\alpha $$-helix structure", "solution": "**Answer:** It has always double stranded $$\\alpha $$-helix structure\n\nRNA has a single standard helix structure.\n

DNA has a double standard helix structure", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 314, "subject": "Chemistry", "question": "Match the following :
\n(i) Riboflavin    (a) Beriberi
\n(ii) Thiamine    (b) Scurvy
\n(iii) Pyridoxine   (c) Cheliosis
\n(iv) Ascorbic acid    (d) Convulsions
", "options": [ { "text": "(i)-(c), (ii)-(d), (iii)-(a), (iv)-(b)" }, { "text": "(i)-(d), (ii)-(b), (iii)-(a), (iv)-(c)" }, { "text": "(i)-c, (ii)-(a), (iii)-(d), (iv)-(b)" }, { "text": "(i)-(a), (ii)-(d), (iii)-(c), (iv)-(b)\n" } ], "answer": "(i)-c, (ii)-(a), (iii)-(d), (iv)-(b)", "solution": "**Answer:** (i)-c, (ii)-(a), (iii)-(d), (iv)-(b)\n\nThiamine (vitamin B1) : Beriberi\n

Riboflavin (vitamin B2) : Cheilosis\n

Pyridoxine (vitamin B6) : Convulsions\n

Ascorbic acid (vitamin C) : Scurvy", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 315, "subject": "Chemistry", "question": "Which of the following will react with CHCl3 +\nalc. KOH?", "options": [ { "text": "Adenine and thymine" }, { "text": "Thymine and proline" }, { "text": "Adenine and lysine" }, { "text": "Adenine and proline" } ], "answer": "Adenine and lysine", "solution": "**Answer:** Adenine and lysine\n\nAdenine and lysine both have primary anline react with CHCl3 + alc. KOH.\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 316, "subject": "Chemistry", "question": "Match List - I and List - II.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
List IList II
(a)Valium(i)Antifertility drug
(b)Morphine(ii)Pernicious anaemia
(c)Norethindrone(iii)Analgesic
(d)Vitamin $${B_{12}}$$(iv)Tranquilizer
", "options": [ { "text": "(a)-(i), (b)-(iii), (c)-(iv), (d)-(ii)" }, { "text": "(a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)" }, { "text": "(a)-(ii), (b)-(iv), (c)-(iii), (d)-(i)" }, { "text": "(a)-(iv), (b)-(iii), (c)-(i), (d)-(ii)" } ], "answer": "(a)-(iv), (b)-(iii), (c)-(i), (d)-(ii)", "solution": "**Answer:** (a)-(iv), (b)-(iii), (c)-(i), (d)-(ii)\n\n

(A) Valium – (4) Tranquilizer

A tranquilizer drug became a standard drug for the treatment of anxiety and one of most commonly prescribed drugs of all time.

\n\n

(B) Morphine – (3) Analgesic

Morphine is effective for both acute and chronic pain and often used before and after surgery.

\n\n

(C) Norethindrone – (1) Antifertility drug

It is a form of progesterone, a female hormone important for regulating ovulation and menstruation.

\n\n

(D) Vitamin B12 – (2) Pernicious anaemia

It is a nutrient that helps to keep our body blood cells and nerve cells healthy and help in making DNA.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 317, "subject": "Chemistry", "question": "Which of the following vitamin is helpful in delaying the blood clotting?", "options": [ { "text": "Vitamin K" }, { "text": "Vitamin C" }, { "text": "Vitamin E" }, { "text": "Vitamin B" } ], "answer": "Vitamin K", "solution": "**Answer:** Vitamin K\n\nVitamin K is helpful in delaying the blood\nclotting.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 318, "subject": "Chemistry", "question": "Which among the following pairs of Vitamins is stored in our body relatively for longer duration?", "options": [ { "text": "Thiamine and Vitamin A" }, { "text": "Ascorbic acid and Vitamin D" }, { "text": "Thiamine and Ascorbic acid" }, { "text": "Vitamin A and Vitamin D" } ], "answer": "Vitamin A and Vitamin D", "solution": "**Answer:** Vitamin A and Vitamin D\n\nVitamins which are soluble in fat and oils but\ninsoluble in water are fat-soluble vitamins, which\nare stored in our body relatively for a longer time.

\ne.g. Vitamin A and Vitamin D

\nThiamine (Vit B1) and Ascorbic acid (Vit C) are\nwater-soluble.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 319, "subject": "Chemistry", "question": "Deficiency of vitamin K causes :", "options": [ { "text": "Cheilosis" }, { "text": "Decrease in blood clotting time" }, { "text": "Increase in blood clotting time" }, { "text": "Increase in fragility of RBC's" } ], "answer": "Increase in blood clotting time", "solution": "**Answer:** Increase in blood clotting time\n\nDeficiency of vitamin K causes increase in blood clotting time.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 320, "subject": "Chemistry", "question": "Thiamine and pyridoxine are also known respectively as :", "options": [ { "text": "Vitamin B2 and Vitamin E" }, { "text": "Vitamin E and Vitamin B2" }, { "text": "Vitamin B6 and Vitamin B2" }, { "text": "Vitamin B1 and Vitamin B6" } ], "answer": "Vitamin B1 and Vitamin B6", "solution": "**Answer:** Vitamin B1 and Vitamin B6\n\nVitamine-B1 is also known as Thiamine while Vitamin B-6 is known as Pyridoxine.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 321, "subject": "Chemistry", "question": "

Sugar moiety in DNA and RNA molecules respectively are

", "options": [ { "text": "$$\\beta$$-D-2-deoxyribose, $$\\beta$$-D-deoxyribose." }, { "text": "$$\\beta$$-D-2-deoxyribose, $$\\beta$$-D-ribose" }, { "text": "$$\\beta$$-D-ribose, $$\\beta$$-D-2-deoxyribose." }, { "text": "$$\\beta$$-D-deoxyribose, $$\\beta$$-D-2-deoxyribose." } ], "answer": "$$\\beta$$-D-2-deoxyribose, $$\\beta$$-D-ribose", "solution": "**Answer:** $$\\beta$$-D-2-deoxyribose, $$\\beta$$-D-ribose\n\nDNA contains $$ \\Rightarrow $$ $$\\beta $$– D – 2 – deoxyribose\n

RNA contains $$ \\Rightarrow $$ $$\\beta $$– D – ribose", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 322, "subject": "Chemistry", "question": "

The number of oxygens present in a nucleotide formed from a base, that is present only in RNA is ___________.

", "options": [], "answer": "9", "solution": "**Answer:** 9\n\nUracil is the base which only present is RNA.

\"JEE
Structure of nucleotides number of 0-9.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 323, "subject": "Chemistry", "question": "

Which one of the following is a water soluble vitamin, that is not excreted easily?

", "options": [ { "text": "Vitamin B2" }, { "text": "Vitamin B1" }, { "text": "Vitamin B6" }, { "text": "Vitamin B12" } ], "answer": "Vitamin B12", "solution": "**Answer:** Vitamin B12\n\nVitamin B12 is water soluble and not excreted easily.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 324, "subject": "Chemistry", "question": "

The sugar produced after complete hydrolysis of DNA is

", "options": [ { "text": "a pentose sugar." }, { "text": "a hexose sugar." }, { "text": "a polysaccharide." }, { "text": "a disaccharide." } ], "answer": "a pentose sugar.", "solution": "**Answer:** a pentose sugar.\n\nDNA mainly consist of a nitrogeneous base, phosphate group and a pentose sugar.\n

\nSo, hydrolysis of DNA produces a pentose sugar.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 325, "subject": "Chemistry", "question": "Which is not true for arginine?", "options": [ { "text": "It has a fairly high melting point." }, { "text": "It is associated with more than one $\\mathrm{pK}_{\\mathrm{a}}$ values." }, { "text": "It has high solubility in benzene." }, { "text": "It is a crystalline solid. " } ], "answer": "It has high solubility in benzene.", "solution": "**Answer:** It has high solubility in benzene.\n\n\"JEE
Arginine is an amino acid that has a basic side chain containing a guanidino group. The other three options are true for arginine:\n

\n(A) It has a fairly high melting point due to the presence of strong ionic interactions between the positively charged guanidino group and the negatively charged carboxylate group in the solid state.

\n(B) It is associated with more than one pKa values because it contains multiple ionizable groups, including the amino group, the carboxylate group, and the guanidino group. These groups can dissociate in a stepwise manner, leading to multiple pKa values.

\n(D) It is a crystalline solid due to the presence of strong ionic interactions between the positively charged guanidino group and the negatively charged carboxylate group, which leads to the formation of a well-ordered crystal lattice.\n

\nHowever, option (C) is not true because arginine is not soluble in benzene. This is because benzene is a nonpolar solvent, and arginine is a polar and charged molecule. Therefore, arginine is more soluble in polar solvents such as water and less soluble in nonpolar solvents such as benzene.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 326, "subject": "Chemistry", "question": "

The naturally occurring amino acid that contains only one basic functional group in its chemical structure is

", "options": [ { "text": "asparagine" }, { "text": "arginine" }, { "text": "histidine" }, { "text": "lysine" } ], "answer": "asparagine", "solution": "**Answer:** asparagine\n\nAmong the given options, arginine, histidine, and lysine are all basic amino acids because they contain amino (-NH2) and imino (-NH=) functional groups that can accept a proton (H+) to become positively charged. \n

\nAsparagine, on the other hand, contains an amide (-CONH2) functional group instead of an imino group. This functional group is not basic because it does not have a proton that can be easily removed. Therefore, the answer is: asparagine.", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 327, "subject": "Chemistry", "question": "

Match List I with List II

\n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
LIST I
Natural amino acid		LIST II
One letter code
A.Glutamic acidI.Q
B.GlutamineII.W
C.TyrosineIII.E
D.TryptophanIV.Y

\n

Choose the correct answer from the options given below:

", "options": [ { "text": "A-III, B-IV, C-I, D-II" }, { "text": "A-IV, B-III, C-I, D-II" }, { "text": "A-II, B-I, C-IV, D-III" }, { "text": "A-III, B-I, C-IV, D-II" } ], "answer": "A-III, B-I, C-IV, D-II", "solution": "**Answer:** A-III, B-I, C-IV, D-II\n\n

The one-letter code for each amino acid is as follows:

\n

Glutamic Acid (A) is represented by the letter E (III).

\nGlutamine (B) is represented by the letter Q (I).

\nTyrosine (C) is represented by the letter Y (IV).

\nTryptophan (D) is represented by the letter W (II).

\n

These one-letter codes are part of the standard notation used in biology and biochemistry to represent the 20 standard amino acids, plus a few others. The codes are designed to be short and easy to remember, while still being distinct from each other.

\n", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 328, "subject": "Chemistry", "question": "

Match List I with List II

\n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
LIST I
Natural Amino Acid		LIST II
One Letter Code
A.ArginineI.D
B.Aspartic acidII.N
C.AsparagineIII.A
D.AlanineIV.R

\n

Choose the correct answer from the options given below:

", "options": [ { "text": "(A) - IV, (B) - I, (C) - III, (D) - II" }, { "text": "(A) - III, (B) - I, (C) - II, (D) - IV" }, { "text": "(A) - IV, (B) - I, (C) - II, (D) - III" }, { "text": "(A) - I, (B) - III, (C) - IV, (D) - II" } ], "answer": "(A) - IV, (B) - I, (C) - II, (D) - III", "solution": "**Answer:** (A) - IV, (B) - I, (C) - II, (D) - III\n\n

In protein chemistry, each amino acid is assigned a one-letter code to simplify notation. The matching for the given amino acids to their one-letter codes is as follows:

\n\n

So, the correct answer is Option C: (A) - IV, (B) - I, (C) - II, (D) - III.

\n", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 329, "subject": "Chemistry", "question": "If one strand of a DNA has the sequence ATGCTTCA, sequence of the bases in complementary strand is :", "options": [ { "text": "CATTAGCT" }, { "text": "TACGAAGT" }, { "text": "ATGCGACT" }, { "text": "GTACTTAC" } ], "answer": "TACGAAGT", "solution": "**Answer:** TACGAAGT\n\nDNA is made up of two strands that are complementary to each other. The base pairing rules state that adenine (A) pairs with thymine (T), and cytosine (C) pairs with guanine (G). Using these rules, we can determine the sequence of the complementary strand of the given DNA sequence ATGCTTCA.\n\nHere's the pairing for the given sequence:\n\n

A (adenine) pairs with T (thymine)

\n

T (thymine) pairs with A (adenine)

\n

G (guanine) pairs with C (cytosine)

\n

C (cytosine) pairs with G (guanine)

\n

T (thymine) pairs with A (adenine)

\n

T (thymine) pairs with A (adenine)

\n

C (cytosine) pairs with G (guanine)

\n

A (adenine) pairs with T (thymine)

\n\nSo, the complementary base sequence to the given DNA strand ATGCTTCA would be:\n\n

1. A pairs with T

\n

2. T pairs with A

\n

3. G pairs with C

\n

4. C pairs with G

\n

5. T pairs with A

\n

6. T pairs with A

\n

7. C pairs with G

\n

8. A pairs with T

\n\nPutting it all together, the complementary sequence is: TACGAAGT. Therefore, the correct answer is:\n\nOption B - TACGAAGT", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 330, "subject": "Chemistry", "question": "

Two nucleotides are joined together by a linkage known as :

", "options": [ { "text": "Phosphodiester linkage\n" }, { "text": "Glycosidic linkage\n" }, { "text": "Disulphide linkage\n" }, { "text": "Peptide linkage" } ], "answer": "Phosphodiester linkage\n", "solution": "**Answer:** Phosphodiester linkage\n\n\n

Phosphodiester linkage

\n

\"JEE

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 331, "subject": "Chemistry", "question": "

From the vitamins $$\\mathrm{A}, \\mathrm{B}_1, \\mathrm{~B}_6, \\mathrm{~B}_{12}, \\mathrm{C}, \\mathrm{D}, \\mathrm{E}$$ and $$\\mathrm{K}$$, the number of vitamins that can be stored in our body is _________.

", "options": [], "answer": "5", "solution": "**Answer:** 5\n\n

Vitamins A, D, E, K and B$$_{12}$$ are stored in liver and adipose tissue.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 332, "subject": "Chemistry", "question": "

The total number of correct statements, regarding the nucleic acids is _________.

\n

A. RNA is regarded as the reserve of genetic information

\n

B. DNA molecule self-duplicates during cell division

\n

C. DNA synthesizes proteins in the cell

\n

D. The message for the synthesis of particular proteins is present in DNA

\n

E. Identical DNA strands are transferred to daughter cells.

", "options": [], "answer": "3", "solution": "**Answer:** 3\n\n

A. RNA is regarded as the reserve of genetic information. (False)

\n

B. DNA molecule self-duplicates during cell division. (True)

\n

C. DNA synthesizes proteins in the cell. (False)

\n

D. The message for the synthesis of particular proteins is present in DNA. (True)

\n

E. Identical DNA strands are transferred to daughter cells. (True)

", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 333, "subject": "Chemistry", "question": "

The $$\\mathrm{F}^{-}$$ ions make the enamel on teeth much harder by converting hydroxyapatite (the enamel on the surface of teeth) into much harder fluoroapatite having the formula.

", "options": [ { "text": "$$\\left[3\\left(\\mathrm{Ca}_3\\left(\\mathrm{PO}_4\\right)_2\\right) \\cdot \\mathrm{CaF}_2\\right]$$\n" }, { "text": "$$\\left[3\\left(\\mathrm{Ca}_3\\left(\\mathrm{PO}_4\\right)_2\\right) \\cdot \\mathrm{Ca}(\\mathrm{OH})_2\\right]$$\n" }, { "text": "$$\\left[3\\left(\\mathrm{Ca}_3\\left(\\mathrm{PO}_4\\right)_3\\right) \\cdot \\mathrm{CaF}_2\\right]$$\n" }, { "text": "$$\\left[3\\left(\\mathrm{Ca}_2\\left(\\mathrm{PO}_4\\right)_2\\right) \\cdot \\mathrm{Ca}(\\mathrm{OH})_2\\right]$$" } ], "answer": "$$\\left[3\\left(\\mathrm{Ca}_3\\left(\\mathrm{PO}_4\\right)_2\\right) \\cdot \\mathrm{CaF}_2\\right]$$\n", "solution": "**Answer:** $$\\left[3\\left(\\mathrm{Ca}_3\\left(\\mathrm{PO}_4\\right)_2\\right) \\cdot \\mathrm{CaF}_2\\right]$$\n\n\n

Hydroxyapatite : $$\\left[3\\left(\\mathrm{Ca}_3\\left(\\mathrm{PO}_4\\right)_2\\right) \\cdot \\mathrm{Ca}(\\mathrm{OH})_2\\right]$$

\n

Fluoroapatite : $$\\left[3\\left(\\mathrm{Ca}_3\\left(\\mathrm{PO}_4\\right)_2\\right) \\cdot \\mathrm{CaF}_2\\right]$$

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 334, "subject": "Chemistry", "question": "The bond dissociation energy of B - F in BF3 is 646 kJ mol-1 whereas that of C - F in CF4 is 515 kj mol-1. The correct reason for higher B - F bond dissociation energy as compared to that of C - F is", "options": [ { "text": "stronger $$\\sigma$$ bond between B and F in BF3 as compared to that between C and F in CF4" }, { "text": "significant $$p\\pi - p\\pi$$ interaction between B and F in BF3 whereas there is no possibility of such interaction between C and F in CF4" }, { "text": "lower degree of $$p\\pi - p\\pi$$ interaction between B and F in BF3 than that between C and F in CF4" }, { "text": "smaller size of B - atom as compared to that of C- atom." } ], "answer": "significant $$p\\pi - p\\pi$$ interaction between B and F in BF3 whereas there is no possibility of such interaction between C and F in CF4", "solution": "**Answer:** significant $$p\\pi - p\\pi$$ interaction between B and F in BF3 whereas there is no possibility of such interaction between C and F in CF4\n\nIn BF3, B is sp2\n hybridised and by Back Bonding method strong p$$\\pi $$-p$$\\pi $$ bond is created between filled p-orbital of F and vacant p-orbital of\nB which leads to shortening of B–F bond length which results in higher bond dissociation energy of\nthe B–F bond. However in CF4, C does not have any vacant\np-orbitals to undergo $$\\pi $$-bonding.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 335, "subject": "Chemistry", "question": "The correct order of bond angles (smallest first) in H2S, NH3, BF3 and SiH4 is ", "options": [ { "text": "H2S < SiH4 < NH3 < BF3 " }, { "text": "H2S < NH3 < BF3 < SiH4" }, { "text": "H2S < NH3 < SiH4 < BF3 " }, { "text": "NH3 < H2S < SiH4 < BF3" } ], "answer": "H2S < NH3 < SiH4 < BF3 ", "solution": "**Answer:** H2S < NH3 < SiH4 < BF3 \n\n1. In H2S molecule, central atom S is a 3rd period element and according to Dragos rule, when a 3rd period or higher period element overlap with a small element whose electronegetivity is low then that over lapping cannot be a effective overlapping, because of higher size of 3rd or higher periods element and smaller size of low electronegative element. \n

Here in H2S, H atom has very low electronegativity and smaller in size so it create more negative charge around sulphur(S) atom and size of S$$-$$2 ion increases, because of this higher size difference efficiency overlapping is not possible. So, any hybridization do not happen in between S and H atom. Lone pair electrons of S atom is present in the pure orbital like s, px, py, pz and it look like this : \n

\"AIEEE\n

As we know the angle between and py is 90o, so bond angle is 90o.\n

2. In NH3, 3 Bond pair(BP) + 1 lone pair (LP) present so angle between bond 107o. \n

\"AIEEE\n

3. $$\\,\\,\\,\\,$$ BF3 has sp2 hybridization. So bond angle is 120o.\n

\"AIEEE\n

4.\"AIEEE\n
Here Si is sp3 hybridised and shape is regular tetrahedral and bond angle 109o 28'", "topic": "Geometry", "subtopic": "Plane Geometry" }, { "id": 336, "subject": "Chemistry", "question": "The bond angle H - X - H is the greatest in the compound :", "options": [ { "text": "CH4" }, { "text": "NH3" }, { "text": "H2O" }, { "text": "PH3" } ], "answer": "CH4", "solution": "**Answer:** CH4\n\nCH4 - sp3 hybridisation - Bond angle (109o28')\n

NH3 - sp3 hybridisation - Bond angle ( 107.8o)\n

H2O - sp3 hybridisation - Bond angle ( 104.5o)\n

PH3 - No hybridisation - Bond angle ( 90o)\n

Note : According to DRAGO RULE, lone pair of electrons are in pure s- orbital of the atom P. That is why PH3 has no hybridisation.", "topic": "Geometry", "subtopic": "Plane Geometry" }, { "id": 337, "subject": "Chemistry", "question": "Chlorine reacts with hot and concentrated NaOH and produces compounds (X) and (Y). Compound\n(X) gives white precipitate with silver nitrate solution. The average bond order between CI and O\natoms in (Y) is.", "options": [], "answer": "1.66to1.67", "solution": "**Answer:** 1.66to1.67\n\n3Cl2 + 6NaOH $$ \\to $$ 5NaCl + NaClO3 + 3H2O\n

NaCl + AgNO3 $$ \\to $$ AgCl- $$ \\downarrow $$ + NaNO3\n

$$ \\therefore $$ X = NaCl then Y = NaClO3\n

Here in anion ClO3- has bond between Cl and O atom.\n

Bond order of Cl–O Bond = \n
Total bond
\n
Total resonating structure
\n.\n

= $${5 \\over 3}$$ = 1.67\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 338, "subject": "Chemistry", "question": "Arrange the following bonds according to their\naverage bond energies in descending order :
\nC–Cl, C–Br, C–F, C–I", "options": [ { "text": "C–Br > C–I > C–Cl > C–F" }, { "text": "C–Cl > C–Br > C–I > C–F" }, { "text": "C–I > C–Br > C–Cl > C–F" }, { "text": "C–F > C–Cl > C–Br > C–I" } ], "answer": "C–F > C–Cl > C–Br > C–I", "solution": "**Answer:** C–F > C–Cl > C–Br > C–I\n\nBond length order in carbon halogen bonds are\n

in the order of C – F $$<$$ C – Cl $$<$$ C – Br $$<$$ C – I\n

Hence, Bond energy order\n

C – F $$>$$ C – Cl $$>$$ C – Br $$>$$ C – I", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 339, "subject": "Chemistry", "question": "Given below are two statements : one is labeled as Assertion A and the other is labelled as Reason R :

Assertion A : The H$$-$$O$$-$$H bond angle in water molecule is 104.5$$^\\circ$$.

Reason R : The lone pair - lone pair repulsion of electrons is higher than the bond pair - bond pair repulsion.

In the light of the above statements, choose the correct answer from the options given below :", "options": [ { "text": "Both A and R are true, and R is the correct explanation of A" }, { "text": "A is true but R is false" }, { "text": "A is false but R is true" }, { "text": "Both A and R are true, but R is not the correct explanation of A" } ], "answer": "Both A and R are true, and R is the correct explanation of A", "solution": "**Answer:** Both A and R are true, and R is the correct explanation of A\n\n\"JEE

\ndue to lp - lp repulsion bond angle decrease.", "topic": "Geometry", "subtopic": "Plane Geometry" }, { "id": 340, "subject": "Chemistry", "question": "

$$\\mathrm{O}-\\mathrm{O}$$ bond length in $$\\mathrm{H}_{2} \\mathrm{O}_{2}$$ is $$\\underline{\\mathrm{X}}$$ than the $$\\mathrm{O}-\\mathrm{O}$$ bond length in $$\\mathrm{F}_{2} \\mathrm{O}_{2}$$. The $$\\mathrm{O}-\\mathrm{H}$$ bond length in $$\\mathrm{H}_{2} \\mathrm{O}_{2}$$ is $$\\underline{Y}$$ than that of the $$\\mathrm{O}-\\mathrm{F}$$ bond in $$\\mathrm{F}_{2} \\mathrm{O}_{2}$$.

\n

Choose the correct option for $$\\underline{X}$$ and $$\\underline{Y}$$ from those given below :

", "options": [ { "text": "$$\\mathrm{X}$$ - shorter, $$\\mathrm{Y}$$ - longer" }, { "text": "$$\\mathrm{X}$$ - longer, $$\\mathrm{Y}$$ - shorter" }, { "text": "$$\\mathrm{X}$$ - shorter, $$\\mathrm{Y}$$ - shorter" }, { "text": "$$\\mathrm{X}$$ - longer, $$\\mathrm{Y}$$ - longer" } ], "answer": "$$\\mathrm{X}$$ - longer, $$\\mathrm{Y}$$ - shorter", "solution": "**Answer:** $$\\mathrm{X}$$ - longer, $$\\mathrm{Y}$$ - shorter\n\n\n\n\n \n \n \n \n\n\n \n \n \n \n \n \n \n \n\n
BondBond Length
O - O$1.48 \\,\\mathop A\\limits^o\\left(\\text { In }{\\mathrm{H}_2 \\mathrm{O}_2}\\right)$
O - O$1.22 \\,\\mathop A\\limits^o\\left(\\text { In }{\\mathrm{O}_2 \\mathrm{F}_2}\\right)$
\n
Thus, $\\mathrm{O}-\\mathrm{O}$ bond length in $\\mathrm{H}_2 \\mathrm{O}_2$ is longer than $\\mathrm{O}-\\mathrm{O}$ bond length in $\\mathrm{O}_2 \\mathrm{~F}_2$.\n

\n\n\n \n \n \n \n\n\n \n \n \n \n \n \n \n \n\n
BondBond Length
O - F$158 \\,\\mathrm{pm}\\left(\\text { In }{\\mathrm{O}_2 \\mathrm{F}_2}\\right)$
O - H$95 \\,\\mathrm{pm}\\left(\\text { In }{\\mathrm{H}_2 \\mathrm{O}_2}\\right)$
\n
Thus, the $\\mathrm{O}-\\mathrm{H}$ bond length in $\\mathrm{H}_2 \\mathrm{O}_2$ is shorter than $\\mathrm{O}-\\mathrm{F}$ bond length in $\\mathrm{O}_2 \\mathrm{~F}_2$.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 341, "subject": "Chemistry", "question": "

For $$\\mathrm{OF}_{2}$$ molecule consider the following :

\n

A. Number of lone pairs on oxygen is 2 .

\n

B. FOF angle is less than $$104.5^{\\circ}$$.

\n

C. Oxidation state of $$\\mathrm{O}$$ is $$-2$$.

\n

D. Molecule is bent '$$\\mathrm{V}$$' shaped.

\n

E. Molecular geometry is linear.

\n

correct options are:

", "options": [ { "text": "A, C, D only" }, { "text": "A, B, D only" }, { "text": "C, D, E only" }, { "text": "B, E, A only" } ], "answer": "A, B, D only", "solution": "**Answer:** A, B, D only\n\n

\"JEE

\n

A : No. of lone pairs on oxygen = 2

\n

B : \"JEE

\n

($$\\theta$$ < Bond angle in H$$_2$$O (104.5$$^\\circ$$)

\n

D : molecule is bent \"v\" shaped

", "topic": "Geometry", "subtopic": "Plane Geometry" }, { "id": 342, "subject": "Chemistry", "question": "

The bond dissociation energy is highest for

", "options": [ { "text": "Cl$$_2$$" }, { "text": "I$$_2$$" }, { "text": "Br$$_2$$" }, { "text": "F$$_2$$" } ], "answer": "Cl$$_2$$", "solution": "**Answer:** Cl$$_2$$\n\nBond energy of $\\mathrm{F}_{2}$ less than $\\mathrm{Cl}_{2}$ due to lone pair lone pair repulsions.\n

\nBond energy order $\\mathrm{Cl}_{2}>\\mathrm{Br}_{2}>\\mathrm{F}_{2}>\\mathrm{I}_{2}$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 343, "subject": "Chemistry", "question": "

Given below are two statements :

\n

Statement I : $$\\mathrm{SO}_{2}$$ and $$\\mathrm{H}_{2} \\mathrm{O}$$ both possess V-shaped structure.

\n

Statement II : The bond angle of $$\\mathrm{SO}_{2}$$ is less than that of $$\\mathrm{H}_{2} \\mathrm{O}$$.

\n

In the light of the above statements, choose the most appropriate answer from the options given below:

", "options": [ { "text": "Both Statement I and Statement II are incorrect" }, { "text": "Both Statement I and Statement II are correct" }, { "text": "Statement I is correct but Statement II is incorrect" }, { "text": "Statement I is incorrect but Statement II is correct" } ], "answer": "Statement I is correct but Statement II is incorrect", "solution": "**Answer:** Statement I is correct but Statement II is incorrect\n\n\"JEE
Statement I: Both SO₂ and H₂O have a V-shaped (or bent) molecular structure. SO₂ has a central sulfur atom with two oxygen atoms and one lone pair, while H₂O has a central oxygen atom with two hydrogen atoms and two lone pairs. In both cases, the repulsion between the lone pairs and the bonded electron pairs leads to a V-shaped structure.\n

\nStatement II: The bond angle in H₂O is approximately 104.5° due to the tetrahedral arrangement of electron pairs (including the lone pairs) around the central oxygen atom. In SO₂, the bond angle is approximately 119°. The presence of a lone pair and the double bonds in SO₂ results in a different electron distribution, leading to a larger bond angle compared to H₂O.", "topic": "Geometry", "subtopic": "Plane Geometry" }, { "id": 344, "subject": "Chemistry", "question": "

Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).

\n

Assertion (A) : There is a considerable increase in covalent radius from $$\\mathrm{N}$$ to $$\\mathrm{P}$$. However from As to Bi only a small increase in covalent radius is observed.

\n

Reason (R) : Covalent and ionic radii in a particular oxidation state increases down the group. In the light of the above statements, choose the most appropriate answer from the options given below:

", "options": [ { "text": "(A) is false but (R) is true" }, { "text": "Both (A) and (R) are true and (R) is the correct explanation of (A)" }, { "text": "Both (A) and (R) are true but (R) is not the correct explanation of (A)" }, { "text": "(A) is true but (R) is false" } ], "answer": "Both (A) and (R) are true but (R) is not the correct explanation of (A)", "solution": "**Answer:** Both (A) and (R) are true but (R) is not the correct explanation of (A)\n\n

According to NCERT,

\n

Statement-I : Factual data,

\n

Statement-II is true.

\n

But correct explanation is presence of completely filled $$d$$ and $$f$$-orbitals of heavier members

", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 345, "subject": "Chemistry", "question": "Which of the following statements is true?", "options": [ { "text": "HF is less polar than HBr" }, { "text": "absolutely pure water does not contain any ions" }, { "text": "chemical bond formation takes place when forces of attraction overcome the forces of repulsion" }, { "text": "in covalency transference of electron takes place" } ], "answer": "chemical bond formation takes place when forces of attraction overcome the forces of repulsion", "solution": "**Answer:** chemical bond formation takes place when forces of attraction overcome the forces of repulsion\n\n(A) Electronegativity of F is 4 and Electronegativity of H is 2.1.\n

$$ \\therefore $$ Electronegativity difference between F and H is = 4 - 2.1 = 1.9\n

Now Electronegativity of Br is 2.8 and Electronegativity of H is 2.1.\n

$$ \\therefore $$ Electronegativity difference between Br and H is = 2.8 - 2.1 = 0.7\n

As in HF Electronegativity difference is more then it's polar nature is also more.\n

(B) At equilibrium pure H2O produce very less H+ and OH- ions.\n

(C) Chemical bond formation takes place when forces of attraction overcome the forces of repulsion.\n

(D) In covalenrt bond, sharing of electons take place. Transfer of electrons take place in ionic bond.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 346, "subject": "Chemistry", "question": "Which of the following compounds contain(s) no covalent bond(s)?\nKCl, PH3, O2, B2H6, H2SO4 ", "options": [ { "text": "KCl, B2H6" }, { "text": "KCl, B2H6, PH3" }, { "text": "KCl, H2SO4" }, { "text": "KCl" } ], "answer": "KCl", "solution": "**Answer:** KCl\n\nHere only KCl is ionic compound all other compound has covalent bond.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 347, "subject": "Chemistry", "question": "The covalent alkaline earth metal halide\n(X = Cl, Br, I) is :", "options": [ { "text": "CaX2" }, { "text": "SrX2" }, { "text": "MgX2" }, { "text": "BeX2" } ], "answer": "BeX2", "solution": "**Answer:** BeX2\n\nAccording to Fajan’s rule, when the size of cation is less then the\ncovalent character would be more for the cation.\n

Since Be2+ has minimum size\namong given cation, therefore, BeX2 would be\nmost covalent among given alkaline with metal\nhalides", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 348, "subject": "Chemistry", "question": "

Arrange the following in increasing order of their covalent character.

\n

A. $$\\mathrm{CaF}_{2}$$

\n

B. $$\\mathrm{CaCl}_{2}$$

\n

C. $$\\mathrm{CaBr}_{2}$$

\n

D. $$\\mathrm{CaI}_{2}$$

\n

Choose the correct answer from the options given below.

", "options": [ { "text": "B < A < C < D" }, { "text": "A < B < C < D" }, { "text": "A < B < D < C" }, { "text": "A < C < B < D" } ], "answer": "A < B < C < D", "solution": "**Answer:** A < B < C < D\n\nFrom Fajan's rule, for a given metal ion, as the size of anion increases, polarizability of anion increases and hence covalent character of the given ionic compound increases.\n

\nHence, the increasing order of covalent character is $\\mathrm{CaF}_{2}<\\mathrm{CaCl}_{2}<\\mathrm{CaBr}_{2}<\\mathrm{Cal}_{2}$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 349, "subject": "Chemistry", "question": "

Given below are two statements.

\n

Statement I: $$\\mathrm{O}_{2}, \\mathrm{Cu}^{2+}$$, and $$\\mathrm{Fe}^{3+}$$ are weakly attracted by magnetic field and are magnetized in the same direction as magnetic field.

\n

Statement II: $$\\mathrm{NaCl}$$ and $$\\mathrm{H}_{2} \\mathrm{O}$$ are weakly magnetized in opposite direction to magnetic field.

\n

In the light of the above statements, choose the most appropriate answer from the options given below.

", "options": [ { "text": "Both Statement I and Statement II are correct." }, { "text": "Both Statement I and Statement II are incorrect." }, { "text": "Statement I is correct but Statement II is incorrect." }, { "text": "Statement I is incorrect but Statement II is correct." } ], "answer": "Both Statement I and Statement II are correct.", "solution": "**Answer:** Both Statement I and Statement II are correct.\n\n$\\mathrm{O}_{2}, \\mathrm{Cu}^{2+}$ and $\\mathrm{Fe}^{3+}$ have 2,1 and 5 unpaired electrons respectively, so these are the paramagnetic species. Hence, they are attracted by magnetic field.\n

\n$\\mathrm{NaCl}$ and $\\mathrm{H}_{2} \\mathrm{O}$ are the diamagnetic species so they are repelled by the magnetic field.", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 350, "subject": "Chemistry", "question": "

Order of Covalent bond;

\n

$$\\mathrm{A.~KF > KI ; LiF > KF}$$

\n

$$\\mathrm{B.~KF < KI ; LiF > KF}$$

\n

$$\\mathrm{C.~SnCl_4 > SnCl_2 ; CuCl > NaCl}$$

\n

$$\\mathrm{D.~LiF > KF ; CuCl < NaCl}$$

\n

$$\\mathrm{E.~KF < KI ; CuCl > NaCl}$$

\n

Choose the correct answer from the options given below :

", "options": [ { "text": "B, C only" }, { "text": "A, B only" }, { "text": "C, E only" }, { "text": "B, C, E only" } ], "answer": "B, C, E only", "solution": "**Answer:** B, C, E only\n\nB is correct $\\mathrm{KF}<\\mathrm{KI}$; $\\mathrm{LiF}>\\mathrm{KF}$\n

\n$\\mathrm{C}$ is correct $\\underset{+4}{\\mathrm{SnCl}_{4}}>\\underset{+2}{\\mathrm{SnCl}_{2}} ; \\mathrm{CuCl}>\\mathrm{NaCl}$\n

\n$\\mathrm{E}$ is correct $\\mathrm{KF}<\\mathrm{KI}$; $\\mathrm{CuCl}>\\mathrm{NaCl}$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 351, "subject": "Chemistry", "question": "

Number of $$\\sigma$$ and $$\\pi$$ bonds present in ethylene molecule is respectively :

", "options": [ { "text": "3 and 1\n" }, { "text": "5 and 1\n" }, { "text": "4 and 1\n" }, { "text": "5 and 2" } ], "answer": "5 and 1\n", "solution": "**Answer:** 5 and 1\n\n\n

Let's analyze the bond structure in an ethylene molecule (C2H4) to determine the count of $$\\sigma$$ and $$\\pi$$ bonds.\n\n

\n\n

Ethylene consists of two carbon atoms double-bonded to each other, with each carbon atom also bonded to two hydrogen atoms. Each single bond is a $$\\sigma$$ bond. The double bond between the carbon atoms is made up of one $$\\sigma$$ bond and one $$\\pi$$ bond. Here's how the bonds break down:

\n\n
    \n\n
  1. Each carbon-hydrogen bond is a $$\\sigma$$ bond. Since there are four C-H bonds, we have 4 $$\\sigma$$ bonds from C-H bonding.
    2. \n\n
  2. The carbon-carbon double bond consists of one $$\\sigma$$ bond and one $$\\pi$$ bond. Thus, we have 1 additional $$\\sigma$$ bond from the C=C double bond.
    3. \n\n
  3. Total $$\\sigma$$ bonds = 4 (from C-H) + 1 (from C=C) = 5.
    4. \n\n
  4. There is 1 $$\\pi$$ bond present in the carbon-carbon double bond.
    5. \n\n
\n\n

Therefore, the correct option is:
\n\n

Option B: 5 and 1

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 352, "subject": "Chemistry", "question": "Which one of the following pairs of molecules will have permanent dipole moments for both members", "options": [ { "text": "NO2 and CO2" }, { "text": "NO2 and O3" }, { "text": "SiF4 abd CO2" }, { "text": "SiF4 abd NO2" } ], "answer": "NO2 and O3", "solution": "**Answer:** NO2 and O3\n\n\"AIEEE\n
Here $$\\mu $$total = $$\\mu $$1 + $$\\mu $$2 $$ \\ne $$ 0\n
\"AIEEE\n
$$\\mu $$total = $$\\mu $$1 $$-$$ $$\\mu $$2 = 0\n
\"AIEEE\n
$$\\mu $$total = $$\\mu $$1 + $$\\mu $$2 $$ \\ne $$ 0\n
\"AIEEE\n
$$\\mu $$1 + $$\\mu $$2 + $$\\mu $$3 = $$\\mu $$4\n

   $$\\mu $$total = 0\n

In NO2 and O3 have permanent dipole moment as both of them are not symmetric molecule. \n

In CO2 and SiF4 each individual bond c = O and S- $$-$$ F have dipole moment but because of symmetric structure individual dipole moment get's cancelled. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 353, "subject": "Chemistry", "question": "The dipole of CCl4, CHCl3 and CH4 are in the order :", "options": [ { "text": "CCl4 < CH4 < CHCl3" }, { "text": "CH4 = CCl4 < CHCl3\n" }, { "text": "CH4 < CCl4 < CHCl3" }, { "text": "CHCl3 < CH4= CCl4" } ], "answer": "CH4 = CCl4 < CHCl3\n", "solution": "**Answer:** CH4 = CCl4 < CHCl3\n\n\nCHCl3 is polar while CH4 and CCl4 are nonpolar.\n

So, dipole moment order is :\n

CH4 = CCl4 < CHCl3\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 354, "subject": "Chemistry", "question": "

Amongst BeF2, BF3, H2O, NH3, CCl4 and HCl, the number of molecules with non-zero net dipole moment is ____________.

", "options": [], "answer": "3", "solution": "**Answer:** 3\n\n$\\mathrm{BeF}_2, \\mathrm{BF}_3$ and $\\mathrm{CCl}_4 \\Rightarrow \\mu_{\\mathrm{net}}=0$

$\\mathrm{H}_2 \\mathrm{O}, \\mathrm{NH}_3$ and $\\mathrm{HCl} \\Rightarrow \\mu_{\\mathrm{net}} \\neq 0$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 355, "subject": "Chemistry", "question": "

Amongst the following, the number of molecule/(s) having net resultant dipole moment is ____________.

\n

NF3, BF3, BeF2, CHCl3, H2S, SiF4, CCl4, PF5

", "options": [], "answer": "3", "solution": "**Answer:** 3\n\nSymmetrical molecules have zero dipole moment like $\\mathrm{BF}_{3}$, $\\mathrm{BeF}_{2}, \\mathrm{SiF}_{4}, \\mathrm{CCl}_{4}$ and $\\mathrm{PF}_{5}$.

\nUnsymmetrical molecules have net diploe moment like $-\\mathrm{NF}_3$, $\\mathrm{CHCl}_3$ and $\\mathrm{H}_2 \\mathrm{S}$

\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 356, "subject": "Chemistry", "question": "

Match List - I with List - II.

\n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
List - IList - II
(A)$$\\psi_{\\mathrm{MO}}=\\psi_{\\mathrm{A}}-\\psi_{\\mathrm{B}}$$(I)Dipole moment
(B)$$\\mu=Q \\times r$$(II)Bonding molecular orbital
(C)$$\\frac{\\mathrm{N}_{\\mathrm{b}}-\\mathrm{N}_{\\mathrm{a}}}{2}$$(III)Anti-bonding molecular orbital
(D)$$\\psi_{\\mathrm{MO}}=\\psi_{\\mathrm{A}}+\\psi_{\\mathrm{B}}$$(IV)Bond order

\n

Choose the correct answer from the options given below :

", "options": [ { "text": "$$(\\mathrm{A})-(\\mathrm{II}),(\\mathrm{B})-(\\mathrm{I}),(\\mathrm{C})-(\\mathrm{IV}),(\\mathrm{D})-(\\mathrm{III})$$" }, { "text": "$$(\\mathrm{A})-(\\mathrm{III}),(\\mathrm{B})-(\\mathrm{IV}),(\\mathrm{C})-(\\mathrm{I}),(\\mathrm{D})-(\\mathrm{II})$$" }, { "text": "$$(\\mathrm{A})-(\\mathrm{III}),(\\mathrm{B})-(\\mathrm{I}),(\\mathrm{C})-(\\mathrm{IV}),(\\mathrm{D})-(\\mathrm{II})$$" }, { "text": "$$(\\mathrm{A})-(\\mathrm{III}),(\\mathrm{B})-(\\mathrm{IV}),(\\mathrm{C})-(\\mathrm{II}),(\\mathrm{D})-(\\mathrm{I})$$" } ], "answer": "$$(\\mathrm{A})-(\\mathrm{III}),(\\mathrm{B})-(\\mathrm{I}),(\\mathrm{C})-(\\mathrm{IV}),(\\mathrm{D})-(\\mathrm{II})$$", "solution": "**Answer:** $$(\\mathrm{A})-(\\mathrm{III}),(\\mathrm{B})-(\\mathrm{I}),(\\mathrm{C})-(\\mathrm{IV}),(\\mathrm{D})-(\\mathrm{II})$$\n\n$\\Psi_{A}-\\Psi_{B}=\\Psi_{M O}$ is anti-boding molecular orbital\n

\n$$\n\\begin{aligned}\n&\\mu=Q \\times r \\text { is dipole moment } \\\\\\\\\n&\\frac{N_{b}-N_{a}}{2}=\\text { bond order } \\\\\\\\\n&\\Psi_{\\mathrm{A}}+\\Psi_{\\mathrm{B}}=\\Psi_{\\mathrm{MO}} \\text { is boding molecular orbital. }\n\\end{aligned}\n$$", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 357, "subject": "Chemistry", "question": "

Statement I : Dipole moment is a vector quantity and by convention it is depicted by a small arrow with tail on the negative centre and head pointing towards the positive centre.

\n

Statement II : The crossed arrow of the dipole moment symbolizes the direction of the shift of charges in the molecules.

\n

In the light of the above statements, choose the most appropriate answer from the options given below :

", "options": [ { "text": "Statement I is correct but Statement II is incorrect" }, { "text": "Statement I is incorrect but Statement II is correct" }, { "text": "Both Statement I and Statement II are correct" }, { "text": "Both Statement I and Statement II are incorrect" } ], "answer": "Statement I is incorrect but Statement II is correct", "solution": "**Answer:** Statement I is incorrect but Statement II is correct\n\nThe dipole moment is a vector quantity and is\ndepicted by an arrow with tail on the positive centre\nand head pointing towards the negative centre. So,\nStatement-I is incorrect.

The crossed arrow of the\ndipole moment symbolizes the direction of the shift\nof charges in the molecules. So, Statement-II is\ncorrect.\n

For example,\n

\"JEE", "topic": "Algebra", "subtopic": "Linear Algebra" }, { "id": 358, "subject": "Chemistry", "question": "Consider the following statements:\n

\n(A) NF3 molecule has a trigonal planar structure.\n

\n(B) Bond length of $\\mathrm{N}_{2}$ is shorter than $\\mathrm{O}_{2}$.\n

\n(C) Isoelectronic molecules or ions have identical bond order.\n

\n(D) Dipole moment of $\\mathrm{H}_{2}\\mathrm{S}$ is higher than that of water molecule.\n

\nChoose the correct answer from the options given below:", "options": [ { "text": "(C) and (D) are correct" }, { "text": "(A) and (B) are correct" }, { "text": "(B) and (C) are correct" }, { "text": "(A) and (D) are correct\n" } ], "answer": "(B) and (C) are correct", "solution": "**Answer:** (B) and (C) are correct\n\nThe correct answer is (B) and (C).\n

\n(A) NF3 molecule has a trigonal pyramidal structure, not a trigonal planar structure. Therefore, statement (A) is false.\n

\n(B) The bond length of $\\mathrm{N}_{2}$ is shorter than that of $\\mathrm{O}_{2}$ due to the smaller atomic radius of nitrogen compared to oxygen. Therefore, statement (B) is true.\n

\n(C) Isoelectronic molecules or ions have the same number of electrons and hence the same bond order. Therefore, statement (C) is true.\n

\n(D) The dipole moment of water ($\\mathrm{H}_{2}\\mathrm{O}$) is higher than that of $\\mathrm{H}_{2}\\mathrm{S}$ due to the greater electronegativity of oxygen compared to sulfur. Therefore, statement (D) is false.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 359, "subject": "Chemistry", "question": "

Choose the polar molecule from the following:

", "options": [ { "text": "$$\\mathrm{CCl}_4$$\n" }, { "text": "$$\\mathrm{CO}_2$$\n" }, { "text": "$$\\mathrm{CH}_2=\\mathrm{CH}_2$$\n" }, { "text": "$$\\mathrm{CHCl}_3$$" } ], "answer": "$$\\mathrm{CHCl}_3$$", "solution": "**Answer:** $$\\mathrm{CHCl}_3$$\n\n

\"JEE

\n

$$\\mu \\neq 0$$

\n

$$\\mathrm{CHCl}_3$$ is polar molecule and rest all molecules are non-polar.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 360, "subject": "Chemistry", "question": "

A diatomic molecule has a dipole moment of $$1.2 \\mathrm{~D}$$. If the bond distance is $$1 \\mathrm{~A}^{\\circ}$$, then fractional charge on each atom is _________ $$\\times 10^{-1}$$ esu.

\n

(Given $$1 \\mathrm{~D}=10^{-18}$$ esucm)

", "options": [], "answer": "0", "solution": "**Answer:** 0\n\n

To find the fractional charge on each atom in a diatomic molecule, we can use the relationship between dipole moment (μ), charge (q), and distance (d):

\n\n

$$ \\mu = q \\times d $$

\n\n

Where:

\n\n\n

First, convert units to be consistent with esu:

\n\n\n

We are given that the bond distance ($ d $) is $ 1 \\mathrm{~A}^{\\circ} $ and that the dipole moment $( \\mu $) is $ 1.2 \\mathrm{~D} $. Using these values, we can calculate the charge ($ q $) as follows:

\n\n

$$ q = \\frac{\\mu}{d} $$

\n\n

Now plug in the given values:

\n\n

$$ q = \\frac{1.2 \\times 10^{-18} \\mathrm{~esucm}}{1 \\times 10^{-8} \\mathrm{~cm}} $$

\n

$$ q = 1.2 \\times 10^{-10} \\mathrm{~esu} $$

\n

$$ q = 1.2 \\times 10^{-9} \\times 10^{-1} \\mathrm{~esu} $$

\n

$$ q = 0.000000012 \\times 10^{-1} \\mathrm{~esu} $$

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 361, "subject": "Chemistry", "question": "

The total number of molecules with zero dipole moment among $$\\mathrm{CH}_4, \\mathrm{BF}_3, \\mathrm{H}_2 \\mathrm{O}, \\mathrm{HF}, \\mathrm{NH}_3, \\mathrm{CO}_2$$ and $$\\mathrm{SO}_2$$ is ________.

", "options": [], "answer": "3", "solution": "**Answer:** 3\n\n

Molecules with zero dipole moment $$=\\mathrm{CO}_2, \\mathrm{CH}_4, \\mathrm{BF}_3$$

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 362, "subject": "Chemistry", "question": "

Given below are two statements:

\n

Statement - I: Since Fluorine is more electronegative than nitrogen, the net dipole moment of $$\\mathrm{NF}_3$$ is greater than $$\\mathrm{NH}_3$$.

\n

Statement - II: In $$\\mathrm{NH}_3$$, the orbital dipole due to lone pair and the dipole moment of $$\\mathrm{NH}$$ bonds are in opposite direction, but in $$\\mathrm{NF}_3$$ the orbital dipole due to lone pair and dipole moments of N-F bonds are in same direction.

\n

In the light of the above statements, choose the most appropriate from the options given below:

", "options": [ { "text": "Statement I is true but Statement II is false.\n" }, { "text": "Both Statement I and Statement II are true.\n" }, { "text": "Both Statement I and Statement II are false.\n" }, { "text": "Statement I is false but Statement II is true." } ], "answer": "Both Statement I and Statement II are false.\n", "solution": "**Answer:** Both Statement I and Statement II are false.\n\n\n

\"JEE

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 363, "subject": "Chemistry", "question": "

Which one of the following molecules has maximum dipole moment?

", "options": [ { "text": "$$\\mathrm{CH}_4$$\n" }, { "text": "$$\\mathrm{NF}_3$$\n" }, { "text": "$$\\mathrm{NH}_3$$\n" }, { "text": "$$\\mathrm{PF}_5$$" } ], "answer": "$$\\mathrm{NH}_3$$\n", "solution": "**Answer:** $$\\mathrm{NH}_3$$\n\n\n

$$\\mathrm{NH}_3 \\text { have more dipole moment than } \\mathrm{NF}_3 \\text {. }$$

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 364, "subject": "Chemistry", "question": "

Number of compounds / species from the following with non-zero dipole moment is _________.

\n

$$\\mathrm{BeCl}_2, \\mathrm{BCl}_3, \\mathrm{NF}_3, \\mathrm{XeF}_4, \\mathrm{CCl}_4, \\mathrm{H}_2 \\mathrm{O}, \\mathrm{H}_2 \\mathrm{~S}, \\mathrm{HBr}, \\mathrm{CO}_2, \\mathrm{H}_2, \\mathrm{HCl}$$

", "options": [], "answer": "5", "solution": "**Answer:** 5\n\n

$$\\mathrm{NF}_3, \\mathrm{H}_2 \\mathrm{O}, \\mathrm{H}_2 \\mathrm{S}, \\mathrm{HBr}, \\mathrm{HCl}$$ have non zero dipole moments.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 365, "subject": "Chemistry", "question": "

Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).

\n

Assertion (A) : $$\\mathrm{NH}_3$$ and $$\\mathrm{NF}_3$$ molecule have pyramidal shape with a lone pair of electrons on nitrogen atom. The resultant dipole moment of $$\\mathrm{NH}_3$$ is greater than that of $$\\mathrm{NF}_3$$.

\n

Reason (R) : In $$\\mathrm{NH}_3$$, the orbital dipole due to lone pair is in the same direction as the resultant dipole moment of the $$\\mathrm{N}-\\mathrm{H}$$ bonds. $$\\mathrm{F}$$ is the most electronegative element.

\n

In the light of the above statements, choose the correct answer from the options given below :

", "options": [ { "text": "(A) is false but (R) is true" }, { "text": "(A) is true but (R) is false" }, { "text": "Both (A) and (R) are true but (R) is NOT the correct explanation of (A)" }, { "text": "Both (A) and (R) are true and (R) is the correct explanation of (A)" } ], "answer": "Both (A) and (R) are true and (R) is the correct explanation of (A)", "solution": "**Answer:** Both (A) and (R) are true and (R) is the correct explanation of (A)\n\n

\"JEE

\n

Dipole moment of $$\\mathrm{NH}_3$$ is higher than $$\\mathrm{NF}_3$$ as $$\\mu$$ of lone-pair is in same direction as the resultant dipole moment of $$\\mathrm{N}-\\mathrm{H}$$ bonds.\n

So, both (A) & (R) are true, & (R) is the correct explanation of (A)

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 366, "subject": "Chemistry", "question": "

Number of compounds from the following with zero dipole moment is _________.

\n

$$\\mathrm{HF}, \\mathrm{H}_2, \\mathrm{H}_2 \\mathrm{~S}, \\mathrm{CO}_2, \\mathrm{NH}_3, \\mathrm{BF}_3, \\mathrm{CH}_4, \\mathrm{CHCl}_3, \\mathrm{SiF}_4, \\mathrm{H}_2 \\mathrm{O}, \\mathrm{BeF}_2$$

", "options": [], "answer": "6", "solution": "**Answer:** 6\n\n

\"JEE

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 367, "subject": "Chemistry", "question": "Hybridisation of underline atom changes in", "options": [ { "text": "$$\\underline {Al} {H_3}$$ changes to $$AlH_4^-$$" }, { "text": "$${H_2}\\underline O $$ changes to $$H_3O^+$$" }, { "text": "$$\\underline N {H_3}$$ changes to $$NH_4^+$$" }, { "text": "in all cases" } ], "answer": "$$\\underline {Al} {H_3}$$ changes to $$AlH_4^-$$", "solution": "**Answer:** $$\\underline {Al} {H_3}$$ changes to $$AlH_4^-$$\n\n(a) $$\\,\\,\\,$$ AlH3 + H$$-$$ $$ \\to $$ AlH$$_4^ - $$\n

Steric number of AlH3 is = $${1 \\over 2}$$ [3 + 3] = 3\n

$$\\therefore\\,\\,\\,$$ AlH3 is sp2 hybridized. \n

Steric number of AlH$$_4^ - $$ = $${1 \\over 2}$$ [3 + 4 +1] = 4 \n

$$\\therefore\\,\\,\\,$$ AlH$$_4^ - $$ is sp3 hybridized. \n

(b) $$\\,\\,\\,$$ H2O + H+ $$ \\to $$ H3O+ \n

Steric no of H2O = $${1 \\over 2}$$ (6+ 2) = 4\n

$$\\therefore\\,\\,\\,\\,$$ H2O s sp3 hybridized.\n

Steric no of H3O+ = $${1 \\over 2}$$ [ 8 + 3 $$-$$1] = 4\n

$$\\therefore\\,\\,\\,$$ H3O+ s also sp3 hybridized.\n

(c) $$\\,\\,\\,$$ NH3 + H+ $$ \\to $$ NH4+ \n

Steric no of NH3 = $${1 \\over 2}$$ [5 + 3] = 4\n

$$\\therefore\\,\\,\\,$$ hybridization of NH3 is sp3 \n

Steric number of NH4+ = $${1 \\over 2}$$ [5 + 4 $$-$$ ] = 4\n

$$\\therefore\\,\\,\\,$$ Hybridization of NH4+ is sp3 ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 368, "subject": "Chemistry", "question": "A square planar complex is formed by hybridisation of which atomic orbitals ?", "options": [ { "text": "s, px, py, dxy" }, { "text": "s, px, py, dx2 - y2" }, { "text": "s, px, py, dz2" }, { "text": "s, px, py, dyz" } ], "answer": "s, px, py, dx2 - y2", "solution": "**Answer:** s, px, py, dx2 - y2\n\nHybridization of square planar complex is dsp2 . \n

\"AIEEE\n

In square planar complex all 4 surrounding atoms and central atom are in the same plane and let this plane is x $$-$$ y plane. As it is dsp2 hybridized, so it has 1 d orbital, 1 s orbital and 2 p orbital.\n

s orbital is non-directional, so we just write it as s. \n

\"AIEEE\n

In px the two lobes are along x-axis and in py two lobes are along y-axis \n

One d orbital is dx2 $$-$$ y2, it means out of four lobes of d orbital two along x-axis and two along y-axis. \n

\"AIEEE ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 369, "subject": "Chemistry", "question": "In which of the following species the interatomic bond angle is 109o28' ?", "options": [ { "text": "NH3, BF4-" }, { "text": "NH4+, BF4-" }, { "text": "NH3, BF3" }, { "text": "NH2-1, BF3" } ], "answer": "NH4+, BF4-", "solution": "**Answer:** NH4+, BF4-\n\nBond angle 109o 28' means Regular Tetrahedral geometry and hybridization is sp3. \n

Steric Number (SN) for sp3 hybridization is 4.\n

In sp3 molecules can have different shapes which is decided by SN number \n

(1) $$\\,\\,\\,\\,$$ 4 bond pair in a molecule then angle between bonds 109o 28' \n

(2) $$\\,\\,\\,\\,$$ 3 bond pair and 1 lone pair in a molecule then angle between bonds 107o.\n

(3) $$\\,\\,\\,\\,$$ 2 bond pair and 2 lone pair in a molecule then angle between bonds 104.5o \n

(4) $$\\,\\,\\,\\,$$ 1 bond pair and 3 lone pair in a molecule then bond angle is undefined as there is only one bond. \n

(1) $$\\,\\,\\,\\,$$ In NH3, 3 Bond pair(BP) +1 lone pair (LP) present so angle between bond 107o. \n

\"AIEEE\n

(2) $$\\,\\,\\,\\,$$ BF$$_4^ - $$, 4 bond pair present so angle is 109o 28'.\n

\"AIEEE\n

(3) In NH$$_4^ + $$, 4 bond pair present so angle between bond is 109o 28'\n

\"AIEEE\n

(4) $$\\,\\,\\,\\,$$ BF3 has sp2 hybridization. So bond angle is 120o.\n

\"AIEEE\n

(5) $$\\,\\,\\,\\,$$ In NH$$_2^ - $$, 2 bond pair and 2 lone pair present, so bond angle is 104.5o \n

\"AIEEE\n

So, $$NH{}^ + $$ and BF$$_4^ - $$ has bond angle 109o 28'. ", "topic": "Geometry", "subtopic": "Plane Geometry" }, { "id": 370, "subject": "Chemistry", "question": "The pair of species having identical shapes for molecules of both species is", "options": [ { "text": "XeF2, CO2" }, { "text": "BF3, PCl3" }, { "text": "PF5, IF5" }, { "text": "CF4, SF4" } ], "answer": "XeF2, CO2", "solution": "**Answer:** XeF2, CO2\n\n\"AIEEE\n
\"AIEEE\n
\"AIEEE\n
\"AIEEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 371, "subject": "Chemistry", "question": "Which one of the following compounds has the smallest bond angle in its molecule?", "options": [ { "text": "OH2" }, { "text": "SH2" }, { "text": "NH3" }, { "text": "SO2" } ], "answer": "SH2", "solution": "**Answer:** SH2\n\n(a)   H2O is sp3 hybridized, and oxygen atom has 2 bond pair and 2 lone pair. So the angle between two O $$-$$ H bond is 104.5o\n

(b)   In H2S molecule, central atom S is a 3rd period element and according to Dragos rule, when a 3rd period or higher period element overlap with a small element whose electronegetivity is low then that over lapping cannot be a effective overlapping, because of higher size of 3rd or higher periods element and smaller size of low electronegative element. \n

Here in H2S, H atom has very low electronegativity and smaller in size so it create more negative charge around sulphur(s) atom and size of S$$-$$2 ion increases, because of this higher size difference efficiency overlapping is not possible. So, any hybridization do not happen in between S and H atom. Lone pair electrons of S atom is present in the pure orbital like s, px, py, pz and it look like this : \n

\"AIEEE\n

As we know the angle between and py is 90o, so bond angle is 90o.\n

(c)   NH3 has sp3 hybridization and N atom has 3 bond pair and one lone pair so bond angle is 107o\n

(d)   SO2 is sp2 hybridized and bond angle is 117o.\n

\"AIEEE", "topic": "Geometry", "subtopic": "Plane Geometry" }, { "id": 372, "subject": "Chemistry", "question": "The states of hybridization of boron and oxygen atoms in boric acid (H3BO3) are respectively ", "options": [ { "text": "sp2 and sp2\n " }, { "text": "sp3 and sp3" }, { "text": "sp3 and sp2" }, { "text": "sp2 and sp3\n " } ], "answer": "sp2 and sp3\n ", "solution": "**Answer:** sp2 and sp3\n \n\nStructure of  H3BO3  is \n

\"AIEEE\n

Here Boron has 3 sigma bond. So it is sp2 hybridised. \n

And oxygen has two sigma bond and two lone pair. So it sp3 hybridised.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 373, "subject": "Chemistry", "question": "Which one of the following has the regular tetrahedral structure?
\n\n(Atomic nos : B = 5, S = 16, Ni = 28, Xe = 54)", "options": [ { "text": "XeF4" }, { "text": "[Ni(CN)4]2- " }, { "text": "$$BF_4^-$$" }, { "text": "SF4" } ], "answer": "$$BF_4^-$$", "solution": "**Answer:** $$BF_4^-$$\n\nRegular Tetrahedral structure is possible in sp3 hybridization where central atom has 4 bond pair and no lone pair.\n

(a)   XeF4  is sp3d2 hybridised and structure is square planar.\n

\"AIEEE\n

(b)   [Ni(CN)4]$$-$$2 is coordinate compound and oxidation number of Ni is +2.\n

Electronic configuration of Ni+2 is $$=$$ [Ar]3d8\n

\"AIEEE\n

But because of CN$$-$$ ion which is a strong field ligand , it can perform pairing of electron.\n

\"AIEEE\n

And the structure of dsp2 hybridization is square planar.\n

\"AIEEE\n

(C) $$\\,\\,\\,\\,$$ BF$$_4^ - $$, 4 bond pair present so angle is 109o 28' and sp3 hybridised. So structure is regular tetrahedral.\n

\"AIEEE\n

(d)   \"AIEEE\n

SF4 is sp3d hybridised and structure is see-saw.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 374, "subject": "Chemistry", "question": "The maximum number of 90° angles between bond pair of electrons is observed in ", "options": [ { "text": "dsp3 hybridization" }, { "text": "sp3d2 hybridization " }, { "text": "dsp2 hybridization " }, { "text": "sp3d hybridization" } ], "answer": "sp3d2 hybridization ", "solution": "**Answer:** sp3d2 hybridization \n\n\"AIEEE\n

Here eight 90o angles between bond pair and bond pair. Those angles are $$\\angle $$1M2, $$\\angle $$2M3, $$\\angle $$3M4, $$\\angle $$4M1, $$\\angle $$5M1, $$\\angle $$5M2, $$\\angle $$5M3, $$\\angle $$5M4.\n\"AIEEE\n

Here twelve 90o angles between bond pair and bond pair. Those angles are $$\\angle $$1M2, $$\\angle $$2M3, $$\\angle $$3M4, $$\\angle $$4M1, $$\\angle $$5M1, $$\\angle $$5M2, $$\\angle $$5M3, $$\\angle $$5M4, $$\\angle $$6M1, $$\\angle $$6M2, $$\\angle $$6M3, $$\\angle $$6M4.\n\"AIEEE\n

Here four 90o angles between bond pair and bond pair. Those angles are $$\\angle $$1M2, $$\\angle $$2M3, $$\\angle $$3M4, $$\\angle $$4M1.\n\"AIEEE\n

Here six 90o angles between bond pair and bond pair. Those angles are $$\\angle $$1M3, $$\\angle $$1M4, $$\\angle $$1M5, $$\\angle $$2M3, $$\\angle $$2M4, $$\\angle $$2M5.", "topic": "Geometry", "subtopic": "Plane Geometry" }, { "id": 375, "subject": "Chemistry", "question": "In which of the following molecules/ions are all the bonds not equal?", "options": [ { "text": "XeF4" }, { "text": "$$BF_4^−$$" }, { "text": "SF4" }, { "text": "SiF4" } ], "answer": "SF4", "solution": "**Answer:** SF4\n\n(a)   XeF4  is sp3d2 hybridised with 4 bond pairs and 1 lone pair and structure is square planar. Here all the bond lengths are equal.\n

\"AIEEE\n

(b) $$\\,\\,\\,\\,$$ BF$$_4^ - $$, 4 bond pair present so angle is 109o 28' and sp3 hybridised. So structure is regular tetrahedral. Here all the bond lengths are equal.\n

\"AIEEE\n

(c)   SF4 is sp3d hybridised with 4 bond pairs and 1 lone pair and its expected trigonal\nbipyramidal geometry gets distorted due to presence of a lone\npair of electrons and it becomes distorted tetrahedral or\nsee-saw with the bond angles equal to < 120o and 179o instead of\nthe expected angles of 120o and 180o respectively. Here axial and equitorial both bonds are presents. And we know axial bonds are longer and weaker.\n

\"AIEEE\n

(d) SiF4\nis sp3\n hybridisation and regular tetrahedral geometry. Here all the bond lengths are equal.\n

\"AIEEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 376, "subject": "Chemistry", "question": "The decreasing values of bond angles from NH3 (106o) to SbH3 (91o\n) down group-15 of the periodic table is due to ", "options": [ { "text": "increasing bp-bp repulsion" }, { "text": "increasing p-orbital character in sp3" }, { "text": "decreasing lp-bp repulsion" }, { "text": "decreasing electronegativity " } ], "answer": "decreasing lp-bp repulsion", "solution": "**Answer:** decreasing lp-bp repulsion\n\nNitrogen(N) atom is smaller in size so it's lone pair is relatively unstable. That is why repulsion between lone pairs present on nitrogen atom and bonded pairs of electrons is relatively high.\n

But size of Sb is more so its lone pair is relatively stable. That is why it does not want to perticipate in bond pair lone pair repulsion and decreases lp-bp repulsion.", "topic": "Geometry", "subtopic": "Plane Geometry" }, { "id": 377, "subject": "Chemistry", "question": "The hybridization of orbitals of N atom in $$NO_3^-$$, $$NO_2^+$$ and $$NH_4^+$$ are respectively :", "options": [ { "text": "sp , sp2, sp3" }, { "text": "sp2, sp , sp3" }, { "text": "sp , sp3, sp2" }, { "text": "sp2, sp3, sp " } ], "answer": "sp2, sp , sp3", "solution": "**Answer:** sp2, sp , sp3\n\n\"AIEEE\n\"AIEEE\n\"AIEEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 378, "subject": "Chemistry", "question": "The structure of IF7 is :", "options": [ { "text": "trigonal bipyramid" }, { "text": "octahedral" }, { "text": "pentagonal bipyramid" }, { "text": "square pyramid" } ], "answer": "pentagonal bipyramid", "solution": "**Answer:** pentagonal bipyramid\n\nThe structure of IF7 pentagonal bipyramidal having\nsp3d3\n hybridisation.\n
\"AIEEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 379, "subject": "Chemistry", "question": "In which of the following pairs the two species are not isostructural ?", "options": [ { "text": "$$CO_3^{2-}$$ and $$NO_3^-$$" }, { "text": "$$PCl_4^{+}$$ and $$SiCl_4$$" }, { "text": "PF5 and BrF5" }, { "text": "$$AlF_6^-$$ and $$SF_6$$" } ], "answer": "PF5 and BrF5", "solution": "**Answer:** PF5 and BrF5\n\nPF5\n $$ \\Rightarrow $$ sp3d hybridized and shape is trigonal bipyramidal.\n

BrF5 $$ \\Rightarrow $$ 5 bond pair and 1 lone pair present so sp3d2 hybridized and shape is square pyramidal.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 380, "subject": "Chemistry", "question": "The group of molecules having identical shape is :", "options": [ { "text": "SF4 , XeF4 , CCl4" }, { "text": "ClF3 , XeOF2 , XeF$$_3^ + $$" }, { "text": "BF3 , PCl3 , XeO3" }, { "text": "PCl5 , IF5 , XeO2F2" } ], "answer": "ClF3 , XeOF2 , XeF$$_3^ + $$", "solution": "**Answer:** ClF3 , XeOF2 , XeF$$_3^ + $$\n\nH = $${1 \\over 2}$$[VE + MA – c + a]\n

ClF3\n, H =\n$${1 \\over 2}$$\n (7 + 3 – 0 + 0) = 5 (sp3d)\n

XeOF2\n, H = $${1 \\over 2}$$(8 + 2 – 0 + 0) = 5 (sp3d)\n

XeF3+\n, H =\n$${1 \\over 2}$$\n (8 + 3 – 1 + 0 ) = 5 (sp3d)\n

All molecules have sp3d hybridization and 3 bond pair + 2 lone pairs. Hence all have identical stucture (T-shape).\n", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 381, "subject": "Chemistry", "question": "The species in which the $\\mathrm{N}$ atom is in a state of $s p$ hybridization is :\n", "options": [ { "text": "$\\mathrm{NO}_2^{+}$" }, { "text": "$\\mathrm{NO}_2^{-}$\n" }, { "text": "$\\mathrm{NO}_3^{-}$\n" }, { "text": "$\\mathrm{NO}_3^{-}$" } ], "answer": "$\\mathrm{NO}_2^{+}$", "solution": "**Answer:** $\\mathrm{NO}_2^{+}$\n\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 382, "subject": "Chemistry", "question": "sp3d2 hybridization is not displayed by :\n", "options": [ { "text": "BrF5" }, { "text": "SF6" }, { "text": "[CrF6]3$$-$$" }, { "text": "PF5" } ], "answer": "PF5", "solution": "**Answer:** PF5\n\nHybridization (X) = $${1 \\over 2}$$ [ VE + MA – c + a ]\n

where, VE = No. of valence electrons of central atom \n
MA = No. of monovalent atoms/groups surrounding the central\natom, \n
c = Charge on the cation, \n
a = Charge on the anion\n

(A) [BrF5] : X = $${1 \\over 2}$$[ 7 + 5 - 0 + 0] = 6 = sp3d2 hybridized\n

(B) [SF6] : X = $${1 \\over 2}$$[ 6 + 6 - 0 + 0] = 6 = sp3d2 hybridized\n

(C) [CrF6]3$$-$$ : Cr+3 = [Ar]3d3\n\"JEE\n

(D) [PF5] : X = $${1 \\over 2}$$[ 5 + 5 - 0 + 0] = 5 = sp3d hybridized\n

Note : [CrF6]3$$-$$ shows d2sp3 hybridization not sp3d2. So option (C) should also be the correct answer.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 383, "subject": "Chemistry", "question": "The group having triangular planar structures is :\n", "options": [ { "text": "BF3, NF3, $$CO_3^{2 - }$$" }, { "text": "$$CO_3^{2 - }$$, $$NO_3^ - $$, SO3" }, { "text": "NH3, SO3, $$CO_3^{2 - }$$" }, { "text": "NCl3, BCl3, SO3" } ], "answer": "$$CO_3^{2 - }$$, $$NO_3^ - $$, SO3", "solution": "**Answer:** $$CO_3^{2 - }$$, $$NO_3^ - $$, SO3\n\n(A) BF3 : sp2 Hybridization : Triangular Planar\n

NF3 : sp3 Hybridization : Tetrahedral\n

$$CO_3^{2 - }$$ : sp2 Hybridization : Triangular Planar\n

(B) $$CO_3^{2 - }$$ : sp2 Hybridization : Triangular Planar\n

$$NO_3^ - $$ : sp2 Hybridization : Triangular Planar\n

SO3 : sp2 Hybridization : Triangular Planar\n

(C) NH3 : sp3 Hybridization : Tetrahedral\n

SO3 : sp2 Hybridization : Triangular Planar\n

$$CO_3^{2 - }$$ : sp2 Hybridization : Triangular Planar\n

(D) NCl3 : sp3 Hybridization : Tetrahedral\n

BCl3 : sp2 Hybridization : Triangular Planar\n

SO3 : sp2 Hybridization : Triangular Planar", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 384, "subject": "Chemistry", "question": "The incorrect geometry is represented by : ", "options": [ { "text": "BF3 - trigonal planar" }, { "text": "H2O - bent" }, { "text": "NF3 - trigonal planar " }, { "text": "AsF5 - trigonal bipyramidal " } ], "answer": "NF3 - trigonal planar ", "solution": "**Answer:** NF3 - trigonal planar \n\n\"JEE\n

NF3 has trigonal pyramidal geometry. Here central atom nitrogen has sp3 hybridization and one lone pair of electron. Because of lone pair there exist repulsion force between lone pair and bond pair of electrons, that is why bond angles are lower than tetrahedral bond angle. ", "topic": "Geometry", "subtopic": "Plane Geometry" }, { "id": 385, "subject": "Chemistry", "question": "Which of the following conversions involves change in both shape and hybridisation ? ", "options": [ { "text": "NH3 $$ \\to $$ NH4+ " }, { "text": "CH4 $$ \\to $$ C2H6" }, { "text": "H2O $$ \\to $$ H3O+" }, { "text": "BF3 $$ \\to $$ BF4$$-$$" } ], "answer": "BF3 $$ \\to $$ BF4$$-$$", "solution": "**Answer:** BF3 $$ \\to $$ BF4$$-$$\n\nBF3  $$ \\to $$  BF4$$-$$\n

\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 386, "subject": "Chemistry", "question": "Identify the pair in which the geometry of the species is $$T$$-shape and square - pyraidal, respectively : ", "options": [ { "text": "ClF3 and $$I{O_4}^ - $$ " }, { "text": "ICl2- and ICl5" }, { "text": "XeOF2 and XeOF4" }, { "text": "IO3- and IO2F2-" } ], "answer": "XeOF2 and XeOF4", "solution": "**Answer:** XeOF2 and XeOF4\n\n\"JEE\n
\"JEE\n
\"JEE\n
\"JEE", "topic": "Geometry", "subtopic": "Plane Geometry" }, { "id": 387, "subject": "Chemistry", "question": "Total number of lone pair of electrons in $${\\rm I}_3^ - $$ ion is ", "options": [ { "text": "3" }, { "text": "9" }, { "text": "6" }, { "text": "12" } ], "answer": "9", "solution": "**Answer:** 9\n\n\"JEE\n

$$\\therefore\\,\\,\\,$$ Total number of lone pair = 9", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 388, "subject": "Chemistry", "question": "The decreasing order of bond angles in BF3, NH3, PF3 and I3- is :", "options": [ { "text": "I3- > NH3 > PF3 > BF3" }, { "text": "I3- > BF3 > NH3 > PF3" }, { "text": "BF3 > I3- > PF3 > NH3" }, { "text": "BF3 > NH3 > PF3 > I3-" } ], "answer": "I3- > BF3 > NH3 > PF3", "solution": "**Answer:** I3- > BF3 > NH3 > PF3\n\n\"JEE\n
\"JEE\n
\"JEE\n
\"JEE", "topic": "Geometry", "subtopic": "Plane Geometry" }, { "id": 389, "subject": "Chemistry", "question": "The type of hybridisation and number of lone pair (s) of electrons of Xe in XeOF4, respectively, are: ", "options": [ { "text": "sp3d and 2" }, { "text": "sp3d2 and 2" }, { "text": "sp3d and 1" }, { "text": "sp3d2 and 1" } ], "answer": "sp3d2 and 1", "solution": "**Answer:** sp3d2 and 1\n\nH = $${1 \\over 2}$$ (V + M - c + a)\n

$$ \\therefore $$ H = $${1 \\over 2}$$(8 + 4) = 6\n
\"JEE\n

From structure, it is clear that it has five bond pairs and one\nlone pair.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 390, "subject": "Chemistry", "question": "The ion that has sp3d2 hybridization for the\ncentral atom, is :", "options": [ { "text": "[ICl2]-" }, { "text": "[ICl4]-" }, { "text": "[IF6]-" }, { "text": "[BrF2]-" } ], "answer": "[ICl4]-", "solution": "**Answer:** [ICl4]-\n\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 391, "subject": "Chemistry", "question": "The correct statement about ICl5 and $$ICl_4^-$$ is", "options": [ { "text": "ICl5 is square pyramidal and $$ICl_4^-$$ is square\nplanar." }, { "text": "ICl5 is trigonal bipyramidal and is\ntetrahedral." }, { "text": "ICl5 is square pyramidal and $$ICl_4^-$$ is\ntetrahedral" }, { "text": "Both are isostructural." } ], "answer": "ICl5 is square pyramidal and $$ICl_4^-$$ is square\nplanar.", "solution": "**Answer:** ICl5 is square pyramidal and $$ICl_4^-$$ is square\nplanar.\n\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 392, "subject": "Chemistry", "question": "The oxoacid of sulphur that does not contain bond between sulphur atoms is ", "options": [ { "text": "H2S2O7" }, { "text": "H2S2O3" }, { "text": "H2S4O6" }, { "text": "H2S2O4" } ], "answer": "H2S2O7", "solution": "**Answer:** H2S2O7\n\n\"JEE\n

Here you can see, H2S2O7 does not have S – S linkage.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 393, "subject": "Chemistry", "question": "The correct statement among the following is :", "options": [ { "text": "(SiH3)3N is planar and more basic than (CH3)3N" }, { "text": "(SiH3)3N is pyramidal and more basic than (CH3)3N" }, { "text": "(SiH3)3N is pyramidal and less basic than (CH3)3N" }, { "text": "(SiH3)3N is planar and less basic than (CH3)3N" } ], "answer": "(SiH3)3N is planar and less basic than (CH3)3N", "solution": "**Answer:** (SiH3)3N is planar and less basic than (CH3)3N\n\n\"JEE\n\n\"JEE\n
In (CH3)3N, one lone pair and 3 sigma bond present. So it's shape is pyramidal.\n

But in (SiH3)3N, due to backbonding lone pair of N shift to vacant 3d orbital of Si. So now N have only 3 sigma bond and hybridization is sp2, that is why it has triangular planar geometry.\n

As in (SiH3)3N. nitrogen(N) don't have any lone pair to donate but in (CH3)3N, lone pair of nitrogen(N) can easily be donated. That is why (CH3)3N is more basic.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 394, "subject": "Chemistry", "question": "The compound that has the largest H–M–H bond angle (M = N, O, S, C) is :", "options": [ { "text": "CH4" }, { "text": "H2S" }, { "text": "NH3" }, { "text": "H2O" } ], "answer": "CH4", "solution": "**Answer:** CH4\n\nCH4 - Bond angle = 109o28'\n

H2S - Bond angle = 92o\n

NH3 - Bond angle = 107o\n

H2O - Bond angle = 104o5'", "topic": "Geometry", "subtopic": "Plane Geometry" }, { "id": 395, "subject": "Chemistry", "question": "The structure of PCl5\n in the solid state is :", "options": [ { "text": "square pyramidal" }, { "text": "tetrahedral [PCl4]+ and octahedral [PCl6]–" }, { "text": "square planar [PCl4]+ and octahedral [PCl6]–" }, { "text": "trigonal bipyramidal" } ], "answer": "tetrahedral [PCl4]+ and octahedral [PCl6]–", "solution": "**Answer:** tetrahedral [PCl4]+ and octahedral [PCl6]–\n\nIn solid state PCl5\n exist in Ionpair i.e. [PCl4]\n+ and [PCl6]\n–\n

[PCl4]+ is tetrahedral.\n
[PCl6]– is octahedral.", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 396, "subject": "Chemistry", "question": "The molecular geometry of SF6 is octahedral.\nWhat is the geometry of SF4 (including lone\npair(s) of electrons, if any)?", "options": [ { "text": "Tetrahedral" }, { "text": "Trigonal bipyramidal" }, { "text": "Square planar" }, { "text": "Pyramidal" } ], "answer": "Trigonal bipyramidal", "solution": "**Answer:** Trigonal bipyramidal\n\n\"JEE\n

Here 4$$\\sigma $$ bonds +1 lone pair\n

Electronic geometry is Trigonal bipyramidal. Moleculer Geometry\nor \nMoleculer Shape is see-saw.\n

Note : The difference between Electronic Geometry and Moleculer Geometry(Moleculer Shape) is that in Electronic Geometry we include lone pair but in Moleculer Geometry(Moleculer Shape) we do not include lone pair.", "topic": "Geometry", "subtopic": "Plane Geometry" }, { "id": 397, "subject": "Chemistry", "question": "If AB4 molecule is a polar molecule, a possible\ngeometry of AB4 is", "options": [ { "text": "Tetrahedral" }, { "text": "see-saw" }, { "text": "Square pyramidal" }, { "text": "Square planar" } ], "answer": "see-saw", "solution": "**Answer:** see-saw\n\n\n\n \n \n \n \n \n \n \n\n\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\n
CompoundElectronic GeometryMoleculer Geometry
or
Moleculer Shape		Lone PairPolarity
AB4TetrahedralTetrahedral0Nonpolar
AB4Trigonal Bipyramidalsee-saw1Polar
AB4Octahedral
(Square Bipyramidal)		Square Planar2Nonpolar
\n

Note : The difference between Electronic Geometry and Moleculer Geometry(Moleculer Shape) is that in Electronic Geometry we include lone pair but in Moleculer Geometry(Moleculer Shape) we do not include lone pair.\n

Here in this question it is not clear which geometry they are talking about, correct answer should be either Trigonal Bipyramidal or see-saw.", "topic": "Geometry", "subtopic": "Plane Geometry" }, { "id": 398, "subject": "Chemistry", "question": "The shape / structure of [XeF5]– and XeO3F2,\nrespectively, are", "options": [ { "text": "Pentagonal planar and trigonal bipyramidal" }, { "text": "Trigonal bipyramidal and pentagonal\nplanar" }, { "text": "Octahedral and square pyramidal" }, { "text": "Trigonal bipyramidal and trigonal\nbipyramidal" } ], "answer": "Pentagonal planar and trigonal bipyramidal", "solution": "**Answer:** Pentagonal planar and trigonal bipyramidal\n\n[XeF5]– $$ \\to $$ 5BP + 2LP $$ \\to $$ sp3d3 hybridisation $$ \\to $$ Pentagonal planar\n

XeO3F2\n$$ \\to $$ 5BP + 0LP $$ \\to $$ sp3d hybridisation $$ \\to $$ Trigonal bipyramidal", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 399, "subject": "Chemistry", "question": "The number of sp2 hybrid orbitals in a molecule\nof benzene is :", "options": [ { "text": "24" }, { "text": "12" }, { "text": "6" }, { "text": "18" } ], "answer": "18", "solution": "**Answer:** 18\n\nEach carbon atom is sp2 hybrid\n

$$ \\therefore $$ 3 sp2 hybrid orbitals are formed by each\ncarbon atom\n

Total sp2 orbitals = 6 × 3 = 18", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 400, "subject": "Chemistry", "question": "Which of the following are isostructural pairs ?

\nA. $$SO_4^{2 - }$$ and $$CrO_4^{2 - }$$

\nB. SiCl4, and TiCl4

\nC. NH3 and NO3-

\nD. BCl3 and BrCl3\n", "options": [ { "text": "B and C only" }, { "text": "C and D only" }, { "text": "A and B only" }, { "text": "A and C only" } ], "answer": "A and B only", "solution": "**Answer:** A and B only\n\n

Isostructural compounds are those compounds which have same structure as well as same hybridisation.\n

\n

Formula for find hybridisation : H = (Lone pair + Sigma bond\n + coordinate bond )

\n\"JEE\n\"JEE\nIf number of sigma bond ($$\\sigma $$), co-ordinate bond and lone pair are same for given pairs, they are isostructural.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 401, "subject": "Chemistry", "question": "The correct shape and $$I - I - I$$ bond angles respectively in $$I_3^ - $$ ion are :", "options": [ { "text": "Linear; 180$$^\\circ$$" }, { "text": "T-shaped; 180$$^\\circ$$ and 90$$^\\circ$$" }, { "text": "Trigonal planar; 120$$^\\circ$$" }, { "text": "Distorted trigonal planar; 135$$^\\circ$$ and 90$$^\\circ$$" } ], "answer": "Linear; 180$$^\\circ$$", "solution": "**Answer:** Linear; 180$$^\\circ$$\n\n

Hybridisation of central $$I$$ in $$I_3^ - $$ is sp3d with 3 lone pair and 2 bond pair.

\n

Shape Linear

\n

\"JEE

\n

Lone pair 3 lone pair

\n

Bond angle 180$$^\\circ$$ (for linear molecule)

\n

\"JEE

", "topic": "Geometry", "subtopic": "Plane Geometry" }, { "id": 402, "subject": "Chemistry", "question": "Which among the following species has unequal bond lengths?", "options": [ { "text": "XeF4" }, { "text": "BF$$_4^ - $$" }, { "text": "SiF4" }, { "text": "SF4" } ], "answer": "SF4", "solution": "**Answer:** SF4\n\n(a)   XeF4  is sp3d2 hybridised with 4 bond pairs and 1 lone pair and structure is square planar. Here all the bond lengths are equal.\n

\"JEE\n

(b) $$\\,\\,\\,\\,$$ BF$$_4^ - $$, 4 bond pair present so angle is 109o 28' and sp3 hybridised. So structure is regular tetrahedral. Here all the bond lengths are equal.\n

\"JEE\n\n

(c) SiF4\nis sp3\n hybridisation and regular tetrahedral geometry. Here all the bond lengths are equal.\n

\"JEE\n

(d)   SF4 is sp3d hybridised with 4 bond pairs and 1 lone pair and its expected trigonal\nbipyramidal geometry gets distorted due to presence of a lone\npair of electrons and it becomes distorted tetrahedral or\nsee-saw with the bond angles equal to < 120o and 179o instead of\nthe expected angles of 120o and 180o respectively. Here axial and equitorial both bonds are presents. And we know axial bonds are longer and weaker.\n

\"JEE\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 403, "subject": "Chemistry", "question": "A central atom in a molecule has two lone pairs of electrons and forms three single bonds. The shape of this molecule is :", "options": [ { "text": "trigonal pyramidal" }, { "text": "T-shaped" }, { "text": "see-saw" }, { "text": "planar triangular" } ], "answer": "T-shaped", "solution": "**Answer:** T-shaped\n\n\"JEE
\nsp3d hybridised
\nT-shaped", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 404, "subject": "Chemistry", "question": "Amongst the following, the linear species is :", "options": [ { "text": "O3" }, { "text": "Cl2O" }, { "text": "N$$_3^ - $$" }, { "text": "NO2" } ], "answer": "N$$_3^ - $$", "solution": "**Answer:** N$$_3^ - $$\n\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 405, "subject": "Chemistry", "question": "The number of species below that have two lone pairs of electrons in their central atom is _________. (Round off to the Nearest Integer).

SF4, BF$$_4^ - $$, ClF3, AsF3, PCl5, BrF5, XeF4, SF6", "options": [], "answer": "2", "solution": "**Answer:** 2\n\n\"JEE\n\"JEE\n\"JEE\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 406, "subject": "Chemistry", "question": "The number of lone pairs of electrons on the central I atom in I$$_3^ - $$ is ____________.", "options": [], "answer": "3", "solution": "**Answer:** 3\n\n

Shape of I3- is :

\n\"JEE\n

The number of lone pairs of electron on the central atom is 3.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 407, "subject": "Chemistry", "question": "The hybridisations of the atomic orbitals of nitrogen in NO$$_2^ - $$, NO$$_2^ + $$ and NH$$_4^ + $$ respectively are.", "options": [ { "text": "sp3, sp2 and sp" }, { "text": "sp, sp2 and sp3" }, { "text": "sp3, sp and sp2" }, { "text": "sp2, sp and sp3" } ], "answer": "sp2, sp and sp3", "solution": "**Answer:** sp2, sp and sp3\n\nHybridisation = Number of sigma bonds + number of lone pair of electrons + number of coordinate bond.

For $$NO_2^ - $$ hybridisation = 2 sigma NO bonds + 1 lone pair on N = 3; so $$NO_2^ - $$ is sp2-hybridised.

\"JEE
For $$NO_2^ + $$ hybridisation = 2 sigma NO bonds = 2; so $$NO_2^ + $$ sp-hybridised.

\"JEE
For $$NO_4^ + $$ Hybridisation = 4 sigma NH bonds = 4; so $$NO_4^ + $$ is sp3-hybridised.

\"JEE
Therefore, hybridisation of N in $$NO_2^ - $$, $$NO_2^ + $$, $$NO_4^ + $$ are sp2, sp, sp3 respectively.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 408, "subject": "Chemistry", "question": "Match List-I with List-II :

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
List-I
(Species)		List-II
(Hybrid Orbitals)
(a)$$S{F_4}$$(i)$$s{p^3}{d^2}$$
(b)$$I{F_5}$$(ii)$${d^2}s{p^3}$$
(c)$$NO_2^ + $$(iii)$$s{p^3}d$$
(d)$$NH_4^ + $$(iv)$$s{p^3}$$
(v)$$sp$$


Choose the correct answer from the options given below :", "options": [ { "text": "(a)-(i), (b)-(ii), (c)-(v) and (d)-(iii)" }, { "text": "(a)-(ii), (b)-(i), (c)-(iv) and (d)-(v)" }, { "text": "(a)-(iii), (b)-(i), (c)-(v) and (d)-(iv)" }, { "text": "(a)-(iv), (b)-(iii), (c)-(ii) and (d)-(v)" } ], "answer": "(a)-(iii), (b)-(i), (c)-(v) and (d)-(iv)", "solution": "**Answer:** (a)-(iii), (b)-(i), (c)-(v) and (d)-(iv)\n\n(a) SF4 - sp3d hybridization

(b) IF5 - sp3d2 hybridization

(c) NO$$_2^ + $$ - sp hybridization

(d) NH$$_4^ + $$ - sp3 hybridization", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 409, "subject": "Chemistry", "question": "AB3 is an interhalogen T-shaped molecule. The number of lone pairs of electrons on A is __________. (Integer answer)", "options": [], "answer": "2", "solution": "**Answer:** 2\n\nT-shaped molecule means 3 sigma bond and 2 lone pairs of electron on central atom.

\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 410, "subject": "Chemistry", "question": "The number of species having non-pyramidal shape among the following is ___________.

(A) SO3

(B) NO$$_3^ - $$

(C) PCl3

(D) CO$$_3^{2 - }$$", "options": [], "answer": "3", "solution": "**Answer:** 3\n\n\"JEE
Hence, non-pyramidal species are SO3, NO$$_3^{- }$$ and CO$$_3^{2 - }$$.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 411, "subject": "Chemistry", "question": "

Consider the species CH4, NH$$_4^ + $$ and BH$$_4^ - $$. Choose the correct option with respect to the these species.

", "options": [ { "text": "They are isoelectronic and only two have tetrahedral structures." }, { "text": "They are isoelectronic and all have tetrahedral structures." }, { "text": "Only two are isoelectronic and all have tetrahedral structures." }, { "text": "Only two are isoelectronic and only two have tetrahedral structures." } ], "answer": "They are isoelectronic and all have tetrahedral structures.", "solution": "**Answer:** They are isoelectronic and all have tetrahedral structures.\n\n\"JEE
All are tetrahedral and each have 10 electrons. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 412, "subject": "Chemistry", "question": "

Number of lone pair(s) of electrons on central atom and the shape BrF3 molecule respectively, are

", "options": [ { "text": "0, triangular planar." }, { "text": "1, pyramidal." }, { "text": "2, bent T-shape." }, { "text": "1, bent T-shape." } ], "answer": "2, bent T-shape.", "solution": "**Answer:** 2, bent T-shape.\n\n\"JEE
\nSteric no. = 5 (sp3d), lone pair = 2

\nBent T shape. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 413, "subject": "Chemistry", "question": "

In the structure of SF4, the lone pair of electrons on S is in.

", "options": [ { "text": "equatorial position and there are two lone pair - bond pair repulsions at 90$$^\\circ$$." }, { "text": "equatorial position and there are three lone pair - bond pair repulsions at 90$$^\\circ$$." }, { "text": "axial position and there are three lone pair - bond pair repulsion at 90$$^\\circ$$." }, { "text": "axial position and there are two lone pair - bond pair repulsion at 90$$^\\circ$$." } ], "answer": "equatorial position and there are two lone pair - bond pair repulsions at 90$$^\\circ$$.", "solution": "**Answer:** equatorial position and there are two lone pair - bond pair repulsions at 90$$^\\circ$$.\n\nSF4 → sp3d hybridisation.

\n\"JEE
\nThe lone pair of electrons on S is in an equatorial position and there are two lone pair-bond pair repulsions at 90°.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 414, "subject": "Chemistry", "question": "

The hybridization of P exhibited in PF5 is spxdy. The value of y is __________.

", "options": [], "answer": "1", "solution": "**Answer:** 1\n\n$\\mathrm{PF}_5 \\Rightarrow \\mathrm{sp}^3 \\mathrm{~d}$ hybridisation\n

(5 sigma bonds, zero lone pair on central atom)\n

Value of $y=1$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 415, "subject": "Chemistry", "question": "

Identify the incorrect statement for PCl5 from the following.

", "options": [ { "text": "In this molecule, orbitals of phosphorous are assumed to undergo sp3d hybridization." }, { "text": "The geometry of PCl5 is trigonal bipyramidal." }, { "text": "PCl5 has two axial bonds stronger than three equatorial bonds." }, { "text": "The three equatorial bonds of PCl5 lie in a plane." } ], "answer": "PCl5 has two axial bonds stronger than three equatorial bonds.", "solution": "**Answer:** PCl5 has two axial bonds stronger than three equatorial bonds.\n\nPCl5

\n\"JEE
\nAll three equatorial bonds in a plane

\nsp3d hybridization

\nTrigonal bipyramidal

\nAxial bonds are weaker than equatorial bonds.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 416, "subject": "Chemistry", "question": "

Based upon VSEPR theory, match the shape (geometry) of the molecules in List-I with the molecules in List-II and select the most appropriate option.

\n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
List - I
(Shape)		List - II
(Molecules)
(A)T-shaped(I)XeF$$_4$$
(B)Trigonal planar(II)SF$$_4$$
(C)Square planar(III)ClF$$_3$$
(D)See-saw(IV)BF$$_3$$

", "options": [ { "text": "(A) - (I), (B) - (II), (C) - (III), (D) - (IV)" }, { "text": "(A) - (III), (B) - (IV), (C) - (I), (D) - (II)" }, { "text": "(A) - (III), (B) - (IV), (C) - (I), (D) - (I)" }, { "text": "(A) - (IV), (B) - (III), (C) - (I), (D) - (II)" } ], "answer": "(A) - (III), (B) - (IV), (C) - (I), (D) - (II)", "solution": "**Answer:** (A) - (III), (B) - (IV), (C) - (I), (D) - (II)\n\n\"JEE", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 417, "subject": "Chemistry", "question": "

Amongst SF4, XeF4, CF4 and H2O, the number of species with two lone pairs of electrons is _____________.

", "options": [], "answer": "1", "solution": "**Answer:** 1\n\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 418, "subject": "Chemistry", "question": "

Number of electron deficient molecules among the following

\n

PH3, B2H6, CCl4, NH3, LiH and BCl3 is

", "options": [ { "text": "0" }, { "text": "1" }, { "text": "2" }, { "text": "3" } ], "answer": "2", "solution": "**Answer:** 2\n\nElectron deficient species have less than 8 electrons (or two electrons for $$\\mathrm{H}$$ ) in their valence (incomplete octet).

\n$$\\mathrm{B}_{2} \\mathrm{H}_{6}, \\mathrm{BCl}_{3}$$ have incomplete octet.\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 419, "subject": "Chemistry", "question": "

Match List I with List II:

\n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
List I
(molecule)		List II
(hybridization ; shape)
(A)XeO$$_3$$(I)sp$$^3$$d ; linear
(B)XeF$$_2$$(II)sp$$^3$$ ; pyramidal
(C)XeOF$$_4$$(III)sp$$^3$$d$$^3$$ ; distorted octahedral
(D)XeF$$_6$$(IV)sp$$^3$$d$$^2$$ ; square pyramidal

\n

Choose the correct answer from the options given below:

", "options": [ { "text": "A-II, B-I, C-IV, D-III" }, { "text": "A-II, B-IV, C-III, D-I" }, { "text": "A-IV, B-II, C-III, D-I" }, { "text": "A-IV, B-II, C-I, D-III" } ], "answer": "A-II, B-I, C-IV, D-III", "solution": "**Answer:** A-II, B-I, C-IV, D-III\n\n$$\\mathrm{XeO}_{3}-s p^{3}$$, Pyramidal\n

\n$$\\mathrm{XeF}_{2}-s p^{3} d$$, linear\n

\n$$\\mathrm{XeOF}_{4}-s p^{3} d^{2}$$, Square Pyramidal\n

\n$$\\mathrm{XeF}_{6}-s p^{3} d^{3}$$, distorted octahedral", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 420, "subject": "Chemistry", "question": "

The sum of number of lone pairs of electrons present on the central atoms of XeO3, XeOF4 and XeF6, is ______________

", "options": [], "answer": "3", "solution": "**Answer:** 3\n\n\"JEE\n\n

\"JEE\n

From structure, it is clear that it has five bond pairs and one\nlone pair.\n\n

\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 421, "subject": "Chemistry", "question": "

Match List-I with List-II :

\n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
List I
(Compound)		List II
(Shape)
(A)BrF$$_5$$(I)bent
(B)[CrF$$_6$$]$$^{3 - }$$(II)square pyramidal
(C)O$$_3$$(III)trigonal bipyramidal
(D)PCl$$_5$$(IV)octahedral

\n

Choose the correct answer from the options given below :

", "options": [ { "text": "$$(\\mathrm{A})-(\\mathrm{I}),(\\mathrm{B})-(\\mathrm{II}),(\\mathrm{C})-(\\mathrm{III}),(\\mathrm{D})-(\\mathrm{IV})$$" }, { "text": "$$(\\mathrm{A})-(\\mathrm{IV}),(\\mathrm{B})-(\\mathrm{III}),(\\mathrm{C})-(\\mathrm{II}),(\\mathrm{D})-(\\mathrm{I})$$" }, { "text": "$$(\\mathrm{A})-(\\mathrm{II}),(\\mathrm{B})-(\\mathrm{IV}),(\\mathrm{C})-(\\mathrm{I}),(\\mathrm{D})-(\\mathrm{III})$$" }, { "text": "$$(\\mathrm{A})-(\\mathrm{III}),(\\mathrm{B})-(\\mathrm{IV}),(\\mathrm{C})-(\\mathrm{II}),(\\mathrm{D})-(\\mathrm{I})$$" } ], "answer": "$$(\\mathrm{A})-(\\mathrm{II}),(\\mathrm{B})-(\\mathrm{IV}),(\\mathrm{C})-(\\mathrm{I}),(\\mathrm{D})-(\\mathrm{III})$$", "solution": "**Answer:** $$(\\mathrm{A})-(\\mathrm{II}),(\\mathrm{B})-(\\mathrm{IV}),(\\mathrm{C})-(\\mathrm{I}),(\\mathrm{D})-(\\mathrm{III})$$\n\n(A) $$\\mathrm{BrF}_{5}-$$ square pyramidal

\n(B) $$\\left[\\mathrm{CrF}_{6}\\right]^{3-}-$$ octahedral

\n(C) $$\\mathrm{O}_{3}-$$ bent

\n(D) $$\\mathrm{PCl}_{5}$$ - trigonal bipyramidal", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 422, "subject": "Chemistry", "question": "

The number of molecule(s) or ion(s) from the following having non-planar structure is ____________.

\n

$$\\mathrm{NO}_{3}^{-}, \\mathrm{H}_{2} \\mathrm{O}_{2}, \\mathrm{BF}_{3}, \\mathrm{PCl}_{3}, \\mathrm{XeF}_{4}, \\mathrm{SF}_{4}, \\mathrm{XeO}_{3}, \\mathrm{PH}_{4}^{+}, \\mathrm{SO}_{3},\\left[\\mathrm{Al}(\\mathrm{OH})_{4}\\right]^{-}$$

", "options": [], "answer": "6", "solution": "**Answer:** 6\n\n$\\mathrm{NO}_{3}^{\\ominus} \\rightarrow$ Trigonal planar (Planar)\n

\n$\\mathrm{H}_{2} \\mathrm{O}_{2} \\rightarrow$ Open book (Non-planar)\n

\n$\\mathrm{BF}_{3} \\rightarrow$ Trigonal planar (Planar)\n

\n$\\mathrm{PCl}_{3} \\rightarrow$ Pyramidal (Non-planar)\n

\n$\\mathrm{XeF}_{4} \\rightarrow$ Square planar (Planar)\n

\n$\\mathrm{SF}_{4} \\rightarrow$ See-Saw (Non-planar) $\\mathrm{XeO}_{3} \\rightarrow$ Pyramidal (Non-planar)\n

\n$\\mathrm{PH}_{4}^{\\oplus} \\rightarrow$ Tetrahedral (Non-planar)\n

\n$\\mathrm{SO}_{3} \\rightarrow$ Trigonal planar (Planar)\n

\n$\\left[\\mathrm{Al}(\\mathrm{OH})_{4}\\right]^{-} \\rightarrow$ Tetrahedral (Non-planar)", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 423, "subject": "Chemistry", "question": "

The number of interhalogens from the following having square pyramidal structure is :

\n

$$\\mathrm{ClF}_{3}, \\mathrm{IF}_{7}, \\mathrm{BrF}_{5}, \\mathrm{BrF}_{3}, \\mathrm{I}_{2} \\mathrm{Cl}_{6}, \\mathrm{IF}_{5}, \\mathrm{ClF}, \\mathrm{ClF}_{5}$$

", "options": [], "answer": "3", "solution": "**Answer:** 3\n\n$\\mathrm{CIF}_{3} \\rightarrow 3 \\sigma$ bond $+2$ lone pair\n

\n$\\mathrm{IF}_{7} \\rightarrow 7 \\sigma$ bond $+0$ lone pair\n

\n$\\mathrm{BrF}_{5} \\rightarrow 5 \\sigma$ bond $+1$ lone pair $\\rightarrow$ Square pyramidal\n

\n$\\mathrm{BrF}_{3} \\rightarrow 3 \\sigma$ bond $+2$ lone pair\n

\n$\\mathrm{I}_{2} \\mathrm{Cl}_{6} \\rightarrow 4 \\sigma$ bond $+2$ lone pair\n

\n$\\mathrm{IF}_{5} \\rightarrow 5 \\sigma$ bond $+1$ lone pair $\\rightarrow$ Square pyramidal\n

\n$\\mathrm{CIF} \\rightarrow 1 \\sigma$ bond $+3$ lone pair\n

\n$\\mathrm{CIF}_{5} \\rightarrow 5 \\sigma$ bond $+1$ lone pair $\\rightarrow$ Square pyramidal", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 424, "subject": "Chemistry", "question": "

Which of the following pair of molecules contain odd electron molecule and an expanded octet molecule?

", "options": [ { "text": "$$\\mathrm{BCl}_{3}$$ and $$\\mathrm{SF}_{6}$$" }, { "text": "$$\\mathrm{NO}$$ and $$\\mathrm{H}_{2} \\mathrm{SO}_{4}$$" }, { "text": "$$\\mathrm{SF}_{6}$$ and $$\\mathrm{H}_{2} \\mathrm{SO}_{4}$$" }, { "text": "$$\\mathrm{BCl}_{3}$$ and $$\\mathrm{NO}$$" } ], "answer": "$$\\mathrm{NO}$$ and $$\\mathrm{H}_{2} \\mathrm{SO}_{4}$$", "solution": "**Answer:** $$\\mathrm{NO}$$ and $$\\mathrm{H}_{2} \\mathrm{SO}_{4}$$\n\n(A) $\\mathrm{BCl}_{3}$ - Even electron molecules\n
       $\\mathrm{SF}_{6}$ - Expended octet molecules\n

\n(B) $\\mathrm{NO}$ - Odd electron molecules
\n      $\\mathrm{H}_{2} \\mathrm{SO}_{4}$ - Expanded octet

\n(C) $\\mathrm{SF}_{6}-$ Even electron molecules
   \n  $\\mathrm{H}_{2} \\mathrm{SO}_{4}$ - Expanded octet\n

\n(D) $\\mathrm{BCl}_{3}-$ Even electron molecules\n
      $\\mathrm{NO}$ - odd electron molecules", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 425, "subject": "Chemistry", "question": "

Number of lone pairs of electrons in the central atom of $$\\mathrm{SCl}_{2}, \\mathrm{O}_{3}, \\mathrm{ClF}_{3}$$ and $$\\mathrm{SF}_{6}$$, respectively, are :

", "options": [ { "text": "0, 1, 2 and 2" }, { "text": "2, 1, 2 and 0" }, { "text": "1, 2, 2 and 0" }, { "text": "2, 1, 0 and 2" } ], "answer": "2, 1, 2 and 0", "solution": "**Answer:** 2, 1, 2 and 0\n\nThe number of lone pair of electrons in the central atom of $\\mathrm{SCl}_{2}, \\mathrm{O}_{3}, \\mathrm{ClF}_{3}$ and $\\mathrm{SF}_{6}$ are 2, 1, 2 and $\\mathrm{0}$ respectively\n

\nTheir structures are as,

\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 426, "subject": "Chemistry", "question": "

Consider, $$\\mathrm{PF}_{5}, \\mathrm{BrF}_{5}, \\mathrm{PCl}_{3}, \\mathrm{SF}_{6},\\left[\\mathrm{ICl}_{4}\\right]^{-}, \\mathrm{ClF}_{3}$$ and $$\\mathrm{IF}_{5}$$.

\n

Amongst the above molecule(s)/ion(s), the number of molecule(s)/ion(s) having $$\\mathrm{sp}^{3}\\mathrm{~d}^{2}$$ hybridisation is __________.

", "options": [], "answer": "4", "solution": "**Answer:** 4\n\n\"JEE\n\"JEE\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 427, "subject": "Chemistry", "question": "Amongst the following, the number of species having the linear shape is _________.\n

\n$\\mathrm{XeF}_{2}, \\mathrm{I}_{3}^{+}, \\mathrm{C}_{3} \\mathrm{O}_{2}, \\mathrm{I}_{3}^{-}, \\mathrm{CO}_{2}, \\mathrm{SO}_{2}, \\mathrm{BeCl}_{2}$ and $\\mathrm{BCl}_{2}^{\\ominus}$", "options": [], "answer": "5", "solution": "**Answer:** 5\n\n\"JEE\n

\"JEE\n

\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 428, "subject": "Chemistry", "question": "

Match List I with List II

\n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
List IList II
A.$$\\mathrm{XeF_4}$$I.See-saw
B.$$\\mathrm{SF_4}$$II.Square-planar
C.$$\\mathrm{NH_{4}^{+}}$$III.Bent T-shaped
D.$$\\mathrm{BrF_3}$$IV.Tetrahedral

\n

Choose the correct answer from the options given below :

", "options": [ { "text": "A - II, B - I, C - III, D - IV" }, { "text": "A - II, B - I, C - IV, D - III" }, { "text": "A - IV, B - I, C - II, D - III" }, { "text": "A - IV, B - III, C - II, D - I" } ], "answer": "A - II, B - I, C - IV, D - III", "solution": "**Answer:** A - II, B - I, C - IV, D - III\n\n\"JEE\n
\"JEE\n
\"JEE\n
\"JEE", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 429, "subject": "Chemistry", "question": "

Match List I with List II

\n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
List I
(molecules/ions)		List II
(No. of lone pairs of e$$^-$$ on central atom)
A.$$\\mathrm{IF_7}$$I.Three
B.$$\\mathrm{ICl}$$$$_4^ - $$II.One
C.$$\\mathrm{XeF_6}$$III.Two
D.$$\\mathrm{XeF_2}$$IV.Zero

\n

Choose the correct answer from the options given below :

", "options": [ { "text": "A - II, B - I, C - IV, D - III" }, { "text": "A - IV, B - I, C - II, D - III" }, { "text": "A - IV, B - III, C - II, D - I" }, { "text": "A - II, B - III, C - IV, D - I" } ], "answer": "A - IV, B - III, C - II, D - I", "solution": "**Answer:** A - IV, B - III, C - II, D - I\n\n

(A) IF$$_7$$ $$-$$ 0 lone pairs

\n

(B) ICI$$_4^ - $$ $$-$$ 2 lone pairs

\n

(C) XeF$$_6$$ $$-$$ 1 lone pair

\n

(D) XeF$$_2$$ $$-$$ 3 lone pairs

\n

\"JEE

", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 430, "subject": "Chemistry", "question": "

The total number of lone pairs of electrons on oxygen atoms of ozone is __________.

", "options": [], "answer": "6", "solution": "**Answer:** 6\n\n

Total no, of lone pairs on oxygen atoms = 6

\n

\"JEE

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 431, "subject": "Chemistry", "question": "The number of $\\mathrm{P}-\\mathrm{O}-\\mathrm{P}$ bonds in $\\mathrm{H}_{4} \\mathrm{P}_{2} \\mathrm{O}_{7},\\left(\\mathrm{HPO}_{3}\\right)_{3}$ and $\\mathrm{P}_{4} \\mathrm{O}_{10}$ are respectively :", "options": [ { "text": "$1,3,6$" }, { "text": "$1,2,4$" }, { "text": "$0,3,4$" }, { "text": "$0,3,6$" } ], "answer": "$1,3,6$", "solution": "**Answer:** $1,3,6$\n\nLet's analyze each compound separately:\n

\n1. $\\mathrm{H}_{4} \\mathrm{P}_{2} \\mathrm{O}_{7}$: This compound is also known as pyrophosphoric acid. Its structure has two phosphorus atoms (P) connected by one oxygen atom (O), forming a P-O-P bond. There are no other P-O-P bonds in its structure.\nNumber of P-O-P bonds = 1\n

\"JEE\n

\n2. $\\left(\\mathrm{HPO}_{3}\\right)_{3}$: This compound is also known as cyclotriphosphoric acid. It has three phosphorus atoms and three oxygen atoms, arranged in a cyclic structure with each P atom connected to two O atoms, forming a six-membered ring. In this structure, there are three P-O-P bonds.\nNumber of P-O-P bonds = 3\n

\"JEE\n

\n3. $\\mathrm{P}_{4} \\mathrm{O}_{10}$: This compound is also known as phosphorus pentoxide. Its structure consists of four phosphorus atoms (P) and ten oxygen atoms (O) connected in a cyclic arrangement. In the structure, each phosphorus atom is connected to four oxygen atoms, with two terminal oxygen atoms and two bridging oxygen atoms. There are six P-O-P bonds in the P4O10 structure.\n

\"JEE\n

\nSo, the number of P-O-P bonds in $\\mathrm{H}_{4} \\mathrm{P}_{2} \\mathrm{O}_{7},\\left(\\mathrm{HPO}_{3}\\right)_{3}$, and $\\mathrm{P}_{4} \\mathrm{O}_{10}$ are 1, 3, and 6, respectively.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 432, "subject": "Chemistry", "question": "$$\\mathrm{ClF}_{5}$$ at room temperature is a:", "options": [ { "text": "Colourless liquid with square pyramidal geometry" }, { "text": "Colourless gas with square pyramidal geometry" }, { "text": "Colourless gas with trigonal bipyramidal geometry." }, { "text": "Colourless liquid with trigonal bipyramidal geometry" } ], "answer": "Colourless liquid with square pyramidal geometry", "solution": "**Answer:** Colourless liquid with square pyramidal geometry\n\n\"JEE
$\\mathrm{ClF}_5$ is actually a colorless liquid at room temperature with a square pyramidal geometry. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 433, "subject": "Chemistry", "question": "

The maximum number of lone pairs of electrons on the central atom from the following species is ____________.

\n

$$\\mathrm{ClO}_{3}{ }^{-}, \\mathrm{XeF}_{4}, \\mathrm{SF}_{4}$$ and $$\\mathrm{I}_{3}{ }^{-}$$

", "options": [], "answer": "3", "solution": "**Answer:** 3\n\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 434, "subject": "Chemistry", "question": "

Match List - I with List - II:

\n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
List - I SpeciesList - II Geometry/Shape
A.$$\\mathrm{H_3O^+}$$I.Tetrahedral
B.Acetylide anionII.Linear
C.$$\\mathrm{NH_4^+}$$III.Pyramidal
D.$$\\mathrm{ClO_2^-}$$IV.Bent

\n

Choose the correct answer from the options given below:

", "options": [ { "text": "A-III, B-I, C-II, D-IV" }, { "text": "A-III, B-II, C-I, D-IV" }, { "text": "A-III, B-IV, C-I, D-II" }, { "text": "A-III, B-IV, C-II, D-I" } ], "answer": "A-III, B-II, C-I, D-IV", "solution": "**Answer:** A-III, B-II, C-I, D-IV\n\n

Let's consider each species in List I and determine their geometries/shapes.

\n

A. $\\mathrm{H_3O^+}$: This ion is formed by the addition of a proton to a water molecule. The central atom (O) is surrounded by three Hydrogen atoms and one lone pair, making its shape pyramidal.

\n

B. Acetylide anion: The acetylide ion ($\\mathrm{C_2H^-}$ or $\\mathrm{C_2^{2-}}$) has a linear geometry as the three atoms are in a straight line.

\n

C. $\\mathrm{NH_4^+}$: The ammonium ion has a central nitrogen atom surrounded by four hydrogen atoms, giving it a tetrahedral geometry.

\n

D. $\\mathrm{ClO_2^-}$: The chlorite ion has a central chlorine atom surrounded by two oxygen atoms and one lone pair, which gives it a bent or V-shaped geometry.

\n

Matching these to List II, we get:

\n

A. $\\mathrm{H_3O^+}$ - III. Pyramidal

\nB. Acetylide anion - II. Linear

\nC. $\\mathrm{NH_4^+}$ - I. Tetrahedral

\nD. $\\mathrm{ClO_2^-}$ - IV. Bent

", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 435, "subject": "Chemistry", "question": "

The number of molecules from the following which contain only two lone pair of electrons is ________

\n

$$\\mathrm{H}_{2} \\mathrm{O}, \\mathrm{N}_{2}, \\mathrm{CO}, \\mathrm{XeF}_{4}, \\mathrm{NH}_{3}, \\mathrm{NO}, \\mathrm{CO}_{2}, \\mathrm{~F}_{2}$$

", "options": [], "answer": "3", "solution": "**Answer:** 3\n\n\"JEE\n\"JEE\n\"JEE
\nThe number of molecules having only 2 lone pair of electrons $=3$

\nWhich are $\\mathrm{H}_2 \\mathrm{O} , \\mathrm{N}_2$ and $\\mathrm{CO}$

\nNote:-

\n$\\mathrm{XeF}_4$ have 2 lps on central atom, but we are asked about lone pair in molecule", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 436, "subject": "Chemistry", "question": "

The compound which does not exist is

", "options": [ { "text": "(NH$$_4$$)$$_2$$BeF$$_4$$" }, { "text": "PbEt$$_4$$" }, { "text": "BeH$$_2$$" }, { "text": "NaO$$_2$$" } ], "answer": "NaO$$_2$$", "solution": "**Answer:** NaO$$_2$$\n\nSodium superoxide (NaO$_2$) is not a stable compound. It is not found in normal conditions due to its high reactivity.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 437, "subject": "Chemistry", "question": "

The number of bent-shaped molecule/s from the following is __________

\n

N$$_3^-$$, NO$$_2^-$$, I$$_3^-$$, O$$_3$$, SO$$_2$$

", "options": [], "answer": "3", "solution": "**Answer:** 3\n\n

A bent-shaped molecule has a molecular geometry with a central atom bonded to two other atoms and one or two pairs of non-bonding electrons. The VSEPR (Valence Shell Electron Pair Repulsion) theory helps us predict the shapes of molecules.

\n

Let's consider the given molecules:

\n
    \n

  1. N$_3^-$: The azide ion has a linear shape, not bent. The nitrogen in the middle is connected to two other nitrogens and there are no lone pairs on the central atom.

    \n

    2. \n

  2. NO$_2^-$: The nitrite ion has a bent shape. The central nitrogen atom is connected to two oxygen atoms and has one lone pair of electrons, giving it a bent geometry.

    \n

    3. \n

  3. I$_3^-$: The triiodide ion has a linear shape, not bent. The central iodine atom is connected to two other iodine atoms and there are no lone pairs on the central atom.

    \n

    4. \n

  4. O$_3$: Ozone has a bent shape. The central oxygen atom is connected to two other oxygen atoms and has one lone pair of electrons, giving it a bent geometry.

    \n

    5. \n

  5. SO$_2$: Sulfur dioxide has a bent shape. The central sulfur atom is connected to two oxygen atoms and has one lone pair of electrons, giving it a bent geometry.

    \n
    6. \n
\n

Therefore, there are 3 bent-shaped molecules: NO$_2^-$, O$_3$, and SO$_2$.

\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 438, "subject": "Chemistry", "question": "

The sum of lone pairs present on the central atom of the interhalogen IF$$_5$$ and IF$$_7$$ is _________

", "options": [], "answer": "1", "solution": "**Answer:** 1\n\n

Interhalogen compounds are the substances that consist of two different halogens. The most common type of interhalogen compounds are binary, containing only two different elements.

\n

In IF$_5$, iodine (I) is the central atom. Iodine has 7 valence electrons, 5 of which are used for bonding with the 5 fluorine (F) atoms. This leaves 2 electrons, or 1 lone pair on the iodine atom.

\n

In IF$_7$, iodine (I) is the central atom again. Iodine has 7 valence electrons, all of which are used for bonding with the 7 fluorine (F) atoms. So, there are no lone pairs on the iodine atom.

\n

Therefore, the sum of lone pairs present on the central atom of the interhalogens IF$_5$ and IF$_7$ is 1.

\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 439, "subject": "Chemistry", "question": "

The number of species from the following carrying a single lone pair on central atom Xenon is ___________.

\n

$$\\mathrm{XeF}_{5}^{+}, \\mathrm{XeO}_{3}, \\mathrm{XeO}_{2} \\mathrm{~F}_{2}, \\mathrm{XeF}_{5}^{-}, \\mathrm{XeO}_{3} \\mathrm{~F}_{2}, \\mathrm{XeOF}_{4}, \\mathrm{XeF}_{4}$$

", "options": [], "answer": "4", "solution": "**Answer:** 4\n\n\"JEE\"JEE\n
So, Answer is 4", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 440, "subject": "Chemistry", "question": "

Match List I with List II

\n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
LIST I
Oxide		LIST II
Type of bond
A.$$\\mathrm{N_2O_4}$$I.1 N = O bond
B.$$\\mathrm{NO_2}$$II.1 N $$-$$ O $$-$$ N bond
C.$$\\mathrm{N_2O_5}$$III.1 N $$-$$ N bond
D.$$\\mathrm{N_2O}$$IV.1 N=N / N $$\\equiv$$ N bond

\n

Choose the correct answer from the options given below:

", "options": [ { "text": "A-II, B-I, C-III, D-IV" }, { "text": "A-III, B-I, C-II, D-IV" }, { "text": "A-III, B-I, C-IV, D-II" }, { "text": "A-II, B-IV, C-III, D-I" } ], "answer": "A-III, B-I, C-II, D-IV", "solution": "**Answer:** A-III, B-I, C-II, D-IV\n\n\"JEE", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 441, "subject": "Chemistry", "question": "

The number of species from the following which have square pyramidal structure is _________

\n

$$\\mathrm{PF}_{5}, \\mathrm{BrF}_{4}^{-}, \\mathrm{IF}_{5}, \\mathrm{BrF}_{5}, \\mathrm{XeOF}_{4}, \\mathrm{ICl}_{4}^{-}$$

", "options": [], "answer": "3", "solution": "**Answer:** 3\n\n

A square pyramidal structure has five bonds and one lone pair, making a total of six electron pairs around the central atom. The geometry of such a molecule can be analyzed using the VSEPR theory.

\n
    \n
  1. $\\mathrm{PF}_{5}$: 5 bond pairs and 0 lone pairs, so its geometry is trigonal bipyramidal, not square pyramidal.
    2. \n
  2. $\\mathrm{BrF}_{4}^{-}$: 4 bond pairs and 2 lone pairs, so its geometry is square planar, not square pyramidal.

    3. \n
  3. $\\mathrm{IF}_{5}$: 5 bond pairs and 1 lone pair, so its geometry is square pyramidal.

    4. \n
  4. $\\mathrm{BrF}_{5}$: 5 bond pairs and 1 lone pair, so its geometry is square pyramidal.
    5. \n
  5. $\\mathrm{XeOF}_{4}$: 5 bond pairs and 1 lone pair, so its geometry is square pyramidal.

    6. \n
  6. $\\mathrm{ICl}_{4}^{-}$: 4 bond pairs and 2 lone pairs, so its geometry is square planar, not square pyramidal.
    7. \n
\n

So the number of species from the given list that have a square pyramidal structure is 3.

\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 442, "subject": "Chemistry", "question": "

The number of species having a square planar shape from the following is __________.

\n

$$\\mathrm{XeF}_{4}, \\mathrm{SF}_{4}, \\mathrm{SiF}_{4}, \\mathrm{BF}_{4}^{-}, \\mathrm{BrF}_{4}^{-},\\left[\\mathrm{Cu}\\left(\\mathrm{NH}_{3}\\right)_{4}\\right]^{2+},\\left[\\mathrm{FeCl}_{4}\\right]^{2-},\\left[\\mathrm{PtCl}_{4}\\right]^{2-}$$

", "options": [], "answer": "4", "solution": "**Answer:** 4\n\n$\\mathrm{XeF}_4 \\rightarrow$ Square planar

\n$\\mathrm{SF}_4 \\rightarrow$ See saw

\n$\\mathrm{SiF}_4 \\rightarrow$ Tetrahedral

\n$\\mathrm{BF}_4^{-} \\rightarrow$ Tetrahedral

\n$\\left[\\mathrm{Cu}\\left(\\mathrm{NH}_3\\right)_4\\right]^{2+} \\rightarrow$ Square planar

\n$\\left[\\mathrm{FeCl}_4\\right]^{2-} \\rightarrow$ Tetrahedral

\n$\\left[\\mathrm{PtCl}_4\\right]^{2-} \\rightarrow$ Square planar

\n$\\mathrm{BrF}_4^{-} \\rightarrow$ Square planar

\nSo, 4 square planer shape compounds are present.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 443, "subject": "Chemistry", "question": "The number of molecules/ion/s having trigonal bipyramidal shape is _______.\n

$\\mathrm{PF}_5, \\mathrm{BrF}_5, \\mathrm{PCl}_5,\\left[\\mathrm{Pt} \\mathrm{Cl}_4\\right]^{2-}, \\mathrm{BF}_3, \\mathrm{Fe}(\\mathrm{CO})_5$", "options": [], "answer": "3", "solution": "**Answer:** 3\n\n\"JEE\n\"JEE\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 444, "subject": "Chemistry", "question": "

The number of non-polar molecules from the following is _________. $$\\mathrm{HF}, \\mathrm{H}_2 \\mathrm{O}, \\mathrm{SO}_2, \\mathrm{H}_2, \\mathrm{CO}_2, \\mathrm{CH}_4, \\mathrm{NH}_3, \\mathrm{HCl}, \\mathrm{CHCl}_3, \\mathrm{BF}_3$$

", "options": [], "answer": "4", "solution": "**Answer:** 4\n\n

To determine whether a molecule is polar or non-polar, we must consider the difference in electronegativity between the atoms and the symmetry of the molecule. Polar molecules occur when there is an electronegativity difference between the bonded atoms. Non-polar molecules either do not have any polar bonds or the polarities cancel each other out because of a symmetrical arrangement. Let's consider each molecule individually:

\n\n\n\n

Given this information, the non-polar molecules from the list are: $$\\mathrm{H}_2, \\mathrm{CO}_2, \\mathrm{CH}_4, \\mathrm{BF}_3$$.

\n\n

Therefore, there are 4 non-polar molecules in the list given.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 445, "subject": "Chemistry", "question": "

The number of species from the following in which the central atom uses $$\\mathrm{sp}^3$$ hybrid orbitals in its bonding is __________.

\n

$$\\mathrm{NH}_3, \\mathrm{SO}_2, \\mathrm{SiO}_2, \\mathrm{BeCl}_2, \\mathrm{CO}_2, \\mathrm{H}_2 \\mathrm{O}, \\mathrm{CH}_4, \\mathrm{BF}_3$$

", "options": [], "answer": "3", "solution": "**Answer:** 3\n\n

To determine the number of species in which the central atom uses $$\\mathrm{sp}^3$$ hybrid orbitals in its bonding, we need to analyze the hybridization of each central atom in the given species.

\n\n

1. $$\\mathrm{NH}_3$$ (Ammonia):

\n\n

The central atom is nitrogen. Ammonia has 3 sigma bonds and 1 lone pair. Thus, the steric number is 4, which corresponds to $$\\mathrm{sp}^3$$ hybridization.

\n\n

2. $$\\mathrm{SO}_2$$ (Sulfur dioxide):

\n\n

The central atom is sulfur. Sulfur dioxide has 2 sigma bonds and 1 lone pair. Thus, the steric number is 3, which corresponds to $$\\mathrm{sp}^2$$ hybridization.

\n\n

3. $$\\mathrm{SiO}_2$$ (Silicon dioxide):

\n\n

The central atom is silicon. Silicon dioxide has a linear structure with double bonds, leading to $$\\mathrm{sp}$$ hybridization.

\n\n

4. $$\\mathrm{BeCl}_2$$ (Beryllium chloride):

\n\n

The central atom is beryllium. Beryllium chloride has 2 sigma bonds and no lone pair. Thus, the steric number is 2, which corresponds to $$\\mathrm{sp}$$ hybridization.

\n\n

5. $$\\mathrm{CO}_2$$ (Carbon dioxide):

\n\n

The central atom is carbon. Carbon dioxide has a linear structure with double bonds, leading to $$\\mathrm{sp}$$ hybridization.

\n\n

6. $$\\mathrm{H}_2\\mathrm{O}$$ (Water):

\n\n

The central atom is oxygen. Water has 2 sigma bonds and 2 lone pairs. Thus, the steric number is 4, which corresponds to $$\\mathrm{sp}^3$$ hybridization.

\n\n

7. $$\\mathrm{CH}_4$$ (Methane):

\n\n

The central atom is carbon. Methane has 4 sigma bonds and no lone pair. Thus, the steric number is 4, which corresponds to $$\\mathrm{sp}^3$$ hybridization.

\n\n

8. $$\\mathrm{BF}_3$$ (Boron trifluoride):

\n\n

The central atom is boron. Boron trifluoride has 3 sigma bonds and no lone pair. Thus, the steric number is 3, which corresponds to $$\\mathrm{sp}^2$$ hybridization.

\n\n

From the above analysis, the species in which the central atom uses $$\\mathrm{sp}^3$$ hybrid orbitals are:

\n\n\n\n

Therefore, the number of species where the central atom uses $$\\mathrm{sp}^3$$ hybrid orbitals is 3.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 446, "subject": "Chemistry", "question": "

Number of compounds with one lone pair of electrons on central atom amongst following is _________.

\n

$$\\mathrm{O}_3, \\mathrm{H}_2 \\mathrm{O}, \\mathrm{SF}_4, \\mathrm{ClF}_3, \\mathrm{NH}_3, \\mathrm{BrF}_5, \\mathrm{XeF}_4$$

", "options": [], "answer": "4", "solution": "**Answer:** 4\n\n

\"JEE

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 447, "subject": "Chemistry", "question": "

The molecule / ion with square pyramidal shape is

", "options": [ { "text": "$$\\mathrm{PCl}_5$$\n" }, { "text": "$$\\left[\\mathrm{Ni}(\\mathrm{CN})_4\\right]^{2-}$$\n" }, { "text": "$$\\mathrm{PF}_5$$\n" }, { "text": "$$\\mathrm{BrF}_5$$" } ], "answer": "$$\\mathrm{BrF}_5$$", "solution": "**Answer:** $$\\mathrm{BrF}_5$$\n\n

$$\\mathrm{BrF_5}$$

\n

\"JEE

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 448, "subject": "Chemistry", "question": "

Aluminium chloride in acidified aqueous solution forms an ion having geometry

", "options": [ { "text": "Square planar\n" }, { "text": "Octahedral\n" }, { "text": "Trigonal bipyramidal\n" }, { "text": "Tetrahedral" } ], "answer": "Octahedral\n", "solution": "**Answer:** Octahedral\n\n\n

$$\\mathrm{AlCl}_3$$ in acidified aqueous solution forms octahedral geometry $$[\\mathrm{Al}(\\mathrm{H}_2 \\mathrm{O})_6]^{3+}$$

", "topic": "Geometry", "subtopic": "Plane Geometry" }, { "id": 449, "subject": "Chemistry", "question": "

Match List I with List II.

\n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
List I
Molecule		List II
Shape
(A)$$\\mathrm{BrF_5}$$(I)T-Shape
(B)$$\\mathrm{H_2O}$$(II)See saw
(C)$$\\mathrm{ClF_3}$$(III)Bent
(D)$$\\mathrm{SF_4}$$(IV)Square pyramidal

\n

Choose the correct answer from the options given below :

", "options": [ { "text": "(A)-(I), (B)-(II), (C)-(IV), (D)-(III)\n" }, { "text": "(A)-(IV), (B)-(III), (C)-(I), (D)-(II)\n" }, { "text": "(A)-(III), (B)-(IV), (C)-(I), (D)-(II)\n" }, { "text": "(A)-(II), (B)-(I), (C)-(III), (D)-(IV)" } ], "answer": "(A)-(IV), (B)-(III), (C)-(I), (D)-(II)\n", "solution": "**Answer:** (A)-(IV), (B)-(III), (C)-(I), (D)-(II)\n\n\n\"JEE\n\"JEE", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 450, "subject": "Chemistry", "question": "

The correct increasing order for bond angles among $$\\mathrm{BF}_3, \\mathrm{PF}_3$$ and $$\\mathrm{ClF}_3$$ is :

", "options": [ { "text": "$$\\mathrm{BF}_3=\\mathrm{PF}_3<\\mathrm{ClF}_3$$\n" }, { "text": "$$\\mathrm{BF}_3<\\mathrm{PF}_3<\\mathrm{ClF}_3$$\n" }, { "text": "$$\\mathrm{ClF}_3<\\mathrm{PF}_3<\\mathrm{BF}_3$$\n" }, { "text": "$$\\mathrm{PF}_3<\\mathrm{BF}_3<\\mathrm{ClF}_3$$" } ], "answer": "$$\\mathrm{ClF}_3<\\mathrm{PF}_3<\\mathrm{BF}_3$$\n", "solution": "**Answer:** $$\\mathrm{ClF}_3<\\mathrm{PF}_3<\\mathrm{BF}_3$$\n\n\n

\"JEE

\n

Order of bond angle is

\n

$$\\mathrm{ClF}_3<\\mathrm{PF}_3<\\mathrm{BF}_3$$

", "topic": "Geometry", "subtopic": "Plane Geometry" }, { "id": 451, "subject": "Chemistry", "question": "

In which one of the following pairs the central atoms exhibit $$\\mathrm{sp}^2$$ hybridization ?

", "options": [ { "text": "$$\\mathrm{BF}_3$$ and $$\\mathrm{NO}_2^{-}$$\n" }, { "text": "$$\\mathrm{NH}_2^{-}$$ and $$\\mathrm{BF}_3$$\n" }, { "text": "$$\\mathrm{NH}_2^{-}$$ and $$\\mathrm{H}_2 \\mathrm{O}$$\n" }, { "text": "$$\\mathrm{H}_2 \\mathrm{O}$$ and $$\\mathrm{NO}_2$$" } ], "answer": "$$\\mathrm{BF}_3$$ and $$\\mathrm{NO}_2^{-}$$\n", "solution": "**Answer:** $$\\mathrm{BF}_3$$ and $$\\mathrm{NO}_2^{-}$$\n\n\n

\"JEE

\n

have central atom with sp$$^2$$ hybridisation NH$$_2^-$$ and H$$_2$$O have sp$$^3$$ hybridised central atom.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 452, "subject": "Chemistry", "question": "

Number of molecules/ions from the following in which the central atom is involved in $$\\mathrm{sp}^3$$ hybridization is ________.

\n

$$\\mathrm{NO}_3^{-}, \\mathrm{BCl}_3, \\mathrm{ClO}_2^{-}, \\mathrm{ClO}_3^{-}$$

", "options": [ { "text": "2" }, { "text": "4" }, { "text": "1" }, { "text": "3" } ], "answer": "2", "solution": "**Answer:** 2\n\n

To determine the number of molecules/ions in which the central atom is involved in $$\\mathrm{sp}^3$$ hybridization, we must analyze the hybridization state for each central atom. Hybridization is typically determined by the number of sigma bonds and lone pairs on the central atom.

\n\n

Let's evaluate each molecule/ion:

\n\n

1. $$\\mathrm{NO}_3^{-}$$:

\n\n

The central atom is nitrogen (N). In $$\\mathrm{NO}_3^{-}$$, nitrogen forms three sigma bonds with oxygen atoms and has no lone pairs. Using the formula for hybridization:

\n\n

Number of hybrid orbitals = number of sigma bonds + number of lone pairs

\n\n

For $$\\mathrm{NO}_3^{-}$$: $$3 + 0 = 3$$ hybrid orbitals. Therefore, nitrogen is $$\\mathrm{sp}^2$$ hybridized.

\n\n

2. $$\\mathrm{BCl}_3$$:

\n\n

The central atom is boron (B). In $$\\mathrm{BCl}_3$$, boron forms three sigma bonds with chlorine atoms and has no lone pairs. Using the same formula:

\n\n

For $$\\mathrm{BCl}_3$$: $$3 + 0 = 3$$ hybrid orbitals. Therefore, boron is $$\\mathrm{sp}^2$$ hybridized.

\n\n

3. $$\\mathrm{ClO}_2^{-}$$:

\n\n

The central atom is chlorine (Cl). In $$\\mathrm{ClO}_2^{-}$$, chlorine forms two sigma bonds with oxygen atoms and has two lone pairs. Therefore,:

\n\n

For $$\\mathrm{ClO}_2^{-}$$: $$2 + 2 = 4$$ hybrid orbitals. Thus, chlorine is $$\\mathrm{sp}^3$$ hybridized.

\n\n

4. $$\\mathrm{ClO}_3^{-}$$:

\n\n

The central atom is chlorine (Cl). In $$\\mathrm{ClO}_3^{-}$$, chlorine forms three sigma bonds with oxygen atoms and has one lone pair. Therefore,:

\n\n

For $$\\mathrm{ClO}_3^{-}$$: $$3 + 1 = 4$$ hybrid orbitals. Thus, chlorine is $$\\mathrm{sp}^3$$ hybridized.

\n\n

Based on the analysis, the molecules/ions $$\\mathrm{ClO}_2^{-}$$ and $$\\mathrm{ClO}_3^{-}$$ have their central atoms involved in $$\\mathrm{sp}^3$$ hybridization.

\n\n

Therefore, the total number is 2.

\n\n

Thus, the correct option is Option A.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 453, "subject": "Chemistry", "question": "

The number of species from the following that have pyramidal geometry around the central atom is _________

\n

$$\\mathrm{S}_2 \\mathrm{O}_3^{2-}, \\mathrm{SO}_4^{2-}, \\mathrm{SO}_3^{2-}, \\mathrm{S}_2 \\mathrm{O}_7^{2-}$$

", "options": [ { "text": "4" }, { "text": "3" }, { "text": "2" }, { "text": "1" } ], "answer": "1", "solution": "**Answer:** 1\n\n

$$\\mathrm{SO}_3^{2-}$$ is the only species with pyramidal geometry.

\n

\"JEE

", "topic": "Geometry", "subtopic": "Plane Geometry" }, { "id": 454, "subject": "Chemistry", "question": "

The shape of carbocation is :

", "options": [ { "text": "tetrahedral\n" }, { "text": "diagonal pyramidal\n" }, { "text": "diagonal\n" }, { "text": "trigonal planar" } ], "answer": "trigonal planar", "solution": "**Answer:** trigonal planar\n\n

Carbocation is $$s p^2$$ hybridised hence it's shape is trigonal planar

\n

\"JEE

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 455, "subject": "Chemistry", "question": "

Match List I with List II

\n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
LIST I
(Molecule)		LIST II
(Shape)
A.$$\\mathrm{NH_3}$$I.Square pyramid
B.\t$$\\mathrm{BrF_5}$$II.Tetrahedral
C.$$\\mathrm{PCl_5}$$III.Trigonal pyramidal
D.$$\\mathrm{CH_4}$$IV.Trigonal bipyramidal

\n

Choose the correct answer from the options given below:

", "options": [ { "text": "A-II, B-IV, C-I, D-III\n" }, { "text": "A-III, B-I, C-IV, D-II\n" }, { "text": "A-IV, B-III, C-I, D-II\n" }, { "text": "A-III, B-IV, C-I, D-II" } ], "answer": "A-III, B-I, C-IV, D-II\n", "solution": "**Answer:** A-III, B-I, C-IV, D-II\n\n\n

To match the molecules in List I with their corresponding shapes in List II, we need to consider the electronic geometry and the VSEPR (Valence Shell Electron Pair Repulsion) theory. Here’s the detailed analysis:

\n\n

$$\\mathrm{NH_3}$$ (Ammonia): Ammonia has a central nitrogen atom bonded to three hydrogen atoms and has one lone pair of electrons. This leads to a trigonal pyramidal shape.

\n\n

$$\\mathrm{BrF_5}$$ (Bromine pentafluoride): Bromine pentafluoride has a central bromine atom bonded to five fluorine atoms and has one lone pair of electrons. This leads to a square pyramidal shape.

\n\n

$$\\mathrm{PCl_5}$$ (Phosphorus pentachloride): Phosphorus pentachloride has a central phosphorus atom bonded to five chlorine atoms and no lone pairs. This results in a trigonal bipyramidal shape.

\n\n

$$\\mathrm{CH_4}$$ (Methane): Methane has a central carbon atom bonded to four hydrogen atoms with no lone pairs, forming a tetrahedral structure.

\n\n

Using the above analyses, we can match the molecules and shapes as follows:

\n\n\n\n

Therefore, the correct matching is option B:

\n\n

Option B: A-III, B-I, C-IV, D-II

", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 456, "subject": "Chemistry", "question": "

Match List I with List II.

\n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
LIST I
LIST II
A.$$
\\mathrm{ICl}
$$		I.T - shape
B.$$
\\mathrm{ICl}_3
$$		II.Square pyramidal
C.$$
\\mathrm{ClF}_5
$$		III.Pentagonal bipyramidal
D.$$
\\mathrm{IF}_7
$$		IV.Linear

\n

Choose the correct answer from the options given below :

", "options": [ { "text": "(A)-(I), (B)-(IV), (C)-(III), (D)-(II)\n" }, { "text": "(A)-(IV), (B)-(I), (C)-(II), (D)-(III)\n" }, { "text": "(A)-(I), (B)-(III), (C)-(II), (D)-(IV)\n" }, { "text": "(A)-(IV), (B)-(III), (C)-(II), (D)-(I)" } ], "answer": "(A)-(IV), (B)-(I), (C)-(II), (D)-(III)\n", "solution": "**Answer:** (A)-(IV), (B)-(I), (C)-(II), (D)-(III)\n\n\n

$$\\begin{aligned}\n& \\mathrm{ICl} \\rightarrow s p^3 \\rightarrow 1 \\mathrm{bp}+3 \\mathrm{lp} \\rightarrow \\text { Linear } \\\\\n& \\mathrm{ICl}_3 \\rightarrow s p^3 d \\rightarrow 3 \\mathrm{bp}+2 \\mathrm{lp} \\rightarrow \\mathrm{T} \\text {-shape } \\\\\n& \\mathrm{CIF}_5 \\rightarrow s p^3 d^2 \\rightarrow 5 \\mathrm{bp}+1 \\mathrm{lp} \\rightarrow \\text { Square Pyramidal } \\\\\n& \\mathrm{IF}_7 \\rightarrow s p^3 d^3 \\rightarrow 7 \\text { bp only } \\rightarrow \\text { Pentagonal bipyramidal } \\\\\n& \\text { (A)-(IV), (B)-(I), (C)-(II), D-(III) }\n\\end{aligned}$$

", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 457, "subject": "Chemistry", "question": "

Total number of species from the following with central atom utilising $$\\mathrm{sp}^2$$ hybrid orbitals for bonding is ________.

$$\\mathrm{NH}_3, \\mathrm{SO}_2, \\mathrm{SiO}_2, \\mathrm{BeCl}_2, \\mathrm{C}_2 \\mathrm{H}_2, \\mathrm{C}_2 \\mathrm{H}_4, \\mathrm{BCl}_3, \\mathrm{HCHO}, \\mathrm{C}_6 \\mathrm{H}_6, \\mathrm{BF}_3, \\mathrm{C}_2 \\mathrm{H}_4 \\mathrm{Cl}_2$$

", "options": [], "answer": "6", "solution": "**Answer:** 6\n\n

$$\\mathrm{SO}_2, \\mathrm{C}_2 \\mathrm{H}_4, \\mathrm{BCl}_3, \\mathrm{HCHO}, \\mathrm{C}_6 \\mathrm{H}_6, \\mathrm{BF}_3$$ are $$s p^2$$ hybridised central atom

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 458, "subject": "Chemistry", "question": "

Match List I with List II

\n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
LIST I
(Compound/Species)		LIST II
(Shape/Geometry)
A.$$\\mathrm{SF_4}$$I.Tetrahedral
B.\t$$\\mathrm{BrF_3}$$II.Pyramidal
C.$$\\mathrm{BrO_3^-}$$III.See saw
D.$$\\mathrm{NH_4^+}$$IV.Bent T-Shape

\n

Choose the correct answer from the options given below:

", "options": [ { "text": "A-III, B-II, C-IV, D-I\n" }, { "text": "A-III, B-IV, C-II, D-I\n" }, { "text": "A-II, B-IV, C-III, D-I\n" }, { "text": "A-II, B-III, C-I, D-IV" } ], "answer": "A-III, B-IV, C-II, D-I\n", "solution": "**Answer:** A-III, B-IV, C-II, D-I\n\n\n

$$\\begin{aligned}\n& \\text { A } \\rightarrow \\text { Sea-saw (III) } \\\\\n& \\text { B } \\rightarrow \\text { Bent T-shape (IV) } \\\\\n& \\text { C } \\rightarrow \\text { Pyramidal (II) } \\\\\n& \\text { D } \\rightarrow \\text { Tetrahedral (I) }\n\\end{aligned}$$

", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 459, "subject": "Chemistry", "question": "

Match List I with List II

\n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
LIST I
(Molecule/Species)		LIST II
(Property/Shape)
A.$$\\mathrm{SO_2Cl_2}$$I.Paramagnetic
B.\t$$\\mathrm{NO}$$II.Diamagnetic
C.$$\\mathrm{NO_2^-}$$III.Tetrahedral
D.$$\\mathrm{I_3^-}$$IV.Linear

\n

Choose the correct answer from the options given below:

", "options": [ { "text": "A-II, B-III, C-I, D-IV\n" }, { "text": "A-III, B-I, C-II, D-IV\n" }, { "text": "A-IV, B-I, C-III, D-II\n" }, { "text": "A-III, B-IV, C-II, D-I" } ], "answer": "A-III, B-I, C-II, D-IV\n", "solution": "**Answer:** A-III, B-I, C-II, D-IV\n\n\n

$$\\begin{aligned}\n& \\mathrm{A} \\rightarrow \\text { Tetrahedral (III) } \\\\\n& \\mathrm{B} \\rightarrow \\text { Paramagnetic (I) } \\\\\n& \\mathrm{C} \\rightarrow \\text { Diamagnetic (II) } \\\\\n& \\mathrm{D} \\rightarrow \\text { Linear (IV) }\n\\end{aligned}$$

", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 460, "subject": "Chemistry", "question": "Which of the following hydrogen bonds is the strongest? ", "options": [ { "text": "O−H…….N" }, { "text": "F−H…….F " }, { "text": "O−H…….O " }, { "text": "O−H…….F " } ], "answer": "F−H…….F ", "solution": "**Answer:** F−H…….F \n\nAmong F, O and N, F is most electronegative so F pulls bond pair of electron in F - H towards itself and develops highly positive charge on H atom. \n

This highly positive charged H atom creates stongest hydrogen bonding by taking lone pair of electron form electronegative atom F/N/O. Hence among the given options F – H ...... F is the strongest bond.\n

Note : F – H ...... N has strongest hydogen bonding among F – H ...... F, F – H ...... O and F – H ...... N because N is least electronegative among F, O and N and can easily donate lone pair of electron.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 461, "subject": "Chemistry", "question": "The intermolecular interaction that is dependent on the inverse cube of distance between the molecule is:", "options": [ { "text": "ion-dipole interaction" }, { "text": "London force" }, { "text": "hydrogen bond" }, { "text": "ion-ion interaction" } ], "answer": "hydrogen bond", "solution": "**Answer:** hydrogen bond\n\nHydrogen bond is a type of strong electrostatic dipole- dipole intersection and dependent on the inverse cube of distance between the molecular ion-dipole -\n

interaction $$ \\propto {1 \\over {{r^3}}}.$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 462, "subject": "Chemistry", "question": "HF has highest boiling point among hydrogen\nhalides, because it has :", "options": [ { "text": "lowest dissociation enthalpy" }, { "text": "strongest hydrogen bonding" }, { "text": "lowest ionic character" }, { "text": "strongest van der Waals' interactions" } ], "answer": "strongest hydrogen bonding", "solution": "**Answer:** strongest hydrogen bonding\n\nDue to strong H-bonding\nbetween HF molecules, HF has highest boiling point among the\nhydrogen halides.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 463, "subject": "Chemistry", "question": "Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R.

Assertion A : Dipole-dipole interactions are the only non-covalent interactions, resulting in hydrogen bond formation.

Reason R : Fluorine is the most electronegative element and hydrogen bonds in HF are symmetrical.

In the light of the above statements, choose the most appropriate answer from the options given below :", "options": [ { "text": "Both A and R are true and R is the correct explanation of A" }, { "text": "A is true but R is false" }, { "text": "A is false but R is true" }, { "text": "Both A and R are true but R is NOT the correct explanation of A" } ], "answer": "A is false but R is true", "solution": "**Answer:** A is false but R is true\n\nDipole - Dipole are not only the interaction\nresponsible for hydrogen bond formation.\nIon-dipole can also be responsible for\nhydrogen bond formation.\n

F is most electronegative element and\nanhydrous HF in solid phase has\nsymmetrical hydrogen bonding.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 464, "subject": "Chemistry", "question": "The number of hydrogen bonded water molecule(s) associated with stoichiometry CuSO4.5H2O is ____________.", "options": [], "answer": "1", "solution": "**Answer:** 1\n\n\"JEE

One hydrogen bonded H2O molecule", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 465, "subject": "Chemistry", "question": "

The correct order of increasing intermolecular hydrogen bond strength is

", "options": [ { "text": "HCN < H2O < NH3" }, { "text": "HCN < CH4 < NH3" }, { "text": "CH4 < HCN < NH3" }, { "text": "CH4 < NH3 < HCN" } ], "answer": "CH4 < HCN < NH3", "solution": "**Answer:** CH4 < HCN < NH3\n\nDue to the high difference in electronegativity of H and\nN the H-bond strength of NH3 is highest. There is\nno H-bond in CH4.

\nCH4 < HCN < NH3", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 466, "subject": "Chemistry", "question": "

Decreasing order of the hydrogen bonding in following forms of water is correctly represented by

\n

A. Liquid water

\n

B. Ice

\n

C. Impure water

\n

Choose the correct answer from the options given below :

", "options": [ { "text": "A = B > C" }, { "text": "B > A > C" }, { "text": "A > B > C" }, { "text": "C > B > A" } ], "answer": "B > A > C", "solution": "**Answer:** B > A > C\n\nExtent of hydrogen bonding :\n

\nIce $>$ liquid water $>$ impure water\n

\n- In ice, 4 molecules of $\\mathrm{H}_{2} \\mathrm{O}$ are connected to $\\mathrm{H}_{2} \\mathrm{O}$ molecule.\n

\n- Impure water will have less hydrogen bonding.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 467, "subject": "Chemistry", "question": "

In an ice crystal, each water molecule is hydrogen bonded to ____________ neighbouring molecules.

", "options": [], "answer": "4", "solution": "**Answer:** 4\n\nIn an ice crystal, each water molecule is hydrogen bonded to four neighbouring molecules.", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 468, "subject": "Chemistry", "question": "Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).

\nAssertion (A): $\\mathrm{PH}_3$ has lower boiling point than $\\mathrm{NH}_3$.

\nReason (R) : In liquid state $\\mathrm{NH}_3$ molecules are associated through vander Waal's forces, but $\\mathrm{PH}_3$ molecules are associated through hydrogen bonding.\n

\nIn the light of the above statements, choose the most appropriate answer from the options given below :\n\n", "options": [ { "text": "Both (A) and (R) are correct and (R) is the correct explanation of (A)" }, { "text": "(A) is not correct but (R) is correct" }, { "text": "(A) is correct but (R) is not correct" }, { "text": "Both $(\\mathbf{A})$ and $(\\mathbf{R})$ are correct but $(\\mathbf{R})$ is not the correct explanation of $(\\mathbf{A})$" } ], "answer": "(A) is correct but (R) is not correct", "solution": "**Answer:** (A) is correct but (R) is not correct\n\n

The correct answer is Option C: (A) is correct but (R) is not correct.

\n\n

Assertion (A): $\\mathrm{PH}_3$ has lower boiling point than $\\mathrm{NH}_3$.

\n\n

This assertion is true. The boiling point of ammonia ($\\mathrm{NH}_3$) is higher than that of phosphine ($\\mathrm{PH}_3$). Ammonia has a boiling point of about -33.34°C, whereas phosphine has a boiling point of about -87.7°C. The reason for the higher boiling point of ammonia as compared to phosphine lies in the strength and type of intermolecular forces present in the substances.

\n\n

Reason (R) : In liquid state $\\mathrm{NH}_3$ molecules are associated through Vander Waals' forces, but $\\mathrm{PH}_3$ molecules are associated through hydrogen bonding.

\n\n

This reason is incorrect. Ammonia ($\\mathrm{NH}_3$) can form hydrogen bonds due to the presence of a highly electronegative nitrogen atom bonded to hydrogen atoms. This allows $\\mathrm{NH}_3$ molecules to strongly associate with each other through hydrogen bonding, which is a much stronger intermolecular force than Van der Waals forces. This hydrogen bonding is responsible for the relatively high boiling point of ammonia.

\n\n

On the other hand, phosphine ($\\mathrm{PH}_3$) does not form hydrogen bonds because the electronegativity difference between phosphorus and hydrogen is not significant enough to enable the formation of hydrogen bonds. Instead, phosphine molecules are associated mainly through weaker Van der Waals forces, leading to a lower boiling point when compared to ammonia.

\n\n

In conclusion, the assertion that $\\mathrm{PH}_3$ has a lower boiling point than $\\mathrm{NH}_3$ is correct, due to the hydrogen bonding present in $\\mathrm{NH}_3$ and absent in $\\mathrm{PH}_3$. However, the reason provided is incorrect because it incorrectly states that $\\mathrm{PH}_3$ forms hydrogen bonds and that $\\mathrm{NH}_3$ is associated through Van der Waals forces. It is actually the other way around, so the correct answer is that (A) is true but (R) is false.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 469, "subject": "Chemistry", "question": "

The correct statement/s about Hydrogen bonding is/are

\n

A. Hydrogen bonding exists when H is covalently bonded to the highly electro negative atom.

\n

B. Intermolecular H bonding is present in $$o$$-nitro phenol

\n

C. Intramolecular $$\\mathrm{H}$$ bonding is present in HF.

\n

D. The magnitude of $$\\mathrm{H}$$ bonding depends on the physical state of the compound.

\n

E. H-bonding has powerful effect on the structure and properties of compounds

\n

Choose the correct answer from the options given below:

", "options": [ { "text": "A, B, C only\n" }, { "text": "A only\n" }, { "text": "A, D, E only\n" }, { "text": "A, B, D only" } ], "answer": "A, D, E only\n", "solution": "**Answer:** A, D, E only\n\n\n

In o-nitrophenol intra molecular hydrogen bonding \nis present.

\n

In HF intermolecular hydrogen bonding is present.

\n

Other statements are correct except B and C.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 470, "subject": "Chemistry", "question": "Among the following, the number of halide(s) which is/are inert to hydrolysis is _________.

(A) BF3

(B) SiCl4

(C) PCl5

(D) SF6", "options": [], "answer": "1", "solution": "**Answer:** 1\n\nBF3 – Shows Partial hydrolysis\n

SiCl4 – Undergoes hydrolysis readily\n

PCl5 – Undergoes hydrolysis by addition–\nelimination mechanism.\n

SF6 – Due to crowding Inert towards hydrolysis.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 471, "subject": "Chemistry", "question": "Lattice energy of an ionic compounds depends upon ", "options": [ { "text": "Charge on the ion only" }, { "text": "Size of the ion only " }, { "text": "Packing of ions only" }, { "text": "Charge on the ion and size of the ion" } ], "answer": "Charge on the ion and size of the ion", "solution": "**Answer:** Charge on the ion and size of the ion\n\n

The lattice energy of an ionic compound is a measure of the strength of the bonds in that ionic compound. Specifically, it is the energy required to separate one mole of an ionic solid into its constituent ions in the gaseous state. The lattice energy depends on several factors, primarily the charge on the ions and the size (or radius) of the ions.

\n\n

1. Charge on the ions: The lattice energy is directly proportional to the product of the charges of the cation and the anion. Higher charges result in stronger electrostatic attraction between the ions, thereby increasing the lattice energy. This relationship can be derived from Coulomb's law, which states that the force of attraction between two charged particles is directly proportional to the product of their charges.

\n\n

Mathematically:

\n\n

$$U \\propto \\frac{q_1 \\cdot q_2}{r}$$

\n\n

where $U$ is the lattice energy, $q_1$ and $q_2$ are the charges on the ions, and $r$ is the distance between the centers of the ions.

\n\n

2. Size of the ions: The lattice energy is inversely proportional to the distance between the ions, which depends on the sum of their ionic radii. Smaller ions can come closer together, increasing the electrostatic attractions between them, thereby increasing the lattice energy. Larger ions, being further apart, result in a lower lattice energy.

\n\n

Considering these two factors, the correct choice is:

\n\n

Option D: Charge on the ion and size of the ion

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 472, "subject": "Chemistry", "question": "The charge/size ratio of a cation determines its polarizing power. Which one of the following\nsequences represents the increasing order of the polarizinig order of the polarizing power of the\ncationic species, K+, Ca2+, Mg2+, Be2+? ", "options": [ { "text": "Mg2+ < Be2+ < K+ < Ca2+" }, { "text": "K+ < Ca2+ < Mg2+ < Be2+" }, { "text": "Be2+ < K+ < Ca2+ < Mg2+" }, { "text": "Ca2+ < Mg2+ < Be2+ \n < K+" } ], "answer": "K+ < Ca2+ < Mg2+ < Be2+", "solution": "**Answer:** K+ < Ca2+ < Mg2+ < Be2+\n\nAs charge/size ratio of a cation determines its polarizing power so high charge and small size of the cations increases\npolarisation.\n

As the size of the given cations decreases as\n

K+\n > Ca2+ > Mg2+ > Be2+\n

Hence, polarising power decreases as\n

K+ < Ca2+ < Mg2+ < Be2+", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 473, "subject": "Chemistry", "question": "Among the following the maximum covalent character is shown by the compound : ", "options": [ { "text": "SnCl2 " }, { "text": "AlCl3 " }, { "text": "MgCl2 " }, { "text": "FeCl2" } ], "answer": "AlCl3 ", "solution": "**Answer:** AlCl3 \n\nCharge of cation/Size of cation is called polarising power.\n

$$ \\therefore $$ (i) Polarising power $$ \\propto $$ charge of cation\n

(ii) Polarising power $$ \\propto $$ \n
1
\n
size of cation
\n\n

Here the AlCl3\n is satisfying the above two conditions i.e., Al\nis in +3 oxidation state and also has small size. So it has more\ncovalent character.\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 474, "subject": "Chemistry", "question": "Match the type of interaction in column A with\nthe distance dependence of their interaction\nenergy in column B\n
\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
AB
(i) ion-ion(a) $${1 \\over r}$$
(ii) dipole-dipole(b) $${1 \\over {{r^2}}}$$
(iii) London dispersion(c) $${1 \\over {{r^3}}}$$
(d) $${1 \\over {{r^6}}}$$
", "options": [ { "text": "(I)-(a), (II)-(b), (III)-(d)" }, { "text": "(I)-(b), (II)-(d), (III)-(c)" }, { "text": "(I)-(a), (II)-(b), (III)-(c)" }, { "text": "(I)-(a), (II)-(c), (III)-(d)" } ], "answer": "(I)-(a), (II)-(c), (III)-(d)", "solution": "**Answer:** (I)-(a), (II)-(c), (III)-(d)\n\nIon-ion interaction energy $$ \\propto $$ $${1 \\over r}$$\n

Dipole-dipole interaction energy $$ \\propto $$ $${1 \\over {{r^3}}}$$\n

London dispersion $$ \\propto $$ $${1 \\over {{r^6}}}$$", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 475, "subject": "Chemistry", "question": "The correct set from the following in which both pairs are in correct order of melting point is :", "options": [ { "text": "LiCl > LiF ; NaCl > MgO" }, { "text": "LiCl > LiF ; MgO > NaCl" }, { "text": "LiF > LiCl ; NaCl > MgO" }, { "text": "LiF > LiCl ; MgO > NaCl" } ], "answer": "LiF > LiCl ; MgO > NaCl", "solution": "**Answer:** LiF > LiCl ; MgO > NaCl\n\n

Correct option is i.e. LiF > LiCl; MgO > NaCl. Melting point is directly proportional to lattice energy. Lattice energy is the energy required to separate a mole of an ionic solid into gaseous ions. It depends upon charge of ions and size of ions.

\n

$$M.P. \\propto L.E. \\propto {{Charge} \\over {Size}}$$

\n

$$\\matrix{\n {Li \\to + 1} & {Li \\to + 1} \\cr \n {F = - 1} & {Cl \\to - 1} \\cr \n\n } $$

\n

Both LiF and LiCl having same charge, so melting point will depend on size.

\n

Larger the size of anion, lesser the lattice energy and hence, melting point order is LiF > LiCl.

\n

Similarly, $$\\matrix{\n {MgO} & {NaCl} \\cr \n {Mg \\to 2 + } & {Na \\to 1 + } \\cr \n {O \\to 2 - } & {Cl \\to 1 - } \\cr \n\n } $$

\n

MgO having + 2 charge which is greater than NaCl (+ 1) charge. So, greater the charge on the ions greater will be lattice energy and hence, melting point order is MgO > NaCl.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 476, "subject": "Chemistry", "question": "

The number of following factors which affect the percent covalent character of the ionic bond is _________

\n

(A) Polarising power of cation

\n

(B) Extent of distortion of anion

\n

(C) Polarisability of the anion

\n

(D) Polarising power of anion

", "options": [], "answer": "3", "solution": "**Answer:** 3\n\n

The covalent character of an ionic bond is largely determined by the polarization of the ions involved in the bond. Polarization refers to the distortion of the electron cloud of an anion by a cation. This distortion leads to a shift in electron density towards the cation, thereby increasing the covalent character of the bond.

\n

Here's a breakdown of the options:

\n

(A) Polarising power of cation: The greater the polarising power of a cation, the greater the distortion of the anion, and therefore, the greater the covalent character of the bond. Thus, this statement is correct.

\n

(B) Extent of distortion of anion: The more an anion is distorted, the more the electron density is shifted towards the cation, and the greater the covalent character of the bond. This statement is also correct.

\n

(C) Polarisability of the anion: The greater the polarisability of an anion, the more it can be distorted, and therefore, the greater the covalent character of the bond. Thus, this statement is also correct.

\n

(D) Polarising power of anion: This statement is incorrect. It is not the polarising power of the anion, but the polarisability of the anion and the polarising power of the cation that influence the covalent character of the bond.

\n

Therefore, three of the factors listed (A, B, C) affect the percent covalent character of the ionic bond.

\n", "topic": "Number Theory", "subtopic": "Divisibility / Factorization" }, { "id": 477, "subject": "Chemistry", "question": "Arrange the bonds in order of increasing ionic character in the molecules. $\\mathrm{LiF}$, $\\mathrm{K}_2 \\mathrm{O}, \\mathrm{N}_2, \\mathrm{SO}_2$ and $\\mathrm{ClF}_3$ :", "options": [ { "text": "$\\mathrm{N}_2<\\mathrm{SO}_2<\\mathrm{ClF}_3<\\mathrm{K}_2 \\mathrm{O}<\\mathrm{LiF}$" }, { "text": "$\\mathrm{ClF}_3<\\mathrm{N}_2<\\mathrm{SO}_2<\\mathrm{K}_2 \\mathrm{O}<\\mathrm{LiF}$" }, { "text": "$\\mathrm{LiF}<\\mathrm{K}_2 \\mathrm{O}<\\mathrm{ClF}_3<\\mathrm{SO}_2<\\mathrm{N}_2$" }, { "text": "$\\mathrm{N}_2<\\mathrm{ClF}_3<\\mathrm{SO}_2<\\mathrm{K}_2 \\mathrm{O}<\\mathrm{LiF}$" } ], "answer": "$\\mathrm{N}_2<\\mathrm{SO}_2<\\mathrm{ClF}_3<\\mathrm{K}_2 \\mathrm{O}<\\mathrm{LiF}$", "solution": "**Answer:** $\\mathrm{N}_2<\\mathrm{SO}_2<\\mathrm{ClF}_3<\\mathrm{K}_2 \\mathrm{O}<\\mathrm{LiF}$\n\nTo determine the order of increasing ionic character in the bonds of the given molecules, we should consider the concept of electronegativity, which is a measure of the tendency of an atom to attract a bonding pair of electrons. The larger the difference in electronegativity between two atoms, the more ionic the bond is. Covalent bonds, on the other hand, have smaller differences in electronegativity.\n\n

We can qualitatively assess the ionic character of the bonds in each molecule as follows :

\n\n- $\\mathrm{N_2}$ : Both nitrogen atoms have the same electronegativity since it is a diatomic molecule of the same element, therefore the bond is purely covalent with no ionic character.\n \n

- $\\mathrm{SO_2}$ : Sulfur and oxygen have different electronegativities, with oxygen being more electronegative than sulfur, but not as drastically different as in ionic compounds. Therefore, the bond has some ionic character but is primarily covalent.\n \n

- $\\mathrm{ClF_3}$ : Chlorine and fluorine have different electronegativities, with fluorine being more electronegative. However, the difference is not as large as between metals and nonmetals, so while this bond has ionic character, it is not as ionic as a bond between a metal and a nonmetal.\n \n

- $\\mathrm{K_2O}$ : Potassium is a metal with low electronegativity, and oxygen is a nonmetal with high electronegativity. Therefore, bonds between potassium and oxygen will have a high degree of ionic character.\n \n

- $\\mathrm{LiF}$ : Lithium is a metal with low electronegativity, and fluorine is a nonmetal with very high electronegativity. The difference in electronegativity is very large, indicating a very high ionic character, even more so than in $\\mathrm{K_2O}$ because fluorine is more electronegative than oxygen.\n\n

Given these considerations, the correct order from least to most ionic character is as follows :

\n\n$$\\mathrm{N_2}<\\mathrm{SO_2}<\\mathrm{ClF_3}<\\mathrm{K_2O}<\\mathrm{LiF}$$\n\n

Therefore, the correct option is :

\n\n

Option A : $$\\mathrm{N_2}<\\mathrm{SO_2}<\\mathrm{ClF_3}<\\mathrm{K_2O}<\\mathrm{LiF}$$

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 478, "subject": "Chemistry", "question": "

Which of the following is least ionic?

", "options": [ { "text": "CoCl$$_2$$" }, { "text": "KCl" }, { "text": "BaCl$$_2$$" }, { "text": "AgCl" } ], "answer": "AgCl", "solution": "**Answer:** AgCl\n\n

$$\\mathrm{AgCl}<\\mathrm{CoCl}_2<\\mathrm{BaCl}_2<\\mathrm{KCl}$$ (ionic character)

\n

Reason: $$\\mathrm{Ag}^{+}$$ has pseudo inert gas configuration.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 479, "subject": "Chemistry", "question": "The number of Cl = O bonds in perchloric acid\nis, \"________\".", "options": [], "answer": "3", "solution": "**Answer:** 3\n\nThe structure of perchloric acid is\n\"JEE\n
The number Cl = O bonds in HClO4 is 3.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 480, "subject": "Chemistry", "question": "

The number of molecules or ions from the following, which do not have odd number of electrons are _________.

\n

(A) NO$$_2$$

\n

(B) ICl$$_4^ - $$

\n

(C) BrF$$_3$$

\n

(D) ClO$$_2$$

\n

(E) NO$$_2^ + $$

\n

(F) NO

", "options": [], "answer": "3", "solution": "**Answer:** 3\n\nOnly $\\mathrm{ICl}_4^{-}, \\mathrm{BrF}_3$ and $\\mathrm{NO}_2^{+}$have even number of electrons.\n

$$\n\\begin{aligned}\n& \\mathrm{NO}_2 \\Rightarrow 23 e^{-} ; \\\\\\\\\n& \\mathrm{ICl}_4^{-} \\Rightarrow 122 e^{-} ; \\\\\\\\\n& \\mathrm{BrF}_3 \\Rightarrow 62 e^{-} ; \\\\\\\\\n& \\mathrm{ClO}_2 \\Rightarrow 33 e^{-} ; \\\\\\\\\n& \\mathrm{NO}_2^{+} \\Rightarrow 22 e^{-} ; \\\\\\\\\n& \\mathrm{NO} \\Rightarrow 15 e^{-}\n\\end{aligned}\n$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 481, "subject": "Chemistry", "question": "

Number of molecules from the following which are exceptions to octet rule is _________.

\n

$$\\mathrm{CO}_2, \\mathrm{NO}_2, \\mathrm{H}_2 \\mathrm{SO}_4, \\mathrm{BF}_3, \\mathrm{CH}_4, \\mathrm{SiF}_4, \\mathrm{ClO}_2, \\mathrm{PCl}_5, \\mathrm{BeF}_2, \\mathrm{C}_2 \\mathrm{H}_6, \\mathrm{CHCl}_3, \\mathrm{CBr}_4$$

", "options": [], "answer": "6", "solution": "**Answer:** 6\n\n

$$\\mathrm{NO}_2, \\mathrm{H}_2 \\mathrm{SO}_4, \\mathrm{BF}_3, \\mathrm{ClO}_2, \\mathrm{PCl}_5, \\mathrm{BeF}_2$$

\n

These are exception of octet rule

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 482, "subject": "Chemistry", "question": "

In the lewis dot structure for $$\\mathrm{NO}_2^{-}$$, total number of valence electrons around nitrogen is _________.

", "options": [], "answer": "8", "solution": "**Answer:** 8\n\n

$$\\text { Lewis dot structure of } \\mathrm{NO}_2^{-} \\text {is: }$$

\n

\"JEE

\n

$$\\text { Total number of valence } \\mathrm{e}^{\\ominus} \\text { around } \\mathrm{N}=8$$

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 483, "subject": "Chemistry", "question": "Which of the following are arranged in an increasing order of their bond strengths?", "options": [ { "text": "$$O_2^-$$ < O2 < $$O_2^+$$ < $$O_2^{2-}$$" }, { "text": "$$O_2^{2-}$$ < $$O_2^-$$ < $$O_2$$ < $$O_2^{+}$$" }, { "text": "$$O_2^-$$ < $$O_2^{2-}$$ < $$O_2$$ < $$O_2^{+}$$" }, { "text": "$$O_2^{+}$$ < $$O_2$$ < $$O_2^-$$ < $$O_2^{2-}$$" } ], "answer": "$$O_2^{2-}$$ < $$O_2^-$$ < $$O_2$$ < $$O_2^{+}$$", "solution": "**Answer:** $$O_2^{2-}$$ < $$O_2^-$$ < $$O_2$$ < $$O_2^{+}$$\n\nNote : \n

(1) $$\\,\\,\\,\\,$$ Bond strength $$ \\propto $$ Bond order \n

(2) $$\\,\\,\\,\\,$$ Bond length $$ \\propto $$ $${1 \\over {Bond\\,\\,order}}$$\n

(3) $$\\,$$ Bond order $$ = {1 \\over 2}$$ [Nb $$-$$ Na] \n

Nb = No of electrons in bonding molecular orbital \n

Na $$=$$ No of electrons in anti bonding molecular orbital \n

(4) $$\\,\\,\\,\\,$$ upto 14 electrons, molecular orbital configuration is \n

\"AIEEE\n

Here Na = Anti bonding electron $$=$$ 4 and Nb = 10\n

(5) $$\\,\\,\\,\\,$$ After 14 electrons to 20 electrons molecular orbital configuration is - - -\n

\"AIEEE\n

Here Na = 10\n

and Nb = 10\n

In O atom 8 electrons present, so in O2, 8 $$ \\times $$ 2 = 16 electrons present. \n

Then in $$O_2^ + $$ no of electrons = 15 \n

in $$O_2^ - $$ no of electrons = 17\n

in $$O_2^{2 - }$$ no of electrons = 18\n

$$\\therefore\\,\\,\\,\\,$$ Molecular orbital configuration of O2 (16 electrons) is \n

$${\\sigma _{1{s^2}}}\\,\\,\\sigma _{1{s^2}}^ * \\,$$ $${\\sigma _{2{s^2}}}\\,\\,\\sigma _{2{s^2}}^ * \\,$$ $${\\sigma _{2p_z^2}}\\,\\,{\\pi _{2p_x^2}} = {\\pi _{2p_y^2}}\\,\\,\\pi _{2p_x^1}^ * \\,\\, = \\pi _{2p_y^1}^ * $$\n

$$\\therefore\\,\\,\\,\\,$$Na = 6\n

Nb = 10\n

$$\\therefore\\,\\,\\,\\,$$ BO = $${1 \\over 2}\\left[ {10 - 6} \\right] = 2$$\n

Molecular orbital configuration of O$$_2^ + $$ (15 electrons) is \n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,{\\sigma _{2{s^2}}}\\,\\sigma _{2{s^2}}^ * \\,{\\sigma _{2p_z^2}}\\,{\\pi _{2p_x^2}}\\, = \\,{\\pi _{2p_y^2}}\\,\\pi _{2p_x^1}^ * \\, = \\,\\pi _{2p_y^o}^ * $$\n

$$\\therefore\\,\\,\\,\\,$$ Nb = 10\n

Na = 5 \n

$$\\therefore\\,\\,\\,\\,$$ BO = $${1 \\over 2}\\left[ {10 - 5} \\right]$$ = 2.5\n

Molecular orbital configuration of $$O_2^ - $$ (17 electrons) is \n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,{\\sigma _{2{s^2}}}\\,\\sigma _{2{s^2}}^ * \\,{\\sigma _{2p_z^2}}\\,{\\pi _{2p_x^2}}\\, = \\,{\\pi _{2p_y^2}}\\,\\pi _{2p_x^2}^ * \\, = \\,\\pi _{2p_y^1}^ * $$\n

$$\\therefore\\,\\,\\,\\,$$ Nb = 10\n

Na = 7\n

$$\\therefore\\,\\,\\,\\,$$ BO = $${1 \\over 2}\\left[ {10 - 7} \\right]$$ = 1.5\n

Molecular orbital configuration of O $$_2^{2 - }$$ (18 electrons) is \n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,{\\sigma _{2{s^2}}}\\,\\sigma _{2{s^2}}^ * \\,{\\sigma _{2p_z^2}}\\,{\\pi _{2p_x^2}}\\, = \\,{\\pi _{2p_y^2}}\\,\\pi _{2p_x^2}^ * \\, = \\,\\pi _{2p_y^2}^ * $$\n

$$\\therefore\\,\\,\\,\\,$$ Nb = 10\n

Na = 8\n

$$\\therefore\\,\\,\\,\\,$$ BO = $${1 \\over 2}$$ [ 10 $$-$$ 8] = 1\n

As Bond strength $$ \\propto $$ Bond order so, correct order is \n

$$O_2^{2 - } < O_2^ - < {O_2} < O_2^ + $$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 484, "subject": "Chemistry", "question": "The bond order in NO is 2.5 while that in NO+ is 3. Which of the following statements is true for these two species? ", "options": [ { "text": "Bond length in NO+ is greater than in NO " }, { "text": "Bond length is unpredictable" }, { "text": "Bond length in NO+ in equal to that in NO" }, { "text": "Bond length in NO is greater than in NO+" } ], "answer": "Bond length in NO is greater than in NO+", "solution": "**Answer:** Bond length in NO is greater than in NO+\n\nMolecular orbital configuration of NO (15 electrons) is \n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,{\\sigma _{2{s^2}}}\\,\\sigma _{2{s^2}}^ * \\,{\\sigma _{2p_z^2}}\\,{\\pi _{2p_x^2}}\\, = \\,{\\pi _{2p_y^2}}\\,\\pi _{2p_x^1}^ * \\, = \\,\\pi _{2p_y^o}^ * $$\n

$$\\therefore\\,\\,\\,\\,$$ Nb = 10\n

Na = 5 \n

$$\\therefore\\,\\,\\,\\,$$ BO = $${1 \\over 2}\\left[ {10 - 5} \\right]$$ = 2.5\n

Similarly Molecular orbital configuration of NO+ (14 electrons) is \n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,{\\sigma _{2{s^2}}}\\,\\sigma _{2{s^2}}^ * \\,{\\sigma _{2p_z^2}}\\,{\\pi _{2p_x^2}}\\, = \\,{\\pi _{2p_y^2}}\\,$$\n

$$\\therefore\\,\\,\\,\\,$$ Nb = 10\n

Na = 4\n

$$\\therefore\\,\\,\\,\\,$$ BO = $${1 \\over 2}\\left[ {10 - 4} \\right]$$ = 3\n

Note :\n
(2) $$\\,\\,\\,\\,$$ Bond length $$ \\propto $$ $${1 \\over {Bond\\,\\,order}}$$\n

So, bond length in NO > NO+", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 485, "subject": "Chemistry", "question": "Which one of the following species is diamagnetic in nature? ", "options": [ { "text": "$$He_2^+$$" }, { "text": "H2" }, { "text": "$$H_2^+$$" }, { "text": "$$H_2^-$$" } ], "answer": "H2", "solution": "**Answer:** H2\n\nTIPS/Formulae : \n

A diamagnetic substance contains no unpaired electron. \n

$${H_2}$$ is diamagnetic as it contains all paired electrons \n

$$\\mathop {{H_2} = \\sigma _b^2}\\limits_{\\left( {diamagnetic} \\right)} \\,\\,,\\,\\,\\mathop {H_2^ + = \\sigma _b^1,}\\limits_{\\left( {paramagnetic} \\right)} \\,\\,\\mathop {H_2^ - = \\sigma _b^2,}\\limits_{\\left( {paramagnetic} \\right)} $$\n

$$\\mathop {\\sigma _a^{ * 1};He_2^ + }\\limits_{\\left( {paramagnetic} \\right)} = \\mathop {\\sigma _b^2,\\sigma _a^{ \\ne 1}}\\limits_{\\left( {paramagnetic} \\right)} $$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 486, "subject": "Chemistry", "question": "Which of the following molecules/ions does not contain unpaired electrons? ", "options": [ { "text": "$$O_2^{2−} $$" }, { "text": "B2" }, { "text": "$$N_2^+$$" }, { "text": "O2" } ], "answer": "$$O_2^{2−} $$", "solution": "**Answer:** $$O_2^{2−} $$\n\n(A) Molecular orbital configuration of O $$_2^{2 - }$$ (18 electrons) is \n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,{\\sigma _{2{s^2}}}\\,\\sigma _{2{s^2}}^ * \\,{\\sigma _{2p_z^2}}\\,{\\pi _{2p_x^2}}\\, = \\,{\\pi _{2p_y^2}}\\,\\pi _{2p_x^2}^ * \\, = \\,\\pi _{2p_y^2}^ * $$\n

So O $$_2^{2 - }$$ has no unpaired electrons.\n\n

(B) Molecular orbital configuration of B2 (10 electrons) is \n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,\\,{\\sigma _{2{s^2}}}\\,\\,\\sigma _{2{s^2}}^ * \\,\\,{\\pi _{2p_x^1}} = {\\pi _{2p_y^1}}$$\n

Here in B2, 2 unpaired electrons present.\n \n

(C) Moleculer orbital configuration of $$N_2^{ + }$$ (13 electrons)\n

= $${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,{\\sigma _{2{s^2}}}\\,\\sigma _{2{s^2}}^ * \\,{\\pi _{2p_x^2}}\\, = \\,{\\pi _{2p_y^2}}\\,{\\sigma _{2p_z^1}}$$\n

Here in $$N_2^{ + }$$, 1 unpaired electron present.\n

(D) Molecular orbital configuration of O2 (16 electrons) is \n

$${\\sigma _{1{s^2}}}\\,\\,\\sigma _{1{s^2}}^ * \\,$$ $${\\sigma _{2{s^2}}}\\,\\,\\sigma _{2{s^2}}^ * \\,$$ $${\\sigma _{2p_z^2}}\\,\\,{\\pi _{2p_x^2}} = {\\pi _{2p_y^2}}\\,\\,\\pi _{2p_x^1}^ * \\,\\, = \\pi _{2p_y^1}^ * $$\n

So O$$_2$$ has 2 unpaired electrons.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 487, "subject": "Chemistry", "question": "Which of the following species exhibits the diamagnetic behaviour? ", "options": [ { "text": "$$O_2^{2−}$$" }, { "text": "NO" }, { "text": "$$O_2^+$$" }, { "text": "O2" } ], "answer": "$$O_2^{2−}$$", "solution": "**Answer:** $$O_2^{2−}$$\n\nThose species which have unpaired electrons are called paramagnetic species.\n

And those species which have no unpaired electrons are called diamagnetic species.\n\n

(a)    $$O_2^{2−}$$ has 18 electrons.\n

Moleculer orbital configuration of $$O_2^{2−}$$ is \n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,{\\sigma _{2{s^2}}}\\,\\sigma _{2{s^2}}^ * \\,{\\sigma _{2p_z^2}}\\,{\\pi _{2p_x^2}}\\, = \\,{\\pi _{2p_y^2}}\\,\\pi _{2p_x^2}^ * \\, = \\,\\pi _{2p_y^2}^ * $$\n

Here is no unpaired electron so it is diamagnetic. \n\n

(b)   NO has 15 electrons. \n

Moleculer orbital configuration of NO is \n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,\\,{\\sigma _{2{s^2}}}\\,\\,\\sigma _{2{s^2}}^ * \\,\\,{\\sigma _{2p_z^2}}\\,\\,{\\pi _{2p_x^2}}\\,= \\,{\\pi _{2p_y^2}}\\,\\pi _{2p_x^1}^ * = \\,\\,\\pi _{2p_y^0}^ * $$\n

Here is 1 unpaired electron, So it is Paramagnetic. \n

(c)   $$O_2$$ has 16 electrons. \n

Moleculer orbital configuration of $$O_2$$ is \n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,\\,{\\sigma _{2{s^2}}}\\,\\,\\sigma _{2{s^2}}^ * \\,\\,{\\sigma _{2p_z^2}}\\,\\,{\\pi _{2p_x^2}}\\,= \\,{\\pi _{2p_y^2}}\\,\\pi _{2p_x^1}^ * = \\,\\,\\pi _{2p_y^1}^ * $$\n

Here 2 unpaired electron present, so it is paramagnetic.\n

(d)   $$O_2^{+}$$ has 15 electrons. \n

Moleculer orbital configuration of $$O_2^{+}$$ is \n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,\\,{\\sigma _{2{s^2}}}\\,\\,\\sigma _{2{s^2}}^ * \\,\\,{\\sigma _{2p_z^2}}\\,\\,{\\pi _{2p_x^2}}\\,= \\,{\\pi _{2p_y^2}}\\,\\pi _{2p_x^1}^ * = \\,\\,\\pi _{2p_y^0}^ * $$\n

Here 1 unpaired electron present, so it is paramagnetic.\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 488, "subject": "Chemistry", "question": "In which of the following ionization processes, the bond order has increased and the magnetic\nbehaviour has changed? ", "options": [ { "text": "$$C_2 \\to C_2^+$$" }, { "text": "$$N_2 \\to N_2^+$$" }, { "text": "$$NO \\to NO^+$$" }, { "text": "$$O_2 \\to O_2^+$$" } ], "answer": "$$NO \\to NO^+$$", "solution": "**Answer:** $$NO \\to NO^+$$\n\n(A) Moleculer orbital configuration of $$C_2$$ (12 electrons)\n

= $${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,{\\sigma _{2{s^2}}}\\,\\sigma _{2{s^2}}^ * \\,{\\pi _{2p_x^2}}\\, = \\,{\\pi _{2p_y^2}}$$\n

$$\\therefore\\,\\,\\,\\,$$Na = 4\n

Nb = 8\n

$$\\therefore\\,\\,\\,\\,$$ BO = $${1 \\over 2}\\left[ {8 - 4} \\right] = 2$$\n

Here no unpaired electron present, so it is diamagnetic.\n

Moleculer orbital configuration of $$C_2^+$$ (11 electrons)\n

= $${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,{\\sigma _{2{s^2}}}\\,\\sigma _{2{s^2}}^ * \\,{\\pi _{2p_x^2}}\\, = \\,{\\pi _{2p_y^1}}$$\n

$$\\therefore\\,\\,\\,\\,$$Na = 4\n

Nb = 7\n

$$\\therefore\\,\\,\\,\\,$$ BO = $${1 \\over 2}\\left[ {7 - 4} \\right] = 1.5$$\n

Here 1 unpaired electron present, so it is paramagnetic.\n

(B) Moleculer orbital configuration of $$N_2$$ (14 electrons)\n

= $${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,{\\sigma _{2{s^2}}}\\,\\sigma _{2{s^2}}^ * \\,{\\pi _{2p_x^2}}\\, = \\,{\\pi _{2p_y^2}}\\,{\\sigma _{2p_z^2}}$$\n

$$\\therefore\\,\\,\\,\\,$$Na = 4\n

Nb = 10\n

$$\\therefore\\,\\,\\,\\,$$ BO = $${1 \\over 2}\\left[ {10 - 4} \\right] = 3$$\n

Here no unpaired electron present, so it is diamagnetic.\n

Moleculer orbital configuration of $$N_2^+$$ (13 electrons)\n

= $${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,{\\sigma _{2{s^2}}}\\,\\sigma _{2{s^2}}^ * \\,{\\pi _{2p_x^2}}\\, = \\,{\\pi _{2p_y^2}}\\,{\\sigma _{2p_z^1}}$$\n

$$\\therefore\\,\\,\\,\\,$$Na = 4\n

Nb = 9\n

$$\\therefore\\,\\,\\,\\,$$ BO = $${1 \\over 2}\\left[ {9 - 4} \\right] = 2.5$$\n

Here 1 unpaired electron present, so it is paramagnetic.\n\n

(C) Moleculer orbital configuration of NO (15 electrons) is \n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,\\,{\\sigma _{2{s^2}}}\\,\\,\\sigma _{2{s^2}}^ * \\,\\,{\\sigma _{2p_z^2}}\\,\\,{\\pi _{2p_x^2}}\\,= \\,{\\pi _{2p_y^2}}\\,\\pi _{2p_x^1}^ * = \\,\\,\\pi _{2p_y^0}^ * $$\n

$$\\therefore\\,\\,\\,\\,$$Na = 5\n

Nb = 10\n

$$\\therefore\\,\\,\\,\\,$$ BO = $${1 \\over 2}\\left[ {10 - 5} \\right] = 2.5$$\n

Here is 1 unpaired electron, So it is paramagnetic. \n

Moleculer orbital configuration of NO+ (14 electrons) is \n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,\\,{\\sigma _{2{s^2}}}\\,\\,\\sigma _{2{s^2}}^ * \\,\\,{\\sigma _{2p_z^2}}\\,\\,{\\pi _{2p_x^2}}\\,= \\,{\\pi _{2p_y^2}}\\,\\pi _{2p_x^0}^ * = \\,\\,\\pi _{2p_y^0}^ * $$\n

$$\\therefore\\,\\,\\,\\,$$Na = 4\n

Nb = 10\n

$$\\therefore\\,\\,\\,\\,$$ BO = $${1 \\over 2}\\left[ {10 - 4} \\right] = 3$$\n

Here is no unpaired electron, So it is diamagnetic. \n\n\n\n\n\n\n

(D) Molecular orbital configuration of O2 (16 electrons) is \n

$${\\sigma _{1{s^2}}}\\,\\,\\sigma _{1{s^2}}^ * \\,$$ $${\\sigma _{2{s^2}}}\\,\\,\\sigma _{2{s^2}}^ * \\,$$ $${\\sigma _{2p_z^2}}\\,\\,{\\pi _{2p_x^2}} = {\\pi _{2p_y^2}}\\,\\,\\pi _{2p_x^1}^ * \\,\\, = \\pi _{2p_y^1}^ * $$\n

$$\\therefore\\,\\,\\,\\,$$Na = 6\n

Nb = 10\n

$$\\therefore\\,\\,\\,\\,$$ BO = $${1 \\over 2}\\left[ {10 - 6} \\right] = 2$$\n

Here 2 unpaired electrons present, so it is paramagnetic.\n

Molecular orbital configuration of O$$_2^ + $$ (15 electrons) is \n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,{\\sigma _{2{s^2}}}\\,\\sigma _{2{s^2}}^ * \\,{\\sigma _{2p_z^2}}\\,{\\pi _{2p_x^2}}\\, = \\,{\\pi _{2p_y^2}}\\,\\pi _{2p_x^1}^ * \\, = \\,\\pi _{2p_y^o}^ * $$\n

$$\\therefore\\,\\,\\,\\,$$ Nb = 10\n

Na = 5 \n

$$\\therefore\\,\\,\\,\\,$$ BO = $${1 \\over 2}\\left[ {10 - 5} \\right]$$ = 2.5\n

Here 1 unpaired electrons present, so it is also paramagnetic.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 489, "subject": "Chemistry", "question": "Which of the following pair of species have the same bond order?", "options": [ { "text": "CN- and NO+" }, { "text": "CN- and CN+" }, { "text": "$$O_2^-$$ and CN-" }, { "text": "NO+ and CN+" } ], "answer": "CN- and NO+", "solution": "**Answer:** CN- and NO+\n\nNumber of electron in\nNO+\n = number of electron in CN–\n = 14 electrons.\n

As both have same number of electrons so their bond order is equal.\n

Moleculer orbital configuration of NO+ (14 electrons) is \n

= $${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,\\,{\\sigma _{2{s^2}}}\\,\\,\\sigma _{2{s^2}}^ * \\,\\,{\\sigma _{2p_z^2}}\\,\\,{\\pi _{2p_x^2}}\\,= \\,{\\pi _{2p_y^2}}\\,\\pi _{2p_x^0}^ * = \\,\\,\\pi _{2p_y^0}^ * $$\n

$$ \\therefore $$ B.O = $${1 \\over 2}\\left[ {10 - 4} \\right]$$ = 3\n\n

Moleculer orbital configuration of CN- (14 electrons) is\n

= $${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,{\\sigma _{2{s^2}}}\\,\\sigma _{2{s^2}}^ * \\,{\\pi _{2p_x^2}}\\, = \\,{\\pi _{2p_y^2}}\\,{\\sigma _{2p_z^2}}$$\n

$$ \\therefore $$ B.O = $${1 \\over 2}\\left[ {10 - 4} \\right]$$ = 3", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 490, "subject": "Chemistry", "question": "Using MO theory, predict which of the following species has the shortest bond length?", "options": [ { "text": "$$O_2^+$$" }, { "text": "$$O_2^-$$" }, { "text": "$$O_2^{2-}$$" }, { "text": "$$O_2^{2+}$$" } ], "answer": "$$O_2^{2+}$$", "solution": "**Answer:** $$O_2^{2+}$$\n\nNote : \n

(1) $$\\,\\,\\,\\,$$ Bond length $$ \\propto $$ $${1 \\over {Bond\\,\\,order}}$$\n

(2) $$\\,$$ Bond order $$ = {1 \\over 2}$$ [Nb $$-$$ Na] \n

Nb = No of electrons in bonding molecular orbital \n

Na $$=$$ No of electrons in anti bonding molecular orbital \n

(4) $$\\,\\,\\,\\,$$ upto 14 electrons, molecular orbital configuration is \n

\"AIEEE\n

Here Na = Anti bonding electron $$=$$ 4 and Nb = 10\n

(5) $$\\,\\,\\,\\,$$ After 14 electrons to 20 electrons molecular orbital configuration is - - -\n

\"AIEEE\n

Here Na = 10\n

and Nb = 10\n

In O atom 8 electrons present, so in O2, 8 $$ \\times $$ 2 = 16 electrons present. \n

Then in $$O_2^ + $$ no of electrons = 15 \n

in $$O_2^ - $$ no of electrons = 17\n

in $$O_2^{2 - }$$ no of electrons = 18\n

$$\\therefore\\,\\,\\,\\,$$ Molecular orbital configuration of O2 (16 electrons) is \n

$${\\sigma _{1{s^2}}}\\,\\,\\sigma _{1{s^2}}^ * \\,$$ $${\\sigma _{2{s^2}}}\\,\\,\\sigma _{2{s^2}}^ * \\,$$ $${\\sigma _{2p_z^2}}\\,\\,{\\pi _{2p_x^2}} = {\\pi _{2p_y^2}}\\,\\,\\pi _{2p_x^1}^ * \\,\\, = \\pi _{2p_y^1}^ * $$\n

$$\\therefore\\,\\,\\,\\,$$Na = 6\n

Nb = 10\n

$$\\therefore\\,\\,\\,\\,$$ BO = $${1 \\over 2}\\left[ {10 - 6} \\right] = 2$$\n

Molecular orbital configuration of O$$_2^ + $$ (15 electrons) is \n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,{\\sigma _{2{s^2}}}\\,\\sigma _{2{s^2}}^ * \\,{\\sigma _{2p_z^2}}\\,{\\pi _{2p_x^2}}\\, = \\,{\\pi _{2p_y^2}}\\,\\pi _{2p_x^1}^ * \\, = \\,\\pi _{2p_y^o}^ * $$\n

$$\\therefore\\,\\,\\,\\,$$ Nb = 10\n

Na = 5 \n

$$\\therefore\\,\\,\\,\\,$$ BO = $${1 \\over 2}\\left[ {10 - 5} \\right]$$ = 2.5\n

Molecular orbital configuration of O$$_2^ {2+ }$$ (14 electrons) is \n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,{\\sigma _{2{s^2}}}\\,\\sigma _{2{s^2}}^ * \\,{\\sigma _{2p_z^2}}\\,{\\pi _{2p_x^2}}\\, = \\,{\\pi _{2p_y^2}}\\,\\pi _{2p_x^0}^ * \\, = \\,\\pi _{2p_y^o}^ * $$\n

$$\\therefore\\,\\,\\,\\,$$ Nb = 10\n

Na = 4 \n

$$\\therefore\\,\\,\\,\\,$$ BO = $${1 \\over 2}\\left[ {10 - 4} \\right]$$ = 3\n

Molecular orbital configuration of $$O_2^ - $$ (17 electrons) is \n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,{\\sigma _{2{s^2}}}\\,\\sigma _{2{s^2}}^ * \\,{\\sigma _{2p_z^2}}\\,{\\pi _{2p_x^2}}\\, = \\,{\\pi _{2p_y^2}}\\,\\pi _{2p_x^2}^ * \\, = \\,\\pi _{2p_y^1}^ * $$\n

$$\\therefore\\,\\,\\,\\,$$ Nb = 10\n

Na = 7\n

$$\\therefore\\,\\,\\,\\,$$ BO = $${1 \\over 2}\\left[ {10 - 7} \\right]$$ = 1.5\n

Molecular orbital configuration of O $$_2^{2 - }$$ (18 electrons) is \n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,{\\sigma _{2{s^2}}}\\,\\sigma _{2{s^2}}^ * \\,{\\sigma _{2p_z^2}}\\,{\\pi _{2p_x^2}}\\, = \\,{\\pi _{2p_y^2}}\\,\\pi _{2p_x^2}^ * \\, = \\,\\pi _{2p_y^2}^ * $$\n

$$\\therefore\\,\\,\\,\\,$$ Nb = 10\n

Na = 8\n

$$\\therefore\\,\\,\\,\\,$$ BO = $${1 \\over 2}$$ [ 10 $$-$$ 8] = 1\n

As Bond length $$ \\propto $$ $${1 \\over {Bond\\,\\,order}}$$\n

So $$O_2^{2+}$$ has the shortest bond length.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 491, "subject": "Chemistry", "question": "In which of the following pairs of molecules/ions, both the species are not likely to exist?", "options": [ { "text": "$$H_2^+$$ , $$He_2^{2-}$$" }, { "text": "$$H_2^-$$ , $$He_2^{2-}$$" }, { "text": "$$H_2^{2+}$$ , $$He_2$$" }, { "text": "$$H_2^-$$ , $$He_2^{2+}$$" } ], "answer": "$$H_2^{2+}$$ , $$He_2$$", "solution": "**Answer:** $$H_2^{2+}$$ , $$He_2$$\n\n$$H_2^{2+}$$ , $$He_2$$\n

Both have zero bond order. Thus, they do not exist.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 492, "subject": "Chemistry", "question": "Stability of the species Li2, $$Li_2^−$$ and $$Li_2^+$$ increases in the order of: ", "options": [ { "text": "Li2 < $$Li_2^+$$ < $$Li_2^-$$" }, { "text": "$$Li_2^+$$ < $$Li_2^-$$ < Li2" }, { "text": "$$Li_2^-$$ <$$Li_2^+$$ < Li2" }, { "text": "$$Li_2^-$$ < Li2 < $$Li_2^+$$" } ], "answer": "$$Li_2^-$$ <$$Li_2^+$$ < Li2", "solution": "**Answer:** $$Li_2^-$$ <$$Li_2^+$$ < Li2\n\nLi2 = $${\\sigma _{1{s^2}}}\\,\\,\\sigma _{1{s^2}}^ * \\,$$ $${\\sigma _{2{s^2}}} \\,$$\n

$$ \\therefore $$ Bond order = $${1 \\over 2}\\left( {4 - 2} \\right)$$ = 1\n

$$Li_2^+$$ = $${\\sigma _{1{s^2}}}\\,\\,\\sigma _{1{s^2}}^ * \\,$$ $${\\sigma _{2{s^1}}} \\,$$\n

$$ \\therefore $$ Bond order = $${1 \\over 2}\\left( {3 - 2} \\right)$$ = 0.5\n

$$Li_2^-$$ = $${\\sigma _{1{s^2}}}\\,\\,\\sigma _{1{s^2}}^ * \\,$$ $${\\sigma _{2{s^2}}} \\,\\,\\sigma _{1{s^1}}^ * \\,$$\n

$$ \\therefore $$ Bond order = $${1 \\over 2}\\left( {4 - 3} \\right)$$ = 0.5\n

The bond order of $$Li_2^+$$ and $$Li_2^-$$ \nis same but $$Li_2^+$$\n is more stable\nthan $$Li_2^-$$ because $$Li_2^+$$\nis smaller in size and\nhas 2 electrons in antibonding orbitals whereas $$Li_2^-$$ has 3 electrons in antibonding orbitals.\nHence $$Li_2^+$$\n is more stable than $$Li_2^-$$.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 493, "subject": "Chemistry", "question": "Which of the following species is not paramagnetic?\n", "options": [ { "text": "CO" }, { "text": "O2" }, { "text": "B2" }, { "text": "NO" } ], "answer": "CO", "solution": "**Answer:** CO\n\nThose species which have unpaired electrons are called paramagnetic species.\n\n

(a)    CO has 14 electrons.\n

Moleculer orbital configuration of CO is \n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,\\,{\\sigma _{2{s^2}}}\\,\\,\\sigma _{2{s^2}}^ * \\,\\,{\\pi _{2p_x^2}} =\\,{\\pi _{2p_y^2}}\\,{\\sigma _{2p_z^2}}$$\n

Here is no unpaired electron so it is diamagnetic. \n

(b)   $$O_2$$ has 16 electrons. \n

Moleculer orbital configuration of $$O_2$$ is \n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,\\,{\\sigma _{2{s^2}}}\\,\\,\\sigma _{2{s^2}}^ * \\,\\,{\\sigma _{2p_z^2}}\\,\\,{\\pi _{2p_x^2}}\\,= \\,{\\pi _{2p_y^2}}\\,\\pi _{2p_x^1}^ * = \\,\\,\\pi _{2p_y^1}^ * $$\n

Here 2 unpaired electron present, so it is paramagnetic.\n

(c)   B2 has 10 electrons.\n

Molecular orbital configuration of B2 is \n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,\\,{\\sigma _{2{s^2}}}\\,\\,\\sigma _{2{s^2}}^ * \\,\\,{\\pi _{2p_x^1}} = {\\pi _{2p_y^1}}$$\n

Here two unpaired electrons present. So it is paramagnetic. \n

(d)   NO has 15 electrons. \n

Moleculer orbital configuration of NO is \n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,\\,{\\sigma _{2{s^2}}}\\,\\,\\sigma _{2{s^2}}^ * \\,\\,{\\sigma _{2p_z^2}}\\,\\,{\\pi _{2p_x^2}}\\,= \\,{\\pi _{2p_y^2}}\\,\\pi _{2p_x^1}^ * = \\,\\,\\pi _{2p_y^0}^ * $$\n

Here is 1 unpaired electron, So it is Paramagnetic. \n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 494, "subject": "Chemistry", "question": "Which of the following is paramagnetic ? ", "options": [ { "text": "NO+ " }, { "text": "CO" }, { "text": "$$O_2^{2 - }$$" }, { "text": "B2" } ], "answer": "B2", "solution": "**Answer:** B2\n\nThose species which have unpaired electrons are called paramagnetic species.\n

(a)   NO+ has 14 electrons. \n

Moleculer orbital configuration of NO+ is \n

$${\\sigma _{1{s^2}}}$$ $$\\sigma _{1{s^2}}^ * $$ $${\\sigma _{2{s^2}}}\\,\\sigma _{2{s^2}}^ * \\,\\,{\\sigma _{2p_z^2}}\\,\\,{\\pi _{2p_x^2}}\\,\\, = \\,\\,{\\pi _{2p_y^2}}$$\n

Here is no unpaired electron, So it is Diamagnetic. \n

(b)    CO has 14 electrons.\n

Moleculer orbital configuration of CO is \n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,\\,{\\sigma _{2{s^2}}}\\,\\,\\sigma _{2{s^2}}^ * \\,\\,{\\pi _{2p_x^2}} =\\,{\\pi _{2p_y^2}}\\,{\\sigma _{2p_z^2}}$$\n

Here is no unpaired electron so it is diamagnetic. \n

(c)   $$O_2^{2 - }$$ has 18 electrons. \n

Moleculer orbital configuration of $$O_2^{2 - }$$ is \n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,\\,{\\sigma _{2{s^2}}}\\,\\,\\sigma _{2{s^2}}^ * \\,\\,{\\sigma _{2p_z^2}}\\,\\,{\\pi _{2p_x^2}}\\,= \\,{\\pi _{2p_y^2}}\\,\\pi _{2p_x^2}^ * = \\,\\,\\pi _{2p_y^2}^ * $$\n

Here also no unpaired electron present, so it is diamagnetic.\n

(d)   B2 has 10 electrons.\n

Molecular orbital configuration of B2 is \n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,\\,{\\sigma _{2{s^2}}}\\,\\,\\sigma _{2{s^2}}^ * \\,\\,{\\pi _{2p_x^1}} = {\\pi _{2p_y^1}}$$\n

Here two unpaired electrons present. \n

So it is paramagnetic. \n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 495, "subject": "Chemistry", "question": "In the molecular orbital diagram for the molecular ion, N2+, the number of electrons in the $$\\sigma $$2pz molecular orbital is : ", "options": [ { "text": "0" }, { "text": "1" }, { "text": "2" }, { "text": "3" } ], "answer": "1", "solution": "**Answer:** 1\n\nElectrons in $$N_2^ + $$ = 7 $$ \\times $$ 2 $$-$$ 1 = 13\n

Moleculer orbital configuration of $$N_2^ + $$ (13)\n

= $${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,{\\sigma _{2{s^2}}}\\,\\sigma _{2{s^2}}^ * \\,{\\pi _{2p_x^2}}\\, = \\,{\\pi _{2p_y^2}}\\,{\\sigma _{2p_z^1}}$$\n

$$\\therefore\\,\\,\\,$$ $$\\,{\\sigma _{2{p_z}}}$$ no. of electrons = 1.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 496, "subject": "Chemistry", "question": "According to molecular orbital theory, which of the following will not be a viable molecule?", "options": [ { "text": "$${\\rm H}e_2^{2 + }$$ " }, { "text": "$${\\rm H}e_2^{ + }$$" }, { "text": "$${\\rm H}_2^{- }$$" }, { "text": "$${\\rm H}_2^{2 - }$$" } ], "answer": "$${\\rm H}_2^{2 - }$$", "solution": "**Answer:** $${\\rm H}_2^{2 - }$$\n\nNote :\n

According to molecules orbital theory, when a molecule have bond order = 0 then that molecule does not exist. \n

(a)$$\\,\\,\\,$$ Configuration of $$He_2^{2 + }$$ (2 electrons) is = $${\\sigma _{1{s^2}}}$$\n

$$\\therefore\\,\\,\\,$$ Bond order = $${1 \\over 2}$$ (2 $$-$$0) = 1\n

(b)$$\\,\\,\\,$$ Configuration of $$He_2^ + $$ (3 electrons) is = $${\\sigma _{1{s^2}}}$$ $$\\sigma _{1{s^1}}^ * $$\n

$$\\therefore\\,\\,\\,$$ Bond order = $${1 \\over 2}$$ (2 $$-$$1) = 0.5\n

(c) $$\\,\\,\\,$$ Configuration of $$H_2^ - $$ (3 electrons) is = $${\\sigma _{1{s^2}}}$$ $$\\sigma _{1{s^1}}^ * $$\n

$$\\therefore$$ Bond order = $${1 \\over 2}$$ (2 $$-$$ 1) = 0.5\n

(d) $$\\therefore\\,\\,\\,$$ Configuration of $$He_2^{2 - }$$ is = $${\\sigma _{1{s^2}}}\\,\\,\\sigma _{1{s^2}}^ * $$ \n

$$\\therefore\\,\\,\\,$$ Bond order = $$ = {1 \\over 2}$$ (2 $$-$$ 2) = 0\n

$$\\therefore\\,\\,\\,$$ $$H_2^{2 - }$$ is not a viable molecule. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 497, "subject": "Chemistry", "question": "According to molecular orbital theory, which of the following is true with respect to Li2+ and Li2$$-$$ ? ", "options": [ { "text": "$$Li_2^ + $$ is unstable and $$Li_2^ - $$ is stable" }, { "text": "$$Li_2^ + $$ is stable and $$Li_2^ - $$ unstable" }, { "text": "Both are stable " }, { "text": "Both are unstable " } ], "answer": "Both are stable ", "solution": "**Answer:** Both are stable \n\n$$Li_2^ + $$  (5 electrons) = $${\\sigma _{1{s^2}}}\\,\\,\\sigma _{1{s^2}}^ * \\,\\,{\\sigma _{2{s^1}}}$$\n

$$Li_2^ - $$  (7 electrons) = $${\\sigma _{1{s^2}}}\\,\\,\\sigma _{1{s^2}}^ * \\,\\,{\\sigma _{2{s^2}}}$$ $$\\,\\sigma _{2{s^1}}^ * \\,$$\n

$$ \\therefore $$   Bond order of $$Li_2^ + $$ = $${{3 - 2} \\over 2}$$ = 0.5\n

Bond order of $$Li_2^ - $$ = $${{4 - 3} \\over 2}$$ = 0.5\n

As both $$Li_2^ + $$ and $$Li_2^ - $$ has non-zero bond order, so both are stable.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 498, "subject": "Chemistry", "question": "During the change of O2 to O2-\n, the incoming electron goes to the orbital :", "options": [ { "text": "$$\\pi 2{p_y}$$" }, { "text": "$${\\sigma ^*}2{p_z}$$" }, { "text": "$${\\pi ^*}2{p_x}$$" }, { "text": "$$\\pi 2{p_x}$$" } ], "answer": "$${\\pi ^*}2{p_x}$$", "solution": "**Answer:** $${\\pi ^*}2{p_x}$$\n\nMolecular orbital configuration of O2 (16 electrons) is \n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,{\\sigma _{2{s^2}}}\\,\\sigma _{2{s^2}}^ * \\,{\\sigma _{2p_z^2}}\\,{\\pi _{2p_x^2}}\\, = \\,{\\pi _{2p_y^2}}\\,\\pi _{2p_x^1}^ * \\, = \\,\\pi _{2p_y^1}^ * $$\n

Molecular orbital configuration of O2- (17 electrons) is \n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,{\\sigma _{2{s^2}}}\\,\\sigma _{2{s^2}}^ * \\,{\\sigma _{2p_z^2}}\\,{\\pi _{2p_x^2}}\\, = \\,{\\pi _{2p_y^2}}\\,\\pi _{2p_x^2}^ * \\, = \\,\\pi _{2p_y^1}^ * $$\n

So, the incoming electron goes to $$\\pi _{2p_x}^ *$$ or $$\\pi _{2p_y}^ *$$", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 499, "subject": "Chemistry", "question": "Among the following species, the diamagnetic\nmolecule is", "options": [ { "text": "CO" }, { "text": "B2" }, { "text": "O2" }, { "text": "NO" } ], "answer": "CO", "solution": "**Answer:** CO\n\nThose species which have unpaired electrons are called paramagnetic species.\n

And those species which have no unpaired electrons are called diamagnetic species.\n\n

(a)    CO has 14 electrons.\n

Moleculer orbital configuration of CO is \n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,\\,{\\sigma _{2{s^2}}}\\,\\,\\sigma _{2{s^2}}^ * \\,\\,{\\pi _{2p_x^2}} =\\,{\\pi _{2p_y^2}}\\,{\\sigma _{2p_z^2}}$$\n

Here is no unpaired electron so it is diamagnetic. \n\n

(b)   B2 has 10 electrons.\n

Molecular orbital configuration of B2 is \n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,\\,{\\sigma _{2{s^2}}}\\,\\,\\sigma _{2{s^2}}^ * \\,\\,{\\pi _{2p_x^1}} = {\\pi _{2p_y^1}}$$\n

Here two unpaired electrons present. So it is paramagnetic.\n

(c)   $$O_2$$ has 16 electrons. \n

Moleculer orbital configuration of $$O_2$$ is \n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,\\,{\\sigma _{2{s^2}}}\\,\\,\\sigma _{2{s^2}}^ * \\,\\,{\\sigma _{2p_z^2}}\\,\\,{\\pi _{2p_x^2}}\\,= \\,{\\pi _{2p_y^2}}\\,\\pi _{2p_x^1}^ * = \\,\\,\\pi _{2p_y^1}^ * $$\n

Here 2 unpaired electron present, so it is paramagnetic.\n

(d)   NO has 15 electrons. \n

Moleculer orbital configuration of NO is \n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,\\,{\\sigma _{2{s^2}}}\\,\\,\\sigma _{2{s^2}}^ * \\,\\,{\\sigma _{2p_z^2}}\\,\\,{\\pi _{2p_x^2}}\\,= \\,{\\pi _{2p_y^2}}\\,\\pi _{2p_x^1}^ * = \\,\\,\\pi _{2p_y^0}^ * $$\n

Here is 1 unpaired electron, So it is Paramagnetic. \n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 500, "subject": "Chemistry", "question": "Among the following, the molecule expected\nto be stabilized by anion formation is :
\nC2, O2, NO, F2", "options": [ { "text": "C2" }, { "text": "NO" }, { "text": "O2" }, { "text": "F2" } ], "answer": "C2", "solution": "**Answer:** C2\n\nC2 has 12 electrons.\n

Moleculer orbital configuration of C2 is \n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,\\,{\\sigma _{2{s^2}}}\\,\\,\\sigma _{2{s^2}}^ * \\,\\,{\\pi _{2p_x^2}} =\\,{\\pi _{2p_y^2}}$$\n

Bond order of C2 = $${{8 - 4} \\over 2}$$ = 2\n

C2- has 13 electrons.\n

Moleculer orbital configuration of C2- is \n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,\\,{\\sigma _{2{s^2}}}\\,\\,\\sigma _{2{s^2}}^ * \\,\\,{\\pi _{2p_x^2}} =\\,{\\pi _{2p_y^2}}$$ $${\\sigma _{2p_z^1}}$$\n

Bond order of C2- = $${{9 - 4} \\over 2}$$ = 2.5\n

As we know, higher is the bond order more will be the stability of the molecule.\n

So we can say, stability of C2 increases when it forms C2- ion.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 501, "subject": "Chemistry", "question": "Among the following molecules / ions,\n$$C_2^{2 - },N_2^{2 - },O_2^{2 - },{O_2}$$\nwhich one is diamagnetic and has the shortest\nbond length?", "options": [ { "text": "$${O_2}$$" }, { "text": "$$C_2^{2 - }$$" }, { "text": "$$N_2^{2 - }$$" }, { "text": "$$O_2^{2 - }$$" } ], "answer": "$$C_2^{2 - }$$", "solution": "**Answer:** $$C_2^{2 - }$$\n\nNote : \n

(1) $$\\,\\,\\,\\,$$ Bond strength $$ \\propto $$ Bond order \n

(2) $$\\,\\,\\,\\,$$ Bond length $$ \\propto $$ $${1 \\over {Bond\\,\\,order}}$$\n

(3) $$\\,$$ Bond order $$ = {1 \\over 2}$$ [Nb $$-$$ Na] \n

Nb = No of electrons in bending molecular orbital \n

Na $$=$$ No of electrons in anti bonding molecular orbital \n

(4) $$\\,\\,\\,\\,$$ upto 14 electrons, molecular orbital configuration is \n

\"JEE\n

Here Na = Anti bonding electron $$=$$ 4 and Nb = 10\n

(5) $$\\,\\,\\,\\,$$ After 14 electrons to 20 electrons molecular orbital configuration is - - -\n

\"JEE\n

Here Na = 10\n

and Nb = 10\n

(A) In O atom 8 electrons present, so in O2, 8 $$ \\times $$ 2 = 16 electrons present. \n\n

Then in $$O_2^{2 - }$$ no of electrons = 18\n

$$\\therefore\\,\\,\\,\\,$$ Molecular orbital configuration of O2 (16 electrons) is \n

$${\\sigma _{1{s^2}}}\\,\\,\\sigma _{1{s^2}}^ * \\,$$ $${\\sigma _{2{s^2}}}\\,\\,\\sigma _{2{s^2}}^ * \\,$$ $${\\sigma _{2p_z^2}}\\,\\,{\\pi _{2p_x^2}} = {\\pi _{2p_y^2}}\\,\\,\\pi _{2p_x^1}^ * \\,\\, = \\pi _{2p_y^1}^ * $$\n

$$\\therefore\\,\\,\\,\\,$$Na = 6\n

Nb = 10\n

$$\\therefore\\,\\,\\,\\,$$ BO = $${1 \\over 2}\\left[ {10 - 6} \\right] = 2$$\n

Here 2 unpaired electrons are present so it is paramagnetic.\n\n

(D) Molecular orbital configuration of O $$_2^{2 - }$$ (18 electrons) is \n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,{\\sigma _{2{s^2}}}\\,\\sigma _{2{s^2}}^ * \\,{\\sigma _{2p_z^2}}\\,{\\pi _{2p_x^2}}\\, = \\,{\\pi _{2p_y^2}}\\,\\pi _{2p_x^2}^ * \\, = \\,\\pi _{2p_y^2}^ * $$\n

$$\\therefore\\,\\,\\,\\,$$ Nb = 10\n

Na = 8\n

$$\\therefore\\,\\,\\,\\,$$ BO = $${1 \\over 2}$$ [ 10 $$-$$ 8] = 1\n

Here no unpaired electrons are present so it is diamagnetic.\n

(B)   $$C_2^{2 - }$$ has 14 electrons. \n

Moleculer orbital configuration of $$C_2^{2 - }$$ is\n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,{\\sigma _{2{s^2}}}\\,\\sigma _{2{s^2}}^ * \\,{\\pi _{2p_x^2}}\\, = \\,{\\pi _{2p_y^2}}\\,{\\sigma _{2p_z^2}}$$\n

$$\\therefore\\,\\,\\,\\,$$Na = 4\n

Nb = 10\n

$$\\therefore\\,\\,\\,\\,$$ BO = $${1 \\over 2}\\left[ {10 - 4} \\right] = 3$$\n

Here no unpaired electron present, so it is diamagnetic.\n

(C)   $$N_2^{2 - }$$ has 16 electrons. \n

Moleculer orbital configuration of $$N_2^{2 - }$$ is\n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,{\\sigma _{2{s^2}}}\\,\\sigma _{2{s^2}}^ * \\,{\\pi _{2p_x^2}}\\, = \\,{\\pi _{2p_y^2}}\\,{\\sigma _{2p_z^2}}\\,\\pi _{2p_x^1}^ * \\, = \\,\\pi _{2p_y^1}^ *$$\n

$$\\therefore\\,\\,\\,\\,$$Na = 6\n

Nb = 10\n

$$\\therefore\\,\\,\\,\\,$$ BO = $${1 \\over 2}\\left[ {10 - 6} \\right] = 2$$\n

Here 2 unpaired electron present, so it is paramagnetic.\n

As Bond length $$ \\propto $$ $${1 \\over {Bond\\,\\,order}}$$\n

so among two diamagnetic ions $$C_2^{2 - }$$ and $$O_2^{2 - }$$, bond order of $$C_2^{2 - }$$ is more so it will have shorter bond length.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 502, "subject": "Chemistry", "question": "Two pi and half sigma bonds are present in : ", "options": [ { "text": "O2" }, { "text": "N$$_2^ + $$" }, { "text": "O$$_2^ + $$" }, { "text": "N2" } ], "answer": "N$$_2^ + $$", "solution": "**Answer:** N$$_2^ + $$\n\nTwo pi and half sigma bonds are presents in molecule with bond order 2.5.\n

Moleculer orbital configuration of $$N_2^+$$ (13 electrons)\n

= $${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,{\\sigma _{2{s^2}}}\\,\\sigma _{2{s^2}}^ * \\,{\\pi _{2p_x^2}}\\, = \\,{\\pi _{2p_y^2}}\\,{\\sigma _{2p_z^1}}$$\n

Bond order = $${1 \\over 2}\\left( {9 - 4} \\right)$$ = 2.5", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 503, "subject": "Chemistry", "question": "In which of the following process, the bond order has increased and paramagnetic character has charged to diamagnetic ? ", "options": [ { "text": "NO $$ \\to $$ NO+" }, { "text": "N2 $$ \\to $$ N2+" }, { "text": "O2 $$ \\to $$ O2+" }, { "text": "O2 $$ \\to $$ O22$$-$$" } ], "answer": "NO $$ \\to $$ NO+", "solution": "**Answer:** NO $$ \\to $$ NO+\n\nMolecular orbital configuration of NO (15 electrons) is \n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,{\\sigma _{2{s^2}}}\\,\\sigma _{2{s^2}}^ * \\,{\\sigma _{2p_z^2}}\\,{\\pi _{2p_x^2}}\\, = \\,{\\pi _{2p_y^2}}\\,\\pi _{2p_x^1}^ * \\, = \\,\\pi _{2p_y^o}^ * $$\n

$$\\therefore\\,\\,\\,\\,$$ Nb = 10\n

Na = 5 \n

$$\\therefore\\,\\,\\,\\,$$ BO = $${1 \\over 2}\\left[ {10 - 5} \\right]$$ = 2.5\n

And in NO one unpaired electron is present , so it is paramagnetic.\n

Similarly Molecular orbital configuration of NO+ (14 electrons) is \n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,{\\sigma _{2{s^2}}}\\,\\sigma _{2{s^2}}^ * \\,{\\sigma _{2p_z^2}}\\,{\\pi _{2p_x^2}}\\, = \\,{\\pi _{2p_y^2}}\\,$$\n

$$\\therefore\\,\\,\\,\\,$$ Nb = 10\n

Na = 4\n

$$\\therefore\\,\\,\\,\\,$$ BO = $${1 \\over 2}\\left[ {10 - 4} \\right]$$ = 3\n

And in NO+ no unpaired electron is present , so it is diamagnetic.\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 504, "subject": "Chemistry", "question": "The bond order and the magnetic characteristics of CN-\n are :", "options": [ { "text": "3, paramagnetic" }, { "text": "$$2{1 \\over 2}$$, paramagnetic" }, { "text": "3, diamagnetic" }, { "text": "$$2{1 \\over 2}$$, diamagnetic" } ], "answer": "3, diamagnetic", "solution": "**Answer:** 3, diamagnetic\n\n    CN- has 14 electrons.\n

Moleculer orbital configuration of CN- is \n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,\\,{\\sigma _{2{s^2}}}\\,\\,\\sigma _{2{s^2}}^ * \\,\\,{\\pi _{2p_x^2}} =\\,{\\pi _{2p_y^2}}\\,{\\sigma _{2p_z^2}}$$\n

Here is no unpaired electron so it is diamagnetic.\n
Nb = 10\n

Na = 4 \n

$$\\therefore\\,\\,\\,\\,$$ BO = $${1 \\over 2}\\left[ {10 - 4} \\right]$$ = 3", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 505, "subject": "Chemistry", "question": "If the magnetic moment of a dioxygen species\nis 1.73 B.M, it may be :", "options": [ { "text": "$$O_2^ - $$ or $$O_2^ + $$" }, { "text": "O2, $$O_2^ - $$ or $$O_2^ + $$" }, { "text": "O2 or $$O_2^ + $$" }, { "text": "O2 or $$O_2^ - $$" } ], "answer": "$$O_2^ - $$ or $$O_2^ + $$", "solution": "**Answer:** $$O_2^ - $$ or $$O_2^ + $$\n\nMagnetic moment = 1.73 BM\n

$$ \\therefore $$ Unpaired electron = 1\n

(1) $$O_2$$ has 16 electrons. \n

Moleculer orbital configuration of $$O_2$$ is \n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,\\,{\\sigma _{2{s^2}}}\\,\\,\\sigma _{2{s^2}}^ * \\,\\,{\\sigma _{2p_z^2}}\\,\\,{\\pi _{2p_x^2}}\\,= \\,{\\pi _{2p_y^2}}\\,\\pi _{2p_x^1}^ * = \\,\\,\\pi _{2p_y^1}^ * $$\n

Here 2 unpaired electron present.\n

(2)   $$O_2^{+}$$ has 15 electrons. \n

Moleculer orbital configuration of $$O_2^{+}$$ is \n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,\\,{\\sigma _{2{s^2}}}\\,\\,\\sigma _{2{s^2}}^ * \\,\\,{\\sigma _{2p_z^2}}\\,\\,{\\pi _{2p_x^2}}\\,= \\,{\\pi _{2p_y^2}}\\,\\pi _{2p_x^1}^ * = \\,\\,\\pi _{2p_y^0}^ * $$\n

Here 1 unpaired electron present.\n

(3)   $$O_2^{-}$$ has 17 electrons. \n

Moleculer orbital configuration of $$O_2^{-}$$ is \n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,\\,{\\sigma _{2{s^2}}}\\,\\,\\sigma _{2{s^2}}^ * \\,\\,{\\sigma _{2p_z^2}}\\,\\,{\\pi _{2p_x^2}}\\,= \\,{\\pi _{2p_y^2}}\\,\\pi _{2p_x^2}^ * = \\,\\,\\pi _{2p_y^1}^ * $$\n

Here 1 unpaired electron present.\n

Hence $$O_2^{-}$$ & $$O_2^{+}$$\n have one unpaired electron. \n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 506, "subject": "Chemistry", "question": "Of the species, NO, NO+, NO2+ and NO-\n, the one with minimum bond strength is :", "options": [ { "text": "NO–" }, { "text": "NO" }, { "text": "NO+" }, { "text": "NO2+" } ], "answer": "NO–", "solution": "**Answer:** NO–\n\nNote : \n

(1) $$\\,\\,\\,\\,$$ Bond strength $$ \\propto $$ Bond order \n

(2) $$\\,\\,\\,\\,$$ Bond length $$ \\propto $$ $${1 \\over {Bond\\,\\,order}}$$\n

(3) $$\\,$$ Bond order $$ = {1 \\over 2}$$ [Nb $$-$$ Na] \n

Nb = No of electrons in bonding molecular orbital \n

Na $$=$$ No of electrons in anti bonding molecular orbital\n

(4) $$\\,\\,\\,\\,$$ upto 14 electrons, molecular orbital configuration is \n

\"JEE\n

Here Na = Anti bonding electron $$=$$ 4 and Nb = 10\n

(5) $$\\,\\,\\,\\,$$ After 14 electrons to 20 electrons molecular orbital configuration is - - -\n

\"JEE\n

Molecular orbital configuration of NO (15 electrons) is \n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,{\\sigma _{2{s^2}}}\\,\\sigma _{2{s^2}}^ * \\,{\\sigma _{2p_z^2}}\\,{\\pi _{2p_x^2}}\\, = \\,{\\pi _{2p_y^2}}\\,\\pi _{2p_x^1}^ * \\, = \\,\\pi _{2p_y^o}^ * $$\n

$$\\therefore\\,\\,\\,\\,$$ Nb = 10\n

Na = 5 \n

$$\\therefore\\,\\,\\,\\,$$ BO = $${1 \\over 2}\\left[ {10 - 5} \\right]$$ = 2.5\n

Similarly Molecular orbital configuration of NO+ (14 electrons) is \n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,{\\sigma _{2{s^2}}}\\,\\sigma _{2{s^2}}^ * \\,{\\sigma _{2p_z^2}}\\,{\\pi _{2p_x^2}}\\, = \\,{\\pi _{2p_y^2}}\\,$$\n

$$\\therefore\\,\\,\\,\\,$$ Nb = 10\n

Na = 4\n

$$\\therefore\\,\\,\\,\\,$$ BO = $${1 \\over 2}\\left[ {10 - 4} \\right]$$ = 3\n

Similarly Molecular orbital configuration of NO2+ (13 electrons) is \n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,{\\sigma _{2{s^2}}}\\,\\sigma _{2{s^2}}^ * \\,{\\sigma _{2p_z^2}}\\,{\\pi _{2p_x^2}}\\, = \\,{\\pi _{2p_y^1}}\\,$$\n

$$\\therefore\\,\\,\\,\\,$$ Nb = 9\n

Na = 4\n

$$\\therefore\\,\\,\\,\\,$$ BO = $${1 \\over 2}\\left[ {9 - 4} \\right]$$ = 2.5\n

Molecular orbital configuration of NO- (16 electrons) is \n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,{\\sigma _{2{s^2}}}\\,\\sigma _{2{s^2}}^ * \\,{\\sigma _{2p_z^2}}\\,{\\pi _{2p_x^2}}\\, = \\,{\\pi _{2p_y^2}}\\,\\pi _{2p_x^1}^ * \\, = \\,\\pi _{2p_y^1}^ * $$\n

$$\\therefore\\,\\,\\,\\,$$ Nb = 10\n

Na = 6\n

$$\\therefore\\,\\,\\,\\,$$ BO = $${1 \\over 2}\\left[ {10 - 6} \\right]$$ = 2\n

As Bond strength $$ \\propto $$ Bond order\n

$$ \\therefore $$ NO– will have minimum bond strength.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 507, "subject": "Chemistry", "question": "According to molecular orbital theory, the species among the following that does not exist is :", "options": [ { "text": "$${O_2}^{2 - }$$" }, { "text": "$$B{e_2}$$" }, { "text": "$$H{e_2}^ - $$" }, { "text": "$$H{e_2}^ + $$" } ], "answer": "$$B{e_2}$$", "solution": "**Answer:** $$B{e_2}$$\n\nNote :\n

According to molecules orbital theory, when a molecule have bond order = 0 then that molecule does not exist.\n

We know, Bond order $$ = {1 \\over 2}$$ [Nb $$-$$ Na] \n

Nb = No of electrons in bonding molecular orbital \n

Na $$=$$ No of electrons in anti bonding molecular orbital \n

(4) $$\\,\\,\\,\\,$$ upto 14 electrons, molecular orbital configuration is \n

\"JEE\n

Here Na = Anti bonding electron $$=$$ 4 and Nb = 10\n

(5) $$\\,\\,\\,\\,$$ After 14 electrons to 20 electrons molecular orbital configuration is - - -\n

\"JEE\n

Here Na = 10\n

and Nb = 10\n

Molecular orbital configuration of O $$_2^{2 - }$$ (18 electrons) is \n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,{\\sigma _{2{s^2}}}\\,\\sigma _{2{s^2}}^ * \\,{\\sigma _{2p_z^2}}\\,{\\pi _{2p_x^2}}\\, = \\,{\\pi _{2p_y^2}}\\,\\pi _{2p_x^2}^ * \\, = \\,\\pi _{2p_y^2}^ * $$\n

$$\\therefore\\,\\,\\,\\,$$ Nb = 10\n

Na = 8\n

$$\\therefore\\,\\,\\,\\,$$ BO = $${1 \\over 2}$$ [ 10 $$-$$ 8] = 1\n \n\n

$$\\,\\,\\,$$ Configuration of $$He_2^ + $$ (3 electrons) is = $${\\sigma _{1{s^2}}}\\,\\,\\sigma _{1{s^1}}^ * $$ \n

$$\\therefore\\,\\,\\,$$ Bond order = $${1 \\over 2}$$ (2 $$-$$1) = 0.5\n

$$\\,\\,\\,$$ Configuration of $$He_2^ - $$ (5 electrons) is = $${\\sigma _{1{s^2}}}\\,\\,\\sigma _{1{s^2}}^ * {\\sigma _{2{s^1}}}\\,$$ \n

$$\\therefore\\,\\,\\,$$ Bond order = $${1 \\over 2}$$ (3 $$-$$2) = 0.5\n\n

$$\\,\\,\\,$$ Configuration of $$Be_2 $$ (4 electrons) is = $${\\sigma _{1{s^2}}}$$ $$\\sigma _{1{s^2}}^ * $$\n

$$\\therefore$$ Bond order = $${1 \\over 2}$$ (2 $$-$$ 2) = 0", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 508, "subject": "Chemistry", "question": "Match List - I with List - II.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
List - I (Molecule)List - II (Bond order)
(a)$$N{e_2}$$(i)1
(b)$${N_2}$$(ii)2
(c)$${F_2}$$(iii)0
(d)$${O_2}$$(iv)3

Choose the correct answer from the options given below :", "options": [ { "text": "(a) $$ \\to $$ (i), (b) $$ \\to $$ (ii), (c) $$ \\to $$ (iii), (d) $$ \\to $$ (iv)" }, { "text": "(a) $$ \\to $$ (iv), (b) $$ \\to $$ (iii), (c) $$ \\to $$ (ii), (d) $$ \\to $$ (i)" }, { "text": "(a) $$ \\to $$ (iii), (b) $$ \\to $$ (iv), (c) $$ \\to $$ (i), (d) $$ \\to $$ (ii)" }, { "text": "(a) $$ \\to $$ (ii), (b) $$ \\to $$ (i), (c) $$ \\to $$ (iv), (d) $$ \\to $$ (iii)" } ], "answer": "(a) $$ \\to $$ (iii), (b) $$ \\to $$ (iv), (c) $$ \\to $$ (i), (d) $$ \\to $$ (ii)", "solution": "**Answer:** (a) $$ \\to $$ (iii), (b) $$ \\to $$ (iv), (c) $$ \\to $$ (i), (d) $$ \\to $$ (ii)\n\nBond order $$ = {1 \\over 2}$$ [Nb $$-$$ Na] \n

Nb = No of electrons in bonding molecular orbital \n

Na $$=$$ No of electrons in anti bonding molecular orbital \n

(4) $$\\,\\,\\,\\,$$ upto 14 electrons, molecular orbital configuration is \n

\"JEE\n

Here Na = Anti bonding electron $$=$$ 4 and Nb = 10\n

(5) $$\\,\\,\\,\\,$$ After 14 electrons to 20 electrons molecular orbital configuration is - - -\n

\"JEE\n

Here Na = 10\n

and Nb = 10\n

(a) Molecular orbital configuration of Ne2 (20 electrons) is \n

$${\\sigma _{1{s^2}}}\\,\\,\\sigma _{1{s^2}}^ * \\,$$ $${\\sigma _{2{s^2}}}\\,\\,\\sigma _{2{s^2}}^ * \\,$$ $${\\sigma _{2p_z^2}}\\,\\,{\\pi _{2p_x^2}} = {\\pi _{2p_y^2}}\\,\\,\\pi _{2p_x^2}^ * \\,\\, = \\pi _{2p_y^2}^ * $$$$\\sigma _{2p_z^2}^*$$\n

$$\\therefore\\,\\,\\,\\,$$Na = 10\n

Nb = 10\n

$$\\therefore\\,\\,\\,\\,$$ BO = $${1 \\over 2}\\left[ {10 - 10} \\right] = 0$$\n

(Note : All inert gases has BO = 0 and it does not exist as molecule. Here Ne is also an inert gas.)\n

(b) Moleculer orbital configuration of $$N_2$$ (14 electrons) is\n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,{\\sigma _{2{s^2}}}\\,\\sigma _{2{s^2}}^ * \\,{\\pi _{2p_x^2}}\\, = \\,{\\pi _{2p_y^2}}\\,{\\sigma _{2p_z^2}}$$\n

$$\\therefore\\,\\,\\,\\,$$Na = 4\n

Nb = 10\n

$$\\therefore\\,\\,\\,\\,$$ BO = $${1 \\over 2}\\left[ {10 - 4} \\right] = 3$$\n

(d) Molecular orbital configuration of F2 (18 electrons) is \n

$${\\sigma _{1{s^2}}}\\,\\,\\sigma _{1{s^2}}^ * \\,$$ $${\\sigma _{2{s^2}}}\\,\\,\\sigma _{2{s^2}}^ * \\,$$ $${\\sigma _{2p_z^2}}\\,\\,{\\pi _{2p_x^2}} = {\\pi _{2p_y^2}}\\,\\,\\pi _{2p_x^2}^ * \\,\\, = \\pi _{2p_y^2}^ * $$\n

$$\\therefore\\,\\,\\,\\,$$Na = 8\n

Nb = 10\n

$$\\therefore\\,\\,\\,\\,$$ BO = $${1 \\over 2}\\left[ {10 - 8} \\right] = 1$$\n\n

(d) Molecular orbital configuration of O2 (16 electrons) is \n

$${\\sigma _{1{s^2}}}\\,\\,\\sigma _{1{s^2}}^ * \\,$$ $${\\sigma _{2{s^2}}}\\,\\,\\sigma _{2{s^2}}^ * \\,$$ $${\\sigma _{2p_z^2}}\\,\\,{\\pi _{2p_x^2}} = {\\pi _{2p_y^2}}\\,\\,\\pi _{2p_x^1}^ * \\,\\, = \\pi _{2p_y^1}^ * $$\n

$$\\therefore\\,\\,\\,\\,$$Na = 6\n

Nb = 10\n

$$\\therefore\\,\\,\\,\\,$$ BO = $${1 \\over 2}\\left[ {10 - 6} \\right] = 2$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 509, "subject": "Chemistry", "question": "AX is a covalent diatomic molecule where A and X are second row elements of periodic table. Based on Molecular orbital theory, the bond order of AX is 2.5. The total number of electrons in AX is __________. (Round off to the Nearest Integer).", "options": [], "answer": "15", "solution": "**Answer:** 15\n\nThe compound AX is NO its bond order is 2.5 and it has total 15 electrons.\n

Note : Total number of electrons equal to 13 will\nalso have the 2.5 bond order. But in this case\nneutral diatomic molecule will not be possible.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 510, "subject": "Chemistry", "question": "In the following the correct bond order sequence is :", "options": [ { "text": "$$O_2^{2 - } > O_2^ + > O_2^ - > {O_2}$$" }, { "text": "$$O_2^ + > O_2^ - > O_2^{2 - } > {O_2}$$" }, { "text": "$$O_2^ + > {O_2} > O_2^ - > O_2^{2 - }$$" }, { "text": "$${O_2} > O_2^ - > O_2^{2 - } > O_2^ + $$" } ], "answer": "$$O_2^ + > {O_2} > O_2^ - > O_2^{2 - }$$", "solution": "**Answer:** $$O_2^ + > {O_2} > O_2^ - > O_2^{2 - }$$\n\nNote : \n

(1) $$\\,\\,\\,\\,$$ Bond strength $$ \\propto $$ Bond order \n

(2) $$\\,\\,\\,\\,$$ Bond length $$ \\propto $$ $${1 \\over {Bond\\,\\,order}}$$\n

(3) $$\\,$$ Bond order $$ = {1 \\over 2}$$ [Nb $$-$$ Na] \n

Nb = No of electrons in bonding molecular orbital \n

Na $$=$$ No of electrons in anti bonding molecular orbital \n

(4) $$\\,\\,\\,\\,$$ upto 14 electrons, molecular orbital configuration is \n

\"JEE\n

Here Na = Anti bonding electron $$=$$ 4 and Nb = 10\n

(5) $$\\,\\,\\,\\,$$ After 14 electrons to 20 electrons molecular orbital configuration is - - -\n

\"JEE\n

Here Na = 10\n

and Nb = 10\n

In O atom 8 electrons present, so in O2, 8 $$ \\times $$ 2 = 16 electrons present. \n

Then in $$O_2^ + $$ no of electrons = 15 \n

in $$O_2^ - $$ no of electrons = 17\n

in $$O_2^{2 - }$$ no of electrons = 18\n

$$\\therefore\\,\\,\\,\\,$$ Molecular orbital configuration of O2 (16 electrons) is \n

$${\\sigma _{1{s^2}}}\\,\\,\\sigma _{1{s^2}}^ * \\,$$ $${\\sigma _{2{s^2}}}\\,\\,\\sigma _{2{s^2}}^ * \\,$$ $${\\sigma _{2p_z^2}}\\,\\,{\\pi _{2p_x^2}} = {\\pi _{2p_y^2}}\\,\\,\\pi _{2p_x^1}^ * \\,\\, = \\pi _{2p_y^1}^ * $$\n

$$\\therefore\\,\\,\\,\\,$$Na = 6\n

Nb = 10\n

$$\\therefore\\,\\,\\,\\,$$ BO = $${1 \\over 2}\\left[ {10 - 6} \\right] = 2$$\n

Molecular orbital configuration of O$$_2^ + $$ (15 electrons) is \n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,{\\sigma _{2{s^2}}}\\,\\sigma _{2{s^2}}^ * \\,{\\sigma _{2p_z^2}}\\,{\\pi _{2p_x^2}}\\, = \\,{\\pi _{2p_y^2}}\\,\\pi _{2p_x^1}^ * \\, = \\,\\pi _{2p_y^o}^ * $$\n

$$\\therefore\\,\\,\\,\\,$$ Nb = 10\n

Na = 5 \n

$$\\therefore\\,\\,\\,\\,$$ BO = $${1 \\over 2}\\left[ {10 - 5} \\right]$$ = 2.5\n

Molecular orbital configuration of $$O_2^ - $$ (17 electrons) is \n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,{\\sigma _{2{s^2}}}\\,\\sigma _{2{s^2}}^ * \\,{\\sigma _{2p_z^2}}\\,{\\pi _{2p_x^2}}\\, = \\,{\\pi _{2p_y^2}}\\,\\pi _{2p_x^2}^ * \\, = \\,\\pi _{2p_y^1}^ * $$\n

$$\\therefore\\,\\,\\,\\,$$ Nb = 10\n

Na = 7\n

$$\\therefore\\,\\,\\,\\,$$ BO = $${1 \\over 2}\\left[ {10 - 7} \\right]$$ = 1.5\n

Molecular orbital configuration of O $$_2^{2 - }$$ (18 electrons) is \n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,{\\sigma _{2{s^2}}}\\,\\sigma _{2{s^2}}^ * \\,{\\sigma _{2p_z^2}}\\,{\\pi _{2p_x^2}}\\, = \\,{\\pi _{2p_y^2}}\\,\\pi _{2p_x^2}^ * \\, = \\,\\pi _{2p_y^2}^ * $$\n

$$\\therefore\\,\\,\\,\\,$$ Nb = 10\n

Na = 8\n

$$\\therefore\\,\\,\\,\\,$$ BO = $${1 \\over 2}$$ [ 10 $$-$$ 8] = 1\n

As Bond strength $$ \\propto $$ Bond order so, correct order is \n

$$O_2^{2 - } < O_2^ - < {O_2} < O_2^ + $$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 511, "subject": "Chemistry", "question": "The difference between bond orders of CO and NO$$^ \\oplus $$ is $${x \\over 2}$$ where x = _____________. (Round off to the Nearest Integer)", "options": [], "answer": "0", "solution": "**Answer:** 0\n\nBond order of CO = 3

Bond order of NO+ = 3

Difference = 0 = $${x \\over 2}$$

$$ \\Rightarrow $$ x = 0\n

Note : \n\n

(1) $$\\,$$ Bond order $$ = {1 \\over 2}$$ [Nb $$-$$ Na] \n

Nb = No of electrons in bending molecular orbital \n

Na $$=$$ No of electrons in anti bonding molecular orbital \n

(4) $$\\,\\,\\,\\,$$ upto 14 electrons, molecular orbital configuration is \n

\"JEE\n

Here Na = Anti bonding electron $$=$$ 4 and Nb = 10\n

(5) $$\\,\\,\\,\\,$$ After 14 electrons to 20 electrons molecular orbital configuration is - - -\n

\"JEE\n

Here Na = 10\n

and Nb = 10\n

(A)    CO has 14 electrons.\n

Moleculer orbital configuration of CO is \n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,\\,{\\sigma _{2{s^2}}}\\,\\,\\sigma _{2{s^2}}^ * \\,\\,{\\pi _{2p_x^2}} =\\,{\\pi _{2p_y^2}}\\,{\\sigma _{2p_z^2}}$$\n

$$\\therefore$$ Nb = 10\n

Na = 4\n

$$\\therefore\\,\\,\\,\\,$$ BO = $${1 \\over 2}$$ [ 10 $$-$$ 4] = 3\n

(B)   NO+ has 14 electrons. \n

Moleculer orbital configuration of NO+ is \n

$${\\sigma _{1{s^2}}}$$ $$\\sigma _{1{s^2}}^ * $$ $${\\sigma _{2{s^2}}}\\,\\sigma _{2{s^2}}^ * \\,\\,{\\sigma _{2p_z^2}}\\,\\,{\\pi _{2p_x^2}}\\,\\, = \\,\\,{\\pi _{2p_y^2}}$$\n

$$\\therefore$$ Nb = 10\n

Na = 4\n

$$\\therefore\\,\\,\\,\\,$$ BO = $${1 \\over 2}$$ [ 10 $$-$$ 4] = 3", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 512, "subject": "Chemistry", "question": "The total number of electrons in all bonding molecular orbitals of $$O_2^{2 - }$$ is ______________.

(Round off to the nearest integer)", "options": [], "answer": "10", "solution": "**Answer:** 10\n\nNb = No of electrons in bonding molecular orbital \n

Na $$=$$ No of electrons in anti bonding molecular orbital \n

(1) $$\\,\\,\\,\\,$$ upto 14 electrons, molecular orbital configuration is \n

\"JEE\n

Here Na = Anti bonding electron $$=$$ 4 and Nb = 10\n

(2) $$\\,\\,\\,\\,$$ After 14 electrons to 20 electrons molecular orbital configuration is - - -\n

\"JEE\n

Here Na = 10\n

and Nb = 10\n

In O atom 8 electrons present, so in O2, 8 $$ \\times $$ 2 = 16 electrons present. \n

Then in $$O_2^ + $$ no of electrons = 15 \n

in $$O_2^ - $$ no of electrons = 17\n

in $$O_2^{2 - }$$ no of electrons = 18\n\n\n

Molecular orbital configuration of O $$_2^{2 - }$$ (18 electrons) is \n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,{\\sigma _{2{s^2}}}\\,\\sigma _{2{s^2}}^ * \\,{\\sigma _{2p_z^2}}\\,{\\pi _{2p_x^2}}\\, = \\,{\\pi _{2p_y^2}}\\,\\pi _{2p_x^2}^ * \\, = \\,\\pi _{2p_y^2}^ * $$\n

$$\\therefore\\,\\,\\,\\,$$ Nb = 10\n

and Na = 8", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 513, "subject": "Chemistry", "question": "The bond order and magnetic behaviour of $$O_2^ - $$ ion are respectively :", "options": [ { "text": "1.5 and paramagnetic " }, { "text": "1.5 and diamagnetic" }, { "text": "2 and diamagnetic" }, { "text": "1 and paramagnetic " } ], "answer": "1.5 and paramagnetic ", "solution": "**Answer:** 1.5 and paramagnetic \n\n$$O_2^ - = {({\\sigma _{1s}})^2}{(\\sigma _{1s}^*)^2}{({\\sigma _{2s}})^2}{(\\sigma _{2s}^*)^2}{({\\sigma _{2{p_z}}})^2}$$$$\\left( {\\pi _{2{p_x}}^2 = \\pi _{2{p_y}}^2} \\right)\\left( {\\pi _{2{p_x}}^{*2} = \\pi _{2{p_y}}^{*1}} \\right)$$

Bond order = $${{10 - 7} \\over 2} = 1.5$$

and paramagnetic.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 514, "subject": "Chemistry", "question": "Match items of List-I with those of List-II :

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
List - I
(Property)		List - II
(Example)
(a)Diamagnetism(i)MnO
(b)Ferrimagnetism(ii)$${O_2}$$
(c)Paramagnetism(iii)NaCl
(d)Antiferromagnetism(iv)$$F{e_3}{O_4}$$


Choose the most appropriate answer from the options given below :", "options": [ { "text": "(a)-(ii), (b)-(i), (c)-(iii), (d)-(iv)" }, { "text": "(a)-(i), (b)-(iii), (c)-(iv), (d)-(ii)" }, { "text": "(a)-(iii), (b)-(iv), (c)-(ii), (d)-(i)" }, { "text": "(a)-(iv), (b)-(ii), (c)-(i), (d)-(iii)" } ], "answer": "(a)-(iii), (b)-(iv), (c)-(ii), (d)-(i)", "solution": "**Answer:** (a)-(iii), (b)-(iv), (c)-(ii), (d)-(i)\n\n

A. NaCl is diamagnetic because all electrons are paired in Na+ and in Cl− . So, it shows diamagnetism.

\n

B. Fe3O4 is ferrimagnetic because of the presence of the unequal alignment of magnetic moment in opposite direction.

\n

C. O2 molecule has two unpaired electrons. So, it is paramagnetic.

\n

D. In MnO, the orientation of the electrons of Mn and O is such that they cancel their effects, hence it is antiferromagnetic.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 515, "subject": "Chemistry", "question": "According to molecular orbital theory, the number of unpaired electron(s) in $$O_2^{2 - }$$ is :", "options": [], "answer": "0", "solution": "**Answer:** 0\n\nMolecular orbital configuration of $$O_2^{2 - }$$ is

$$\\sigma _{1s}^2\\sigma _{1s}^{*2}\\sigma _{2s}^2\\sigma _{2s}^{*2}\\left( {\\pi 2p_x^2 = \\pi 2p_y^2} \\right)\\left( {\\pi _{2px}^{*2} = \\pi _{2py}^{*2}} \\right)$$

Zero unpaired electron", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 516, "subject": "Chemistry", "question": "Number of paramagnetic oxides among the following given oxides is ____________.

Li2O, CaO, Na2O2, KO2, MgO and K2O", "options": [ { "text": "1" }, { "text": "2" }, { "text": "3" }, { "text": "0" } ], "answer": "1", "solution": "**Answer:** 1\n\nLi2O $$\\Rightarrow$$ 2Li+ O2$$-$$

CaO $$\\Rightarrow$$ Ca2+ O2$$-$$

Na2O2 $$\\Rightarrow$$ 2Na+ O$$_2^{2 - }$$

KO2 $$\\Rightarrow$$ O$$_2^{- }$$

MgO $$\\Rightarrow$$ Mg2+ O2$$-$$

K2O $$\\Rightarrow$$ 2K+ O2$$-$$

O2$$-$$ $$\\Rightarrow$$ Complete octet, diamagnetic

$$O_2^{2−}$$ has 18 electrons.\n

Moleculer orbital configuration of $$O_2^{2−}$$ is \n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,{\\sigma _{2{s^2}}}\\,\\sigma _{2{s^2}}^ * \\,{\\sigma _{2p_z^2}}\\,{\\pi _{2p_x^2}}\\, = \\,{\\pi _{2p_y^2}}\\,\\pi _{2p_x^2}^ * \\, = \\,\\pi _{2p_y^2}^ * $$\n

Here is no unpaired electron so it is diamagnetic.\n

$$O_2^{−}$$ has 17 electrons.\n

Moleculer orbital configuration of $$O_2^{−}$$ is \n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,{\\sigma _{2{s^2}}}\\,\\sigma _{2{s^2}}^ * \\,{\\sigma _{2p_z^2}}\\,{\\pi _{2p_x^2}}\\, = \\,{\\pi _{2p_y^2}}\\,\\pi _{2p_x^2}^ * \\, = \\,\\pi _{2p_y^1}^ * $$\n

Here is 1 unpaired electron so it is paramagnetic. \n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 517, "subject": "Chemistry", "question": "The spin-only magnetic moment value of $$B_2^ + $$ species is _____________ $$\\times$$ 10$$-$$2 BM. (Nearest integer) [Given : $$\\sqrt 3 $$ = 1.73]", "options": [], "answer": "173", "solution": "**Answer:** 173\n\n$$B_2^ + \\Rightarrow \\sigma _{1s}^2\\sigma _{1s}^{*2}\\sigma _{2s}^2\\sigma _{2s}^{*2}\\pi _{2py}^1 \\simeq \\pi _{2pz}^0$$

It has one\nunpaired electron.

Spin - only magnetic moment = $$\\mu \n$$\n

= $$\\sqrt {n\\left( {n + 1} \\right)} $$\n

n = Number of unpaired electrons\n

$$= \\sqrt {1(1 + 2)} = \\sqrt 3 $$ BM

= 1.73 BM

= 1.73 $$\\times$$ 10$$-$$2 BM", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 518, "subject": "Chemistry", "question": "

Consider the ions/molecule

\n

O$$_2^ + $$, O2, O$$_2^ - $$, O$$_2^ {2-} $$

\n

For increasing bond order the correct option is :

", "options": [ { "text": "O$$_2^ {2-} $$ < O$$_2^ - $$ < O2 < O$$_2^ + $$" }, { "text": "O$$_2^ - $$ < O$$_2^ {2-} $$ < O2 < O$$_2^ + $$" }, { "text": "O$$_2^ - $$ < O$$_2^ {2-} $$ < O$$_2^ + $$ < O2" }, { "text": "O$$_2^ - $$ < O$$_2^ + $$ < O$$_2^ {2-} $$ < O2" } ], "answer": "O$$_2^ {2-} $$ < O$$_2^ - $$ < O2 < O$$_2^ + $$", "solution": "**Answer:** O$$_2^ {2-} $$ < O$$_2^ - $$ < O2 < O$$_2^ + $$\n\nNote : \n

(1) $$\\,\\,\\,\\,$$ Bond strength $$ \\propto $$ Bond order \n

(2) $$\\,\\,\\,\\,$$ Bond length $$ \\propto $$ $${1 \\over {Bond\\,\\,order}}$$\n

(3) $$\\,$$ Bond order $$ = {1 \\over 2}$$ [Nb $$-$$ Na] \n

Nb = Number of electrons in bonding molecular orbital \n

Na $$=$$ Number of electrons in anti bonding molecular orbital \n

(4) $$\\,\\,\\,\\,$$ upto 14 electrons, molecular orbital configuration is \n

\"JEE\n

Here Na = Anti bonding electron $$=$$ 4 and Nb = 10\n

(5) $$\\,\\,\\,\\,$$ After 14 electrons to 20 electrons molecular orbital configuration is - - -\n

\"JEE\n

Here Na = 10\n

and Nb = 10\n

In O atom 8 electrons present, so in O2, 8 $$ \\times $$ 2 = 16 electrons present. \n

Then in $$O_2^ + $$ no of electrons = 15 \n

in $$O_2^ - $$ no of electrons = 17\n

in $$O_2^{2 - }$$ no of electrons = 18\n

$$\\therefore\\,\\,\\,\\,$$ Molecular orbital configuration of O2 (16 electrons) is \n

$${\\sigma _{1{s^2}}}\\,\\,\\sigma _{1{s^2}}^ * \\,$$ $${\\sigma _{2{s^2}}}\\,\\,\\sigma _{2{s^2}}^ * \\,$$ $${\\sigma _{2p_z^2}}\\,\\,{\\pi _{2p_x^2}} = {\\pi _{2p_y^2}}\\,\\,\\pi _{2p_x^1}^ * \\,\\, = \\pi _{2p_y^1}^ * $$\n

$$\\therefore\\,\\,\\,\\,$$Na = 6\n

Nb = 10\n

$$\\therefore\\,\\,\\,\\,$$ BO = $${1 \\over 2}\\left[ {10 - 6} \\right] = 2$$\n

Molecular orbital configuration of O$$_2^ + $$ (15 electrons) is \n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,{\\sigma _{2{s^2}}}\\,\\sigma _{2{s^2}}^ * \\,{\\sigma _{2p_z^2}}\\,{\\pi _{2p_x^2}}\\, = \\,{\\pi _{2p_y^2}}\\,\\pi _{2p_x^1}^ * \\, = \\,\\pi _{2p_y^o}^ * $$\n

$$\\therefore\\,\\,\\,\\,$$ Nb = 10\n

Na = 5 \n

$$\\therefore\\,\\,\\,\\,$$ BO = $${1 \\over 2}\\left[ {10 - 5} \\right]$$ = 2.5\n

Molecular orbital configuration of $$O_2^ - $$ (17 electrons) is \n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,{\\sigma _{2{s^2}}}\\,\\sigma _{2{s^2}}^ * \\,{\\sigma _{2p_z^2}}\\,{\\pi _{2p_x^2}}\\, = \\,{\\pi _{2p_y^2}}\\,\\pi _{2p_x^2}^ * \\, = \\,\\pi _{2p_y^1}^ * $$\n

$$\\therefore\\,\\,\\,\\,$$ Nb = 10\n

Na = 7\n

$$\\therefore\\,\\,\\,\\,$$ BO = $${1 \\over 2}\\left[ {10 - 7} \\right]$$ = 1.5\n

Molecular orbital configuration of O $$_2^{2 - }$$ (18 electrons) is \n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,{\\sigma _{2{s^2}}}\\,\\sigma _{2{s^2}}^ * \\,{\\sigma _{2p_z^2}}\\,{\\pi _{2p_x^2}}\\, = \\,{\\pi _{2p_y^2}}\\,\\pi _{2p_x^2}^ * \\, = \\,\\pi _{2p_y^2}^ * $$\n

$$\\therefore\\,\\,\\,\\,$$ Nb = 10\n

Na = 8\n

$$\\therefore\\,\\,\\,\\,$$ BO = $${1 \\over 2}$$ [ 10 $$-$$ 8] = 1\n

So, correct order of Bond order is\n

$$O_2^{2 - } < O_2^ - < {O_2} < O_2^ + $$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 519, "subject": "Chemistry", "question": "

Bonding in which of the following diatomic molecule(s) become(s) stronger, on the basis of MO Theory, by removal of an electron?

\n

(A) NO

\n

(B) N2

\n

(C) O2

\n

(D) C2

\n

(E) B2

\n

Choose the most appropriate answer from the options given below :

", "options": [ { "text": "(A), (B), (C) only" }, { "text": "(B), (C), (E) only" }, { "text": "(A), (C) only" }, { "text": "(D) only" } ], "answer": "(A), (C) only", "solution": "**Answer:** (A), (C) only\n\nIf an electron is removed from the anti-bonding\norbital, then it will tend to increase the bond order.\nThe HOMO in NO and O2 is antibonding molecular\norbital .

\nHence, in NO and O2 bond order will increase on\nloss of electron.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 520, "subject": "Chemistry", "question": "

The correct order of bond orders of $${C_2}^{2 - }$$, $${N_2}^{2 - }$$ and $${O_2}^{2 - }$$

is, respectively", "options": [ { "text": "$${C_2}^{2 - }$$ < $${N_2}^{2 - }$$ < $${O_2}^{2 - }$$" }, { "text": "$${O_2}^{2 - }$$ < $${N_2}^{2 - }$$ < $${C_2}^{2 - }$$" }, { "text": "$${C_2}^{2 - }$$ < $${O_2}^{2 - }$$ < $${N_2}^{2 - }$$" }, { "text": "$${N_2}^{2 - }$$ < $${C_2}^{2 - }$$ < $${O_2}^{2 - }$$" } ], "answer": "$${O_2}^{2 - }$$ < $${N_2}^{2 - }$$ < $${C_2}^{2 - }$$", "solution": "**Answer:** $${O_2}^{2 - }$$ < $${N_2}^{2 - }$$ < $${C_2}^{2 - }$$\n\nNote : \n\n

(1) $$\\,$$ Bond order $$ = {1 \\over 2}$$ [Nb $$-$$ Na] \n

Nb = No of electrons in bonding molecular orbital \n

Na $$=$$ No of electrons in anti bonding molecular orbital \n

(2) $$\\,\\,\\,\\,$$ upto 14 electrons, molecular orbital configuration is \n

\"JEE\n

Here Na = Anti bonding electron $$=$$ 4 and Nb = 10\n

(3) $$\\,\\,\\,\\,$$ After 14 electrons to 20 electrons molecular orbital configuration is - - -\n

\"JEE\n

Here Na = 10\n

and Nb = 10\n

In O atom 8 electrons present, so in O2, 8 $$ \\times $$ 2 = 16 electrons present. \n\n

in $$O_2^{2 - }$$ no of electrons = 18\n\n

(A) Molecular orbital configuration of O $$_2^{2 - }$$ (18 electrons) is \n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,{\\sigma _{2{s^2}}}\\,\\sigma _{2{s^2}}^ * \\,{\\sigma _{2p_z^2}}\\,{\\pi _{2p_x^2}}\\, = \\,{\\pi _{2p_y^2}}\\,\\pi _{2p_x^2}^ * \\, = \\,\\pi _{2p_y^2}^ * $$\n

$$\\therefore\\,\\,\\,\\,$$ Nb = 10\n

Na = 8\n

$$\\therefore\\,\\,\\,\\,$$ BO = $${1 \\over 2}$$ [ 10 $$-$$ 8] = 1\n

(B)   $$C_2^{2 - }$$ has 14 electrons. \n

Moleculer orbital configuration of $$C_2^{2 - }$$ is\n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,{\\sigma _{2{s^2}}}\\,\\sigma _{2{s^2}}^ * \\,{\\pi _{2p_x^2}}\\, = \\,{\\pi _{2p_y^2}}\\,{\\sigma _{2p_z^2}}$$\n

$$\\therefore\\,\\,\\,\\,$$Na = 4\n

Nb = 10\n

$$\\therefore\\,\\,\\,\\,$$ BO = $${1 \\over 2}\\left[ {10 - 4} \\right] = 3$$\n\n

(C)   $$N_2^{2 - }$$ has 16 electrons. \n

Moleculer orbital configuration of $$N_2^{2 - }$$ is\n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,{\\sigma _{2{s^2}}}\\,\\sigma _{2{s^2}}^ * \\,{\\pi _{2p_x^2}}\\, = \\,{\\pi _{2p_y^2}}\\,{\\sigma _{2p_z^2}}\\,\\pi _{2p_x^1}^ * \\, = \\,\\pi _{2p_y^1}^ *$$\n

$$\\therefore\\,\\,\\,\\,$$Na = 6\n

Nb = 10\n

$$\\therefore\\,\\,\\,\\,$$ BO = $${1 \\over 2}\\left[ {10 - 6} \\right] = 2$$\n

The correct order of bond orders of $${C_2}^{2 - }$$, $${N_2}^{2 - }$$ and $${O_2}^{2 - }$$\n

$${O_2}^{2 - }$$ < $${N_2}^{2 - }$$ < $${C_2}^{2 - }$$\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 521, "subject": "Chemistry", "question": "

Among the following species

\n

$$\\mathrm{N}_{2}, \\mathrm{~N}_{2}^{+}, \\mathrm{N}_{2}^{-}, \\mathrm{N}_{2}^{2-}, \\mathrm{O}_{2}, \\mathrm{O}_{2}^{+}, \\mathrm{O}_{2}^{-}, \\mathrm{O}_{2}^{2-}$$

\n

the number of species showing diamagnesim is _______________.

", "options": [], "answer": "2", "solution": "**Answer:** 2\n\nThose species which have unpaired electrons are called paramagnetic species.\n

And those species which have no unpaired electrons are called diamagnetic species.\n\n

(1)   $$N_2$$ has 14 electrons. \n

Moleculer orbital configuration of $$N_2$$\n

= $${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,{\\sigma _{2{s^2}}}\\,\\sigma _{2{s^2}}^ * \\,{\\pi _{2p_x^2}}\\, = \\,{\\pi _{2p_y^2}}\\,{\\sigma _{2p_z^2}}$$\n

Here no unpaired electron present, so it is diamagnetic.\n\n

(2) Moleculer orbital configuration of $$N_2^{ + }$$ (13 electrons)\n

= $${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,{\\sigma _{2{s^2}}}\\,\\sigma _{2{s^2}}^ * \\,{\\pi _{2p_x^2}}\\, = \\,{\\pi _{2p_y^2}}\\,{\\sigma _{2p_z^1}}$$\n

Here in $$N_2^{ + }$$, 1 unpaired electron present, so it is paramagnetic.\n\n

(3)   $$\\mathrm{N}_{2}^{2-}$$ has 16 electrons. \n

Moleculer orbital configuration of $$\\mathrm{N}_{2}^{2-}$$ is \n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,\\,{\\sigma _{2{s^2}}}\\,\\,\\sigma _{2{s^2}}^ * \\,\\,{\\sigma _{2p_z^2}}\\,\\,{\\pi _{2p_x^2}}\\,= \\,{\\pi _{2p_y^2}}\\,\\pi _{2p_x^1}^ * = \\,\\,\\pi _{2p_y^1}^ * $$\n

Here 2 unpaired electron present, so it is paramagnetic.\n

(4)   $$\\mathrm{N}_{2}^{-}$$ has 15 electrons. \n

Moleculer orbital configuration of $$\\mathrm{N}_{2}^{-}$$ is \n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,\\,{\\sigma _{2{s^2}}}\\,\\,\\sigma _{2{s^2}}^ * \\,\\,{\\sigma _{2p_z^2}}\\,\\,{\\pi _{2p_x^2}}\\,= \\,{\\pi _{2p_y^2}}\\,\\pi _{2p_x^1}^ * = \\,\\,\\pi _{2p_y^0}^ * $$\n

Here 1 unpaired electron present, so it is paramagnetic.\n\n\n\n

(a)    $$O_2^{2−}$$ has 18 electrons.\n

Moleculer orbital configuration of $$O_2^{2−}$$ is \n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,{\\sigma _{2{s^2}}}\\,\\sigma _{2{s^2}}^ * \\,{\\sigma _{2p_z^2}}\\,{\\pi _{2p_x^2}}\\, = \\,{\\pi _{2p_y^2}}\\,\\pi _{2p_x^2}^ * \\, = \\,\\pi _{2p_y^2}^ * $$\n

Here is no unpaired electron so it is diamagnetic. \n\n

(b)    $$O_2^{−}$$ has 17 electrons.\n

Moleculer orbital configuration of $$O_2^{2−}$$ is \n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,{\\sigma _{2{s^2}}}\\,\\sigma _{2{s^2}}^ * \\,{\\sigma _{2p_z^2}}\\,{\\pi _{2p_x^2}}\\, = \\,{\\pi _{2p_y^2}}\\,\\pi _{2p_x^2}^ * \\, = \\,\\pi _{2p_y^1}^ * $$\n

Here 1 unpaired electron present, so it is paramagnetic.\n

(c)   $$O_2$$ has 16 electrons. \n

Moleculer orbital configuration of $$O_2$$ is \n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,\\,{\\sigma _{2{s^2}}}\\,\\,\\sigma _{2{s^2}}^ * \\,\\,{\\sigma _{2p_z^2}}\\,\\,{\\pi _{2p_x^2}}\\,= \\,{\\pi _{2p_y^2}}\\,\\pi _{2p_x^1}^ * = \\,\\,\\pi _{2p_y^1}^ * $$\n

Here 2 unpaired electron present, so it is paramagnetic.\n

(d)   $$O_2^{+}$$ has 15 electrons. \n

Moleculer orbital configuration of $$O_2^{+}$$ is \n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,\\,{\\sigma _{2{s^2}}}\\,\\,\\sigma _{2{s^2}}^ * \\,\\,{\\sigma _{2p_z^2}}\\,\\,{\\pi _{2p_x^2}}\\,= \\,{\\pi _{2p_y^2}}\\,\\pi _{2p_x^1}^ * = \\,\\,\\pi _{2p_y^0}^ * $$\n

Here 1 unpaired electron present, so it is paramagnetic.\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 522, "subject": "Chemistry", "question": "

Amongst the following, the number of oxide(s) which are paramagnetic in nature is

\n

$$\\mathrm{Na}_{2} \\mathrm{O}, \\mathrm{KO}_{2}, \\mathrm{NO}_{2}, \\mathrm{~N}_{2} \\mathrm{O}, \\mathrm{ClO}_{2}, \\mathrm{NO}, \\mathrm{SO}_{2}, \\mathrm{Cl}_{2} \\mathrm{O}$$

", "options": [], "answer": "4", "solution": "**Answer:** 4\n\nParamagnetic species: $\\mathrm{KO}_{2}, \\mathrm{NO}_{2}, \\mathrm{ClO}_{2}, \\mathrm{NO}$\n

\nDiamagnetic species are: $\\mathrm{Na}_{2} \\mathrm{O}, \\mathrm{N}_{2} \\mathrm{O}, \\mathrm{SO}_{2}, \\mathrm{Cl}_{2} \\mathrm{O}$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 523, "subject": "Chemistry", "question": "

According to MO theory, number of species/ions from the following having identical bond order is ________.

\n

$$\\mathrm{CN}^{-}, \\mathrm{NO}^{+}, \\mathrm{O}_{2}, \\mathrm{O}_{2}^{+}, \\mathrm{O}_{2}^{2+}$$

", "options": [], "answer": "3", "solution": "**Answer:** 3\n\n$\\mathrm{CN}^{-}, \\mathrm{NO}^{+}$and $\\mathrm{O}_{2}^{2+}$ have bond order of 3 \n

\n$\\mathrm{O}_{2}$ has bond order of 2\n

\n$\\mathrm{O}_{2}^{+}$ has bond order of $2.5$\n

\n$\\therefore 3$ species have similar bond order.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 524, "subject": "Chemistry", "question": "

The number of paramagnetic species among the following is ___________.

\n

$$\\mathrm{B}_{2}, \\mathrm{Li}_{2}, \\mathrm{C}_{2}, \\mathrm{C}_{2}^{-}, \\mathrm{O}_{2}^{2-}, \\mathrm{O}_{2}^{+}$$ and $$\\mathrm{He}_{2}^{+}$$

", "options": [], "answer": "4", "solution": "**Answer:** 4\n\nThose species which have unpaired electrons are called paramagnetic species.\n

And those species which have no unpaired electrons are called diamagnetic species.\n

B2 has 10 electrons.\n

Molecular orbital configuration of B2 is \n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,\\,{\\sigma _{2{s^2}}}\\,\\,\\sigma _{2{s^2}}^ * \\,\\,{\\pi _{2p_x^1}} = {\\pi _{2p_y^1}}$$\n

Here two unpaired electrons present. So it is paramagnetic. \n

$$O_2^{2−}$$ has 18 electrons.\n

Moleculer orbital configuration of $$O_2^{2−}$$ is \n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,{\\sigma _{2{s^2}}}\\,\\sigma _{2{s^2}}^ * \\,{\\sigma _{2p_z^2}}\\,{\\pi _{2p_x^2}}\\, = \\,{\\pi _{2p_y^2}}\\,\\pi _{2p_x^2}^ * \\, = \\,\\pi _{2p_y^2}^ * $$\n

Here is no unpaired electron so it is diamagnetic. \n\n

$$O_2^{+}$$ has 15 electrons. \n

Moleculer orbital configuration of $$O_2^{+}$$ is \n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,\\,{\\sigma _{2{s^2}}}\\,\\,\\sigma _{2{s^2}}^ * \\,\\,{\\sigma _{2p_z^2}}\\,\\,{\\pi _{2p_x^2}}\\,= \\,{\\pi _{2p_y^2}}\\,\\pi _{2p_x^1}^ * = \\,\\,\\pi _{2p_y^0}^ * $$\n

Here 1 unpaired electron present, so it is paramagnetic.\n

$$C_2$$ has 12 electrons. \n

Moleculer orbital configuration of $$C_2$$\n

= $${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,{\\sigma _{2{s^2}}}\\,\\sigma _{2{s^2}}^ * \\,{\\pi _{2p_x^2}}\\, = \\,{\\pi _{2p_y^2}}$$\n

Here no unpaired electron present, so it is diamagnetic.\n

$$C_2^{ - }$$ has 13 electrons. \n

Moleculer orbital configuration of $$C_2^{ - }$$ is\n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,{\\sigma _{2{s^2}}}\\,\\sigma _{2{s^2}}^ * \\,{\\pi _{2p_x^2}}\\, = \\,{\\pi _{2p_y^2}}\\,{\\sigma _{2p_z^1}}$$\n

Here 1 unpaired electron present, so it is paramagnetic.\n

Li2 has 6 electrons.\n

Li2 = $${\\sigma _{1{s^2}}}\\,\\,\\sigma _{1{s^2}}^ * \\,$$ $${\\sigma _{2{s^2}}} \\,$$\n

Here no unpaired electron present, so it is diamagnetic.\n

Configuration of $$He_2^ + $$ (3 electrons) is = $${\\sigma _{1{s^2}}}$$ $$\\sigma _{1{s^1}}^ * $$\n

Here 1 unpaired electron present, so it is paramagnetic.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 525, "subject": "Chemistry", "question": "

According to MO theory the bond orders for $$\\mathrm{O}$$$$_2^{2 - }$$, $$\\mathrm{CO}$$ and $$\\mathrm{NO^+}$$ respectively, are

", "options": [ { "text": "1, 3 and 3" }, { "text": "2, 3 and 3" }, { "text": "1, 2 and 3" }, { "text": "1, 3 and 2" } ], "answer": "1, 3 and 3", "solution": "**Answer:** 1, 3 and 3\n\n

\n\n\n\n\n\n\n \n \n \n \n\n\n \n \n \n \n \n \n \n \n \n \n \n \n\n
SpeciesB.O.
O$$_2^{2 - }$$1
CO3
NO$$^ \\oplus $$3

\nNote : \n\n

(1) $$\\,$$ Bond order $$ = {1 \\over 2}$$ [Nb $$-$$ Na] \n

Nb = No of electrons in bending molecular orbital \n

Na $$=$$ No of electrons in anti bonding molecular orbital \n

(4) $$\\,\\,\\,\\,$$ upto 14 electrons, molecular orbital configuration is \n

\"JEE\n

Here Na = Anti bonding electron $$=$$ 4 and Nb = 10\n

(5) $$\\,\\,\\,\\,$$ After 14 electrons to 20 electrons molecular orbital configuration is - - -\n

\"JEE\n

Here Na = 10\n

and Nb = 10\n

(A) In O atom 8 electrons present, so in O2, 8 $$ \\times $$ 2 = 16 electrons present. \n\n

Then in $$O_2^{2 - }$$ no of electrons = 18\n

$$ \\therefore $$ Molecular orbital configuration of O $$_2^{2 - }$$ (18 electrons) is \n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,{\\sigma _{2{s^2}}}\\,\\sigma _{2{s^2}}^ * \\,{\\sigma _{2p_z^2}}\\,{\\pi _{2p_x^2}}\\, = \\,{\\pi _{2p_y^2}}\\,\\pi _{2p_x^2}^ * \\, = \\,\\pi _{2p_y^2}^ * $$\n

$$\\therefore$$ Nb = 10\n

Na = 8\n

$$\\therefore\\,\\,\\,\\,$$ BO = $${1 \\over 2}$$ [ 10 $$-$$ 8] = 1\n

(B)    CO has 14 electrons.\n

Moleculer orbital configuration of CO is \n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,\\,{\\sigma _{2{s^2}}}\\,\\,\\sigma _{2{s^2}}^ * \\,\\,{\\pi _{2p_x^2}} =\\,{\\pi _{2p_y^2}}\\,{\\sigma _{2p_z^2}}$$\n

$$\\therefore$$ Nb = 10\n

Na = 4\n

$$\\therefore\\,\\,\\,\\,$$ BO = $${1 \\over 2}$$ [ 10 $$-$$ 4] = 3\n

(C)   NO+ has 14 electrons. \n

Moleculer orbital configuration of NO+ is \n

$${\\sigma _{1{s^2}}}$$ $$\\sigma _{1{s^2}}^ * $$ $${\\sigma _{2{s^2}}}\\,\\sigma _{2{s^2}}^ * \\,\\,{\\sigma _{2p_z^2}}\\,\\,{\\pi _{2p_x^2}}\\,\\, = \\,\\,{\\pi _{2p_y^2}}$$\n

$$\\therefore$$ Nb = 10\n

Na = 4\n

$$\\therefore\\,\\,\\,\\,$$ BO = $${1 \\over 2}$$ [ 10 $$-$$ 4] = 3", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 526, "subject": "Chemistry", "question": "

The magnetic behaviour of $$\\mathrm{Li_2O,Na_2O_2}$$ and $$\\mathrm{KO_2}$$, respectively, are :

", "options": [ { "text": "paramagnetic, paramagnetic and diamagnetic" }, { "text": "paramagnetic, diamagnetic and paramagnetic" }, { "text": "diamagnetic, paramagnetic and diamagnetic" }, { "text": "diamagnetic, diamagnetic and paramagnetic" } ], "answer": "diamagnetic, diamagnetic and paramagnetic", "solution": "**Answer:** diamagnetic, diamagnetic and paramagnetic\n\n$\\mathrm{Li}_{2} \\mathrm{O} \\rightarrow \\mathrm{O}^{2-} \\rightarrow$ diamagnetic\n

\n$\\mathrm{Na}_{2} \\mathrm{O}_{2} \\rightarrow \\mathrm{O}_{2}^{2-} \\rightarrow$ diamagnetic\n

\n$\\mathrm{KO}_{2} \\rightarrow \\mathrm{O}_{2}^{-} \\rightarrow$ paramagnetic", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 527, "subject": "Chemistry", "question": "

What is the number of unpaired electron(s) in the highest occupied molecular orbital of the following species : $$\\mathrm{{N_2};N_2^ + ;{O_2};O_2^ + }$$ ?

", "options": [ { "text": "0, 1, 0, 1" }, { "text": "2, 1, 0, 1" }, { "text": "0, 1, 2, 1" }, { "text": "2, 1, 2, 1" } ], "answer": "0, 1, 2, 1", "solution": "**Answer:** 0, 1, 2, 1\n\n\n\n\n \n \n \n \n\n\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\n
Molecule$\\begin{gathered}\\text { No. of unpaired electron in } \\\\\\text { highest occupied } \\\\\\text { molecular orbital }\\end{gathered}$
$\\mathrm{N}_{2}$0
$\\mathrm{N}_{2}^{\\oplus}$1
$\\mathrm{O}_{2}$2
$\\mathrm{O}_{2}^{\\oplus}$1
", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 528, "subject": "Chemistry", "question": "

In which of the following processes, the bond order increases and paramagnetic character changes to diamagnetic one ?

", "options": [ { "text": "$$\\mathrm{O}_{2} \\rightarrow \\mathrm{O}_{2}^{2-}$$" }, { "text": "$$\\mathrm{N}_{2} \\rightarrow \\mathrm{N}_{2}^{+}$$" }, { "text": "$$\\mathrm{NO} \\rightarrow \\mathrm{NO}^{+}$$" }, { "text": "$$\\mathrm{O}_{2} \\rightarrow \\mathrm{O}_{2}^{+}$$" } ], "answer": "$$\\mathrm{NO} \\rightarrow \\mathrm{NO}^{+}$$", "solution": "**Answer:** $$\\mathrm{NO} \\rightarrow \\mathrm{NO}^{+}$$\n\nLet's analyze each option:\n

\nOption A: $\\mathrm{O}_{2} \\rightarrow \\mathrm{O}_{2}^{2-}$

\nThe bond order of $\\mathrm{O}_{2}$ is 2 and it is paramagnetic. When two electrons are added to form $\\mathrm{O}_{2}^{2-}$, the bond order becomes 1 (decreases) and it becomes diamagnetic. This option does not meet the required conditions.\n

\nOption B: $\\mathrm{N}_{2} \\rightarrow \\mathrm{N}_{2}^{+}$

\nThe bond order of $\\mathrm{N}_{2}$ is 3 and it is diamagnetic. When one electron is removed to form $\\mathrm{N}_{2}^{+}$, the bond order becomes 2.5 (decreases) and it remains diamagnetic. This option does not meet the required conditions.\n

\nOption C: $\\mathrm{NO} \\rightarrow \\mathrm{NO}^{+}$

\nThe bond order of $\\mathrm{NO}$ is 2.5 and it is paramagnetic. When one electron is removed to form $\\mathrm{NO}^{+}$, the bond order becomes 3 (increases) and it becomes diamagnetic. This option meets the required conditions.\n

\nOption D: $\\mathrm{O}_{2} \\rightarrow \\mathrm{O}_{2}^{+}$

\nThe bond order of $\\mathrm{O}_{2}$ is 2 and it is paramagnetic. When one electron is removed to form $\\mathrm{O}_{2}^{+}$, the bond order becomes 2.5 (increases) and it remains paramagnetic. This option does not meet the required conditions.\n

\nTherefore, the correct answer is $\\mathrm{NO} \\rightarrow \\mathrm{NO}^{+}$.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 529, "subject": "Chemistry", "question": "

The bond order and magnetic property of acetylide ion are same as that of

", "options": [ { "text": "$$\\mathrm{O}_{2}^{+}$$" }, { "text": "$$\\mathrm{O}_{2}^{-}$$" }, { "text": "$$\\mathrm{N}_{2}^{+}$$" }, { "text": "$$\\mathrm{NO}^{+}$$" } ], "answer": "$$\\mathrm{NO}^{+}$$", "solution": "**Answer:** $$\\mathrm{NO}^{+}$$\n\n

The acetylide ion has the formula $$\\mathrm{C}_{2}^{2-}$$. To determine its bond order, we need to first write the molecular orbital (MO) diagram for this ion.

\n

The MO diagram for $$\\mathrm{C}_{2}^{2-}$$ is:

\n

$$\\mathrm{σ_{1s}^{2}}\\mathrm{σ_{1s}^{2}}\\mathrm{σ_{2s}^{2}}\\mathrm{σ_{2s}^{2}}\\mathrm{π_{2p}^{4}}$$

\n

The bond order is given by the difference between the number of bonding electrons and the number of antibonding electrons, divided by 2. In this case, there are 4 bonding electrons and 2 antibonding electrons, so the bond order is:

\n

$$\\frac{4-2}{2}=1$$

\n

Therefore, the bond order of $$\\mathrm{C}_{2}^{2-}$$ is 1.

\n

Now, we need to determine the magnetic property of $$\\mathrm{C}_{2}^{2-}$$. Since the bond order is 1, we know that the ion has a single unpaired electron. Therefore, it is paramagnetic.

\n

Among the given options, the only ion with a bond order of 1 and paramagnetic property is $$\\mathrm{NO}^{+}$$.

\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 530, "subject": "Chemistry", "question": "Given below are two statements :

\nStatement (I) : A $\\pi$ bonding MO has lower electron density above and below the inter-nuclear axis.\n

\nStatement (II) : The $\\pi^*$ antibonding MO has a node between the nuclei.

\nIn the light of the above statements, choose the most appropriate answer from the options given below :", "options": [ { "text": "Both Statement I and Statement II are true" }, { "text": "Both Statement I and Statement II are false" }, { "text": "Statement I is true but Statement II is false" }, { "text": "Statement I is false but Statement II is true" } ], "answer": "Statement I is false but Statement II is true", "solution": "**Answer:** Statement I is false but Statement II is true\n\n

Let's analyze both statements:

\n\n

Statement (I): \"A $\\pi$ bonding MO has lower electron density above and below the inter-nuclear axis.\"

\n\n

This statement is false. In molecular orbital (MO) theory, a pi bond ($\\pi$ bond) is formed by the sideways overlap of p-orbitals from two adjacent atoms. The characteristic electron density of a $\\pi$ bonding molecular orbital is concentrated above and below the inter-nuclear axis, not lower. These regions of electron density are where the p-orbitals overlap and electrons are likely to be found. This concentration of electron density above and below the inter-nuclear axis is what allows the $\\pi$ bond to provide additional stability to the molecule, alongside any sigma ($\\sigma$) bonds that may be present.

\n\n

Statement (II): \"The $\\pi^*$ antibonding MO has a node between the nuclei.\"

\n\n

This statement is true. In a $\\pi^*$ (pi-star) antibonding molecular orbital, there is indeed a nodal plane between the nuclei where the probability of finding electrons is essentially zero. This node arises from the out-of-phase combination of the p-orbitals from adjacent atoms, which results in destructive interference and a region of zero electron density in the bonding region. The presence of this nodal plane is what makes the $\\pi^*$ orbital antibonding—the electrons in this orbital actually serve to destabilize the bond between the two atoms.

\n\n

Therefore, the correct answer is:

\n

Option D: Statement I is false but Statement II is true.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 531, "subject": "Chemistry", "question": "

Sum of bond order of CO and NO$$^+$$ is ________.

", "options": [], "answer": "6", "solution": "**Answer:** 6\n\n

$$\\begin{array}{lcl}\n\\mathrm{CO} \\Rightarrow & \\overline{\\mathrm{C}} \\equiv \\stackrel{+}{\\mathrm{O}} & : \\mathrm{BO}=3 \\\\\n\\mathrm{NO}^{+} \\Rightarrow & \\mathrm{N} \\equiv \\mathrm{O}^{+} & : \\mathrm{BO}=3\n\\end{array}$$

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 532, "subject": "Chemistry", "question": "

The linear combination of atomic orbitals to form molecular orbitals takes place only when the combining atomic orbitals

\n

A. have the same energy

\n

B. have the minimum overlap

\n

C. have same symmetry about the molecular axis

\n

D. have different symmetry about the molecular axis

\n

Choose the most appropriate from the options given below:

", "options": [ { "text": "B, C, D only" }, { "text": "A, B, C only" }, { "text": "B and D only" }, { "text": "A and C only" } ], "answer": "A and C only", "solution": "**Answer:** A and C only\n\n

* Molecular orbital should have maximum overlap

\n

* Symmetry about the molecular axis should be similar

", "topic": "Discrete Mathematics", "subtopic": "Combinatorics" }, { "id": 533, "subject": "Chemistry", "question": "

The number of species from the following which are paramagnetic and with bond order equal to one is _________.

\n

$$\\mathrm{H}_2, \\mathrm{He}_2^{+}, \\mathrm{O}_2^{+}, \\mathrm{N}_2^{2-}, \\mathrm{O}_2^{2-}, \\mathrm{F}_2, \\mathrm{Ne}_2^{+}, \\mathrm{B}_2$$

", "options": [], "answer": "1", "solution": "**Answer:** 1\n\n

\n\n\n\n\n\n\n\n \n \n \n \n \n\n\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\n
Sol.Magnetic behaviourBond order
$$\\mathrm{H}_2$$Diamagnetic$$1$$
$$\\mathrm{He_2^+}$$Paramagnetic$$0.5$$
$$\\mathrm{O_2^+}$$Paramagnetic$$2.5$$
$$\\mathrm{N_2^{2-}}$$Paramagnetic$$2$$
$$\\mathrm{O_2^{2-}}$$Diamagnetic$$1$$
$$\\mathrm{F_2}$$Diamagnetic$$1$$
$$\\mathrm{Ne_2^+}$$Paramagnetic$$0.5$$
$$\\mathrm{B_2}$$Paramagnetic$$1$$

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 534, "subject": "Chemistry", "question": "

The total number of anti bonding molecular orbitals, formed from $$2 s$$ and $$2 p$$ atomic orbitals in a diatomic molecule is _______.

", "options": [], "answer": "4", "solution": "**Answer:** 4\n\n

Antibonding molecular orbital from $$2 \\mathrm{~s}=1$$

\n

Antibonding molecular orbital from $$2 p=3$$

\n

Total $$=4$$

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 535, "subject": "Chemistry", "question": "

The total number of molecular orbitals formed from $$2 \\mathrm{s}$$ and $$2 \\mathrm{p}$$ atomic orbitals of a diatomic molecule is __________.

", "options": [], "answer": "8", "solution": "**Answer:** 8\n\n

Two molecular orbitals $$\\sigma 2 \\mathrm{s}$$ and $$\\sigma * 2 \\mathrm{s}$$.

\n

Six molecular orbitals $$\\sigma 2 \\mathrm{p}_z$$ and $$\\sigma * 2 \\mathrm{p}_{\\mathrm{z}}$$.

\n

$$\\pi 2 \\mathrm{p}_{\\mathrm{x}}, \\pi 2 \\mathrm{p}_{\\mathrm{y}}$$ and $$\\pi * 2 \\mathrm{p}_{\\mathrm{x}}, \\pi^* 2 \\mathrm{p}_{\\mathrm{y}}$$

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 536, "subject": "Chemistry", "question": "

Total number of electrons present in $$\\left(\\pi^*\\right)$$ molecular orbitals of $$\\mathrm{O}_2, \\mathrm{O}_2^{+}$$ and $$\\mathrm{O}_2^{-}$$ is ________.

", "options": [], "answer": "6", "solution": "**Answer:** 6\n\n

$$\\begin{aligned}\n& \\mathrm{O}_2(16 \\mathrm{e}):\\left(\\sigma_{1 \\mathrm{~s}}\\right)^2\\left(\\sigma_{1 \\mathrm{~s}}^*\\right)^2\\left(\\sigma_{2 \\mathrm{~s}}\\right)^2\\left(\\sigma_{2 \\mathrm{~s}}^*\\right)^2 \\\\\n& \\left(\\sigma_{2 \\mathrm{p}}\\right)^2\\left[\\left(\\pi_{2 \\mathrm{p}}\\right)^2=\\left(\\pi_{2 \\mathrm{p}}\\right)^2\\right],\\left[\\left(\\pi_{2 \\mathrm{p}}^*\\right)^1=\\left(\\pi_{2 \\mathrm{p}}^*\\right)^1\\right]\n\\end{aligned}$$

\n

Number of $$\\mathrm{e}^{-}$$ present in $$\\left(\\pi^*\\right)$$ of $$\\mathrm{O}_2=2$$

\n

Number of $$\\mathrm{e}^{-}$$ present in $$\\left(\\pi^*\\right)$$ of $$\\mathrm{O}_2^{+}=1$$

\nNumber of $$\\mathrm{e}^{-}$$ present in $$\\left(\\pi^*\\right)$$ of $$\\mathrm{O}_2^{-}=3$$

\n

So total $$\\mathrm{e}^{-}$$ in $$\\left(\\pi^*\\right)=2+1+3=6$$

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 537, "subject": "Chemistry", "question": "

The total number of species from the following in which one unpaired electron is present, is _______.

\n

$$\\mathrm{N}_2, \\mathrm{O}_2, \\mathrm{C}_2^{-}, \\mathrm{O}_2^{-}, \\mathrm{O}_2^{2-}, \\mathrm{H}_2^{+}, \\mathrm{CN}^{-}, \\mathrm{He}_2^{+}$$

", "options": [], "answer": "4", "solution": "**Answer:** 4\n\n

\n\n\n\n\n\n\n \n \n \n \n\n\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\n
SpeciesUnpaired e
$$
\\mathrm{N}_2
$$		0
$$
\\mathrm{O}_2
$$		2
$$
\\mathrm{C}_2^{-}
$$		1
$$
\\mathrm{O}_2^{-}
$$		1
$$
\\mathrm{O}_2^{2-}
$$		0
$$
\\mathrm{H}_2^{+}
$$		1
$$
\\mathrm{CN}^{-}
$$		0
$$
\\mathrm{He}_2^{+}
$$		1

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 538, "subject": "Chemistry", "question": "

Number of molecules/species from the following having one unpaired electron is ________.

\n

$$\\mathrm{O}_2, \\mathrm{O}_2^{-1}, \\mathrm{NO}, \\mathrm{CN}^{-1}, \\mathrm{O}_2^{2-}$$

", "options": [], "answer": "2", "solution": "**Answer:** 2\n\n

$$\\mathrm{O}_2^{-} \\text {and } \\mathrm{NO} \\text { have } 1 \\text { unpaired electron. }$$

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 539, "subject": "Chemistry", "question": "

When $$\\psi_{\\mathrm{A}}$$ and $$\\psi_{\\mathrm{B}}$$ are the wave functions of atomic orbitals, then $$\\sigma^*$$ is represented by :

", "options": [ { "text": "$$\\psi_A+2 \\psi_B$$\n" }, { "text": "$$\\psi_{\\mathrm{A}}-\\psi_{\\mathrm{B}}$$\n" }, { "text": "$$\\psi_A-2 \\psi_B$$\n" }, { "text": "$$\\psi_{\\mathrm{A}}+\\psi_{\\mathrm{B}}$$" } ], "answer": "$$\\psi_{\\mathrm{A}}-\\psi_{\\mathrm{B}}$$\n", "solution": "**Answer:** $$\\psi_{\\mathrm{A}}-\\psi_{\\mathrm{B}}$$\n\n\n

In Molecular Orbital Theory, molecular orbitals are formed by the linear combination of atomic orbitals (LCAO). The wave functions of atomic orbitals $\\psi_A$ and $\\psi_B$ can combine in two ways:

\n\n
    \n
  1. Constructive Interference: This leads to the formation of a bonding molecular orbital ($\\sigma$):
    2. \n
\n

$ \\sigma = \\psi_A + \\psi_B $

\n\n
    \n
  1. Destructive Interference: This leads to the formation of an antibonding molecular orbital ($\\sigma^*$):
    2. \n
\n

$ \\sigma^* = \\psi_A - \\psi_B $

\n\n

Conclusion

\n\n

The antibonding molecular orbital $\\sigma^*$ is represented by:

\n\n

$ \\sigma^* = \\psi_A - \\psi_B $

\n\n

Therefore, the correct answer is:

\n\n

Option B: $\\psi_{\\mathrm{A}} - \\psi_{\\mathrm{B}}$

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 540, "subject": "Chemistry", "question": "

Number of molecules having bond order 2 from the following molecules is _________.

$$\\mathrm{C}_2, \\mathrm{O}_2, \\mathrm{Be}_2, \\mathrm{Li}_2, \\mathrm{Ne}_2, \\mathrm{~N}_2, \\mathrm{He}_2$$

", "options": [], "answer": "2", "solution": "**Answer:** 2\n\n

To determine the number of molecules with a bond order of 2 from the given molecules, we need to first calculate the bond order for each molecule. The bond order can be determined using Molecular Orbital Theory (MOT). The bond order is given by the formula:

\n\n

\n\n

$$\\text{Bond Order} = \\frac{\\text{Number of bonding electrons} - \\text{Number of anti-bonding electrons}}{2}$$

\n\n

\n\n

Let's evaluate each molecule individually:

\n\n

1. $$\\mathrm{C}_2$$:

\n\n

For $$\\mathrm{C}_2$$, the total number of electrons is 12. The molecular orbital configuration will be $$\\left( \\sigma_{1s} \\right)^2 \\left( \\sigma_{1s}^* \\right)^2 \\left( \\sigma_{2s} \\right)^2 \\left( \\sigma_{2s}^* \\right)^2 \\left( \\pi_{2p_x} \\right)^2 \\left( \\pi_{2p_y} \\right)^2$$. There are 8 bonding electrons and 4 anti-bonding electrons:

\n\n

\n\n

$$\\text{Bond Order} = \\frac{8 - 4}{2} = 2$$

\n\n

\n\n

2. $$\\mathrm{O}_2$$:

\n\n

For $$\\mathrm{O}_2$$, the total number of electrons is 16. The molecular orbital configuration will be $$\\left( \\sigma_{1s} \\right)^2 \\left( \\sigma_{1s}^* \\right)^2 \\left( \\sigma_{2s} \\right)^2 \\left( \\sigma_{2s}^* \\right)^2 \\left( \\sigma_{2p_z} \\right)^2 \\left( \\pi_{2p_x} \\right)^2 \\left( \\pi_{2p_y} \\right)^2 \\left( \\pi_{2p_x}^* \\right)^1 \\left( \\pi_{2p_y}^* \\right)^1$$. There are 10 bonding electrons and 6 anti-bonding electrons:

\n\n

\n\n

$$\\text{Bond Order} = \\frac{10 - 6}{2} = 2$$

\n\n

\n\n

3. $$\\mathrm{Be}_2$$:

\n\n

For $$\\mathrm{Be}_2$$, the total number of electrons is 8. The molecular orbital configuration will be $$\\left( \\sigma_{1s} \\right)^2 \\left( \\sigma_{1s}^* \\right)^2 \\left( \\sigma_{2s} \\right)^2 \\left( \\sigma_{2s}^* \\right)^2$$. There are 4 bonding electrons and 4 anti-bonding electrons:

\n\n

\n\n

$$\\text{Bond Order} = \\frac{4 - 4}{2} = 0$$

\n\n

\n\n

4. $$\\mathrm{Li}_2$$:

\n\n

For $$\\mathrm{Li}_2$$, the total number of electrons is 6. The molecular orbital configuration will be $$\\left( \\sigma_{1s} \\right)^2 \\left( \\sigma_{1s}^* \\right)^2 \\left( \\sigma_{2s} \\right)^2$$. There are 4 bonding electrons and 2 anti-bonding electrons:

\n\n

\n\n

$$\\text{Bond Order} = \\frac{4 - 2}{2} = 1$$

\n\n

\n\n

5. $$\\mathrm{Ne}_2$$:

\n\n

For $$\\mathrm{Ne}_2$$, the total number of electrons is 20. The molecular orbital configuration will be $$\\left( \\sigma_{1s} \\right)^2 \\left( \\sigma_{1s}^* \\right)^2 \\left( \\sigma_{2s} \\right)^2 \\left( \\sigma_{2s}^* \\right)^2 \\left( \\sigma_{2p_z} \\right)^2 \\left( \\pi_{2p_x} \\right)^2 \\left( \\pi_{2p_y} \\right)^2 \\left( \\pi_{2p_x}^* \\right)^2 \\left( \\pi_{2p_y}^* \\right)^2 \\left( \\sigma_{2p_z}^* \\right)^2$$. There are 10 bonding electrons and 10 anti-bonding electrons:

\n\n

\n\n

$$\\text{Bond Order} = \\frac{10 - 10}{2} = 0$$

\n\n

\n\n

6. $$\\mathrm{N}_2$$:

\n\n

For $$\\mathrm{N}_2$$, the total number of electrons is 14. The molecular orbital configuration will be $$\\left( \\sigma_{1s} \\right)^2 \\left( \\sigma_{1s}^* \\right)^2 \\left( \\sigma_{2s} \\right)^2 \\left( \\sigma_{2s}^* \\right)^2 \\left( \\sigma_{2p_z} \\right)^2 \\left( \\pi_{2p_x} \\right)^2 \\left( \\pi_{2p_y} \\right)^2$$. There are 10 bonding electrons and 4 anti-bonding electrons:

\n\n

\n\n

$$\\text{Bond Order} = \\frac{10 - 4}{2} = 3$$

\n\n

\n\n

7. $$\\mathrm{He}_2$$:

\n\n

For $$\\mathrm{He}_2$$, the total number of electrons is 4. The molecular orbital configuration will be $$\\left( \\sigma_{1s} \\right)^2 \\left( \\sigma_{1s}^* \\right)^2$$. There are 2 bonding electrons and 2 anti-bonding electrons:

\n\n

\n\n

$$\\text{Bond Order} = \\frac{2 - 2}{2} = 0$$

\n\n

\n\n

Thus, the molecules with a bond order of 2 from the given list are $$\\mathrm{C}_2$$ and $$\\mathrm{O}_2$$. Therefore, the number of molecules having bond order 2 is 2.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 541, "subject": "Chemistry", "question": "Identify the species having one $$\\pi$$-bond and maximum number of canonical forms from the following :", "options": [ { "text": "SO3" }, { "text": "O2" }, { "text": "SO2" }, { "text": "CO$$_3^{2 - }$$" } ], "answer": "CO$$_3^{2 - }$$", "solution": "**Answer:** CO$$_3^{2 - }$$\n\nAmong SO3, O2, SO2 and CO$$_3^{2 - }$$, only O2 and CO$$_3^{2 - }$$ has only one $$\\pi$$-bond.

\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 542, "subject": "Chemistry", "question": "

The difference in energy between the actual structure and the lowest energy resonance structure for the given compound is

", "options": [ { "text": "electromeric energy\n" }, { "text": "resonance energy\n" }, { "text": "ionization energy\n" }, { "text": "hyperconjugation energy" } ], "answer": "resonance energy\n", "solution": "**Answer:** resonance energy\n\n\n

The difference in energy between the actual structure and the lowest energy resonance structure for the given compound is known as resonance energy.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 543, "subject": "Chemistry", "question": "The pair that contains two P – H bonds in each of the oxoacids is ", "options": [ { "text": "H3PO2 and H4P2O5" }, { "text": "H4P2O5 and H4P2O6" }, { "text": "H4P2O5 and H3PO3" }, { "text": "H3PO3 and H3PO2" } ], "answer": "H3PO2 and H4P2O5", "solution": "**Answer:** H3PO2 and H4P2O5\n\n\"JEE\n\"JEE\n\"JEE\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 544, "subject": "Chemistry", "question": "The lowest oxidation number of an atom in a compound $\\mathrm{A}_2 \\mathrm{B}$ is -2 . The number of electrons in its valence shell is _______.", "options": [], "answer": "6", "solution": "**Answer:** 6\n\n$\\mathrm{A}_2 \\mathrm{~B} \\rightarrow 2 \\mathrm{~A}^{+}+\\mathrm{B}^{-2}$\n

When an atom has the lowest oxidation number of -2 in a compound, it means the atom has gained two electrons beyond its neutral state to achieve this oxidation state. This is because gaining electrons makes the oxidation number more negative. In a neutral atom, the electrons in the outermost shell are known as valence electrons, which play a crucial role in chemical reactions and bonding.

In the given compound $\\mathrm{A}_2\\mathrm{B}$, atom B has an oxidation state of -2, indicating it has gained two electrons. For an atom to gain two electrons to complete its octet implies that its neutral state had six valence electrons. Therefore, before gaining electrons, the atom B would have had six electrons in its valence shell to begin with. As it gains two more electrons, it reaches an oxidation number of -2, implying it now has a complete octet or eight electrons in its valence shell, but the original count of valence electrons in its neutral state was six.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 545, "subject": "Chemistry", "question": "The relative strength of interionic/intermolecular forces in decreasing order is :\n", "options": [ { "text": "dipole-dipole $$>$$ ion-dipole $$>$$ ion-ion" }, { "text": "ion-dipole $$>$$ dipole-dipole $$>$$ ion-ion" }, { "text": "ion-dipole $$>$$ ion-ion $$>$$ dipole-dipole" }, { "text": "ion-ion $$>$$ ion-dipole $$>$$ dipole-dipole" } ], "answer": "ion-ion $$>$$ ion-dipole $$>$$ dipole-dipole", "solution": "**Answer:** ion-ion $$>$$ ion-dipole $$>$$ dipole-dipole\n\nIonic interactions are stronger as compared to\nvan der waal interactions.\n

So, correct order is\n

ion-ion $$>$$ ion-dipole $$>$$ dipole-dipole", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 546, "subject": "Chemistry", "question": "For the reaction CO (g) + (1/2) O2 (g) $$\\leftrightharpoons$$ CO2 (g), Kp/Kc is :", "options": [ { "text": "RT" }, { "text": "(RT)-1" }, { "text": "(RT)-1/2" }, { "text": "(RT)1/2" } ], "answer": "(RT)-1/2", "solution": "**Answer:** (RT)-1/2\n\n$${K_p} = {K_c}{\\left( {RT} \\right)^{\\Delta n}};$$ \n

$$\\Delta n = 1 - \\left( {1 + {1 \\over 2}} \\right)$$\n

$$ = 1 - {3 \\over 2} = - {1 \\over 2}.$$\n

$$\\therefore$$ $$\\,\\,\\,\\,{{{K_p}} \\over {{K_c}}} = {\\left( {RT} \\right)^{ - 1/2}}$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 547, "subject": "Chemistry", "question": "For the reaction equilibrium
\nN2O4 (g) $$\\leftrightharpoons$$ 2NO2 (g)
\nthe concentrations of N2O4 and NO2 at equilibrium are 4.8 $$\\times$$ 10-2 and 1.2 $$\\times$$ 10-2 mol L-1 respectively. The value of Kc for the reaction is", "options": [ { "text": "3 $$\\times$$ 10-1 mol L-1" }, { "text": "3 $$\\times$$ 10-3 mol L-1" }, { "text": "3 $$\\times$$ 103 mol L-1" }, { "text": "3.3 $$\\times$$ 102 mol L-1" } ], "answer": "3 $$\\times$$ 10-3 mol L-1", "solution": "**Answer:** 3 $$\\times$$ 10-3 mol L-1\n\n$${K_c} = {{{{\\left[ {N{O_2}} \\right]}^2}} \\over {\\left[ {{N_2}{O_4}} \\right]}} = {{\\left[ {1.2 \\times {{10}^{ - 2}}} \\right]^2} \\over {\\left[ {4.8 \\times {{10}^{ - 2}}} \\right]}}$$\n

$$ = 3 \\times {10^{ - 3}}\\,mol/L$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 548, "subject": "Chemistry", "question": "For the reaction, CO(g) + Cl2(g) $$\\leftrightharpoons$$ COCl2(g) the $${{{K_p}} \\over {{K_c}}}$$ is equal to :", "options": [ { "text": "$$\\sqrt {RT} $$" }, { "text": "RT" }, { "text": "1/RT" }, { "text": "1.0" } ], "answer": "1/RT", "solution": "**Answer:** 1/RT\n\n$${K_p} = {K_c}{\\left( {RT} \\right)^{\\Delta n}};$$ \n

Here $$\\Delta n = 1 - 2 = - 1$$ \n

$$\\therefore$$ $${{{k_p}} \\over {{K_c}}} = {1 \\over {RT}}$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 549, "subject": "Chemistry", "question": "What is the equilibrium expression for the reaction
\nP4 (s) + 5O2 $$\\leftrightharpoons$$ P4O10 (s)?", "options": [ { "text": "Kc = [P4O10] / 5[P4] [O2] " }, { "text": "Kc = 1/[O2]5" }, { "text": "Kc = [P4O10] / [P4] [O2]5" }, { "text": "Kc = [O2]5" } ], "answer": "Kc = 1/[O2]5", "solution": "**Answer:** Kc = 1/[O2]5\n\nFor $${P_4}\\left( s \\right) + 5{O_2}\\left( g \\right)\\,\\rightleftharpoons\\,{P_4}{O_{10}}\\left( 8 \\right)$$\n

$${K_c} = {1 \\over {{{\\left( {{O_2}} \\right)}^5}}}.$$ The solids have concentration unity", "topic": "Algebra", "subtopic": "Algebraic Expressions" }, { "id": 550, "subject": "Chemistry", "question": "The equilibrium constant for the reaction N2(g) + O2(g) $$\\leftrightharpoons$$ 2NO(g) at temperature T is\n4 $$\\times$$ 10-4. The value of Kc for the reaction NO(g) $$\\leftrightharpoons$$ $$1 \\over 2$$N2 (g) + $$1 \\over 2$$O2 (g) at the same temperature is :", "options": [ { "text": "2.5 $$\\times$$ 102" }, { "text": "4 $$\\times$$ 10-4" }, { "text": "50" }, { "text": "0.02" } ], "answer": "50", "solution": "**Answer:** 50\n\n$${K_c} = {{{{\\left[ {NO} \\right]}^2}} \\over {\\left[ {{N_2}} \\right]\\left[ {{O_2}} \\right]}} = 4 \\times {10^{ - 4}}$$ \n

$$K{'_c} = {{{{\\left[ {{N_2}} \\right]}^{1/2}}{{\\left[ {{Q_2}} \\right]}^{1/2}}} \\over {\\left[ {NO} \\right]}}$$\n

$$ = {1 \\over {\\sqrt {{K_c}} }}$$\n

$$ = {1 \\over {\\sqrt {4 \\times {{10}^{ - 4}}} }}$$\n

$$ = 50$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 551, "subject": "Chemistry", "question": "For the reaction\n2NO2 (g) $$\\leftrightharpoons$$ 2NO (g) + O2 (g), (Kc = 1.8 $$\\times$$ 10-6 at 184oC) (R = 0.0831 kJ/(mol. K))
\nWhen Kp and Kc are compared at 184oC , it is found that :", "options": [ { "text": "Kp is greater than Kc" }, { "text": "Kp is less than Kc" }, { "text": "Kp = Kc" }, { "text": "Whether Kp is greater than, less than or equal to Kc depends upon the total gas\npressure " } ], "answer": "Kp is greater than Kc", "solution": "**Answer:** Kp is greater than Kc\n\nFor the reaction : - \n

$$2N{O_2}\\left( g \\right)\\rightleftharpoons2NO\\left( g \\right) + {O_2}\\left( g \\right)$$ \n

Given $${K_c} = 1.8 \\times {10^{ - 6}}\\,\\,$$ at $$\\,\\,{184^ \\circ }C$$ \n

$$R=0.0831$$ $$\\,\\,kJ/mol.k$$\n

$${K_p} = 1.8 \\times {10^{ - 6}} \\times 0.0831 \\times 457$$\n

$$ = 6.836 \\times {10^{ - 6}}$$ \n

$$\\left[ {} \\right.$$ as $$\\,\\,\\,\\,\\,{184^ \\circ }C = \\left( {273 + 184} \\right) = 457\\,k,\\,\\,$$ \n

$$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\left. {\\Delta n = \\left( {2 + 1, - 1} \\right) = 1} \\right]$$\n

Hence it is clear that $${K_p} > {K_c}$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 552, "subject": "Chemistry", "question": "An amount of solid NH4HS is placed in a flask already containing ammonia gas at a\ncertain temperature and 0.50 atm. Pressure. Ammonium hydrogen sulphide\ndecomposes to yield NH3 and H2S gases in the flask. When the decomposition\nreaction reaches equilibrium, the total pressure in the flask rises to 0.84 atm. The\nequilibrium constant for NH4HS decomposition at this temperature is :", "options": [ { "text": "0.30" }, { "text": "0.11" }, { "text": "0.17" }, { "text": "0.18" } ], "answer": "0.11", "solution": "**Answer:** 0.11\n\n$$\\mathop {N{H_4}HS\\left( s \\right)\\,\\rightleftharpoons\\,}\\limits_{\\matrix{\n {start} \\cr \n {At\\,\\,equib.} \\cr \n\n } } \\,\\,\\mathop {N{H_3}\\left( g \\right)}\\limits_{\\matrix{\n {0.5\\,\\,atm} \\cr \n {0.5 + x\\,\\,atm} \\cr \n\n } } \\,\\, + \\mathop {{H_2}S\\left( g \\right)}\\limits_{\\matrix{\n {0\\,\\,atm} \\cr \n {x\\,\\,atm} \\cr \n\n } } $$\n

Then $$0.5 + x + x = 2x + 0.5 = 0.84\\,\\,$$ (Given)\n

$$ \\Rightarrow x = 0.17\\,\\,atm.$$\n

$${P_{N{H_3}}} = 0.5 + 0.17 = 0.67\\,atm;$$\n

$$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,$$ $${P_H}_{_2S} = 0.17\\,atm$$\n

$$K = {P_{N{H_3}}} \\times {P_{{H_2}S}}$$ \n

$$ = 0.67 \\times 0.17\\,at{m^2}$$\n

$$ = 0.1139 = 0.11$$ ", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 553, "subject": "Chemistry", "question": "Phosphorus pentachloride dissociates as follows, in a closed reaction vessel
\nPCl5 (g) $$\\leftrightharpoons$$ PCl3 (g) + Cl2 (g)
\nIf total pressure at equilibrium of the reaction mixture is P and degree of dissociation of PCl5 is x, the\npartial pressure of PCl3 will be ", "options": [ { "text": "$$\\left( {{x \\over {x + 1}}} \\right)P$$ " }, { "text": "$$\\left( {{2x \\over {1 - x}}} \\right)P$$ " }, { "text": "$$\\left( {{x \\over {x - 1}}} \\right)P$$ " }, { "text": "$$\\left( {{x \\over {1 - x}}} \\right)P$$ " } ], "answer": "$$\\left( {{x \\over {x + 1}}} \\right)P$$ ", "solution": "**Answer:** $$\\left( {{x \\over {x + 1}}} \\right)P$$ \n\n$$\\mathop {PC{l_5}\\left( g \\right)}\\limits_{1 - x} \\mathop {\\,\\rightleftharpoons\\,PC{l_3}\\left( g \\right)}\\limits_x \\,\\, + \\,\\,\\mathop {C{l_2}\\left( g \\right)}\\limits_x $$\n

Total moles after dissociation \n

$$1 - x + x + x = 1 + x$$\n

$${P_{PC{l_3}}} = $$ mole fraction of $$PCl{}_3 \\times $$ Total pressure \n

$$ = \\left( {{x \\over {1 + x}}} \\right)P$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 554, "subject": "Chemistry", "question": "The equilibrium constant for the reaction
\nSO3 (g) $$\\leftrightharpoons$$ SO2 (g) + $$1 \\over 2$$ O2 (g)
\nis Kc = 4.9 $$\\times$$ 10–2. The value of Kc for the reaction
\n2SO2 (g) + O2 (g) $$\\leftrightharpoons$$ 2SO3 (g) will be :", "options": [ { "text": "416" }, { "text": "9.8 $$\\times$$ 10-2" }, { "text": "4.9 $$\\times$$ 10-2" }, { "text": "2.40 $$\\times$$ 10-3" } ], "answer": "416", "solution": "**Answer:** 416\n\n$$S{O_3}\\left( g \\right)\\,\\rightleftharpoons\\,S{O_2}\\left( g \\right)\\,\\, + \\,\\,{1 \\over 2}{O_2}\\left( g \\right)$$\n

$${K_c} = {{\\left[ {S{O_2}} \\right]{{\\left[ {{O_2}} \\right]}^{{\\raise0.5ex\\hbox{$\\scriptstyle 1$}\n\\kern-0.1em/\\kern-0.15em\n\\lower0.25ex\\hbox{$\\scriptstyle 2$}}}}} \\over {\\left[ {S{O_3}} \\right]}}$$\n

$$ = 4.9 \\times {10^{ - 2}};$$\n

On taking the square of the above reaction \n

$${{{{\\left[ {S{O_2}} \\right]}^2}\\left[ {{O_2}} \\right]} \\over {{{\\left[ {S{O_3}} \\right]}^2}}}$$\n

$$ = 24.01 \\times {10^{ - 4}}$$\n

now $$K{'_C}$$ \n

for $$\\,\\,2S{O_2}\\left( g \\right)\\,\\, + \\,\\,{O_2}\\left( g \\right)\\,\\rightleftharpoons\\,2S{O_3}$$\n

$$ = {{{{\\left[ {S{O_3}} \\right]}^2}} \\over {{{\\left[ {S{O_2}} \\right]}^2}\\left[ {{O_2}} \\right]}}$$\n

$$ = {1 \\over {24.01 \\times {{10}^{ - 4}}}}$$\n

$$ = 416$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 555, "subject": "Chemistry", "question": "The equilibrium constants KP1 and KP2 for the reactions X $$\\leftrightharpoons$$ 2Y and Z $$\\leftrightharpoons$$ P + Q, respectively are in\nthe ratio of 1 : 9. If the degree of dissociation of X and Z be equal then the ratio of total pressure at\nthese equilibria is :", "options": [ { "text": "1 : 36" }, { "text": "1 : 1" }, { "text": "1 : 3" }, { "text": "1 : 9" } ], "answer": "1 : 36", "solution": "**Answer:** 1 : 36\n\nLet the initial moles of $$X$$ be $$'a'$$ \n

and that of $$Z$$ be $$'b'$$ the for the given reactions, \n

we have $$X\\,\\,\\,\\,\\,\\,\\,\\,\\,\\rightleftharpoons\\,\\,\\,\\,\\,\\,\\,\\,\\,2Y$$ \n

$$\\eqalign{\n & Initial\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,a\\,\\,moles\\,\\,\\,\\,\\,\\,\\,0 \\cr \n & At\\,\\,equi.\\,\\,\\,\\,\\,\\,\\,a\\left( {1 - \\alpha } \\right)\\,\\,\\,\\,\\,\\,\\,2a\\alpha \\cr \n & (moles) \\cr} $$\n

Total no. of moles $$ = a\\left( {1 - \\alpha } \\right) + 2a\\alpha $$\n

$$ = a - a\\alpha + 2a\\alpha $$ $$ = a\\left( {1 + \\alpha } \\right)$$\n

Now, $$\\,\\,\\,{K_{{p_1}}} = {{{{\\left( {{n_y}} \\right)}^2}} \\over {{n_x}}} \\times {\\left( {{{{P_{{T_1}}}} \\over {\\sum n }}} \\right)^{\\Delta n}}$$\n

or, $$\\,\\,\\,{K_{{p_1}}} = {{{{\\left( {2a\\alpha } \\right)}^2}.{P_{{T_1}}}} \\over {\\left[ {a\\left( {1 - \\alpha } \\right)} \\right]\\left[ {a\\left( {1 + \\alpha } \\right)} \\right]}}$$\n

$$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,$$ $$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,$$ $$Z\\,\\,\\,\\,\\,\\,\\,\\,\\,\\rightleftharpoons\\,\\,\\,\\,\\,\\,\\,\\,\\,P+Q$$ \n

$$\\eqalign{\n & Initial\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,b\\,\\,moles\\,\\,\\,\\,\\,\\,\\,\\,\\,0\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,0 \\cr \n & Ar\\,\\,equi.\\,\\,\\,\\,\\,\\,b\\left( {1 - \\alpha } \\right)\\,\\,\\,\\,\\,\\,\\,\\,b\\alpha \\,\\,\\,\\,\\,\\,\\,b\\alpha \\cr \n & (moles) \\cr} $$\n

Total no. of moles \n

$$ = b\\left( {1 - \\alpha } \\right) + b\\alpha + b\\alpha $$\n

$$ = b - b\\alpha + b\\alpha + b\\alpha $$\n

$$ = b\\left( {1 + \\alpha } \\right)$$\n

Now $$\\,\\,\\,{K_{{P_2}}} = {{{n_Q} \\times {n_P}} \\over {{n_z}}} \\times {\\left[ {{{{P_{{T_2}}}} \\over {\\sum\\nolimits_n \\, }}} \\right]^{\\Delta n}}$$\n

or$$\\,\\,\\,{K_{{P_2}}} = {{\\left( {b\\alpha } \\right)\\left( {b\\alpha } \\right).{P_{{T_2}}}} \\over {\\left[ {b\\left( {1 - \\alpha } \\right)} \\right]\\left[ {b\\left( {1 + \\alpha } \\right)} \\right]}}$$\n

or$$\\,\\,\\,$$ $$\\,\\,\\,{{{K_{{P_1}}}} \\over {{K_{P2}}}} = {{4{\\alpha ^2}.{P_{{T_1}}}} \\over {\\left( {1 - {\\alpha ^2}} \\right)}} \\times {{{{\\left( {1 - \\alpha } \\right)}^2}} \\over {{P_{{T_2}}}.{\\alpha ^2}}}$$ \n

$$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,$$ $$ = {{4{P_{{T_1}}}} \\over {{P_{{T_2}}}}}$$\n
or$$\\,\\,\\,{{{P_{{T_1}}}} \\over {{P_{T2}}}} = {1 \\over 9}\\,\\,\\,\\,\\,$$ [ as $$\\,\\,\\,{{{K_{{P_1}}}} \\over {{K_{{P_1}}}}} = {1 \\over 9}\\,\\,$$ given ]\n

or$$\\,\\,\\,{{{P_{{T_1}}}} \\over {{P_{T2}}}} = {1 \\over {36}}$$ \n

or$$\\,\\,\\,1:36$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 556, "subject": "Chemistry", "question": "For the following three reactions a, b and c, equilibrium constants are given:
\na. CO (g) + H2O (g) $$\\leftrightharpoons$$ CO2(g) + H2 (g) ; K1
\nb. CH4 (g) + H2O (g) $$\\leftrightharpoons$$ CO(g) + 3H2 (g) ; K2
\nc. CH4 (g) + 2H2O (g) $$\\leftrightharpoons$$ CO2(g) + 4H2 (g) ; K3
", "options": [ { "text": "$${K_1}\\sqrt {{K_2}} = {K_3}$$ " }, { "text": "K2K3 = K1" }, { "text": "K3 = K1K2" }, { "text": "K3.$$K_2^3$$ = $$K_1^2$$" } ], "answer": "K3 = K1K2", "solution": "**Answer:** K3 = K1K2\n\nReaction $$(c)$$ can be obtained by adding reactions $$(a)$$ and $$(b)$$ therefore $${K_3} = {K_1}.{K_2}$$\n

Hence $$(c)$$ is the correct answer. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 557, "subject": "Chemistry", "question": "In aqueous solution the ionization constants for carbonic acid are
\nK1 = 4.2 x 10–7 and K2 = 4.8 x 10–11
\nSelect the correct statement for a saturated 0.034 M solution of the carbonic acid.", "options": [ { "text": "The concentration of\n$$CO_3^{2−}$$ is 0.034 M." }, { "text": "The concentration of\n$$CO_3^{2−}$$ is greater than that of $$HCO_3^{−}$$" }, { "text": "The concentration of H+ and $$HCO_3^−$$ are approximately equal." }, { "text": "The concentration of H+ is double that of $$CO_3^−$$." } ], "answer": "The concentration of H+ and $$HCO_3^−$$ are approximately equal.", "solution": "**Answer:** The concentration of H+ and $$HCO_3^−$$ are approximately equal.\n\n$$\\mathop {{H_2}C{O_3}\\left( {aq} \\right)}\\limits_{0.034 - x} \\, + \\,{H_2}O\\left( l \\right)\\,\\rightleftharpoons\\,\\mathop {HCO_3^ - \\left( {aq} \\right)}\\limits_x \\, + \\,\\mathop {{H_3}{O^ + }\\left( {aq} \\right)}\\limits_x $$\n

$${K_1} = {{\\left[ {HCO_3^ - } \\right]\\left[ {{H_3}{O^ + }} \\right]} \\over {\\left[ {{H_2}C{O_3}} \\right]}}$$\n

$$ = {{x \\times x} \\over {0.034 - x}}$$\n

$$ \\Rightarrow 4.2 \\times {10^{ - 7}} = {{{x^2}} \\over {0.034}}$$\n

$$ \\Rightarrow x = 1.195 \\times {10^{ - 4}}$$\n

As $${H_2}C{O_3}$$ is a weak acid so the concentration of \n

$${H_2}C{O_3}$$ will remain $$0.034\\,\\,$$ as $$0.034 > > x.$$ \n

$$x = \\left[ {{H^ + }} \\right] = \\left[ {HCO_3^ - } \\right]$$ \n

$$ = 1.195 \\times {10^{ - 4}}$$ \n

Now, $$\\mathop {HCO_3^ - }\\limits_{x - y} \\left( {aq} \\right) + {H_2}O\\left( l \\right)\\,\\rightleftharpoons\\,\\mathop {CO_3^{2 - }\\left( {aq} \\right)}\\limits_y \\, + \\,\\mathop {{H_3}{O^ + }\\left( {aq} \\right)}\\limits_y $$ \n

As $$HCO_3^ - \\,$$ is again a weak acid (weaker than $${H_2}C{O_3})$$ \n

with $$x > > y.$$ \n

$${K_2} = {{\\left[ {CO_3^{{2^ - }}} \\right]\\left[ {{H_3}{O^ + }} \\right]} \\over {\\left[ {HCO_3^ - } \\right]}}$$ \n

$$ = {{y \\times \\left( {x + y} \\right)} \\over {\\left( {x - y} \\right)}}$$ \n

Note : $$\\left[ {{H_3}{O^ + }} \\right] = {H^ + }\\,\\,$$ from first step $$(x)$$ \n

and from second step $$\\left( y \\right) = \\left( {x + y} \\right)$$\n

[As $$\\,\\,\\,x > > y\\,\\,$$ so $$\\,\\,\\,x + y \\simeq x\\,\\,\\,$$ and $$x - y \\simeq x$$]\n

So, $$\\,\\,\\,{K_2} \\simeq {{y \\times x} \\over x} = y$$ \n

$$ \\Rightarrow {K_2} = 4.8 \\times {10^{ - 11}}$$ \n

$$ = y = \\left[ {CO_3^{{2^ - }}} \\right]$$ \n

So the concentration of $$\\,\\,\\,\\left[ {{H^ + }} \\right] = \\left[ {HCO_3^ - } \\right] = \\,\\,\\,$$ concentrations obtained from the first step. As the dissociation will be very low in second step so there will be no change in these concentrations. \n

Thus the final concentrations are \n

$$\\left[ {{H^ + }} \\right] = \\left[ {HCO_3^ - } \\right] = 1.195 \\times {10^{ - 4}}$$ \n

$$\\& \\,\\,\\,\\left[ {CO_3^{2 - }} \\right] = 4.8 \\times {10^{ - 11}}$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 558, "subject": "Chemistry", "question": "A vessel at 1000 K contains CO2 with a pressure of 0.5 atm. Some of the CO2 is converted into CO on\nthe addition of graphite. If the total pressure at equilibrium is 0.8 atm, the value of K is :", "options": [ { "text": "3 atm " }, { "text": "0.3 atm " }, { "text": "0.18 atm" }, { "text": "1.8 atm " } ], "answer": "1.8 atm ", "solution": "**Answer:** 1.8 atm \n\n$$C{O_2} + {C_{\\left( {grapnite} \\right)}}\\,\\rightleftharpoons\\,2CO$$\n

$${P_{initial}}\\,\\,0.5atm\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,0$$\n

$${P_{final}}\\,\\,\\,\\left( {0.5 - x} \\right)atm\\,\\,\\,\\,\\,\\,\\,2x\\,atm$$\n

Total $$P$$ at equilibrium \n

$$ = 0.5 - x + 2x = 0.5 + x\\,atm$$\n

$$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,0.8 = 0.5 + x$$\n

$$\\therefore$$ $$\\,\\,\\,x = 0.8 - 0.5 = 0.3\\,atm$$\n

Now $$\\,\\,\\,{k_p} = {\\left( {{P_{CO}}} \\right)^2}/{P_{C{O_2}}}$$\n

$$ = {{{{\\left( {2 \\times 0.3} \\right)}^2}} \\over {\\left( {0.5 - 0.3} \\right)}} = {{{{\\left( {0.6} \\right)}^2}} \\over {\\left( {0.2} \\right)}}$$\n

$$ = 1.8\\,atm$$ ", "topic": "Discrete Mathematics", "subtopic": "Other" }, { "id": 559, "subject": "Chemistry", "question": "The equilibrium constant (KC) for the reaction N2(g) + O2(g) $$\\to$$ 2NO(g) at temperature T is 4 $$\\times$$ 10–4. The value of KC for the reaction, NO(g) $$\\to$$ 1/2N2(g) + 1/2O2(g) at the same temperature is : ", "options": [ { "text": "0.02" }, { "text": "2.5 $$\\times$$ 102" }, { "text": "4 $$\\times$$ 10-4" }, { "text": "50.0" } ], "answer": "50.0", "solution": "**Answer:** 50.0\n\nFor the reaction\n

$${N_2} + {O_2} \\to 2NO$$ \n

$$\\,K = 4 \\times {10^{ - 4}}$$\n

Hence for the reaction \n

$$NO \\to {1 \\over 2}{N_2} + {1 \\over 2}{O_2}$$\n

$$K' = {1 \\over {\\sqrt K }}$$\n

$$ = {1 \\over {\\sqrt {4 \\times {{10}^{ - 4}}} }}$$\n

$$ = 50$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 560, "subject": "Chemistry", "question": "For the reaction SO2 (g) + $${1 \\over 2} O_2(g) \\leftrightharpoons$$ SO3(g).
\n if KP = KC(RT)x where the symbols have usual meaning then\nthe value of x is: (assuming ideality)", "options": [ { "text": "-1" }, { "text": "-1/2" }, { "text": "1/2" }, { "text": "1" } ], "answer": "-1/2", "solution": "**Answer:** -1/2\n\n$$S{O_2}\\left( g \\right) + {1 \\over 2}{O_2}\\left( g \\right)\\,\\rightleftharpoons\\,S{O_3}\\left( g \\right)$$\n

$${K_p} = {K_C}{\\left( {RT} \\right)^x}$$ \n

where $$x = \\Delta {n_g} = $$ number of gaseous moles in product \n

$$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,$$$$-$$ number of gaseous moles in reactant \n

$$ = 1 - \\left( {1 + {1 \\over 2}} \\right)$$ \n

$$ = 1 - {3 \\over 2} = - {1 \\over 2}$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 561, "subject": "Chemistry", "question": "The standard Gibbs energy change at 300 K for the reaction 2A $$\\leftrightharpoons$$ B + C is 2494.2 J. At a given time,\nthe composition of the reaction mixture is [A] = 1/2, [B] = 2 and [C] = 1/2. The reaction proceeds in the: [R = 8.314 J/K/mol, e = 2.718]", "options": [ { "text": "reverse direction because Q > Kc" }, { "text": "forward direction because Q < Kc" }, { "text": "reverse direction because Q < Kc" }, { "text": "forward direction because Q > Kc" } ], "answer": "reverse direction because Q > Kc", "solution": "**Answer:** reverse direction because Q > Kc\n\n$$\\Delta {G^ \\circ } = 2494.2J$$ \n

$$2A\\,\\rightleftharpoons\\,B + C$$\n

$$R = 8.314\\,J/K/mol.$$\n

$$e = 2.718$$ \n

$$\\left[ A \\right] = {1 \\over 2},\\left[ B \\right] = 2,$$ \n

$$\\left[ C \\right] = {1 \\over 2};Q = {{\\left[ B \\right]\\left[ C \\right]} \\over {{{\\left[ A \\right]}^2}}}$$\n

$$ = {{2 \\times 1/2} \\over {{{\\left( {{1 \\over 2}} \\right)}^2}}} = 4$$ \n

$$\\Delta {G^ \\circ } = - 2.303\\,\\,RT\\,\\log \\,{K_c}.$$\n

$$2494.2J = - 2.303 \\times \\left( {8.314J/K/mol} \\right)$$ \n

$$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, \\times \\left( {300K} \\right)\\log {K_c}$$ \n

$$ \\Rightarrow \\log \\,{K_c}$$ \n

$$ = - {{2494.2\\,J} \\over {2.303 \\times 8.314\\,J/K/mol \\times 300\\,K}}$$\n

$$ \\Rightarrow \\log \\,{K_c} = - 0.4341;\\,\\,{K_c} = 0.37;\\,\\,Q > {K_c}.$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 562, "subject": "Chemistry", "question": "The equilibrium constant at 298 K for a reaction A + B $$\\leftrightharpoons$$ C + D is 100. If the initial concentration of\nall the four species were 1M each, then equilibrium concentration of D (in mol L–1) will be: ", "options": [ { "text": "0.818" }, { "text": "1.818" }, { "text": "1.182" }, { "text": "0.182" } ], "answer": "1.818", "solution": "**Answer:** 1.818\n\nGiven, \n

\"JEE \n

$$\\therefore$$ $$\\,\\,\\,{K_c} = {\\left( {{{1 + a} \\over {1 - a}}} \\right)^2} = 100$$\n

$$\\therefore$$ $$\\,\\,\\,{{1 + a} \\over {1 - a}} = 10$$ \n

On solving\n

$$a=0.81$$\n

$${\\left[ D \\right]_{At\\,eq}} = 1 + a = 1 + 0.81 = 1.81$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 563, "subject": "Chemistry", "question": "A solid XY kept in an evacuated sealed container undergoes decomposition to form a mixture of gases X and Y at temperature T. The equilibrium pressure is 10 bar in this vessel. Kp for this reaction is :\n", "options": [ { "text": "5" }, { "text": "10" }, { "text": "25" }, { "text": "100" } ], "answer": "25", "solution": "**Answer:** 25\n\n

To determine the equilibrium constant, $K_p$, for the decomposition reaction of the solid compound XY into its gaseous components X and Y, we need to consider the following reaction:

\n\n

\n\n

$$ \\text{XY (s)} \\rightleftharpoons \\text{X (g)} + \\text{Y (g)} $$

\n\n

\n\n

For a reaction where a solid decomposes into gases, the equilibrium constant $K_p$ is defined in terms of the partial pressures of the gaseous products. Since XY is a solid, its activity is considered to be 1 and does not appear in the equilibrium expression. Hence the expression for $K_p$ is:

\n\n

\n\n

$$ K_p = P_X \\cdot P_Y $$

\n\n

\n\n

Where $P_X$ and $P_Y$ are the partial pressures of the gases X and Y, respectively.

\n\n

Given that the total pressure at equilibrium is 10 bar, and assuming that X and Y are present in equal amounts due to the stoichiometry of the decomposition reaction (i.e., 1:1), we can write:

\n\n

\n\n

$$ P_X = P_Y = \\frac{10}{2} = 5 \\, \\text{bar} $$

\n\n

\n\n

Substituting these partial pressures into the expression for $K_p$ gives:

\n\n

\n\n

$$ K_p = P_X \\cdot P_Y = 5 \\, \\text{bar} \\cdot 5 \\, \\text{bar} = 25 \\, \\text{bar}^2 $$

\n\n

\n\n

Therefore, the equilibrium constant $K_p$ for this reaction is 25. Thus, the correct answer is:

\n\n

Option C - 25

", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 564, "subject": "Chemistry", "question": "At a certain temperature in a $$5$$ $$L$$ vessel, 2 moles of carbon monoxide and 3 moles of chlorine were allowed to reach equilibrium according to the reaction, \n
         CO + Cl2 $$\\rightleftharpoons$$ COCl2\n
At equilibrium, if one mole of CO is present then equilibrium constant (Kc) for the reaction is : ", "options": [ { "text": "2" }, { "text": "2.5" }, { "text": "3" }, { "text": "4" } ], "answer": "2.5", "solution": "**Answer:** 2.5\n\n\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n
CO+Cl2$$\\rightleftharpoons$$COCl2
Initially number of moles230
At equilibrium number of moles121
\n

The equilibrium constant,\n

Kc = $${{\\left[ {COC{l_2}} \\right]} \\over {\\left[ {CO} \\right]\\left[ {C{l_2}} \\right]}}$$\n

=  $${{\\left( {{1 \\over 5}} \\right)} \\over {\\left( {{1 \\over 5}} \\right) \\times \\left( {{2 \\over 5}} \\right)}}$$\n

=   $${5 \\over 2}$$\n

=  2.5", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 565, "subject": "Chemistry", "question": "In which one of the following equilibria, Kp $$ \\ne $$ KC ?", "options": [ { "text": "2NO(g) ⇋ N2(g) + O2(g)" }, { "text": "2C(s) + O2(g) ⇋ 2CO(g)" }, { "text": "2HI(g) ⇋ H2(g) + I2(g)" }, { "text": "NO2(g) + SO2(g) ⇋ NO(g) + SO3(g)" } ], "answer": "2C(s) + O2(g) ⇋ 2CO(g)", "solution": "**Answer:** 2C(s) + O2(g) ⇋ 2CO(g)\n\nWe know,\n
Kp = KC(RT)$$\\Delta $$ng\n

Kp = KC when $$\\Delta $$ng = 0\n
Kp $$ \\ne $$ KC when $$\\Delta $$ng $$ \\ne $$ 0 and T $$ \\ne $$ 12 K\n

In this reaction, 2C(s) + O2(g) ⇋ 2CO(g)\n
$$\\Delta $$ng = 1, so Kp $$ \\ne $$ KC", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 566, "subject": "Chemistry", "question": "For the following reactions, equilibrium\nconstants are given :
\n
S(s) + O2(g) ⇋ SO2(g); K1 = 1052
\n
2S(s) + 3O2(g) ⇋ 2SO3(g); K2 = 10129
\n
The equilibrium constant for the reaction,
\n
2SO2(g) + O2(g) ⇋ 2SO3(g) is :", "options": [ { "text": "10181" }, { "text": "1025" }, { "text": "1077" }, { "text": "10154" } ], "answer": "1025", "solution": "**Answer:** 1025\n\nS(s) + O2(g) ⇋ SO2(g); K1 = 1052\n

By reversing the equation, we get\n

SO2(g); ⇋ S(s) + O2(g) ; $${1 \\over {{K_1}}} = {1 \\over {{{10}^{52}}}}$$ ..............(1)\n

2S(s) + 3O2(g) ⇋ 2SO3(g); K2 = 10129 ....................(2)\n

Performing (2) - 2 $$ \\times $$ (1), we get\n

2SO2(g) + O2(g) ⇋ 2SO3(g)\n

$$ \\therefore $$ Equilibrium constant of this reaction is\n

= $${{{K_2}} \\over {K_1^2}}$$ = $${{{{10}^{129}}} \\over {{{\\left( {{{10}^{52}}} \\right)}^2}}}$$ = 1025", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 567, "subject": "Chemistry", "question": "Consider the reaction \n
N2(g) + 3H2(g) $$\\rightleftharpoons$$ 2NH3(g)\n

The equilibrium constant of the above reaction is Kp. If pure ammonia is left to dissociate, the partial pressure of ammonia at equilibrium is given by (Assume that PNH3 << Ptotal at equilibrium) ", "options": [ { "text": "$${{{3^{{3 \\over 2}}}{K_P^{{1 \\over 2}}}{P^2}} \\over 4}$$" }, { "text": "$${{K_P^{{1 \\over 2}}{P^2}} \\over 4}$$" }, { "text": "$${{{3^{{3 \\over 2}}}{K_P^{{1 \\over 2}}}{P^2}} \\over 16}$$" }, { "text": "$${{K_P^{{1 \\over 2}}{P^2}} \\over 16}$$" } ], "answer": "$${{{3^{{3 \\over 2}}}{K_P^{{1 \\over 2}}}{P^2}} \\over 16}$$", "solution": "**Answer:** $${{{3^{{3 \\over 2}}}{K_P^{{1 \\over 2}}}{P^2}} \\over 16}$$\n\nN2(g) + 3H2(g) $$\\rightleftharpoons$$ 2NH3(g) ; Keq = Kp\n

Write this equation reverse way,\n

2NH3(g)  $$\\rightleftharpoons$$ N2(g) + 3H2(g) ; Keq = $${1 \\over {{K_p}}}$$\n

\n\n\n \n \n \n \n \n \n \n \n\n\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\n
2NH3(g)⇌N2(g)+3H2(g)
At t = 0Po00
At t = teqPNH3p3p
\n
At equillibrium\n

PTotal = PNH3 + PN2 + PH2\n

= PNH3 + p + 3p\n

(As PNH3 << Ptotal so we can ignore PNH3)\n

$$ \\therefore $$ PTotal = 4p\n

$$ \\Rightarrow $$ p = $${{{{P_{total}}} \\over 4}}$$\n

Formula of\n
Keq = $${{{p_{{N_2}}} \\times {{\\left( {{p_{{H_2}}}} \\right)}^3}} \\over {{{\\left( {{p_{N{H_3}}}} \\right)}^2}}}$$ = $${1 \\over {{K_p}}}$$\n

$$ \\Rightarrow $$ $${1 \\over {{K_p}}}$$ = $${{p \\times 27{p^3}} \\over {{{\\left( {{p_{N{H_3}}}} \\right)}^2}}}$$\n

$$ \\Rightarrow $$ $${{{\\left( {{p_{N{H_3}}}} \\right)}^2}}$$ = Kp $$ \\times $$ 27 $$ \\times $$ $${{{\\left( {{{{P_{total}}} \\over 4}} \\right)}^4}}$$\n

$$ \\Rightarrow $$ PNH3 = $$\\sqrt {{K_p}} \\times {\\left( {27} \\right)^{{1 \\over 2}}} \\times {\\left( {{{{P_{Total}}} \\over 4}} \\right)^{{4 \\over 2}}}$$\n

= $${{{3^{{3 \\over 2}}}K_p^{{1 \\over 2}}P_{Total}^2} \\over {16}}$$\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 568, "subject": "Chemistry", "question": "5.1 g NH4SH is introduced in 3.0 L evacuated flask at 327ºC. 30% of the solid NH4SH decomposed to NH3 and H2S as gases . The Kp of the reaction at 327oC is (R = 0.082 L atm mol–1 K–1, Molar mass of S = 32 g mol–1 molar mass of N = 14 g mol–1) \n", "options": [ { "text": "0.242 $$ \\times $$ 10$$-$$4 atm2" }, { "text": "1 $$ \\times $$ 10–4 atm2\n" }, { "text": "4.9 $$ \\times $$ 10$$-$$3 atm2" }, { "text": "0.242 atm2" } ], "answer": "0.242 atm2", "solution": "**Answer:** 0.242 atm2\n\n   NH4SH(s)  $$\\rightleftharpoons$$    NH3(g) + H2S(g)\n

$$n = {{5.1} \\over {51}} = .1\\,mole\\,\\,\\,\\,\\,\\,\\,\\,0\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,0\\,\\,$$\n

$$.1\\left( { - 1 - \\alpha } \\right)\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,.1\\alpha \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,.1\\alpha $$\n

$$\\alpha \\,\\, = \\,\\,30\\% = .3$$\n

so number of moles at equilibrium\n

$$\\,\\,\\,\\,\\,\\,\\,.1\\,(1 - .3)\\,\\,\\,\\,\\,.1\\,\\, \\times \\,\\,.3\\,\\,\\,\\,\\,\\,\\,\\,\\,.1\\,\\, \\times \\,.3$$\n

$$ = \\,\\,\\,\\,.07\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, = .03\\;\\,\\,\\,\\,\\,\\,\\,\\,\\, = .03$$\n

Now use PV = nRT at equilibrium\n

Ptotal $$ \\times $$ 3 lit = (.03 + .03) $$ \\times $$ .082 $$ \\times $$ 600\n

Ptotal = .984 atm\n

At equilibrium\n

PNH3 = PH2S = $${{{P_{total}}} \\over 2}$$ = .492\n

So kp = PNH3 . PH2S = (.492) (.492)\n

kp = .242 atm2", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 569, "subject": "Chemistry", "question": "The values of Kp/Kc for the following reactions at 300 K are, respectively : (At 300 K, RT = 24.62 dm3 atm mol–1) \n

N2(g) + O2(g)  $$\\rightleftharpoons$$ 2 NO(g)\n

N2O4(g)  $$\\rightleftharpoons$$   2 NO(g)\n

N2(g) + 3H2(g) $$\\rightleftharpoons$$ 2 NH3(g)", "options": [ { "text": "24.62 dm3 atm mol–1, 606.0 dm6 atm2 mol–2 1.65 $$ \\times $$ 10–3 dm–6 atm–2 mol2" }, { "text": "1,4.1 $$ \\times $$ 10–2 dm–3 atm–1 mol, 606 dm6 atm2 mol–2" }, { "text": "1,24.62 dm3 atm mol–1 , 1.65 $$ \\times $$ 10–3 dm–6 atm –2 mol2" }, { "text": "1, 24.62 dm3 atm mol–1, 606.0 dm6 atm2 mol–2\n" } ], "answer": "1,24.62 dm3 atm mol–1 , 1.65 $$ \\times $$ 10–3 dm–6 atm –2 mol2", "solution": "**Answer:** 1,24.62 dm3 atm mol–1 , 1.65 $$ \\times $$ 10–3 dm–6 atm –2 mol2\n\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 570, "subject": "Chemistry", "question": "For a reaction X + Y ⇌ 2Z , 1.0 mol of X, 1.5 mol
of Y and 0.5 mol of Z were taken in a 1 L\nvessel and
allowed to react. At equilibrium, the concentration
of Z was 1.0 mol L–1. The equilibrium constant of reaction
is $${x \\over {15}}$$. The value of x is _________.", "options": [], "answer": "16", "solution": "**Answer:** 16\n\n\"JEE\n

Keq = $${{{{\\left( 1 \\right)}^2}} \\over {{3 \\over 4} \\times {5 \\over 4}}}$$ = $${{16} \\over {15}}$$\n

$$ \\therefore $$ x = 16", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 571, "subject": "Chemistry", "question": "The value of KC is 64 at 800 K for the reaction\n

N2(g) + 3H2(g) ⇌ 2NH3(g)\n

The value of KC for the following reaction is :\n

NH3(g) ⇌ $${1 \\over 2}$$N2(g) + $${3 \\over 2}$$H2(g)", "options": [ { "text": "8" }, { "text": "$${1 \\over 8}$$" }, { "text": "$${1 \\over 4}$$" }, { "text": "$${1 \\over {64}}$$" } ], "answer": "$${1 \\over 8}$$", "solution": "**Answer:** $${1 \\over 8}$$\n\nN2(g) + 3H2(g) ⇌ 2NH3(g) ; KC\n

2NH3(g) ⇌ N2(g) + 3H2(g) ; $${1 \\over {{K_C}}}$$\n

Multiplying by $${1 \\over 2}$$, reaction becomes\n

NH3(g) ⇌ $${1 \\over 2}$$N2(g) + $${3 \\over 2}$$H2(g) ; \n

$$ \\therefore $$ New KC = $${\\left( {{1 \\over {{K_C}}}} \\right)^{{1 \\over 2}}}$$ = $${\\left( {{1 \\over {64}}} \\right)^{{1 \\over 2}}}$$ = $${1 \\over 8}$$\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 572, "subject": "Chemistry", "question": "The variation of equilibrium constant with\ntemperature is given below :\n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
TemperatureEquilibrium Constant
T1 = 25oCK1 = 10
T2 = 100oCK2 = 100
\n
The values of $$\\Delta $$Ho, $$\\Delta $$Go at
T1 and $$\\Delta $$Go at T2\n(in kJ mol–1) respectively, are close to :
[Use\nR = 8.314 J K–1 mol–1]", "options": [ { "text": "28.4, –5.71 and –14.29" }, { "text": "0.64, –7.14 and –5.71" }, { "text": "28.4, –7.14 and –5.71" }, { "text": "0.64, –5.71 and –14.29" } ], "answer": "28.4, –5.71 and –14.29", "solution": "**Answer:** 28.4, –5.71 and –14.29\n\nln $$\\left[ {{{{k_2}} \\over {{k_1}}}} \\right]$$ = $${{\\Delta H^\\circ } \\over R}\\left\\{ {{1 \\over {{T_1}}} - {1 \\over {{T_2}}}} \\right\\}$$\n

$$ \\Rightarrow $$ ln(10) = $${{\\Delta H^\\circ } \\over R}\\left\\{ {{1 \\over {298}} - {1 \\over {373}}} \\right\\}$$\n

$$ \\Rightarrow $$ $${\\Delta H^\\circ }$$ = 28.37 kJ/mol\n

$$\\Delta $$Go = –RT ln K\n

T1 = 25oC K1 = 10\n

$$\\Delta $$Go at T1 = –8.314 × 298 × 2.303 × log 10\n

= –5.71 kJ/mol\n

$$\\Delta $$Go at T2 = –8.314 × 373 × 2.303 × log(100)\n

= –14.29 kJ/mol\n", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 573, "subject": "Chemistry", "question": "For the reaction\n

Fe2N(s) + $${3 \\over 2}$$H2(g) ⇌ 2Fe(s) + NH3(g)", "options": [ { "text": "KC = Kp(RT)1/2" }, { "text": "KC = Kp(RT)-1/2" }, { "text": "KC = Kp(RT)" }, { "text": "KC = Kp(RT)3/2" } ], "answer": "KC = Kp(RT)1/2", "solution": "**Answer:** KC = Kp(RT)1/2\n\nKP = KC(RT)$$\\Delta $$ng\n

= KC(RT)$$1 - {3 \\over 2}$$\n

= KC(RT)$$ - {1 \\over 2}$$\n

$$ \\Rightarrow $$ KC = Kp(RT)1/2", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 574, "subject": "Chemistry", "question": "Consider the following reaction:\n

N2O4(g) ⇌ 2NO2(g); $$\\Delta $$Ho = +58 kJ\n

For each of the following cases (a, b), the direction in which the equilibrium shifts is :\n
(a) Temperature is decreased.\n
(b) Pressure is increased by adding N2\n at constant T.", "options": [ { "text": "(a) towards reactant, (b) towards product" }, { "text": "(a) towards reactant, (b) no change" }, { "text": "(a) towards product, (b) towards reactant" }, { "text": "(a) towards product, (b) no change" } ], "answer": "(a) towards reactant, (b) no change", "solution": "**Answer:** (a) towards reactant, (b) no change\n\n$$ \\because $$ Given reaction is endothermic.\n

$$ \\therefore $$ On decreasing temperature backward\nreaction will be favoured.\n

On adding N2, pressure is increased at\nconstant T, and volume would also be constant\nso no change is observed.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 575, "subject": "Chemistry", "question": "If the equilibrium constant for
A ⇌ B + C is $$K_{eq}^{(1)}$$ and that of\n
B + C ⇌ P is $$K_{eq}^{(2)}$$, the equilibrium\n
constant for A ⇌ P is :", "options": [ { "text": "$${{K_{eq}^{(1)}} \\over {K_{eq}^{(2)}}}$$" }, { "text": "$${K_{eq}^{(1)}}$$ + $${K_{eq}^{(2)}}$$" }, { "text": "$${K_{eq}^{(2)}}$$ - $${K_{eq}^{(1)}}$$" }, { "text": "$${K_{eq}^{(1)}}$$ $${K_{eq}^{(2)}}$$" } ], "answer": "$${K_{eq}^{(1)}}$$ $${K_{eq}^{(2)}}$$", "solution": "**Answer:** $${K_{eq}^{(1)}}$$ $${K_{eq}^{(2)}}$$\n\nA ⇌ B + C $$K_{eq}^{(1)}$$ = $${{\\left[ B \\right]\\left[ C \\right]} \\over {\\left[ A \\right]}}$$ ....(1)\n

B + C ⇌ P $$K_{eq}^{(2)}$$ = $${{\\left[ P \\right]} \\over {\\left[ B \\right]\\left[ C \\right]}}$$ ....(2)\n

For A ⇌ P    Keq = $${{\\left[ P \\right]} \\over {\\left[ A \\right]}}$$\n

Multiplying equation (1) & (2),\n

$$K_{eq}^{(1)}$$ $$ \\times $$ $$K_{eq}^{(2)}$$ = $${{\\left[ P \\right]} \\over {\\left[ A \\right]}}$$ = Keq", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 576, "subject": "Chemistry", "question": "An open beaker of water in equilibrium with\nwater vapour is in a sealed container. When a\nfew grams of glucose are added to the beaker\nof water, the rate at which water molecules :", "options": [ { "text": "leaves the solution increases" }, { "text": "leaves the vapour increases" }, { "text": "leaves the vapour decreases" }, { "text": "leaves the solution decreases" } ], "answer": "leaves the solution decreases", "solution": "**Answer:** leaves the solution decreases\n\nWith addition of solute in solvent, surface area for vapourisation decreases causes lowering in vapour\npressure.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 577, "subject": "Chemistry", "question": "At 1990 K and 1 atm pressure, there are equal number of Cl2, molecules and Cl atoms in the\nreaction mixture. The value of Kp for the reaction Cl2 (g) $$ \\rightleftharpoons $$ 2Cl(g) under the above conditions\nis x $$ \\times $$ 10-1.
\nThe value of x is _______. (Rounded off to the nearest integer)", "options": [], "answer": "5", "solution": "**Answer:** 5\n\nCl2(g) $$\\rightleftharpoons$$ 2Cl(g)

Let moles of both of Cl2 and Cl molecule be x.

Partial pressure of Cl is, $${p_{Cl}} = {x \\over {2x}} \\times 1 = {1 \\over 2}$$

Partial pressure of Cl2 is, $${p_{C{l_2}}} = {x \\over {2x}} \\times 1 = {1 \\over 2}$$

Now, $${K_p} = {{{{({p_{Cl}})}^2}} \\over {{p_{C{l_2}}}}} $$\n

$$\\Rightarrow {K_p} = {{{{(1/2)}^2}} \\over {1/2}} = {1 \\over 2} = 0.5$$

= 5 $$\\times$$ 10$$-$$1

Hence, x $$\\times$$ 10$$-$$1

x = 5", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 578, "subject": "Chemistry", "question": "For the reaction A(g) $$ \\to $$ B(g) the value of the equilibrium constant at 300 K and 1 atm is equal to 100.0. The value of $$\\Delta $$rG for the reaction at 300 K and 1 atm in J mol-1 is – xR, where x is _______. (Rounded off to the nearest integer)
\n[R = 8.31 J mol–1K-1 and ln 10 = 2.3)", "options": [], "answer": "1380", "solution": "**Answer:** 1380\n\nFor a reaction, A(g) $$\\to$$ B(g)

Given, Kp (equilibrium constant) = 100

Temperature = 300 K

Pressure = 1 atm

Formula used, $$\\Delta$$G$$^\\circ$$ = $$-$$ RT ln Kp .... (i)

Here, $$\\Delta$$G$$^\\circ$$ = standard Gibb's free energy

R = gas constant = 8.31 J mol$$-$$1 K$$-$$1

Put value in Eq. (i), we get

$$\\Delta$$G$$^\\circ$$ = $$-$$ R (300) ln 100

$$\\Delta$$G$$^\\circ$$ = $$-$$ R (300) (2) ln (10)

$$\\because$$ ln (10) = 2.3

$$\\Delta$$G$$^\\circ$$ = $$-$$ R(300) (2) (2.3)

$$\\Delta$$G$$^\\circ$$ = $$-$$ 1380 R

Hence, $$\\Delta$$G$$^\\circ$$ = $$-$$ xR

$$ \\therefore $$ x = 1380", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 579, "subject": "Chemistry", "question": "A homogeneous ideal gaseous reaction $$A{B_{2(g)}} \\rightleftharpoons {A_{(g)}} + 2{B_{(g)}}$$ is carried out in a 25 litre flask at 27$$^\\circ$$C. The initial amount of AB2 was 1 mole and the equilibrium pressure was 1.9 atm. The value of Kp is x $$\\times$$ 10$$-$$2. The value of x is _________. (Integer answer)

[R = 0.08206 dm3atm K$$-$$1mol$$-$$1]", "options": [], "answer": "72TO75", "solution": "**Answer:** 72TO75\n\n

\"JEE

\n

[p = Total pressure at equilibrium = 1.9 atm]

\n

Now, at equilibrium pV = (1 + 2x)RT

\n

$$ \\Rightarrow 1 + 2x = {{pV} \\over {RT}} = {{1.9 \\times 25} \\over {0.082 \\times 300}} = 1.93$$

\n

[V = 25 L, R = 0.082 L atm mol$$-$$1 K$$-$$1 T = 300 K]

\n

$$ \\Rightarrow x = {{1.93 - 1} \\over 2} = 0.465$$

\n

$$ \\Rightarrow {K_p} = {{{p_A} \\times p_B^2} \\over {{p_{A{B_2}}}}} \\Rightarrow {{\\left( {{x \\over {1 + 2x}}p} \\right) \\times {{\\left( {{{2x} \\over {1 + 2x}}p} \\right)}^2}} \\over {\\left( {{{1 - x} \\over {1 + 2x}}p} \\right)}}$$

\n

$$ = {{4{x^3} \\times {p^3}} \\over {{{(1 + 2x)}^3}}} \\times {{(1 + 2x)} \\over {(1 - x) \\times p}} = {{4{x^3} \\times {p^2}} \\over {{{(1 + 2x)}^2} \\times (1 - x)}}$$

\n

$$ = {{4 \\times {{(0.465)}^3} \\times {{(1.9)}^2}} \\over {{{(1 + 2 \\times 0.465)}^2} \\times (1 - 0.465)}} = 0.7285$$ atm

\n

$$ = 72.85 \\times {10^{ - 2}}$$ atm $$ \\simeq 73 \\times {10^{ - 2}} = x \\times {10^{ - 2}}$$

\n

$$\\therefore$$ $$x = 73$$

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 580, "subject": "Chemistry", "question": "For the reaction $$A(g) \\rightleftharpoons B(g)$$ at 495 K, $$\\Delta$$rG$$^\\circ$$ = $$-$$9.478 kJ mol$$-$$1.
If we start the reaction in a closed container at 495 K with 22 millimoles of A, the amount of B in the equilibrium mixture is ____________ millimoles.
(Round off to the Nearest Integer). [R = 8.314 J mol$$-$$1 K$$-$$1; ln 10 = 2.303]", "options": [], "answer": "20", "solution": "**Answer:** 20\n\n$$\\Delta$$Go = $$-$$RT ln Keq

$$-$$9.478 $$\\times$$ 103 = $$-$$495 $$\\times$$ 8.314 ln Keq

ln Keq = 2.303 = ln 10

So, Keq = 10

Now, A(g) $$\\rightleftharpoons$$ B(g)

$$\\matrix{\n {t = 0} & {22} & 0 \\cr \n {t = t} & {22 - x} & x \\cr \n\n } $$

$$Keq = {{[B]} \\over {[A]}} = {x \\over {(22 - x)}} = 10$$

x = 20

So, millimoles of B = 20", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 581, "subject": "Chemistry", "question": "0.01 moles of a weak acid HA (Ka = 2.0 $$\\times$$ 10$$-$$6) is dissolved in 1.0 L of 0.1 M HCl solution. The degree of dissociation of HA is __________ $$\\times$$ 10$$-$$5 (Round off to the Nearest Integer).

[Neglect volume change on adding HA. Assume degree of dissociation <<1 ]", "options": [], "answer": "2", "solution": "**Answer:** 2\n\n\"JEE

Now,

$${K_a} = {{[{H^ + }][{A^ - }]} \\over {[HA]}}$$

$$ \\Rightarrow 2 \\times {10^{ - 6}} = {{(0.1)({{10}^{ - 2}}\\alpha )} \\over {{{10}^{ - 2}}}}$$

$$ \\Rightarrow \\alpha = 2 \\times {10^{ - 5}}$$", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 582, "subject": "Chemistry", "question": "Consider the reaction

$$N_{2}O_{4}\\left( g\\right) \\rightleftharpoons 2NO_{2}\\left( g\\right) $$

The temperature at which KC = 20.4 and KP = 600.1, is ____________ K. (Round off to the Nearest Integer). [Assume all gases are ideal and R = 0.0831 L bar K$$-$$1 mol$$-$$1]", "options": [], "answer": "354", "solution": "**Answer:** 354\n\nN2O4(g) $$\\rightleftharpoons$$ 2NO2(g)

$$\\Delta$$ng = 2 $$-$$ 1 = 1

KP = KC(RT)$$\\Delta$$ng

600.1 = 20.4 (0.0831 $$\\times$$ T)1

$$ \\Rightarrow $$ T = $${{600.1} \\over {20.4 \\times 0.0831}}$$ = 354 K", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 583, "subject": "Chemistry", "question": "The gas phase reaction $$2A(g) \\rightleftharpoons {A_2}(g)$$ at 400 K has $$\\Delta$$Go = + 25.2 kJ mol-1.

The equilibrium constant KC for this reaction is ________ $$\\times$$ 10$$-$$2. (Round off to the Nearest Integer).

[Use : R = 8.3 J mol$$-$$1 K$$-$$1, ln 10 = 2.3 log10 2 = 0.30, 1 atm = 1 bar]

[antilog ($$-$$0.3) = 0.501]", "options": [], "answer": "2", "solution": "**Answer:** 2\n\nUsing formula,

$$\\Delta$$G$$^\\circ$$ = $$-$$ RTln(Kp)

$$ \\Rightarrow $$ 25.2 $$\\times$$ 103 = $$-$$8.3 $$\\times$$ 400 $$\\times$$ 2.3 log (Kp)

$$ \\Rightarrow $$ Kp = 10$$-$$3.3

= 10$$-$$3 $$\\times$$ 0.501

= 5.01 $$\\times$$ 10$$-$$4 Bar$$-$$1

Also,

$${{{K_p}} \\over {{K_c}}} = {(RT)^{\\Delta {n_g}}}$$

$$ \\Rightarrow {{{K_p}} \\over {{K_c}}} = {(RT)^{ - 1}}$$

$$ \\Rightarrow {K_c} = {K_p}(RT)$$

$$ = 5.01 \\times {10^{ - 4}} \\times 8.3 \\times 400$$

$$ = 1.66 \\times {10^{ - 5}}$$ m3/mole

$$ = 1.66 \\times {10^{ - 2}}$$ L/mol", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 584, "subject": "Chemistry", "question": "2SO2(g) + O2(g) $$\\rightleftharpoons$$ 2SO3(g)

In an equilibrium mixture, the partial pressures are

PSO3 = 43 kPa; PO2 = 530 Pa and PSO2 = 45 kPa. The equilibrium constant KP = ___________ $$\\times$$ 10$$-$$2. (Nearest integer)", "options": [], "answer": "172", "solution": "**Answer:** 172\n\nOn reaction, 2SO2(g) + O2(g) $$\\rightarrow$$ 2SO3(g)

Given values are : pSO3 = 45kPa, pSO2 = 530 Pa = 0.53 kPa

pSO2 = 43 kPa

Now, $${K_p} = {{{{[{p_{S{O_3}(g)}}]}^2}} \\over {{{[{p_{S{O_2}(g)}}]}^2} \\times [{p_{{O_2}}}]}}$$

On putting given values, we get

$$ \\Rightarrow {K_p} = {{{{(43)}^2}} \\over {{{(45)}^2} \\times 0.53}}$$

$$ = {{1849} \\over {2025 \\times 0.53}} = {{1849} \\over {1073.25}}$$

$$ = 1.7228$$

$$ = {{1.7228 \\times {{10}^2}} \\over {{{10}^2}}} = 172.28 \\times {10^{ - 2}} = 172$$

Hence, the equilibrium constant, Kp = 172.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 585, "subject": "Chemistry", "question": "Value of KP for the equilibrium reaction

N2O4(g) $$\\rightleftharpoons$$ 2NO2(g) at 288 K is 47.9. The KC for this reaction at same temperature is ____________. (Nearest integer)

(R = 0.083 L bar K$$-$$1 mol$$-$$1)", "options": [], "answer": "2", "solution": "**Answer:** 2\n\n$${K_C} = {{{K_P}} \\over {RT}} = {{47.9} \\over {0.083 \\times 288}} = 2$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 586, "subject": "Chemistry", "question": "For the reaction

A + B $$\\rightleftharpoons$$ 2C

the value of equilibrium constant is 100 at 298 K. If the initial concentration of all the three species is 1 M each, then the equilibrium concentration of C is x $$\\times$$ 10$$-$$1 M. The value of x is ____________. (Nearest integer)", "options": [], "answer": "25", "solution": "**Answer:** 25\n\n\"JEE

$$K = {{[C]_{eq}^2} \\over {{{[A]}_{eq}}{{[B]}_{eq}}}} = {{{{(1 + 2x)}^2}} \\over {(1 - x)(1 - x)}}$$

$$100 = {\\left( {{{1 + 2x} \\over {1 - x}}} \\right)^2}$$

$$\\left( {{{1 + 2x} \\over {1 - x}}} \\right) = 10$$

$$x = {3 \\over 4}$$

$$[C]{e_{q.}} = 1 + 2x$$

$$ = 1 + 2\\left( {{3 \\over 4}} \\right)$$

= 2.5 M

= 25 $$\\times$$ 10-1 M", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 587, "subject": "Chemistry", "question": "Assuming that Ba(OH)2 is completely ionised in aqueous solution under the given conditions the concentration of H3O+ ions in 0.005 M aqueous solution of Ba(OH)2 at 298 K is ______________ $$\\times$$ 10$$-$$12 mol L$$-$$1. (Nearest integer)", "options": [], "answer": "1", "solution": "**Answer:** 1\n\n$$Ba{(OH)_2} \\to B{a^{ + 2}} + 2O{H^ - } \\downarrow 2 \\times 0.005 = 0.01 = {10^{ - 2}}$$

At 298 K : in aq. solution $$[{H_3}{O^ + }][O{H^ - }] = {10^{ - 14}}$$

$$[{H_3}{O^ + }] = {{{{10}^{ - 14}}} \\over {{{10}^{ - 2}}}} = {10^{ - 12}}$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 588, "subject": "Chemistry", "question": "PCl5 $$\\rightleftharpoons$$ PCl3 + Cl2

Kc = 1.844

3.0 moles of PCl5 is introduced in a 1 L closed reaction vessel at 380 K. The number of moles of PCl5 at equilibrium is ______________ $$\\times$$ 10$$-$$3. (Round off to the Nearest Integer)", "options": [], "answer": "1400", "solution": "**Answer:** 1400\n\nPCl5(g) $$\\rightleftharpoons$$ PCl3(g) + Cl2(g) K2 = 1.844

t = 0 3moles

t = $$\\infty$$ x x

$$ \\Rightarrow {{[PC{l_3}][C{l_2}]} \\over {[PC{l_5}]}} = {{{x^2}} \\over {3 - x}} = 1.844$$

$$ \\Rightarrow {x^2} + 1.844 - 5.532 = 0$$

$$ \\Rightarrow x = {{ - 1.844 + \\sqrt {{{(1.844)}^2} + 4 \\times 5.532} } \\over 2}$$

$$ \\cong 1.604$$

$$\\Rightarrow$$ Moles of PCl5 = 3 $$-$$ 1.604 $$\\cong$$ 1.396", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 589, "subject": "Chemistry", "question": "The equilibrium constant for the reaction

A(s) $$\\rightleftharpoons$$ M(s) + $${1 \\over 2}$$O2(g)

is Kp = 4. At equilibrium, the partial pressure of O2 is _________ atm. (Round off to the nearest integer)", "options": [], "answer": "16", "solution": "**Answer:** 16\n\nkp = Po$$_2^{1/2}$$ = 4

$$\\therefore$$ Po2 = 16 bar = 16 atm", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 590, "subject": "Chemistry", "question": "The OH$$-$$ concentration in a mixture of 5.0 mL of 0.0504 M NH4Cl and 2 mL of 0.0210 M NH3 solution is x $$\\times$$ 10$$-$$6 M. The value of x is ___________. (Nearest integer)

[Given Kw = 1 $$\\times$$ 10$$-$$14 and Kb = 1.8 $$\\times$$ 10$$-$$5]", "options": [], "answer": "3", "solution": "**Answer:** 3\n\n$$\\left[ {NH_4^ + } \\right]$$ = 0.0504 & [NH3] = 0.0210

So, $${K_b} = {{[NH_4^ + ][H{O^ - }]} \\over {[N{H_3}]}}$$

$$[H{O^ - }] = {{{K_b} \\times [N{H_3}]} \\over {[NH_4^ + ]}} = 1.8 \\times {10^{ - 5}} \\times {2 \\over 5} \\times {{210} \\over {504}} = 3 \\times {10^{ - 6}}$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 591, "subject": "Chemistry", "question": "The equilibrium constant Kc at 298 K for the reaction A + B $$\\rightleftharpoons$$ C + D is 100. Starting with an equimolar solution with concentrations of A, B, C and D all equal to 1M, the equilibrium concentration of D is ___________ $$\\times$$ 10$$-$$2 M. (Nearest integer)", "options": [], "answer": "182", "solution": "**Answer:** 182\n\n

\"JEE

\n

$$\\therefore$$ $${K_C} = {\\left( {{{1 + x} \\over {1 - x}}} \\right)^2}$$

\n

$$100 = {\\left( {{{1 + x} \\over {1 - x}}} \\right)^2}$$

\n

$${{1 + x} \\over {1 - x}} = 10$$

\n

$$x = {9 \\over {11}}$$

\n

Moles of D = 1 + x

\n

$$ = 1 + {9 \\over {11}} = {{20} \\over {11}}$$

\n

$$ = 1.818 = 181.8 \\times {10^{ - 2}} = 181.8 \\times {10^{ - 2}}$$

\n

$$ \\cong 182 \\times {10^{ - 2}}$$ M

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 592, "subject": "Chemistry", "question": "The reaction rate for the reaction

[PtCl4]2$$-$$ + H2O $$\\rightleftharpoons$$ [Pt(H2O)Cl3]$$-$$ + Cl$$-$$

was measured as a function of concentrations of different species. It was observed that $${{ - d\\left[ {{{\\left[ {PtC{l_4}} \\right]}^{2 - }}} \\right]} \\over {dt}} = 4.8 \\times {10^{ - 5}}\\left[ {{{\\left[ {PtC{l_4}} \\right]}^{2 - }}} \\right] - 2.4 \\times {10^{ - 3}}\\left[ {{{\\left[ {Pt({H_2}O)C{l_3}} \\right]}^ - }} \\right]\\left[ {C{l^ - }} \\right]$$.

where square brackets are used to denote molar concentrations. The equilibrium constant Kc = ____________ . (Nearest integer)", "options": [], "answer": "0", "solution": "**Answer:** 0\n\nThe rate equation provided in your question describes the rates of the forward and reverse reactions for this chemical system. The equilibrium constant, $K_c$, is the ratio of the forward rate constant ($k_f$) to the reverse rate constant ($k_r$). At equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction, so the rate equation simplifies to:\n\n

$$k_f[\\text{{PtCl}}_4]^{2-} = k_r[\\text{{Pt(H}}_2\\text{{O)Cl}}_3]^-[\\text{{Cl}}^-]$$\n\n

Given the rate equation:\n\n

$$-\\frac{d[\\text{{PtCl}}_4]^{2-}}{dt} = 4.8 \\times 10^{-5} [\\text{{PtCl}}_4]^{2-} - 2.4 \\times 10^{-3} [\\text{{Pt(H}}_2\\text{{O)Cl}}_3]^-[\\text{{Cl}}^-]$$\n\n

At equilibrium, $-\\frac{d[\\text{{PtCl}}_4]^{2-}}{dt} = 0$, so:\n\n

$$0 = 4.8 \\times 10^{-5} [\\text{{PtCl}}_4]^{2-} - 2.4 \\times 10^{-3} [\\text{{Pt(H}}_2\\text{{O)Cl}}_3]^-[\\text{{Cl}}^-]$$\n\n

Rearranging terms, we find:\n\n

$$4.8 \\times 10^{-5} [\\text{{PtCl}}_4]^{2-} = 2.4 \\times 10^{-3} [\\text{{Pt(H}}_2\\text{{O)Cl}}_3]^-[\\text{{Cl}}^-]$$\n\n

Now, the equilibrium constant $K_c$ is defined as the ratio of the concentrations of the products to the reactants, each raised to the power of their stoichiometric coefficients. For the reaction in question, we have:\n\n

$$K_c = \\frac{[\\text{{Pt(H}}_2\\text{{O)Cl}}_3]^-[\\text{{Cl}}^-]}{[\\text{{PtCl}}_4]^{2-}}$$\n\n

Dividing both sides of our rate equation by $[\\text{{PtCl}}_4]^{2-}$, we find that:\n\n

$$K_c = \\frac{4.8 \\times 10^{-5}}{2.4 \\times 10^{-3}} = \\frac{1}{50}$$ = 0.02\n\n

So, the equilibrium constant $K_c$ for this reaction is approximately 0, when rounded to the nearest integer.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 593, "subject": "Chemistry", "question": "The number of moles of NH3, that must be added to 2L of 0.80 M AgNO3 in order to reduce the concentration of Ag+ ions to 5.0 $$\\times$$ 10$$-$$8 M (Kformation for [Ag(NH3)2]+ = 1.0 $$\\times$$ 108) is ____________. (Nearest integer)

[Assume no volume change on adding NH3]", "options": [], "answer": "4", "solution": "**Answer:** 4\n\nLet moles added = a

\"JEE
$${{0.8} \\over {(5 \\times {{10}^{ - 8}})\\left( {{a \\over 2} - 1.6} \\right)}} = {10^8}$$

$$\\Rightarrow$$ $${{a \\over 2}}$$ $$-$$ 1.6 = 0.4 $$\\Rightarrow$$ a = 4", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 594, "subject": "Chemistry", "question": "When 5.1 g of solid NH4HS is introduced into a two litre evacuated flask at 27$$^\\circ$$C, 20% of the solid decomposes into gaseous ammonia and hydrogen sulphide. The Kp for the reaction at 27$$^\\circ$$C is x $$\\times$$ 10$$-$$2. The value of x is _____________. (Integer answer) [Given R = 0.082 L atm K$$-$$1 mol$$-$$]", "options": [], "answer": "6", "solution": "**Answer:** 6\n\nmoles of NH4HS initially taken = $${{5.1g} \\over {51g/mol}}$$ = 0.1 mol

volume of vessel = 2 litre

\"JEE

$$\\Rightarrow$$ partial pressure of each component

$$P = {{nRT} \\over V} = {{0.1 \\times 0.2 \\times 0.082 \\times 300} \\over 2}$$

= 0.246 atm

$$\\Rightarrow$$ kP = P$$N{H_3}$$ $$\\times$$ P$${H_2}S$$ = (0.246)2 = 0.060516

= 6.05 $$\\times$$ 10$$-$$2

$$ \\therefore $$ x = 6", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 595, "subject": "Chemistry", "question": "

4.0 moles of argon and 5.0 moles of PCl5 are introduced into an evacuated flask of 100 litre capacity at 610 K. The system is allowed to equilibrate. At equilibrium, the total pressure of mixture was found to be 6.0 atm. The Kp for the reaction is :

\n

[Given : R = 0.082 L atm K$$-$$1 mol$$-$$1]

", "options": [ { "text": "2.25" }, { "text": "6.24" }, { "text": "12.13" }, { "text": "15.24" } ], "answer": "2.25", "solution": "**Answer:** 2.25\n\n

\"JEE

\n

Here 4 moles of inert gas argon also present.

\n

$$\\therefore$$ Total moles of mixture present at equilibrium,

\n

nT = 5 + x + 4

\n

= 9 + x

\n

At equilibrium, total pressure (pT) = 6 atm

\n

Volume (v) = 100 L

\n

Temperature = 610 K

\n

$$\\therefore$$ Using ideal gas equation,

\n

$${P_T}V = {n_T}RT$$

\n

$$ \\Rightarrow 6 \\times 100 = (9 + x) \\times 0.082 \\times 610$$

\n

$$ \\Rightarrow x = 3$$

\n

Now,

\n

$${K_P} = {{{P_{PC{l_3}}} \\times {P_{C{l_2}}}} \\over {{P_{PC{l_5}}}}}$$

\n

$$ = {{\\left[ {{3 \\over {9 + 3}} \\times 6} \\right] \\times \\left[ {{3 \\over {9 + 3}} \\times 6} \\right]} \\over {\\left[ {{{5 - 3} \\over {9 + 3}} \\times 6} \\right]}}$$

\n

$$ = {{27} \\over {12}}$$

\n

$$ = {9 \\over 4}$$

\n

= 2.25 atm

\n

Note :

\n

Inert gas always contribute to total mole and pressure calculation.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 596, "subject": "Chemistry", "question": "

A box contains 0.90 g of liquid water in equilibrium with water vapour at 27$$^\\circ$$C. The equilibrium vapour pressure of water at 27$$^\\circ$$C is 32.0 Torr. When the volume of the box is increased, some of the liquid water evaporates to maintain the equilibrium pressure. If all the liquid water evaporates, then the volume of the box must be __________ litre. [nearest integer]

\n

(Given : R = 0.082 L atm K$$-$$1 mol$$-$$1)

\n

(Ignore the volume of the liquid water and assume water vapours behave as an ideal gas.)

", "options": [], "answer": "29", "solution": "**Answer:** 29\n\n

We know, 760 Torr = 1 atm

\n

$$\\therefore$$ 32 Torr = $${{32} \\over {760}}$$ atm

\n

As all the liquid water evaporates so entire water is in gaseous state.

\n

$$\\therefore$$ Weight of water vapour = 0.9 g

\n

$$\\therefore$$ Moles of water vapour (n) = $${{0.9} \\over {18}}$$

\n

Pressure (P) = $${{32} \\over {760}}$$ atm

\n

Temperature (T) = (27 + 273) K = 300 K

\n

R = 0.082 L atm K$$-$$1 mol$$-$$1

\n

Given water vapour act as an ideal gas, so we can apply ideal gas equation.

\n

From ideal gas equation,

\n

PV = nRT

\n

$$ \\Rightarrow {{32} \\over {760}} \\times v = {{0.9} \\over {18}} \\times 0.082 \\times 300$$

\n

$$ \\Rightarrow v = 29$$ L

", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 597, "subject": "Chemistry", "question": "

2NOCl(g) $$\\rightleftharpoons$$ 2NO(g) + Cl2(g)

\n

In an experiment, 2.0 moles of NOCl was placed in a one-litre flask and the concentration of NO after equilibrium established, was found to be 0.4 mol/L. The equilibrium constant at 30$$^\\circ$$C is ______________ $$\\times$$ 10$$-$$4.

", "options": [], "answer": "125", "solution": "**Answer:** 125\n\n

\"JEE

\n

Given that at equilibrium, concentration of NO = 0.4 mol/L

\n

$$\\therefore$$ 2x = 0.4

\n

$$\\Rightarrow$$ x = 0.2

\n

$$\\therefore$$ Concentration of NOCl at equilibrium,

\n

[NOCl]eq = 2 $$-$$ 2 $$\\times$$ 0.2 = 1.6

\n

and [NO]eq = 0.4

\n

and [Cl2]eq = 0.2

\n

We know,

\n

$${K_C} = {{{{[NO]}^2}[C{l_2}]} \\over {{{[NOCl]}^2}}}$$

\n

$$ = {{{{[0.4]}^2}[0.2]} \\over {{{[1.6]}^2}}}$$

\n

$$ \\Rightarrow {K_C} = 12.5 \\times {10^{ - 3}}$$

\n

$$ \\Rightarrow {K_C} = 125 \\times {10^{ - 4}}$$

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 598, "subject": "Chemistry", "question": "

The standard free energy change ($$\\Delta$$G$$^\\circ$$) for 50% dissociation of N2O4 into NO2 at 27$$^\\circ$$C and 1 atm pressure is $$-$$ x J mol$$-$$1. The value of x is ___________. (Nearest Integer)

\n

[Given : R = 8.31 J K$$-$$1 mol$$-$$1, log 1.33 = 0.1239 ln 10 = 2.3]

", "options": [], "answer": "710", "solution": "**Answer:** 710\n\n\"JEE
\n$\\mathrm{k}_{\\mathrm{P}}=\\frac{\\left(\\frac{1}{1.5} \\times 1\\right)^2}{\\left(\\frac{0.5}{1.5} \\times 1\\right)}=\\frac{1}{0.75}=\\frac{100}{75}$

\n$=1.33$

\n$\\Delta \\mathrm{G}^0=-\\mathrm{RT} \\ell \\mathrm{nk}_{\\mathrm{P}}$

\n$=-8.31 \\times 300 \\times \\ln (1.33)=-710.45 \\mathrm{~J} / \\mathrm{mol}$

\n$=-710 \\mathrm{~J} / \\mathrm{mol}$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 599, "subject": "Chemistry", "question": "

PCl5 dissociates as

\n

PCl5(g) $$\\rightleftharpoons$$ PCl3(g) + Cl2(g)

\n

5 moles of PCl5 are placed in a 200 litre vessel which contains 2 moles of N2 and is maintained at 600 K. The equilibrium pressure is 2.46 atm. The equilibrium constant Kp for the dissociation of PCl5 is __________ $$\\times$$ 10$$-$$3. (nearest integer)

\n

(Given : R = 0.082 L atm K$$-$$1 mol$$-$$1; Assume ideal gas behaviour)

", "options": [], "answer": "1107", "solution": "**Answer:** 1107\n\n

\"JEE

\n

Here 2 moles of N2 also present that is why 2 moles always have to add in total mole calculation.

\n

At equilibrium,

\n

Pressure (P) = 2.46 atm

\n

Volume (V) = 200 L

\n

Temperature (T) = 600 K

\n

$$\\therefore$$ Applying ideal gas equation,

\n

PV = nRT

\n

$$\\Rightarrow$$ 2.46 $$\\times$$ 200 = (7 + x) $$\\times$$ 0.082 $$\\times$$ 600

\n

$$\\Rightarrow$$ x = 3

\n

Now,

\n

$${K_P} = {{{P_{PC{l_3}}} \\times {P_{C{l_2}}}} \\over {{P_{PC{l_5}}}}}$$

\n

$$ = {{\\left[ {{3 \\over {7 + 3}} \\times 2.46} \\right]\\left[ {{3 \\over {7 + 3}} \\times 2.46} \\right]} \\over {\\left[ {{{5 - 3} \\over {7 + 3}} \\times 2.46} \\right]}}$$

\n

$$ = {{{3 \\over {10}} \\times {3 \\over {10}} \\times {{(2.46)}^2}} \\over {{2 \\over {10}} \\times 2.46}}$$

\n

$$ = {9 \\over {20}} \\times 2.46$$

\n

$$ = 1107 \\times {10^{ - 3}}$$ atm

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 600, "subject": "Chemistry", "question": "

For a reaction at equilibrium

\n

A(g) $$\\rightleftharpoons$$ B(g) + $${1 \\over 2}$$ C(g)

\n

the relation between dissociation constant (K), degree of dissociation ($$\\alpha$$) and equilibrium pressure (p) is given by :

", "options": [ { "text": "$$K = {{{\\alpha ^{{1 \\over 2}}}{p^{{3 \\over 2}}}} \\over {{{\\left( {1 + {3 \\over 2}\\alpha } \\right)}^{{1 \\over 2}}}(1 - \\alpha )}}$$" }, { "text": "$$K = {{{\\alpha ^{{3 \\over 2}}}{p^{{1 \\over 2}}}} \\over {{{\\left( {2 + \\alpha } \\right)}^{{1 \\over 2}}}(1 - \\alpha )}}$$" }, { "text": "$$K = {{{{(\\alpha \\,p)}^{{3 \\over 2}}}} \\over {{{\\left( {1 + {3 \\over 2}\\alpha } \\right)}^{{1 \\over 2}}}(1 - \\alpha )}}$$" }, { "text": "$$K = {{{{(\\alpha \\,p)}^{{3 \\over 2}}}} \\over {{{\\left( {1 + \\alpha } \\right)}}{{(1 - \\alpha )}^{{1 \\over 2}}}}}$$" } ], "answer": "$$K = {{{\\alpha ^{{3 \\over 2}}}{p^{{1 \\over 2}}}} \\over {{{\\left( {2 + \\alpha } \\right)}^{{1 \\over 2}}}(1 - \\alpha )}}$$", "solution": "**Answer:** $$K = {{{\\alpha ^{{3 \\over 2}}}{p^{{1 \\over 2}}}} \\over {{{\\left( {2 + \\alpha } \\right)}^{{1 \\over 2}}}(1 - \\alpha )}}$$\n\n

\"JEE

\n

Now,

\n

$${K_P}$$ or $$K = {{{P_B} \\times {{\\left( {{P_C}} \\right)}^{{1 \\over 2}}}} \\over {{P_A}}}$$

\n

$$ = {{\\left( {{\\alpha \\over {1 + {\\alpha \\over 2}}}} \\right)P \\times {{\\left[ {\\left( {{{{\\alpha \\over 2}} \\over {1 + {\\alpha \\over 2}}}} \\right)P} \\right]}^{{1 \\over 2}}}} \\over {\\left( {{{1 - \\alpha } \\over {1 + {\\alpha \\over 2}}}} \\right)P}}$$

\n

$$ = {{\\left( {{{2\\alpha } \\over {2 + \\alpha }}} \\right)P \\times {{\\left[ {\\left( {{\\alpha \\over {2 + \\alpha }}} \\right)P} \\right]}^{{1 \\over 2}}}} \\over {\\left( {{{2(1 - \\alpha )} \\over {2 + \\alpha }}} \\right)P}}$$

\n

$$= {\\alpha \\over {1 - \\alpha }} \\times {\\left( {{{\\alpha P} \\over {2 + \\alpha }}} \\right)^{{1 \\over 2}}}$$

\n

$$ = {{{\\alpha ^{{3 \\over 2}}}\\,.\\,{P^{{1 \\over 2}}}} \\over {(1 - \\alpha ){{(2 + \\alpha )}^{{1 \\over 2}}}}}$$

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 601, "subject": "Chemistry", "question": "

2O3(g) $$\\rightleftharpoons$$ 3O2(g)

\n

At 300 K, ozone is fifty percent dissociated. The standard free energy change at this temperature and 1 atm pressure is ($$-$$) ____________ J mol$$-$$1. (Nearest integer)

\n

[Given : ln 1.35 = 0.3 and R = 8.3 J K$$-$$1 mol$$-$$1]

", "options": [], "answer": "747", "solution": "**Answer:** 747\n\n$\\underset{1-x}{2 \\mathrm{O}_3(\\mathrm{~g})} \\rightleftharpoons \\underset{\\frac{3 \\mathrm{x}}{2}}{3 \\mathrm{O}_2(\\mathrm{~g})}$

\nGiven, $x=0.5$\n

\n$\\therefore \\mathrm{k}_{\\mathrm{p}}=\\frac{[3(0.5)]^{3} \\times 1}{[2]^{3} \\times(0.5)^{2} \\times 1.25}$\n

\n$\\therefore \\mathrm{k}_{\\mathrm{p}}=\\frac{27}{8} \\times \\frac{0.5}{1.25}=1.35$\n

\n$$\n\\begin{aligned}\n\\Delta \\mathrm{G}^{\\circ} &=-2.303 \\mathrm{RT} \\log \\mathrm{k}_{\\mathrm{p}} \\\\\\\\\n&=-2.303 \\times 8.3 \\times 300 \\log 1.35 \\\\\\\\\n&=-8.3 \\times 300 \\ln (1.35) \\\\\\\\\n&=-747 \\mathrm{~J} \\mathrm{~mol}^{-1}\n\\end{aligned}\n$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 602, "subject": "Chemistry", "question": "

The equilibrium constant for the reversible reaction

\n

2A(g) $$\\rightleftharpoons$$ 2B(g) + C(g) is K1

\n

$${3 \\over 2}$$A(g) $$\\rightleftharpoons$$ $${3 \\over 2}$$B(g) + $${3 \\over 4}$$C(g) is K2.

\n

K1 and K2 are related as :

", "options": [ { "text": "$${K_1} = \\sqrt {{K_2}} $$" }, { "text": "$${K_2} = \\sqrt {{K_1}} $$" }, { "text": "$${K_2} = K_1^{3/4}$$" }, { "text": "$${K_1} = K_2^{3/4}$$" } ], "answer": "$${K_2} = K_1^{3/4}$$", "solution": "**Answer:** $${K_2} = K_1^{3/4}$$\n\n$2 \\mathrm{A}(\\mathrm{g})=2 \\mathrm{B}(\\mathrm{~g})+\\mathrm{C}(\\mathrm{g}) \\quad K_{1} \\quad\\quad ...(i)$\n

\n$$\n\\frac{3}{2} \\mathrm{A}(\\mathrm{g})=\\frac{3}{2} \\mathrm{B}(\\mathrm{g})+\\frac{3}{4} \\mathrm{C}(\\mathrm{g}) \\quad K_{2}\\quad\\quad...(ii)\n$$\n

\neq. (ii) is $\\frac{3}{4}$ times of eq. (i), hence,\n

\n$K_{2}=\\left(K_{1}\\right)^{\\frac{3}{4}}$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 603, "subject": "Chemistry", "question": "

At $$298 \\mathrm{~K}$$, the equilibrium constant is $$2 \\times 10^{15}$$ for the reaction :

\n

$$\\mathrm{Cu}(\\mathrm{s})+2 \\mathrm{Ag}^{+}(\\mathrm{aq}) \\rightleftharpoons \\mathrm{Cu}^{2+}(\\mathrm{aq})+2 \\mathrm{Ag}(\\mathrm{s})$$

\n

The equilibrium constant for the reaction

\n

$$\n\\frac{1}{2} \\mathrm{Cu}^{2+}(\\mathrm{aq})+\\mathrm{Ag}(\\mathrm{s}) \\rightleftharpoons \\frac{1}{2} \\mathrm{Cu}(\\mathrm{s})+\\mathrm{Ag}^{+}(\\mathrm{aq})\n$$

\n

is $$x \\times 10^{-8}$$. The value of $$x$$ is _____________. (Nearest Integer)

", "options": [], "answer": "2", "solution": "**Answer:** 2\n\n$\\mathrm{Cu}(\\mathrm{s})+2 \\mathrm{Ag}^{+}(\\mathrm{aq}) \\rightleftharpoons \\mathrm{Cu}^{2+}(\\mathrm{aq})+2 \\mathrm{Ag}(\\mathrm{s})$\n

\n$k=2 \\times 10^{15}$\n

\n$\\frac{1}{2} \\mathrm{Cu}(\\mathrm{s})+\\mathrm{Ag}^{+}(\\mathrm{aq}) \\rightleftharpoons \\mathrm{Cu}^{+2}(\\mathrm{aq})+2 \\mathrm{Ag}(\\mathrm{s})$\n

\n$\\mathrm{K}^{\\prime}=\\frac{1}{(\\mathrm{~K})^{1 / 2}}=\\frac{1}{\\left(2 \\times 10^{15}\\right)^{1 / 2}}$\n

\n$=2.23 \\times 10^{-8}$\n

\n$x \\simeq 2$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 604, "subject": "Chemistry", "question": "

At $$600 \\mathrm{~K}, 2 \\mathrm{~mol}$$ of $$\\mathrm{NO}$$ are mixed with $$1 \\mathrm{~mol}$$ of $$\\mathrm{O}_{2}$$.

\n

$$2 \\mathrm{NO}_{(\\mathrm{g})}+\\mathrm{O}_{2}(\\mathrm{g}) \\rightleftarrows 2 \\mathrm{NO}_{2}(\\mathrm{g})$$

\n

The reaction occurring as above comes to equilibrium under a total pressure of 1 atm. Analysis of the system shows that $$0.6 \\mathrm{~mol}$$ of oxygen are present at equilibrium. The equilibrium constant for the reaction is ________. (Nearest integer)

", "options": [], "answer": "2", "solution": "**Answer:** 2\n\n\"JEE
Partial pressure of $\\mathrm{NO}(\\mathrm{g})=\\frac{1.2}{2.6} \\times 1$\n

\nPartial pressure of $\\mathrm{O}_{2}(\\mathrm{~g})=\\frac{0.6}{2.6}$\n

\nPartial pressure of $\\mathrm{NO}_{2}(\\mathrm{~g})=\\frac{0.8}{2.6}$\n

\n$$\n\\begin{aligned}\n\\mathrm{K}_{\\mathrm{p}}=\\frac{\\left(\\mathrm{P}_{\\mathrm{NO}_{2}}\\right)^{2}}{\\left(\\mathrm{P}_{\\mathrm{NO}}\\right)^{2}\\left(\\mathrm{P}_{\\mathrm{O}_{2}}\\right)} &=\\frac{0.8 \\times 0.8 \\times 2.6}{1.2 \\times 1.2 \\times 0.6} \\\\\\\\\n&=1.925 \\\\\\\\\n& \\approx 2\n\\end{aligned}\n$$", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 605, "subject": "Chemistry", "question": "

(i) $$\\mathrm{X}(\\mathrm{g}) \\rightleftharpoons \\mathrm{Y}(\\mathrm{g})+\\mathrm{Z}(\\mathrm{g}) \\quad \\mathrm{K}_{\\mathrm{p} 1}=3$$

\n

(ii) $$\\mathrm{A}(\\mathrm{g}) \\rightleftharpoons 2 \\mathrm{~B}(\\mathrm{g}) \\quad \\mathrm{K}_{\\mathrm{p} 2}=1$$

\n

If the degree of dissociation and initial concentration of both the reactants $$\\mathrm{X}(\\mathrm{g})$$ and $$\\mathrm{A}(\\mathrm{g})$$ are equal, then the ratio of the total pressure at equilibrium $$\\left(\\frac{p_{1}}{p_{2}}\\right)$$ is equal to $$\\mathrm{x}: 1$$. The value of $$\\mathrm{x}$$ is _____________ (Nearest integer)

", "options": [], "answer": "12", "solution": "**Answer:** 12\n\n\"JEE\n

$$\n\\begin{aligned}\n& \\mathrm{k}_{\\mathrm{p}_1}=\\frac{\\left(\\frac{\\alpha}{1+\\alpha} \\times \\mathrm{p}_1\\right)^2}{\\frac{1-\\alpha}{1+\\alpha} \\mathrm{p}_1} \\\\\\\\\n& 3=\\frac{\\alpha^2 \\times \\mathrm{p}_1}{1-\\alpha^2}\n\\end{aligned}\n$$\n

\"JEE\n

$$\n\\begin{aligned}\n& \\mathrm{k}_{\\mathrm{p}_2}=\\frac{\\left(\\frac{2 \\alpha}{1+\\alpha} \\times \\mathrm{p}_2\\right)^2}{\\frac{1-\\alpha}{1+\\alpha} \\times \\mathrm{p}_2} \\\\\\\\\n& 1=\\frac{4 \\alpha^2 \\times \\mathrm{p}_2}{1-\\alpha^2} \\\\\\\\\n& \\frac{\\mathrm{k}_{\\mathrm{p}_1}}{\\mathrm{k}_{\\mathrm{p}_2}}=\\frac{\\mathrm{p}_1}{4 \\mathrm{p}_2} \\\\\\\\\n& \\frac{3}{1}=\\frac{\\mathrm{p}_1}{4 \\mathrm{p}_2} \\\\\\\\\n& {\\mathrm{p}_1} : {\\mathrm{p}_2} = 12 : 1 \\\\\\\\\n& \\therefore \\mathrm{x}=12\n\\end{aligned}\n$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 606, "subject": "Chemistry", "question": "

For reaction : $$\\mathrm{SO}_{2}(\\mathrm{~g})+\\frac{1}{2} \\mathrm{O}_{2}(\\mathrm{~g}) \\rightleftharpoons \\mathrm{SO}_{3}(\\mathrm{~g})$$

\n

$$\\mathrm{K}_{\\mathrm{p}}=2 \\times 10^{12}$$ at $$27^{\\circ} \\mathrm{C}$$ and $$1 \\mathrm{~atm}$$ pressure. The $$\\mathrm{K}_{\\mathrm{c}}$$ for the same reaction is ____________ $$\\times 10^{13}$$. (Nearest integer)

\n

(Given $$\\mathrm{R}=0.082 \\mathrm{~L} \\mathrm{~atm} \\mathrm{~K}^{-1} \\mathrm{~mol}^{-1}$$)

", "options": [], "answer": "1", "solution": "**Answer:** 1\n\n$$\n\\begin{aligned}\n& \\mathrm{SO}_{2(\\mathrm{~g})}+\\frac{1}{2} \\mathrm{O}_{2(\\mathrm{~g})} \\rightleftharpoons \\mathrm{SO}_{3(\\mathrm{~g})} \\\\\\\\\n& \\mathrm{K}_{\\mathrm{P}}=2 \\times 10^{12} \\text { at } 300 \\mathrm{~K} \\\\\\\\\n& \\mathrm{~K}_{\\mathrm{P}}=\\mathrm{K}_{\\mathrm{C}} \\times(\\mathrm{RT})^{\\Delta \\mathrm{n}_{\\mathrm{g}}} \\\\\\\\\n& \\Rightarrow 2 \\times 10^{12}=\\mathrm{K}_{\\mathrm{C}} \\times(0.082 \\times 300)^{-1 / 2} \\\\\\\\\n& \\Rightarrow \\mathrm{~K}_{\\mathrm{C}}=9.92 \\times 10^{12} \\\\\\\\\n& \\Rightarrow \\mathrm{~K}_{\\mathrm{C}}=0.992 \\times 10^{13}\n\\end{aligned}\n$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 607, "subject": "Chemistry", "question": "

Water decomposes at 2300 K

\n

$$\\mathrm{H_2O(g)\\to H_2(g)+\\frac{1}{2}O_2(g)}$$

\n

The percent of water decomposing at 2300 K and 1 bar is ___________ (Nearest integer).

\n

Equilibrium constant for the reaction is $$2\\times10^{-3}$$ at 2300 K.

", "options": [], "answer": "2", "solution": "**Answer:** 2\n\n$\\mathrm{H}_{2} \\mathrm{O} \\longrightarrow \\mathrm{H}_{2}+\\frac{1}{2} \\mathrm{O}_{2}$\n

\n$$\n\\begin{aligned}\n& \\mathrm{K}_{\\mathrm{p}}=\\frac{\\mathrm{P}_{\\mathrm{T}}^{\\frac{3}{2}}\\left(1+\\frac{\\alpha}{2}\\right) \\alpha^{\\frac{3}{2}}}{2^{\\frac{1}{2}} \\mathrm{P}_{\\mathrm{T}}\\left(1+\\frac{\\alpha}{2}\\right)^{\\frac{3}{2}}(1-\\alpha)} \\\\\\\\\n& 2 \\times 10^{-3}=\\frac{\\alpha^{\\frac{3}{2}}}{2^{\\frac{1}{2}}}\\left[\\text { as } \\alpha<<1 \\text { and let } \\mathrm{P}_{\\mathrm{T}}=1\\right]\n\\end{aligned}\n$$\n

\n$$\n\\begin{aligned}\n& \\alpha^{\\frac{3}{2}}=2^{\\frac{3}{2}} \\times 10^{-3} \\\\\\\\\n& \\approx 2 \\times 10^{-2}\n\\end{aligned}\n$$\n

\n$\\therefore \\%$ of water decomposition $=2 \\%$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 608, "subject": "Chemistry", "question": "

Consider the following reaction approaching equilibrium at 27$$^\\circ$$C and 1 atm pressure

\n

$$\\mathrm{A+B}$$ $$\\mathrel{\\mathop{\\kern0pt\\rightleftharpoons}\n\\limits_{{k_r} = {{10}^2}}^{{k_f} = {{10}^3}}} $$ $$\\mathrm{C+D}$$

\n

The standard Gibb's energy change $$\\mathrm{(\\Delta_r G^\\theta)}$$ at 27$$^\\circ$$C is ($$-$$) ___________ kJ mol$$^{-1}$$ (Nearest integer).

\n

(Given : $$\\mathrm{R=8.3~J~K^{-1}~mol^{-1}}$$ and $$\\mathrm{\\ln 10=2.3}$$)

", "options": [], "answer": "6", "solution": "**Answer:** 6\n\n$\\because \\Delta \\mathrm{G}^{0}=-\\mathrm{RT} \\ln \\mathrm{K}_{\\mathrm{eq}}$\n

\nand $\\mathrm{K}_{\\mathrm{eq}}=\\frac{\\mathrm{K}_{\\mathrm{f}}}{\\mathrm{K}_{\\mathrm{b}}}$\n

\n$\\therefore \\mathrm{K}_{\\mathrm{eq}}=\\frac{10^{3}}{10^{2}}=10$\n

\n$\\therefore \\Delta \\mathrm{G}=-\\mathrm{RT} \\ln 10$\n

\n$\\Rightarrow-(8.3 \\times 300 \\times 2.3)=-5.7 \\mathrm{~kJ} \\mathrm{~mole}^{-1} \\approx 6 \\mathrm{~kJ}$ $\\mathrm{mole}^{-1}$ (nearest integer)", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 609, "subject": "Chemistry", "question": "

4.5 moles each of hydrogen and iodine is heated in a sealed ten litre vessel. At equilibrium, 3 moles of $$\\mathrm{HI}$$ were found. The equilibrium constant for $$\\mathrm{H}_{2}(\\mathrm{~g})+\\mathrm{I}_{2}(\\mathrm{~g}) \\rightleftharpoons 2 \\mathrm{HI}(\\mathrm{g})$$ is _________.

", "options": [], "answer": "1", "solution": "**Answer:** 1\n\n\"JEE\n

$$\n\\mathrm{K}_{\\mathrm{c}}=\\frac{[\\mathrm{HI}]^2}{\\left[\\mathrm{H}_2\\right]\\left[\\mathrm{I}_2\\right]}=\\frac{(3)^2}{3 \\times 3}=\\frac{9}{9}=1\n$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 610, "subject": "Chemistry", "question": "

A mixture of 1 mole of $$\\mathrm{H}_{2} \\mathrm{O}$$ and 1 mole of $$\\mathrm{CO}$$ is taken in a 10 litre container and heated to $$725 \\mathrm{~K}$$. At equilibrium $$40 \\%$$ of water by mass reacts with carbon monoxide according to the equation : \n

$$\\mathrm{CO}(\\mathrm{g})+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g}) \\rightleftharpoons \\mathrm{CO}_{2}(\\mathrm{~g})+\\mathrm{H}_{2}(\\mathrm{~g})$$.\n

The equilibrium constant $$\\mathrm{K}_{\\mathrm{c}} \\times 10^{2}$$ for the reaction is ____________. (Nearest integer)

", "options": [], "answer": "44", "solution": "**Answer:** 44\n\nThe given reaction is : \n\n

$$\\mathrm{CO(g)} + \\mathrm{H_{2}O(g)} \\rightleftharpoons \\mathrm{CO_{2}(g)} + \\mathrm{H_{2}(g)} $$\n\n

We are given that $40\\%$ of water by mass reacts with carbon monoxide.\n\n

The molecular weight of water (H2O) is approximately $18 \\, \\mathrm{g/mol}$, so $1 \\, \\mathrm{mole}$ of water weighs $18 \\, \\mathrm{g}$. Therefore, the mass of the reacted water is $0.40 \\times 18 \\, \\mathrm{g} = 7.2 \\, \\mathrm{g}$.\n\n

Since from the stoichiometry of the reaction we can see that $1 \\, \\mathrm{mole}$ of CO reacts with $1 \\, \\mathrm{mole}$ of H2O to form $1 \\, \\mathrm{mole}$ of CO2 and $1 \\, \\mathrm{mole}$ of H2, this means that $7.2 \\, \\mathrm{g}$ of H2O is equivalent to $7.2 \\, \\mathrm{g} / 18 \\, \\mathrm{g/mol} = 0.4 \\, \\mathrm{mole}$.\n\n

We start with $1 \\, \\mathrm{mole}$ of CO and $1 \\, \\mathrm{mole}$ of H2O. At equilibrium, we have :\n\n

- CO: $1 \\, \\mathrm{mole} - 0.4 \\, \\mathrm{mole} = 0.6 \\, \\mathrm{mole}$\n

- H2O : $1 \\, \\mathrm{mole} - 0.4 \\, \\mathrm{mole} = 0.6 \\, \\mathrm{mole}$\n

- CO2 : $0 \\, \\mathrm{mole} + 0.4 \\, \\mathrm{mole} = 0.4 \\, \\mathrm{mole}$\n

- H2 : $0 \\, \\mathrm{mole} + 0.4 \\, \\mathrm{mole} = 0.4 \\, \\mathrm{mole}$\n\n

The volume of the container is $10 \\, \\mathrm{litres}$. Therefore, we can convert the moles to concentrations (in M = moles/L) as :\n\n

- [CO] = $0.6 \\, \\mathrm{M}$\n

- [H2O] = $0.6 \\, \\mathrm{M}$\n

- [CO2] = $0.4 \\, \\mathrm{M}$\n

- [H2] = $0.4 \\, \\mathrm{M}$\n\n

The equilibrium constant $K_{c}$ for the reaction can be calculated as :\n\n

$$K_{c} = \\frac{[\\mathrm{CO_{2}}][\\mathrm{H_{2}}]}{[\\mathrm{CO}][\\mathrm{H_{2}O}]}$$\n\n

So, substituting the values we get :\n\n

$$K_{c} = \\frac{(0.4 \\times 0.4)}{(0.6 \\times 0.6)} = 0.44$$\n\n

As per the question, we need to calculate the value of $K_{c} \\times 10^{2}$ :\n\n

$$K_{c} \\times 10^{2} = 0.44 \\times 10^{2} = 44 \\, (\\text{rounded to the nearest\n\n integer})$$\n\n

Therefore, $K_{c} \\times 10^{2} = 44$.", "topic": "Algebra", "subtopic": "Equations and Inequalities" }, { "id": 611, "subject": "Chemistry", "question": "

$$\\mathrm{A}(g) \\rightleftharpoons 2 \\mathrm{~B}(g)+\\mathrm{C}(g)$$

\n

For the given reaction, if the initial pressure is $$450 \\mathrm{~mm} ~\\mathrm{Hg}$$ and the pressure at time $$\\mathrm{t}$$ is $$720 \\mathrm{~mm} ~\\mathrm{Hg}$$ at a constant temperature $$\\mathrm{T}$$ and constant volume $$\\mathrm{V}$$. The fraction of $$\\mathrm{A}(\\mathrm{g})$$ decomposed under these conditions is $$x \\times 10^{-1}$$. The value of $$x$$ is ___________ (nearest integer)

", "options": [], "answer": "3", "solution": "**Answer:** 3\n\n

Given the reaction:

\n

$$\\mathrm{A}_{(\\mathrm{g})} \\rightleftharpoons 2 \\mathrm{~B}_{(\\mathrm{g})}+\\mathrm{C}_{(\\mathrm{g})}$$

\n

At the beginning of the reaction (at time = 0), the total pressure is solely due to A, hence it is 450 mmHg.

\n

As the reaction progresses, let's denote 'x' as the pressure decrease due to the decomposition of A. Correspondingly, the pressure increases by '2x' and 'x' for B and C respectively, following the stoichiometry of the reaction.

\n

At time 't', the total pressure (P(T)) is the sum of the pressures of A, B, and C, which is given to be 720 mmHg.

\n

This gives us the equation:

\n

$$450 \\text{ mmHg} - x \\text{ mmHg (decrease in A's pressure)} + 2x \\text{ mmHg (increase in B's pressure)} + x \\text{ mmHg (increase in C's pressure)} = 720 \\text{ mmHg (total pressure at time t)}$$

\n

Solving for 'x' gives:

\n

$$x = 135 \\text{ mmHg}$$

\n

The fraction of A decomposed would then be this change in pressure divided by the initial pressure:

\n

$$\\text{Fraction decomposed} = \\frac{x}{\\text{Initial pressure}} = \\frac{135 \\text{ mmHg}}{450 \\text{ mmHg}} = 0.3 = 3 \\times 10^{-1}$$

\n", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 612, "subject": "Chemistry", "question": "

For a concentrated solution of a weak electrolyte ($$\\mathrm{K}_{\\text {eq }}=$$ equilibrium constant) $$\\mathrm{A}_{2} \\mathrm{B}_{3}$$ of concentration '$$c$$', the degree of dissociation '$$\\alpha$$' is :

", "options": [ { "text": "$$\\left(\\frac{K_{e q}}{25 c^{2}}\\right)^{\\frac{1}{5}}$$" }, { "text": "$$\\left(\\frac{K_{e q}}{108 c^{4}}\\right)^{\\frac{1}{5}}$$" }, { "text": "$$\\left(\\frac{K_{e q}}{5 c^{4}}\\right)^{\\frac{1}{5}}$$" }, { "text": "$$\\left(\\frac{K_{e q}}{6 c^{5}}\\right)^{\\frac{1}{5}}$$" } ], "answer": "$$\\left(\\frac{K_{e q}}{108 c^{4}}\\right)^{\\frac{1}{5}}$$", "solution": "**Answer:** $$\\left(\\frac{K_{e q}}{108 c^{4}}\\right)^{\\frac{1}{5}}$$\n\n$$\n\\begin{aligned}\n& \\mathrm{A}_2 \\mathrm{~B}_3(\\mathrm{aq} .) \\rightleftharpoons 2 \\mathrm{~A}_{\\text {(aq.) }}^{3+}+3 \\mathrm{~B}_{\\text {(aq) }}^{2-} \\\\\\\\\n& \\mathrm{c}(1-\\alpha) \\quad\\quad\\quad 2 \\mathrm{c} \\alpha\\quad\\quad\\quad 3 \\mathrm{c}\\alpha\\\\\\\\\n& \\mathrm{K}_{\\mathrm{eq}}=\\frac{\\left[\\mathrm{A}^{3+}\\right]^2\\left[\\mathrm{~B}^{2-}\\right]^3}{\\left[\\mathrm{~A}_2 \\mathrm{~B}_3\\right]}=\\frac{4 \\mathrm{c}^2 \\alpha^2 \\times 27 \\mathrm{c}^3 \\alpha^3}{\\mathrm{c}(1-\\alpha)} \\\\\\\\\n& \\mathrm{K}_{\\mathrm{eq}}=\\frac{108 \\mathrm{c}^5 \\alpha^5}{\\mathrm{c}} \\alpha=\\left(\\frac{\\mathrm{K}_{\\mathrm{eq}}}{108 \\mathrm{c}^4}\\right)^{\\frac{1}{5}}\n\\end{aligned}\n$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 613, "subject": "Chemistry", "question": "

The equilibrium composition for the reaction $$\\mathrm{PCl}_{3}+\\mathrm{Cl}_{2} \\rightleftharpoons \\mathrm{PCl}_{5}$$ at $$298 \\mathrm{~K}$$ is given below:

\n

$$\\left[\\mathrm{PCl}_{3}\\right]_{\\mathrm{eq}}=0.2 \\mathrm{~mol} \\mathrm{~L}^{-1},\\left[\\mathrm{Cl}_{2}\\right]_{\\mathrm{eq}}=0.1 \\mathrm{~mol} \\mathrm{~L}^{-1},\\left[\\mathrm{PCl}_{5}\\right]_{\\mathrm{eq}}=0.40 \\mathrm{~mol} \\mathrm{~L}^{-1}$$

\n

If $$0.2 \\mathrm{~mol}$$ of $$\\mathrm{Cl}_{2}$$ is added at the same temperature, the equilibrium concentrations of $$\\mathrm{PCl}_{5}$$ is __________ $$\\times 10^{-2} \\mathrm{~mol} \\mathrm{~L}^{-1}$$

\n

Given : $$\\mathrm{K}_{\\mathrm{c}}$$ for the reaction at $$298 \\mathrm{~K}$$ is 20

", "options": [], "answer": "49", "solution": "**Answer:** 49\n\n

The initial equilibrium concentrations are given as:

\n$[\\mathrm{PCl}_{3}]_{\\text{eq}} = 0.2 \\, \\mathrm{mol/L}$, $[\\mathrm{Cl}_{2}]_{\\text{eq}} = 0.1 \\, \\mathrm{mol/L}$, $[\\mathrm{PCl}_{5}]_{\\text{eq}} = 0.4 \\, \\mathrm{mol/L}$.

\n

After adding $0.2 \\, \\mathrm{mol}$ of $\\mathrm{Cl_2}$, the new concentration of $\\mathrm{Cl_2}$ becomes $0.1 + 0.2 = 0.3 \\, \\mathrm{mol/L}$.

\n

Let the change in concentration be $x$, so at the new equilibrium, the concentrations are:

\n$[\\mathrm{PCl}_{3}] = 0.2 - x \\, \\mathrm{mol/L}$, $[\\mathrm{Cl}_{2}] = 0.3 - x \\, \\mathrm{mol/L}$, $[\\mathrm{PCl}_{5}] = 0.4 + x \\, \\mathrm{mol/L}$.

\n

Using the equilibrium constant expression $K_c = 20$, we get:

\n$20 = \\frac{0.4+x}{(0.2-x)(0.3-x)}$.

\n

Solving for $x$, we find $x \\approx 0.086$.

\n

Therefore, the new equilibrium concentration of $\\mathrm{PCl}_5$ is:

\n$[\\mathrm{PCl}_5]_{\\text{new eq}} = 0.4 + 0.086 = 0.486 \\, \\mathrm{mol/L} = 48.6 \\times 10^{-2} \\, \\mathrm{mol/L}$.

\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 614, "subject": "Chemistry", "question": "

$$\\mathrm{A}_{(\\mathrm{g})} \\rightleftharpoons \\mathrm{B}_{(\\mathrm{g})}+\\frac{\\mathrm{C}}{2}(\\mathrm{g})$$ The correct relationship between $$\\mathrm{K}_{\\mathrm{P}}, \\alpha$$ and equilibrium pressure $$\\mathrm{P}$$ is

", "options": [ { "text": "$$K_P=\\frac{\\alpha^{1 / 2} P^{3 / 2}}{(2+\\alpha)^{3 / 2}}$$\n" }, { "text": "$$K_P=\\frac{\\alpha^{3 / 2} P^{1 / 2}}{(2+\\alpha)^{1 / 2}(1-\\alpha)}$$\n" }, { "text": "$$K_P=\\frac{\\alpha^{1 / 2} P^{1 / 2}}{(2+\\alpha)^{3 / 2}}$$\n" }, { "text": "$$K_P=\\frac{\\alpha^{1 / 2} P^{1 / 2}}{(2+\\alpha)^{1 / 2}}$$" } ], "answer": "$$K_P=\\frac{\\alpha^{3 / 2} P^{1 / 2}}{(2+\\alpha)^{1 / 2}(1-\\alpha)}$$\n", "solution": "**Answer:** $$K_P=\\frac{\\alpha^{3 / 2} P^{1 / 2}}{(2+\\alpha)^{1 / 2}(1-\\alpha)}$$\n\n\n

$$\\begin{aligned}\n& \\mathrm{A}_{(\\mathrm{g})} \\rightleftharpoons \\mathrm{B}_{(\\mathrm{g})}+\\frac{\\mathrm{C}}{2}(\\mathrm{~g}) \\\\\n& \\mathrm{t}=\\mathrm{t}_{\\mathrm{eq}} \\quad(1-\\alpha) \\quad \\alpha \\quad \\frac{\\alpha}{2}\n\\end{aligned}$$

\n

$$\\mathrm{P}_{\\mathrm{B}}=\\frac{\\alpha}{\\left(1+\\frac{\\alpha}{2}\\right)} \\cdot \\mathrm{P}, \\mathrm{P}_{\\mathrm{A}}=\\frac{(1-\\alpha)}{\\left(1+\\frac{\\alpha}{2}\\right)} \\cdot \\mathrm{P}, \\mathrm{P}_{\\mathrm{C}}=\\frac{\\frac{\\alpha}{2}}{\\left(1+\\frac{\\alpha}{2}\\right)} . \\mathrm{P}$$

\n

$$\\mathrm{K_P=\\frac{P_B \\cdot P_C^{\\frac{1}{2}}}{P_A}}$$

\n

$$\\mathrm{=\\frac{(\\alpha)^{\\frac{3}{2}}(P)^{\\frac{1}{2}}}{(1-\\alpha)(2+\\alpha)^{\\frac{1}{2}}}}$$

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 615, "subject": "Chemistry", "question": "

For the given reaction, choose the correct expression of $$\\mathrm{K}_{\\mathrm{C}}$$ from the following :-

\n

$$\\mathrm{Fe}_{(\\mathrm{aq})}^{3+}+\\mathrm{SCN}_{(\\mathrm{aq})}^{-} \\rightleftharpoons(\\mathrm{FeSCN})_{(\\mathrm{aq})}^{2+}$$

", "options": [ { "text": "$$\\mathrm{K}_{\\mathrm{C}}=\\frac{\\left[\\mathrm{Fe}^{3+}\\right]\\left[\\mathrm{SCN}^{-}\\right]}{\\left[\\mathrm{FeSCN}^{2+}\\right]}$$\n" }, { "text": "$$\\mathrm{K}_{\\mathrm{C}}=\\frac{\\left[\\mathrm{FeSCN}^{2+}\\right]}{\\left[\\mathrm{Fe}^{3+}\\right]\\left[\\mathrm{SCN}^{-}\\right]}$$\n" }, { "text": "$$\\mathrm{K}_{\\mathrm{C}}=\\frac{\\left[\\mathrm{FeSCN}^{2+}\\right]^2}{\\left[\\mathrm{Fe}^{3+}\\right]\\left[\\mathrm{SCN}^{-}\\right]}$$\n" }, { "text": "$$\\mathrm{K}_{\\mathrm{C}}=\\frac{\\left[\\mathrm{FeSCN}^{2+}\\right]}{\\left[\\mathrm{Fe}^{3+}\\right]^2\\left[\\mathrm{SCN}^{-}\\right]^2}$$" } ], "answer": "$$\\mathrm{K}_{\\mathrm{C}}=\\frac{\\left[\\mathrm{FeSCN}^{2+}\\right]}{\\left[\\mathrm{Fe}^{3+}\\right]\\left[\\mathrm{SCN}^{-}\\right]}$$\n", "solution": "**Answer:** $$\\mathrm{K}_{\\mathrm{C}}=\\frac{\\left[\\mathrm{FeSCN}^{2+}\\right]}{\\left[\\mathrm{Fe}^{3+}\\right]\\left[\\mathrm{SCN}^{-}\\right]}$$\n\n\n

$$\\begin{aligned}\n& \\mathrm{K}_{\\mathrm{C}}=\\frac{\\text { Products ion conc. }}{\\text { Reactants ion conc. }} \\\\\n& \\mathrm{K}_{\\mathrm{C}}=\\frac{\\left[\\mathrm{FeSCN}^{2+}\\right]}{\\left[\\mathrm{Fe}^{3+}\\right]\\left[\\mathrm{SCN}^{-}\\right]}\n\\end{aligned}$$

", "topic": "Algebra", "subtopic": "Algebraic Expressions" }, { "id": 616, "subject": "Chemistry", "question": "

For the reaction $$\\mathrm{N}_2 \\mathrm{O}_{4(\\mathrm{~g})} \\rightleftarrows 2 \\mathrm{NO}_{2(\\mathrm{~g})}, \\mathrm{K}_{\\mathrm{p}}=0.492 \\mathrm{~atm}$$ at $$300 \\mathrm{~K} . \\mathrm{K}_{\\mathrm{c}}$$ for the reaction at same temperature is _________ $$\\times 10^{-2}$$.

\n

(Given : $$\\mathrm{R}=0.082 \\mathrm{~L} \\mathrm{~atm} \\mathrm{~mol}^{-1} \\mathrm{~K}^{-1}$$)

", "options": [], "answer": "2", "solution": "**Answer:** 2\n\n

$$\\begin{aligned}\n& \\mathrm{K}_{\\mathrm{P}}=\\mathrm{K}_{\\mathrm{C}} \\cdot(\\mathrm{RT})^{\\Delta \\mathrm{n}_{\\mathrm{g}}} \\\\\n& \\Delta \\mathrm{n}_{\\mathrm{g}}=1 \\\\\n& \\Rightarrow \\mathrm{K}_{\\mathrm{c}}=\\frac{\\mathrm{K}_{\\mathrm{P}}}{\\mathrm{RT}}=\\frac{0.492}{0.082 \\times 300}=2 \\times 10^{-2}\n\\end{aligned}$$

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 617, "subject": "Chemistry", "question": "

The following concentrations were observed at $$500 \\mathrm{~K}$$ for the formation of $$\\mathrm{NH}_3$$ from $$\\mathrm{N}_2$$ and $$\\mathrm{H}_2$$. At equilibrium ; $$\\left[\\mathrm{N}_2\\right]=2 \\times 10^{-2} \\mathrm{M},\\left[\\mathrm{H}_2\\right]=3 \\times 10^{-2} \\mathrm{M}$$ and $$\\left[\\mathrm{NH}_3\\right]=1.5 \\times 10^{-2} \\mathrm{M}$$. Equilibrium constant for the reaction is ________.

", "options": [], "answer": "417", "solution": "**Answer:** 417\n\n

$$\\begin{aligned}\n& \\mathrm{K}_{\\mathrm{C}}=\\frac{\\left[\\mathrm{NH}_3\\right]^2}{\\left[\\mathrm{~N}_2\\right]\\left[\\mathrm{H}_2\\right]^3} \\\\\n& \\mathrm{~K}_{\\mathrm{C}}=\\frac{\\left(1.5 \\times 10^{-2}\\right)^2}{\\left(2 \\times 10^{-2}\\right) \\times\\left(3 \\times 10^{-2}\\right)^3} \\\\\n& \\mathrm{~K}_{\\mathrm{C}}=417\n\\end{aligned}$$

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 618, "subject": "Chemistry", "question": "

The equilibrium constant for the reaction

\n

$$\\mathrm{SO}_3(\\mathrm{~g}) \\rightleftharpoons \\mathrm{SO}_2(\\mathrm{~g})+\\frac{1}{2} \\mathrm{O}_2(\\mathrm{~g})$$

\n

is $$\\mathrm{K}_{\\mathrm{c}}=4.9 \\times 10^{-2}$$. The value of $$\\mathrm{K}_{\\mathrm{c}}$$ for the reaction given below is $$2 \\mathrm{SO}_2(\\mathrm{~g})+\\mathrm{O}_2(\\mathrm{~g}) \\rightleftharpoons 2 \\mathrm{SO}_3(\\mathrm{~g})$$ is :

", "options": [ { "text": "49" }, { "text": "416" }, { "text": "41.6" }, { "text": "4.9" } ], "answer": "416", "solution": "**Answer:** 416\n\n

The reaction

\n

$$2 \\mathrm{SO}_2+\\mathrm{O}_2 \\rightleftharpoons 2 \\mathrm{SO}_3$$

\n

can be formed from the given reaction by reverting it and multiplying coefficients by 2.

\n

Thus,

\n

$$\\begin{aligned}\n\\mathrm{K}_c^{\\prime} & =\\mathrm{K}_{\\mathrm{c}}^{-2}=\\frac{1}{\\mathrm{~K}_{\\mathrm{c}}^2}=\\left(\\frac{1}{4.9 \\times 10^{-2}}\\right)^2 \\\\\n& =416\n\\end{aligned}$$

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 619, "subject": "Chemistry", "question": "

For the given hypothetical reactions, the equilibrium constants are as follows :

\n

$$\\begin{aligned}\n& \\mathrm{X} \\rightleftharpoons \\mathrm{Y} ; \\mathrm{K}_1=1.0 \\\\\n& \\mathrm{Y} \\rightleftharpoons \\mathrm{Z} ; \\mathrm{K}_2=2.0 \\\\\n& \\mathrm{Z} \\rightleftharpoons \\mathrm{W} ; \\mathrm{K}_3=4.0\n\\end{aligned}$$

\n

The equilibrium constant for the reaction $$\\mathrm{X} \\rightleftharpoons \\mathrm{W}$$ is

", "options": [ { "text": "12.0" }, { "text": "8.0" }, { "text": "6.0" }, { "text": "7.0" } ], "answer": "8.0", "solution": "**Answer:** 8.0\n\n

To find the equilibrium constant for the overall reaction $$\\mathrm{X} \\rightleftharpoons \\mathrm{W}$$, we need to combine the equilibrium constants for the individual reactions given. Let's analyze this step-by-step:

\n\n

The given reactions and their equilibrium constants are:

\n\n

$$\\begin{aligned} & \\mathrm{X} \\rightleftharpoons \\mathrm{Y} \\quad K_1 = 1.0 \\\\ & \\mathrm{Y} \\rightleftharpoons \\mathrm{Z} \\quad K_2 = 2.0 \\\\ & \\mathrm{Z} \\rightleftharpoons \\mathrm{W} \\quad K_3 = 4.0 \\end{aligned}$$

\n\n

The equilibrium constant for the overall reaction $$\\mathrm{X} \\rightleftharpoons \\mathrm{W}$$ is the product of the equilibrium constants for the individual steps. This is because the equilibrium constant for a composite reaction is the product of the equilibrium constants for the sequential reactions that lead to the composite reaction.

\n\n

So, we have:

\n\n

$$K_{\\text{overall}} = K_1 \\cdot K_2 \\cdot K_3$$

\n\n

Now, substituting the given values:

\n\n

$$K_{\\text{overall}} = 1.0 \\cdot 2.0 \\cdot 4.0 = 8.0$$

\n\n

Therefore, the equilibrium constant for the reaction $$\\mathrm{X} \\rightleftharpoons \\mathrm{W}$$ is 8.0, which corresponds to Option B.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 620, "subject": "Chemistry", "question": "

Given below are two statements :

\n

Statement I : On passing $$\\mathrm{HCl}_{(\\mathrm{g})}$$ through a saturated solution of $$\\mathrm{BaCl}_2$$, at room temperature white turbidity appears.

\n

Statement II : When $$\\mathrm{HCl}$$ gas is passed through a saturated solution of $$\\mathrm{NaCl}$$, sodium chloride is precipitated due to common ion effect.

\n

In the light of the above statements, choose the most appropriate answer from the options given below :

", "options": [ { "text": "Both Statement I and Statement II are correct\n" }, { "text": "Statement I is incorrect but Statement II is correct\n" }, { "text": "Both Statement I and Statement II are incorrect\n" }, { "text": "Statement I is correct but Statement II is incorrect" } ], "answer": "Statement I is correct but Statement II is incorrect", "solution": "**Answer:** Statement I is correct but Statement II is incorrect\n\n

To determine the correctness of the statements, we'll analyze each one individually by applying principles of solubility, common ion effect, and chemical equilibria.

\n
\n

Statement I:

\n

On passing $\\mathrm{HCl}_{(g)}$ through a saturated solution of $\\mathrm{BaCl}_2$ at room temperature, white turbidity appears.

\n

Analysis:

\n\n

Dissolution of HCl Gas:

\n

When $\\mathrm{HCl}_{(g)}$ is bubbled through water, it dissolves and dissociates completely:

\n

$ \\mathrm{HCl}_{(aq)} \\rightarrow \\mathrm{H}^+_{(aq)} + \\mathrm{Cl}^-_{(aq)} $

\n

This increases the concentration of $\\mathrm{Cl}^-$ ions in the solution.

\n

Effect on BaCl₂ Solubility:

\n

The solubility equilibrium of $\\mathrm{BaCl}_2$ in water is:

\n

$ \\mathrm{BaCl}_2 \\leftrightarrow \\mathrm{Ba}^{2+}_{(aq)} + 2\\mathrm{Cl}^-_{(aq)} $

\n

Adding more $\\mathrm{Cl}^-$ shifts the equilibrium to the left (Le Chatelier's Principle), causing $\\mathrm{BaCl}_2$ to precipitate.

\n

The precipitation of $\\mathrm{BaCl}_2$ manifests as a white turbidity.

\n\n

Conclusion:

\n\n

Statement I is correct.

\n\n
\n

Statement II:

\n

When $\\mathrm{HCl}$ gas is passed through a saturated solution of $\\mathrm{NaCl}$, sodium chloride is precipitated due to common ion effect.

\n

Analysis:

\n\n

Dissolution of HCl Gas:

\n

Similar to before, $\\mathrm{HCl}$ increases $\\mathrm{Cl}^-$ ion concentration.

\n

Effect on NaCl Solubility:

\n

The solubility equilibrium of $\\mathrm{NaCl}$ is:

\n

$ \\mathrm{NaCl} \\leftrightarrow \\mathrm{Na}^+_{(aq)} + \\mathrm{Cl}^-_{(aq)} $

\n

However, $\\mathrm{NaCl}$ is highly soluble in water, and its solubility is not significantly affected by the common ion effect from $\\mathrm{Cl}^-$.

\n

The solubility product ($K_{sp}$) of $\\mathrm{NaCl}$ is large, and the addition of $\\mathrm{Cl}^-$ ions does not cause $\\mathrm{NaCl}$ to precipitate under normal conditions.

\n

No precipitation occurs; the solution remains clear.

\n\n

Conclusion:

\n\n

Statement II is incorrect.

\n\n
\n

Final Answer:

\n\n

Statement I is correct but Statement II is incorrect.

\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 621, "subject": "Chemistry", "question": "

The ratio $$\\frac{K_P}{K_C}$$ for the reaction :

\n

$$\\mathrm{CO}_{(\\mathrm{g})}+\\frac{1}{2} \\mathrm{O}_{2(\\mathrm{~g})} \\rightleftharpoons \\mathrm{CO}_{2(\\mathrm{~g})}$$ is :

", "options": [ { "text": "1" }, { "text": "$$\n(\\mathrm{RT})^{1 / 2}\n$$" }, { "text": "RT" }, { "text": "$$\\mathrm{\n\\frac{1}{\\sqrt{R T}}}\n$$" } ], "answer": "$$\\mathrm{\n\\frac{1}{\\sqrt{R T}}}\n$$", "solution": "**Answer:** $$\\mathrm{\n\\frac{1}{\\sqrt{R T}}}\n$$\n\n

To solve this problem, we need to understand the relationship between the equilibrium constant in terms of pressure, $$K_P$$, and the equilibrium constant in terms of concentration, $$K_C$$. The relationship between these two constants for a general reaction is given by:\n\n

$$ K_P = K_C (RT)^{\\Delta n} $$

\n\n

where:

\n\n

\n\n

\n\n

$$\\Delta n$$ is the change in the number of moles of gas (moles of gaseous products minus moles of gaseous reactants),

\n\n

\n\n

\n\n

$$R$$ is the ideal gas constant, and

\n\n

\n\n

\n\n

$$T$$ is the temperature in Kelvin.

\n\n

\n\n

For the given reaction:\n\n

$$ \\mathrm{CO}_{(\\mathrm{g})}+\\frac{1}{2} \\mathrm{O}_{2(\\mathrm{~g})} \\rightleftharpoons \\mathrm{CO}_{2(\\mathrm{~g})} $$

\n\n

we need to calculate $$\\Delta n$$.

\n\n

On the reactants side, we have:

\n\n\n\n

So, the total number of moles of reactants is:

\n\n

$$ 1 + \\frac{1}{2} = \\frac{3}{2} = 1.5 $$

\n\n

On the products side, we have:

\n\n\n\n

The total number of moles of products is:

\n\n

$$ 1 $$

\n\n

Therefore, $$\\Delta n$$ is:

\n\n

$$ \\Delta n = \\text{moles of products} - \\text{moles of reactants} = 1 - 1.5 = -0.5 $$

\n\n

Now, we can use the relationship between $$K_P$$ and $$K_C$$:

\n\n

$$ K_P = K_C (RT)^{\\Delta n} $$

\n\n

Substituting $$\\Delta n = -0.5$$, we get:

\n\n

$$ K_P = K_C (RT)^{-0.5} $$

\n\n

or equivalently:

\n\n

$$ \\frac{K_P}{K_C} = (RT)^{-0.5} $$

\n\n

Therefore, the correct option is:

\n\n

Option D: $$\\frac{1}{\\sqrt{R T}}$$

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 622, "subject": "Chemistry", "question": "

At $$-20^{\\circ} \\mathrm{C}$$ and $$1 \\mathrm{~atm}$$ pressure, a cylinder is filled with equal number of $$\\mathrm{H}_2, \\mathrm{I}_2$$ and $$\\mathrm{HI}$$ molecules for the reaction\n$$\\mathrm{H}_2(\\mathrm{~g})+\\mathrm{I}_2(\\mathrm{~g}) \\rightleftharpoons 2 \\mathrm{HI}(\\mathrm{g})$$, the $$\\mathrm{K}_{\\mathrm{p}}$$ for the process is $$x \\times 10^{-1}$$.

\n

$$\\mathrm{x}=$$ __________.

\n

[Given : $$\\mathrm{R}=0.082 \\mathrm{~L} \\mathrm{~atm} \\mathrm{~K}^{-1} \\mathrm{~mol}^{-1}$$]

", "options": [ { "text": "2" }, { "text": "1" }, { "text": "10" }, { "text": "0.01" } ], "answer": "10", "solution": "**Answer:** 10\n\n

At $$-20^{\\circ} \\mathrm{C}$$ and a pressure of $$1 \\mathrm{~atm}$$, a cylinder contains equal quantities of $$\\mathrm{H}_2, \\mathrm{I}_2,$$ and $$\\mathrm{HI}$$ molecules. The equilibrium constant for the reaction

\n\n

$$\\mathrm{H}_2(\\mathrm{~g})+\\mathrm{I}_2(\\mathrm{~g}) \\rightleftharpoons 2 \\mathrm{HI}(\\mathrm{g})$$

\n\n

denoted as $$\\mathrm{K}_{\\mathrm{p}}$$, is given by the expression $$x \\times 10^{-1}$$.

\n\n

To solve for $$x$$, use the provided equilibrium expression:

\n\n

$$\\mathrm{K_{P} = \\mathrm{K_C}} =\\frac{\\left[\\mathrm{HI}\\right]^2}{\\left[\\mathrm{H}_2\\right]\\left[\\mathrm{I}_2\\right]}=1 = 10 \\times 10^{-1}$$

\n\n

Therefore, $$x = 10$$.

", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 623, "subject": "Chemistry", "question": "Change in volume of the system does not alter which of the following equilibria?", "options": [ { "text": "N2(g) + O2(g) $$\\leftrightharpoons$$ 2NO (g)" }, { "text": "PCl5(g) $$\\leftrightharpoons$$ PCl3 (g) + Cl2 (g)" }, { "text": "N2(g) + 3H2(g) $$\\leftrightharpoons$$ 2NH3 (g)" }, { "text": "SO2Cl2 (g) $$\\leftrightharpoons$$ SO2 (g) + Cl2 (g)" } ], "answer": "N2(g) + O2(g) $$\\leftrightharpoons$$ 2NO (g)", "solution": "**Answer:** N2(g) + O2(g) $$\\leftrightharpoons$$ 2NO (g)\n\nIn this reaction the ratio of number of moles of reactants to products is same $$i.e.\\,\\,2:2,$$ hence change in volume will not alter the number of moles. ", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 624, "subject": "Chemistry", "question": "Consider the reaction equilibrium
\n2 SO2 (g) + O2 (g) $$\\leftrightharpoons$$ 2 SO3 (g); $$\\Delta H^o$$ = -198 kJ
\nOne the basis of Le Chatelier's principle, the condition favourable for the forward reaction is :", "options": [ { "text": "increasing temperature as well as pressure" }, { "text": "lowering the temperature and increasing the pressure" }, { "text": "any value of temperature and pressure" }, { "text": "lowering temperature as well as pressure" } ], "answer": "lowering the temperature and increasing the pressure", "solution": "**Answer:** lowering the temperature and increasing the pressure\n\nDue to exothermicity of reaction low or optimum temperature will be required. Since $$3$$ moles are changing to $$2$$ moles.\n

$$\\therefore$$ High pressure will be required. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 625, "subject": "Chemistry", "question": "The exothermic formation of ClF3 is represented by the equation:
\nCl2 (g) + 3F2 (g) $$\\leftrightharpoons$$ 2ClF3 (g); $$\\Delta H$$ = -329 kJ
\nWhich of the following will increase the quantity of ClF3 in an equilibrium mixture of\nCl2, F2 and ClF3? ", "options": [ { "text": "Increasing the temperature" }, { "text": "Removing Cl2" }, { "text": "Increasing the volume of the container " }, { "text": "Adding F2" } ], "answer": "Adding F2", "solution": "**Answer:** Adding F2\n\nThe reaction given is an exothermic reaction thus accordingly to Lechatalier's principle lowering of temperature, addition of $${F_2}$$ and or $$C{l_2}$$ favour the for ward direction and hence the production of $$Cl{F_3}.$$ ", "topic": "Algebra", "subtopic": "Equations and Inequalities" }, { "id": 626, "subject": "Chemistry", "question": "The following reaction occurs in the Blast Furnace where iron ore is reduced to iron\nmetal :\n

Fe2O3(s) + 3 CO(g) $$\\rightleftharpoons$$ 2 Fe(1) + 3 CO2(g)\n

Using the Le Chatelier’s principle, predict which one of the following will not disturb the equilibrium ?", "options": [ { "text": "Removal of CO " }, { "text": "Removal of CO2 " }, { "text": "Addition of CO2" }, { "text": "Addition of Fe2O3" } ], "answer": "Addition of Fe2O3", "solution": "**Answer:** Addition of Fe2O3\n\n

Addition of a solid component to a system at constant\n\npressure has no effect on the equilibrium. Therefore,\n\naddition of Fe2O3 will not disturb the equilibrium.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 627, "subject": "Chemistry", "question": "In which of the following reactions, an increase in the volume of the container will favour the formation of products? ", "options": [ { "text": "2NO2(g) $$\\rightleftharpoons$$ 2NO(g) + O2(g)" }, { "text": "H2(g) + I2(g) $$\\rightleftharpoons$$ 2HI(g)" }, { "text": "4NH3(g) + 5O2(g) $$\\rightleftharpoons$$ 4NO(g) + 6H2O(1)" }, { "text": "3O2 (g) $$\\rightleftharpoons$$ 2O3(g)" } ], "answer": "2NO2(g) $$\\rightleftharpoons$$ 2NO(g) + O2(g)", "solution": "**Answer:** 2NO2(g) $$\\rightleftharpoons$$ 2NO(g) + O2(g)\n\nFrom Boyle's law, we know when volume increase then pressure decreases, and to keep the pressure on original value the reaction should proceeds in the direction where the number of moles of gases increases. \n

In this reaction, only satisfy this. \n

2 NO2(g) $$\\rightleftharpoons$$ 2 NO(g) + O2 (g) \n

$$\\therefore\\,\\,\\,\\,$$ $$\\Delta $$ng = (2 + 1) $$-$$ 2 = 1", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 628, "subject": "Chemistry", "question": "The gas phase reaction 2NO2(g) $$ \\to $$ N2O4(g) is an exothermic reaction. The decomposition of N2O4, in equilibrium mixture of NO2(g) and N2O4(g), can be increased by : ", "options": [ { "text": "lowering the temperature." }, { "text": "increasing the pressure." }, { "text": "addition of an inert gas at constant volume. " }, { "text": "addition of an inert gas at constant pressure. " } ], "answer": "addition of an inert gas at constant pressure. ", "solution": "**Answer:** addition of an inert gas at constant pressure. \n\n2NO2 (g) $$\\buildrel \\, \\over\n \\longrightarrow $$ N2O4 (g); $$\\Delta $$H = $$-$$ ve \n

At equilibrium, N2O4 $$\\rightleftharpoons$$ 2NO2 ; $$\\Delta $$H = + ve\n

On adding the inert gas at constant pressure, the number of moles per unit volume of various reactants and products will decrease. Hence, the equilibrium will shift towards the direction in which there is increase in number of moles of gases. Therefore, in above case, reaction will move towards forward direction which will lead to the decomposition of dinitrogen tetraoxide (N2O4).", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 629, "subject": "Chemistry", "question": "For the reaction,\n
2SO2(g) + O2(g) = 2SO3(g), $$\\Delta $$H = –57.2 kJ mol–1\n and KC = 1.7 × 1016\n
Which of the following statement is incorrect ? ", "options": [ { "text": "The equilibrium will shift in forward direction as the pressure increase." }, { "text": "The addition of inert gas at constant volume will be not affect the equilibrium constant." }, { "text": "The equilibrium constant is large suggestive of reaction going to completion and so no catalyst is\nrequired." }, { "text": "The equilibrium constant decreases as the temperature increase." } ], "answer": "The equilibrium constant is large suggestive of reaction going to completion and so no catalyst is\nrequired.", "solution": "**Answer:** The equilibrium constant is large suggestive of reaction going to completion and so no catalyst is\nrequired.\n\nEquilibrium constant has no relation with\ncatalyst.\n

Catalyst only affects the rate with\nwhich a reaction proceeds.\n

Here we use catalyst V2O5 to\nspeed up the reaction.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 630, "subject": "Chemistry", "question": "The INCORRECT match in the following is :", "options": [ { "text": "$$\\Delta $$Go\n = 0, K = 1" }, { "text": "$$\\Delta $$Go\n < 0, K < 1" }, { "text": "$$\\Delta $$Go\n > 0, K < 1" }, { "text": "$$\\Delta $$Go\n < 0, K > 1" } ], "answer": "$$\\Delta $$Go\n < 0, K < 1", "solution": "**Answer:** $$\\Delta $$Go\n < 0, K < 1\n\nWe know, $$\\Delta {G^o} = - RT\\ln K$$\n
Case 1 : \n
If $$\\Delta $$Go < 0\n
$$ \\Rightarrow $$ $$- RT\\ln K$$ < 0\n
$$ \\Rightarrow $$ $$\\ln K$$ > 0\n
$$ \\Rightarrow $$ K > 1\n

Case 2 : \n
If $$\\Delta $$Go > 0\n
$$ \\Rightarrow $$ $$- RT\\ln K$$ > 0\n
$$ \\Rightarrow $$ $$\\ln K$$ < 0\n
$$ \\Rightarrow $$ K < 1\n

Case 2 : \n
If $$\\Delta $$Go = 0\n
$$ \\Rightarrow $$ $$- RT\\ln K$$ = 0\n
$$ \\Rightarrow $$ $$\\ln K$$ = 0\n
$$ \\Rightarrow $$ K = 1", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 631, "subject": "Chemistry", "question": "

40% of HI undergoes decomposition to H2 and I2 at 300 K. $$\\Delta$$G$$^\\Theta $$ for this decomposition reaction at one atmosphere pressure is __________ J mol$$-$$1. [nearest integer]

\n

(Use R = 8.31 J K$$-$$1 mol$$-$$1 ; log 2 = 0.3010, ln 10 = 2.3, log 3 = 0.477)

", "options": [], "answer": "2735", "solution": "**Answer:** 2735\n\n

\"JEE

\n

$$\\therefore$$ $$K = {{{{\\left( {{\\alpha \\over 2}} \\right)}^{1/2}} \\times {{\\left( {{\\alpha \\over 2}} \\right)}^{1/2}}} \\over {(1 - \\alpha )}}$$

\n

Given $$\\alpha = {{40} \\over {100}} = 0.4$$

\n

$$\\therefore$$ $$K = {{{{\\left( {{{0.4} \\over 2}} \\right)}^{1/2}} \\times {{\\left( {{{0.4} \\over 2}} \\right)}^{1/2}}} \\over {(1 - 0.4)}}$$

\n

$$ = {1 \\over 3}$$

\n

We know,

\n

$$\\Delta G^\\circ = - RT\\ln K$$

\n

$$ = - RT\\ln \\left( {{1 \\over 3}} \\right)$$

\n

$$ = + RT\\ln 3$$

\n

$$ = + 8.314 \\times 300 \\times \\ln 3$$

\n

= 2735 J/mol

", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 632, "subject": "Chemistry", "question": " Consider the following equation:\n

\n$2 \\mathrm{SO}_{2}(g)+\\mathrm{O}_{2}(g) \\rightleftharpoons 2 \\mathrm{SO}_{3}(g), \\Delta H=-190 \\mathrm{~kJ}$\n

\nThe number of factors which will increase the yield of $\\mathrm{SO}_{3}$ at equilibrium from the following is _______.

\nA. Increasing temperature

\nB. Increasing pressure

\nC. Adding more $\\mathrm{SO}_{2}$

\nD. Adding more $\\mathrm{O}_{2}$

\nE. Addition of catalyst", "options": [], "answer": "3", "solution": "**Answer:** 3\n\n

$$\\mathrm{2SO_2+O_2\\rightleftharpoons 2SO_3~~\\Delta H=-190~kJ}$$

\n

It is an exothermic reaction

\n

$$\\therefore$$ factor B, C, D will increase the amount of SO$$_3$$.

", "topic": "Number Theory", "subtopic": "Divisibility / Factorization" }, { "id": 633, "subject": "Chemistry", "question": "

At 298 K

\n

$$\\mathrm{N_2~(g)+3H_2~(g)\\rightleftharpoons~2NH_3~(g),~K_1=4\\times10^5}$$

\n

$$\\mathrm{N_2~(g)+O_2~(g)\\rightleftharpoons~2NO~(g),~K_2=1.6\\times10^{12}}$$

\n

$$\\mathrm{H_2~(g)+\\frac{1}{2}O_2~(g)\\rightleftharpoons~H_2O~(g),~K_3=1.0\\times10^{-13}}$$

\n

Based on above equilibria, then equilibrium constant of the reaction, $$\\mathrm{2NH_3(g)+\\frac{5}{2}O_2~(g)\\rightleftharpoons~2NO~(g)+3H_2O~(g)}$$ is ____________ $$\\times10^{-33}$$ (Nearest integer).

", "options": [], "answer": "4", "solution": "**Answer:** 4\n\n

$$\\mathrm{2NH_3(g)\\frac{5}{2}O_2(g)\\rightleftharpoons 2NO(g)+3H_2O(g)}$$

\n

Clearly, $$\\mathrm{{K_{eq}} = {1 \\over {{k_1}}} \\times {k_2} \\times k_3^3}$$

\n

$$ = {{1.6 \\times {{10}^{12}} \\times {{10}^{ - 39}}} \\over {4 \\times {{10}^5}}}$$

\n

$$ = 0.4 \\times {10^{ - 32}}$$

\n

$$ = 4 \\times {10^{ - 33}}$$

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 634, "subject": "Chemistry", "question": "

The number of correct statement/s involving equilibria in physical processes from the following is ________

\n

(A) Equilibrium is possible only in a closed system at a given temperature.

\n

(B) Both the opposing processes occur at the same rate.

\n

(C) When equilibrium is attained at a given temperature, the value of all its parameters became equal.

\n

(D) For dissolution of solids in liquids, the solubility is constant at a given temperature.

", "options": [], "answer": "3", "solution": "**Answer:** 3\n\n

(A) Equilibrium is possible only in a closed system at a given temperature.

\n\n

(B) Both the opposing processes occur at the same rate.

\n\n

(C) When equilibrium is attained at a given temperature, the value of all its parameters became equal.

\n\n

(D) For dissolution of solids in liquids, the solubility is constant at a given temperature.

\n\n

Therefore, there are 3 correct statements: (A), (B), and (D).

\n", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 635, "subject": "Chemistry", "question": "

The following reaction occurs in the Blast furnance where iron ore is reduced to iron metal

\n

$$\\mathrm{Fe}_2 \\mathrm{O}_{3(s)}+3 \\mathrm{CO}_{(g)} \\rightleftharpoons \\mathrm{Fe}_{(j)}+3 \\mathrm{CO}_{2(g)}$$

\n

Using the Le-chatelier's principle, predict which one of the following will not disturb the equilibrium.

", "options": [ { "text": "Addition of $$\\mathrm{CO}_2$$\n" }, { "text": "Removal of $$\\mathrm{CO}$$\n" }, { "text": "Addition of $$\\mathrm{Fe}_2 \\mathrm{O}_3$$\n" }, { "text": "Removal of $$\\mathrm{CO}_2$$" } ], "answer": "Addition of $$\\mathrm{Fe}_2 \\mathrm{O}_3$$\n", "solution": "**Answer:** Addition of $$\\mathrm{Fe}_2 \\mathrm{O}_3$$\n\n\n

For the reaction :

\n

$$\\mathrm{Fe}_2 \\mathrm{O}_{3(\\mathrm{~s})}+3 \\mathrm{CO}_{(\\mathrm{g})} \\rightleftharpoons \\mathrm{Fe}_{(\\mathrm{l})}+3 \\mathrm{CO}_{2(\\mathrm{~g})}$$

\n

Addition or removal of $$\\mathrm{Fe}_2 \\mathrm{O}_{3(\\mathrm{s})}$$ and/or $$\\mathrm{Fe}_{\\text {(l) }}$$ will not affect the equilibrium quotient and the equilibrium.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 636, "subject": "Chemistry", "question": "In respect of the equation k = Ae-Ea/RT in chemical kinetics, which one of the following statements is correct?", "options": [ { "text": "A is adsorption factor" }, { "text": "Ea is energy of activation" }, { "text": "R is Rydberg’s constant" }, { "text": "k is equilibrium constant" } ], "answer": "Ea is energy of activation", "solution": "**Answer:** Ea is energy of activation\n\nIn equation $$K = A{e^{ - {E_a}/RT}};A = $$ Frequency factor\n

$$K = $$ velocity constant, $$R=$$ gas constant and \n

$${E_b} = $$ energy of activation ", "topic": "Algebra", "subtopic": "Equations and Inequalities" }, { "id": 637, "subject": "Chemistry", "question": "Rate of a reaction can be expressed by Arrhenius equation as:
\n$$$k = A\\,{e^{ - E/RT}}$$$\nIn this equation, E represents ", "options": [ { "text": "the energy above which all the colliding molecules will react" }, { "text": "the energy below which colliding molecules will not react " }, { "text": "the total energy of the reacting molecules at a temperature, T " }, { "text": "the fraction of molecules with energy greater than the activation energy of the reaction" } ], "answer": "the energy above which all the colliding molecules will react", "solution": "**Answer:** the energy above which all the colliding molecules will react\n\nIn Arrhenius equation $$K = A\\,{e^{ - E/RT}},\\,\\,E$$ is the energy of activation, which is required by the colliding molecules to react resulting in the formation of products. ", "topic": "Algebra", "subtopic": "Equations and Inequalities" }, { "id": 638, "subject": "Chemistry", "question": "The energies of activation for forward and reverse reactions for A2 + B2 $$\\leftrightharpoons$$ 2AB are 180 kJ mol−1\nand 200 kJ mol−1 respectively. The presence of catalyst lowers the activation energy of both (forward and reverse) reactions by 100 kJ mol−1. The enthalpy change of the reaction ( A2 + B2 $$\\to$$ 2AB) in the presence of catalyst will be (in kJ mol−1) ", "options": [ { "text": "300" }, { "text": "120" }, { "text": "200" }, { "text": "20" } ], "answer": "20", "solution": "**Answer:** 20\n\n$$\\Delta {H_R} = {E_f} - {E_b} = 180 - 200 = - 20kJ/mol$$ \n

The nearest correct answer given in choices may be obtained by neglecting sign. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 639, "subject": "Chemistry", "question": "The rate of a reaction doubles when its temperature changes from 300K to 310K. Activation energy of such\na reaction will be:(R = 8.314 JK–1 mol–1 and log 2 = 0.301) ", "options": [ { "text": "48.6 kJ mol–1 " }, { "text": "58.5 kJ mol–1" }, { "text": "60.5 kJ mol–1 " }, { "text": "53.6 kJ mol–1" } ], "answer": "53.6 kJ mol–1", "solution": "**Answer:** 53.6 kJ mol–1\n\nActivation energy can be calculated from the equation \n

$${{\\log \\,{k_2}} \\over {\\log \\,{k_1}}} = {{{E_a}} \\over {2.303R}}\\left( {{1 \\over {{T_1}}} - {1 \\over {{T_2}}}} \\right)$$ \n

given $${{{k_2}} \\over {{k_1}}} = 2\\,\\,{T_2} = 310\\,K\\,\\,{T_1} = 300\\,K$$\n

$$ = \\log 2 = {{ - {E_a}} \\over {2.303 \\times 8.314}}\\left( {{1 \\over {310}} - {1 \\over {300}}} \\right)$$\n

$${E_a} = 53598.6J/mol$$\n

$$ = 53.6kJ/mol.$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 640, "subject": "Chemistry", "question": "Two reactions R1 and R2 have identical pre-exponential factors. Activation energy of R1 exceeds that of R2 by 10 kJ mol–1. If k1 and k2 are rate constants for reactions R1 and R2 respectively at 300 K, then ln(k2/k1) is equal to :
\n(R = 8.314 J mol–1 K–1) ", "options": [ { "text": "12" }, { "text": "6" }, { "text": "4" }, { "text": "8" } ], "answer": "4", "solution": "**Answer:** 4\n\nWe know, from arrhenius equation,\n

k = A.$${e^{{{ - {E_a}} \\over {RT}}}}$$\n

$$ \\therefore $$ k1 = A.$${e^{{{ - {E_{{a_1}}}} \\over {RT}}}}$$ ......(1)\n

k2 = A.$${e^{{{ - {E_{{a_2}}}} \\over {RT}}}}$$ ......(2)\n

On dividing equation (2) by (1), we get\n

$${{{k_2}} \\over {{k_1}}} = {e^{{{\\left( {{E_{{a_1}}} - {E_{{a_2}}}} \\right)} \\over {RT}}}}$$\n

$$ \\Rightarrow $$ $$\\ln \\left( {{{{k_2}} \\over {{k_1}}}} \\right) = {{\\left( {{E_{{a_1}}} - {E_{{a_2}}}} \\right)} \\over {RT}}$$ = $${{10,000} \\over {8.314 \\times 300}}$$ = 4", "topic": "Number Theory", "subtopic": "Divisibility / Factorization" }, { "id": 641, "subject": "Chemistry", "question": "The rate of a reaction A doubles on increasing the temperature from 300 to 310 K. By how much, the temperature of reaction B should be Increased from 300 K so that rate doubles if activation energy of the reaction B is twice to that of reaction A.", "options": [ { "text": "9.84 K" }, { "text": "4.92 K" }, { "text": "2.45 K" }, { "text": "19.67 K" } ], "answer": "4.92 K", "solution": "**Answer:** 4.92 K\n\nFor reaction A, T1 = 300 K, T2 = 310 K, k2 = 2 k1\n

$$\\log {{{k_2}} \\over {{k_1}}} = {{{E_{{a_1}}}} \\over {2.303R}}\\left[ {{1 \\over {{T_1}}} - {1 \\over {{T_2}}}} \\right]$$\n

$$ \\therefore $$ $$\\log {{2{k_1}} \\over {{k_1}}} = \\log 2 = {{{E_{{a_1}}}} \\over {2.303R}}\\left[ {{1 \\over {300}} - {1 \\over {310}}} \\right]$$ ...(1)\n

For reaction B, T1 = 300 K, T2 = ?, k2 = 2k1, Ea2 = 2Ea1\n

$$\\log {{2{k_1}} \\over {{k_1}}} = \\log 2 = {{2{E_{{a_1}}}} \\over {2.303R}}\\left[ {{1 \\over {300}} - {1 \\over {{T_2}}}} \\right]$$\n

From eq. (i) and (ii), we get\n

$${{2{E_{{a_1}}}} \\over {2.303R}}\\left[ {{1 \\over {300}} - {1 \\over {{T_2}}}} \\right] = {{{E_{{a_1}}}} \\over {2.303R}}\\left[ {{1 \\over {300}} - {1 \\over {310}}} \\right]$$\n

$$ \\Rightarrow $$ $$2\\left[ {{1 \\over {300}} - {1 \\over {{T_2}}}} \\right] = \\left[ {{1 \\over {300}} - {1 \\over {310}}} \\right]$$\n

$$ \\Rightarrow $$ T2 = $${{{300 \\times 310} \\over {610}}}$$ $$ \\times $$ 2 = 304.92 K\n

$$ \\therefore $$ Increased temperature = (304.92 – 300) = 4.92 K", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 642, "subject": "Chemistry", "question": "The rate of a reaction quadruples when the temperature changes from 300 to 310 K. The activation energy of this reaction is :\n

(Assume activation energy and preexponential factor are independent of\ntemperature; ln 2 = 0.693; R = 8.314 J mol−1 K−1)\n", "options": [ { "text": "107.2 kJ mol$$-$$1" }, { "text": "53.6 kJ mol$$-$$1" }, { "text": "26.8 kJ mol$$-$$1" }, { "text": "214.4 kJ mol$$-$$1" } ], "answer": "107.2 kJ mol$$-$$1", "solution": "**Answer:** 107.2 kJ mol$$-$$1\n\n

According to Arrhenius equation,

\n

$$\\log {{{k_2}} \\over {{k_1}}} = {{{E_a}} \\over {2.303R}}\\left( {{{{T_2} - {T_1}} \\over {{T_1}{T_2}}}} \\right)$$

\n

Substituting the given values in the equation, we get

\n

$$\\log 4 = {{{E_a}} \\over {2.303 \\times 8.314\\,J\\,{K^{ - 1}}mo{l^{ - 1}}}}\\left( {{{310 - 300} \\over {300 \\times 310}}} \\right)$$

\n

$${E_a} = {{0.602 \\times 2.303 \\times 8.314 \\times 300 \\times 310} \\over {10}}$$

\n

= 107197.12 J/mol$$-$$1 or 107.2 kJ/mol$$-$$1

", "topic": "Number Theory", "subtopic": "Divisibility / Factorization" }, { "id": 643, "subject": "Chemistry", "question": "If a reaction follows the Arrhenius equation, the plot ln k vs $${1 \\over {\\left( {RT} \\right)}}$$ gives straight line with a gradient ($$-$$ y) unit. \n

The energy required to active the reactant is : ", "options": [ { "text": "y unit" }, { "text": "y/R unit" }, { "text": "yR unit" }, { "text": "$$-$$y unit" } ], "answer": "y unit", "solution": "**Answer:** y unit\n\nAccording to Arrhenius equation,\n
k = A$${e^{ - {{{E_a}} \\over {RT}}}}$$\n

or ln k = ln A - $${{{{E_a}} \\over {RT}}}$$\n

Comparing the above equation\nwith straight line equation,\n

y = mx + c,\n

we get, slope or gradient (m) = –Ea\n
and Intercept (c) = ln A\n

Also given that slope or gradient (m) = -y\n

$$ \\therefore $$ -y = –Ea\n

$$ \\Rightarrow $$ Ea = y\n

So the activation energy of the reactant, Ea = y unit", "topic": "Algebra", "subtopic": "Equations and Inequalities" }, { "id": 644, "subject": "Chemistry", "question": "For the reaction of H2 with I2, the rate constant is 2.5 × 10–4 dm3\n mol–1s–1\n at 327°C and 1.0 dm3\n mol–1\n at\n527°C. The activation energy for the reaction, in kJ mole–1\n is : (R = 8.314 JK–1\n mol–1\n)", "options": [ { "text": "59" }, { "text": "166" }, { "text": "72" }, { "text": "150" } ], "answer": "166", "solution": "**Answer:** 166\n\nFrom Arrhenius equation, we get\n

$$\\log {{{K_2}} \\over {{K_1}}} = {{{E_a}} \\over {2.303R}}\\left( {{1 \\over {{T_1}}} - {1 \\over {{T_2}}}} \\right)$$\n

$$ \\Rightarrow $$ $$\\log {1 \\over {2.5 \\times {{10}^{ - 4}}}} = {{{E_a}} \\over {2.303 \\times 8.314}}\\left( {{1 \\over {600}} - {1 \\over {800}}} \\right)$$\n

$$ \\Rightarrow $$ 3.6 = $${{{E_a}} \\over {2.303 \\times 8.314}} \\times {{200} \\over {600 \\times 800}}$$\n

$$ \\Rightarrow $$ E$$a$$ = 166 kJ/mol", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 645, "subject": "Chemistry", "question": "The rate of a certain biochemical reaction at physiological temperature (T) occurs 106 times faster with\nenzyme than without. The change in the activation energy upon adding enzyme is :", "options": [ { "text": "+ 6RT" }, { "text": "– 6 (2.303)RT" }, { "text": "– 6RT" }, { "text": "+ 6(2.303)RT" } ], "answer": "– 6 (2.303)RT", "solution": "**Answer:** – 6 (2.303)RT\n\nThe rate constant of a reaction without catalyst is\n

$$k = A{e^{ - {{{E_a}} \\over {RT}}}}$$\n

The rate constant in presence of catalyst is\ngiven by\n

$$k' = A{e^{ - {{E{'_a}} \\over {RT}}}}$$\n

$$ \\therefore $$ $${{k'} \\over k} = {e^{ - {{\\left( {E{'_a} - {E_a}} \\right)} \\over {RT}}}}$$\n

$$ \\Rightarrow $$ 106 = $${e^{ - {{\\left( {E{'_a} - {E_a}} \\right)} \\over {RT}}}}$$\n

$$ \\Rightarrow $$ ln 106 = $${ - {{\\left( {E{'_a} - {E_a}} \\right)} \\over {RT}}}$$\n

$$ \\Rightarrow $$ $${E{'_a} - {E_a}}$$ = -RTln 106\n

$$ \\Rightarrow $$ $${E{'_a} - {E_a}}$$ = -6RT$$ \\times $$2.303", "topic": "Discrete Mathematics", "subtopic": "Logic" }, { "id": 646, "subject": "Chemistry", "question": "The rate of a reaction decreased by 3.555\ntimes when the temperature was changed\nfrom 40oC to 30oC. The activation energy\n(in kJ mol–1) of the reaction is _______. \n

Take;\nR = 8.314 J mol–1 K–1 ln 3.555 = 1.268", "options": [], "answer": "100", "solution": "**Answer:** 100\n\nk = A$${e^{ - {{{E_a}} \\over {RT}}}}$$\n

$$ \\therefore $$ $$\\ln {{{k_2}} \\over {{k_1}}} = {{{E_a}} \\over R}\\left( {{1 \\over {{T_1}}} - {1 \\over {{T_2}}}} \\right)$$\n

$$ \\Rightarrow $$ ln (3.555) = $${{{E_a}} \\over {8.314}}\\left( {{1 \\over {303}} - {1 \\over {313}}} \\right)$$\n

$$ \\Rightarrow $$ Ea = $${{1.268 \\times 8.314 \\times 3.3 \\times 313} \\over {10}}$$\n

= 99980.7 = 99.98 kJ/mol $$ \\simeq $$ 100 kJ/mol", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 647, "subject": "Chemistry", "question": "A flask contains a mixture of compounds A and\nB. Both compounds decompose by first-order\nkinetics. The half-lives for A and B are 300 s\nand 180 s, respectively. If the concentrations\nof A and B are equal initially, the time required\nfor the concentration of A to be four times that\nof B(in s) : \n
(Use ln 2 = 0.693)", "options": [ { "text": "180" }, { "text": "120" }, { "text": "300" }, { "text": "900" } ], "answer": "900", "solution": "**Answer:** 900\n\nAt = A0.e-k1t\n

Bt = B0.e-k2t\n

k1 = $${{\\ln 2} \\over {300}}$$\n

k2 = $${{\\ln 2} \\over {180}}$$\n

Given, A0 = B0\n

and At\n and Bt\n are related as [A] = 4[B]\n

$$ \\therefore $$ A0.e-k1t = 4B0.e-k2t\n

$$ \\Rightarrow $$ $${e^{\\left( {{{\\ln 2} \\over {180}} - {{\\ln 2} \\over {300}}} \\right)t}}$$ = 4\n

$$ \\Rightarrow $$ $${\\left( {{{\\ln 2} \\over {180}} - {{\\ln 2} \\over {300}}} \\right)t}$$ = ln 4 = 2ln 2\n

$$ \\Rightarrow $$ $${\\left( {{1 \\over {180}} - {{\\mathop{\\rm l}\\nolimits} \\over {300}}} \\right)t}$$ = 2\n

$$ \\Rightarrow $$ t = $${{2 \\times 180 \\times 300} \\over {120}}$$ = 900 sec", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 648, "subject": "Chemistry", "question": "The number of molecules with energy greater\nthan the threshold energy for a reaction\nincreases five fold by a rise of temperature\nfrom 27oC to 42oC. Its energy of activation in\nJ/mol is _____.\n
(Take ln 5 = 1.6094; R = 8.314 J mol–1 K–1)\n", "options": [], "answer": "84297.47to84297.48", "solution": "**Answer:** 84297.47to84297.48\n\n$$ \\because $$ k = A$${e^{ - {{{E_a}} \\over {RT}}}}$$\n

T1 = 300K, T2 = 315K\n

As per question KT2 = 5KT2 as molecules activated are increased five times so K will increases five time.\n

$$\\ln \\left( {{{{K_{{T_2}}}} \\over {{K_{{T_1}}}}}} \\right)$$ = $${{{E_a}} \\over R}\\left( {{1 \\over {{T_1}}} - {1 \\over {{T_2}}}} \\right)$$\n

$$ \\Rightarrow $$ ln 5 = $${{{E_a}} \\over R}\\left( {{{15} \\over {300 \\times 315}}} \\right)$$\n

$$ \\Rightarrow $$ Ea = $${{1.6094 \\times 8.314 \\times 300 \\times 315} \\over {15}}$$\n

= 84297.47 Joules/mole", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 649, "subject": "Chemistry", "question": "A sample of milk splits after 60 min. at 300 K\nand after 40 min. at 400 K when the population\nof lactobacillus acidophilus in it doubles. The\nactiva tion energy (in kJ/ mol) for this process\nis closest to__________.

\n(Given, R = 8.3 J mol–1 K–1, $$\\ln \\left( {{3 \\over 2}} \\right) = 0.4$$,\ne–3 = 4.0)\n", "options": [], "answer": "3.98TO3.99", "solution": "**Answer:** 3.98TO3.99\n\nUsing Arrehenius equation\n

K = A$${e^{ - {{{E_a}} \\over {RT}}}}$$\n

$$\\ln \\left( {{{{k_{400}}} \\over {{k_{300}}}}} \\right) = {{{E_a}} \\over R}\\left( {{1 \\over {300}} - {1 \\over {400}}} \\right)$$\n

$$ \\Rightarrow $$ $$\\ln \\left( {{{60} \\over {40}}} \\right) = {{{E_a}} \\over R}\\left( {{{100} \\over {300 \\times 400}}} \\right)$$\n

$$ \\Rightarrow $$ $$\\ln \\left( {{3 \\over 2}} \\right) = {{{E_a}} \\over {1200R}}$$\n

$$ \\therefore $$ Ea = 0.4 × 1200 × 8.3 = 3984 J/mol\n

Ea = 3.984 kJ/mol\n= 3.98 kJ/mol\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 650, "subject": "Chemistry", "question": "For the following reactions
\n$$A\\buildrel {700K} \\over\n \\longrightarrow {\\mathop{\\rm Product}\\nolimits} $$
\n$$A\\mathrel{\\mathop{\\kern0pt\\longrightarrow}\n\\limits_{catalyst}^{500K}} {\\mathop{\\rm Product}\\nolimits} $$
\nit was found that Ea is decreased by 30 kJ/mol\nin the presence of catalyst.
\nIf the rate remains unchanged, the activation\nenergy for catalysed reaction is (Assume pre\nexponential factor is same):", "options": [ { "text": "198 kJ/mol" }, { "text": "135 kJ/mol" }, { "text": "105 kJ/mol" }, { "text": "75 kJ/mol" } ], "answer": "75 kJ/mol", "solution": "**Answer:** 75 kJ/mol\n\nK1 = A$${e^{ - {{{E_a}} \\over {R \\times 700}}}}$$\n

K2 = A$${e^{ - {{\\left( {{E_a} - 30} \\right)} \\over {R \\times 500}}}}$$\n

$$ \\because $$Rate is same\n

$$ \\therefore $$ Rate constant will also be same\n

K1 = K2\n

A$${e^{ - {{{E_a}} \\over {R \\times 700}}}}$$ = A$${e^{ - {{\\left( {{E_a} - 30} \\right)} \\over {R \\times 500}}}}$$\n

$$ \\Rightarrow $$ $${{{\\left( {{E_a} - 30} \\right)} \\over {R \\times 500}}}$$ = $${{{{E_a}} \\over {R \\times 700}}}$$\n

$$ \\Rightarrow $$ 5Ea = 7Ea – 210\n

$$ \\Rightarrow $$ 210 = 2Ea\n

$$ \\Rightarrow $$ Ea = 105 kJ/mole\n

$$ \\therefore $$ Activation energy in the presence of\ncatalyst = 105 – 30 = 75 kJ/mol", "topic": "Number Theory", "subtopic": "Divisibility / Factorization" }, { "id": 651, "subject": "Chemistry", "question": "The rate constant of a reaction increases by five times on increase in temperature from 27$$^\\circ$$C to 52$$^\\circ$$C. The value of activation energy in kJ mol$$-$$1 is _________. (Rounded off to the nearest integer)

[R = 8.314 J K$$-$$1mol$$-$$1]", "options": [], "answer": "52", "solution": "**Answer:** 52\n\nT1 = (273 + 27) = 300 K, T2 = (273 + 52) = 325 K

Given, temperature coefficient of the reaction,

$${\\alpha _T} = {{{K_{325}}} \\over {{K_{300}}}} = 5$$

$$\\log {{{K_{{T_2}}}} \\over {{K_{{T_1}}}}} = {{{E_a}} \\over {2.303R}} \\times \\left( {{{{T_2} - {T_1}} \\over {{T_1}{T_2}}}} \\right)$$

$$\\log {{{K_{325}}} \\over {{K_{300}}}} = {{{E_a}} \\over {2.303 \\times 8.314}}\\left( {{{325 - 300} \\over {300 \\times 325}}} \\right)$$

$$\\log 5 = {{{E_a}} \\over {2.303 \\times 8.319}} \\times {{25} \\over {300 \\times 325}}$$

Ea = 52194.78 J mol$$-$$

= 52.194 kJ mol$$-$$1

$$\\simeq$$ 52 kJ mol$$-$$1", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 652, "subject": "Chemistry", "question": "If the activation energy of a reaction is 80.9 kJ mol$$-$$1, the fraction of molecules at 700 K, having enough energy to react to form
products is e$$-$$x. The value of x is __________. (Rounded off to the nearest integer) [Use R = 8.31 J K$$-$$1 mol$$-$$1]", "options": [], "answer": "14", "solution": "**Answer:** 14\n\nEa = 80.9 kJ/mol

Fraction of molecules able to cross energy barrier = e$$-$$Ea/RT = e$$-$$x

x = $${{{E_a}} \\over {RT}} = {{80.9 \\times 1000} \\over {8.31 \\times 700}} = 13.91$$

$$ \\Rightarrow $$ x $$ \\simeq $$ 14", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 653, "subject": "Chemistry", "question": "The decomposition of formic acid on gold surface follows first order kinetics. If the rate constant at 300 K is 1.0 $$\\times$$ 10$$-$$3 s$$-$$1 and the activation energy Ea = 11.488 kJ mol$$-$$1, the rate constant at 200 K is ____________ $$\\times$$ 10$$-$$5 s$$-$$1. (Round off to the Nearest Integer). (Given : R = 8.314 J mol$$-$$1 K$$-$$1)", "options": [], "answer": "10", "solution": "**Answer:** 10\n\n$$\\log {{{k_2}} \\over {{k_1}}} = {{{E_a}} \\over {2.303R}}\\left[ {{1 \\over {{T_1}}} - {1 \\over {{T_2}}}} \\right]$$

k1 (at 200 K) = ?

k2 (at 300 K) = $$1 \\times {10^{ - 3}}{s^{ - 1}}$$

$$\\log {{1 \\times {{10}^{ - 3}}} \\over {{k_1}}} = {{11.488 \\times {{10}^3}} \\over {2.303 \\times 8.314}}\\left[ {{1 \\over {600}}} \\right] = 1$$

$$ \\Rightarrow $$ $${{1 \\times {{10}^{ - 3}}} \\over {{k_1}}} = 10$$

$$ \\Rightarrow $$ $${k_1} = 10 \\times {10^{ - 5}}{s^{ - 1}}$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 654, "subject": "Chemistry", "question": "A and B decompose via first order kinetics with half-lives 54.0 min and 18.0 min respectively. Starting from an equimolar non reactive mixture of A and B, the time taken for the concentration of A to become 16 times that of B is _________ min. (Round off to the Nearest Integer).", "options": [], "answer": "108", "solution": "**Answer:** 108\n\nInitially : $$\\left[ {{A_0}} \\right] = \\left[ {{B_0}} \\right] = a$$

After time 't' min : $$\\left[ A \\right] = 16\\left[ B \\right]$$

$$\\left[ A \\right] = \\left[ {{A_0}} \\right]{e^{ - {k_A}t}}$$

$$\\left[ B \\right] = \\left[ {{B_0}} \\right]{e^{ - {k_B}t}}$$

$$ \\Rightarrow a\\,.\\,{e^{ - {k_A}t}} = 16a{e^{ - {k_B}t}}$$

$$ \\Rightarrow {e^{ - \\left( {{k_A} - {k_B}} \\right)t}} = 16$$

$$ \\Rightarrow \\left( {{k_B} - {k_A}} \\right)t = \\ln 16$$

$$ \\Rightarrow \\ln 2\\left( {{1 \\over {18}} - {1 \\over {54}}} \\right)t = 4\\ln 2$$

$$ \\Rightarrow t = {{54 \\times 18 \\times 4} \\over {36}} = 108$$ min", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 655, "subject": "Chemistry", "question": "The first order rate constant for the decomposition of CaCO3 at 700 K is 6.36 $$\\times$$ 10$$-$$3s$$-$$1 and activation energy is 209 kJ mol$$-$$1. Its rate constant (in s$$-$$1) at 600 K is x $$\\times$$ 10$$-$$6. The value of x is ___________. (Nearest integer)

[Given R = 8.31 J K$$-$$ mol$$-$$1; log 6.36 $$\\times$$ 10$$-$$3 = $$-$$2.19, 10$$-$$4.79 = 1.62 $$\\times$$ 10$$-$$5]", "options": [], "answer": "16", "solution": "**Answer:** 16\n\nK700 = 6.36 $$\\times$$ 10$$-$$3s$$-$$1;

K600 = x $$\\times$$ 10$$-$$6s$$-$$1

Ea = 209 kJ/mol

Applying;

$$\\log \\left( {{{{K_{{T_2}}}} \\over {{K_{{T_1}}}}}} \\right) = {{ - {E_a}} \\over {2.303R}}\\left( {{1 \\over {{T_2}}} - {1 \\over {{T_1}}}} \\right)$$

$$\\log \\left( {{{{K_{700}}} \\over {{K_{600}}}}} \\right) = {{ - {E_a}} \\over {2.303R}}\\left( {{1 \\over {700}} - {1 \\over {600}}} \\right)$$

$$\\log \\left( {{{6.36 \\times {{10}^{ - 3}}} \\over {{K_{600}}}}} \\right) = {{ + 209 \\times 1000} \\over {2.303 \\times 8.31}}\\left( {{{100} \\over {700 \\times 600}}} \\right)$$

log(6.36 $$\\times$$ 10$$-$$3) $$-$$ logK600 = 2.6

$$\\Rightarrow$$ logK600 = $$-$$2.19 $$-$$ 2.6 = $$-$$4.79

$$\\Rightarrow$$ K600 = 10$$-$$4.79 = 1.62 $$\\times$$ 10$$-$$5

= 1.62 $$\\times$$ 10$$-$$6

= x $$\\times$$ 10$$-$$6

$$\\Rightarrow$$ x = 16", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 656, "subject": "Chemistry", "question": "For the reaction A $$\\to$$ B, the rate constant k(in s$$-$$1) is given by

$${\\log _{10}}k = 20.35 - {{(2.47 \\times {{10}^3})} \\over T}$$

The energy of activation in kJ mol$$-$$1 is ____________. (Nearest integer) [Given : R = 8.314 J K$$-$$1 mol$$-$$1]", "options": [], "answer": "47", "solution": "**Answer:** 47\n\nGiven,

$$\\log K = 20.35 - {{2.47 \\times {{10}^3}} \\over T}$$

We know

$$\\log K = \\log A - {{{E_a}} \\over {2.303RT}}$$

$$ \\Rightarrow {{{E_a}} \\over {2.303RT}} = 2.47 \\times {10^3}$$

$${E_a} = 2.47 \\times {10^3} \\times 2.303 \\times {{8.314} \\over {1000}}$$ KJ/mole

= 47.29 = 47 (Nearest integer)", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 657, "subject": "Chemistry", "question": "

The activation energy of one of the reactions in a biochemical process is 532611 J mol$$-$$1. When the temperature falls from 310 K to 300 K, the change in rate constant observed is k300 = x $$\\times$$ 10$$-$$3 k310. The value of x is ____________.

\n

[Given : $$\\ln 10 = 2.3$$, R = 8.3 J K$$-$$1 mol$$-$$1]

", "options": [], "answer": "1", "solution": "**Answer:** 1\n\n$$\n\\begin{aligned}\n& \\ln \\left(\\frac{\\mathrm{K}_2}{\\mathrm{~K}_1}\\right)=\\frac{\\mathrm{E}_{\\mathrm{a}}}{\\mathrm{R}}\\left(\\frac{1}{\\mathrm{~T}_1}-\\frac{1}{\\mathrm{~T}_2}\\right) \\\\\\\\\n& \\ln \\left(\\frac{\\mathrm{K}_2}{\\mathrm{~K}_1}\\right)=\\frac{532611}{8.3} \\times\\left(\\frac{10}{310 \\times 300}\\right)\n\\end{aligned}\n$$

\nwhere $\\mathrm{K}_2$ is at $310 \\mathrm{~K} \\,\\& \\mathrm{~K}_1$ is at $300 \\mathrm{~K}$

\n$$\n\\begin{aligned}\n& \\ln \\left(\\frac{\\mathrm{K}_2}{\\mathrm{~K}_1}\\right)=6.9 \\\\\\\\\n& =3 \\times \\ln 10 \\\\\\\\\n& \\ln \\frac{\\mathrm{K}_2}{\\mathrm{~K}_1}=\\ln 10^3 \\\\\\\\\n& \\mathrm{~K}_2=\\mathrm{K}_1 \\times 10^3 \\\\\\\\\n& \\mathrm{~K}_1=\\mathrm{K}_2 \\times 10^3\n\\end{aligned}\n$$

\nSo K = 1", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 658, "subject": "Chemistry", "question": "

The equation

k = (6.5 $$\\times$$ 1012s$$-$$1)e$$-$$26000K/T

is followed for the decomposition of compound A. The activation energy for the reaction is ________ kJ mol$$-$$1. [nearest integer]

\n

(Given : R = 8.314 J K$$-$$1 mol$$-$$1)

", "options": [], "answer": "216", "solution": "**Answer:** 216\n\n $k=A e^{\\frac{-E_{a}}{R T}}$\n

\n$$\n\\begin{aligned}\n\\frac{E_{a}}{R T} &=\\frac{26000}{T} \\\\\\\\\nE_{a} &=26000 \\times 8.314 \\\\\\\\\n&=216164 \\mathrm{~J} \\\\\\\\\n&=216 \\mathrm{~kJ}\n\\end{aligned}\n$$", "topic": "Algebra", "subtopic": "Equations and Inequalities" }, { "id": 659, "subject": "Chemistry", "question": "

It has been found that for a chemical reaction with rise in temperature by 9 K the rate constant gets doubled. Assuming a reaction to be occurring at 300 K, the value of activation energy is found to be ____________ kJ mol$$-$$1. [nearest integer]

\n

(Given ln10 = 2.3, R = 8.3 J K$$-$$1 mol$$-$$1, log 2 = 0.30)

", "options": [], "answer": "59", "solution": "**Answer:** 59\n\n$T_{1}=300 \\mathrm{~K}$\n

\n(Rate constant)\n

\n$\\mathrm{K}_{2}=2 \\mathrm{~K}_{1}$, on increase temperature by $9 \\mathrm{~K}$\n

\n$\\mathrm{T}_{2}=309 \\mathrm{~K}$\n

\n$\\mathrm{Ea}=?$\n

\n$\\log \\frac{\\mathrm{K}_{2}}{\\mathrm{~K}_{1}}=\\frac{\\mathrm{Ea}}{2.3 \\mathrm{R}}\\left[\\frac{\\mathrm{T}_{2}-\\mathrm{T}_{1}}{\\mathrm{~T}_{2} \\cdot \\mathrm{T}_{1}}\\right]$\n

\n$\\log 2=\\frac{\\mathrm{Ea}}{2.3 \\times 8.3}\\left[\\frac{9}{309 \\times 300}\\right]$\n

\n$\\mathrm{Ea}=\\frac{0.3 \\times 309 \\times 300 \\times 2.3 \\times 8.3}{9}$\n

\n$=58988.1 \\mathrm{~J} / \\mathrm{mole}$\n

\n$\\simeq 59 \\mathrm{~kJ} / \\mathrm{mole}$", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 660, "subject": "Chemistry", "question": "

The rate constant for a first order reaction is given by the following equation:

\n

$$\\ln k = 33.24 - {{2.0 \\times {{10}^4}\\,K} \\over T}$$

\n

The activation energy for the reaction is given by ____________ kJ mol$$-$$1. (In nearest integer) (Given : R = 8.3 J K$$-$$1 mol$$-$$1)

", "options": [], "answer": "166", "solution": "**Answer:** 166\n\n$\\ln \\mathrm{k}=\\ln \\mathrm{A}-\\frac{\\mathrm{E}_{\\mathrm{A}}}{\\mathrm{RT}}$

\nGiven: $\\ln k=33.24-\\frac{2.0 \\times 10^4}{\\mathrm{~T}}$

\n$$\n\\begin{aligned}\n&\\therefore \\text { on comparing } \\frac{\\mathrm{E}_{\\mathrm{A}}}{\\mathrm{R}}=2.0 \\times 10^4 \\\\\\\\\n&\\therefore \\mathrm{E}_{\\mathrm{A}}=2.0 \\times 10^4 \\times \\mathrm{R} \\\\\\\\\n&\\Rightarrow \\mathrm{E}_{\\mathrm{A}}=2.0 \\times 10^4 \\times 8.3 \\mathrm{~J} \\\\\\\\\n&\\Rightarrow \\mathrm{E}_{\\mathrm{A}}=16.6 \\times 10^4 \\mathrm{~J}=166 \\mathrm{~kJ}\n\\end{aligned}\n$$", "topic": "Algebra", "subtopic": "Equations and Inequalities" }, { "id": 661, "subject": "Chemistry", "question": "

A flask is filled with equal moles of A and B. The half lives of A and B are 100 s and 50 s respectively and are independent of the initial concentration. The time required for the concentration of A to be four times that of B is ___________ s.

\n

(Given : ln 2 = 0.693)

", "options": [], "answer": "200", "solution": "**Answer:** 200\n\n$\\mathrm{k}_{\\mathrm{A}}=\\frac{\\ln 2}{100} ; \\mathrm{k}_{\\mathrm{B}}=\\frac{\\ln 2}{50}$

\n$\\mathrm{A}_{\\mathrm{t}}=\\mathrm{A}_0 \\times \\mathrm{e}^{-\\mathrm{k}_{\\mathrm{A}} \\mathrm{t}}$

\n$\\mathrm{A}_{\\mathrm{t}}=\\mathrm{A}_0 \\times \\mathrm{e}^{\\left(\\frac{-\\ln 2}{100} \\times \\mathrm{t}\\right)}$

\n$\\mathrm{B}_{\\mathrm{t}}=\\mathrm{B}_0 \\times \\mathrm{e}^{\\left(\\frac{-\\ln 2}{50} \\times \\mathrm{t}\\right)}$

\n$\\mathrm{A}_0=\\mathrm{B}_0$

\n$\\& \\mathrm{~A}_{\\mathrm{t}}=4 \\mathrm{~B}_{\\mathrm{t}}$

\n$\\mathrm{e}^{-\\frac{\\ln 2}{100} \\times \\mathrm{t}}=4 \\times \\mathrm{e}^{-\\frac{\\ln 2}{50} \\times \\mathrm{t}}$

\n$\\mathrm{e}^{\\frac{\\ln 2}{100} \\times \\mathrm{t}}=4$

\n$\\mathrm{e}^{\\frac{\\ln 2}{100} \\times \\mathrm{t}}=4$

\n$\\frac{\\ln 2}{100} \\times \\mathrm{t}=\\ln 4=2 \\ln 2$

\n$\\mathrm{t}=200 ~ \\mathrm{sec}$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 662, "subject": "Chemistry", "question": "

Catalyst A reduces the activation energy for a reaction by 10 kJ mol$$-$$1 at 300 K. The ratio of rate constants, $${{{}^kT,\\,Catalysed} \\over {{}^kT,\\,Uncatalysed}}$$ is ex. The value of x is ___________. [nearest integer]

\n

[Assume that the pre-exponential factor is same in both the cases. Given R = 8.31 J K$$-$$1 mol$$-$$1]

", "options": [], "answer": "4", "solution": "**Answer:** 4\n\nActivation Energy : It is the minimum amount of energy required to activate the molecules/ atoms so that they can undergo the chemical reaction.\n\n

A catalyst increases the rate of a reaction by lowering the activation energy so that more reactant molecules collide with enough energy to surmount the smaller energy barrier.\n

In the Question,\n

Given : $\\mathrm{E}_{\\text {cat }}-\\mathrm{E}_{\\text {uncat }}=10 \\mathrm{~kJ} / \\mathrm{mol}$\n

$$\n\\mathrm{T}=300 \\mathrm{~K}\n$$\n

According to Arrhenius Equation,\n

$$\n\\begin{aligned}\n& \\mathrm{K}=A \\mathrm{e}^{-\\mathrm{Ea} / \\mathrm{RT}} \\\\\\\\\n& \\frac{\\mathrm{E}_{\\text {cat }}}{\\mathrm{E}_{\\text {uncat }}}=e^{\\frac{E a-E_a^1}{R T}}\n\\end{aligned}\n$$\n

$$\n\\begin{aligned}\n& =e^{\\frac{10 \\times 10^3}{8.21 \\times 300}} \\\\\\\\\n& =e^4 \\\\\\\\\n& =4\n\\end{aligned}\n$$", "topic": "Number Theory", "subtopic": "Divisibility / Factorization" }, { "id": 663, "subject": "Chemistry", "question": "

For the reaction P $$\\to$$ B, the values of frequency factor A and activation energy EA are 4 $$\\times$$ 1013 s$$-$$1 and 8.3 kJ mol$$-$$1 respectively. If the reaction is of first order, the temperature at which the rate constant is 2 $$\\times$$ 10$$-$$6 s$$-$$1 is _____________ $$\\times$$ 10$$-$$1 K.

\n

(Given : ln 10 = 2.3, R = 8.3 J K$$-$$1 mol$$-$$1, log2 = 0.30)

", "options": [], "answer": "225", "solution": "**Answer:** 225\n\n$$\n\\begin{aligned}\n&k=A \\mathrm{e}^{\\frac{-E_{a}}{R T}} \\\\\\\\\n&\\ln k=\\ln A-\\frac{E_{a}}{R T} \\\\\\\\\n&\\Rightarrow \\log \\left(2 \\times 10^{-6}\\right)=\\log \\left(4 \\times 10^{13}\\right)-\\frac{8.3 \\times 10^{3}}{8.3 \\times T \\times 2.3} \\\\\\\\\n&\\Rightarrow \\log (2)-6=2 \\times \\log (2)+13-\\frac{8.3 \\times 10^{3}}{8.3 \\times T \\times 2.3} \\\\\\\\\n&\\Rightarrow-19.3=-\\frac{10^{3}}{T \\times 2.3} \\Rightarrow T=225 \\times 10^{-1} \\mathrm{~K}\n\\end{aligned}\n$$", "topic": "Number Theory", "subtopic": "Divisibility / Factorization" }, { "id": 664, "subject": "Chemistry", "question": "For a reversible reaction $\\mathrm{A} \\rightleftharpoons \\mathrm{B}$, the $\\Delta \\mathrm{H}_{\\text{forward reaction}} = 20 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$. The activation energy of the uncatalysed forward reaction is $300 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$. When the reaction is catalysed keeping the reactant concentration same, the rate of the catalysed forward reaction at $27^{\\circ} \\mathrm{C}$ is found to be same as that of the uncatalysed reaction at $327^{\\circ} \\mathrm{C}$.

The activation energy of the catalysed backward reaction is ___________ $\\mathrm{kJ}~ \\mathrm{mol}^{-1}$.", "options": [], "answer": "130", "solution": "**Answer:** 130\n\nUsing the Arrhenius equation:\n

\n$$k = Ae^{-E_a/RT}$$\n

\nwhere $k$ is the rate constant, $A$ is the pre-exponential factor, $E_a$ is the activation energy, $R$ is the gas constant, and $T$ is the temperature in Kelvin.\n

\nThe equation is used for both the uncatalyzed forward reaction and the catalyzed backward reaction. By setting the two equations equal to each other and cancelling out the pre-exponential factor, we get:\n

\n$$e^{\\frac{300 \\times 10^3}{600 \\times R}} = e^{\\frac{-E_a}{300 \\times R}}$$\n

\nSimplifying the equation, we get:\n

\n$$\\frac{300 \\times 10^3}{600 \\times R} = \\frac{E_a}{300 \\times R}$$\n

\nSolving for $E_a$, we get:\n

\n$$E_a = \\frac{10^3}{2}\\times 300 = 150 \\times 10^3\\ \\mathrm{J~mol^{-1}} = 150\\ \\mathrm{kJ~mol^{-1}}$$\n

\nThis gives us the activation energy for the uncatalyzed forward reaction.\n

\nTo find the activation energy for the catalyzed backward reaction, we use the relationship:\n

\n$$E_{\\text{rev,catalysed}} = E_{\\text{fwd,uncat}} - \\Delta H_{\\text{forward reaction}}$$\n

\nSubstituting the given values, we get:\n

\n$$E_{\\text{rev,catalysed}} = 150\\ \\mathrm{kJ~mol^{-1}} - 20\\ \\mathrm{kJ~mol^{-1}} = 130\\ \\mathrm{kJ~mol^{-1}}$$\n

\nTherefore, the activation energy for the catalyzed backward reaction is $\\boxed{130\\ \\mathrm{kJ~mol^{-1}}}$.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 665, "subject": "Chemistry", "question": "

The number of given statement/s which is/are correct is __________.

\n

(A) The stronger the temperature dependence of the rate constant, the higher is the activation energy.

\n

(B) If a reaction has zero activation energy, its rate is independent of temperature.

\n

(C) The stronger the temperature dependence of the rate constant, the smaller is the activation energy.

\n

(D) If there is no correlation between the temperature and the rate constant then it means that the reaction has negative activation energy.

", "options": [], "answer": "2", "solution": "**Answer:** 2\n\n

Clearly, if $\\mathrm{E}_a=0, \\mathrm{~K}$ is temperature independent if $\\mathrm{E}_a>0, \\mathrm{~K}$ increase with increase in temperature if $\\mathrm{E}_a<0, \\mathrm{~K}$ decrease with increase in temperature

\n\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 666, "subject": "Chemistry", "question": "

For a reaction taking place in three steps at same temperature, overall rate constant $$\\mathrm{K}=\\frac{\\mathrm{K}_1 \\mathrm{~K}_2}{\\mathrm{~K}_3}$$. If $$\\mathrm{Ea}_1, \\mathrm{Ea}_2$$ and $$\\mathrm{Ea}_3$$ are 40, 50 and $$60 \\mathrm{~kJ} / \\mathrm{mol}$$ respectively, the overall $$\\mathrm{Ea}$$ is ________ $$\\mathrm{kJ} / \\mathrm{mol}$$.

", "options": [], "answer": "30", "solution": "**Answer:** 30\n\n

$$\\begin{aligned} & \\mathrm{K}=\\frac{\\mathrm{K}_1 \\cdot \\mathrm{K}_2}{\\mathrm{~K}_3}=\\frac{\\mathrm{A}_1 \\cdot \\mathrm{A}_2}{\\mathrm{~A}_3} \\cdot \\mathrm{e}^{-\\frac{\\left(\\mathrm{E}_{\\mathrm{a}_1}+\\mathrm{E}_{\\mathrm{a}_2}-\\mathrm{E}_{\\mathrm{a}_3}\\right)}{R T}} \\\\ & \\mathrm{~A} \\cdot \\mathrm{e}^{-\\mathrm{E}_{\\mathrm{a}} / \\mathrm{RT}}=\\frac{\\mathrm{A}_1 \\mathrm{~A}_2}{\\mathrm{~A}_3} \\cdot \\mathrm{e}^{-\\frac{\\left(\\mathrm{E}_{\\mathrm{a}_1}+\\mathrm{E}_{\\mathrm{a}_2}-\\mathrm{E}_{\\mathrm{a}_3}\\right)}{R T}}\\end{aligned}$$

\n

$$\\mathrm{E}_{\\mathrm{a}}=\\mathrm{E}_{\\mathrm{a}_1}+\\mathrm{E}_{\\mathrm{a}_2}-\\mathrm{E}_{\\mathrm{a}_3}=40+50-60=30 \\mathrm{~kJ} / \\mathrm{mole} .$$

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 667, "subject": "Chemistry", "question": "Consider the reaction, 2A + B $$\\to$$ products. When concentration of B alone was doubled, the half-life did not change. When the concentration of A\nalone was doubled, the rate increased by two times. The unit of rate constant for this reaction is ", "options": [ { "text": "L mol-1 s-1 " }, { "text": "no unit" }, { "text": "mol L-1 s-1 " }, { "text": "s-1 " } ], "answer": "L mol-1 s-1 ", "solution": "**Answer:** L mol-1 s-1 \n\nRate = k [A]x[B]y\n

When [B] is doubled, keeping [A] constant half-life of the reaction\ndoes not change.\n\n

For a first order reaction $${t_{1/2}} = {{0.693} \\over K}\\,\\,$$ i.e. for a first order reaction $${t_{1/2}}$$ does not depend up on the concentration. Hence the reaction is first order with respect to B. Now when\n[A] is doubled, keeping [B] constant, the rate also doubles.\nHence the reaction is first order with respect to A.\n

$$\\therefore$$ $$\\,\\,\\,\\,\\,$$\nRate = k [A]1[B]1\n

Order of reaction $$=1+1=2$$ \n

Now for a nth order reaction, unit of rate constant is\n
(L)n–1 (mol)1–n\n s–1 when n = 2, unit of rate constant
is $$L\\,\\,mo{l^{ - 1}}\\,{\\sec ^{ - 1}}.$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 668, "subject": "Chemistry", "question": "At 518oC the rate of decomposition of a sample of gaseous acetaldehyde initially at a pressure of 363 Torr,\nwas 1.00 Torr s–1 when 5% had reacted and 0.5 Torr s–1 when 33% had reacted. The order of the reaction is ", "options": [ { "text": "0" }, { "text": "2" }, { "text": "3" }, { "text": "1" } ], "answer": "2", "solution": "**Answer:** 2\n\nFor a nth order reaction, the rate of reaction at time t ,\n

Rate = K [Pt] n \n

Here Pt = pressure at time t, k = constant.\n

Note : \n

Here instead of concentration of product, pressure of product is given. \n

When 5% is reacted at a rate 1 Toss S$$-$$1 \n

Then un-reacted is 95%..\n

As initial pressure is 363 Torr \n

then after 5% reaction completed the pressure will be \n

= 363 $$ \\times $$ $${{95} \\over {100}}$$ Torr. \n

$$\\therefore\\,\\,\\,$$ 1 = K $${\\left[ {363 \\times {{95} \\over {100}}} \\right]^n}$$ . . . . . . . .(1)\n

When 33% is reacted at a rate 0.5 , Torr S$$-$$1 then un-reacted is 67%\n

So, after 33% reaction completion, the pressure is = $$363 \\times {{67} \\over {100}}$$ Torr. \n

$$\\therefore\\,\\,\\,$$ 0.5 = K $${\\left[ {363 \\times {{67} \\over {100}}} \\right]^n}.......$$ (2)\n

Dividing (1) by (2), we get \n

$${1 \\over {0.5}} = {{{{\\left[ {363 \\times {{95} \\over {100}}} \\right]}^n}} \\over {{{\\left[ {363 \\times {{67} \\over {100}}} \\right]}^n}}}$$ \n

$$ \\Rightarrow \\,\\,2 = {\\left[ {{{95} \\over {67}}} \\right]^n}$$\n

$$ \\Rightarrow \\,\\,$$ 2 = $${\\left[ {1.41} \\right]^n}$$ \n

$$ \\Rightarrow \\,\\,$$ 2 = $${\\left[ {\\sqrt 2 } \\right]^n}$$ \n

$$ \\Rightarrow \\,\\,\\,\\,2 = {2^{{n \\over 2}}}$$ \n

$$\\therefore\\,\\,\\,$$ $${n \\over 2}$$ = 1 \n

$$ \\Rightarrow \\,\\,\\,\\,$$ n = 2\n

This is a 2nd order reaction. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 669, "subject": "Chemistry", "question": "For the reaction, 2A + B $$ \\to $$ products, when the concentrations of A and B both were doubled, the rate of the reaction increased from 0.3 mol L$$-$$1s$$-$$1 to 2.4 mol L$$-$$1s$$-$$1. When the concentration of A alone is doubled, the rate increased from 0.3 mol L$$-$$1s$$-$$1 to 0.6 mol L$$-$$1s$$-$$1. ", "options": [ { "text": "Total order of the reaction is 4" }, { "text": "Order of the reaction with respect to B is 2" }, { "text": "Order of the reaction with respect to B is 1" }, { "text": "Order of the reaction with respect to A is 2" } ], "answer": "Order of the reaction with respect to B is 2", "solution": "**Answer:** Order of the reaction with respect to B is 2\n\n$$r = K{\\left[ A \\right]^x}{\\left[ B \\right]^y}$$\n

$$ \\Rightarrow 8 = {2^3} = {2^{x + y}}$$\n

$$ \\Rightarrow x + y = 3\\,...(1)$$\n

$$ \\Rightarrow 2 = {2^x}$$\n

$$ \\Rightarrow x = 1,y = 2$$\n

Order w.r.t. A = 1\n

Order w.r.t. B = 2", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 670, "subject": "Chemistry", "question": "2NO(g) + Cl2(g) $$\\rightleftharpoons $$ 2NOCl(s)

This reaction was studied at $$-$$10$$^\\circ$$ and the following data was obtained

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
Run$${[NO]_0}$$$${[C{l_2}]_0}$$$${r_0}$$
10.100.100.18
20.100.200.35
30.200.201.40


$${[NO]_0}$$ and $${[C{l_2}]_0}$$ are the initial concentrations and r0 is the initial reaction rate.

The overall order of the reaction is __________. (Round off to the Nearest Integer).", "options": [], "answer": "3", "solution": "**Answer:** 3\n\nWe know,

$$Rate = K{\\left[ A \\right]^x}{\\left[ B \\right]^y}$$

Here from,

Exp 1 : $$0.18 = K{\\left[ {0.1} \\right]^x}{\\left[ {0.1} \\right]^y}$$ ..... (1)

Exp. 2 : $$0.35 = K{\\left[ {0.1} \\right]^x}{\\left[ {0.2} \\right]^y}$$ .... (2)

Exp. 3 : $$1.40 = K{\\left[ {0.2} \\right]^x}{\\left[ {0.2} \\right]^y}$$ .... (3)

(2) $$\\div$$ (3)

$${{0.35} \\over {1.40}} = {{K \\times {{\\left[ {0.1} \\right]}^x}{{\\left[ {0.2} \\right]}^y}} \\over {K \\times {{\\left[ {0.2} \\right]}^x}{{\\left[ {0.2} \\right]}^y}}}$$

$$ \\Rightarrow {1 \\over 4} = {\\left( {{1 \\over 2}} \\right)^x}$$

$$ \\Rightarrow x = 2$$

(1) $$\\div$$ (2)

$$\\left( {{1 \\over 2}} \\right) = {\\left( {{1 \\over 2}} \\right)^y}$$

$$ \\Rightarrow y = 1$$\n

$$ \\therefore $$ x + y = 3", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 671, "subject": "Chemistry", "question": "

At 345 K, the half life for the decomposition of a sample of a gaseous compound initially at 55.5 kPa was 340 s. When the pressure was 27.8 kPa, the half life was found to be 170 s. The order of the reaction is ____________. [integer answer]

", "options": [], "answer": "0", "solution": "**Answer:** 0\n\n$t_{1 / 2} \\propto \\frac{1}{\\left[P_{0}\\right]^{n-1}}$\n

\n$$\n\\begin{aligned}\n&\\frac{\\left(\\mathrm{t}_{1 / 2}\\right)_{1}}{\\left(\\mathrm{t}_{1 / 2}\\right)_{2}}=\\frac{\\left[\\mathrm{P}_{0}\\right]_{2}^{\\mathrm{n}-1}}{\\left[\\mathrm{P}_{0}\\right]_{1}^{\\mathrm{n}-1}} \\\\\\\\\n&\\frac{340}{170}=\\left(\\frac{27.8}{55.5}\\right)^{\\mathrm{n}-1} \\\\\\\\\n&2=\\left(\\frac{1}{2}\\right)^{\\mathrm{n}-1} \\\\\\\\\n&2=(2)^{1-n} \\\\\\\\\n&1-\\mathrm{n}=1 \\\\\\\\\n&\\mathrm{n}=0\n\\end{aligned}\n$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 672, "subject": "Chemistry", "question": "

The half life for the decomposition of gaseous compound $$\\mathrm{A}$$ is $$240 \\mathrm{~s}$$ when the gaseous pressure was 500 Torr initially. When the pressure was 250 Torr, the half life was found to be $$4.0$$ min. The order of the reaction is ______________. (Nearest integer)

", "options": [], "answer": "1", "solution": "**Answer:** 1\n\n$$\\left(\\mathrm{t}_{1 / 2}\\right)_{\\mathrm{A}}=240 \\mathrm{~s}$$ when $$\\mathrm{P}=500$$ torr\n

\n$$\\left(t_{1 / 2}\\right)_{A}=4 \\min =4 \\times 60=240$$ sec when $$P=250$$ torr\n

\nIf means half-life is independent of concentration of reactant present.\n

\n$\\therefore$ Order of reaction $=1$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 673, "subject": "Chemistry", "question": "

$$2 \\mathrm{NO}+2 \\mathrm{H}_{2} \\rightarrow \\mathrm{N}_{2}+2 \\mathrm{H}_{2} \\mathrm{O}$$

\n

The above reaction has been studied at $$800^{\\circ} \\mathrm{C}$$. The related data are given in the table below

\n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
Reaction serial numberInitial Pressure of $${H_2}/kPa$$Initial Pressure of $$NO/kPa$$Initial rate $$\\left( {{{ - dp} \\over {dt}}} \\right)/(kPa/s)$$
165.640.00.135
265.620.10.033
338.665.60.214
419.265.60.106

\n

The order of the reaction with respect to NO is ___________.

", "options": [], "answer": "2", "solution": "**Answer:** 2\n\nLet the rate of reaction ( $r$ ) is as\n

\n$$\n\\mathrm{r}=\\mathrm{K}[\\mathrm{NO}]^{n}\\left[\\mathrm{H}_{2}\\right]^{\\mathrm{m}}\n$$\n

\nFrom $1^{\\text {st }}$ data\n

\n$$\n0.135=\\mathrm{K}[40]^{\\mathrm{n}} \\cdot(65.6)^{\\mathrm{m}}\\quad\\quad...(1)\n$$\n

\nFrom $2^{\\text {nd }}$ data\n

\n$$\n0.033=\\mathrm{K}(20.1)^{\\mathrm{n}} \\cdot(65.6)^{\\mathrm{m}}\\quad\\quad...(2)\n$$\n

\nOn dividing equation (1) by equation (2)\n

\n$$\n\\begin{aligned}\n&\\frac{0.135}{0.033}=\\left(\\frac{40}{20.1}\\right)^{n} \\\\\\\\\n&4=(2)^{n} \\\\\\\\\n&\\therefore n=2 \\\\\\\\\n&\\therefore \\text { Order of reaction w.r.t. NO is } 2 .\n\\end{aligned}\n$$\n", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 674, "subject": "Chemistry", "question": "

For kinetic study of the reaction of iodide ion with $$\\mathrm{H}_{2} \\mathrm{O}_{2}$$ at room temperature :

\n

(A) Always use freshly prepared starch solution.

\n

(B) Always keep the concentration of sodium thiosulphate solution less than that of KI solution.

\n

(C) Record the time immediately after the appearance of blue colour.

\n

(D) Record the time immediately before the appearance of blue colour.

\n

(E) Always keep the concentration of sodium thiosulphate solution more than that of KI solution.

\n

Choose the correct answer from the options given below :

", "options": [ { "text": "$$(\\mathrm{A}),(\\mathrm{B}),(\\mathrm{C})$$ only" }, { "text": "$$(\\mathrm{A}),(\\mathrm{D}),(\\mathrm{E})$$ only" }, { "text": "$$(\\mathrm{D}),(\\mathrm{E})$$ only" }, { "text": "$$(\\mathrm{A}),(\\mathrm{B}),(\\mathrm{E})$$ only" } ], "answer": "$$(\\mathrm{A}),(\\mathrm{B}),(\\mathrm{C})$$ only", "solution": "**Answer:** $$(\\mathrm{A}),(\\mathrm{B}),(\\mathrm{C})$$ only\n\nTo minimize contamination, use freshly prepared starch solution to determine end point. As $\\mathrm{KI}$ is used in excess to consume all the $\\mathrm{H}_{2} \\mathrm{O}_{2}$ the concentration of sodium thiosulphate solution is less than $\\mathrm{KI}$ solution. After appearance of blue colour record the time immediately.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 675, "subject": "Chemistry", "question": "

The reaction between X and Y is first order with respect to X and zero order with respect to Y.

\n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
Experiment$${{[X]} \\over {mol\\,{L^{ - 1}}}}$$$${{[Y]} \\over {mol\\,{L^{ - 1}}}}$$$${{Initial\\,rate} \\over {mol\\,{L^{ - 1}}\\,{{\\min }^{ - 1}}}}$$
I0.10.1$$2 \\times {10^{ - 3}}$$
IL0.2$$4 \\times {10^{ - 3}}$$
III0.40.4$$M \\times {10^{ - 3}}$$
IV0.10.2$$2 \\times {10^{ - 3}}$$

\n

Examine the data of table and calculate ratio of numerical values of M and L. (Nearest Integer)

", "options": [], "answer": "40", "solution": "**Answer:** 40\n\n$$r=k[X][Y]^{0}=k[X]$$\n

\nUsing I & II\n

\n$\\frac{4 \\times 10^{-3}}{2 \\times 10^{-3}}=\\left(\\frac{L}{0.1}\\right) \\quad \\Rightarrow \\quad \\mathrm{L}=0.2$\n

\nUsing I & III\n

\n$\\frac{M \\times 10^{-3}}{2 \\times 10^{-3}}=\\frac{0.4}{0.1} \\quad \\Rightarrow \\quad \\mathrm{M}=8$\n

\n$\\frac{M}{L}=\\frac{8}{0.2}=40$", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 676, "subject": "Chemistry", "question": "

For conversion of compound A $$\\to$$ B, the rate constant of the reaction was found to be $$\\mathrm{4.6\\times10^{-5}~L~mol^{-1}~s^{-1}}$$. The order of the reaction is ____________.

", "options": [], "answer": "2", "solution": "**Answer:** 2\n\n

As unit is L mol$$^{-1}$$ s$$^{-1}$$, order of the reaction is 2.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 677, "subject": "Chemistry", "question": "

A student has studied the decomposition of a gas AB$$_3$$ at 25$$^\\circ$$C. He obtained the following data.

\n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
p (mm Hg)50100200400
relative t$$_{1/2}$$ (s)4210.5

\n

The order of the reaction is

", "options": [ { "text": "2" }, { "text": "0.5" }, { "text": "1" }, { "text": "0 (zero)" } ], "answer": "2", "solution": "**Answer:** 2\n\n$$\n\\begin{aligned}\n& \\mathrm{t}_{1 / 2} \\propto\\left(\\mathrm{P}_{\\mathrm{o}}\\right)^{1-\\mathrm{n}} \\\\\\\\\n& \\frac{\\left(\\mathrm{t}_{1 / 2}\\right)_1}{\\left(\\mathrm{t}_{1 / 2}\\right)_2}=\\frac{\\left(\\mathrm{P}_o\\right)_1^{1-\\mathrm{n}}}{\\left(\\mathrm{P}_{o_2}\\right)_2^{1-\\mathrm{n}}} \\\\\\\\\n& \\Rightarrow\\left(\\frac{4}{2}\\right)=\\left(\\frac{50}{100}\\right)^{1-\\mathrm{n}} \\\\\\\\\n& \\Rightarrow 2=\\left(\\frac{1}{2}\\right)^{1-\\mathrm{n}} \\\\\\\\\n& \\Rightarrow 2=(2)^{\\mathrm{n}-1} \\\\\\\\\n& \\Rightarrow \\mathrm{n}-1=1 \\\\\\\\\n& \\Rightarrow \\mathrm{n}=2\n\\end{aligned}\n$$", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 678, "subject": "Chemistry", "question": "

The reaction $$2 \\mathrm{NO}+\\mathrm{Br}_{2} \\rightarrow 2 \\mathrm{NOBr}$$

\n

takes places through the mechanism given below:

\n

$$\\mathrm{NO}+\\mathrm{Br}_{2} \\Leftrightarrow \\mathrm{NOBr}_{2}$$ (fast)

\n

$$\\mathrm{NOBr}_{2}+\\mathrm{NO} \\rightarrow 2 \\mathrm{NOBr}$$ (slow)

\n

The overall order of the reaction is ___________.

", "options": [], "answer": "3", "solution": "**Answer:** 3\n\n

The overall order of a reaction is determined by the slow (rate-determining) step.

\n

Here, the slow step is:\n$ \\text{NOBr}_2 + \\text{NO} \\rightarrow 2 \\text{NOBr} $

\n

This is a second-order reaction: first-order with respect to $\\text{NOBr}_2$ and first-order with respect to NO.

\n

However, $\\text{NOBr}_2$ is not a reactant in the overall reaction. It's an intermediate. So, we need to express it in terms of the initial reactants.

\n

From the first (fast) step, we get: \n$ \\text{NO} + \\text{Br}_2 \\rightleftharpoons \\text{NOBr}_2 $

\n

In the steady state approximation, the rate of formation of $\\text{NOBr}_2$ equals the rate of its consumption. We can write this as:

\n

$ k_1[\\text{NO}][\\text{Br}_2] = k_{-1}[\\text{NOBr}_2] + k_2[\\text{NOBr}_2][\\text{NO}] $

\n

Since the second reaction (slow step) is much slower than the first one, $k_2[\\text{NOBr}_2][\\text{NO}]$ term is negligible in comparison to $k_{-1}[\\text{NOBr}_2]$.

\n

So, we can simplify to:

\n$ k_1[\\text{NO}][\\text{Br}_2] \\approx k_{-1}[\\text{NOBr}_2] $

\n

Solving for $[\\text{NOBr}_2]$, we get:\n$ [\\text{NOBr}_2] \\approx \\frac{k_1}{k_{-1}} [\\text{NO}][\\text{Br}_2] $

\n

Substitute $[\\text{NOBr}_2]$ into the rate equation for the slow step:

\n$ \\text{Rate} = k_2[\\text{NO}][\\text{NOBr}_2] $

\n$ \\text{Rate} = k_2[\\text{NO}] \\left( \\frac{k_1}{k_{-1}} [\\text{NO}][\\text{Br}_2] \\right) $

\n

So the overall reaction order is 3 (2 with respect to NO and 1 with respect to $\\text{Br}_2$).

\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 679, "subject": "Chemistry", "question": "

For a chemical reaction $$\\mathrm{A}+\\mathrm{B} \\rightarrow$$ Product, the order is 1 with respect to $$\\mathrm{A}$$ and $$\\mathrm{B}$$.

\n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
$$\\mathrm{Rate}$$
$$\\mathrm{mol~L^{-1}~S^{-1}}$$		$$\\mathrm{[A]}$$
$$\\mathrm{mol~L^{-1}}$$		$$\\mathrm{[B]}$$
$$\\mathrm{mol~L^{-1}}$$
0.10200.5
0.40$$x$$0.5
0.8040$$y$$

\n

What is the value of $$x$$ and $$y$$ ?

", "options": [ { "text": "80 and 4" }, { "text": "160 and 4" }, { "text": "80 and 2" }, { "text": "40 and 4" } ], "answer": "80 and 2", "solution": "**Answer:** 80 and 2\n\n$$\n\\begin{aligned}\n& \\mathrm{r}=\\mathrm{K}[\\mathrm{A}]^1[\\mathrm{~B}]^1 \\\\\\\\\n& 0.1=\\mathrm{K}(20)^1(0.5)^1 ........(i) \\\\\\\\\n& 0.40=\\mathrm{K}(\\mathrm{x})^1(0.5)^1 ........(ii) \\\\\\\\\n& 0.80=\\mathrm{K}(40)^1(\\mathrm{y})^1 ........(iii)\n\\end{aligned}\n$$\n

From (i) and (ii)\n

$$\nx=80\n$$\n

From (i) and (iii)\n

$$\ny=2\n$$", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 680, "subject": "Chemistry", "question": "

Consider the following first order gas phase reaction at constant temperature $$\n\\mathrm{A}(\\mathrm{g}) \\rightarrow 2 \\mathrm{B}(\\mathrm{~g})+\\mathrm{C}(\\mathrm{g})$$

\n

If the total pressure of the gases is found to be 200 torr after 23 $$\\mathrm{sec}$$. and 300 torr upon the complete decomposition of A after a very long time, then the rate constant of the given reaction is ________ $$\\times 10^{-2} \\mathrm{~s}^{-1}$$ (nearest integer)

\n

[Given : $$\\log _{10}(2)=0.301$$]

", "options": [], "answer": "3", "solution": "**Answer:** 3\n\n

To find the rate constant of the first-order reaction indicated, let us first understand the reaction dynamics based on the total pressure change over time. The reaction is:

\n\n

$$\\mathrm{A(g)} \\rightarrow 2\\mathrm{B(g)} + \\mathrm{C(g)}$$

\n\n

Initially, only A is present. As the reaction progresses, A decreases, and B and C are formed. For each mole of A that reacts, a total of 3 moles of gas are produced (2 moles of B and 1 mole of C), leading to an increase in the total pressure of the system if the volume remains constant.

\n\n

At the start: Total pressure due to A only.

\n

Finally, when A has completely decomposed: Total pressure is due to 3 times the initial amount of A, as no A is left and for every mole of A decomposed, 3 moles of gas (2B + 1C) are produced.

\n\n

The total pressure change thus directly relates to the extent of reaction, i.e., how much A has decomposed into B and C.

\n\n

The total pressure at 23 sec is 200 torr, and the final total pressure is 300 torr after a very long time, indicating that the pressure increases by 100 torr due to the complete decomposition of A.

\n\n

To use this information, we need to understand the relationship between pressure change and concentration in a closed system, especially for a first-order reaction. For a first-order reaction:

\n\n

$$\\ln\\left(\\frac{[A]_0}{[A]}\\right) = kt$$

\n\n

Where:\n$$[A]_0$$ is the initial concentration (or in this context, partial pressure),\n$$[A]$$ is the concentration (or partial pressure) at time t,\n$$k$$ is the rate constant, and\n$$t$$ is the time.

\n\n

Let's denote the initial total pressure due to A as $$P_0$$. At complete decomposition, the pressure increase is from the conversion of A to 2B + C which increases the total pressure to 300 torr, indicating that initially, $$P_0$$ must have been 100 torr since the pressure increase is due to the tripling effect of the reaction, completing to 300 torr.

\n\n

At 23 sec, the total pressure is at 200 torr. This means 100 torr is due to the unreacted A, and the additional 100 torr is from the formation of 2B + C. Given the nature of the reaction (1:3 stoichiometry of A to the total gas), we can deduce that at 23 seconds, half of A has reacted because the pressure due to the products matches the pressure due to the unreacted A.

\n\n

Hence, at 23 sec, $$[A] = \\frac{1}{2}[A]_0$$. Plugging this into the first-order reaction formula:

\n\n

$$\\ln\\left(\\frac{[A]_0}{\\frac{1}{2}[A]_0}\\right) = kt$$

\n

$$\\ln(2) = k \\times 23$$

\n\n

Given that $$\\log_{10}(2) = 0.301$$, and knowing that $$\\ln(2) = \\log_{e}(2)$$ (where $$\\log_{e}(2) \\approx 0.693$$, for conversion from base 10 to e we can use the natural logarithm of 2 directly), we have:

\n\n

$$0.693 = k \\times 23$$

\n

$$k = \\frac{0.693}{23} = 0.0301 \\, \\mathrm{s}^{-1}$$

\n\n

When expressed in the format requested, $$k = 3.01 \\times 10^{-2} \\, \\mathrm{s}^{-1}$$. Rounding to the nearest integer, $$k = 3 \\times 10^{-2} \\, \\mathrm{s}^{-1}$$.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 681, "subject": "Chemistry", "question": "

Consider the following reaction

\n

$$\\mathrm{A}+\\mathrm{B} \\rightarrow \\mathrm{C}$$

\n

The time taken for A to become $$1 / 4^{\\text {th }}$$ of its initial concentration is twice the time taken to become $$1 / 2$$ of the same. Also, when the change of concentration of B is plotted against time, the resulting graph gives a straight line with a negative slope and a positive intercept on the concentration axis.

\n

The overall order of the reaction is ________.

", "options": [], "answer": "1", "solution": "**Answer:** 1\n\n

To determine the overall order of the reaction given by $$\\mathrm{A} + \\mathrm{B} \\rightarrow \\mathrm{C}$$, we can derive the information based on the given details about the kinetics of reactant A and the graphical behavior of reactant B.

\n\n

The first piece of information tells us that the time for the concentration of A to reduce to $1/4$ of its initial concentration ($[A]_0/4$) is twice the time it takes to reduce to half of its initial concentration ($[A]_0/2$). This characteristic is a hallmark of first-order reactions. In first-order reactions, the time it takes for the concentration of the reactant to reduce to half of its initial value (known as the half-life, $$t_{1/2}$$) is constant and does not depend on the initial concentration. Specifically, the fact that the concentration decreases by a factor of 4 (to $1/4$ its initial value) in twice the time it takes to decrease by a factor of 2 (to $1/2$ its initial value) indicates a constant half-life, consistent with first-order kinetics for reactant A.

\n\n

As for reactant B, the information provided is that a plot of its concentration change over time is a straight line with a negative slope and a positive intercept on the concentration axis. This description matches the behavior of a reactant in a reaction of zero-order kinetics with respect to B, where the rate of reaction is constant and independent of the concentration of B. In zero-order reactions, the concentration of the reactant decreases linearly over time, since the rate of decrease is constant.

\n\n

Therefore, considering the kinetics of both A and B:

\n\n\n\n

The overall order of the reaction is the sum of the individual orders with respect to each reactant, which gives us:

\n\n

$$\\text{Overall order} = \\text{Order with respect to A} + \\text{Order with respect to B} = 1 (A) + 0 (B) = 1.$$

\n\n

Hence, the overall order of the reaction is 1.

", "topic": "Discrete Mathematics", "subtopic": "Other" }, { "id": 682, "subject": "Chemistry", "question": "

During Kinetic study of reaction $$\\mathrm{2 A+B \\rightarrow C+D}$$, the following results were obtained :

\n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
$$\\mathrm{A [M]}$$$$\\mathrm{B [M]}$$initial rate of formation of $$\\mathrm{D}$$
I0.10.1$$6.0\\times10^{-3}$$
II0.30.2$$7.2\\times10^{-2}$$
III0.30.4$$2.88\\times10^{-1}$$
IV0.40.1$$2.40\\times10^{-2}$$

\n

Based on above data, overall order of the reaction is _________.

", "options": [], "answer": "3", "solution": "**Answer:** 3\n\n

$$\\begin{aligned}\n& \\text { Rate }=k[A]^x[B]^y \\\\\n& \\frac{6 \\times 10^{-}}{2.4 \\times 10^{-}}=\\left(\\frac{0.1}{0.4}\\right)^x \\Rightarrow x=1 \\\\\n& \\frac{7.2 \\times 10^{-}}{2.88 \\times 10^{-}}=\\left(\\frac{0.2}{0.4}\\right)^y \\Rightarrow y=2\n\\end{aligned}$$

", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 683, "subject": "Chemistry", "question": "If half-life of a substance is 5 yrs, then the total amount of substance left after 15 years, when initial amount is 64 grams is", "options": [ { "text": "16 grams" }, { "text": "2 grams" }, { "text": "32 grams" }, { "text": "8 grams" } ], "answer": "8 grams", "solution": "**Answer:** 8 grams\n\n$${t_{1/2}} = 5\\,\\,$$ years, $$T=15$$ years hence -\n

total number of half life periods $$ = {{15} \\over 3} = 3.$$ \n

$$\\therefore$$ Amount left $$ = {{64} \\over {{{\\left( 2 \\right)}^3}}} = 8g$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 684, "subject": "Chemistry", "question": "The integrated rate equation is Rt = log C0 - log Ct . The straight line graph is obtained by\nplotting", "options": [ { "text": "time vs log Ct" }, { "text": "$${1 \\over {time}}$$ vs Ct" }, { "text": "time vs Ct" }, { "text": "$${1 \\over {time}}$$ vs $${1 \\over {{C_t}}}$$ " } ], "answer": "time vs log Ct", "solution": "**Answer:** time vs log Ct\n\n$$Rt = \\log {C_o} - {{\\mathop{\\rm logC}\\nolimits} _t}$$ \n

It is clear from the equation that if we plot a graph between $$\\log \\,{C_t}$$ and time, a straight line with a slope equal to $$ - {k \\over {2.303}}$$ and intercept equal to $$\\log \\,\\left[ {{A_o}} \\right]$$ will be obtained. ", "topic": "Discrete Mathematics", "subtopic": "Other" }, { "id": 685, "subject": "Chemistry", "question": "Units of rate constant of first and zero order reactions in terms of molarity M unit are respectively", "options": [ { "text": "sec-1, Msec-1" }, { "text": "sec-1, M" }, { "text": "Msec-1, sec-1" }, { "text": "M, sec-1" } ], "answer": "sec-1, Msec-1", "solution": "**Answer:** sec-1, Msec-1\n\nFor a zero order reaction.\n

rate $$ = k{\\left[ A \\right]^ \\circ }\\,\\,\\,$$ i.e. rate $$=k$$ \n

hence unit of $$k = M.{\\sec ^{ - 1}}$$\n

For a first order reaction. \n

rate $$ = k\\left[ A \\right]$$\n

$$k=M.$$ $${\\sec ^{ - 1}}/M = {\\sec ^{ - 1}}$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 686, "subject": "Chemistry", "question": "For the reaction system: \n

2NO(g) + O2(g) $$\\to$$ 2NO2(g) volume is suddenly reduce to half its value by increasing the pressure on it. If the reaction is of first order with respect to O2 and second order with respect to NO, the rate of reaction will", "options": [ { "text": "diminish to one-eighth of its initial value" }, { "text": "increase to eight times of its initial value" }, { "text": "increase to four times of its initial value" }, { "text": "diminish to one-fourth of its initial value" } ], "answer": "increase to eight times of its initial value", "solution": "**Answer:** increase to eight times of its initial value\n\n$$r = k\\left[ {{O_2}} \\right]{\\left[ {NO} \\right]^2}.$$ \n

When the volume is reduced to $$1/2,$$ the conc. will double\n

$$\\therefore$$ $$\\,\\,\\,\\,\\,$$ New rate $$ = k\\left[ {2{O_2}} \\right]{\\left[ {2NO} \\right]^2} = 8k\\left[ {{O^2}} \\right]{\\left[ {NO} \\right]^2}$$\n

The new rate increases to eight times of its initial. ", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 687, "subject": "Chemistry", "question": "In a first order reaction, the concentration of the reactant decreases from 0.8 M to 0.4 M in 15\nminutes. The time taken for the concentration to change from 0.1 M to 0.025 M is", "options": [ { "text": "30 minutes" }, { "text": "60 minutes" }, { "text": "7.5 minutes" }, { "text": "15 minutes" } ], "answer": "30 minutes", "solution": "**Answer:** 30 minutes\n\nAs the concentration of reactant decreases from $$0.8$$ to $$0.4$$ in $$15$$ minutes hence the $${t_{1/2}}$$ is $$15$$ minutes. To fall the concentration from $$0.1$$ to $$0.025$$ we need two half lives i.e., $$30$$ minutes. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 688, "subject": "Chemistry", "question": "A reaction involving two different reactants can never be ", "options": [ { "text": "Unimolecular reaction " }, { "text": "First order reaction" }, { "text": "second order reaction" }, { "text": "Bimolecular reaction " } ], "answer": "Unimolecular reaction ", "solution": "**Answer:** Unimolecular reaction \n\nThe molecularity of reaction is the number of reactant molecules taking part in a single step of the reaction. \n

NOTE : The reaction involving two different reactant can never be unimolecular. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 689, "subject": "Chemistry", "question": "t1/4 can be taken as the time taken for the concentration of a reactant to drop to $$3 \\over 4$$ of its initial value. If the rate constant for a first order reaction is K, the t1/4 can be written as ", "options": [ { "text": "0.10 / K" }, { "text": "0.29 / K" }, { "text": "0.69 / K " }, { "text": "0.75 / K" } ], "answer": "0.29 / K", "solution": "**Answer:** 0.29 / K\n\n$${t_{1/4}} = {{2.303} \\over K}\\log {1 \\over {3/4}}$$\n

$$ = {{2.303} \\over K}\\log {4 \\over 3}$$\n

$$ = {{2.303} \\over K}\\left( {\\log \\,4 - \\log 3} \\right)$$\n

$$ = {{2.303} \\over K}\\left( {2{{\\log }^2} - \\log 3} \\right)$$\n

$$ = {{2.303} \\over K}\\left( {2 \\times 0.301 - 0.4771} \\right)$$\n

$$ = {{0.29} \\over K}$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 690, "subject": "Chemistry", "question": "The following mechanism has been proposed for the reaction of NO with Br2 to form NOBr:
\nNO(g) + Br2 (g) $$\\leftrightharpoons$$ NOBr2 (g)
\nNOBr2 (g) + NO (g) $$\\to$$ 2NOBr (g)
\nIf the second step is the rate determining step, the order of the reaction with respect to NO(g) is ", "options": [ { "text": "1" }, { "text": "0" }, { "text": "3" }, { "text": "2" } ], "answer": "2", "solution": "**Answer:** 2\n\n$$\\left( i \\right)\\,\\,\\,\\,\\,\\,\\,\\,\\,NO\\left( g \\right) + B{r_2}\\left( g \\right)\\,\\,\\,\\rightleftharpoons\\,NOB{r_2}\\left( g \\right)$$ \n

$$\\left( {ii} \\right)\\,\\,\\,\\,\\,\\,\\,\\,\\,NOB{r_2}\\left( g \\right) + NO\\left( g \\right)\\,\\buildrel \\, \\over\n \\longrightarrow 2NOBr\\left( g \\right)$$ \n

Rate law equation $$ = k\\left[ {NOB{r_2}} \\right]\\left[ {NO} \\right]$$ \n

But $$NOB{r_2}$$ is intermediate and \n

must not appear in the rate law equation\n

from $$1$$st step $${K_C} = {{\\left[ {NOB{r_2}} \\right]} \\over {\\left[ {NO} \\right]\\left[ {B{r_2}} \\right]}}$$ \n

$$\\therefore$$ $$\\,\\,\\,\\,\\,\\left[ {NOB{r_2}} \\right] = {K_C}\\left[ {NO} \\right]\\left[ {B{r_2}} \\right]$$ \n

$$\\therefore$$ $$\\,\\,\\,\\,$$ Rate law equation $$ = k.{K_C}{\\left[ {NO} \\right]^2}\\left[ {B{r_2}} \\right]$$\n

hence order of reaction is $$2$$ w.r.t. $$NO.$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 691, "subject": "Chemistry", "question": "The half life period of a first order chemical reaction is 6.93 minutes. The time required for the\ncompletion of 99% of the chemical reaction will be (log 2=0.301) :", "options": [ { "text": "230.3 minutes" }, { "text": "23.03 minutes " }, { "text": "46.06 minutes " }, { "text": "460.6 minutes " } ], "answer": "46.06 minutes ", "solution": "**Answer:** 46.06 minutes \n\nFor first order reaction\n

$$k = {{2.303} \\over t}\\log {{100} \\over {100 - 99}}$$\n

$${{0.693} \\over {6.93}} = {{2.303} \\over t}\\log {{100} \\over 1}$$\n

$${{0.693} \\over {6.93}} = {{2.303 \\times 2} \\over t}$$\n

$$ \\Rightarrow t = 46.06\\min $$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 692, "subject": "Chemistry", "question": "The time for half life period of a certain reaction A $$\\to$$ products is 1 hour. When the initial\nconcentration of the reactant ‘A’, is 2.0 mol L–1, how much time does it take for its concentration to\ncome from 0.50 to 0.25 mol L–1 if it is a zero order reaction ?", "options": [ { "text": "4 h" }, { "text": "0.5 h" }, { "text": "0.25 h" }, { "text": "1 h" } ], "answer": "0.25 h", "solution": "**Answer:** 0.25 h\n\nFor the reaction\n

$$A\\,\\, \\to \\,\\,$$ Product ; given $${t_{1/2}} = 1\\,\\,$$ hour\n

for a zero order reaction \n

$${t_{completion}}\\,\\, = {{\\left[ {{A_0}} \\right]} \\over k} = {{initial\\,\\,conc.} \\over {rate\\,\\,cons\\tan t}}$$\n

$$\\therefore$$ $$\\,\\,\\,\\,\\,{t_{1/2}} = {{\\left[ {{A_0}} \\right]} \\over {2K}}$$\n

or $$\\,\\,\\,\\,\\,k = {{\\left[ {{A_0}} \\right]} \\over {2{t_{1/2}}}} = {2 \\over {2 \\times 1}} = 1\\,\\,mol\\,li{t^{ - 1}}\\,h{r^{ - 1}}$$\n

Further for a zero order reaction \n

$$k = {{dx} \\over {dt}} = {{change\\,\\,\\,in\\,\\,\\,concentration} \\over {time}}$$\n

$$I = {{0.50 - 0.25} \\over {time}}\\,\\,\\,$$\n

$$\\therefore$$ $$\\,\\,\\,\\,\\,$$ time $$ = 0.25\\,\\,hr.$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 693, "subject": "Chemistry", "question": "For a first order reaction, (A) $$\\to$$ products, the concentration of A changes from 0.1 M to 0.025 M in 40\nminutes. The rate of reaction when the concentration of A is 0.01 M is :", "options": [ { "text": "1.73 x 10–5 M/ min" }, { "text": "3.47 x 10–4 M/min" }, { "text": "3.47 x 10–5 M/min" }, { "text": "1.73 x 10–4 M/min" } ], "answer": "3.47 x 10–4 M/min", "solution": "**Answer:** 3.47 x 10–4 M/min\n\nFor a first order reaction\n

$$k = {{2.0303} \\over t}\\,\\log \\,{a \\over {a - x}}$$\n

$$ = {{2.303} \\over {40}}\\log {{0.1} \\over {0.025}}$$\n

$$ = {{2.303} \\over {40}}\\log 4$$\n

$$ = {{2.303 \\times 0.6020} \\over {40}}$$\n

$$ = 3.47 \\times {10^{ - 2}}$$\n

$$R = K{\\left( A \\right)^1} = 3.47 \\times {10^{ - 2}} \\times 0.01$$\n

$$ = 3.47 \\times {10^{ - 4}}$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 694, "subject": "Chemistry", "question": "Higher order (>3) reactions are rare due to", "options": [ { "text": "increase in entropy and activation energy as more molecules are involved" }, { "text": "shifting of equilibrium towards reactants due to elastic collisions" }, { "text": "loss of active species on collision" }, { "text": "low probability of simultaneous collision of all the reacting species" } ], "answer": "low probability of simultaneous collision of all the reacting species", "solution": "**Answer:** low probability of simultaneous collision of all the reacting species\n\nReactions of higher order $$\\left( { > 3} \\right)$$ are very rate due to very less chances of many molecules to undergo effective collisions. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 695, "subject": "Chemistry", "question": "Decomposition of H2O2 follows a first order reaction. In fifty minutes the concentration of H2O2 decreases\nfrom 0.5 to 0.125 M in one such decomposition. When the concentration of H2O2 reaches 0.05 M, the rate of formation of O2 will be :", "options": [ { "text": "6.93 $$\\times$$ 10-4 mol min-1" }, { "text": "6.96 $$\\times$$ 10-2 mol min-1" }, { "text": "1.34 $$\\times$$ 10-2 mol min-1" }, { "text": "2.66 L min–1 at STP" } ], "answer": "6.93 $$\\times$$ 10-4 mol min-1", "solution": "**Answer:** 6.93 $$\\times$$ 10-4 mol min-1\n\n$${H_2}{O_2}\\left( {aq} \\right) \\to {H_2}O\\left( {aq} \\right) + {1 \\over 2}{O_2}\\left( g \\right)$$ \n

For a first order reaction\n

$$k = {{2.303} \\over t}\\log {a \\over {\\left( {a - x} \\right)}}$$\n

Given $$a = 0.5,\\left( {a - x} \\right) = 0.125,\\,t = 50\\,\\,$$ min\n

$$\\therefore$$ $$\\,\\,\\,\\,\\,k = {{2.303} \\over {50}}\\log {{0.5} \\over {0.125}}$$\n

$$ = 2.78 \\times {10^{ - 2}}\\,{\\min ^{ - 1}}$$\n

$$r = k\\left[ {{H_2}{O_2}} \\right] = 2.78 \\times {10^{ - 2}} \\times 0.05$$ \n

$$ = 1.386 \\times {10^{ - 3}}\\,\\,mol\\,{\\min ^{ - 1}}$$\n

Now\n

$$ - {{d\\left[ {{H_2}{O_2}} \\right]} \\over {dt}} = {{d\\left[ {{H_2}O} \\right]} \\over {dt}} = {{2d\\left[ {{O_2}} \\right]} \\over {dt}}$$\n

$$\\therefore$$ $$\\,\\,\\,\\,\\,{{2d\\left[ {{O_2}} \\right]} \\over {dt}} - {{d\\left[ {{H_2}{O_2}} \\right]} \\over {dt}}$$\n

$$\\therefore$$ $$\\,\\,\\,\\,\\,$$ $${{d\\left[ {{O_2}} \\right]} \\over {dt}} = {1 \\over 2} \\times {{d\\left[ {{H_2}{O_2}} \\right]} \\over {dt}}$$ \n

$$ = {{1.386 \\times {{10}^{ - 3}}} \\over 2} = 6.93 \\times {10^{ - 4}}\\,mol\\,{\\min ^{ - 1}}$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 696, "subject": "Chemistry", "question": "N2O5 decomposes to NO2 and O2 and follows first order kinetics. After 50 minutes, the pressure inside the vessel increases from 50 mmHg to 87.5 mmHg. The pressure of the gaseous mixture after 100 minute at constant temperature will be :", "options": [ { "text": "175.0 mmHg" }, { "text": "116.25 mmHg" }, { "text": "136.25 mmHg" }, { "text": "106.25 mmHg" } ], "answer": "106.25 mmHg", "solution": "**Answer:** 106.25 mmHg\n\n\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n
N2O5$$\\rightleftharpoons$$2NO2+ $${1 \\over 2}$$O2
At t = 0 5000
At t = 50 min50 $$-$$ P12P1$${{{P_1}} \\over 2}$$
\n

Total pressure after 50 min, \n

= 50 $$-$$ P1 + 2P1 + $${{{P_1}} \\over 2}$$ = 87.5\n

$$\\therefore\\,\\,\\,$$ 50 + $${{3{P_1}} \\over 2}$$ = 87.5\n

$$ \\Rightarrow $$ $$\\,\\,\\,$$ P1 = 25 \n

So, 50 min is the Half $$-$$ life period of the reaction. \n

Then 100 min is two half life \n

\n\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n
N2O5$$\\rightleftharpoons$$ 2NO2+ $${1 \\over 2}$$O2
At t = 100 min50 $$-$$ P22P2$${{{P_2}} \\over 2}$$
\n

$$\\therefore\\,\\,\\,$$ 50 $$-$$ P2 = $${{25} \\over 2}$$\n

P2 = 37.5\n

$$\\therefore\\,\\,\\,$$ Total pressure at t = 100 min\n

= 50 $$-$$ P2 + 2P2 + $${{{P_2}} \\over 2}$$\n

= 50 + $${3 \\over 2}$$ $$ \\times $$ 37.5\n

= 106.25 mm of Hg", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 697, "subject": "Chemistry", "question": "For a first order reaction, A $$ \\to $$ P, t1/2 (half-life) is 10 days The time required for $${1 \\over 4}$$th conversion of A (in days) is : (ln 2 = 0.693,    ln 3 = 1.1)", "options": [ { "text": "5" }, { "text": "3.2" }, { "text": "4.1" }, { "text": "2.5" } ], "answer": "4.1", "solution": "**Answer:** 4.1\n\nFor first order reaction, \n

The half life, t$${1 \\over 2}$$ = $${{0.693} \\over k}$$\n

Here given t$${1 \\over 2}$$ = 10 days\n

$$\\therefore\\,\\,\\,$$ k = $${{0.693} \\over {10}}$$ = 0.0693 days$$-$$1\n

Now, the time required for $${1 \\over 4}$$th conversion of A is , \n

t = $${{2.303} \\over k}$$ log10 $$\\left( {{a \\over {a - x}}} \\right)$$\n

= $${{2.303} \\over {0.0693}}$$ log $$\\left( {{1 \\over {1 - {1 \\over 4}}}} \\right)$$\n

= $${{2.303} \\over {0.0693}}$$ log $$\\left( {{4 \\over 3}} \\right)$$\n

= 4.1 days.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 698, "subject": "Chemistry", "question": "If 50% of a reaction occurs in 100 second and 75% of the reaction occurs in 200 secod, the order of this reaction is : ", "options": [ { "text": "Zero " }, { "text": "1" }, { "text": "2" }, { "text": "3" } ], "answer": "1", "solution": "**Answer:** 1\n\nAssume initial concentration of the reactant = 1 M\n

After 100 second, concentration becomes of the reactant \n

= 1 $$ \\times $$ $${{50} \\over {100}}$$ = 0.5 M\n

After 200 second, concentration of the reactant becomes \n

= 1 $$ \\times $$ $${{25} \\over {100}}$$ = 0.25 M.\n

So, in first 100 second reactant concentration becomes half of initial and in the second 100 second concentration becomes 0.5 M to 0.25 M, which is also half of 0.5 M.\n

So, after each 100 second period, concentration of reactant becomes half of initial concentration. \n

So, 100 second is the half life period and it is independent of concentration of reactant. This is characteristic of first order reaction. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 699, "subject": "Chemistry", "question": "The reaction 2X $$ \\to $$ B is a zeroth order reaction. If the initial concentration of X is 0.2 M, the half-life is 6 h. When the initial concentration of X is 0.5 M, the time required to reach its final concentration of 0.2 M will\nbe: \n", "options": [ { "text": "18.0 h" }, { "text": "9.0 h" }, { "text": "7.2 h" }, { "text": "12.0 h" } ], "answer": "18.0 h", "solution": "**Answer:** 18.0 h\n\nFor zero order reaction,\n

t1/2 = $${{{a_0}} \\over {2k}}$$\n

$$ \\Rightarrow $$ k = $${{{a_0}} \\over {2{t_{1/2}}}}$$ = $${{0.2} \\over {2 \\times 6}}$$ = 1.67 $$ \\times $$ 10-2 mol L–1h–1\n

For zero order reaction,\n

A0 - At = kt\n

$$ \\Rightarrow $$ 0.5 - 0.2 = 1.67 $$ \\times $$ 10-2 t\n

$$ \\Rightarrow $$ t = $${{0.3} \\over {1.67 \\times {{10}^{ - 2}}}}$$ = 18 h", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 700, "subject": "Chemistry", "question": "Decomposition of X exhibits a rate constant of 0.05 $$\\mu $$g/year. How many year are required for the decomposition of 5$$\\mu $$g of X into 2.5 $$\\mu $$g? \n", "options": [ { "text": "50" }, { "text": "20" }, { "text": "25" }, { "text": "40" } ], "answer": "50", "solution": "**Answer:** 50\n\nAccording to unit of rate constant it is a zero order\nreaction, then half life of zero order reaction.\n

t1/2 = $${{{a_0}} \\over {2k}} = {5 \\over {2 \\times 0.05}}$$ = 50 years", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 701, "subject": "Chemistry", "question": "It is true that :", "options": [ { "text": "A first order reaction is always a single step reaction" }, { "text": "A zero order reaction is a multistep reaction" }, { "text": "A zero order reaction is a single step reaction" }, { "text": "A second order reaction is always a multistep reaction" } ], "answer": "A zero order reaction is a multistep reaction", "solution": "**Answer:** A zero order reaction is a multistep reaction\n\nZero order reaction has complex mechanism.\n
Zero order reaction is a multistep reaction.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 702, "subject": "Chemistry", "question": "If 75% of a first order reaction was completed\nin 90 minutes, 60% of the same reaction would\nbe completed in approximately (in minutes)\n_______.\n

(Take : log 2 = 0.30; log 2.5 = 0.40)", "options": [], "answer": "60", "solution": "**Answer:** 60\n\nt75% = 90 min = 2 × t1/2\n

$$ \\Rightarrow $$ t1/2 = 45 min\n

Rate constant, K = $${{0.693} \\over {45}}$$ min-1\n

Time for completion of 60% of the reaction,\n

t60% = $${{2.303} \\over K}\\log {{10} \\over 4}$$\n

= $${{2.303 \\times 45} \\over {0.693}}\\log 2.5$$\n

= 60 min ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 703, "subject": "Chemistry", "question": "Gaseous cyclobutene isomerises to butadiene in a first order process which has a 'k' value of 3.3 $$ \\times $$ 10-4 s-1 at 153°C. The time in minutes it takes for the isomerization to proceed 40% to completion at this temperature is ______.
\n(Rounded off to the nearest integer)", "options": [], "answer": "26", "solution": "**Answer:** 26\n\n\"JEE
It is a first order isomerisation reaction. Integrated rate law for 1st order reaction is

$$kt = \\ln {{{{[A]}_0}} \\over {{{[A]}_t}}}$$ ...(i)

Here,

k = rate constant

$${[A]_0}$$ = initial concentration

$${[A]_t}$$ = concentration at time 't'

Given, k = 3.3 $$\\times$$ 10$$-$$4 s$$-$$1

$${[A]_0} = 100 \\Rightarrow {[A]_t} = 100 - 40 = 60$$

Put values in Eq. (i), we get

$$3.3 \\times {10^{ - 4}}{s^{ - 1}} \\times t = \\ln {{100} \\over {60}}$$

t = 1547.95 s = 25.79 min (1 min = 60 s or 1s = $${1 \\over {60}}$$ min) = 26 minutes", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 704, "subject": "Chemistry", "question": "Sucrose hydrolyses in acid solution into glucose and fructose following first order rate law with a half-life of 3.33 h at 25$$^\\circ$$C. After 9 h, the fraction of sucrose remaining is f. The value of $${\\log _{10}}\\left( {{1 \\over f}} \\right)$$ is ________ $$\\times$$ 10$$-$$2. (Rounded off to the nearest integer)

[Assume : ln 10 = 2.303, ln 2 = 0.693]", "options": [], "answer": "81", "solution": "**Answer:** 81\n\nGiven, $$\\mathop {{C_{12}}{H_{22}}{O_{11}}}\\limits_{Sucrose} + {H_2}O\\buildrel {1st\\,order} \\over\n \\longrightarrow \\mathop {{C_6}{H_{12}}{O_6}}\\limits_{Glu\\cos e} + \\mathop {{C_6}{H_{12}}{O_6}}\\limits_{Fructose} $$

$${t_{1/2}} = {{10} \\over 3}h$$

At t = 0, a = [A]0      (initial conc.)

t = 9h, a $$-$$ x = [A]t         [conc. at time t]

For using 1st order equation,

$$K = {{2.303} \\over t}\\log {{{{[A]}_0}} \\over {{{[A]}_t}}} $$\n

$$\\Rightarrow {{K \\times t} \\over {2.303}} = \\log {{{{[A]}_0}} \\over {{{[A]}_t}}}$$

$${{\\ln 2 \\times 9} \\over {10/3 \\times 2.303}} = \\log \\left( {{1 \\over F}} \\right) \\Rightarrow \\log \\left( {{1 \\over F}} \\right) = 0.8124$$ ($$\\because$$ $$k = {{\\ln 2} \\over {{t_{1/2}}}}$$)

$$\\log \\left( {{1 \\over F}} \\right) = 81.24 \\times {10^{ - 2}}$$

x = 81.24 or x $$\\approx$$ 81", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 705, "subject": "Chemistry", "question": "For a certain first order reaction 32% of the reactant is left after 570s. The rate constant of this reaction is _________ $$\\times$$ 10$$-$$3 s$$-$$1. (Round off to the Nearest Integer). [Given : log102 = 0.301, ln10 = 2.303]", "options": [], "answer": "2", "solution": "**Answer:** 2\n\n$$k = {1 \\over t}\\ln \\left[ {{a \\over {a - x}}} \\right]$$

$$k = {{2.303} \\over {570}}\\log \\left( {{{100} \\over {32}}} \\right)$$

$$k = {{2.303} \\over {570}}\\left[ {\\log ({{10}^2}) - \\log {2^5}} \\right]$$

$$k = {{2.303} \\over {570}} \\times 0.5$$

$$k = 2 \\times {10^{ - 3}}{s^{ - 1}}$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 706, "subject": "Chemistry", "question": "A reaction has a half life of 1 min. The time required for 99.9% completion of the reaction is _________ min. (Round off to the Nearest Integer). [Use : ln 2 = 0.69; ln 10 = 2.3]", "options": [], "answer": "10", "solution": "**Answer:** 10\n\nGiven t1/2 = 1 min

$$ \\therefore $$ $$K = {{\\ln 2} \\over {{t_{1/2}}}} = {{\\ln 2} \\over 1}$$

From formula we know,

$$Kt = \\ln \\left( {{{100} \\over {100 - x}}} \\right)$$

$$ \\Rightarrow {{\\ln 2} \\over 1} = {1 \\over t}\\ln \\left( {{{100} \\over {0.1}}} \\right)$$

$$ \\Rightarrow 0.3 = {1 \\over t}\\ln \\left( {{{10}^3}} \\right)$$

$$ \\Rightarrow 0.3 = {1 \\over t} \\times 3$$

$$ \\Rightarrow t = 10$$ min", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 707, "subject": "Chemistry", "question": "The inactivation rate of a viral preparation is proportional to the amount of virus. In the first minute after preparation, 10% of the virus is inactivated. The rate constant for viral inactivation is ___________ $$\\times$$ 10$$-$$3 min$$-$$1. (Nearest integer)

[Use : ln 10 = 2.303; log10 3 = 0.477; property
of logarithm : log xy = y log x]", "options": [], "answer": "106", "solution": "**Answer:** 106\n\nUnit of rate constant is min$$-$$1, so it must be a first order reaction. For first order reaction,

k $$\\times$$ t = 2.303 log$${{{A_0}} \\over {{A_t}}}$$

k is the rate constant

t is the time

A0 is the initial conc.

At is the conc. at time, t

Using formula,

A0 = 100, At = 90 min 1 min

So, K $$\\times$$ 1 = 2.303 $$\\times$$ log$${{100} \\over {90}}$$

= 2.303 $$\\times$$ (log 10 $$-$$ 2 log 3)

= 2.303 $$\\times$$ (1 $$-$$ 2 $$\\times$$ 0.477)

= 0.10593

= 105.93 $$\\times$$ 10$$-$$3

$$\\approx$$ 106

Hence, the rate constant for viral inactivation is 106.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 708, "subject": "Chemistry", "question": "$$PC{l_5}(g) \\to PC{l_3}(g) + C{l_2}(g)$$

In the above first order reaction the concentration of PCl5 reduces from initial concentration 50 mol L$$-$$1 to 10 mol L$$-$$1 in 120 minutes at 300 K. The rate constant for the reaction at 300 K is x $$\\times$$ 10$$-$$2 min$$-$$1. The value of x is __________. [Given log5 = 0.6989]", "options": [], "answer": "1", "solution": "**Answer:** 1\n\n$$PC{l_5}(g)\\mathrel{\\mathop{\\kern0pt\\longrightarrow}\n\\limits_{300K}^{I\\,order}} PC{l_3}(g) + C{l_2}(g)$$

t = 0

50M

t = 120min

10 M

$$ \\Rightarrow K = {{2.303} \\over t}\\log {{[{A_0}]} \\over {[{A_t}]}}$$

$$ \\Rightarrow K = {{2.303} \\over {120}}\\log {{50} \\over {10}}$$

$$ \\Rightarrow K = {{2.303} \\over {120}} \\times 0.6989 = 0.013413$$ min$$-$$1

$$ = 1.3413 \\times {10^{ - 2}}$$ min$$-$$1

1.34 $$\\Rightarrow$$ Nearest integer = 1", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 709, "subject": "Chemistry", "question": "$${N_2}{O_{5(g)}} \\to 2N{O_{2(g)}} + {1 \\over 2}{O_{2(g)}}$$

In the above first order reaction the initial concentration of N2O5 is 2.40 $$\\times$$ 10$$-$$2 mol L$$-$$1 at 318 K. The concentration of N2O5 after 1 hour was 1.60 $$\\times$$ 10$$-$$2 mol L$$-$$1. The rate constant of the reaction at 318 K is ______________ $$\\times$$ 10$$-$$3 min$$-$$1. (Nearest integer)

[Given : log 3 = 0.477, log 5 = 0.699]", "options": [], "answer": "7", "solution": "**Answer:** 7\n\n$$K = {{2.303} \\over t}\\log {{{{[{N_2}{O_5}]}_0}} \\over {{{[{N_2}{O_5}]}_t}}}$$

$$ = {{2.303} \\over {60}}\\log {{2.4} \\over {1.6}} = 6.76 \\times {10^{ - 3}}$$ min$$-$$1 $$ \\approx $$ 7 $$\\times$$ 10$$-$$3 min$$-$$1", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 710, "subject": "Chemistry", "question": "For the first order reaction A $$\\to$$ 2B, 1 mole of reactant A gives 0.2 moles of B after 100 minutes. The half life of the reaction is __________ min. (Round off to the nearest integer).

[Use : ln 2 = 0.69, ln 10 = 2.3]

Properties of logarithms : ln xy = y ln x;

$$\\ln \\left( {{x \\over y}} \\right) = \\ln x - \\ln y$$

(Round off to the nearest integer)", "options": [], "answer": "600to700", "solution": "**Answer:** 600to700\n\n\"JEE
Now, $$t = {{{t_{1/2}}} \\over {\\ln 2}} \\times {{[{A_0}]} \\over {[{A_t}]}}$$

$$100 = {{{t_{1/2}}} \\over {\\ln 2}} \\times \\ln {1 \\over {0.9}} \\Rightarrow {t_{1/2}} = 690$$ min (taking ln 3 = 1.11)

Ans. 600 to 700", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 711, "subject": "Chemistry", "question": "For a first order reaction, the ratio of the time for 75% completion of a reaction to the time for 50% completion is ____________. (Integer answer)", "options": [], "answer": "2", "solution": "**Answer:** 2\n\n$$k = {{2.303} \\over t}\\log {a \\over {a - x}}$$

$${{2.303} \\over {{t_{50\\% }}}}\\log {{100} \\over {100 - 50}} = {{2.303} \\over {{t_{75\\% }}}}\\log {{100} \\over {100 - 75}}$$

$$ \\Rightarrow $$ $${t_{75\\% }} = 2{t_{50\\% }}$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 712, "subject": "Chemistry", "question": "

A radioactive element has a half life of 200 days. The percentage of original activity remaining after 83 days is ___________. (Nearest integer)

\n

(Given : antilog 0.125 = 1.333, antilog 0.693 = 4.93)

", "options": [], "answer": "75", "solution": "**Answer:** 75\n\n$\\lambda=\\frac{2.303}{t} \\log \\frac{A_{0}}{A}$\n

\n$$\n\\begin{aligned}\n&\\frac{0.693}{200}=\\frac{2.303}{83} \\log \\frac{A_{0}}{A} \\\\\\\\\n&\\frac{A}{A_{0}}=0.75\n\\end{aligned}\n$$\n

\nHence, percentage of original activity remaining after 83 days is $75 \\%$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 713, "subject": "Chemistry", "question": "

For a first order reaction A $$\\to$$ B, the rate constant, k = 5.5 $$\\times$$ 10$$-$$14 s$$-$$1. The time required for 67% completion of reaction is x $$\\times$$ 10$$-$$1 times the half life of reaction. The value of x is _____________ (Nearest integer)

\n

(Given : log 3 = 0.4771)

", "options": [], "answer": "16", "solution": "**Answer:** 16\n\n$$\n\\begin{aligned}\n&\\because \\mathrm{kt}=\\ln \\frac{\\mathrm{A}_{0}}{\\mathrm{~A}} \\\\\\\\\n&\\frac{\\ln 2}{\\mathrm{t}_{\\frac{1}{2}}} \\mathrm{t}_{67 \\%}=\\ln \\frac{\\mathrm{A}_{0}}{0.33 \\mathrm{~A}_{0}} \\\\\\\\\n&\\frac{\\log 2}{\\mathrm{t}_{\\frac{1}{2}}} \\mathrm{t}_{67 \\%}=\\log \\frac{1}{0.33} \\\\\\\\\n&\\mathrm{t}_{67 \\%}=1.566 \\mathrm{t}_{1 / 2} \\\\\\\\\n&\\mathrm{x}=15.66\n\\end{aligned}\n$$\n

\nNearest integer $=16$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 714, "subject": "Chemistry", "question": "

For a first order reaction, the time required for completion of 90% reaction is 'x' times the half life of the reaction. The value of 'x' is

\n

(Given : ln 10 = 2.303 and log 2 = 0.3010)

", "options": [ { "text": "1.12" }, { "text": "2.43" }, { "text": "3.32" }, { "text": "33.31" } ], "answer": "3.32", "solution": "**Answer:** 3.32\n\n$$\\mathrm{A} \\rightarrow$$ Products

\nFor a first order reaction,

\n$$\n\\mathrm{t}_{1 / 2}=\\frac{\\ln 2}{\\mathrm{k}}=\\frac{0.693}{\\mathrm{k}}\n$$

\nTime for $$90 \\%$$ conversion,

\n$$t_{90 \\%}=\\frac{1}{k} \\ln \\frac{100}{10}=\\frac{\\ln 10}{k}=\\frac{2.303}{k}$$

\n$$t_{90\\%}=\\frac{2.303}{0.693} t_{1 / 2}=3.32 t_{1 / 2}$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 715, "subject": "Chemistry", "question": "

For a reaction $$\\mathrm{A} \\rightarrow 2 \\mathrm{~B}+\\mathrm{C}$$ the half lives are $$100 \\mathrm{~s}$$ and $$50 \\mathrm{~s}$$ when the concentration of reactant $$\\mathrm{A}$$ is $$0.5$$ and $$1.0 \\mathrm{~mol} \\mathrm{~L}^{-1}$$ respectively. The order of the reaction is ______________ . (Nearest Integer)

", "options": [], "answer": "2", "solution": "**Answer:** 2\n\n$t_{1 / 2} \\propto \\frac{1}{\\left(a_{0}\\right)^{n-1}}$\n

\n$$\n\\begin{array}{ll}\n\\mathrm{t}_{1 / 2}=100 \\,\\mathrm{sec} & \\mathrm{a}_{0}=0.5 \\\\\n\\mathrm{t}_{1 / 2}=50 \\,\\mathrm{sec} & \\mathrm{a}_{0}=1\n\\end{array}\n$$\n

\n$$\\frac{100}{50}=\\left(\\frac{1}{0 \\cdot 5}\\right)^{n-1}$$\n

\n$$(2)=(2)^{n-1}$$\n

\n$$\\mathrm{n}-1=1$$\n

\n$$n=2$$\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 716, "subject": "Chemistry", "question": "

At $$30^{\\circ} \\mathrm{C}$$, the half life for the decomposition of $$\\mathrm{AB}_{2}$$ is $$200 \\mathrm{~s}$$ and is independent of the initial concentration of $$\\mathrm{AB}_{2}$$. The time required for $$80 \\%$$ of the $$\\mathrm{AB}_{2}$$ to decompose is

\n

Given: $$\\log 2=0.30$$ $$\\quad \\log 3=0.48$$

", "options": [ { "text": "200 s" }, { "text": "323 s" }, { "text": "467 s" }, { "text": "532 s" } ], "answer": "467 s", "solution": "**Answer:** 467 s\n\nSince, half-life is independent of the initial concentration of $$A B_{2}$$. Hence, the reaction is \"First Order\".\n

\n$$k=\\frac{2.303 \\log 2}{t_{1 / 2}}$$\n

\n$$\\frac{2.303 \\log 2}{t_{1 / 2}}=\\frac{2.303}{t} \\log \\frac{100}{(100-80)}$$\n

\n$$\\frac{2.303 \\times 0.3}{200}=\\frac{2.303}{t} \\log 5$$\n

\n$$t=467 \\mathrm{~s}$$\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 717, "subject": "Chemistry", "question": "

$$\\matrix{\n {[A]} & \\to & {[B]} \\cr \n {{\\mathop{\\rm Reactant}\\nolimits} } & {} & {{\\mathop{\\rm Product}\\nolimits} } \\cr \n\n } $$

\n

If formation of compound $$[\\mathrm{B}]$$ follows the first order of kinetics and after 70 minutes the concentration of $$[\\mathrm{A}]$$ was found to be half of its initial concentration. Then the rate constant of the reaction is $$x \\times 10^{-6} \\mathrm{~s}^{-1}$$. The value of $$x$$ is ______________. (Nearest Integer)

", "options": [], "answer": "165", "solution": "**Answer:** 165\n\n$\\mathrm{K}=\\frac{0.693}{\\mathrm{t}_{1 / 2}}=\\frac{0.693}{70 \\times 60}$

\n$=\\frac{6930}{7 \\times 6} \\times 10^{-6}$

\n$=165 \\times 10^{-6} \\mathrm{~s}^{-1}$", "topic": "Algebra", "subtopic": "Linear Algebra" }, { "id": 718, "subject": "Chemistry", "question": "

For the given first order reaction

\n

$$\\mathrm{A} \\rightarrow \\mathrm{B}$$

\n

the half life of the reaction is $$0.3010 \\mathrm{~min}$$. The ratio of the initial concentration of reactant to the concentration of reactant at time $$2.0 \\mathrm{~min}$$ will be equal to ___________. (Nearest integer)

", "options": [], "answer": "100", "solution": "**Answer:** 100\n\n$\\mathrm{t}_{1 / 2}=\\frac{0.693}{\\mathrm{~K}} \\quad\\quad \\mathrm{t}_{1 / 2}$ given $=0.3010$\n

\n$$\n\\begin{aligned}\n&K=\\frac{0.693}{0.3010} \\\\\\\\\n&K=2.30\n\\end{aligned}\n$$\n

\n$\\mathrm{K}=\\frac{2.303}{\\mathrm{t}} \\log \\frac{\\left(\\mathrm{A}_{0}\\right)}{\\left(\\mathrm{A}_{\\mathrm{t}}\\right)}$\n

\n$A_{0} \\rightarrow$ initial concentration of reactant\n

\n$A_{t} \\rightarrow$ concentration of reactant at time $t$\n

\n$2.303=\\frac{2.303}{2} \\log \\frac{\\left(A_{0}\\right)}{\\left(A_{t}\\right)}$\n

\n$2=\\log \\frac{\\left(A_{0}\\right)}{\\left(A_{t}\\right)}$\n

\n$100=\\frac{A_{0}}{A_{t}}$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 719, "subject": "Chemistry", "question": "

Assuming $$1 \\,\\mu \\mathrm{g}$$ of trace radioactive element X with a half life of 30 years is absorbed by a growing tree. The amount of X remaining in the tree after 100 years is ______ $$\\times\\, 10^{-1} \\mu \\mathrm{g}$$.

\n

[Given : ln 10 = 2.303; log 2 = 0.30]

", "options": [], "answer": "1", "solution": "**Answer:** 1\n\n$$t=\\frac{1}{\\lambda} \\ln \\left(\\frac{a}{a-x}\\right)$$\n

\n$$\n\\begin{aligned}\n&\\Rightarrow100=\\left(\\frac{30}{\\ln 2}\\right)\\left[\\ln \\left(\\frac{1}{w}\\right)\\right] \\\\\n&\\Rightarrow{\\left[\\frac{100 \\times \\log 2}{30}\\right]=\\log \\left(\\frac{1}{w}\\right)} \\\\\n&\\Rightarrow1=\\log \\left(\\frac{1}{w}\\right) \\\\\n&\\Rightarrow\\frac{1}{w}=10 \\\\\n&\\text { So } w=0.1 \\mu g\n\\end{aligned}\n$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 720, "subject": "Chemistry", "question": "

$$\\mathrm{A}$$ $$\\rightarrow \\mathrm{B}$$

\n

The above reaction is of zero order. Half life of this reaction is $$50 \\mathrm{~min}$$. The time taken for the concentration of $$\\mathrm{A}$$ to reduce to one-fourth of its initial value is ____________ min. (Nearest integer)

", "options": [], "answer": "75", "solution": "**Answer:** 75\n\n\"JEE\n

$ \\mathrm{a}-\\mathrm{x}=\\frac{\\mathrm{a}}{4} \\Rightarrow \\mathrm{x}=\\frac{3 \\mathrm{a}}{4} $\n

$ \\mathrm{t}_{1 / 2}=\\frac{\\mathrm{a}}{2 \\mathrm{~K}}=50 \\mathrm{~min} .$\n

$ \\Rightarrow \\frac{\\mathrm{a}}{\\mathrm{K}}=100 \\mathrm{~min} . $\n

$ \\mathrm{t}=\\frac{\\mathrm{x}}{\\mathrm{K}}=\\frac{3 \\mathrm{a}}{4 \\mathrm{~K}}=75 \\mathrm{~min} $", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 721, "subject": "Chemistry", "question": "The rate constant for a first order reaction is $20 \\mathrm{~min}^{-1}$. The time required for the initial concentration of the reactant to reduce to its $\\frac{1}{32}$ level is _______ $\\times 10^{-2} \\mathrm{~min}$. (Nearest integer)

\n(Given : $\\ln 10=2.303$ and $$ \\log 2=0.3010 \\text { )}$$", "options": [], "answer": "17", "solution": "**Answer:** 17\n\n$K=20 \\mathrm{~min}^{-1}$\n\n

$$\n\\mathrm{t}_{1 / 2}=\\frac{0.6932}{20}=\\frac{\\ln 2}{20}\n$$\n\n

Required time $=\\mathrm{n} \\times \\mathrm{t}_{1 / 2}$\n\n

$$C = {{{C_0}} \\over {{2^n}}} = {{{C_0}} \\over {32}}$$\n

$$ \\Rightarrow $$ $${2^n} = 32$$ = $${2^5}$$\n

$$ \\Rightarrow $$ n = 5\n\n

$$\n\\begin{aligned}\n\\text { Required time } & =\\frac{5 \\times 0.6932}{20} \\\\\\\\\n& =0.173 \\mathrm{~min} \\\\\\\\\n& =17.3 \\times 10^{-2} \\mathrm{~min}\n\\end{aligned}\n$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 722, "subject": "Chemistry", "question": "

A and B are two substances undergoing radioactive decay in a container. The half life of A is 15 min and that of B is 5 min. If the initial concentration of B is 4 times that of A and they both start decaying at the same time, how much time will it take for the concentration of both of them to be same? _____________ min.

", "options": [], "answer": "15", "solution": "**Answer:** 15\n\n\"JEE\n

After $15 \\min ,[A]=[B]=\\frac{Co}{2}$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 723, "subject": "Chemistry", "question": "An organic compound undergoes first-order decomposition. If the time taken for the $60 \\%$ decomposition is $540 \\mathrm{~s}$, then the time required for $90 \\%$ decomposition will be ________ s. (Nearest integer).\n

\nGiven: $\\ln 10=2.3 ; \\log 2=0.3$", "options": [], "answer": "1350", "solution": "**Answer:** 1350\n\nFor the first order reaction,\n

$$\nk=\\frac{1}{t} \\ln \\frac{a}{a-x} $$\n

$$ \\Rightarrow $$ $$ t=\\frac{1}{k} \\ln \\frac{a}{a-x}\n$$\n

When reaction is $60 \\%$ completed,\n

$$\nx=\\frac{60}{100} a=0.6 a, t=540 \\text { seconds } ; $$\n

$k= \\frac{1}{t_1} \\ln \\frac{a}{a-0.6 a}$\n

$$ \\therefore $$ $$t_1=\\frac{1}{k} \\ln \\frac{a}{0.4 a}\n$$\n

When reaction is $90 \\%$ completed, i.e., $x=0.9 a$\n

$$\n\\begin{aligned}\nk= \\frac{1}{t_2} \\ln \\frac{a}{a-0.9 a} \\\\\\\\\n\\Rightarrow t_2=\\frac{1}{k} \\ln \\frac{a}{0.1 a} \\\\\\\\\n \\therefore \\frac{t_1}{t_2} =\\frac{\\frac{1}{k} \\ln \\frac{a}{0.4 a}}{\\frac{1}{k} \\ln \\frac{a}{0.1 a}} \\\\\\\\\n\\Rightarrow \\frac{540}{t_2} =\\frac{\\ln \\frac{10}{4}}{\\ln 10}\n\n\\end{aligned}\n$$\n

$$ \\Rightarrow $$ $\\frac{540}{t_2} =\\frac{2.3\\log 10-2.3\\log 4}{2.3\\log 10}$\n

$$ \\Rightarrow $$ $\\frac{540}{t_2} =\\frac{2.3-2.3(0.6)}{2.3}$\n

$$ \\Rightarrow $$ $\\frac{540}{t_2} =0.4$\n

$$ \\Rightarrow $$ $$\nt_2=\\frac{540}{0.4}=1350 \n$$ sec", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 724, "subject": "Chemistry", "question": "

If compound A reacts with B following first order kinetics with rate constant $$2.011 \\times 10^{-3} \\mathrm{~s}^{-1}$$. The time taken by $$\\mathrm{A}$$ (in seconds) to reduce from $$7 \\mathrm{~g}$$ to $$2 \\mathrm{~g}$$ will be ___________. (Nearest Integer)

\n

$$[\\log 5=0.698, \\log 7=0.845, \\log 2=0.301]$$

", "options": [], "answer": "623", "solution": "**Answer:** 623\n\n

$$t = {{2.303} \\over k}\\log {{{C_0}} \\over {{C_t}}}$$

\n

$$ = {{2.303} \\over {2.011 \\times {{10}^{ - 3}}}}\\log {7 \\over 2}$$

\n

$$ = {{2.303 \\times {{10}^3}} \\over {2.011}}(.845 - .301)$$

\n

$$ = 622.99$$

\n

$$ \\approx 623$$ sec.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 725, "subject": "Chemistry", "question": "

A first order reaction has the rate constant, $$\\mathrm{k=4.6\\times10^{-3}~s^{-1}}$$. The number of correct statement/s from the following is/are __________

\n

Given : $$\\mathrm{\\log3=0.48}$$

\n

A. Reaction completes in 1000 s.

\n

B. The reaction has a half-life of 500 s.

\n

C. The time required for 10% completion is 25 times the time required for 90% completion.

\n

D. The degree of dissociation is equal to ($$\\mathrm{1-e^{-kt}}$$)

\n

E. The rate and the rate constant have the same unit.

", "options": [], "answer": "1", "solution": "**Answer:** 1\n\n(A) $\\underset{1-\\alpha}{\\mathrm{A}} \\longrightarrow$ Products\n

\n$\\mathrm{k}=4.6 \\times 10^{-3} \\mathrm{~s}^{-1}$\n

\n$\\mathrm{kt}=\\ln \\frac{1}{1-\\alpha}$\n

\n$\\alpha=1-\\mathrm{e}^{-\\mathrm{kt}}$\n

\nReaction completes at infinite time.\n

(B) For first order reaction,\n

\nHalf-life $=\\frac{0.693}{4.6 \\times 10^{-3}}=150.65 \\mathrm{~s}$\n

\n(C) For $10 \\%$ completion, $t=t_1$\n

$$\n\\begin{aligned}\n& a=100, a-x=100-10=90 \\\\\\\\\n& k=\\frac{2.303}{t_1} \\log \\frac{100}{90}=4.6 \\times 10^{-3} \\\\\\\\\n& t_1=\\frac{2.303}{4.6 \\times 10^{-3}} \\times 0.04575\n\\end{aligned}\n$$\n

For $90 \\%$ completion $t=t_2, a=100, a-x=100-90=10$\n

$$\n\\begin{aligned}\n& k=\\frac{2.303}{t_2} \\log \\frac{100}{10}=4.6 \\times 10^{-3} \\\\\\\\\n& \\frac{2.303}{t_2} \\times 1=4.6 \\times 10^{-3} \\Rightarrow t_2=\\frac{2.303}{4.6 \\times 10^{-3}} \\\\\\\\\n& \\frac{t_2}{t_1}=\\frac{1}{0.04575}=21.85 \\quad \\text { (Incorrect) }\n\\end{aligned}\n$$\n

\n(E) Unit of rate = mol L–1s–1\n

Unit of rate constant = s–1\n

Both have different units.\n

\n$\\therefore$ Number of correct statements $=1$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 726, "subject": "Chemistry", "question": "

For the first order reaction A $$\\to$$ B, the half life is 30 min. The time taken for 75% completion of the reaction is _________ min. (Nearest integer)

\n

Given : log 2 = 0.3010

\n

log 3 = 0.4771

\n

log 5 = 0.6989

", "options": [], "answer": "60", "solution": "**Answer:** 60\n\nTime taken for $75 \\%$ completion\n

\n$=2 \\times \\mathrm{t}_{1 / 2}$\n

\n$=2 \\times 30$\n

\n$=60 \\mathrm{~min}$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 727, "subject": "Chemistry", "question": "

A(g) $$\\to$$ 2B(g) + C(g) is a first order reaction. The initial pressure of the system was found to be 800 mm Hg which increased to 1600 mm Hg after 10 min. The total pressure of the system after 30 min will be _________ mm Hg. (Nearest integer)

", "options": [], "answer": "2200", "solution": "**Answer:** 2200\n\nAt 10 minutes, the pressure increased by 2x:\n

\n$$800 + 2x = 1600$$

\n$$2x = 800$$

\n$$x = 400$$\n

\nThe rate constant (k) can be found as:\n

\n$$k = \\frac{2.303}{10} \\log \\frac{800}{400} = \\frac{2.303 \\times \\log 2}{10}$$\n

\nFor 30 minutes, we can set up the equation:\n

\n$$k = \\frac{2.303}{30} \\log \\frac{800}{800 - y}$$

\n$$\\frac{2.303 \\times \\log 2}{10} = \\frac{2.303}{30} \\log \\left(\\frac{800}{800 - y}\\right)$$\n

\nSolving for y:\n

\n$$\\left(\\frac{800}{800 - y}\\right) = 8$$

\n$$800 - y = 100$$

\n$$y = 700$$\n

\nNow we can find the total pressure after 30 minutes:\n

\nTotal pressure = (800 - y) + 2y + y

\nTotal pressure = 100 + 1400 + 700

\nTotal pressure = 2200 mm Hg\n

\nSo, the total pressure of the system after 30 minutes is 2200 mm Hg.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 728, "subject": "Chemistry", "question": "

$$\\mathrm{t}_{87.5}$$ is the time required for the reaction to undergo $$87.5 \\%$$ completion and $$\\mathrm{t}_{50}$$ is the time required for the reaction to undergo $$50 \\%$$ completion. The relation between $$\\mathrm{t}_{87.5}$$ and $$\\mathrm{t}_{50}$$ for a first order reaction is $$\\mathrm{t}_{87.5}=x \\times \\mathrm{t}_{50}$$ The value of $$x$$ is ___________. (Nearest integer)

", "options": [], "answer": "3", "solution": "**Answer:** 3\n\nFor a first-order reaction, the relation between the reaction rate constant (k) and time (t) for a given percentage of completion (p) is:\n

\n$$\\ln\\left(\\frac{1}{1-p}\\right) = kt$$\n

\nFor t₅₀ (50% completion), p = 0.5:\n

\n$$\\ln\\left(\\frac{1}{1-0.5}\\right) = k\\cdot t_{50}$$

\n$$\\ln\\left(\\frac{1}{0.5}\\right) = k\\cdot t_{50}$$

\n$$\\ln(2) = k\\cdot t_{50}$$\n

\nFor t₈₇.₅ (87.5% completion), p = 0.875:\n

\n$$\\ln\\left(\\frac{1}{1-0.875}\\right) = k\\cdot t_{87.5}$$

\n$$\\ln\\left(\\frac{1}{0.125}\\right) = k\\cdot t_{87.5}$$\n$$\\ln(8) = k\\cdot t_{87.5}$$\n

\nNow, we need to find the relationship between t₈₇.₅ and t₅₀:\n

\n$$\\frac{k\\cdot t_{87.5}}{k\\cdot t_{50}} = \\frac{\\ln(8)}{\\ln(2)}$$\n

\nSince the k's cancel out, we have:\n

\n$$\\frac{t_{87.5}}{t_{50}} = \\frac{\\ln(8)}{\\ln(2)}$$\n

\nUsing the property of logarithms, we get:\n

\n$$\\frac{t_{87.5}}{t_{50}} = \\frac{\\ln(2^3)}{\\ln(2)}$$\n

\n$$\\frac{t_{87.5}}{t_{50}} = 3$$\n

\nSo, the value of x is 3.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 729, "subject": "Chemistry", "question": "

The number of incorrect statement/s from the following is ___________

\n

A. The successive half lives of zero order reactions decreases with time.

\n

B. A substance appearing as reactant in the chemical equation may not affect the rate of reaction

\n

C. Order and molecularity of a chemical reaction can be a fractional number

\n

D. The rate constant units of zero and second order reaction are $$\\mathrm{mol} ~\\mathrm{L}^{-1} \\mathrm{~s}^{-1}$$ and $$\\mathrm{mol}^{-1} \\mathrm{~L} \\mathrm{~s}^{-1}$$ respectively

", "options": [], "answer": "1", "solution": "**Answer:** 1\n\n

A. The successive half lives of zero order reactions decreases with time. This statement is correct. The half-life of a zero-order reaction is given by the formula $t_{1/2} = [A]_0/2k$, where $[A]_0$ is the initial concentration and k is the rate constant. As the concentration decreases, the half-life also decreases.

\n

B. A substance appearing as reactant in the chemical equation may not affect the rate of reaction. This statement is correct. For example, in a zero-order reaction, the rate of reaction does not depend on the concentration of reactants.

\n

C. Order and molecularity of a chemical reaction can be a fractional number. This statement is incorrect. The order of a reaction can be zero, fractional or negative. However, molecularity, which refers to the number of reactant molecules taking part in an elementary reaction, can only be a whole number.

\n

D. The rate constant units of zero and second order reaction are $$\\mathrm{mol} ~\\mathrm{L}^{-1} \\mathrm{~s}^{-1}$$ and $$\\mathrm{mol}^{-1} \\mathrm{~L} \\mathrm{~s}^{-1}$$ respectively. This statement is correct. The units of the rate constant for a zero-order reaction are M/s or mol·L−1·s−1 . The units of the rate constant for a second-order reaction are 1/(M·s) or L·mol−1·s−1 .

\n", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 730, "subject": "Chemistry", "question": "

A molecule undergoes two independent first order reactions whose respective half lives are 12 min and 3 min. If both the reactions are occurring then the time taken for the 50% consumption of the reactant is ___________ min. (Nearest integer)

", "options": [], "answer": "2", "solution": "**Answer:** 2\n\n

In the given problem, a molecule is undergoing two separate, simultaneous first-order reactions. Each of these reactions has its own rate constant and half-life. The half-life ($T$) of a first-order reaction is related to its rate constant ($k$) by the equation:

\n

$$T = \\frac{\\ln(2)}{k}$$

\n

When two or more first-order reactions are occurring independently and simultaneously (also known as parallel or competing reactions), the overall rate constant for the disappearance of the reactant is the sum of the rate constants for the individual reactions:

\n

$$k_{\\text{total}} = k_1 + k_2$$

\n

Conversely, the overall half-life for the disappearance of the reactant is determined by the reciprocal of the sum of the reciprocals of the individual half-lives:

\n

$$\\frac{1}{T_{\\text{total}}} = \\frac{1}{T_1} + \\frac{1}{T_2}$$

\n

This equation reflects the fact that the reactant is disappearing more quickly than it would due to any single reaction, because it is being consumed by two reactions at once.

\n

Substituting the given half-lives of $12$ minutes and $3$ minutes into this equation gives:

\n

$$\\begin{align}\n\\frac{1}{T_{\\text{total}}} &= \\frac{1}{12\\text{ min}} + \\frac{1}{3\\text{ min}} \\\\\\\\\n&= \\frac{5}{12\\text{ min}}\n\\end{align}$$

\n

Solving for $T_{\\text{total}}$ then gives:

\n

$$T_{\\text{total}} = \\frac{12}{5}\\text{ min} = 2.4\\text{ min}$$

\n

This is the time taken for $50\\%$ of the reactant to be consumed when both reactions are occurring. Rounding to the nearest integer gives an answer of $2$ minutes.

\n", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 731, "subject": "Chemistry", "question": "The following data were obtained during the first order thermal decomposition of a gas A at constant volume :

\n$\\mathrm{A}(\\mathrm{g}) \\rightarrow 2 \\mathrm{~B}(\\mathrm{~g})+\\mathrm{C}(\\mathrm{g})$

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
S.No.Time /sTotal pressure /(atm)
1.00.1
2.1150.28
\n
\nThe rate constant of the reaction is ________ $\\times 10^{-2} \\mathrm{~s}^{-1}$ (nearest integer)", "options": [], "answer": "2", "solution": "**Answer:** 2\n\n\"JEE\n
$\\begin{aligned} & 0.1+2 \\mathrm{x}=0.28 \\\\\\\\ & 2 \\mathrm{x}=0.18 \\\\\\\\ & \\mathrm{x}=0.09 \\\\\\\\ & \\mathrm{~K}=\\frac{1}{115} \\ln \\frac{0.1}{0.1-0.09} \\\\\\\\ & =0.0200 \\mathrm{sec}^{-1} \\\\\\\\ & =2 \\times 10^{-2} \\mathrm{sec}^{-1}\\end{aligned}$", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 732, "subject": "Chemistry", "question": "

Time required for completion of $$99.9 \\%$$ of a First order reaction is ________ times of half life $$\\left(t_{1 / 2}\\right)$$ of the reaction.

", "options": [], "answer": "10", "solution": "**Answer:** 10\n\n

$$\\frac{\\mathrm{t}_{99.9 \\%}}{\\mathrm{t}_{1 / 2}}=\\frac{\\frac{2.303}{\\mathrm{k}}\\left(\\frac{\\mathrm{a}}{\\mathrm{a}-\\mathrm{x}}\\right)}{\\frac{2.303}{\\mathrm{k}} \\log 2}=\\frac{\\log \\left(\\frac{100}{100-99.9}\\right)}{\\log 2}=\\frac{\\log 10^3}{\\log 2}=\\frac{3}{0.3}=10$$

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 733, "subject": "Chemistry", "question": "

$$\\mathrm{r}=\\mathrm{k}[\\mathrm{A}]$$ for a reaction, $$50 \\%$$ of $$\\mathrm{A}$$ is decomposed in 120 minutes. The time taken for $$90 \\%$$ decomposition of $$\\mathrm{A}$$ is _________ minutes.

", "options": [], "answer": "399", "solution": "**Answer:** 399\n\n

$$\\mathrm{r}=\\mathrm{k}[\\mathrm{A}]$$

\n

So, order of reaction $$=1$$

\n

$$\\mathrm{t}_{1 / 2}=120 \\mathrm{~min}$$

\n

For $$90 \\%$$ completion of reaction

\n

$$\\begin{aligned}\n& \\Rightarrow \\mathrm{k}=\\frac{2.303}{\\mathrm{t}} \\log \\left(\\frac{\\mathrm{a}}{\\mathrm{a}-\\mathrm{x}}\\right) \\\\\n& \\Rightarrow \\frac{0.693}{\\mathrm{t}_{1 / 2}}=\\frac{2.303}{\\mathrm{t}} \\log \\frac{100}{10} \\\\\n& \\therefore \\mathrm{t}=399 \\mathrm{~min} .\n\\end{aligned}$$

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 734, "subject": "Chemistry", "question": "

Integrated rate law equation for a first order gas phase reaction is given by (where $$\\mathrm{P}_{\\mathrm{i}}$$ is initial pressure and $$\\mathrm{P}_{\\mathrm{t}}$$ is total pressure at time $$t$$)

", "options": [ { "text": "$$k=\\frac{2.303}{t} \\times \\log \\frac{P_i}{\\left(2 P_i-P_t\\right)}$$\n" }, { "text": "$$\\mathrm{k}=\\frac{2.303}{\\mathrm{t}} \\times \\log \\frac{\\left(2 \\mathrm{P}_{\\mathrm{i}}-\\mathrm{P}_{\\mathrm{t}}\\right)}{\\mathrm{P}_{\\mathrm{i}}}$$\n" }, { "text": "$$k=\\frac{2.303}{t} \\times \\frac{P_i}{\\left(2 P_i-P_t\\right)}$$\n" }, { "text": "$$\\mathrm{k}=\\frac{2.303}{\\mathrm{t}} \\times \\log \\frac{2 \\mathrm{P}_{\\mathrm{i}}}{\\left(2 \\mathrm{P}_{\\mathrm{i}}-\\mathrm{P}_{\\mathrm{t}}\\right)}$$" } ], "answer": "$$k=\\frac{2.303}{t} \\times \\log \\frac{P_i}{\\left(2 P_i-P_t\\right)}$$\n", "solution": "**Answer:** $$k=\\frac{2.303}{t} \\times \\log \\frac{P_i}{\\left(2 P_i-P_t\\right)}$$\n\n\n

$$\\begin{array}{llll}\n\\mathrm{A} \\rightarrow & \\mathrm{B} & + & \\mathrm{C} \\\\\n\\mathrm{P}_{\\mathrm{i}} & 0 & & 0 \\\\\n\\mathrm{P}_{\\mathrm{i}}-\\mathrm{x} & \\mathrm{x} & & \\mathrm{x}\n\\end{array}$$

\n

$$\\begin{aligned}\n& \\mathrm{P_t=P_i+x} \\\\\n& \\mathrm{P_i-x=P_i-P_t+P_i} \\\\\n& \\mathrm{=2 P_i-P_t} \\\\\n& \\mathrm{K=\\frac{2.303}{t} \\log \\frac{P_i}{2 P_i-P_t}}\n\\end{aligned}$$

", "topic": "Algebra", "subtopic": "Equations and Inequalities" }, { "id": 735, "subject": "Chemistry", "question": "

The rate of First order reaction is $$0.04 \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{~s}^{-1}$$ at 10 minutes and $$0.03 \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{~s}^{-1}$$ at 20 minutes after initiation. Half life of the reaction is _______ minutes.

(Given $$\\log 2=0.3010, \\log 3=0.4771$$)

", "options": [], "answer": "24", "solution": "**Answer:** 24\n\n

$$\\begin{aligned}\n& 0.04=\\mathrm{k}[\\mathrm{A}]_0 \\mathrm{e}^{-\\mathrm{k} \\times 10 \\times 60} \\quad \\text{..... (1)}\\\\\n& 0.03=\\mathrm{k}[\\mathrm{A}]_0 \\mathrm{e}^{-\\mathrm{k} \\times 20 \\times 60} \\quad \\text{..... (2)}\n\\end{aligned}$$

\n

$$\\begin{aligned}\n\\frac{4}{3} & =\\mathrm{e}^{600 \\mathrm{k}(2-1)} /(2) \\\\\n\\frac{4}{3} & =\\mathrm{e}^{600 \\mathrm{k}} \\\\\n\\ln \\frac{4}{3} & =600 \\mathrm{k} \\\\\n\\ln \\frac{4}{3} & =600 \\times \\frac{\\ln 2}{\\mathrm{t}_{1 / 2}} \\\\\n\\mathrm{t}_{1 / 2} & =600 \\frac{\\ln 2}{\\ln \\frac{4}{3}} \\sec \\\\\n\\mathrm{t}_{1 / 2} & =600 \\times \\frac{\\log 2}{\\log 4-\\log 3} \\mathrm{sec} .=10 \\times \\frac{0.3010}{0.6020-0.477} \\mathrm{~min} \\\\\n\\mathrm{t}_{1 / 2} & =24.08 \\mathrm{~min}\n\\end{aligned}$$

\n

Ans. 24

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 736, "subject": "Chemistry", "question": "

Consider the following reaction, the rate expression of which is given below

\n

$$\\begin{aligned}\n& \\mathrm{A}+\\mathrm{B} \\rightarrow \\mathrm{C} \\\\\n& \\text { rate }=\\mathrm{k}[\\mathrm{A}]^{1 / 2}[\\mathrm{~B}]^{1 / 2}\n\\end{aligned}$$

\n

The reaction is initiated by taking $$1 \\mathrm{~M}$$ concentration of $$\\mathrm{A}$$ and $$\\mathrm{B}$$ each. If the rate constant $$(\\mathrm{k})$$ is $$4.6 \\times 10^{-2} \\mathrm{~s}^{-1}$$, then the time taken for $$\\mathrm{A}$$ to become $$0.1 \\mathrm{~M}$$ is _________ sec.\n(nearest integer)

", "options": [], "answer": "50", "solution": "**Answer:** 50\n\n

$$\\begin{aligned}\n& A+B \\rightarrow C \\\\\n& \\frac{-d[A]}{d t}=k[A]^{1 / 2}[B]^{1 / 2}\n\\end{aligned}$$

\n

Since, $$[A]=[B]$$

\n

$$\\begin{aligned}\n& \\Rightarrow \\quad \\frac{-d[A]}{d t}=k[A] \\\\\n& \\Rightarrow \\quad k t=\\ln \\frac{[A]_0}{[A]} \\\\\n& \\Rightarrow \\quad t=\\frac{1}{4.6 \\times 10^{-2}} \\times \\ln \\left(\\frac{1}{0.1}\\right) \\\\\n& =\\frac{2.303}{4.6} \\times 100 \\approx 50\n\\end{aligned}$$

", "topic": "Algebra", "subtopic": "Algebraic Expressions" }, { "id": 737, "subject": "Chemistry", "question": "

Consider the two different first order reactions given below

\n

$$\\begin{aligned}\n& \\mathrm{A}+\\mathrm{B} \\rightarrow \\mathrm{C} \\text { (Reaction 1) } \\\\\n& \\mathrm{P} \\rightarrow \\mathrm{Q} \\text { (Reaction 2) }\n\\end{aligned}$$

\n

The ratio of the half life of Reaction 1 : Reaction 2 is $$5: 2$$ If $$t_1$$ and $$t_2$$ represent the time taken to complete $$2 / 3^{\\text {rd }}$$ and $$4 / 5^{\\text {th }}$$ of Reaction 1 and Reaction 2 , respectively, then the value of the ratio $$t_1: t_2$$ is _________ $$\\times 10^{-1}$$ (nearest integer).\n[Given : $$\\log _{10}(3)=0.477$$ and $$\\log _{10}(5)=0.699$$]

", "options": [], "answer": "17", "solution": "**Answer:** 17\n\n

$$\\mathrm{A+B \\rightarrow C}$$ Reaction .... (1)

\n

$$\\mathrm{P} \\rightarrow \\mathrm{Q} \\quad$$ Reaction .... (2)

\n

$$\\frac{\\left(t_{\\frac{1}{2}}\\right)_1}{\\left(t_{\\frac{1}{2}}\\right)_2}=\\frac{k}{k_1}= \\frac{5}{2}$$

\n

$$\\begin{aligned}\n& \\frac{\\frac{t_2}{3}}{t_{\\frac{4}{5}}^5}=\\frac{k_2}{k_1} \\frac{\\log 3}{\\log 5} \\\\\n& =\\left(\\frac{5}{2}\\right) \\frac{0.477}{0.699}=1.7 \\\\\n& =17 \\times 10^{-1} \\\\\n& =x=17\n\\end{aligned}$$

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 738, "subject": "Chemistry", "question": "

Time required for $$99.9 \\%$$ completion of a first order reaction is _________ times the time required for completion of $$90 \\%$$ reaction.(nearest integer)

", "options": [], "answer": "3", "solution": "**Answer:** 3\n\n

To determine the time required for a certain level of completion ($x\\%$) of a first-order reaction, we can use the formula that relates the time $ t $, the rate constant $ k $, and the concentration of the reactant. The formula for a first-order reaction, when expressed in terms of the initial concentration $[A]_0$ and the concentration at time $t$, $[A]_t$, is given by the integrated rate law for first-order reactions:

\n\n

$$\\ln\\left(\\frac{[A]_0}{[A]_t}\\right) = kt$$

\n\n

For a given percentage of completion, we use the fact that $[A]_t = [A]_0(1-\\frac{x}{100})$, where $x$ is the percentage completion. Thus, the equation becomes:

\n\n

$$\\ln\\left(\\frac{[A]_0}{[A]_0(1-\\frac{x}{100})}\\right) = kt$$

\n\n

Simplifying, we get:

\n\n

$$\\ln\\left(\\frac{1}{1-\\frac{x}{100}}\\right) = kt$$

\n\n

Let's calculate the time required for $99.9\\%$ and $90\\%$:

\n\n\n\n

$$\\ln\\left(\\frac{1}{1-\\frac{99.9}{100}}\\right) = kt_{99.9}$$

\n\n

$$\\ln\\left(\\frac{1}{0.001}\\right) = kt_{99.9}$$

\n\n

$$\\ln(1000) = kt_{99.9}$$

\n\n\n\n

$$\\ln\\left(\\frac{1}{1-\\frac{90}{100}}\\right) = kt_{90}$$

\n\n

$$\\ln\\left(\\frac{1}{0.1}\\right) = kt_{90}$$

\n\n

$$\\ln(10) = kt_{90}$$

\n\n

Since $k$ is a constant for a given reaction at a constant temperature, we can compare the times $t_{99.9}$ and $t_{90}$ directly by comparing the logarithmic values:

\n\n

$$\\frac{t_{99.9}}{t_{90}} = \\frac{\\ln(1000)}{\\ln(10)}$$

\n\n

Using the property of logarithms, we know that $\\ln(1000) = 3\\ln(10)$ because $1000 = 10^3$, so:

\n\n

$$\\frac{t_{99.9}}{t_{90}} = \\frac{3\\ln(10)}{\\ln(10)} = 3$$

\n\n

Therefore, the time required for $99.9\\%$ completion of a first-order reaction is $3$ times the time required for completion of $90\\%$ of the reaction. The nearest integer to this value is $3$.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 739, "subject": "Chemistry", "question": "$$\\beta$$ - particle is emitted in radioactivity by", "options": [ { "text": "conversion of proton to neutron" }, { "text": "from outermost orbit" }, { "text": "conversion of neutron to proton" }, { "text": "$$\\beta$$ -particle is not emitted" } ], "answer": "conversion of neutron to proton", "solution": "**Answer:** conversion of neutron to proton\n\n$${}_0{n^1} \\to {}_{ + 1}{p^1} + {}_{ - 1}{e^{}}$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 740, "subject": "Chemistry", "question": "The radionucleide $${}_{90}^{234}Th$$ undergoes two successive $$\\beta$$ -decays followed by one $$\\alpha$$-decay. The atomic number\nand the mass number respectively of the resulting radionucleide are", "options": [ { "text": "94 and 230" }, { "text": "90 and 230" }, { "text": "92 and 230" }, { "text": "92 and 234" } ], "answer": "90 and 230", "solution": "**Answer:** 90 and 230\n\n$${}_{90}^{234}Th\\buildrel { - \\beta } \\over\n \\longrightarrow {}_{91}^{234}X\\buildrel {\\, - \\beta } \\over\n \\longrightarrow {}_{92}^{234}Th\\buildrel { - \\alpha } \\over\n \\longrightarrow {}_{90}^{230}Th$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 741, "subject": "Chemistry", "question": "The half-life of a radioactive isotope is three hours. If the initial mass of the isotope were 256 g, the mass of\nit remaining undecayed after 18 hours would be", "options": [ { "text": "8.0 g" }, { "text": "12.0 g" }, { "text": "16.0 g" }, { "text": "4.0 g" } ], "answer": "4.0 g", "solution": "**Answer:** 4.0 g\n\n$${t_{1/2}} = 3\\,$$ hrs. $$T=18$$ hours \n

as $$\\,\\,\\,\\,T = n \\times {t_{1/2}}$$ \n

$$\\therefore$$ $$\\,\\,\\,\\,n = {{18} \\over 3} = 6$$\n

Initial mass $$\\left( {{C_0}} \\right) = 256\\,g$$ \n

$$\\therefore$$ $$\\,\\,\\,\\,{C_n} = {{{C_0}} \\over {{2^n}}} = {{256} \\over {{{\\left( 2 \\right)}^6}}} = {{256} \\over {64}} = 4g.$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 742, "subject": "Chemistry", "question": "Consider the following nuclear reactions
\n$${}_{92}^{238}M \\to {}_Y^XN + 2{}_2^4He$$
\n$${}_Y^XN \\to {}_B^AL + 2{\\beta ^ + }$$
\nThe number of neutrons in the element L is ", "options": [ { "text": "140" }, { "text": "144" }, { "text": "142" }, { "text": "146" } ], "answer": "144", "solution": "**Answer:** 144\n\nWe know that, $\\alpha$ - decay formula is -

\n$$\n{ }_Z^A X \\stackrel{-\\alpha}{\\rightarrow}{ }_{Z-2}^{A-4} Y\n$$

\nAnd the formula for $\\beta$ is -

\n$$\n{ }_Z^A X \\stackrel{-\\beta}{\\rightarrow}{ }_{Z-1}^A Y\n$$

\nNow for the given above question we see that, -

\n$$\n\\begin{aligned}\n&{ }_{92}^{238} M \\stackrel{-2 \\alpha}{\\rightarrow}{ }_{92-4}^{238-8} N \\\\\\\\\n&{ }_Y^X X \\stackrel{-2 \\beta^{+}}{\\rightarrow}{ }_{86}^{230} L\n\\end{aligned}\n$$

\nTherefore we can now calculate the number of neutrons in $L$, -

\n$$\nL=230-86\n$$

\n$$\nL=144\n$$

\nSo, as we can see the no. of neutrons present in $\\mathrm{L}$ is 144 .", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 743, "subject": "Chemistry", "question": "The half – life of a radioisotope is four hours. If the initial mass of the isotope was 200 g, the\nmass remaining after 24 hours undecayed is", "options": [ { "text": "1.042 g " }, { "text": "4.167 g" }, { "text": "3.125 g" }, { "text": "2.084 g" } ], "answer": "3.125 g", "solution": "**Answer:** 3.125 g\n\n$${N_t} = {N_0}{\\left( {{1 \\over 2}} \\right)^n}$$ where $$n$$ is number of half life periods. \n

$$n = {{Total\\,\\,time} \\over {half\\,\\,life}} = {{24} \\over 4} = 6$$\n

$$\\therefore$$ $$\\,\\,\\,\\,\\,{N_t} = 200{\\left( {{1 \\over 2}} \\right)^6} = 3.125g.$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 744, "subject": "Chemistry", "question": "Hydrogen bomb is based on the principle of ", "options": [ { "text": "Nuclear fission " }, { "text": "Natural radioactivity " }, { "text": "Nuclear fusion" }, { "text": "Artificial radioactivity" } ], "answer": "Nuclear fusion", "solution": "**Answer:** Nuclear fusion\n\n

Option C

\n

Nuclear fusion

\n

Explanation :

\n

A hydrogen bomb, also known as a thermonuclear bomb, uses the principle of nuclear fusion. In a fusion reaction, two lighter atomic nuclei combine to form a heavier nucleus, and a substantial amount of energy is released in the process. In the case of a hydrogen bomb, isotopes of hydrogen (such as deuterium and tritium) fuse together to form helium, releasing a large amount of energy.

\n

It's worth noting that a hydrogen bomb usually involves a two-stage process. The first stage is a fission bomb (like those used in Hiroshima and Nagasaki) that creates the conditions necessary for the fusion reaction in the second stage. Despite this, the majority of the energy in a hydrogen bomb comes from fusion, which is why it is categorized as a fusion weapon.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 745, "subject": "Chemistry", "question": "The photon of hard gamma radiation knocks a proton out of $${}_{12}^{24}Mg$$ nucleus to form ", "options": [ { "text": "the isotope of parent nucleus" }, { "text": "the isobar of parent nucleus " }, { "text": "the nuclide $${}_{11}^{23}Na$$" }, { "text": "the isobar of $${}_{11}^{23}Na$$" } ], "answer": "the nuclide $${}_{11}^{23}Na$$", "solution": "**Answer:** the nuclide $${}_{11}^{23}Na$$\n\n$${}_{12}M{g^{24}}\\buildrel \\, \\over\n \\longrightarrow {}_{11}N{a^{23}} + {}_1{H^1}.$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 746, "subject": "Chemistry", "question": "In the transformation of $${}_{92}^{238}U$$ to $${}_{92}^{234}U$$, if one emission is an α-particle, what should be the other\nemission(s)? ", "options": [ { "text": "Two $$\\beta^-$$" }, { "text": "Two $$\\beta^-$$ and one $$\\beta^+$$" }, { "text": "One $$\\beta^-$$ and one $$\\gamma$$" }, { "text": "One $$\\beta^+$$ and One $$\\beta^-$$" } ], "answer": "Two $$\\beta^-$$", "solution": "**Answer:** Two $$\\beta^-$$\n\n$${}_{92}^{238}U\\buildrel { - \\alpha } \\over\n \\longrightarrow {}_{90}^{234}Th\\buildrel {\\, - 2\\beta } \\over\n \\longrightarrow {}_{92}^{234}U$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 747, "subject": "Chemistry", "question": "A radioactive element gets spilled over the floor of a room. Its half-life period is 30 days. If the initial\nactivity is ten times the permissible value, after how many days will it be safe to enter the room? ", "options": [ { "text": "1000 days" }, { "text": "300 days " }, { "text": "10 days" }, { "text": "100 days" } ], "answer": "100 days", "solution": "**Answer:** 100 days\n\nSuppose activity of safe working $$=A$$\n

Given $${A_0} = 10A$$\n

$$\\lambda = {{0.693} \\over {{t_{1/2}}}} = {{0.693} \\over {30}}$$\n

$${t_{{\\raise0.5ex\\hbox{$\\scriptstyle 1$}\n\\kern-0.1em/\\kern-0.15em\n\\lower0.25ex\\hbox{$\\scriptstyle 2$}}}} = {{2.303} \\over \\lambda }\\log {{{A_0}} \\over A}$$\n

$$ = {{2.303} \\over {0.693/30}}\\log {{10A} \\over A}$$\n

$$ = {{2.303 \\times 30} \\over {0.693}} \\times \\log 10$$\n

$$ = 100$$ days.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 748, "subject": "Chemistry", "question": "Which of the following nuclear reactions will generate an isotope?", "options": [ { "text": "neutron particle emission" }, { "text": "positron emission " }, { "text": "$$\\alpha$$-particle emission" }, { "text": "$$\\beta$$-particle emission" } ], "answer": "neutron particle emission", "solution": "**Answer:** neutron particle emission\n\nNOTE : Isotopes are atoms of same element having same atomic number but different atomic masses. Neutron has atomic number $$0$$ and atomic mass $$1.$$ So loss of neutron will generate isotope. $$e.g.,$$\n
$$${}_{92}{U^{238}} + {}_0{n^1} \\to {}_{92}{U^{239}}$$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 749, "subject": "Chemistry", "question": "During the nuclear explosion, one of the products is 90Sr with half life of 6.93 years. If 1 $$\\mu $$ g of 90Sr was absorbed in the bones of newly born baby in placed of Ca, how much time, in years, is required to reduce much time, in year, is required to reduce it by 90% if it not lost metabolically.", "options": [], "answer": "23to23.03", "solution": "**Answer:** 23to23.03\n\nAll nuclear decays follow first order kinetics\n

t = $${1 \\over k}\\ln {{\\left[ {{A_0}} \\right]} \\over {\\left[ A \\right]}}$$\n

= $${{\\left( {{t_{1/2}}} \\right)} \\over {0.693}} \\times 2.303{\\log _{10}}10$$\n

= 10 × 2.303 × 1\n

= 23.03 years\n

Shortcut Method :\n

t90% = $${{10} \\over 3} \\times {t_{50\\% }}$$\n

= $${{10} \\over 3} \\times 6.93$$ = 23.1", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 750, "subject": "Chemistry", "question": "Isotope(s) of hydrogen which emits low energy $$\\beta$$$$-$$ particles with t1/2 value > 12 years is/are", "options": [ { "text": "Protium" }, { "text": "Tritium" }, { "text": "Deuterium" }, { "text": "Deuterium and Tritium" } ], "answer": "Tritium", "solution": "**Answer:** Tritium\n\n$$_1^1$$H and $$_1^2$$H are stable while $$_1^3$$H is radioactive.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 751, "subject": "Chemistry", "question": "The ratio of $\\frac{{ }^{14} \\mathrm{C}}{{ }^{12} \\mathrm{C}}$ in a piece of wood is $\\frac{1}{8}$ part that of atmosphere. If half life of ${ }^{14} \\mathrm{C}$ is 5730 years, the age of wood sample is ________ years.", "options": [], "answer": "17190", "solution": "**Answer:** 17190\n\n

The given problem involves calculating the age of a wood sample using the carbon-14 dating method. Carbon-14 ($^{14}C$) is a radioactive isotope of carbon that decays over time, and its ratio compared to carbon-12 ($^{12}C$) can be used to date organic materials. The ratio of $\\frac{^{14}C}{^{12}C}$ in the wood is $\\frac{1}{8}$th of that in the atmosphere, suggesting the $^{14}C$ has decayed. The half-life of $^{14}C$ is given as 5730 years, which is the time for half the $^{14}C$ to decay.

To solve for the age of the wood, we use the formula for radioactive decay:

$$N = N_0 e^{-\\lambda t},$$

where $N$ is the remaining amount of $^{14}C$, $N_0$ is the original amount of $^{14}C$, $\\lambda$ is the decay constant, and $t$ is the time (age of the wood).

Since the ratio $\\frac{N}{N_0} = \\frac{1}{8}$, we infer that $N = \\frac{N_0}{8}$. Substituting into the decay equation and solving for $t$, first we find the decay constant $\\lambda$:

$$\\lambda = \\frac{0.693}{t_{1/2}} = \\frac{0.693}{5730}.$$

Substitute $\\lambda$ and rearrange to solve for $t$:

$$t = \\frac{\\ln(8)}{\\lambda} = \\frac{\\ln(8)}{0.693} \\times 5730 = 3 \\times 5730 = 17190 \\text{ years}.$$

This calculation shows that the wood sample is 17190 years old.

", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 752, "subject": "Chemistry", "question": "

The half-life of radioisotope bromine - 82 is 36 hours. The fraction which remains after one day is ________ $$\\times 10^{-2}$$.

\n

(Given antilog $$0.2006=1.587$$)

", "options": [], "answer": "63", "solution": "**Answer:** 63\n\n

Half life of bromine $$-82=36$$ hours

\n

$$\\begin{aligned}\n& \\mathrm{t}_{1 / 2}=\\frac{0.693}{\\mathrm{~K}} \\\\\n& \\mathrm{~K}=\\frac{0.693}{36}=0.01925 \\mathrm{~hr}^{-1} \\\\\n& 1^{\\text {st }} \\text { order rxn kinetic equation } \\\\\n& \\mathrm{t}=\\frac{2.303}{\\mathrm{~K}} \\log \\frac{\\mathrm{a}}{\\mathrm{a}-\\mathrm{x}} \\\\\n& \\log \\frac{\\mathrm{a}}{\\mathrm{a}-\\mathrm{x}}=\\frac{\\mathrm{t} \\times \\mathrm{K}}{2.303}(\\mathrm{t}=1 \\text { day }=24 \\mathrm{~hr}) \\\\\n& \\log \\frac{\\mathrm{a}}{\\mathrm{a}-\\mathrm{x}}=\\frac{24 \\mathrm{hr} \\times 0.01925 \\mathrm{hr}^{-1}}{2.303} \\\\\n& \\log \\frac{\\mathrm{a}}{\\mathrm{a}-\\mathrm{x}}=0.2006 \\\\\n& \\frac{\\mathrm{a}}{\\mathrm{a}-\\mathrm{x}}=\\mathrm{anti} \\log (0.2006) \\\\\n& \\frac{\\mathrm{a}}{\\mathrm{a}-\\mathrm{x}}=1.587 \\\\\n& \\text { If }=1 \\\\\n& \\frac{1}{1-\\mathrm{x}}=1.587 \\Rightarrow 1-\\mathrm{x}=0.6301=\\text { Fraction remain } \\\\\n& \\mathrm{after} \\text { one day }\n\\end{aligned}$$

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 753, "subject": "Chemistry", "question": "For the reaction A + 2B $$\\to$$ C, rate is given by R = [A] [B]2 then the order of the reaction is", "options": [ { "text": "3" }, { "text": "6" }, { "text": "5" }, { "text": "7" } ], "answer": "3", "solution": "**Answer:** 3\n\nNOTE : Order is the sum of the power of the concentrations terms in rate law expression. \n

Hence the order of reaction is $$=1+2=3$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 754, "subject": "Chemistry", "question": "The rate law for a reaction between the substances A and B is given by Rate = k[A]n [B]m On doubling the concentration of A and halving the concentration of B, the ratio of the new rate to the earlier rate of the reaction will be as", "options": [ { "text": "(m + n)" }, { "text": "(n - m) " }, { "text": "2( n - m)" }, { "text": "$${1 \\over {{2^{(m + n)}}}}$$ " } ], "answer": "2( n - m)", "solution": "**Answer:** 2( n - m)\n\n$$Rat{e_1} = k{\\left[ A \\right]^n}{\\left[ B \\right]^m};$$ \n

$$Rat{e_2} = k{\\left[ {2A} \\right]^n}{\\left[ {{\\raise0.5ex\\hbox{$\\scriptstyle 1$}\n\\kern-0.1em/\\kern-0.15em\n\\lower0.25ex\\hbox{$\\scriptstyle 2$}}B} \\right]^m}$$\n

$$\\therefore$$ $$\\,\\,\\,\\,{{Rat{e_2}} \\over {Rat{e_1}}} = {{k{{\\left[ {2A} \\right]}^n}{{\\left[ {{\\raise0.5ex\\hbox{$\\scriptstyle 1$}\n\\kern-0.1em/\\kern-0.15em\n\\lower0.25ex\\hbox{$\\scriptstyle 2$}}B} \\right]}^m}} \\over {k{{\\left[ A \\right]}^n}{{\\left[ B \\right]}^m}}}$$\n

$$ = {\\left[ 2 \\right]^n}{\\left[ {{\\raise0.5ex\\hbox{$\\scriptstyle 1$}\n\\kern-0.1em/\\kern-0.15em\n\\lower0.25ex\\hbox{$\\scriptstyle 2$}}} \\right]^m} = {2^n}{.2^{ - m}} = {2^{n - m}}$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 755, "subject": "Chemistry", "question": "The rate equation for the reaction 2A + B $$\\to$$ C is found to be: rate k[A][B]. The correct\nstatement in relation to this reaction is that the ", "options": [ { "text": "unit of K must be s-1" }, { "text": "values of k is independent of the initial concentration of A and B" }, { "text": "rate of formation of C is twice the rate of disappearance of A " }, { "text": "t1/2 is a constant" } ], "answer": "values of k is independent of the initial concentration of A and B", "solution": "**Answer:** values of k is independent of the initial concentration of A and B\n\nThe velocity constant depends on temperature only. It is independent of concentration of reactants. ", "topic": "Algebra", "subtopic": "Equations and Inequalities" }, { "id": 756, "subject": "Chemistry", "question": "A reaction was found to be second order with respect to the concentration of carbon monoxide. If the\nconcentration of carbon monoxide is doubled, with everything else kept the same, the rate of reaction\nwill", "options": [ { "text": "remain unchanged " }, { "text": "triple" }, { "text": "increase by a factor of 4" }, { "text": "double" } ], "answer": "increase by a factor of 4", "solution": "**Answer:** increase by a factor of 4\n\nSince the reaction is $$2$$nd order w.r.t CO. Thus, rate law is given as. \n

$$r = k{\\left[ {CO} \\right]^2}$$ \n

Let initial concentration of $$CO$$ is a i.e. $$\\left[ {CO} \\right] = a$$\n

$$\\therefore$$ $$\\,\\,\\,\\,{r_1} = k{\\left( a \\right)^2} = k{a^2}$$ \n

when concentration becomes doubled, i.e. $$\\left[ {CO} \\right] = 2a$$\n

$$\\therefore$$ $$\\,\\,\\,\\,\\,{r_2} = k{\\left( {2a} \\right)^2} = 4k{a^2}$$ \n

$$\\therefore$$ $$\\,\\,\\,\\,\\,{r_2} = 4{r_1}$$ \n

So, the rate of reaction becomes $$4$$ times.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 757, "subject": "Chemistry", "question": "

Consider the reaction :
\n
Cl2(aq) + H2S(aq) → S(s) + 2H+ (aq) + 2Cl– (aq)
\n
The rate equation for this reaction is rate = k [Cl2] [H2S]
\n
Which of these mechanisms is/are consistent with this rate equation?


\n

(A) Cl2 + H2S $$\\to$$ H+ + Cl– + Cl+ + HS– (slow)
\n
Cl+ + HS– $$\\to$$ H+ + Cl– + S (fast)

\n

(B) H2S $$ \\Leftrightarrow $$ H+ + HS– (fast equilibrium)
\n
Cl2 + HS– $$\\to$$ 2Cl– + H+ + S (slow)

", "options": [ { "text": "B only" }, { "text": "Both A and B" }, { "text": "Neither A nor B" }, { "text": "A only" } ], "answer": "A only", "solution": "**Answer:** A only\n\nSince the slow step is the rate determining step hence-\n

if we consider option $$(1)$$ we find\n

Rate $$ = k\\left[ {C{l_2}} \\right]\\left[ {{H_2}S} \\right]$$\n

Now if we consider option $$(2)$$ we find \n

Rate $$ = k\\left[ {C{l_2}} \\right]\\left[ {H{S^ - }} \\right]\\,\\,\\,\\,\\,...\\left( 1 \\right)$$\n

From equation $$(i)$$\n

$$k = {{\\left[ {{H^ + }} \\right]\\left[ {H{S^ - }} \\right]} \\over {{H_2}S}}$$ $$\\,\\,\\,\\,\\,$$ or\n

$$\\left[ {H{S^ - }} \\right] = {{k\\left[ {{H_2}S} \\right]} \\over {{H^ + }}}$$\n

Substituting this value in equation $$(1)$$ we find \n

Rate $$ = k\\left[ {C{l_2}} \\right]K{{\\left[ {{H_2}S} \\right]} \\over {{H^ + }}}$$\n

$$ = k'{{\\left[ {C{l_2}} \\right]\\left[ {{H_2}S} \\right]} \\over {\\left[ {{H^ + }} \\right]}}$$\n

hence only, mechanism $$(1)$$ is consistent with the given rate equation.", "topic": "Algebra", "subtopic": "Equations and Inequalities" }, { "id": 758, "subject": "Chemistry", "question": "For the non – stoichiometre reaction 2A + B $$\\to$$ C + D, the following kinetic data were obtained in three\nseparate experiments, all at 298 K.\n
\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
Initial Concentration (A)Initial Concentration (B)Initial rate of formation of C (mol L-1 s-1)
0.1 M0.1 M1.2 x 10-3
0.1 M0.2 M1.2 x 10-3
0.2 M0.1 M2.4 x 10-3
\n
\nThe rate law for the formation of C is:", "options": [ { "text": "$${{dc} \\over {dt}} = k[A]{[B]^2}$$ " }, { "text": "$${{dc} \\over {dt}} = k[A]$$ " }, { "text": "$${{dc} \\over {dt}} = k[A]{[B]}$$" }, { "text": "$${{dc} \\over {dt}} = k[A^2]{[B]}$$ " } ], "answer": "$${{dc} \\over {dt}} = k[A]$$ ", "solution": "**Answer:** $${{dc} \\over {dt}} = k[A]$$ \n\nLet rate of reaction $$ = {{d\\left[ C \\right]} \\over t} = k{\\left[ A \\right]^x}{\\left[ B \\right]^y}$$\n

Now from the given data\n

$$1.2 \\times {10^{ - 3}} = k{\\left[ {0.1} \\right]^x}{\\left[ {0.1} \\right]^y}\\,\\,\\,\\,\\,...\\left( i \\right)$$\n

$$1.2 \\times {10^{ - 3}} = k{\\left[ {0.1} \\right]^x}{\\left[ {0.2} \\right]^y}\\,\\,\\,\\,\\,...\\left( {ii} \\right)$$\n

$$2.4 \\times {10^{ - 3}} = k{\\left[ {0.2} \\right]^x}{\\left[ {0.1} \\right]^y}\\,\\,\\,\\,...\\left( {iii} \\right)$$\n

Dividing equation $$(i)$$ by $$(ii)$$\n

$$ \\Rightarrow \\,\\,\\,\\,{{1.2 \\times {{10}^{ - 3}}} \\over {1.2 \\times {{10}^{ - 3}}}} = {{k{{\\left[ {0.1} \\right]}^x}{{\\left[ {0.1} \\right]}^y}} \\over {k{{\\left[ {0.1} \\right]}^x}{{\\left[ {0.2} \\right]}^y}}}$$\n

We find, $$y=0$$ \n

Now dividing equation $$(i)$$ by $$(iii)$$ \n

$$ \\Rightarrow \\,\\,\\,\\,{{1.2 \\times {{10}^{ - 3}}} \\over {2.4 \\times {{10}^{ - 3}}}} = {{k{{\\left[ {0.1} \\right]}^x}{{\\left[ {0.1} \\right]}^y}} \\over {k{{\\left[ {0.2} \\right]}^x}{{\\left[ {0.1} \\right]}^y}}}$$\n

We find, $$x=1$$ \n

Hence $${{d\\left[ C \\right]} \\over {dt}} = k{\\left[ A \\right]^1}{\\left[ B \\right]^0}$$ ", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 759, "subject": "Chemistry", "question": "The reaction of ozone with oxygen atoms in the presence of chlorine atoms can occur\nby a two step process shown below :\n

O3(g) + Cl$${^ \\bullet }$$ (g) $$ \\to $$ O2(g) + ClO$${^ \\bullet }$$ (g) . . . . . .(i)\n

ki = 5.2 × 109 L mol−1 s−1\n

ClO$${^ \\bullet }$$(g) + O$${^ \\bullet }$$(g) $$ \\to $$ O2(g) + Cl$${^ \\bullet }$$ (g) . . . . . . (ii)\n

kii = 2.6 × 1010 L mol−1 s−1\n

The closest rate constant for the overall reaction O3(g) + O$${^ \\bullet }$$ (g) $$ \\to $$ 2 O2(g) is : ", "options": [ { "text": "5.2 × 109 L mol−1 s−1" }, { "text": "2.6 × 1010 L mol−1 s−1" }, { "text": "3.1 × 1010 L mol−1 s−1" }, { "text": "1.4 × 1020 L mol−1 s−1" } ], "answer": "1.4 × 1020 L mol−1 s−1", "solution": "**Answer:** 1.4 × 1020 L mol−1 s−1\n\n

We have

\n

$${O_3}(g) + C{l^ \\bullet }(g) \\to {O_2}(g) + Cl{O^ \\bullet }(g)$$ ...... (1)

\n

$$Cl{O^ \\bullet }(g) + {O^ \\bullet }(g) \\to {O_2}(g) + C{l^ \\bullet }(g)$$ ...... (2)

\n

On adding Eq. (1) and Eq. (2), we get the required equation for overall reaction

\n

$${O_3}(g) + {O^ \\bullet }(g) \\to 2{O_2}(g)$$

\n

Therefore,

\n

$${k_{overall}} = {k_1} \\times {k_2}$$

\n

$$ = 5.2 \\times {10^9} \\times 2.6 \\times {10^{10}}$$

\n

$$ = 1.4 \\times {10^{20}}$$ L mol$$-$$1 s$$-$$1

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 760, "subject": "Chemistry", "question": "The rate law for the reaction below is given by the expression k [A] [B]\n

A + B $$ \\to $$ Product\n

If the concentration of B is increased from 0.1 to 0.3 mole, keeping the value of A at 0.1 mole, the rate constant will be :\n", "options": [ { "text": "k" }, { "text": "k/3" }, { "text": "3k" }, { "text": "9k" } ], "answer": "k", "solution": "**Answer:** k\n\nRate constant only depends on temperature only and it is independed of concentration of reactants.", "topic": "Algebra", "subtopic": "Algebraic Expressions" }, { "id": 761, "subject": "Chemistry", "question": "For the reaction 2A + B $$ \\to $$ C, the values of initial rate at diffrent reactant concentrations are\ngiven in the table below. The rate law for the reaction is :\n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
[A] (mol L-1) [B] (mol L-1)Initial Rate (mol L-1s-1)
0.050.050.045
0.100.050.090
0.200.100.72
", "options": [ { "text": "Rate = k[A][B]2" }, { "text": "Rate = k[A]2[B]2" }, { "text": "Rate = k[A]2[B]" }, { "text": "Rate = k[A][B]" } ], "answer": "Rate = k[A][B]2", "solution": "**Answer:** Rate = k[A][B]2\n\nRate law for the reaction,\n

2A + B $$ \\to $$ C\n

Rate law (R) = k[A]x[B]y\n

From experiment 1 :\n

R1 = 0.045 = k[0.05]x [0.05]y ............(i)\n

From experiment 2 :\n

R2 = 0.090 = k[0.1]x [0.05]y ...............(ii)\n

From experiment 3 :\n

R3 = 0.72 = k[0.2]x [0.1]y ...................(iii)\n

Divide equation (ii) by equation (i),\n

$${{{R_2}} \\over {{R_1}}} = {{0.09} \\over {0.045}} = {{k{{\\left[ {0.1} \\right]}^x}{{\\left[ {0.05} \\right]}^y}} \\over {k{{\\left[ {0.05} \\right]}^x}{{\\left[ {0.05} \\right]}^y}}}$$\n

$$ \\Rightarrow $$ 2 = (2)x\n

$$ \\Rightarrow $$ x = 1\n

Divide equation (iii) by equation (ii),\n

$${{{R_3}} \\over {{R_2}}} = {{0.72} \\over {0.09}} = {{k{{\\left[ {0.2} \\right]}^x}{{\\left[ {0.1} \\right]}^y}} \\over {k{{\\left[ {0.1} \\right]}^x}{{\\left[ {0.05} \\right]}^y}}}$$\n

$$ \\Rightarrow $$ 8 = 2x 2y\n

$$ \\Rightarrow $$ 8 = 21 2y [as x = 1]\n

$$ \\Rightarrow $$ 2y = 4\n

$$ \\Rightarrow $$ y = 2\n

$$ \\therefore $$ Rate law (R) = k[A]1[B]2\n\n\n", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 762, "subject": "Chemistry", "question": "The following results were obtained during kinetic studies of the reaction ; \n

2A + B $$ \\to $$ Products\n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
Experiment[A] (in mol L$$-$$1)[b] (in mol L$$-$$1)Initial Rate of reaction
(In mol L$$-$$1 min$$-$$1)
I0.100.206.93 G 10$$-$$3
II0.100.256.93 G 10$$-$$3
III0.200.301.386 G 10$$-$$2
\n

The time (in minutes) required to consume half of A is : ", "options": [ { "text": "5" }, { "text": "10" }, { "text": "1" }, { "text": "100" } ], "answer": "10", "solution": "**Answer:** 10\n\n

To determine the time required to consume half of A, we must first determine the rate law for the reaction based on the provided kinetic data.

\n\n

From the data, we can write the general rate law for the reaction as:

\n\n

$$ \\text{Rate} = k [A]^m [B]^n $$

\n\n

We can now use the experimental data to find the values of the exponents $ m $ and $ n $, as well as the rate constant $ k $.

\n\n

From Experiment I and II, we can see that the initial concentration of A remains constant at 0.10 M while the concentration of B changes.

\n\n

$$ \\begin{align*} \\text{Experiment I:} \\quad \\text{Rate} = k [0.10]^m [0.20]^n &= 6.93 \\times 10^{-3} \\, \\text{mol L}^{-1} \\text{min}^{-1} \\\\ \\text{Experiment II:} \\quad \\text{Rate} = k [0.10]^m [0.25]^n &= 6.93 \\times 10^{-3} \\, \\text{mol L}^{-1} \\text{min}^{-1} \\end{align*} $$

\n\n

By comparing the rates from Experiment I and II:

\n\n

$$ \\frac{[0.10]^m [0.25]^n}{[0.10]^m [0.20]^n} = \\frac{6.93 \\times 10^{-3}}{6.93 \\times 10^{-3}} $$

\n\n

$$ \\frac{[0.25]^n}{[0.20]^n} = 1 \\implies \\left(\\frac{0.25}{0.20}\\right)^n = 1 \\implies n = 0 $$

\n\n

Thus, the reaction is zero order with respect to B.

\n\n

Next, we compare Experiment I and III to determine the order with respect to A:

\n\n

$$ \\begin{align*} \\text{Experiment I:} \\quad \\text{Rate} = k [0.10]^m [0.20]^0 &= 6.93 \\times 10^{-3} \\, \\text{mol L}^{-1} \\text{min}^{-1} \\\\ \\text{Experiment III:} \\quad \\text{Rate} = k [0.20]^m [0.30]^0 &= 1.386 \\times 10^{-2} \\, \\text{mol L}^{-1} \\text{min}^{-1} \\end{align*} $$

\n\n

$$ \\frac{k [0.20]^m}{k [0.10]^m} = \\frac{1.386 \\times 10^{-2}}{6.93 \\times 10^{-3}} $$

\n\n

$$ \\left( \\frac{0.20}{0.10} \\right)^m = 2 \\implies (2)^m = 2 \\implies m = 1 $$

\n\n

Therefore, the reaction is first-order with respect to A.

\n\n

Now, we can write the rate law as:

\n\n

$$ \\text{Rate} = k [A]^1 $$

\n\n

Let's calculate the rate constant $ k $ using data from any experiment (say Experiment I):

\n\n

$$ 6.93 \\times 10^{-3} = k \\times 0.10 $$

\n\n

$$ k = \\frac{6.93 \\times 10^{-3}}{0.10} = 6.93 \\times 10^{-2} \\, \\text{min}^{-1} $$

\n\n

To determine the half-life (time required to consume half of A) for a first-order reaction, we use the formula:

\n\n

$$ t_{1/2} = \\frac{0.693}{k} $$

\n\n

Substituting the value of $ k $:

\n\n

$$ t_{1/2} = \\frac{0.693}{6.93 \\times 10^{-2}} = 10 \\, \\text{minutes} $$

\n\n

Therefore, the time required to consume half of A is:

\n\n

Option B: 10

", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 763, "subject": "Chemistry", "question": "For the reaction
2H2(g) + 2NO(g) $$ \\to $$ N2(g) + 2H2O(g)
the observed rate expression is, rate = Kf[NO]2[H2]. The rate expression for the reverse reaction is :", "options": [ { "text": "Kb[N2][H2O]" }, { "text": "Kb[N2][H2O]2/[H2]" }, { "text": "Kb[N2][H2O]2/[NO] " }, { "text": "Kb[N2][H2O]2" } ], "answer": "Kb[N2][H2O]2/[H2]", "solution": "**Answer:** Kb[N2][H2O]2/[H2]\n\nKeq = $${{{k_f}} \\over {{k_b}}}$$ = $${{\\left[ {{N_2}} \\right]{{\\left[ {{H_2}O} \\right]}^2}} \\over {{{\\left[ {{H_2}} \\right]}^2}{{\\left[ {NO} \\right]}^2}}}$$\n\n

$${k_f}\\left[ {{H_2}} \\right]{\\left[ {NO} \\right]^2} = {k_b}{{\\left[ {{N_2}} \\right]{{\\left[ {{H_2}O} \\right]}^2}} \\over {\\left[ {{H_2}} \\right]}}$$\n

At equilibrium rf = rb\n

Given rf = $${k_f}\\left[ {{H_2}} \\right]{\\left[ {NO} \\right]^2}$$\n

$$ \\therefore $$ rb = $${k_b}{{\\left[ {{N_2}} \\right]{{\\left[ {{H_2}O} \\right]}^2}} \\over {\\left[ {{H_2}} \\right]}}$$", "topic": "Algebra", "subtopic": "Algebraic Expressions" }, { "id": 764, "subject": "Chemistry", "question": "For a reaction of order n, the unit of the rate constant is :", "options": [ { "text": "mol1$$-$$n L1$$-$$n s" }, { "text": "mol1$$-$$n L2n s$$-$$1" }, { "text": "mol1$$-$$n Ln$$-$$1 s$$-$$1" }, { "text": "mol1$$-$$n L1$$-$$n s$$-$$1" } ], "answer": "mol1$$-$$n Ln$$-$$1 s$$-$$1", "solution": "**Answer:** mol1$$-$$n Ln$$-$$1 s$$-$$1\n\nRate = k[A]n

comparing units

$${{(mol/l)} \\over {\\sec }} = k{\\left( {{{mol} \\over l}} \\right)^n}$$

$$ \\Rightarrow k = mo{l^{(l - n)}}{l^{(n - 1)}}{s^{ - 1}}$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 765, "subject": "Chemistry", "question": "

A $$\\to$$ B

\n

The rate constants of the above reaction at 200 K and 300 K are 0.03 min$$^{-1}$$ and 0.05 min$$^{-1}$$ respectively. The activation energy for the reaction is ___________ J (Nearest integer)

\n

(Given : $$\\mathrm{ln10=2.3}$$

\n

$$\\mathrm{R=8.3~J~K^{-1}~mol^{-1}}$$\n

$$\\mathrm{\\log5=0.70}$$

\n

$$\\mathrm{\\log3=0.48}$$

\n

$$\\mathrm{\\log2=0.30}$$)

", "options": [], "answer": "2520", "solution": "**Answer:** 2520\n\n$$\n\\begin{aligned}\n& \\log \\frac{\\mathrm{k}_2}{\\mathrm{k}_1}=\\frac{\\mathrm{E}_{\\mathrm{a}}}{2.3 \\times 8.3}\\left(\\frac{1}{200}-\\frac{1}{300}\\right) \\\\\\\\\n& \\log \\frac{0.05}{0.03}=\\frac{E_a}{2.3 \\times 8.3}\\left(\\frac{1}{600}\\right) \\\\\\\\\n& \\begin{aligned}\n(0.70-0.48)=\\frac{E_a}{2.3 \\times 8.3} \\times \\frac{1}{600}\n\\end{aligned} \\\\\\\\\n& \\begin{aligned}\n& \\Rightarrow 0.22 =\\frac{E_a}{2.3 \\times 8.3} \\times \\frac{1}{600} \\\\\\\\\n& \\Rightarrow {E_a} = 2.3 \\times 8.3 \\times 600 \\times 0.22 \\\\\\\\\n& =2519.88 \\\\\\\\\n& \\approx 2520 \\mathrm{~J}\n\\end{aligned}\n\\end{aligned}\n$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 766, "subject": "Chemistry", "question": "

The number of correct statement/s from the following is __________

\n

A. Larger the activation energy, smaller is the value of the rate constant.

\n

B. The higher is the activation energy, higher is the value of the temperature coefficient.

\n

C. At lower temperatures, increase in temperature causes more change in the value of k than at higher temperature

\n

D. A plot of $$\\mathrm{\\ln k}$$ vs $$\\frac{1}{T}$$ is a straight line with slope equal to $$-\\frac{E_a}{R}$$

", "options": [], "answer": "3", "solution": "**Answer:** 3\n\n(A) $k=A e^{-\\frac{E_{a}}{R T}}\\left(E_{a} \\uparrow k \\downarrow\\right)$\n

\n(B) $\\ln k=\\ln A-\\frac{E_{a}}{R T}$\n

\n$$\n\\frac{1}{\\mathrm{k}} \\cdot \\frac{\\mathrm{dk}}{\\mathrm{dT}}=\\frac{+\\mathrm{E}_{\\mathrm{a}}}{\\mathrm{RT}^{2}}\n$$\n

\n$\\mathrm{E}_{\\mathrm{a}} \\uparrow$ temp. coefficient $\\uparrow$

\n(C)
\n\"JEE

Option (C ) is wrong. $\\Delta$k may be greater or lesser\ndepending on temperature.

\n(D) $\\ln k=\\ln A-\\frac{E_{a}}{R T}$\n

\nSlope of $\\ln k$ vs $\\frac{1}{T}$ is $\\left(\\frac{-E_{a}}{R}\\right)$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 767, "subject": "Chemistry", "question": "

For a reaction $$A \\xrightarrow{\\mathrm{K}_1} \\mathrm{~B} \\xrightarrow{\\mathrm{K}_2} \\mathrm{C}$$\nIf the rate of formation of B is set to be zero then the concentration of B is given by :

", "options": [ { "text": "$$(\\mathrm{K}_1-\\mathrm{K}_2)[\\mathrm{A}]$$\n" }, { "text": "$$(\\mathrm{K}_1+\\mathrm{K}_2)[\\mathrm{A}]$$\n" }, { "text": "$$\\mathrm{K}_1 \\mathrm{K}_2[\\mathrm{~A}]$$\n" }, { "text": "$$(\\mathrm{K}_1 / \\mathrm{K}_2)[\\mathrm{A}]$$" } ], "answer": "$$(\\mathrm{K}_1 / \\mathrm{K}_2)[\\mathrm{A}]$$", "solution": "**Answer:** $$(\\mathrm{K}_1 / \\mathrm{K}_2)[\\mathrm{A}]$$\n\n

To determine the concentration of $ \\mathrm{B} $ when the rate of formation of $ \\mathrm{B} $ is set to zero, let's consider the reaction sequence:

\n\n

$$ \\mathrm{A} \\xrightarrow{\\mathrm{K}_1} \\mathrm{B} \\xrightarrow{\\mathrm{K}_2} \\mathrm{C} $$

\n\n

The rate of formation of $ \\mathrm{B} $ from $ \\mathrm{A} $ is given by:

\n\n

$$ \\text{Rate of formation of } \\mathrm{B} = \\mathrm{K}_1[\\mathrm{A}] $$

\n\n

The rate of consumption of $ \\mathrm{B} $ to form $ \\mathrm{C} $ is given by:

\n\n

$$ \\text{Rate of consumption of } \\mathrm{B} = \\mathrm{K}_2[\\mathrm{B}] $$

\n\n

At steady state, where the rate of formation of $ \\mathrm{B} $ is set to zero, the rate of formation of $ \\mathrm{B} $ is equal to the rate of its consumption. Therefore, we have:

\n\n

$$ \\mathrm{K}_1[\\mathrm{A}] = \\mathrm{K}_2[\\mathrm{B}] $$

\n\n

Solving for $ [\\mathrm{B}] $, we get:

\n\n

$$ [\\mathrm{B}] = \\frac{\\mathrm{K}_1}{\\mathrm{K}_2} [\\mathrm{A}] $$

\n\n

Therefore, the correct concentration of $ \\mathrm{B} $ is:

\n\n

Option D: $$ \\left(\\frac{\\mathrm{K}_1}{\\mathrm{K}_2}\\right)[\\mathrm{A}] $$

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 768, "subject": "Chemistry", "question": "

Consider the following single step reaction in gas phase at constant temperature.

\n

$$2 \\mathrm{~A}_{(\\mathrm{g})}+\\mathrm{B}_{(\\mathrm{g})} \\rightarrow \\mathrm{C}_{(\\mathrm{g})}$$

\n

The initial rate of the reaction is recorded as $$\\mathrm{r}_1$$ when the reaction starts with $$1.5 \\mathrm{~atm}$$ pressure of $$\\mathrm{A}$$ and $$0.7 \\mathrm{~atm}$$ pressure of B. After some time, the rate $$r_2$$ is recorded when the pressure of C becomes $$0.5 \\mathrm{~atm}$$. The ratio $$\\mathrm{r}_1: \\mathrm{r}_2$$ is _________ $$\\times 10^{-1}$$. (Nearest integer)

", "options": [], "answer": "315", "solution": "**Answer:** 315\n\n

$$2 \\mathrm{~A}(\\mathrm{~g})+\\mathrm{B}(\\mathrm{g}) \\rightarrow \\mathrm{C}(\\mathrm{g})$$

\n

As this is single step reaction,

\n

$$\\mathrm{r}=\\mathrm{k}_{\\mathrm{f}}[\\mathrm{A}]^2[\\mathrm{~B}]^2$$

\n

$$\\begin{array}{lllll} \n& 2 \\mathrm{~A}(\\mathrm{~g})+ & \\mathrm{B}(\\mathrm{g}) \\rightarrow & \\mathrm{C}(\\mathrm{g}) \\\\\n\\mathrm{t}=0 & 1.5 & 0.7 & 0 \\\\\n\\mathrm{t}=\\mathrm{t}^{\\prime} & 1.5-2 \\mathrm{x} \\quad & 0.7-\\mathrm{x} & \\mathrm{x}\n\\end{array}$$

\n

$$\\begin{aligned}\n& \\Rightarrow x=0.5 \\mathrm{~atm} \\\\\n& \\mathrm{r}_1=\\mathrm{k}_{\\mathrm{f}}(1.5)^2(0.7) \\\\\n& \\mathrm{r}_2=\\mathrm{k}_{\\mathrm{f}}(0.5)^2(0.2) \\\\\n& \\frac{r_1}{r_2}=\\frac{9 \\times 7}{2}=31.5 \\\\\n& =315 \\times 10^{-1}\n\\end{aligned}$$

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 769, "subject": "Chemistry", "question": "The differential rate law for the reaction H2 + I2 $$\\to$$ 2HI is", "options": [ { "text": "$$ - {{d\\left[ {{H_2}} \\right]} \\over {dt}}$$ = $$ - {{d\\left[ {{I_2}} \\right]} \\over {dt}}$$ = $$ - {{d\\left[ {{HI}} \\right]} \\over {dt}}$$" }, { "text": "$$ {{d\\left[ {{H_2}} \\right]} \\over {dt}}$$ = $$ {{d\\left[ {{I_2}} \\right]} \\over {dt}}$$ = $$ {{d\\left[ {{HI}} \\right]} \\over {dt}}$$" }, { "text": "$${1 \\over 2}{{d\\left[ {{H_2}} \\right]} \\over {dt}}$$ = $${1 \\over 2}{{d\\left[ {{I_2}} \\right]} \\over {dt}}$$ = $$ - {{d\\left[ {{HI}} \\right]} \\over {dt}}$$" }, { "text": "$$ - 2{{d\\left[ {{H_2}} \\right]} \\over {dt}}$$ = $$ - 2{{d\\left[ {{I_2}} \\right]} \\over {dt}}$$ = $${{d\\left[ {{HI}} \\right]} \\over {dt}}$$" } ], "answer": "$$ - 2{{d\\left[ {{H_2}} \\right]} \\over {dt}}$$ = $$ - 2{{d\\left[ {{I_2}} \\right]} \\over {dt}}$$ = $${{d\\left[ {{HI}} \\right]} \\over {dt}}$$", "solution": "**Answer:** $$ - 2{{d\\left[ {{H_2}} \\right]} \\over {dt}}$$ = $$ - 2{{d\\left[ {{I_2}} \\right]} \\over {dt}}$$ = $${{d\\left[ {{HI}} \\right]} \\over {dt}}$$\n\nrate of appearance of $$HI = {1 \\over 2}{{d\\left[ {HI} \\right]} \\over {dt}}$$ \n

rate of formation of $${H_2} = {{ - d\\left[ {{H_2}} \\right]} \\over {dt}}$$ \n

rate of formation of $${I_2} = {{ - d\\left[ {{I_2}} \\right]} \\over {dt}}$$\n

hence $${{ - d\\left[ {{H_2}} \\right]} \\over {dt}} = - {{ - d\\left[ {{I_2}} \\right]} \\over {dt}} = {1 \\over 2}{{d\\left[ {HI} \\right]} \\over {dt}}$$ \n

or $$\\,\\,\\,\\,$$ $$ - {{2d\\left[ {{H_2}} \\right]} \\over {dt}} = - {{2d\\left[ {{I_2}} \\right]} \\over {dt}} = {{d\\left[ {HI} \\right]} \\over {dt}}$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 770, "subject": "Chemistry", "question": "For a reaction $${1 \\over 2}A \\to 2B$$ rate of disappearance of ‘A’ is related to the rate of appearance of ‘B’ by the\nexpression ", "options": [ { "text": "$$ - {{d[A]} \\over {dt}}$$ = $${1 \\over 2}{{d[B]} \\over {dt}}$$" }, { "text": "$$ - {{d[A]} \\over {dt}}$$ = $${1 \\over 4}{{d[B]} \\over {dt}}$$" }, { "text": "$$ - {{d[A]} \\over {dt}}$$ = $${{d[B]} \\over {dt}}$$ " }, { "text": "$$ - {{d[A]} \\over {dt}}$$ = $$4{{d[B]} \\over {dt}}$$ " } ], "answer": "$$ - {{d[A]} \\over {dt}}$$ = $${1 \\over 4}{{d[B]} \\over {dt}}$$", "solution": "**Answer:** $$ - {{d[A]} \\over {dt}}$$ = $${1 \\over 4}{{d[B]} \\over {dt}}$$\n\nThe rates of reactions for the reaction\n

$${1 \\over 2}A \\to 2B$$\n

can be written either as \n

$$ - 2{d \\over {dt}}\\left[ A \\right]\\,\\,$$ with respect to $$'A'$$ \n

or $$\\,\\,\\,\\,\\,$$ $${1 \\over 2}{d \\over {dt}}\\left[ B \\right]\\,\\,\\,\\,$$ with respect to $$'B'$$ \n

From the above, we have\n

$$ - 2{d \\over {dt}}\\left[ A \\right] = {1 \\over 2}{d \\over {dt}}\\left[ B \\right]$$\n

or $$\\,\\,\\,\\,\\,$$ $$ - {d \\over {dt}}\\left[ A \\right] = {1 \\over 4}{d \\over {dt}}\\left[ B \\right]$$ \n

i.e., correct answer is (b)", "topic": "Algebra", "subtopic": "Algebraic Expressions" }, { "id": 771, "subject": "Chemistry", "question": "The rate of a chemical reaction doubles for every 10oC rise of temperature. If the temperature is raised\nby 50oC , the rate of the reaction increases by about :", "options": [ { "text": "24 times " }, { "text": "32 times" }, { "text": "64 times" }, { "text": "10 times " } ], "answer": "32 times", "solution": "**Answer:** 32 times\n\nSince for every $${10^ \\circ }C$$ rise in temperature rate doubles for $${50^ \\circ }C$$ rise in temp increase in reaction rate $$ = {2^5} = 32$$ times", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 772, "subject": "Chemistry", "question": "For a reaction scheme $$A\\buildrel {{k_1}} \\over\n \\longrightarrow B\\buildrel {{k_2}} \\over\n \\longrightarrow C$$,\n

if\nthe rate of formation of B is set to be zero then\nthe concentration of B is given by :", "options": [ { "text": "$${k_1}{k_2}[A]$$" }, { "text": "$$\\left( {{{{k_1}} \\over {{k_2}}}} \\right)[A]$$" }, { "text": "$$({k_1} + {k_2})[A]$$" }, { "text": "$$({k_1} - {k_2})[A]$$" } ], "answer": "$$\\left( {{{{k_1}} \\over {{k_2}}}} \\right)[A]$$", "solution": "**Answer:** $$\\left( {{{{k_1}} \\over {{k_2}}}} \\right)[A]$$\n\nGiven,\nthe rate of formation of B is set to be zero.\n

$$ \\therefore $$ $${{d\\left[ B \\right]} \\over {dt}} = 0$$\n

For this reaction,\n

$$A\\mathrel{\\mathop{\\kern0pt\\longrightarrow}\n\\limits_{{R_1}}^{{k_1}}} B$$\n

Rate of reaction for this reaction (R1) = $${k_1}\\left[ A \\right]$$\n

For this reaction,\n

$$B\\mathrel{\\mathop{\\kern0pt\\longrightarrow}\n\\limits_{{R_2}}^{{k_2}}} C$$\n

Rate of reaction for this reaction (R2) = $${k_2}\\left[ B \\right]$$\n

Net rate of formation of B = $${{d\\left[ B \\right]} \\over {dt}}$$ = R1 - R2\n

As $${{d\\left[ B \\right]} \\over {dt}} = 0$$\n

$$ \\therefore $$ R1 - R2 = 0\n

$$ \\Rightarrow $$ $${k_1}\\left[ A \\right]$$ - $${k_2}\\left[ B \\right]$$ = 0\n

$$ \\Rightarrow $$ $$\\left[ B \\right] = {{{k_1}} \\over {{k_2}}}\\left[ A \\right]$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 773, "subject": "Chemistry", "question": "In the following reaction; xA $$ \\to $$ yB\n
$${\\log _{10}}\\left[ { - {{d\\left[ A \\right]} \\over {dt}}} \\right] = {\\log _{10}}\\left[ {{{d\\left[ B \\right]} \\over {dt}}} \\right] + 0.3010$$\n
'A' and 'B' respectively can be :", "options": [ { "text": "n-Butane and Iso-butane" }, { "text": "C2H4 and C4H8" }, { "text": "C2H4 and C6H6" }, { "text": "N2O4 and NO2" } ], "answer": "C2H4 and C4H8", "solution": "**Answer:** C2H4 and C4H8\n\nxA $$ \\to $$ yB\n

$${1 \\over x}\\left\\{ { - {{d\\left[ A \\right]} \\over {dt}}} \\right\\} = {1 \\over y}\\left\\{ {{{d\\left[ B \\right]} \\over {dt}}} \\right\\}$$\n

$$ \\Rightarrow $$ $$ - {{d\\left[ A \\right]} \\over {dt}} = {x \\over y}\\left\\{ {{{d\\left[ B \\right]} \\over {dt}}} \\right\\}$$\n

Taking log both sides, we get\n

$$ \\Rightarrow $$ $${\\log _{10}}\\left[ { - {{d\\left[ A \\right]} \\over {dt}}} \\right] = {\\log _{10}}\\left[ {{{d\\left[ B \\right]} \\over {dt}}} \\right] + {\\log _{10}}\\left( {{x \\over y}} \\right)$$\n

Given that,\n

$${\\log _{10}}\\left[ { - {{d\\left[ A \\right]} \\over {dt}}} \\right] = {\\log _{10}}\\left[ {{{d\\left[ B \\right]} \\over {dt}}} \\right] + 0.3010$$\n

So, by comparing both of them we get\n

$${\\log _{10}}\\left( {{x \\over y}} \\right)$$ = 0.3010 = $${\\log _{10}}\\left( 2 \\right)$$\n

$$ \\therefore $$ $${{x \\over y}}$$ = 2\n

$$ \\Rightarrow $$ x = 2y\n

If x = 2 then y = 1\n

The reaction is of type 2A $$ \\to $$ B\n

So possible reaction is\n

2C2H4 $$ \\to $$ C4H8", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 774, "subject": "Chemistry", "question": "NO2 required for a reaction is produced by the decomposition of N2O5 in CCl4 as per the equation,\n
2N2O5(g) $$ \\to $$ 4NO2(g) + O2(g).\n
The initial concentration of N2O5 is 3.00 mol L–1\n and it is 2.75 mol L–1\n after 30 minutes. The rate of\nformation of NO2 is :", "options": [ { "text": "2.083 × 10–3\n mol L–1\n min–1" }, { "text": "8.333 × 10–3\n mol L–1\n min–1" }, { "text": "4.167 × 10–3\n mol L–1\n min–1" }, { "text": "1.667 × 10–2\n mol L–1\n min–1" } ], "answer": "1.667 × 10–2\n mol L–1\n min–1", "solution": "**Answer:** 1.667 × 10–2\n mol L–1\n min–1\n\nRate of disappearance of N2O5 = $$ - {{\\Delta \\left[ {{N_2}{O_5}} \\right]} \\over {\\Delta \\left[ t \\right]}}$$ = $$ - {{\\left[ {2.75 - 3} \\right]} \\over {30}}$$ = $$ + {1 \\over {120}}$$ mol L-1 min-1\n

Rate of formation of NO2 = $$ + {{\\Delta \\left[ {N{O_2}} \\right]} \\over {\\Delta \\left[ t \\right]}}$$\n

As $$ - {1 \\over 2}{{\\Delta \\left[ {{N_2}{O_5}} \\right]} \\over {\\Delta \\left[ t \\right]}} = + {1 \\over 4}{{\\Delta \\left[ {N{O_2}} \\right]} \\over {\\Delta \\left[ t \\right]}}$$\n
$$ \\Rightarrow $$ $${{\\Delta \\left[ {N{O_2}} \\right]} \\over {\\Delta \\left[ t \\right]}}$$ = $$ + {1 \\over {120}} \\times {4 \\over 2}$$ = 0.01667 = 1.667 × 10–2\n mol L–1\n min–1", "topic": "Algebra", "subtopic": "Equations and Inequalities" }, { "id": 775, "subject": "Chemistry", "question": "For the reaction
2A + 3B +\n$${3 \\over 2}$$C\n$$ \\to $$ 3P, which\nstatement is correct ?", "options": [ { "text": "$${{d{n_A}} \\over {dt}} = {{d{n_B}} \\over {dt}} = {{d{n_C}} \\over {dt}}$$" }, { "text": "$${{d{n_A}} \\over {dt}} = {2 \\over 3}{{d{n_B}} \\over {dt}} = {3 \\over 4}{{d{n_C}} \\over {dt}}$$" }, { "text": "$${{d{n_A}} \\over {dt}} = {3 \\over 2}{{d{n_B}} \\over {dt}} = {3 \\over 4}{{d{n_C}} \\over {dt}}$$" }, { "text": "$${{d{n_A}} \\over {dt}} = {2 \\over 3}{{d{n_B}} \\over {dt}} = {4 \\over 3}{{d{n_C}} \\over {dt}}$$" } ], "answer": "$${{d{n_A}} \\over {dt}} = {2 \\over 3}{{d{n_B}} \\over {dt}} = {4 \\over 3}{{d{n_C}} \\over {dt}}$$", "solution": "**Answer:** $${{d{n_A}} \\over {dt}} = {2 \\over 3}{{d{n_B}} \\over {dt}} = {4 \\over 3}{{d{n_C}} \\over {dt}}$$\n\n2A + 3B +\n$${3 \\over 2}$$C\n$$ \\to $$ 3P\n

rate = $$ - {1 \\over 2}{{d\\left[ A \\right]} \\over {dt}} = - {1 \\over 3}{{d\\left[ B \\right]} \\over {dt}} = - {2 \\over 3}{{d\\left[ C \\right]} \\over {dt}} = {1 \\over 3}{{d\\left[ P \\right]} \\over {dt}}$$\n

$$ \\Rightarrow $$ $${{d\\left[ A \\right]} \\over {dt}} = {2 \\over 3}{{d\\left[ B \\right]} \\over {dt}} = {4 \\over 3}{{d\\left[ C \\right]} \\over {dt}}$$$$ = - {2 \\over 3}{{d\\left[ P \\right]} \\over {dt}}$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 776, "subject": "Chemistry", "question": "The reaction 2A + B2 $$ \\to $$ 2AB is an elementary reaction.

For a certain quantity of reactants, if the volume of the reaction vessel is reduced by a factor of 3, the rate of the reaction increases by a factor of ____________. (Round off to the Nearest Integer).", "options": [], "answer": "27", "solution": "**Answer:** 27\n\nLet initial concentration of A & B is a & b

Rate = $$K{[A]^2}{[{B_2}]^1}$$

$${r_1} = K{\\left( {{a \\over V}} \\right)^2}{\\left( {{b \\over V}} \\right)^1}$$

$${r_2} = K{\\left( {{{3a} \\over V}} \\right)^2}{\\left( {{{3b} \\over V}} \\right)^1}$$

$$\\left( {{{{r_2}} \\over {{r_1}}}} \\right) = 27$$", "topic": "Number Theory", "subtopic": "Divisibility / Factorization" }, { "id": 777, "subject": "Chemistry", "question": "For a chemical reaction A $$\\to$$ B, it was found that concentration of B is increased by 0.2 mol L$$-$$ in 30 min. The average rate of the reaction is ____________ $$\\times$$ 10$$-$$1 mol L$$-$$1 h$$-$$1. (in nearest integer)", "options": [], "answer": "4", "solution": "**Answer:** 4\n\nA $$\\to$$ B

$$\\matrix{\n {t = 0} & 0 \\cr \n {t = 30\\,\\min } & {0.2M} \\cr \n\n } $$

Av. rate of reaction = $$ - {{\\Delta [A]} \\over {\\Delta t}} = {{\\Delta [B]} \\over {\\Delta t}} = {{(0.2 - 0)} \\over {{1 \\over 2}}}$$

$$ = 0.4 = 4 \\times {10^{ - 1}}$$ mol / L $$\\times$$ hr", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 778, "subject": "Chemistry", "question": "The reaction that occurs in a breath analyser, a device used to determine the alcohol level in a person's blood stream is

$$2{K_2}C{r_2}{O_7} + 8{H_2}S{O_4} + 3{C_2}{H_6}O \\to 2C{r_2}{(S{O_4})_3} + 3{C_2}{H_4}{O_2} + 2{K_2}S{O_4} + 11{H_2}O$$

If the rate of appearance of Cr2(SO4)3 is 2.67 mol min$$-$$1 at a particular time, the rate of disappearance of C2H6O at the same time is _____________ mol min$$-$$1. (Nearest integer)", "options": [], "answer": "4", "solution": "**Answer:** 4\n\n$$\\left( {{{Rate\\,of\\,disappearance\\,of\\,{C_2}{H_6}O} \\over 3}} \\right)$$

$$ = \\left( {{{Rate\\,of\\,disappearance\\,of\\,C{r_2}{{(S{O_4})}_3}} \\over 2}} \\right)$$

$$ \\Rightarrow \\left( {{{2.67\\,mol/\\min \\, \\times 3} \\over 2}} \\right) = $$ rate of disappearance of C2H6O.

$$\\Rightarrow$$ Rate of disappearance of C2H6O = 4.005 mol/min.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 779, "subject": "Chemistry", "question": "

For a given chemical reaction

\n

$$\\gamma$$1A + $$\\gamma$$2B $$\\to$$ $$\\gamma$$3C + $$\\gamma$$4D

\n

Concentration of C changes from 10 mmol dm$$-$$3 to 20 mmol dm$$-$$3 in 10 seconds. Rate of appearance of D is 1.5 times the rate of disappearance of B which is twice the rate of disappearance A. The rate of appearance of D has been experimentally determined to be 9 mmol dm$$-$$3 s$$-$$1. Therefore, the rate of reaction is _____________ mmol dm$$-$$3 s$$-$$1. (Nearest Integer)

", "options": [], "answer": "1", "solution": "**Answer:** 1\n\nRate $=\\frac{1}{r_{1}}\\left(\\frac{-d[A]}{d t}\\right)=\\frac{1}{r_{2}}\\left(\\frac{-d[B]}{d t}\\right)=\\frac{1}{r_{3}}\\left(\\frac{-d[C]}{d t}\\right)$\n

\n$$\n=\\frac{1}{r_{4}}\\left(\\frac{d[D]}{d t}\\right)\n$$\n

\n$\\frac{\\mathrm{d}[\\mathrm{D}]}{\\mathrm{dt}}=\\frac{\\mathrm{r}_{4}}{\\mathrm{r}_{2}}\\left(\\frac{-\\mathrm{d}[\\mathrm{B}]}{\\mathrm{dt}}\\right)$\n

\n$$\n\\frac{r_{4}}{r_{2}}=\\frac{3}{2}\n$$\n

\n$\\frac{-\\mathrm{d}[\\mathrm{B}]}{\\mathrm{dt}}=\\frac{\\mathrm{r}_{2}}{\\mathrm{r}_{1}}\\left(\\frac{-\\mathrm{d}[\\mathrm{A}]}{\\mathrm{dt}}\\right) \\Rightarrow \\frac{\\mathrm{r}_{2}}{\\mathrm{r}_{1}}=2$\n

\n$r_{2}=2 r_{1}$

\n$$\n\\begin{aligned}\n&\\mathrm{r}_{4}=1.5 \\mathrm{r}_{2}=3 \\mathrm{r}_{1} \\\\\\\\\n&\\frac{\\mathrm{d}[\\mathrm{C}]}{\\mathrm{dt}}=1 \\mathrm{~m} \\cdot \\mathrm{mol} ~\\mathrm{dm}^{-3} \\mathrm{sec}^{-1} \\\\\\\\\n&\\frac{\\mathrm{d}[\\mathrm{D}]}{\\mathrm{dt}}=9 \\mathrm{~m} \\cdot \\mathrm{mol}~ \\mathrm{dm}^{-3} \\mathrm{sec}^{-1} \\\\\\\\\n&\\frac{\\mathrm{d}[\\mathrm{D}]}{\\mathrm{dt}}=\\frac{\\mathrm{r}_{4}}{r_{3}} \\cdot \\frac{\\mathrm{d}[\\mathrm{C}]}{\\mathrm{dt}} \\Rightarrow \\frac{\\mathrm{r}_{4}}{\\mathrm{r}_{3}}=9 \\\\\\\\\n&\\mathrm{r}_{4}=9 \\mathrm{r}_{3}=3 \\mathrm{r}_{1} \\\\\\\\\n&\\Rightarrow \\mathrm{r}_{1}=3 \\mathrm{r}_{3} \\\\\\\\\n&\\begin{array}{rr}\n3 \\mathrm{r}_{3} \\mathrm{~A}+6 \\mathrm{r}_{3} \\mathrm{~B} \\rightarrow \\mathrm{r}_{3} \\mathrm{C}+9 \\mathrm{r}_{3} \\mathrm{D}\n\\end{array} \\\\\\\\\n&\\begin{aligned}\n\\therefore \\text { rate of reaction } &=\\frac{1}{9} \\times 9 \\mathrm{~m} \\cdot \\mathrm{mol}~ \\mathrm{dm}^{-3} \\mathrm{sec}^{-1} \\\\\\\\\n&=1 \\mathrm{~m} \\cdot \\mathrm{mol}~ \\mathrm{dm}^{-3} \\mathrm{sec}^{-1}\n\\end{aligned}\n\\end{aligned}\n$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 780, "subject": "Chemistry", "question": "

$$\\mathrm{KClO}_{3}+6 \\mathrm{FeSO}_{4}+3 \\mathrm{H}_{2} \\mathrm{SO}_{4} \\rightarrow \\mathrm{KCl}+3 \\mathrm{Fe}_{2}\\left(\\mathrm{SO}_{4}\\right)_{3}+3 \\mathrm{H}_{2} \\mathrm{O}$$

\n

The above reaction was studied at $$300 \\mathrm{~K}$$ by monitoring the concentration of $$\\mathrm{FeSO}_{4}$$ in which initial concentration was $$10 \\mathrm{M}$$ and after half an hour became 8.8 M. The rate of production of $$\\mathrm{Fe}_{2}\\left(\\mathrm{SO}_{4}\\right)_{3}$$ is _________ $$\\times 10^{-6} \\mathrm{~mol} \\mathrm{~L} \\mathrm{~s}^{-1}$$ (Nearest integer)

", "options": [], "answer": "333", "solution": "**Answer:** 333\n\nThe Rate of Reaction (ROR) for a reaction can be calculated based on the changes in concentrations of the reactants or the products over time. In the given reaction,\n\n

$$\\mathrm{KClO}_{3}+6 \\mathrm{FeSO}_{4}+3 \\mathrm{H}_{2} \\mathrm{SO}_{4} \\rightarrow \\mathrm{KCl}+3 \\mathrm{Fe}_{2}\\left(\\mathrm{SO}_{4}\\right)_{3}+3 \\mathrm{H}_{2} \\mathrm{O}$$\n\n

The rates of reaction can be written as :\n\n

$$\n\\begin{aligned}\n& \\mathrm{ROR}=-\\frac{\\Delta\\left[\\mathrm{KClO}_3\\right]}{\\Delta \\mathrm{t}}=\\frac{-1}{6} \\frac{\\Delta\\left[\\mathrm{FeSO}_4\\right]}{\\Delta \\mathrm{t}} \\\\\\\\\n&=\\frac{+1}{3} \\frac{\\Delta\\left[\\mathrm{Fe}_2\\left(\\mathrm{SO}_4\\right)_3\\right]}{\\Delta \\mathrm{t}} \\\\\n\\end{aligned}\n$$\n\n

From this, we can express the rate of formation of $\\mathrm{Fe}_2\\left(\\mathrm{SO}_4\\right)_3$ in terms of the change in concentration of $\\mathrm{FeSO}_4$ :\n\n

$$ \\frac{\\Delta\\left[\\mathrm{Fe}_2\\left(\\mathrm{SO}_4\\right)_3\\right]}{\\Delta \\mathrm{t}}=\\frac{1}{2} \\frac{-\\Delta\\left[\\mathrm{FeSO}_4\\right]}{\\Delta \\mathrm{t}} $$\n\n

We know that the initial concentration of $\\mathrm{FeSO}_4$ was 10 M and after 30 minutes (or 1800 seconds), it became 8.8 M. Substituting these values in, we get :\n\n

$$=\\frac{1}{2} \\frac{(10-8.8)}{30 \\times 60} = 0.333 \\times 10^{-3} $$\n\n

To express the rate in terms of $10^{-6} \\, \\mathrm{mol \\, L^{-1} \\, s^{-1}}$, we multiply the rate by $10^{3}$ :\n\n

$$ =333 \\times 10^{-6} \\, \\mathrm{mol \\, L^{-1} \\, s^{-1}}$$\n\n

Therefore, the rate of production of $\\mathrm{Fe}_{2}\\left(\\mathrm{SO}_{4}\\right)_{3}$ is $333 \\times 10^{-6} \\, \\mathrm{mol \\, L^{-1} \\, s^{-1}}$.", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 781, "subject": "Chemistry", "question": "

$$\\mathrm{NO}_2$$ required for a reaction is produced by decomposition of $$\\mathrm{N}_2 \\mathrm{O}_5$$ in $$\\mathrm{CCl}_4$$ as by equation

\n

$$2 \\mathrm{~N}_2 \\mathrm{O}_{5(\\mathrm{~g})} \\rightarrow 4 \\mathrm{NO}_{2(\\mathrm{~g})}+\\mathrm{O}_{2(\\mathrm{~g})}$$

\n

The initial concentration of $$\\mathrm{N}_2 \\mathrm{O}_5$$ is $$3 \\mathrm{~mol} \\mathrm{~L}^{-1}$$ and it is $$2.75 \\mathrm{~mol} \\mathrm{~L}^{-1}$$ after 30 minutes.

\n

The rate of formation of $$\\mathrm{NO}_2$$ is $$\\mathrm{x} \\times 10^{-3} \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{~min}^{-1}$$, value of $$\\mathrm{x}$$ is _________. (nearest integer)

", "options": [], "answer": "17", "solution": "**Answer:** 17\n\n

Rate of reaction (ROR)

\n

$$\\begin{aligned}\n& =-\\frac{1}{2} \\frac{\\Delta\\left[\\mathrm{N}_2 \\mathrm{O}_5\\right]}{\\Delta \\mathrm{t}}=\\frac{1}{4} \\frac{\\left[\\mathrm{NO}_2\\right]}{\\Delta \\mathrm{t}}=\\frac{\\Delta\\left[\\mathrm{O}_2\\right]}{\\Delta \\mathrm{t}} \\\\\n& \\mathrm{ROR}=-\\frac{1}{2} \\frac{\\Delta\\left[\\mathrm{N}_2 \\mathrm{O}_5\\right]}{\\Delta \\mathrm{t}}=-\\frac{1}{2} \\frac{(2.75-3)}{30} \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{~min}^{-1} \\\\\n& \\mathrm{ROR}=-\\frac{1}{2} \\frac{(-0.25)}{30} \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{~min}^{-1} \\\\\n& \\text { ROR }=\\frac{1}{240} \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{~min}^{-1}\n\\end{aligned}$$

\n

Rate of formation of $$\\mathrm{NO}_2=\\frac{\\Delta\\left[\\mathrm{NO}_2\\right]}{\\Delta \\mathrm{t}}=4 \\times \\mathrm{ROR}$$

\n

$$=\\frac{4}{240}=16.66 \\times 10^{-3} \\mathrm{molL}^{-1} \\mathrm{~min}^{-1} \\simeq 17 \\times 10^{-3} \\text {. }$$

", "topic": "Algebra", "subtopic": "Equations and Inequalities" }, { "id": 782, "subject": "Chemistry", "question": "The artificial sweetener that has the highest sweetness value in comparison to cane\nsugar is :", "options": [ { "text": "Aspartane " }, { "text": "Saccharin " }, { "text": "Sucralose" }, { "text": "Alitame" } ], "answer": "Alitame", "solution": "**Answer:** Alitame\n\n

Alitame is 2000 times sweeter than sucrose.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 783, "subject": "Chemistry", "question": "The number of sp2 hybridised carbons present in \"Aspartame\" is", "options": [], "answer": "9", "solution": "**Answer:** 9\n\nStrcuture of Aspartame is:\n
\"JEE\n

Number of sp2\n hybridised carbon atom is = 9\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 784, "subject": "Chemistry", "question": "A chemist has 4 samples of artificial sweetener\nA, B, C and D. To identify these samples, he\nperformed certain experiments and noted the\nfollowing observations :

\n(i) A and D both form blue-violet colour with\nninhydrin.

\n(ii) Lassaigne extract of C gives positive\nAgNO3 test and negative Fe4[Fe(CN)6]3 test.

\n(iii) Lassaigne extract of B and D gives positive\nsodium nitroprusside test

\nBased on these observations which option is\ncorrect ?", "options": [ { "text": "A : Aspartame ; B : Alitame ;\nC : Saccharin ; D ; Sucralose" }, { "text": "A : Alitame ; B : Saccharin ;\nC : Aspartame ; D ; Sucralose" }, { "text": "A : Aspartame ; B : Saccharin ;\nC : Sucralose ; D ; Alitame" }, { "text": "A : Saccharin ; B : Alitame ;\nC : Sucralose ; D ; Aspartame" } ], "answer": "A : Aspartame ; B : Saccharin ;\nC : Sucralose ; D ; Alitame", "solution": "**Answer:** A : Aspartame ; B : Saccharin ;\nC : Sucralose ; D ; Alitame\n\n(i) A and D give positive test with ninhydrin because both have free carboxylic and amine group. Both aspartame and alitame contains\namino acids.\n

(ii) Lassaigne extract give +ve test with\nAgNO3. So Cl is present, –ve test with\nFe4[Fe(CN)6]3 means N is absent. So it\ncan't be Aspartame or Saccharine or\nAlitame, so C is sucralose.\n

(iii) Lassaigne solution of B and D given +ve\nsodium nitroprusside test, so it is a sulphur containing compound,\nSaccharine and Alitame also contains sulphur.\n\n

$$ \\therefore $$ A = Aspartame, B = Saccharin\n

C = Sucralose, D = Alitame.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 785, "subject": "Chemistry", "question": "Match List - I with List - II.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
List - I
Chemical Compound		List - II
Used as
(a)Sucralose(i)Synthetic detergent
(b)Glyceryl ester of stearic acid(ii)Artificial sweetener
(c)Sodium benzoate(iii)Antiseptic
(d)Bithionol(iv)Food preservative


Choose the correct match :", "options": [ { "text": "(a)-(i), (b)-(ii), (c)-(iv), (d)-(iii)" }, { "text": "(a)-(iii), (b)-(ii), (c)-(iv), (d)-(i)" }, { "text": "(a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)" }, { "text": "(a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)" } ], "answer": "(a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)", "solution": "**Answer:** (a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)\n\n(a) Sucralose $$ \\to $$ Artificial sweetener

\n(b) Glyceryl ester of stearic acid $$ \\to $$ Synthetic detergent

\n(c) Sodium benzoate $$ \\to $$ Food preservative

\n(d) Bithionol $$ \\to $$ Antiseptic", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 786, "subject": "Chemistry", "question": "

Which amongst the following is not a pesticide?

", "options": [ { "text": "DDT" }, { "text": "Organophosphates" }, { "text": "Dieldrin" }, { "text": "Sodium arsenite" } ], "answer": "Sodium arsenite", "solution": "**Answer:** Sodium arsenite\n\nNaAsO2 (Sodium arsenite) is an insecticide, antibacterial agent herbicide.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 787, "subject": "Chemistry", "question": "

Match List I with List II

\n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
List - IList -II
A.ZymaseI.Stomach
B.DiastaseII.Yeast
C.UreaseIII.Malt
D.PepsinIV.Soyabean

\n

Choose the correct answer from the options given below :

", "options": [ { "text": "A-II, B-III, C-I, D-IV" }, { "text": "A-II, B-III, C-IV, D-I" }, { "text": "A-III, B-II, C-IV, D-I" }, { "text": "A-III, B-II, C-I, D-IV" } ], "answer": "A-II, B-III, C-IV, D-I", "solution": "**Answer:** A-II, B-III, C-IV, D-I\n\nZymase naturally occurs in yeast.

\n Diastase is found in malt.

\n Urease is found in soyabean

\n Pepsin is found in stomach ", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 788, "subject": "Chemistry", "question": "

Which one of the following is an example of artificial sweetner?

", "options": [ { "text": "Bithional" }, { "text": "Alitame" }, { "text": "Salvarsan" }, { "text": "Lactose" } ], "answer": "Alitame", "solution": "**Answer:** Alitame\n\nAlitame is an aspartic acid-containing sweetener.\n

$$\n\\mathrm{C}_{14} \\mathrm{H}_{25} \\mathrm{O}_4 \\mathrm{N}_3 \\mathrm{S}-\\text { Alitame }\n$$\n

First generation artificial sweeteners - Saccharin, cyclamate and aspartame\n

Second generation artificial sweeteners - Sucralose, alitame and neotame", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 789, "subject": "Chemistry", "question": "

Which of the following artificial sweeteners has the highest sweetness value in comparison to cane sugar ?

", "options": [ { "text": "Alitame" }, { "text": "Sucralose" }, { "text": "Saccharin" }, { "text": "Aspartame" } ], "answer": "Alitame", "solution": "**Answer:** Alitame\n\nSweetness value order wrt cane sugar\n

Alitame > Sucralose > Saccharin > Aspartame\n

Sucralose = 600\n

Aspartame = 100\n

Saccharin = 550\n

Alitame = 2000", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 790, "subject": "Chemistry", "question": "

Match List I with List II:

\n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
LIST ILIST II
A.SaccharinI.High potency sweetener
B.AspartameII.First artificial sweetening agent
C.AlitameIII.Stable at cooking temperature
D.SucraloseIV.Unstable at cooking temperature

\n

Choose the correct answer from the options given below:

", "options": [ { "text": "A-II, B-III, C-IV, D-I" }, { "text": "A-II, B-IV, C-I, D-III" }, { "text": "A-II, B-IV, C-III, D-I" }, { "text": "A-IV, B-III, C-I, D-II" } ], "answer": "A-II, B-IV, C-I, D-III", "solution": "**Answer:** A-II, B-IV, C-I, D-III\n\nA. Saccharin is first popular artificial sweetening agent. It is 550 times as sweet as cane sugar. It is used as intake by diabetic person.

\nB. Aspartame is 100 times as sweet as cane sugar. It is used in cold foods and soft drinks because it is unstable at cooking temperature.

\nC. Alitame is 2000 times as sweet compared to cane sugar. It has excellent stability at high temperature. So, can be used in cooking and baking.

\nD. Sucralose is 600 times as sweet as cane sugar. It is heat stable and finds its use in baked goods.", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 791, "subject": "Chemistry", "question": "

Match List I with List II

\n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
LIST I
Vitamin		LIST II
Deficiency disease
A.Vitamin AI.Beri-Beri
B.ThiamineII.Cheilosis
C.Ascorbic acidIII.Xeropthalmia
D.RiboflavinIV.Scurvy

\n

Choose the correct answer from the options given below:

", "options": [ { "text": "A-III, B-I, C-IV, D-II" }, { "text": "A-IV, B-I, C-III, D-II" }, { "text": "A-IV, B-II, C-III, D-I" }, { "text": " A-III, B-II, C-IV, D-I" } ], "answer": "A-III, B-I, C-IV, D-II", "solution": "**Answer:** A-III, B-I, C-IV, D-II\n\n

We can match the vitamins with their deficiency diseases as follows:

\n\n

So the correct answer is:
\n
A-III, B-I, C-IV, D-II.

\n", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 792, "subject": "Chemistry", "question": "

Match List I with List II

\n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
LIST I
(Compound)		LIST II
(Uses)
A.IdoformI.Fire extinguisher
B.\tCarbon tetrachlorideII.Insecticide
C.CFCIII.Antiseptic
D.DDTIV.Refrigerants

\n

Choose the correct answer from the options given below:

", "options": [ { "text": "A-II, B-IV, C-I, D-III\n" }, { "text": "A-III, B-I, C-IV, D-II\n" }, { "text": "A-III, B-II, C-IV, D-I\n" }, { "text": "A-I, B-II, C-III, D-IV" } ], "answer": "A-III, B-I, C-IV, D-II\n", "solution": "**Answer:** A-III, B-I, C-IV, D-II\n\n\n

The matching for each compound with its respective use is based on their known applications:

\n\n\n\n

Thus, the correct option is Option B: A-III, B-I, C-IV, D-II.

", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 793, "subject": "Chemistry", "question": "Match List - I with List - II.\n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
List I (Compound)List II (Use)
(A) Carbon tetrachloride(I) Paint remover
(B) Methylene chloride(II) Refrigerators and air conditioners
(C) DDT(III) Fire extinguisher
(D) Freons(IV) Non Biodegradable insecticide
\n
\nChoose the correct answer from the options given below :", "options": [ { "text": "(A)-(II), (B)-(III), (C)-(I), (D)-(IV)" }, { "text": "(A)-(III), (B)-(I), (C)-(IV), (D)-(II)" }, { "text": "(A)-(I), (B)-(II), (C)-(III), (D)-(IV)" }, { "text": "(A)-(IV), (B)-(III), (C)-(II), (D)-(I)" } ], "answer": "(A)-(III), (B)-(I), (C)-(IV), (D)-(II)", "solution": "**Answer:** (A)-(III), (B)-(I), (C)-(IV), (D)-(II)\n\n

Each of the compounds listed in List I has distinct uses. Let's match them correctly with the uses mentioned in List II:

\n\n\n\n

Considering these associations, we can match the compounds with their correct uses:

\n\n\n\n

So, the correct match according to the given options will be:

\n\n

Option B:
\n(A)-(III), (B)-(I), (C)-(IV), (D)-(II)

", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 794, "subject": "Chemistry", "question": "Which one of the following methods is neither meant for the synthesis nor for\nseparation of amines?", "options": [ { "text": "Hinsberg method" }, { "text": "Hofmann method " }, { "text": "Wurtz reaction" }, { "text": "Curtius reaction" } ], "answer": "Wurtz reaction", "solution": "**Answer:** Wurtz reaction\n\nWurtz reaction is for the preparation of hydrocarbons from alkyl halide\n

$$RX + 2Na + XR\\buildrel \\, \\over\n \\longrightarrow R - R + 2NaX$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 795, "subject": "Chemistry", "question": "Which one of the following is the strongest base in aqueous solution? ", "options": [ { "text": "Trimethylamine " }, { "text": "Aniline" }, { "text": "Dimethylamine" }, { "text": "Methylamine" } ], "answer": "Dimethylamine", "solution": "**Answer:** Dimethylamine\n\nNOTE : Aromatic amines are less basic than aliphatic amines. Among aliphatic amines the order of basicity is $${2^ \\circ } > {1^ \\circ } > {3^ \\circ }.$$ The electron density is decreased in $${3^ \\circ }$$ amine due to crowing of alkyl group over $$N$$ atom which makes the approach and bonding by a proton relatively difficult. Therefore the basicity decreases. Further Phenyl group show $$ - {\\rm I}$$ effect, thus decreases the electron density on nitrogen atom and hence the basicity.\n

$$\\therefore$$ dimethylamine ($${2^ \\circ }$$ aliphatic amine) is strongest base among gtiven choices.\n

$$\\therefore$$ The correct order of basic strength is Dimethyl amine $$>$$ Methyl amine $$>$$ Trimethyl-amine $$>$$ Aniline.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 796, "subject": "Chemistry", "question": "Considering the basic strength of amines in aqueous solution, which one has the smallest pKb value?", "options": [ { "text": "(CH3)3N" }, { "text": "C6H5NH2" }, { "text": "(CH3)2NH" }, { "text": "CH3NH2" } ], "answer": "(CH3)2NH", "solution": "**Answer:** (CH3)2NH\n\nArylamines are less basic than alkyl amines and even ammonia. This is due to resonance. In aryl amines the lone pair of electrons on $$N$$ is partly shared with the ring and is thus less available for sharing with a portion.\n

In alkylamines, the electron releasing alkyl group increases the electron density on nitrogen atom and thus also increases the ability of amine for protonation. Hence more the no. of alkyl groups higher should be the basicity of amine. But a slight discrepancy occurs in case of trimethyl amines due to steric effect. Hence the correct order is \n$$${\\left( {CH{}_3} \\right)_2}NH > C{H_3}N{H_2} > \\left( {C{H_3}} \\right){}_3N > {C_6}{H_5}N{H_2}$$$", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 797, "subject": "Chemistry", "question": "On heating an aliphatic primary amine with chloroform and ethanolic potassium hydroxide, the organic\ncompound formed is:", "options": [ { "text": "an alkyl cyanide" }, { "text": "an alkyl isocyanide" }, { "text": "an alkanol" }, { "text": "an alkanediol" } ], "answer": "an alkyl isocyanide", "solution": "**Answer:** an alkyl isocyanide\n\n\"JEE ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 798, "subject": "Chemistry", "question": "In the Hofmann bromamide degradation reaction, the number of moles of NaOH and Br2 used per mole of\namine produced are:", "options": [ { "text": "Four moles of NaOH and two moles of Br2." }, { "text": "Two moles of NaOH and two moles of Br2" }, { "text": "Four moles of NaOH and one mole of Br2" }, { "text": "One mole of NaOH and one mole of Br2" } ], "answer": "Four moles of NaOH and one mole of Br2", "solution": "**Answer:** Four moles of NaOH and one mole of Br2\n\n$$4$$ moles of $$NaOH$$ and one mole of $$B{r_2}$$ is required during production of one mole of amine during Hoffmann's bromamide degradation reaction.\n

\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 799, "subject": "Chemistry", "question": "The test to distinguish primary, secondary and tertiary amines is :", "options": [ { "text": "Carbylamine reaction" }, { "text": "C6H5SO2Cl" }, { "text": "Sandmeyer’s reaction" }, { "text": "Mustard oil test" } ], "answer": "C6H5SO2Cl", "solution": "**Answer:** C6H5SO2Cl\n\nHinsberg’s reagent (C6H5SO2Cl) forms monoalkyl\nsulphonamide with 1o amines which is soluble in KOH. With 2o\namines it gives dialkyl sulphonamide which is insoluble in KOH\nand with 3o amines there is no reaction.\n

In mustard oil test, 1o amines on action of CS2\n and HgCl2\n give\nalkyl isothiocyanate having mustard oil smell. 2o amines react\nwith CS2\n but not with HgCl2\n while 3o amines give no reaction.\nHowever, this test is not able to distinguish 2o and 3o amines.\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 800, "subject": "Chemistry", "question": "Which of the following is not a correct method of the preparation of benzylamine from cyanobenzene ?", "options": [ { "text": "(i) SnCl2 + HCl(gas)          (ii) NaBH4" }, { "text": "H2/Ni\n" }, { "text": "(i) LiAlH4                           (ii) H3O+" }, { "text": "(i) HCl/H2O                        (ii) NaBH4" } ], "answer": "(i) HCl/H2O                        (ii) NaBH4", "solution": "**Answer:** (i) HCl/H2O                        (ii) NaBH4\n\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 801, "subject": "Chemistry", "question": "Ethylamine (C2H5NH2) can be obtained from N-ethylphatalimide on treatment with :", "options": [ { "text": "CaH2" }, { "text": "H2O" }, { "text": "NaBH4" }, { "text": "NH2NH2" } ], "answer": "NH2NH2", "solution": "**Answer:** NH2NH2\n\nN-ethyl phthalimide on treatment with\nNH2—NH2 gives ethylamine.\n

It is the final step of Galbriel phatalimide synthesis reaction.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 802, "subject": "Chemistry", "question": "Which of the following amines can be prepared by Gabriel phthalimide reaction ?", "options": [ { "text": "neo-pentylamine" }, { "text": "n-butylamine" }, { "text": "t-butylamine" }, { "text": "triethylamine" } ], "answer": "n-butylamine", "solution": "**Answer:** n-butylamine\n\nGabriel phthalimide synthesis is used to prepare 1o aliphatic amine.\n\"JEE\n\"JEE\n\"JEE\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 803, "subject": "Chemistry", "question": "A compound 'X' on treatment with Br2/NaOH, provided C3H9N, which gives positive carbylamine test. Compound 'X' is : ", "options": [ { "text": "CH3CH2CH2CONH2" }, { "text": "CH3CON(CH3)2" }, { "text": "CH3CH2COCH2NH" }, { "text": "CH3COCH2NHCH3" } ], "answer": "CH3CH2CH2CONH2", "solution": "**Answer:** CH3CH2CH2CONH2\n\nBr2/NaOH(Hoffman Hypobromide reagent) converts amide into primary amine having one carbon atom less, which gives\ncarbylamine test. In the Amide there should be
-CONH2 group to produce primary amine.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 804, "subject": "Chemistry", "question": "The most appropriate reagent for conversion of\nC2H5CN into CH3CH2CH2NH2 is", "options": [ { "text": "NaBH4" }, { "text": "CaH2" }, { "text": "Na(CN)BH3" }, { "text": "LiAlH4" } ], "answer": "LiAlH4", "solution": "**Answer:** LiAlH4\n\nCH3–CH2–C$$ \\equiv $$N $$\\buildrel {LiAl{H_4}} \\over\n \\longrightarrow $$ CH3–CH2–CH2–NH2", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 805, "subject": "Chemistry", "question": "Which of the following is least basic?", "options": [ { "text": "$${(C{H_3}CO)_2}\\mathop N\\limits^{..} H$$" }, { "text": "$$(C{H_3}CO)\\mathop N\\limits^{..} H{C_2}{H_5}$$" }, { "text": "$${({C_2}{H_5})_3}\\mathop N\\limits^{..} $$" }, { "text": "$${({C_2}{H_5})_2}\\mathop N\\limits^{..} H$$" } ], "answer": "$${(C{H_3}CO)_2}\\mathop N\\limits^{..} H$$", "solution": "**Answer:** $${(C{H_3}CO)_2}\\mathop N\\limits^{..} H$$\n\nBasic strength $$ \\propto $$ availability of lone pair.

\n\"JEE
\nIn this case lone pair of $$\\mathop N\\limits^{ \\bullet \\bullet } $$ is highly participating in\nresonance.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 806, "subject": "Chemistry", "question": "Ammonolysis of Alkyl halides followed by the treatment with NaOH solution can be used to prepare primary, secondary and tertiary amines. The purpose of NaOH in the reaction is :", "options": [ { "text": "to remove acidic impurities" }, { "text": "to remove basic impurities" }, { "text": "to activate NH3 used in the reaction" }, { "text": "to increase the reactivity of alkyl halide" } ], "answer": "to remove acidic impurities", "solution": "**Answer:** to remove acidic impurities\n\n\"JEE
\nDuring the reaction HX (acid) is formed

\nHence, we use NaOH to remove these acidic impurities", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 807, "subject": "Chemistry", "question": "Primary, secondary and tertiary amines can be separated using:", "options": [ { "text": "Chloroform and KOH" }, { "text": "Acetyl amide" }, { "text": "Benzene sulphonic acid" }, { "text": "para-Toluene sulphonyl chloride" } ], "answer": "para-Toluene sulphonyl chloride", "solution": "**Answer:** para-Toluene sulphonyl chloride\n\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 808, "subject": "Chemistry", "question": "In the reaction of hypobromite with amide, the carbonyl carbon is lost as :", "options": [ { "text": "CO2" }, { "text": "CO" }, { "text": "HCO$$_3^ - $$" }, { "text": "CO$$_3^{2 - }$$" } ], "answer": "CO$$_3^{2 - }$$", "solution": "**Answer:** CO$$_3^{2 - }$$\n\n\"JEE
\nCarbonyl carbon is lost as $$CO_3^{2 - }$$.\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 809, "subject": "Chemistry", "question": "Compound A is converted to B on reaction with CHCl3 and KOH. The compound B is toxic and can be decomposed by C. A, B and C respectively are :", "options": [ { "text": "primary amine, nitrile compound, conc. HCl" }, { "text": "secondary amine, isonitrile compound, conc. NaOH" }, { "text": "primary amine, isonitrile compound, conc. HCl" }, { "text": "secondary amine, nitrile compound, conc. NaOH" } ], "answer": "primary amine, isonitrile compound, conc. HCl", "solution": "**Answer:** primary amine, isonitrile compound, conc. HCl\n\n

Primary amine reacts in presence of chloroform gives isonitrile which is toxic in nature which further reacts in presence of HCl/H3O+ to give primary amine and acid.

\n

\"JEE

\n

Hence, A-primary amine, B-isonitrile compound and C-conc. HCl.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 810, "subject": "Chemistry", "question": "The number of nitrogen atoms in a semicarbazone molecule of acetone is ______________.", "options": [], "answer": "3", "solution": "**Answer:** 3\n\n

Three nitrogen atoms are present as per structure below

\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 811, "subject": "Chemistry", "question": "In gaseous triethyl amine the \"$$-$$C$$-$$N$$-$$C$$-$$\" bond angle is _________ degree.", "options": [], "answer": "108", "solution": "**Answer:** 108\n\nIn gaseous triethyl amine the \"$$-$$C$$-$$N$$-$$C$$-$$\" bond angle is 108 degree.", "topic": "Geometry", "subtopic": "Plane Geometry" }, { "id": 812, "subject": "Chemistry", "question": "Which of the following is not a correct statement for primary aliphatic amines?", "options": [ { "text": "The intermolecular association in primary amines is less than the intermolecular association in secondary amines." }, { "text": "Primary amines on treating with nitrous acid solution from corresponding alcohols except mythyl amine." }, { "text": "Primary amines are less basic than the secondary amines." }, { "text": "Primary amines can be prepared by the Gabriel phthalimide synthesis." } ], "answer": "The intermolecular association in primary amines is less than the intermolecular association in secondary amines.", "solution": "**Answer:** The intermolecular association in primary amines is less than the intermolecular association in secondary amines.\n\nThe intermolecular association is more prominent in case of primary amines as compared to secondary, due to the availability of two hydrogen atom.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 813, "subject": "Chemistry", "question": "

Given below are two statements :

\n

Statement I : In Hofmann degradation reaction, the migration of only an alkyl group takes place from carbonyl carbon of the amide to the nitrogen atom.

\n

Statement II : The group is migrated in Hofmann degradation reaction to electron deficient atom.

\n

In the light of the above statements, choose the most appropriate answer from the options given below :

", "options": [ { "text": "Both Statement I and Statement II are correct." }, { "text": "Both Statement I and Statement II are incorrect." }, { "text": "Statement I is correct but Statement II is incorrect." }, { "text": "Statement I is incorrect but Statement II is correct." } ], "answer": "Statement I is incorrect but Statement II is correct.", "solution": "**Answer:** Statement I is incorrect but Statement II is correct.\n\nHofmann bromamide degradation\n

\nIn this degradation, the migration of the alkyl/aryl group occurs to the electron-deficient nitrogen (nitrene).\n

\nStatement (I) is not absolutely correct as it mentions only the alkyl group, whereas migration of aryl groups may also occur depending on migratory aptitude.\n

\nStatement (II) is correct as the migration occurs to the electron-deficient atom.", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 814, "subject": "Chemistry", "question": "

Which statement is NOT correct for p-toluenesulphonyl chloride?

", "options": [ { "text": "It is known as Hinsberg's reagent." }, { "text": "It is used to distinguish primary and secondary amines." }, { "text": "On treatment with secondary amine, it leads to a product, that is soluble in alkali." }, { "text": "It doesn't react with tertiary amines." } ], "answer": "On treatment with secondary amine, it leads to a product, that is soluble in alkali.", "solution": "**Answer:** On treatment with secondary amine, it leads to a product, that is soluble in alkali.\n\n\"JEE\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 815, "subject": "Chemistry", "question": "

The conversion of propan-1-ol to n-butylamine involves the sequential addition of reagents. The correct sequential order of reagents is

", "options": [ { "text": "(i) SOCl2 (ii) KCN (iii) H2/Ni, Na(Hg)/C2H5OH" }, { "text": "(i) HCl (ii) H2/Ni, Na(Hg)/C2H5OH" }, { "text": "(i) SOCl2 (ii) KCN (iii) CH3NH2" }, { "text": "(i) HCl (ii) CH3NH2" } ], "answer": "(i) SOCl2 (ii) KCN (iii) H2/Ni, Na(Hg)/C2H5OH", "solution": "**Answer:** (i) SOCl2 (ii) KCN (iii) H2/Ni, Na(Hg)/C2H5OH\n\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 816, "subject": "Chemistry", "question": "

The number of sp3 hybridised carbons in an acyclic neutral compound with molecular formula C4H5N is ___________.

", "options": [], "answer": "1", "solution": "**Answer:** 1\n\n$\\mathrm{C}_{4} \\mathrm{H}_{5} \\mathrm{~N}$\n

\n$$\n\\begin{aligned}\n\\mathrm{DBE} & =(\\mathrm{C}+1)-\\left(\\frac{\\mathrm{H}+\\mathrm{X}-\\mathrm{N}}{2}\\right) \\\\\\\\\n& =4+1-\\left(\\frac{5-1}{2}\\right)=5-2=3\n\\end{aligned}\n$$\n

\n3 double bond equivalent are present in compound

\n\"JEE
\nOnly 1 $s p^{3}$ hybridised carbon is there\n

\n(Keeping compound as acyclic)", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 817, "subject": "Chemistry", "question": "

Match List I with List II.

\n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
List IList II
(A)Benzenesulphonyl chloride(I)Test for primary amines
(B)Hoffmann bromamide reaction(II)Anti Saytzeff
(C)Carbylamine reaction(III)Hinsberg reagent
(D)Hoffmann orientation(IV)Known reaction of Isocyanates.

\n

Choose the correct answer from the options given below:

", "options": [ { "text": "A-IV, B-III, C-II, D-I" }, { "text": "A-IV, B-II, C-I, D-III" }, { "text": "A-III, B-IV, C-I, D-II" }, { "text": "A-IV, B-III, C-I, D-II" } ], "answer": "A-III, B-IV, C-I, D-II", "solution": "**Answer:** A-III, B-IV, C-I, D-II\n\n(A) Benzene sulphonyl chloride is also known as Hinsberg reagent.\n

\n(B) Hoffmann bromamide reaction involves conversion of amide to amine having one $\\mathrm{C}$ atom less. This reaction involves isocyanate as intermediate.\n

\n(C) Carbylamine reaction is a test given by all primary amines.\n

\n(D) Hoffmann orientation refers to the addition of molecules to unsymmetrical alkenes according to anti Saytzeff's rule.\n

\nCorrect match is\n

\nA-III, B-IV, C-I, D-II", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 818, "subject": "Chemistry", "question": "

Reaction of propanamide with $$\\mathrm{Br_2/KOH(aq)}$$ produces :

", "options": [ { "text": "Propanenitrile" }, { "text": "Propylamine" }, { "text": "Ethylnitrile" }, { "text": "Ethylamine" } ], "answer": "Ethylamine", "solution": "**Answer:** Ethylamine\n\n

Propanamide $$\\mathrm{\\buildrel {B{r_2}/KOH} \\over\n \\longrightarrow}$$ Ethylamine

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 819, "subject": "Chemistry", "question": "

An amine $$(\\mathrm{X})$$ is prepared by ammonolysis of benzyl chloride. On adding p-toluenesulphonyl chloride to it the solution remains clear. Molar mass of the amine $$(\\mathrm{X})$$ formed is _________ $$\\mathrm{g} \\mathrm{mol}^{-1}$$.

\n

(Given molar mass in $$\\mathrm{gmol}^{-1} \\mathrm{C}: 12, \\mathrm{H}: 1, \\mathrm{O}: 16, \\mathrm{~N}: 14$$)

", "options": [], "answer": "287", "solution": "**Answer:** 287\n\n

\"JEE

\n

Molar mass of amine $$=287 \\mathrm{~g} / \\mathrm{mol}$$

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 820, "subject": "Chemistry", "question": "A. Phenyl methanamine

B. N,N-Dimethylaniline

C. N-Methyl aniline

D. Benzenamine

Choose the correct order of basic nature of the above amines.", "options": [ { "text": "D > C > B > A" }, { "text": "A > C > B > D" }, { "text": "D > B > C > A" }, { "text": "A > B > C > D" } ], "answer": "A > B > C > D", "solution": "**Answer:** A > B > C > D\n\n

In phenyl methanamine, the lone pair on nitrogen of -NH2\ngroup is localised and does not undergoes resonance.\n

\n\n\"JEE\n\n

(B), (C) and (D) are aromatic amines in which lone pair of electrons\nof N-atoms goes in resonance\n(+R effect) with the benzene ring. So, Lewis basicity or donation of\nlone of electrons of these amines will be decreased in comparison to\n(A).\n

\n\n\"JEE\n\n

+ I effects of two -CH3\ngroups increases electron density on N atom while lone pair of N atom take part in resonance with the benzene ring and decreases electron density on N atom. Both this effect try to compensate each other.\n

\n\n\"JEE\n\n

+ I effects of one -CH3\ngroups increases electron density on N atom while lone pair of N atom take part in resonance with the benzene ring and decreases electron density on N atom. Both this effect try to compensate each other.

\n\n\"JEE\n\n

It has no + I-effect on N-atom to increase electron density on the N atom which was decreased due to lone pair of N atom take part in resonance with the benzene ring.\nSo, A is purely aliphatic 1°-amine. B is aromatic.\n3°-amine with more aliphatic nature (for two -CH3\ngroups). C is\naromatic 2°-amine with less aliphatic nature (for one -CH3\ngroup).\n
D is purely aromatic 1°-amine.\n

Hence basicity order is A > B > C > D.\n

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 821, "subject": "Chemistry", "question": "Given below are two statements, one is labelled as Assertion (A) and other is labelled as Reason (R).

Assertion (A) : Gabriel phthalimide synthesis cannot be used to prepare aromatic primary amines.

Reason (R) : Aryl halides do not undergo nucleophilic substitution reaction.

In the light of the above statements, choose the correct answer from the options given below :", "options": [ { "text": "Both (A) and (R) true but (R) is not the correct explanation of (A)" }, { "text": "(A) is false but (R) is true." }, { "text": "Both (A) and (R) true and (R) is correct explanation of (A)." }, { "text": "(A) is true but (R) is false." } ], "answer": "Both (A) and (R) true and (R) is correct explanation of (A).", "solution": "**Answer:** Both (A) and (R) true and (R) is correct explanation of (A).\n\nGabriel phthalimide synthesis

\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 822, "subject": "Chemistry", "question": "Given below are two statements :

Statement I : Aniline is less basic than acetamide.

Statement II : In aniline, the lone pair of electrons on nitrogen atom is delocalized over benzene ring due to resonance and hence less available to a proton.

Choose the most appropriate option;", "options": [ { "text": "Statement I is true but statement II is false." }, { "text": "Statement I is false but statement II is true." }, { "text": "Both statement I and statement II are true." }, { "text": "Both statement I and statement II are false." } ], "answer": "Statement I is false but statement II is true.", "solution": "**Answer:** Statement I is false but statement II is true.\n\nExplanation : aniline is more basic than acetamide because in acetamide, lone pair of nitrogen is delocalized to more electronegative element oxygen.

In Aniline lone pair of nitrogen delocalized over benzene ring.", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 823, "subject": "Chemistry", "question": "$$R - CN\\mathrel{\\mathop{\\kern0pt\\longrightarrow}\n\\limits_{(ii){H_2}O}^{(i)DIBAL - H}} R - Y$$

Consider the above reaction and identify \"Y\"", "options": [ { "text": "$$-$$CH2NH2" }, { "text": "$$-$$CONH2" }, { "text": "$$-$$CHO" }, { "text": "$$-$$COOH" } ], "answer": "$$-$$CHO", "solution": "**Answer:** $$-$$CHO\n\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 824, "subject": "Chemistry", "question": "The total number of reagents from those given blow, that can convert nitrobenzene into aniline is ___________. (Integer answer)

I. Sn $$-$$ HCl

II. Sn $$-$$ NH4OH

III. Fe $$-$$ HCl

IV. Zn $$-$$ HCl

V. H2 $$-$$ Pd

VI. H2 $$-$$ Raney Nickel", "options": [], "answer": "5", "solution": "**Answer:** 5\n\n\"JEE

(i) Sn + HCl

(ii) Fe + HCl

(iii) Zn + HCl

(iv) H2 $$-$$ Pd

(v) H2 (Raney Ni)", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 825, "subject": "Chemistry", "question": "

In Friedel-Crafts alkylation of aniline, one gets

", "options": [ { "text": "alkylated product with ortho and para substitution." }, { "text": "secondary amine after acidic treatment." }, { "text": "an amide product." }, { "text": "positively charged nitrogen at benzene ring." } ], "answer": "positively charged nitrogen at benzene ring.", "solution": "**Answer:** positively charged nitrogen at benzene ring.\n\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 826, "subject": "Chemistry", "question": "

Decarboxylation of all six possible forms of diaminobenzoic acids C6H3(NH2)2COOH yields three products A, B and C. Three acids give a product 'A', two acids gives a product 'B' and one acid give a product 'C'. The melting point of product 'C' is

", "options": [ { "text": "63$$^\\circ$$C" }, { "text": "90$$^\\circ$$C" }, { "text": "104$$^\\circ$$C" }, { "text": "142$$^\\circ$$C" } ], "answer": "142$$^\\circ$$C", "solution": "**Answer:** 142$$^\\circ$$C\n\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 827, "subject": "Chemistry", "question": "

Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).

\n

Assertion (A) : Experimental reaction of $$\\mathrm{CH}_{3} \\mathrm{Cl}$$ with aniline and anhydrous $$\\mathrm{AlCl}_{3}$$ does not give $$o$$ and $$p$$-methylaniline.

\n

Reason (R): The $$-\\mathrm{NH}_{2}$$ group of aniline becomes deactivating because of salt formation with anhydrous $$\\mathrm{AlCl}_{3}$$ and hence yields $$m$$-methyl aniline as the product.

\n

In the light of the above statements, choose the most appropriate answer from the options given below :

", "options": [ { "text": "Both A and R are true and (R) is the correct explanation of (A)." }, { "text": "Both A and R are true but (R) is not the correct explanation of (A)." }, { "text": "(A) is true, but (R) is false." }, { "text": "(A) is false, but (R) is true." } ], "answer": "(A) is true, but (R) is false.", "solution": "**Answer:** (A) is true, but (R) is false.\n\n

\"JEE

Aniline does not undergo Friedel Craft reaction\nbecause the reagent AlCl3 being electron deficient\nacts as a Lewis acid.

", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 828, "subject": "Chemistry", "question": "

Given below are two statements: one is labelled as Assertion $$\\mathbf{A}$$ and the other is labelled as Reason $$\\mathbf{R}$$

\n

Assertion A : Aniline on nitration yields ortho, meta & para nitro derivatives of aniline.

\n

Reason $$\\mathrm{R}$$ : Nitrating mixture is a strong acidic mixture.

\n

In the light of the above statements, choose the correct answer from the options given below

", "options": [ { "text": "Both A and R are true and R is the correct explanation of A" }, { "text": "Both $$\\mathrm{A}$$ and $$\\mathrm{R}$$ are true but $$\\mathrm{R}$$ is NOT the correct explanation of $$\\mathrm{A}$$" }, { "text": "A is true but R is false" }, { "text": "A is false but R is true" } ], "answer": "Both A and R are true and R is the correct explanation of A", "solution": "**Answer:** Both A and R are true and R is the correct explanation of A\n\n\"JEE
\nDue to formation of anilinium ion in acidic\nmedium meta product is also obtained in\nsignificant amount ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 829, "subject": "Chemistry", "question": "

Match List I with List II

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
List IList II
ReactionReagents
(A)Hoffmann Degradation(I)$$\\mathrm{Conc. KOH}$$, $$\\Delta$$
(B)Clemenson reduction(II)$$\\mathrm{CHCl_3,NaOH/H_3O}$$$$^ \\oplus $$
(C)Cannizaro reaction(III)$$\\mathrm{Br_2,NaOH}$$
(D)Reimer-Tiemann Reaction(IV)$$\\mathrm{Zn-Hg/HCl}$$
\n

Choose the correct answer from the options given below :

", "options": [ { "text": "(A) - III, (B) - IV, (C) - I, (D) - II" }, { "text": "(A) - II, (B) - IV, (C) - I, (D) - III" }, { "text": "(A) - III, (B) - IV, (C) - II, (D) - I" }, { "text": "(A) - II, (B) - I, (C) - III, (D) - IV" } ], "answer": "(A) - III, (B) - IV, (C) - I, (D) - II", "solution": "**Answer:** (A) - III, (B) - IV, (C) - I, (D) - II\n\n\n\n\n \n \n \n \n \n \n\n\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\n
ReactionReagents
(A)$\\begin{aligned}  &  \\text { Hoffmann } \\\\ &  \\text { degradation }\\end{aligned}$$\\longrightarrow$$\\mathrm{Br}_{2}, \\mathrm{NaOH}$
(B)$\\begin{aligned}  &  \\text { Clemenson } \\\\ &  \\text { reduction }\\end{aligned}$$\\longrightarrow$$\\mathrm{Zn}-\\mathrm{Hg} / \\mathrm{HCl}$
(C)$\\begin{aligned}  &  \\text { Cannizaro- } \\\\ &  \\text { reaction }\\end{aligned}$$\\longrightarrow$Conc. $\\mathrm{KOH}, \\Delta$
(D)$\\begin{aligned}  &  \\text { Reimer-Tiemann } \\\\ &  \\text { reaction }\\end{aligned}$$\\longrightarrow$$\\begin{aligned}  &  \\mathrm{CHCl}_{3}, \\\\ &  \\mathrm{NaOH} / \\mathrm{H}_{3} \\mathrm{O}^{+}\\end{aligned}$

\n$\\therefore$ Correct match is : (A)-III, (B) -IV, (C) -I, (D)-II", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 830, "subject": "Chemistry", "question": "

Match List I with List II

\n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
List I (Amines)List II ($$\\mathrm{pK_b}$$)
A.AnilineI.3.25
B.EthanamineII.3.00
C.N-EthylethanamineIII.9.38
D.N, N-DiethylethanamineIV.3.29

\n

Choose the correct answer from the options given below :

", "options": [ { "text": "A-III, B-II, C-I, D-IV" }, { "text": "A-III, B-II, C-IV, D-I" }, { "text": "A-III, B-IV, C-II, D-I" }, { "text": "A-I, B-IV, C-II, D-III" } ], "answer": "A-III, B-IV, C-II, D-I", "solution": "**Answer:** A-III, B-IV, C-II, D-I\n\n

Aromatic amines are less basic than aliphatic amines. Among given aliphatic amines, $2^{\\circ}$ amine is most basic, followed by $3^{\\circ}$ amine and $1^{\\circ}$ amine. Therefore the correct basic strength $\\left(\\mathrm{K}_{\\mathrm{b}}\\right)$ order of the given amines is

\n

$$\n\\underset{(C)}{\\mathrm{CH}_3 \\mathrm{CH}_2 \\mathrm{NHCH}_2 \\mathrm{CH}_3}>\\underset{(D)}{\\left(\\mathrm{CH}_3 \\mathrm{CH}_2\\right)_3 \\mathrm{~N}}>\\underset{(B)}{\\mathrm{CH}_3 \\mathrm{CH}_2 \\mathrm{NH}_2}>\\underset{(A)}{\\mathrm{C}_6 \\mathrm{H}_5 \\mathrm{NH}_2}\n$$

\n

The $\\mathrm{pK}_{\\mathrm{b}}$ order of the given amines will be just the opposite of their basic strength order. The correct matching is\n

\n$A-I I I, B-I V, C-I I, D-I$

", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 831, "subject": "Chemistry", "question": "

Match List I with List II

\n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
List I
Isomeric pairs		List II
Type of isomers
A.Propanamine and N-MethylethanamineI.Metamers
B.Hexan-2-one and Hexan-3-oneII.Positional isomers
C.Ethanamide and HydroxyethanimineIII.Functional isomers
D.o-nitrophenol and p-nitrophenolIV.Tautomers

\n

Choose the correct answer from the options given below :

", "options": [ { "text": "A-III, B-I, C-IV, D-II" }, { "text": "A-III, B-IV, C-I, D-II" }, { "text": "A-II, B-III, C-I, D-IV" }, { "text": "A-IV, B-III, C-I, D-II" } ], "answer": "A-III, B-I, C-IV, D-II", "solution": "**Answer:** A-III, B-I, C-IV, D-II\n\n

A. Propanamine      N–Methylethanamine

\"JEE

\n

B. Hexan–2–one      Hexan–3–one

\n

\"JEE

\n

C. Ethanamide     Hydroxyethanimine

\n

\"JEE

\n

D. o–Nitrophenol    p–nitrophenol

\n

\"JEE

", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 832, "subject": "Chemistry", "question": "

Given below are two statements:

\n

Statement I : Pure Aniline and other arylamines are usually colourless.

\n

Statement II : Arylamines get coloured on storage due to atmospheric reduction.

\n

In the light of the above statements, choose the most appropriate answer from the options given below :

", "options": [ { "text": "Both Statement I and Statement II are correct" }, { "text": "Statement I is correct but Statement II is incorrect" }, { "text": "Both Statement I and Statement II are incorrect" }, { "text": "Statement I is incorrect but Statement I is correct" } ], "answer": "Statement I is correct but Statement II is incorrect", "solution": "**Answer:** Statement I is correct but Statement II is incorrect\n\nThe most appropriate answer is Statement-I is correct but Statement-II is incorrect.\n

\nPure Aniline and other arylamines are usually colorless, which is in agreement with Statement-I. However, Statement-II is incorrect as arylamines do not get colored on storage due to atmospheric reduction. Instead, get\ncoloured on storage due to atmospheric oxidation and become darker in color upon exposure to air.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 833, "subject": "Chemistry", "question": "

Number of isomeric aromatic amines with molecular formula $$\\mathrm{C}_{8} \\mathrm{H}_{11} \\mathrm{~N}$$, which can be synthesized by Gabriel Phthalimide synthesis is ____________.

", "options": [], "answer": "6", "solution": "**Answer:** 6\n\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 834, "subject": "Chemistry", "question": "Given below are two statements :

\nStatement (I) : The $\\mathrm{NH}_2$ group in Aniline is ortho and para directing and a powerful activating group.\n

\nStatement (II) : Aniline does not undergo Friedel-Craft's reaction (alkylation and acylation).

\nIn the light of the above statements, choose the most appropriate answer from the options given below :\n", "options": [ { "text": "Both Statement I and Statement II are correct" }, { "text": " Both Statement I and Statement II are incorrect" }, { "text": "Statement I is correct but Statement II is incorrect" }, { "text": "Statement I is incorrect but Statement II is correct" } ], "answer": "Both Statement I and Statement II are correct", "solution": "**Answer:** Both Statement I and Statement II are correct\n\n

The given statements pertain to aniline, a primary amine where the amino group ($\\mathrm{NH}_2$) is directly attached to a benzene ring. Let's analyze the statements:

\n\n

Statement (I): The $\\mathrm{NH}_2$ group in Aniline is ortho and para directing and a powerful activating group.

\n\n

This statement is correct. The amino group ($\\mathrm{NH}_2$) in aniline is an electron-donating group due to the lone pair of electrons on the nitrogen atom. It increases the electron density on the benzene ring, particularly at the ortho and para positions. This in turn makes the ortho and para positions more reactive toward electrophilic aromatic substitution reactions. Therefore, the amino group is considered to be an ortho and para director and is one of the most powerful activating groups in the context of electrophilic aromatic substitution reactions.

\n\n

Statement (II): Aniline does not undergo Friedel-Craft's reaction (alkylation and acylation).

\n\n

This statement is also correct. Aniline does not undergo Friedel-Crafts alkylation or acylation reactions. The reason for this is twofold: Firstly, the Lewis acid catalysts used in Friedel-Crafts reactions (such as AlCl$_3$) react with the amino group to form a complex that deactivates the benzene ring toward further reaction. Secondly, the strong interaction between the Lewis acid and the lone pair on the nitrogen can lead to a salt formation rather than the desired alkylation or acylation of the benzene ring. Moreover, the acidic conditions can lead to protonation of the amino group, converting it into an $\\mathrm{NH}_3^+$ group, which is meta-directing and deactivating, further inhibiting the reaction.

\n\n

Given this analysis, the correct option is:

\n\n

Option A: Both Statement I and Statement II are correct.

", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 835, "subject": "Chemistry", "question": "Given below are two statements :

\nStatement (I) : Aminobenzene and aniline are same organic compounds.

\nStatement (II) : Aminobenzene and aniline are different organic compounds.

\nIn the light of the above statements, choose the most appropriate answer from the options given below :", "options": [ { "text": "Both Statement I and Statement II are correct" }, { "text": "Statement I is incorrect but Statement II is correct" }, { "text": "Statement I is correct but Statement II is incorrect" }, { "text": "Both Statement I and Statement II are incorrect" } ], "answer": "Statement I is correct but Statement II is incorrect", "solution": "**Answer:** Statement I is correct but Statement II is incorrect\n\n

Statement (I) says that aminobenzene and aniline are the same organic compounds. Statement (II) says that aminobenzene and aniline are different organic compounds. To evaluate these statements, let's examine what each name refers to.

\n\n

Aminobenzene and aniline both refer to the same compound, which has the chemical formula $$C_6H_5NH_2$$. This molecule consists of a benzene ring with an amino group ($$NH_2$$) attached to it. Aminobenzene is a name that describes the structure based on its functional groups - an amino group attached to a benzene ring. Aniline is the common name for this compound and is recognized by IUPAC as an acceptable name.

\n\n

Given this information:

\n

Statement I is correct because aminobenzene and aniline refer to the same compound, that is, $$C_6H_5NH_2$$.

\n

Statement II is incorrect because it states that aminobenzene and aniline are different organic compounds, which is not true.

\n\n

Therefore, the most appropriate answer from the given options is:

\n

Option C\nStatement I is correct but Statement II is incorrect.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 836, "subject": "Chemistry", "question": "

Given below are two statements :

\n

Statement I : Aniline reacts with con. $$\\mathrm{H}_2 \\mathrm{SO}_4$$, followed by heating at $$453-473 \\mathrm{~K}$$ gives $$\\mathrm{p}$$-aminobenzene sulphonic acid, which gives blood red colour in the 'Lassaigne's test'.

\n

Statement II : In Friedel - Craft's alkylation and acylation reactions, aniline forms salt with the $$\\mathrm{AlCl}_3$$ catalyst. Due to this, nitrogen of aniline aquires a positive charge and acts as deactivating group.

\n

In the light of the above statements, choose the correct answer from the options given below :

", "options": [ { "text": "Both statement I and statement II are true\n" }, { "text": "Statement I is true but statement II is false\n" }, { "text": "Statement I is false but statement II is true\n" }, { "text": "Both statement I and statement II are false" } ], "answer": "Both statement I and statement II are true\n", "solution": "**Answer:** Both statement I and statement II are true\n\n\n

\"JEE

", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 837, "subject": "Chemistry", "question": "

A compound $$(x)$$ with molar mass $$108 \\mathrm{~g} \\mathrm{~mol}^{-1}$$ undergoes acetylation to give product with molar mass $$192 \\mathrm{~g} \\mathrm{~mol}^{-1}$$. The number of amino groups in the compound $$(x)$$ is ___________.

", "options": [], "answer": "2", "solution": "**Answer:** 2\n\n

\"JEE

\n

Gain in molecular weight after acylation with one $$-\\mathrm{NH}_2$$ group is 42.

\n

Total increase in molecular weight $$=84$$

\n

$$\\therefore$$ Number of amino group in $$x=\\frac{84}{42}=2$$

", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 838, "subject": "Chemistry", "question": "

Given below are two statements :

\n

Statement (I) : All the following compounds react with p-toluenesulfonyl chloride.

\n

$$\\mathrm{C}_6 \\mathrm{H}_5 \\mathrm{NH}_2 \\quad\\left(\\mathrm{C}_6 \\mathrm{H}_5\\right)_2 \\mathrm{NH} \\quad\\left(\\mathrm{C}_6 \\mathrm{H}_5\\right)_3 \\mathrm{N}$$

\n

Statement (II) : Their products in the above reaction are soluble is aqueous $$\\mathrm{NaOH}$$.

\n

In the light of the above statements, choose the correct answer from the options given below :

", "options": [ { "text": "Statement I is false but Statement II is true\n" }, { "text": "Both Statement I and Statement II are true\n" }, { "text": "Both Statement I and Statement II are false\n" }, { "text": "Statement I is true but Statement II is false" } ], "answer": "Both Statement I and Statement II are false\n", "solution": "**Answer:** Both Statement I and Statement II are false\n\n\n

Let's analyze both statements one by one.

\n\n

Statement (I): All the following compounds react with p-toluenesulfonyl chloride.

\n\n

The compounds given are:

\n\n

$$\\mathrm{C}_6 \\mathrm{H}_5 \\mathrm{NH}_2 \\quad (\\mathrm{Aniline})$$
\n\n

$$\\left(\\mathrm{C}_6 \\mathrm{H}_5\\right)_2 \\mathrm{NH} \\quad (\\mathrm{Diphenylamine})$$

\n\n

$$\\left(\\mathrm{C}_6 \\mathrm{H}_5\\right)_3 \\mathrm{N} \\quad (\\mathrm{Triphenylamine})$$

\n\n

P-toluenesulfonyl chloride (TsCl) reacts with amines to form sulfonamides. The general reaction is:

\n\n

$$\\mathrm{RNH}_2 + \\mathrm{TsCl} \\rightarrow \\mathrm{RNHTs} + \\mathrm{HCl}$$

\n\n

For this reaction to occur, the amine must have at least one hydrogen attached to the nitrogen. Let's check the provided compounds:

\n\n

1. $$\\mathrm{C}_6 \\mathrm{H}_5 \\mathrm{NH}_2$$ (Aniline) - primary amine, has 2 hydrogens, so it reacts.

\n\n

2. $$\\left(\\mathrm{C}_6 \\mathrm{H}_5\\right)_2 \\mathrm{NH}$$ (Diphenylamine) - secondary amine, has 1 hydrogen, so it reacts.

\n\n

3. $$\\left(\\mathrm{C}_6 \\mathrm{H}_5\\right)_3 \\mathrm{N}$$ (Triphenylamine) - tertiary amine, has no hydrogen, so it does not react.

\n\n

Hence, Statement I is false, since $$\\left(\\mathrm{C}_6 \\mathrm{H}_5\\right)_3 \\mathrm{N}$$ does not react with p-toluenesulfonyl chloride.

\n\n

Statement (II): Their products in the above reaction are soluble in aqueous $$\\mathrm{NaOH}$$.

\n\n

The products of the reaction of aniline and diphenylamine with p-toluenesulfonyl chloride (if it were to happen for the sake of analysis) would be sulfonamides:

\n\n

$$\\mathrm{RNH}_2 + \\mathrm{TsCl} \\rightarrow \\mathrm{RNHTs}$$

\n\n

Sulfonamides (RNHTs) are generally not soluble in aqueous NaOH because they are neutral compounds and NaOH is a strong base. Only aniline itself is soluble in aqueous NaOH due to its basic nature, but its product with TsCl is not.

\n\n

Thus, Statement II is also false, as the products would not be soluble in aqueous NaOH.

\n\n

Conclusion:

\n\n

Both Statement I and Statement II are false. Therefore, the correct answer is:

\n\n

Option C

\n\n

Both Statement I and Statement II are false.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 839, "subject": "Chemistry", "question": "

If $$279 \\mathrm{~g}$$ of aniline is reacted with one equivalent of benzenediazonium chloride, the maximum amount of aniline yellow formed will be ________ g. (nearest integer)

\n

(consider complete conversion).

", "options": [], "answer": "591", "solution": "**Answer:** 591\n\n

\"JEE

\n

Mole of aniline $$=\\frac{279}{93}=3 \\mathrm{~mol}$$

\n

Mole of aniline yellow formed $$=3 \\mathrm{~mol}$$

\n

$$\\begin{gathered}\n\\text { Mass }=3 \\times 197 \\\\\n=591 \\mathrm{~g}\n\\end{gathered}$$

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 840, "subject": "Chemistry", "question": "

$$9.3 \\mathrm{~g}$$ of pure aniline is treated with bromine water at room temperature to give a white precipitate of the product '$$\\mathrm{P}$$'. The mass of product '$$\\mathrm{P}$$' obtained is $$26.4 \\mathrm{~g}$$. The percentage yield is ________ %.

", "options": [], "answer": "80", "solution": "**Answer:** 80\n\n

\"JEE

\n

moles of aniline taken $$=\\frac{9.3}{93}=0.1$$

\n

$$\\% \\text { yield }=\\frac{26.4}{0.1 \\times 330} \\times 100=80 \\%$$

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 841, "subject": "Chemistry", "question": "Fluorination of an aromatic ring is easily accomplished by treating a diazonium salt with HBF4. Which of the following conditions is correct about this reaction ?", "options": [ { "text": "Only heat" }, { "text": "NaNO2/Cu" }, { "text": "Cu2O/H2O" }, { "text": "NaF/Cu" } ], "answer": "Only heat", "solution": "**Answer:** Only heat\n\n\"JEE", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 842, "subject": "Chemistry", "question": "Hinsberg's reagent is :", "options": [ { "text": "C6H5COCl" }, { "text": "C6H5SO2Cl" }, { "text": "SOCl2" }, { "text": "(COCl)2" } ], "answer": "C6H5SO2Cl", "solution": "**Answer:** C6H5SO2Cl\n\n

The correct answer is Option B: C6H5SO2Cl.

\n\n

Hinsberg's reagent is benzenesulfonyl chloride (C6H5SO2Cl). It is used to differentiate primary, secondary, and tertiary amines.

\n\n

Here's why the other options are incorrect :

\n\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 843, "subject": "Chemistry", "question": "The mass percentage of nitrogen in histamine is _____.", "options": [], "answer": "37.80to38.20", "solution": "**Answer:** 37.80to38.20\n\n\"JEE\n

Chemical formula of Histamine : C5H9N3\n

$$ \\therefore $$ % by mas of N = $${{3 \\times 14} \\over {5 \\times 12 + 1 \\times 9 + 3 \\times 14}} \\times 100$$\n

= $${{42} \\over {111}} \\times 100$$\n

= 37.84%", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 844, "subject": "Chemistry", "question": "A reaction of benzonitrile with one equivalent CH3MgBr followed by hydrolysis produces a yellow liquid \"P\". The compound \"P\" will give positive _________.", "options": [ { "text": "Iodoform test" }, { "text": "Schiff's test" }, { "text": "Ninhydrin's test" }, { "text": "Tollen's test" } ], "answer": "Iodoform test", "solution": "**Answer:** Iodoform test\n\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 845, "subject": "Chemistry", "question": "

A primary aliphatic amine on reaction with nitrous acid in cold (273 K) and there after raising temperature of reaction mixture to room temperature (298 K), gives a/an

", "options": [ { "text": "nitrile" }, { "text": "alcohol" }, { "text": "diazonium salt" }, { "text": "secondary amine" } ], "answer": "alcohol", "solution": "**Answer:** alcohol\n\n$\\mathrm{R}-\\mathrm{NH}_2 \\underset{+\\mathrm{HCl}}{\\stackrel{\\mathrm{NaNO}_2}{\\longrightarrow}} \\mathrm{R}-\\mathrm{N}_2^{+} \\rightarrow \\mathrm{R}^{+} \\underset{\\mathrm{H}_2 \\mathrm{O}}{\\longrightarrow} \\mathrm{R}-\\mathrm{OH}$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 846, "subject": "Chemistry", "question": "

Match List I with List II

\n

1 - Bromopropane is reacted with reagents in List I to give product in List II

\n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
LIST I - ReagentLIST II - Product
A.$$\\mathrm{KOH}$$ (alc)I.Nitrile
B.$$\\mathrm{KCN}$$ (alc)II.Ester
C.$$\\mathrm{AgNO_2}$$III.Alkene
D.$$\\mathrm{H_3CCOOAg}$$IV.Nitroalkane

\n

Choose the correct answer from the options given below:

", "options": [ { "text": "A-I, B-II, C-III, D-IV" }, { "text": "A-IV, B-III, C-II, D-I" }, { "text": "A-III, B-I, C-IV, D-II" }, { "text": "A-I, B-III, C-IV, D-II" } ], "answer": "A-III, B-I, C-IV, D-II", "solution": "**Answer:** A-III, B-I, C-IV, D-II\n\nFor the reactions of bromopropane with the reagents in List I, the products formed in List II are as follows:\n

\nA. $$\\mathrm{KOH}$$ (alc) - Alcoholic KOH is a strong base and will cause elimination of HBr, forming an alkene as the product. So, A-III.\n

\nB. $$\\mathrm{KCN}$$ (alc) - Potassium cyanide (KCN) will replace the bromine atom in the alkyl halide with a cyanide group, forming a nitrile. So, B-I.\n

\nC. $$\\mathrm{AgNO_2}$$ - Silver nitrite (AgNO2) will replace the bromine atom with a nitro group, forming a nitroalkane. So, C-IV.\n

\nD. $$\\mathrm{H_3CCOOAg}$$ - Silver acetate (H3CCOOAg) will replace the bromine atom with an acetate group, forming an ester. So, D-II.\n

\nThe correct answer is Option C: A-III, B-I, C-IV, D-II.", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 847, "subject": "Chemistry", "question": "

The compound formed by the reaction of ethanal with semicarbazide contains _________ number of nitrogen atoms.

", "options": [], "answer": "3", "solution": "**Answer:** 3\n\n\"JEE\n

When ethanal reacts with semicarbazide, semi-carbazone is formed via condensation, number of nitrogen atoms present in product is 3.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 848, "subject": "Chemistry", "question": "

Number of compounds from the following which cannot undergo Friedel-Crafts reactions is: _________

\n

toluene, nitrobenzene, xylene, cumene, aniline, chlorobenzene, $$m$$-nitroaniline, $$m$$-dinitrobenzene

", "options": [], "answer": "4", "solution": "**Answer:** 4\n\n

Compounds which can not undergo Friedel Crafts reaction are

\n

\"JEE

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 849, "subject": "Chemistry", "question": "

Which of the following nitrogen containing compound does not give Lassaigne's test ?

", "options": [ { "text": "Glycene\n" }, { "text": "Phenyl hydrazine\n" }, { "text": "Urea\n" }, { "text": "Hydrazine" } ], "answer": "Hydrazine", "solution": "**Answer:** Hydrazine\n\n

Hydrazine $$(\\mathrm{N}_2 \\mathrm{H}_4)$$ doesn't contain any carbon atom and hence doesn't give Lassaigne test.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 850, "subject": "Chemistry", "question": "

Given below are two statements :

\n

Statement (I) : Kjeldahl method is applicable to estimate nitrogen in pyridine.

\n

Statement (II) : The nitrogen present in pyridine can easily be converted into ammonium sulphate in Kjeldahl method.

\n

In the light of the above statements, choose the correct answer from the options given below :

", "options": [ { "text": "Both Statement I and Statement II are true\n" }, { "text": "Both Statement I and Statement II are false\n" }, { "text": "Statement I is true but Statement II is false\n" }, { "text": "Statement I is false but Statement II is true" } ], "answer": "Both Statement I and Statement II are false\n", "solution": "**Answer:** Both Statement I and Statement II are false\n\n\n

In order to determine the correct option, let’s analyze both statements in the context of the Kjeldahl method.

\n\n

Statement I: Kjeldahl method is applicable to estimate nitrogen in pyridine.

\n\n

The Kjeldahl method is a widely used procedure for estimating the nitrogen content in organic compounds. However, this method is generally not applicable to aromatic nitrogen compounds like pyridine because the nitrogen in these compounds is not easily converted to ammonium form. Therefore, Statement I is false.

\n\n

Statement II: The nitrogen present in pyridine can easily be converted into ammonium sulphate in the Kjeldahl method.

\n\n

The nitrogen in pyridine is part of the aromatic ring structure, making it more stable and less reactive compared to nitrogen in aliphatic amines or amides. This means it cannot be easily converted into ammonium compounds such as ammonium sulphate via the Kjeldahl method. Therefore, Statement II is also false.

\n\n

Thus, the correct option is:

\n\n

Option B

\n\n

Both Statement I and Statement II are false.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 851, "subject": "Chemistry", "question": "

$$\\mathrm{X} \\mathrm{~g}$$ of ethanamine was subjected to reaction with $$\\mathrm{NaNO}_2 / \\mathrm{HCl}$$ followed by hydrolysis to liberate $$\\mathrm{N}_2$$ and $$\\mathrm{HCl}$$. The $$\\mathrm{HCl}$$ generated was completely neutralised by 0.2 moles of $$\\mathrm{NaOH} . \\mathrm{X}$$ is _________ g.

", "options": [], "answer": "9", "solution": "**Answer:** 9\n\n

$$\\begin{aligned}\n& \\mathrm{HCl}+\\mathrm{NaOH} \\rightarrow \\mathrm{NaCl}+\\mathrm{H}_2 \\mathrm{O} \\\\\n& \\mathrm{mol} \\text { of } \\mathrm{HCl}=0.2 \\mathrm{~mol}\n\\end{aligned}$$

\n

$$\\mathrm{CH}_3 \\mathrm{CH}_2 \\mathrm{NH}_2 \\xrightarrow[+\\mathrm{HCl}]{\\mathrm{NaNO}_2} \\mathrm{CH}_3 \\mathrm{CH}_2 \\mathrm{N}_2^{+} \\mathrm{Cl}^{-} \\xrightarrow{\\mathrm{H}_2 \\mathrm{O}} \\mathrm{CH}_3 \\mathrm{CH}_2 \\mathrm{OH}+\\mathrm{N}_2+\\mathrm{HCl}$$

\n

$$0.2 \\mathrm{~mol} \\mathrm{~HCl}$$ would be generated by $$0.2 \\mathrm{~mol}$$ $$\\mathrm{C}_2 \\mathrm{H}_5 \\mathrm{NH}_2$$

\n

$$\\mathrm{x}=0.2 \\times 45=9 \\mathrm{~g}$$

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 852, "subject": "Chemistry", "question": "

Given below are two statements :

\n

Statement I : Piciric acid is 2,4,6 - trinitrotoluene.

\n

Statement II : Phenol - 2,4 - disulphonic acid is treated with Conc. $$\\mathrm{HNO}_3$$ to get picric acid.

\n

In the light of the above statements, choose the most appropriate answer from the options given below :

", "options": [ { "text": "Statement I is incorrect but Statement II is correct\n" }, { "text": "Both Statement I and Statement II are incorrect\n" }, { "text": "Both Statement I and Statement II are correct\n" }, { "text": "Statement I is correct but Statement II is incorrect" } ], "answer": "Statement I is incorrect but Statement II is correct\n", "solution": "**Answer:** Statement I is incorrect but Statement II is correct\n\n\n

S-I is incorrect as picric acid is 2,4,6-trinitrophenol

\n

S-II is correct

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 853, "subject": "Chemistry", "question": "

$$9.3 \\mathrm{~g}$$ of pure aniline upon diazotisation followed by coupling with phenol gives an orange dye. The mass of orange dye produced (assume 100% yield/conversion) is ________ g. (nearest integer)

", "options": [], "answer": "20", "solution": "**Answer:** 20\n\n

$$\\begin{aligned}\n& \\text { Moles of aniline }=0.1 \\\\\n& \\text { Moles of dye }=0.1 \\\\\n& \\text { Mass of dye }=0.1 \\times 198=19.8 \\mathrm{~gm} \\\\\n& \\approx 20 \\\\\n&\n\\end{aligned}$$

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 854, "subject": "Chemistry", "question": "CH3 - Mg - Br is an organometallic compound due to :", "options": [ { "text": "Mg - Br bond" }, { "text": "C - Mg bond" }, { "text": "C - Br bond" }, { "text": "C - H bond" } ], "answer": "C - Mg bond", "solution": "**Answer:** C - Mg bond\n\nCompounds that contain at least one carbon metal bond are known as organometallic compounds. In $$C{H_3}$$$$ - Mg - Br$$ (Grignard's reagent) a bond is present between carbon and $$Mg$$ (Metal) hence it is an organometallic compound. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 855, "subject": "Chemistry", "question": "Coordination compound have great importance in biological systems. In this context which of\nthe following statements is incorrect?", "options": [ { "text": "Chlorophylls are green pigments in plants and contains calcium" }, { "text": "Carboxypeptidase – A is an enzyme and contains zinc " }, { "text": "Cyanocobalamin is B12 and contains cobalt " }, { "text": "Haemoglobin is the red pigment of blood and contains iron" } ], "answer": "Chlorophylls are green pigments in plants and contains calcium", "solution": "**Answer:** Chlorophylls are green pigments in plants and contains calcium\n\nThe chlorophyll molecule plays an important role in photosynthesis, contain porphyrin ring and the metal $$Mg$$ not $$Ca.$$", "topic": "Discrete Mathematics", "subtopic": "Logic" }, { "id": 856, "subject": "Chemistry", "question": "Match the metals (column I) with the coordination compound(s)/enzyme(s) (column II) \n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
(Column I) Metals
(Column II) Coordination
compounds(s) enzyme(s)
(A)Co(i)Wilkinson catalyst
(B)Zn(ii)Chlorophyl
(C)Rh(iii)Vitamin B12
(D)Mg(iv)Carbonic anhydrase
", "options": [ { "text": "(A)-(iii); (B)-(iv); (C)-(i); (D)-(ii)" }, { "text": "(A)-(iv); (B)-(iii); (C)-(i); (D)-(ii)" }, { "text": "(A)-(i); (B)-(ii); (C)-(iii); (D)-(iv)" }, { "text": "(A)-(ii); (B)-(i); (C)-(iv); (D)-(iii)" } ], "answer": "(A)-(iii); (B)-(iv); (C)-(i); (D)-(ii)", "solution": "**Answer:** (A)-(iii); (B)-(iv); (C)-(i); (D)-(ii)\n\nCo $$ \\to $$ Vitamin B12\n

Zn $$ \\to $$ Carbonic anhydrase\n

Rh $$ \\to $$ Wilkinson catalyst\n

Mg $$ \\to $$ Chlorophyll", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 857, "subject": "Chemistry", "question": "The number of bridging CO ligand(s) and Co-Co bond(s) in Co2(CO)8, respectively are : ", "options": [ { "text": "2 and 0" }, { "text": "0 and 2" }, { "text": "4 and 0" }, { "text": "2 and 1" } ], "answer": "2 and 1", "solution": "**Answer:** 2 and 1\n\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 858, "subject": "Chemistry", "question": "Mn2(CO)10 is an organometallic compound due to the presence of - ", "options": [ { "text": "C–O bond" }, { "text": "Mn – Mn bond " }, { "text": "Mn – O bond" }, { "text": "Mn – C bond " } ], "answer": "Mn – C bond ", "solution": "**Answer:** Mn – C bond \n\nCompounds that contain at least one carbon-metal\nbond are called organometallic compounds.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 859, "subject": "Chemistry", "question": "The compound that inhibits the growth of\ntumors is :", "options": [ { "text": "cis-[Pd(Cl)2(NH3)2]" }, { "text": "trans-[Pd(Cl)2(NH3)2]" }, { "text": "cis-[Pt(Cl)2(NH3)2]" }, { "text": "trans-[Pt(Cl)2(NH3)2]" } ], "answer": "cis-[Pt(Cl)2(NH3)2]", "solution": "**Answer:** cis-[Pt(Cl)2(NH3)2]\n\nCis-platin or cis-[Pt(Cl)2(NH3)2] is used as an anti-cancer drug.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 860, "subject": "Chemistry", "question": "The compound used in the treatment of lead poisoning is :", "options": [ { "text": "desferrioxime B" }, { "text": "Cis-platin" }, { "text": "D-penicillamine" }, { "text": "EDTA" } ], "answer": "EDTA", "solution": "**Answer:** EDTA\n\nEDTA is used in treatment of lead poisoning.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 861, "subject": "Chemistry", "question": "The potassium ferrocyanide solution gives a Prussian blue colour, when added to :", "options": [ { "text": "CoCl3" }, { "text": "FeCl2" }, { "text": "CoCl2" }, { "text": "FeCl3" } ], "answer": "FeCl3", "solution": "**Answer:** FeCl3\n\nFeCl3 + K4[Fe(CN)6] $$\\to$$ Fe4[Fe(CN)6]3 (Prussian blue)", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 862, "subject": "Chemistry", "question": "

Number of complexes which will exhibit synergic bonding amongst, $$[Cr{(CO)_6}]$$, $$[Mn{(CO)_5}]$$ and $$[M{n_2}{(CO)_{10}}]$$ is ___________.

", "options": [], "answer": "3", "solution": "**Answer:** 3\n\nCarbonyl complex compounds have tendency to show synergic bonding.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 863, "subject": "Chemistry", "question": "

Which statement is not true with respect to nitrate ion test?

", "options": [ { "text": "A dark brown ring is formed at the junction of two solutions." }, { "text": "Ring is formed due to nitroferrous sulphate complex." }, { "text": "The brown complex is [Fe(H2O)5 (NO)]SO4." }, { "text": "Heating the nitrate salt with conc. H2SO4, light brown fumes are evolved." } ], "answer": "Ring is formed due to nitroferrous sulphate complex.", "solution": "**Answer:** Ring is formed due to nitroferrous sulphate complex.\n\nBrown ring test\n

\n$$\n\\begin{aligned}\n&\\mathrm{NO}_{3}^{-}+3 \\mathrm{Fe}^{+2}+4 \\mathrm{H}^{+} \\rightarrow \\mathrm{NO}+3 \\mathrm{Fe}^{+3}+2 \\mathrm{H}_{2} \\mathrm{O} \\\\\\\\\n&{\\left[\\mathrm{Fe}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}\\right]^{2+} \\mathrm{NO} \\rightarrow \\underset{\\text{Brown ring}}{\\left[\\mathrm{Fe}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{5} \\mathrm{NO}]^{2+} \\mathrm{H}_{2} \\mathrm{O}\\right.}}\n\\end{aligned}\n$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 864, "subject": "Chemistry", "question": "

Correct formula of the compound which gives a white precipitate with BaCl2 solution, but not with AgNO3 solution, is :

", "options": [ { "text": "[Co(NH3)5Br]SO4" }, { "text": "[Co(NH3)5SO4]Br" }, { "text": "[Pt(NH3)4Cl2]Br2" }, { "text": "[Pt(NH3)4Br2]Cl2" } ], "answer": "[Co(NH3)5Br]SO4", "solution": "**Answer:** [Co(NH3)5Br]SO4\n\nOnly the counter ions are replaced easily by other reagents.

$\\left[\\mathrm{Co}\\left(\\mathrm{NH}_{3}\\right)_{5} \\mathrm{Br}\\right] \\mathrm{SO}_{4}+\\mathrm{BaCl}_{2} \\rightarrow\\left[\\mathrm{Co}\\left(\\mathrm{NH}_{3}\\right)_{5} \\mathrm{Br}\\right] \\mathrm{Cl}_{2}+\\underset{\\text{White ppt.}}{\\mathrm{BaSO}_{4}{\\downarrow}}$\n

\nWhite ppt. It does not have chloride ion as counter ion so it will not give white ppt with $\\mathrm{AgNO}_{3}$.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 865, "subject": "Chemistry", "question": "

$$\\mathrm{Fe}^{3+}$$ cation gives a prussian blue precipitate on addition of potassium ferrocyanide solution due to the formation of :

", "options": [ { "text": "$$\\left[\\mathrm{Fe}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}\\right]_{2}\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}\\right]$$" }, { "text": "$$\\mathrm{Fe}_{2}\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}\\right]_{2}$$" }, { "text": "$$\\mathrm{Fe}_{3}\\left[\\mathrm{Fe}(\\mathrm{OH})_{2}(\\mathrm{CN})_{4}\\right]_{2}$$" }, { "text": "$$\\mathrm{Fe}_{4}\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}\\right]_{3}$$" } ], "answer": "$$\\mathrm{Fe}_{4}\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}\\right]_{3}$$", "solution": "**Answer:** $$\\mathrm{Fe}_{4}\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}\\right]_{3}$$\n\n$4 \\mathrm{Fe}^{3+}+3\\left[\\mathrm{Fe}(\\mathrm{CN})_6\\right]^{-4} \\longrightarrow\\underset{\\text{Prussian Blue}} {\\mathrm{Fe}_4\\left[\\mathrm{Fe}(\\mathrm{CN})_6\\right]_3}$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 866, "subject": "Chemistry", "question": "

A solution of $$\\mathrm{FeCl_3}$$ when treated with $$\\mathrm{K_4[Fe(CN)_6]}$$ gives a prussium blue precipitate due to the formation of :

", "options": [ { "text": "$$\\mathrm{Fe[Fe(CN)_{6}]}$$" }, { "text": "$$\\mathrm{Fe_{4}[Fe(CN)_{6}]_{3}}$$" }, { "text": "$$\\mathrm{Fe_{3}[Fe(CN)_{6}]_{2}}$$" }, { "text": "$$\\mathrm{K[Fe_{2}(CN)_{6}]}$$" } ], "answer": "$$\\mathrm{Fe_{4}[Fe(CN)_{6}]_{3}}$$", "solution": "**Answer:** $$\\mathrm{Fe_{4}[Fe(CN)_{6}]_{3}}$$\n\n$\\mathrm{Fe}^{3+}+\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}\\right]^{4-} \\longrightarrow \\mathrm{Fe}_{4}\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}\\right]_{3}$(Prussium Blue)", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 867, "subject": "Chemistry", "question": "

To inhibit the growth of tumours, identify the compounds used from the following :

\n

A. EDTA

\n

B. Coordination Compounds of Pt

\n

C. D - Penicillamine

\n

D. Cis - Platin

\n

Choose the correct answer from the option given below :

", "options": [ { "text": "A and B Only" }, { "text": "C and D Only" }, { "text": "B and D Only" }, { "text": "A and C Only" } ], "answer": "B and D Only", "solution": "**Answer:** B and D Only\n\n

Cis-platin is [Pt(NH$$_3$$)$$_2$$Cl$$_2$$]; cis platin and other complexes of pt are used to inhibit the growth of tumours.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 868, "subject": "Chemistry", "question": "

The sum of bridging carbonyls in $$\\mathrm{W(CO)_6}$$ and $$\\mathrm{Mn_2(CO)_{10}}$$ is ____________.

", "options": [], "answer": "0", "solution": "**Answer:** 0\n\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 869, "subject": "Chemistry", "question": "

The mismatched combinations are

\n

A. Chlorophyll - Co

\n

B. Water hardness - EDTA

\n

C. Photography $$-\\left[\\mathrm{Ag}(\\mathrm{CN})_{2}\\right]^{-}$$

\n

D. Wilkinson catalyst $$-\\left[\\left(\\mathrm{Ph}_{3} \\mathrm{P}\\right)_{3} \\mathrm{RhCl}\\right]$$

\n

E. Chelating ligand - D-Penicillamine

\n

Choose the correct answer from the options given below :

", "options": [ { "text": "A and E Only" }, { "text": "D and E Only" }, { "text": "A and C Only" }, { "text": "A, C, and E Only" } ], "answer": "A and C Only", "solution": "**Answer:** A and C Only\n\nLet's analyze each combination :\n

\nA. Chlorophyll - Co : Mismatched. Chlorophyll has a magnesium (Mg) ion at its center, not cobalt (Co).\n

\nB. Water hardness - EDTA : Correct match. EDTA (ethylenediaminetetraacetic acid) is used to treat water hardness by chelating metal ions like Ca²⁺ and Mg²⁺.\n

\nC. Photography - [Ag(CN)₂]⁻ : Mismatched. Silver halides (AgX, where X = Cl, Br, I) are used in photography, not [Ag(CN)₂]⁻.\n

\nD. Wilkinson catalyst - [(Ph₃P)₃RhCl] : Correct match. Wilkinson's catalyst is a homogeneous hydrogenation catalyst with the formula [(Ph₃P)₃RhCl].\n

\nE. Chelating ligand - D-Penicillamine : Correct match. D-Penicillamine is a chelating agent that can bind to metal ions through multiple coordination sites.\n

\nThe mismatched combinations are A and C. So, the correct answer is :\n

\nA and C Only.", "topic": "Discrete Mathematics", "subtopic": "Combinatorics" }, { "id": 870, "subject": "Chemistry", "question": "

Given below are two statements, one is labelled as Assertion A and the other is labelled as Reason R.

\n

Assertion A : $$\\left[\\mathrm{CoCl}\\left(\\mathrm{NH}_{3}\\right)_{5}\\right]^{2+}$$ absorbs at lower wavelength of light with respect to $$\\left[\\mathrm{Co}\\left(\\mathrm{NH}_{3}\\right)_{5}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)\\right]^{3+}$$

\n

Reason R : It is because the wavelength of the light absorbed depends on the oxidation state of the metal ion.

\n

In the light of the above statements, choose the correct answer from the options given below:

", "options": [ { "text": "$$\\mathrm{A}$$ is false but $$\\mathrm{R}$$ is true" }, { "text": "Both $$\\mathrm{A}$$ and $$\\mathrm{R}$$ are true but $$\\mathrm{R}$$ is NOT the correct explanation of $$A$$" }, { "text": "Both $$\\mathrm{A}$$ and $$\\mathrm{R}$$ are true and $$\\mathrm{R}$$ is the correct explanation of $$\\mathrm{A}$$" }, { "text": "$$\\mathrm{A}$$ is true but $$\\mathrm{R}$$ is false" } ], "answer": "$$\\mathrm{A}$$ is false but $$\\mathrm{R}$$ is true", "solution": "**Answer:** $$\\mathrm{A}$$ is false but $$\\mathrm{R}$$ is true\n\n

Assertion A : $$\\left[\\mathrm{CoCl}\\left(\\mathrm{NH}_{3}\\right)_{5}\\right]^{2+}$$ absorbs at a lower wavelength of light with respect to $$\\left[\\mathrm{Co}\\left(\\mathrm{NH}_{3}\\right)_{5}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)\\right]^{3+}$$.

\n

Given that H2O is a stronger field ligand than Cl-, it would result in a larger splitting of the d-orbitals and hence a larger energy difference. This would lead to absorption at a lower wavelength. Therefore, the Assertion A is false.

\n

Reason R : It is because the wavelength of the light absorbed depends on the oxidation state of the metal ion.

\n

While it's true that the oxidation state can influence the absorption wavelength by affecting the energy levels of the d-orbitals, the difference in the absorption wavelength in this specific case is not primarily due to the oxidation state (the oxidation state of Co in both complexes is the same (+3)), but rather due to the difference in ligand field strength (H2O vs Cl-). Therefore, the Reason R is not providing the correct explanation for Assertion A.

\n

So, the correct answer would be:

\n

Option A : A is false but R is true.

\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 871, "subject": "Chemistry", "question": "

Match List I with List II

\n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
LIST I
Complex		LIST II
Colour
A.$$Mg(N{H_4})P{O_4}$$I.brown
B.$${K_3}[Co{(N{O_2})_6}]$$II.white
C.$$MnO{(OH)_2}$$III.yellow
D.$$F{e_4}{[Fe{(CN)_6}]_3}$$IV.blue

\n

Choose the correct answer from the options given below :

", "options": [ { "text": "A-II, B-IV, C-I, D-III" }, { "text": "A-II, B-III, C-I, D-IV" }, { "text": "A-III, B-IV, C-II, D-I" }, { "text": "A-II, B-III, C-IV, D-I" } ], "answer": "A-II, B-III, C-I, D-IV", "solution": "**Answer:** A-II, B-III, C-I, D-IV\n\n

In this problem, we are matching chemical compounds with their corresponding color. Let's go over each one :

\n

A. $$Mg(NH_4)PO_4$$ - Magnesium ammonium phosphate is white in color, so A matches with II.

\n

B. $$K_3[Co(NO_2)_6]$$ - Potassium hexanitritocobaltate(III) is a yellow complex, so B matches with III.

\n

C. $$MnO(OH)_2$$ - Manganese hydroxide is brown in color, so C matches with I.

\n

D. $$Fe_4[Fe(CN)_6]_3$$ - Prussian blue or Ferric ferrocyanide is deep blue in color, so D matches with IV.

\n

So, the correct option is Option B: A-II, B-III, C-I, D-IV.

\n", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 872, "subject": "Chemistry", "question": "

The complex that dissolves in water is :

", "options": [ { "text": "$$\\mathrm{Fe}_{4}\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}\\right]_{3}$$" }, { "text": "$$\\left(\\mathrm{NH}_{4}\\right)_{3}\\left[\\mathrm{As}\\left(\\mathrm{Mo}_{3} \\mathrm{O}_{10}\\right)_{4}\\right]$$" }, { "text": "$$\\mathrm{K}_{3}\\left[\\mathrm{Co}\\left(\\mathrm{NO}_{2}\\right)_{6}\\right]$$" }, { "text": "$$\\left[\\mathrm{Fe}_{3}(\\mathrm{OH})_{2}(\\mathrm{OAc})_{6}\\right] \\mathrm{Cl}$$" } ], "answer": "$$\\left[\\mathrm{Fe}_{3}(\\mathrm{OH})_{2}(\\mathrm{OAc})_{6}\\right] \\mathrm{Cl}$$", "solution": "**Answer:** $$\\left[\\mathrm{Fe}_{3}(\\mathrm{OH})_{2}(\\mathrm{OAc})_{6}\\right] \\mathrm{Cl}$$\n\nOption A : $$\\mathrm{Fe}_{4}\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}\\right]_{3}$$ (Prussian Blue) \n

Prussian blue is a very stable and intensely colored compound, but it is not soluble in water. The individual cyanide ligands and the iron ions are bonded together strongly, which results in a crystalline structure that does not easily dissociate into its components in water.\n\n

Option B : $$\\left(\\mathrm{NH}_{4}\\right)_{3}\\left[\\mathrm{As}\\left(\\mathrm{Mo}_{3} \\mathrm{O}_{10}\\right)_{4}\\right]$$ (Ammonium Arseno Molybdate) \n

In this compound, the arsenic and molybdenum atoms are coordinated in a complex polyatomic anion. Despite the presence of ammonium ions, which are generally soluble, the overall complex has a low solubility due to the strong interactions between arsenic, molybdenum, and oxygen within the anion.\n\n

Option C : $$\\mathrm{K}_{3}\\left[\\mathrm{Co}\\left(\\mathrm{NO}_{2}\\right)_{6}\\right]$$ \n

This is a potassium salt of an anionic complex. Although potassium salts are generally soluble in water, the nitrite ligands attached to the cobalt in the anion form a tightly bonded structure, resulting in very poor solubility in water.\n\n

Option D : $$\\left[\\mathrm{Fe}_{3}(\\mathrm{OH})_{2}(\\mathrm{OAc})_{6}\\right] \\mathrm{Cl}$$ \n

This complex is composed of iron atoms coordinated with hydroxide and acetate ligands, and a chloride counterion. Hydroxide and acetate ions are both polar and capable of forming hydrogen bonds with water molecules, which is a key factor in solubility. The chloride ion is also soluble in water. The presence of these ions, combined with the likely overall charge of the complex (since it's paired with a counterion), makes it soluble in water.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 873, "subject": "Chemistry", "question": "Given below are two statements :

\nStatement (I) : Dimethyl glyoxime forms a six-membered covalent chelate when treated with $\\mathrm{NiCl}_2$ solution in presence of $\\mathrm{NH}_4 \\mathrm{OH}$.\n

\nStatement (II) : Prussian blue precipitate contains iron both in $(+2)$ and $(+3)$ oxidation states.

\nIn the light of the above statements, choose the most appropriate answer from the options given below :", "options": [ { "text": "Statement I is false but Statement II is true" }, { "text": "Both Statement I and Statement II are true" }, { "text": "Statement I is true but Statement II is false" }, { "text": "Both Statement I and Statement II are false" } ], "answer": "Statement I is false but Statement II is true", "solution": "**Answer:** Statement I is false but Statement II is true\n\n

Let's analyze each statement separately:

\n\n

Statement (I): \"JEE\n\n

So, Statement I is false.

\n\n

Statement (II): Prussian blue, also known as ferric ferrocyanide, is a famous pigment that contains iron in both +2 and +3 oxidation states. Its chemical formula can be represented as $\\mathrm{Fe}_4\\mathrm{[Fe(CN)_6]_3}$. In this formula, the iron within the square brackets, $\\mathrm{[Fe(CN)_6]^{4-}}$, is in the +2 oxidation state (ferrocyanide ion), while the iron outside the square brackets is in the +3 oxidation state (ferric ion). The compound is a complex salt that arises from the reaction of ferric and ferrous ions with cyanide ions. Therefore, Statement II is also true.

\n\n

With both statements being true, the correct answer is:

\n\n

Option A : Statement I is false but Statement II is true.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 874, "subject": "Chemistry", "question": "

Identity the incorrect pair from the following :

", "options": [ { "text": "Haber process - Iron\n" }, { "text": "Polythene preparation - $$\\mathrm{TiCl}_4, \\mathrm{Al}\\left(\\mathrm{CH}_3\\right)_3$$\n" }, { "text": "Photography - AgBr\n" }, { "text": "Wacker process - $$\\mathrm{Pt} \\mathrm{Cl}_2$$" } ], "answer": "Wacker process - $$\\mathrm{Pt} \\mathrm{Cl}_2$$", "solution": "**Answer:** Wacker process - $$\\mathrm{Pt} \\mathrm{Cl}_2$$\n\n

The catalyst used in Wacker's process is $$\\mathrm{PdCl_2}$$

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 875, "subject": "Chemistry", "question": "

Match List I with List II

\n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
List - I
(Substances)		List - II
(Element Present)
(A)Ziegler catalyst(I)Rhodium
(B)Blood Pigment(II)Cobalt
(C)Wilkinson catalyst(III)Iron
(D)Vitamin $$\\mathrm{B_{12}}$$(IV)Titanium

\n

Choose the correct answer from the options given below:

", "options": [ { "text": "A-IV, B-III, C-I, D-II\n" }, { "text": "A-II, B-IV, C-I, D-III\n" }, { "text": "A-III, B-II, C-IV, D-I\n" }, { "text": "A-II, B-III, C-IV, D-I" } ], "answer": "A-IV, B-III, C-I, D-II\n", "solution": "**Answer:** A-IV, B-III, C-I, D-II\n\n\n

Ziegler catalyst $$\\rightarrow$$ Titanium

\n

Blood pigment $$\\rightarrow$$ Iron

\n

Wilkinson catalyst $$\\rightarrow$$ Rhodium

\n

Vitamin $$\\mathrm{B}_{12} \\rightarrow$$ Cobalt

", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 876, "subject": "Chemistry", "question": "

In which one of the following metal carbonyls, $$\\mathrm{CO}$$ forms a bridge between metal atoms?

", "options": [ { "text": "$$\\left[\\mathrm{Os}_3(\\mathrm{CO})_{12}\\right]$$\n" }, { "text": "$$\\left[\\mathrm{Ru}_3(\\mathrm{CO})_{12}\\right]$$\n" }, { "text": "$$\\left[\\mathrm{Mn}_2(\\mathrm{CO})_{10}\\right]$$\n" }, { "text": "$$\\left[\\mathrm{Co}_2(\\mathrm{CO})_8\\right]$$" } ], "answer": "$$\\left[\\mathrm{Co}_2(\\mathrm{CO})_8\\right]$$", "solution": "**Answer:** $$\\left[\\mathrm{Co}_2(\\mathrm{CO})_8\\right]$$\n\n\"JEE\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 877, "subject": "Chemistry", "question": "

A reagent which gives brilliant red precipitate with Nickel ions in basic medium is

", "options": [ { "text": "dimethyl glyoxime\n" }, { "text": "sodium nitroprusside\n" }, { "text": "meta-dinitrobenzene\n" }, { "text": "neutral $$\\mathrm{FeCl}_3$$" } ], "answer": "dimethyl glyoxime\n", "solution": "**Answer:** dimethyl glyoxime\n\n\n

$$\\mathrm{Ni}^{2+}+2 \\mathrm{dmg}^{-} \\rightarrow\\left[\\mathrm{Ni}(\\mathrm{dmg})_2\\right]$$

\n

Rosy red/Bright Red precipitate

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 878, "subject": "Chemistry", "question": "

Choose the correct statements from the following :

\n

(A) Ethane-1, 2-diamine is a chelating ligand.

\n

(B) Metallic aluminium is produced by electrolysis of aluminium oxide in presence of cryolite.

\n

(C) Cyanide ion is used as ligand for leaching of silver.

\n

(D) Phosphine act as a ligand in Wilkinson catalyst.

\n

(E) The stability constants of $$\\mathrm{Ca}^{2+}$$ and $$\\mathrm{Mg}^{2+}$$ are similar with EDTA complexes.

\n

Choose the correct answer from the options given below :

", "options": [ { "text": "(B), (C), (E) only" }, { "text": "(A), (D), (E) only" }, { "text": "(C), (D), (E) only" }, { "text": "(A), (B), (C) only" } ], "answer": "(A), (B), (C) only", "solution": "**Answer:** (A), (B), (C) only\n\n

Let's examine each statement for correctness:

\n\n

(A) Ethane-1, 2-diamine is a chelating ligand.

\n

Ethane-1,2-diamine, also known as ethylenediamine (en), has two nitrogen atoms that can coordinate to a metal ion, forming a ring structure in the process. Because it can form these two bonds, it can \"chelate\" a metal ion, thus it is correctly identified as a chelating ligand.

\n\n

(B) Metallic aluminium is produced by electrolysis of aluminium oxide in presence of cryolite.

\n

Aluminium is indeed produced industrially by the Hall-Héroult process, which involves the electrolysis of aluminium oxide ($$\\mathrm{Al_2O_3}$$) dissolved in molten cryolite ($$\\mathrm{Na_3AlF_6}$$). Cryolite acts as a solvent for the aluminium oxide and reduces the melting point of the mixture, thus decreasing energy consumption during electrolysis. This statement is correct.

\n\n

(C) Cyanide ion is used as a ligand for leaching of silver.

\n

Gold and silver are often extracted from their ores via a leaching process using a cyanide solution. The cyanide ion ($$\\mathrm{CN^-}$$) complexes with the metal ions to form soluble complexes like [Ag(CN)$_2$]$^-$, enabling the separation of silver from the ore. So, this statement is correct as well.

\n\n

(D) Phosphine act as a ligand in Wilkinson's catalyst.

\n

Wilkinson's catalyst is $$\\mathrm{RhCl(PPh_3)_3}$$, where PPh$_3$ stands for triphenylphosphine, a type of phosphine ligand. Phosphines are indeed ligands in Wilkinson's catalyst, and they play an important role in its catalytic activity, particularly in hydrogenation reactions. Therefore, this statement is correct.

\n\n

(E) The stability constants of $$\\mathrm{Ca}^{2+}$$ and $$\\mathrm{Mg}^{2+}$$ are similar with EDTA complexes.

\n

EDTA (ethylenediaminetetraacetic acid) forms strong complexes with many metal ions including $$\\mathrm{Ca}^{2+}$$ and $$\\mathrm{Mg}^{2+}$$. However, the stability constants of their complexes with EDTA are not similar; the stability constant for the calcium complex is notably higher than that for the magnesium complex. Thus, this statement is incorrect.

\n\n

With all the information above, we can conclude:

\n

(A) is correct, (B) is correct, (C) is correct, (D) is correct, and (E) is incorrect.

\n

Therefore, the correct statements are (A), (B), (C), and (D), making Option D—(A), (B), (C) only—the correct choice.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 879, "subject": "Chemistry", "question": "

Given below are two statements:

\n

Statement I: $$\\mathrm{N}\\left(\\mathrm{CH}_3\\right)_3$$ and $$\\mathrm{P}\\left(\\mathrm{CH}_3\\right)_3$$ can act as ligands to form transition metal complexes.

\n

Statement II: As N and P are from same group, the nature of bonding of $$\\mathrm{N}\\left(\\mathrm{CH}_3\\right)_3$$ and $$\\mathrm{P}\\left(\\mathrm{CH}_3\\right)_3$$ is always same with transition metals.

\n

In the light of the above statements, choose the most appropriate answer from the options given below:

", "options": [ { "text": "Both Statement I and Statement II are incorrect.\n" }, { "text": "Both Statement I and Statement II are correct.\n" }, { "text": "Statement I is incorrect but Statement II is correct.\n" }, { "text": "Statement I is correct but Statement II is incorrect." } ], "answer": "Statement I is correct but Statement II is incorrect.", "solution": "**Answer:** Statement I is correct but Statement II is incorrect.\n\n

Answer: Option D - Statement I is correct but Statement II is incorrect.

\n\n

Explanation:

\n\n

Statement I: $$\\mathrm{N}\\left(\\mathrm{CH}_3\\right)_3$$ (trimethylamine) and $$\\mathrm{P}\\left(\\mathrm{CH}_3\\right)_3$$ (trimethylphosphine) can indeed act as ligands to form transition metal complexes. Ligands are molecules or ions that can donate a pair of electrons to a metal atom to form a coordinate bond. Both nitrogen and phosphorus have lone pairs of electrons that can be donated to transition metals, making $$\\mathrm{N}\\left(\\mathrm{CH}_3\\right)_3$$ and $$\\mathrm{P}\\left(\\mathrm{CH}_3\\right)_3$$ suitable ligands. Therefore, Statement I is correct.

\n\n

Statement II: While N (Nitrogen) and P (Phosphorus) are both from group 15 of the periodic table and share some similar properties, the nature of their bonding with transition metals is not always the same. Nitrogen compounds typically form stronger bonds with metals compared to phosphorus compounds. This difference is due to several factors, including the difference in atomic sizes, electronegativity, and the extent of orbital overlap. Nitrogen, being smaller and more electronegative, can form more effective overlap and stronger bonds with metals compared to phosphorus. Therefore, Statement II is incorrect.

\n\n

Given this analysis, the most appropriate answer is Option D: Statement I is correct but Statement II is incorrect.

", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 880, "subject": "Chemistry", "question": "

Match List I with List II

\n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
LIST I
(Compound)		LIST II
(Colour]
A.$$\\mathrm{Fe}_4\\left[\\mathrm{Fe}(\\mathrm{CN})_6\\right]_3 \\cdot \\mathrm{xH_2O}$$I.Violet
B.$$\\left[\\mathrm{Fe}(\\mathrm{CN})_5 \\mathrm{NOS}\\right]^{4-}$$II.Blood Red
C.$$[\\mathrm{Fe}(\\mathrm{SCN})]^{2+}$$III.Prussian Blue
D.$$\\left(\\mathrm{NH}_4\\right)_3 \\mathrm{PO}_4\\cdot12 \\mathrm{MoO}_3$$IV.Yellow

\n

Choose the correct answer from the options given below:

", "options": [ { "text": "A-IV, B-I, C-II, D-III\n" }, { "text": "A-I, B-II, C-III, D-IV\n" }, { "text": "A-II, B-III, C-IV, D-I\n" }, { "text": "A-III, B-I, C-II, D-IV" } ], "answer": "A-III, B-I, C-II, D-IV", "solution": "**Answer:** A-III, B-I, C-II, D-IV\n\n

To match List I with List II correctly, we need to identify the colors associated with each compound.

\n\n

Let's analyze each compound:

\n\n

1. $$\\mathrm{Fe}_4\\left[\\mathrm{Fe}(\\mathrm{CN})_6\\right]_3 \\cdot \\mathrm{xH_2O}$$ - This is known as Prussian Blue.

\n\n

2. $$\\left[\\mathrm{Fe}(\\mathrm{CN})_5 \\mathrm{NOS}\\right]^{4-}$$ - This complex is known to have a violet color.

\n\n

3. $$[\\mathrm{Fe}(\\mathrm{SCN})]^{2+}$$ - This ion is known for its blood red color.

\n\n

4. $$\\left(\\mathrm{NH}_4\\right)_3 \\mathrm{PO}_4\\cdot12 \\mathrm{MoO}_3$$ - This is commonly known as Ammonium phosphomolybdate, which is yellow.

\n\n

Matching each item, we get:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
LIST I
(Compound)		LIST II
(Colour)
A.$$\\mathrm{Fe}_4\\left[\\mathrm{Fe}(\\mathrm{CN})_6\\right]_3 \\cdot \\mathrm{xH_2O}$$III.Prussian Blue
B.$$\\left[\\mathrm{Fe}(\\mathrm{CN})_5 \\mathrm{NOS}\\right]^{4-}$$I.Violet
C.$$[\\mathrm{Fe}(\\mathrm{SCN})]^{2+}$$II.Blood Red
D.$$\\left(\\mathrm{NH}_4\\right)_3 \\mathrm{PO}_4\\cdot12 \\mathrm{MoO}_3$$IV.Yellow
\n\n

Based on the analysis, the correct matching is:

\n\n

Option D: A-III, B-I, C-II, D-IV

", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 881, "subject": "Chemistry", "question": "One mole of the complex compound Co(NH3)5Cl3, gives 3 moles of ions on dissolution in water. One mole of the same complex reacts with two moles of AgNO3 solution to yield two moles of AgCl (s). The structure of the complex is :", "options": [ { "text": "[Co(NH3)3Cl3]. 2NH3" }, { "text": "[Co(NH3)4Cl2] Cl. NH3" }, { "text": "[Co(NH3)4Cl] Cl2. NH3" }, { "text": "[Co(NH3)5Cl] Cl2" } ], "answer": "[Co(NH3)5Cl] Cl2", "solution": "**Answer:** [Co(NH3)5Cl] Cl2\n\n$$Co\\,{\\left( {N{H_3}} \\right)_5}C{l_3}\\,\\,\\leftrightharpoons\\,\\,{\\left[ {Co{{\\left( {N{H_3}} \\right)}_5}Cl} \\right]^{ + 2}} + 2C{l^ - }$$\n

$$\\therefore$$ $$\\,\\,\\,\\,$$ Structure is $$\\,\\,\\,\\left[ {Co{{\\left( {N{H_3}} \\right)}_5}Cl} \\right]C{l_2}$$\n

Now $$\\,\\,\\,\\left[ {Co{{\\left( {N{H_3}} \\right)}_5}Cl} \\right]C{l_2} + 2AgN{O_3}$$\n

$$ \\to \\,\\,\\,\\left[ {Co{{\\left( {N{H_3}} \\right)}_5}Cl} \\right]{\\left( {N{O_3}} \\right)_2} + 2AgCl$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 882, "subject": "Chemistry", "question": "Ammonia forms the complex ion [Cu(NH3)4]2+ with copper ions in alkaline solutions but not in acidic solutions. What is the reason for it? ", "options": [ { "text": "In acidic solutions protons coordinate with ammonia molecules forming $$NH^+_4$$ ions and NH3\n molecules are not available" }, { "text": "In alkaline solutions insoluble Cu(OH)2\n is precipitated which is soluble in excess of any alkali" }, { "text": "Copper hydroxide is an amphoteric substance" }, { "text": "In acidic solutions hydration protects copper ions." } ], "answer": "In acidic solutions protons coordinate with ammonia molecules forming $$NH^+_4$$ ions and NH3\n molecules are not available", "solution": "**Answer:** In acidic solutions protons coordinate with ammonia molecules forming $$NH^+_4$$ ions and NH3\n molecules are not available\n\n\"AIEEE ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 883, "subject": "Chemistry", "question": "In the coordination compound, K4[Ni(CN)4], the oxidation state of nickel is :", "options": [ { "text": "0" }, { "text": "+1" }, { "text": "+2" }, { "text": "-1" } ], "answer": "0", "solution": "**Answer:** 0\n\nLet the $$O.$$ No of $$Ni$$ in \n

$$\\,\\,{K_4}\\left[ {Fe{{\\left( {CN} \\right)}_6}} \\right]\\,\\,$$ $$be$$ $$=x$$ then\n

$$4\\left( { + 1} \\right) + x + \\left( { - 1} \\right) \\times 4 = 0$$\n

$$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, \\Rightarrow 4 + x - 4 = 0$$ \n

$$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,$$ $$x=0$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 884, "subject": "Chemistry", "question": "The coordination number of central metal atom in a complex is determined by :", "options": [ { "text": "the number of ligands around a metal ion bonded by sigma bonds" }, { "text": "the number of only anionic ligands bonded to the metal ion" }, { "text": "the number of ligands around a metal ion bonded by sigma and pi- bonds both " }, { "text": "the number of ligands around a metal ion bonded by pi-bonds" } ], "answer": "the number of ligands around a metal ion bonded by sigma bonds", "solution": "**Answer:** the number of ligands around a metal ion bonded by sigma bonds\n\nThe coordination number of central metal atom in a complex is equal to number of monovalent ligands, twice the number of bidentate ligands and so on, around the metal ion bonded by coordinate bonds. Hence coordination number $$=$$ no. of $$\\sigma $$ bonds formed by metals with ligands", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 885, "subject": "Chemistry", "question": "The oxidation state of Cr in [Cr(NH3)4Cl2]+ is :", "options": [ { "text": "+3 " }, { "text": "+2" }, { "text": "+1" }, { "text": "0" } ], "answer": "+3 ", "solution": "**Answer:** +3 \n\nOxidation state of $$Cr$$ in $${\\left[ {Cr{{\\left( {N{H_3}} \\right)}_4}C{l_2}} \\right]^ + }.$$ \n

Let it be $$x,\\,\\,1 \\times x + 4 \\times 0 + 2 \\times \\left( { - 1} \\right) = 1$$\n

Therefore $$\\,\\,\\,$$ $$x=3.$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 886, "subject": "Chemistry", "question": "How many EDTA (ethylenediaminetetraacetic acid) molecules are required to make an octahedral\ncomplex with a Ca2+ ion?", "options": [ { "text": "Six" }, { "text": "Three" }, { "text": "One" }, { "text": "Two" } ], "answer": "One", "solution": "**Answer:** One\n\n$$EDTA$$ has hexadentate four donar $$O$$ atoms and $$2$$ donar $$N$$ atoms and for the formation of octahedral complex one molecule is required", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 887, "subject": "Chemistry", "question": "The coordination number and the oxidation state of the element ‘E’ in the complex [E(en)2(C2O4)]NO2 (where (en) is ethylene diamine) are, respectively,", "options": [ { "text": "6 and 2 " }, { "text": "4 and 2" }, { "text": "4 and 3" }, { "text": "6 and 3" } ], "answer": "6 and 3", "solution": "**Answer:** 6 and 3\n\nIn the given complex we have two bidentate ligands (i.e. $$en$$ and $${C_2}{O_4}$$ ), so coordination number of $$E$$ is $$6$$ $$\\left( {2 \\times 2 + 1 \\times 2 = 6} \\right)$$ \n

Let the oxidation state of $$E$$ in complex be $$x,$$ \n

then $$\\left[ {x + \\left( { - 2} \\right) = 1} \\right]$$ or $$x - 2 = 1$$ \n

or $$\\,\\,\\,\\,$$ $$x=+3,$$ so its oxidation state is $$+3$$ \n

Thus option $$(d)$$ is correct.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 888, "subject": "Chemistry", "question": "A solution containing 2.675g of CoCl3. 6NH3 (molar mass = 267.5 g mol–1) is passed through a cation exchanger. The chloride ions obtained in solution were treated with excess of AgNO3 to give\n4.78 g of AgCl (molar mass = 143.5 g mol–1). The formula of the complex is : (At. Mass of Ag = 108 u)", "options": [ { "text": "[Co(NH3)6]Cl3 " }, { "text": "[CoCl2(NH3)4]Cl " }, { "text": "[CoCl3(NH3)3] " }, { "text": "[CoCl(NH3)5]Cl2 " } ], "answer": "[Co(NH3)6]Cl3 ", "solution": "**Answer:** [Co(NH3)6]Cl3 \n\n$$\\mathop {CoC{l_3}.6N{H_3}}\\limits_{2.675g} \\buildrel \\, \\over\n \\longrightarrow xC{l^ - }$$\n

$$xC{l^ - } + AgN{O_3}\\buildrel \\, \\over\n \\longrightarrow \\mathop {x\\,AgCl \\downarrow }\\limits_{4.78g} $$\n

Number of moles of the complex \n

$$ = {{2.675} \\over {267.5}} = 0.01$$ moles\n

Number of moles of $$AgCl$$ obtained \n

$$ = {{4.78} \\over {143.5}} = 0.03$$ moles \n

$$\\therefore$$ $$\\,\\,\\,$$ Number of moles of $$AgCl$$ obtained \n

$$ = 3 \\times \\,\\,$$ No. of moles of complex\n

$$\\therefore$$ $$\\,\\,\\,$$ $$n = {{0.03} \\over {0.01}} = 3$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 889, "subject": "Chemistry", "question": "The equation which is balanced and represents the correct product(s) is :", "options": [ { "text": "[Mg (H2O)6 ]2+ + (EDTA)4- $$\\buildrel {excess\\,NaOH} \\over\n \\longrightarrow $$ [Mg (EDTA) ]2+ + 6H2O" }, { "text": "CuSO4 + KCN $$\\to$$ K2 [Cu (CN)4] + K2SO4 " }, { "text": "Li2O + 2KCl $$\\to$$ 2LiCl + K2O " }, { "text": "[CoCl(NH3)5]+ + 5H+ $$\\to$$ Co2+ + $$5NH_4^+$$ + Cl-" } ], "answer": "[CoCl(NH3)5]+ + 5H+ $$\\to$$ Co2+ + $$5NH_4^+$$ + Cl-", "solution": "**Answer:** [CoCl(NH3)5]+ + 5H+ $$\\to$$ Co2+ + $$5NH_4^+$$ + Cl-\n\n

(A) [Mg (H2O)6 ]2+ + (EDTA)4- $$\\buildrel {excess\\,NaOH} \\over\n \\longrightarrow $$ [Mg (EDTA) ]2+ + 6H2O

\n

The above equation is incorrect because the product formed would be [Mg (EDTA)]2− .

\n

(B) CuSO4 + KCN $$\\to$$ K2[Cu (CN)4] + K2SO4

\n

The above equation is incorrect. Thus, the correct equation is

\n

2KCN + CuSO4 $$\\to$$ K2SO4 + Cu + (CN)2(Cyanogen Gas)

\n

(C) Li2O + 2KCl $$\\to$$ 2LiCl + K2O

\n

The above equation is incorrect because K2O, a stronger base cannot be generated by a weaker base, Li2O.

\n

(D) [CoCl(NH3)5]+ + 5H+ $$\\to$$ Co2+ + $$5NH_4^+$$ + Cl-

\n

The above equation is correct because amine complexes decomposes under acidic medium. Thus, the complex [CoCl(NH3)5] decomposes under acidic conditions to give ammonium ions.

", "topic": "Algebra", "subtopic": "Equations and Inequalities" }, { "id": 890, "subject": "Chemistry", "question": "Which of the following is an example of homoleptic complex?", "options": [ { "text": "[Co(NH3)6]Cl3" }, { "text": "[Pt(NH3)2Cl2]\n" }, { "text": "[Co(NH3)4Cl2]" }, { "text": "[Co(NH3)5Cl]Cl2\n" } ], "answer": "[Co(NH3)6]Cl3", "solution": "**Answer:** [Co(NH3)6]Cl3\n\n

Homoleptic complexes are those compounds in which all the ligands bound to the central metal are identical. Thus, these types of complexes have only one type of ligands. In the complex [Co(NH3)6]Cl3 , central atom Co has ammonia ligands as all six.

\n\n

Heteroleptic complexes are those compounds in which central transition metal has more than one type of ligands. So, rest of the options are examples of heteroleptic complexes.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 891, "subject": "Chemistry", "question": "[Co2(CO)8] displays :", "options": [ { "text": "one Co−Co bond, six terminal CO and two bridging CO" }, { "text": "one Co−Co bond, four terminal CO and four bridging CO" }, { "text": "no Co−Co bond, six terminal CO and two bridging CO" }, { "text": "no Co−Co bond, four terminal CO and four bridging CO\n" } ], "answer": "one Co−Co bond, six terminal CO and two bridging CO", "solution": "**Answer:** one Co−Co bond, six terminal CO and two bridging CO\n\n

The structure of [Co2(CO)8] is

\n\"JEE\n

From the structure, we can see that there is one CO−CO bond, six terminals CO and two bridging CO.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 892, "subject": "Chemistry", "question": "The oxidation states of Cr in [Cr(H2O)6]Cl3, [Cr(C6H6)2] and K2[Cr(CN)2(O)2(O2)(NH3)] respectively are :", "options": [ { "text": "+3, 0 and +4" }, { "text": "+3, +4 and +6 " }, { "text": "+3, +2 and +4" }, { "text": "+3, 0 and +6" } ], "answer": "+3, 0 and +6", "solution": "**Answer:** +3, 0 and +6\n\nAssume oxidation state of Cr in all the compounds = x \n

(i)$$\\,\\,\\,$$ In [Cr(H2O)6] Cl3 oxidation state of Cr is \n

x + 0 $$ \\times $$ 6 + ($$-$$1 $$ \\times $$ ) = O\n

$$ \\Rightarrow \\,\\,\\,\\,\\,$$ x + 0 $$-$$ 3 = O\n

$$ \\Rightarrow \\,\\,\\,\\,$$ x = + 3\n

(ii)$$\\,\\,\\,$$ [Cr (C6 H6)2] oxidation state of Cr is \n

x + 0 $$ \\times $$ 2 = 0\n

$$ \\Rightarrow \\,\\,\\,\\,$$ x = 0\n

(iii) $$\\,\\,\\,$$ In K2 [ Cr(CN)2 (O)2 (O2) (NH3)] oxidation state of Cr is \n

1 $$ \\times $$ 2 + x + ($$-$$ 1 $$ \\times $$ 2) + ($$-$$2 $$ \\times $$ 2) + ($$-$$2) + 0 = 0\n

$$ \\Rightarrow \\,\\,\\,\\,$$ x = + 6\n

$$\\therefore\\,\\,\\,$$ + 3, 0 and + 6 is the correct answer. \n

Note : \n

O2 molecule can have 0, $$-$$ 1, $$-$$ 2 oxidation state but in K2 [ Cr (CN)2 (O)2 (O2) NH3 ] if we choose zero as the oxidation state of O2 then for Cr oxidation state will be $$+$$ 4. But + 4 oxidation state of Cr is unstable and $$+$$ 6 is most stable that is why we choose $$-$$ 2 oxidation state of O2. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 893, "subject": "Chemistry", "question": "The coordination number of Th in K4[Th(C2O4)4(OH2)2] is :\n
(C2O$${_4^{2 - }}$$ = Oxalato)", "options": [ { "text": "14" }, { "text": "10" }, { "text": "8" }, { "text": "6" } ], "answer": "10", "solution": "**Answer:** 10\n\nOxalato (C2O42–) is a bidentate and H2O is unidentate\nligand.\n

4C2O42– creates 8 covalent bonds.\n

2H2O creates 2 covalent bonds.\n

$$ \\therefore $$ Around Th 10 coordinate covalent bonds will be present.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 894, "subject": "Chemistry", "question": "The coordination numbers of Co and Al in [Co(Cl)(en)2]Cl and K3[Al(C2O4)3], respectively, are :\n(en = ethane-1, 2-diamine) ", "options": [ { "text": "3 and 3" }, { "text": "6 and 6" }, { "text": "5 and 3" }, { "text": "5 and 6" } ], "answer": "5 and 6", "solution": "**Answer:** 5 and 6\n\nHere in [Co(Cl)(en)2]Cl\n
'Cl' is monodentate so one coordinate linkage will be made with Co.\n
'en' is bidentate so two coordinate linkage will be made with Co. There are two 'en ' present so 4 coordinate linkage will be made with Co.\n
$$ \\therefore $$ Total 5 coordinate linkage will be made with Co by Cl and en. So C.N. of Co is 5.\n

Here in K3[Al(C2O4)3]\n
C2O4-2 is bidentate so two coordinate linkage will be made with Al. There are three C2O4-2' present so 6 coordinate linkage will be made with Al.\n
$$ \\therefore $$ So C.N. of Al is 6.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 895, "subject": "Chemistry", "question": "Complexes (ML5) of metals Ni and Fe have\nideal square pyramidal and trigonal\nbipyramidal grometries, respectively. The sum\nof the 90°, 120° and 180° L-M-L angles in the\ntwo complexes is ________.", "options": [], "answer": "20", "solution": "**Answer:** 20\n\n\"JEE\n
$$\\angle $$90o = 6\n

$$\\angle $$120o = 3\n

$$\\angle $$180o = 1\n

Total = 10\n\"JEE\n
$$\\angle $$90o = 8\n

$$\\angle $$180o = 2\n

Total = 10\n

$$ \\therefore $$ Total number of 180o, 90o and 120o L-M-L bond\nangles = 10 + 10 = 20", "topic": "Geometry", "subtopic": "Plane Geometry" }, { "id": 896, "subject": "Chemistry", "question": "The oxidation states of iron atoms in\ncompounds (A), (B) and (C), respectively, are x,\ny and z. The sum of x, y and z is ________.\n

Na4[Fe(CN)5(NOS)]\n
       (A)\n

Na4[FeO4]\n
       (B)\n

[Fe2(CO)9]\n
       (C)", "options": [], "answer": "6", "solution": "**Answer:** 6\n\nNa4[Fe(CN)5(NOS)]\n

Let the O.S. of Fe be x\n

OS of CN = –1\n
OS of NOS = –1\n

$$ \\therefore $$ (+1)4 + x + (–1)5 + (–1)1 = 0\n

$$ \\Rightarrow $$ x = +2\n

Na4[FeO4]\n

Let O.S. of Fe be y\n

(+1)4 + y + (–2)4 = 0\n

$$ \\Rightarrow $$ y = +4\n

[Fe2(CO)9]\n

Let O.S. of Fe be z\n

2z + 0 × 9 = 0\n

$$ \\Rightarrow $$ z = 0\n

so (x + y + z) = +2 + 4 + 0 = 6", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 897, "subject": "Chemistry", "question": "The total number of coordination sites
in ethylenediaminetetraacetate (EDTA4–) is _____.", "options": [], "answer": "6", "solution": "**Answer:** 6\n\n[EDTA]4– is ethylenediaminetetraacetate anion.\nIt is a hexadentate ligand.\n\"JEE\n

It has six co-ordination sites.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 898, "subject": "Chemistry", "question": "Number of bridging CO ligands in [Mn2(CO)10] is __________.", "options": [], "answer": "0", "solution": "**Answer:** 0\n\n\"JEE\n

$$ \\therefore $$ Number of bridging CO ligands = 0.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 899, "subject": "Chemistry", "question": "The total number of unpaired electrons present in the complex K3[Cr(oxalate)3] is _____________.", "options": [], "answer": "3", "solution": "**Answer:** 3\n\nIn $${K_3}\\left[ {Cr\\left( {{C_2}{O_4}} \\right)} \\right]$$, oxidation number of Cr :

$$3( + 1) + x + 3( - 2) = 0$$

$$ \\Rightarrow x = + 3$$

$$ \\therefore $$ $${}_{24}C{r^{ + 3}} = \\left[ {Ar} \\right]3{d^3}$$

$$ \\therefore $$ Number of unpaired electrons = 3", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 900, "subject": "Chemistry", "question": "The secondary valency and the number of hydrogen bonded water molecule(s) in CuSO4 . 5H2O, respectively, are :", "options": [ { "text": "5 and 1" }, { "text": "4 and 1" }, { "text": "6 and 5" }, { "text": "6 and 4" } ], "answer": "4 and 1", "solution": "**Answer:** 4 and 1\n\nCuSO4.5H2O $$ \\Rightarrow $$ [Cu(H2O)4]SO4.H2O

\n In CuSO4.5H2O, Cu is co-ordinated with 4 water molecule and two more oxygen atom from sulphate ion and fifth water molecule is hydrogen bonded.

\n So secondary valency = 4

\n No. of hydrogen bond per molecule = 1", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 901, "subject": "Chemistry", "question": "Three moles of AgCl get precipitated when one mole of an octahedral co-ordination compound with empirical formula CrCl3.3NH3.3H2O reacts with excess of silver nitrate. The number of chloride ions satisfying the secondary valency of the metal ion is ______________.", "options": [], "answer": "0", "solution": "**Answer:** 0\n\nMole of AgCl precipitated is equal to mole of Cl- present in ionization sphere.

\"JEE

Since none of Cl- is present in the co-ordination sphere. Therefore answer is zero.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 902, "subject": "Chemistry", "question": "Which one of the following metal complexes is most stable?", "options": [ { "text": "[Co(en)(NH3)4]Cl2" }, { "text": "[Co(en)3]Cl2" }, { "text": "[Co(en)2(NH3)2]Cl2" }, { "text": "[Co(NH3)6]Cl2" } ], "answer": "[Co(en)3]Cl2", "solution": "**Answer:** [Co(en)3]Cl2\n\nComplex [Co(en)3]Cl2 is most stable complex among the given complex compounds because more number of chelate rings are present in this complex as compare to others.

(1) [Co(en)(NH3)4]Cl2 1 chelate ring

(2) [Co(en)3]Cl2 3 chelate ring

(3) [Co(en)2(NH3)2]Cl2 2 chelate ring

(4) [Co(NH3)6]Cl2 0 chelate ring ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 903, "subject": "Chemistry", "question": "3 moles of metal complex with formula Co(en)2Cl3 gives 3 moles of silver chloride on treatment with excess of silver nitrate. The secondary valency of Co in the complex is ___________.

(Round off to the nearest integer)", "options": [], "answer": "6", "solution": "**Answer:** 6\n\n$$3\\left[ {Co{{(en)}_2}C{l_2}} \\right]Cl + \\mathop {AgN{o_3}}\\limits_{(excess)} \\to \\mathop {3AgCl}\\limits_{(white\\,ppt.)} $$

Secondary valency of Co = 6

(C. N.)", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 904, "subject": "Chemistry", "question": "The ratio of number of water molecules in Mohr's salt and potash alum is ____________ $$\\times$$ 10$$-$$1. (Integer answer)", "options": [], "answer": "5", "solution": "**Answer:** 5\n\nMohr's salt : (NH4)2 Fe(SO4)2 . 6H2O

The number of water molecules in Mohr's salt = 6

Potash alum : KAl(SO4)2 . 12H2O

The number of water molecules in potash alum = 12

So ratio of number of water molecules in Mohr's salt and potash alum

$$ = {6 \\over {12}}$$

$$ = {1 \\over 2}$$

= 0.5

= 5 $$\\times$$ 10$$-$$1", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 905, "subject": "Chemistry", "question": "The denticity of an organic ligand, biuret is :", "options": [ { "text": "2" }, { "text": "4" }, { "text": "3" }, { "text": "6" } ], "answer": "2", "solution": "**Answer:** 2\n\nBiuret ligand is\n
\"JEE\n
It forms complexes like\n

\"JEE\n

The denticity of organic ligand is 2.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 906, "subject": "Chemistry", "question": "The sum of oxidation states of two silver ions in [Ag(NH3)2] [Ag(CN)2] complex is _____________.", "options": [], "answer": "2", "solution": "**Answer:** 2\n\n[Ag(NH3)2][Ag(CN)2] complex dissociates into [Ag(NH3)2]+ and [Ag(CN)2].

Oxidation of Ag in [Ag(NH3)2]+

Ag + 0 $$\\times$$ 2 = + 1

Ag = + 1

Oxidation state of Ag in [Ag(CN)2]$$-$$

Ag + ($$-$$1) $$\\times$$ 2 = $$-$$ 1

Ag $$-$$ 2 = $$-$$ 1

$$\\Rightarrow$$ Ag = + 1

$$\\therefore$$ Sum of oxidation states of two silver ions in [Ag(NH3)2][Ag(CN)2] complex is 2.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 907, "subject": "Chemistry", "question": "

Given below are two statements.

\n

$$\\bullet$$ Statement I : In CuSO4 . 5H2O, Cu-O bonds are present.

\n

$$\\bullet$$ Statement II : In CuSO4 . 5H2O, ligands coordinating with Cu(II) ion are O-and S-based ligands.

\n

In the light of the above statements, choose the correct answer from the options given below.

", "options": [ { "text": "Both Statement I and Statement II are correct." }, { "text": "Both Statement I and Statement II are incorrect." }, { "text": "Statement I is correct but Statement II is incorrect." }, { "text": "Statement I is incorrect but Statement II is correct." } ], "answer": "Statement I is correct but Statement II is incorrect.", "solution": "**Answer:** Statement I is correct but Statement II is incorrect.\n\n

Statement I is true but statement II is false. Only\noxygen atom forms a Co-ordinate bond with Cu+2 in\nCuSO4.5H2O

\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 908, "subject": "Chemistry", "question": "

In the cobalt-carbonyl complex : [Co2(CO)8], number of Co-Co bonds is \"X\" and terminal CO ligands is \"Y\". X + Y = ___________.

", "options": [], "answer": "7", "solution": "**Answer:** 7\n\n

\"JEE

\n

x = 1

\n

y = 6

\n

$$\\therefore$$ x + y = 7

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 909, "subject": "Chemistry", "question": "

The conductivity of a solution of complex with formula $$\\mathrm{CoCl}_{3}\\left(\\mathrm{NH}_{3}\\right)_{4}$$ corresponds to 1 : 1 electrolyte, then the primary valency of central metal ion is __________.

", "options": [], "answer": "3", "solution": "**Answer:** 3\n\nIn 1: 1 type of electrolyte the ions have $+1$ and $-1$ charge on them

\n$\\therefore$ Possible compound is $\\rightarrow\\left[\\mathrm{Co}\\left(\\mathrm{NH}_{3}\\right)_{4} \\mathrm{Cl}_{2}\\right]^{+} \\mathrm{Cl}^{-}$\n

Oxidation state of central atom represents the total number of primary valency\n

\n$\\therefore$ Primary valency will be 3 .", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 910, "subject": "Chemistry", "question": "

Low oxidation state of metals in their complexes are common when ligands :

", "options": [ { "text": "have good $$\\pi$$-accepting character" }, { "text": "have good $$\\sigma$$-donor character" }, { "text": "are having good $$\\pi$$-donating ability" }, { "text": "are having poor $$\\sigma$$-donating ability" } ], "answer": "have good $$\\pi$$-accepting character", "solution": "**Answer:** have good $$\\pi$$-accepting character\n\nLigands like : CO, are sigma donor and $\\pi$-acceptor and they make stronger bond with lower oxidation state metal ion, in this case back bonding is more effective", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 911, "subject": "Chemistry", "question": "

Which of the following are the example of double salt?

\n

A. $$\\mathrm{FeSO}_{4} \\cdot\\left(\\mathrm{NH}_{4}\\right)_{2} \\mathrm{SO}_{4} \\cdot 6 \\mathrm{H}_{2} \\mathrm{O}$$

\n

B. $$\\mathrm{CuSO}_{4}\\cdot 4 \\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}$$

\n

C. $$\\mathrm{K}_{2} \\mathrm{SO}_{4} \\cdot \\mathrm{Al}_{2}\\left(\\mathrm{SO}_{4}\\right)_{3} \\cdot 24 \\mathrm{H}_{2} \\mathrm{O}$$

\n

D. $$\\mathrm{Fe}(\\mathrm{CN})_{2}\\cdot4 \\mathrm{KCN}$$

\n

Choose the correct answer :

", "options": [ { "text": "A, B and D only" }, { "text": "B and D only" }, { "text": "A and B only" }, { "text": "A and C only" } ], "answer": "A and C only", "solution": "**Answer:** A and C only\n\n$A=\\mathrm{FeSO}_{4} \\cdot\\left(\\mathrm{NH}_{4}\\right)_{2} \\mathrm{SO}_{4} \\cdot 6 \\mathrm{H}_{2} \\mathrm{O}$ - double salt\n

B. $\\mathrm{CuSO}_{4} \\cdot 4 \\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}$\n$=\\left[\\mathrm{Cu}\\left(\\mathrm{NH}_{3}\\right)_{4}\\right] \\mathrm{SO}_{4} \\cdot \\mathrm{H}_{2} \\mathrm{O} $ - complex salt\n

C. $\\mathrm{K}_{2} \\mathrm{SO}_{4} \\cdot \\mathrm{Al}_{2}\\left(\\mathrm{SO}_{4}\\right)_{3} \\cdot 24 \\mathrm{H}_{2} \\mathrm{O}$ - double salt\n

D. $\\mathrm{Fe}(\\mathrm{CN})_{2} \\cdot 4 \\mathrm{KCN}$ = \n$\\mathrm{K}_{4}\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}\\right]$\n- complex salt\n

Option A, $\\mathrm{FeSO}_{4} \\cdot\\left(\\mathrm{NH}_{4}\\right)_{2} \\mathrm{SO}_{4} \\cdot 6 \\mathrm{H}_{2} \\mathrm{O}$, is a double salt because it is formed by the combination of two different salts: iron(II) sulfate ($\\mathrm{FeSO}_{4}$) and ammonium sulfate ($\\left(\\mathrm{NH}_{4}\\right)_{2} \\mathrm{SO}_{4}$).\n

Option B, $\\mathrm{CuSO}_{4}\\cdot 4 \\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}$, is a complex salt because it contains a central copper ion coordinated to several ammonia molecules.\n\n

In this compound, the copper(II) ion (Cu2+) acts as the central metal ion, while the four ammonia molecules (NH3) act as ligands, coordinating to the copper ion through their lone pairs of electrons. The sulfate ion (SO42-) and water molecules (H2O) are not involved in the coordination sphere and simply crystallize along with the complex.\n

Option C, $\\mathrm{K}_{2} \\mathrm{SO}_{4} \\cdot \\mathrm{Al}_{2}\\left(\\mathrm{SO}_{4}\\right)_{3} \\cdot 24 \\mathrm{H}_{2} \\mathrm{O}$, is also a double salt, formed by the combination of two different salts: potassium sulfate ($\\mathrm{K}_{2} \\mathrm{SO}_{4}$) and aluminum sulfate ($\\mathrm{Al}_{2}\\left(\\mathrm{SO}_{4}\\right)_{3}$).\n\n

The 24 water molecules in the formula are not directly involved in the salt formation but instead function to stabilize the crystal lattice by forming hydrogen bonds with the sulfate ions and other water molecules.\n\n

Option D, $\\mathrm{Fe}(\\mathrm{CN})_{2}\\cdot4 \\mathrm{KCN}$, is not an example of a double salt. It is a complex salt, formed by the coordination of iron(II) ions to cyanide ligands and potassium ions.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 912, "subject": "Chemistry", "question": "

The primary and secondary valencies of cobalt respectively in $$\\mathrm{[Co(NH_3)_5Cl]Cl_2}$$ are :

", "options": [ { "text": "2 and 8" }, { "text": "2 and 6" }, { "text": "3 and 5" }, { "text": "3 and 6" } ], "answer": "3 and 6", "solution": "**Answer:** 3 and 6\n\n

$$\\mathrm{[Co(NH_3)_5Cl]Cl_2}$$

\n

x + 5 (0) – 1 = + 2 $$ \\Rightarrow $$ x = + 3

\n

Secondary valency is the number of\nligands in the complex which is equal to 6.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 913, "subject": "Chemistry", "question": "

The set which does not have ambidentate ligand(s) is :

", "options": [ { "text": "$$\\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-}, \\mathrm{NO}_{2}{ }^{-}, \\mathrm{NCS}^{-}$$" }, { "text": "$$\\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-}$$, ethylene diammine, $$\\mathrm{H}_{2} \\mathrm{O}$$" }, { "text": "$$\\mathrm{NO}_{2}^{-}, \\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-}$$, EDTA $$^{4-}$$" }, { "text": "$$\\mathrm{EDTA}^{4-}, \\mathrm{NCS}^{-}, \\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-}$$" } ], "answer": "$$\\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-}$$, ethylene diammine, $$\\mathrm{H}_{2} \\mathrm{O}$$", "solution": "**Answer:** $$\\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-}$$, ethylene diammine, $$\\mathrm{H}_{2} \\mathrm{O}$$\n\nOption A :\n

- $$\\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-}$$ (Oxalate) is a bidentate ligand. It always binds through its two oxygen atoms.\n

- $$\\mathrm{NO}_{2}{ }^{-}$$ (Nitrite) is an ambidentate ligand. It can coordinate either through nitrogen or oxygen.\n

- $$\\mathrm{NCS}^{-}$$ (Thiocyanate) can also bind through either nitrogen or sulfur, making it an ambidentate ligand.\n\n

Option B :\n

- $$\\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-}$$ (Oxalate) as mentioned earlier, is a bidentate ligand.\n

- Ethylene diamine (en) is also a bidentate ligand, as it always binds through its two nitrogen atoms.\n

- $$\\mathrm{H}_{2} \\mathrm{O}$$ (Water) is a monodentate ligand, binding only through one oxygen atom.\n\n

Option C :\n

- $$\\mathrm{NO}_{2}^{-}$$ as mentioned, is an ambidentate ligand.\n

- $$\\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-}$$ as mentioned, is a bidentate ligand.\n

- $$\\mathrm{EDTA}^{4-}$$ is a hexadentate ligand. It always binds through six donor atoms and is not ambidentate.\n\n

Option D :\n

- $$\\mathrm{EDTA}^{4-}$$ as mentioned, is a hexadentate ligand.\n

- $$\\mathrm{NCS}^{-}$$ as mentioned, is an ambidentate ligand.\n

- $$\\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-}$$ as mentioned, is a bidentate ligand.\n\n

From the above analysis, it's clear that Option B is the one that does not contain any ambidentate ligands. So, the correct answer is Option B.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 914, "subject": "Chemistry", "question": "

Number of ambidentate ligands in a representative metal complex $$\\left[\\mathrm{M}(\\mathrm{en})(\\mathrm{SCN})_{4}\\right]$$ is ___________.

\n

[en = ethylenediamine]

", "options": [], "answer": "4", "solution": "**Answer:** 4\n\n

Ambidentate ligands are ligands that can bond to a metal atom through two different atoms. They can attach through one site or the other, but not both at the same time.

\n

In the given complex $$[\\mathrm{M}(\\mathrm{en})(\\mathrm{SCN})_{4}]$$:

\n\n

Since there are four $$\\mathrm{SCN}^{-}$$ ligands in the given complex, the number of ambidentate ligands is 4.

\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 915, "subject": "Chemistry", "question": "

Yellow compound of lead chromate gets dissolved on treatment with hot $$\\mathrm{NaOH}$$ solution. The product of lead formed is a :

", "options": [ { "text": "Tetraanionic complex with coordination number six\n" }, { "text": "Neutral complex with coordination number four\n" }, { "text": "Dianionic complex with coordination number six\n" }, { "text": "Dianionic complex with coordination number four" } ], "answer": "Dianionic complex with coordination number four", "solution": "**Answer:** Dianionic complex with coordination number four\n\n

Lead chromate (\n$$\\mathrm{PbCrO}_4$$\n) dissolves in hot NaOH solution to form products due to a chemical reaction. The reaction involves the formation of a compound where lead (\n$$\\mathrm{Pb}$$\n) is coordinated by hydroxide ions (\n$$\\mathrm{OH}^-$$\n).

\n

The balanced chemical reaction is:

\n

\n$$\\mathrm{PbCrO}_4 + 4\\mathrm{NaOH} (hot, excess) \\rightarrow \\left[\\mathrm{Pb}(\\mathrm{OH})_4\\right]^{2-} + \\mathrm{Na}_2\\mathrm{CrO}_4$$\n

\n

In this reaction, lead forms a dianionic complex (\n$$\\left[\\mathrm{Pb}(\\mathrm{OH})_4\\right]^{2-}$$\n) with hydroxide ions. This indicates that the lead is in the center of a complex with a coordination number of four, meaning it is directly bound to four hydroxide ions. Therefore, the correct description of the product formed with lead is a dianionic complex with a coordination number of four.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 916, "subject": "Chemistry", "question": "

The coordination geometry around the manganese in decacarbonyldimanganese $$(0)$$ is

", "options": [ { "text": "Trigonal bipyramidal\n" }, { "text": "Square pyramidal\n" }, { "text": "Square planar\n" }, { "text": "Octahedral" } ], "answer": "Octahedral", "solution": "**Answer:** Octahedral\n\n

$$\\mathrm{Mn_2(CO)_{10}}$$

\n

\"JEE

", "topic": "Geometry", "subtopic": "Plane Geometry" }, { "id": 917, "subject": "Chemistry", "question": "

An octahedral complex with the formula $$\\mathrm{CoCl}_3 \\cdot \\mathrm{nNH}_3$$ upon reaction with excess of $$\\mathrm{AgNO}_3$$ solution gives 2 moles of $$\\mathrm{AgCl}$$. Consider the oxidation state of $$\\mathrm{Co}$$ in the complex is '$$x$$'. The value of \"$$x+n$$\" is __________.

", "options": [ { "text": "6" }, { "text": "5" }, { "text": "3" }, { "text": "8" } ], "answer": "8", "solution": "**Answer:** 8\n\n

To solve this problem, we need to determine the oxidation state of cobalt ($$x$$) and the number of ammonia molecules ($$n$$) in the complex $$\\mathrm{CoCl}_3 \\cdot \\mathrm{nNH}_3$$, given that it produces 2 moles of $$\\mathrm{AgCl}$$ upon reaction with excess $$\\mathrm{AgNO}_3$$.

\n\n

First, let's write the reaction between the complex and $$\\mathrm{AgNO}_3$$.

\n\n

The precipitate formation indicates that some chloride ions are free (not coordinated to the metal). Since 2 moles of $$\\mathrm{AgCl}$$ are formed, it indicates that there are 2 chloride ions that are free to react with $$\\mathrm{AgNO}_3$$.

\n\n

Therefore, we can conclude that in the complex, 1 chloride ion is coordinated to the cobalt, and 2 chloride ions are free. This can be represented as:

\n\n

$$\\mathrm{[Co(NH_3)_{n}Cl]Cl_2}$$

\n\n

Now, let's determine the oxidation state of cobalt (Co). The charge on the entire complex should be zero, so we can write the charge balance equation as follows:

\n\n

The sum of the charges is:

\n\n

$$x + (0 \\cdot n) + (-1 \\cdot 1) + (-1 \\cdot 2) = 0$$

\n\n

Solving this, we get:

\n\n

$$x - 1 - 2 = 0$$

\n\n

$$x - 3 = 0$$

\n\n

$$x = 3$$

\n\n

So, the oxidation state of Co is 3.

\n\n

Now, we need to find the value of $$n$$. Since the coordination number of Co in an octahedral complex is typically 6 and we have 1 chloride ion coordinated to Co, the number of ammonia molecules coordinated to Co would be:

\n\n

$$n = 6 - 1 = 5$$

\n\n

Finally, we calculate $$x + n$$:

\n\n

$$x + n = 3 + 5 = 8$$

\n\n

So, the value of \"$$x + n$$\" is 8.

\n\n

Therefore, the correct option is:

\n\n

Option D: 8

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 918, "subject": "Chemistry", "question": "The most stable ion is :", "options": [ { "text": "[Fe(OH)3]3-" }, { "text": "[Fe(Cl)6]3-" }, { "text": "[Fe(CN)6]3-" }, { "text": "[Fe(H2O)6]3+" } ], "answer": "[Fe(CN)6]3-", "solution": "**Answer:** [Fe(CN)6]3-\n\nThe cyano and hydroxo complexes are far more stable than those formed by halide ion. This is due to the fact that $$C{N^ - }$$ and $$O{H^ - }$$ are strong lewis bases (nucleophiles). Further $${\\left[ {Fe{{\\left( {OH} \\right)}_5}} \\right]^{3 - }}$$ is not formed. hence most stable ion is $${\\left[ {Fe{{\\left( {CN} \\right)}_6}} \\right]^{3 - }}$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 919, "subject": "Chemistry", "question": "The value of the ‘spin only’ magnetic moment for one of the following configurations\nis 2.84 BM. The correct one is :", "options": [ { "text": "d4 (in strong ligand field)" }, { "text": "d4 (in weak ligand field) " }, { "text": "d3 (in weak as well as in strong fields)" }, { "text": "d5 (in strong ligand field) " } ], "answer": "d4 (in strong ligand field)", "solution": "**Answer:** d4 (in strong ligand field)\n\n\"AIEEE ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 920, "subject": "Chemistry", "question": "In which of the following octahedral complexes of Co (at. no. 27), will the magnitude of $$\\Delta _o$$ be the highest?", "options": [ { "text": "[Co(CN)6]3− " }, { "text": "[Co(C2O4)3]3− " }, { "text": "[Co(H2O)6]3+ " }, { "text": "[Co(NH3)6]3+" } ], "answer": "[Co(CN)6]3− ", "solution": "**Answer:** [Co(CN)6]3− \n\nIn octahedral complex the magnitude of $${\\Delta _o}$$ will be highest in a complex having strongest ligand. Of the given ligands $$C{N^ - }$$ is strongest so $${\\Delta _o}$$ will be highest for $${\\left( {Co{{\\left( {CN} \\right)}_6}} \\right)^{3 - }}.$$ Thus option $$(a)$$ is correct.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 921, "subject": "Chemistry", "question": "The octahedral complex of a metal ion M3+ with four monodentate ligands L1, L2, L3 and L4 absorb wavelengths in the region of red, green, yellow and blue, respectively. The increasing order of ligand strength of the four ligands is :", "options": [ { "text": "L3 < L2 < L4 < L1" }, { "text": "L1 < L2 < L4 < L3" }, { "text": "L4 < L3 < L2 < L1" }, { "text": "L1 < L3 < L2 < L4" } ], "answer": "L1 < L3 < L2 < L4", "solution": "**Answer:** L1 < L3 < L2 < L4\n\n\"JEE \n

For a given metal ion, weak filed ligands create a complex with smaller $$\\Delta $$, which will absorbs light of longer $$\\lambda $$ and thus lower frequency. Conservely, stronger filled ligands create a larger $$\\Delta ,$$ absorb light of shorter $$\\lambda $$ and thus higher $$v.i.e.$$ higher energy.\n

$$\\mathop {{\\mathop{\\rm Re}\\nolimits} d}\\limits_{\\lambda = 650\\,nm} \\,\\,\\mathop { < Yellow}\\limits_{570\\,\\,nm} \\,\\,\\mathop { < Green}\\limits_{490\\,nm} < \\mathop {Blue}\\limits_{450\\,nm} $$ \n

So order of ligand strength is \n

$${L_1} < {L_3} < {L_2} < {L_4}$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 922, "subject": "Chemistry", "question": "Which of the following compounds is not colored yellow?", "options": [ { "text": "K3[Co(NO2)6]" }, { "text": "(NH4)3[As(Mo3O10)4]" }, { "text": "BaCrO4" }, { "text": "Zn2[Fe(CN)6]" } ], "answer": "Zn2[Fe(CN)6]", "solution": "**Answer:** Zn2[Fe(CN)6]\n\nZn2[Fe(CN)6] is white in color as it does not have unpaired\nelectrons.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 923, "subject": "Chemistry", "question": "Identify the correct trend given below : (Atomic No.=Ti : 22, Cr : 24 and Mo : 42)\n", "options": [ { "text": "$$\\Delta $$o of [Cr(H2O)6]2+ >\n

[Mo(H2O)6]2+ and\n

$$\\Delta $$o of [Ti(H2O)6]3+ > [Ti(H2O)6]2+" }, { "text": "$$\\Delta $$o of [Cr(H2O)6]2+ >\n

[Mo(H2O)6]2+ and\n

$$\\Delta $$o of [Ti(H2O)6]3+ < [Ti(H2O)6]2+" }, { "text": "$$\\Delta $$o of [Cr(H2O)6]2+ \n

< [Mo(H2O)6]2+ and\n

$$\\Delta $$o of [Ti(H2O)6]3+ > [Ti(H2O)6]2+" }, { "text": "$$\\Delta $$o of [Cr(H2O)6]2+ \n

< [Mo(H2O)6]2+ and\n

$$\\Delta $$o of [Ti(H2O)6]3+ < [Ti(H2O)6]2+" } ], "answer": "$$\\Delta $$o of [Cr(H2O)6]2+ \n

< [Mo(H2O)6]2+ and\n

$$\\Delta $$o of [Ti(H2O)6]3+ > [Ti(H2O)6]2+", "solution": "**Answer:** $$\\Delta $$o of [Cr(H2O)6]2+ \n

< [Mo(H2O)6]2+ and\n

$$\\Delta $$o of [Ti(H2O)6]3+ > [Ti(H2O)6]2+\n\n

$$\\Delta$$0 of complex [Mo(H2O)6]2+ is greater than that of [Cr(H2O)6]2+. It is due to the fact that Mo2+ belongs to second row transition element while Cr2+ is first row transition element.

\n

Magnitude of crystal field splitting energy ($$\\Delta$$0) depends on the charge on the metal ion. Greater the charge, greater is the crystal field splitting energy. Therefore, [Ti(H2O)6]3+ > [Ti(H2O)6]2+.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 924, "subject": "Chemistry", "question": "Two complexes [Cr(H2O)6]Cl3 (A) and [Cr(NH3)6]Cl3 (B) are violet and yellow coloured, respectively. The incorrect statement regarding them is : ", "options": [ { "text": "$$\\Delta $$0 values of (A) and (B) are calculated from the energies of violet and yellow light, respectively " }, { "text": "both are paramagnetic with three unpaired electrons" }, { "text": "both absorb energies corresponding to their complementry colors. " }, { "text": "$$\\Delta $$0 value for (A) is less than that of (B). " } ], "answer": "$$\\Delta $$0 values of (A) and (B) are calculated from the energies of violet and yellow light, respectively ", "solution": "**Answer:** $$\\Delta $$0 values of (A) and (B) are calculated from the energies of violet and yellow light, respectively \n\nNH3 is a strong field ligand.\n

H2O is weak field ligand.\n

Let $$\\Delta $$violet is the crystal field splitting energy of complex A and $$\\Delta $$yellow is for complex B. \n

$$ \\therefore $$   $$\\Delta $$yellow > $$\\Delta $$violet.\n

We know, $$\\Delta $$ = $${{hc} \\over {{\\lambda _{absorb}}}}$$\n

Violet color will absorb yellow color and yellow color will absorb violet color. \n

$$ \\therefore $$   $$\\Delta $$yellow = $${{hc} \\over {{\\lambda _{violet}}}}$$\n

and $$\\Delta $$violet = $${{hc} \\over {{\\lambda _{yellow}}}}$$\n

So, $$\\Delta $$o of A is calculated from energy of yellow light and $$\\Delta $$o of B is calculated from energy of violet light. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 925, "subject": "Chemistry", "question": "The crystal field stabilization energy (CFSE) of [Fe(H2O)6]Cl2 and K2[NiCl4] respectively, are : ", "options": [ { "text": "– 0.4 $$\\Delta $$0 and – 0.8 $$\\Delta $$t" }, { "text": "– 0.6 $$\\Delta $$0 and – 0.8 $$\\Delta $$t" }, { "text": "– 2.4 $$\\Delta $$0 and – 1.2 $$\\Delta $$t" }, { "text": "– 0.4 $$\\Delta $$0 and – 1.2 $$\\Delta $$t" } ], "answer": "– 0.4 $$\\Delta $$0 and – 0.8 $$\\Delta $$t", "solution": "**Answer:** – 0.4 $$\\Delta $$0 and – 0.8 $$\\Delta $$t\n\nCN of [Fe(H2O)6]Cl2 = 6\n

For CN = 6, CFSE = $$\\left[ {{3 \\over 5} \\times n - {2 \\over 5} \\times {n_1}} \\right]{\\Delta _0}$$\n

Here, n = number of electron in eg and \n
n1 = number of electron in t2g\n

Electronic configuration of Fe+2 according to crystal field theory = $$t_{2g}^4e_g^2$$\n

$$ \\therefore $$ CFSE = $$\\left[ {{3 \\over 5} \\times 2 - {2 \\over 5} \\times 4} \\right]{\\Delta _0}$$ = - 0.4 $${\\Delta _0}$$\n

CN of K2[NiCl4] = 4\n

For CN = 4, CFSE = $$\\left[ {{2 \\over 5} \\times n - {3 \\over 5} \\times {n_1}} \\right]{\\Delta _t}$$\n

Here, n = number of electron in t2g and \n
n1 = number of electron in eg\n

Electronic configuration of Ni+2 according to crystal field theory = $$e_g^4t_{2g}^4$$\n

$$ \\therefore $$ CFSE = $$\\left[ {{2 \\over 5} \\times 4 - {3 \\over 5} \\times 4} \\right]{\\Delta _t}$$ = - 0.8 $${\\Delta _t}$$", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 926, "subject": "Chemistry", "question": "Three complexes,\n
[CoCl(NH3)5]\n2+(I),\n
[Co(NH3)5H2O]3+ (II) and\n
[Co(NH3)6]\n3+(III)\n
absorb light in the visible region. The correct order of the wavelength of light absorbed by them is :", "options": [ { "text": "(III) > (I) > (II)" }, { "text": "(III) > (II) > (I)" }, { "text": "(I) > (II) > (III)" }, { "text": "(II) > (I) > (III)" } ], "answer": "(I) > (II) > (III)", "solution": "**Answer:** (I) > (II) > (III)\n\nAs in a co-ordination compound, the strong field\nligand causes higher splitting of the d-orbitals\n

Also we know,\n

strength of ligand $$ \\propto $$ $${1 \\over {{\\lambda _{absorbed}}}}$$\n

Order of strength of ligand\n

NH3 > H2O > Cl-\n

Therefore decreasing order of wavelength\nabsorbed is (I) > (II) > (III)\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 927, "subject": "Chemistry", "question": "The degenerate orbitals of [Cr(H2O)6]3+ are :", "options": [ { "text": "dx2 and dxz" }, { "text": "dxz and dyz" }, { "text": "dyz and dz2" }, { "text": "dx2 - y2 and dxy" } ], "answer": "dxz and dyz", "solution": "**Answer:** dxz and dyz\n\nDegenerate orbitals means orbitals of equal energy.\n

Cr3+ forms an\noctahedral inner orbitals complex and its d orbital get splitted into two differnt energy level.\n

The two set of degenerate orbitals are \n

(1) dxy, dyz and\ndxz\n

(2) dx2 - y2 and dz2", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 928, "subject": "Chemistry", "question": "The correct order of the spin-only magnetic moment of metal ions in the following low-spin\ncomplexes, [V(CN)6]4–,[Fe(CN)6]4–, [Ru(NH3)6]3+, and [Cr(NH3)6]2+ , is :", "options": [ { "text": "V2+ > Ru3+ > Cr2+ > Fe2+" }, { "text": "V2+ > Cr2+ > Ru3+ > Fe2+" }, { "text": "Cr2+ > V2+ > Ru3+ > Fe2+" }, { "text": "Cr2+ > Ru3+ > Fe2+ > V2+" } ], "answer": "V2+ > Cr2+ > Ru3+ > Fe2+", "solution": "**Answer:** V2+ > Cr2+ > Ru3+ > Fe2+\n\n\"JEE\n
Here number of unpaired electrons = 3\n

$$ \\therefore $$ Spin only magnetic moment ($$\\mu $$) = $$\\sqrt {3\\left( {3 + 2} \\right)} = \\sqrt {15} $$ B.M\n

Note : Energy of t2g is less than eg. As electrons always go to the lower energy level orbitals first, that is why electrons goes to the t2g orbital first.\n\"JEE\n
Here number of unpaired electrons = 0\n

$$ \\therefore $$ Spin only magnetic moment ($$\\mu $$) = 0 B.M\n

Note : (1) As CN- is a strong field ligand so Energy gap between eg and t2g orbital is very high.\n

(2) [Fe(CN)6]4– is an octahedral complex. And for octahedral complex Energy gap between eg and t2g orbital is called $$\\Delta $$0 or Crystal Field splitting Energy.\n

(3) Energy required to pair up the electron in same orbital is called Pairing Energy(P).\n

(4) For strong field ligand, $$\\Delta $$0 is very high and for weak field ligand, $$\\Delta $$0 is very low.\n

(5) For strong field ligand, $$\\Delta $$0 > P, so when electron gets energy, pairing of electrons happens as for pairing of electrons very low energy requied.\n

(6) For weak field ligand, $$\\Delta $$0 < P, so when electron gets energy, pairing of electrons does not happens as the energy required to enter into the eg orbital is less than pairing energy. That is why electron go to eg orbital first.\n

(7) As CN- is a strong field ligand so pairing of electron occurs.\n\"JEE\n
Here number of unpaired electrons = 1\n

$$ \\therefore $$ Spin only magnetic moment ($$\\mu $$) = $$\\sqrt {1\\left( {1 + 2} \\right)} = \\sqrt {3} $$ B.M\n

Note : (1) In [Ru(NH3)6]3+ complex, NH3 is a strong field ligand so Energy gap between eg and t2g orbital is very high. That is why pairing of electrons occurs.\n\"JEE\n
Here number of unpaired electrons = 2\n

$$ \\therefore $$ Spin only magnetic moment ($$\\mu $$) = $$\\sqrt {2\\left( {2 + 2} \\right)} = \\sqrt {8} $$ B.M\n

$$ \\therefore $$ Correct order of the spin-only magnetic moment of metal ions\n

V2+ > Cr2+ > Ru3+ > Fe2+", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 929, "subject": "Chemistry", "question": "The pair of metal ions that can give a spin only magnetic moment of 3.9 BM for the complex [M(H2O)6]Cl2, is - \n", "options": [ { "text": "V2+ and Fe2+ " }, { "text": "V2+ and Co2+ " }, { "text": "Co2+ and Fe2+ " }, { "text": "Cr2+ and Mn2+" } ], "answer": "V2+ and Co2+ ", "solution": "**Answer:** V2+ and Co2+ \n\nMagnetic moment = $$\\sqrt {n\\left( {n + 2} \\right)} $$ BM\n

$$ \\therefore $$ $$\\sqrt {n\\left( {n + 2} \\right)} $$ = 3.9\n

$$ \\Rightarrow $$ n = 3\n

$$ \\therefore $$ no. of unpaired electron = 3\n

H2O is weak field ligand so no pairing of electrons happens.\n

Fe2+ = t2g4eg2\n

Co2+ = t2g5eg2\n

V2+ = t2g3eg0\n

$$ \\therefore $$ M is V, Co.\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 930, "subject": "Chemistry", "question": "Homoleptic octahedral complexes of a metal ion 'M3+' with three monodenate ligands L1, L2 and L3 adsorb wavelenths in the region of green, blue and red respectively. The increasing order of the ligands strength is : ", "options": [ { "text": "L3 < L1 < L2" }, { "text": "L3 < L2 < L1" }, { "text": "L1 < L2 < L3" }, { "text": "L2 < L1 < L3" } ], "answer": "L3 < L1 < L2", "solution": "**Answer:** L3 < L1 < L2\n\nStronger the ligand, absorption of light having lower wavelength is more.\n

Order of $$\\lambda $$ : Red $$>$$ Green $$>$$ Blue\n\n

Hence, ligand strength is L3 $$<$$ L1 $$<$$ L2", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 931, "subject": "Chemistry", "question": "The complex that has highest crystal field splitting energy ($$\\Delta $$), is : ", "options": [ { "text": "[Co(NH3)5(H2O)]Cl3" }, { "text": "K2[CoCl4]" }, { "text": "[Co(NH3)5Cl]Cl2" }, { "text": "K3[Co(CN)6]" } ], "answer": "K3[Co(CN)6]", "solution": "**Answer:** K3[Co(CN)6]\n\nAs complex K3[Co(CN)6] have CN$$-$$ ligand which is strongfield ligand amongst the given ligands in other complexes. ", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 932, "subject": "Chemistry", "question": "For a d4 metal ion in an octahedral field, the\ncorrect electronic configuration is :", "options": [ { "text": "$$t_{2g}^4e_g^0$$ when $$\\Delta $$0 < P" }, { "text": "$$t_{2g}^3e_g^1$$ when $$\\Delta $$0 > P" }, { "text": "$$e_g^2t_{2g}^2$$ when $$\\Delta $$0 < P" }, { "text": "$$t_{2g}^3e_g^1$$ when $$\\Delta $$0 < P" } ], "answer": "$$t_{2g}^3e_g^1$$ when $$\\Delta $$0 < P", "solution": "**Answer:** $$t_{2g}^3e_g^1$$ when $$\\Delta $$0 < P\n\n\"JEE\n

back pairing is not possible
because pairing energy > $$\\Delta $$0.", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 933, "subject": "Chemistry", "question": "Considering that $$\\Delta $$0\n > P, the magnetic moment
(in BM) of [Ru(H2O)6]2+ would be _________.", "options": [], "answer": "0", "solution": "**Answer:** 0\n\nRu(44) : [Kr] 4d75s1\n

Ru+2 = [Kr]4d6\n

As $$\\Delta $$0\n > P,\n

$$ \\therefore $$ Pairing of e–s will take place.\n

No. of unpaired e–s = 0\n

$$ \\therefore $$ Magnetic moment = 0 B.M", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 934, "subject": "Chemistry", "question": "The values of the crystal field stabilization\nenergies for a high spin d6 metal ion in\noctahedral and tetrahedral fields, respectively,\nare :", "options": [ { "text": "–0.4$$\\Delta $$0\n and –0.27$$\\Delta $$t" }, { "text": "–1.6$$\\Delta $$0\n and –0.4$$\\Delta $$t" }, { "text": "–0.4$$\\Delta $$0\n and –0.6$$\\Delta $$t" }, { "text": "–2.4$$\\Delta $$0\n and –0.27$$\\Delta $$t" } ], "answer": "–0.4$$\\Delta $$0\n and –0.6$$\\Delta $$t", "solution": "**Answer:** –0.4$$\\Delta $$0\n and –0.6$$\\Delta $$t\n\nIn octahedral\n\"JEE\n

CFSE = [-0.4nt2g + 0.6neg]$$\\Delta $$0\n

= [-0.4$$ \\times $$4 + 0.6$$ \\times $$2]$$\\Delta $$0\n

= -0.4 $$\\Delta $$0\n

In tetrahedral fields\n\"JEE\n

CFSE = [-0.6neg + 0.4nt2g]$$\\Delta $$t\n

= (– 0.6 × 3 + 0.4 × 3)$$\\Delta $$t\n

= (–1.8 + 1.2)$$\\Delta $$t\n

= –0.6$$\\Delta $$t", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 935, "subject": "Chemistry", "question": "The one that can exhibit highest paramagnetic\nbehaviour among the following is :\n

gly = glycinato; bpy = 2, 2'-bipyridine", "options": [ { "text": "[Fe(en)(bpy)(NH3)2]2+" }, { "text": "[Pd(gly)2]" }, { "text": "[Co(OX)2(OH)2]– ($$\\Delta $$0 > P)" }, { "text": "[Ti(NH3)6]3+" } ], "answer": "[Co(OX)2(OH)2]– ($$\\Delta $$0 > P)", "solution": "**Answer:** [Co(OX)2(OH)2]– ($$\\Delta $$0 > P)\n\n[Co(OX)2(OH)2]– \n

(As here $$\\Delta $$0 > P so those ligands act as strong field ligand and pairing happens.)\n

27Co+5 = [18Ar] 3d4 4s0( $$t_{2g}^{2,1,1},e_g^{0,0}$$ )\n

It has highest number of unpaired e– s. so it is most paramagnetic.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 936, "subject": "Chemistry", "question": "The Crystal Field Stabilization Energy\n
(CFSE) of [CoF3(H2O)3] ($$\\Delta $$0 < P) is :", "options": [ { "text": "-0.8 $$\\Delta $$0" }, { "text": "-0.4 $$\\Delta $$0" }, { "text": "-0.8 $$\\Delta $$0 + 2P" }, { "text": "-0.4 $$\\Delta $$0 + P" } ], "answer": "-0.4 $$\\Delta $$0", "solution": "**Answer:** -0.4 $$\\Delta $$0\n\n\"JEE\n

As $$\\Delta $$0 < P, so all ligands behaves as weak field ligands.\n

For octahedral\n

Crystal field stabilization energy (CFSE) \n

= (-0.4$$\\Delta $$0) $$ \\times $$ nt2g + (+0.6$$\\Delta $$0) $$ \\times $$ neg + np\n

nt2g = number of electrons in t2g orbital\n

neg = number of electrons in eg orbital\n

n = number of extra pairs\n

p = Pairing energy\n

Here nt2g = 1\n

neg = 0\n

n = 0 ( For weak field ligands n always zero. Here H2O is weak field ligand)\n

$$ \\therefore $$ Crystal field stabilization energy (CFSE) \n

= (-0.4$$\\Delta $$0) $$ \\times $$ 4 + (+0.6$$\\Delta $$0) $$ \\times $$ 2 + 0$$ \\times $$P\n

= -1.6$$\\Delta $$0 + 1.2$$\\Delta $$0\n

= -0.4$$\\Delta $$0", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 937, "subject": "Chemistry", "question": "The d-electron configuration of [Ru(en)3\n]Cl2\n
and [Fe(H2O)6]Cl2\n, respectively are :", "options": [ { "text": "$$t_{2g}^4e_g^2$$ and $$t_{2g}^6e_g^0$$" }, { "text": "$$t_{2g}^6e_g^0$$ and $$t_{2g}^6e_g^0$$" }, { "text": "$$t_{2g}^6e_g^0$$ and $$t_{2g}^4e_g^2$$" }, { "text": "$$t_{2g}^4e_g^2$$ and $$t_{2g}^4e_g^2$$" } ], "answer": "$$t_{2g}^6e_g^0$$ and $$t_{2g}^4e_g^2$$", "solution": "**Answer:** $$t_{2g}^6e_g^0$$ and $$t_{2g}^4e_g^2$$\n\n[Ru(en)3\n]Cl2 :\n

Here CN = 6 so octahedral splitting happens.\n

'en' is strong field ligand so $$\\Delta $$0 > P(pairing energy). That is why pairing of electrons happens.\n

$$\\Delta $$0 = Energy gap between eg and t2g orbital.\n
\"JEE\n

[Fe(H2O)6]Cl2 :\n

Here CN = 6 so octahedral splitting happens.\n

H2O is weak field ligand so $$\\Delta $$0 < P(pairing energy). That is why no pairing of electrons happens.\n

$$\\Delta $$0 = Energy gap between eg and t2g orbital.\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 938, "subject": "Chemistry", "question": "Consider that a d6 metal ion (M2+) forms a\ncomplex with aqua ligands, and the spin only\nmagnetic moment of the complex is 4.90 BM.\nThe geometry and the crystal field stabilization\nenergy of the complex is", "options": [ { "text": "tetrahedral and – 1.6 $$\\Delta $$t\n + 1P" }, { "text": "octahedral and –2.4 $$\\Delta $$0 + 2P" }, { "text": "tetrahedral and –0.6 $$\\Delta $$t" }, { "text": "octahedral and –1.6 $$\\Delta $$0" } ], "answer": "tetrahedral and –0.6 $$\\Delta $$t", "solution": "**Answer:** tetrahedral and –0.6 $$\\Delta $$t\n\nSpin only magnetic moment = 4.9 = $$\\sqrt {n\\left( {n + 2} \\right)} $$\n

From this, n = 4 (unpaired electrons)\n

(A) In octahedral complex: [M(H2O)6]\n2+\n\"JEE\n

C.F.S.E. = (–0.4 $$\\Delta $$0) × 4 + (+0.6 $$\\Delta $$0) × 2 + 0 × P\n
= –0.4 $$\\Delta $$0\n

(B) In tetrahedral complex: [M(H2O)4]\n2+\n

\"JEE\n
C.F.S.E. = (–0.6 $$\\Delta $$t) × 3 + (+0.4 $$\\Delta $$t) × 3 + 0 × P\n
= –0.6 $$\\Delta $$t\n", "topic": "Geometry", "subtopic": "Plane Geometry" }, { "id": 939, "subject": "Chemistry", "question": "The correct order of the spin-only magnetic\nmoments of the following complexes is :\n
(I) [Cr(H2O)6]Br2\n
(II) Na4[Fe(CN)6]\n
(III) Na3[Fe(C2O4)3] ($$\\Delta $$0 $$>$$ P)\n
(IV) (Et4N)2[CoCl4]", "options": [ { "text": "(III) > (I) > (II) > (IV)" }, { "text": "(II) $$ \\approx $$ (I) > (IV) > (III)" }, { "text": "(III) > (I) > (IV) > (II)" }, { "text": "(I) > (IV) > (III) > (II)" } ], "answer": "(I) > (IV) > (III) > (II)", "solution": "**Answer:** (I) > (IV) > (III) > (II)\n\n(I) [Cr(H2O)6]Br2\n

H2O is weak field ligand so it can not pair up all the electrons.\n

Cr2+ : [Ar] 4s03d4 ($$t_{2g}^3e{g^1}$$)\n

Unpaired e– = 4\n

Magnetic moment = $$\\sqrt {24} $$ = 4.89 BM\n

(II) Na4[Fe(CN)6]\n

CN- is strong field ligand so it pair up all the electrons.\n

Fe2+ = [Ar] 4s03d6 ($$t_{2g}^6e{g^0}$$)\n

Unpaired e– = 0\n

Magnetic moment = 0 BM\n

(III) Na3[Fe(C2O4)3] \n

As $$\\Delta $$0 $$>$$ P, so pairing of electrons happens.\n

Fe3+ = [Ar] 4s03d5 ($$t_{2g}^5e{g^0}$$)\n

Unpaired e– = 1\n

Magnetic moment = $$\\sqrt 3 $$ = 1.73 BM\n

(IV) (Et4N)2[CoCl4]\n

Cl-is weak field ligand so it can not pair up all the electrons.\n

Co2+ = [Ar] 4s03d7 ($$e{g^4}t_{2g}^3$$)\n

Unpaired e– = 3\n

Magnetic moment = $$\\sqrt {15} $$ = 3.87 BM\n

Hence order of magnetic moment is\n
I > IV > III > II", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 940, "subject": "Chemistry", "question": "[Pd(F)(Cl)(Br)(I)]2– has n number of\ngeometrical isomers. Then, the spin-only\nmagnetic moment and crystal field stabilisation\nenergy [CFSE] of [Fe(CN)6]n–6, respectively,\nare:
\n[Note : Ignore the pairing energy]", "options": [ { "text": "1.73 BM and –2.0 $$\\Delta $$0" }, { "text": "5.92 BM and 0" }, { "text": "2.84 BM and –1.6 $$\\Delta $$0" }, { "text": "0 BM and –2.4 $$\\Delta $$0" } ], "answer": "1.73 BM and –2.0 $$\\Delta $$0", "solution": "**Answer:** 1.73 BM and –2.0 $$\\Delta $$0\n\nComplex [Pd(F)(Cl)(Br)(I)]2– (square planar\ngeometry)-has 3 geometrical isomers.\n

$$ \\therefore $$ n = 3\n

[Fe(CN)6]n–6 becomes [Fe(CN)6]3-\n

Here Oxidation state of Fe = +3\n

$$ \\therefore $$ Fe+3 = [Ar]3d54s0\n

CN- is strong field ligand so it pairing of electrons happens.\n

E.C. according to CFT = ($$t_{2g}^5e{g^0}$$)\n

$$ \\therefore $$ No. of unpaired e- = 1\n

Then Magnetic moment = $$\\sqrt {n\\left( {n + 2} \\right)} $$ = $$\\sqrt 3 $$ = 1.73 B.M\n

CFSE = [(–0.4 × 5) + (0.6 × 0)] $$\\Delta $$0\n= –2.0 $$\\Delta $$0", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 941, "subject": "Chemistry", "question": "Among the statements(a)-(d) the incorrect ones are :
\n(a) Octahedral CO(III) complexes with strong fields ligands have very high magnetic moments.
\n(b) When $$\\Delta $$0 < P, the d-electron configuration of Co(III) in an octahedral complex is $$t_{eg}^4e_g^2$$
\n(c) Wavelength of light absorbed by [Co(en)3]3+ is lower than that of [CoF6]3-
\n(d) If the $$\\Delta $$0 for an octahedral complex of CO(III) is 18,000 cm-1, the $$\\Delta $$t for its tetrahedral complex with the same ligand be 16,000 cm-1\n", "options": [ { "text": "(a) and (b) only" }, { "text": "(b) and (c) only " }, { "text": "(c) and (d) only " }, { "text": "(a) and (d) only" } ], "answer": "(a) and (d) only", "solution": "**Answer:** (a) and (d) only\n\n(a) Co3+ with strong field complex forms low\nmagnetic moment complex.\n

(b) If $$\\Delta $$0 < P configuration of Co3+ will be $$t_{eg}^4e_g^2$$.\n

(c) Splitting power of ethylenediamine (en) is\ngreater than fluoride (F–) ligand therefore more\nenergy absorbed by [Co(en)3]3+ as compared to\n[CoF6]3–.\n

So wave length of light absorbed by [Co(en)3]3+\nis lower than that of [CoF6]3–\n

(d) $${\\Delta _t} = {4 \\over 9}{\\Delta _0}$$ = $${4 \\over 9} \\times 18000$$ = 8000 cm-1\n

$$ \\therefore $$ Statement (a) and (d) are incorrect.", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 942, "subject": "Chemistry", "question": "The theory that can completely/properly explain the nature of bonding in [Ni(Co)4] is :", "options": [ { "text": "Wemer's theory" }, { "text": "Valence bond theory" }, { "text": "Crystal field theory" }, { "text": "Molecular orbital theory" } ], "answer": "Molecular orbital theory", "solution": "**Answer:** Molecular orbital theory\n\nIn complex [Ni(CO)4] decrease in Ni–C bond\nlength and increase in C–O bond length as well\nas it's magnetic property is explained by MOT.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 943, "subject": "Chemistry", "question": "The electronic spectrum of [Ti(H2O)6]3+ shows a single broad peak with a maximum at 20,300 cm-1\n.\nThe crystal field stabilization energy (CFSE) of the complex ion, in kJ mol-1, is :", "options": [ { "text": "83.7" }, { "text": "242.5" }, { "text": "145.5" }, { "text": "97" } ], "answer": "97", "solution": "**Answer:** 97\n\n[Ti(H2O)6]3+\n

Ti = [Ar]3d24s2\n

Ti+3 = [Ar]3d1\n
\"JEE\n

For octahedral\n

Crystal field stabilization energy (CFSE) \n

= (-0.4$$\\Delta $$0) $$ \\times $$ nt2g + (+0.6$$\\Delta $$0) $$ \\times $$ neg + np\n

nt2g = number of electrons in t2g orbital\n

neg = number of electrons in eg orbital\n

n = number of extra pairs\n

p = Pairing energy\n

Here nt2g = 1\n

neg = 0\n

n = 0 ( For weak field ligands n always zero. Here H2O is weak field ligand)\n

$$ \\therefore $$ Crystal field stabilization energy (CFSE) \n

= (-0.4$$\\Delta $$0) $$ \\times $$ 1 + (+0.6$$\\Delta $$0) $$ \\times $$ 0 + 0$$ \\times $$P\n

= -0.4$$\\Delta $$0 = –0.4 × 20300 = –8120 cm–1\n

CFSE (in kJ) = $${{8120} \\over {83.7}}$$ = 97 kJ/mol\n", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 944, "subject": "Chemistry", "question": "The calculated magnetic moments (spin only value) for species $${[FeC{l_4}]^{2 - }}$$, $${[Co{({C_2}{O_4})_3}]^{3 - }}$$ and $$MnO_4^{2 - }$$ respectively are :", "options": [ { "text": "5.92, 4.90 and 0 BM" }, { "text": "4.90, 0 and 1.73 BM" }, { "text": "5.82, 0 and 0 BM" }, { "text": "4.90, 0 and 2.83 BM" } ], "answer": "4.90, 0 and 1.73 BM", "solution": "**Answer:** 4.90, 0 and 1.73 BM\n\n(i) $${[FeC{l_4}]^{2 - }}\\buildrel {} \\over\n \\longrightarrow F{e^{2 + }}\\buildrel {} \\over\n \\longrightarrow [Ar]3{d^6}$$

$$\\therefore$$ Cl is weak field ligand so does not pairing occur

\"JEE
So, magnetic moment $$(\\mu ) = \\sqrt {n(n + 2)} BM$$

$$ = \\sqrt {4(4 + 2)} BM$$ (n = Number of total unpaired e$$-$$ = 4)

$$ = \\sqrt {24} BM = 4.90\\,BM$$

(ii) $${[Co{({C_2}{O_4})_3}]^{3 - }}\\mathrel{\\mathop{\\kern0pt\\longrightarrow}\n\\limits_{}} C{o^{3 + }}\\mathrel{\\mathop{\\kern0pt\\longrightarrow}\n\\limits_{}} [Ar]3{d^6}$$

C2O4 is strong field ligand so pairing occur.

\"JEE
All electrons are paired, n = 0

hence, $$\\mu$$ = 0

(iii) $$MnO_4^{2 - }\\mathrel{\\mathop{\\kern0pt\\longrightarrow}\n\\limits_{}} M{n^{ + 6}}\\mathrel{\\mathop{\\kern0pt\\longrightarrow}\n\\limits_{}} [Ar]3{d^1}$$

$$n = 1$$

$$ = \\sqrt {n(n + 2)} BM$$

$$ = \\sqrt {1(1 + 2)} BM$$

$$ = \\sqrt 3 BM$$

$$ = 1.73\\,BM$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 945, "subject": "Chemistry", "question": "[Ti(H2O)6]3+ absorbs light of wavelength 498 nm during a d $$-$$ d transition. The octahedral splitting energy for the above complex is ____________ $$\\times$$ 10$$-$$19 J. (Round off to the Nearest Integer). h = 6.626 $$\\times$$ 10$$-$$34 Js; c = 3 $$\\times$$ 108 ms$$-$$1", "options": [], "answer": "4", "solution": "**Answer:** 4\n\nOctahedral splitting energy = $${{hc} \\over \\lambda }$$

$$ = {{6.626 \\times {{10}^{ - 34}} \\times 3 \\times {{10}^8}} \\over {498 \\times {{10}^{ - 9}}}}$$

$$ = 3.99 \\times {10^{ - 19}}J$$

$$ \\approx 4 \\times {10^{ - 19}}$$", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 946, "subject": "Chemistry", "question": "The correct order of intensity of colors of the compounds is :", "options": [ { "text": "$${[Ni{(CN)_4}]^{2 - }} > {[NiC{l_4}]^{2 - }} > {[Ni{({H_2}O)_6}]^{2 + }}$$" }, { "text": "$${[Ni{({H_2}O)_6}]^{2 + }} > {[NiC{l_4}]^{2 - }} > {[Ni{(CN)_4}]^{2 - }}$$" }, { "text": "$${[NiC{l_4}]^{2 - }} > {[Ni{({H_2}O)_6}]^{2 + }} > {[Ni{(CN)_4}]^{2 - }}$$" }, { "text": "$${[NiC{l_4}]^{2 - }} > {[Ni{(CN)_4}]^{2 - }} > {[Ni{({H_2}O)_6}]^{2 + }}$$" } ], "answer": "$${[NiC{l_4}]^{2 - }} > {[Ni{({H_2}O)_6}]^{2 + }} > {[Ni{(CN)_4}]^{2 - }}$$", "solution": "**Answer:** $${[NiC{l_4}]^{2 - }} > {[Ni{({H_2}O)_6}]^{2 + }} > {[Ni{(CN)_4}]^{2 - }}$$\n\nCorrect order of intensity of colours of the compounds is

[NiCl4]2$$-$$ > [Ni(H2O)6]2+ > [Ni(CN)4]2$$-$$

Ni is in +2 oxidation state in all complexes. The intensity of colour depends on the strength of the ligand attached with the central metal atom because more strong ligand more splitting energy, less is intensity of colour. Strength of ligand is in the order CN$$-$$ > H2O > Cl$$-$$.

Splitting energy order [NiCl4]2$$-$$ < [Ni(H2O)6]2+ < [Ni(CN)4]2$$-$$

$$\\therefore$$ Intensity of colour of compound

[NiCl4]2$$-$$ > [Ni(H2O)6]2+ > [Ni(CN)4]2$$-$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 947, "subject": "Chemistry", "question": "The total number of unpaired electrons present in [Co(NH3)6]Cl2 and [Co(NH3)6]Cl3 is :", "options": [], "answer": "3", "solution": "**Answer:** 3\n\n[Co(NH3)6]Cl2

Co2+ : [Ar]3d74s04p0

For this complex $$\\Delta$$0 < P.E., so pairing of electron does not take place.

sp3d2 hybridisation

Total 3 unpaired electrons are present.

[Co(NH3)6]Cl3

Co3+ : [Ar]3d6 4s0 4p0

d2sp3 hybridistion

NH3 acts as SFL because $$\\Delta$$0 > P.E.

So, here all electrons becomes paired.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 948, "subject": "Chemistry", "question": "Arrange the following Cobalt complexes in the order of increasing Crystal Field Stabilization Energy (CFSE) value.

Complexes : $$\\mathop {{{[Co{F_6}]}^{3 - }}}\\limits_A ,\\mathop {{{[Co{{({H_2}O)}_6}]}^{2 + }}}\\limits_B ,\\mathop {{{[Co{{(N{H_3})}_6}]}^{3 + }}}\\limits_C and \\mathop {{{[Co{{({en})}_3}]}^{3 + }}}\\limits_D $$

Choose the correct option :", "options": [ { "text": "A < B < C < D" }, { "text": "B < A < C < D" }, { "text": "B < C < D < A" }, { "text": "C < D < B < A" } ], "answer": "B < A < C < D", "solution": "**Answer:** B < A < C < D\n\n(i) CFSE $$\\propto$$ charge or oxidation no. of central metal ion.

(ii) CFSE $$\\propto$$ strength of ligand

en > NH3 > H2O > F$$-$$

$$\\therefore$$ order of CFSE

$$\\mathop {{{[Co{{(en)}_3}]}^{ + 3}}}\\limits^{III} > \\mathop {{{[Co{{(N{H_3})}_6}]}^{ + 3}}}\\limits^{III} > \\mathop {{{[Co{F_6}]}^{ - 3}}}\\limits^{III} > \\mathop {{{[Co{{({H_2}O)}_6}]}^{ + 2}}}\\limits^{III} $$", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 949, "subject": "Chemistry", "question": "The Crystal Field Stabilization Energy (CFSE) and magnetic moment (spin-only) of an octahedral aqua complex of a metal ion (Mz+) are $$-$$0.8 $$\\Delta$$0 and 3.87 BM, respectively. Identify (Mz+) :", "options": [ { "text": "V3+" }, { "text": "Cr3+" }, { "text": "Mn4+" }, { "text": "Co2+" } ], "answer": "Co2+", "solution": "**Answer:** Co2+\n\nCo2+ = [Ar] 3d74so\n

in [Co(H2O)6]2+, H2O will behave as weak field\nligand\n

Co2+ = t2g5\n, eg2\n

\"JEE\n
CFSE = (–0.4 × 5 + 2 × 0.6) Δ0\n = –0.8 Δ0\n\n

Magnetic moment (spin only) can be calculated as :\n

$$\\mu = \\sqrt {n\\left( {n + 2} \\right)} $$\n

where, $$\\mu $$ = spin only magnetic moment and\n

n = number of unpaired electrons = 3\n

= $$\\sqrt {3\\left( {3 + 2} \\right)} = \\sqrt {15} $$ = 3.87 BM", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 950, "subject": "Chemistry", "question": "

Which of the following will have maximum stabilization due to crystal field?

", "options": [ { "text": "[Ti(H2O)6]3+" }, { "text": "[Co(H2O)6]2+" }, { "text": "[Co(CN)6]3$$-$$" }, { "text": "[Cu(NH3)4]2+" } ], "answer": "[Co(CN)6]3$$-$$", "solution": "**Answer:** [Co(CN)6]3$$-$$\n\nThe given complexes are:\n

\n$\\left[\\mathrm{Ti}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}\\right]^{3+},\\left[\\mathrm{Co}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}\\right]^{2+},\\left[\\mathrm{Co}(\\mathrm{CN})_{6}\\right]^{-3},\\left[\\mathrm{Cu}\\left(\\mathrm{NH}_{3}\\right)_{4}\\right]^{2+}$\n

\n$\\mathrm{CN}^{-}$is the strongest ligand among the given complexes CFSE value for the $\\left[\\mathrm{Co}(\\mathrm{CN})_{6}\\right]^{-3}$ complex will be highest as it has $d^{6}$ configuration with a CFSE value of $-2.40 \\,\\Delta_{0}+2 \\mathrm{P}$, where $P$ represents pairing energy and $\\Delta_{0}$ represents splitting energy in octahedral field.\n

\nThe value of $\\Delta_{0}$ is high for cyanide complexes.", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 951, "subject": "Chemistry", "question": "

The spin-only magnetic moment value of an octahedral complex among CoCl3.4NH3, NiCl2.6H2O and PtCl4.2HCl, which upon reaction with excess of AgNO3 gives 2 moles of AgCl is ___________ B.M. (Nearest integer)

", "options": [], "answer": "3", "solution": "**Answer:** 3\n\n

From the given information, we are looking for a complex which, upon reaction with excess of AgNO₃, gives 2 moles of AgCl. This implies that the complex has 2 chloride ions involved.

\n

Considering the complexes :

\n
    \n
  1. CoCl₃.4NH₃ : This complex has 3 chloride ions, so it's not the one we are looking for.
    2. \n
  2. NiCl₂.6H₂O : This complex has 2 chloride ions, so it's a potential candidate.
    3. \n
  3. PtCl₄.2HCl : This complex has 4 chloride ions, so it's not the one we are looking for.
    4. \n
\n

Therefore, the complex we are interested in is NiCl₂.6H₂O.

\n$\\mathrm{CoCl}_{3} \\cdot 4 \\mathrm{NH}_{3} \\underset{\\text { excess }}{\\stackrel{\\mathrm{AgNO}_{3}}{\\longrightarrow}}\\left[\\mathrm{Co}\\left(\\mathrm{NH}_{3}\\right)_{4} \\cdot \\mathrm{Cl}_{2}\\right]+\\mathrm{AgCl}$\n

\n$$\n\\left.\\left.\\mathrm{NiCl}_{2} \\cdot 6 \\mathrm{H}_{2} \\mathrm{O} \\underset{\\text { excess }}{\\stackrel{\\mathrm{AgNO}_{3}}{\\longrightarrow}}\\right[ \\mathrm{Ni}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}\\right]^{2+}+2 \\mathrm{AgCl}\n$$\n

\n$\\mathrm{PtCl}_{4} \\cdot 2 \\mathrm{HCl} \\longrightarrow\\left[\\mathrm{PtCl}_{6}\\right]^{4-}+\\mathrm{No} ~\\mathrm{AgCl} ~\\mathrm{ppt}$\n

The next step is to find the oxidation state of the nickel ion. The nickel ion must have a charge of +2 to balance the -2 charge from the two chloride ions, thus it is Ni2+.

\n

In the case of Ni²⁺, the electron configuration is [Ar]3d8. For an octahedral complex, the d-orbitals split into two sets under the influence of ligands: the $e_g$ set which includes d(x²-y²) and d(z²) orbitals, and the $t_{2g}$ set which includes the d(xy), d(xz), and d(yz) orbitals. Electrons will occupy the lower energy $t_{2g}$ orbitals first.

\n

The 3d8 electron configuration implies there are 8 electrons in the 3d orbitals. The first six electrons pair up in the three $t_{2g}$ orbitals, and the next two electrons will go into the two $e_g$ orbitals, with each one having one unpaired electron.

\n

The spin-only magnetic moment (μ) can be calculated using the formula :

\n

$$\n\\mu = \\sqrt{n(n+2)} \\, \\text{B.M.}\n$$

\n

where n is the number of unpaired electrons. In this case, n = 2, so

\n

$$\n\\mu = \\sqrt{2 \\times (2+2)} = \\sqrt{8} \\, \\text{B.M.}\n$$

\n

Rounding to the nearest integer, the spin-only magnetic moment is approximately 3 B.M.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 952, "subject": "Chemistry", "question": "

Reaction of [Co(H2O)6]2+ with excess ammonia and in the presence of oxygen results into a diamagnetic product. Number of electrons present in t2g-orbitals of the product is ___________.

", "options": [], "answer": "6", "solution": "**Answer:** 6\n\n$\\left[\\mathrm{Co}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}\\right]^{+2} \\underset{\\mathrm{O}_{2}}{\\stackrel{\\mathrm{NH}_{3}}{\\longrightarrow}}\\left[\\mathrm{Co}\\left(\\mathrm{NH}_{3}\\right)_{6}\\right]^{+3}+e^{-}$\n

\n$$\n\\mathrm{Co}^{+3} \\longrightarrow 3 \\mathrm{~d}^{6}\n$$\n

\n$\\mathrm{NH}_{3}$ is a strong field ligand.\n

\n$$\n3 \\mathrm{~d}^{6} \\longrightarrow \\mathrm{t}_{2 \\mathrm{~g}}^{6} ~\\mathrm{eg}^{\\circ}\n$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 953, "subject": "Chemistry", "question": "

If [Cu(H2O)4]2+ absorbs a light of wavelength 600 nm for d-d transition, then the value of octahedral crystal field splitting energy for [Cu(H2O)6]2+ will be ____________ $$\\times$$ 10$$-$$21 J. [Nearest Integer]

\n

(Given : h = 6.63 $$\\times$$ 10$$-$$34 Js and c = 3.08 $$\\times$$ 108 ms$$-$$1)

", "options": [], "answer": "766", "solution": "**Answer:** 766\n\n$\\left[\\mathrm{Cu}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{4}\\right]^{2+}$ is tetrahedral\n

\n$\\left[\\mathrm{Cu}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}\\right]^{2+}$ is octahedral\n

\n$$\n\\because \\Delta_{\\mathrm{t}}=\\frac{4}{9} \\times \\Delta_{0}\n$$\n

\n$$\n\\begin{aligned}\n&\\Delta_{\\mathrm{t}}=\\frac{6.63 \\times 10^{-34} \\times 3.08 \\times 10^{8}}{600 \\times 10^{-9}} \\\\\\\\\n&\\Delta_{0}=\\frac{9}{4} \\times \\frac{6.63 \\times 10^{-34} \\times 3.08 \\times 10^{8}}{600 \\times 10^{-9}} \\approx 765.7 \\times 10^{-21} \\mathrm{~J}\n\\end{aligned}\n$$", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 954, "subject": "Chemistry", "question": "

Transition metal complex with highest value of crystal field splitting ($$\\Delta$$0) will be :

", "options": [ { "text": "[Cr(H2O)6]3+" }, { "text": "[Mo(H2O)6]3+" }, { "text": "[Fe(H2O)6]3+" }, { "text": "[Os(H2O)6]3+" } ], "answer": "[Os(H2O)6]3+", "solution": "**Answer:** [Os(H2O)6]3+\n\nCrystal field splitting $$\\left(\\Delta_{0}\\right)$$ for octahedral complexes depends on oxidation state of the metal as well as to which transition series the metal belongs. For the same oxidation state, the crystal field splitting $$\\left(\\Delta_{0}\\right)$$ increases as we move from $$3 d \\rightarrow 4 d \\rightarrow 5 d $$

$$\\mathrm{Cr}^{3+}$$ and $$\\mathrm{Fe}^{3+}$$ belong to $$3 d$$ series, $$\\mathrm{Mo}^{3+}$$ belongs to $$4 d$$ series and $$\\mathrm{Os}^{3+}$$ belongs to $$5 d$$ series.

Therefore crystal field splitting $$\\left(\\Delta_{0}\\right)$$ is highest for $$\\left[\\mathrm{Os}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}\\right]^{3+}$$.", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 955, "subject": "Chemistry", "question": "

Consider the following metal complexes :

\n

$$\\left[\\mathrm{Co}\\left(\\mathrm{NH}_{3}\\right)_{6}\\right]^{3+}$$

\n

$$\\left[\\mathrm{CoCl}\\left(\\mathrm{NH}_{3}\\right)_{5}\\right]^{2+}$$

\n

$$\\left[\\mathrm{Co}(\\mathrm{CN})_{6}\\right]^{3-}$$

\n

$$\\left[\\mathrm{Co}\\left(\\mathrm{NH}_{3}\\right)_{5}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)\\right]^{3+}$$

\n

The spin-only magnetic moment value of the complex that absorbes light with shortest wavelength is _____________ B. M. (Nearest integer)

", "options": [], "answer": "0", "solution": "**Answer:** 0\n\nIn all complexes, Co is present in $+3$ oxidation state and all complexes are low spin or inner orbital complex.\n

\nThe stronger the ligand, the higher the crystal field splitting.\n

\nSo, the order of crystal field splitting is\n

\n$\\left[\\mathrm{Co}(\\mathrm{CN})_{6}\\right]^{3-}>\\left[\\mathrm{Co}\\left(\\mathrm{NH}_{3}\\right)_{6}\\right]^{3+}>\\left[\\mathrm{Co}\\left(\\mathrm{NH}_{3}\\right)_{5}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)\\right]^{3+}>$ $\\left[\\mathrm{CoCl}\\left(\\mathrm{NH}_{3}\\right)_{5}\\right]^{2+}$

Shortest wavelength is shown by complex having maximum crystal field splitting.

\n\"JEE
Spin only magnetic moment $=\\sqrt{0(0+2)}=0$ B.M", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 956, "subject": "Chemistry", "question": "

The correct order of energy of absorption for the following metal complexes is :

\n

A : [Ni(en)3]2+ , B : [Ni(NH3)6]2+ , C : [Ni(H2O)6]2+

", "options": [ { "text": "C < B < A" }, { "text": "B < C < A" }, { "text": "C < A < B" }, { "text": "A < C < B" } ], "answer": "C < B < A", "solution": "**Answer:** C < B < A\n\nStronger is ligand attached to metal ion, greater will be the splitting between $$\\mathrm{t}_{2} \\mathrm{g}$$ and $e_g$ (hence greater will be $$\\Delta \\mathrm{U})$$

$$ \\therefore$$ Greater will be absorption of energy.

Hence correct order\n

\n$$\\left[\\mathrm{Ni}(\\mathrm{en})_{3}\\right]^{2+}>\\left[\\mathrm{Ni}\\left(\\mathrm{NH}_{3}\\right)_{6}\\right]^{2+}>\\left[\\mathrm{Ni}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}\\right]^{2+}$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 957, "subject": "Chemistry", "question": "

$$\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}\\right]^{3-}$$ should be an inner orbital complex. Ignoring the pairing energy, the value of crystal field stabilization energy for this complex is $$(-)$$ ____________ $$\\Delta_{0}$$. (Nearest integer)

", "options": [], "answer": "2", "solution": "**Answer:** 2\n\nIn $\\left[\\mathrm{Fe}(\\mathrm{CN})^{6}\\right]^{3-}$, $\\mathrm{Fe}$ is present in (+3) oxidation state $\\mathrm{Fe}(\\mathrm{III}) \\Rightarrow$ inner orbital complex $\\Rightarrow d^{5}$ (with pairing)\n

\nConfiguration $\\Rightarrow t_{2 g}^{5}$\n

\nCFSE $=5 \\times \\frac{-2}{5} \\Delta_{0}=-2 \\Delta_{0}$", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 958, "subject": "Chemistry", "question": "

Octahedral complexes of copper(II) undergo structural distortion (Jahn-Teller). Which one of the given copper (II) complexes will show the maximum structural distortion? (en - ethylenediamine; $$\\mathrm{H}_{2} \\mathrm{~N}_{-} \\mathrm{CH}_{2}-\\mathrm{CH}_{2}-\\mathrm{NH}_{2}$$)

", "options": [ { "text": "$$\\left[\\mathrm{Cu}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}\\right] \\mathrm{SO}_{4}$$" }, { "text": "$$\\left[\\mathrm{Cu}(\\mathrm{en})\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{4}\\right] \\mathrm{SO}_{4}$$" }, { "text": "cis-$$\\left[\\mathrm{Cu}\\left(\\mathrm{en})_{2} \\mathrm{Cl}_{2}\\right]\\right.$$" }, { "text": "trans-$$\\left[\\mathrm{Cu}\\left(\\mathrm{en})_{2} \\mathrm{Cl}_{2}\\right]\\right.$$" } ], "answer": "$$\\left[\\mathrm{Cu}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}\\right] \\mathrm{SO}_{4}$$", "solution": "**Answer:** $$\\left[\\mathrm{Cu}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}\\right] \\mathrm{SO}_{4}$$\n\n

According to John teller any nonlinear molecular system in a degenerate electronic state will be unstable and will undergo some kind of distortion which will lower its symmetry and energy and split the degenerate state.

\n

In case of octahedral $$d^{9}$$ configuration, the last electron may occupy either $$d^{2}$$ or $$d_{x^{2} y^{2}}$$ orbitals of $$e_{g}$$ set.

If it occupies $$\\mathrm{dz}^{2}$$ orbital most of the electron density will be concentrated between the metal and the two ligands on the $$\\mathrm{z}$$ axis. Thus there will be greater electrostatic repulsion associated with these ligands than with the other four on xy plane.

\n\n

The Jahn Teller effect is mostly observed in octahedral environments. The considerable distortions are usually observed in high spin $$\\mathrm{d}^{4}$$, low spin $$\\mathrm{d}^{7}$$ and $$\\mathrm{d}^{9}$$ configuration.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 959, "subject": "Chemistry", "question": "If the CFSE of $\\left[\\mathrm{Ti}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}\\right]^{3+}$ is $-96.0 \\mathrm{~kJ} / \\mathrm{mol}$, this complex will absorb maximum at wavelength ___________ $\\mathrm{nm}$. (nearest integer)\n

\nAssume Planck's constant (h) $=6.4 \\times 10^{-34} \\mathrm{Js}$, Speed of light $(\\mathrm{c})=3.0 \\times 10^{8} \\mathrm{~m} / \\mathrm{s}$ and Avogadro's\n

\nConstant $\\left(\\mathrm{N}_{\\mathrm{A}}\\right)=6 \\times 10^{23} / \\mathrm{mol}$", "options": [], "answer": "480", "solution": "**Answer:** 480\n\n$$\n\\begin{aligned}\n& {\\left[\\mathrm{Ti}\\left(\\mathrm{H}_2 \\mathrm{O}\\right)_6\\right]^{3+},} \\\\\\\\\n& \\mathrm{CFSE}=-0.4 \\Delta_0 =-96.0 \\mathrm{~kJ} \\mathrm{~mol}^{-1} \\\\\\\\\n& \\therefore \\Delta_0=\\frac{-96.0}{-0.4} \\\\\\\\\n& \\therefore \\Delta_0=240 \\mathrm{~kJ} \\mathrm{~mol}^{-1} \\\\\\\\\n\\end{aligned}\n$$\n

$$\n\\begin{aligned}\n\\frac{\\mathrm{hc}}{\\lambda} & =\\frac{96 \\times 10^3}{0.4 \\times 6 \\times 10^{23}} \\\\\\\\\n\\lambda & =\\frac{0.4 \\times 6 \\times 10^{23} \\times 6.4 \\times 10^{-34} \\times 3 \\times 10^8}{96 \\times 10^3} \\\\\\\\\n& =0.48 \\times 10^{-6} \\mathrm{~m} \\\\\\\\\n& =480 \\times 10^{-9} \\mathrm{~m} \\\\\\\\\n& =480 \\mathrm{~nm}\n\\end{aligned}\n$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 960, "subject": "Chemistry", "question": "

Which of the following complex will show largest splitting of d-orbitals?

", "options": [ { "text": "$$[\\mathrm{Fe(C_2O_4)_3]^{3-}}$$" }, { "text": "$$[\\mathrm{Fe(CN)_6]^{3-}}$$" }, { "text": "$$[\\mathrm{Fe(NH_3)_6]^{3+}}$$" }, { "text": "$$[\\mathrm{FeF_6]^{3-}}$$" } ], "answer": "$$[\\mathrm{Fe(CN)_6]^{3-}}$$", "solution": "**Answer:** $$[\\mathrm{Fe(CN)_6]^{3-}}$$\n\nCN–\nis strongest field ligand among given ligands. So maximum splitting \nin d-orbitals take place.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 961, "subject": "Chemistry", "question": "

Which of the following is correct order of ligand field strength?

", "options": [ { "text": "$$\\mathrm{S}^{2-}<\\mathrm{C}_{2} \\mathrm{O}_{4}^{2-}<\\mathrm{NH}_{3} < $$ en $$<\\mathrm{CO}$$" }, { "text": "$$\\mathrm{CO}<\\mathrm{en}<\\mathrm{NH}_{3}<\\mathrm{C}_{2} \\mathrm{O}_{4}^{2}<\\mathrm{S}^{2-}$$" }, { "text": "$$\\mathrm{S}^{2-}<\\mathrm{NH}_{3}<$$ en $$<\\mathrm{CO}<\\mathrm{C}_{2} \\mathrm{O}_{4}^{2}$$" }, { "text": "$$\\mathrm{NH}_{3}<\\mathrm{en}<\\mathrm{CO}<\\mathrm{S}^{2-}<\\mathrm{C}_{2} \\mathrm{O}_{4}^{2}$$" } ], "answer": "$$\\mathrm{S}^{2-}<\\mathrm{C}_{2} \\mathrm{O}_{4}^{2-}<\\mathrm{NH}_{3} < $$ en $$<\\mathrm{CO}$$", "solution": "**Answer:** $$\\mathrm{S}^{2-}<\\mathrm{C}_{2} \\mathrm{O}_{4}^{2-}<\\mathrm{NH}_{3} < $$ en $$<\\mathrm{CO}$$\n\n

Ligand field strength

\n

$$\\mathrm{{S^{2 - }} < {C_2}O_4^{2 - } < N{H_3} < en < CO}$$

", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 962, "subject": "Chemistry", "question": "

Correct order of spin only magnetic moment of the following complex ions is :

\n

(Given At.no. Fe : 26, Co : 27)

", "options": [ { "text": "$$\\mathrm{[CoF_6]^{3-} > [FeF_6]^{3-} > [Co(C_2O_4)_3]^{3-}}$$" }, { "text": "$$\\mathrm{[FeF_6]^{3-} > [Co(C_2O_4)_3]^{3-} > [CoF_6]^{3-}}$$" }, { "text": "$$\\mathrm{[FeF_6]^{3-} > [CoF_6]^{3-} > [Co(C_2O_4)_3]^{3-}}$$" }, { "text": "$$\\mathrm{[Co(C_2O_4)_3]^{3-} > [CoF_6]^{3-} > [FeF_6]^{3-}}$$" } ], "answer": "$$\\mathrm{[FeF_6]^{3-} > [CoF_6]^{3-} > [Co(C_2O_4)_3]^{3-}}$$", "solution": "**Answer:** $$\\mathrm{[FeF_6]^{3-} > [CoF_6]^{3-} > [Co(C_2O_4)_3]^{3-}}$$\n\n

$${\\left[ {Fe{F_6}} \\right]^{3 - }}\\buildrel {} \\over\n \\longrightarrow 5$$ unpaired electrons

\n

$${\\left[ {Co{{\\left( {{C_2}{O_4}} \\right)}_3}} \\right]^{3 - }}\\buildrel {} \\over\n \\longrightarrow 0$$ unpaired electron

\n

$${\\left[ {Co{F_6}} \\right]^{3 - }}\\buildrel {} \\over\n \\longrightarrow 4$$ unpaired electrons

\n

So, correct answer is : (3)

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 963, "subject": "Chemistry", "question": "

Match List I with List II

\n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
List I
Coordination entity		List II
Wavelength of light absorbed in nm
A.$$\\mathrm{[CoCl(NH_3)_5]^{2+}}$$I.310
B.$$\\mathrm{[Co(NH_3)_6]^{3+}}$$II.475
C.$$\\mathrm{[Co(CN)_6]^{3-}}$$III.535
D.$$\\mathrm{[Cu(H_2O)_4]^{2+}}$$IV.600

\n

Choose the correct answer from the options given below :

", "options": [ { "text": "A-IV, B-I, C-III, D-II" }, { "text": "A-III, B-II, C-I, D-IV" }, { "text": "A-III, B-I, C-II, D-IV" }, { "text": "A-II, B-III, C-IV, D-I" } ], "answer": "A-III, B-II, C-I, D-IV", "solution": "**Answer:** A-III, B-II, C-I, D-IV\n\n

Co-ordination compounds absorb a particular wavelength following certain rules.

\n

$$\n\\begin{aligned}\n\\text{Wavelength of light absorbed} & \\propto \\frac{1}{\\text { Oxidation state of metal ion }} \\\\\\\\\n& \\propto \\frac{1}{\\text { Strength of ligand }}\n\\end{aligned}\n$$

\n\n

Ligand field strength : $\\mathrm{CN}^{-}>\\mathrm{NH}_{3}>\\mathrm{H}_{2} \\mathrm{O}>\\mathrm{Cl}^{-}$

\n

\n\n\n \n \n \n \n \n \n\n\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\n
C.$\\left[\\mathrm{Co}^{||||}(\\mathrm{CN})_{6}\\right]^{3-}$I.$310 \\mathrm{~nm}$
B.$\\left[\\mathrm{Co}^{||||}\\left(\\mathrm{NH}_{3}\\right)_{6}\\right]^{3+}$II.$475 \\mathrm{~nm}$
A.$\\left[\\mathrm{Co}^{||||}\\mathrm{Cl}\\left(\\mathrm{NH}_{3}\\right)_{5}\\right]^{2+}$III.$535 \\mathrm{~nm}$
D.$\\left[\\mathrm{Co}^{||}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{4}\\right]^{2+}$IV.$600 \\mathrm{~nm}$

", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 964, "subject": "Chemistry", "question": "

Which of the following cannot be explained by crystal field theory?

", "options": [ { "text": "The order of spectrochemical series" }, { "text": "Stability of metal complexes" }, { "text": "Magnetic properties of transition metal complexes" }, { "text": "Colour of metal complexes" } ], "answer": "The order of spectrochemical series", "solution": "**Answer:** The order of spectrochemical series\n\nCFT does not explain the order of spectrochemical\nseries because as per CFT, anionic ligands should\nexert greatest splitting effect. However, they lie on\nlower end of the spectrochemical series.", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 965, "subject": "Chemistry", "question": "

The d-electronic configuration of $$\\mathrm{[CoCl_4]^{2-}}$$ in tetrahedral crystal field in $${e^mt_2^n}$$. Sum of \"m\" and \"number of unpaired electrons\" is ___________

", "options": [], "answer": "7", "solution": "**Answer:** 7\n\n$$\n\\mathrm{Co}^{2+}: 3 \\mathrm{~d}^7 4 \\mathrm{~s}^0, \\mathrm{Cl}^{-}: \\mathrm{WFL}\n$$

\n\"JEE
\nConfiguration $\\mathrm{e}^4 \\mathrm{t}_2{ }^3: \\mathrm{m}=4$

\nNumber of unpaired electrons $=3$

\nSo, answer $=7$", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 966, "subject": "Chemistry", "question": " The complex with highest magnitude of crystal field splitting energy $\\left(\\Delta_{0}\\right)$ is :", "options": [ { "text": "$\\left[\\mathrm{Fe}\\left(\\mathrm{OH}_{2}\\right)_{6}\\right]^{3+}$" }, { "text": "$\\left[\\mathrm{Mn}\\left(\\mathrm{OH}_{2}\\right)_{6}\\right]^{3+}$" }, { "text": "$\\left[\\mathrm{Cr}\\left(\\mathrm{OH}_{2}\\right)_{6}\\right]^{3+}$" }, { "text": "$\\left[\\mathrm{Ti}\\left(\\mathrm{OH}_{2}\\right)_{6}\\right]^{3+}$" } ], "answer": "$\\left[\\mathrm{Cr}\\left(\\mathrm{OH}_{2}\\right)_{6}\\right]^{3+}$", "solution": "**Answer:** $\\left[\\mathrm{Cr}\\left(\\mathrm{OH}_{2}\\right)_{6}\\right]^{3+}$\n\nThe crystal field splitting energy ($\\Delta_0$) is proportional to the electrostatic interaction between the metal ion and the ligands. This interaction is affected by the size of the metal ion, as well as the size of the ligands.\n

\nThe size of the metal ion can be estimated by its ionic radius, which is the distance from the nucleus to the outermost electron shell. A smaller ionic radius corresponds to a greater electrostatic interaction with the ligands, resulting in a larger $\\Delta_0$ value.\n

\nOut of the given options, the ionic radii of the metal ions are:\n

\n- $\\mathrm{Ti}^{3+}$: 67 pm

\n- $\\mathrm{Cr}^{3+}$: 62 pm

\n- $\\mathrm{Mn}^{3+}$: 65 pm

\n- $\\mathrm{Fe}^{3+}$: 65 pm\n

\nThe smaller ionic radius of $\\mathrm{Cr}^{3+}$ compared to the other metal ions indicates a stronger electrostatic interaction with the ligands, resulting in a higher crystal field splitting energy ($\\Delta_0$).\n

\nTherefore, the correct answer is option (C) $\\left[\\mathrm{Cr}\\left(\\mathrm{OH}_2\\right)_6\\right]^{3+}$, as it has the highest tendency to attract ligands due to its smaller ionic radius.", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 967, "subject": "Chemistry", "question": "The homoleptic and octahedral complex of $\\mathrm{Co}^{2+}$ and $\\mathrm{H}_{2} \\mathrm{O}$ has ___________ unpaired electron(s) in the $t_{2\\mathrm{g}}$ set of orbitals.", "options": [], "answer": "1", "solution": "**Answer:** 1\n\nWhen the water ligands coordinate to the $\\mathrm{Co}^{2+}$ ion in $\\mathrm{[Co(H_2O)6]^{2+}}$, the $d$ orbitals of the $\\mathrm{Co}^{2+}$ ion split into two sets of orbitals: the $t{2g}$ set, which consists of the $d_{xy}$, $d_{xz}$, and $d_{yz}$ orbitals, and the $e_g$ set, which consists of the $d_{z^2}$ and $d_{x^2-y^2}$ orbitals.

The $t_{2g}$ orbitals are lower in energy than the $e_g$ orbitals.\n

\nThe $\\mathrm{Co}^{2+}$ ion has a d$^7$ electron configuration, with three electrons in the $t_{2g}$ set and two electrons in the $e_g$ set. In an octahedral crystal field, the three $t_{2g}$ orbitals will be lower in energy than the two $e_g$ orbitals. Therefore, the three electrons in the $t_{2g}$ set will occupy all three $t_{2g}$ orbitals, and the two electrons in the $e_g$ set will occupy the higher energy $e_g$ orbitals. As a result, there is only $\\boxed{1}$ unpaired electron in the $t_{2g}$ set of orbitals.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 968, "subject": "Chemistry", "question": "

Match List I with List II

\n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
LIST I
Complex		LIST II
CFSE ($$\\Delta_0$$)
A.$$\\mathrm{[Cu(NH_3)_6]^{2+}}$$I.$$-0.6$$
B.$$\\mathrm{[Ti(H_2O)_6]^{3+}}$$II.$$-2.0$$
C.$$\\mathrm{[Fe(CN)_6]^{3-}}$$III.$$-1.2$$
D.$$\\mathrm{[NiF_6]^{4-}}$$IV.$$-0.4$$

\n

Choose the correct answer from the options given below:

", "options": [ { "text": "A-I, B-IV, C-II, D-III" }, { "text": "A-III, B-IV, C-I, D-II" }, { "text": "A-II, B-III, C-I, D-IV" }, { "text": "A-I, B-II, C-IV, D-III" } ], "answer": "A-I, B-IV, C-II, D-III", "solution": "**Answer:** A-I, B-IV, C-II, D-III\n\n

In order to match these complexes with their respective crystal field splitting energies (CFSE), we need to understand the coordination number, the geometry of the complex, and the nature of the ligand.

\n

The crystal field splitting energy, often denoted as Δ₀ or Δ, is the energy difference between the higher-energy and lower-energy sets of d-orbitals in a transition metal ion when in a crystal field created by a set of ligands.

\n

The sign of CFSE is often taken as negative because it represents a lowering in energy due to the field of the ligands. The magnitude of CFSE depends on the nature of the ligand, with ligands causing larger splitting being known as strong-field ligands and those causing smaller splitting being known as weak-field ligands.

\n

Based on spectrochemical series, we know the strength of ligands as follows: \nF⁻ < H₂O < NH₃ < CN⁻

\n

The crystal field splitting energy also depends on the configuration of the d electrons in the transition metal ion. For octahedral complexes, the d orbitals split into two sets with an energy difference Δ₀: a lower-energy set (dxy, dyz, dxz) called t₂g, and a higher-energy set (dz², dx²-y²) called eg.

\n

Now let's analyze each complex:

\n
    \n

  1. [Cu(NH₃)₆]²⁺: Cu²⁺ ion has a d⁹ configuration. It is an octahedral complex with a strong field ligand (NH₃), hence the splitting will be large.

    \n
    2. \n

  2. [Ti(H₂O)₆]³⁺: Ti³⁺ ion has a d¹ configuration. It is an octahedral complex with a weak field ligand (H₂O), hence the splitting will be small.

    \n
    3. \n

  3. [Fe(CN)₆]³⁻: Fe³⁺ ion has a d⁵ configuration. It is an octahedral complex with a very strong field ligand (CN⁻), hence the splitting will be very large.

    \n
    4. \n

  4. [NiF₆]⁴⁻: Ni²⁺ ion has a d⁸ configuration. It is an octahedral complex with a very weak field ligand (F⁻), hence the splitting will be very small.

    \n
    5. \n
\n

So, the correct order of CFSE values for these complexes should be:

\n

[NiF₆]⁴⁻ < [Ti(H₂O)₆]³⁺ < [Cu(NH₃)₆]²⁺ < [Fe(CN)₆]³⁻

\n

Therefore, the correct answer is:

\n

A-I, B-IV, C-II, D-III

\n", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 969, "subject": "Chemistry", "question": "

The ratio of spin-only magnetic moment values $$\\mu_{\\text {eff }}\\left[\\mathrm{Cr}(\\mathrm{CN})_{6}\\right]^{3-} / \\mu_{\\text {eff }}\\left[\\mathrm{Cr}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}\\right]^{3+}$$ is _________.

", "options": [], "answer": "1", "solution": "**Answer:** 1\n\nIn both given complexes, namely, $$[\\mathrm{Cr}(\\mathrm{CN})_{6}]^{3-}$$ and $$[\\mathrm{Cr}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}]^{3+}$$, the chromium atom is in the +3 oxidation state. This results in a $$d^3$$ configuration for the chromium ion.\n\n

Despite the difference in ligands (CN- and H2O), both complexes have a low-spin configuration because CN- is a strong field ligand and H2O is a weak field ligand. This configuration is denoted as $$\\mathrm{t}_{2g}^3 \\mathrm{e}_{\\mathrm{g}}^0$$, indicating that there are three unpaired electrons in the t2g orbital and no unpaired electrons in the eg orbital.\n\n

Using the spin-only magnetic moment formula, $$\\mu = \\sqrt{n(n+2)}$$, where n is the number of unpaired electrons, we find that :\n\n

For $$[\\mathrm{Cr}(\\mathrm{CN})_{6}]^{3-}$$, the spin-only magnetic moment is $$\\mu_{1}=\\sqrt{3(3+2)} = \\sqrt{15}$$ BM.\n\n

And for $$[\\mathrm{Cr}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}]^{3+}$$, the spin-only magnetic moment is also $$\\mu_{2}=\\sqrt{3(3+2)} = \\sqrt{15}$$ BM.\n\n

Therefore, the ratio of the spin-only magnetic moments is $$\\frac{\\mu_{1}}{\\mu_{2}} = \\frac{\\sqrt{15}}{\\sqrt{15}} = 1$$.\n

So, the spin-only magnetic moment of $$[\\mathrm{Cr}(\\mathrm{CN})_{6}]^{3-}$$ is equal to the spin-only magnetic moment of $$[\\mathrm{Cr}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}]^{3+}$$.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 970, "subject": "Chemistry", "question": "

The correct order of the number of unpaired electrons in the given complexes is

\n

A. $$\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}\\right]^{3-}$$

\n

B. $$\\left[\\mathrm{Fe} \\mathrm{F}_{6}\\right]^{3-}$$

\n

C. $$\\left[\\mathrm{CoF}_{6}\\right]^{3-}$$

\n

D. $$\\left.[\\mathrm{Cr} \\text { (oxalate})_{3}\\right]^{3-}$$

\n

E. $$\\left[\\mathrm{Ni}(\\mathrm{CO})_{4}\\right]$$

\n

Choose the correct answer from the options given below:

", "options": [ { "text": "E < A < B < D < C" }, { "text": "A < E < D < C < B" }, { "text": "A < E < C < B < D" }, { "text": "E < A < D < C < B" } ], "answer": "E < A < D < C < B", "solution": "**Answer:** E < A < D < C < B\n\n

The number of unpaired electrons in a complex can be determined by the oxidation state of the central metal atom and the ligand field splitting energy. In general, a complex with a high-spin configuration will have more unpaired electrons than a complex with a low-spin configuration.

\n

In the given complexes, the oxidation states of the central metal atoms are:

\n\n

The ligand field splitting energy of a complex depends on the type of ligand. In general, ligands that are strong field ligands will split the d orbitals more than ligands that are weak field ligands.

\n

The following table shows the number of unpaired electrons in each complex :

\n\n\n\n \n \n \n \n \n \n\n\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\n
ComplexOxidation stateLigand field splitting energyNumber of unpaired electrons
[Fe(CN)6]3-Fe(II)Low-spin1
[FeF6]3-Fe(III)High-spin5
[CoF6]3-Co(III)High-spin4
[Cr(ox)3]3-Cr(III)High-spin3
[Ni(CO)4]Ni(0)None0
\n

Therefore, the correct order of the number of unpaired electrons in the given complexes is :

\n
E < A < D < C < B
", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 971, "subject": "Chemistry", "question": "

Match List I with List II

\n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
List - I
Complex		List - II
Crystal Field splitting energy ($$\\Delta_0$$)
A.$${[Ti{({H_2}O)_6}]^{2 + }}$$I.$$-1.2$$
B.$${[V{({H_2}O)_6}]^{2 + }}$$II.$$-0.6$$
C.$${[Mn{({H_2}O)_6}]^{3 + }}$$III.0
D.$${[Fe{({H_2}O)_6}]^{3 + }}$$IV.$$-0.8$$

\n

Choose the correct answer from the options given below:

", "options": [ { "text": "A-IV, B-I, C-II, D-III" }, { "text": "A-II, B-IV, C-III, D-I" }, { "text": "A-II, B-IV, C-I, D-III" }, { "text": "A-IV, B-I, C-III, D-II" } ], "answer": "A-IV, B-I, C-II, D-III", "solution": "**Answer:** A-IV, B-I, C-II, D-III\n\n$$\n\\begin{aligned}\n& \\text{(A)} ~~ {\\left[\\mathrm{Ti}\\left(\\mathrm{H}_2 \\mathrm{O}\\right)_6\\right]^{2+}} \\\\\\\\\n& \\mathrm{Ti}^{2+} \\Rightarrow 3 \\mathrm{~d}^2 4 \\mathrm{~s}^0 \\\\\\\\\n& \\mathrm{t}_{2 \\mathrm{~g}} \\mathrm{e}^{-}=2 \\\\\\\\\n& \\mathrm{e}_{\\mathrm{g}} \\mathrm{e}^{-}=0 \\\\\\\\\n& \\mathrm{CFSE}=[-0.4 \\times 2+0.6 \\times 0] \\Delta_0 \\\\\\\\\n& \\quad=-0.8 \\Delta\n\\end{aligned}\n$$\n

\n$$\n\\begin{aligned}\n& \\text{(B)} ~~ {\\left[\\mathrm{V}\\left(\\mathrm{H}_2 \\mathrm{O}\\right)_6\\right]^{2+}} \\\\\\\\\n& \\mathrm{V}^{2+} \\Rightarrow 3 \\mathrm{~d}^3 4 \\mathrm{~s}^0 \\\\\\\\\n& \\mathrm{t}_{2 \\mathrm{~g}} \\mathrm{e}^{-}=3 \\\\\\\\\n& \\mathrm{e}_{\\mathrm{g}} \\mathrm{e}^{-}=0 \\\\\\\\\n& \\mathrm{CFSE}=[-0.4 \\times 3+0.6 \\times 0] \\Delta_0 \\\\\\\\\n& =-1.2 \\Delta_0\n\\end{aligned}\n$$\n

\n$$\n\\begin{aligned}\n& \\text{(C)} ~~{\\left[\\mathrm{Mn}\\left(\\mathrm{H}_2 \\mathrm{O}\\right)_6\\right]^{3+}} \\\\\\\\\n& \\mathrm{Mn}^{3+} \\Rightarrow 3 \\mathrm{~d}^4 4 \\mathrm{~s}^0 \\\\\\\\\n& \\mathrm{t}_{2 \\mathrm{~g}} \\mathrm{e}^{-}=3 \\\\\\\\\n& \\mathrm{e}_{\\mathrm{g}} \\mathrm{e}^{-}=1 \\\\\\\\\n& \\mathrm{CFSE}=[-0.4 \\times 3+0.6 \\times 1] \\Delta_0 \\\\\\\\\n& =-0.6 \\Delta_0\n\\end{aligned}\n$$

\n$$\n\\begin{aligned}\n& \\text{(D)} ~~{\\left[\\mathrm{Fe}\\left(\\mathrm{H}_2 \\mathrm{O}\\right)_6\\right]^{3+}} \\\\\\\\\n& \\mathrm{Fe}^{3+} \\Rightarrow 3 \\mathrm{~d}^5 4 \\mathrm{~s}^0 \\\\\\\\\n& \\mathrm{t}_{2 \\mathrm{~g}} \\mathrm{e}^{-}=3 \\quad \\mathrm{e}_{\\mathrm{g}}=2 \\\\\\\\\n& \\mathrm{CFSE}=[-0.4 \\times 3+0.6 \\times 2] \\Delta_0 \\\\\\\\\n& \\quad=0 \\Delta_0\n\\end{aligned}\n$$", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 972, "subject": "Chemistry", "question": "

For a metal ion, the calculated magnetic moment is $$4.90 ~\\mathrm{BM}$$. This metal ion has ___________ number of unpaired electrons.

", "options": [], "answer": "4", "solution": "**Answer:** 4\n\n

Using the equation $\\mu = \\sqrt{n(n+2)}$, where $\\mu$ is the magnetic moment and $n$ is the number of unpaired electrons, we can substitute $\\mu = 4.90$:

\n

$4.90 = \\sqrt{n(n+2)}$

\n

Squaring both sides of the equation:

\n

$(4.90)^2 = n(n+2)$

\n

$24 = n^2 + 2n$

\n

Rearranging the equation:

\n

$n^2 + 2n - 24 = 0$

\n

Solving the quadratic equation, we find:

\n

$n =4$

\n

Therefore, the metal ion has 4 unpaired electrons.

\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 973, "subject": "Chemistry", "question": "

The octahedral diamagnetic low spin complex among the following is :

", "options": [ { "text": "[Co(NH$$_3$$)$$_6$$]$$^{3+}$$" }, { "text": "[NiCl$$_4$$]$$^{2-}$$" }, { "text": "[CoCl$$_6$$]$$^{3-}$$" }, { "text": "[CoF$$_6$$]$$^{3-}$$" } ], "answer": "[Co(NH$$_3$$)$$_6$$]$$^{3+}$$", "solution": "**Answer:** [Co(NH$$_3$$)$$_6$$]$$^{3+}$$\n\n

In the complex [Co(NH$_3$)$_6$]$^{3+}$, Co is in the +3 oxidation state. This means it has a configuration of $d^6$. NH$_3$ is a strong field ligand, which causes the d-electrons to pair up. Therefore, this complex is a low-spin (also referred to as spin-paired) complex.

\n

The term "low-spin" means that the electrons prefer to pair up in the lower energy d-orbitals rather than occupy the higher energy orbitals. When all the electrons are paired, the complex is diamagnetic, which means it's not attracted to an external magnetic field. The hybridization of this complex is $d^2sp^3$, also known as inner orbital complex, because the inner d-orbitals are used in the hybridization.

\n

The complexes [CoF$_6$]$^{3-}$ and [CoCl$_6$]$^{3-}$ are both octahedral, but they do have unpaired electrons because $F^-$ and $Cl^-$ are not strong enough field ligands to cause pairing of all the electrons in the d-orbitals. Therefore, they are not low-spin or diamagnetic.

\n

Lastly, the complex [NiCl$_4$]$^{2-}$ is not octahedral. Because $Cl^-$ is a weak field ligand, the electrons in the d-orbitals do not pair up, and the $Ni^{2+}$ ion, with a $d^8 $ configuration, undergoes sp^3 hybridization, resulting in a tetrahedral complex.

\n

Therefore, the only octahedral, diamagnetic, and low-spin complex in the options provided is [Co(NH$_3$)$_6$]$^{3+}$ (Option A).

\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 974, "subject": "Chemistry", "question": "

In potassium ferrocyanide, there are ________ pairs of electrons in the $$t_{2g}$$ set of orbitals.

", "options": [], "answer": "3", "solution": "**Answer:** 3\n\n$\\mathrm{K}_4\\left[\\mathrm{Fe}(\\mathrm{CN})_6\\right]$

\n\"JEE
\n$$\n\\begin{aligned}\n& \\mathrm{Fe}^{+2}=[\\mathrm{Ar}] 3 \\mathrm{~d}^6 \\\\\\\\\n& \\mathrm{CN}^{-}=\\mathrm{SFL}\n\\end{aligned}\n$$

\n$\\mathrm{t}_{2 \\mathrm{~g}}$ contain 6 electron so it become 3 pairs", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 975, "subject": "Chemistry", "question": "

Match List I with List II

\n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
LIST I
Coordination Complex		LIST II
Number of unpaired electrons
A.$$\\left[\\mathrm{Cr}(\\mathrm{CN})_{6}\\right]^{3-}$$I.0
B.$$\\left[\\mathrm{Fe}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}\\right]^{2+}$$II.3
C.$$\\left[\\mathrm{Co}\\left(\\mathrm{NH}_{3}\\right)_{6}\\right]^{3+}$$III.2
D.$$\\left[\\mathrm{Ni}\\left(\\mathrm{NH}_{3}\\right)_{6}\\right]^{2+}$$IV.4

\n

Choose the correct answer from the options given below:

", "options": [ { "text": "A-IV, B-III, C-II, D-I" }, { "text": "A-II, B-I, C-IV, D-III" }, { "text": "A-II, B-IV, C-I, D-III" }, { "text": "A-III, B-IV, C-I, D-II" } ], "answer": "A-II, B-IV, C-I, D-III", "solution": "**Answer:** A-II, B-IV, C-I, D-III\n\nFor option (A)

\n$$\n\\begin{aligned}\n& \\mathrm{Cr}^{+3}: 3 \\mathrm{~d}^3 \\\\\\\\\n& \\mathrm{CN}^{-} \\rightarrow \\mathrm{SFL}\n\\end{aligned}\n$$

\nNo. of unpaired electrons $=3$

\nFor option (B)

\n$$\n\\begin{aligned}\n& \\mathrm{Fe}^{+2}: 3 \\mathrm{~d}^6 \\\\\\\\\n& \\mathrm{H}_2 \\mathrm{O}: \\mathrm{WFL}\n\\end{aligned}\n$$

\nNo. of unpaired electrons $=4$

\nFor option (C)

\n$$\n\\begin{aligned}\n& \\mathrm{Co}^{+3}: 3 \\mathrm{~d}^6 \\\\\\\\\n& \\mathrm{NH}_3: \\mathrm{SFL}\n\\end{aligned}\n$$

\nNo. of unpaired electrons $=0$

\nFor option (D)

\n$$\n\\begin{aligned}\n& \\mathrm{Ni}^{+2}: 3 \\mathrm{~d}^8 \\\\\\\\\n& \\mathrm{NH}_3: \\mathrm{SFL}\n\\end{aligned}\n$$

\nNo. of unpaired electrons $=2$", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 976, "subject": "Chemistry", "question": "

The observed magnetic moment of the complex $$\\left.\\left[\\operatorname{Mn}(\\underline{N} C S)_{6}\\right)\\right]^{x^{-}}$$ is $$6.06 ~\\mathrm{BM}$$. The numerical value of $$x$$ is __________.

", "options": [], "answer": "4", "solution": "**Answer:** 4\n\n

The complex is given as $[\\mathrm{Mn}(\\mathrm{NCS})_{6}]^{x-}$. Here, $\\mathrm{NCS}^{-}$ acts as a ligand. Each $\\mathrm{NCS}^{-}$ has a charge of -1 and since there are six of them, they contribute a total charge of -6 to the complex.

\n

Manganese (Mn) in this complex is in the +2 oxidation state. We know this because in its ground state, Mn has 5 electrons in its 3d orbitals and 2 electrons in its 4s orbital. But in the Mn²⁺ cation, the 2 electrons from the 4s orbital have been removed, leaving 5 unpaired electrons in the 3d orbitals. This matches the given magnetic moment of 6.06 BM, which corresponds to 5 unpaired electrons.

\n

So, the overall charge of the complex is -4: the Mn²⁺ ion contributes a charge of +2 and the six $\\mathrm{NCS}^{-}$ ligands contribute a total charge of -6. When you add these together, you get -4. Hence, $x = -4$.

\n

In summary, the $[\\mathrm{Mn}(\\mathrm{NCS})_{6}]^{x-}$ complex has an overall charge of -4, meaning x = -4.

So the numerical value of x will be 4.\n

\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 977, "subject": "Chemistry", "question": "

Which of the following complex is octahedral, diamagnetic and the most stable?

", "options": [ { "text": "$$\\mathrm{Na}_{3}\\left[\\mathrm{CoCl}_{6}\\right]$$" }, { "text": "$$\\left[\\mathrm{Co}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}\\right] \\mathrm{Cl}_{2}$$" }, { "text": "$$\\mathrm{K}_{3}\\left[\\mathrm{Co}(\\mathrm{CN})_{6}\\right]$$" }, { "text": "$$\\left[\\mathrm{Ni}\\left(\\mathrm{NH}_{3}\\right)_{6}\\right] \\mathrm{Cl}_{2}$$" } ], "answer": "$$\\mathrm{K}_{3}\\left[\\mathrm{Co}(\\mathrm{CN})_{6}\\right]$$", "solution": "**Answer:** $$\\mathrm{K}_{3}\\left[\\mathrm{Co}(\\mathrm{CN})_{6}\\right]$$\n\n

Option A : \n$$\\mathrm{Na}_{3}\\left[\\mathrm{CoCl}_{6}\\right]$$

\n

This is an octahedral complex, but it is not diamagnetic. Co in this complex is in +3 oxidation state, and the d-electron configuration is $t_{2g}^6 e_{g}^0$, leading to two unpaired electrons.

\n

Option B : \n$$\\left[\\mathrm{Co}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}\\right] \\mathrm{Cl}_{2}$$

\n

This is also an octahedral complex, but it is not diamagnetic. The Co ion is in +2 oxidation state with a d-electron configuration of $t_{2g}^5 e_{g}^2$, leading to 3 unpaired electrons.

\n

Option C : \n$$\\mathrm{K}_{3}\\left[\\mathrm{Co}(\\mathrm{CN})_{6}\\right]$$

\n

This complex is both octahedral and diamagnetic. Co in this complex is in +3 oxidation state and the d-electron configuration is $t_{2g}^6 e_{g}^0$, which means no unpaired electrons. This complex is more stable due to a strong field ligand (CN-) which leads to a greater splitting of d-orbitals and more pairing of electrons.

\n

Option D : \n$$\\left[\\mathrm{Ni}\\left(\\mathrm{NH}_{3}\\right)_{6}\\right] \\mathrm{Cl}_{2}$$

\n

This is an octahedral complex, and it is diamagnetic. The Ni ion is in +2 oxidation state with a d-electron configuration of $t_{2g}^8 e_{g}^0$, leading to no unpaired electrons. However, NH3 is a weaker field ligand compared to CN-, so the complex is less stable than Option C.

\n

Hence, the complex that is octahedral, diamagnetic, and the most stable among the given options is $$\\mathrm{K}_{3}\\left[\\mathrm{Co}(\\mathrm{CN})_{6}\\right]$$.

\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 978, "subject": "Chemistry", "question": "

The correct order of spin only magnetic moments for the following complex ions is

", "options": [ { "text": "$$\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}\\right]^{3-}<\\left[\\mathrm{CoF}_{6}\\right]^{3-}<\\left[\\mathrm{MnBr}_{4}\\right]^{2-}<\\left[\\mathrm{Mn}(\\mathrm{CN})_{6}\\right]^{3-}$$" }, { "text": "$$\\left[\\mathrm{MnBr}_{4}\\right]^{2-}<\\left[\\mathrm{CoF}_{6}\\right]^{3-}<\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}\\right]^{3-}<\\left[\\mathrm{Mn}(\\mathrm{CN})_{6}\\right]^{3-}$$" }, { "text": "$$\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}\\right]^{3-}<\\left[\\mathrm{Mn}(\\mathrm{CN})_{6}\\right]^{3-}<\\left[\\mathrm{CoF}_{6}\\right]^{3-}<\\left[\\mathrm{MnBr}_{4}\\right]^{2-}$$" }, { "text": "$$\\left[\\mathrm{CoF}_{6}\\right]^{3-}<\\left[\\mathrm{MnBr}_{4}\\right]^{2-}<\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}\\right]^{3-}<\\left[\\mathrm{Mn}(\\mathrm{CN})_{6}\\right]^{3-}$$" } ], "answer": "$$\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}\\right]^{3-}<\\left[\\mathrm{Mn}(\\mathrm{CN})_{6}\\right]^{3-}<\\left[\\mathrm{CoF}_{6}\\right]^{3-}<\\left[\\mathrm{MnBr}_{4}\\right]^{2-}$$", "solution": "**Answer:** $$\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}\\right]^{3-}<\\left[\\mathrm{Mn}(\\mathrm{CN})_{6}\\right]^{3-}<\\left[\\mathrm{CoF}_{6}\\right]^{3-}<\\left[\\mathrm{MnBr}_{4}\\right]^{2-}$$\n\n

The spin-only magnetic moment of a complex ion is calculated using the formula $\\mu = \\sqrt{n(n+2)}$, where $n$ is the number of unpaired electrons in the complex ion. In the case of transition metal complexes, the number of unpaired electrons and hence the magnetic moment can be influenced by the field strength of the ligands (the Ligand Field Theory).

\n

Here, CN$^-$ is a strong-field ligand (or high-spin ligand) that can cause pairing of electrons, while F$^-$ and Br$^-$ are considered weak-field ligands (or low-spin ligands) that do not promote electron pairing.

\n
    \n

  1. $\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}\\right]^{3-}$: In this complex, Fe is in +3 oxidation state and has 5 d-electrons. Due to the strong field ligand, CN$^-$, all the 5 electrons are paired, and there are 0 unpaired electrons. The spin-only magnetic moment is $\\mu = \\sqrt{0(0+2)} = 0$.

    \n

    2. \n

  2. $\\left[\\mathrm{CoF}_{6}\\right]^{3-}$: Here, Co is in +3 oxidation state and has 6 d-electrons. As F$^-$ is a weak field ligand, no electron pairing occurs, and hence there are 4 unpaired electrons. The spin-only magnetic moment is $\\mu = \\sqrt{4(4+2)} = \\sqrt{24}$.

    \n

    3. \n

  3. $\\left[\\mathrm{MnBr}_{4}\\right]^{2-}$: Here, Mn is in +2 oxidation state and has 5 d-electrons. Since Br$^-$ is a weak field ligand, no electron pairing occurs, and hence there are 5 unpaired electrons. The spin-only magnetic moment is $\\mu = \\sqrt{5(5+2)} = \\sqrt{35}$.

    \n

    4. \n

  4. $\\left[\\mathrm{Mn}(\\mathrm{CN})_{6}\\right]^{3-}$: In this complex, Mn is in +3 oxidation state and has 4 d-electrons. Due to the strong field ligand, CN$^-$, all the 4 electrons are paired, and there are 0 unpaired electrons. The spin-only magnetic moment is $\\mu = \\sqrt{0(0+2)} = 0$.

    \n
    5. \n
\n

So the correct order of spin-only magnetic moments is:

\n$$\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}\\right]^{3-}<\\left[\\mathrm{Mn}(\\mathrm{CN})_{6}\\right]^{3-}<\\left[\\mathrm{CoF}_{6}\\right]^{3-}<\\left[\\mathrm{MnBr}_{4}\\right]^{2-}$$

\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 979, "subject": "Chemistry", "question": "

Given below are two statements, one is labelled as Assertion $$\\mathbf{A}$$ and the other is labelled as Reason $$\\mathbf{R}$$.

\n

Assertion A: The spin only magnetic moment value for $$\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}\\right]^{3-}$$ is $$1.74 \\mathrm{BM}$$, whereas for $$\\left[\\mathrm{Fe}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}\\right]^{3+}$$ is $$5.92 \\mathrm{BM}$$.

\n

Reason $$\\mathbf{R}$$ : In both complexes, $$\\mathrm{Fe}$$ is present in +3 oxidation state.

\n

In the light of the above statements, choose the correct answer from the options given below:

", "options": [ { "text": "Both A and R are true and R is the correct explanation of A" }, { "text": "A is true but R is false" }, { "text": "A is false but R is true" }, { "text": "Both A and R are true but R is NOT the correct explanation of A" } ], "answer": "Both A and R are true but R is NOT the correct explanation of A", "solution": "**Answer:** Both A and R are true but R is NOT the correct explanation of A\n\n\"JEE
\nUnpaired electron $=1$

\n$$\n\\mu=\\sqrt{\\mathrm{n}(\\mathrm{n}+2)}=\\sqrt{1 \\times 3}=1.74 \\text { B.M. }\n$$

\n$\\left[\\mathrm{Fe}\\left(\\mathrm{H}_2 \\mathrm{O}\\right)_6\\right]^{3+}$ No pairing because $\\mathrm{H}_2 \\mathrm{O}$ is WFL

\nNumber of unpaired electrons $=5, \\mu=5.92 \\mathrm{BM}$

\nAssertion is true, Reason is true but not correct explanation.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 980, "subject": "Chemistry", "question": "$\\left[\\mathrm{Co}\\left(\\mathrm{NH}_3\\right)_6\\right]^{3+}$ and $\\left[\\mathrm{CoF}_6\\right]^{3-}$ are respectively known as :", "options": [ { "text": "Inner orbital Complex, Spin paired Complex" }, { "text": "Spin paired Complex, Spin free Complex" }, { "text": "Spin free Complex, Spin paired Complex" }, { "text": "Outer orbital Complex, Inner orbital Complex" } ], "answer": "Spin paired Complex, Spin free Complex", "solution": "**Answer:** Spin paired Complex, Spin free Complex\n\n$$\n\\left[\\mathrm{Co}\\left(\\mathrm{NH}_3\\right)_6\\right]^{3+}\n$$\n

$\\mathrm{Co}^{3+}($ strong field ligand $) \\Rightarrow 3 \\mathrm{~d}^6\\left(\\mathrm{t}_{2 \\mathrm{~g}}^6, \\mathrm{e}_{\\mathrm{g}}^0\\right)$,\n

Hybridisation : $\\mathrm{d}^2 \\mathrm{sp}^3$\n

Inner obital complex(spin paired complex)\n

Pairing will take place.\n

$$\n\\left[\\mathrm{CoF}_6\\right]^{3-}\n$$\n

$\\mathrm{Co}^{3+}($ weak field ligand $) \\Rightarrow 3 \\mathrm{~d}^6\\left(\\mathrm{t}_{2 \\mathrm{~g}}^4, \\mathrm{e}_{\\mathrm{g}}^2\\right)$\n

Hybridisation : $\\mathrm{sp}^3 \\mathrm{~d}^2$\n

Outer orbital complex (spin free complex) \n

No pairing will take place", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 981, "subject": "Chemistry", "question": "Which of the following compounds show colour due to d-d transition?", "options": [ { "text": "$\\mathrm{K}_2 \\mathrm{Cr}_2 \\mathrm{O}_7$" }, { "text": "$\\mathrm{CuSO}_4 \\cdot 5 \\mathrm{H}_2 \\mathrm{O}$" }, { "text": "$\\mathrm{KMnO}_4$" }, { "text": "$\\mathrm{K}_2 \\mathrm{CrO}_4$" } ], "answer": "$\\mathrm{CuSO}_4 \\cdot 5 \\mathrm{H}_2 \\mathrm{O}$", "solution": "**Answer:** $\\mathrm{CuSO}_4 \\cdot 5 \\mathrm{H}_2 \\mathrm{O}$\n\n

Among the given options, the compounds that show color due to d-d transitions are those that have partially filled d-orbitals when in the form of complex ions or compounds.\n\n

Let's examine each option:\n\n

Option A: $\\mathrm{K}_2 \\mathrm{Cr}_2 \\mathrm{O}_7$\n\n

Chromium in $\\mathrm{K}_2 \\mathrm{Cr}_2 \\mathrm{O}_7$ exists in the +6 oxidation state as the dichromate ion ($\\mathrm{Cr}_2\\mathrm{O}_7^{2-}$). In this oxidation state, chromium does not have any electrons in the d-orbitals (it has a $\\mathrm{[Ar]}\\;3\\mathrm{d}^0$ configuration) so there is no possibility for d-d transitions. Instead, the color is due to charge transfer transitions.\n\n

Option B: $\\mathrm{CuSO}_4 \\cdot 5 \\mathrm{H}_2 \\mathrm{O}$\n\n

Copper(II) sulfate pentahydrate, $\\mathrm{CuSO}_4 \\cdot 5 \\mathrm{H}_2 \\mathrm{O}$, contains copper in a +2 oxidation state. In this state, copper has a $\\mathrm{[Ar]}\\;3\\mathrm{d}^9$ electron configuration, which means there is one unpaired electron in the d-orbitals. Due to the presence of this unpaired electron, d-d transitions can occur, and this is why $\\mathrm{CuSO}_4 \\cdot 5 \\mathrm{H}_2 \\mathrm{O}$ is blue in color.\n\n

Option C: $\\mathrm{KMnO}_4$\n\n

In potassium permanganate ($\\mathrm{KMnO}_4$), manganese is in the +7 oxidation state, and it has a $\\mathrm{[Ar]}\\;3\\mathrm{d}^0$ electron configuration. Like chromium in the +6 state, manganese in the +7 state does not have any d-electrons available for d-d transitions. The deep purple color of $\\mathrm{KMnO}_4$ is also due to charge transfer transitions.\n\n

Option D: $\\mathrm{K}_2 \\mathrm{CrO}_4$\n\n

Potassium chromate ($\\mathrm{K}_2 \\mathrm{CrO}_4$) contains chromium in the +6 oxidation state as the chromate ion ($\\mathrm{CrO}_4^{2-}$). Like dichromate, chromium does not have d-electrons and therefore cannot undergo d-d transitions in this compound. The color is due to charge transfer transitions.\n\n

So the only compound from the options listed that shows color due to d-d transitions is Option B: $\\mathrm{CuSO}_4 \\cdot 5 \\mathrm{H}_2 \\mathrm{O}$.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 982, "subject": "Chemistry", "question": "Given below are two statements :

\nStatement (I) : A solution of $\\left[\\mathrm{Ni}\\left(\\mathrm{H}_2 \\mathrm{O}\\right)_6\\right]^{2+}$ is green in colour.

\nStatement (II) : A solution of $\\left[\\mathrm{Ni}(\\mathrm{CN})_4\\right]^{2-}$ is colourless.

\nIn the light of the above statements, choose the most appropriate answer from the options given below :\n\n", "options": [ { "text": "Statement I is incorrect but Statement II is correct" }, { "text": "Both Statement I and Statement II are correct" }, { "text": "Both Statement I and Statement II are incorrect" }, { "text": "Statement I is correct but Statement II is incorrect" } ], "answer": "Both Statement I and Statement II are correct", "solution": "**Answer:** Both Statement I and Statement II are correct\n\n

In order to determine the correctness of the given statements, we need to analyze the nature of the mentioned complexes.

\n\n

Statement (I) asserts that a solution of $\\left[\\mathrm{Ni}\\left(\\mathrm{H}_2 \\mathrm{O}\\right)_6\\right]^{2+}$ is green in color. This complex involves a nickel(II) cation coordinated with six water molecules. Nickel(II) has the electronic configuration $[Ar]3d^8$. In an octahedral field created by water ligands, the d-orbitals split into two sets, $t_{2g}$ and $e_g$, due to crystal field splitting. The presence of unpaired electrons allows for d-d transitions when light is absorbed, which is responsible for the characteristic color of the complex. Since typical $\\left[\\mathrm{Ni}\\left(\\mathrm{H}_2 \\mathrm{O}\\right)_6\\right]^{2+}$ complexes are indeed green in color due to such transitions, Statement (I) is correct.

\n\n

Statement (II) states that a solution of $\\left[\\mathrm{Ni}(\\mathrm{CN})_4\\right]^{2-}$ is colorless. In this complex, nickel(II) is surrounded by four cyanide ligands. Cyanide is a strong field ligand, which leads to a large crystal field splitting that exceeds the pairing energy of the electrons. Hence, all the electrons in the nickel(II) ion are paired, and there are no unpaired electrons available for d-d transitions. As there are no d-d transitions to absorb light in the visible region, the complex appears colorless. Therefore, Statement (II) is also correct.

\n\n

Given that both statements are correct, the most appropriate answer would be :

\n\n

Option B : Both Statement I and Statement II are correct.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 983, "subject": "Chemistry", "question": "

Consider the following complex ions

\n

$$\\begin{aligned}\n& \\mathrm{P}=\\left[\\mathrm{FeF}_6\\right]^{3-} \\\\\n& \\mathrm{Q}=\\left[\\mathrm{V}\\left(\\mathrm{H}_2 \\mathrm{O}\\right)_6\\right]^{2+} \\\\\n& \\mathrm{R}=\\left[\\mathrm{Fe}\\left(\\mathrm{H}_2 \\mathrm{O}\\right)_6\\right]^{2+}\n\\end{aligned}$$

\n

The correct order of the complex ions, according to their spin only magnetic moment values (in B.M.) is :

", "options": [ { "text": "R < Q < P" }, { "text": "R < P < Q" }, { "text": "Q < R < P" }, { "text": "Q < P < R" } ], "answer": "Q < R < P", "solution": "**Answer:** Q < R < P\n\n

$$\\left[\\mathrm{FeF}_6\\right]^{3-}: \\mathrm{Fe}^{+3}:[\\mathrm{Ar}] 3 \\mathrm{~d}^5$$

\n

F : Weak field Ligand

\n

\"JEE

\n

No. of unpaired electron's $$=5$$

\n

$$\\begin{aligned}\n& \\mu=\\sqrt{5(5+2)} \\\\\n& \\mu=\\sqrt{35} \\mathrm{~BM} \\\\\n& {\\left[\\mathrm{V}\\left(\\mathrm{H}_2 \\mathrm{O}\\right)_6\\right]^{+2}: \\mathrm{V}^{+2}: 3 \\mathrm{~d}^3}\n\\end{aligned}$$

\n

\"JEE

\n

No. of unpaired electron's $$=3$$

\n

$$\\begin{aligned}\n& \\mu=\\sqrt{3(3+2)} \\\\\n& \\mu=\\sqrt{15} \\mathrm{~BM} \\\\\n& {\\left[\\mathrm{Fe}\\left(\\mathrm{H}_2 \\mathrm{O}\\right)_6\\right]^{+2}: \\mathrm{Fe}^{+2}: 3 \\mathrm{~d}^6}\n\\end{aligned}$$

\n

H$$_2$$O : Weak field Ligand

\n

\"JEE

\n

No. of unpaired electron's $$=4$$

\n

$$\\begin{aligned}\n& \\mu=\\sqrt{4(4+2)} \\\\\n& \\mu=\\sqrt{24} \\mathrm{~BM}\n\\end{aligned}$$

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 984, "subject": "Chemistry", "question": "

The Spin only magnetic moment value of square planar complex $$\\left[\\mathrm{Pt}\\left(\\mathrm{NH}_3\\right)_2 \\mathrm{Cl}\\left(\\mathrm{NH}_2 \\mathrm{CH}_3\\right)\\right] \\mathrm{Cl}$$ is _________ B.M. (Nearest integer)

\n

(Given atomic number for $$\\mathrm{Pt}=78$$)

", "options": [], "answer": "0", "solution": "**Answer:** 0\n\n

$$\\mathrm{Pt^2+ (d^8)}$$

\n

\"JEE

\n

$$\\mathrm{Pt}^{2+} \\rightarrow \\mathrm{dsp}^2$$ hybridization and have no unpaired $$\\mathrm{e}^{-} \\mathrm{s}$$.

\n

$$\\therefore$$ Magnetic moment $$=0$$

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 985, "subject": "Chemistry", "question": "

Select the option with correct property -

", "options": [ { "text": "$$\\left[\\mathrm{Ni}(\\mathrm{CO})_4\\right]$$ and $$\\left[\\mathrm{NiCl}_4\\right]^{2-}$$ both Paramagnetic\n" }, { "text": "$$\\left[\\mathrm{Ni}(\\mathrm{CO})_4\\right]$$ and $$\\left[\\mathrm{NiCl}_4\\right]^{2-}$$ both Diamagnetic\n" }, { "text": "$$\\left[\\mathrm{NiCl}_4\\right]^{2-}$$ Diamagnetic, $$\\left[\\mathrm{Ni}(\\mathrm{CO})_4\\right]$$ Paramagnetic\n" }, { "text": "$$\\left[\\mathrm{Ni}(\\mathrm{CO})_4\\right]$$ Diamagnetic, $$\\left[\\mathrm{NiCl}_4\\right]^{2-}$$ Paramagnetic" } ], "answer": "$$\\left[\\mathrm{Ni}(\\mathrm{CO})_4\\right]$$ Diamagnetic, $$\\left[\\mathrm{NiCl}_4\\right]^{2-}$$ Paramagnetic", "solution": "**Answer:** $$\\left[\\mathrm{Ni}(\\mathrm{CO})_4\\right]$$ Diamagnetic, $$\\left[\\mathrm{NiCl}_4\\right]^{2-}$$ Paramagnetic\n\n

$$\\left[\\mathrm{Ni}(\\mathrm{CO})_4\\right] \\rightarrow$$ diamagnetic, $$\\mathrm{sp}^3$$ hybridisation, number of unpaired electrons $$=0$$

\n

$$\\left[\\mathrm{NiCl}_4\\right]^{2-}, \\rightarrow$$ paramagnetic, $$\\mathrm{sp}^3$$ hybridisation, number of unpaired electrons $$=2$$

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 986, "subject": "Chemistry", "question": "

Match List I with List II

\n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
List - I
(Complex ion)		List - II
(Electronic Configuration)
(A)$$\\mathrm{[Cr(H_2O)_6]^{3+}}$$(I)$$t_{2 g}{ }^2 e_g^0$$
(B)$$\\left[\\mathrm{Fe}\\left(\\mathrm{H}_2 \\mathrm{O}\\right)_6\\right]^{3+}$$(II)$$t_{2 g}{ }^3 e_g{ }^0$$
(C)$$\\left[\\mathrm{Ni}\\left(\\mathrm{H}_2 \\mathrm{O}\\right)_6\\right]^{2+}$$(III)$$t_{2 g}{ }^3 e_g{ }^2$$
(D)$$\\left[\\mathrm{V}\\left(\\mathrm{H}_2 \\mathrm{O}\\right)_6\\right]^{3+}$$(IV)$$t_{2 g}{ }^6 e_g^2$$

\n

Choose the correct answer from the options given below:

", "options": [ { "text": "A-IV, B-I, C-II, D-III\n" }, { "text": "A-III, B-II, C-IV, D-I\n" }, { "text": "A-II, B-III, C-IV, D-I\n" }, { "text": "A-IV, B-III, C-I, D-II" } ], "answer": "A-II, B-III, C-IV, D-I\n", "solution": "**Answer:** A-II, B-III, C-IV, D-I\n\n\n

$$\\begin{aligned}\n& {\\left[\\mathrm{Cr}\\left(\\mathrm{H}_2 \\mathrm{O}\\right)_6\\right]^{3+} \\text { Contains } \\mathrm{Cr}^{3+}:[\\mathrm{Ar}] 3 \\mathrm{~d}^3: \\mathrm{t}_{2 \\mathrm{~g}}^3 \\mathrm{e}_{\\mathrm{g}}^{\\mathrm{o}}} \\\\\n& {\\left[\\mathrm{Fe}\\left(\\mathrm{H}_2 \\mathrm{O}\\right)_6\\right]^{3+} \\text { Contains } \\mathrm{Fe}^{3+}:[\\mathrm{Ar}] 3 \\mathrm{~d}^5: \\mathrm{t}_{2 \\mathrm{~g}}^3 \\mathrm{e}_{\\mathrm{g}}^2} \\\\\n& {\\left[\\mathrm{Ni}\\left(\\mathrm{H}_2 \\mathrm{O}\\right)_6\\right]^{2+} \\text { Contains } \\mathrm{Ni}^{2+}:[\\mathrm{Ar}] 3 \\mathrm{~d}^8: \\mathrm{t}_{2 \\mathrm{~g}}^6 \\mathrm{e}_{\\mathrm{g}}^2} \\\\\n& {\\left[\\mathrm{~V}\\left(\\mathrm{H}_2 \\mathrm{O}\\right)_6\\right]^{3+} \\text { Contains } \\mathrm{V}^{3+}:[\\mathrm{Ar}] 3 \\mathrm{~d}^2: \\mathrm{t}_{2 \\mathrm{~g}}^2 \\mathrm{eg}^{\\mathrm{o}}}\n\\end{aligned}$$

", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 987, "subject": "Chemistry", "question": "

The 'Spin only' Magnetic moment for $$\\left[\\mathrm{Ni}\\left(\\mathrm{NH}_3\\right)_6\\right]^{2+}$$ is _________ $$\\times 10^{-1} \\mathrm{~BM}$$. (given $$=$$ Atomic number of $$\\mathrm{Ni}: 28$$)

", "options": [], "answer": "28", "solution": "**Answer:** 28\n\n

$$\\mathrm{NH}_3$$ act as WFL with $$\\mathrm{Ni}^{2+}$$

\n

$$\\mathrm{Ni}^{2+}=3 \\mathrm{~d}^8$$

\n

\"JEE

\n

No. of unpaired electron $$=2$$

\n

$$\\begin{aligned}\n\\mu=\\sqrt{\\mathrm{n}(\\mathrm{n}+2)} & =\\sqrt{8}=2.82 \\mathrm{~BM} \\\\\n& =28.2 \\times 10^{-1} \\mathrm{~BM} \\\\\n\\mathrm{x} & =28\n\\end{aligned}$$

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 988, "subject": "Chemistry", "question": "

Number of complexes from the following with even number of unpaired \"$$\\mathrm{d}$$\" electrons is ________ $$[\\mathrm{V}(\\mathrm{H}_2 \\mathrm{O})_6]^{3+},[\\mathrm{Cr}(\\mathrm{H}_2 \\mathrm{O})_6]^{2+},[\\mathrm{Fe}(\\mathrm{H}_2 \\mathrm{O})_6]^{3+},[\\mathrm{Ni}(\\mathrm{H}_2 \\mathrm{O})_6]^{3+},[\\mathrm{Cu}(\\mathrm{H}_2 \\mathrm{O})_6]^{2+}$$ [Given atomic numbers: $$\\mathrm{V}=23, \\mathrm{Cr}=24, \\mathrm{Fe}=26, \\mathrm{Ni}=28 \\mathrm{Cu}=29$$]

", "options": [ { "text": "1" }, { "text": "5" }, { "text": "2" }, { "text": "4" } ], "answer": "2", "solution": "**Answer:** 2\n\n

$$\\left[\\mathrm{V}\\left(\\mathrm{H}_2 \\mathrm{O}\\right)_6\\right]^{3+} \\rightarrow 2$$ unpaired electrons

\n

$$\\left[\\mathrm{Cr}\\left(\\mathrm{H}_2 \\mathrm{O}\\right)_6\\right]^{2+} \\rightarrow 4$$ unpaired electrons

\n

Above 2 complex have even number of unpaired electrons.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 989, "subject": "Chemistry", "question": "

The correct sequence of ligands in the order of decreasing field strength is :

", "options": [ { "text": "$$\\mathrm{NCS}^{-}>\\mathrm{EDTA}^{4-}>\\mathrm{CN}^{-}>\\mathrm{CO}$$\n" }, { "text": "$$\\mathrm{CO}>\\mathrm{H}_2 \\mathrm{O}>\\mathrm{F}^{-}>\\mathrm{S}^{2-}$$\n" }, { "text": "$$\\mathrm{S}^{2-}>{ }^{-} \\mathrm{OH}>\\mathrm{EDTA}^{4-}>\\mathrm{CO}$$\n" }, { "text": "$${ }^{-} \\mathrm{OH}>\\mathrm{F}^{-}>\\mathrm{NH}_3>\\mathrm{CN}^{-}$$" } ], "answer": "$$\\mathrm{CO}>\\mathrm{H}_2 \\mathrm{O}>\\mathrm{F}^{-}>\\mathrm{S}^{2-}$$\n", "solution": "**Answer:** $$\\mathrm{CO}>\\mathrm{H}_2 \\mathrm{O}>\\mathrm{F}^{-}>\\mathrm{S}^{2-}$$\n\n\n

$$\n\\text { Field strength order : } \\mathrm{CO}>\\mathrm{H}_2 \\mathrm{O}>\\mathrm{F}^{-}>\\mathrm{S}^{2-}\n$$

", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 990, "subject": "Chemistry", "question": "

If an iron (III) complex with the formula $$\\left[\\mathrm{Fe}\\left(\\mathrm{NH}_3\\right)_x(\\mathrm{CN})_y\\right]^-$$ has no electron in its $$e_g$$ orbital, then the value of $$x+y$$ is

", "options": [ { "text": "6" }, { "text": "4" }, { "text": "3" }, { "text": "5" } ], "answer": "6", "solution": "**Answer:** 6\n\n

Balancing charges,

\n

$$\\begin{aligned}\n& 3-y=-1 \\\\\n& \\Rightarrow y=4 \\\\\n& x=2 \\\\\n& x+y=6 \\\\\n&\n\\end{aligned}$$

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 991, "subject": "Chemistry", "question": "

The number of unpaired d-electrons in $$\\left[\\mathrm{Co}\\left(\\mathrm{H}_2 \\mathrm{O}\\right)_6\\right]^{3+}$$ is ________.

", "options": [ { "text": "0" }, { "text": "2" }, { "text": "1" }, { "text": "4" } ], "answer": "0", "solution": "**Answer:** 0\n\n

$$\\left[\\mathrm{Co}\\left(\\mathrm{H}_2 \\mathrm{O}\\right)_6\\right]^{3+}$$ is an inner orbital complex.

\n

So, electronic configuration is $$\\mathrm{t}_{2 \\mathrm{~g}}^6 \\mathrm{e}_{\\mathrm{g}}^0$$.

\n

All electrons are paired.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 992, "subject": "Chemistry", "question": "

A first row transition metal in its +2 oxidation state has a spin-only magnetic moment value of $$3.86 \\mathrm{~BM}$$. The atomic number of the metal is

", "options": [ { "text": "22" }, { "text": "23" }, { "text": "26" }, { "text": "25" } ], "answer": "23", "solution": "**Answer:** 23\n\n

To determine the atomic number of the metal based on the provided spin-only magnetic moment value, we use the formula for calculating the spin-only magnetic moment:

\n\n

$ \\mu = \\sqrt{n(n+2)} \\mu_{B} $

\n\n

where $ \\mu $ is the magnetic moment in Bohr magnetons ($ \\mu_{B} $), and $ n $ is the number of unpaired electrons.

\n\n

Given that the spin-only magnetic moment value is $ 3.86 \\, \\mathrm{BM} $, we can set this value equal to the formula to solve for $ n $:

\n\n

$ 3.86 = \\sqrt{n(n+2)} $

\n\n

Squaring both sides gives:

\n\n

$ 14.8996 = n(n+2) $

\n\n

Since $ n $ must be an integer and this equation doesn't solve neatly for an integer $ n $, we look for a value of $ n $ that would give a product close to $ 14.8996 $ when plugged into $ \\sqrt{n(n+2)} $.

\n\n

Practically, we can test the square root values near $ 3.86 $ for different $ n $ to see which one gives a close match. Considering the usual spin-only magnetic moments for common numbers of unpaired electrons:

\n\n\n

The value $ n = 3 $ gives a magnetic moment of approximately $ 3.87 \\mu_B $, which is very close to the given value, $ 3.86 \\mu_B $. Therefore, the metal has 3 unpaired electrons in its d-orbital configuration.

\n\n

In the +2 oxidation state, metals lose electrons from the s-orbital before the d-orbital. Given that the element is a first row transition metal, starting with scandium (Sc) at atomic number 21 which has an electronic configuration of $[Ar] 3d^1 4s^2$ in its neutral state, we can deduce the configurations:

\n\n\n

So, the atomic number of the metal with a +2 oxidation state and a spin-only magnetic moment value of $3.86 \\mathrm{~BM}$ (corresponding to 3 unpaired electrons) is 23, which identifies the metal as Vanadium (V). Therefore,

\n\n

Option B (23) is the correct answer.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 993, "subject": "Chemistry", "question": "

Match List I with List II

\n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
LIST I
(Complex ion)		LIST II
(Spin only magnetic moment in B.M.)
A.$$
\\left[\\mathrm{Cr}\\left(\\mathrm{NH}_3\\right)_6\\right]^{3+}
$$		I.4.90
B.\t$$
\\left[\\mathrm{NiCl}_4\\right]^{2-}
$$
II.3.87
C.$$
\\left[\\mathrm{CoF}_6\\right]^{3-}
$$
III.0.0
D.$$
\\left[\\mathrm{Ni}(\\mathrm{CN})_4\\right]^{2-}
$$
IV.2.83

\n

Choose the correct answer from the options given below :

", "options": [ { "text": "(A)-(IV), (B)-(III), (C)-(I), (D)-(II)\n" }, { "text": "(A)-(II), (B)-(III), (C)-(I), (D)-(IV)\n" }, { "text": "(A)-(I), (B)-(IV), (C)-(II), (D)-(III)\n" }, { "text": "(A)-(II), (B)-(IV), (C)-(I), (D)-(III)" } ], "answer": "(A)-(II), (B)-(IV), (C)-(I), (D)-(III)", "solution": "**Answer:** (A)-(II), (B)-(IV), (C)-(I), (D)-(III)\n\n

\"JEE

", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 994, "subject": "Chemistry", "question": "

Total number of unpaired electrons in the complex ions $$[\\mathrm{Co}(\\mathrm{NH}_3)_6]^{3+}$$ and $$[\\mathrm{NiCl}_4]^{2-}$$ is ________.

", "options": [], "answer": "2", "solution": "**Answer:** 2\n\n

$$\\left[\\mathrm{Co}\\left(\\mathrm{NH}_3\\right)_6\\right]^{3+}: \\mathrm{Co}^{3+}: \\mathrm{t}_{2 \\mathrm{~g}}^6 \\mathrm{e}_{\\mathrm{g}}^0: \\mathrm{n}=0$$

\n

$$\\left[\\mathrm{NiCl}_4\\right]^{2-}: \\mathrm{Ni}^{2+}: t_{2 \\mathrm{~g}}^6 \\mathrm{e}_{\\mathrm{g}}^2: \\mathrm{n}=2$$

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 995, "subject": "Chemistry", "question": "

Number of Complexes with even number of electrons in $$\\mathrm{t_{2 g}}$$ orbitals is -

\n

$$\\left[\\mathrm{Fe}\\left(\\mathrm{H}_2 \\mathrm{O}\\right)_6\\right]^{2+},\\left[\\mathrm{Co}\\left(\\mathrm{H}_2 \\mathrm{O}\\right)_6\\right]^{2+},\\left[\\mathrm{Co}\\left(\\mathrm{H}_2 \\mathrm{O}\\right)_6\\right]^{3+},\\left[\\mathrm{Cu}\\left(\\mathrm{H}_2 \\mathrm{O}\\right)_6\\right]^{2+},\\left[\\mathrm{Cr}\\left(\\mathrm{H}_2 \\mathrm{O}\\right)_6\\right]^{2+}$$

", "options": [ { "text": "3" }, { "text": "2" }, { "text": "1" }, { "text": "5" } ], "answer": "3", "solution": "**Answer:** 3\n\n

To determine the number of complexes with an even number of electrons in the $$\\mathrm{t_{2g}}$$ orbitals, we first need to determine the electronic configuration of the metal ions in each complex and then find the distribution of electrons among the $$\\mathrm{t_{2g}}$$ and $$\\mathrm{e_{g}}$$ orbitals.

\n\n

Let's consider each complex one by one:

\n\n

1. $$\\left[\\mathrm{Fe}\\left(\\mathrm{H}_2 \\mathrm{O}\\right)_6\\right]^{2+}$$

\n\n

The oxidation state of Fe is +2. The electronic configuration of $$\\mathrm{Fe}^{2+}$$ is $$\\left[\\mathrm{Ar}\\right]3d^6$$.

\n\n

In an octahedral field, the splitting pattern for the 3d orbitals is such that: $$t_{2g}$$ (lower energy) and $$e_g$$ (higher energy).

\n\n

So, the electronic configuration in octahedral field will be: $$t_{2g}^4 e_g^2$$. Therefore, there are 4 electrons in the $$t_{2g}$$ orbitals (even).

\n\n

2. $$\\left[\\mathrm{Co}\\left(\\mathrm{H}_2 \\mathrm{O}\\right)_6\\right]^{2+}$$

\n\n

The oxidation state of Co is +2. The electronic configuration of $$\\mathrm{Co}^{2+}$$ is $$\\left[\\mathrm{Ar}\\right]3d^7$$.

\n\n

In an octahedral field, the splitting pattern for the 3d orbitals will be: $$t_{2g}^5 e_g^2$$. Therefore, there are 5 electrons in the $$t_{2g}$$ orbitals (odd).

\n\n

3. $$\\left[\\mathrm{Co}\\left(\\mathrm{H}_2 \\mathrm{O}\\right)_6\\right]^{3+}$$

\n\n

The oxidation state of Co is +3. The electronic configuration of $$\\mathrm{Co}^{3+}$$ is $$\\left[\\mathrm{Ar}\\right]3d^6$$.

\n\n

In an octahedral field, the splitting pattern for the 3d orbitals will be: $$t_{2g}^4 e_g^2$$. Therefore, there are 4 electrons in the $$t_{2g}$$ orbitals (even).

\n\n

4. $$\\left[\\mathrm{Cu}\\left(\\mathrm{H}_2 \\mathrm{O}\\right)_6\\right]^{2+}$$

\n\n

The oxidation state of Cu is +2. The electronic configuration of $$\\mathrm{Cu}^{2+}$$ is $$\\left[\\mathrm{Ar}\\right]3d^9$$.

\n\n

In an octahedral field, the splitting pattern for the 3d orbitals will be: $$t_{2g}^6 e_g^3$$. Therefore, there are 6 electrons in the $$t_{2g}$$ orbitals (even).

\n\n

5. $$\\left[\\mathrm{Cr}\\left(\\mathrm{H}_2 \\mathrm{O}\\right)_6\\right]^{2+}$$

\n\n

The oxidation state of Cr is +2. The electronic configuration of $$\\mathrm{Cr}^{2+}$$ is $$\\left[\\mathrm{Ar}\\right]3d^4$$.

\n\n

In an octahedral field, the splitting pattern for the 3d orbitals will be: $$t_{2g}^3 e_g^1$$. Therefore, there are 3 electrons in the $$t_{2g}$$ orbitals (odd).

\n\n

So, the complexes with an even number of electrons in the $$\\mathrm{t_{2g}}$$ orbitals are:

\n\n

1. $$\\left[\\mathrm{Fe}\\left(\\mathrm{H}_2 \\mathrm{O}\\right)_6\\right]^{2+}$$ (4 electrons)
\n\n

    \n\n
  1. $$\\left[\\mathrm{Co}\\left(\\mathrm{H}_2 \\mathrm{O}\\right)_6\\right]^{3+}$$ (4 electrons)

    2. \n\n
  2. $$\\left[\\mathrm{Cu}\\left(\\mathrm{H}_2 \\mathrm{O}\\right)_6\\right]^{2+}$$ (6 electrons)

    3. \n\n
\n\n

Hence, there are 3 complexes with an even number of electrons in the $$\\mathrm{t_{2g}}$$ orbitals. Therefore, the correct answer is option A.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 996, "subject": "Chemistry", "question": "

The 'spin only' magnetic moment value of $$\\mathrm{MO}_4{ }^{2-}$$ is ________ BM. (Where M is a metal having least metallic radii. among $$\\mathrm{Sc}, \\mathrm{Ti}, \\mathrm{V}, \\mathrm{Cr}, \\mathrm{Mn}$$ and $$\\mathrm{Zn}$$ ).

\n

(Given atomic number: $$\\mathrm{Sc}=21, \\mathrm{Ti}=22, \\mathrm{~V}=23, \\mathrm{Cr}=24, \\mathrm{Mn}=25$$ and $$\\mathrm{Zn}=30$$)

", "options": [], "answer": "0", "solution": "**Answer:** 0\n\n

The given ion is $$\\mathrm{MO}_4{ }^{2-}$$ where M is the metal ion and needs to be identified from Sc, Ti, V, Cr, Mn, and Zn based on the least metallic radii. Among these metals, Zn has the least metallic radii. Therefore, M is identified as Zn.

\n\n

Zinc (Zn) has an atomic number of 30. In a neutral state, zinc has the electronic configuration:

\n\n

$$\\mathrm{Zn: [Ar] 3d^{10} 4s^2}$$

\n\n

Upon forming the ion $$\\mathrm{Zn}^{2+}$$, it loses two electrons from the 4s orbital, resulting in the electronic configuration:

\n\n

$$\\mathrm{Zn^{2+}: [Ar] 3d^{10}}$$

\n\n

In the $$\\mathrm{Zn}^{2+}$$ ion, all d-orbitals are fully filled with electrons. Therefore, there are no unpaired electrons present in the $$\\mathrm{Zn}^{2+}$$ ion.

\n\n

The ‘spin-only’ magnetic moment is given by the formula:

\n\n

$$\\mu = \\sqrt{n(n+2)} \\, \\text{BM}$$

\n\n

where $$n$$ is the number of unpaired electrons.

\n\n

For $$\\mathrm{Zn}^{2+}$$, $$n = 0$$.

\n\n

Substituting $$n = 0$$ in the formula we get:

\n\n

$$\\mu = \\sqrt{0(0+2)} = \\sqrt{0} = 0 \\, \\text{BM}$$

\n\n

Therefore, the 'spin-only' magnetic moment value of $$\\mathrm{MO}_4{ }^{2-}$$ where M is Zn is 0 BM.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 997, "subject": "Chemistry", "question": "

The correct order of ligands arranged in increasing field strength.

", "options": [ { "text": "$$\\mathrm{F}^{-}<\\mathrm{Br}^{-}<\\mathrm{I}^{-}<\\mathrm{NH}_3$$\n" }, { "text": "$$\\mathrm{H}_2 \\mathrm{O}<\\mathrm{^{-}OH}<\\mathrm{CN}^{-}<\\mathrm{NH}_3$$\n" }, { "text": "$$\\mathrm{Br}^{-}<\\mathrm{F}^{-}<\\mathrm{H}_2 \\mathrm{O}<\\mathrm{NH}_3$$\n" }, { "text": "$$\\mathrm{Cl}^{-}<\\mathrm{^{-}OH}<\\mathrm{Br}^{-}<\\mathrm{CN}^{-}$$" } ], "answer": "$$\\mathrm{Br}^{-}<\\mathrm{F}^{-}<\\mathrm{H}_2 \\mathrm{O}<\\mathrm{NH}_3$$\n", "solution": "**Answer:** $$\\mathrm{Br}^{-}<\\mathrm{F}^{-}<\\mathrm{H}_2 \\mathrm{O}<\\mathrm{NH}_3$$\n\n\n

The correct answer is Option C:\n\n

$$ \\mathrm{Br}^{-}<\\mathrm{F}^{-}<\\mathrm{H}_2 \\mathrm{O}<\\mathrm{NH}_3 $$.

\n\n

\n\n

Here's why:

\n\n

The spectrochemical series is a list of ligands arranged in order of increasing field strength. This means that ligands higher on the series cause a larger splitting of the d-orbitals in a transition metal complex, resulting in a larger crystal field stabilization energy (CFSE).

\n\n

Here's a breakdown of the factors influencing the spectrochemical series:

\n\n\n\n

Let's analyze why the other options are incorrect:

\n\n\n\n

Therefore, the correct order in increasing field strength is: $$ \\mathrm{Br}^{-}<\\mathrm{F}^{-}<\\mathrm{H}_2 \\mathrm{O}<\\mathrm{NH}_3 $$.

\n\n

", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 998, "subject": "Chemistry", "question": "

Which one of the following complexes will exhibit the least paramagnetic behaviour ?\n[Atomic number, $$\\mathrm{Cr}=24, \\mathrm{Mn}=25, \\mathrm{Fe}=26, \\mathrm{Co}=27$$]

", "options": [ { "text": "$$\\left[\\mathrm{Fe}\\left(\\mathrm{H}_2 \\mathrm{O}\\right)_6\\right]^{2+}$$\n" }, { "text": "$$\\left[\\mathrm{Mn}\\left(\\mathrm{H}_2 \\mathrm{O}\\right)_6\\right]^{2+}$$\n" }, { "text": "$$\\left[\\mathrm{Co}\\left(\\mathrm{H}_2 \\mathrm{O}\\right)_6\\right]^{2+}$$\n" }, { "text": "$$\\left[\\mathrm{Cr}\\left(\\mathrm{H}_2 \\mathrm{O}\\right)_6\\right]^{2+}$$" } ], "answer": "$$\\left[\\mathrm{Co}\\left(\\mathrm{H}_2 \\mathrm{O}\\right)_6\\right]^{2+}$$\n", "solution": "**Answer:** $$\\left[\\mathrm{Co}\\left(\\mathrm{H}_2 \\mathrm{O}\\right)_6\\right]^{2+}$$\n\n\n

\"JEE

\n

So, the complex with minimum number of unpaired $$\\mathrm{e}^{\\ominus}$$ is option (2) $$\\left[\\mathrm{Co}\\left(\\mathrm{H}_2 \\mathrm{O}\\right)_6\\right]^{2+}$$.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 999, "subject": "Chemistry", "question": "

The spin-only magnetic moment value of the ion among $$\\mathrm{Ti}^{2+}, \\mathrm{V}^{2+}, \\mathrm{Co}^{3+}$$ and $$\\mathrm{Cr}^{2+}$$, that acts as strong oxidising agent in aqueous solution is _________ BM (Near integer).

\n

(Given atomic numbers : $$\\mathrm{Ti}: 22, \\mathrm{~V}: 23, \\mathrm{Cr}: 24, \\mathrm{Co}: 27$$)

", "options": [], "answer": "5", "solution": "**Answer:** 5\n\n

The ion which acts as strong oxidising agent in aqueous solution is $$\\mathrm{Cr}^{2+}:[\\mathrm{Ar}] 4 s^{\\circ} 3 d^4$$

\n

$$\\mu=\\sqrt{4(4+2)}=4.89 \\Rightarrow 5$$

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1000, "subject": "Chemistry", "question": "

The number of complexes from the following with no electrons in the $$t_2$$ orbital is ______.

$$\\mathrm{TiCl}_4,\\left[\\mathrm{MnO}_4\\right]^{-},\\left[\\mathrm{FeO}_4\\right]^{2-},\\left[\\mathrm{FeCl}_4\\right]^{-},\\left[\\mathrm{CoCl}_4\\right]^{2-}$$

", "options": [ { "text": "4" }, { "text": "2" }, { "text": "3" }, { "text": "1" } ], "answer": "3", "solution": "**Answer:** 3\n\n

Identifying the electronic configuration and geometry :

\n

The condition \"no electrons in the $ t_2 $ orbital\" typically applies to a tetrahedral crystal field splitting pattern. In a tetrahedral field:

\n\n\n

The $ d $-orbitals split into two sets: $ e $ (lower energy, 2 orbitals) and $ t_2 $ (higher energy, 3 orbitals).

\n

Electrons fill the lower-energy $ e $ set before occupying the $ t_2 $ set.

\n

Thus, to have no electrons in $ t_2 $, either:

\n

The metal has a $ d^0 $ configuration (no $ d $-electrons at all), or

\n

The $ d $-electron count is so low that all electrons can fit into the $ e $ orbitals without needing to occupy $ t_2 $.

\n\n\n

Analyzing each complex :

\n

(i) $\\mathrm{TiCl}_4$ :

\n\n\n

Ti in TiCl₄ is in the +4 oxidation state (since TiCl₄ is neutral and each Cl is -1).

\n

Ti (Z = 22) neutral : $ 3d^2 4s^2 $. As Ti⁴⁺, it loses all 4 valence electrons (2 from 4s and 2 from 3d), resulting in $ d^0 $.

\n

With $ d^0 $, there are no electrons to occupy $ t_2 $ orbitals.

\n

No electrons in $ t_2 $.

\n

(ii) $[\\mathrm{MnO}_4]^{-}$ (permanganate) :

\n

Mn in permanganate ($\\mathrm{MnO_4}^-$) is in the +7 oxidation state.

\n

Mn (Z = 25) neutral is $ 3d^5 4s^2 $. Mn⁷⁺ means removing all 7 valence electrons, leaving $ d^0 $.

\n

With $ d^0 $, no electrons in $ t_2 $.

\n

No electrons in $ t_2 $.

\n

(iii) $[\\mathrm{FeO}_4]^{2-}$ (ferrate) :

\n

Fe in $\\mathrm{FeO_4}^{2-}$: Oxygen contributes -8 total. The ion is -2. Thus, Fe + (-8) = -2 → Fe = +6 oxidation state.

\n

Fe (Z = 26) neutral is $ 3d^6 4s^2 $. Fe⁶⁺ means removing 6 electrons from the valence shell. After losing 2 from 4s and 4 from 3d, we get $ d^2 $.

\n

In a tetrahedral field, $ d^2 $ fills the lower $ e $ orbitals (2 electrons into e orbitals).

\n

No need to occupy $ t_2 $ orbitals since both electrons fit into e.

\n

No electrons in $ t_2 $.

\n

(iv) $[\\mathrm{FeCl}_4]^{-}$ :

\n

Fe in $\\mathrm{FeCl_4}^- $: Cl total charge = -4, complex = -1, so Fe = +3.

\n

Fe(III) is $ d^5 $.

\n

In a tetrahedral field, with 5 $ d $-electrons, after filling the 2 $ e $ orbitals, we have 3 more electrons that must go into $ t_2 $.

\n

Has electrons in $ t_2 $.

\n

(v) $[\\mathrm{CoCl}_4]^{2-}$:

\n

Co in $\\mathrm{CoCl_4}^{2-}$ : Cl total = -4, complex = -2, so Co = +2.

\n

Co(II) is $ d^7 $.

\n

For $ d^7 $, even after filling the $ e $ set (2 orbitals), we have 5 more electrons left, which must occupy the $ t_2 $ orbitals.

\n

Has electrons in $ t_2 $.

\n\n\n

Counting the complexes with no electrons in $ t_2 $ :

\n\n\n

TiCl₄: No electrons in $ t_2 $.

\n

[MnO₄]⁻: No electrons in $ t_2 $.

\n

[FeO₄]²⁻: No electrons in $ t_2 $.

\n

[FeCl₄]⁻: Has electrons in $ t_2 $.

\n

[CoCl₄]²⁻: Has electrons in $ t_2 $.

\n\n

Total with no electrons in $ t_2 $ = 3.

\n

Answer :

\n

3

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1001, "subject": "Chemistry", "question": "

Match List I with List II.

\n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
LIST I
Tetrahedral Complex		LIST II
Electronic configuration
A.$$
\\mathrm{TiCl}_4
$$
I.$$
\\mathrm{e}^2, \\mathrm{t}_2^0
$$
B.\t$$
\\left[\\mathrm{FeO}_4\\right]^{2-}
$$
II.$$
\\mathrm{e^4, t_2^3}
$$
C.$$
\\left[\\mathrm{FeCl}_4\\right]^{-}
$$
III.$$
\\mathrm{e}^0, \\mathrm{t}_2^0
$$
D.$$
\\left[\\mathrm{CoCl}_4\\right]^{2-}
$$
IV.$$
\\mathrm{e}^2, \\mathrm{t}_2^3
$$

\n

Choose the correct answer from the options given below :

", "options": [ { "text": "(A)-(IV), (B)-(III), (C)-(I), (D)-(II)\n" }, { "text": "(A)-(III), (B)-(I), (C)-(IV), (D)-(II)\n" }, { "text": "(A)-(I), (B)-(III), (C)-(IV), (D)-(II)\n" }, { "text": "(A)-(III), (B)-(IV), (C)-(II), (D)-(I)" } ], "answer": "(A)-(III), (B)-(I), (C)-(IV), (D)-(II)\n", "solution": "**Answer:** (A)-(III), (B)-(I), (C)-(IV), (D)-(II)\n\n\n

(A) $$\\mathrm{TiCl}_4: \\mathrm{Ti}^{4+}: 3 d^0 4 s^0$$ or $$\\mathrm{e}^0 \\mathrm{t}_2^0$$

\n

(B) $$\\left[\\mathrm{FeO}_4\\right]^{2-}: \\mathrm{Fe}^{6+}: 3 d^2 4 s^0$$ or $$\\mathrm{e}^2 \\mathrm{t}_2^0$$

\n

(C) $$\\left[\\mathrm{FeCl}_4\\right]^{-}: \\mathrm{Fe}^{3+}: 3 d^5 4 s^0$$ or $$\\mathrm{e}^2 \\mathrm{t}_2^3$$

\n

(D) $$\\left[\\mathrm{CoCl}_4\\right]^{2-}: \\mathrm{Co}^{2+}: 3 d^7 4 s^0$$ or $$\\mathrm{e}^4 \\mathrm{t}_2^3$$

\n

$$\\therefore$$ Correct matching of tetrahedral complexes given in List-I with their electronic configuration given in List-II is

\n

(A)-(III); (B)-(I); (C)-(IV); (D)-(II)

", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1002, "subject": "Chemistry", "question": "

Consider the following complexes

\n

(A) $$\\left[\\mathrm{CoCl}\\left(\\mathrm{NH}_3\\right)_5\\right]^{2+}$$, (B) $$\\left[\\mathrm{Co}(\\mathrm{CN})_6\\right]^{3-}$$, (C) $$\n\\left[\\mathrm{Co}\\left(\\mathrm{NH}_3\\right)_5\\left(\\mathrm{H}_2 \\mathrm{O}\\right)\\right]^{3+}\n$$, (D) $$\\left[\\mathrm{Cu}\\left(\\mathrm{H}_2 \\mathrm{O}\\right)_4\\right]^{2+}$$

\n

The correct order of A, B, C and D in terms of wavenumber of light absorbed is :

", "options": [ { "text": "D < A < C < B" }, { "text": "B < C < A < D" }, { "text": "C < D < A < B" }, { "text": "A < C < B < D" } ], "answer": "D < A < C < B", "solution": "**Answer:** D < A < C < B\n\n

Wavenumber $$=\\frac{1}{\\lambda} \\propto$$ Frequency $$\\propto \\Delta_0$$

\n

Wavenumber order : $$\\mathrm{D}<\\mathrm{A}<\\mathrm{C}<\\mathrm{B}$$

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1003, "subject": "Chemistry", "question": "

The difference in the 'spin-only' magnetic moment values of $$\\mathrm{KMnO}_4$$ and the manganese product formed during titration of $$\\mathrm{KMnO}_4$$ against oxalic acid in acidic medium is ________ $$\\mathrm{BM}$$. (nearest integer)

", "options": [], "answer": "6", "solution": "**Answer:** 6\n\n

The 'spin-only' magnetic moment for $$\\mathrm{Mn}^{7+}$$ is $$0$$ BM. During the titration of $$\\mathrm{KMnO}_4$$ with oxalic acid in an acidic medium, manganese is reduced to $$\\mathrm{Mn}^{2+}$$, which has a 'spin-only' magnetic moment of $$5.91$$ BM.

\n\n

Thus, the difference in magnetic moment values is calculated as follows:

\n\n

$$0 \\, \\text{BM}$$ (for $$\\mathrm{Mn}^{7+}$$) subtracted from $$5.91 \\, \\text{BM}$$ (for $$\\mathrm{Mn}^{2+}$$), which results in $$5.91 \\, \\text{BM}$$.

\n\n

Rounding this to the nearest integer, we get a difference of approximately $$6$$ BM.

", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 1004, "subject": "Chemistry", "question": "The type isomerism present in nitropentammine chromium $$\\left( {{\\rm I}{\\rm I}{\\rm I}} \\right)$$ chloride is :", "options": [ { "text": "optical " }, { "text": "linkage" }, { "text": "ionization" }, { "text": "polymerisation." } ], "answer": "linkage", "solution": "**Answer:** linkage\n\nThe chemical formula of nitropentammine chromium $$\\left( {{\\rm I}{\\rm I}{\\rm I}} \\right)$$ chloride is \n

$$\\left[ {Cr{{\\left( {N{H_3}} \\right)}_5}\\,N{O_2}} \\right]C{l_2}$$\n

It can exist in following two structures \n

$$\\left[ {Cr{{\\left( {N{H_3}} \\right)}_5}N{O_2}} \\right]C{l_2}\\,\\,$$ and \n

nitropentammine chromium $$\\left( {{\\rm I}{\\rm I}{\\rm I}} \\right)$$ chloride\n

$$\\left[ {Cr{{\\left( {N{H_3}} \\right)}_5}ONO} \\right]C{l_2}$$ \n

Nitropentammine chromium $$\\left( {{\\rm I}{\\rm I}{\\rm I}} \\right)$$ chloride\n

Therefore the type of isomerism found in this compound is linkage isomerism as nitro group is linked through $$N$$ as $$ - N{O_2}$$ or through $$O$$ as $$ - ONO.$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1005, "subject": "Chemistry", "question": "Which one the following has largest number of isomers? (R = alkyl group, en = ethylenediamine) ", "options": [ { "text": "[Ru(NH3)4Cl2]+" }, { "text": "[Co(en)2Cl2]+" }, { "text": "[Ir(PR3)2 H(CO)]2+ " }, { "text": "[Co(NH3)5Cl]2+" } ], "answer": "[Co(en)2Cl2]+", "solution": "**Answer:** [Co(en)2Cl2]+\n\nIsomers\n

\"AIEEE ", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 1006, "subject": "Chemistry", "question": "Which of the following compounds shows optical isomerism? ", "options": [ { "text": "[Cu(NH3)4]+2 " }, { "text": "[ZnCl4]-2" }, { "text": "[Cr(C2O4)3]-3" }, { "text": "[Co(CN)6]-3" } ], "answer": "[Cr(C2O4)3]-3", "solution": "**Answer:** [Cr(C2O4)3]-3\n\n\"AIEEE ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1007, "subject": "Chemistry", "question": "Which of the following pairs represents linkage isomers ? ", "options": [ { "text": "[Pd (P Ph3)2 (NCS)2] and [Pd (P Ph3)2 (SCN)2]" }, { "text": "[Co(NH3)5 NO3]SO4 and [Co(NH3)5 SO4]NO3" }, { "text": "[Pt Cl2(NH3)4] Br2 and [Pt Br2(NH3)4] Cl2 " }, { "text": "[Cu (NH3)4] [Pt Cl4] and [Pt (NH3)4] [CuCl4]" } ], "answer": "[Pd (P Ph3)2 (NCS)2] and [Pd (P Ph3)2 (SCN)2]", "solution": "**Answer:** [Pd (P Ph3)2 (NCS)2] and [Pd (P Ph3)2 (SCN)2]\n\nThe $$SC{N^ - }$$ ion can coordinate through $$S$$ or $$N$$ atom giving rise to linkage isomerism\n

$$M \\leftarrow SCN\\,\\,\\,$$ thiocyanto\n

$$M \\leftarrow NCS\\,\\,\\,$$ isothiocyanato.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1008, "subject": "Chemistry", "question": "Which of the following has an optical isomer ? \n", "options": [ { "text": "[Co(en)(NH3)2]2+" }, { "text": "[Co(H2O)4(en)]3+" }, { "text": "[Co(en)2(NH3)2]3+" }, { "text": "[Co(NH3)3Cl]+" } ], "answer": "[Co(en)2(NH3)2]3+", "solution": "**Answer:** [Co(en)2(NH3)2]3+\n\n\"AIEEE ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1009, "subject": "Chemistry", "question": "Which one of the following has an optical isomer ?
\n(en = ethylenediamine)", "options": [ { "text": "[Zn (en) (NH3)2]2+ " }, { "text": "[Co (en)3]3+ " }, { "text": "[Co (H2O)4 en]3+ " }, { "text": "[Zn (en)2]2+" } ], "answer": "[Co (en)3]3+ ", "solution": "**Answer:** [Co (en)3]3+ \n\nFor a substance to be optical isomers following conditions should be fulfilled \n

$$(a)\\,\\,\\,$$ A coordination compound which can rotate the plane of polarized light is said to be optically active. \n

$$(b)\\,\\,\\,$$ When the coordination compounds have same formula but differ in their abilities to rotate directions of the plane of polarized light are said to exhibit optical isomerism and the molecules are optical isomers. The optical isomers are pair of molecules which are non-super-imposable mirror images of each other.\n

$$(c)\\,\\,\\,$$ This is due to the absence of elements of symmetry in the complex. \n

$$(d)\\,\\,\\,$$ Optical isomerism is expected in tetrahedral complexes of the type $$Mabcd.$$ \n

Based on this only option $$(2)$$ shows optical isomerism $${\\left[ {Co{{\\left( {en} \\right)}_3}} \\right]^{3 + }}$$ \n

\"AIEEE \n

Complexes of $$Z{n^{ + + }}\\,\\,$$ cannot show optical isomerism as they are tetrahedral complexes with plane of symmetry. \n

$${\\left[ {Co{{\\left( {{H_2}O} \\right)}_4}\\left( {en} \\right)} \\right]^{3 + }}\\,\\,$$ have two planes of symmetry hence it is also optically inactive. \n

Hence the formula of the complex is $$\\left[ {Co{{\\left( {N{H_3}} \\right)}_6}} \\right]C{l_3}$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1010, "subject": "Chemistry", "question": "Which of the following complex species is not expected to exhibit optical isomerism?", "options": [ { "text": "[Co (en)2 Cl2]-" }, { "text": "[Co (NH3)3 Cl3]" }, { "text": "[Co (en) (NH3)2 Cl2]+" }, { "text": "[Co (en)3]3+" } ], "answer": "[Co (NH3)3 Cl3]", "solution": "**Answer:** [Co (NH3)3 Cl3]\n\nOctahedral coordination entities of the type $$M{a_3}{b_3}$$ exhibit Geometrical isomerism. The compound exists both as facial and meridional isomers. \n

\"JEE ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1011, "subject": "Chemistry", "question": "The number of geometric isomers that can exist for square planar [Pt (Cl) (py) (NH3) (NH2OH)]+ is (py = pyridine) :", "options": [ { "text": "3" }, { "text": "4" }, { "text": "6" }, { "text": "2" } ], "answer": "3", "solution": "**Answer:** 3\n\nSquare planar complexes of type $$M\\left[ {ABCD} \\right]\\,$$ form three isomers. Their position may be obtained by fixing the position of one ligand and placing at the trans position any one of the remaining three ligands one by one. \n

\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1012, "subject": "Chemistry", "question": "Which one of the following complexes shows optical isomerism?
\n(en = ethylenediamine)", "options": [ { "text": "cis[Co(en)2Cl2]Cl" }, { "text": "trans[Co(en)2Cl2]Cl" }, { "text": "[Co(NH3)4Cl2]Cl" }, { "text": "[Co(NH3)3Cl3]" } ], "answer": "cis[Co(en)2Cl2]Cl", "solution": "**Answer:** cis[Co(en)2Cl2]Cl\n\nOptical isomerism occurs when a molecule is non-superimposable with its mirror image hence the complex $$cis - \\left[ {Co{{\\left( {en} \\right)}_2}C{l_2}} \\right]Cl$$ is optically active. \n

\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1013, "subject": "Chemistry", "question": "The total number of possible isomers for square-planar [Pt(Cl)(NO2)(NO3)(SCN)]2- is :", "options": [ { "text": "8" }, { "text": "12" }, { "text": "16" }, { "text": "24" } ], "answer": "12", "solution": "**Answer:** 12\n\n

The complexes with formula [Mabcd]n± can have three geometrical isomers. As NO2− and SCN− are ambidentate ligands, therefore 4 × 3 = 12 geometrical isomers will be possible.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1014, "subject": "Chemistry", "question": "Which of the following complexes will shows geometrical isomerism ? ", "options": [ { "text": "aquachlorobis (ethylenediamine) cobalt (II) chloride" }, { "text": "pentaaquachlorohromium (III) chloride" }, { "text": "potassium amminetrichloropltinate (II) " }, { "text": "potassium tris(oxalato) chromate (III)" } ], "answer": "aquachlorobis (ethylenediamine) cobalt (II) chloride", "solution": "**Answer:** aquachlorobis (ethylenediamine) cobalt (II) chloride\n\n(a)   [Co(H2O) Cl(en)2]Cl  $$ \\to $$  it shows gemetrical isomerism.\n

\"JEE\n

(b)   [Cr(H2O)5Cl]Cl2  :   This complex is [MA5B] type. It does not show geometrical isomerism.\n

(c)   K[Pt(NH3)Cl3]  :  Thiscomplex is [MAB3] type. It does not show geometrical isomerism. \n

(d)   K3[Cr(OX)3]  :  This complex is [M(AA)3] type. It does not show geometrical isomerism. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1015, "subject": "Chemistry", "question": "Consider the following reaction and statements:
\n[Co(NH3)4Br2]+ + Br- $$\\to$$ [Co(NH3)3Br3] + NH3
\n(I) Two isomers are produced if the reactant complex ion is a cis-isomer
\n(II) Two isomers are produced if the reactant complex ion is a trans-isomer
\n(III) Only one isomer is produced if the reactant complex ion is a trans-isomer
\n(IV) Only one isomer is produced if the reactant complex ion is a cis – isomer
\nThe correct statements are
", "options": [ { "text": "(II) and (IV)" }, { "text": "(I) and (II)" }, { "text": "(I) and (III)" }, { "text": "(III) and (IV)" } ], "answer": "(I) and (III)", "solution": "**Answer:** (I) and (III)\n\nWhen reactant is cis isomer then following reaction takes place. \n

\"JEE\n
So, if reactant is cis - isomer then two isomers are produced. \n

When reactant is trans isomer then following reaction takes place.\n

\"JEE\n

So, if the reactant is trans isomer then only one isomer is produced.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1016, "subject": "Chemistry", "question": "The total number of isomers for a square planar complex
[M(F)(Cl)(SCN)(NO2)] is", "options": [ { "text": "16" }, { "text": "8" }, { "text": "12" }, { "text": "4" } ], "answer": "12", "solution": "**Answer:** 12\n\n\"JEE\n\"JEE\n

So, The total number of isomers for a square planar complex [M(F)(Cl)(SCN)(NO2\n)] is 12", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1017, "subject": "Chemistry", "question": "A reaction of cobalt (III) chloride and ethylenediamine in a 1 : 2 mole ratio generates two isomeric products A (violet coloured) and B (green coloured). A can show optical activity, but B is optically inactive. What\ntype of isomers does A and B represcent? ", "options": [ { "text": "Ionisation isomers" }, { "text": "Linkage isomers " }, { "text": "Coordination isomers " }, { "text": "Geometrical isomers " } ], "answer": "Geometrical isomers ", "solution": "**Answer:** Geometrical isomers \n\nCoCl3\n + en $$ \\to $$ [Co(en)2Cl2]Cl\n
   1      :   2\n\"JEE\n

A and B are Geometrical isomers.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1018, "subject": "Chemistry", "question": "The species that can have a trans-isomer is :\n
(en = ehane-1, 2-diamine, ox = oxalate)", "options": [ { "text": "[Cr(en)2(ox)]+" }, { "text": "[Pt(en)Cl2]" }, { "text": "[Pt(en)2Cl2]2+" }, { "text": "[Zn(en)Cl2]" } ], "answer": "[Pt(en)2Cl2]2+", "solution": "**Answer:** [Pt(en)2Cl2]2+\n\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1019, "subject": "Chemistry", "question": "The complex that can show optical activity is :", "options": [ { "text": "cis-[Fe(NH3)2(CN)4]–" }, { "text": "trans-[Cr(Cl2)(ox)2]3–" }, { "text": "trans-[Fe(NH3)2(CN)4]–" }, { "text": "cis-[CrCl2(ox)2]3– (ox = oxalate)" } ], "answer": "cis-[CrCl2(ox)2]3– (ox = oxalate)", "solution": "**Answer:** cis-[CrCl2(ox)2]3– (ox = oxalate)\n\n\"JEE\n

It does not have symmetry, so, optically active.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1020, "subject": "Chemistry", "question": "Consider the complex ions,\n
trans-[Co(en)2Cl2]+ (A) and
cis-[Co(en)2Cl2]+ (B).\n
The correct statement regarding them is :", "options": [ { "text": "both (A) and (B) can be optically active." }, { "text": "both (A) and (B) can not be optically active." }, { "text": "(A) can not be optically active, but (B) can\nbe optically active." }, { "text": "(A) can be optically active, but (B) can not\nbe optically active." } ], "answer": "(A) can not be optically active, but (B) can\nbe optically active.", "solution": "**Answer:** (A) can not be optically active, but (B) can\nbe optically active.\n\ntrans-[Co(en)2Cl2]+ contain a plane of\nsymmetry. It cannot be optically active.\n

cis-[Co(en)2Cl2]+ does not contain\nany plane of symmetry. Hence the compound (B) is\noptically active.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1021, "subject": "Chemistry", "question": "The number of isomers possible for\n
[Pt(en)(NO2)2] is :", "options": [ { "text": "3" }, { "text": "1" }, { "text": "4" }, { "text": "2" } ], "answer": "3", "solution": "**Answer:** 3\n\n\"JEE\n

Total 3 geometrical isomers are possible.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1022, "subject": "Chemistry", "question": "The one that is not expected to show isomerism\nis :", "options": [ { "text": "[Pt(NH3)2Cl2]" }, { "text": "[Ni(NH3)4(H2O)2]2+\n" }, { "text": "[Ni(en)3]2+" }, { "text": "[Ni(NH3)2Cl2]" } ], "answer": "[Ni(NH3)2Cl2]", "solution": "**Answer:** [Ni(NH3)2Cl2]\n\n[Pt(NH3)2Cl2] is dsp2 hybridisation and shows geometrical\nisomerism.\n

[Ni(NH3)4(H2O)2]2+ is Octahedral, show\ngeometrical isomerism.\n

[Ni(en)3]2+ is Octahedral and shows optical\nisomerism.\n

[Ni(NH3)2Cl2] is sp3 hybridisation and tetrahedral complex, therefore does not show geometrical and optical isomerism and structural isomerism.\n

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1023, "subject": "Chemistry", "question": "Complex X of composition Cr(H2O)6Cln\nhas a spin only magnetic moment of 3.83\nBM. It reacts with AgNO3 and shows\ngeometrical isomerism. The IUPAC\nnomenclature of X is :", "options": [ { "text": "Hexaaqua chromium (III) chloride" }, { "text": "Tetraaquadichlorido chromium(IV)\nchloride dihydrate" }, { "text": "Tetraaquadichlorido chromium (III)\nchloride dihydrate" }, { "text": "Dichloridotetraaqua chromium (IV)\nchloride dihydrate" } ], "answer": "Tetraaquadichlorido chromium (III)\nchloride dihydrate", "solution": "**Answer:** Tetraaquadichlorido chromium (III)\nchloride dihydrate\n\nGiven complex Cr(H2O)6Cln\n

As the magnetic mement is 3.83 BM then \n

Spin only magnetic moment = $$\\sqrt {n\\left( {n + 2} \\right)} $$ BM = 3.83\n

$$ \\Rightarrow $$ n = 3\n

Chromium\nin +3 oxidation state so molecular formula is\nCr(H2O)6Cln.\n

$$ \\therefore $$ This formula have following isomers\n

(a) [Cr(H2O)6]Cl3 : react with AgNO3 but does\nnot show geometrical isomerism.\n

(b) [Cr(H2O)5Cl]Cl2.H2O react with AgNO3 but\ndoes not show geometrical isomerism.\n

(c) [Cr(H2O)4Cl2]Cl.2H2O react with AgNO3 &\nshow geometrical isomerism.\n

(d) [Cr(H2O)3Cl3].3H2O does not react with\nAgNO3 & show geometrical isomerism.\n

Compound will be\n[Cr(H2O)4Cl2] Cl.2H2O\n

IUPAC NAME : Tetraaquadichlorido chromium(III) chloride dihydrate \n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1024, "subject": "Chemistry", "question": "Among (a) – (d) the complexes that can display\ngeometrical isomerism are :
\n(a) [Pt(NH3)3Cl]+
\n(b) [Pt(NH3)Cl5]–
\n(c) [Pt(NH3)2Cl(NO2)]
\n (d) [Pt(NH3)4ClBr]2+", "options": [ { "text": "(a) and (b)" }, { "text": "(c) and (d)" }, { "text": "(d) and (a)" }, { "text": "(b) and (c)" } ], "answer": "(c) and (d)", "solution": "**Answer:** (c) and (d)\n\n[Pt(NH3)2Cl(NO2)] and [Pt(NH3)4ClBr]2+ can display geometrical isomerism.\n
\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1025, "subject": "Chemistry", "question": "The complex that can show fac- and mer-isomers is :", "options": [ { "text": "[CoCl2(en)2]" }, { "text": "[Pt(NH3)2Cl2]" }, { "text": "[Co(NH3)3(NO2)3]" }, { "text": "[Co(NH3)4Cl2]+" } ], "answer": "[Co(NH3)3(NO2)3]", "solution": "**Answer:** [Co(NH3)3(NO2)3]\n\n[Ma3b3] type complex shows fac and mer\nisomerism.\n

So [Co(NH3)3(NO2)3] is correct answer.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1026, "subject": "Chemistry", "question": "The number of possible optical isomers for the complexes MA2B2 with sp3 and dsp2 hydridized metal atom. respectively, is :
\nNote : A and B are unidentate netural and unidentate monoanionic ligands, respectively.", "options": [ { "text": "0 and 2\n" }, { "text": "0 and 0" }, { "text": "0 and 1" }, { "text": "2 and 2" } ], "answer": "0 and 0", "solution": "**Answer:** 0 and 0\n\n(a) If the complex MA2B2 is sp3 hybridised then\nthe shape of this complex is tetrahedral this\nstructure is opticaly inactive due to the presence\nof plane of symmetry.\n\"JEE\n

(b) If the complex MA2B2 is dsp2 hybridised then\nthe shape of this complex is square planar.\n\"JEE\n
Both isomers are optically inactive due to the\npresence of plane of symmetry.\n

$$ \\therefore $$ Total number of optical isomer is zero in both\nthe cases.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1027, "subject": "Chemistry", "question": "The isomer(s) of [Co(NH3)4Cl2] that has/have\na Cl–Co–Cl angle of 90°, is/are :", "options": [ { "text": "cis only" }, { "text": "cis and trans" }, { "text": "meridional and trans" }, { "text": "trans only" } ], "answer": "cis only", "solution": "**Answer:** cis only\n\n[Co(NH3)4Cl2] has 2 geometrical isomers.\n\"JEE\n
Among cis and trans isomers, cis isomer has Cl–Co–Cl angle of 90o.", "topic": "Geometry", "subtopic": "Plane Geometry" }, { "id": 1028, "subject": "Chemistry", "question": "The number of stereoisomers possible for [Co(ox)2(Br)(NH3)]2$$-$$ is ___________. [ox = oxalate]", "options": [], "answer": "3", "solution": "**Answer:** 3\n\n

The coordination compound [Co(ox)2(Br)(NH3)]2$$-$$\nof general\nformula [M(A - A)2BC]\ncan show both geometrical and optical\nisomerism (stereoisomerism).

\n

\"JEE

\n

The cis-form produces non-superimposable mirror images, i.e.\nenantiomeric pairs (optically active)

\n

\"JEE

\n

The trans-form is optically inactive.\n

So, total number of stereoisomers possible\n= cis( $$ \\pm $$ ) + trans = 3\n

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1029, "subject": "Chemistry", "question": "The equivalents of ethylene diamine required to replace the neutral ligands from the coordination sphere of the trans-complex of CoCl3 . 4NH3 is _________. (Round off to the Nearest Integer).", "options": [], "answer": "2", "solution": "**Answer:** 2\n\n

$$[Co{(N{H_3}]_4}C{l_2}]Cl + 2en \\to [Co{(en)_2}C{l_2}] + 4N{H_3}$$

\n

NH3 is the neutral monodentate ligand. Ethylene diamine is a neutral didentate ligand.

\n

$${H_2}\\mathop N\\limits^{ \\bullet \\,\\, \\bullet } - C{H_2} - C{H_2} - \\mathop N\\limits^{ \\bullet \\,\\, \\bullet } {H_2}(en)$$

\n

So, two ethylene diamine are equivalent to four 'NH3' ligand.

", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1030, "subject": "Chemistry", "question": "Match List - I with List - II :

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
List - I
List - II
(a)$$[Co{(N{H_3})_6}][Cr{(CN)_6}]$$(i)Linkage isomerism
(b)$$[Co{(N{H_3})_3}{(N{O_2})_3}]$$(ii)Solvate isomerism
(c)$$[Cr{({H_2}O)_6}C{l_3}$$(iii)Co-ordination isomerism
(d)$$cis - {[CrC{l_2}{(ox)_2}]^{3 - }}$$(iv)Optical isomerism


Choose the correct answer from the options given below :", "options": [ { "text": "(a)-(i), (b)-(ii), (c)-(iii), (d)-(iv)" }, { "text": "(a)-(iii), (b)-(i), (c)-(ii), (d)-(iv)" }, { "text": "(a)-(ii), (b)-(i), (c)-(iii), (d)-(iv)" }, { "text": "(a)-(iv), (b)-(ii), (c)-(iii), (d)-(i)" } ], "answer": "(a)-(iii), (b)-(i), (c)-(ii), (d)-(iv)", "solution": "**Answer:** (a)-(iii), (b)-(i), (c)-(ii), (d)-(iv)\n\n(a) [Co(NH3)6][Cr(CN)6] $$ \\to $$ (iii) Coordination isomerism

\n(b) [Co(NH3)3(NO2)3] $$ \\to $$ (i) Linkage isomerism

\n(c) [Cr(H2O)6]Cl3 $$ \\to $$ (ii) Solvate isomerism

\n(d) cis-[CrCl2(ox)2]3– $$ \\to $$ (iv) Optical isomerism", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1031, "subject": "Chemistry", "question": "The number of geometrical isomers found in the metal
complexes [PtCl2(NH3)2], [Ni(CO)4], [Ru(H2O)3Cl3 and [CoCl2(NH3)4]+ respectively, are :", "options": [ { "text": "1, 1, 1, 1" }, { "text": "2, 1, 2, 2" }, { "text": "2, 0, 2, 2" }, { "text": "2, 1, 2, 1" } ], "answer": "2, 0, 2, 2", "solution": "**Answer:** 2, 0, 2, 2\n\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1032, "subject": "Chemistry", "question": "The number of geometrical isomers possible in triamminetrinitrocobalt (III) is X and in trioxalatochromate (III) is Y. Then the value of X + Y is _______________.", "options": [], "answer": "2", "solution": "**Answer:** 2\n\nTriamminetrinitrocobalt (III) $$\\to$$ [Co(NO2)3(NH3)3]

trioxalatochromate (III) ion $$\\to$$ [Cr(C2O4)3]3$$-$$[Co(NO2)3(NH3)3]

\"JEE

X + Y = 2 + 0 = 2.0", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1033, "subject": "Chemistry", "question": "Indicate the complex/complex ion which did not show any geometrical isomerism :", "options": [ { "text": "[CoCl2(en)2]" }, { "text": "[Co(CN)5(NC)]3$$-$$" }, { "text": "[Co(NH3)3(NO2)3]" }, { "text": "[Co(NH3)4Cl2]+" } ], "answer": "[Co(CN)5(NC)]3$$-$$", "solution": "**Answer:** [Co(CN)5(NC)]3$$-$$\n\n(a) [CoCl2(en)2] show Cis-trans isomerism

(b) [Co(CN)5(NC)]3$$-$$ can't show G.I.

(3) [Co(NH3)3(NO2)3] show fac & mer isomerism

(d) [Co(NH3)4Cl2]+ show cis & trans isomerism", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1034, "subject": "Chemistry", "question": "The number of optical isomers possible for [Cr(C2O4)3]3$$-$$ is ____________.", "options": [], "answer": "2", "solution": "**Answer:** 2\n\nThe number of optical isomers for [Cr(C2O4)3]3$$-$$ is two.

\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1035, "subject": "Chemistry", "question": "

(a) CoCl3.4NH3,    (b) CoCl3.5NH3,      (c) CoCl3.6NH3    and    (d) CoCl(NO3)2.5NH3.

\n

Number of complex(es) which will exist in cis-trans form is/are _______________.

", "options": [], "answer": "1", "solution": "**Answer:** 1\n\n$\\mathrm{CoCl}_{3} \\cdot 4 \\mathrm{NH}_{3} \\Rightarrow\\left[\\mathrm{Co}\\left(\\mathrm{NH}_{3}\\right)_{4} \\mathrm{Cl}_{2}\\right] \\mathrm{Cl}$\n

\n$\\mathrm{CoCl}_{3} \\cdot 5 \\mathrm{NH}_{3} \\Rightarrow\\left[\\mathrm{Co}\\left(\\mathrm{NH}_{3}\\right)_{5} \\mathrm{Cl}\\right] \\mathrm{Cl}_{2}$\n

\n$\\mathrm{CoCl}_{3} \\cdot 6 \\mathrm{NH}_{3} \\Rightarrow\\left[\\mathrm{Co}\\left(\\mathrm{NH}_{3}\\right)_{6}\\right] \\mathrm{Cl}_{3}$\n

\nOnly $\\left[\\mathrm{Co}\\left(\\mathrm{NH}_{3}\\right)_{4} \\mathrm{Cl}_{2}\\right]$ can show geometrical isomerism. Hence can exist in cis-trans form.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1036, "subject": "Chemistry", "question": "

Total number of relatively more stable isomer(s) possible for octahedral complex $$\\left[\\mathrm{Cu}(\\mathrm{en})_{2}(\\mathrm{SCN})_{2}\\right]$$ will be _________.

", "options": [], "answer": "3", "solution": "**Answer:** 3\n\n[Cu(en)2(SCN)2]

\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1037, "subject": "Chemistry", "question": "

The complex cation which has two isomers is :

", "options": [ { "text": "$$\\left[\\mathrm{Co}\\left(\\mathrm{NH}_{3}\\right)_{5} \\mathrm{Cl}\\right]^{+}$$" }, { "text": "$$\\left[\\mathrm{Co}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}\\right]^{3+}$$" }, { "text": "$$\\left[\\mathrm{Co}\\left(\\mathrm{NH}_{3}\\right)_{5} \\mathrm{NO}_{2}\\right]^{2+}$$" }, { "text": "$$\\left[\\mathrm{Co}\\left(\\mathrm{NH}_{3}\\right)_{5} \\mathrm{Cl}\\right]^{2+}$$" } ], "answer": "$$\\left[\\mathrm{Co}\\left(\\mathrm{NH}_{3}\\right)_{5} \\mathrm{NO}_{2}\\right]^{2+}$$", "solution": "**Answer:** $$\\left[\\mathrm{Co}\\left(\\mathrm{NH}_{3}\\right)_{5} \\mathrm{NO}_{2}\\right]^{2+}$$\n\nComplex $$\\left[\\mathrm{Co}\\left(\\mathrm{NH}_{3}\\right)_{5} \\mathrm{NO}_{2}\\right]^{2+}$$ will have two isomer one \nlinked through N (Nitro) and one through O (Nitrite).\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1038, "subject": "Chemistry", "question": "The $\\mathrm{Cl}-\\mathrm{Co}-\\mathrm{Cl}$ bond angle values in a fac- $\\left[\\mathrm{Co}\\left(\\mathrm{NH}_{3}\\right)_{3} \\mathrm{Cl}_{3}\\right]$ complex is/are :", "options": [ { "text": "$90^{\\circ} ~\\&~ 120^{\\circ}$" }, { "text": "$90^{\\circ}$" }, { "text": "$180^{\\circ}$" }, { "text": "$90^{\\circ} ~\\& ~180^{\\circ}$" } ], "answer": "$90^{\\circ}$", "solution": "**Answer:** $90^{\\circ}$\n\n

\"JEE

\n

All the Cl$$-$$Co$$-$$Cl bond angles are of 90$$^\\circ$$.

", "topic": "Geometry", "subtopic": "Plane Geometry" }, { "id": 1039, "subject": "Chemistry", "question": "

Chiral complex from the following is :

\n

Here en = ethylene diamine

", "options": [ { "text": "$$\\mathrm{cis-[PtCl_2(en)_2]^{2+}}$$" }, { "text": "$$\\mathrm{trans-[Co(NH_3)_4Cl_2]^{+}}$$" }, { "text": "$$\\mathrm{trans-[PtCl_2(en)_2]^{2+}}$$" }, { "text": "$$\\mathrm{cis-[PtCl_2(NH_3)_2]}$$" } ], "answer": "$$\\mathrm{cis-[PtCl_2(en)_2]^{2+}}$$", "solution": "**Answer:** $$\\mathrm{cis-[PtCl_2(en)_2]^{2+}}$$\n\n

\"JEE

\nThis is chiral complex form.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1040, "subject": "Chemistry", "question": "

The total number of stereoisomers for the complex $$\\left[\\mathrm{Cr}(o x)_{2} \\mathrm{ClBr}\\right]^{3-}$$ (where $$o x=$$ oxalate) is :

", "options": [ { "text": "1" }, { "text": "3" }, { "text": "2" }, { "text": "4" } ], "answer": "3", "solution": "**Answer:** 3\n\n$\\left[\\mathrm{Cr}(\\mathrm{Ox})_2 \\mathrm{ClBr}\\right]^{-3}$

\n- No. of isomers -\n

\n\"JEE
\n- This structure has plane of symmetry, So no optical isomerism will be shown.

\"JEE
\n- This structure does not contain plane of symmetry, So two forms $d$ as well as 1 will be shown.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1041, "subject": "Chemistry", "question": "

If $$\\mathrm{Ni}^{2+}$$ is replaced by $$\\mathrm{Pt}^{2+}$$ in the complex $$\\left[\\mathrm{NiCl}_{2} \\mathrm{Br}_{2}\\right]^{2-}$$, which of the following properties are expected to get changed ?

\n

A. Geometry

\n

B. Geometrical isomerism

\n

C. Optical isomerism

\n

D. Magnetic properties

", "options": [ { "text": "A and D" }, { "text": "A, B and D" }, { "text": "A, B and C" }, { "text": "B and C" } ], "answer": "A, B and D", "solution": "**Answer:** A, B and D\n\nThe complex $\\left[\\mathrm{NiCl}_{2} \\mathrm{Br}_{2}\\right]^{2-}$ is a tetrahedral complex due to the fact that Nickel (Ni) in this complex is in its +2 oxidation state, and thus adopts a dsp² hybridization, resulting in a tetrahedral geometry.\n\n

If $\\mathrm{Ni}^{2+}$ is replaced by $\\mathrm{Pt}^{2+}$, the new complex would be $\\left[\\mathrm{PtCl}_{2} \\mathrm{Br}_{2}\\right]^{2-}$. Platinum (Pt) in its +2 oxidation state typically forms square planar complexes due to its preference for dsp² hybridization.\n\n\n

    \n

  1. Geometry :

    \n\n
    2. \n

  2. Geometrical Isomerism :

    \n\n
    3. \n

  3. Optical Isomerism :

    \n\n
    4. \n

  4. Magnetic Properties :

    \n\n
    5. \n
\n

Based on this information, the properties expected to change when $\\mathrm{Ni}^{2+}$ is replaced by $\\mathrm{Pt}^{2+}$ are:

\n\n

Optical isomerism remains unchanged (inactive in both cases).

\n

Therefore, the correct answer is :

\n

Option B: A, B, and D (Geometry, Geometrical Isomerism, and Magnetic Properties).

\n", "topic": "Geometry", "subtopic": "Plane Geometry" }, { "id": 1042, "subject": "Chemistry", "question": "Which of the following complex is homoleptic?", "options": [ { "text": "$\\left[\\mathrm{Ni}\\left(\\mathrm{NH}_3\\right)_2 \\mathrm{Cl}_2\\right]$" }, { "text": "$\\left[\\mathrm{Co}\\left(\\mathrm{NH}_3\\right)_4 \\mathrm{Cl}_2\\right]^{+}$" }, { "text": "$\\left[\\mathrm{Fe}\\left(\\mathrm{NH}_3\\right)_4 \\mathrm{Cl}_2\\right]^{+}$" }, { "text": "$\\left[\\mathrm{Ni}(\\mathrm{CN})_4\\right]^{2-}$" } ], "answer": "$\\left[\\mathrm{Ni}(\\mathrm{CN})_4\\right]^{2-}$", "solution": "**Answer:** $\\left[\\mathrm{Ni}(\\mathrm{CN})_4\\right]^{2-}$\n\n

A homoleptic complex is one in which a central metal atom or ion is surrounded by only one kind of donor groups, ligands or ions. On the other hand, a heteroleptic complex contains a central metal surrounded by more than one kind of donor groups, ligands or ions.

\n\n

Let's examine the given options :

\n\n

Option A: $\\left[\\mathrm{Ni}\\left(\\mathrm{NH}_3\\right)_2 \\mathrm{Cl}_2\\right]$
\n
This complex contains two different ligands, ammonia ($\\mathrm{NH}_3$) and chloride (Cl). Therefore, it is a heteroleptic complex.

\n\n

Option B: $\\left[\\mathrm{Co}\\left(\\mathrm{NH}_3\\right)_4 \\mathrm{Cl}_2\\right]^{+}$
\n
Just like Option A, this complex has two types of ligands, ammonia ($\\mathrm{NH}_3$) and chloride (Cl), making it a heteroleptic complex.

\n\n

Option C: $\\left[\\mathrm{Fe}\\left(\\mathrm{NH}_3\\right)_4 \\mathrm{Cl}_2\\right]^{+}$
\n
Again, this complex has ammonia ($\\mathrm{NH}_3$) and chloride (Cl) ligands, so it is a heteroleptic complex.

\n\n

Option D: $\\left[\\mathrm{Ni}(\\mathrm{CN})_4\\right]^{2-}$
\n
This complex has only cyanide (CN) ligands surrounding the nickel ion, with no other types of ligands present. Thus, this is a homoleptic complex because it is coordinated by a single type of ligand.

\n\n

Therefore, Option D is the correct answer as it represents a homoleptic complex.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1043, "subject": "Chemistry", "question": "

Number of complexes which show optical isomerism among the following is ________.

\n

$$\\text { cis- }\\left[\\mathrm{Cr}(\\mathrm{ox})_2 \\mathrm{Cl}_2\\right]^{3-},\\left[\\mathrm{Co}(\\text {en})_3\\right]^{3+}, \\text { cis- }\\left[\\mathrm{Pt}(\\text {en})_2 \\mathrm{Cl}_2\\right]^{2+}, \\text { cis- }\\left[\\mathrm{Co}(\\text {en})_2 \\mathrm{Cl}_2\\right]^{+}, \\text {trans- }\\left[\\mathrm{Pt}(\\text {en})_2 \\mathrm{Cl}_2\\right]^{2+}, \\text { trans- }\\left[\\mathrm{Cr}(\\mathrm{ox})_2 \\mathrm{Cl}_2\\right]^{3-}$$

", "options": [], "answer": "4", "solution": "**Answer:** 4\n\n

cis $$-\\left[\\mathrm{Cr}(\\mathrm{ox})_2 \\mathrm{Cl}_2\\right]^{3-} \\rightarrow$$ can show optical isomerism (no POS & COS)

\n

$$\\left[\\mathrm{Co}(\\mathrm{en})_3\\right]^{3+} \\rightarrow \\text { can show (no POS & COS) }$$

\n

$$\\text { cis }-\\left[\\mathrm{Pt}(\\mathrm{en})_2 \\mathrm{Cl}_2\\right]^{2+} \\rightarrow$$ can show (no POS & COS)

\n

$$\\text { cis }-\\left[\\mathrm{Co}(\\mathrm{en})_2 \\mathrm{Cl}_2\\right]^{+} \\rightarrow$$ can show (no POS & COS)

\n

$$\\operatorname{trans}-\\left[\\mathrm{Pt}(\\mathrm{en})_2 \\mathrm{Cl}_2\\right]^{2+} \\rightarrow$$ can't show (contains POS & COS)

\n

$$\\text { trans }-\\left[\\mathrm{Cr}(\\mathrm{ox})_2 \\mathrm{Cl}_2\\right]^{3-} \\rightarrow$$ can't show (contains POS & COS)

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1044, "subject": "Chemistry", "question": "

Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).

\n

Assertion (A): The total number of geometrical isomers shown by $$[\\mathrm{Co}(\\mathrm{en})_2 \\mathrm{Cl}_2]^{+}$$ complex ion is three.

\n

Reason (R): $$[\\mathrm{Co}(\\mathrm{en})_2 \\mathrm{Cl}_2]^{+}$$ complex ion has an octahedral geometry.

\n

In the light of the above statements, choose the most appropriate answer from the options given below :

", "options": [ { "text": "(A) is correct but (R) is not correct" }, { "text": "(A) is not correct but (R) is correct" }, { "text": "Both (A) and (R) are correct but (R) is not the correct explanation of (A)" }, { "text": "Both (A) and (R) are correct and (R) is the correct explanation of (A)" } ], "answer": "(A) is not correct but (R) is correct", "solution": "**Answer:** (A) is not correct but (R) is correct\n\n

To evaluate the assertion and the reason, let's first analyze the given complex ion $$[\\mathrm{Co}(\\mathrm{en})_2 \\mathrm{Cl}_2]^{+}$$.

\n\n

1. Assertion (A): The total number of geometrical isomers shown by $$[\\mathrm{Co}(\\mathrm{en})_2 \\mathrm{Cl}_2]^{+}$$ complex ion is three.

\n\n

2. Reason (R): $$[\\mathrm{Co}(\\mathrm{en})_2 \\mathrm{Cl}_2]^{+}$$ complex ion has an octahedral geometry.

\n\n

First, we know that $$[\\mathrm{Co}(\\mathrm{en})_2 \\mathrm{Cl}_2]^{+}$$ is a coordination complex with $$\\mathrm{Co}$$ (Cobalt) in the center coordinated to two ethylenediamine (en) ligands and two chlorides (Cl). Ethylenediamine is a bidentate ligand, meaning it binds through two donor atoms, giving the overall complex an octahedral geometry around the central cobalt ion.

\n\n

To verify the assertion about the geometrical isomers:

\n\n

In an octahedral complex, the two chlorides and the two $$\\mathrm{en}$$ ligands can have different spatial arrangements relative to each other. Specifically, for $$[\\mathrm{Co}(\\mathrm{en})_2 \\mathrm{Cl}_2]^{+}$$, the possible geometrical isomers are:

\n\n\n\n

These two configurations are the typical geometrical isomers for this type of complex. Thus, the assertion that there are three geometrical isomers appears incorrect, as the common understanding is that there are only two geometrical isomers for this type of octahedral complex.

\n\n

Now, let’s consider the reason:

\n\n

The reason states that $$[\\mathrm{Co}(\\mathrm{en})_2 \\mathrm{Cl}_2]^{+}$$ has an octahedral geometry. This statement is indeed correct because the coordination number of 6 (from the four donor atoms of the two $$\\mathrm{en}$$ ligands and the two chloride ions) leads to an octahedral geometry.

\n\n

Therefore, the most appropriate answer is:

\n\n

Option B: (A) is not correct but (R) is correct

", "topic": "Geometry", "subtopic": "Plane Geometry" }, { "id": 1045, "subject": "Chemistry", "question": "

The metal atom present in the complex MABXL (where A, B, X and L are unidentate ligands and $$\\mathrm{M}$$ is metal) involves $$\\mathrm{sp}^3$$ hybridization. The number of geometrical isomers exhibited by the complex is :

", "options": [ { "text": "0" }, { "text": "2" }, { "text": "3" }, { "text": "4" } ], "answer": "0", "solution": "**Answer:** 0\n\n

To determine the number of geometrical isomers for the given complex $$\\mathrm{MABXL}$$, where $$\\mathrm{M}$$ is a metal and $$\\mathrm{A}$$, $$\\mathrm{B}$$, $$\\mathrm{X}$$, and $$\\mathrm{L}$$ are unidentate ligands, we need to consider the geometry implied by the $\\mathrm{sp}^3$ hybridization of the metal. The $\\mathrm{sp}^3$ hybridization suggests a tetrahedral geometry for the metal complex.

\n\n

In a tetrahedral complex, geometrical isomerism does not arise. This is because geometrical isomerism (also known as cis-trans isomerism) typically occurs in square planar or octahedral complexes where ligands can be positioned differently around the central metal atom to give distinct isomers. In a tetrahedral complex, every position is equivalent due to the symmetric arrangement of the ligands around the metal center, making it impossible to distinguish between different sides of the molecule as you would with a square planar or octahedral complex.

\n\n

Given that the complex follows tetrahedral geometry because of $\\mathrm{sp}^3$ hybridization and that tetrahedral complexes do not exhibit geometrical isomerism, the number of geometrical isomers exhibited by the complex is:

\n\n

Option A: 0

\n\n

This is because in a tetrahedral arrangement of ligands around a metal center, no geometrical isomers can be formed due to the lack of cis or trans arrangements possible within such a symmetric geometry.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1046, "subject": "Chemistry", "question": "The IUPAC name of the coordination compound K3[Fe(CN)6] is", "options": [ { "text": "Potassium hexacyanoferrate (II)" }, { "text": "Potassium hexacyanoferrate (III)" }, { "text": "Potassium hexacyanoiron (II)" }, { "text": "Tripotassium hexcyanoiron (II) " } ], "answer": "Potassium hexacyanoferrate (III)", "solution": "**Answer:** Potassium hexacyanoferrate (III)\n\n$${K_3}\\left[ {Fe{{\\left( {CN} \\right)}_6}} \\right]\\,\\,\\,$$ is potassium hexacyano ferrate $$\\left( {{\\rm I}{\\rm I}{\\rm I}} \\right).$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1047, "subject": "Chemistry", "question": "The IUPAC name for the complex [Co(NO2)(NH3)5]Cl2 is ", "options": [ { "text": "nitrito-N-pentaamminecobalt (III) chloride" }, { "text": "nitrito-N-pentaamminecobalt (II) chloride" }, { "text": "pentaammine nitrito-N-cobalt (II) chloride" }, { "text": "pentaammine nitrito-N-cobalt (III) chloride " } ], "answer": "pentaammine nitrito-N-cobalt (III) chloride ", "solution": "**Answer:** pentaammine nitrito-N-cobalt (III) chloride \n\n

The IUPAC name for the complex $[\\text{Co(NO}_2)(\\text{NH}_3)_5]\\text{Cl}_2$ is:\nPentaammine(nitrito-$\\text{N,O}$)cobalt(III) chloride

\n\n

Here's how to arrive at the name:\n

\n

The complex is a cobalt complex, and cobalt has a +3 charge in this compound (since it has one negatively charged NO$_2$ ligand and one negatively charged Cl ion, for a total of -2, to balance the +3 charge on cobalt).

\n

The ligands are NH$_3$ (ammonia) and NO$_2$ (nitrito).

\n

Ammonia is a neutral ligand, so its name is just \"ammine\" and it is designated by the prefix \"penta\" because there are five of them.

\n

Nitrito is a negatively charged ligand, so its name is \"nitrito-$\\text{N,O}$\" (the \"-$\\text{N,O}$\" indicates that the NO$_2$ ligand is bound through both the nitrogen and oxygen atoms).

\n

Finally, the compound has a chloride ion, so its name is \"chloride\" and it is designated by the prefix \"di\" because there are two of them.

\n

Putting it all together, we get the name \"pentaammine(nitrito-$\\text{N,O}$)cobalt(III) chloride\".

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1048, "subject": "Chemistry", "question": "Which among the following will be named as dibromidobis (ethylene diamine) chromium(III) bromide?", "options": [ { "text": "[Cr (en)3 ] Br3 " }, { "text": "[Cr (en)2 Br2 ] Br" }, { "text": "[Cr (en) Br4 ]-" }, { "text": "[Cr (en) Br2 ] Br" } ], "answer": "[Cr (en)2 Br2 ] Br", "solution": "**Answer:** [Cr (en)2 Br2 ] Br\n\n$$\\left[ {Cr{{\\left( {en} \\right)}_2}B{r_2}} \\right]Br$$\n

dibromidobis (ethylenediamine) chromium $$\\left( {{\\rm I}{\\rm I}{\\rm I}} \\right)$$ Bromide.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1049, "subject": "Chemistry", "question": "The IUPAC name of complex [Pt(NH3)2Cl(NH2CH3)]Cl is:\n", "options": [ { "text": "Diamminechlorido (methanamine) platinum(II)chloride." }, { "text": "Biasmmine (methanamine)chlorido platinum(II)chloride." }, { "text": "Diamminechlorido (amminomethane) platinum(II) chloride." }, { "text": "Diammine (methanamine) chlorido platinum(II)Chloride." } ], "answer": "Diamminechlorido (methanamine) platinum(II)chloride.", "solution": "**Answer:** Diamminechlorido (methanamine) platinum(II)chloride.\n\n[Pt(NH3)2Cl(NH2CH3)]Cl\n

x + 0 – 1 + 0 = +1\n

x = +2\n

So Pt present in Pt+2 form.\n

$$ \\therefore $$ Correct IUPAC form\n

Diamminechlorido (methanamine) platinum(II)chloride.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1050, "subject": "Chemistry", "question": "Complex A has a composition of H12O6Cl3Cr. If the complex on treatment with conc.H2SO4\n loses\n13.5% of its original mass, the correct molecular formula of A is :\n
[Given: atomic mass of Cr = 52 amu and Cl = 35 amu]", "options": [ { "text": "[Cr(H2O)5Cl]Cl2.H2O" }, { "text": "[Cr(H2O)4Cl2]Cl.2H2O" }, { "text": "[Cr(H2O)3Cl3].3H2O" }, { "text": "[Cr(H2O)6]Cl3" } ], "answer": "[Cr(H2O)4Cl2]Cl.2H2O", "solution": "**Answer:** [Cr(H2O)4Cl2]Cl.2H2O\n\nLet x molecule of water are lost then\n

13.5 = $$\\left[ {{{x \\times 18} \\over {6 \\times 18 + 3 \\times 35 + 52}}} \\right] \\times 100$$\n

$$ \\Rightarrow $$ x = 1.99 $$ \\simeq $$ 2\n

$$ \\therefore $$ Around two moles of water are lost during\nheating.\n

$$ \\therefore $$ Formula of complex could be\n

[Cr(H2O)4Cl2]Cl.2H2O ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1051, "subject": "Chemistry", "question": "

The IUPAC name of $$\\mathrm{K}_{3}\\left[\\mathrm{Co}\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}\\right)_{3}\\right]$$ is:-

", "options": [ { "text": "Potassium tris(oxalato)cobalt(III)" }, { "text": "Potassium tris(oxalato)cobaltate(III)" }, { "text": "Potassium trioxalatocobaltate(III)" }, { "text": "Potassium trioxalatocobalt(III)" } ], "answer": "Potassium trioxalatocobaltate(III)", "solution": "**Answer:** Potassium trioxalatocobaltate(III)\n\n

To determine the IUPAC name of the compound K₃[Co(C₂O₄)₃], let's follow the nomenclature rules for coordination compounds:

\n\n

Identify the Cation and Anion: The compound consists of potassium ions (K⁺) and a complex anion [Co(C₂O₄)₃]³⁻.

\n

Name the Cation First: The cation is named first. So, we start with \"Potassium.\"

\n

Determine the Ligand Name: The ligand C₂O₄²⁻ is called \"oxalato.\"

\n

Count the Number of Ligands: There are three oxalato ligands. Since \"oxalato\" does not contain any numerical prefixes (like di-, tri-), we use the prefixes \"di-\", \"tri-\", etc., for simple ligands. So, we use \"trioxalato.\"

\n

Name the Central Metal Atom: Since the complex ion is an anion, the metal name ends with the suffix \"-ate.\" Therefore, \"cobalt\" becomes \"cobaltate.\"

\n

Specify the Oxidation State: The oxidation state of cobalt in this complex can be calculated:

\n\n\n

Let the oxidation state of Co be x.

\n

Each oxalato ligand has a charge of -2.

\n

The overall charge of the complex ion is -3.

\n

So, x + 3(-2) = -3 ⇒ x -6 = -3 ⇒ x = +3.

\n

Therefore, we indicate the oxidation state as (III).

\n\n

Combining all these, the IUPAC name is Potassium trioxalatocobaltate(III).

\n

Answer: Option C

\n

Potassium trioxalatocobaltate(III)

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1052, "subject": "Chemistry", "question": "

Number of ambidentate ligands among the following is _________.

\n

$$\\mathrm{NO}_2^{-}, \\mathrm{SCN}^{-}, \\mathrm{C}_2 \\mathrm{O}_4^{2-}, \\mathrm{NH}_3, \\mathrm{CN}^{-}, \\mathrm{SO}_4^{2-}, \\mathrm{H}_2 \\mathrm{O} \\text {. }$$

", "options": [], "answer": "3", "solution": "**Answer:** 3\n\n

Ambidentate ligands are ligands that can attach to a central metal atom through two different atoms. Let’s analyze the given ligands one by one:

\n\n\n\n

Based on this analysis, the ambidentate ligands are $$\\mathrm{NO}_2^{-}$$, $$\\mathrm{SCN}^{-}$$, and $$\\mathrm{CN}^{-}$$.

\n\n

Therefore, the number of ambidentate ligands among the given options is 3.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1053, "subject": "Chemistry", "question": "

The correct IUPAC name of $$[\\mathrm{PtBr}_2(\\mathrm{PMe}_3)_2]$$ is :

", "options": [ { "text": "bis(trimethylphosphine)dibromoplatinum(II)\n" }, { "text": "dibromodi(trimethylphosphine)platinum(II)\n" }, { "text": "dibromobis(trimethylphosphine)platinum(II)\n" }, { "text": "bis[bromo(trimethylphosphine)]platinum(II)" } ], "answer": "dibromobis(trimethylphosphine)platinum(II)\n", "solution": "**Answer:** dibromobis(trimethylphosphine)platinum(II)\n\n\n

The correct IUPAC name of $$[\\mathrm{PtBr}_2(\\mathrm{PMe}_3)_2]$$ is dibromobis(trimethylphosphine)platinum(II).

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1054, "subject": "Chemistry", "question": "The correct order of magnetic moments (spin only values in B.M.) among is :
\n(Atomic numbers: Mn = 25; Fe = 26, Co =27) ", "options": [ { "text": "[MnCl4]2- > [CoCl4]-2 > [Fe(CN)6]-4" }, { "text": "[Fe(CN)6]-4 > [CoCl4]2- > [MnCl4]2-" }, { "text": "[Fe(CN)6]4- > [MnCl4]2- > [CoCl4]2-" }, { "text": "[MnCl4]2- > [Fe(CN)6]4- > [CoCl4]2-" } ], "answer": "[MnCl4]2- > [CoCl4]-2 > [Fe(CN)6]-4", "solution": "**Answer:** [MnCl4]2- > [CoCl4]-2 > [Fe(CN)6]-4\n\n\"AIEEE \n

NOTE : The greater the number of unpaired electrons, greater the magnitude of magnetic moment. Hence the correct order will be \n$$${\\left[ {MnC{l_4}} \\right]^{ - - }} > {\\left[ {CoC{l_4}} \\right]^{ - - }} > {\\left[ {Fe{{\\left( {CN} \\right)}_6}} \\right]^{4 - }}$$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1055, "subject": "Chemistry", "question": "Which one of the following complexes in an outer orbital complex? ", "options": [ { "text": "[Fe(CN)6]4- " }, { "text": "[Ni(NH3)6]2+" }, { "text": "[Co(NH3)6]3+ " }, { "text": "[Mn(CN)6]4-" } ], "answer": "[Ni(NH3)6]2+", "solution": "**Answer:** [Ni(NH3)6]2+\n\nHybridisation\n

$$\\mathop {{{\\left[ {Fe{{\\left( {CN} \\right)}_6}} \\right]}^{4 - }},}\\limits_{{d^2}s{p^3}} \\,\\,\\mathop {{{\\left[ {Mn{{\\left( {CN} \\right)}_6}} \\right]}^{4 - }},}\\limits_{{d^2}s{p^3}} $$ \n

$$\\mathop {{{\\left[ {Co{{\\left( {N{H_3}} \\right)}_3}} \\right]}^{3 + }},}\\limits_{{d^2}s{p^3}} \\,\\,\\mathop {{{\\left[ {Ni{{\\left( {N{H_3}} \\right)}_6}} \\right]}^{2 + }}}\\limits_{s{p^3}{d^2}} $$\n

Hence $${\\left[ {Ni{{\\left( {N{H_3}} \\right)}_6}} \\right]^{2 + }}$$ is outer orbital complex. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1056, "subject": "Chemistry", "question": "Among the properties (a) reducing (b) oxidising (c) complexing, the set of properties shown by CN– ion\ntowards metal species is :", "options": [ { "text": "c, a" }, { "text": "b, c" }, { "text": "a, b" }, { "text": "a, b, c" } ], "answer": "c, a", "solution": "**Answer:** c, a\n\n$$C{N^ - }$$ ion acts good complexing as well as reducing agent. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1057, "subject": "Chemistry", "question": "Which one of the following cyano complexes would exhibit the lowest value of\nparamagnetic behaviour?
\n(At. No. Cr = 24, Mn = 25, Fe = 26, Co = 27) ", "options": [ { "text": "[Cr(CN)6]-3 " }, { "text": "[Mn(CN)6]-3" }, { "text": "[Fe(CN)6]-3 " }, { "text": "[Co(CN)6]-3" } ], "answer": "[Co(CN)6]-3", "solution": "**Answer:** [Co(CN)6]-3\n\n\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n
No. of unpaired electron
a)Co3+0
b)Fe3+1
c)Mn3+2
d)Cr3+3
\n

The effective magnetic moment is given by the number of unpaired electrons in a substance, the lesser the number of unpaired electrons lower is its magnetic moment in Bohr $$-$$ Magneton and lower shall be its paramagnetism", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1058, "subject": "Chemistry", "question": "Nickel (Z = 28) combines with a uninegative monodentate ligand X– to form a paramagnetic complex [NiX4]2− . The number of unpaired electron(s) in the nickel and geometry of this complex ion are,\nrespectively ", "options": [ { "text": "one, tetrahedral" }, { "text": "two, tetrahedral " }, { "text": "one, square planar" }, { "text": "two, square planar" } ], "answer": "two, tetrahedral ", "solution": "**Answer:** two, tetrahedral \n\n$${\\left[ {Ni{X_4}} \\right]^{2 - }},$$ the electronic configuration of $$N{i^{2 + }}$$ is \n

\"AIEEE \n

It contains two unpaired electrons and the hybridisation is $$s{p^3}$$ (tetrahedral).", "topic": "Geometry", "subtopic": "Plane Geometry" }, { "id": 1059, "subject": "Chemistry", "question": "In Fe(CO)5, the Fe – C bond possesses :", "options": [ { "text": "π-character only" }, { "text": "both $$\\sigma$$ and π characters " }, { "text": "ionic character" }, { "text": "$$\\sigma$$-character" } ], "answer": "both $$\\sigma$$ and π characters ", "solution": "**Answer:** both $$\\sigma$$ and π characters \n\nDue to some back-bonding by side-wise overlapping of between $$d$$-orbitals of metal and $$p$$-orbital of carbon, the $$Fe-C$$ bond in $$Fe{\\left( {CO} \\right)_5}$$ has both $$\\sigma $$ and $$\\pi $$ character.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1060, "subject": "Chemistry", "question": "Which one of the following has a square planar geometry? ", "options": [ { "text": "[CoCl4]2-" }, { "text": "[FeCl4]2-" }, { "text": "[NiCl4]2-" }, { "text": "[PtCl4]2-" } ], "answer": "[PtCl4]2-", "solution": "**Answer:** [PtCl4]2-\n\nComplexes with $$ds{p^2}$$ hybridisation are square planar. So $$\\,\\,{\\left[ {PtC{l_4}} \\right]^{2 - }}\\,\\,$$ is square planar in shape.", "topic": "Geometry", "subtopic": "Plane Geometry" }, { "id": 1061, "subject": "Chemistry", "question": "Which of the following facts about the complex [Cr (NH3)6 ]Cl3 is wrong?", "options": [ { "text": "The complex is paramagnetic " }, { "text": "The complex is an outer orbital complex " }, { "text": "The complex gives white precipitate with silver nitrate solution" }, { "text": "The complex involves d2sp3 hybridization and is octahedral in shape." } ], "answer": "The complex is an outer orbital complex ", "solution": "**Answer:** The complex is an outer orbital complex \n\n$$\\left[ {Cr{{\\left( {N{H_3}} \\right)}_6}} \\right]C{l_3}\\,\\,$$ is an inner orbital complex, because in this complex $$d-$$orbital used is of lower quantum number i.e. $$(n-1).$$ It results from $${d^2}\\,\\,\\,$$ $$s{p^3}$$ (inner orbital) hybridization.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1062, "subject": "Chemistry", "question": "The magnetic moment (spin only) of [NiCl4]2− is", "options": [ { "text": "5.46 BM " }, { "text": "2.82 BM" }, { "text": "1.41 BM " }, { "text": "1.82 BM" } ], "answer": "2.82 BM", "solution": "**Answer:** 2.82 BM\n\n$${\\left[ {NiC{l_4}} \\right]^{2 - }}\\,\\left( {{d^8}} \\right)$$\n

\"AIEEE \n

i.e. the number of unpaired electrons in $${\\left[ {NiCl{}_4} \\right]^{2 - }}\\,\\,$$ is $$2.$$\n

$$n = \\sqrt {n\\left( {n + 2} \\right)} = \\sqrt {2\\left( 4 \\right)} $$\n

$$ = 2\\sqrt 2 = 2 \\times 1.41$$\n

$$ = 2.82\\,BM$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1063, "subject": "Chemistry", "question": "The pair having the same magnetic moment is :
\n[At. No.: Cr = 24, Mn = 25, Fe = 26, Co = 27] ", "options": [ { "text": "[Cr(H2O)6]2+ and [Fe(H2O)6]2+ " }, { "text": "[Mn(H2O)6]2+ and [Cr(H2O)6]2+" }, { "text": "[CoCl4]2– and [Fe(H2O)6]2+ " }, { "text": "[Cr(H2O)6]2+ and [CoCl4]2– " } ], "answer": "[Cr(H2O)6]2+ and [Fe(H2O)6]2+ ", "solution": "**Answer:** [Cr(H2O)6]2+ and [Fe(H2O)6]2+ \n\n\"JEE\n
Since $$(a)$$ and $$(b),$$ each has $$4$$ unpaired electrons, they will have same magnetic moment. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1064, "subject": "Chemistry", "question": "Which one of the following complexes will consume more equivalents of aqueous\nsolution of Ag(NO3)?", "options": [ { "text": "Na3[CrCl6]\n" }, { "text": "[Cr(H2O)5Cl]Cl2" }, { "text": "[Cr(H2O)6]Cl3" }, { "text": "Na2[CrCl5(H2O)]" } ], "answer": "[Cr(H2O)6]Cl3", "solution": "**Answer:** [Cr(H2O)6]Cl3\n\n

[Cr(H2O)6]Cl3 has the highest primary valency among the given complexes, that is, three, therefore, it will consume three moles of AgNO3 and precipitate three moles of AgCl.

\n

[Cr(H2O)6]Cl3 + 3AgNo3 $$\\to$$ 3AgCl + [Cr(H2O)6](NO)3

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1065, "subject": "Chemistry", "question": "On treatment of 100 mL of 0.1 M solution of CoCl3. 6H2O with excess AgNO3; 1.2 $$\\times$$ 1022 ions are precipitated. The complex is :", "options": [ { "text": "[Co(H2O)3Cl3].3H2O " }, { "text": "[Co(H2O)6]Cl3" }, { "text": "[Co(H2O)5Cl]Cl2.H2O " }, { "text": "[Co(H2O)4Cl2]Cl.2H2O " } ], "answer": "[Co(H2O)5Cl]Cl2.H2O ", "solution": "**Answer:** [Co(H2O)5Cl]Cl2.H2O \n\nMoles of complex = $${{Molarity \\times Volume\\left( {mL} \\right)} \\over {1000}}$$\n

= $${{100 \\times 0.1} \\over {1000}}$$ = 0.01 mole\n

Moles of ions precipitated with excess of AgNO3\n

= $${{1.2 \\times {{10}^{22}}} \\over {6.023 \\times {{10}^{23}}}}$$ = 0.02 mole\n

So 0.01 × n = 0.02\n

$$ \\Rightarrow $$ n = 2\n

It means 2Cl– ions present in ionization sphere.\n

$$ \\therefore $$ The formula of complex is [Co(H2O)5Cl]Cl2.H2O ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1066, "subject": "Chemistry", "question": "The correct combination is : ", "options": [ { "text": "[Ni(CN)4]2- $$-$$ tetrahedral ;\n

[Ni(CO)4] $$-$$ paramagnetic" }, { "text": "[Ni(Cl)4]2- $$-$$ paramagnetic ;\n

[Ni(CO)4] $$-$$ tetrahedral" }, { "text": "[Ni(Cl)4]2- $$-$$ square-planar ;\n

[Ni(CN)4] 2-$$-$$ paramagnetic" }, { "text": "[Ni(Cl)4]2- $$-$$ diamagnetic ;\n

[Ni(CO)4] $$-$$square-planar" } ], "answer": "[Ni(Cl)4]2- $$-$$ paramagnetic ;\n

[Ni(CO)4] $$-$$ tetrahedral", "solution": "**Answer:** [Ni(Cl)4]2- $$-$$ paramagnetic ;\n

[Ni(CO)4] $$-$$ tetrahedral\n\n[NiCl4]2– : Oxidation state of Ni in [NiCl4]2– = + 2 \n
\"JEE\n

Cl– is a weak field ligand and cannot take part in pairing of electrons.\n\"JEE\n
Hence, the complex is tetrahedral and paramagnetic with two unpaired electrons.\n

[Ni(CO)4] : Oxidation state of Ni in [Ni(CO)4] is zero. CO is a strong field ligand thus pairing of electrons takes place in d-orbitals.\n\"JEE\n
Hence, the complex is tetrahedral and diamagnetic.", "topic": "Discrete Mathematics", "subtopic": "Combinatorics" }, { "id": 1067, "subject": "Chemistry", "question": "The correct order of spin-only magnetic moments among the following is : \n
(Atomic number : Mn = 25, Co = 27, Ni = 28, Zn = 30)", "options": [ { "text": "[ZnCl4]2- > [NiCl4]2- > [CoCl4]2- > [MnCl4]2-" }, { "text": "[CoCl4]2- > [MnCl4]2- > [NiCl4]2- > [ZnCl4]2-" }, { "text": "[NiCl4]2- > [CoCl4]2- > [MnCl4]2- > [ZnCl4]2-" }, { "text": "[MnCl4]2- > [CoCl4]2- > [NiCl4]2- > [ZnCl4]2-" } ], "answer": "[MnCl4]2- > [CoCl4]2- > [NiCl4]2- > [ZnCl4]2-", "solution": "**Answer:** [MnCl4]2- > [CoCl4]2- > [NiCl4]2- > [ZnCl4]2-\n\nWe know, \n

Spin only magnetic moment ($$\\mu $$) = $$\\sqrt {n\\left( {n + 2} \\right)} $$   B.M\n

Where, n is the number of unpaired electrons.\n

So, the complex having higher number of unpaired electrons will have higher value of spin only magnetic moment. \n

(1)   Zn+2 : [Ar] 3d10\n
Here 0 unpaired electrons present \n

(2)   Ni+2 : [Ar] 3d8\n
Here 2 unpaired electrons present.\n

(3)   Co+2 : [Ar] 3d7\n
Here 3 unpaired electrons present.\n

(4)   Mn+2 : [Ar] 3d5\n
Here 5 unpaired electrons present.\n

So, correct order is $$ \\to $$\n
    [MnCl4]2$$-$$ > [CoCl4]2$$-$$ > [NiCl4]2$$-$$ > [ZnCl4]2$$-$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1068, "subject": "Chemistry", "question": "In Wilkinson's catalyst, the hybridization of central metal ion and its shape are respectively : ", "options": [ { "text": "sp3d, trigonal bipyramidal " }, { "text": "sp3, tetrahedral" }, { "text": "dsp2, square planar " }, { "text": "d2sp3, octahedral " } ], "answer": "dsp2, square planar ", "solution": "**Answer:** dsp2, square planar \n\nWilkinson's catalyst is [RhCl(PPh3)3]\n

Here central atom is Rh.\n

Oxidation state of Rh here is = + 1\n

$$ \\therefore $$   Electronic configuration of Rh+ = [Kr]4d8\n

\"JEE\n
And the complex is square planar\n

\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1069, "subject": "Chemistry", "question": "The calculated spin-only magnetic moments\n(BM) of the anionic and cationic species of\n[Fe(H2O)6]2 and [Fe(CN)6], respectively, are :", "options": [ { "text": "2.84 and 5.92" }, { "text": "4.9 and 0" }, { "text": "0 and 4.9" }, { "text": "0 and 5.92" } ], "answer": "4.9 and 0", "solution": "**Answer:** 4.9 and 0\n\nCompount is Fe(H2O)6]2 [Fe(CN)6]\n

Cation is Fe(H2O)6]2+\n

Anion is [Fe(CN)6]4-\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1070, "subject": "Chemistry", "question": "The correct statements among I to III are :
\n(I) Valence bond theory cannot explain the\ncolor exhibited by transition metal\ncomplexes.
\n(II) Valence bond theory can predict\nquantitatively the magnetic properties of\ntranstition metal complexes.
\n(III) Valence bond theory cannot distinguish\nligands as weak and strong field ones.", "options": [ { "text": "(II) and (III) only" }, { "text": "(I) and (II) only" }, { "text": "(I), (II) and (III)" }, { "text": "(I) and (III) only" } ], "answer": "(I) and (III) only", "solution": "**Answer:** (I) and (III) only\n\n

To determine which statements are correct, let's analyze each one individually in the context of Valence Bond Theory (VBT) as it applies to transition metal complexes.

\n
\n

Statement (I):

\n

\"Valence bond theory cannot explain the color exhibited by transition metal complexes.\"

\n

Analysis:

\n\n

Color in Transition Metal Complexes:

\n

The colors of transition metal complexes arise from electronic transitions between different energy levels of the d-orbitals, specifically d-d transitions.

\n

These transitions occur when an electron absorbs light energy and moves from a lower-energy d-orbital to a higher-energy d-orbital.

\n\n

Valence Bond Theory Limitations:

\n

VBT focuses on the hybridization of atomic orbitals to form covalent bonds.

\n

It does not account for the splitting of d-orbitals into different energy levels in the presence of ligands (known as crystal field splitting).

\n

Therefore, VBT cannot explain the origin of color in these complexes because it doesn't address the electronic transitions responsible for color.

\n\n

Conclusion:

\n\n

Statement (I) is correct.

\n\n
\n

Statement (II):

\n

\"Valence bond theory can predict quantitatively the magnetic properties of transition metal complexes.\"

\n

Analysis:

\n\n

Magnetic Properties:

\n

The magnetic behavior of a complex depends on the number of unpaired electrons in the metal ion.

\n

Quantitative prediction requires calculating the magnetic moment, often using the formula:

\n

$ \\mu = \\sqrt{n(n+2)} \\ \\text{Bohr Magnetons (BM)} $

\n

where $ n $ is the number of unpaired electrons.

\n

Valence Bond Theory Capabilities:

\n

VBT can provide a qualitative idea about the magnetic properties by indicating whether a complex is paramagnetic (unpaired electrons present) or diamagnetic (no unpaired electrons).

\n

However, VBT does not offer the tools to quantitatively predict the exact magnetic moment.

\n

Accurate quantitative predictions require more advanced theories like Crystal Field Theory (CFT) or Ligand Field Theory (LFT).

\n\n

Conclusion:

\n\n

Statement (II) is incorrect.

\n\n
\n

Statement (III):

\n

\"Valence bond theory cannot distinguish ligands as weak and strong field ones.\"

\n

Analysis:

\n\n

Weak and Strong Field Ligands:

\n

Ligands are classified based on their ability to split the d-orbitals of the metal ion, influencing the pairing of electrons.

\n

Strong field ligands cause a large splitting, often leading to low-spin complexes.

\n

Weak field ligands cause small splitting, leading to high-spin complexes.

\n

Valence Bond Theory Limitations:

\n

VBT does not address the energy splitting of d-orbitals.

\n

It assumes that all bonds are formed via overlap of orbitals without considering the effect of ligands on d-orbital energies.

\n

Therefore, VBT cannot distinguish between weak and strong field ligands because it doesn't involve the spectrochemical series or orbital splitting concepts.

\n\n

Conclusion:

\n\n

Statement (III) is correct.

\n\n
\n

Final Answer:

\n\n

Statements (I) and (III) are correct.

\n

Statement (II) is incorrect.

\n\n
\n

Answer: Option D

\n

(I) and (III) only

", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 1071, "subject": "Chemistry", "question": "The metal d-orbitals that are directly facing the ligands in K3[Co(CN)6] are - ", "options": [ { "text": "dx2$$-$$y2 and dz2" }, { "text": "dxy, dxz and dyz " }, { "text": "dxz, dyz and dz2 " }, { "text": "dxy and dx2 $$-$$ y2" } ], "answer": "dx2$$-$$y2 and dz2", "solution": "**Answer:** dx2$$-$$y2 and dz2\n\nDue to presence of strong field ligand (CN–) pairing occurs\nin which two d-orbitals i.e., dx2–y2 and dz2 directly face the CN-\nligand.\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1072, "subject": "Chemistry", "question": "The difference in the number of unpaired electrons of a metal ion in its high spin and low-spin octahedral complexes is two. The metal ion is :", "options": [ { "text": "Mn2+ " }, { "text": "Ni2+" }, { "text": "Co2+" }, { "text": "Fe2+" } ], "answer": "Co2+", "solution": "**Answer:** Co2+\n\n\"JEE\n
$$ \\therefore $$ The difference in the number of unpaired electrons = 3 - 1 = 2", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1073, "subject": "Chemistry", "question": "The magnetic moment of an octahedral homoleptic Mn(II) complex is 5.9 BM. The suitable ligand for this complex is - ", "options": [ { "text": "CN$$-$$ " }, { "text": "ethylenediamine " }, { "text": "NCS–" }, { "text": "CO" } ], "answer": "NCS–", "solution": "**Answer:** NCS–\n\nGiven, $$\\mu $$ = 5.9 BM\n

$$ \\therefore $$ n = number of unpaired electron = 5\n

\"JEE\n
For Mn+2 with 5 unpaired electronic configuration only possible when ligand is weak field ligand.\n

Here weak field ligand is NCS$$-$$.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1074, "subject": "Chemistry", "question": "The correct order of the calculated spin-only\nmagnetic moments of complexs (A) to (D) is:
\n(A) Ni(CO)4
\n(B) [Ni(H2O)6]Cl2
\n(C) Na2[Ni(CN)4]
\n(D) PdCl2(PPh3)2", "options": [ { "text": "(C) < (D) < (B) < (A)" }, { "text": "(C) $$ \\approx $$ (D) < (B) < (A)" }, { "text": "(A) $$ \\approx $$ (C) $$ \\approx $$ (D) < (B)" }, { "text": "(A) $$ \\approx $$ (C) < (B) $$ \\approx $$ (D)" } ], "answer": "(A) $$ \\approx $$ (C) $$ \\approx $$ (D) < (B)", "solution": "**Answer:** (A) $$ \\approx $$ (C) $$ \\approx $$ (D) < (B)\n\n(A) Ni(CO)4\n

Ni = 3d84s2\n

CO is strong field ligand. So pairing of elections happens.\n

$$ \\therefore $$ Number of unpaired electrons = 0\n

$$ \\therefore $$ $$\\mu $$spin = 0\n

(B) [Ni(H2O)6]Cl2\n

Ni+2 = 3d84s0\n

H2O is weak field ligand. So no pairing of electrons happens.\n

Number of unpaired electron = 2\n

$$ \\therefore $$ $$\\mu $$spin = $$\\sqrt {n\\left( {n + 2} \\right)} $$ = $$\\sqrt {2\\left( {2 + 2} \\right)} $$ = $$\\sqrt 8 $$ B.M\n

(C) Na2[Ni(CN)4]\n

Ni+2 = 3d84s0\n

CN- is strong field ligand. So pairing of electrons happens.\n

Number of unpaired electron = 0\n

$$ \\therefore $$ $$\\mu $$spin = 0\n

(D) PdCl2(PPh3)2\n

Pd2+ = 4d8\n

This is dsp2 complex. And shape is square planar.\n

$$ \\therefore $$ $$\\mu $$spin = 0", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1075, "subject": "Chemistry", "question": "For octahedral Mn(II) and tetrahedral Ni(II)\ncomplexes, consider the following statements:\n

(I) both the complexes can be high spin.\n
(II) Ni(II) complex can very rarely be low spin.\n
(III) with strong field ligands, Mn(II) complexes\ncan be low spin.\n
(IV)aqueous solution of Mn(II) ions is yellow in\ncolour.\n

The correct statements are :", "options": [ { "text": "(I), (III) and (IV) only" }, { "text": "(I) and (II) only" }, { "text": "(II), (III) and (IV) only" }, { "text": "(I), (II) and (III) only" } ], "answer": "(I), (II) and (III) only", "solution": "**Answer:** (I), (II) and (III) only\n\n(I) Under weak field ligand, octahedral Mn(II) and tetrahedral Ni(II) both the complexes are high spin\ncomplex.\n

(II) Tetrahedral Ni(II) complex can very rarely be low spin because square planar (under strong ligand)\ncomplexes of Ni(II) are low spin complexes.\n

(III)With strong field ligands Mn (II) complexes can be low spin because they have less number of unpaired\nelectron (unpaired electron = 1) While with weak field ligands Mn(II) complexes can be high spin\nbecause they have more number of unpaired electron (unpaired electron = 5)\n

(IV) Aqueous solution of Mn(II) ions is pink in colour.", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 1076, "subject": "Chemistry", "question": "The pair in which both the species have the\nsame magnetic moment (spin only) is :", "options": [ { "text": "[Cr(H2O)6]2+ and [CoCl4]2–" }, { "text": "[Co(OH)4]2– and [Fe(NH3)6]2+" }, { "text": "[Mn(H2O)6]2+ and [Cr(H2O)]2+" }, { "text": "[Cr(H2O)6]2+ and [Fe(H2O)6]2+" } ], "answer": "[Cr(H2O)6]2+ and [Fe(H2O)6]2+", "solution": "**Answer:** [Cr(H2O)6]2+ and [Fe(H2O)6]2+\n\nSpecies with same number of unpaired\nelectrons have equal magnetic moment.\n

\n\n\n \n \n \n \n \n \n\n\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\n
ComplexLigandLigand TypeNumber of unpaired electrons
[Mn(H2O)6]2+H2OWeak Field Ligand5
[Cr(H2O)]2+H2OWeak Field Ligand4
[CoCl4]2–ClWeak Field Ligand3
[Fe(H2O)6]2+H2OWeak Field Ligand4
[Co(OH)4]2–OHWeak Field Ligand3
[Fe(NH3)6]2+NH3Weak Field Ligand4
", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1077, "subject": "Chemistry", "question": "The molecule in which hybrid MOs involve only\none d-orbital of the central atom is :", "options": [ { "text": "XeF4" }, { "text": "[Ni(CN)4]2–" }, { "text": "[CrF6]3–" }, { "text": "BrF5" } ], "answer": "[Ni(CN)4]2–", "solution": "**Answer:** [Ni(CN)4]2–\n\nXeF4 = sp3d2\n

[Ni(CN)4]2– = dsp2\n

[CrF6]3– = d2sp2\n

BrF5 = sp3d2", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1078, "subject": "Chemistry", "question": "The species that has a spin-only magnetic\nmoment of 5.9 BM, is :\n
(Td = tetrahedral)", "options": [ { "text": "[MnBr4]2– (Td)" }, { "text": "[NiCl4]2– (Td)" }, { "text": "Ni(CO)4 (Td)" }, { "text": "[Ni(CN)4]2– (square planar)" } ], "answer": "[MnBr4]2– (Td)", "solution": "**Answer:** [MnBr4]2– (Td)\n\n\"JEE\n

$$ \\therefore $$ Unpaired electrons = 5\n

$$ \\therefore $$ $$\\mu $$ = $$\\sqrt {5\\left( {5 + 2} \\right)} $$ = 5.9 BM", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1079, "subject": "Chemistry", "question": "The hybridization and magnetic nature of $${[Mn{(CN)_6}]^{4 - }}$$ and $${[Fe{(CN)_6}]^{3 - }}$$, respectively are :", "options": [ { "text": "sp3d2 and diamagnetic" }, { "text": "d2sp3 and paramagnetic" }, { "text": "sp3d2 and paramagnetic " }, { "text": "d2sp3 and diamagnetic" } ], "answer": "d2sp3 and paramagnetic", "solution": "**Answer:** d2sp3 and paramagnetic\n\n$${[Mn{(CN)_6}]^{4 - }}$$\n

Mn2+ = 3d5 4s0\n

CN– is a strong field ligand.\n

$$ \\therefore $$ Pairing will occur.\n
\"JEE\n

$${[Fe{(CN)_6}]^{3 - }}$$\n

Fe3+ = 3d54s0\n

CN– is a strong field ligand.\n

$$ \\therefore $$ Pairing will occur.\n
\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1080, "subject": "Chemistry", "question": "In which of the following order the given complex ions are arranged correctly with respect to their decreasing spin only magnetic moment?

(i) [FeF6]3$$-$$

(ii) [Co(NH3)6]3+

(iii) [NiCl4]2$$-$$

(iv) [Cu(NH3)4]2+", "options": [ { "text": "(iii) > (iv) > (ii) > (i)" }, { "text": "(ii) > (iii) > (i) > (iv)" }, { "text": "(i) > (iii) > (iv) > (ii)" }, { "text": "(ii) > (i) > (iii) > (iv)" } ], "answer": "(i) > (iii) > (iv) > (ii)", "solution": "**Answer:** (i) > (iii) > (iv) > (ii)\n\n

Spin only magnetic moment, $$\\mu = \\sqrt {n(n + 2)} BM$$

\n

where, n = number of unpaired electrons and $$\\mu$$ $$\\propto$$ n.

\n

(i) $${[Fe{F_6}]^{3 - }} \\Rightarrow F{e^{3 + }} = (3{d^5}) , {F^ - }$$ (weak field ligand).

\n

Thus, pairing of electron does not take place.

\n

\"JEE

\n

(ii) $${[Co{(N{H_3})_6}]^{3 + }} \\Rightarrow C{o^{3 + }} = (3{d^6}), N{H_3}$$ (strong field ligand).

\n

Thus, pairing of electron takes place.

\n

\"JEE

\n

(iii) $${[NiC{l_4}]^{2 - }} \\Rightarrow N{i^{2 + }} = (3{d^8}), C{l^ - }$$ (weak field ligand).

\n

Thus, pairing of electron takes place.

\n

\"JEE

\n

(iv) $${[Cu{(N{H_3})_4}]^{2 + }} \\Rightarrow C{u^{2 + }}(3{d^9}), N{H_3}$$ (strong field ligand).

\n

Thus, pairing of electron takes place.

\n

\"JEE

\n

So, the decreasing order of $$\\mu$$ is

\n

$$\\mathop {{\\mu _i}}\\limits_{(n = 5)} > \\mathop {{\\mu _{iii}}}\\limits_{(n = 2)} > \\mathop {{\\mu _{iv}}}\\limits_{(n = 1)} > \\mathop {{\\mu _{ii}}}\\limits_{(n = 0)} $$

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1081, "subject": "Chemistry", "question": "The spin only magnetic moment of a divalent ion in aqueous solution (atomic number 29) is _________ BM.", "options": [], "answer": "2", "solution": "**Answer:** 2\n\n

Z = 29 [Cu] $$\\buildrel { - 2{e^ - }} \\over\n \\longrightarrow $$ Cu2+ = [Ar] 3d9

\n

\"JEE

\n

Number of unpaired electron, n = 1

\n

$$\\therefore$$ Spin only magnetic moment,

\n

$$\\mu = \\sqrt {n(n + 2)} BM = \\sqrt {1(1 + 2)} BM = \\sqrt 3 BM$$

\n

= 1.73 BM $$ \\simeq $$ 2 BM

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1082, "subject": "Chemistry", "question": "On complete reaction of FeCl3 with oxalic acid in aqueous solution containing KOH, resulted in the formation of product A. The secondary valency of Fe in the product A is __________. (Round off to the Nearest Integer).", "options": [], "answer": "6", "solution": "**Answer:** 6\n\nFeCl3 + 3H2C2O4 + 6KOH $$ \\to $$ K3[Fe(C2O4)3] + 3KCl + 6H2O\n\"JEE\n
Coordination number = 6\n

Secondary valency is 6", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1083, "subject": "Chemistry", "question": "According to the valence bond theory the hybridization of central metal atom is dsp2 for which one of the following compounds?", "options": [ { "text": "$$NiC{l_2}.6{H_2}O$$" }, { "text": "$${K_2}[Ni{(CN)_4}]$$" }, { "text": "$$[Ni{(CO)_4}]$$" }, { "text": "$$N{a_2}[NiC{l_4}]$$" } ], "answer": "$${K_2}[Ni{(CN)_4}]$$", "solution": "**Answer:** $${K_2}[Ni{(CN)_4}]$$\n\nAccording to VBT i.e. valence bond theory,

Electronic configuration of Ni = [Ar]3d84s2.

(a) NiCl2 . 6H2O

NiCl2 . 6H2O $$\\rightarrow$$ NiCl2 + 6H2O

Oxidation number of Ni(x) = x + 2($$-$$1) = 0; where x, $$-$$1 and 2 are the oxidation number of Ni, oxidation number of Cl and number of Cl atoms respectively.

NiCl2 $$\\Rightarrow$$ x + ($$-$$2) = 0 $$\\Rightarrow$$ x = 2

Electronic configuration of Ni2+ = [Ar]3d84s0

Cl$$-$$ is a weak field ligand. So, no pairing of electrons occurs.

For C.N. = 6

\"JEE
(b) K2[Ni(CN)4]

K2[Ni(CN)4] $$\\rightarrow$$ 2K+ + [Ni(CN)4]2$$-$$

x + 4($$-$$1) $$-$$ ($$-$$2) = 0; where x, 4, $$-$$1 and $$-$$2 are the oxidation number of Ni, number of CN ligands, charge on one CN and charge on complex.

[Ni(CN)4]2$$-$$ $$\\Rightarrow$$ x $$-$$ 4 + 2 = 0 $$\\Rightarrow$$ x = +2

Electronic configuration of Ni2+ $$\\Rightarrow$$ [Ar]3d84s0

CN$$-$$ is a strong field ligand. So, pairing of electrons occur. For C.N. = 4

\"JEE
(c) Ni(CO)4

CO is neutral and strong field ligand. So, pairing of electrons occur.

Oxidation number of Ni is zero

Electronic configuration of Ni = [Ar]3d104s0

For C.N. = 4

\"JEE
(d) Na2[NiCl4]

Na2[NiCl4] $$\\rightarrow$$ 2Na + [NiCl4]2$$-$$

x + 4($$-$$1) $$-$$ ($$-$$2) = 0; where x, 4, $$-$$1 and $$-$$2 are the oxidation number of Ni, number of CN ligands, charge on one CN and charge on complex.

[NiCl4]2$$-$$ $$\\Rightarrow$$ x $$-$$ 4 + 2 = 0 $$\\Rightarrow$$ x = +2

Electronic configuration of Ni2+ = [Ar]3d84s0

For C.N. = 4

\"JEE
Hence, only K2[Ni(CN)4] has dsp2 hybridization.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1084, "subject": "Chemistry", "question": "The spin-only magnetic moment value for the complex [Co(CN)6]4$$-$$ is __________ BM.

[At. no. of Co = 27]", "options": [], "answer": "2", "solution": "**Answer:** 2\n\n[Co(CN)6]4$$-$$

x + 6 $$\\times$$ ($$-$$1) = $$-$$4

where, x, 6, $$-$$1 and $$-$$4 are the oxidation number of Co, number of CN ligands, charge on one CN and charge on complex.

x = +2, i.e. Co2+

Electronic configuration of Co2+ : [Ar]3d7 and CN$$-$$ is a strong field ligand which can pair electron of central atom.

\"JEE
It has one unpaired electron (n) in 4d-subshell.

So, spin only magnetic moment ($$\\mu$$) = $$\\sqrt {n(n + 2)} $$ BM = $$\\sqrt {1(1 + 2)} $$ BM = $$\\sqrt 3 $$ BM

where, n = number of unpaired electrons

$$\\mu$$ = $$\\sqrt 3 $$ BM = 1.73 BM

Nearest integer = 2", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1085, "subject": "Chemistry", "question": "Spin only magnetic moment of an octahedral complex of Fe2+ in the presence of a strong field ligand in BM is :", "options": [ { "text": "4.89" }, { "text": "2.82" }, { "text": "0" }, { "text": "3.46" } ], "answer": "0", "solution": "**Answer:** 0\n\nIn presence of SFL $$\\Delta$$0 > P means pairing occurs therefore

For Fe+2 $$\\to$$ 3d6

\"JEE

$$\\therefore$$ No of unpaired e-(s) = 0

$$\\therefore$$ $$\\mu$$ = $$\\sqrt {n(n + 2)} $$BM = 0

[n = No of unpaired e$$-$$(s)]

In NiCl2, Ni+2 is having configuration 3d8

$$\\therefore$$ Number of unpaired electron = 2

After formation of oxidised product

[Ni(CN)6]$$-$$2$$ \\Rightarrow $$ Ni+4 is obtained

Ni+4 $$\\Rightarrow$$ 3d6 and CN$$-$$ is strong field ligand

$$\\therefore$$ Number of unpaired electrons = 0

$$\\therefore$$ The charge is 2 $$-$$ 0 = 2", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 1086, "subject": "Chemistry", "question": "Which one of the following species doesn't have a magnetic moment of 1.73 BM, (spin only value)?", "options": [ { "text": "O$$_2^ + $$" }, { "text": "CuI" }, { "text": "[Cu(NH3)4]Cl2" }, { "text": "O$$_2^ - $$" } ], "answer": "CuI", "solution": "**Answer:** CuI\n\nSpecies does not have a magnetic moment of 1.7 BM means species must not contain single unpaired electron.

(a) Molecular orbital configuration of O$$_2^ + $$ (15 electrons) is \n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,{\\sigma _{2{s^2}}}\\,\\sigma _{2{s^2}}^ * \\,{\\sigma _{2p_z^2}}\\,{\\pi _{2p_x^2}}\\, = \\,{\\pi _{2p_y^2}}\\,\\pi _{2p_x^1}^ * \\, = \\,\\pi _{2p_y^o}^ * $$

Unpaired electron = 1

$$\\mu = \\sqrt {n(n + 2)} = \\sqrt {1(1 + 2)} = \\sqrt 3 = 1.73$$

where n = no. of unpaired e$$-$$

$$\\therefore$$ $$\\mu$$ = 1.73 BM

(b) Cul $$\\Rightarrow$$ Cu+ $$-$$[Ar]3d10, Unpaired electron = 0

I$$-$$ $$\\to$$ [Xe], unpaired electron = 0

Therefore, $$\\mu$$ = 0

(c) [Cu(NH3)4]Cl2

Cu2+ $$\\to$$ [Ar]3d9

Unpaired electron = 1,

Therefore, $$\\mu$$ = 1.73 BM

(d) Molecular orbital configuration of $$O_2^ - $$ (17 electrons) is \n

$${\\sigma _{1{s^2}}}\\,\\sigma _{1{s^2}}^ * \\,{\\sigma _{2{s^2}}}\\,\\sigma _{2{s^2}}^ * \\,{\\sigma _{2p_z^2}}\\,{\\pi _{2p_x^2}}\\, = \\,{\\pi _{2p_y^2}}\\,\\pi _{2p_x^2}^ * \\, = \\,\\pi _{2p_y^1}^ * $$

Unpaired electron = 1

Therefore, $$\\mu$$ = 1.73 BM

Therefore, Cul does not have magnetic moment of 1.73 BM.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1087, "subject": "Chemistry", "question": "An aqueous solution of NiCl2 was heated with excess sodium cyanide in presence of strong oxidizing agent to form [Ni(CN)6]2$$-$$. The total change in number of unpaired electrons on metal centre is _______________.", "options": [], "answer": "2", "solution": "**Answer:** 2\n\n[Ni(CN)6]2-

Ni+4 $$\\to$$ d6, CN- strong field ligand. So pairing will happen.

\"JEE

Here zero unpaired electron

NiCl2 $$\\to$$ Ni2+ $$\\to$$ d8

\"JEECl- Weak field ligand.$$\\to$$ two unpaired e-

$$ \\therefore $$ Change = 2.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1088, "subject": "Chemistry", "question": "Which one of the following species responds to an external magnetic field?", "options": [ { "text": "[Fe(H2O)6]3+" }, { "text": "[Ni(CN)4]2$$-$$" }, { "text": "[Co(CN)6]3$$-$$" }, { "text": "[Ni(CO)4]" } ], "answer": "[Fe(H2O)6]3+", "solution": "**Answer:** [Fe(H2O)6]3+\n\n1. $${[\\mathop {Fe}\\limits^{ + 3} {({H_2}O)_6}]^{3 + }}$$

$$F{e^{3 + }}:[Ar]3{d^5}$$

Hybridisation : $$s{p^3}{d^2}$$

Magnetic nature : Paramagnetic (so this complex response to external magnetic field)

2. $${[Ni{(CN)_4}]^{2 - }}$$

$$N{i^{2 + }}:[Ar]3{d^8}$$

Hybridisation : dsp2

Magnetic nature : diamagnetic

3. $${[Co{(CN)_6}]^{3 - }}$$

$$C{o^{3 + }}:[Ar]3{d^6}$$

Hybridisation : $${d^2}s{p^3}$$

Magnetic nature : diamagnetic

4. $$[Ni{(CO)_4}]$$

$$Ni:[Ar]3{d^8}4{s^2}$$

Hybridisation : $$s{p^3}$$

Magnetic nature : diamagnetic", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 1089, "subject": "Chemistry", "question": "The type of hybridisation and magnetic property of the complex [MnCl6]3$$-$$, respectively, are :", "options": [ { "text": "sp3d2 and diamagnetic" }, { "text": "d2sp3 and diamagnetic" }, { "text": "d2sp3 and paramagnetic" }, { "text": "sp3d2 and paramagnetic" } ], "answer": "sp3d2 and paramagnetic", "solution": "**Answer:** sp3d2 and paramagnetic\n\n[MnCl6]3$$-$$

\"JEE

Paramagnetic and having 4 unpaired electrons.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1090, "subject": "Chemistry", "question": "Given below are two statements :

Statement I : $${[Mn{(CN)_6}]^{3 - }}$$, $${[Fe{(CN)_6}]^{3 - }}$$ and $${[Co{({C_2}{O_4})_3}]^{3 - }}$$ are d2sp3 hybridised.

Statement II : $${[MnCl)_6}{]^{3 - }}$$ and $${[Fe{F_6}]^{3 - }}$$ are paramagnetic and have 4 and 5 unpaired electrons, respectively.

In the light of the above statements, choose the correct answer from the options given below :", "options": [ { "text": "Statement I is correct but statement II is false" }, { "text": "Both statement I and Statement II are false" }, { "text": "Statement I is incorrect but statement II is true" }, { "text": "Both statement I and statement II are true" } ], "answer": "Both statement I and statement II are true", "solution": "**Answer:** Both statement I and statement II are true\n\n

Hybridization of all the given complexes are given in following table.

\n

$${[Mn{(CN)_6}]^{3 - }} \\Rightarrow M{n^{3 + }}$$

\n

CN$$-$$ is a strong field ligand.

\n

$$\\therefore$$ Pairing of electron takes place.

\n

\"JEE

\n

$${[Fe{(CN)_6}]^{3 - }} \\Rightarrow F{e^{3 + }}$$

\n

CN$$-$$ is a strong ligand.

\n

$$\\therefore$$ Pairing of electron takes place.

\n

\"JEE

\n

$${[Co{({C_2}{O_4})_3}]^{3 - }} \\Rightarrow C{o^{3 + }}$$

\n

$${C_2}O_4^{2 - }$$ acts as strong field ligand with Co3+.

\n

$$\\therefore$$ Pairing of electrons takes place.

\n

\"JEE

\n

$${[Fe{F_6}]^{3 - }} \\Rightarrow F{e^{3 + }}$$

\n

F is a weak field ligand.

\n

$$\\therefore$$ Pairing of electron does not occurs.

\n

\"JEE

\n

$${[MnC{l_6}]^{3 - }} \\Rightarrow M{n^{3 + }}$$

\n

Cl$$-$$ is a weak field ligand.

\n

$$\\therefore$$ Pairing of electron does not occurs.

\n

\"JEE

\n

$${[Fe{F_6}]^{3 - }}$$ have 5 unpaired electrons and $${[MnC{l_6}]^{3 - }}$$ have 4 unpaired electrons.

\n

So, both statement I and statement II are true.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1091, "subject": "Chemistry", "question": "The overall stability constant of the complex ion [Cu(NH3)4]2+ is 2.1 $$\\times$$ 1013. The overall dissociations constant is y $$\\times$$ 10$$-$$14. Then y is __________. (Nearest integer)", "options": [], "answer": "5", "solution": "**Answer:** 5\n\n\"JEE\n

Given ks = 2.1 $$\\times$$ 1013

Kd = $${1 \\over {{k_s}}}$$ = 4.7 $$\\times$$ 10$$-$$14

$$\\therefore$$ y = 4.7 $$ \\approx $$ 5", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1092, "subject": "Chemistry", "question": "1 mol of an octahedral metal complex with formula MCl3 . 2L on reaction with excess of AgNO3 gives 1 mol of AgCl. The denticity of Ligand L is ____________. (Integer answer)", "options": [], "answer": "2", "solution": "**Answer:** 2\n\nMCl3 . 2L octahedral

$$\\mathop {MC{l_3}.2L}\\limits_{1\\,mole} \\buildrel {Ex.\\,AgN{O_3}} \\over\n \\longrightarrow $$ 1 mole of AgCl

Its means that one Cl$$-$$ ion present in ionization sphere.

$$\\therefore$$ formula = [MCl2L2]Cl

For octahedral complex coordination no. is 6

$$\\therefore$$ L act as bidentate ligand", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1093, "subject": "Chemistry", "question": "Spin only magnetic moment in BM of [Fe(CO)4(C2O4)]+ is :", "options": [ { "text": "5.92" }, { "text": "0" }, { "text": "1" }, { "text": "1.73" } ], "answer": "1.73", "solution": "**Answer:** 1.73\n\n[Fe(CO)4(C2O4)]+

\"JEE

One unpaired electron Spin only magnetic moment = $$\\sqrt 3 $$ B.M. = 1.73 BM", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1094, "subject": "Chemistry", "question": "

$${[Fe{(CN)_6}]^{4 - }}$$

\n

$${[Fe{(CN)_6}]^{3 - }}$$

\n

$${[Ti{(CN)_6}]^{3 - }}$$

\n

$${[Ni{(CN)_4}]^{2 - }}$$

\n

$${[Co{(CN)_6}]^{3 - }}$$

\n

Among the given complexes, number of paramagnetic complexes is ____________.

", "options": [], "answer": "2", "solution": "**Answer:** 2\n\n$\\left[\\mathrm{Fe}(\\mathrm{CN})_6\\right]^{4-} \\quad$ Diamagnetic

\n$\\left[\\mathrm{Fe}(\\mathrm{CN})_6\\right]^{3-} \\quad$ Paramagnetic (1 unpaired electron)

\n$\\left[\\mathrm{Ti}(\\mathrm{CN})_6\\right]^{3-} \\quad$ Paramagnetic (1 unpaired electron)

\n$\\left[\\mathrm{Ni}(\\mathrm{CN})_4\\right]^{2-} \\quad$ Diamagnetic

\n$\\left[\\mathrm{Co}(\\mathrm{CN})_6\\right]^{3-} \\quad$ Diamagnetic", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1095, "subject": "Chemistry", "question": "

Given below are two statements :

\n

Statement I : [Ni(CN)4]2$$-$$ is square planar and diamagnetic complex, with dsp2 hybridization for Ni but [Ni(CO)4] is tetrahedral, paramagnetic and with sp3-hybridication for Ni.

\n

Statement II : [NiCl4]2$$-$$ and [Ni(CO)4] both have same d-electron configuration have same geometry and are paramagnetic.

\n

In light the above statements, choose the correct answer from the options given below :

", "options": [ { "text": "Both Statement I and Statement II are true." }, { "text": "Both Statement I and Statement II are false." }, { "text": "Statement I is correct but Statement II is false." }, { "text": "Statement I is false but Statement II is correct." } ], "answer": "Both Statement I and Statement II are false.", "solution": "**Answer:** Both Statement I and Statement II are false.\n\n[Ni(CN)4]2– : d8 configuration, SFL, sq. planar splitting (dsp2), diamagnetic.

\n[Ni(CO)4] : d10 config (after excitation), SFL, tetrahedral splitting (sp3), diamagnetic.

\n[NiCl4]2– : d8 config, WFL, tetrahedral splitting (sp3), paramagnetic(2 unpaired e–).", "topic": "Geometry", "subtopic": "Plane Geometry" }, { "id": 1096, "subject": "Chemistry", "question": "

Arrange the following coordination compounds in the increasing order of magnetic moments. (Atomic numbers : Mn = 25; Fe = 26)

\n

A. [FeF6]3$$-$$

\n

B. [Fe(CN)6]3$$-$$

\n

C. [MnCl6]3$$-$$ (high spin)

\n

D. [Mn(CN)6]3$$-$$

\n

Choose the correct answer from the options given below :

", "options": [ { "text": "A < B < D < C" }, { "text": "B < D < C < A" }, { "text": "A < C < D < B" }, { "text": "B < D < A < C" } ], "answer": "B < D < C < A", "solution": "**Answer:** B < D < C < A\n\n\n\n\n \n \n \n \n \n\n\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\n
Coordination CompoundNumber of unpaired e– (n)Magnetic moment (µ) (B.M)
A. [FeF6]3$$-$$ - d555.91
B. [Fe(CN)6]3$$-$$ - d511.73
C. [MnCl6]3$$-$$ - d444.89
D. [Mn(CN)6]3$$-$$ - d422.82

\nHence, correct order of magnetic moment is $2 < 4 < 3 < 1$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1097, "subject": "Chemistry", "question": "

Acidified potassium permanganate solution oxidises oxalic acid. The spin-only magnetic moment of the manganese product formed from the above reaction is ____________ B.M. (Nearest integer)

", "options": [], "answer": "6", "solution": "**Answer:** 6\n\n$\\mathrm{KMnO}_{4}$ (acidic medium) $+\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4} \\rightarrow \\mathrm{CO}_{2}+\\mathrm{Mn}^{+2}$

$\\mathrm{Mn}^{+2}$ has 5 unpaired electrons\n

\n$\\therefore $ Spin only magnetic moment $=\\sqrt{5(5+2)}$\n

\n$$\n\\begin{aligned}\n&=\\sqrt{5 \\times 7} \\\\\\\\\n&=\\sqrt{35} \\\\\\\\\n&\\simeq 5.92 \\text { B.M. } \\\\\\\\\n&\\simeq 6 \\text { B.M. }\n\\end{aligned}\n$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1098, "subject": "Chemistry", "question": "

Amongst FeCl3.3H2O, K3[Fe(CN)6] and [Co(NH3)6]Cl3, the spin-only magnetic moment value of the inner-orbital complex that absorbs light at shortest wavelength is ____________ B.M. [nearest integer]

", "options": [], "answer": "2", "solution": "**Answer:** 2\n\n$\\left[\\mathrm{Fe}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{3} \\mathrm{Cl}_{3}\\right] \\rightarrow$ Outer-orbital complex\n

\n$$\n\\mathrm{K}_{3}\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}\\right] \\rightarrow \\text { Inner-orbital complex }\n$$\n

\n$\\left[\\mathrm{Co}\\left(\\mathrm{NH}_{3}\\right)_{6}\\right] \\mathrm{Cl}_{3} \\rightarrow$ Inner-orbital complex\n

\nSince $\\mathrm{CN}^{-}$is a strong field ligand than $\\mathrm{NH}_{3}$. Hence $\\mathrm{K}_{3}\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}\\right]$ is the inner-orbital complex that absorbs light at shortest wavelength.\n

\n$\\mathrm{Fe}(\\text{III}) \\rightarrow$ valence shell configuration $3 \\mathrm{~d}^{5}$\n

\nSince $\\mathrm{CN}^{-}$will do pairing, so unpaired electron $=1$\n

\n$$\n\\mu=\\sqrt{1(1+2)}=\\sqrt{3} \\mathrm{BM} \\simeq 2 \\mathrm{BM}\n$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1099, "subject": "Chemistry", "question": "

Match List - I with List - II :

\n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
List - IList -II
(A)$${[PtC{l_4}]^{2 - }}$$(I)$$s{p^3}d$$
(B)$$Br{F_5}$$(II)$${d^2}s{p^3}$$
(C)$$PC{l_5}$$(III)$$ds{p^2}$$
(D)$${[Co{(N{H_3})_6}]^{3 + }}$$(IV)$$s{p^3}{d^2}$$

\n

Choose the most appropriate answer from the options given below :

", "options": [ { "text": "(A)-(II), (B)-(IV), (C)-(I), (D)-(III)" }, { "text": "(A)-(III), (B)-(IV), (C)-(I), (D)-(II)" }, { "text": "(A)-(III), (B)-(I), (C)-(IV), (D)-(II)" }, { "text": "(A)-(II), (B)-(I), (C)-(IV), (D)-(III)" } ], "answer": "(A)-(III), (B)-(IV), (C)-(I), (D)-(II)", "solution": "**Answer:** (A)-(III), (B)-(IV), (C)-(I), (D)-(II)\n\n\n\n\n\n\n\n\n\n\n \n \n \n \n \n \n\n\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\n
List - IList -II
(A)$${[PtC{l_4}]^{2 - }}$$(III)$$ds{p^2}$$
(B)$$Br{F_5}$$(IV)$$s{p^3}d^2$$
(C)$$PC{l_5}$$(I)$$s{p^3}d$$
(D)$${[Co{(N{H_3})_6}]^{3 + }}$$(II)$$d^2s{p^3}$$
", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1100, "subject": "Chemistry", "question": "

In the following brown complex, the oxidation state of iron is +_____________.

\n

$${[Fe{({H_2}O)_6}]^{2 + }} + NO \\to \\mathop {{{[Fe{{({H_2}O)}_5}(NO)]}^{2 + }}}\\limits_{\\text{Brown complex}} + {H_2}O$$

", "options": [], "answer": "1", "solution": "**Answer:** 1\n\n

To determine the oxidation state of iron in the brown complex ${[Fe({H_2}O)_5(NO)]^{2+}}$, we need to analyze the ligands and their charges, as well as the overall charge of the complex.

\n

Step 1: Understanding the Complex

\n

The complex is:

\n

$ {[Fe({H_2}O)_5(NO)]^{2+}} $

\n\n

Iron (Fe): The central metal atom whose oxidation state we need to find.

\n

Ligands:

\n

5 Water Molecules ($H_2O$): Neutral ligands (charge = 0).

\n

1 Nitrosyl Group (NO): Can have different charges depending on its mode of bonding.

\n\n

Step 2: Assigning Charges to Ligands

\n\n

Water ($H_2O$): Neutral ligand, so it contributes 0 to the overall charge.

\n

Nitrosyl (NO): Can act as:

\n

NO$^+$ (nitrosonium ion) with a +1 charge.

\n

NO (neutral molecule) with 0 charge.

\n

NO$^-$ (nitroxide ion) with a –1 charge.

\n\n

Step 3: Setting Up the Oxidation State Equation

\n

Let $ x $ be the oxidation state of Fe.

\n

The sum of the oxidation states of all components equals the overall charge of the complex:

\n

$ x + (5 \\times 0) + (\\text{Charge of NO}) = +2 $

\n

Simplifying:

\n

$ x + (\\text{Charge of NO}) = +2 $

\n

Step 4: Determining the Charge of NO in the Complex

\n

In the brown ring complex, experimental evidence shows that the nitrosyl ligand acts as NO$^+$. This is because:

\n\n

The NO ligand forms a linear bond with Fe, characteristic of NO$^+$.

\n

The complex is known to involve a reduction of the oxidation state of Fe to an unusual value.

\n\n

Thus, Charge of NO = +1.

\n

Step 5: Calculating the Oxidation State of Fe

\n

Substitute the charge of NO into the equation:

\n

$ x + (+1) = +2 $

\n

Solving for $ x $:

\n

$ x = +2 - (+1) = +1 $

\n

Therefore, the oxidation state of Fe in the brown complex is $ +1 $.

\n

Conclusion

\n

The oxidation state of iron in ${[Fe({H_2}O)_5(NO)]^{2+}}$ is +1.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1101, "subject": "Chemistry", "question": "

Spin only magnetic moment ($$\\mu$$s) of $${K_3}[Fe{(CN)_6}]$$ is ____________ B.M.

\n

(Nearest integer)

", "options": [], "answer": "2", "solution": "**Answer:** 2\n\n$\\mathrm{K}_{3}\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}\\right]$\n

\n$\\mathrm{CN}^{-}$ have $-1$ oxidation State.\n

\nLet oxidation state of $\\mathrm{Fe}$ in the complex is $x$.\n

\nSo, $x-6=-3 \\Rightarrow x=+3$\n

\nNow, $\\mathrm{CN}^{-}$ion is a strong field ligand.\n

\nFor $Fe^{3+}: -3d^54s^0$ \"JEE

\nUnpaired electron $=1$\n

\n$\\mu_{\\mathrm{B}}=\\sqrt{n(n+2)}=\\sqrt{1 \\times(1+2)}=1.7$\n

\n$\\mu_{\\mathrm{B}}=2$ B.M\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1102, "subject": "Chemistry", "question": "

The difference between spin only magnetic moment values of $$\\left[\\mathrm{Co}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}\\right]\\mathrm{Cl}_{2}$$ and $$\\left[\\mathrm{Cr}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}\\right] \\mathrm{Cl}_{3}$$ is ___________.

", "options": [], "answer": "0", "solution": "**Answer:** 0\n\n$\\mathrm{Co} \\rightarrow 4 s^{2} 3 d^{7}$\n

\n$\\mathrm{H}_{2} \\mathrm{O}$ is weak field ligand.\n

\n$\\mathrm{Co}^{+2} \\rightarrow 3 d^{7}$\n

\n$$\n\\begin{aligned}\n\\mathrm{n}=3 \\quad \\mu_{1} &=\\sqrt{n(\\mathrm{n}+2)} \\\\\n&=\\sqrt{15} \\text { B.M. }\n\\end{aligned}\n$$\n

\n$\\mathrm{Cr} \\rightarrow 4 s^{1} 3 d^{5}$\n

\n$\\mathrm{Co}^{+3} \\rightarrow 3 d^{3}$\n

\n$\\mathrm{n}=3 \\quad \\mu_{2}=\\sqrt{15}$ B.M.\n

\n$\\mu_{1}-\\mu_{2}=0$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1103, "subject": "Chemistry", "question": "

The metal complex that is diamagnetic is (Atomic number: $$\\mathrm{Fe}, 26 ; \\mathrm{Cu}, 29)$$

", "options": [ { "text": "$$\\mathrm{K}_{3}\\left[\\mathrm{Cu}(\\mathrm{CN})_{4}\\right]$$" }, { "text": "$$\\mathrm{K}_{2}\\left[\\mathrm{Cu}(\\mathrm{CN})_{4}\\right]$$" }, { "text": "$$\\mathrm{K}_{3}\\left[\\mathrm{Fe}(\\mathrm{CN})_{4}\\right]$$" }, { "text": "$$\\mathrm{K}_{4}\\left[\\mathrm{FeCl}_{6}\\right]$$" } ], "answer": "$$\\mathrm{K}_{3}\\left[\\mathrm{Cu}(\\mathrm{CN})_{4}\\right]$$", "solution": "**Answer:** $$\\mathrm{K}_{3}\\left[\\mathrm{Cu}(\\mathrm{CN})_{4}\\right]$$\n\n$$\\Rightarrow \\mathrm{K}_{3}\\left[\\mathrm{Cu}(\\mathrm{CN})_{4}\\right]$$ is diamagnetic\n

\n$$\\mathrm{Cu}(\\mathrm{I}) \\Rightarrow \\mathrm{d}^{10}$$ configuration $$\\Rightarrow$$ No unpaired electrons.\n

\n$$\\Rightarrow \\mathrm{K}_{2}\\left[\\mathrm{Cu}(\\mathrm{CN})_{4}\\right], \\mathrm{K}_{3}\\left[\\mathrm{Fe}(\\mathrm{CN})_{4}\\right]$$ and $$\\mathrm{K}_{4}\\left[\\mathrm{FeCl}_{6}\\right]$$ are paramagnetic in nature", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1104, "subject": "Chemistry", "question": "

Match List I with List II

\n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
List - I (Complex)List - II (Hybridization)
(A)$$Ni{(CO)_4}$$(I)$$s{p^3}$$
(B)$${[Ni{(CN)_4}]^{2 - }}$$(II)$$s{p^3}{d^2}$$
(C)$${[Co{(CN)_6}]^{3 - }}$$(III)$${d^2}s{p^3}$$
(D)$${[Co{F_6}]^{3 - }}$$(IV)$$ds{p^2}$$

\n

Choose the correct answer from the options given below :

", "options": [ { "text": "A-IV, B-I, C-III, D-II" }, { "text": "A-I, B-IV, C-III, D-II" }, { "text": "A-I, B-IV, C-II, D-III" }, { "text": "A-IV, B-I, C-II, D-III" } ], "answer": "A-I, B-IV, C-III, D-II", "solution": "**Answer:** A-I, B-IV, C-III, D-II\n\nNi(CO)4 Hybridisation sp3\n

\n[Ni(CN)4]2– Hybridisation dsp2

\n [Co(CN)6]3– Hybridisation d2sp3

\n[Co(F)6]3– Hybridisation sp3d2", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1105, "subject": "Chemistry", "question": "

The spin only magnetic moment of $$\\left[\\mathrm{Mn}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}\\right]^{2+}$$ complexes is _________ B.M. (Nearest integer)

\n

(Given : Atomic no. of Mn is 25)

", "options": [], "answer": "6", "solution": "**Answer:** 6\n\n

The spin only magnetic moment of the $$\\left[\\mathrm{Mn}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}\\right]^{2+}$$ complex can be calculated using the formula:

\n\n

μ = $$\\sqrt {n\\left( {n + 2} \\right)} $$ Bohr magnetons

\n\n

Where n is the number of unpaired electrons in the complex. To find the number of unpaired electrons in $$\\left[\\mathrm{Mn}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}\\right]^{2+}$$, we can use the electron configuration of the Mn ion:

\n\n

For Mn2+ : 1s2 2s2 2p6 3s2 3p6 3d5

\n\n

From the electron configuration, we can see that Mn2+ has 5 unpaired electrons. Thus, the spin only magnetic moment of $$\\left[\\mathrm{Mn}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}\\right]^{2+}$$ is:

\n\n

μ = $$\\sqrt {5\\left( {5 + 2} \\right)} $$ = $$\\sqrt {35} $$ = 5.9 Bohr Magnetons

\n\n

The nearest integer to 5.9 is 6.

\n

Hence, the spin only magnetic moment of $$\\left[\\mathrm{Mn}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}\\right]^{2+}$$ is 6 Bohr magnetons.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1106, "subject": "Chemistry", "question": "

Cobalt chloride when dissolved in water forms pink colored complex $$\\underline{\\mathrm{X}}$$ which has\noctahedral geometry. This solution on treating with conc $$\\mathrm{HCl}$$ forms deep blue complex, $$\\underline{\\mathrm{Y}}$$ which has a $$\\underline{\\mathrm{Z}}$$ geometry. $$\\mathrm{X}, \\mathrm{Y}$$ and $$\\mathrm{Z}$$, respectively, are

", "options": [ { "text": "$$\\mathrm{X}=\\left[\\mathrm{Co}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}\\right]^{2+}, \\mathrm{Y}=\\left[\\mathrm{CoCl}_{4}\\right]^{2-}, \\mathrm{Z}=$$ Tetrahedral" }, { "text": "$$\\mathrm{X}=\\left[\\mathrm{Co}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}\\right]^{2+}, \\mathrm{Y}=\\left[\\mathrm{CoCl}_{6}\\right]^{3-}, \\mathrm{Z}=$$ Octahedral" }, { "text": "$$\\mathrm{X}=\\left[\\mathrm{Co}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{4} \\mathrm{Cl}_{2}\\right]^{+}, \\mathrm{Y}=\\left[\\mathrm{CoCl}_{4}\\right]^{2-}, \\mathrm{Z}=$$ Tetrahedral" }, { "text": "$$\\mathrm{X}=\\left[\\mathrm{Co}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}\\right]^{3+}, \\mathrm{Y}=\\left[\\mathrm{CoCl}_{6}\\right]^{3-}, \\mathrm{Z}=$$ Octahedral" } ], "answer": "$$\\mathrm{X}=\\left[\\mathrm{Co}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}\\right]^{2+}, \\mathrm{Y}=\\left[\\mathrm{CoCl}_{4}\\right]^{2-}, \\mathrm{Z}=$$ Tetrahedral", "solution": "**Answer:** $$\\mathrm{X}=\\left[\\mathrm{Co}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}\\right]^{2+}, \\mathrm{Y}=\\left[\\mathrm{CoCl}_{4}\\right]^{2-}, \\mathrm{Z}=$$ Tetrahedral\n\n\"JEE", "topic": "Geometry", "subtopic": "Plane Geometry" }, { "id": 1107, "subject": "Chemistry", "question": "

Match List I with List II:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
List I (Complexes) List II (Hybridisation)
A.$\\left[\\mathrm{Ni}(\\mathrm{CO})_{4}\\right]$I.$\\mathrm{sp}^{3}$
B.$\\left[\\mathrm{Cu}\\left(\\mathrm{NH}_{3}\\right)_{4}\\right]^{2+}$II.dsp$^{2}$
C.$\\left[\\mathrm{Fe}\\left(\\mathrm{NH}_{3}\\right)_{6}\\right]^{2+}$III.$\\mathrm{sp}^{3}\\mathrm{d}^{2}$
D.$\\left[\\mathrm{Fe}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}\\right]^{2+}$IV.$\\mathrm{d}^{2} \\mathrm{sp}^{3}$
", "options": [ { "text": "A-I, B-II, C-III, D-IV" }, { "text": "A-II, B-I, C-III, D-IV" }, { "text": "A-I, B-II, C-IV, D-III" }, { "text": "A-II, B-I, C-IV, D-III" } ], "answer": "A-I, B-II, C-IV, D-III", "solution": "**Answer:** A-I, B-II, C-IV, D-III\n\n

$$\\mathrm{Ni{(CO)_4}\\buildrel {} \\over\n \\longrightarrow s{p^3}}$$

\n

$$\\mathrm{{\\left[ {Cu{{(N{H_3})}_4}} \\right]^{2 + }}\\buildrel {} \\over\n \\longrightarrow ds{p^2}}$$

\n

$$\\mathrm{{\\left[ {Fe{{(N{H_3})}_6}} \\right]^{2 + }}\\buildrel {} \\over\n \\longrightarrow {d^2}s{p^3}}$$

\n

$$\\mathrm{{\\left[ {Fe{{({H_2}O)}_6}} \\right]^{2 + }}\\buildrel {} \\over\n \\longrightarrow s{p^3}{d^2}}$$

", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1108, "subject": "Chemistry", "question": "

Total number of moles of AgCl precipitated on addition of excess of AgNO$$_3$$ to one mole each of the following complexes $$\\mathrm{[Co(NH_3)_4Cl_2]Cl,[Ni(H_2O)_6]Cl_2,[Pt(NH_3)_2Cl_2]}$$ and $$\\mathrm{[Pd(NH_3)_4]Cl_2}$$ is ___________.

", "options": [], "answer": "5", "solution": "**Answer:** 5\n\n$\\left[\\mathrm{Co}\\left(\\mathrm{NH}_{3}\\right)_{4} \\mathrm{Cl}_{2}\\right] \\mathrm{Cl} \\stackrel{\\mathrm{AgNO}_{3}}{\\longrightarrow} \\mathrm{AgCl}$\n

\n$$\n\\begin{aligned}\n& {\\left[\\mathrm{Ni}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}\\right] \\mathrm{Cl}_{2} \\stackrel{\\mathrm{AgNO}_{3}}{\\longrightarrow} 2 \\mathrm{AgCl}} \\\\\\\\\n& {\\left[\\mathrm{Pd}\\left(\\mathrm{NH}_{3}\\right)_{4}\\right] \\mathrm{Cl}_{2} \\stackrel{\\mathrm{AgNO}_{3}}{\\longrightarrow} 2 \\mathrm{AgCl}}\n\\end{aligned}\n$$\n

\nTotal moles of $\\mathrm{AgCl}$ precipitated $=5$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1109, "subject": "Chemistry", "question": "

The number of paramagnetic species from the following is _____________.

\n

$$\\mathrm{{[Ni{(CN)_4}]^{2 - }},[Ni{(CO)_4}],{[NiC{l_4}]^{2 - }}}$$

\n

$$\\mathrm{{[Fe{(CN)_6}]^{4 - }},{[Cu{(N{H_3})_4}]^{2 + }}}$$

\n

$$\\mathrm{{[Fe{(CN)_6}]^{3 - }}\\,and\\,{[Fe{({H_2}O)_6}]^{2 + }}}$$

", "options": [], "answer": "4", "solution": "**Answer:** 4\n\n$$\n\\begin{array}{ll}\n\\text { Species } & \\text { Magnetic property } \\\\\\\\\n{\\left[\\mathrm{Ni}(\\mathrm{CN})_4\\right]^{2-}} & \\text { Diamagnetic } \\\\\\\\\n{\\left[\\mathrm{Ni}(\\mathrm{CO})_4\\right]} & \\text { Diamagnetic } \\\\\\\\\n\\left.[\\mathrm{NiCl}]_4\\right]^{2-} & \\text { Paramagnetic } \\\\\\\\\n\\left.[\\mathrm{FeCN})_6\\right]^{4-} & \\text { Diamagnetic } \\\\\\\\\n{\\left[\\mathrm{Fe}(\\mathrm{CN})_6\\right]^{3-}} & \\text { Paramagnetic } \\\\\\\\\n\\left.\\mathrm{Fe}\\left(\\mathrm{H}_2 \\mathrm{O}\\right)_6\\right]^{2+} & \\text { Paramagnetic } \\\\\\\\\n{\\left[\\mathrm{Cu}\\left(\\mathrm{NH}_3\\right)_4\\right]^{2+}} & \\text { Paramagnetic }\n\\end{array}\n$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1110, "subject": "Chemistry", "question": "

The hybridization and magnetic behaviour of cobalt ion in $$\\mathrm{[Co(NH_3)_6]^{3+}}$$ complex, respectively is :

", "options": [ { "text": "$$\\mathrm{sp^3d^2}$$ and paramagnetic" }, { "text": "$$\\mathrm{sp^3d^2}$$ and diamagnetic" }, { "text": "$$\\mathrm{d^2sp^3}$$ and diamagnetic" }, { "text": "$$\\mathrm{d^2sp^3}$$ and paramagnetic" } ], "answer": "$$\\mathrm{d^2sp^3}$$ and diamagnetic", "solution": "**Answer:** $$\\mathrm{d^2sp^3}$$ and diamagnetic\n\n$\\left[\\mathrm{Co}\\left(\\mathrm{NH}_{3}\\right)_{6}\\right]^{3+}$ is diamagnetic with $\\mathrm{d}^{2} \\mathrm{sp}^{3}$ hybridisation of $\\mathrm{Co}^{+3}$.\n

\nThis is because $\\mathrm{NH}_{3}$ is a strong field ligand and forces electrons to pair up in a $\\mathrm{d}^{6}$ configuration.\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1111, "subject": "Chemistry", "question": "

Which of the following complexes will exhibit maximum attraction to an applied magnetic field?

", "options": [ { "text": "$$\\left[\\mathrm{Ni}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}\\right]^{2+}$$" }, { "text": "$$\\left[\\mathrm{Co}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}\\right]^{2+}$$" }, { "text": "$$\\left[\\mathrm{Co}(\\mathrm{en})_{3}\\right]^{3+}$$" }, { "text": "$$\\left[\\mathrm{Zn}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}\\right]^{2+}$$" } ], "answer": "$$\\left[\\mathrm{Co}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}\\right]^{2+}$$", "solution": "**Answer:** $$\\left[\\mathrm{Co}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}\\right]^{2+}$$\n\nTo determine which complex exhibits maximum attraction to an applied magnetic field, we need to look at the number of unpaired electrons in each complex. The more unpaired electrons a complex has, the greater the attraction to the magnetic field.\n

\n1. Option A: [Ni(H₂O)₆]²⁺\nNi²⁺ has a 3d⁸ electron configuration. In an octahedral crystal field, two of the electrons will occupy the lower energy t₂g orbitals, while the remaining six electrons will fill the higher energy e_g orbitals. There are 2 unpaired electrons in this complex.\n

\n2. Option B: [Co(H₂O)₆]²⁺\nCo²⁺ has a 3d⁷ electron configuration. In an octahedral crystal field, two of the electrons will occupy the lower energy t₂g orbitals, while the remaining five electrons will fill the higher energy e_g orbitals. There are 3 unpaired electrons in this complex.\n

\n3. Option C: [Co(en)₃]³⁺\nCo³⁺ has a 3d⁶ electron configuration. In an octahedral crystal field, all six electrons will occupy the lower energy t₂g orbitals. There are no unpaired electrons in this complex.\n

\n4. Option D: [Zn(H₂O)₆]²⁺\nZn²⁺ has a 3d¹⁰ electron configuration, with all orbitals being completely filled. There are no unpaired electrons in this complex.\n

\nBased on the number of unpaired electrons, the complex with the maximum attraction to an applied magnetic field is [Co(H₂O)₆]²⁺ (Option B), as it has 3 unpaired electrons.", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 1112, "subject": "Chemistry", "question": "

The magnetic moment is measured in Bohr Magneton (BM).

\n

Spin only magnetic moment of $$\\mathrm{Fe}$$ in $$\\left[\\mathrm{Fe}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}\\right]^{3+}$$ and $$\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}\\right]^{3-}$$ complexes respectively is :

", "options": [ { "text": "3.87 B.M. and 1.732 B.M." }, { "text": "5.92 B.M. and 1.732 B.M." }, { "text": "6.92 B.M. in both" }, { "text": "4.89 B.M. and 6.92 B.M." } ], "answer": "5.92 B.M. and 1.732 B.M.", "solution": "**Answer:** 5.92 B.M. and 1.732 B.M.\n\n

For spin-only magnetic moment, we use the formula:

\n$$\\mathrm{Magnetic ~moment} = \\sqrt{n(n+2)} \\cdot \\mathrm{BM}$$

\nwhere $n$ is the total number of unpaired electrons.

\n

For $$\\left[\\mathrm{Fe}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}\\right]^{3+}$$, the electronic configuration of $\\mathrm{Fe^{3+}}$ is $\\mathrm{d^5}$. Here, all the five electrons will be unpaired due to the high spin nature of Fe$^{3+}$. Hence, $n=5$ and magnetic moment

$$= \\sqrt{5(5+2)} \\cdot \\mathrm{BM} \\approx 5.92 \\mathrm{BM}$$

\n

For $$\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}\\right]^{3-}$$, the $\\mathrm{Fe^{3+}}$ ion has electronic configuration $\\mathrm{d^5}$. In this case, the strong field ligand cyanide ($\\mathrm{CN}^{-}$) will pair up the electrons in the $\\mathrm{d}$ orbital. So, there will be only one unpaired electron. Hence, $n=1$ and magnetic moment

$$= \\sqrt{1(1+2)} \\cdot \\mathrm{BM} \\approx 1.732 \\mathrm{BM}$$

\n

Therefore, the answer is 5.92 B.M. and 1.732 B.M.

\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1113, "subject": "Chemistry", "question": "

Identify from the following species in which $$\\mathrm{d}^2 \\mathrm{sp}^3$$ hybridization is shown by central atom :

", "options": [ { "text": "$$\\left[\\mathrm{Co}\\left(\\mathrm{NH}_3\\right)_6\\right]^{3+}$$\n" }, { "text": "$$\\mathrm{SF}_6$$\n" }, { "text": "$$\\left[\\mathrm{Pt}\\left(\\mathrm{Cl}_4\\right)\\right]^{2-}$$\n" }, { "text": "$$\\mathrm{BrF}_5$$" } ], "answer": "$$\\left[\\mathrm{Co}\\left(\\mathrm{NH}_3\\right)_6\\right]^{3+}$$\n", "solution": "**Answer:** $$\\left[\\mathrm{Co}\\left(\\mathrm{NH}_3\\right)_6\\right]^{3+}$$\n\n\n

$$\\left[\\mathrm{Co}\\left(\\mathrm{NH}_3\\right)_6\\right]^{+3}-\\mathrm{d}^2 \\mathrm{sp}^3$$ hybridization

\n

$$\\mathrm{BrF}_5-\\mathrm{sp}^3 \\mathrm{d}^2$$ hybridization

\n

$$\\left[\\mathrm{PtCl}_4\\right]^{-2}-\\mathrm{dsp}^2$$ hybridization

\n

$$\\mathrm{SF}_6-\\mathrm{sp}^3 \\mathrm{d}^2$$ hybridization

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1114, "subject": "Chemistry", "question": "

The correct statements from following are:

\n

A. The strength of anionic ligands can be explained by crystal field theory.

\n

B. Valence bond theory does not give a quantitative interpretation of kinetic stability of coordination compounds.

\n

C. The hybridization involved in formation of $$\\left[\\mathrm{Ni}(\\mathrm{CN})_4\\right]^{2-}$$ complex is $$\\mathrm{dsp}^2$$.

\n

D. The number of possible isomer(s) of cis- $$\\left[\\mathrm{PtCl}_2(\\mathrm{en})_2\\right]^{2+}$$ is one

\n

Choose the correct answer from the options given below:

", "options": [ { "text": "B, C only" }, { "text": "B, D only" }, { "text": "A, C only" }, { "text": "A, D only" } ], "answer": "B, C only", "solution": "**Answer:** B, C only\n\n

(A) Strength of anionic ligands cannot be explained by CFT instead LFT i.e., ligand field theory explains the strength of ligands.

\n\n

(B) VBT does not give a quantitative interpretation of kinetic stability of coordination compounds.

\n\n

(C) $\\left[\\mathrm{Ni}(\\mathrm{CN})_4\\right]^{2-} \\rightarrow \\mathrm{Ni}^{2+}$ in presence of $\\mathrm{CN}^{-}$ligand.

\n\n
\"JEE\n\n

Square planar geometry since $\\rightarrow 4, d s p^2$ hybridised orbitals created.

\n\n

(D) cis- $\\left[\\mathrm{PtCl}_2(\\mathrm{en})_2\\right] \\Rightarrow$ It has two possible isomers

\n
\"JEE", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 1115, "subject": "Chemistry", "question": "

Match List I with List II

\n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
LIST ILIST II
A.$$\\mathrm{K}_2\\left[\\mathrm{Ni}(\\mathrm{CN})_4\\right]$$I.$$sp^3$$
B.\t$$\\left[\\mathrm{Ni}(\\mathrm{CO})_4\\right]$$II.$$sp^3d^2$$
C.$$\\left[\\mathrm{Co}\\left(\\mathrm{NH}_3\\right)_6\\right] \\mathrm{Cl}_3$$III.$$dsp^2$$
D.$$\\mathrm{Na}_3\\left[\\mathrm{CoF}_6\\right]$$IV.$$d^2sp^3$$

\n

Choose the correct answer from the options given below:

", "options": [ { "text": "A-III, B-II, C-IV, D-I\n" }, { "text": "A-III, B-I, C-II, D-IV\n" }, { "text": "A-I, B-III, C-II, D-IV\n" }, { "text": "A-III, B-I, C-IV, D-II" } ], "answer": "A-III, B-I, C-IV, D-II", "solution": "**Answer:** A-III, B-I, C-IV, D-II\n\n

$$\\begin{aligned}\n& \\stackrel{+2}{\\text { (A) } \\mathrm{K}_2\\left[\\mathrm{Ni}(\\mathrm{CN})_4\\right]} \\\\\n& \\mathrm{Ni}^{2+}:[\\mathrm{Ar}] 3 \\mathrm{~d}^8 4 \\mathrm{~s}^{\\circ},\\left(\\mathrm{CN}^{-} \\text {is } \\mathrm{S} . \\mathrm{F} . \\mathrm{L}\\right) \\\\\n& \\text { Pre hybridization state of } \\mathrm{Ni}^{+2}\n\\end{aligned}$$

\n

\"JEE

\n

(B) $$[\\mathrm{Ni}(\\mathrm{CO})_4]$$

\n

$$\\mathrm{Ni}:[\\mathrm{Ar}] 3 \\mathrm{~d}^8 4 \\mathrm{~s}^2$$

\n

$$\\mathrm{CO}$$ is S.F.L , so pairing occur

\n

Pre hybridization state of $$\\mathrm{Ni}$$

\n

\"JEE

\n

(C) $$[\\mathrm{Co}(\\mathrm{NH}_3)_6] \\mathrm{Cl}_3$$

\n

$$\\mathrm{Co}^{+3}:[\\mathrm{Ar}] 3 \\mathrm{~d}^6 4 \\mathrm{~s}^0$$

\n

With $$\\mathrm{Co}^{3+}, \\mathrm{NH}_3$$ act as S.F.L

\n

\"JEE

\n

(D) $$\\mathrm{Na}_3[\\mathrm{CoF}_6]$$

\n

$$\\mathrm{Co}^{3+}:[\\mathrm{Ar}] 3 \\mathrm{~d}^6\\left(\\mathrm{~F}^{\\ominus}: \\text { W.F.L }\\right)$$

\n

\"JEE

", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1116, "subject": "Chemistry", "question": "

The coordination environment of $$\\mathrm{Ca}^{2+}$$ ion in its complex with $$\\mathrm{EDTA}^{4-}$$ is :

", "options": [ { "text": "trigonal prismatic\n" }, { "text": "octahedral\n" }, { "text": "square planar\n" }, { "text": "tetrahedral" } ], "answer": "octahedral\n", "solution": "**Answer:** octahedral\n\n\n

The coordination environment of the $$\\mathrm{Ca}^{2+}$$ ion when it forms a complex with $$\\mathrm{EDTA}^{4-}$$ (ethylenediaminetetraacetic acid) is octahedral. Ethylenediaminetetraacetic acid (EDTA) is a hexadentate ligand, which means it has six donor atoms that can bind to a central metal ion. In the case of EDTA, these donor atoms are four oxygen atoms from its four carboxyl groups and two nitrogen atoms from its two amine groups.

\n\n

When $$\\mathrm{Ca}^{2+}$$ forms a complex with $$\\mathrm{EDTA}^{4-}$$, all six donor atoms from EDTA coordinate with the calcium ion. As a result, the coordination number of the calcium ion is 6, leading to an octahedral geometry. This is because the octahedral geometry is the most common and energetically favorable arrangement for a coordination number of 6. In such a geometry, the six ligands are placed at equal distances from the central ion and at 90° angles relative to adjacent ligands, maximizing the distance between all ligands to minimize repulsion.

\n\n

The correct answer is Option B, octahedral.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1117, "subject": "Chemistry", "question": "

Consider the following test for a group-IV cation.

\n

$$\\mathrm{M}^{2+}+\\mathrm{H}_2 \\mathrm{S} \\rightarrow \\mathrm{A} \\text { (Black precipitate)+ byproduct }$$

\n

$$\\mathrm{A}+\\text { aqua regia } \\rightarrow \\mathrm{B}+\\mathrm{NOCl}+\\mathrm{S}+\\mathrm{H}_2 \\mathrm{O}$$

\n

$$\\mathrm{B}+\\mathrm{KNO}_2+\\mathrm{CH}_3 \\mathrm{COOH} \\rightarrow \\mathrm{C}+\\text { byproduct }$$

\n

The spin-only magnetic moment value of the metal complex $$\\mathrm{C}$$ is _________ $$\\mathrm{BM}$$ (Nearest integer)

", "options": [], "answer": "0", "solution": "**Answer:** 0\n\n

\"JEE

\n

$$\\begin{aligned}\n& \\text { In } \\mathrm{K}_3\\left[\\mathrm{Co}\\left(\\mathrm{NO}_2\\right)_6\\right], \\mathrm{Co}^{+3}: 3 \\mathrm{~d}^6 4 \\mathrm{~s}^0 \\\\\n& \\mathrm{Co}^{3+}: \\mathrm{d}^2 \\mathrm{sp}^3 \\text { Hybridisation } \\\\\n& \\text { Number of unpaired } \\mathrm{e}^{-}=0 \\\\\n& \\text { Magnetic moment }=\\sqrt{n(n+2)}=0 \\text { B.M } \\\\\n&\n\\end{aligned}$$

", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 1118, "subject": "Chemistry", "question": "

Given below are two statements :

\n

Statement I: $$\\mathrm{PF}_5$$ and $$\\mathrm{BrF}_5$$ both exhibit $$\\mathrm{sp}^3 \\mathrm{~d}$$ hybridisation.

\n

Statement II: Both $$\\mathrm{SF}_6$$ and $$[\\mathrm{Co}(\\mathrm{NH}_3)_6]^{3+}$$ exhibit $$\\mathrm{sp}^3 \\mathrm{~d}^2$$ hybridisation.

\n

In the light of the above statements, choose the correct answer from the options given below :

", "options": [ { "text": "Both Statement I and Statement II are false\n" }, { "text": "Statement I is true but Statement II is false\n" }, { "text": "Statement I is false but Statement II is true\n" }, { "text": "Both Statement I and Statement II are true" } ], "answer": "Both Statement I and Statement II are false\n", "solution": "**Answer:** Both Statement I and Statement II are false\n\n\n

Hybridisation of $$\\mathrm{P}$$ in $$\\mathrm{PF}_5$$ is $$s p^3 d$$ but the hybridisation of $$\\mathrm{Br}$$ in $$\\mathrm{BrF}_5$$ is $$s p^3 d^2$$. So, statement-I is false.

\n

Hybridisation of $$\\mathrm{S}$$ in $$\\mathrm{SF}_6$$ is $$s p^3 d^2$$ but the hybridisation of $$\\mathrm{Co}^{3+}$$ in $$[\\mathrm{Co}(\\mathrm{NH}_3)_6]^{3+}$$ is $$d^2 s p^3$$ as $$\\mathrm{NH}_3$$ is a strong field ligand forcing the unpaired electrons to pair up. So, statement-II is false.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1119, "subject": "Chemistry", "question": "

Match List I with List II

\n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
LIST I
(Hybridization)		LIST II
(Orientation in Shape)
A.sp$$^3$$I.Trigonal bipyramidal
B.\tdsp$$^2$$II.Octahedral
C.sp$$^3$$dIII.Tetrahedral
D.sp$$^3$$d$$^2$$IV.Square planar

\n

Choose the correct answer from the options given below:

", "options": [ { "text": "A-III, B-I, C-IV, D-II\n" }, { "text": "A-III, B-IV, C-I, D-II\n" }, { "text": "A-IV, B-III, C-I, D-II\n" }, { "text": "A-II, B-I, C-IV, D-III" } ], "answer": "A-III, B-IV, C-I, D-II\n", "solution": "**Answer:** A-III, B-IV, C-I, D-II\n\n\n

A $$\\rightarrow$$ Tetrahedral (III)

\n

B $$\\rightarrow$$ Square planar (IV)

\n

C $$\\rightarrow$$ Trigonal bipyramidal (I)

\n

D $$\\rightarrow$$ Octahedral (II)

", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1120, "subject": "Chemistry", "question": "When KMnO4 acts as an oxidising agent and ultimately forms [MnO4]-2, MnO2, Mn2O3, Mn+2\nthen the number of electrons transferred in each case respectively is :", "options": [ { "text": "4, 3, 1, 5" }, { "text": "1, 5, 3, 7" }, { "text": "1, 3, 4, 5" }, { "text": "3, 5, 7, 1" } ], "answer": "1, 3, 4, 5", "solution": "**Answer:** 1, 3, 4, 5\n\n\"AIEEE ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1121, "subject": "Chemistry", "question": "How do we differentiate between Fe3+ and Cr3+ in group III?", "options": [ { "text": "by taking excess of NH4OH solution" }, { "text": "by increasing $$NH_4^+$$ ion concentration" }, { "text": "by decreasing OH- ion concentration" }, { "text": "both (b) and (c)" } ], "answer": "by increasing $$NH_4^+$$ ion concentration", "solution": "**Answer:** by increasing $$NH_4^+$$ ion concentration\n\nFeCl3 + 3NH4OH $$ \\to $$ Fe(OH)3 ↓(reddish brown) + 3NH4Cl\n

CrCl3 + 3NH4OH $$ \\to $$ Cr(OH)3 (Bluish green) + 3NH4Cl", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 1122, "subject": "Chemistry", "question": "What would happen when a solution of potassium chromate is treated with an excess of dilute nitric acid?", "options": [ { "text": "$$Cr_2O^{2-}_7$$ and H2O are formed" }, { "text": "$$CrO^{2-}_4$$ is reduced to +3 state of Cr" }, { "text": "$$CrO^{2-}_4$$ is oxidized to +7 state of Cr" }, { "text": "Cr3+ and $$Cr_2O^{2-}_7$$ are formed" } ], "answer": "$$Cr_2O^{2-}_7$$ and H2O are formed", "solution": "**Answer:** $$Cr_2O^{2-}_7$$ and H2O are formed\n\nWhen a solution of potassium chromate is treated with an excess of dilute nitric acid. Potassium dichromate and $${H_2}O$$ are formed.\n

$$2{K_2}Cr{O_4} + 2HN{O_3} \\to {K_2}C{r_2}{O_7} + 2KN{O_3} + {H_2}O$$ \n

Hence $$C{r_2}{O_7}^ - $$ and $${H_2}O$$ are formed. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1123, "subject": "Chemistry", "question": "Which one of the following nitrates will leave behind a metal on strong heating?", "options": [ { "text": "Copper nitrate" }, { "text": "Manganese nitrate" }, { "text": "Silver nitrate" }, { "text": "Ferric nitrate" } ], "answer": "Silver nitrate", "solution": "**Answer:** Silver nitrate\n\n$$AgN{O_3}$$ on heating till red hot decomposes as follows: \n

$$AgN{O_3} \\to Ag + N{O_2} + {1 \\over 2}{O_2}$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1124, "subject": "Chemistry", "question": "Excess of KI reacts with CuSO4 solution and then Na2S2O3 solution is added to it. Which of the statements is incorrect for this reaction?", "options": [ { "text": "Cu2I2 is reduced" }, { "text": "Evolved I2 is reduced " }, { "text": "Na2S2O3 is oxidized " }, { "text": "CuI2 is formed" } ], "answer": "CuI2 is formed", "solution": "**Answer:** CuI2 is formed\n\n$$4\\mathop {KI}\\limits^{ - 1} + 2CuS{O_4} \\to \\mathop {{I_2}}\\limits^0 + C{u_2}{I_2} + 2{K_2}S{O_4}$$\n

$$\\mathop {{I_2}}\\limits^0 + 2N{a_2}\\mathop {{S_2}}\\limits^{2 + } {O_3} \\to N{a_2}\\mathop {{S_4}}\\limits^{ + 2.5} {O_6} + \\mathop {2NaI}\\limits^{ - 1} $$\n

In this $$Cu{I_2}$$ is not formed", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1125, "subject": "Chemistry", "question": "The oxidation state of chromium in the final product formed the reaction between $$K{\\rm I}$$ and acidified potassium dichromate solution is :", "options": [ { "text": "$$+3$$ " }, { "text": "$$+2$$ " }, { "text": "$$+6$$ " }, { "text": "$$+4$$ " } ], "answer": "$$+3$$ ", "solution": "**Answer:** $$+3$$ \n\n$$C{r_2}O_7^{2 - } + 6{{\\rm I}^ - } + 14{H^ + }\\buildrel \\, \\over\n \\longrightarrow 3{{\\rm I}_2} + 7{H_2}O + 2C{r^{3 + }}$$ \n

oxidation state of $$Cr$$ is $$3+.$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1126, "subject": "Chemistry", "question": "Amount of oxalic acid present in a solution can be determined by its titration with $$KMn{O_4}$$ solution in the presence of $${H_2}S{O_4}.$$ The titration gives unsatisfactory result when carried out in the presence of $$HCl,$$ because $$HCl$$ :", "options": [ { "text": "gets oxidised by oxalic acid to chlorine" }, { "text": "furnishes $${H^ + }$$ ions in addition to those from oxalic acid " }, { "text": "reduces permanganate to $$M{n^{2 + }}$$ " }, { "text": "oxaidises oxalic acid to carbon doxide and water " } ], "answer": "reduces permanganate to $$M{n^{2 + }}$$ ", "solution": "**Answer:** reduces permanganate to $$M{n^{2 + }}$$ \n\nThe titration of oxalic acid with $$KMnO{}_4$$ in presence of $$HCl$$ gives unsatisfactory result because of the fact that $$KMn{O_4}$$ can also oxidise $$HCl$$ along with oxalic acid. $$HCl$$ on oxidation gives $$C{l_2}$$ and $$HCl$$ reduces $$KMnO{}_4$$ to $$M{n^{2 + }}$$ thus the correct answer is $$(c).$$ ", "topic": "Number Theory", "subtopic": "Divisibility / Factorization" }, { "id": 1127, "subject": "Chemistry", "question": "The colour of KMnO4 is due to :", "options": [ { "text": "d – d transition" }, { "text": "L $$\\to$$ M charge transfer transition" }, { "text": "$$\\sigma$$ – $$\\sigma$$* transition" }, { "text": "M $$\\to$$ L charge transfer transition" } ], "answer": "L $$\\to$$ M charge transfer transition", "solution": "**Answer:** L $$\\to$$ M charge transfer transition\n\n$$L \\to M$$ charge transfer spectra. $$KMn{O_4}$$ is colored because it absorbs light in the visible range of electromagnetic radiation. The permanganate ion is the source of color, as a ligand to metal, $$\\left( {L \\to M} \\right)$$ charge transfer takes place between oxygen's $$p$$ orbitals and the empty $$d$$-orbitals on the metal. This charge transfer takes place when a photon of light is absorbed, which leads to the purple color of the compound. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1128, "subject": "Chemistry", "question": "Wilkinson catalyst is : (Et = C2H5)", "options": [ { "text": "[(Ph3P)3 RhCl]" }, { "text": "[(Et3P)3RhCl] \n" }, { "text": "[(Et3P)3IrCl] " }, { "text": "[(Ph3P)3IrCl]" } ], "answer": "[(Ph3P)3 RhCl]", "solution": "**Answer:** [(Ph3P)3 RhCl]\n\nWilkinsion catalyst is [(Ph3P)3RhCl]", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1129, "subject": "Chemistry", "question": "$$\\underline A \\,\\,\\buildrel {4KOH,{O_2}} \\over\n \\longrightarrow \\,\\,\\mathop {2\\underline B }\\limits_{\\left( {Green} \\right)} \\,\\, + \\,\\,2{H_2}O$$\n

$$3\\underline B \\,\\,\\buildrel {4HCl} \\over\n \\longrightarrow \\,\\,\\mathop {2\\underline C }\\limits_{\\left( {Purple} \\right)} \\,\\, + \\,\\,Mn{O_2} + 2{H_2}O$$\n

$$2\\underline C \\,\\,\\buildrel {{H_2}O.KI} \\over\n \\longrightarrow \\,\\,\\mathop {2\\underline A }\\limits_{\\left( {Purple} \\right)} \\,\\, + \\,\\,2KOH\\,\\, + \\,\\,\\underline D $$\n

In the above sequence of reactions, $${\\underline A }$$ and $${\\underline D }$$, respectively, are : ", "options": [ { "text": "Kl and KMnO4" }, { "text": "Kl and K2MnO4" }, { "text": "KlO3 and MnO2" }, { "text": "MnO2 and KlO3" } ], "answer": "MnO2 and KlO3", "solution": "**Answer:** MnO2 and KlO3\n\n$$Mn{O_2}(A) \\,\\,\\buildrel {4KOH,{O_2}} \\over\n \\longrightarrow \\,\\,\\mathop {2K_2MnO_4(B) }\\limits_{\\left( {Green} \\right)} \\,\\, + \\,\\,2{H_2}O$$\n

$$3K_2MnO_4(B) \\,\\,\\buildrel {4HCl} \\over\n \\longrightarrow \\,\\,\\mathop {2K_2MnO_4(C) }\\limits_{\\left( {Purple} \\right)} \\,\\, + \\,\\,Mn{O_2} + 2{H_2}O$$\n

$$2K_2MnO_4(C) \\,\\,\\buildrel {{H_2}O.KI} \\over\n \\longrightarrow \\,\\,\\mathop {2 Mn{O_2}(A) }\\limits_{\\left( {Purple} \\right)} \\,\\, + \\,\\,2KOH\\,\\, + \\,\\,KIO_3(D) $$\n

$$ \\therefore $$ A $$ \\to $$ MnO2\n

D $$ \\to $$ KIO3", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1130, "subject": "Chemistry", "question": "Consider the following reactions :

\n$$NaCl{\\rm{ }} + {\\rm{ }}{K_2}C{r_2}{O_7} + \\mathop {{H_2}S{O_4}}\\limits_{(conc.)} $$$$ \\to $$ (A) + side products\n

\n(A) + NaOH $$ \\to $$ (B) + Side products

\n$$\\left( B \\right){\\rm{ }} + \\mathop {{H_2}S{O_4}}\\limits_{(dilute)} + {\\rm{ }}{H_2}{O_2}$$ $$ \\to $$ (C) + Side products

\nThe sum of the total number of atoms in one molecule each of (A) and (B) and (C) is\n\n", "options": [], "answer": "18", "solution": "**Answer:** 18\n\n$$NaCl{\\rm{ }} + {\\rm{ }}{K_2}C{r_2}{O_7} + \\mathop {{H_2}S{O_4}}\\limits_{(conc.)} $$$$ \\to $$
2CrO2Cl2(A) + 4NaHSO4 + 2KHSO4 + 3H2O\n

CrO2Cl2(A) + 4NaOH $$ \\to $$ Na2CrO4(B) + 2NaCl + 2H2O\n

Na2CrO4(B) + 2H2SO4 + 2H2O2 $$ \\to $$\n
CrO5(C) + 2NaHSO4 + 3H2O\n

A = CrO2Cl2\n

B = Na2CrO4\n

C = CrO5\n

Total number of atom in A + B + C = 18", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1131, "subject": "Chemistry", "question": "5 g of zinc is treated separately with an excess\nof

\n(a) dilute hydrochloric acid and
\n(b) aqueous sodium hydroxide.

\nThe ratio of the volumes of H2 evolved in these\ntwo reactions is :", "options": [ { "text": "1 : 2" }, { "text": "1 : 1" }, { "text": "1 : 4" }, { "text": "2 : 1" } ], "answer": "1 : 1", "solution": "**Answer:** 1 : 1\n\nZn + 2dil. HCl $$ \\to $$ ZnCl2 + H2\n

Zn + 2NaOH $$ \\to $$ Na2ZnO2 + H2\n

From one mole of Zn, 1 mol of H2 is produced\nby both NaOH and HCl.\n

The ratio of the volume of H2 is 1 : 1", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1132, "subject": "Chemistry", "question": "The sum of the total number of bonds between\nchromium and oxygen atoms in chromate and\ndichromate ions is ____________.", "options": [], "answer": "18", "solution": "**Answer:** 18\n\n\"JEE\n
Note : Here if we consider only $$\\sigma $$ bonds so the answer\nwould be 12 but there are 6$$\\pi $$ bonds also, if we consider them then\nthe total number of Cr – O bonds will be 18.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1133, "subject": "Chemistry", "question": "The incorrect statement is :", "options": [ { "text": "In manganate and permanganate ions, the p-bonding takes place by overlap of p-orbitals\nof oxygen and d-orbitals of manganese" }, { "text": "Manganate ion is green in colour and\npermanganate ion in purple in colour" }, { "text": "Manganate and permanganate ions are\nparamagnetic" }, { "text": "Manganate and permanganate ions are\ntetrahedral" } ], "answer": "Manganate and permanganate ions are\nparamagnetic", "solution": "**Answer:** Manganate and permanganate ions are\nparamagnetic\n\n

To identify the incorrect statement, let's analyze each option :

\n

Option A : "In manganate and permanganate ions, the π-bonding takes place by overlap of p-orbitals of oxygen and d-orbitals of manganese." \nThis statement is true. In both manganate ${MnO}_4^{2-} $ and permanganate ${MnO}_4^- $ ions, the manganese atom is in a high oxidation state (+6 and +7, respectively) and forms bonds with oxygen atoms involving the overlap of p-orbitals of oxygen and d-orbitals of manganese.

\n

Option B : "Manganate ion is green in colour and permanganate ion in purple in colour."\nThis statement is true. Manganate ions are indeed green and permanganate ions are purple in aqueous solution.

\n

Option C : "Manganate and permanganate ions are paramagnetic."\nThis statement is incorrect. Both manganate and permanganate ions are diamagnetic, not paramagnetic. In these ions, manganese is in +6 and +7 oxidation states, respectively, and has no unpaired electrons.

\nManganate($$\\mathop {Mn}\\limits^{ + 6} $$O4\n2–)\n,\n

Mn(25) = [Ar]3d64s2\n

Mn+6(25) = [Ar]3d1 $$ \\to $$ Paramagnetic\n

Permanganate ($$\\mathop {Mn}\\limits^{ + 7} $$O4-)\n

Mn(25) = [Ar]3d64s2\n

Mn+7(25) = [Ar]3d0 $$ \\to $$ Diamagnetic\n

Option D : "Manganate and permanganate ions are tetrahedral."\nThis statement is true. Both manganate and permanganate ions have a tetrahedral geometry, with the manganese atom at the center and the four oxygen atoms surrounding it.

\n

Therefore, the incorrect statement is option C.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1134, "subject": "Chemistry", "question": "The incorrect statement among the following is :", "options": [ { "text": "RuO4 is an oxidizing agent" }, { "text": "Cr2O3 is an amphoteric oxide." }, { "text": "VOSO4 is a reducing agent" }, { "text": "Red colour of ruby is due to the presence of Co3+" } ], "answer": "Red colour of ruby is due to the presence of Co3+", "solution": "**Answer:** Red colour of ruby is due to the presence of Co3+\n\nRed colour of ruby is due to presence of ${Cr}^{3+} $ (chromium ions), not ${Co}^{3+} $ (cobalt ions). in Al2O3.\nChromium is the trace element that causes ruby’s red colour, which\nranges from an orange red to a publish red. The strength of ruby’s\nred depends on how much chromium is present.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1135, "subject": "Chemistry", "question": "In mildly alkaline medium, thiosulphate ion is oxidized by $$MnO_4^ - $$ to \"A\". The oxidation state of sulphur in \"A\" is __________.", "options": [], "answer": "6", "solution": "**Answer:** 6\n\n\"JEE\n

Oxidation number of sulphur in SO42- (A) is + 6.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1136, "subject": "Chemistry", "question": "Fex2 and Fey3 are known when x and y are :", "options": [ { "text": "x = F, Cl, Br, I and y = F, Cl, Br, I" }, { "text": "x = F, Cl, Br and y = F, Cl, Br, I" }, { "text": "x = F, Cl, Br, I and y = F, Cl, Br" }, { "text": "x = Cl, Br, I and y = F, Cl, Br, I" } ], "answer": "x = F, Cl, Br, I and y = F, Cl, Br", "solution": "**Answer:** x = F, Cl, Br, I and y = F, Cl, Br\n\nFeI3 does not exist as I– reduces Fe3+ to Fe2+.

\n2Fe3+ + 2I\n– $$ \\to $$ 2Fe2+ + I2. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1137, "subject": "Chemistry", "question": "Given below are two statements :

Statement I : Potassium permanganate on heating at 573K forms potassium manganate.

Statement II : Both potassium permanganate and potassium manganate are tetrahedral and paramagnetic in nature.

In the light of the above statements, choose the most appropriate answer from the options given below :", "options": [ { "text": "Statement I is true but statement II is false" }, { "text": "Statement I is false but statement II is true" }, { "text": "Both statement I and statement II are false" }, { "text": "Both statement I and statement II are true" } ], "answer": "Statement I is true but statement II is false", "solution": "**Answer:** Statement I is true but statement II is false\n\nStatement-1 : KMnO4 $$\\buildrel {573K} \\over\n \\longrightarrow $$ K2MnO4 + MnO2 + O2

\nStatement-2 :
KMnO4 $$ \\Rightarrow $$ Mn7+ $$ \\Rightarrow $$ Diamagnetic.
\n KMnO4 $$ \\Rightarrow $$ Mn6+ $$ \\Rightarrow $$ Paramagnetic.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1138, "subject": "Chemistry", "question": "Match List - I with List - II

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
List - I
List - II
(a)Chlorophyll(i)Ruthenium
(b)Vitamin-$${B_{12}}$$(ii)Platinum
(c)Anticancer drug(iii)Cobalt
(d)Grubbs catalyst(iv)Magnesium


Choose the most appropriate answer from the options given below :", "options": [ { "text": "(a) - (iii), (b) - (ii), (c) - (iv), (d) - (i)" }, { "text": "(a) - (iv), (b) - (iii), (c) - (ii), (d) - (i)" }, { "text": "(a) - (iv), (b) - (iii), (c) - (i), (d) - (ii)" }, { "text": "(a) - (iv), (b) - (ii), (c) - (iii), (d) - (i)" } ], "answer": "(a) - (iv), (b) - (iii), (c) - (ii), (d) - (i)", "solution": "**Answer:** (a) - (iv), (b) - (iii), (c) - (ii), (d) - (i)\n\nChlorophyll is a coordination compound of\nmagnesium.

\nVitamin B-12, cyanocobalamine is a\ncoordination compound of cobalt.

\nCisplatin is used as an anti-cancer drug and\nis a coordination compound of platinum.\nGrubbs catalyst is a compound of Ruthenium.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1139, "subject": "Chemistry", "question": "The oxide that shows magnetic property is :", "options": [ { "text": "MgO" }, { "text": "SiO2" }, { "text": "Mn3O4" }, { "text": "Na2O" } ], "answer": "Mn3O4", "solution": "**Answer:** Mn3O4\n\nMn3O4 is paramagnetic due to presence of unpaired electrons. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1140, "subject": "Chemistry", "question": "The correct order of following 3d metal oxides, according to their oxidation numbers is :

(a) CrO3, (b) Fe2O3, (c) MnO2, (d) V2O5, (e) Cu2O", "options": [ { "text": "(d) > (a) > (b) > (c) > (e)" }, { "text": "(a) > (c) > (d) > (b) > (e)" }, { "text": "(a) > (d) > (c) > (b) > (e)" }, { "text": "(c) > (a) > (d) > (e) > (b)" } ], "answer": "(a) > (d) > (c) > (b) > (e)", "solution": "**Answer:** (a) > (d) > (c) > (b) > (e)\n\n(a) $$\\mathop C\\limits^{ + 6} r{O_3}$$ : Since there are three oxygen atoms each with an oxidation number of -2, the total oxidation number from the oxygen atoms is -6. To make the compound neutral, Chromium (Cr) must therefore have an oxidation number of +6.\n\n

(b) $$\\mathop {F{e_2}}\\limits^{ + 3} {O_3}$$ : Here, there are three oxygen atoms, contributing a total oxidation number of -6. To balance this, the total oxidation number for the two iron (Fe) atoms must be +6, meaning each Fe atom has an oxidation number of +3.

(c) $$\\mathop {Mn}\\limits^{ + 4} {O_2}$$ : With two oxygen atoms, the total oxidation number is -4. Thus, the manganese (Mn) atom must have an oxidation number of +4 to balance this.

(d) $$\\mathop {{V_2}}\\limits^{ + 5} {O_5}$$ : In this case, there are five oxygen atoms for a total oxidation number of -10. To balance this, the total oxidation number for the two vanadium (V) atoms must be +10, meaning each V atom has an oxidation number of +5.

(e) $$\\mathop {C{u_2}}\\limits^{ + 1} O$$ : Here, there's one oxygen atom with an oxidation number of -2. To balance this, the total oxidation number for the two copper (Cu) atoms must be +2, meaning each Cu atom has an oxidation number of +1.\n\n

So, the order of the oxidation numbers from highest to lowest is : \n

(a) CrO$_3$ > (d) V$_2$O$_5$ > (c) MnO$_2$ > (b) Fe$_2$O$_3$ > (e) Cu$_2$O.\n\n

Therefore, the answer is Option C : (a) > (d) > (c) > (b) > (e).", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1141, "subject": "Chemistry", "question": "Which one of the following metals forms interstitial hydride easily?", "options": [ { "text": "Cr" }, { "text": "Fe" }, { "text": "Mn" }, { "text": "Co" } ], "answer": "Cr", "solution": "**Answer:** Cr\n\nElements of group 7, 8, 9 do not form hydrides thus Cr will only form hydride among the given elements (Fe, Mn, Co).", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1142, "subject": "Chemistry", "question": "The addition of dilute NaOH to Cr3+ salt solution will give :", "options": [ { "text": "a solution of [Cr(OH)4]$$-$$" }, { "text": "precipitate of Cr2O3(H2O)n" }, { "text": "precipitate of [Cr(OH)6]3$$-$$" }, { "text": "precipitate of Cr(OH)3" } ], "answer": "precipitate of Cr2O3(H2O)n", "solution": "**Answer:** precipitate of Cr2O3(H2O)n\n\n$$C{r^{3 + }} + \\mathop {NaOH}\\limits_{dil.} \\buildrel {} \\over\n \\longrightarrow \\mathop {C{r_2}{O_3}.{{({H_2}O)}_n}}\\limits_{precipitate} $$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1143, "subject": "Chemistry", "question": "Potassium permanganate on heating at 513 K gives a product which is :", "options": [ { "text": "paramagnetic and colourless" }, { "text": "diamagnetic and green" }, { "text": "diamagnetic and colourless" }, { "text": "paramagnetic and green" } ], "answer": "paramagnetic and green", "solution": "**Answer:** paramagnetic and green\n\n$$2KMn{O_4}\\mathrel{\\mathop{\\kern0pt\\longrightarrow}\n\\limits_{200^\\circ C}^\\Delta } \\mathop {{K_2}Mn{O_4}}\\limits_{Green} + \\mathop {Mn{O_2}}\\limits_{Black} + {O_2}$$

In K2MnO4, manganese oxidation state is +6 and hence it has one unpaired e$$-$$.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1144, "subject": "Chemistry", "question": "In the structure of the dichromate ion, there is a :", "options": [ { "text": "linear symmetrical Cr-O-Cr bond." }, { "text": "non-linear symmetrical Cr-O-Cr bond." }, { "text": "linear unsymmetrical Cr-O-Cr bond." }, { "text": "non-linear unsymmetrical Cr-O-Cr bond." } ], "answer": "non-linear symmetrical Cr-O-Cr bond.", "solution": "**Answer:** non-linear symmetrical Cr-O-Cr bond.\n\n\"JEE

Dichromate ion contain non-linear symmetrical Cr-O-Cr bond.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1145, "subject": "Chemistry", "question": "In the given chemical reaction, colors of the Fe2+ and Fe3+ ions, are respectively :

5Fe2+ + MnO$$_4^ - $$ + 8H+ $$\\to$$ Mn2+ + 4H2O + 5Fe3+", "options": [ { "text": "Yellow, Orange" }, { "text": "Yellow, Green" }, { "text": "Green, Orange" }, { "text": "Green, Yellow" } ], "answer": "Green, Yellow", "solution": "**Answer:** Green, Yellow\n\nColour of Fe2+ is observed green and Fe3+ is yellow.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1146, "subject": "Chemistry", "question": "

The number of terminal oxygen atoms present in the product B obtained from the following reaction is _____________.

\n

FeCr2O4 + Na2CO3 + O2 $$\\to$$ A + Fe2O3 + CO2

\n

A + H+ $$\\to$$ B + H2O + Na+

", "options": [], "answer": "6", "solution": "**Answer:** 6\n\n$\\mathrm{FeCr}_{2} \\mathrm{O}_{4}+\\mathrm{Na}_{2} \\mathrm{CO}_{3}+\\mathrm{O}_{2} \\longrightarrow \\mathrm{Fe}_{2} \\mathrm{O}_{3}+\\mathrm{CO}_{2}$ $+\\mathrm{Na}_{2} \\mathrm{CrO}_{4}$\n\n(A)\n

\n$$\n\\mathrm{Na}_{2} \\mathrm{CrO}_{4}+\\mathrm{H}^{+} \\longrightarrow \\mathrm{Cr}_{2} \\mathrm{O}_{7}^{-2}+\\mathrm{H}_{2} \\mathrm{O}+\\overset{+}{\\mathrm{Na}}\n$$

\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1147, "subject": "Chemistry", "question": "

An acidified manganate solution undergoes disproportionation reaction. The spin-only magnetic moment value of the product having manganese in higher oxidation state is _____________ B.M. (Nearest integer)

", "options": [], "answer": "0", "solution": "**Answer:** 0\n\n$$\n\\begin{aligned}\n& 3 \\mathrm{MnO}_4^{2-}+4 \\mathrm{H}^{+} \\longrightarrow 2 \\overset{+7}{\\mathrm{MnO}_4^{-}}+\\overset{+4}{\\mathrm{MnO}_2}+2 \\mathrm{H}_2 \\mathrm{O} \\\\\\\\\n& \\overset{+7}{\\mathrm{Mn}}=\\text { no. of unpaired electrons is '0' } \\\\\\\\\n& \\mu=0 \\text { B.M. }\n\\end{aligned}\n$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1148, "subject": "Chemistry", "question": "

The number of statements correct from the following for Copper (at. no. 29) is/are ____________.

\n

(A) Cu(II) complexes are always paramagnetic.

\n

(B) Cu(I) complexes are generally colourless

\n

(C) Cu(I) is easily oxidized

\n

(D) In Fehling solution, the active reagent has Cu(I)

", "options": [], "answer": "3", "solution": "**Answer:** 3\n\n(A) $\\mathrm{Cu}$ (II) complexes are always paramagnetic as they have one unpaired electron due to $d^{9}$ configuration of $\\mathrm{Cu}(\\mathrm{II})$\n

\n(B) $\\mathrm{Cu}(\\mathrm{I})$ complexes are generally colourless due to $d^{10}$ configuration.\n

\n(C) $\\mathrm{Cu}(\\mathrm{I})$ is easily oxidised to $\\mathrm{Cu}^{+2}$ in aqueous solution\n

\n$2 \\mathrm{Cu}^{+} \\rightarrow \\mathrm{Cu}^{+2}+\\mathrm{Cu}$\n

\n$\\mathrm{Cu}^{+1}$ disproportionates to $\\mathrm{Cu}^{+2}$ and $\\mathrm{Cu}$\n

\n$\\left(E_{\\text {cell }}^{\\circ}>0\\right.$ for this cell reaction in aqueous solution)\n

\nIn Fehling's solution, active reagent has $\\mathrm{Cu}(\\mathrm{II})$ which is reduced to $\\mathrm{Cu}(\\mathrm{I})$ on reaction with aldehydes.\n

\nHence (D) statement is incorrect", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1149, "subject": "Chemistry", "question": "

The spin-only magnetic moment value of the most basic oxide of vanadium among V2O3, V2O4 and V2O5 is _____________ B.M. (Nearest integer)

", "options": [], "answer": "3", "solution": "**Answer:** 3\n\nThe most basic oxide among $\\mathrm{V}_{2} \\mathrm{O}_{3}, \\mathrm{~V}_{2} \\mathrm{O}_{4}$ and $\\mathrm{V}_{2} \\mathrm{O}_{5}$ is $\\mathrm{V}_{2} \\mathrm{O}_{3}$\n

\n$$\n\\mathrm{V}_{2} \\mathrm{O}_{3}=\\mathrm{V}^{+3}\\left(\\mathrm{~d}^{2}\\right)\n$$\n

\nMagnetic moment $=\\sqrt{2(2+2)}=\\sqrt{8}$\n

\n$$\n=2.83 \\approx 3\n$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1150, "subject": "Chemistry", "question": "

Manganese (VI) has ability to disproportionate in acidic solution. The difference in oxidation states of two ions it forms in acidic solution is ____________.

", "options": [], "answer": "3", "solution": "**Answer:** 3\n\nManganese $(\\mathrm{VI})$ disproportionates in acidic medium as\n

\n$3 \\mathrm{MnO}_{4}^{2-}+4 \\mathrm{H}^{+} \\longrightarrow 2 \\mathrm{MnO}_{4}^{-}+\\mathrm{MnO}_{2}+2 \\mathrm{H}_{2} \\mathrm{O}$\n

\nThe difference in oxidation states of $\\mathrm{Mn}$ in the products formed $=7-4=3$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1151, "subject": "Chemistry", "question": "

The difference in oxidation state of chromium in chromate and dichromate salts is ___________.

", "options": [], "answer": "0", "solution": "**Answer:** 0\n\nChromate ion $\\rightarrow \\mathrm{CrO}_{4}^{2-}$, oxidation state of $\\mathrm{Cr}=+6$

Dichromate ion $\\rightarrow \\mathrm{Cr_2O}_{7}^{2-}$, oxidation state of $\\mathrm{Cr}=+6$

$\\therefore $ Difference in oxidation state $=$ zero", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1152, "subject": "Chemistry", "question": "

The dark purple colour of $$\\mathrm{KMnO}_{4}$$ disappears in the titration with oxalic acid in acidic medium. The overall change in the oxidation number of manganese in the reaction is :

", "options": [ { "text": "5" }, { "text": "1" }, { "text": "7" }, { "text": "2" } ], "answer": "5", "solution": "**Answer:** 5\n\n$2 \\overset{+7}{\\mathrm{KMnO}_{4}}+5 \\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}+3 \\mathrm{H}_{2} \\mathrm{SO}_{4} \\rightarrow$\n\n$$\n\\mathrm{K}_{2} \\mathrm{SO}_{4}+2 \\overset{+2}{\\mathrm{MnSO}_{4}}+10 \\mathrm{CO}_{2}+8 \\mathrm{H}_{2} \\mathrm{O}\n$$\n

\nChange is oxidation state $M n$ is 5.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1153, "subject": "Chemistry", "question": "

Given below are two statements.

\n

Statement I : Iron (III) catalyst, acidified $$\\mathrm{K}_{2} \\mathrm{Cr}_{2} \\mathrm{O}_{7}$$ and neutral $$\\mathrm{KMnO}_{4}$$ have the ability to oxidise $$\\mathrm{I}^{-}$$ to $$\\mathrm{I}_{2}$$ independently.

\n

Statement II : Manganate ion is paramagnetic in nature and involves $$\\mathrm{p} \\pi-\\mathrm{p} \\pi$$ bonding.

\n

In the light of the above statements, choose the correct answer from the options given below.

", "options": [ { "text": "Both Statement I and Statement II are true." }, { "text": "Both Statement I and Statement II are false." }, { "text": "Statement I is true but Statement II is false." }, { "text": "Statement I is false but Statement II is true." } ], "answer": "Both Statement I and Statement II are false.", "solution": "**Answer:** Both Statement I and Statement II are false.\n\nManganate ion $\\mathrm{MnO}_{4}^{2-}$ has tetrahedral structure

\n\"JEE\n
\nhas only $\\mathrm{d} \\pi-\\mathrm{p} \\pi\\,\\, \\pi$-bonds.\n

\n$\\mathrm{Fe}^{3+}$ is not used as a catalyst in the conversion of I- to $\\mathrm{I}_{2}$ by $\\mathrm{K}_{2} \\mathrm{Cr}_{2} \\mathrm{O}_{7} \\cdot \\mathrm{K}_{2} \\mathrm{Cr}_{2} \\mathrm{O}_{7}$ oxidise $\\mathrm{I}^{-}$ in acidic medium easily", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1154, "subject": "Chemistry", "question": "

The total number of $$\\mathrm{Mn}=\\mathrm{O}$$ bonds in $$\\mathrm{Mn}_{2} \\mathrm{O}_{7}$$ is __________.

", "options": [ { "text": "4" }, { "text": "5" }, { "text": "6" }, { "text": "3" } ], "answer": "6", "solution": "**Answer:** 6\n\nStructure of $\\mathrm{Mn}_{2} \\mathrm{O}_{7}$ is as :

\n\"JEE\n
\n$\\therefore$ There are total $6\\, \\mathrm{M}=\\mathrm{O}$ bonds are present in $\\mathrm{Mn}_{2} \\mathrm{O}_{7}$ compound.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1155, "subject": "Chemistry", "question": "

In neutral or alkaline solution, $$\\mathrm{MnO}_{4}^{-}$$ oxidises thiosulphate to :

", "options": [ { "text": "$$\\mathrm{S}_{2} \\mathrm{O}_{7}^{2-}$$" }, { "text": "$$\\mathrm{S}_{2} \\mathrm{O}_{8}^{2-}$$" }, { "text": "$$\\mathrm{SO}_{3}^{2-}$$" }, { "text": "$$\\mathrm{SO}_{4}^{2-}$$" } ], "answer": "$$\\mathrm{SO}_{4}^{2-}$$", "solution": "**Answer:** $$\\mathrm{SO}_{4}^{2-}$$\n\n$8 \\mathrm{MnO}_{4}^{-}+3 \\mathrm{~S}_{2} \\mathrm{O}_{3}^{2-}+ \\mathrm{H}_{2} \\mathrm{O}\\rightarrow 8 \\mathrm{MnO}_{2}+6 \\mathrm{SO}_{4}^{2-}+$ $2 \\mathrm{OH}^{-}$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1156, "subject": "Chemistry", "question": "

Match List - I with List - II, match the gas evolved during each reaction.

\n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
List - IList - II
(A)$$\\left(\\mathrm{NH}_{4}\\right)_{2} \\mathrm{Cr}_{2} \\mathrm{O}_{7} \\stackrel{\\Delta}{\\longrightarrow}$$(I)$$\\mathrm{H}_{2}$$
(B)$$\\mathrm{KMnO}_{4}+\\mathrm{HCl} \\rightarrow$$(II)$$\\mathrm{N}_{2}$$
(C)$$\\mathrm{Al}+\\mathrm{NaOH}+\\mathrm{H}_{2} \\mathrm{O} \\rightarrow$$(III)$$\\mathrm{O}_{2}$$
(D)$$\\mathrm{NaNO}_{3} \\stackrel{\\Delta}{\\longrightarrow}$$(IV)$$\\mathrm{Cl}_{2}$$

\n

Choose the correct answer from the options given below :

", "options": [ { "text": "$$(\\mathrm{A})-(\\mathrm{II}),(\\mathrm{B})-(\\mathrm{III}),(\\mathrm{C})-(\\mathrm{I}),(\\mathrm{D})-(\\mathrm{IV})$$" }, { "text": "$$(\\mathrm{A})-(\\mathrm{III}),(\\mathrm{B})-(\\mathrm{I}),(\\mathrm{C})-(\\mathrm{IV}),(\\mathrm{D})-(\\mathrm{II})$$" }, { "text": "$$(\\mathrm{A})-(\\mathrm{II}),(\\mathrm{B})-(\\mathrm{IV}),(\\mathrm{C})-(\\mathrm{I}),(\\mathrm{D})-(\\mathrm{III})$$" }, { "text": "$$(\\mathrm{A})-(\\mathrm{III}),(\\mathrm{B})-(\\mathrm{IV}),(\\mathrm{C})-(\\mathrm{I}),(\\mathrm{D})-(\\mathrm{II})$$" } ], "answer": "$$(\\mathrm{A})-(\\mathrm{II}),(\\mathrm{B})-(\\mathrm{IV}),(\\mathrm{C})-(\\mathrm{I}),(\\mathrm{D})-(\\mathrm{III})$$", "solution": "**Answer:** $$(\\mathrm{A})-(\\mathrm{II}),(\\mathrm{B})-(\\mathrm{IV}),(\\mathrm{C})-(\\mathrm{I}),(\\mathrm{D})-(\\mathrm{III})$$\n\n$\\left(\\mathrm{NH}_{4}\\right)_{2} \\mathrm{Cr}_{2} \\mathrm{O}_{7} \\stackrel{\\Delta}{\\longrightarrow} \\mathrm{N}_{2}+4 \\mathrm{H}_{2} \\mathrm{O}+\\mathrm{Cr}_{2} \\mathrm{O}_{3}$\n

\n$\\mathrm{KMnO}_{4}+\\mathrm{HCl} \\longrightarrow \\mathrm{KCl}+\\mathrm{MnCl}_{2}+\\mathrm{Cl}_{2}+\\mathrm{H}_{2} \\mathrm{O}$\n

\n$\\mathrm{Al}+\\mathrm{NaOH}+\\mathrm{H}_{2} \\mathrm{O} \\longrightarrow \\mathrm{Na}\\left(\\mathrm{Al}(\\mathrm{OH})_{4}\\right)+\\mathrm{H}_{2}$\n

\n$2 \\mathrm{NaNO}_{3}(\\mathrm{~s}) \\stackrel{\\Delta}{\\longrightarrow} 2 \\mathrm{NaNO}_{2}(\\mathrm{~s})+\\mathrm{O}_{2}$", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1157, "subject": "Chemistry", "question": "

The disproportionation of $$\\mathrm{MnO}_{4}^{2-}$$ in acidic medium resulted in the formation of two manganese compounds $$\\mathrm{A}$$ and $$\\mathrm{B}$$. If the oxidation state of $$\\mathrm{Mn}$$ in $$\\mathrm{B}$$ is smaller than that of A, then the spin-only magnetic moment $$(\\mu)$$ value of B in BM is __________. (Nearest integer)

", "options": [], "answer": "4", "solution": "**Answer:** 4\n\n$3 \\overset{+6}{\\mathrm{MnO}_{4}^{-2}}+4 \\mathrm{H}^{+} \\rightarrow \\overset{+4}{\\mathrm{MnO}_{2}}+\\overset{+7}{\\mathrm{MnO}_{4}^{-}}$\n

\n$\\mathrm{Mn} \\rightarrow 4 s^{2} 3 d^{5}$\n

\n$\\mathrm{Mn}^{+4} \\rightarrow 3 d^{3}$\n

\n$$\n\\mathrm{n}=3\n$$\n

\n$$\n\\begin{aligned}\n\\mu &=\\sqrt{n(n+2)} \\\\\\\\\n&=\\sqrt{3(5)} \\\\\\\\\n&=\\sqrt{15} \\\\\\\\\n&=3.87 \\approx 4 \\text { B.M. }\n\\end{aligned}\n$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1158, "subject": "Chemistry", "question": "

In neutral or faintly alkaline medium, $$\\mathrm{KMnO}_{4}$$ being a powerful oxidant can oxidize, thiosulphate almost quantitatively, to sulphate. In this reaction overall change in oxidation state of manganese will be :

", "options": [ { "text": "5" }, { "text": "1" }, { "text": "0" }, { "text": "3" } ], "answer": "3", "solution": "**Answer:** 3\n\nIn neutral or Faintly alkaline medium, thiosulphate is oxidised almost quantitatively to sulphate ion according to reaction given below,

$8 \\mathrm{MnO}_{4}^{-}+3 \\mathrm{~S}_{2} \\mathrm{O}_{3}^{2-}+\\mathrm{H}_{2} \\mathrm{O} \\rightarrow 8 \\mathrm{MnO}_{2}+6 \\mathrm{SO}_{4}^{2-}+2 \\mathrm{OH}^{-}$

Here the $\\mathrm{Mn}$ changes from $\\mathrm{Mn}^{+7}$ to $\\mathrm{Mn}^{+4}$\n

\nThus overall change in its oxidation number would be of 3 .\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1159, "subject": "Chemistry", "question": "

Which element is not present in Nessler's reagent?

", "options": [ { "text": "Mercury" }, { "text": "Iodine" }, { "text": "Potassium" }, { "text": "Oxygen" } ], "answer": "Oxygen", "solution": "**Answer:** Oxygen\n\n

Nessler’s reagent is K2[HgI4]\n

\n

Therefore, among the options given, oxygen is the only element that is not present in Nessler's reagent.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1160, "subject": "Chemistry", "question": "

Among following compounds, the number of those present in copper matte is ___________.

\n

A. $$\\mathrm{CuCO_{3}}$$

\n

B. $$\\mathrm{Cu_{2}S}$$

\n

C. $$\\mathrm{Cu_{2}O}$$

\n

D. $$\\mathrm{FeO}$$

", "options": [], "answer": "1", "solution": "**Answer:** 1\n\n

Copper matte is a mixture of copper sulfide and iron sulfide formed as a result of smelting of copper ore. It contains mainly copper sulfide (Cu2S) as the major component and iron sulfide (FeS) as a minor component.

\n

\nThe compounds A. CuCO3, C. Cu2O, and D. FeO are not present in copper matte, but can be formed as a result of further processing or refining of the matte.

\n\n

Therefore, among the given compounds, only \"B. Cu2S\" is present in copper matte.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1161, "subject": "Chemistry", "question": "

When $$\\mathrm{Cu}^{2+}$$ ion is treated with $$\\mathrm{KI}$$, a white precipitate, $$\\mathrm{X}$$ appears in solution. The solution is titrated with sodium thiosulphate, the compound $$\\mathrm{Y}$$ is formed. $$\\mathrm{X}$$ and $$\\mathrm{Y}$$ respectively are :

", "options": [ { "text": "\n\n\n\n\n\n\n \n \n \n \n\n
$$\\mathrm{X=CuI_2}$$$$\\mathrm{Y=Na_2S_4O_6}$$
" }, { "text": "\n\n\n\n\n\n\n \n \n \n \n\n
$$\\mathrm{X=Cu_2I_2}$$$$\\mathrm{Y=Na_2S_4O_6}$$
" }, { "text": "\n\n\n\n\n\n\n \n \n \n \n\n
$$\\mathrm{X=CuI_2}$$$$\\mathrm{Y=Na_2S_2O_3}$$
" }, { "text": "\n\n\n\n\n\n\n \n \n \n \n\n
$$\\mathrm{X=Cu_2I_2}$$$$\\mathrm{Y=Na_2S_4O_5}$$
" } ], "answer": "\n\n\n\n\n\n\n \n \n \n \n\n
$$\\mathrm{X=Cu_2I_2}$$$$\\mathrm{Y=Na_2S_4O_6}$$
", "solution": "**Answer:** \n\n\n\n\n\n\n \n \n \n \n\n
$$\\mathrm{X=Cu_2I_2}$$$$\\mathrm{Y=Na_2S_4O_6}$$
\n\n$2 \\mathrm{Cu}^{2+}+4 \\mathrm{KI} \\longrightarrow \\underset{\\text { White ppt. }}{\\mathrm{Cu}_{2} \\mathrm{I}_{2}}+\\mathrm{I}_{2}$\n\n

$\\mathrm{I}_{2}+\\mathrm{Na}_{2} \\mathrm{~S}_{2} \\mathrm{O}_{3} \\longrightarrow 2 \\mathrm{Nal}+\\mathrm{Na}_{2} \\mathrm{~S}_{4} \\mathrm{O}_{6}$\n\n

$\\mathrm{X}=\\mathrm{Cu}_{2} \\mathrm{I}_{2}$\n\n

$\\mathrm{Y}=\\mathrm{Na}_{2} \\mathrm{~S}_{4} \\mathrm{O}_{6}$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1162, "subject": "Chemistry", "question": "Formulae for Nessler's reagent is :", "options": [ { "text": "$\\mathrm{KHgI}_3$" }, { "text": "$\\mathrm{HgI}_2$" }, { "text": "$\\mathrm{KHg}_2 \\mathrm{I}_2$" }, { "text": "$\\mathrm{K}_2 \\mathrm{HgI}_4$" } ], "answer": "$\\mathrm{K}_2 \\mathrm{HgI}_4$", "solution": "**Answer:** $\\mathrm{K}_2 \\mathrm{HgI}_4$\n\n

Nessler's reagent is $$\\mathrm{K_2H_g I_4}$$

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1163, "subject": "Chemistry", "question": "$\\mathrm{KMnO}_4$ oxidises $\\mathrm{I}^{-}$ in acidic and neutral/faintly alkaline solutions, respectively, to :", "options": [ { "text": "$\\mathrm{IO}_3^{-} ~\\&~ \\mathrm{IO}_3^{-}$" }, { "text": "$\\mathrm{I}_2 ~\\&~ \\mathrm{I}_2$" }, { "text": "$\\mathrm{I}_2 ~\\&~ \\mathrm{IO}_3^{-}$" }, { "text": "$\\mathrm{IO}_3^{-} ~\\&~ \\mathrm{I}_2$" } ], "answer": "$\\mathrm{I}_2 ~\\&~ \\mathrm{IO}_3^{-}$", "solution": "**Answer:** $\\mathrm{I}_2 ~\\&~ \\mathrm{IO}_3^{-}$\n\n

Potassium permanganate ($\\mathrm{KMnO}_4$) is a strong oxidizing agent that can oxidize iodide ions ($\\mathrm{I}^{-}$) to different products depending on the pH of the solution.

\n\n

In acidic solutions, $\\mathrm{KMnO}_4$ oxidizes $\\mathrm{I}^{-}$ to molecular iodine ($\\mathrm{I}_2$) according to the following equation:

\n\n$$\n2\\mathrm{MnO}_4^{-} + 16\\mathrm{H}^{+} + 10\\mathrm{I}^{-} \\rightarrow 2\\mathrm{Mn}^{2+} + 8\\mathrm{H}_2\\mathrm{O} + 5\\mathrm{I}_2\n$$\n\n

In neutral or faintly alkaline solutions, the oxidation product is different because the $\\mathrm{I}^{-}$ is oxidized to iodate ion ($\\mathrm{IO}_3^{-}$). The reaction in a neutral or slightly alkaline solution is as follows:

\n\n$$\n2\\mathrm{MnO}_4^{-} + 2\\mathrm{H}_2\\mathrm{O} + 10\\mathrm{I}^{-} \\rightarrow 2\\mathrm{MnO}_2 + 4\\mathrm{OH}^{-} + 5\\mathrm{IO}_3^{-}\n$$\n\n

So according to the reactions above, in acidic solutions, $\\mathrm{I}^{-}$ is oxidized to $\\mathrm{I}_2$ while in neutral or faintly alkaline solutions, $\\mathrm{I}^{-}$ is oxidized to $\\mathrm{IO}_3^{-}$. Therefore, the correct answer is:

\n\n

Option C: $\\mathrm{I}_2 ~\\&~ \\mathrm{IO}_3^{-}$

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1164, "subject": "Chemistry", "question": "

During the qualitative analysis of $$\\mathrm{SO}_{3}^{2-}$$ using dilute $$\\mathrm{H}_{2} \\mathrm{SO}_{4}, \\mathrm{SO}_{2}$$ gas is evolved which turns $$\\mathrm{K}_{2} \\mathrm{Cr}_{2} \\mathrm{O}_{7}$$ solution (acidified with dilute $$\\mathrm{H}_{2} \\mathrm{SO}_{4}$$) :

", "options": [ { "text": "blue" }, { "text": "black" }, { "text": "red" }, { "text": "green" } ], "answer": "green", "solution": "**Answer:** green\n\n

$$\\mathrm{S{O_2} + C{r_2}O_7^{2 - }\\mathrel{\\mathop{\\kern0pt\\longrightarrow}\n\\limits_{}} \\mathop {C{r^{3 + }}}\\limits_{(green)} + SO_4^{2 - }}$$

", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 1165, "subject": "Chemistry", "question": "

The number of electrons involved in the reduction of permanganate to manganese dioxide in acidic medium is _____________.

", "options": [], "answer": "3", "solution": "**Answer:** 3\n\n

$$\\mathrm{3{e^ - } + 4{H^ + } + MnO_4^ - \\buildrel {} \\over\n \\longrightarrow Mn{O_2} + 2{H_2}O}$$

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1166, "subject": "Chemistry", "question": "

An indicator 'X' is used for studying the effect of variation in concentration of iodide on the rate of reaction of iodide ion with $$\\mathrm{H_2O_2}$$ at room temp. The indicator 'X' forms blue coloured complex with compound 'A' present in the solution. The indicator 'X' and compound 'A' respectively are :

", "options": [ { "text": "Methyl orange and $$\\mathrm{H_2O_2}$$" }, { "text": "Methyl orange and iodine" }, { "text": "Starch and $$\\mathrm{H_2O_2}$$" }, { "text": "Starch and iodine" } ], "answer": "Starch and iodine", "solution": "**Answer:** Starch and iodine\n\n

Starch forms blue coloured complex with iodine.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1167, "subject": "Chemistry", "question": "

A chloride salt solution acidified with dil.HNO$$_3$$ gives a curdy white precipitate, [A], on addition of AgNO$$_3$$. [A] on treatment with NH$$_4$$OH gives a clear solution B. A and B are respectively :

", "options": [ { "text": "$$\\mathrm{H[AgCl_3]~\\&~(NH_4)[Ag(OH)_2]}$$" }, { "text": "$$\\mathrm{AgCl~\\&~[Ag(NH_3)_2]Cl}$$" }, { "text": "$$\\mathrm{H[AgCl_3]~\\&~[Ag(NH_3)_2]Cl}$$" }, { "text": "$$\\mathrm{AgCl~\\&~(NH_4)[Ag(OH)_2]}$$" } ], "answer": "$$\\mathrm{AgCl~\\&~[Ag(NH_3)_2]Cl}$$", "solution": "**Answer:** $$\\mathrm{AgCl~\\&~[Ag(NH_3)_2]Cl}$$\n\n$\\mathrm{Cl}^{-}(\\mathrm{aq})+\\mathrm{AgNO}_{3}(\\mathrm{aq}) \\stackrel{\\text { dil } \\mathrm{HNO}_{3}}{\\longrightarrow} \\underset{(\\mathrm{A})}{\\mathrm{AgCl}} \\downarrow$\n

\n$$\n\\mathrm{AgCl} \\downarrow+2 \\mathrm{NH}_4 \\mathrm{OH}(\\mathrm{aq}) \\rightarrow\\underset{(B)}{\\left[\\mathrm{Ag}\\left(\\mathrm{NH}_3\\right)_2\\right]} \\mathrm{Cl}(\\mathrm{aq})+2 \\mathrm{H}_2 \\mathrm{O}\n$$\n

\n$\\therefore(\\mathrm{A})$ is $\\mathrm{AgCl}$ and $(\\mathrm{B})$ is $\\left[\\mathrm{Ag}\\left(\\mathrm{NH}_{3}\\right)_{2}\\right] \\mathrm{Cl}$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1168, "subject": "Chemistry", "question": "

An ammoniacal metal salt solution gives a brilliant red precipitate on addition of dimethylglyoxime. The metal ion is :

", "options": [ { "text": "Co$$^{2+}$$" }, { "text": "Fe$$^{2+}$$" }, { "text": "Ni$$^{2+}$$" }, { "text": "Cu$$^{2+}$$" } ], "answer": "Ni$$^{2+}$$", "solution": "**Answer:** Ni$$^{2+}$$\n\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1169, "subject": "Chemistry", "question": "

Prolonged heating is avoided during the preparation of ferrous ammonium sulphate to :

", "options": [ { "text": "prevent hydrolysis" }, { "text": "prevent breaking" }, { "text": "prevent oxidation" }, { "text": "prevent reduction" } ], "answer": "prevent oxidation", "solution": "**Answer:** prevent oxidation\n\n

Prolonged heating is avoided during the preparation of ferrous ammonium sulfate to prevent oxidation. Ferrous ammonium sulfate is a green crystalline solid that is used as a reducing agent in various chemical reactions. When heated, it can undergo oxidation to form ferric ammonium sulfate, which is a brown crystalline solid.

\n

Here is a chemical equation for the oxidation of ferrous ammonium sulfate:

\n$$\\begin{equation}\n\\ce{FeSO4(NH4)2SO4.6H2O + O2 -> Fe2(SO4)3(NH4)2SO4.12H2O}\n\\end{equation}$$

As you can see, the oxidation of ferrous ammonium sulfate results in the formation of ferric ammonium sulfate, which is a brown crystalline solid. This is why prolonged heating is avoided during the preparation of ferrous ammonium sulfate.

\n

Here are the other options and why they are incorrect:

\n\n\n", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 1170, "subject": "Chemistry", "question": "

Which halogen is known to cause the reaction given below :

\n

$$2 \\mathrm{Cu}^{2+}+4 \\mathrm{X}^{-} \\rightarrow \\mathrm{Cu}_{2} \\mathrm{X}_{2}(\\mathrm{s})+\\mathrm{X}_{2}$$

", "options": [ { "text": "All halogens" }, { "text": "Only Iodine" }, { "text": "Only Bromine" }, { "text": "Only Chlorine" } ], "answer": "Only Iodine", "solution": "**Answer:** Only Iodine\n\n

The given reaction describes the formation of a copper(II) halide and the corresponding diatomic halogen molecule. Not all halogens will cause this reaction to occur.

\n

This reaction is known to occur with iodine (I) because the formation of $\\mathrm{I}_{2}$ (a diatomic iodine molecule) is easier due to the relatively low bond energy of the $\\mathrm{I}-\\mathrm{I}$ bond.

\n

On the other hand, the bond energies of $\\mathrm{Cl}-\\mathrm{Cl}$ and $\\mathrm{Br}-\\mathrm{Br}$ are higher, making it more difficult for these diatomic halogen molecules to form. Thus, these halogens would not typically cause this reaction to occur.

\n

Therefore, the correct answer is:

\n

Option B : Only Iodine.

\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1171, "subject": "Chemistry", "question": "

Element not present in Nessler's reagent is :

", "options": [ { "text": "I" }, { "text": "$$\\mathrm{N}$$" }, { "text": "$$\\mathrm{K}$$" }, { "text": "$$\\mathrm{Hg}$$" } ], "answer": "$$\\mathrm{N}$$", "solution": "**Answer:** $$\\mathrm{N}$$\n\nNessler's reagent is a 2% solution of potassium tetraiodomercurate(II) in aqueous potassium hydroxide. Its chemical formula is $$\\mathrm{K}_2[\\mathrm{HgI}_4]$$. The elements present in it are Potassium (K), Mercury (Hg), and Iodine (I). Nitrogen (N) is not present in Nessler's reagent.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1172, "subject": "Chemistry", "question": "

$$\\mathrm{NaCl}$$ reacts with conc. $$\\mathrm{H}_2 \\mathrm{SO}_4$$ and $$\\mathrm{K}_2 \\mathrm{Cr}_2 \\mathrm{O}_7$$ to give reddish fumes (B), which react with $$\\mathrm{NaOH}$$ to give yellow solution (C). (B) and (C) respectively are ;

", "options": [ { "text": "$$\\mathrm{CrO}_2 \\mathrm{Cl}_2, \\mathrm{Na}_2 \\mathrm{CrO}_4$$\n" }, { "text": "$$\\mathrm{Na}_2 \\mathrm{CrO}_4, \\mathrm{CrO}_2 \\mathrm{Cl}_2$$\n" }, { "text": "$$\\mathrm{CrO}_2 \\mathrm{Cl}_2, \\mathrm{KHSO}_4$$\n" }, { "text": "$$\\mathrm{CrO}_2 \\mathrm{Cl}_2, \\mathrm{Na}_2 \\mathrm{Cr}_2 \\mathrm{O}_7$$" } ], "answer": "$$\\mathrm{CrO}_2 \\mathrm{Cl}_2, \\mathrm{Na}_2 \\mathrm{CrO}_4$$\n", "solution": "**Answer:** $$\\mathrm{CrO}_2 \\mathrm{Cl}_2, \\mathrm{Na}_2 \\mathrm{CrO}_4$$\n\n\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1173, "subject": "Chemistry", "question": "

Choose the correct statements from the following

\n

A. $$\\mathrm{Mn}_2 \\mathrm{O}_7$$ is an oil at room temperature

\n

B. $$\\mathrm{V}_2 \\mathrm{O}_4$$ reacts with acid to give $$\\mathrm{VO}_2{ }^{2+}$$

\n

C. $$\\mathrm{CrO}$$ is a basic oxide

\n

D. $$\\mathrm{V}_2 \\mathrm{O}_5$$ does not react with acid

\n

Choose the correct answer from the options given below :

", "options": [ { "text": "A, B and D only" }, { "text": "A, B and C only" }, { "text": "A and C only" }, { "text": "B and C only" } ], "answer": "A and C only", "solution": "**Answer:** A and C only\n\n

(A) $$\\mathrm{Mn}_2 \\mathrm{O}_7$$ is green oil at room temperature.

\n

(B) $$\\mathrm{V}_2 \\mathrm{O}_4$$ dissolve in acids to give $$\\mathrm{VO}^{2+}$$ salts.

\n

(C) $$\\mathrm{CrO}$$ is basic oxide

\n

(D) $$\\mathrm{V}_2 \\mathrm{O}_5$$ is amphoteric it reacts with acid as well as base.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1174, "subject": "Chemistry", "question": "

Identify correct statements from below:

\n

A. The chromate ion is square planar.

\n

B. Dichromates are generally prepared from chromates.

\n

C. The green manganate ion is diamagnetic.

\n

D. Dark green coloured $$\\mathrm{K}_2 \\mathrm{MnO}_4$$ disproportionates in a neutral or acidic medium to give permanganate.

\n

E. With increasing oxidation number of transition metal, ionic character of the oxides decreases.

\n

Choose the correct answer from the options given below:

", "options": [ { "text": "B, D, E only" }, { "text": "A, B, C only" }, { "text": "A, D, E only" }, { "text": "B, C, D only" } ], "answer": "B, D, E only", "solution": "**Answer:** B, D, E only\n\n

A. $$\\mathrm{CrO}_4{ }^{2-}$$ is tetrahedral

\n

B. $$2 \\mathrm{Na}_2 \\mathrm{CrO}_4+2 \\mathrm{H}^{+} \\rightarrow \\mathrm{Na}_2 \\mathrm{Cr}_2 \\mathrm{O}_7+2 \\mathrm{Na}^{+}+\\mathrm{H}_2 \\mathrm{O}$$

\n

C. As per NCERT, green manganate is paramagnetic with 1 unpaired electron.

\n

D. Statement is correct

\n

E. Statement is correct

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1175, "subject": "Chemistry", "question": "

The metals that are employed in the battery industries are

\n

A. $$\\mathrm{Fe}$$

\n

B. $$\\mathrm{Mn}$$

\n

C. $$\\mathrm{Ni}$$

\n

D. $$\\mathrm{Cr}$$

\n

E. $$\\mathrm{Cd}$$

\n

Choose the correct answer from the options given below:

", "options": [ { "text": "A, B, C, D and E" }, { "text": "A, B, C and D only" }, { "text": "B, D and E only" }, { "text": "B, C and E only" } ], "answer": "B, C and E only", "solution": "**Answer:** B, C and E only\n\n

Mn, Ni and Cd metals used in battery industries.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1176, "subject": "Chemistry", "question": "

$$\\mathrm{KMnO}_4$$ decomposes on heating at $$513 \\mathrm{~K}$$ to form $$\\mathrm{O}_2$$ along with

", "options": [ { "text": "$$\\mathrm{K}_2 \\mathrm{MnO}_4 ~\\& \\mathrm{~Mn}$$" }, { "text": "$$\\mathrm{MnO}_2 ~\\& \\mathrm{~K}_2 \\mathrm{O}_2$$\n" }, { "text": "\n$$\\mathrm{K}_2 \\mathrm{MnO}_4 ~\\& \\mathrm{~MnO}_2$$\n" }, { "text": "$$\\mathrm{Mn} ~\\& \\mathrm{~KO}_2$$" } ], "answer": "\n$$\\mathrm{K}_2 \\mathrm{MnO}_4 ~\\& \\mathrm{~MnO}_2$$\n", "solution": "**Answer:** \n$$\\mathrm{K}_2 \\mathrm{MnO}_4 ~\\& \\mathrm{~MnO}_2$$\n\n\n

$$\\mathrm{KMnO}_4 \\xrightarrow{\\Delta} \\mathrm{K}_2 \\mathrm{MnO}_4+\\mathrm{MnO}_2+\\mathrm{O}_2$$

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1177, "subject": "Chemistry", "question": "

The correct IUPAC name of $$\\mathrm{K}_2 \\mathrm{MnO}_4$$ is

", "options": [ { "text": "Potassium tetraoxidomanganate (VI)\n" }, { "text": "Dipotassium tetraoxidomanganate (VII)\n" }, { "text": "Potassium tetraoxopermanganate (VI)\n" }, { "text": "Potassium tetraoxidomanganese (VI)" } ], "answer": "Potassium tetraoxidomanganate (VI)\n", "solution": "**Answer:** Potassium tetraoxidomanganate (VI)\n\n\n

$$\\begin{aligned}\n& \\mathrm{K}_2 \\mathrm{MnO}_4 \\\\\n& 2+\\mathrm{x}-8=0 \\\\\n& \\Rightarrow \\mathrm{x}=+6\n\\end{aligned}$$

\n

$$\\mathrm{O.S.}$$ of $$\\mathrm{Mn}=+6$$\n

IUPAC Name $$=$$

\n

Potassium tetraoxidomanganate(VI)

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1178, "subject": "Chemistry", "question": "

The orange colour of $$\\mathrm{K}_2 \\mathrm{Cr}_2 \\mathrm{O}_7$$ and purple colour of $$\\mathrm{KMnO}_4$$ is due to\n

", "options": [ { "text": "Charge transfer transition in both.\n" }, { "text": "$$\\mathrm{d} \\rightarrow \\mathrm{d}$$ transitions in $$\\mathrm{KMnO}_4$$ and charge transfer transitions in $$\\mathrm{K}_2 \\mathrm{Cr}_2 \\mathrm{O}_7$$.\n" }, { "text": "$$\\mathrm{d} \\rightarrow \\mathrm{d}$$ transitions in both\n" }, { "text": "$$\\mathrm{d} \\rightarrow \\mathrm{d}$$ transitions in $$\\mathrm{K}_2 \\mathrm{Cr}_2 \\mathrm{O}_7$$ and charge transfer transitions in $$\\mathrm{KMnO}_4$$." } ], "answer": "Charge transfer transition in both.\n", "solution": "**Answer:** Charge transfer transition in both.\n\n\n

$$\\left.\\begin{array}{l}\n\\mathrm{K}_2 \\mathrm{Cr}_2 \\mathrm{O}_7 \\rightarrow \\mathrm{Cr}^{+6} \\rightarrow \\text { No d - d transition } \\\\\n\\mathrm{KMnO}_4 \\rightarrow \\mathrm{Mn}^{7+} \\rightarrow \\text { No d - d transition }\n\\end{array}\\right\\} \\text { Charge transfer }$$

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1179, "subject": "Chemistry", "question": "

Consider the following reaction

\n

$$\\mathrm{MnO}_2+\\mathrm{KOH}+\\mathrm{O}_2 \\rightarrow \\mathrm{A}+\\mathrm{H}_2 \\mathrm{O} \\text {. }$$

\n

Product '$$\\mathrm{A}$$' in neutral or acidic medium disproportionate to give products '$$\\mathrm{B}$$' and '$$\\mathrm{C}$$' along with water. The sum of spin-only magnetic moment values of $$\\mathrm{B}$$ and $$\\mathrm{C}$$ is ________ BM. (nearest integer) (Given atomic number of $$\\mathrm{Mn}$$ is 25)

", "options": [], "answer": "4", "solution": "**Answer:** 4\n\n

$$\\mathrm{A}$$ is $$\\mathrm{K}_2 \\mathrm{MnO}_4$$

\n

$$\\mathrm{B}$$ and $$\\mathrm{C}$$ are $$\\mathrm{KMnO}_4$$ and $$\\mathrm{MnO}_2$$

\n

$$\\mathrm{KMnO}_4(\\mu=0)$$

\n

$$\\mathrm{MnO}_2(\\mathrm{Mn}^{4+})(\\mu=3.87)$$

\n

Sum $$=3.87=4$$ (Nearest integer)

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1180, "subject": "Chemistry", "question": "

While preparing crystals of Mohr's salt, dil $$\\mathrm{H}_2 \\mathrm{SO}_4$$ is added to a mixture of ferrous sulphate and ammonium sulphate, before dissolving this mixture in water, dil $$\\mathrm{H_2SO_4}$$ is added here to :

", "options": [ { "text": "prevent the hydrolysis of ferrous sulphate\n" }, { "text": "prevent the hydrolysis of ammonium sulphate\n" }, { "text": "increase the rate of formation of crystals\n" }, { "text": "make the medium strongly acidic" } ], "answer": "prevent the hydrolysis of ferrous sulphate\n", "solution": "**Answer:** prevent the hydrolysis of ferrous sulphate\n\n\n

While preparing crystal of Mohr's salt, dil. $$\\mathrm{H}_2 \\mathrm{SO}_4$$ is added to prevent hydrolysis of ferrous sulphate.

", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 1181, "subject": "Chemistry", "question": "

The fusion of chromite ore with sodium carbonate in the presence of air leads to the formation of products $$\\mathrm{A}$$ and $$\\mathrm{B}$$ along with the evolution of $$\\mathrm{CO}_2$$. The sum of spin-only magnetic moment values of A and B is _________ B.M. (Nearest integer)

\n

[Given atomic number : $$\\mathrm{C}: 6, \\mathrm{Na}: 11, \\mathrm{O}: 8, \\mathrm{Fe}: 26, \\mathrm{Cr}: 24$$]

", "options": [], "answer": "6", "solution": "**Answer:** 6\n\n

To determine the spin-only magnetic moments of products A and B formed from the fusion of chromite ore with sodium carbonate in the presence of air, we first need to examine the given reaction and the properties of the products.

\n\n

The reaction provided is:

\n\n

$4 \\mathrm{FeCr}_2 \\mathrm{O}_4 + 8 \\mathrm{Na}_2 \\mathrm{CO}_3 + 7 \\mathrm{O}_2 \\rightarrow 8 \\mathrm{Na}_2 \\mathrm{CrO}_4 (\\mathrm{A}) + 2 \\mathrm{Fe}_2 \\mathrm{O}_3 (\\mathrm{B}) + 8 \\mathrm{CO}_2$

\n\nIdentifying the Products:\n

\n
\nDetermining the Magnetic Moments:\n

\n
    \n
  1. For Product A ($\\mathrm{Na}_2 \\mathrm{CrO}_4$):
    2. \n

\n\n

$ \\mu = \\sqrt{n(n+2)} \\text{ BM} = \\sqrt{0(0+2)} \\text{ BM} = 0 \\text{ BM} $

\n\n
    \n
  1. For Product B ($\\mathrm{Fe}_2 \\mathrm{O}_3$):
    2. \n
\n\n

\n\n

$ \\mu = \\sqrt{n(n+2)} \\text{ BM} = \\sqrt{5(5+2)} \\text{ BM} = \\sqrt{35} \\text{ BM} \\approx 5.9 \\text{ BM} $

\n\nSum of Magnetic Moments:\n\n\n

Therefore, the sum of the spin-only magnetic moments of A and B is:

\n\n

$ 0 + 5.9 \\approx 6 \\text{ BM}$

\n\nConclusion:\n\n

The sum of the spin-only magnetic moment values of A and B is approximately 6 B.M. (to the nearest integer).

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1182, "subject": "Chemistry", "question": "Most common oxidation states of Ce (cerium) are :", "options": [ { "text": "+3, +4" }, { "text": "+2, +3" }, { "text": "+2, +4" }, { "text": "+3, +5" } ], "answer": "+3, +4", "solution": "**Answer:** +3, +4\n\nCommon oxidation states of $$Ce$$(Cerium) are $$+3$$ and $$+4$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1183, "subject": "Chemistry", "question": "The radius of La3+ (Atomic number of La = 57) is 1.06 Å. Which one of the following given values will be\nclosest to the radius of Lu3+ (Atomic number of Lu = 71) ?", "options": [ { "text": "1.60 Å" }, { "text": "1.40 Å" }, { "text": "0.85 Å" }, { "text": "1.60 Å" } ], "answer": "0.85 Å", "solution": "**Answer:** 0.85 Å\n\nIonic radili $$ \\propto {1 \\over z}$$\n

Thus, $${{{z_2}} \\over {{z_1}}} \\Rightarrow {{1.06} \\over {\\left( {Ionic\\,\\,\\,radii\\,\\,\\,of\\,\\,L{u^{3 + }}} \\right)}} = {{71} \\over {57}}$$ \n

$$ \\Rightarrow \\,\\,\\,$$ Ionic radii of $$L{u^{3 + }} = 0.85\\,\\mathop A\\limits^ \\circ $$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1184, "subject": "Chemistry", "question": "Cerium (Z = 58) is an important member of the lanthanoids. Which of the following\nstatements about cerium is incorrect? ", "options": [ { "text": "The common oxidation states of cerium are +3 and +4 " }, { "text": "Cerium (IV) acts as an oxidizing agent" }, { "text": "The +4 oxidation state of cerium is not known in solutions" }, { "text": "The +3 oxidation state of cerium is more stable than the +4 oxidation state " } ], "answer": "The +4 oxidation state of cerium is not known in solutions", "solution": "**Answer:** The +4 oxidation state of cerium is not known in solutions\n\nThe $$+4$$ oxidation state of cerium is also known in solution. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1185, "subject": "Chemistry", "question": "The lanthanide contraction is responsible for the fact that :", "options": [ { "text": "Zr and Y have about the same radius" }, { "text": "Zr and Nb have similar oxidation\nstate" }, { "text": "Zr and Hf have about the same radius" }, { "text": "Zr and Zn have the same oxidation" } ], "answer": "Zr and Hf have about the same radius", "solution": "**Answer:** Zr and Hf have about the same radius\n\n\"AIEEE\n
In the entire d-block,\n

Only in group 3, the size increase from top to bottom due to increase of no of shells in an atom. That is why size of Sc < Y < La. And shell size in group 3 is 3d < 4d < 5d.\n

But from group 4 to 12, the size of 3d < 4d but 4d $$ \\cong $$ 5d due to lanthanide contraction. That is why Zr and Hf have about the same radius.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1186, "subject": "Chemistry", "question": "Which of the following factors may be regarded as the main cause of lanthanide\ncontraction?", "options": [ { "text": "Poor shielding of one of 4f electron by another in the subshell" }, { "text": "Effective shielding of one of 4f electrons by another in the subshell " }, { "text": "Poorer shielding of 5d electrons by 4f electrons" }, { "text": "Greater shielding of 5d electrons by 4f electrons" } ], "answer": "Poorer shielding of 5d electrons by 4f electrons", "solution": "**Answer:** Poorer shielding of 5d electrons by 4f electrons\n\nThe main reason of lanthanide contraction is, poor shielding of 4f electrons on the outer most shell electrons.\n

In lanthanides, there is poorer shielding of $$5d$$ electrons by $$4f$$ electrons resulting in greater attraction of the nucleus over $$5d$$ electrons and reduction of the atomic radii.", "topic": "Number Theory", "subtopic": "Divisibility / Factorization" }, { "id": 1187, "subject": "Chemistry", "question": "Lanthanoid contraction is caused due to :", "options": [ { "text": "the appreciable shielding on outer electrons by 4f electrons from the nuclear charge " }, { "text": "the appreciable shielding on outer electrons by 5d electrons from the nuclear charge " }, { "text": "the same effective nuclear charge from Ce to Lu " }, { "text": "the imperfect shielding on outer electrons by 4f electrons from the nuclear charge " } ], "answer": "the imperfect shielding on outer electrons by 4f electrons from the nuclear charge ", "solution": "**Answer:** the imperfect shielding on outer electrons by 4f electrons from the nuclear charge \n\nThe main reason of lanthanide contraction is, poor shielding of 4f electrons on the outer most shell electrons. As f orbital is defused nature so shielding of 4f electron is imperfect that is why nucleus can attract outer most shell electron easily and the size of atom decreases.\n

In lanthanides, there is poorer shielding of $$5d$$ electrons by $$4f$$ electrons resulting in greater attraction of the nucleus over $$5d$$ electrons and reduction of the atomic radii.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1188, "subject": "Chemistry", "question": "Identify the incorrect statement among the following :", "options": [ { "text": "d-Block elements show irregular and erratic chemical properties among themselves " }, { "text": "La and Lu have partially filled d orbitals and no other partially filled orbitals " }, { "text": "The chemistry of various lanthanoids is very similar " }, { "text": "4f and 5f orbitals are equally shielded " } ], "answer": "4f and 5f orbitals are equally shielded ", "solution": "**Answer:** 4f and 5f orbitals are equally shielded \n\n$$4f$$ orbital is nearer to nucleus as compared to $$5f$$ orbital therefore, shielding of $$4f$$ is more than $$5f.$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1189, "subject": "Chemistry", "question": "The actinoids exhibits more number of oxidation states in general than the lanthanoids. This is\nbecause :", "options": [ { "text": "the 5f orbitals are more buried than the 4f orbitals " }, { "text": "there is a similarity between 4f and 5f orbitals in their angular part of the wave function " }, { "text": "the actinoids are more reactive than the lanthanoids" }, { "text": "the 5f orbitals extend further from the nucleus than the 4f orbitals " } ], "answer": "the 5f orbitals extend further from the nucleus than the 4f orbitals ", "solution": "**Answer:** the 5f orbitals extend further from the nucleus than the 4f orbitals \n\nNOTE : More the distance between nucleus and outer orbitals, lesser will be force of attraction on them. Distance between nucleus and $$5f$$ orbitals is more as compared to distance between $$4f$$ orbital and nucleus. So actinoids exhibit more number of oxidation states in general than the lanthanoids. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1190, "subject": "Chemistry", "question": "Larger number of oxidation states are exhibited by the actinoids than those by the lanthanoids, the\nmain reason being :", "options": [ { "text": "4f orbitals more diffused than the 5f orbitals " }, { "text": "lesser energy difference between 5f and 6d than between 4f and 5d orbitals " }, { "text": "more energy difference between 5f and 6d than between 4f and 5d orbitals" }, { "text": "more reactive nature of the actinoids than the lanthanoids" } ], "answer": "lesser energy difference between 5f and 6d than between 4f and 5d orbitals ", "solution": "**Answer:** lesser energy difference between 5f and 6d than between 4f and 5d orbitals \n\nNOTE : The main reason for exhibiting larger number of oxidation states by actinoids as compared to lanthanoids is lesser energy difference between $$5f$$ and $$6d$$ orbitals as compared to that between $$4f$$ and $$5d$$ orbitals. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1191, "subject": "Chemistry", "question": "Knowing that the Chemistry of lanthanoids (Ln) is dominated by its +3 oxidation state, which of the\nfollowing statements is incorrect?", "options": [ { "text": "Because of the large size of the Ln (III) ions the bonding in its compounds is predominantly ionic in\n character." }, { "text": "The ionic sizes of Ln (III) decrease in general with increasing atomic number. " }, { "text": "Ln (III) compounds are generally colourless" }, { "text": "Ln (III) hydroxides are mainly basic in character." } ], "answer": "Ln (III) compounds are generally colourless", "solution": "**Answer:** Ln (III) compounds are generally colourless\n\nMost of the $$L{n^{3 + }}\\,\\,$$ compounds except $$L{a^{3 + }}$$ and $$L{u^{3 + }}$$ are coloured due to the presence of $$f$$-electrons. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1192, "subject": "Chemistry", "question": "In context of the lanthanoids, which of the following statements is not correct?", "options": [ { "text": "All the members exhibit +3 oxidation state " }, { "text": "Because of similar properties the separation of lanthanoids is not easy. " }, { "text": "Availability of 4f electrons results in the formation of compounds in +4 state for all the members\n of the series." }, { "text": "There is a gradual decrease in the radii of the members with increasing atomic number in the\n series. " } ], "answer": "Availability of 4f electrons results in the formation of compounds in +4 state for all the members\n of the series.", "solution": "**Answer:** Availability of 4f electrons results in the formation of compounds in +4 state for all the members\n of the series.\n\nThe number stable oxidation states of lanthanides is $$+3$$.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1193, "subject": "Chemistry", "question": "The outer electron configuration of $$Gd$$ (Atomic No. $$64$$) is : ", "options": [ { "text": "$$4{f^3}\\,5{d^5}\\,6{s^2}$$ " }, { "text": "$$4{f^8}\\,5{d^0}\\,6{s^2}$$ " }, { "text": "$$4{f^4}\\,5{d^4}\\,6{s^2}$$ " }, { "text": "$$4{f^7}\\,5{d^1}\\,6{s^2}$$ " } ], "answer": "$$4{f^7}\\,5{d^1}\\,6{s^2}$$ ", "solution": "**Answer:** $$4{f^7}\\,5{d^1}\\,6{s^2}$$ \n\nThe configuration of $$Gd$$ is $$4{f^7}\\,5{d^1}\\,6{s^2}.$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1194, "subject": "Chemistry", "question": "The 71st electron of an element X with an atomic number of 71 enters into the orbital :", "options": [ { "text": "4f" }, { "text": "6s" }, { "text": "6p" }, { "text": "5d" } ], "answer": "5d", "solution": "**Answer:** 5d\n\nIn lutetium (71Lu) the 71st electron enters in 5d-orbital.\n

Electron configuration is [Xe]4f145d16s2", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1195, "subject": "Chemistry", "question": "The highest possible oxidation states of uranium and plutonium, respectively are :", "options": [ { "text": "4 and 6" }, { "text": "6 and 4" }, { "text": "7 and 6" }, { "text": "6 and 7" } ], "answer": "6 and 7", "solution": "**Answer:** 6 and 7\n\n

The highest possible oxidation states of uranium and plutonium can be identified by considering the electronic configurations of these elements and noting the number of valence electrons available for bonding.

\n

Uranium (U, atomic number 92) has the electronic configuration: \n$ [Rn] 5f^3 6d^1 7s^2 $

\n

Plutonium (Pu, atomic number 94) has the electronic configuration: \n$ [Rn] 5f^6 7s^2 $

\n

To find the highest oxidation state, we sum the number of $d$, $f$, and $s$ electrons in the valence shell.

\n

For uranium :\n

$ 3 (f) + 1 (d) + 2 (s) = 6 $\n

So the highest oxidation state of uranium is +6.

\n

For plutonium :\n

$ 6 (f) + 2 (s) = 8 $\n

So the highest oxidation state of plutonium is +8. However, in practice, plutonium is known to exhibit a highest oxidation state of +7.

\n

Comparing with the given options, the correct answer is :

\n

Option D : 6 and 7.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1196, "subject": "Chemistry", "question": "The maximum number of possible oxidation\nstates of actinoides are shown by :", "options": [ { "text": "nobelium (No) and lawrencium (Lr)" }, { "text": "actinium (Ac) and thorium (Th)" }, { "text": "neptunium (Np) and plutonium (Pu)" }, { "text": "berkelium (Bk) and californium (Cf)" } ], "answer": "neptunium (Np) and plutonium (Pu)", "solution": "**Answer:** neptunium (Np) and plutonium (Pu)\n\nActinoid 'Th' shows oxidation state = +4\n

Actinoid 'Ac' shows oxidation state = +3\n

Actinoid 'Pu' shows oxidation states = +3, +4, +5, +6, +7\n

Actinoid 'Np' shows oxidation states = +3, +4, +5, +6, +7\n

Actinoid 'Bk' shows oxidation states = +3, +4\n

Actinoid 'Cm' shows oxidation states = +3, +4\n

Actinoid 'Lr' shows oxidation states = +3\n

$$ \\therefore $$ Maximum oxidation state is shown by Np\nand Pu.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1197, "subject": "Chemistry", "question": "The correct order of atomic radii is :", "options": [ { "text": "Ho > N > Eu > Ce " }, { "text": "Ce > Eu > Ho > N " }, { "text": "N > Ce > Eu > Ho" }, { "text": "Eu > Ce > Ho > N " } ], "answer": "Eu > Ce > Ho > N ", "solution": "**Answer:** Eu > Ce > Ho > N \n\nIn lanthanide series from left to right atomic radius decreases but the exception is Europiun (Eu) and Ytterbium (Yb).\n

So the correct order is : Eu > Ce > Ho > N", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1198, "subject": "Chemistry", "question": "The effect of lanthanoid contraction in the lanthanoid series of elements by and large means : ", "options": [ { "text": "increase in atomic radii and decrease in ionic radii \n" }, { "text": "increase in both atomic and ionic radii\n" }, { "text": "decrease in both atomic and ionic radii " }, { "text": "decrease in atomic radii and increase in ionic radii" } ], "answer": "decrease in both atomic and ionic radii ", "solution": "**Answer:** decrease in both atomic and ionic radii \n\nDue to Lanthanoid contraction both atomic radii and ionic radii decreases gradually in the lanthanoid series. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1199, "subject": "Chemistry", "question": "The atomic radius of Ag is closed to :", "options": [ { "text": "Au" }, { "text": "Cu" }, { "text": "Hg" }, { "text": "Ni" } ], "answer": "Au", "solution": "**Answer:** Au\n\nAtomic radius of Ag and Au is nearly same due\nto lanthanide contraction.\n

In periodic table in group 1, 2, 3 from top to bottom radius increases but from group 4 to 12 size increases for 3d to 4d series elements but size of 4d and 5d series elements are approximately same.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1200, "subject": "Chemistry", "question": "The elements with atomic numbers 101 and 104\nbelong to, respectively :", "options": [ { "text": "Group 6 and Actinoids" }, { "text": "Actinoids and Group 4" }, { "text": "Group 11 and Group 4" }, { "text": "Actinoids and Group 6" } ], "answer": "Actinoids and Group 4", "solution": "**Answer:** Actinoids and Group 4\n\nActinoids contains 14 elements with atomic\nnumber 90 to 103. Hence element with atomic\nnumber 101 is Actinoids.\n

Element with atomic number 104 is a d-block\nelement of group 4.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1201, "subject": "Chemistry", "question": "The correct electronic configuration and spin-only
magnetic moment (BM) of Gd3+ (Z = 64),\nrespectively, are :", "options": [ { "text": "[Xe]5f7 and 8.9" }, { "text": "[Xe]4f7 and 7.9" }, { "text": "[Xe]5f7 and 7.9" }, { "text": "[Xe]4f7 and 8.9" } ], "answer": "[Xe]4f7 and 7.9", "solution": "**Answer:** [Xe]4f7 and 7.9\n\nGd3+ (Z = 64) = [Xe] 4f7\n

Magnetic moment ($$\\mu $$) = $$\\sqrt {n\\left( {n + 2} \\right)} $$ B.M\n

= $$\\sqrt {7\\left( {7 + 2} \\right)} $$\n

= 7.9 B.M\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1202, "subject": "Chemistry", "question": "In the sixth period, the orbitals that are filled are :", "options": [ { "text": "6s, 5d, 5f, 6p" }, { "text": "6s, 4f, 5d, 6p" }, { "text": "6s, 6p, 6d, 6f" }, { "text": "6s, 5f, 6d, 6p" } ], "answer": "6s, 4f, 5d, 6p", "solution": "**Answer:** 6s, 4f, 5d, 6p\n\nAs per (n + l) rule in 6th period, order of orbitals filling is 6s, 4f, 5d, 6p.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1203, "subject": "Chemistry", "question": "The lanthanoid that does NOT show +4 oxidation\nstate is :", "options": [ { "text": "Tb" }, { "text": "Dy" }, { "text": "Ce" }, { "text": "Eu" } ], "answer": "Eu", "solution": "**Answer:** Eu\n\nEuropium (Eu)\n
Atomic No = 63\n
Electronic configuration = [Xe]4f76s2\n
Can show only + 2 and + 3 oxidation state.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1204, "subject": "Chemistry", "question": "Mischmetal is an alloy consisting mainly of :", "options": [ { "text": "lanthanoid and actinoid metals" }, { "text": "actinoid and transition metals" }, { "text": "lanthanoid metals" }, { "text": "actinoid metals" } ], "answer": "lanthanoid metals", "solution": "**Answer:** lanthanoid metals\n\nMisch metal is an alloy consisting mainly of\nlanthanoid metals.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1205, "subject": "Chemistry", "question": "Given below are two statements :

Statement I : CeO2 can be used for oxidation of aldehydes and ketones.

Statement II : Aqueous solution of EuSO4 is a strong reducing agent.

In the light of the above statements, choose the correct answer from the options given below :", "options": [ { "text": "Statement I is true but Statement II is false." }, { "text": "Statement I is false but Statement II is true" }, { "text": "Both Statement I and Statement II are false" }, { "text": "Both Statement I and Statement II are true" } ], "answer": "Both Statement I and Statement II are true", "solution": "**Answer:** Both Statement I and Statement II are true\n\nThe +3 oxidation state of lanthanide is most\nstable and therefore lanthanide in +4 oxidation\nstate has strong tendence to gain e–\n and\nconverted into +3 and therefore act as strong\noxidizing agent.\neg Ce+4\n
And therefore CeO2\n is used to oxidized alcohol\naldehyde and ketones.\n

Lanthanide in +2 oxidation state has strong\ntendency to loss e–\n and converted into +3\noxidation state therefore act as strong reducing\nagent.\n

$$ \\therefore $$ EuSO4 act as a strong reducing agent.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1206, "subject": "Chemistry", "question": "Which one of the following lanthanoids does not form MO2? [M is lanthanoid metal]", "options": [ { "text": "Nd" }, { "text": "Dy" }, { "text": "Yb" }, { "text": "Pr" } ], "answer": "Yb", "solution": "**Answer:** Yb\n\nNd (60) = 4f4 6s2

\nPr (59) = 4f3 6s2

\nDy (66) = 4f10 6s2

\nYb (70) = 4f14 6s2

\nYb+2 has fully-filled 4f orbital, it will require very\nlarge amount of energy to reach +4 oxidation\nstate.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1207, "subject": "Chemistry", "question": "Given below are two statements :

Statement I : The Eo value for Ce4+ / Ce3+ is + 1.74 V.

Statement II : Ce is more stable in Ce4+ state than Ce3+ state.

In the light of the above statements, choose the most appropriate answer from the options given below :", "options": [ { "text": "Statement I is incorrect but statement II is correct" }, { "text": "Both statement I and statement II are correct" }, { "text": "Both statement I and statement II are incorrect" }, { "text": "Statement I is correct but statement II is incorrect" } ], "answer": "Statement I is correct but statement II is incorrect", "solution": "**Answer:** Statement I is correct but statement II is incorrect\n\nCe4+ $$\\buildrel {{e^ - }} \\over\n \\longrightarrow $$ Ce3+   E° = +1.74 V

\nPositive SRP and higher SRP means greater\noxidising power. So, Ce4+ wants to reduce to Ce3+.\nIndicates Ce4+ is less stable than Ce3+.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1208, "subject": "Chemistry", "question": "Given below are two statement : one is labelled as Assertion A and the other is labelled as Reason R :

Assertion A : Size of Bk3+ ion is less than Np3+ ion.

Reason R : The above is a consequence of the lanthanoid contraction.

In the light of the above statements, choose the correct answer from the options given below :", "options": [ { "text": "Both A and B are true and R is the correct explanation of A" }, { "text": "A is false but R is true" }, { "text": "A is true but R is false" }, { "text": "Both A and B are true but R is not the correct explanation of A" } ], "answer": "A is true but R is false", "solution": "**Answer:** A is true but R is false\n\nSize of Bk3+ is 98 pm

\nSize of Np3+ is 101 pm

\nSo size of Np3+ is more than Bk3+ ion.

\nthere is a gradual decrease in the size of M3+ ions\nacross the series. This may be referred to as the\nactinoid contraction.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1209, "subject": "Chemistry", "question": "Arrange the following metal complex/compounds in the increasing order of spin only magnetic moment. Presume all the three, high spin system.

(Atomic numbers Ce = 58, Gd = 64 and Eu = 63)

(a) (NH4)2[Ce(NO3)6]

(b) Gd(NO3)3 and

(c) Eu(NO3)3", "options": [ { "text": "(a) < (b) < (c)" }, { "text": "(b) < (a) < (c)" }, { "text": "(a) < (c) < (b)" }, { "text": "(c) < (a) < (b)" } ], "answer": "(a) < (c) < (b)", "solution": "**Answer:** (a) < (c) < (b)\n\n

(A) $${}_{58}Ce \\to [Xe]4{f^2}5{d^0}6{s^2}$$

\n

In complex $$C{e^{ + 4}} \\to [Xe]4{f^0}5{d^0}6{s^0}$$

\n

There is no unpaired electron, so, $$\\mu$$m = 0 BM

\n

(B) $${}_{64}Gd \\to [Xe]4{f^7}5{d^1}6{s^2}$$

\n

In complex, $${}_{64}G{d^{2 + }} \\to [Xe]4{f^7}5{d^1}$$

\n

There are 8 unpaired electrons.

\n

$$\\therefore$$ $${\\mu _m} = \\sqrt {8(8 + 2)} = \\sqrt {80} = 8.94BM$$

\n

(C) $${}_{63}Eu \\to [Xe]4{f^9}5{d^0}6{s^0}$$

\n

In complex $${}_{63}E{u^{ + 3}} \\to [Xe]4{f^6}5{d^0}6{s^0}$$

\n

contain six unpaired electrons.

\n

So, $${\\mu _m} = \\sqrt {6(6 + 2)} = \\sqrt {48} BM = 6.93BM$$

\n

Hence, order of spin only magnetic moment

\n

$$B > C > A$$.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1210, "subject": "Chemistry", "question": "Number of electrons present in 4f orbital of Ho3+ ion is ____________. (Given Atomic No. of Ho = 67)", "options": [], "answer": "10", "solution": "**Answer:** 10\n\nHo = [Xe]4f116s2

Ho3+ = [Xe] 4f10

So number of e$$-$$ present in 4f is 10.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1211, "subject": "Chemistry", "question": "The number of 4f electrons in the ground statement electronic configuration of Gd2+ is ___________. [Atomic number of Gd = 64]", "options": [], "answer": "7", "solution": "**Answer:** 7\n\nThe electronic configuration of

$$_{64}Gd:[Xe]4{f^7}5{d^1}6{s^2}$$

So, the electronic configuration of

$$_{64}G{d^{2 + }}:[Xe]4{f^7}5{d^1}6{s^0}$$

i.e. the number of 4f electrons in the ground state electronic configuration of $$G{d^{2 + }}$$ is 7.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1212, "subject": "Chemistry", "question": "The number of f electrons in the ground state electronic configuration of Np (Z = 93) is ___________. (Nearest integer)", "options": [], "answer": "18", "solution": "**Answer:** 18\n\nNp = 1s2 22 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2

4f14 5d10 6p6 7s2 5f4 6d1

Total No. of 'f' electron = 14e$$-$$ + 4e$$-$$ = 18", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1213, "subject": "Chemistry", "question": "Which one of the following lanthanides exhibits +2 oxidation state with diamagnetic nature? (Given Z for Nd = 60, Yb = 70, La = 57, Ce = 58)", "options": [ { "text": "Nd" }, { "text": "Yb" }, { "text": "La" }, { "text": "Ce" } ], "answer": "Yb", "solution": "**Answer:** Yb\n\nYb (70) = 4f14 6s2\n

Yb+2 = 4f14 6s0\n

$$ \\because $$ All the electrons are paired hence Yb+2 is\ndiamagnetic.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1214, "subject": "Chemistry", "question": "The Eu2+ ion is a strong reducing agent in spite of its ground state electronic configuration (outermost) : [Atomic number of Eu = 63]", "options": [ { "text": "4f76s2" }, { "text": "4f6" }, { "text": "4f7" }, { "text": "4f66s2" } ], "answer": "4f7", "solution": "**Answer:** 4f7\n\nEu $$\\to$$ [Xe]4f76s2

Eu2+ $$\\to$$ [Xe]4f7", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1215, "subject": "Chemistry", "question": "

Which one of the lanthanoids given below is the most stable in divalent form?

", "options": [ { "text": "Ce (Atomic Number 58)" }, { "text": "Sm (Atomic number 62)" }, { "text": "Eu (Atomic Number 63)" }, { "text": "Yb (Atomic Number 70)" } ], "answer": "Eu (Atomic Number 63)", "solution": "**Answer:** Eu (Atomic Number 63)\n\n

The stability of a divalent state can often be attributed to the stability of the electron configuration of that state. Among the options given, we can evaluate the electron configurations of each element in their divalent state to find out which is the most stable. The electronic configurations of the neutral atoms are :

\n\n

In their divalent state (+2 oxidation state), the electron configurations would be :

\n\n

Among these, Eu in its divalent state has a half-filled (4f) shell, which offers extra stability due to symmetrical electron distribution. Therefore, Eu (Atomic Number 63) is the most stable in the divalent form.

\n

So the correct answer is Option C : Eu (Atomic Number 63).

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1216, "subject": "Chemistry", "question": "The 'f' orbitals are half and completely filled, respectively in lanthanide ions :\n

[Given : Atomic no. Eu, 63; Sm, 62; Tm, 69; Tb, 65; Yb, 70; Dy, 66]

", "options": [ { "text": "Eu2+ and Tm2+" }, { "text": "Sm2+ and Tm3+" }, { "text": "Tb4+ and Yb2+" }, { "text": "Dy3+ and Yb3+" } ], "answer": "Tb4+ and Yb2+", "solution": "**Answer:** Tb4+ and Yb2+\n\nTb $$ \\to $$ 4f9 6s2

\nTb+4 $$ \\to $$ 4f7

\nYb $$ \\to $$ 4f14 6s2

\nYb+2 $$ \\to $$ 4f14

\nHence, the pair Tb+4 and Yb+2 have half filled and\ncompletely filled f subshells respectively.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1217, "subject": "Chemistry", "question": "

The most common oxidation state of Lanthanoid elements is +3. Which of the following is likely to deviate easily from +3 oxidation state?

", "options": [ { "text": "Ce (At. No. 58)" }, { "text": "La (At. No. 57)" }, { "text": "Lu (At. No. 71)" }, { "text": "Gd (At. No. 64)" } ], "answer": "Ce (At. No. 58)", "solution": "**Answer:** Ce (At. No. 58)\n\nCe → [Xe]4f1 5d1 6s2

\nCe+4 → [xe] 4f° 5d° 6s°

\nCerium in +4 oxidation state acquires inert gas\nconfiguration.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1218, "subject": "Chemistry", "question": "

Cerium (IV) has a noble gas configuration. Which of the following is correct statement about it?

", "options": [ { "text": "It will not prefer to undergo redox reactions." }, { "text": "It will prefer to gain electron and act as an oxidizing agent." }, { "text": "It will prefer to give away an electron and behave as reducing agent." }, { "text": "It acts as both, oxidizing and reducing agent." } ], "answer": "It will prefer to gain electron and act as an oxidizing agent.", "solution": "**Answer:** It will prefer to gain electron and act as an oxidizing agent.\n\nCerium exists in two different oxidation state $$+3$$, $$+4$$

\n$$\n\\begin{array}{ll}\n\\mathrm{Ce}^{+4}+\\mathrm{e}^{-} \\rightarrow \\mathrm{Ce}^{3+} & \\mathrm{E}^{0}=+1.61 \\mathrm{~V} \\\\\n\\mathrm{Ce}^{+3}+3 \\mathrm{e}^{-} \\rightarrow \\mathrm{Ce} & \\mathrm{E}^{0}=-2.336 \\mathrm{~V}\n\\end{array}\n$$

\nIt shows $$\\mathrm{Ce}^{+4}$$ acts as a strong oxidising agent & accepts electron.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1219, "subject": "Chemistry", "question": "

Which one amongst the following are good oxidizing agents?

\n

A. Sm$$^{2+}$$

\n

B. Ce$$^{2+}$$

\n

C. Ce$$^{4+}$$

\n

D. Tb$$^{4+}$$

\n

Choose the most appropriate answer from the options given below :

", "options": [ { "text": "C and D only" }, { "text": "A and B only" }, { "text": "D only" }, { "text": "C only" } ], "answer": "C and D only", "solution": "**Answer:** C and D only\n\nThe most appropriate answer is $C$ and $D$ only. $Ce^{4+}$ and $Tb^{4+}$ are good oxidizing agents because they have a high oxidation state and a strong tendency to gain electrons. $Sm^{2+}$ and $Ce^{2+}$ are reducing agents because they have a low oxidation state and a tendency to lose electrons.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1220, "subject": "Chemistry", "question": "

The pair of lanthanides in which both elements have high third - ionization energy is :

", "options": [ { "text": "Dy, Gd" }, { "text": "Lu, Yb" }, { "text": "Eu, Yb" }, { "text": "Eu, Gd" } ], "answer": "Eu, Yb", "solution": "**Answer:** Eu, Yb\n\n1. Eu²⁺: [Xe] 4f⁷

\n2. Yb²⁺: [Xe] 4f¹⁴\n

\nEu²⁺ has a half-filled 4f orbital, while Yb²⁺ has a fully-filled 4f orbital. Both of these configurations lead to increased stability due to the symmetrical distribution of electrons in the 4f subshell, resulting in higher ionization energies for both Europium (Eu) and Ytterbium (Yb).\n

\nThus, the correct answer is Eu (Europium) and Yb (Ytterbium), as both elements have high third ionization energies due to their half-filled and fully-filled 4f subshell configurations in their divalent cation forms.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1221, "subject": "Chemistry", "question": "

Given below are two statement: one is labelled as Assertion $$\\mathbf{A}$$ and the other is labelled as Reason $$\\mathbf{R}$$

\n

Assertion A: $$5 \\mathrm{f}$$ electrons can participate in bonding to a far greater extent than $$4 \\mathrm{f}$$ electrons

\n

Reason R: $$5 \\mathrm{f}$$ orbitals are not as buried as $$4 \\mathrm{f}$$ orbitals

\n

In the light of the above statements, choose the correct answer from the options given below

", "options": [ { "text": "Both A and R are true and R is the correct explanation of A" }, { "text": "Both A and R are true but R is NOT the correct explanation of A" }, { "text": "A is true but R is false" }, { "text": "A is false but R is true" } ], "answer": "Both A and R are true and R is the correct explanation of A", "solution": "**Answer:** Both A and R are true and R is the correct explanation of A\n\n

Assertion A: 5f electrons can participate in bonding to a far greater extent than 4f electrons

\n

This is correct. In actinides (elements in which 5f orbitals are being filled), the 5f electrons can participate in bonding. This contrasts with lanthanides (elements in which 4f orbitals are being filled), where the 4f electrons largely do not participate in bonding.

\n

Reason R: 5f orbitals are not as buried as 4f orbitals

\n

This is also correct. The term "buried" refers to the degree to which an orbital is shielded from the outside environment by other electrons. The 5f orbitals are less shielded, or less "buried," than the 4f orbitals. This means they are more exposed to the outside environment and can more readily interact with other atoms to form bonds. This is the reason why 5f electrons can participate in bonding to a greater extent than 4f electrons.

\n

Therefore, both A and R are true, and R is the correct explanation of A.

\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1222, "subject": "Chemistry", "question": "

Strong reducing and oxidizing agents among the following, respectively, are :

", "options": [ { "text": "$$\\mathrm{Ce}^{4+}$$ and $$\\mathrm{Eu}^{2+}$$" }, { "text": "$$\\mathrm{Eu}^{2+}$$ and $$\\mathrm{Ce}^{4+}$$" }, { "text": "$$\\mathrm{Ce}^{4+}$$ and $$\\mathrm{Tb}^{4+}$$" }, { "text": "$$\\mathrm{Ce}^{3+}$$ and $$\\mathrm{Ce}^{4+}$$" } ], "answer": "$$\\mathrm{Eu}^{2+}$$ and $$\\mathrm{Ce}^{4+}$$", "solution": "**Answer:** $$\\mathrm{Eu}^{2+}$$ and $$\\mathrm{Ce}^{4+}$$\n\n

Strong reducing agents are those that readily donate electrons, and strong oxidizing agents are those that readily accept electrons.

\n\n

So the correct answer is: $\\mathrm{Eu}^{2+}$ and $\\mathrm{Ce}^{4+}$.

\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1223, "subject": "Chemistry", "question": "

The electronic configuration for Neodymium is:

\n

[Atomic Number for Neodymium 60]

", "options": [ { "text": "$$[\\mathrm{Xe}] 4 \\mathrm{f}^4 6 \\mathrm{~s}^2$$\n" }, { "text": "$$[\\mathrm{Xe}] 5 \\mathrm{f}^4 7 \\mathrm{~s}^2$$\n" }, { "text": "$$[\\mathrm{Xe}] 4 \\mathrm{f}^6 6 \\mathrm{~s}^2$$\n" }, { "text": "$$[\\mathrm{Xe}] 4 \\mathrm{f}^1 5 \\mathrm{~d}^1 6 \\mathrm{~s}^2$$" } ], "answer": "$$[\\mathrm{Xe}] 4 \\mathrm{f}^4 6 \\mathrm{~s}^2$$\n", "solution": "**Answer:** $$[\\mathrm{Xe}] 4 \\mathrm{f}^4 6 \\mathrm{~s}^2$$\n\n\n

The electronic configuration of an element is determined by adding electrons to atomic orbitals following the Aufbau principle, the Pauli exclusion principle, and Hund's rule. For atoms with atomic numbers higher than that of xenon (Xe, atomic number 54), the electrons start filling the outer orbitals following the xenon core. In this case, we're looking at neodymium (Nd), which has an atomic number of 60.

\n\n

To find the electronic configuration of neodymium, we first note the configuration of xenon, which serves as the core, and then add the additional electrons beyond xenon to the relevant orbitals.

\n\n

Neodymium has 60 - 54 = 6 electrons more than xenon. These additional electrons will fill the 4f and 6s orbitals. Based on its position in the periodic table, neodymium’s electrons start filling the 4f orbitals before the 5d. Since the 6s orbital is at a lower energy level than the 4f orbital, it gets filled before the 4f. Neodymium will have 2 electrons in the 6s orbital, and the remaining 4 electrons will go into the 4f orbital.

\n\n

Therefore, the correct electronic configuration for neodymium is:

\n$$[\\mathrm{Xe}] 4 \\mathrm{f}^4 6 \\mathrm{~s}^2$$\n\n

So, Option A is the correct answer.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1224, "subject": "Chemistry", "question": "

Given below are two statements :

\n

Statement (I) : In the Lanthanoids, the formation $$\\mathrm{Ce}^{+4}$$ is favoured by its noble gas configuration.

\n

Statement (II) : $$\\mathrm{Ce}^{+4}$$ is a strong oxidant reverting to the common +3 state.

\n

In the light of the above statements, choose the most appropriate answer from the options given below :

", "options": [ { "text": "Both Statement I and Statement II are false\n" }, { "text": "Statement I is true but Statement II is false\n" }, { "text": "Both Statement I and Statement II are true\n" }, { "text": "Statement I is false but Statement II is true" } ], "answer": "Both Statement I and Statement II are true\n", "solution": "**Answer:** Both Statement I and Statement II are true\n\n\n

Statement (1) is true, $$\\mathrm{Ce}^{+4}$$ has noble gas electronic configuration.

\n

Statement (2) is also true due to high reduction potential for $$\\mathrm{Ce}^{4+} / \\mathrm{Ce}^{3+}(+1.74 \\mathrm{~V})$$, and stability of $$\\mathrm{Ce}^{3+}, \\mathrm{Ce}^{4+}$$ acts as strong oxidizing agent.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1225, "subject": "Chemistry", "question": "

Total number of ions from the following with noble gas configuration is _________.

$$\\mathrm{Sr}^{2+}(z=38), \\mathrm{Cs}^{+}(z=55), \\mathrm{La}^{2+}(z=57), \\mathrm{Pb}^{2+}(z=82), \\mathrm{Yb}^{2+}(z=70)$$ and $$\\mathrm{Fe}^{2+}(z=26)$$

", "options": [], "answer": "2", "solution": "**Answer:** 2\n\n

To determine which of these ions have a noble gas configuration, we must consider what the phrase \"noble gas configuration\" means. Atoms or ions with a noble gas configuration have completely filled electron shells, similar to the electron configuration of the noble gases, which are the elements found in Group 18 of the periodic table. Noble gases are helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), and radon (Rn).

\n\n

Let's examine each ion given:

\n\n\n
${\\left[\\mathrm{Fe}^{2+}\\right]=[\\mathrm{Ar}] 3 \\mathrm{~d}^6}$\n\n

Therefore, the ions $$\\mathrm{Sr}^{2+}$$ and $$\\mathrm{Cs}^{+}$$ have a noble gas configuration. Altogether, there are 2 ions with a noble gas configuration.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1226, "subject": "Chemistry", "question": "

Which of the following acts as a strong reducing agent? (Atomic number: $$\\mathrm{Ce}=58, \\mathrm{Eu}=63, \\mathrm{Gd}=64, \\mathrm{Lu}=71$$)

", "options": [ { "text": "$$\\mathrm{Eu}^{2+}$$\n" }, { "text": "$$\\mathrm{Gd}^{3+}$$\n" }, { "text": "$$\\mathrm{Lu}^{3+}$$\n" }, { "text": "$$\\mathrm{Ce}^{4+}$$" } ], "answer": "$$\\mathrm{Eu}^{2+}$$\n", "solution": "**Answer:** $$\\mathrm{Eu}^{2+}$$\n\n\n$\\mathrm{Eu}^{2+}$ is a strong reducing agent changing to the common +3 state while all other are provided at their higher oxidation states.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1227, "subject": "Chemistry", "question": "

Diamagnetic Lanthanoid ions are :

", "options": [ { "text": "$$\\mathrm{La}^{3+} \\& \\mathrm{~Ce}^{4+}$$\n" }, { "text": "$$\\mathrm{Nd}^{3+} \\& \\mathrm{~Ce}^{4+}$$\n" }, { "text": "$$\\mathrm{Lu}^{3+} \\& \\mathrm{~Eu}^{3+}$$\n" }, { "text": "$$\\mathrm{Nd}^{3+} \\& \\mathrm{~Eu}^{3+}$$" } ], "answer": "$$\\mathrm{La}^{3+} \\& \\mathrm{~Ce}^{4+}$$\n", "solution": "**Answer:** $$\\mathrm{La}^{3+} \\& \\mathrm{~Ce}^{4+}$$\n\n\n

$$\\mathrm{Ce}:[\\mathrm{Xe}] 4 \\mathrm{f}^1 5 \\mathrm{~d}^1 6 \\mathrm{~s}^2 ; \\mathrm{Ce}^{4+}$$ diamagnetic

\n

$$\\mathrm{La}:[\\mathrm{Xe}] 4 \\mathrm{f}^0 5 \\mathrm{~d}^1 6 \\mathrm{~s}^2 ; \\mathrm{La}^{3+}$$ diamagnetic

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1228, "subject": "Chemistry", "question": "

The electronic configuration of Einsteinium is :

\n

(Given atomic number of Einsteinium $$=99$$)

", "options": [ { "text": "$$[\\mathrm{Rn}] 5 \\mathrm{f}^{13} 6 \\mathrm{~d}^{\\circ} 7 \\mathrm{~s}^2$$\n" }, { "text": "$$[\\mathrm{Rn}] 5 \\mathrm{f}^{12} 6 \\mathrm{~d}^{\\circ} 7 \\mathrm{~s}^2$$\n" }, { "text": "$$[\\mathrm{Rn}] 5 \\mathrm{f}^{11} 6 \\mathrm{~d}^{\\circ} 7 \\mathrm{~s}^2$$\n" }, { "text": "$$[\\mathrm{Rn}] 5 \\mathrm{f}^{10} 6 \\mathrm{~d}^{\\circ} 7 \\mathrm{~s}^2$$" } ], "answer": "$$[\\mathrm{Rn}] 5 \\mathrm{f}^{11} 6 \\mathrm{~d}^{\\circ} 7 \\mathrm{~s}^2$$\n", "solution": "**Answer:** $$[\\mathrm{Rn}] 5 \\mathrm{f}^{11} 6 \\mathrm{~d}^{\\circ} 7 \\mathrm{~s}^2$$\n\n\n

The correct option for the electronic configuration of Einsteinium (atomic number 99) is:

\n\nOption C: [Rn] 5f11 6d0 7s2\n\n

Here's why:

\n\n
    \n
  1. Atomic Number and Electrons: Einsteinium has an atomic number of 99, which means it has 99 protons in its nucleus. Since atoms are neutral, it also has 99 electrons surrounding the nucleus.

    2. \n
  2. Electron Shells: Electrons fill up energy shells around the nucleus. The order of filling follows the Aufbau principle, which generally fills lower energy shells before moving to higher ones.

    3. \n
  3. Noble Gas Core Notation: The notation [Rn] represents the filled electron configuration of Radon (atomic number 86). This is a shorthand to avoid writing out the entire configuration of the inner shells, as they are not involved in the chemical behavior of Einsteinium.

    4. \n
  4. Filling the f-orbital: Einsteinium is an actinide element, located in the f-block of the periodic table. The 5f subshell fills up after the 6s subshell.

    5. \n
  5. Aufbau Principle: According to the Aufbau principle, the 5f subshell can hold up to 14 electrons. In Einsteinium, 11 electrons fill the 5f subshell (5f11).

    6. \n
  6. Empty d-orbital: The 6d subshell has a higher energy level than the 5f subshell. Since the 5f subshell isn't completely filled, the 6d subshell remains empty (6d0).

    7. \n
  7. Outermost Electrons: The remaining two electrons fill the outermost 7s subshell (7s2).
    8. \n
\n

Therefore, the complete electronic configuration of Einsteinium is:

\n\n[Rn] 5f11 6d0 7s2", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1229, "subject": "Chemistry", "question": "

Number of colourless lanthanoid ions among the following is __________.

$$\\mathrm{Eu}^{3+}, \\mathrm{Lu}^{3+}, \\mathrm{Nd}^{3+}, \\mathrm{La}^{3+}, \\mathrm{Sm}^{3+}$$

", "options": [], "answer": "2", "solution": "**Answer:** 2\n\n

The color of lanthanoid ions in solutions is mainly due to the electronic transitions within the 4f subshell. The lanthanoid ions are more likely to be colorless when they have fully filled (with 14 electrons) or completely empty (with 0 electrons) f orbitals because, in these cases, there are no electrons to undergo f-f transitions, and as a result, no absorption of visible light occurs leading to colorlessness.

\n\n

Let's consider the electronic configurations of the lanthanoid ions provided:

\n\n

- $$\\mathrm{Eu}^{3+}$$: Europium (Eu) has an atomic number of 63. Neutral europium ([Xe]4f7 6s^2) loses three electrons to form Eu3+, leaving it with an electronic configuration equivalent to [Xe]4f6. With 6 electrons in the f orbital, it can undergo f-f transitions, thus it is not colorless.

\n\n

- $$\\mathrm{Lu}^{3+}$$: Lutetium (Lu) has an atomic number of 71. In its 3+ ionic state, lutetium has lost its 6s and 5d electrons and is left with a completely filled 4f orbital ([Xe] 4f14). This configuration cannot allow for any f-f transitions, as there are no available energy levels within the f orbital for an electron to jump to, making Lu3+ colorless.

\n\n

- $$\\mathrm{Nd}^{3+}$$: Neodymium (Nd) has an atomic number of 60. In its 3+ state ([Xe] 4f3), it clearly has partially filled f orbitals, which can absorb visible light for f-f transitions, so it is not colorless.

\n\n

- $$\\mathrm{La}^{3+}$$: Lanthanum (La) has an atomic number of 57. In its 3+ ionic state, it has a configuration of [Xe], meaning that its 4f orbital is completely empty. Since there are no electrons in the f orbital to undergo f-f transitions, La3+ is colorless.

\n\n

- $$\\mathrm{Sm}^{3+}$$: Samarium (Sm) has an atomic number of 62. As a 3+ ion ([Xe] 4f5), it too has electrons in the f orbital capable of undergoing f-f transitions, so it is not colorless.

\n\n

From the analysis, the colorless lanthanoid ions among the ones listed are $$\\mathrm{Lu}^{3+}$$ and $$\\mathrm{La}^{3+}$$.

\n\n

Therefore, the number of colorless lanthanoid ions among the given options is 2.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1230, "subject": "Chemistry", "question": "

The number of element from the following that do not belong to lanthanoids is $$\\mathrm{Eu}, \\mathrm{Cm}, \\mathrm{Er}, \\mathrm{Tb}, \\mathrm{Yb}$$ and $$\\mathrm{Lu}$$

", "options": [ { "text": "3" }, { "text": "1" }, { "text": "4" }, { "text": "5" } ], "answer": "1", "solution": "**Answer:** 1\n\n

To determine the number of elements from the given list that do not belong to the lanthanoids, let's first define what lanthanoids are. The lanthanoids consist of the 15 chemical elements in the periodic table from Lanthanum ($\\mathrm{La}$) with atomic number 57 to Lutetium ($\\mathrm{Lu}$) with atomic number 71. Thus, lanthanoids include:

\n\n\n\n

Given the list of elements: Eu (Europium), Cm (Curium), Er (Erbium), Tb (Terbium), Yb (Ytterbium), and Lu (Lutetium), we can now check which of these are not part of the lanthanoids:

\n\n\n\n

From the given list, only Cm (Curium) does not belong to the lanthanoids, which means the correct answer is:

\n\n

Option B: 1

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1231, "subject": "Chemistry", "question": "Which of the following ions has the maximum magnetic moment?", "options": [ { "text": "Mn2+" }, { "text": "Fe2+" }, { "text": "Ti2+" }, { "text": "Cr2+" } ], "answer": "Mn2+", "solution": "**Answer:** Mn2+\n\n$$M{n^{ + + }} - 5\\,\\,$$ unpaired electrons\n

$$F{e^{ + + }} - 4\\,\\,$$ unpaired electrons\n

$$T{i^{ + + }}$$$$ - 2\\,\\,$$ unpaired electrons \n

$$C{r^{ + + }} - 4\\,\\,$$ unpaired electrons\n

hence maximum no. of unpaired electron is present in $$M{n^{ + + }}.$$ \n

NOTE : Magnetic moment $$ \\propto $$ number of unpaired electrons", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1232, "subject": "Chemistry", "question": "For making good quality mirrors, plates of float glass are used. These are obtained by floating molten glass\nover a liquid metal which does not solidify before glass. The metal used can be :", "options": [ { "text": "tin" }, { "text": "sodium" }, { "text": "magnesium" }, { "text": "mercury" } ], "answer": "mercury", "solution": "**Answer:** mercury\n\nIt is mercury because it exists as liquid at room temperature.", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1233, "subject": "Chemistry", "question": "The number of d-electrons retained in Fe2+ (At no of Fe = 26) ion is :", "options": [ { "text": "4" }, { "text": "5" }, { "text": "6" }, { "text": "3" } ], "answer": "6", "solution": "**Answer:** 6\n\nFe (Z = 26) $$\\to$$ e- = 26

\nFe2+ (Z = 26) $$\\to$$ e- = 24

\nElectronic configuration:

\nFe (Z = 26) $$\\to$$ 1s2 2s2 2p6 3s2 3p6 3d6 4s2

\nFe2+ (Z = 26) $$\\to$$ 1s2 2s2 2p6 3s2 3p6 3d6 4s0

\nas 4s is the outermost shell of Fe that's why e- will always removed first from 4s, so retained e- in d-shell is 6 ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1234, "subject": "Chemistry", "question": "A red solid is insolvable in water. However it becomes soluble if some $$K{\\rm I}$$ is added to water. Heating the red solid in a test tube results in liberation of some violet colored fumes and droplets of a metal appear on the cooler parts of the test tube. The red solid is :", "options": [ { "text": "$$Hg{{\\rm I}_2}$$ " }, { "text": "$$HgO$$ " }, { "text": "$$P{b_3}{O_4}$$ " }, { "text": "$${\\left( {N{H_4}} \\right)_2}C{r_2}{O_7}$$ " } ], "answer": "$$Hg{{\\rm I}_2}$$ ", "solution": "**Answer:** $$Hg{{\\rm I}_2}$$ \n\nWhen $$K{\\rm I}$$ is added to mercuric iodide it dissolve in it and form complex.\n

$$\\mathop {Hg{{\\rm I}_2}}\\limits_{red,solid(inso{\\mathop{\\rm lub}} le)} \\,\\, + \\,\\,\\,K{\\rm I} \\to \\mathop {{K_2}\\left[ {Hg{{\\rm I}_4}} \\right]}\\limits_{(soluble)} $$ \n

On heating $$Hg{{\\rm I}_2}$$ decomposes as \n

$$Hg{{\\rm I}_2}\\,\\,\\,\\,Hg\\,\\,\\rightleftharpoons\\, + \\mathop {{{\\rm I}_2}}\\limits_{(violetvapours)} $$", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1235, "subject": "Chemistry", "question": "Which one of the following statements is correct?", "options": [ { "text": "From a mixed precipitate of AgCl and AgI, ammonia solution dissolves only AgCl" }, { "text": "Ferric ions give a deep green precipitate on adding potassium ferrocyanide solution" }, { "text": "On boiling a solution having K+, Ca2+ and HCO3 ions we get a precipitate of K2Ca(CO3)2." }, { "text": "Manganese salts give a violet borax bead test in the reducing flame" } ], "answer": "From a mixed precipitate of AgCl and AgI, ammonia solution dissolves only AgCl", "solution": "**Answer:** From a mixed precipitate of AgCl and AgI, ammonia solution dissolves only AgCl\n\nBetween AgCl and AgI, AgI is less soluble, hence ammonia can dissolve ppt. of AgCl only\ndue to formation of complex as given below:\n

AgCl + 2NH3 $$ \\to $$ [Ag(NH3)2]Cl", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1236, "subject": "Chemistry", "question": "The atomic numbers of vanadium (V), Chromium (Cr), manganese (Mn) and iron (Fe) are respectively 23, 24, 25 and 26. Which one of these may be expected to have the highest second ionization enthalpy?", "options": [ { "text": "Cr" }, { "text": "Mn" }, { "text": "Fe" }, { "text": "V" } ], "answer": "Cr", "solution": "**Answer:** Cr\n\n

Due to stable electronic configuration of Cr+1 ion (3d5)

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1237, "subject": "Chemistry", "question": "Of the following outer electronic configurations of atoms, the highest oxidation state is achieved by which one of them?", "options": [ { "text": "$$\\left( {n - 1} \\right){d^3}n{s^2}$$ " }, { "text": "$$\\left( {n - 1} \\right){d^5}n{s^1}$$ " }, { "text": "$$\\left( {n - 1} \\right){d^8}n{s^2}$$ " }, { "text": "$$\\left( {n - 1} \\right){d^5}n{s^2}$$ " } ], "answer": "$$\\left( {n - 1} \\right){d^5}n{s^2}$$ ", "solution": "**Answer:** $$\\left( {n - 1} \\right){d^5}n{s^2}$$ \n\n$$\\left( {n - 1} \\right){d^5}n{s^2}\\,\\,$$ attains the maximum $$O.S.$$ of $$+7$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1238, "subject": "Chemistry", "question": "On heating mixture of $$C{u_2}O$$ and $$C{u_2}S$$ will give :", "options": [ { "text": "$$C{u_2}S{O_3}$$ " }, { "text": "$$CuO + CuS$$ " }, { "text": "$$Cu + S{O_3}$$" }, { "text": "$$Cu + S{O_2}$$ " } ], "answer": "$$Cu + S{O_2}$$ ", "solution": "**Answer:** $$Cu + S{O_2}$$ \n\n$$2C{u_2}O + C{u_2}S\\buildrel \\, \\over\n \\longrightarrow 6Cu + S{O_2}$$ self reduction \n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1239, "subject": "Chemistry", "question": "Calomel $$\\left( {H{g_2}C{l_2}} \\right)$$ on reaction with ammonium hydroxide gives :", "options": [ { "text": "$$HgO$$ " }, { "text": "$$H{g_2}O$$ " }, { "text": "$$N{H_2} - Hg - Hg - Cl$$ " }, { "text": "$$HgN{H_2}Cl$$ " } ], "answer": "$$HgN{H_2}Cl$$ ", "solution": "**Answer:** $$HgN{H_2}Cl$$ \n\n$$H{g_2}C{l_2} + 2N{H_4}OH\\buildrel \\, \\over\n \\longrightarrow HgN{H_2}Cl + N{H_4}Cl + 2{H_2}O$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1240, "subject": "Chemistry", "question": "The “spin-only” magnetic moment [in units of Bohr magneton, (µB)] of Ni2+ in aqueous solution would\nbe : (Atomic number of Ni = 28) ", "options": [ { "text": "2.84" }, { "text": "4.90" }, { "text": "0 " }, { "text": "1.73" } ], "answer": "2.84", "solution": "**Answer:** 2.84\n\nThe number of unpaired electrons in $$N{i^{2 + }}\\left( {aq} \\right) = 2$$ \n

Water is weak ligand hence no pairing will take place spin magnetic moment \n

$$\\eqalign{\n & = \\sqrt {n\\left( {n + 2} \\right)} = \\sqrt {2\\left( {2 + 2} \\right)} \\cr \n & = \\sqrt 8 = 2.82 \\cr} $$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1241, "subject": "Chemistry", "question": "A metal, $$M$$ forms chlorides in its $$+2$$ and $$+4$$ oxidation states. Which of the following statements about these chlorides is correct?", "options": [ { "text": "$$MC{l_2}$$ is more ionic than $$MC{l_4}$$" }, { "text": "$$MC{l_2}$$ is more easily hydrolysed than $$MC{l_4}$$ " }, { "text": "$$MC{l_2}$$ is more volatile than $$MC{l_4}$$" }, { "text": "$$MC{l_2}$$ is more soluble in anhydrous ethanol than $$MC{l_4}$$." } ], "answer": "$$MC{l_2}$$ is more ionic than $$MC{l_4}$$", "solution": "**Answer:** $$MC{l_2}$$ is more ionic than $$MC{l_4}$$\n\nMetal atom in the lower oxidation state forms the ionic bond and in the higher oxidation state the covalent bond. Because higher oxidation state means small size and great polarizing power and hence greater the covalent character. Hence $$MC{l_2}$$ is more ionic than $$MC{l_4}.$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1242, "subject": "Chemistry", "question": "In context with the transition elements, which of the following statements is incorrect ? ", "options": [ { "text": "In addition to the normal oxidation states, the zero oxidation state is also shown by these elements in complexes." }, { "text": "In the highest oxidation states, the transition metal show basic character and form cationic complexes." }, { "text": "In the highest oxidation states of the first five transition elements (Sc to Mn), all the 4s and 3d electrons are used for bonding" }, { "text": "Once the d5 configuration is exceeded, the tendency to involve all the 3d electrons in bonding\n decreases. " } ], "answer": "In the highest oxidation states, the transition metal show basic character and form cationic complexes.", "solution": "**Answer:** In the highest oxidation states, the transition metal show basic character and form cationic complexes.\n\nLower oxidation state of an element forms more basic oxide and hydroxide, while the higher oxidation state will form more acidic oxide/hydroxide. For example, \n

\"AIEEE ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1243, "subject": "Chemistry", "question": "Iron exhibits + 2 and +3 oxidation states. Which of the following statements about iron is incorrect?", "options": [ { "text": "Ferrous oxide is more basic in nature than the ferric oxide" }, { "text": "Ferrous compounds are relatively more ionic than the corresponding ferric compounds" }, { "text": "Ferrous compounds are less volatile than the corresponding ferric compounds" }, { "text": "Ferrous compounds are more easily hydrolysed than the corresponding ferric compounds." } ], "answer": "Ferrous compounds are more easily hydrolysed than the corresponding ferric compounds.", "solution": "**Answer:** Ferrous compounds are more easily hydrolysed than the corresponding ferric compounds.\n\n$$F{e^{3 + }}$$ is easily hydrolysed than $$F{e^{2 + }}$$ due to more positive charge.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1244, "subject": "Chemistry", "question": "Which of the following arrangements does not represent the correct order of the property stated against it? ", "options": [ { "text": "Ni2+ < Co2+ < Fe2+ < Mn2+ : ionic size " }, { "text": "Co3+ < Fe3+ < Cr3+ < Sc3+ : stability in aqueous solution " }, { "text": "Sc < Ti < Cr < Mn : number of oxidation states" }, { "text": "V2+ < Cr2+ < Mn2+ < Fe2+ : paramagnetic behaviour" } ], "answer": "V2+ < Cr2+ < Mn2+ < Fe2+ : paramagnetic behaviour", "solution": "**Answer:** V2+ < Cr2+ < Mn2+ < Fe2+ : paramagnetic behaviour\n\n(1)\n

\n\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n
V = 3d34s2V2+ = 3d3= 3 unpaired electron
Cr = 3d54s1Cr2+ = 3d4= 4 unpaired electron
Mn = 3d54s2Mn2+ = 3d5= 5 unpaired electron
Fe = 3d64s2Fe2+ = 3d6= 4 unpaired electron
\n

hence the correct order of paramagnetic behaviour \n

$${V^{2 + }} < Cr{}^{2 + } = F{e^{2 + }} < M{n^{2 + }}$$ \n

(2) $$\\,\\,\\,\\,$$ \n

For the same oxidation state, the-ionic radii generally $$dc$$-creases as the atomic number increases in a particular transition series.\n

Hence the order is \n

$$M{n^{ + + }} > F{e^{ + + }} > C{o^{ + + }} > N{i^{ + + }}$$ \n

(3) \n

In solution, the stability of the compounds depends upon electrode potentials, $$SEP$$ of the transitions metal ions are given as \n

$$C{o^{3 + }}/Co = + 1.97,\\,\\,\\,F{e^{3 + }}/Fe = + 0.77;$$\n

$$C{r^{3 + }}/C{r^{2 + }} = - 0.41,\\,\\,S{c^{3 + }}$$ is highly stable as it does not show $$+2$$ $$O.$$ $$S.$$\n

(4) \n

$$Sc - \\left( { + 2} \\right),\\left( { + 3} \\right)$$\n

$$Ti - \\left( { + 2} \\right),\\left( { + 3} \\right),\\left( { + 4} \\right)$$\n

$$Cr - \\left( { + 1} \\right),\\left( { + 2} \\right),\\left( { + 3} \\right),\\left( { + 4} \\right),\\left( { + 5} \\right),\\left( { + 6} \\right)$$\n

$$Mn - \\left( { + 2} \\right),\\left( { + 3} \\right),\\left( { + 4} \\right),\\left( { + 5} \\right),\\left( { + 6} \\right),\\left( { + 7} \\right)$$\n

$$i.e.\\,Sc < Ti < Cr = Mn$$\n

$${E_a} = 53598.6J/mol$$\n

$$ = 53.6\\,kJ/mol.$$ \n

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1245, "subject": "Chemistry", "question": "Four successive members of the first row transition elements are listed below with atomic numbers. Which\none of them is expected to have the highest $$E_{{M^{3 + }}/{M^{2 + }}}^0$$ value? ", "options": [ { "text": "Mn (Z = 25)" }, { "text": "Fe (Z = 26) " }, { "text": "Co (Z = 27) " }, { "text": "Cr (Z = 24) " } ], "answer": "Co (Z = 27) ", "solution": "**Answer:** Co (Z = 27) \n\n\"JEE ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1246, "subject": "Chemistry", "question": "Which series of reactions correctly represents chemical relations related to iron and its compound?", "options": [ { "text": "Fe $$\\buildrel {dil.\\,{H_2}S{O_4}} \\over\n \\longrightarrow $$ FeSO4 $$\\buildrel {{H_2}S{O_4},{O_2}} \\over\n \\longrightarrow $$ Fe2(SO4)3 $$\\buildrel {heat} \\over\n \\longrightarrow $$ Fe" }, { "text": "Fe $$\\buildrel {{O_2} \\, {, heat}} \\over\n \\longrightarrow $$ FeO $$\\buildrel {dil.\\,{H_2}S{O_4}} \\over\n \\longrightarrow $$ FeSO4 $$\\buildrel {heat} \\over\n \\longrightarrow $$ Fe " }, { "text": "Fe $$\\buildrel {{Cl_2} \\, {, heat}} \\over\n \\longrightarrow $$ FeCl3 $$\\buildrel {heat, \\, air} \\over\n \\longrightarrow $$ FeCl2 $$\\buildrel {Zn} \\over\n \\longrightarrow $$ Fe " }, { "text": "Fe $$\\buildrel {{O_2},\\,heat} \\over\n \\longrightarrow $$ Fe3O4 $$\\buildrel {CO,\\,{{600}^o}C} \\over\n \\longrightarrow $$ FeO $$\\buildrel {CO,\\,{{700}^o}C} \\over\n \\longrightarrow $$ Fe" } ], "answer": "Fe $$\\buildrel {{O_2},\\,heat} \\over\n \\longrightarrow $$ Fe3O4 $$\\buildrel {CO,\\,{{600}^o}C} \\over\n \\longrightarrow $$ FeO $$\\buildrel {CO,\\,{{700}^o}C} \\over\n \\longrightarrow $$ Fe", "solution": "**Answer:** Fe $$\\buildrel {{O_2},\\,heat} \\over\n \\longrightarrow $$ Fe3O4 $$\\buildrel {CO,\\,{{600}^o}C} \\over\n \\longrightarrow $$ FeO $$\\buildrel {CO,\\,{{700}^o}C} \\over\n \\longrightarrow $$ Fe\n\nIn equation \n

(i) $$\\,\\,\\,$$ $$F{e_2}{\\left( {S{O_4}} \\right)_3}\\,\\,\\,$$ \n

and in equation \n

(ii) $$\\,\\,\\,$$ $$F{e_2}{\\left( {S{O_4}} \\right)_3}$$ \n

on decomposing will form oxide instead of $$Fe.$$ \n

The correct sequence of reactions is \n

$$Fe\\buildrel {{O_2},heat} \\over\n \\longrightarrow \\,\\,F{e_3}{O_4}\\buildrel {Co,{{600}^ \\circ }C} \\over\n \\longrightarrow $$\n

$$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,F{e_2}{\\left( {S{O_4}} \\right)_3}\\buildrel \\Delta \\over\n \\longrightarrow Fe$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1247, "subject": "Chemistry", "question": "Match the catalysts to the correct processes :\n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
CatalystProcess
(A)TiCl3(i)Wacker process
(B)PdCl2(ii)Ziegler - Natta
polymerization
(C)CuCl2(iii)Contact process
(D)V2O5(iv)Deacon's process
", "options": [ { "text": "$$\\left( A \\right) - \\left( {ii} \\right),\\,\\,\\left( B \\right) - \\left( {iii} \\right),\\,\\,\\left( C \\right) - \\left( {iv} \\right),\\,\\,\\left( D \\right) - \\left( i \\right)$$ " }, { "text": "$$\\left( A \\right) - \\left( {iii} \\right),\\,\\,\\left( B \\right) - \\left( i \\right),\\,\\,\\left( C \\right) - \\left( {ii} \\right),\\,\\,\\left( D \\right) - \\left( {iv} \\right)$$ " }, { "text": "$$\\left( A \\right) - \\left( {iii} \\right),\\,\\,\\left( B \\right) - \\left( {ii} \\right),\\,\\,\\left( C \\right) - \\left( {iv} \\right),\\,\\,\\left( D \\right) - \\left( i \\right)$$ " }, { "text": "$$\\left( A \\right) - \\left( {ii} \\right),\\,\\,\\left( B \\right) - \\left( i \\right),\\,\\,\\left( C \\right) - \\left( {iv} \\right),\\,\\,\\left( D \\right) - \\left( {iii} \\right)$$ " } ], "answer": "$$\\left( A \\right) - \\left( {ii} \\right),\\,\\,\\left( B \\right) - \\left( i \\right),\\,\\,\\left( C \\right) - \\left( {iv} \\right),\\,\\,\\left( D \\right) - \\left( {iii} \\right)$$ ", "solution": "**Answer:** $$\\left( A \\right) - \\left( {ii} \\right),\\,\\,\\left( B \\right) - \\left( i \\right),\\,\\,\\left( C \\right) - \\left( {iv} \\right),\\,\\,\\left( D \\right) - \\left( {iii} \\right)$$ \n\n

To match the catalysts to the correct processes :

\n
    \n

  1. $TiCl_3$ is well-known as a catalyst in Ziegler-Natta polymerization which is used for the production of polymers of olefins. So, $(A) - (ii)$.

    \n
    2. \n

  2. $PdCl_2$ is used in the Wacker process, which is used to convert ethene to acetaldehyde. Hence, $(B) - (i)$.

    \n
    3. \n

  3. $V_2O_5$ is used in the Contact process to manufacture sulfuric acid by catalyzing the oxidation of sulfur dioxide to sulfur trioxide. Therefore, $(D) - (iii)$.

    \n
    4. \n

  4. $CuCl_2$ is used in Deacon's process for the oxidation of hydrochloric acid to chlorine. Hence, $(C) - (iv)$.

    \n
    5. \n
\n

Comparing these pairings with the options given, we find that the correct option is :

\n

$$\n\\left( A \\right) - \\left( {ii} \\right),\\,\\,\\left( B \\right) - \\left( {i} \\right),\\,\\,\\left( C \\right) - \\left( {iv} \\right),\\,\\,\\left( D \\right) - \\left( {iii} \\right)\n$$

\n", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1248, "subject": "Chemistry", "question": "The transition metal ions responsible for color in ruby and emerald are,\nrespectively :", "options": [ { "text": "Cr3+   and   Co3+" }, { "text": "Co3+    and    Cr3+" }, { "text": "Co3+    and    Co3+" }, { "text": "Cr3+    and    Cr3+" } ], "answer": "Cr3+    and    Cr3+", "solution": "**Answer:** Cr3+    and    Cr3+\n\n

On addition of impurities to the colourless gemstones, brilliant colours are produced. On addition of chromium to colourless corundum (Al2O3), a red ruby is formed. Red colour ruby is obtained when Cr3+ ions replace Al3+ ions in the octahedral sites of corundum. A green emerald emerges when chromium is added to colourless beryl (Be3Al2(SiO3)6). Green colour emerald is obtained when Cr3+ ions replace Al3+ in the octahedral site of beryl.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1249, "subject": "Chemistry", "question": "Which one of the following species is stable in aqueous solution?", "options": [ { "text": "Cr2+" }, { "text": "Cu+" }, { "text": "MnO$$_4^{3 - }$$" }, { "text": "MnO$$_4^{2 - }$$" } ], "answer": "MnO$$_4^{2 - }$$", "solution": "**Answer:** MnO$$_4^{2 - }$$\n\n

MnO42− disproportionate as (in neutral or acidic solution)

\n

3MnO42− + 4H+ → 2MnO42− + MnO2 + 2H2O

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1250, "subject": "Chemistry", "question": "Which of the following compounds is metallic and ferromagnetic ?", "options": [ { "text": "CrO2" }, { "text": "VO2" }, { "text": "MnO2" }, { "text": "TiO2" } ], "answer": "CrO2", "solution": "**Answer:** CrO2\n\nOut of all the four given metallic oxides $$Cr{O_2}$$ is attracted by magnetic field very strongly. The effect persists even when the magnetic field is removed. Thus $$Cr{O_2}$$ is metallic and ferromagnetic in nature. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1251, "subject": "Chemistry", "question": "The reaction of zinc with dilute and concentrated nitric acid, respectively, produces :", "options": [ { "text": "NO2 and NO" }, { "text": "NO and N2O" }, { "text": "NO2 and N2O" }, { "text": "N2O and NO2" } ], "answer": "N2O and NO2", "solution": "**Answer:** N2O and NO2\n\n\"JEE ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1252, "subject": "Chemistry", "question": "The pair of compounds having metals in their highest oxidation state is :", "options": [ { "text": "MnO2 and CrO2Cl2 " }, { "text": "[NiCl4]2$$-$$ and [CoCl4]2$$-$$ " }, { "text": "[Fe(CN)6]3$$-$$ and [Cu(CN)4]2$$-$$" }, { "text": "[FeCl4]$$-$$ and Co2O3 " } ], "answer": "MnO2 and CrO2Cl2 ", "solution": "**Answer:** MnO2 and CrO2Cl2 \n\n

\n\n\n\n\n\n\n\n \n \n \n \n \n\n\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\n
ComplexMetalOxidation state
$${[Fe{(CN)_6}]^{2 - }}$$ and $${[Cu{(CN)_4}]^{2 - }}$$Fe and Cu+4 and +2
$${[FeC{l_4}]^ - }$$ and $$C{o_2}{O_3}$$Fe and Co+3 and +3
$${[NiC{l_4}]^{2 - }}$$ and $${[CoC{l_4}]^{2 - }}$$Ni and Co+2 and +2
$$Mn{O_2}$$ and $$Cr{O_2}C{l_2}$$Mn and Cr+4 and +6

\n

In order to determine which pair of compounds have metals in their highest oxidation states, we need to consider the common oxidation states of the metals involved, as well as the rules for determining oxidation states in compounds and complex ions. Let's evaluate each option:

\n\n

Option A: MnO2 and CrO2Cl2

\n\n

In MnO2, manganese (Mn) has an oxidation state of +4. This is not the highest oxidation state manganese can achieve; it can go up to +7, as seen in KMnO4 (potassium permanganate).

\n\n

In CrO2Cl2, chromium (Cr) has an oxidation state of +6, which is indeed its highest oxidation state.

\n\n

Thus, this option contains one metal in its highest oxidation state (Cr) but not the other (Mn).

\n\n

Option B: [NiCl4]2- and [CoCl4]2-

\n\n

For both [NiCl4]2- and [CoCl4]2-, the nickel (Ni) and cobalt (Co) are in a +2 oxidation state. Neither of these represents the highest oxidation state for these metals. Nickel can have a +3 state, and cobalt can go up to +4 in very rare cases.

\n\n

Option C: [Fe(CN)6]3- and [Cu(CN)4]2-

\n\n

In [Fe(CN)6]3-, the iron (Fe) has an oxidation state of +3, which is a common high oxidation state for iron, though iron can also exist in a +2 state commonly and +6 in very rare conditions. Therefore, +3 is considered a high oxidation state but not the absolute highest.

\n\n

In [Cu(CN)4]2-, copper (Cu) has an oxidation state of +2, which is the highest stable oxidation state for copper in most of its compounds.

\n\n

Option D: [FeCl4]- and Co2O3

\n\n

In [FeCl4]-, iron (Fe) has an oxidation state of +3. As mentioned, +3 is a high state for iron, but not the absolute highest oxidation state it can achieve.

\n\n

In Co2O3, cobalt (Co) has an oxidation state of +3, as evidenced by the formula, where two Co atoms interact with three O atoms (each oxygen providing a -2 charge, for a total of -6, requiring each Co to be +3 to balance the charge). +3 is indeed among the highest common oxidation states for cobalt, although, as mentioned, Co can technically reach +4.

\n\n

Conclusion: Based on above analysis, while none of the options perfectly fits the criteria of both elements being in their absolute highest oxidation states (especially considering the rare or less common states), Option A and Option D come closest, with chromium and cobalt being in their common high oxidation states in the provided compounds. However, the question asks for the highest oxidation state, and strictly speaking, none of the options perfectly satisfy the condition for both elements included. Among the choices, Option A is the closest because chromium is in its highest oxidation state of +6 in CrO2Cl2, and manganese in MnO2 is in a +4 state, which while not its absolute highest, represents a commonly encountered high oxidation state. Given the context and conventional exams' focus on common oxidation states, Option A would be the best fit despite the slight discrepancy with the question's phrasing.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1253, "subject": "Chemistry", "question": "Which of the following ions does not liberate hydrogen gas on reaction with dilute acids ?", "options": [ { "text": "Ti2+ " }, { "text": "V2+ " }, { "text": "Cr2+ " }, { "text": "Mn2+ " } ], "answer": "Mn2+ ", "solution": "**Answer:** Mn2+ \n\nThe third ionisation energy of Mn is too high due to\nstable half filled 3d-orbital. Hence, it cannot further get oxidised\nto liberate hydrogen.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1254, "subject": "Chemistry", "question": "In the following reactions, ZnO is respectively acting as a/an :

\n(a) ZnO + Na2O $$\\to$$ Na2ZnO2

\n(b) ZnO + CO2 $$\\to$$ ZnCO3", "options": [ { "text": "base and base " }, { "text": "acid and acid " }, { "text": "acid and base" }, { "text": "base and acid" } ], "answer": "acid and base", "solution": "**Answer:** acid and base\n\nZinc oxide (ZnO) when react with Na2O it act as acid while with CO2 it act as base. Therefore, it is an amphoteric oxide.\n

(a) ZnO + Na2O $$\\to$$ Na2ZnO2\n
     Acid      Base          Salt

\n(b) ZnO + CO2 $$\\to$$ ZnCO3\n
     Base      Acid       Salt", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1255, "subject": "Chemistry", "question": "The incorrect statement is : ", "options": [ { "text": "Cu2+ salts give red coloured borax bead test in reducing flame. " }, { "text": "Cu2+ and Ni2+ ions give black precipitate with H2S in presence of HCl solution. " }, { "text": "Ferric ion gives blood red color with potasium thiocyanate." }, { "text": "Cu2+ ion gives chocolate coloured preciitate with potassium ferrocyanide solution. " } ], "answer": "Cu2+ and Ni2+ ions give black precipitate with H2S in presence of HCl solution. ", "solution": "**Answer:** Cu2+ and Ni2+ ions give black precipitate with H2S in presence of HCl solution. \n\n

Cu2+ and Ni2+ ions belong to Group II and Group IV respectively.

\n\"JEE\n

Group II sulphides precipitated out even under low concentration of sulphides, while Group IV sulphides require higher concentration of sulphides. To fulfill this reaction condition, dil. HCl is chosen in Group II reagents and NH4OH is chosen in Group IV reagents.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1256, "subject": "Chemistry", "question": "When XO2 is fused with an alkali metal hydroxide in presence of an oxidizing agent such as KNO3 ; a dark green product is formed which disproportioates in acidic solution to afforda dark purple solution. X is : ", "options": [ { "text": "Ti" }, { "text": "V" }, { "text": "Cr" }, { "text": "Mn" } ], "answer": "Mn", "solution": "**Answer:** Mn\n\nX is Mn.\n

MnO2  +  2KOH  +  KNO3  $$ \\to $$  K2MnO4  +  KNO2  +  H2O\n

Here   K2MnO4  is dark green.\n

3MnO4$$-$$2  +  4H+  $$ \\to $$  2MnO4$$-$$  +  MnO2  +  2H2O\n

In acidic solution, K2MnO4 changes to dark purple solution of KMnO4.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1257, "subject": "Chemistry", "question": "The pair that has similar atomic radii is :", "options": [ { "text": "Mo and W" }, { "text": "Mn and Fe" }, { "text": "Ti and Hf" }, { "text": "Sc and Ni" } ], "answer": "Mo and W", "solution": "**Answer:** Mo and W\n\nElectron configuration of Mo(Molybdenum) = [Kr] 4d5 5s1. So it belongs to 4d series.\n

Electron configuration of W(Tungsten) = [Xe] 4f14 5d4 6s2. So it belongs to 5d series.\n

Size of 3d < 4d = 5d. Due to lanthanoid contraction size of 4d and 5d series elements are almost same.\n
", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1258, "subject": "Chemistry", "question": "The correct order of the first ionization enthalpies is :", "options": [ { "text": "Ti < Mn < Zn < Ni" }, { "text": "Zn < Ni < Mn < Ti" }, { "text": "Mn < Ti < Zn < Ni" }, { "text": "Ti < Mn < Ni < Zn" } ], "answer": "Ti < Mn < Ni < Zn", "solution": "**Answer:** Ti < Mn < Ni < Zn\n\nIn a period, from left to right ionization enthalpy increses.\n

So in period 4, correct order for those 4 elements is\n

Ti < Mn < Ni < Zn\n

Exceptions in period 4 are\n

(1) V(Vanadium) < Ti(Titanium)\n

(2) Ni(Titanium) < Co(Cobalt)\n

(3) Ga(Gallium) < Zn(Zinc)\n

(4) Se(Selenium) < As(Arsenic)", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1259, "subject": "Chemistry", "question": "The INCORRECT statement is :", "options": [ { "text": "the color of [CoCl(NH3)5]2+ is violet as it absorbs the yellow light." }, { "text": "the gemstone, ruby, has Cr3+ ions occupying the octahedral sites of beryl." }, { "text": "the spin-only magnetic moment of [Ni(NH3)4(H2O)2]2+ is 2.83 BM." }, { "text": "the spin-only magnetic moments of [Fe(H2O)6]2+ and [Cr(H2O)6]2+ are nearly similar. " } ], "answer": "the gemstone, ruby, has Cr3+ ions occupying the octahedral sites of beryl.", "solution": "**Answer:** the gemstone, ruby, has Cr3+ ions occupying the octahedral sites of beryl.\n\n(A) Complementarty color of yellow is violet. As [CoCl(NH3)5]2+ absorbs yellow so it will show complementary color violet.\n

(B)The building block of ruby is aluminium oxide (Al2O3) containing about 0.5 – 1% Cr3+ ions which are randomly\ndistributed in the position normally occupied by\nAl3+ ions.\n

(C) Total number of unpaired electons in [Fe(H2O)6]2+ = 4\n
and total number of unpaired electons in [Cr(H2O)6]2+ = 4\n
So, the spin-only magnetic moments are same in both cases.\n

(D) Total number of unpaired electons in [Ni(NH3)4(H2O)2]2+ = 2\n
So, $$\\mu $$ = $$\\sqrt {2\\left( {2 + 2} \\right)} $$ BM = 2.83 BM", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1260, "subject": "Chemistry", "question": "The statement that is INCORRECT about the\ninterstitial compounds is :", "options": [ { "text": "They are chemically reactive" }, { "text": "They have metallic conductivity" }, { "text": "They have high melting points" }, { "text": "They are very hard" } ], "answer": "They are chemically reactive", "solution": "**Answer:** They are chemically reactive\n\nInterstitial compound are –\n

(i) very hard\n

(ii) chemically inert\n

(iv) high melting point.\n

As interstitial compounds are chamically inert.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1261, "subject": "Chemistry", "question": "The element that usually does NOT show variable oxidation states is : ", "options": [ { "text": "V" }, { "text": "Cu" }, { "text": "Sc" }, { "text": "ti" } ], "answer": "Sc", "solution": "**Answer:** Sc\n\nSc3+ has noble gas configuration hence only +3 exists.\n

Most common oxidation states of :\n

(i) Sc : +3\n

(ii) V : +2, +3, +4, +5\n

(iii) Ti : +2, +3, +4\n

(iv) Cu : +1, +2", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1262, "subject": "Chemistry", "question": "The transition element that has lowest enthalpy of atomisation, is : ", "options": [ { "text": "Fe" }, { "text": "Cu" }, { "text": "V" }, { "text": "Zn" } ], "answer": "Cu", "solution": "**Answer:** Cu\n\nSince Zn is not a transition element so transition element having lowest atomisation energy out of Cu, V, Fe is Cu.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1263, "subject": "Chemistry", "question": "Consider the hydrated ions of Ti2+, V2+, Ti3+, and Sc3+. The correct order of their spin-only magnetic\nmoments is : ", "options": [ { "text": "Sc3+ < Ti3+ < V2+ < Ti2+" }, { "text": "Sc3+ < Ti3+ < Ti2+ < V2+ " }, { "text": "Ti3+ < Ti2+ < Sc3+ < V2+ " }, { "text": "V2+ < Ti2+ < Ti3+ < Sc3+" } ], "answer": "Sc3+ < Ti3+ < Ti2+ < V2+ ", "solution": "**Answer:** Sc3+ < Ti3+ < Ti2+ < V2+ \n\nAs we know that\n

$$\\mu = \\sqrt {n\\left( {n + 2} \\right)} $$\nwhere n = no. of unpaired electrons i.e. greater the no. of unpaired electron more will be the spin-only magnetic moments.\n

. Electronic configuration of the given transition\nmetal ions are\n

Sc3+ (Z = 21) = 1s22s22p63s23p6\n

$$ \\therefore $$ n = 0\n

Ti2+ (Z = 22) = 1s22s22p63s23p63d2\n

$$ \\therefore $$ n = 2\n

Ti3+ (Z = 22) = 1s22s22p63s23p63d1\n

$$ \\therefore $$ n = 1\n

V2+ (Z = 23) = 1s22s22p63s23p63d3\n

$$ \\therefore $$ n = 3\n

So the correct\nincreasing order of magnetic moment is\n

Sc3+ < Ti3+ < Ti2+ < V2+ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1264, "subject": "Chemistry", "question": "The incorrect statement(s) among (a) - (c) is\n(are)\n

(a) W(VI) is more stable than Cr(VI).\n

(b) In the presence of HCl, permanganate\ntitrations provide satisfactory results.\n

(c) Some lanthanoid oxides can be used as\nphosphors.

", "options": [ { "text": "(a) and (b) only" }, { "text": "(a) only" }, { "text": "(b) only" }, { "text": "(b) and (c) only" } ], "answer": "(b) only", "solution": "**Answer:** (b) only\n\nW(VI) is more stable than Cr(VI)\n

Permanganate titrations in presence of HCl\nare unsatisfactory as HCl is oxidised to Cl2\n

Lanthanoid oxides are used as phosphors.", "topic": "Number Theory", "subtopic": "Divisibility / Factorization" }, { "id": 1265, "subject": "Chemistry", "question": "The set that contains atomic numbers of only\ntransition elements, is :", "options": [ { "text": "21, 32, 53, 64" }, { "text": "9, 17, 34, 38" }, { "text": "37, 42, 50, 64" }, { "text": "21, 25, 42, 72" } ], "answer": "21, 25, 42, 72", "solution": "**Answer:** 21, 25, 42, 72\n\nElements with atomic number 21, 25, 42 and 72\nbelongs to transition metals.\n

Tranition elements \n
= 21 to 30\n
37 to 48\n
57 & 72 to 80", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1266, "subject": "Chemistry", "question": "Given below are two statements :

\nStatement I : Colourless cupric metaborate is reduced to cuprous metaborate in a luminous\nflame.

\nStatement II : Cuprous metaborate is obtained by heating boric anhydride and copper\nsulphate in a non-luminous flame.

\nIn the light of the above statements, choose the most appropriate answer from the options\ngiven below.", "options": [ { "text": "Statement I is true but Statement II is false" }, { "text": "Statement I is false but Statement II is true" }, { "text": "Both Statement I and Statement II are false" }, { "text": "Both Statement I and Statement II are true" } ], "answer": "Both Statement I and Statement II are false", "solution": "**Answer:** Both Statement I and Statement II are false\n\nIn presence of luminous flame, blue cupric metaborate is\nreduced to colourless cuprous metaborate.\n

\"JEE\n
If non-luminous flame is present in reaction, cupric metaborate is\nobtained by heating boric anhydride with copper sulphate.\n

\"JEE\n
So, both statements are false.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1267, "subject": "Chemistry", "question": "What is the correct order of the following elements with respect to their density?", "options": [ { "text": "Cr < Zn < Co < Cu < Fe" }, { "text": "Cr < Fe < Co < Cu < Zn" }, { "text": "Zn < Cr < Fe < Co < Cu" }, { "text": "Zn < Cu < Co < Fe < Cr" } ], "answer": "Zn < Cr < Fe < Co < Cu", "solution": "**Answer:** Zn < Cr < Fe < Co < Cu\n\n

Generally, due to decrease in metallic radius and increase in atomic mass density increase across the period from left to right.

\n

\n\n\n\n\n\n \n \n \n \n\n\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\n
MetalDensity $$(g/c{m^3})$$
Zn7.13
Cr7.19
Fe7.8
Co8.7
Cu8.9

\n

Correct order is Cu > Co > Fe > Cr > Zn.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1268, "subject": "Chemistry", "question": "In which of the following pairs, the outer most electronic configuration will be the same?", "options": [ { "text": "Cr+ and Mn2+" }, { "text": "Ni2+ and Cu+" }, { "text": "V2+ and Cr+" }, { "text": "Fe2+ and Co+" } ], "answer": "Cr+ and Mn2+", "solution": "**Answer:** Cr+ and Mn2+\n\nCr+ $$ \\to $$ [Ar]3d5\n

Mn2+ $$ \\to $$ [Ar]3d5", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1269, "subject": "Chemistry", "question": "What is the spin-only magnetic moment value (BM) of a divalent metal ion with atomic number 25, in it's aqueous solution?", "options": [ { "text": "zero" }, { "text": "5.26" }, { "text": "5.0" }, { "text": "5.92" } ], "answer": "5.92", "solution": "**Answer:** 5.92\n\n25Mn = [Ar]3d5 4s2
\nMn2+ = [Ar]3d5 4s0

\nNo. of unpaired electron (n) = 5

\n$$\\mu = \\sqrt {n\\left( {n + 2} \\right)} \\,BM$$

\n$$ = \\sqrt {5 \\times 7} {\\rm{ }} = \\sqrt {35} {\\rm{ }} = {\\rm{ }}5.92{\\rm{ }}$$ BM\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1270, "subject": "Chemistry", "question": "The common positive oxidation states for an element with atomic number 24, are :", "options": [ { "text": "+1 and +3" }, { "text": "+2 to +6" }, { "text": "+1 to +6" }, { "text": "+1 and +3 to +6" } ], "answer": "+2 to +6", "solution": "**Answer:** +2 to +6\n\n24Cr = [Ar]3d54s1
\n

The element with atomic number 24 is Chromium (Cr). Chromium commonly exhibits several positive oxidation states, which are due to the loss of electrons from both the 4s and 3d orbitals.

\n

Here are the common oxidation states for Chromium :

\n
    \n
  1. +2 (as in ${Cr}^{2+} $) : This is observed where the electron configuration is $[{Ar}] 3d^4$
    2. \n
  2. +3 (as in ${Cr}^{3+} )$ : This is the most stable state, with the electron configuration $[ {Ar} ] 3d^3$
    3. \n
  3. +6 (as in ${CrO}_4^{2-} $ and ${Cr}_2\\text{O}_7^{2-} )$ : In these cases, chromium exhibits an oxidation state of +6.
    4. \n
\n

Hence, the common positive oxidation states for Chromium are +2, +3, and +6.

Therefore, the correct option is :

\n

Option B : +2 to +6.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1271, "subject": "Chemistry", "question": "In the ground state of atomic Fe(Z = 26), the spin-only magnetic moment is ____________ $$\\times$$ 10$$-$$1 BM. (Round off to the Nearest Integer). [Given : $$\\sqrt 3 $$ = 1.73, $$\\sqrt 2 $$ = 1.41 ]", "options": [], "answer": "49", "solution": "**Answer:** 49\n\n$${}_{26}Fe = [Ar]3{d^6}4{s^2}$$

No. of unpaired electrons = 4

$$\\mu = \\sqrt {n(n + 2)} BM$$

$$ = \\sqrt {4(4 + 2)} = \\sqrt {24} $$

$$ = 2\\sqrt {3 \\times 2} = 2[1.73 \\times 1.41]$$

= 4.8786 BM

= $$48.78 \\times {10^{ - 1}}$$ BM", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1272, "subject": "Chemistry", "question": "The set having ions which are coloured and paramagnetic both is :", "options": [ { "text": "Cu2+, Cr3+, Sc+" }, { "text": "Cu2+, Zn2+, Mn4+" }, { "text": "Sc3+, V5+, Ti4+" }, { "text": "Ni2+, Mn7+, Hg2+" } ], "answer": "Cu2+, Cr3+, Sc+", "solution": "**Answer:** Cu2+, Cr3+, Sc+\n\n$$\\left. \\matrix{\n C{u^{2 + }}:[Ar]3{d^9}4{s^0} \\hfill \\cr \n C{r^{3 + }}:[Ar]3{d^3}4{s^0} \\hfill \\cr \n S{c^ + }:[Ar]3{d^1}4{s^1} \\hfill \\cr} \\right]$$All are coloured and paramagnetic due to presence of unpaired electrons", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1273, "subject": "Chemistry", "question": "Number of electrons that Vanadium (Z = 23) has in p-orbitals is equal to _____________", "options": [], "answer": "12", "solution": "**Answer:** 12\n\n$${}_{23}V:1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^3}4{s^2}$$

Number of electrons in p-orbitals is equal to 12.00.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1274, "subject": "Chemistry", "question": "The spin only magnetic moments (in BM) for free Ti3+, V2+ and Sc3+ ions respectively are :

(At.No. Sc : 21, Ti : 22, V : 23)", "options": [ { "text": "3.87, 1.73, 0" }, { "text": "1.73, 3.87, 0" }, { "text": "1.73, 0, 3.87" }, { "text": "0, 3.87, 1.73" } ], "answer": "1.73, 3.87, 0", "solution": "**Answer:** 1.73, 3.87, 0\n\n$$\\mu = \\sqrt {n(n + 2)} BM$$

$$\\matrix{\n {T{i^{ + 3}} = [Ar]3{d^1}} & {n = 1} & {\\mu = 1.73\\,BM} \\cr \n {{V^{ + 2}} = [Ar]3{d^3}} & {n = 3} & {\\mu = 3.87\\,BM} \\cr \n {S{c^{ + 3}} = [Ar]3{d^0}4{s^0}} & {n = 0} & {\\mu = 0} \\cr \n\n } $$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1275, "subject": "Chemistry", "question": "Which one of the following when dissolved in water gives coloured solution in nitrogen atmosphere?", "options": [ { "text": "CuCl2" }, { "text": "AgCl" }, { "text": "ZnCl2" }, { "text": "Cu2Cl2" } ], "answer": "CuCl2", "solution": "**Answer:** CuCl2\n\n(a) CuCl2 + nH2O $$\\to$$ Cu$$_{(aq.)}^{ + 2}$$ (blue colour)

(b) AgCl + nH2O $$\\to$$ Insoluble

(c) ZnCl2 + nH2O $$\\to$$ Zn$$_{(aq.)}^{ + 2}$$ Colourless

(d) Cu2Cl2 + nH2O $$\\to$$ Insoluble", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1276, "subject": "Chemistry", "question": "The nature of oxides V2O3 and CrO is indexed as 'X' and 'Y' type respectively. The correct set of X and Y is :", "options": [ { "text": "\n\n \n \n \n \n \n\n
X = basicY = amphoteric
" }, { "text": "\n\n \n \n \n \n \n\n
X = amphotericY = basic
" }, { "text": "\n\n \n \n \n \n \n\n
X = acidicY = acidic
" }, { "text": "\n\n \n \n \n \n \n\n
X = basicY = basic
" } ], "answer": "\n\n \n \n \n \n \n\n
X = basicY = basic
", "solution": "**Answer:** \n\n \n \n \n \n \n\n
X = basicY = basic
\n\nV2O3 basic

CrO basic", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1277, "subject": "Chemistry", "question": "The value of magnetic quantum number of the outermost electron of Zn+ ion is ______________.", "options": [], "answer": "0", "solution": "**Answer:** 0\n\nZn+ $$\\to$$ 1s22s22p63s23p63d104s1

Outermost electron is in 4s subshell

m = 0", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1278, "subject": "Chemistry", "question": "Identify the element for which electronic configuration in +3 oxidation state is [Ar]3d5 :", "options": [ { "text": "Ru" }, { "text": "Mn" }, { "text": "Co" }, { "text": "Fe" } ], "answer": "Fe", "solution": "**Answer:** Fe\n\nMn(25) = [Ar]3d54s2\n

Mn+3 = [Ar]3d44s0\n

Ru-belongs to 4d transition series\n

Co (27) = [Ar]3d74s2\n

Co+3 = [Ar]3d64s0\n

Fe(26) = [Ar]3d64s2\n

Fe+3 = [Ar]3d54s0", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1279, "subject": "Chemistry", "question": "

The electronic configuration of Pt (atomic number 78) is :

", "options": [ { "text": "[Xe] 4f14 5d9 6s1" }, { "text": "[Kr] 4f14 5d10" }, { "text": "[Xe] 4f14 5d10" }, { "text": "[Xe] 4f14 5d8 6s2" } ], "answer": "[Xe] 4f14 5d9 6s1", "solution": "**Answer:** [Xe] 4f14 5d9 6s1\n\nAtomic number of Pt is 78

\nElectronic configuration is – 78Pt $$ \\to $$ [Xe] 4 f14 5d9 6s1 (Exceptional electronic\nconfiguration)", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1280, "subject": "Chemistry", "question": "

Spin only magnetic moment of [MnBr6]4$$-$$ is _________ B.M. (round off to the closest integer)

", "options": [], "answer": "6", "solution": "**Answer:** 6\n\n$\\begin{aligned} & {\\left[\\mathrm{MnBr}_6\\right]^{4-}} \\\\\\\\ & x-6=-4\\end{aligned}$\n

$$\n\\begin{aligned}\nx & =+2 \\\\\\\\\n\\mathrm{Mn} & =[\\mathrm{Ar}] 3 d^5 4 s^2 \\\\\\\\\n\\mathrm{Mn}^{2+} & =[\\mathrm{Ar}] 3 d^5 4 s^0\n\\end{aligned}\n$$\n

\"JEE\n
So, number of unpaired electrons $(n)=5$\n

$$\n\\begin{aligned}\n\\mu & =\\sqrt{n(n+2)} \\\\\\\\\n\\mu & =\\sqrt{5(5+2)} \\\\\\\\\n& =\\sqrt{35} \\\\\\\\\n& =5.91 \\text { B.M. } \\\\\\\\\n& \\approx 6 \\text { B.M. }\n\\end{aligned}\n$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1281, "subject": "Chemistry", "question": "

For the reaction given below :

\n

CoCl3 . xNH3 + AgNO3 (aq) $$\\to$$

\n

If two equivalents of AgCl precipitate out, then the value of x will be _____________.

", "options": [], "answer": "5", "solution": "**Answer:** 5\n\n$$\n\\begin{aligned}\n&\\mathrm{CoCl}_3 \\cdot \\mathrm{xNH}_3+\\mathrm{AgNO}_3 \\rightarrow \\underset{\\text{2 mol}}{\\mathrm{AgCl} \\downarrow}\\\\\\\\\n&\\left[\\mathrm{Co}\\left(\\mathrm{NH}_3\\right)_5 \\mathrm{Cl}\\right] \\mathrm{Cl}_2+\\mathrm{AgNO}_3 \\rightarrow \\underset{\\text{2 mol}}{\\mathrm{AgCl}} \\downarrow\\\\\\\\\n&x=5\n\\end{aligned}\n$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1282, "subject": "Chemistry", "question": "

The metal ion (in gaseous state) with lowest spin-only magnetic moment value is :

", "options": [ { "text": "V2+" }, { "text": "Ni2+" }, { "text": "Cr2+" }, { "text": "Fe2+" } ], "answer": "Ni2+", "solution": "**Answer:** Ni2+\n\n\n\n \n \n \n \n \n\n\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\n
Valence shell configurationUnpaired electrons
V2+3d34s0n = 3
Ni2+3d84s0n = 2
Cr2+3d44s0n = 4
Fe2+3d64s0n = 4


\nSince Ni2+ has least number of unpaired electrons.\nHence Ni2+ will have lowest spin only magnetic\nmoment Value.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1283, "subject": "Chemistry", "question": "

Among the following, which is the strongest oxidizing agent?

", "options": [ { "text": "Mn3+" }, { "text": "Fe3+" }, { "text": "Ti3+" }, { "text": "Cr3+" } ], "answer": "Mn3+", "solution": "**Answer:** Mn3+\n\nStrongest oxidising agent have highest reduction potential value

\n$$\\mathrm{E}_{\\mathrm{Fe}^{3+} \\mid \\mathrm{Fe}^{2+}}^{\\circ}= + 0.77 \\mathrm{~V} \\quad \\mathrm{E}_{\\mathrm{Ti}^{3+} \\mid \\mathrm{Ti}^{2+}}^{\\circ}=-0.37 \\mathrm{~V}$$

\n$$\\mathrm{E}^{\\circ}_{\\mathrm{Mn}^{+3} \\mid \\mathrm{Mn}^{+2}}=+1.57 \\mathrm{~V} \\quad \\mathrm{E}^{\\circ}_{\\mathrm{Cr}^{3+}\\mid\\mathrm{Cr}^{2+}} = -0.41 \\mathrm{~V}$$

\n$$\\mathrm{Mn}^{+3}$$ is the best oxidising agent among the given series.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1284, "subject": "Chemistry", "question": "

Metals generally melt at very high temperature. Amongst the following, the metal with the highest melting point will be :

", "options": [ { "text": "Hg" }, { "text": "Ag" }, { "text": "Ga" }, { "text": "Cs" } ], "answer": "Ag", "solution": "**Answer:** Ag\n\nMelting points of the given metals

\nHg : –38.83° C

\nAg : 961.8° C

\nGa : 29.76° C

\nCs : 28.44° C

\n$$ \\therefore $$ Metal having highest melting point is Ag.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1285, "subject": "Chemistry", "question": "

Among Co3+, Ti2+, V2+ and Cr2+ ions, one if used as a reagent cannot liberate H2 from dilute mineral acid solution, its spin-only magnetic moment in gaseous state is ___________ B.M. (Nearest integer)

", "options": [], "answer": "5", "solution": "**Answer:** 5\n\n$$\\mathrm{Co}^{3+}$$ will not liberate $$\\mathrm{H}_{2}$$ gas an reaction with dilute acid\n

\n$$\\mathrm{E}_{\\mathrm{Co}^{3+}/\\mathrm{Co}^{2+}}^{\\mathrm{O}}=+1.97$$\n

\nAnd $$\\mathrm{Co}^{3+}$$ has electronic configuration $$=[\\operatorname{Ar}] 3 d^{6}$$\n

\n$$\\therefore 4$$ unpaired $$\\mathrm{e}^{-}$$ are present in it\n

\n$$\\therefore$$ Spin-only magnetic moment $$=\\sqrt{4(4+2)} =4.92 \\approx 5$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1286, "subject": "Chemistry", "question": "

The spin-only magnetic moment value of the compound with strongest oxidizing ability among $$\\mathrm{MnF}_{4}, \\mathrm{MnF}_{3}$$ and $$\\mathrm{MnF}_{2}$$ is ____________ B.M. [nearest integer]

", "options": [], "answer": "5", "solution": "**Answer:** 5\n\n$\\mathrm{MnF}_{3}$ has the strongest oxidizing ability\n

\n$$\n\\left[\\begin{array}{l}\n\\mathrm{E}_{\\mathrm{Mn}^{+3} / \\mathrm{Mn}^{+2}}^{\\circ} \\simeq 1.57 \\mathrm{~V} \\\\\n\\& \\,\\,\\mathrm{E}_{\\mathrm{Mn}^{+4} / \\mathrm{Mn}^{+2}}^{\\circ} \\simeq 1.2 \\mathrm{~V}\n\\end{array}\\right]\n$$\n

\nSo, spin only magnetic moment\n

\n$$\n=\\sqrt{4(4+2)}=\\sqrt{24} \\text { B.M. }\n$$\n

\n$\\simeq 5$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1287, "subject": "Chemistry", "question": "

Which of the following has least tendency to liberate $$\\mathrm{H}_{2}$$ from mineral acids?

", "options": [ { "text": "Cu" }, { "text": "Mn" }, { "text": "Ni" }, { "text": "Zn" } ], "answer": "Cu", "solution": "**Answer:** Cu\n\nThe metal atom whose oxidation potential is less than that of hydrogen can release $\\mathrm{H}_{2}$ from mineral acids.

\n$\\mathrm{E}_{\\mathrm{Zn} / \\mathrm{Zn}^{+2}}=0.76$

\n$\\mathrm{E}_{\\mathrm{Ni/Ni}^{+ 2}}=0.25$

\n$E_{M n / M n^{+2}}=1.18$

\n$\\mathrm{E}_{\\mathrm{Cu}^{\\circ} / \\mathrm{Cu}^{+2}}=-0.34$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1288, "subject": "Chemistry", "question": "

The reaction of zinc with excess of aqueous alkali, evolves hydrogen gas and gives :

", "options": [ { "text": "$$\\mathrm{Zn}(\\mathrm{OH})_{2}$$" }, { "text": "$$\\mathrm{ZnO}$$" }, { "text": "$$\\left[\\mathrm{Zn}(\\mathrm{OH})_{4}\\right]^{2-}$$" }, { "text": "$$\\left[\\mathrm{ZnO}_{2}\\right]^{2-}$$" } ], "answer": "$$\\left[\\mathrm{Zn}(\\mathrm{OH})_{4}\\right]^{2-}$$", "solution": "**Answer:** $$\\left[\\mathrm{Zn}(\\mathrm{OH})_{4}\\right]^{2-}$$\n\nZinc dissolves in excess of aqueous alkali.\n

\n$$\\mathrm{Zn}+2 \\mathrm{OH}^{-}+2 \\mathrm{H}_{2} \\mathrm{O} \\longrightarrow\\left[\\mathrm{Zn}(\\mathrm{OH})_{4}\\right]^{2-}+\\mathrm{H}_{2} \\uparrow$$ (Tetrahydroxozincate(II) ion)\n

\nHowever, this reaction in NCERT is given as\n

\n$$\\mathrm{Zn}+2 \\mathrm{NaOH} \\longrightarrow \\mathrm{Na}_{2} \\mathrm{ZnO}_{2}+\\mathrm{H}_{2} \\uparrow$$\n

\n$$\\mathrm{ZnO}_{2}^{2-}$$ is anhydrous form of $$\\left[\\mathrm{Zn}(\\mathrm{OH})_{4}\\right]^{2-}$$.\n

\n$$\\mathrm{ZnO}_{2}^{2-}+2 \\mathrm{H}_{2} \\mathrm{O} \\rightleftharpoons\\left[\\mathrm{Zn}(\\mathrm{OH})_{4}\\right]^{2-}$$\n

\nSo in aqueous medium best answer of this question is $\\left[\\mathrm{Zn}(\\mathrm{OH})_{4}\\right]^{2-}$.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1289, "subject": "Chemistry", "question": "

In following pairs, the one in which both transition metal ions are colourless is :

", "options": [ { "text": "$$\\mathrm{Sc}^{3+}, \\mathrm{Zn}^{2+}$$" }, { "text": "$$\\mathrm{Ti}^{4+}, \\mathrm{Cu}^{2+}$$" }, { "text": "$$\\mathrm{V}^{2+}, \\mathrm{Ti}^{3+}$$" }, { "text": "$$\\mathrm{Zn}^{2+}, \\mathrm{Mn}^{2+}$$" } ], "answer": "$$\\mathrm{Sc}^{3+}, \\mathrm{Zn}^{2+}$$", "solution": "**Answer:** $$\\mathrm{Sc}^{3+}, \\mathrm{Zn}^{2+}$$\n\n$\\mathrm{Sc}^{+3}$ and $\\mathrm{Zn}^{+2}$ are colourless as they contain no unpaired electron. Whereas the transition metal ions $\\mathrm{Cu}^{+2}, \\mathrm{Ti}^{+3}, \\mathrm{~V}^{+2}$ and $\\mathrm{Mn}^{+2}$ are coloured as they contain unpaired electrons.

The unpaired electron from lower energy $d$ orbital gets excited to a higher energy $d$ orbital on absorbing light of frequency which lies in visible region. The colour complementary to light absorbed is observed.\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1290, "subject": "Chemistry", "question": "

Which of the following $$3\\mathrm{d}$$-metal ion will give the lowest enthalpy of hydration $$\\left(\\Delta_{\\text {hyd }} \\mathrm{H}\\right)$$ when dissolved in water ?

", "options": [ { "text": "$$\\mathrm{Cr}^{2+}$$" }, { "text": "$$\\operatorname{Mn}^{2+}$$" }, { "text": "$$\\mathrm{Fe}^{2+}$$" }, { "text": "$$\\mathrm{Co}^{2+}$$" } ], "answer": "$$\\operatorname{Mn}^{2+}$$", "solution": "**Answer:** $$\\operatorname{Mn}^{2+}$$\n\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1291, "subject": "Chemistry", "question": "

Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).

\n

Assertion (A) : $$\\mathrm{Cu^{2+}}$$ in water is more stable than $$\\mathrm{Cu^{+}}$$.

\n

Reason (R) : Enthalpy of hydration for $$\\mathrm{Cu^{2+}}$$ is much less than that of $$\\mathrm{Cu^{+}}$$.

\n

In the light of the above statements, choose the correct answer from the options given below :

", "options": [ { "text": "(A) is correct but (R) is not correct" }, { "text": "Both (A) and (R) are correct but (R) is not the correct explanation of (A)" }, { "text": "(A) is not correct but (R) is correct" }, { "text": "Both (A) and (R) are correct and (R) is the correct explanation of (A)" } ], "answer": "(A) is correct but (R) is not correct", "solution": "**Answer:** (A) is correct but (R) is not correct\n\nCu2+ ion is more stable than Cu+ ion due to high charge density on Cu2+ ion, it has higher hydration energy when it bonds to water molecules thus it forms stronger bonds.\n

$$ \\therefore $$ (A) is correct but (R) is not correct.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1292, "subject": "Chemistry", "question": "

Highest oxidation state of Mn is exhibited in $$\\mathrm{Mn_2O_7}$$. The correct statements about $$\\mathrm{Mn_2O_7}$$ are

\n

(A) Mn is tetrahedrally surrounded by oxygen atoms.

\n

(B) Mn is octahedrally surrounded by oxygen atoms.

\n

(C) Contains Mn-O-Mn bridge.

\n

(D) Contains Mn-Mn bond.

\n

Choose the correct answer from the options given below :

", "options": [ { "text": "A and C only" }, { "text": "A and D only" }, { "text": "B and C only" }, { "text": "B and D only" } ], "answer": "A and C only", "solution": "**Answer:** A and C only\n\n

The correct statements about $$\\mathrm{Mn_2O_7}$$ are:

\n\n

(A) Mn is tetrahedrally surrounded by oxygen atoms.

\n

(C) Contains Mn-O-Mn bridge.

\n\n

\"JEE

\n\n

Manganese (Mn) in $$\\mathrm{Mn_2O_7}$$ exhibits its highest oxidation state of +7. The oxidation state is the charge an atom would have if the electrons in a chemical bond were shared equally.\n

\n

In $$\\mathrm{Mn_2O_7}$$, the manganese atoms are surrounded by oxygen atoms in a tetrahedral arrangement. This means that each manganese atom is bonded to four oxygen atoms arranged in a pyramid shape. This tetrahedral arrangement of oxygen atoms results in the highest oxidation state for manganese in $$\\mathrm{Mn_2O_7}$$.

\n\n

$$\\mathrm{Mn_2O_7}$$ also contains Mn-O-Mn bridges, which are bonds between two manganese atoms through a shared oxygen atom. These bonds help to reinforce the structure of the molecule and contribute to its stability.

\n\n

It is incorrect to state that Mn is octahedrally surrounded by oxygen atoms or that $$\\mathrm{Mn_2O_7}$$ contains a Mn-Mn bond. An octahedral arrangement would consist of six oxygen atoms surrounding a central manganese atom, and a Mn-Mn bond would involve two manganese atoms sharing electrons without an oxygen atom in between. Neither of these are present in $$\\mathrm{Mn_2O_7}$$.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1293, "subject": "Chemistry", "question": "

$$\\mathrm{Nd^{2+}}$$ = __________

", "options": [ { "text": "$$\\mathrm{4f^4 6s^2}$$" }, { "text": "$$\\mathrm{4f^3}$$" }, { "text": "$$\\mathrm{4f^4}$$" }, { "text": "$$\\mathrm{4f^2 6s^2}$$" } ], "answer": "$$\\mathrm{4f^4}$$", "solution": "**Answer:** $$\\mathrm{4f^4}$$\n\nNd(60) = [Xe] 4f4 5d0 6s2\n

Nd2+\n= [Xe] 4f4 5d0 5s0", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1294, "subject": "Chemistry", "question": "

The correct order of basicity of oxides of vanadium is :

", "options": [ { "text": "$$\\mathrm{V_2O_3 > V_2O_5 > V_2O_4}$$" }, { "text": "$$\\mathrm{V_2O_3 > V_2O_4 > V_2O_5}$$" }, { "text": "$$\\mathrm{V_2O_5 > V_2O_4 > V_2O_3}$$" }, { "text": "$$\\mathrm{V_2O_4 > V_2O_3 > V_2O_5}$$" } ], "answer": "$$\\mathrm{V_2O_3 > V_2O_4 > V_2O_5}$$", "solution": "**Answer:** $$\\mathrm{V_2O_3 > V_2O_4 > V_2O_5}$$\n\n$\\mathrm{V}_{2} \\mathrm{O}_{3}>\\mathrm{V}_{2} \\mathrm{O}_{4}>\\mathrm{V}_{2} \\mathrm{O}_{5}$\n\n

As positive oxidation state increases acidic nature increases and basic nature decreases.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1295, "subject": "Chemistry", "question": "

Given below are two statements :

\n

Statement I : Nickel is being used as the catalyst for producing syn gas and edible fats.

\n

Statement II : Silicon forms both electron rich and electron deficient hydrides.

\n

In the light of the above statements, choose the most appropriate answer from the options given below :

", "options": [ { "text": "Statement I is correct but statement II is incorrect" }, { "text": "Both the statements I and II are correct" }, { "text": "Both the statements I and II are incorrect" }, { "text": "Statement I is incorrect but statement II is correct" } ], "answer": "Statement I is correct but statement II is incorrect", "solution": "**Answer:** Statement I is correct but statement II is incorrect\n\n

Statement I is correct, Statement II is incorrect as Si forms electron precise hydride.

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1296, "subject": "Chemistry", "question": "

The set of correct statements is :

\n

(i) Manganese exhibits +7 oxidation state in its oxide.

\n

(ii) Ruthenium and Osmium exhibit +8 oxidation in their oxides.

\n

(iii) Sc shows +4 oxidation state which is oxidizing in nature.

\n

(iv) Cr shows oxidising nature in +6 oxidation state.

", "options": [ { "text": "(ii), (iii) and (iv)" }, { "text": "(ii) and (iii)" }, { "text": "(i) and (iii)" }, { "text": "(i), (ii) and (iv)" } ], "answer": "(i), (ii) and (iv)", "solution": "**Answer:** (i), (ii) and (iv)\n\n

(i) $$\\mathrm{Mn_2O_7\\to Mn}$$ in (+7) oxidation state

\n

(ii) It is also correct

\n

(iii) Sc only shows +3 oxidation state

\n

(iv) $$\\mathrm{Cr^{+6}}$$ is oxidising in nature

", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1297, "subject": "Chemistry", "question": "

How many of the following metal ions have similar value of spin only magnetic moment in gaseous state? ______________

\n

(Given : Atomic number V, 23; Cr, 24; Fe, 26; Ni, 28)

\n

V$$^{3+}$$, Cr$$^{3+}$$, Fe$$^{2+}$$, Ni$$^{3+}$$

", "options": [], "answer": "2", "solution": "**Answer:** 2\n\n$\\mu_{\\mathrm{s}}=\\sqrt{\\mathrm{n}(\\mathrm{n}+2)} \\mathrm{BM} \\quad(\\mathrm{n}=\\mathrm{no}$ of unpaired electrons)

\n$$\n\\begin{array}{ll}\n & n \\\\\n\\mathrm{V}^{3+}:[\\mathrm{Ar}] 3 \\mathrm{~d}^2 4 \\mathrm{~s}^0 & 2 \\\\\\\\\n\\mathrm{Cr}^{3+}:[\\mathrm{Ar}] 3 \\mathrm{~d}^3 4 \\mathrm{~s}^0 & 3 \\\\\\\\\n\\mathrm{Fe}^{2+}:[\\mathrm{Ar}] 3 \\mathrm{~d}^6 4 \\mathrm{~s}^0 & 4 \\\\\\\\\n\\mathrm{Ni}^{3+}:[\\mathrm{Ar}] 3 \\mathrm{~d}^7 4 \\mathrm{~s}^0 & 3\n\\end{array}\n$$

\n$\\mathrm{Cr}^{3+} ~\\&~ \\mathrm{Ni}^{3+}$ have same value of $\\mu_{\\mathrm{s}}$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1298, "subject": "Chemistry", "question": "In Chromyl chloride, the oxidation state of chromium is $(+)$ _______________.", "options": [], "answer": "6", "solution": "**Answer:** 6\n\nIn chromyl chloride, $\\mathrm{CrO}_2\\mathrm{Cl}_2$, the oxidation state of chromium can be calculated by assigning oxidation states to the other atoms in the molecule and then using the fact that the sum of the oxidation states in a molecule must be equal to zero.\n

\nThe oxidation state of oxygen is usually -2 and the oxidation state of chlorine is usually -1.

Therefore, we can write:\n

\n$$\\mathrm{Cr} + 2\\mathrm{O}(-2) + 2\\mathrm{Cl}(-1) = 0$$\n

\nSimplifying the equation, we get:\n

\n$$\\mathrm{Cr} - 4 - 2 = 0$$\n

\n$$\\mathrm{Cr} - 6 = 0$$\n

\n$$\\mathrm{Cr} = +6$$\n

\nTherefore, the oxidation state of chromium in chromyl chloride is $(+6)$.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1299, "subject": "Chemistry", "question": "The total change in the oxidation state of manganese involved in the reaction of $\\mathrm{KMnO}_{4}$ and potassium iodide in the acidic medium is ____________.", "options": [], "answer": "5", "solution": "**Answer:** 5\n\nThe reaction of $\\mathrm{KMnO}_4$ and potassium iodide ($\\mathrm{KI}$) in acidic medium can be represented as follows:\n

\n$$\\mathrm{MnO_4^-} + \\mathrm{I^-} + \\mathrm{H^+} \\rightarrow \\mathrm{Mn^{2+}} + \\mathrm{I_2} + \\mathrm{H_2O}$$\n

\nIn this reaction, $\\mathrm{KMnO}_4$ acts as an oxidizing agent, while $\\mathrm{KI}$ acts as a reducing agent. The oxidation state of manganese changes from +7 in $\\mathrm{KMnO_4}$ to +2 in $\\mathrm{Mn^{2+}}$, which is a reduction of 5 units.

Therefore, the total change in the oxidation state of manganese involved in the reaction is $\\boxed{5}$.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1300, "subject": "Chemistry", "question": "

Which of the following statements are correct?

\n

(A) The M$$^{3+}$$/M$$^{2+}$$ reduction potential for iron is greater than manganese.

\n

(B) The higher oxidation states of first row d-block elements get stabilized by oxide ion.

\n

(C) Aqueous solution of Cr$$^{2+}$$ can liberate hydrogen from dilute acid.

\n

(D) Magnetic moment of V$$^{2+}$$ is observed between 4.4 - 5.2 BM.

\n

Choose the correct answer from the options given below :

", "options": [ { "text": "(B), (C) only" }, { "text": "(A), (B) only" }, { "text": "(A), (B), (D) only" }, { "text": "(C), (D) only" } ], "answer": "(B), (C) only", "solution": "**Answer:** (B), (C) only\n\nThe reduction electrode potential of $\\mathrm{Mn}^{3+} / \\mathrm{Mn}^{2+}$ is $+1.57 \\mathrm{~V}$ while that of $\\mathrm{Fe}^{3+} / \\mathrm{Fe}^{2+}$ is $+0.77 \\mathrm{~V}$, hence $\\mathrm{A}$ is wrong. Higher oxidation state of smaller $d$-block elements is stabilized (or say form compounds) with smaller anion oxide that can be explained by stearic reason hence $B$ is correct.\n

\nThe oxidation electrode potential of $\\mathrm{Cr}^{2+} / \\mathrm{Cr}^{3+}$ is $+0.41 \\mathrm{~V}$ hence it can reduce $\\mathrm{H}^{+}$and so liberate $\\mathrm{H}_2$.\nThe unpaired electrons in $\\mathrm{V}^{2+}$ are 3 hence the magnetic moment of $\\mathrm{V}^{2+}$ will be lesser than 4.4 BM.

Hence, only $B$ and $C$ are correct.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1301, "subject": "Chemistry", "question": "

In chromyl chloride, the number of d-electrons present on chromium is same as in

(Given at no. of $$\\mathrm{Ti}: 22, \\mathrm{~V}: 23, \\mathrm{Cr}: 24, \\mathrm{Mn}: 25, \\mathrm{Fe}: 26$$ )

", "options": [ { "text": "Mn (VII)" }, { "text": "Ti (III)" }, { "text": "Fe (III)" }, { "text": "V (IV)" } ], "answer": "Mn (VII)", "solution": "**Answer:** Mn (VII)\n\nWhen a mixture containing chloride ion is heated with $\\mathrm{K}_2 \\mathrm{Cr}_2 \\mathrm{O}_7$ and concentrated $\\mathrm{H}_2 \\mathrm{SO}_{4^{\\prime}}$ deep orange-red fumes of chromyl chloride $\\left(\\mathrm{CrO}_2 \\mathrm{Cl}_2\\right)$ are formed

\n$$\n\\begin{aligned}\n& \\mathrm{K}_2 \\mathrm{Cr}_2 \\mathrm{O}_7+4 \\mathrm{NaCl}+6 \\mathrm{H}_2 \\mathrm{SO}_4 \\rightarrow 2 \\mathrm{KHSO}_4+4 \\mathrm{NaHSO}_4 +2 \\mathrm{CrO}_2 \\mathrm{Cl}_2 \\uparrow+3 \\mathrm{H}_2 \\mathrm{O}\n\\end{aligned}\n$$

\nOrange-red fumes

\nSo in this case, $\\mathrm{X}$ in $\\mathrm{CrO}_2 \\mathrm{Cl}_2$.

\nOxidation state of $\\mathrm{Cl}=-1, \\mathrm{O}=-2, \\mathrm{Cr}=x$

\n$$\nx+2 \\times(-2)+2 \\times(-1)=0 \\Rightarrow x=+6\n$$

\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1302, "subject": "Chemistry", "question": "

Given below are two statements: one is labelled as \"Assertion A\" and the other is labelled as \"Reason R\"

\n

Assertion A : In the complex $$\\mathrm{Ni}(\\mathrm{CO})_{4}$$ and $$\\mathrm{Fe}(\\mathrm{CO})_{5}$$, the metals have zero oxidation state.

\n

Reason R : Low oxidation states are found when a complex has ligands capable of $$\\pi$$-donor character in addition to the $\\sigma$-bonding.

\n

In the light of the above statements, choose the most appropriate answer from the options given below

", "options": [ { "text": "A is not correct but R is correct" }, { "text": "A is correct but R is not correct" }, { "text": "Both A and R are correct but R is NOT the correct explanation of A" }, { "text": "Both A and R are correct and R is the correct explanation of A" } ], "answer": "A is correct but R is not correct", "solution": "**Answer:** A is correct but R is not correct\n\n\n
  • Option B: "A is correct but R is incorrect" is the right answer. Here is the reasoning:
    • \n\n

    Assertion A is correct. In the complex $$\\mathrm{Ni(CO)_4}$$, if we assume the oxidation state of Ni to be x, we can set up the equation:

    \n

    $\nx + (0 \\times 4) = 0\n$

    \n

    From which we find that (x = 0).

    \n

    Similarly, in $$\\mathrm{Fe(CO)_5}$$, if we let the oxidation state of Fe be y, the equation becomes:

    \n

    $\ny + (0 \\times 5) = 0\n$

    \n

    Solving this gives (y = 0).

    \n

    Therefore, the oxidation states of both Ni and Fe are zero in their respective complexes, verifying that Assertion A is true.

    \n

    Regarding Reason R, it is incorrect. The low oxidation states in these complexes are stabilized through synergic bonding, which involves the ligand being a $$\\pi$$-acceptor, not a $$\\pi$$-donor. The CO ligand in these complexes accepts electron density from the metal into its antibonding molecular orbitals, characterizing a $$\\pi$$-acid ligand, and not a $$\\pi$$-donor ligand as stated. Consequently, Reason R does not hold true.

    \n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1303, "subject": "Chemistry", "question": "

    During the reaction of permanganate with thiosulphate, the change in oxidation of manganese occurs by value of 3. Identify which of the below medium will favour the reaction.

    ", "options": [ { "text": "aqueous neutral" }, { "text": "both aqueous acidic and faintly alkaline" }, { "text": "both aqueous acidic and neutral" }, { "text": "aqueous acidic" } ], "answer": "aqueous neutral", "solution": "**Answer:** aqueous neutral\n\n

    In a neutral or weakly alkaline solution, permanganate ions are reduced to MnO₂, with a change in oxidation state from +7 to +4, which is a change of 3 units.

    \n

    The balanced redox reaction is :

    \n

    3MnO₄⁻ + 5S₂O₃²⁻ + 2H₂O → 3MnO₂ + 5SO₄²⁻ + 10OH⁻

    ", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 1304, "subject": "Chemistry", "question": "Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).

    \nAssertion (A) : In aqueous solutions $\\mathrm{Cr}^{2+}$ is reducing while $\\mathrm{Mn}^{3+}$ is oxidising in nature.

    \nReason (R) : Extra stability to half filled electronic configuration is observed than incompletely filled electronic configuration.\n

    \nIn the light of the above statements, choose the most appropriate answer from the options given below :", "options": [ { "text": "(A) is true but (R) is false" }, { "text": "Both (A) and (R) are true and (R) is the correct explanation of (A)" }, { "text": "Both (A) and (R) are true but (R) is not the correct explanation of (A)" }, { "text": "(A) is false but (R) is true" } ], "answer": "Both (A) and (R) are true and (R) is the correct explanation of (A)", "solution": "**Answer:** Both (A) and (R) are true and (R) is the correct explanation of (A)\n\n

    Let us analyze both the Assertion (A) and the Reason (R) to determine the correct answer.

    \n\n

    Assertion (A): In aqueous solutions, $\\mathrm{Cr}^{2+}$ is reducing while $\\mathrm{Mn}^{3+}$ is oxidising in nature. This assertion is true. The $\\mathrm{Cr}^{2+}$ ion has a tendency to be oxidized to $\\mathrm{Cr}^{3+}$ because $\\mathrm{Cr}^{3+}$ has a more stable electronic configuration. Therefore, $\\mathrm{Cr}^{2+}$ acts as a reducing agent. On the other hand, $\\mathrm{Mn}^{3+}$ tends to stabilize by being reduced to $\\mathrm{Mn}^{2+}$, which is a more stable electronic configuration due to having a half-filled d5 subshell ($\\mathrm{Mn}^{2+}$: [Ar] 3d5). Thus, $\\mathrm{Mn}^{3+}$ acts as an oxidizing agent.

    \n\n

    Reason (R): Extra stability to half filled electronic configuration is observed than incompletely filled electronic configuration. This reason is also true. Half-filled and fully filled electron configurations, like those found in $\\mathrm{Mn}^{2+}$ and $\\mathrm{Cr}^{0}$ respectively, are particularly stable due to symmetry and exchange energy considerations. This is a reason why certain ions like $\\mathrm{Mn}^{3+}$ want to be reduced to $\\mathrm{Mn}^{2+}$, and why $\\mathrm{Cr}^{0}$ (with a [Ar] 3d5 4s1 configuration) is not as common as $\\mathrm{Cr}^{3+}$ (which has a [Ar] 3d3 configuration, and while not half-filled, is favored due to crystal field stabilization energy considerations in octahedral complexes).

    \n\n

    If we evaluate these statements together, we see that Assertion (A) is directly related to the electronic configurations and their stability, as asserted by Reason (R). The stability of half-filled and full orbitals contributes to the observed redox behavior of $\\mathrm{Cr}^{2+}$ and $\\mathrm{Mn}^{3+}$ in aqueous solutions.

    \n\n

    Therefore, the correct answer is Option B: Both (A) and (R) are true and (R) is the correct explanation of (A).

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1305, "subject": "Chemistry", "question": "

    Which of the following electronic configuration would be associated with the highest magnetic moment?

    ", "options": [ { "text": "$$[\\mathrm{Ar}] 3 \\mathrm{~d}^7$$\n" }, { "text": "$$[\\mathrm{Ar}] 3 \\mathrm{~d}^8$$\n" }, { "text": "$$[\\mathrm{Ar}] 3 \\mathrm{~d}^3$$\n" }, { "text": "$$[\\mathrm{Ar}] 3 \\mathrm{~d}^6$$" } ], "answer": "$$[\\mathrm{Ar}] 3 \\mathrm{~d}^6$$", "solution": "**Answer:** $$[\\mathrm{Ar}] 3 \\mathrm{~d}^6$$\n\n

    The magnetic moment of an atom or ion with unpaired electrons is given by the formula $$\\mu = \\sqrt{n(n+2)}$$ Bohr magnetons (BM), where $$n$$ is the number of unpaired electrons. The magnetic moment depends on the number of unpaired electrons: more unpaired electrons lead to a higher magnetic moment.

    Looking at the given options with respect to their electron configurations and calculating their respective magnetic moments based on their unpaired electrons:

    Hence, the electron configuration associated with the highest magnetic moment is $$[\\mathrm{Ar}] 3 \\mathrm{~d}^6$$.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1306, "subject": "Chemistry", "question": "

    Choose the correct option having all the elements with $$\\mathrm{d}^{10}$$ electronic configuration from the following :

    ", "options": [ { "text": "$${ }^{46} \\mathrm{Pd},{ }^{28} \\mathrm{Ni},{ }^{26} \\mathrm{Fe},{ }^{24} \\mathrm{Cr}$$\n" }, { "text": "$${ }^{29} \\mathrm{Cu},{ }^{30} \\mathrm{Zn},{ }^{48} \\mathrm{Cd},{ }^{47} \\mathrm{Ag}$$\n" }, { "text": "$${ }^{27} \\mathrm{Co},{ }^{28} \\mathrm{Ni},{ }^{26} \\mathrm{Fe},{ }^{24} \\mathrm{Cr}$$\n" }, { "text": "$${ }^{28} \\mathrm{Ni},{ }^{24} \\mathrm{Cr},{ }^{26} \\mathrm{Fe},{ }^{29} \\mathrm{Cu}$$" } ], "answer": "$${ }^{29} \\mathrm{Cu},{ }^{30} \\mathrm{Zn},{ }^{48} \\mathrm{Cd},{ }^{47} \\mathrm{Ag}$$\n", "solution": "**Answer:** $${ }^{29} \\mathrm{Cu},{ }^{30} \\mathrm{Zn},{ }^{48} \\mathrm{Cd},{ }^{47} \\mathrm{Ag}$$\n\n\n

    $$\\begin{aligned}\n& {[\\mathrm{Cr}]=[\\mathrm{Ar}] 4 \\mathrm{~s}^1 3 \\mathrm{~d}^5} \\\\\n& {[\\mathrm{Cd}]=[\\mathrm{Kr}] 5 \\mathrm{~s}^2 4 \\mathrm{~d}^{10}} \\\\\n& {[\\mathrm{Cu}]=[\\mathrm{Ar}] 4 \\mathrm{~s}^1 3 \\mathrm{~d}^{10}} \\\\\n& {[\\mathrm{Ag}]=[\\mathrm{Kr}] 5 \\mathrm{~s}^1 4 \\mathrm{~d}^{10}} \\\\\n& {[\\mathrm{Zn}]=[\\mathrm{Ar}] 4 \\mathrm{~s}^2 3 \\mathrm{~d}^{10}}\n\\end{aligned}$$

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1307, "subject": "Chemistry", "question": "

    Which of the following statements are correct about $$\\mathrm{Zn}, \\mathrm{Cd}$$ and $$\\mathrm{Hg}$$ ?

    \n

    A. They exhibit high enthalpy of atomization as the d-subshell is full.

    \n

    B. $$\\mathrm{Zn}$$ and $$\\mathrm{Cd}$$ do not show variable oxidation state while $$\\mathrm{Hg}$$ shows $$+\\mathrm{I}$$ and $$+\\mathrm{II}$$.

    \n

    C. Compounds of $$\\mathrm{Zn}, \\mathrm{Cd}$$ and $$\\mathrm{Hg}$$ are paramagnetic in nature.

    \n

    D. $$\\mathrm{Zn}, \\mathrm{Cd}$$ and $$\\mathrm{Hg}$$ are called soft metals.

    \n

    Choose the most appropriate from the options given below:

    ", "options": [ { "text": "C, D only" }, { "text": "B, C only" }, { "text": "A, D only" }, { "text": "B, D only" } ], "answer": "B, D only", "solution": "**Answer:** B, D only\n\n

    (A) $$\\mathrm{Zn}, \\mathrm{Cd}, \\mathrm{Hg}$$ exhibit lowest enthalpy of atomization in respective transition series.

    \n

    (C) Compounds of $$\\mathrm{Zn}, \\mathrm{Cd}$$ and $$\\mathrm{Hg}$$ are diamagnetic in nature.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1308, "subject": "Chemistry", "question": "

    Number of metal ions characterized by flame test among the following is ________.

    \n

    $$\\mathrm{Sr}^{2+}, \\mathrm{Ba}^{2+}, \\mathrm{Ca}^{2+}, \\mathrm{Cu}^{2+}, \\mathrm{Zn}^{2+}, \\mathrm{Co}^{2+}, \\mathrm{Fe}^{2+}$$

    ", "options": [], "answer": "4", "solution": "**Answer:** 4\n\n

    All the following metal ions will respond to flame test.

    \n

    $$\\mathrm{Sr}^{2+}, \\mathrm{Ba}^{2+}, \\mathrm{Ca}^{2+}, \\mathrm{Cu}^{2+}$$

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1309, "subject": "Chemistry", "question": "

    Give below are two statements :

    \n

    Statement I : The higher oxidation states are more stable down the group among transition elements unlike p-block elements.

    \n

    Statement II : Copper can not liberate hydrogen from weak acids.

    \n

    In the light of the above statements, choose the correct answer from the options given below :

    ", "options": [ { "text": "Statement I is true but Statement II is false\n" }, { "text": "Both Statement I and Statement II are false\n" }, { "text": "Both Statement I and Statement II are true\n" }, { "text": "Statement I is false but Statement II is true" } ], "answer": "Both Statement I and Statement II are true\n", "solution": "**Answer:** Both Statement I and Statement II are true\n\n\n

    To evaluate the given statements, let's analyze each statement in the context of chemical principles:

    \n\n

    Statement I: The higher oxidation states are more stable down the group among transition elements unlike p-block elements.

    \n\n

    This statement is true. In transition elements, as we move down the group, the electrons are being added to the n-1 d orbitals, where 'n' represents the principal quantum number. The ability to exhibit higher oxidation states stems from the fact that not only are the s electrons of the outermost shell used in bonding, but the d electrons are also involved. Moreover, down the group, the effective nuclear charge decreases, and the shielding effect increases, so electrons from both s and d orbitals can be easily used for bonding, leading to stable higher oxidation states. This trend is opposite in the case of p-block elements, where the stability of higher oxidation states generally decreases down the group due to the inert pair effect, which makes it hard to remove electrons from the s orbital as we move down the group.

    \n\n

    Statement II: Copper can not liberate hydrogen from weak acids.

    \n\n

    This statement is also true. Copper (Cu) is less reactive and lies below hydrogen in the electrochemical series, meaning it has a higher reduction potential than hydrogen. For a metal to displace hydrogen from an acid, it must have a lower reduction potential than hydrogen. This is not the case with copper. Therefore, copper does not react with weak acids (like acetic acid) to liberate hydrogen gas. This behavior can be explained through the electrochemical series, where metals placed above hydrogen are capable of displacing hydrogen from acids (exemplified by metals such as zinc or magnesium), whereas copper does not have this ability due to its relative placement in the series.

    \n\n

    Based on the analysis, Option C is the correct answer: Both Statement I and Statement II are true.

    ", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 1310, "subject": "Chemistry", "question": "

    A transition metal '$$\\mathrm{M}$$' among $$\\mathrm{Sc}, \\mathrm{Ti}, \\mathrm{V}, \\mathrm{Cr}, \\mathrm{Mn}$$ and $$\\mathrm{Fe}$$ has the highest second ionisation enthalpy. The spin-only magnetic moment value of $$\\mathrm{M}^{+}$$ ion is _______ BM (Near integer)

    \n

    (Given atomic number $$\\mathrm{Sc}: 21, \\mathrm{Ti}: 22, \\mathrm{~V}: 23, \\mathrm{Cr}: 24, \\mathrm{Mn}: 25, \\mathrm{Fe}: 26$$)

    ", "options": [], "answer": "6", "solution": "**Answer:** 6\n\n

    To solve this problem, we need to understand what factors influence the second ionisation enthalpy of transition metals and how to calculate the spin-only magnetic moment of their ions. The second ionization enthalpy refers to the energy required to remove an electron from the singly charged cation of an element.

    \n\n

    Generally, for transition metals, the removal of an electron from a d-orbital requires less energy compared to removing an electron from an s-orbital of a valence shell due to the penetration effect and the shield. However, among the given options, the highest second ionisation enthalpy would correspond to the element whose cation (after losing one electron) attains a half-filled or completely filled d-orbital configuration due to the extra stability associated with such electronic arrangements.

    \n\n

    Let's examine the electronic configurations of the cations formed after the first ionization (this helps in understanding the stability and hence the energy required for the second ionization):

    \n\n\n

    Among these configurations, $$\\mathrm{Mn}^+$$ ([Ar] $$3d^5$$) has a half-filled d-orbital, lending it extra stability due to exchange energy and symmetry. Consequently, $$\\mathrm{Mn}$$ has the highest second ionization enthalpy among the given elements, as it would require more energy to remove an electron from a stable $$\\mathrm{Mn}^+$$ ion.

    \n\n

    Now, to find the spin-only magnetic moment of $$\\mathrm{Mn}^+$$, we use the formula:

    \n

    $$\\mu = \\sqrt{n(n + 2)}\\, \\text{B.M.}$$

    \n

    where $$n$$ is the number of unpaired electrons. In the case of $$\\mathrm{Mn}^+$$, the electronic configuration is [Ar] $$3d^5$$, which means all five d-electrons are unpaired.

    \n

    Thus, $$n = 5$$, and substituting in the formula for magnetic moment:

    \n

    $$\\mu = \\sqrt{5(5 + 2)} = \\sqrt{35} \\approx 5.92\\, \\text{B.M.}$$

    \n

    Therefore, the spin-only magnetic moment value of $$\\mathrm{Mn}^+$$ ion is approximately 5.92 BM, and the nearest integer would be 6 BM.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1311, "subject": "Chemistry", "question": "

    The electronic configuration of $$\\mathrm{Cu}(\\mathrm{II})$$ is $$3 \\mathrm{~d}^9$$ whereas that of $$\\mathrm{Cu}(\\mathrm{I})$$ is $$3 \\mathrm{~d}^{10}$$. Which of the following is correct?

    ", "options": [ { "text": "$$\\mathrm{Cu}(\\mathrm{II})$$ is more stable\n" }, { "text": "$$\\mathrm{Cu}$$ (II) is less stable\n" }, { "text": "$$\\mathrm{Cu}(\\mathrm{I})$$ and $$\\mathrm{Cu}(\\mathrm{II})$$ are equally stable\n" }, { "text": "Stability of $$\\mathrm{Cu}(\\mathrm{I})$$ and $$\\mathrm{Cu}(\\mathrm{II})$$ depends on nature of copper salts" } ], "answer": "$$\\mathrm{Cu}(\\mathrm{II})$$ is more stable\n", "solution": "**Answer:** $$\\mathrm{Cu}(\\mathrm{II})$$ is more stable\n\n\n

    $$\\mathrm{Cu}^{2+}$$ is more stable.

    \n

    $$\\mathrm{Cu}^{+} \\longrightarrow \\mathrm{Cu}^{2+}+\\mathrm{Cu}$$ disproportionation reaction

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1312, "subject": "Chemistry", "question": "

    The element which shows only one oxidation state other than its elemental form is :

    ", "options": [ { "text": "Nickel\n" }, { "text": "Titanium\n" }, { "text": "Cobalt\n" }, { "text": "Scandium" } ], "answer": "Scandium", "solution": "**Answer:** Scandium\n\n

    The element among the given options that shows only one oxidation state other than its elemental form is Scandium.

    \n\n

    Explanation:

    \n\n\n\n

    Therefore, the correct answer is Option D: Scandium.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1313, "subject": "Chemistry", "question": "

    A first row transition metal with highest enthalpy of atomisation, upon reaction with oxygen at high temperature forms oxides of formula $$\\mathrm{M}_2 \\mathrm{O}_{\\mathrm{n}}$$ (where $$\\mathrm{n}=3,4,5$$). The 'spin-only' magnetic moment value of the amphoteric oxide from the above oxides is _________ $$\\mathrm{BM}$$ (near integer)

    \n

    (Given atomic number: $$\\mathrm{Sc}: 21, \\mathrm{Ti}: 22, \\mathrm{~V}: 23, \\mathrm{Cr}: 24, \\mathrm{Mn}: 25, \\mathrm{Fe}: 26, \\mathrm{Co}: 27, \\mathrm{Ni}: 28, \\mathrm{Cu}: 29, \\mathrm{Zn}: 30$$)

    ", "options": [], "answer": "0", "solution": "**Answer:** 0\n\n

    Vanadium has highest enthalpy of atomization among first row transition elements.

    \n

    $$\\mathrm{V}_2 \\mathrm{O}_5$$ is amphoteric

    \n

    In $$\\mathrm{V}^{5+}$$ there are no unpaired electrons.

    \n

    Thus, $$\\mu=0$$

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1314, "subject": "Chemistry", "question": "

    Iron (III) catalyses the reaction between iodide and persulphate ions, in which

    \n

    A. $$\\mathrm{Fe}^{3+}$$ oxidises the iodide ion

    \n

    B. $$\\mathrm{Fe}^{3+}$$ oxidises the persulphate ion

    \n

    C. $$\\mathrm{Fe}^{2+}$$ reduces the iodide ion

    \n

    D. $$\\mathrm{Fe}^{2+}$$ reduces the persulphate ion

    \n

    Choose the most appropriate answer from the options given below:

    ", "options": [ { "text": "B and C only\n" }, { "text": "A only\n" }, { "text": "A and D only\n" }, { "text": "B only" } ], "answer": "A and D only\n", "solution": "**Answer:** A and D only\n\n\n

    To determine the correct answer to this question, let's analyze the catalysis mechanism involving Iron (III) in the reaction between iodide ($$\\mathrm{I^-}$$) and persulfate ($$\\mathrm{S_2O_8^{2-}}$$) ions.

    \n\n

    Iron (III), or $$\\mathrm{Fe}^{3+}$$, can act as a catalyst by undergoing redox reactions, cycling between $$\\mathrm{Fe}^{2+}$$ and $$\\mathrm{Fe}^{3+}$$. Here’s how it typically works:

    \n\n

    1. $$\\mathrm{Fe}^{3+}$$ can oxidize iodide ions to iodine. The reaction will proceed as follows:

    \n\n

    $$\\mathrm{2Fe^{3+} + 2I^- \\rightarrow 2Fe^{2+} + I_2}$$

    \n\n

    2. $$\\mathrm{Fe}^{2+}$$, now formed, can be oxidized back to $$\\mathrm{Fe}^{3+}$$ by reducing persulfate ions. The reaction will be:

    \n\n

    $$\\mathrm{2Fe^{2+} + S_2O_8^{2-} \\rightarrow 2Fe^{3+} + 2SO_4^{2-}}$$

    \n\n

    Based on this mechanism, $$\\mathrm{Fe}^{3+}$$ is responsible for oxidizing the iodide ions, and $$\\mathrm{Fe}^{2+}$$ is responsible for reducing the persulfate ions. Thus, we can conclude:

    \n\n

    A. $$\\mathrm{Fe}^{3+}$$ oxidises the iodide ion: This statement is correct.

    \n\n

    B. $$\\mathrm{Fe}^{3+}$$ oxidises the persulphate ion: This statement is incorrect because $$\\mathrm{Fe}^{3+}$$ does not oxidize persulfate ions.

    \n\n

    C. $$\\mathrm{Fe}^{2+}$$ reduces the iodide ion: This statement is incorrect because $$\\mathrm{Fe}^{2+}$$ does not reduce iodide ions.

    \n\n

    D. $$\\mathrm{Fe}^{2+}$$ reduces the persulphate ion: This statement is correct.

    \n\n

    Therefore, the most appropriate answer is:

    \n\n

    Option C: A and D only

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1315, "subject": "Chemistry", "question": "

    The metal that shows highest and maximum number of oxidation state is :

    ", "options": [ { "text": "Co" }, { "text": "Mn" }, { "text": "Fe" }, { "text": "Ti" } ], "answer": "Mn", "solution": "**Answer:** Mn\n\n

    The metal that shows the highest and maximum number of oxidation states among the options given is manganese (Mn), which is option B. Manganese exhibits a wide range of oxidation states from +2 to +7, which is more diverse than the other metals mentioned. To further detail:

    \n\n\n\n

    Therefore, the correct answer is Option B: Mn (Manganese), which can show the highest and maximum number of oxidation states among the options provided.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1316, "subject": "Chemistry", "question": "

    The number of ions from the following that have the ability to liberate hydrogen from a dilute acid is _________.

    \n

    $$\\mathrm{Ti}^{2+}, \\mathrm{Cr}^{2+} \\text { and } \\mathrm{V}^{2+}$$

    ", "options": [ { "text": "1" }, { "text": "2" }, { "text": "3" }, { "text": "0" } ], "answer": "3", "solution": "**Answer:** 3\n\n

    To determine the ability of the ions to liberate hydrogen from a dilute acid, we need to understand the standard electrode potentials of these ions in comparison to hydrogen. The standard electrode potential of hydrogen ($$\\mathrm{H}^{+}/\\mathrm{H}_{2}$$) is set at 0 volts. Any metal ion with a negative standard reduction potential compared to hydrogen has the ability to liberate hydrogen gas from dilute acids because they are more likely to donate electrons to the $$\\mathrm{H}^{+}$$ ions found in acids, reducing them to $$\\mathrm{H}_{2}$$ gas.

    \n\n

    The standard reduction potentials for the ions mentioned are as follows:

    \n\n\n\n

    Due to their negative standard reduction potentials relative to hydrogen, all three ions ($$\\mathrm{Ti}^{2+}, \\mathrm{Cr}^{2+}, \\mathrm{V}^{2+}$$) have the capacity to liberate hydrogen from a dilute acid. Therefore, the correct answer is:

    \n\n

    Option C: 3

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1317, "subject": "Chemistry", "question": "

    Arrange the following elements in the increasing order of number of unpaired electrons in it.

    \n

    (A) $$\\mathrm{Sc}$$

    \n

    (B) $$\\mathrm{Cr}$$

    \n

    (C) $$\\mathrm{V}$$

    \n

    (D) $$\\mathrm{Ti}$$

    \n

    (E) $$\\mathrm{Mn}$$

    \n

    Choose the correct answer from the options given below :

    ", "options": [ { "text": "$$\n(\\mathrm{A})<(\\mathrm{D})<(\\mathrm{C})<(\\mathrm{B})<(\\mathrm{E})\n$$" }, { "text": "$$\n(\\mathrm{A})<(\\mathrm{D})<(\\mathrm{C})<(\\mathrm{E})<(\\mathrm{B})\n$$" }, { "text": "$$\n(\\mathrm{B})<(\\mathrm{C})<(\\mathrm{D})<(\\mathrm{E})<(\\mathrm{A})\n$$" }, { "text": "$$\n(\\mathrm{C})<(\\mathrm{E})<(\\mathrm{B})<(\\mathrm{A})<(\\mathrm{D})\n$$" } ], "answer": "$$\n(\\mathrm{A})<(\\mathrm{D})<(\\mathrm{C})<(\\mathrm{E})<(\\mathrm{B})\n$$", "solution": "**Answer:** $$\n(\\mathrm{A})<(\\mathrm{D})<(\\mathrm{C})<(\\mathrm{E})<(\\mathrm{B})\n$$\n\n

    Electronic configuration of the given metals

    \n

    (A) $$\\mathrm{Sc} \\Rightarrow 3 \\mathrm{~d}^1 4 \\mathrm{~s}^2 \\Rightarrow \\mathrm{n}=1$$

    \n

    (B) $$\\mathrm{Cr} \\Rightarrow 3 \\mathrm{~d}^5 4 \\mathrm{~s}^1 \\Rightarrow \\mathrm{n}=6$$

    \n

    (C) $$\\mathrm{V} \\Rightarrow 3 \\mathrm{~d}^3 4 \\mathrm{~s}^2 \\Rightarrow \\mathrm{n}=3$$

    \n

    (D) $$\\mathrm{Ti} \\Rightarrow 3 \\mathrm{~d}^2 4 \\mathrm{~s}^2 \\Rightarrow \\mathrm{n}=2$$

    \n

    (E) $$\\mathrm{M n \\Rightarrow 3 d^5 4 s^2 \\Rightarrow n=5}$$

    \n

    $$\\therefore \\quad$$ Correct order of increasing number of unpaired electrons.

    \n

    $$\\text { A }<\\text { D }<\\text { C }<\\mathrm{E}<\\mathrm{B}$$

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1318, "subject": "Chemistry", "question": "

    Among $$\\mathrm{VO}_2^{+}, \\mathrm{MnO}_4^{-}$$ and $$\\mathrm{Cr}_2 \\mathrm{O}_7^{2-}$$, the spin-only magnetic moment value of the species with least oxidising ability is __________ BM (Nearest integer).

    \n

    (Given atomic member $$\\mathrm{V}=23, \\mathrm{Mn}=25, \\mathrm{Cr}=24$$)

    ", "options": [], "answer": "0", "solution": "**Answer:** 0\n\n

    In order to determine the spin-only magnetic moment value of the species with the least oxidizing ability, we first need to analyze their oxidation states and electron configurations.

    \n\n

    1. $$\\mathrm{VO}_2^{+}$$:

    \n\n

    For vanadium in $$\\mathrm{VO}_2^{+}$$, the oxidation state is +5. The electronic configuration of V is $$[Ar] \\, 3d^3 \\, 4s^2$$. Thus, in +5 oxidation state, vanadium will have zero d-electrons (since it loses 5 electrons) and, consequently, $$\\mathrm{VO}_2^{+}$$ is diamagnetic (since zero unpaired electrons).

    \n\n

    2. $$\\mathrm{MnO}_4^{-}$$:

    \n\n

    For manganese in $$\\mathrm{MnO}_4^{-}$$, the oxidation state is +7. The electronic configuration of Mn is $$[Ar] \\, 3d^5 \\, 4s^2$$. Thus, in +7 oxidation state, manganese will have zero d-electrons (since it loses 7 electrons) and, consequently, $$\\mathrm{MnO}_4^{-}$$ is also diamagnetic (since zero unpaired electrons).

    \n\n

    3. $$\\mathrm{Cr}_2 \\mathrm{O}_7^{2-}$$:

    \n\n

    For chromium in $$\\mathrm{Cr}_2 \\mathrm{O}_7^{2-}$$, we have two chromium atoms. Each chromium is in the +6 oxidation state. The electronic configuration of Cr is $$[Ar] \\, 3d^5 \\, 4s^1$$. Thus, in +6 oxidation state, each chromium will have zero d-electrons (since it loses 6 electrons) and, consequently, $$\\mathrm{Cr}_2 \\mathrm{O}_7^{2-}$$ is also diamagnetic (since zero unpaired electrons).

    \n\n

    From these observations, we note that all the species we analyzed are diamagnetic. However, to identify the species with the least oxidizing ability, we look at their standard reduction potentials (E° values).\n\n

    Typically, the oxidizing ability increases with increasing positive E° values. Given the nature of these species, we can infer that:

    \n\n\n\n

    Therefore, among the given species, $$\\mathrm{VO}_2^{+}$$ has the least oxidizing ability.

    \n\n

    As mentioned previously, $$\\mathrm{VO}_2^{+}$$ has zero unpaired electrons, making it diamagnetic. The spin-only magnetic moment is given by the formula:

    \n\n

    $$\\mu = \\sqrt{n(n+2)}$$

    \n\n

    where $$n$$ is the number of unpaired electrons.

    \n\n

    For $$\\mathrm{VO}_2^{+}$$, $$n=0$$, thus:

    \n\n

    $$\\mu = \\sqrt{0(0+2)} = 0 \\ \\text{BM}$$

    \n\n

    Given the options, the nearest integer value of the spin-only magnetic moment for the species with the least oxidizing ability (which is $$\\mathrm{VO}_2^{+}$$) is indeed:

    \n\n

    0 BM

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1319, "subject": "Chemistry", "question": "

    Among $$\\mathrm{CrO}, \\mathrm{Cr}_2 \\mathrm{O}_3$$ and $$\\mathrm{CrO}_3$$, the sum of spin-only magnetic moment values of basic and amphoteric oxides is _________ $$10^{-2} \\mathrm{BM}$$ (nearest integer).

    \n

    (Given atomic number of $$\\mathrm{Cr}$$ is 24 )

    ", "options": [], "answer": "877", "solution": "**Answer:** 877\n\n

    First, we need to understand the oxidation states of chromium in the given compounds and determine their magnetic moments based on their electronic configurations.

    \n\n

    1. $$\\mathrm{CrO}$$: In $$\\mathrm{CrO}$$, the oxidation state of chromium is +2. The electronic configuration of chromium (Cr) is $$1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^1$$. In the +2 oxidation state, two electrons are removed, typically from the 4s and one of the 3d orbitals, leaving us with the configuration $$3d^4$$.

    \n\n

    To find the spin-only magnetic moment, we use the formula:

    \n\n

    $$\\mu = \\sqrt{n(n+2)} \\mathrm{BM}$$

    \n\n

    where n is the number of unpaired electrons. For $$\\mathrm{Cr}^{2+}$$, we have 4 unpaired electrons in the 3d orbitals.

    \n\n

    Thus,

    \n\n

    $$\\mu = \\sqrt{4(4+2)} = \\sqrt{24} \\approx 4.90 \\mathrm{BM}$$

    \n\n

    2. $$\\mathrm{Cr}_2 \\mathrm{O}_3$$: In $$\\mathrm{Cr}_2 \\mathrm{O}_3$$, the oxidation state of chromium is +3. The electronic configuration of $$\\mathrm{Cr}^{3+}$$ is $$3d^3$$ after losing three electrons.

    \n\n

    For $$\\mathrm{Cr}^{3+}$$, there are 3 unpaired electrons.

    \n\n

    Thus,

    \n\n

    $$\\mu = \\sqrt{3(3+2)} = \\sqrt{15} \\approx 3.87 \\mathrm{BM}$$

    \n\n

    3. $$\\mathrm{CrO}_3$$: In $$\\mathrm{CrO}_3$$, the oxidation state of chromium is +6. The electronic configuration of $$\\mathrm{Cr}^{6+}$$ is $$3d^0$$ after losing six electrons. There are no unpaired electrons for $$\\mathrm{Cr}^{6+}$$.

    \n\n

    Since $$\\mathrm{Cr}^{6+}$$ has no unpaired electrons, its spin-only magnetic moment is 0 BM.

    \n\n

    The magnetic moment values for each compound are:

    \n\n\n\n

    The sum of the spin-only magnetic moments of the basic and amphoteric oxides is:

    \n\n

    $$4.90 + 3.87 = 8.77 \\mathrm{BM}$$

    \n\n

    Expressing in terms of $$10^{-2} \\mathrm{BM}$$,

    \n\n

    $$8.77 \\mathrm{BM} \\times 100 = 877 (\\times 10^{-2} \\mathrm{BM})$$

    \n\n

    Thus, the sum of spin-only magnetic moment values of basic and amphoteric oxides is approximately 877 $$\\times 10^{-2} \\mathrm{BM}$$ (nearest integer).

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1320, "subject": "Chemistry", "question": "Several blocks of magnesium are fixed to the bottom of a ship to :", "options": [ { "text": "make the ship lighter" }, { "text": "prevent action of water and salt" }, { "text": "prevent puncturing by under-sea rocks" }, { "text": "keep away the sharks" } ], "answer": "prevent action of water and salt", "solution": "**Answer:** prevent action of water and salt\n\nMagnesium provides cathodic protection and prevent rusting or corrosion. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1321, "subject": "Chemistry", "question": "In a hydrogen – oxygen fuel cell, combustion of hydrogen occurs to :", "options": [ { "text": "generate heat " }, { "text": "remove adsorbed oxygen from electrode surfaces " }, { "text": "produce high purity water" }, { "text": "create potential difference between the two electrodes" } ], "answer": "create potential difference between the two electrodes", "solution": "**Answer:** create potential difference between the two electrodes\n\nIn $${H_2} - {O_2}$$ fuel cell, the combustion of $${H_2}$$ occurs to create potential difference between the two electrodes ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1322, "subject": "Chemistry", "question": "In a fuel cell methanol is used as fuel and oxygen gas is used as an oxidizer. The reaction is
    \nCH3OH(l) + 3/2O2 $$\\to$$ CO2 (g) + 2H2O (l)
    \nAt 298K standard Gibb’s energies of formation for CH3OH(l), H2O(l) and CO2 (g) are -166.2, -237.2 and -394.4 kJ mol−1 respectively. If standard enthalpy of combustion of methanol is -726 kJ mol−1, efficiency of the fuel cell will be ", "options": [ { "text": "87%" }, { "text": "90%" }, { "text": "97%" }, { "text": "80%" } ], "answer": "97%", "solution": "**Answer:** 97%\n\n$$C{H_3}OH\\left( l \\right) + {3 \\over 2}{O_2}\\left( g \\right)$$\n

    $$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, \\to C{O_2}\\left( g \\right) + 2{H_2}O\\left( l \\right)$$\n

    $$\\Delta {G_r} = \\Delta {G_f}\\left( {C{O_2},g} \\right) + 2\\Delta {G_f}\\left( {{H_2}O,l} \\right) - $$\n

    $$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\Delta {G_f}\\left( {C{H_3}OH,l} \\right) - {3 \\over 2}\\Delta {G_f}\\left( {{O_2},g} \\right)$$\n

    $$ = - 394.4 + 2\\left( { - 237.2} \\right) - \\left( { - 166.2} \\right) - 0$$\n

    $$ = - 394.4 - 474.4 + 166.2$$\n

    $$ = - 702.6\\,kJ$$\n

    $$\\% \\,\\,$$ efficiency $$\\,\\, = {{702.6} \\over {726}} \\times 100$$ $$ = 97\\% $$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1323, "subject": "Chemistry", "question": "Galvanization is applying a coating of :", "options": [ { "text": "Cr" }, { "text": "Cu" }, { "text": "Zn" }, { "text": "Pb" } ], "answer": "Zn", "solution": "**Answer:** Zn\n\nGalvanization is the process by which zinc is coated over corrosive (easily rusted) metals to prevent them from corrosion. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1324, "subject": "Chemistry", "question": "Identify the correct statement :", "options": [ { "text": "Iron corrodes in oxygen-free water." }, { "text": "Iron corrodes more rapidly in salt water because its electrochemical\npotential is higher." }, { "text": "Corrosion of iron can be minimized by forming a contact with another\nmetal with a higher reduction potential." }, { "text": "Corrosion of iron can be minimized by forming an impermeable barrier\nat its surface." } ], "answer": "Corrosion of iron can be minimized by forming an impermeable barrier\nat its surface.", "solution": "**Answer:** Corrosion of iron can be minimized by forming an impermeable barrier\nat its surface.\n\n

    The two conditions necessary for the corrosion of iron to take place are presence of moisture and oxygen. Factors that catalyse the process of rusting are the presence of carbon dioxide, acids and impurities. It can be minimised by introducing of a barrier film between surface of iron and atmosphere. This can be done by

    \n\n

    (i) painting the surface.

    \n\n

    (ii) coating the surface with oil or grease.

    \n\n

    (iii) electroplating iron with non-corrosive metals such as nickel or chromium.

    \n\n

    (iv) covering the surface of iron with layer of more active metal with higher oxidation potential like zinc.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1325, "subject": "Chemistry", "question": "Compound A used as a strong oxidizing agent is amphoteric in nature. It is the part of lead storage batteries. Compound A is :", "options": [ { "text": "PbO" }, { "text": "PbO2" }, { "text": "PbSO4" }, { "text": "Pb3O4" } ], "answer": "PbO2", "solution": "**Answer:** PbO2\n\nPbO2 is strong oxidizing agent because Pb+4 is\nnot stable and can be easily reduced to Pb+2.

    \nPbO2 is used in lead storage batteries. It is\nalso amphoteric in nature.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1326, "subject": "Chemistry", "question": "

    Match List - I with List - II.

    \n

    \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    List - IList - II
    (A)$$Cd(s) + 2Ni{(OH)_3}(s) \\to CdO(s) + 2Ni{(OH)_2}(s) + {H_2}O(l)$$(I)Primary battery
    (B)$$Zn(Hg) + HgO(s) \\to ZnO(s) + Hg(l)$$(II)Discharging of secondary battery
    (C)$$2PbS{O_4}(s) + 2{H_2}O(l) \\to Pb(s) + Pb{O_2}(s) + 2{H_2}S{O_4}(aq)$$(III)Fuel cell
    (D)$$2{H_2}(g) + {O_2}(g) \\to 2{H_2}O(l)$$(IV)Charging of secondary battery

    \n

    Choose the correct answer from the options given below:

    ", "options": [ { "text": "$$(\\mathrm{A})-(\\mathrm{I}),(\\mathrm{B})-(\\mathrm{II}),(\\mathrm{C})-(\\mathrm{III}),(\\mathrm{D})-(\\mathrm{IV})$$" }, { "text": "$$(\\mathrm{A})-(\\mathrm{IV}),(\\mathrm{B})-(\\mathrm{I}),(\\mathrm{C})-(\\mathrm{II}),(\\mathrm{D})-(\\mathrm{III})$$" }, { "text": "$$(\\mathrm{A})-(\\mathrm{II}),(\\mathrm{B})-(\\mathrm{I}),(\\mathrm{C})-(\\mathrm{IV}),(\\mathrm{D})-(\\mathrm{III})$$" }, { "text": "$$(\\mathrm{A})-(\\mathrm{II}),(\\mathrm{B})-(\\mathrm{I}),(\\mathrm{C})-(\\mathrm{III}),(\\mathrm{D})-(\\mathrm{IV})$$" } ], "answer": "$$(\\mathrm{A})-(\\mathrm{II}),(\\mathrm{B})-(\\mathrm{I}),(\\mathrm{C})-(\\mathrm{IV}),(\\mathrm{D})-(\\mathrm{III})$$", "solution": "**Answer:** $$(\\mathrm{A})-(\\mathrm{II}),(\\mathrm{B})-(\\mathrm{I}),(\\mathrm{C})-(\\mathrm{IV}),(\\mathrm{D})-(\\mathrm{III})$$\n\n(a) $\\mathrm{Cd}(\\mathrm{s})+2 \\mathrm{Ni}(\\mathrm{OH})_3(\\mathrm{~s}) \\rightarrow \\mathrm{CdO}(\\mathrm{s})+2 \\mathrm{Ni}(\\mathrm{OH})_2(\\mathrm{~s})$ $+\\mathrm{H}_2 \\mathrm{O}(l)$\nDischarge of secondary Battery

    \n(b) $\\mathrm{Zn}(\\mathrm{Hg})+\\mathrm{HgO}$ (s) $\\rightarrow \\mathrm{ZnO}(\\mathrm{s})+\\mathrm{Hg}(l)$\n(Primary Battery Mercury cell)

    \n(c) $2 \\mathrm{PbSO}_4(\\mathrm{~s})+2 \\mathrm{H}_2 \\mathrm{O}(l) \\rightarrow \\mathrm{Pb}(\\mathrm{s})+\\mathrm{PbO}_2(\\mathrm{~s})+$\n$2 \\mathrm{H}_2 \\mathrm{SO}_4(\\mathrm{aq})$\n(Charging of secondary Battery)

    \n(d) $2 \\mathrm{H}_2(\\mathrm{~g})+\\mathrm{O}_2(\\mathrm{~g}) \\rightarrow 2 \\mathrm{H}_2 \\mathrm{O}(l)$ (Fuel cell)", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1327, "subject": "Chemistry", "question": "

    For lead storage battery pick the correct statements

    \n

    A. During charging of battery, $$\\mathrm{PbSO}_{4}$$ on anode is converted into $$\\mathrm{PbO}_{2}$$

    \n

    B. During charging of battery, $$\\mathrm{PbSO}_{4}$$ on cathode is converted into $$\\mathrm{PbO}_{2}$$

    \n

    C. Lead storage battery consists of grid of lead packed with $$\\mathrm{PbO}_{2}$$ as anode

    \n

    D. Lead storage battery has $$\\sim 38 \\%$$ solution of sulphuric acid as an electrolyte

    \n

    Choose the correct answer from the options given below:

    ", "options": [ { "text": "A, B, D only" }, { "text": "B, C only" }, { "text": "B, C, D only" }, { "text": "B, D only" } ], "answer": "B, D only", "solution": "**Answer:** B, D only\n\n

    Let's review each statement:

    \n

    A. During charging of battery, PbSO₄ on anode is converted into PbO₂.

    \n\n

    B. During charging of battery, PbSO₄ on cathode is converted into PbO₂.

    \n\n

    C. Lead storage battery consists of grid of lead packed with PbO₂ as anode.

    \n\n

    D. Lead storage battery has ∼38% solution of sulphuric acid as an electrolyte.

    \n\n

    Therefore, the correct answer is:

    \n

    B, D only.

    \n", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 1328, "subject": "Chemistry", "question": "

    Which of the following statements is not correct about rusting of iron?

    ", "options": [ { "text": "Rusting of iron is envisaged as setting up of electrochemical cell on the surface of iron object.\n" }, { "text": "Dissolved acidic oxides $$\\mathrm{SO}_2, \\mathrm{NO}_2$$ in water act as catalyst in the process of rusting.\n" }, { "text": "Coating of iron surface by tin prevents rusting, even if the tin coating is peeling off.\n" }, { "text": "When $$\\mathrm{pH}$$ lies above 9 or 10, rusting of iron does not take place." } ], "answer": "Coating of iron surface by tin prevents rusting, even if the tin coating is peeling off.\n", "solution": "**Answer:** Coating of iron surface by tin prevents rusting, even if the tin coating is peeling off.\n\n\n

    The statement that is not correct about the rusting of iron is Option C. Let's examine each option one by one:

    \n\n

    Option A: Rusting of iron is envisaged as setting up of an electrochemical cell on the surface of iron object. This statement is correct. Rusting of iron is an electrochemical process that occurs when iron (Fe) reacts with water (H2O) and oxygen (O2) to form hydrated iron(III) oxide, which we commonly know as rust. Iron acts as the anode, the part where oxidation occurs as iron loses electrons. Oxygen, often dissolved in water, acts as the cathode where reduction occurs, completing the electrochemical cell.

    \n\n

    Option B: Dissolved acidic oxides such as $$\\mathrm{SO}_2$$ and $$\\mathrm{NO}_2$$ in water act as catalyst in the process of rusting. This statement is correct. Acidic oxides like $$\\mathrm{SO}_2$$ and $$\\mathrm{NO}_2$$ can dissolve in water to form acidic solutions which can lead to the acceleration of the rate of rusting as they may lower the pH of water, thereby increasing the availability of hydrogen ions (H+) which may facilitate the electrochemical reactions involved in rusting.

    \n\n

    Option C: Coating of iron surface by tin prevents rusting, even if the tin coating is peeling off. This statement is incorrect. Tin (Sn) is used as a protective coating for iron to prevent it from rusting; however, if the tin coating starts to peel off, it exposes the iron underneath to the atmosphere, which will then be susceptible to rusting. Once the protective layer is compromised, the iron is at risk, especially if the exposed parts are more anodic than tin. Tin as a more cathodic metal can even act to accelerate the corrosion of the exposed iron—this is known as a bimetallic corrosion or galvanic corrosion.

    \n\n

    Option D: When $$\\mathrm{pH}$$ lies above 9 or 10, rusting of iron does not take place. This statement is generally correct. Rusting of iron is lessened in alkaline conditions, i.e., when the pH is above 9 or 10 because the increased availability of hydroxide ions (OH-) can reduce the solubility of iron(II) and iron(III) ions, leading to their precipitation as hydroxides rather than participating in the rusting process.

    \n\n

    Therefore, the statement from the given options that is not correct about the rusting of iron is Option C, which falsely claims that a peeling tin surface can still prevent rusting of iron.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1329, "subject": "Chemistry", "question": "

    Match List I with List II

    \n

    \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    LIST I
    (Cell)    		LIST II
    (Use/Property/Reaction)
    A.Leclanche cellI.Converts energy of combustion into electrical energy
    B.\tNi - Cd cellII.Does not involve any ion in solution and is used in hearing aids
    C.Fuel cellIII.Rechargeable
    D.Mercury cellIV.Reaction at anode $$\\mathrm{Zn} \\rightarrow \\mathrm{Zn}^{2+}+2 \\mathrm{e}^{-}$$

    \n

    Choose the correct answer from the options given below :

    ", "options": [ { "text": "A-III, B-I, C-IV, D-II\n" }, { "text": "A-I, B-II, C-III, D-IV\n" }, { "text": "A-IV, B-III, C-I, D-II\n" }, { "text": "A-II, B-III, C-IV, D-I" } ], "answer": "A-IV, B-III, C-I, D-II\n", "solution": "**Answer:** A-IV, B-III, C-I, D-II\n\n\n

    To find the correct match between List I (Cell) and List II (Use/Property/Reaction), let's first understand each item and its corresponding match:

    \n\n

    List I (Cell):

    \n\n\n

    List II (Use/Property/Reaction):

    \n\n\n

    Given these explanations, the correct matches would be:

    \n\n\n

    Therefore, the correct answer is Option C: A-IV, B-III, C-I, D-II.

    ", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1330, "subject": "Chemistry", "question": "

    Fuel cell, using hydrogen and oxygen as fuels,

    \n

    A. has been used in spaceship

    \n

    B. has as efficiency of $$40 \\%$$ to produce electricity

    \n

    C. uses aluminum as catalysts

    \n

    D. is eco-friendly

    \n

    E. is actually a type of Galvanic cell only

    \n

    Choose the correct answer from the options given below:

    ", "options": [ { "text": "A, D, E only\n" }, { "text": "A, B, C only\n" }, { "text": "A, B, D only\n" }, { "text": "A, B, D, E only" } ], "answer": "A, D, E only\n", "solution": "**Answer:** A, D, E only\n\n\n

    Fuel cells produce electricity with an efficiency of about $$70 \\%$$.

    \n

    Fuel cells are pollution free, thus, eco-friendly.

    \n

    Fuel cells are type of Galvanic cells only.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1331, "subject": "Chemistry", "question": "

    The reaction at cathode in the cells commonly used in clocks involves.

    \n", "options": [ { "text": "oxidation of $$\\mathrm{Mn}$$ from +2 to +7\n" }, { "text": "reduction of $$\\mathrm{Mn}$$ from +4 to +3\n" }, { "text": "oxidation of $$\\mathrm{Mn}$$ from +3 to +4\n" }, { "text": "reduction of $$\\mathrm{Mn}$$ from +7 to +2" } ], "answer": "reduction of $$\\mathrm{Mn}$$ from +4 to +3\n", "solution": "**Answer:** reduction of $$\\mathrm{Mn}$$ from +4 to +3\n\n\n

    In the cells commonly used in clocks, specifically alkaline batteries, the reaction at the cathode involves the reduction of manganese dioxide (MnO2). The manganese in MnO2 is initially in the +4 oxidation state, and it gets reduced to the +3 oxidation state. Thus, the correct reaction occurring at the cathode can be represented as follows:

    \n\n

    The correct answer is:

    \n\n

    Option B
    reduction of $$\\mathrm{Mn}$$ from +4 to +3

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1332, "subject": "Chemistry", "question": "Conductivity (Seimen’s S) is directly proportional to area of the vessel and the concentration\nof the solution in it and is inversely proportional to the length of the vessel then, then constant of proportionality is expressed in :", "options": [ { "text": "Sm mol-1" }, { "text": "Sm2 mol-1" }, { "text": "S-2m2 mol" }, { "text": "S2m2 mol-2" } ], "answer": "Sm2 mol-1", "solution": "**Answer:** Sm2 mol-1\n\nGiven $$S \\propto {{area\\, \\times \\,conc} \\over \\ell } = {{\\kappa {m^2}mol} \\over {m \\times {m^3}}}$$\n

    $$\\therefore$$ $$\\,\\,\\,\\,\\kappa = S{m^2}mo{l^{ - 1}}$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1333, "subject": "Chemistry", "question": "When during electrolysis of a solution of AgNO3, 9650 coulombs of charge pass through the electroplating\nbath, the mass of silver deposited on the cathode will be :", "options": [ { "text": "10.8 g" }, { "text": "21.6 g" }, { "text": "108 g" }, { "text": "1.08 g" } ], "answer": "10.8 g", "solution": "**Answer:** 10.8 g\n\nWhen $$96500$$ coulomb of electricity is passed through the electroplating bath the amount of Ag deposited $$=108g$$\n

    $$\\therefore$$ when $$9650$$ coulomb of electricity is passed deposited Ag. \n

    $$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, = {{108} \\over {96500}} \\times 9650 = 10.8\\,g$$ ", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 1334, "subject": "Chemistry", "question": "The limiting molar conductivities Λ° for NaCl, KBr and KCl are 126, 152 and 150 S cm2 mol-1 respectively. The Λ° for NaBr is ", "options": [ { "text": "128 S cm2 mol-1" }, { "text": "278 S cm2 mol-1" }, { "text": "176 S cm2 mol-1" }, { "text": "302 S cm2 mol-1" } ], "answer": "128 S cm2 mol-1", "solution": "**Answer:** 128 S cm2 mol-1\n\n$${A^ \\circ }Nacl = {\\lambda ^ \\circ }N{a^ + } + \\lambda C{l^ - }\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,...\\left( i \\right)$$\n

    $${A^ \\circ }KBr = {\\lambda ^ \\circ }{K^ + } + \\lambda Br\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,...\\left( {ii} \\right)$$\n

    $${A^ \\circ }KCl = {\\lambda ^ \\circ }{K^ + } + \\lambda C{l^ - }\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,...\\left( {iii} \\right)$$\n

    operating $$(i)+(ii)-(iii)$$\n

    $${A^ \\circ }NaBr = {\\lambda ^ \\circ }N{a^ + } + {\\lambda ^ \\circ }B{r^ - }$$\n

    $$ = 126 + 152 - 150$$\n

    $$ = 128\\,\\,\\,S\\,c{m^2}\\,mo{l^{ - 1}}$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1335, "subject": "Chemistry", "question": "Aluminium oxide may be electrolysed at 1000oC to furnish aluminium metal (Atomic\nmass = 27 amu; 1 Faraday = 96,500 Coulombs). The cathode reaction is Al3+ + 3e- $$\\to$$ Alo To prepare 5.12 kg of aluminium metal by this method would require ", "options": [ { "text": "5.49 $$\\times$$ 107 C of electricity " }, { "text": "1.83 $$\\times$$ 107 C of electricity " }, { "text": "5.49 $$\\times$$ 104 C of electricity " }, { "text": "5.49 $$\\times$$ 101 C of electricity " } ], "answer": "5.49 $$\\times$$ 107 C of electricity ", "solution": "**Answer:** 5.49 $$\\times$$ 107 C of electricity \n\n$$1$$ mole of $${e^ - } = 1F = 96500\\,C$$ \n

    $$27g$$ of $$Al$$ is deposited by $$3 \\times 96500\\,C$$\n

    $$5120$$ $$g$$ of $$Al$$ will be deposited by \n

    $$ = {{3 \\times 96500 \\times 5120} \\over {27}}$$\n

    $$ = 5.49 \\times {10^7}C$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1336, "subject": "Chemistry", "question": "The highest electrical conductivity of the following aqueous solutions is of :", "options": [ { "text": "0.1 M acetic acid" }, { "text": "0.1 M chloroacetic acid" }, { "text": "0.1 M fluoroacetic acid " }, { "text": "0.1 M difluoroacetic acid " } ], "answer": "0.1 M difluoroacetic acid ", "solution": "**Answer:** 0.1 M difluoroacetic acid \n\nThus difluoro acetic acid being strongest acid will furnish maximum number of ions showing highest electrical conductivity. The decreasing acidic strength of the carboxylic acids given is difluoro acetic acid $$>$$ fluoro acetic acid $$>$$ chloro acitic acid $$>$$ acetic acid. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1337, "subject": "Chemistry", "question": "
    \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    Electrolyte:KClKNO3HClNaOAcNaCl
    $${ \\wedge ^\\infty }(Sc{m^2}mo{l^{ - 1}}):$$
    		149.9145426.291126.5
    \n
    \nCalculate $$ \\wedge _{HOAc}^\\infty $$ Using appropriate molar conductances of the electrolytes listed\nabove at infinite dilution in H2O at 25oC", "options": [ { "text": "517.2 " }, { "text": "552.7 " }, { "text": "390.7" }, { "text": "217.5 " } ], "answer": "390.7", "solution": "**Answer:** 390.7\n\n$$A_{HCl}^\\infty = 426.2\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,...\\left( i \\right)$$ \n

    $$A_{AcONa}^\\infty = 91.0\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,...\\left( {ii} \\right)$$\n

    $$A_{NaCl}^\\infty = 126.5\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,...\\left( {iii} \\right)$$\n

    $$A_{AcOH}^\\infty = \\left( i \\right) + \\left( {ii} \\right) - \\left( {iii} \\right)$$\n

    $$ = \\left[ {426.2 + 91.0 - 126.5} \\right]$$\n

    $$ = 390.7$$ ", "topic": "Discrete Mathematics", "subtopic": "Graph Theory" }, { "id": 1338, "subject": "Chemistry", "question": "The molar conductivities $$ \\wedge _{NaOAc}^o$$ and $$ \\wedge _{HCl}^o$$ and at infinite dilution in water at 25oC are 91.0 and 426.2 Scm2/mol respectively. To calculate $$ \\wedge _{HOAc}^o$$ , the additional value required is ", "options": [ { "text": "$$ \\wedge _{{H_2}O}^o$$" }, { "text": "$$ \\wedge _{KCl}^o$$ " }, { "text": "$$ \\wedge _{NaOH}^o$$" }, { "text": "$$ \\wedge _{NaCl}^o$$" } ], "answer": "$$ \\wedge _{NaCl}^o$$", "solution": "**Answer:** $$ \\wedge _{NaCl}^o$$\n\n$$\\Lambda _{C{H_3}COOH}^o$$ is given by the following equation\n

    $$\\Lambda _{C{H_3}COOH}^o = \\left( {\\Lambda _{C{H_3}COONa}^o + \\Lambda _{HCl}^o} \\right) - \\left( {\\Lambda _{NaCl}^o} \\right)$$\n

    Hence $$\\Lambda _{NaCl}^ \\circ $$ is required.", "topic": "Discrete Mathematics", "subtopic": "Graph Theory" }, { "id": 1339, "subject": "Chemistry", "question": "Resistance of a conductivity cell filled with a solution of an electrolyte of concentration 0.1 M is 100$$\\Omega $$.\nThe conductivity of this solution is 1.29 S m–1. Resistance of the same cell when filled with 0.2 M of\nthe same solution is 520 $$\\Omega $$, The molar conductivity of 0.02 M solution of the electrolyte will be ", "options": [ { "text": "124 $$\\times$$ 10–4 S m2 mol–1" }, { "text": "1240 $$\\times$$ 10–4 S m2 mol–1" }, { "text": "1.24 $$\\times$$ 10–4 S m2 mol–1" }, { "text": "12.4 $$\\times$$ 10–4 S m2 mol–1" } ], "answer": "12.4 $$\\times$$ 10–4 S m2 mol–1", "solution": "**Answer:** 12.4 $$\\times$$ 10–4 S m2 mol–1\n\n$$R = 100\\Omega ,\\kappa = {1 \\over R}\\left( {{l \\over a}} \\right),{l \\over a}$$ (cell constant) \n

    $$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,$$ $$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,$$ $$ = 1.29 \\times 100{m^{ - 1}}$$\n

    Given, $$R = 520\\Omega ,C = 0.2M,\\,\\mu $$ (molar conductivity) $$=$$ ?\n

    $$\\mu = \\kappa \\times V\\,\\,\\,\\,\\,\\,\\,\\,$$ ($$\\kappa $$ can be calculated as $$\\kappa = {1 \\over R}\\left( {{1 \\over a}} \\right)$$\n

    now cell constant is known.\n

    Hence, $$\\mu = {1 \\over {520}} \\times 129 \\times {{1000} \\over {0.2}} \\times {10^{ - 6}}{m^3}$$ \n

    $$ = 12.4 \\times {10^{ - 4}}\\,S{m^2}mo{l^{ - 1}}$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1340, "subject": "Chemistry", "question": "The equivalent conductances of two strong electrolytes at infinite dilution in H2O (where ions move\nfreely through a solution) at 25oC are given below:
    \n$$ \\wedge _{C{H_3}COONa}^o$$ = 91.0 S cm2/equiv
    \n$$ \\wedge _{HCl}^o$$ = 426.2 S cm2/equiv
    \nWhat additional information/quantity one needs to calculate $$ \\wedge ^o$$ of an aqueous solution of acetic acid? ", "options": [ { "text": "$$ \\wedge ^o$$ of chloroacetic acid (C/CH2COOH)" }, { "text": "$$ \\wedge ^o$$ of NaCl" }, { "text": "$$ \\wedge ^o$$ of CH3COOK" }, { "text": "The limiting equivalent conductance of $${H^ + }( \\wedge _{{H^ + }}^o)$$" } ], "answer": "$$ \\wedge ^o$$ of NaCl", "solution": "**Answer:** $$ \\wedge ^o$$ of NaCl\n\nNOTE : According to Kohlrausch's law, molar conductivity of weak electrolyte acetic acid $$\\left( {C{H_3}COOH} \\right)$$ can be calculated as follows: \n

    $${\\Lambda ^o}_{C{H_3}COOH} = \\left( {{\\Lambda ^o}_{C{H_3}COONa} + {\\Lambda ^o}_{HCl}} \\right) - {\\Lambda ^o}_{NaCl}$$\n

    $$\\therefore$$ Value of $${\\Lambda ^o}_{NaCl}$$ should also be known \n

    for calculating value of $${\\Lambda ^o}_{C{H_3}COOH}$$", "topic": "Discrete Mathematics", "subtopic": "Graph Theory" }, { "id": 1341, "subject": "Chemistry", "question": "Resistance of 0.2 M solution of an electrolyte is 50 $$\\Omega$$. The specific conductance of the solution is 1.4 S m-1. The resistance of 0.5 M solution of the same electrolyte is 280 $$\\Omega$$. The molar conductivity of 0.5 M solution of the electrolyte in S m2 mol-1 is :", "options": [ { "text": "5 × 103" }, { "text": "5 × 102" }, { "text": "5 × 10-4" }, { "text": "5 × 10-3" } ], "answer": "5 × 10-4", "solution": "**Answer:** 5 × 10-4\n\nGiven for $$0.2$$ $$M$$ solution\n

    $$R = 50\\Omega $$\n

    $$\\kappa = 1.45\\,S\\,{m^{ - 1}} = 1.4 \\times {10^{ - 2}}\\,S\\,c{m^{ - 1}}$$\n

    Now, $$R = \\rho {\\ell \\over a} = {1 \\over \\kappa } \\times {\\ell \\over a}$$\n

    $$ \\Rightarrow {\\ell \\over a} = R \\times \\kappa = 50 \\times 1.4 \\times {10^{ - 2}}$$\n

    For $$0.5$$ $$M$$ solution\n

    $$R = 280\\,\\Omega ;\\,\\kappa = ?$$\n

    $${\\ell \\over a} = 50 \\times 1.4 \\times {10^{ - 2}}$$\n

    $$ \\Rightarrow R = \\rho {\\ell \\over a} = {1 \\over \\kappa } \\times {\\ell \\over a}$$\n

    $$ \\Rightarrow \\,\\,\\,\\,\\kappa = {1 \\over {280}} \\times 50 \\times 1.4 \\times {10^{ - 2}}$$ \n

    $$ = {1 \\over {280}} \\times 70 \\times {10^{ - 2}} = 2.5 \\times {10^{ - 3}}\\,S\\,c{m^{ - 1}}$$ \n

    Now, $${\\Lambda _m} = {{\\kappa \\times 1000} \\over M} = {{2.5 \\times {{10}^{ - 3}} \\times 1000} \\over {0.5}}$$\n

    $$ = 5\\,S\\,c{m^2}mo{l^{ - 1}} = 5 \\times {10^{ - 4}}\\,S\\,{m^2}\\,mo{l^{ - 1}}$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1342, "subject": "Chemistry", "question": "The equivalent conductance of NaCl at concentration C and at infinite dilution are $${\\lambda _C}$$ and $${\\lambda _\\infty }$$, respectively. The correct relationship between $${\\lambda _C}$$ and $${\\lambda _\\infty }$$ is given as:\n(where the constant B is positive)", "options": [ { "text": "$${\\lambda _C} = {\\lambda _\\infty } + (B)C$$" }, { "text": "$${\\lambda _C} = {\\lambda _\\infty } - (B)C$$" }, { "text": "$${\\lambda _C} = {\\lambda _\\infty } - (B)\\sqrt C$$" }, { "text": "$${\\lambda _C} = {\\lambda _\\infty } + (B)\\sqrt C$$" } ], "answer": "$${\\lambda _C} = {\\lambda _\\infty } - (B)\\sqrt C$$", "solution": "**Answer:** $${\\lambda _C} = {\\lambda _\\infty } - (B)\\sqrt C$$\n\nAccording to Debye Huckle onsager equation,\n

    $$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,{\\lambda _C} = {\\lambda _\\infty } - B\\sqrt C $$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1343, "subject": "Chemistry", "question": "Two Faraday of electricity is passed through a solution of CuSO4. The mass of copper deposited at the\ncathode is: (at. mass of Cu = 63.5 amu)", "options": [ { "text": "63.5 g" }, { "text": "2 g" }, { "text": "127 g" }, { "text": "0 g" } ], "answer": "63.5 g", "solution": "**Answer:** 63.5 g\n\n$$C{u^{2 + }} + 2{e^ - }\\buildrel \\, \\over\n \\longrightarrow Cu$$\n

    $$2F\\,\\,i.e.\\,\\,2 \\times 96500\\,C$$ deposit $$Cu=1$$ mol $$=63.5$$ $$g$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1344, "subject": "Chemistry", "question": "Oxidation of succinate ion produces ethylene and carbon dioxide gases. On passing 0.2 Faraday electricity through an aqueous solution of potassium succinate, the total volume of gases (at both cathode and anode) at STP (1 atm and 273 K) is :", "options": [ { "text": "2.24 L" }, { "text": "4.48 L" }, { "text": "6.72 L " }, { "text": "8.96 L " } ], "answer": "8.96 L ", "solution": "**Answer:** 8.96 L \n\n

    The reaction is

    \n

    $$2C{H_3}COOK + 2{H_2}O\\buildrel {Electrolysis} \\over\n \\longrightarrow C{H_3} - C{H_3} + 2C{O_2} + {H_2} + 2KOH$$

    \n

    At the anode (Oxidation):

    \n

    \"JEE

    \n

    At the cathode (Reduction):

    \n

    $$2{H_2}O\\buildrel { + 2{e^ - }} \\over\n \\longrightarrow 2O{H^ - } + 2\\mathop H\\limits^ \\bullet $$

    \n

    $$2\\mathop H\\limits^ \\bullet \\buildrel {} \\over\n \\longrightarrow {H_2}$$

    \n

    Total number of moles of gases

    \n

    = moles of C2H6 + moles of CO2 + moles of H2

    \n

    $$n = {{0.2} \\over 2} + {{0.2} \\over 1} + {{0.2} \\over 2} = 0.4$$

    \n

    $$V = {{nRT} \\over p}$$

    \n

    $$ = {{(0.4 \\times 0.0821 \\times 273)} \\over 1} = 8.96$$ L

    ", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1345, "subject": "Chemistry", "question": "How long (approximate) should water be electrolysed by passing through 100 amperes current so that the\noxygen released can completely burn 27.66 g of diborane?
    \n(Atomic weight of B = 10.8 u)", "options": [ { "text": "1.6 hours" }, { "text": "6.4 hours" }, { "text": "0.8 hours" }, { "text": "3.2 hours" } ], "answer": "3.2 hours", "solution": "**Answer:** 3.2 hours\n\nRequired reaction : \n

    B2H6 + 3O2 $$ \\to $$ B2 O3 + 3 H2 O\n

    Here molar mass of B2H6 =10.8 $$ \\times $$ 2 + 6 = 27.6 gm\n

    Given weight of B2H6 = 27.66 g\n

    $$\\therefore\\,\\,\\,\\,$$No of moles of B2H6 = $${{27.6} \\over {27.66}} \\simeq 1$$ mole.\n

    For combustion of 1 mole B2H6 3 moles O2 required. \n

    This 3 mole of O2 is obtained by electrolysis of H2O.\n

    2H2O($$l$$) $$ \\to $$ O2 (g) + 4 H+ (aq) + 4 e$$-$$ \n

    From Faradays law of electrolysis, \n

    moles $$ \\times $$ nf = $${{It} \\over {96500}}$$\n

    Here moles of O2 = 3.\n

    Nf of O2 = 4 (in H2 change of O = $$-$$2 \n

    and in O2 change of 0 = O. \n

    So change in charge = 2 .\n

    for two atoms of O2 change in charge = 2 $$ \\times $$ 2 = 4)\n

    $$\\therefore\\,\\,\\,\\,$$ 3 $$ \\times $$ 4 = $${{100 \\times t} \\over {96500}}$$\n

    $$ \\Rightarrow \\,\\,\\,\\,$$ t = 12 $$ \\times $$ 965 sec. \n

    $$ \\Rightarrow \\,\\,\\,\\,$$ t = $${{12 \\times 965} \\over {60 \\times 60}}$$ hr\n

    = 3.2 hr", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1346, "subject": "Chemistry", "question": "When an electric currents passed through acidified water, 112 mL of hydrogen gas at N.T.P. was collected at the cathode in 965 seconds. The current passed, in ampere, is : ", "options": [ { "text": "1.0" }, { "text": "0.5" }, { "text": "0.1" }, { "text": "2.0" } ], "answer": "1.0", "solution": "**Answer:** 1.0\n\nReaction at cathode : \n

    2H+ + 2e$$-$$ $$ \\to $$ H2\n

    We know, \n

    $$\\omega $$ = zIt = $${{EIt} \\over {96500}}$$\n
    <
    no. of moles of H2 = $${{112} \\over {22400}}$$\n

    $$\\therefore\\,\\,\\,$$ mass (w) of H2 = $${{112} \\over {22400}}$$ $$ \\times $$ 2\n

    $$\\therefore\\,\\,\\,\\,$$ $${{112} \\over {22400}}$$ $$ \\times $$ 2 = $${{1 \\times I \\times 965} \\over {96500}}$$\n

    $$ \\Rightarrow $$$$\\,\\,\\,\\,$$ $$I$$ = 1 A", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1347, "subject": "Chemistry", "question": "When 9.65 ampere current was passed for 1.0 hour into nitrobenzene in acidic medium, the amount of p-aminophenol produced is : ", "options": [ { "text": "9.81 g" }, { "text": "10.9 g" }, { "text": "98.1 g" }, { "text": "109.0 g" } ], "answer": "9.81 g", "solution": "**Answer:** 9.81 g\n\n\"JEE\n
    Moler mass of p$$-$$aminophenol \n

    =   6 $$ \\times $$ 12 + 7 + 14 + 16\n

    = 109 gmol$$-$$1\n

    Eq. weight (E) = $${W \\over Q}$$ $$ \\times $$ 96500\n

    $$ \\Rightarrow $$$$\\,\\,\\,$$ W = $${{E\\,Q} \\over {96500}}$$\n

    $$ \\Rightarrow $$$$\\,\\,\\,$$ W = $${{E\\,I\\,t} \\over {96500}}$$\n

    $$ \\Rightarrow $$$$\\,\\,\\,$$ W = $${{109} \\over 4} \\times {{9.65 \\times 3600} \\over {96500}}$$\n

    $$ \\Rightarrow $$$$\\,\\,\\,$$ W = 9.81 gm", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1348, "subject": "Chemistry", "question": "Consider the statements S1 and S2\n

    S1 : Conductivity always increases with decrease in the concentration of electrolyte.\n

    S2 : Molar conductivity always increases with decrease in the concentration of electrolyte.\n

    The correct option among the following is :


    ", "options": [ { "text": "Both S1 and S2 are wrong" }, { "text": "S1 is correct and S2 is wrong" }, { "text": "Both S1 and S2 are correct" }, { "text": "S1 is wrong and S2 is correct " } ], "answer": "S1 is wrong and S2 is correct ", "solution": "**Answer:** S1 is wrong and S2 is correct \n\nWe know conductivity (k) = $${G \\over V}$$\n

    V = volume\n

    When concentration decreases volume increases and when volume increases then conductivity (k) decreases.\n

    So, we can say S1 is incorrect.\n

    We know that,\n

    $${\\lambda _m} = {k \\over c}$$\n

    where\n

    $$\\lambda $$m = molar conductivity\n
    k = conductivity\n
    c = concentration\n

    So, when concentration decreases molar conductivity increases.\n

    So, we can say S2 is correct.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1349, "subject": "Chemistry", "question": "$$ \\wedge _m^ \\circ $$ for NaCl, HCl and NaA are 126.4, 425.9 and 100.5 S cm2 mol–1, respectively. If the conductivity of 0.001 M HA is- \n

    5 $$ \\times $$ 10–5 S cm–1, degree of dissociation of HA is - ", "options": [ { "text": "0.50" }, { "text": "0.125" }, { "text": "0.25" }, { "text": "0.75" } ], "answer": "0.125", "solution": "**Answer:** 0.125\n\n$$ \\wedge _m^ \\circ $$ (HA) = $$ \\wedge _m^ \\circ $$ (HCl) + $$ \\wedge _m^ \\circ $$ (NaA) $$-$$ $$ \\wedge _m^ \\circ $$ (NaCl)\n

    = 425.9 + 100.5 $$-$$ 126.4\n

    = 400 S cm2 . mol$$-$$1\n

        $$ \\wedge _m^ c $$ = $${{K \\times 1000} \\over M}$$\n

    = $${{5 \\times {{10}^{ - 5}} \\times 1000} \\over {{{10}^{ - 3}}}}$$\n

    = 50 S cm2 mol$$-$$1\n

    $$\\alpha $$ = $${{\\Lambda _m^c} \\over {\\Lambda _m^o}}$$ = $${{50} \\over {400}}$$ = 0.125", "topic": "Discrete Mathematics", "subtopic": "Graph Theory" }, { "id": 1350, "subject": "Chemistry", "question": "A solution of Ni(NO3)2 is electrolysed between\nplatinum electrodes using 0.1 Faraday\nelectricity. How many mole of Ni will be\ndeposited at the cathode?", "options": [ { "text": "0.10" }, { "text": "0.15" }, { "text": "0.20" }, { "text": "0.05" } ], "answer": "0.05", "solution": "**Answer:** 0.05\n\nCathode reaction :\n

    Ni+2 + 2e- $$ \\to $$ Ni(s)\n

    $$ \\therefore $$ From 2 mole of electrons 1 mole of Ni is deposited at the cathode.\n

    So from 0.1 F or 0.1 mole of electrons $${1 \\over 2} \\times 0.1$$ = 0.05 mole of Ni is deposited at the cathode.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1351, "subject": "Chemistry", "question": "Potassium chlorate is prepared by the\nelectrolysis of KCl in basic solution\n

    6OH- + Cl- $$ \\to $$ ClO3- + 3H2O + 6e-\n

    If only 60% of the current is utilized in the\nreaction, the time (rounded to the nearest hour)\nrequired to produce 10 g of KClO3 using a\ncurrent of 2 A is_________.\n

    (Given : F = 96,500 C mol–1; molar mass of\nKCIO3 = 122 g mol–1)", "options": [], "answer": "11", "solution": "**Answer:** 11\n\nFor synthesis of 1 mole of ClO3- , 6F of charge\nis required.\n

    $$ \\therefore $$ To synthesise\n$${{10} \\over {122}}$$ moles of KClO3,\n

    Charge required = $${{10} \\over {122}} \\times 6$$ F\n

    $${{10} \\over {122}} \\times 6 = {{2 \\times t(hr) \\times 3600} \\over {96500}} \\times {{60} \\over {100}}$$\n

    $$ \\Rightarrow $$ t(hr) = $${{965 \\times 100} \\over {122 \\times 2 \\times 36}}$$ = 10.98 hr $$ \\simeq $$ 11 Hr", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1352, "subject": "Chemistry", "question": "250 mL of a waste solution obtained from the\nworkshop of a goldsmith contains 0.1 M AgNO3\nand 0.1 M AuCl. The solution was electrolyzed\nat 2V by passing a current of 1A for 15\nminutes. The metal/metals electrodeposited will\nbe\n

    [ $$E_{A{g^ + }/Ag}^0$$ = 0.80 V, $$E_{A{u^ + }/Au}^0$$ = 1.69 V ]", "options": [ { "text": "Silver and gold in equal mass proportion" }, { "text": "Silver and gold in proportion to their atomic\nweights" }, { "text": "Only gold" }, { "text": "Only silver" } ], "answer": "Only gold", "solution": "**Answer:** Only gold\n\nMillimoles of Au+ = 0.1 × 250 = 25\n

    Mole of Au+ = $${{25} \\over {1000}}$$ = $${1 \\over {40}}$$\n

    Charge passed = I × t = 1 × 15 × 60 = 900 C\n

    moles of e– passed = $${{900} \\over {96500}}$$ = $${9 \\over {965}}$$\n

    Only gold will be deposited as quantity of\ncharge passed is less than the amount of Au+\npresent.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1353, "subject": "Chemistry", "question": "An acidic solution of dichromate is electrolyzed\nfor 8 minutes using 2A current. As per the\nfollowing equation\n
    Cr2O72-\n + 14H+ + 6e– $$ \\to $$ 2Cr3+ + 7H2O\n
    The amount of Cr3+ obtained was 0.104 g. The\nefficiency of the process(in%) is\n(Take : F = 96000 C, At. mass of chromium = 52)\n______.\n", "options": [], "answer": "60", "solution": "**Answer:** 60\n\nCr2O72-\n + 14H+ + 6e– $$ \\to $$ 2Cr3+ + 7H2O\n

    I = 2 A, t = 8 min\n

    From 1st law of faraday,\n

    wCr+3 = z $$ \\times $$ i $$ \\times $$ t\n

    $$ \\Rightarrow $$ wCr+3 = $${{52} \\over {96000 \\times 3}} \\times 2 \\times 8 \\times 60$$\n

    = $${{52} \\over {300}}$$\n

    Mass of Cr3+ ions actually obtained = 0.104 gm\n

    % efficiency = \n
    Actual obtained Amt
    \n
    Theo. obtained Amt
    \n     $$ \\times $$ 100\n

    = $${{0.104} \\over {{{52} \\over {300}}}}$$ $$ \\times $$ 100\n

    = 60 %", "topic": "Algebra", "subtopic": "Equations and Inequalities" }, { "id": 1354, "subject": "Chemistry", "question": "Let CNaCl \nand CBaSO4 be the conductances (in S) measured for saturated aqueous solutions of NaCl\nand BaSO4, respectively, at a temperature T.\nWhich of the following is false?\n", "options": [ { "text": "Ionic mobilities of ions from both salts increase with T." }, { "text": "CNaCl(T2) > CNaCl(T1) for T2 > T1" }, { "text": "CBaSO4(T2) > CBaSO4(T1) for T2 > T1" }, { "text": "CNaCl >> CBaSO4 at a given T" } ], "answer": "CNaCl >> CBaSO4 at a given T", "solution": "**Answer:** CNaCl >> CBaSO4 at a given T\n\nBaSO4 is sparingly\nsoluble\nsalt but NaCl is completely soluble salt so it will produce more number of ions. That is why\n
    Conductance (NaCl) > Conductance (BaSO4)", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1355, "subject": "Chemistry", "question": "108 g of silver (molar mass 108 g mol–1) is\ndeposited at cathode from AgNO3(aq) solution\nby a certain quantity of electricity. The\nvolume (in L) of oxygen gas produced at\n273 K and 1 bar pressure from water by the\nsame quantity of electricity is _______.", "options": [], "answer": "5.66to5.68", "solution": "**Answer:** 5.66to5.68\n\nCathode : Ag+(aq) + e- $$ \\to $$ Ag(s)\n

    Moles of Ag deposited = $${{108} \\over {108}}$$ = 1 mole\n

    Anode : 2H2O $$ \\to $$ O2 + 4H+ + 4e-\n

    Here we have to find volume of O2 evolved.\n

    Equivalance of Ag = Equivalance of O2\n

    $$ \\Rightarrow $$ 1 $$ \\times $$ 1 = nO2 $$ \\times $$ 4\n

    $$ \\Rightarrow $$ nO2 = $${1 \\over 4}$$ mol\n

    $$ \\therefore $$ Volume of O2 evolved \n

    = $${1 \\over 4}$$ $$ \\times $$ 22.4\n

    = 5.6 lit", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1356, "subject": "Chemistry", "question": "The equation that is incorrect is :", "options": [ { "text": "$${\\left( {\\Lambda _m^0} \\right)_{KCl}} - {\\left( {\\Lambda _m^0} \\right)_{NaCl}} = {\\left( {\\Lambda _m^0} \\right)_{KBr}} - {\\left( {\\Lambda _m^0} \\right)_{NaBr}}$$" }, { "text": "$${\\left( {\\Lambda _m^0} \\right)_{NaBr}} - {\\left( {\\Lambda _m^0} \\right)_{NaI}} = {\\left( {\\Lambda _m^0} \\right)_{KBr}} - {\\left( {\\Lambda _m^0} \\right)_{NaBr}}$$" }, { "text": "$${\\left( {\\Lambda _m^0} \\right)_{NaBr}} - {\\left( {\\Lambda _m^0} \\right)_{NaCl}} = {\\left( {\\Lambda _m^0} \\right)_{KBr}} - {\\left( {\\Lambda _m^0} \\right)_{KCl}}$$" }, { "text": "$${\\left( {\\Lambda _m^0} \\right)_{{H_2}O}} = {\\left( {\\Lambda _m^0} \\right)_{HCl}} + {\\left( {\\Lambda _m^0} \\right)_{NaOH}} - {\\left( {\\Lambda _m^0} \\right)_{NaCl}}$$" } ], "answer": "$${\\left( {\\Lambda _m^0} \\right)_{NaBr}} - {\\left( {\\Lambda _m^0} \\right)_{NaI}} = {\\left( {\\Lambda _m^0} \\right)_{KBr}} - {\\left( {\\Lambda _m^0} \\right)_{NaBr}}$$", "solution": "**Answer:** $${\\left( {\\Lambda _m^0} \\right)_{NaBr}} - {\\left( {\\Lambda _m^0} \\right)_{NaI}} = {\\left( {\\Lambda _m^0} \\right)_{KBr}} - {\\left( {\\Lambda _m^0} \\right)_{NaBr}}$$\n\nLeft hand side :\n
    $${\\left( {\\Lambda _m^0} \\right)_{NaBr}} - {\\left( {\\Lambda _m^0} \\right)_{NaI}}$$ = $${\\left( {\\Lambda _m^0} \\right)_{Br^-}} - {\\left( {\\Lambda _m^0} \\right)_{I^-}}$$ ....(1)\n

    Right hand side :\n

    $${\\left( {\\Lambda _m^0} \\right)_{KBr}} - {\\left( {\\Lambda _m^0} \\right)_{NaBr}}$$ = $${\\left( {\\Lambda _m^0} \\right)_{K^+}} - {\\left( {\\Lambda _m^0} \\right)_{Na^+}}$$ ....(2)\n

    According to Kohlrausch's law option (B) is incorrect as (1) and (2) are not equal.", "topic": "Algebra", "subtopic": "Equations and Inequalities" }, { "id": 1357, "subject": "Chemistry", "question": "Consider the following reaction

    $$MnO_4^ - + 8{H^ + } + 5{e^ - } \\to M{n^{ + 2}} + 4{H_2}O,{E^o} = 1.51V$$.

    The quantity of electricity required in Faraday to reduce five moles of $$MnO_4^ - $$ is ___________. (Integer answer)", "options": [], "answer": "25", "solution": "**Answer:** 25\n\n$$MnO_4^ - + 8{H^ + } + 5{e^ - } \\to M{n^{ + 2}} + 4{H_2}O,{E^o} = 1.51V$$\n

    1 mole of MnO4- required 5 moles of\nelectrons or 5 F electricity.\n

    $$ \\therefore $$ 5 moles of MnO4- required 25 F electricity.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1358, "subject": "Chemistry", "question": "A 5.0 m mol dm$$-$$3 aqueous solution of KCl has a conductance of 0.55 mS when measured in a cell of cell constant 1.3 cm$$-$$1. The molar conductivity of this solution is ___________ mSm2 mol$$-$$1. (Round off to the Nearest Integer).", "options": [], "answer": "14", "solution": "**Answer:** 14\n\nConductance = $${{Conductivity} \\over {Cell\\,cons\\tan t}}$$

    $$ \\therefore $$ Conductivity = 0.55 $$\\times$$ 10$$-$$3 $$\\times$$ 1.3 S cm$$-$$1

    Molar conductivity = $${{Conductivity\\,(S\\,c{m^{ - 1}}) \\times 1000} \\over {Molarity\\,(mol/L)}}$$

    $$ = {{0.55 \\times {{10}^{ - 3}} \\times 1.3 \\times 100} \\over {5 \\times {{10}^{ - 3}}}}$$

    = 143 S cm2 mol$$-$$1

    = 14.3 mS m2 mol$$-$$1

    $$ \\approx $$ 14 mS m2 mol$$-$$1", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1359, "subject": "Chemistry", "question": "A KCl solution of conductivity 0.14 S m$$-$$1 shows a resistance of 4.19$$\\Omega$$ in a conductivity cell. If the same cell is filled with an HCl solution, the resistance drops to 1.03$$\\Omega$$. The conductivity of the HCl solution is ____________ $$\\times$$ 10$$-$$2 S m$$-$$1. (Round off to the Nearest Integer).", "options": [], "answer": "57", "solution": "**Answer:** 57\n\nFor KCl solution,

    $$R = \\left( {{1 \\over K}} \\right)\\left( {{l \\over A}} \\right) \\Rightarrow {l \\over A} = R \\times K = 4.19 \\times 0.14$$

    = 0.58

    For HCl solution,

    $$R = \\left( {{1 \\over K}} \\right)\\left( {{l \\over A}} \\right)$$

    $$ \\Rightarrow K = {{(l/A)} \\over R} = {{0.58} \\over {1.03}} = 0.56 = 56 \\times {10^{ - 2}}$$ Sm$$-$$1", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1360, "subject": "Chemistry", "question": "Potassium chlorate is prepared by electrolysis of KCl in basic solution as shown by following equation.

    6OH$$-$$ + Cl$$-$$ $$\\to$$ ClO3$$-$$ + 3H2O + 6e$$-$$

    A current of xA has to be passed for 10h to produce 10.0g of potassium chlorate. The value of x is ____________. (Nearest integer)

    (Molar mass of KClO3 = 122.6 g mol$$-$$1, F = 96500 C)", "options": [], "answer": "1", "solution": "**Answer:** 1\n\n$$W = {E \\over F} \\times I \\times t$$

    $$10 = {{122.6} \\over {96500 \\times 6}} \\times x \\times 10 \\times 3600$$

    $$x = 1.311$$

    Ans. (1)", "topic": "Algebra", "subtopic": "Equations and Inequalities" }, { "id": 1361, "subject": "Chemistry", "question": "The conductivity of a weak acid HA of concentration 0.001 mol L$$-$$1 is 2.0 $$\\times$$ 10$$-$$5 S cm$$-$$1. If $$\\Lambda _m^o$$(HA) = 190 S cm2 mol$$-$$1, the ionization constant (Ka) of HA is equal to ______________ $$\\times$$ 10$$-$$6. (Round off to the Nearest Integer)", "options": [], "answer": "12", "solution": "**Answer:** 12\n\n$$\\Lambda _m^{} = 1000 \\times {\\kappa \\over M}$$

    $$ = 1000 \\times {{2 \\times {{10}^{ - 5}}} \\over {0.001}} = 20$$ S cm2 mol$$-$$1

    $$ \\Rightarrow \\alpha = {{\\Lambda _m^{}} \\over {\\Lambda _m^\\infty }} = {{20} \\over {190}} = \\left( {{2 \\over {19}}} \\right)$$

    HA $$\\rightleftharpoons$$ H+ + A$$-$$

    $$0.001(1 - \\alpha )0.001\\alpha 0.001\\alpha $$

    $$ \\Rightarrow {k_a} = 0.001\\left( {{{{\\alpha ^2}} \\over {1 - \\alpha }}} \\right) = {{0.001 \\times {{\\left( {{2 \\over {19}}} \\right)}^2}} \\over {1 - \\left( {{2 \\over {19}}} \\right)}}$$

    $$ = 12.3 \\times {10^{ - 6}}$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1362, "subject": "Chemistry", "question": "Given below are two statements :

    Statement I : The limiting molar conductivity of KCl (strong electrolyte) is higher compared to that of CH3COOH (weak electrolyte).

    Statement II : Molar conductivity decreases with decrease in concentration of electrolyte.

    In the light of the above statements, choose the most appropriate answer from the options given below :", "options": [ { "text": "Statement I is true but Statement II is false." }, { "text": "Statement I is false but Statement II is true." }, { "text": "Both Statement I and Statement II are true." }, { "text": "Both Statement I and Statement II are false." } ], "answer": "Both Statement I and Statement II are false.", "solution": "**Answer:** Both Statement I and Statement II are false.\n\n\n\n \n \n \n \n \n \n \n\n\n \n \n \n \n \n \n \n\n
    Ion$${H^ + }$$$${K^ + }$$$$C{l^ - }$$$$C{H_3}CO{O^ - }$$
    $$\\Lambda _{m\\,Sc{m^2}/mole}^\\infty $$349.873.576.340.9


    So,

    $$\\Lambda _{m\\,\\,C{H_3}COOH}^\\infty = \\Lambda _{m\\,\\,({H^ + })}^\\infty + \\Lambda _{m\\,\\,C{H_3}CO{O^ - }}^\\infty $$

    = 349.8 + 40.9 = 390.7 Scm2/mole

    $$\\Lambda _{m\\,\\,KCl}^\\infty = \\Lambda _{m\\,\\,({K^ + })}^\\infty + \\Lambda _{m\\,\\,(C{l^ - })}^\\infty $$

    = 73.5 + 76.3 = 149.3 Scm2/mole

    So, Statement I is wrong or false.

    As the concentration decreases, the dilution increases which increases the degree of dissociation, thus increasing the no. of ions, which increases the molar conductance.

    So Statement II is false.

    Image", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1363, "subject": "Chemistry", "question": "The resistance of a conductivity cell with cell constant 1.14 cm$$-$$1, containing 0.001 M KCl at 298 K is 1500 $$\\Omega$$. The molar conductivity of 0.001 M KCl solution at 298 K in S cm2 mol$$-$$1 is ____________. (Integer answer)", "options": [], "answer": "760", "solution": "**Answer:** 760\n\n$$K = {1 \\over R} \\times {l \\over A} = \\left( {\\left( {{1 \\over {1500}}} \\right) \\times 1.14} \\right)$$ S cm$$-$$1

    $$ \\Rightarrow { \\wedge _m} = 1000 \\times {{\\left( {{{1.14} \\over {1500}}} \\right)} \\over {0.001}}$$ S cm2 mol$$-$$1

    = 760 S cm2 mol$$-$$1

    $$\\Rightarrow$$ 760", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1364, "subject": "Chemistry", "question": "Match List - I with List - II

    \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    List - I
    (Parmeter)    		List - II
    (Unit)
    (a)Cell constant(i)$$S\\,c{m^2}mo{l^{ - 1}}$$
    (b)Molar conductivity(ii)Dimensionless
    (c)Conductivity(iii)$${m^{ - 1}}$$
    (d)Degree of dissociation of electrolyte(iv)$${\\Omega ^{ - 1}}{m^{ - 1}}$$


    Choose the most appropriate answer from the options given below :", "options": [ { "text": "(a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)" }, { "text": "(a)-(iii), (b)-(i), (c)-(ii), (d)-(iv)" }, { "text": "(a)-(i), (b)-(iv), (c)-(iii), (d)-(ii)" }, { "text": "(a)-(ii), (b)-(i), (c)-(iii), (d)-(iv)" } ], "answer": "(a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)", "solution": "**Answer:** (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)\n\nCell constant = $$\\left( {{l \\over A}} \\right)$$ $$\\Rightarrow$$ Units = m$$-$$1

    Molar conductivity ($$\\Lambda $$m) $$\\Rightarrow$$ Units = Sm2 mole$$-$$1

    Conductivity (K) $$\\Rightarrow$$ Units = S m$$-$$1

    Degree of dissociation ($$\\alpha$$ $$\\to$$ Dimensionless

    $$\\therefore$$ (a) - (iii), (b) - (i), (c) - (iv), (d) - (ii)", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1365, "subject": "Chemistry", "question": "If the conductivity of mercury at 0$$^\\circ$$C is 1.07 $$\\times$$ 106 S m$$-$$1 and the resistance of a cell containing mercury is 0.243$$\\Omega$$, then the cell constant of the cell is x $$\\times$$ 104 m$$-$$1. The value of x is ____________. (Nearest integer)", "options": [], "answer": "26", "solution": "**Answer:** 26\n\nConductance (G) is reciprocal of resistance (R)

    $$R = {1 \\over G}$$ or $$G = {1 \\over R} = {1 \\over {0.243\\Omega }} = 4.115{\\Omega ^{ - 1}}$$

    Relation between conductance (G),

    conductivity ($$\\kappa $$) and cell constant $$\\left( {{l \\over A}} \\right)$$ is given as

    $$\\kappa = {{Gl} \\over A}$$

    $$ \\Rightarrow {l \\over A} = {\\kappa \\over G} = {{1.07 \\times {{10}^6}S{m^{ - 1}}} \\over {4.115{\\Omega ^{ - 1}}}} = 26 \\times {10^4}{m^{ - 1}} \\Rightarrow x = 26$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1366, "subject": "Chemistry", "question": "

    A dilute solution of sulphuric acid is electrolysed using a current of 0.10 A for 2 hours to produce hydrogen and oxygen gas. The total volume of gases produced a STP is _____________ cm3. (Nearest integer)

    \n

    [Given : Faraday constant F = 96500 C mol$$-$$1 at STP, molar volume of an ideal gas is 22.7 L mol$$-$$1]

    ", "options": [], "answer": "127", "solution": "**Answer:** 127\n\n$2 \\mathrm{~F}$ produces $=\\frac{3}{2}$ mole of gas\n

    \n$0.10 \\times 2 \\times 3600$ coulomb produces\n

    \n$$\n\\begin{aligned}\n& =\\frac{\\frac{3}{2} \\times 0.1 \\times 2 \\times 3600}{2 \\times 96500} \\\\\\\\\n& =0.0056 \\text { moles of gas }\n\\end{aligned}\n$$

    \nVolume of gas produced $=0.0056 \\times 22.7 \\mathrm{~L}$\n

    \n$$\n\\begin{aligned}\n& \\simeq 0.127 \\mathrm{~L} \\\\\\\\\n& =127 \\mathrm{~mL}\n\\end{aligned}\n$$", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1367, "subject": "Chemistry", "question": "

    The quantity of electricity in Faraday needed to reduce 1 mol of Cr2O$$_7^{2 - }$$ to Cr3+ is ____________.

    ", "options": [], "answer": "6", "solution": "**Answer:** 6\n\n$$\n\\overset{+6}{\\mathrm{Cr}_2} \\mathrm{O}_7^{2-} \\longrightarrow 2 \\mathrm{Cr}^{3+}\n$$

    \n$\\because$ Each $\\mathrm{Cr}$ is converting from $+6$ to $+3$

    \n$\\therefore 6$ faradays of charge is required", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1368, "subject": "Chemistry", "question": "

    The limiting molar conductivities of NaI, NaNO3 and AgNO3 are 12.7, 12.0 and 13.3 mS m2 mol$$-$$1, respectively (all at 25$$^\\circ$$C). The limiting molar conductivity of AgI at this temperature is ____________ mS m2 mol$$-$$1.

    ", "options": [], "answer": "14", "solution": "**Answer:** 14\n\nGiven

    \n(1) $\\lambda_m^{\\infty}(\\mathrm{NaI})=12.7 \\,\\mathrm{mS} \\mathrm{m}^2 \\mathrm{~mol}^{-1}$

    \n(2) $\\lambda_{\\mathrm{m}}^{\\infty}\\left(\\mathrm{NaNO}_3\\right)=12.0 \\,\\mathrm{mS} \\mathrm{m}^2 \\mathrm{~mol}^{-1}$

    \n(3) $\\lambda_{\\mathrm{m}}^{\\infty}\\left(\\mathrm{AgNO}_3\\right)=13.3 \\,\\mathrm{mS} \\mathrm{m}^2 \\mathrm{~mol}^{-1}$

    \n$\\lambda_{\\mathrm{m}}^{\\infty}(\\mathrm{Ag} \\mathrm{I})=(1)+(3)-(2)$

    \n$$\n=12.7+13.3-12.0\n$$

    \n$$\n=26.0-12.0\n$$

    \n$\\lambda_{\\mathrm{m}}^{\\infty}(\\mathrm{Ag} \\mathrm{I})=14.0$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1369, "subject": "Chemistry", "question": "

    A solution of Fe2(SO4)3 is electrolyzed for 'x' min with a current of 1.5 A to deposit 0.3482 g of Fe. The value of x is ___________. [nearest integer]

    \n

    Given : 1 F = 96500 C mol$$-$$1

    \n

    Atomic mass of Fe = 56 g mol$$-$$1

    ", "options": [], "answer": "20", "solution": "**Answer:** 20\n\n$\\mathrm{Fe}^{3+}+3 \\mathrm{e}^{-} \\rightarrow \\mathrm{Fe}$\n

    \n$$\n\\text {Moles of Fe deposited }=\\frac{0.3482}{56}=6.2 \\times 10^{-3}\n$$\n

    \nFor 1 mole $\\mathrm{Fe}$, charge required is $3 \\mathrm{~F}$\n

    \nFor $6.2 \\times 10^{-3}$ mole $\\mathrm{Fe}$, charge required is $3 \\times 6.2 \\times 10^{-3} \\mathrm{~F}$\n

    \nSince, charge required $=18.6 \\times 10^{-3} \\times 96500 \\mathrm{C}$\n

    \n$$\n=1794.9 \\mathrm{C}\n$$\n

    \nAnd,\n

    \n$$\n1.5 \\times \\mathrm{t}=1794.9\n$$\n

    \n$t=\\frac{1794.9}{1.5 \\times 60} \\min$\n

    \n$$\n\\mathrm{t} \\simeq 20 \\mathrm{~min}\n$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1370, "subject": "Chemistry", "question": "

    The resistance of a conductivity cell containing 0.01 M KCl solution at 298 K is 1750 $$\\Omega$$. If the conductivity of 0.01 M KCl solution at 298 K is 0.152 $$\\times$$ 10$$-$$3 S cm$$-$$1, then the cell constant of the conductivity cell is ____________ $$\\times$$ 10$$-$$3 cm$$-$$1.

    ", "options": [], "answer": "266", "solution": "**Answer:** 266\n\nMolarity of $\\mathrm{KCl}$ solution $=0.1 ~\\mathrm{M}$\n

    \n$$\n\\begin{array}{ll}\n\\text { Resistance } & =1750 ~\\mathrm{ohm} \\\\\\\\\n\\text { Conductivity } & =0.152 \\times 10^{-3} \\mathrm{~S} \\mathrm{~cm}^{-1} \\\\\\\\\n\\text { Conductivity } & =\\frac{\\text { Cell constant }}{\\text { Resistance }} \\\\\\\\\n\\therefore \\text { Cell constant } & =0.152 \\times 10^{-3} \\times 1750 \\\\\\\\\n& =266 \\times 10^{-3} \\mathrm{~cm}^{-1}\n\\end{array}\n$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1371, "subject": "Chemistry", "question": "

    The molar conductivity of a conductivity cell filled with 10 moles of 20 mL NaCl solution is $${\\Lambda _{m1}}$$ and that of 20 moles another identical cell heaving 80 mL NaCl solution is $${\\Lambda _{m2}}$$. The conductivities exhibited by these two cells are same. The relationship between $${\\Lambda _{m2}}$$ and $${\\Lambda _{m1}}$$ is

    ", "options": [ { "text": "$${\\Lambda _{m2}}$$ = 2$${\\Lambda _{m1}}$$" }, { "text": "$${\\Lambda _{m2}}$$ = $${\\Lambda _{m1}}$$ / 2" }, { "text": "$${\\Lambda _{m2}}$$ = $${\\Lambda _{m1}}$$" }, { "text": "$${\\Lambda _{m2}}$$ = 4$${\\Lambda _{m1}}$$" } ], "answer": "$${\\Lambda _{m2}}$$ = 2$${\\Lambda _{m1}}$$", "solution": "**Answer:** $${\\Lambda _{m2}}$$ = 2$${\\Lambda _{m1}}$$\n\n$$\\Lambda_{\\mathrm{m}_{1}}=\\frac{\\mathrm{k}_{1} \\times 1000}{\\mathrm{M}_{1}}=\\frac{\\mathrm{k} \\times 1000}{\\frac{10}{0.02}}$$\n

    \n$$\n\\Lambda_{\\mathrm{m}_{2}}=\\frac{\\mathrm{k}_{2} \\times 1000}{\\frac{20}{0.08}}\n$$\n

    \nIt is given that $$\\mathrm{k}_{1}=\\mathrm{k}_{2}$$\n

    \n$$\n\\mathrm{k}_{1}=\\frac{\\Lambda_{\\mathrm{m}_{1}}}{2} \\quad \\quad \\mathrm{k}_{2}=\\frac{\\Lambda_{\\mathrm{m}_{2}}}{4}\n$$\n

    \nApplying the given condition on conductivity.\n

    \n$$\n\\begin{gathered}\n\\frac{\\Lambda_{\\mathrm{m}_{1}}}{2}=\\frac{\\Lambda_{\\mathrm{m}_{2}}}{4} \\\\\n\\Lambda_{\\mathrm{m}_{2}}=2 \\Lambda_{\\mathrm{m}_{1}}\n\\end{gathered}\n$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1372, "subject": "Chemistry", "question": "

    The amount of charge in $$\\mathrm{F}$$ (Faraday) required to obtain one mole of iron from $$\\mathrm{Fe}_{3} \\mathrm{O}_{4}$$ is ___________. (Nearest Integer)

    ", "options": [], "answer": "3", "solution": "**Answer:** 3\n\nFor $$\\mathrm{Fe}_{3} \\mathrm{O}_{4}$$,\n

    \n$$\nx=\\frac{+8}{3}\n$$\n

    \nwhere x is oxidation state of Fe.\n

    \n$$\n\\mathrm{Fe}_{3} \\mathrm{O}_{4}+8 \\mathrm{H}^{+}+8 \\mathrm{e}^{-} \\longrightarrow 3 \\mathrm{Fe}+4 \\mathrm{H}_{2} \\mathrm{O}\n$$\n

    \nCharge required $$=\\frac{8}{3} \\times \\mathrm{F}=\\frac{8 \\mathrm{~F}}{3} \\simeq 3 \\mathrm{~F}$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1373, "subject": "Chemistry", "question": "

    Given below are two statements :

    \n

    Statement I : For KI, molar conductivity increases steeply with dilution

    \n

    Statement II : For carbonic acid, molar conductivity increases slowly with dilution

    \n

    In the light of the above statements, choose the correct answer from the options given below :

    ", "options": [ { "text": "Both Statement I and Statement II are true" }, { "text": "Both Statement I and Statement II are false" }, { "text": "Statement I is true but Statement II is false" }, { "text": "Statement I is false but Statement II is true" } ], "answer": "Both Statement I and Statement II are false", "solution": "**Answer:** Both Statement I and Statement II are false\n\nStatement I is false : KI is a strong electrolyte\nwhich gets completely dissociate on dissolution.\nOn dilution, the molar conductivity almost\nremain constant.\n

    Statement II is false : carbonic acid is a weak\nelectrolyte which do not gets completely\ndissociate on dissolution. On dilution the molar\nconductivity of weak electrolyte increases\nsharply.

    \n\"JEE

    \n$$ \\therefore $$ Both Statements are false.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1374, "subject": "Chemistry", "question": "

    Resistance of a conductivity cell (cell constant $$129 \\mathrm{~m}^{-1}$$) filled with $$74.5 \\,\\mathrm{ppm}$$ solution of $$\\mathrm{KCl}$$ is $$100 \\,\\Omega$$ (labelled as solution 1). When the same cell is filled with $$\\mathrm{KCl}$$ solution of $$149 \\,\\mathrm{ppm}$$, the resistance is $$50 \\,\\Omega$$ (labelled as solution 2). The ratio of molar conductivity of solution 1 and solution 2 is i.e. $$\\frac{\\wedge_{1}}{\\wedge_{2}}=x \\times 10^{-3}$$. The value of $$x$$ is __________. (Nearest integer)

    \n

    Given, molar mass of $$\\mathrm{KCl}$$ is $$74.5 \\mathrm{~g} \\mathrm{~mol}^{-1}$$.

    ", "options": [], "answer": "1000", "solution": "**Answer:** 1000\n\n$$\\frac{l}{A}=129 \\mathrm{~m}^{-1}$$\n

    \n$$\\mathrm{KCl}$$ solution $$1 \\Rightarrow 74.5 \\,\\mathrm{ppm}, \\mathrm{R}_{1}=100 \\Omega$$\n

    \n$$\\mathrm{KCl}$$ solution $$2 \\Rightarrow 149 \\,\\mathrm{ppm}, \\mathrm{R}_{2}=50 \\Omega$$

    \n$$\n\\begin{aligned}\n&\\text { Here, } \\frac{p p m_{1}}{p p m_{2}}=\\frac{M_{1}}{M_{2}}=\\left(\\frac{w_{1 / M_{0}}}{V} \\times \\frac{V}{w_{2 / M_{0}}}\\right) \\\\\n&\\frac{\\Lambda_{1}}{\\Lambda_{2}}=\\frac{k_{1} \\times \\frac{1000}{M_{1}}}{k_{2} \\times \\frac{1000}{M_{2}}} \\\\\n&=\\frac{k_{1}}{k_{2}} \\times \\frac{M_{1}}{M_{2}} \\\\\n&=\\frac{50}{100} \\times 2 \\\\\n&=\\frac{\\Lambda_{1}}{\\Lambda_{2}}=1000 \\times 10^{-3} \\\\\n&=1000\n\\end{aligned}\n$$", "topic": "Discrete Mathematics", "subtopic": "Graph Theory" }, { "id": 1375, "subject": "Chemistry", "question": "

    Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R)

    \n

    Assertion (A) : An aqueous solution of $$\\mathrm{KOH}$$ when used for volumetric analysis, its concentration should be checked before the use.

    \n

    Reason (R) : On aging, $$\\mathrm{KOH}$$ solution absorbs atmospheric $$\\mathrm{CO}_{2}$$.

    \n

    In the light of the above statements, choose the correct answer from the options given below :

    ", "options": [ { "text": "(A) is correct but (R) is not correct" }, { "text": "(A) is not correct but (R) is correct" }, { "text": "Both (A) and (R) are correct but (R) is not the correct explanation of (A)" }, { "text": "Both (A) and (R) are correct and (R) is the correct explanation of (A)" } ], "answer": "Both (A) and (R) are correct and (R) is the correct explanation of (A)", "solution": "**Answer:** Both (A) and (R) are correct and (R) is the correct explanation of (A)\n\n

    In volumetric analysis, the concentration of a solution is a crucial factor in determining the accuracy of the results. In the case of an aqueous solution of KOH, the concentration can change over time due to the absorption of atmospheric CO2. This occurs because KOH is a basic (alkaline) solution and reacts with carbon dioxide (CO2) from the air to form potassium carbonate (K2CO3).

    \n\n

    The reaction between KOH and CO2 is given by :

    \n\n

    KOH + CO2 $$ \\to $$ K2CO3

    \n\n

    This reaction will change the concentration of KOH in the solution, which in turn will impact the accuracy of the results obtained in volumetric analysis. For example, if the concentration of KOH is lower than it was when the solution was prepared, then more solution will be needed to reach the endpoint in a reaction, and the results will be inaccurate.

    \n\n

    That's why it is important to check the concentration of the KOH solution before using it for volumetric analysis, as stated in the assertion (A). And the reason (R) provides the explanation for why the concentration should be checked. Hence, both the assertion and the reason are correct and the correct answer is (D). Both (A) and (R) are correct and (R) is the correct explanation of (A).

    ", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1376, "subject": "Chemistry", "question": "

    $$1 \\times 10^{-5} ~\\mathrm{M} ~\\mathrm{AgNO}_{3}$$ is added to $$1 \\mathrm{~L}$$ of saturated solution of $$\\mathrm{AgBr}$$. The conductivity of this solution at $$298 \\mathrm{~K}$$ is _____________ $$\\times 10^{-8} \\mathrm{~S} \\mathrm{~m}^{-1}$$.

    \n

    [Given : $$\\mathrm{K}_{\\mathrm{SP}}(\\mathrm{AgBr})=4.9 \\times 10^{-13}$$ at $$298 \\mathrm{~K}$$

    \n

    $$\n\\begin{aligned}\n& \\lambda_{\\mathrm{Ag}^{+}}^{0}=6 \\times 10^{-3} \\mathrm{~S} \\mathrm{~m}^{2} \\mathrm{~mol}^{-1} \\\\\n& \\lambda_{\\mathrm{Br}^{-}}^{0}=8 \\times 10^{-3} \\mathrm{~S} \\mathrm{~m}^{2} \\mathrm{~mol}^{-1} \\\\\n& \\left.\\lambda_{\\mathrm{NO}_{3}^{-}}^{0}=7 \\times 10^{-3} \\mathrm{~S} \\mathrm{~m}^{2} \\mathrm{~mol}^{-1}\\right]\n\\end{aligned}\n$$

    ", "options": [], "answer": "13039", "solution": "**Answer:** 13039\n\n$$\n\\begin{aligned}\n& \\operatorname{AgBr}(\\mathrm{S}) \\rightleftharpoons \\underset{\\left(10^{-5}+\\mathrm{x}\\right)}{\\rightleftharpoons \\mathrm{Ag}^{+}}(\\mathrm{aq})+\\mathrm{Br}_{\\mathrm{x}}^{-}(\\mathrm{aq}) \\\\\\\\\n& x\\left(x+10^{-5}\\right)=4.9 \\times 10^{-13} \\\\\\\\\n& x \\simeq 4.9 \\times 10^{-8} \\mathrm{M} \\\\\\\\\n& \\lambda_{\\mathrm{Ag}^{+}}^{\\circ}=6 \\times 10^{-3} \\mathrm{~S} \\mathrm{~cm}^2 \\mathrm{~mol}^{-1} \\\\\\\\\n& \\lambda_{\\mathrm{Br}^{-}}^0=8 \\times 10^{-3} \\mathrm{~S} \\mathrm{~cm}^2 \\mathrm{~mol}^{-1} \\\\\\\\\n& \\mathrm{~K}_{\\text {solution }}=\\mathrm{K}_{\\mathrm{Ag}^{+}}+\\mathrm{K}_{\\mathrm{Br}^{-}}+\\mathrm{K}_{\\mathrm{NO}_3^{-}} \\\\\\\\\n& =6 \\times 10^{-3} \\times 10^{-5} \\times 10^3+8 \\times 10^{-3} \\times 4.9 \\times 10^{-8} \\times 10^3 \\\\\\\\\n& +7 \\times 10^{-3} \\times 10^{-5} \\times 10^3 \\\\\\\\\n& =(6000+39.2+7000) \\times 10^{-8} \\\\\\\\\n& =13039.2 \\times 10^{-8} ~\\mathrm{Sm}^{-1}\n\\end{aligned}\n$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1377, "subject": "Chemistry", "question": " The resistivity of a $0.8 \\mathrm{M}$ solution of an electrolyte is $5 \\times 10^{-3} \\Omega~ \\mathrm{cm}$.

    Its molar conductivity is _________ $\\times 10^{4}~ \\Omega^{-1} \\mathrm{~cm}^{2} \\mathrm{~mol}^{-1}$. (Nearest integer)", "options": [], "answer": "25", "solution": "**Answer:** 25\n\n$$\n\\begin{aligned}\n\\text { Molar conductivity } & =\\frac{\\mathrm{k} \\times 1000}{\\mathrm{C}} \\\\\\\\\n& =\\frac{\\frac{1}{5 \\times 10^{-3}} \\times 1000}{0.8} \\\\\\\\\n& =\\frac{10^6}{4}=0.25 \\times 10^6 \\\\\\\\\n& =25 \\times 10^4 \\Omega^{-1} \\mathrm{~cm}^2 \\mathrm{~mol}^{-1}\n\\end{aligned}\n$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1378, "subject": "Chemistry", "question": "

    Which one of the following statements is correct for electrolysis of brine solution?

    ", "options": [ { "text": "$$\\mathrm{O}_{2}$$ is formed at cathode" }, { "text": "$$\\mathrm{H}_{2}$$ is formed at anode" }, { "text": "$$\\mathrm{Cl}_{2}$$ is formed at cathode" }, { "text": "$$\\mathrm{OH}^{-}$$ is formed at cathode" } ], "answer": "$$\\mathrm{OH}^{-}$$ is formed at cathode", "solution": "**Answer:** $$\\mathrm{OH}^{-}$$ is formed at cathode\n\nCathode : $\\mathrm{H}_2 \\mathrm{O}_{(l)}+e^{-} \\longrightarrow \\frac{1}{2} \\mathrm{H}_{2(g)}+\\mathrm{OH}_{(a q)}^{-}$\n

    Anode $: \\mathrm{Cl}_{(a q)}^{-} \\longrightarrow \\frac{1}{2} \\mathrm{Cl}_{2(g)}+e^{-}$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1379, "subject": "Chemistry", "question": "The number of correct statements from the following is _______.

    \n(A) Conductivity always decreases with decrease in concentration for both strong and weak electrolytes.\n

    \n(B) The number of ions per unit volume that carry current in a solution increases on dilution.\n

    \n(C) Molar conductivity increases with decrease in concentration\n

    \n(D) The variation in molar conductivity is different for strong and weak electrolytes\n

    \n(E) For weak electrolytes, the change in molar conductivity with dilution is due to decrease in degree of dissociation.", "options": [], "answer": "3", "solution": "**Answer:** 3\n\nLet us evaluate each statement one by one to determine whether it is true or false:\n

    \n(A) Conductivity always decreases with decrease in concentration for both strong and weak electrolytes.\n

    \nThis statement is true. Conductivity is defined as the ability of a solution to conduct electricity, and it depends on the concentration of ions in the solution. As the concentration of ions decreases, the conductivity of the solution also decreases, regardless of whether the electrolyte is strong or weak.\n

    \n(B) The number of ions per unit volume that carry current in a solution increases on dilution.\n

    \nThis statement is false. The number of ions per unit volume in a solution decreases on dilution, because the total number of ions in the solution remains the same, but the volume increases.\n

    \n(C) Molar conductivity increases with decrease in concentration.\n

    \nThis statement is generally true. Molar conductivity is a measure of the ability of an electrolyte to conduct electricity, and it depends on both the concentration of the electrolyte and the mobility of the ions. As the concentration of the electrolyte decreases, the molar conductivity increases because the ions become more separated from each other, and the electrostatic interactions between them become weaker, allowing them to move more freely in the solution.\n

    \n(D) The variation in molar conductivity is different for strong and weak electrolytes.\n

    \nThis statement is true. The variation in molar conductivity with concentration is different for strong and weak electrolytes. Strong electrolytes dissociate completely in solution, so their molar conductivity increases rapidly with dilution, while weak electrolytes dissociate only partially, so their molar conductivity increases more slowly with dilution.\n

    \n(E) For weak electrolytes, the change in molar conductivity with dilution is due to decrease in degree of dissociation.\n

    \nThis statement is false. For weak electrolytes, the change in molar conductivity with dilution is due to an increase in the degree of dissociation, not a decrease. \n

    \nTherefore, there are $\\boxed{3}$ correct statements in the given options, namely (A), (C), (D).", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1380, "subject": "Chemistry", "question": "

    A metal surface of $$100 \\mathrm{~cm}^{2}$$ area has to be coated with nickel layer of thickness $$0.001 \\mathrm{~mm}$$. A current of $$2 \\mathrm{~A}$$ was passed through a solution of $$\\mathrm{Ni}\\left(\\mathrm{NO}_{3}\\right)_{2}$$ for '$$\\mathrm{x}$$' seconds to coat the desired layer. The value of $$\\mathrm{x}$$ is __________. (Nearest integer) ( $$\\rho_{\\mathrm{Ni}}$$ (density of Nickel) is $$10 \\mathrm{~g} \\mathrm{~mL}$$, Molar mass of Nickel is $$60 \\mathrm{~g} \\mathrm{~mol}^{-1}$$ $$\\left.\\mathrm{F}=96500 ~\\mathrm{C} ~\\mathrm{mol}^{-1}\\right)$$

    ", "options": [], "answer": "161", "solution": "**Answer:** 161\n\nUsing the Faraday's law of electrolysis, we can directly relate the amount of substance deposited (in this case, the nickel layer) with the electric charge passed through the electrolyte. \n

    \nThe formula for Faraday's law of electrolysis is:\n

    \n$$W = z \\times i \\times t$$\n

    \nwhere W is the amount of substance deposited (in grams), z is the electrochemical equivalent (grams per coulomb), i is the current (in amperes), and t is the time (in seconds).\n

    \nBy relating the density and volume of the nickel layer to the electric charge passed through the electrolyte, we can calculate the time needed for the deposition:\n

    \n$$10 \\times 100 \\times 0.0001 = \\frac{\\left(\\frac{\\text { atomic wt. }}{\\text { v.f }}\\right) \\times 2 \\times x}{96500}$$\n

    \nwhere v.f is the valence factor for the reaction (in this case, 2).\n

    \nSolving for x, we get:\n

    \n$$x = 161 \\, \\mathrm{sec}$$\n

    \nSo, the value of x is 161 seconds.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1381, "subject": "Chemistry", "question": "

    The specific conductance of $$0.0025 ~\\mathrm{M}$$ acetic acid is $$5 \\times 10^{-5} \\mathrm{~S} \\mathrm{~cm}^{-1}$$ at a certain temperature. The dissociation constant of acetic acid is __________ $$\\times ~10^{-7}$$ (Nearest integer)

    \n

    Consider limiting molar conductivity of $$\\mathrm{CH}_{3} \\mathrm{COOH}$$ as $$400 \\mathrm{~S} \\mathrm{~cm}^{2} \\mathrm{~mol}^{-1}$$

    ", "options": [], "answer": "66", "solution": "**Answer:** 66\n\n

    Given that the specific conductance, $k$, is $5 \\times 10^{-5} ~S~cm^{-1}$ and the concentration, $C$, is $0.0025~M$, we can find the molar conductivity, $\\lambda_m$, as follows:

    \n

    $$\\lambda_m = \\frac{k}{C} \\times 1000 = \\frac{5 \\times 10^{-5} \\times 10^3}{0.0025} = \\frac{5 \\times 10^{-2}}{2.5 \\times 10^{-3}} = 20 ~S~cm^2~mol^{-1}$$

    \n

    Next, we find the degree of dissociation, $\\alpha$, by dividing $\\lambda_m$ by the limiting molar conductivity, $\\lambda_m^o$:

    \n

    $$\\alpha = \\frac{20}{400} = \\frac{1}{20}$$

    \n

    Finally, we use the formula for the dissociation constant of a weak acid, $K_a$:

    \n

    $$K_a = \\frac{C \\alpha^{2}}{1-\\alpha} = \\frac{0.0025 \\times \\frac{1}{20} \\times \\frac{1}{20}}{\\frac{19}{20}} = \\frac{0.0025}{19 \\times 20} = 6.6 \\times 10^{-6} = 66 \\times 10^{-7}$$

    \n

    So, the correct answer is 66.

    \n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1382, "subject": "Chemistry", "question": "The amount of electricity in Coulomb required for the oxidation of $1 \\mathrm{~mol}$ of $\\mathrm{H}_2 \\mathrm{O}$ to $\\mathrm{O}_2$ is __________ $\\times 10^5 \\mathrm{C}$.", "options": [], "answer": "2", "solution": "**Answer:** 2\n\n$\\begin{aligned} & 2 \\mathrm{H}_2 \\mathrm{O} \\rightarrow \\mathrm{O}_2+4 \\mathrm{H}^{+}+4 \\mathrm{e}^{-} \\\\\\\\ & \\frac{\\mathrm{W}}{\\mathrm{E}}=\\frac{\\mathrm{Q}}{96500} \\\\\\\\ & \\text { mole } \\times \\text { n-factor }=\\frac{\\mathrm{Q}}{96500}\\end{aligned}$\n

    $\\begin{aligned} & 1 \\times 2=\\frac{Q}{96500} \\\\\\\\ & Q=2 \\times 96500 \\mathrm{C} \\\\\\\\ & =1.93 \\times 10^5 \\mathrm{C}\\end{aligned}$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1383, "subject": "Chemistry", "question": "

    The mass of silver (Molar mass of $$\\mathrm{Ag}: 108 \\mathrm{~gmol}^{-1}$$ ) displaced by a quantity of electricity which displaces $$5600 \\mathrm{~mL}$$ of $$\\mathrm{O}_2$$ at S.T.P. will be ______ g.

    ", "options": [], "answer": "108", "solution": "**Answer:** 108\n\n

    First, we need to determine the amount of $$\\mathrm{O}_2$$ (oxygen gas) in moles that is displaced by the quantity of electricity mentioned. To do this, we'll use the molar volume of a gas at Standard Temperature and Pressure (S.T.P.), which is approximately $$22.4 \\mathrm{~L/mol}$$ (or $$22400 \\mathrm{~mL/mol}$$).

    \n\n

    The volume of $$\\mathrm{O}_2$$ is given as $$5600 \\mathrm{~mL}$$. Now, we convert this volume to moles:

    \n$$\n\\text{moles of } \\mathrm{O}_2 = \\frac{\\text{volume of } \\mathrm{O}_2}{\\text{molar volume}} = \\frac{5600 \\mathrm{~mL}}{22400 \\mathrm{~mL/mol}} = 0.25 \\mathrm{~mol}\n$$\n\n

    Next, we'll use Faraday's laws of electrolysis to relate the moles of $$\\mathrm{O}_2$$ to the moles of silver being displaced. In electrolysis, silver is deposited at the cathode according to the following half-reaction:

    \n$$\n\\mathrm{Ag}^+ + e^- \\rightarrow \\mathrm{Ag}\n$$\n\n

    This tells us that for each mole of $$\\mathrm{Ag}^+$$ ions, only 1 mole of electrons is required to reduce it to silver metal $$\\mathrm{Ag}$$. However, the liberation of oxygen gas involves the following half-reaction:

    \n$$\n2\\mathrm{H_2O} (l) \\rightarrow \\mathrm{O}_2 (g) + 4H^+ (aq) + 4e^-\n$$ \n\n

    From this reaction, we can see that 1 mole of $$\\mathrm{O}_2$$ gas requires 4 moles of electrons to be produced. Therefore, the number of moles of electrons associated with the $$0.25 \\mathrm{~mol}$$ of $$\\mathrm{O}_2$$ will be:

    \n$$\n\\text{moles of electrons for } \\mathrm{O}_2 = 0.25 \\mathrm{~mol} \\times 4 = 1 \\mathrm{~mol}\n$$\n\n

    Now, since it takes 1 mole of electrons to reduce 1 mole of $$\\mathrm{Ag}^+$$. The moles of electrons required is equal to the moles of $$\\mathrm{Ag}$$ produced. Hence, we have 1 mole of $$\\mathrm{Ag}$$ being deposited.

    \n\n

    Finally, to calculate the mass of this silver, we use the molar mass of silver:

    \n$$\n\\text{Mass of Ag} = (\\text{moles of Ag}) \\times (\\text{Molar mass of Ag}) = 1 \\mathrm{~mol} \\times 108 \\mathrm{~gmol}^{-1} = 108 \\mathrm{~g}\n$$\n\n

    So, the mass of silver displaced by the quantity of electricity that displaces 5600 mL of $$\\mathrm{O}_2$$ at S.T.P. will be $$108 \\mathrm{~g}$$.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1384, "subject": "Chemistry", "question": "The values of conductivity of some materials at $298.15 \\mathrm{~K}^{-1} ~\\text{in} ~\\mathrm{Sm}^{-1}$ are $2.1 \\times 10^3$,

    $1.0 \\times 10^{-16}, 1.2 \\times 10,3.91,1.5 \\times 10^{-2}, 1 \\times 10^{-7}, 1.0 \\times 10^3$.

    The number of conductors among the materials is _____________.", "options": [], "answer": "4", "solution": "**Answer:** 4\n\n

    Conductivity (S m$$^{-1}$$)

    \n

    $$\\left.\\begin{array}{l}\n2.1 \\times 10^3 \\\\\n1.2 \\times 10 \\\\\n3.91 \\\\\n1 \\times 10^3\n\\end{array}\\right\\} \\text { conductors at } 298.15 \\mathrm{~K}$$

    \n

    $$1 \\times 10^{-16} \\text { Insulator at } 298.15 \\mathrm{~K}$$

    \n

    $$\\left.\\begin{array}{l}\n1.5 \\times 10^{-2} \\\\\n1 \\times 10^{-7}\n\\end{array}\\right\\} \\text { Semiconductor at } 298.15 \\mathrm{~K}$$

    \n

    Therefore number of conductors is 4.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1385, "subject": "Chemistry", "question": "

    Identify the factor from the following that does not affect electrolytic conductance of a solution.

    ", "options": [ { "text": "The nature of solvent used.\n" }, { "text": "The nature of the electrolyte added.\n" }, { "text": "The nature of the electrode used.\n" }, { "text": "Concentration of the electrolyte." } ], "answer": "The nature of the electrode used.\n", "solution": "**Answer:** The nature of the electrode used.\n\n\n

    Conductivity of electrolytic cell is affected by concentration of electrolyte, nature of electrolyte and nature of solvent.

    ", "topic": "Number Theory", "subtopic": "Divisibility / Factorization" }, { "id": 1386, "subject": "Chemistry", "question": "

    Number of alkanes obtained on electrolysis of a mixture of $$\\mathrm{CH}_3 \\mathrm{COONa}$$ and $$\\mathrm{C}_2 \\mathrm{H}_5 \\mathrm{COONa}$$ is ________.

    ", "options": [], "answer": "3", "solution": "**Answer:** 3\n\n

    $$\\mathrm{CH}_3 \\mathrm{COONa} \\rightarrow \\dot{\\mathrm{C}} \\mathrm{H}_3$$

    \n

    $$\\mathrm{C}_2 \\mathrm{H}_5 \\mathrm{COONa} \\rightarrow \\dot{\\mathrm{C}}_2 \\mathrm{H}_5$$

    \n

    $$2 \\dot{\\mathrm{C}}_2 \\mathrm{H}_5 \\rightarrow \\mathrm{CH}_3-\\mathrm{CH}_2-\\mathrm{CH}_2-\\mathrm{CH}_3$$

    \n

    $$2 \\dot{\\mathrm{CH}}_3 \\rightarrow \\mathrm{CH}_3-\\mathrm{CH}_3$$

    \n

    $$\\dot{\\mathrm{C}} \\mathrm{H}_3+\\dot{\\mathrm{C}}_2 \\mathrm{H}_5 \\rightarrow \\mathrm{CH}_3-\\mathrm{CH}_2-\\mathrm{CH}_3$$

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1387, "subject": "Chemistry", "question": "

    One Faraday of electricity liberates $$x \\times 10^{-1}$$ gram atom of copper from copper sulphate. $$x$$ is ________.

    ", "options": [], "answer": "5", "solution": "**Answer:** 5\n\n

    To find the value of x when one Faraday of electricity liberates x times $10^{-1}$ gram atom of copper from copper sulphate, we need to understand how electricity interacts with copper ions in solution. The key reaction is:

    $$\\mathrm{Cu}^{2+} + 2 \\mathrm{e}^{-} \\rightarrow \\mathrm{Cu}$$

    This shows that copper ions (Cu2+) gain two electrons (e-) to become copper metal (Cu). From electrochemistry, we know that:

    This means x equals 5.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1388, "subject": "Chemistry", "question": "

    The mass of zinc produced by the electrolysis of zine sulphate solution with a steady current of $$0.015 \\mathrm{~A}$$ for 15 minutes is _________ $$\\times 10^{-4} \\mathrm{~g}$$.

    \n

    (Atomic mass of zinc $$=65.4 \\mathrm{~amu}$$)

    ", "options": [], "answer": "46", "solution": "**Answer:** 46\n\n

    $$\\begin{aligned} & \\mathrm{Zn}^{+2}+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Zn} \\\\ & \\mathrm{W}=\\mathrm{Z} \\times \\mathrm{i} \\times \\mathrm{t} \\\\ & =\\frac{65.4}{2 \\times 96500} \\times 0.015 \\times 15 \\times 60 \\\\ & =45.75 \\times 10^{-4} \\mathrm{gm}\\end{aligned}$$

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1389, "subject": "Chemistry", "question": "

    A constant current was passed through a solution of $$\\mathrm{AuCl}_4^{-}$$ ion between gold electrodes. After a period of 10.0 minutes, the increase in mass of cathode was $$1.314 \\mathrm{~g}$$. The total charge passed through the solution is _______ $$\\times 10^{-2} \\mathrm{~F}$$.

    \n

    (Given atomic mass of $$\\mathrm{Au}=197$$)

    ", "options": [], "answer": "2", "solution": "**Answer:** 2\n\n

    $$\\begin{aligned}\n& \\frac{\\mathrm{W}}{\\mathrm{E}}=\\frac{\\text { charge }}{1 \\mathrm{~F}} \\\\\n& \\frac{1.314}{\\frac{197}{3}}=\\frac{\\mathrm{Q}}{1 \\mathrm{F}} \\\\\n& \\mathrm{Q}=2 \\times 10^{-2} \\mathrm{~F}\n\\end{aligned}$$

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1390, "subject": "Chemistry", "question": "

    Alkaline oxidative fusion of $$\\mathrm{MnO}_2$$ gives \"A\" which on electrolytic oxidation in alkaline solution produces B. A and B respectively are

    ", "options": [ { "text": "$$\\mathrm{Mn}_2 \\mathrm{O}_3$$ and $$\\mathrm{MnO}_4^{2-}$$\n" }, { "text": "$$\\mathrm{Mn}_2 \\mathrm{O}_7$$ and $$\\mathrm{MnO}_4^{-}$$\n" }, { "text": "$$\\mathrm{MnO}_4^{2-}$$ and $$\\mathrm{MnO}_4^{-}$$\n" }, { "text": "$$\\mathrm{MnO}_4^{2-}$$ and $$\\mathrm{Mn}_2 \\mathrm{O}_7$$" } ], "answer": "$$\\mathrm{MnO}_4^{2-}$$ and $$\\mathrm{MnO}_4^{-}$$\n", "solution": "**Answer:** $$\\mathrm{MnO}_4^{2-}$$ and $$\\mathrm{MnO}_4^{-}$$\n\n\n

    Alkaline oxidative fusion of $$\\mathrm{MnO}_2$$ :

    \n

    $$2 \\mathrm{MnO}_2+4 \\mathrm{OH}^{-}+\\mathrm{O}_2 \\rightarrow 2 \\mathrm{MnO}_4^{2-}+2 \\mathrm{H}_2 \\mathrm{O}$$

    \n

    Electrolytic oxidation of $$\\mathrm{MnO}_4^{2-}$$ in alkaline medium.

    \n

    $$\\mathrm{MnO}_4^{2-} \\rightarrow \\mathrm{MnO}_4^{-}+e^{-}$$

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1391, "subject": "Chemistry", "question": "

    Which out of the following is a correct equation to show change in molar conductivity with respect to concentration for a weak electrolyte, if the symbols carry their usual meaning :

    ", "options": [ { "text": "$$\\Lambda_{\\mathrm{m}}-\\Lambda_{\\mathrm{m}}^{\\circ}+\\mathrm{AC}^{\\frac{1}{2}}=0$$" }, { "text": "$$\\Lambda_{\\mathrm{m}}^2 \\mathrm{C}+\\mathrm{K}_{\\mathrm{a}} \\Lambda_{\\mathrm{m}}^{\\mathrm{o}^2}-\\mathrm{K}_{\\mathrm{a}} \\Lambda_{\\mathrm{m}} \\Lambda_{\\mathrm{m}}^{\\circ}=0$$" }, { "text": "$$\\Lambda_{\\mathrm{m}}-\\Lambda_{\\mathrm{m}}^{\\circ}-\\mathrm{AC}^{\\frac{1}{2}}=0$$" }, { "text": "$$\\Lambda_{\\mathrm{m}}^2 \\mathrm{C}-\\mathrm{K}_{\\mathrm{a}} \\Lambda_{\\mathrm{m}}^{\\circ 2}+\\mathrm{K}_{\\mathrm{a}} \\Lambda_{\\mathrm{m}} \\Lambda_{\\mathrm{m}}^{\\circ}=0$$" } ], "answer": "$$\\Lambda_{\\mathrm{m}}^2 \\mathrm{C}-\\mathrm{K}_{\\mathrm{a}} \\Lambda_{\\mathrm{m}}^{\\circ 2}+\\mathrm{K}_{\\mathrm{a}} \\Lambda_{\\mathrm{m}} \\Lambda_{\\mathrm{m}}^{\\circ}=0$$", "solution": "**Answer:** $$\\Lambda_{\\mathrm{m}}^2 \\mathrm{C}-\\mathrm{K}_{\\mathrm{a}} \\Lambda_{\\mathrm{m}}^{\\circ 2}+\\mathrm{K}_{\\mathrm{a}} \\Lambda_{\\mathrm{m}} \\Lambda_{\\mathrm{m}}^{\\circ}=0$$\n\n\n

    The dissociation of a weak electrolyte ($\\mathrm{HA}$) in solution can be represented as:

    \n\n

    $ \\mathrm{HA}(\\mathrm{aq}) \\rightleftharpoons \\mathrm{H}^{+}(\\mathrm{aq}) + \\mathrm{A}^{-}(\\mathrm{aq}) $

    \n\n

    The degree of dissociation ($\\alpha$) for a weak electrolyte is given by:

    \n\n

    $ \\alpha = \\frac{\\Lambda_{\\mathrm{m}}}{\\Lambda_{\\mathrm{m}}^{\\circ}} $

    \n\n

    Where $\\Lambda_{\\mathrm{m}}$ is the molar conductivity at a given concentration $C$, and $\\Lambda_{\\mathrm{m}}^{\\circ}$ is the molar conductivity at infinite dilution.

    \n\n

    The dissociation constant ($K_{\\mathrm{a}}$) is described by:

    \n\n

    $ K_{\\mathrm{a}} = \\frac{\\alpha^2 C}{1 - \\alpha} $

    \n\n

    Rewriting this equation in terms of $\\Lambda_{\\mathrm{m}}$ and $\\Lambda_{\\mathrm{m}}^{\\circ}$:

    \n\n

    $ K_{\\mathrm{a}} = \\frac{\\left(\\frac{\\Lambda_{\\mathrm{m}}}{\\Lambda_{\\mathrm{m}}^{\\circ}}\\right)^2 C}{1 - \\frac{\\Lambda_{\\mathrm{m}}}{\\Lambda_{\\mathrm{m}}^{\\circ}}} $

    \n\n

    By multiplying out and simplifying the above expression:

    \n\n

    $ \\left( \\frac{\\Lambda_{\\mathrm{m}}}{\\Lambda_{\\mathrm{m}}^{\\circ}} \\right)^2 C + K_{\\mathrm{a}} \\left( \\frac{\\Lambda_{\\mathrm{m}}}{\\Lambda_{\\mathrm{m}}^{\\circ}} \\right) = K_{\\mathrm{a}} $

    \n\n

    Substitute $\\alpha = \\frac{\\Lambda_{\\mathrm{m}}}{\\Lambda_{\\mathrm{m}}^{\\circ}}$:

    \n\n

    $ \\left(\\frac{\\Lambda_{\\mathrm{m}}}{\\Lambda_{\\mathrm{m}}^{\\circ}}\\right)^2 C + K_{\\mathrm{a}} \\left(\\frac{\\Lambda_{\\mathrm{m}}}{\\Lambda_{\\mathrm{m}}^{\\circ}}\\right) - K_{\\mathrm{a}} = 0 $

    \n\n

    Rewriting by multiplying through by $\\left( \\Lambda_{\\mathrm{m}}^{\\circ} \\right)^2$:

    \n\n

    $ \\Lambda_{\\mathrm{m}}^2 C + K_{\\mathrm{a}} \\Lambda_{\\mathrm{m}} \\Lambda_{\\mathrm{m}}^{\\circ} - K_{\\mathrm{a}} \\left( \\Lambda_{\\mathrm{m}}^{\\circ} \\right)^2 = 0 $

    \n\n

    Thus, the correct equation that describes the change in molar conductivity with respect to concentration for a weak electrolyte is given by:

    \n\n

    Option D:

    \n\n

    $ \\Lambda_{\\mathrm{m}}^2 C - K_{\\mathrm{a}} \\Lambda_{\\mathrm{m}}^{\\circ 2} + K_{\\mathrm{a}} \\Lambda_{\\mathrm{m}} \\Lambda_{\\mathrm{m}}^{\\circ} = 0 $

    ", "topic": "Algebra", "subtopic": "Equations and Inequalities" }, { "id": 1392, "subject": "Chemistry", "question": "

    For a strong electrolyte, a plot of molar conductivity against (concentration) $${ }^{1 / 2}$$ is a straight line, with a negative slope, the correct unit for the slope is

    ", "options": [ { "text": "$$\\mathrm{S} \\mathrm{cm}^2 \\mathrm{~mol}^{-3 / 2} \\mathrm{~L}$$\n" }, { "text": "$$\\mathrm{S} \\mathrm{cm}{ }^2 \\mathrm{~mol}^{-3 / 2} \\mathrm{~L}^{-1 / 2}$$\n" }, { "text": "$$\\mathrm{S} \\mathrm{cm}{ }^2 \\mathrm{~mol}^{-1} \\mathrm{~L}^{1 / 2}$$\n" }, { "text": "$$\\mathrm{S} \\mathrm{cm}^2 \\mathrm{~mol}^{-3 / 2} \\mathrm{~L}^{1 / 2}$$" } ], "answer": "$$\\mathrm{S} \\mathrm{cm}^2 \\mathrm{~mol}^{-3 / 2} \\mathrm{~L}^{1 / 2}$$", "solution": "**Answer:** $$\\mathrm{S} \\mathrm{cm}^2 \\mathrm{~mol}^{-3 / 2} \\mathrm{~L}^{1 / 2}$$\n\n

    The molar conductivity ($\\Lambda_m$) of a strong electrolyte depends on the concentration ($c$) according to Kohlrausch's law, which can be mathematically expressed as $\\Lambda_m = \\Lambda_m^0 - k\\sqrt{c}$, where $\\Lambda_m^0$ is the molar conductivity at infinite dilution, and $k$ is a constant. The graph of molar conductivity ($\\Lambda_m$) against the square root of concentration ($c^{1/2}$) is a straight line with a negative slope.

    \n\n

    The unit of molar conductivity ($\\Lambda_m$) is Siemens meter squared per mole ($\\text{S cm}^2 \\text{mol}^{-1}$). Concentration ($c$) has the unit moles per liter ($\\text{mol L}^{-1}$), and thus its square root ($c^{1/2}$) has the unit $\\text{mol}^{1/2} \\text{L}^{-1/2}$.

    \n\n

    The slope of the plot of $\\Lambda_m$ against $c^{1/2}$ is derived from the expression $k$ in $k\\sqrt{c}$. Therefore, to find the correct unit of the slope, we look at the division of the unit of $\\Lambda_m$ by the unit of $c^{1/2}$:

    \n\n

    $\\frac{\\text{S cm}^2 \\text{mol}^{-1}}{\\text{mol}^{1/2} \\text{L}^{-1/2}} = \\text{S cm}^2 \\text{mol}^{-1-1/2} \\text{L}^{1/2} = \\text{S cm}^2 \\text{mol}^{-3/2} \\text{L}^{1/2}$

    \n\n

    Thus, the correct unit for the slope of a plot of molar conductivity against the square root of concentration for a strong electrolyte is $\\text{S cm}^2 \\text{mol}^{-3/2} \\text{L}^{1/2}$, which matches with Option D.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1393, "subject": "Chemistry", "question": "

    Given below are two statements :

    \n

    Statement (I) : Fusion of $$\\mathrm{MnO}_2$$ with $$\\mathrm{KOH}$$ and an oxidising agent gives dark green $$\\mathrm{K}_2 \\mathrm{MnO}_4$$.

    \n

    Statement (II) : Manganate ion on electrolytic oxidation in alkaline medium gives permanganate ion.

    \n

    In the light of the above statements, choose the correct answer from the options given below :

    ", "options": [ { "text": "Statement I is false but Statement II is true\n" }, { "text": "Statement I is true but Statement II is false\n" }, { "text": "Both Statement I and Statement II are false\n" }, { "text": "Both Statement I and Statement II are true" } ], "answer": "Both Statement I and Statement II are true", "solution": "**Answer:** Both Statement I and Statement II are true\n\n

    Statement I is indeed true. When $$\\mathrm{MnO}_2$$ (manganese dioxide) is fused with $$\\mathrm{KOH}$$ (potassium hydroxide) and an oxidising agent such as $$\\mathrm{KNO}_3$$ (potassium nitrate), it forms dark green potassium manganate ($$\\mathrm{K}_2\\mathrm{MnO}_4$$). The reaction can be represented as follows:

    \n\n

    $$\\mathrm{2MnO}_2 + 4KOH + \\mathrm{O}_2 \\rightarrow \\mathrm{2K}_2\\mathrm{MnO}_4 + 2H_2\\mathrm{O}$$

    \n\n

    This process involves the oxidation of manganese dioxide to manganate ion ($$\\mathrm{MnO}_4^{2-}$$) in an alkaline medium.

    \n\n

    Statement II is also true. The manganate ion ($$\\mathrm{MnO}_4^{2-}$$), when subjected to electrolytic oxidation in an alkaline medium, can indeed be converted into permanganate ion ($$\\mathrm{MnO}_4^-$$), which is characterized by a deep purple color. This conversion is an example of an electron-loss (oxidation) process at the anode of an electrolytic cell. The reaction can be represented as follows:

    \n\n

    $$\\mathrm{MnO}_4^{2-} + \\mathrm{H}_2\\mathrm{O} \\rightarrow \\mathrm{MnO}_4^- + 2\\mathrm{OH}^- + 2e^-$$

    \n\n

    Thus, both statements I and II are accurate, making the correct answer:

    \n\n

    Option D: Both Statement I and Statement II are true.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1394, "subject": "Chemistry", "question": "

    Molar ionic conductivities of divalent cation and anion are $$57 \\mathrm{~S~cm}^2 \\mathrm{~mol}^{-1}$$ and $$73 \\mathrm{~S~cm}^2 \\mathrm{~mol}^{-1}$$ respectively. The molar conductivity of solution of an electrolyte with the above cation and anion will be:

    ", "options": [ { "text": "$$187 \\mathrm{~S} \\mathrm{~cm}^2 \\mathrm{~mol}^{-1}$$\n" }, { "text": "$$260 \\mathrm{~S} \\mathrm{~cm}^2 \\mathrm{~mol}^{-1}$$\n" }, { "text": "$$65 \\mathrm{~S} \\mathrm{~cm}^2 \\mathrm{~mol}^{-1}$$\n" }, { "text": "$$130 \\mathrm{~S} \\mathrm{~cm}^2 \\mathrm{~mol}^{-1}$$" } ], "answer": "$$130 \\mathrm{~S} \\mathrm{~cm}^2 \\mathrm{~mol}^{-1}$$", "solution": "**Answer:** $$130 \\mathrm{~S} \\mathrm{~cm}^2 \\mathrm{~mol}^{-1}$$\n\n

    The compound with divalent cation $$\\left(\\mathrm{A}^{2+}\\right)$$ and anion $$\\left(B^{2-}\\right)$$ will be $$A B$$.

    \n

    Molar conductivity of its solution will be

    \n

    $$57+73=130 \\mathrm{~S} \\mathrm{~cm}^2 \\mathrm{~mol}^{-1}$$

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1395, "subject": "Chemistry", "question": "

    The quantity of silver deposited when one coulomb charge is passed through $$\\mathrm{AgNO}_3$$ solution :

    ", "options": [ { "text": "$$0.1 \\mathrm{~g}$$ atom of silver\n" }, { "text": "1 chemical equivalent of silver\n" }, { "text": "$$1 \\mathrm{~g}$$ of silver\n" }, { "text": "1 electrochemical equivalent of silver" } ], "answer": "1 electrochemical equivalent of silver", "solution": "**Answer:** 1 electrochemical equivalent of silver\n\n

    To determine the quantity of silver deposited when one coulomb of charge is passed through $$\\mathrm{AgNO}_3$$ solution, we need to refer to Faraday's laws of electrolysis, especially to the concept of the electrochemical equivalent. The electrochemical equivalent (ECE) of a substance is the amount of the substance that is deposited or dissolved at an electrode during electrolysis by one coulomb of charge.

    \n\n

    The electrochemical equivalent of a substance can be calculated using the formula:

    \n\n

    $$E = \\frac{M}{nF}$$

    \n\n

    where:

    \n\n\n\n

    In the case of silver ($\\mathrm{Ag}$) being deposited from $ \\mathrm{AgNO}_3 $, when silver ion ($\\mathrm{Ag}^+$) is reduced to metallic silver ($\\mathrm{Ag}$), the reaction involves one electron ($n = 1$). The molar mass of silver (Ag) is approximately $107.868 \\, \\mathrm{g/mol}$. Using this information:

    \n\n

    $$E = \\frac{107.868 \\, \\mathrm{g/mol}}{1 \\cdot 96485 \\, \\mathrm{C/mol}} \\approx 0.001118 \\, \\mathrm{g/C}$$

    \n\n

    This means that for every coulomb of charge passed through the solution, $0.001118 \\, \\mathrm{g}$ of silver is deposited.

    \n\n

    Given the options:

    \n\n\n\n

    Therefore, the correct answer is Option D: 1 electrochemical equivalent of silver.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1396, "subject": "Chemistry", "question": "For the following cell with hydrogen electrodes at two different pressure p1 and p2. What will be the emf for the given cell :\n

    $$\\eqalign{\n & Pt({H_2})|{H^ + }(aq)|Pt({H_2}) \\cr \n & \\,\\,\\,\\,\\,{p_1}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,1M\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,{p_2} \\cr} $$", "options": [ { "text": "$${{RT} \\over F}{\\log _e}{{{P_1}} \\over {{P_2}}}$$ " }, { "text": "$${{RT} \\over 2F}{\\log _e}{{{P_1}} \\over {{P_2}}}$$" }, { "text": "$${{RT} \\over F}{\\log _e}{{{P_2}} \\over {{P_1}}}$$" }, { "text": "none of these" } ], "answer": "$${{RT} \\over 2F}{\\log _e}{{{P_1}} \\over {{P_2}}}$$", "solution": "**Answer:** $${{RT} \\over 2F}{\\log _e}{{{P_1}} \\over {{P_2}}}$$\n\nOxidation half cell : -\n

    $${H_2}\\left( g \\right)\\buildrel \\, \\over\n \\longrightarrow 2{H^ + }\\left( {1M} \\right) + 2{e^ - }\\,\\,\\,...{P_1}$$\n

    Reduction half cell\n

    $$2{H^ + }\\left( {1M} \\right) + 2{e^ - }\\buildrel \\, \\over\n \\longrightarrow {H_2}\\left( g \\right)\\,\\,\\,...{P_2}$$ \n

    The net cell reaction\n

    $$\\mathop {{H_2}}\\limits_{{P_1}} \\left( g \\right)\\buildrel \\, \\over\n \\longrightarrow \\mathop {{H_2}}\\limits_{{P_2}} \\left( g \\right)$$\n

    $$E_{cell}^ \\circ = 0.00\\,V$$ $$\\,\\,\\,\\,\\,$$ $$n=2$$ \n

    $$\\therefore$$ $$\\,\\,\\,\\,$$ $${E_{cell}} = E_{cell}^ \\circ - {{RT} \\over {nF}}{\\log _c}K$$\n

    $$ = 0 - {{RT} \\over {nF}}{\\log _e}{{{P_2}} \\over {{P_1}}}$$ \n

    or $$\\,\\,\\,\\,{E_{cell}} = {{RT} \\over {2F}}{\\log _e}{{{P_1}} \\over {{P_2}}}$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1397, "subject": "Chemistry", "question": "EMF of a cell in terms of reduction potential of its left and right electrodes is :", "options": [ { "text": "E = Eleft - Eright" }, { "text": "E = Eleft + Eright" }, { "text": "E = Eright - Eleft" }, { "text": "E = -(Eright + Eleft)" } ], "answer": "E = Eright - Eleft", "solution": "**Answer:** E = Eright - Eleft\n\n$${E_{cell}} = \\,\\,$$ Reduction potential of cathode (right) \n

    $$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,$$ $$-$$ Reduction potential of anode (left)\n

    $$ = {E_{right}} - {E_{left}}.$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1398, "subject": "Chemistry", "question": "Which of the following reaction is possible at anode?", "options": [ { "text": "2Cr3+ + 7H2O $$\\to Cr_2O_7^{2-}$$ + 14H+" }, { "text": "F2 $$\\to$$ 2F-" }, { "text": "(1/2) O2 + 2H+ $$\\to$$ H2O " }, { "text": "none of these" } ], "answer": "2Cr3+ + 7H2O $$\\to Cr_2O_7^{2-}$$ + 14H+", "solution": "**Answer:** 2Cr3+ + 7H2O $$\\to Cr_2O_7^{2-}$$ + 14H+\n\n$$2C{r^{3 + }} + 7{H_2}O \\to C{r_2}O_7^{2 - } + 14{H^ + }$$\n

    $$O.S.\\,\\,$$ of $$Cr$$ changes from $$+3$$ to $$+6$$ by loss of electrons. \n

    At anode oxidation takes place. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1399, "subject": "Chemistry", "question": "When the sample of copper with zinc impurity is to be purified by electrolysis, the appropriate\nelectrodes are :", "options": [ { "text": "cathode = pure zinc, anode = pure copper" }, { "text": "cathode = impure sample, anode = pure copper" }, { "text": "cathode = impure zinc, anode = impure sample" }, { "text": "cathode = pure copper, anode = impure sample" } ], "answer": "cathode = pure copper, anode = impure sample", "solution": "**Answer:** cathode = pure copper, anode = impure sample\n\nPure metal always deposits at cathode.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1400, "subject": "Chemistry", "question": "For a cell reaction involving a two-electron change, the standard e.m.f. of the cell is found to be 0.295 V at 25oC. The equilibrium constant of the reaction at 25oC will be", "options": [ { "text": "29.5 $$\\times$$ 10-2" }, { "text": "10" }, { "text": "1 $$\\times$$ 1010" }, { "text": "1 $$\\times$$ 10-10" } ], "answer": "1 $$\\times$$ 1010", "solution": "**Answer:** 1 $$\\times$$ 1010\n\nThe equlibrium constant is released to the standard emf of cell by the expression\n

    $$\\log \\,K = {E^ \\circ }_{cell} \\times {n \\over {0.059}}$$ \n

    $$ = 0.295 \\times {2 \\over {0.059}}$$\n

    $$\\log \\,K = {{590} \\over {59}} = 10$$ \n

    or $$\\,\\,\\,\\,\\,K = 1 \\times {10^{10}}$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1401, "subject": "Chemistry", "question": "For the redox reaction Zn(s) + Cu2+(0.1 M) $$\\to$$ Zn2+(1M) + Cu(s) taking place in a cell, $$E_{cell}^o$$ is 1.10 volt. Ecell for the cell will be ($$2.303{{RT} \\over F}$$ = 0.0591)", "options": [ { "text": "1.80 volt" }, { "text": "1.07 volt" }, { "text": "0.82 volt" }, { "text": "2.14 volt" } ], "answer": "1.07 volt", "solution": "**Answer:** 1.07 volt\n\n$${E_{cell}} = {E^ \\circ }_{cell} + {{0.059} \\over n}\\log {{\\left[ {C{u^{ + 2}}} \\right]} \\over {\\left[ {Z{n^{ + 2}}} \\right]}}$$\n

    $$ = 1.10 + {{0.059} \\over 2}\\log \\left[ {0.1} \\right]$$ \n

    $$ = 1.10 - 0.0295$$\n

    $$ = 1.07V$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1402, "subject": "Chemistry", "question": "Standard reduction electrode potentials of three metals A,B&C are respectively +0.5 V, -3.0 V & -1.2 V. The\nreducing, powers of these metals are", "options": [ { "text": "A > B > C" }, { "text": "C > B > A" }, { "text": "A > C > B" }, { "text": "B > C > A" } ], "answer": "B > C > A", "solution": "**Answer:** B > C > A\n\n$$\\matrix{\n A & B & C \\cr \n { + 0.5C} & { - 3.0V} & { - 1.2V} \\cr \n\n } $$\n

    NOTE : The higher the negative value of reduction potential, the more is the reducing power. \n

    Hence $$B > C > A.$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1403, "subject": "Chemistry", "question": "Consider the following Eo values
    \n
    $$E_{F{e^{3 + }}/F{e^{2 + }}}^o$$ = 0.77 V;
    \n
    $$E_{S{n^{2 + }}/S{n}}^o$$ = -0.14 V
    \n
    Under standard conditions the potential for the reaction
    \n
    Sn(s) + 2Fe3+(aq) $$\\to$$ 2Fe2+(aq) + Sn2+(aq) is :



    ", "options": [ { "text": "1.68 V" }, { "text": "0.63 V " }, { "text": "0.91 V " }, { "text": "1.40 V " } ], "answer": "0.91 V ", "solution": "**Answer:** 0.91 V \n\n$$F{e^{3 + }} + {e^ - } \\to F{e^{2 + }}\\Delta {G^ \\circ }$$\n

    $$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, = - 1 \\times F \\times 0.77$$\n

    $$S{n^{2 + }} + 2{e^ - } \\to Sn\\left( s \\right)\\Delta {G^ \\circ }$$\n

    $$ = - 2 \\times F\\left( { - 0.14} \\right)$$ \n

    $$Sn\\left( s \\right) + 2F{e^{3 + }}\\left( {aq} \\right)$$\n

    $$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, \\to 2F{e^{2 + }}\\left( {aq} \\right) + S{n^{2 + }}\\left( {aq} \\right)$$\n

    $$\\therefore$$ Standard potential for the given reaction\n

    or $$E_{cell}^o = E_{Sn/S{n^{2 + }}}^o + E_{F{e^{3 + }}/F{e^{2 + }}}^0$$ \n

    $$ = 0.14 + 0.77 = 0.91\\,V$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1404, "subject": "Chemistry", "question": "The standard e.m.f of a cell, involving one electron change is found to be 0.591 V at 25oC.\nThe equilibrium constant of the reaction is (F = 96,500 C mol-1: R = 8.314 JK-1 mol-1) ", "options": [ { "text": "1.0 $$\\times$$ 101" }, { "text": "1.0 $$\\times$$ 1030" }, { "text": "1.0 $$\\times$$ 1010" }, { "text": "1.0 $$\\times$$ 105" } ], "answer": "1.0 $$\\times$$ 1010", "solution": "**Answer:** 1.0 $$\\times$$ 1010\n\n$$E_{cell}^o = E_{cell}^o - {{0.059} \\over n}\\log \\,{K_c}$$ \n

    or $$\\,\\,\\,\\,\\,\\,$$ $$0 = 0.591 - {{0.0591} \\over 1}\\log {K_c}$$\n

    or $$\\,\\,\\,\\,\\,\\,$$ $$\\log \\,{K_c} = {{0.591} \\over {0.0591}} = 10$$\n

    or $$\\,\\,\\,\\,\\,\\,$$ $${K_c} = 1 \\times {10^{10}}$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1405, "subject": "Chemistry", "question": "In a cell that utilises the reaction Zn(s) + 2H+\n(aq) $$\\to$$ Zn2+(aq) + H2(g) addition of H2SO4 to cathode compartment, will ", "options": [ { "text": "lower the E and shift equilibrium to the left " }, { "text": "increases the E and shift equilibrium to the left" }, { "text": "increase the E and shift equilibrium to the right " }, { "text": "Lower the E and shift equilibrium to the right " } ], "answer": "increase the E and shift equilibrium to the right ", "solution": "**Answer:** increase the E and shift equilibrium to the right \n\n$$Zn\\left( s \\right) + 2{H^ + } + \\left( {aq} \\right)\\,\\,\\rightleftharpoons\\,\\,Z{n^{2 + }}\\left( {aq} \\right) + {H_2}\\left( g \\right)$$\n

    $${E_{cell}} = E_{cell}^ \\circ - {{0.059} \\over 2}\\log {{\\left[ {Z{n^{2 + }}} \\right]\\left[ {{H_2}} \\right]} \\over {{{\\left[ {{H^ + }} \\right]}^2}}}$$ \n

    Addition of $${H_2}S{O_4}$$ will increases $$\\left[ {{H^ + }} \\right]$$ and $${E_{cell}}$$ will also increase and the equilibrium will shift towards $$RHS$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1406, "subject": "Chemistry", "question": "The $$E_{{M^{3 + }}/{M^{2 + }}}^o$$ values for Cr, Mn, Fe and Co are – 0.41, +1.57, + 0.77 and +1.97 V\nrespectively. For which one of these metals the change in oxidation state form +2 to +3 is\neasiest? ", "options": [ { "text": "Fe" }, { "text": "Mn" }, { "text": "Cr" }, { "text": "Co" } ], "answer": "Cr", "solution": "**Answer:** Cr\n\nThe given values show that $$Cr$$ has maximum oxidation potential, therefore its oxidation will be easiest. (Change the sign to get the oxidation values)", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1407, "subject": "Chemistry", "question": "For a spontaneous reaction the ∆G , equilibrium constant (K) and $$E_{cell}^o$$ will be respectively", "options": [ { "text": "-ve, >1, +ve" }, { "text": "+ve, >1, -ve " }, { "text": "-ve, <1, -ve " }, { "text": "-ve, >1, -ve " } ], "answer": "-ve, >1, +ve", "solution": "**Answer:** -ve, >1, +ve\n\nNOTE : For spontaneous reaction $$\\Delta G$$ should be negative. Equilibrium constant should be more than one\n

    $$\\left( {\\Delta G = - 2.303\\,RT\\,\\log {K_c},\\,\\,} \\right.$$ \n

    If $${K_c} = 1\\,\\,$$ then $$\\,\\,\\,\\Delta G = 0;\\,\\,$$\n

    If $${K_c} < 1$$ then $$\\left. {\\Delta G = + ve} \\right).\\,\\,$$\n

    Again$$\\,\\,\\,\\Delta G = - nFE_{cell}^ \\circ $$\n

    $$E_{cell}^ \\circ \\,\\,\\,$$ must be $$+ve$$ to have $$\\Delta G - ve.$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1408, "subject": "Chemistry", "question": "Given the data at 25oC,
    \nAg + I- $$\\to$$ AgI + e- , Eo = 0.152 V
    \nAg $$\\to$$ Ag+ + e-, Eo = -0.800 V
    \nWhat is the value of log Ksp for AgI? (2.303 RT/F = 0.059 V)", "options": [ { "text": "–8.12" }, { "text": "+8.612" }, { "text": "–37.83" }, { "text": "–16.13" } ], "answer": "–16.13", "solution": "**Answer:** –16.13\n\n$$\\left( i \\right)\\,\\,\\,Ag \\to A{g^ + } + {e^ - }\\,\\,$$\n

    $$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,{E^ \\circ } = - 0.800\\,V$$\n

    $$\\left( {ii} \\right)\\,\\,\\,Ag + {{\\rm I}^ - } \\to Ag{\\rm I} + {e^ - }$$\n

    $$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,{E^ \\circ } = 0.152\\,V$$\n

    From $$(i)$$ and $$(ii)$$ we have, \n

    $$Ag{\\rm I} \\to A{g^ + } + {{\\rm I}^ - }$$\n

    $$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,{E^ \\circ } = - 0.952\\,\\,V$$\n

    $$E_{cell}^ \\circ = {{0.059} \\over n}\\log \\,K$$\n

    $$\\therefore$$ $$ - 0.952 = {{0.059} \\over 1}\\,\\log \\left[ {A{g^ + }} \\right]\\left[ {{{\\rm I}^ - }} \\right]$$\n
    [ as $$\\,\\,\\,k = \\left[ {A{g^ + }} \\right]\\left[ {{{\\rm I}^ - }} \\right]$$\n

    or $$\\,\\,\\, - {{0.952} \\over {0.059}} = \\log \\,{K_{sp}}$$ \n

    or $$\\,\\,\\, - 16.13 = \\log \\,{K_{sp}}$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1409, "subject": "Chemistry", "question": "The cell, Zn | Zn2+ (1M) || Cu2+ (1M) | Cu($$E_{cell}^o$$ = 1.10V) was allowed to be completely discharged at 298 K. The relative concentration of Zn2+ to Cu2+ $$\\left[ {{{\\left[ {Z{n^{2 + }}} \\right]} \\over {\\left[ {C{u^{2 + }}} \\right]}}} \\right]$$ is", "options": [ { "text": "antilog (24.08) " }, { "text": "37.3" }, { "text": "1037.3" }, { "text": "9.65 $$\\times$$ 104\n " } ], "answer": "1037.3", "solution": "**Answer:** 1037.3\n\n$${E_{cell}} = 0;\\,\\,$$ when cell is completely discharged. \n

    $${E_{cell}} = {E^ \\circ }_{cell} - {{0.059} \\over 2}\\log \\left( {{{\\left[ {Z{n^{2 + }}} \\right]} \\over {\\left[ {C{u^{2 + }}} \\right]}}} \\right)$$ \n

    or$$\\,\\,\\,\\,0 - 1.1 - {{0.059} \\over 2}\\log \\left( {{{\\left[ {Z{n^{2 + }}} \\right]} \\over {\\left[ {C{u^{2 + }}} \\right]}}} \\right)$$\n

    $$\\log \\left( {{{\\left[ {Z{n^{2 + }}} \\right]} \\over {\\left[ {C{u^{2 + }}} \\right]}}} \\right) = {{2 \\times 1.1} \\over {0.059}} = 37.3$$\n

    $$\\therefore$$ $$\\,\\,\\,\\,\\left( {{{\\left[ {Z{n^{2 + }}} \\right]} \\over {\\left[ {C{u^{2 + }}} \\right]}}} \\right) = {10^{37.3}}$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1410, "subject": "Chemistry", "question": "Given $$E_{C{r^{3 + }}/Cr}^o$$ = -0.72 V; $$E_{Fe^{2+}/Fe}^o$$ = -0.42V, The potential for the cell Cr | Cr3+ (0.1M) || Fe2+ (0.01 M) | Fe is", "options": [ { "text": "0.26 V" }, { "text": "0.399 V" }, { "text": "−0.339 V" }, { "text": "−0.26 V" } ], "answer": "0.26 V", "solution": "**Answer:** 0.26 V\n\nFrom the given representation of the cell, $${E_{cell}}$$ can be found as follows. \n

    $${E_{cell}} = E_{F{e^{2 + }}/Fe}^ \\circ - E_{C{r^{3 + }}/Cr}^ \\circ - {{0.059} \\over 6}\\log {{{{\\left[ {C{r^{3 + }}} \\right]}^2}} \\over {{{\\left[ {F{e^{2 + }}} \\right]}^3}}}$$ $$\\left[ {} \\right.$$ Nernst - Equ. $$\\left. {} \\right]$$\n

    $$ = - 0.42 - \\left( { - 0.72} \\right) - {{0.059} \\over 6}\\log {{{{\\left( {0.1} \\right)}^2}} \\over {{{\\left( {0.01} \\right)}^3}}}$$\n

    $$ = - 0.42 + 0.72 - {{0.059} \\over 6}\\log {{0.1 \\times 0.1} \\over {0.01 \\times 0.01 \\times 0.01}}$$\n

    $$ = 0.3 - {{0.059} \\over 6}\\log {{{{10}^{ - 2}}} \\over {{{10}^{ - 6}}}}$$\n

    $$ = 0.3 - {{0.059} \\over 6} \\times 4$$\n

    $$ = 0.30 - 0.0393 = 0.26\\,V$$\n

    Hence option (d) is correct answer. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1411, "subject": "Chemistry", "question": "Given : $$E_{F{e^{3 + }}/Fe}^o$$ = -0.036V; $$E_{F{e^{2 + }}/Fe}^o$$ = -0.439 V
    \nThe value of standard electrode potential for the change,
    \nFe3+ (aq) + e- $$\\to$$ Fe2+ (aq) will be ", "options": [ { "text": "-0.072 V" }, { "text": "0.385 V" }, { "text": "0.770 V " }, { "text": "0.270" } ], "answer": "0.770 V ", "solution": "**Answer:** 0.770 V \n\nGiven \n

    $$F{e^{3 + }} + 3{e^ - } \\to Fe,\\,\\,{E^ \\circ }_{F{e^{3 + }}/Fe}$$\n

    $$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, = - 0.036V\\,\\,\\,...\\left( i \\right)$$\n

    $$F{e^{2 + }} + 2{e^ - } \\to Fe,\\,\\,{E^ \\circ }_{F{e^{2 + }}/Fe}$$\n

    $$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, = - 0.439V\\,\\,\\,...\\left( {ii} \\right)$$\n

    we have to calculate\n

    $$F{e^{3 + }} + {e^ - } \\to F{e^{2 + }},\\,\\,\\Delta G = ?$$\n

    To obtain this equation subtract equ $$(ii)$$ from $$(i)$$ we get \n

    $$F{e^{3 + }} + {e^ - } \\to F{e^{2 + }}\\,\\,\\,...\\left( {iii} \\right)$$\n

    As we know that $$\\Delta G = - nFE$$\n

    Thus for reaction $$(iii)$$ \n

    $$\\Delta G = \\Delta {G_1} - \\Delta G;\\,$$\n

    $$\\, - nF{E^ \\circ } = - nF{E_1} - \\left( { - nF{E_2}} \\right)$$\n

    $$\\, - nF{E^ \\circ } = nF{E_2} - nF{E_1}$$\n

    $$ - 1F{E^ \\circ } = 2 \\times 0.439F - 3 \\times 0.036\\,F$$\n

    $$ - 1F{E^ \\circ } = 0.770\\,F$$\n

    $$\\therefore$$ $$\\,\\,\\,\\,{E^ \\circ } = - 0.770V$$\n

    $${O^{ - - }} > {F^ - } > N{a^ + } > M{g^{ + + }} > A{l^{3 + }}$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1412, "subject": "Chemistry", "question": "The correct order of $$E_{{M^{2 + }}/M}^o$$ values with negative sign for the four successive elements Cr, Mn, Fe\nand Co is :", "options": [ { "text": "Mn > Cr > Fe > Co" }, { "text": "Cr > Fe > Mn > Co" }, { "text": "Fe > Mn > Cr > Co" }, { "text": "Cr > Mn > Fe > Co" } ], "answer": "Mn > Cr > Fe > Co", "solution": "**Answer:** Mn > Cr > Fe > Co\n\nThe value of $$E_{{M^{2 + }}/M}^ \\circ $$ for given metal ions are \n

    $$E_{M{n^{2 + }}/Mn}^ \\circ = - 1.18\\,V,\\,\\,$$ \n

    $$E_{C{r^{2 + }}/Cr}^ \\circ = - 0.9V,$$ \n

    $$E_{F{e^{2 + }}/Fe}^ \\circ = - 0.44\\,V$$ and \n

    $$E_{C{o^{2 + }}/Co}^ \\circ = - 0.28\\,V.$$\n

    So when we consider these values with a negative sign (i.e., taking the magnitude of the negative potentials), we get the values as :

    \n\n

    Therefore, the correct order of $E_{M^{2+}/M}^\\circ$ values with the negative sign will be :

    \n

    Mn > Cr > Fe > Co

    \n

    This matches with option A.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1413, "subject": "Chemistry", "question": "The Gibbs energy for the decomposition of Al2O3 at 500oC is as follows :
    \n
    $${2 \\over 3}A{l_2}{O_3}$$ $$\\to$$ $${4 \\over 3}Al + {O_2}$$, $${\\Delta _r}G$$ = + 966 kJ mol–1
    \n
    The potential difference needed for electrolytic reduction of Al2O3 at 500oC is at least :

    ", "options": [ { "text": "4.5 V" }, { "text": "3.0 V" }, { "text": "2.5 V" }, { "text": "5.0 V" } ], "answer": "2.5 V", "solution": "**Answer:** 2.5 V\n\n$$\\Delta G = - nFE$$ \n

    or $$E = {{\\Delta G} \\over { - nF}} = {{966 \\times {{10}^3}} \\over {4 \\times 36500}}$$ \n

    $$ = - 2.5\\,\\,V$$\n

    $$\\therefore$$ The potential difference needed for the reduction $$=2.5$$ $$V$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1414, "subject": "Chemistry", "question": "The reduction potential of hydrogen half cell will be negative if :", "options": [ { "text": "p(H2) = 1 atm and [H+] = 1.0 M" }, { "text": "p(H2) = 1 atm and [H+] = 2.0 M" }, { "text": "p(H2) = 2 atm and [H+] =1.0 M" }, { "text": "p(H2) = 2 atm and [H+] =2.0 M" } ], "answer": "p(H2) = 2 atm and [H+] =1.0 M", "solution": "**Answer:** p(H2) = 2 atm and [H+] =1.0 M\n\n$${H^ + } + {e^ - }\\buildrel \\, \\over\n \\longrightarrow {1 \\over 2}{H_2};$$\n

    $$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,$$ $$ E = {E^ \\circ } - {{0.059} \\over 1}\\log {{P_{{H_2}}^{{\\raise0.5ex\\hbox{$\\scriptstyle 1$}\n\\kern-0.1em/\\kern-0.15em\n\\lower0.25ex\\hbox{$\\scriptstyle 2$}}}} \\over {\\left[ {{H^ + }} \\right]}}$$\n

    Now if $${P_{{H_2}}} = 2\\,\\,$$ atm and $$\\left[ {{H^ + }} \\right] = 1M$$\n

    then $$E = 0 - {{0.059} \\over 1}\\log {{{2^{1/2}}} \\over 1} = - 2$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1415, "subject": "Chemistry", "question": "The standard reduction potentials for Zn2+/ Zn, Ni2+/ Ni, and Fe2+/ Fe are –0.76, –0.23 and –0.44 V respectively. The reaction

    X + Y2+ $$\\to$$ X2+ + Y will be spontaneous when :", "options": [ { "text": "X = Ni, Y = Fe" }, { "text": "X = Ni, Y = Zn" }, { "text": "X = Fe, Y = Zn" }, { "text": "X = Zn, Y = Ni" } ], "answer": "X = Zn, Y = Ni", "solution": "**Answer:** X = Zn, Y = Ni\n\nFor a spontaneous reaction $$\\Delta G$$ must be $$-ve$$\n

    Since $$\\Delta G = - nF{E^ \\circ }$$\n

    Hence for $$\\Delta G$$ to be $$-ve$$ $$\\Delta {E^ \\circ }$$ has to be positive. \n

    Which is possible when $$X = Zn,Y = Ni$$\n

    $$Zn + N{i^{ + + }}\\buildrel \\, \\over\n \\longrightarrow Z{n^{ + + }} + Ni$$\n

    $$E_{Zn/Z{n^{ + 2}}}^ \\circ + E_{N{i^{2 + }}/Ni}^ \\circ $$\n

    $$ = 0.76 + \\left( { - 0.23} \\right) = + 0.53$$ (positive).", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1416, "subject": "Chemistry", "question": "Given
    \n
    $$E_{C{r^{2 + }}/Cr}^o$$ = -0.74 V; $$E_{MnO_4^ - /M{n^{2 + }}}^o$$ = 1.51 V
    \n
    $$E_{C{r_2}O_7^{2 - }/C{r^{3 + }}}^o$$ = 1.33 V; $$E_{Cl/C{l^ - }}^o$$ = 1.36 V
    \n
    Based on the data given above, strongest oxidising agent will be :


    ", "options": [ { "text": "Cr3+" }, { "text": "Mn2+" }, { "text": "$$MnO_4^ - $$ " }, { "text": "Cl-" } ], "answer": "$$MnO_4^ - $$ ", "solution": "**Answer:** $$MnO_4^ - $$ \n\n

    In electrochemistry, the strongest oxidizing agent will be the one with the highest standard electrode potential (E°), because a higher E° value means a greater tendency to gain electrons, i.e., get reduced. An oxidizing agent gains electrons and in doing so, oxidizes another species.

    \n

    From the provided data, the species with the highest standard electrode potential (E°) is MnO₄⁻, with an E° of 1.51 V. This means that MnO₄⁻ has the greatest tendency to gain electrons and thus is the strongest oxidizing agent.

    \n

    Therefore, the correct answer is :

    \n

    Option C : $$MnO_4^ - $$ .

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1417, "subject": "Chemistry", "question": "Given below are the half-cell reactions:
    \n
    Mn2+ + 2e- $$\\to$$ Mn; Eo = -1.18 V
    \n
    2(Mn3+ + e- $$\\to$$ Mn2+); Eo = +1.51 V
    \n
    The Eo for 3Mn2+ $$\\to$$ Mn + 2Mn3+ will be :


    ", "options": [ { "text": "– 0.33 V; the reaction will not occur" }, { "text": "– 0.33 V; the reaction will occur" }, { "text": "– 2.69 V; the reaction will not occur " }, { "text": "– 2.69 V; the reaction will occur" } ], "answer": "– 2.69 V; the reaction will not occur ", "solution": "**Answer:** – 2.69 V; the reaction will not occur \n\n(a)$$\\,\\,\\,\\,\\,$$ $$M{n^{2 + }} + 2{e^ - } \\to Mn;\\,\\,{E^ \\circ } = - 1.18V;\\,...\\left( i \\right)$$\n

    (b) $$\\,\\,\\,\\,\\,$$ $$M{n^{3 + }} + e \\to M{n^{2 + }};\\,\\,{E^ \\circ } = - 1.51V;\\,...\\left( {ii} \\right)$$\n

    Now multiplying equation $$(ii)$$ by two and subtracting from equation $$(i)$$\n

    $$3M{n^{2 + }} \\to M{n^ + } + 2M{n^{3 + }};$$\n

    $${E^ \\circ } = {E_{{O_X}}} + {E_{{\\mathop{\\rm Re}\\nolimits} d.}} = - 1.18 + \\left( { - 1.51} \\right) = - 2.69V$$ \n

    [ $$-ve$$ value of $$EMF$$ (i.e., $$\\Delta G = + ve$$) shows that the reaction is non-spontaneous ]", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1418, "subject": "Chemistry", "question": "What will occur if a block of copper metal is dropped into a beaker containing a\nsolution of 1M ZnSO4?", "options": [ { "text": "The copper metal will dissolve and zinc metal will be deposited." }, { "text": "The copper metal will dissolve with evolution of hydrogen gas." }, { "text": "The copper metal will dissolve with evolution of oxygen gas." }, { "text": "No reaction will occur." } ], "answer": "No reaction will occur.", "solution": "**Answer:** No reaction will occur.\n\nIn the reaction of copper with zinc ions, zinc is more active (easily oxidised) than copper; thus, no reaction will take place.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1419, "subject": "Chemistry", "question": "Given
    \n$$E_{C{l_2}/C{l^ - }}^o$$ = 1.36 V, $$E_{C{r^{3 + }}/Cr}^o$$ = - 0.74 V
    \n$$E_{C{r_2}{O_7}^{2 - }/C{r^{3 + }}}^o$$ = 1.33 V, $$E_{Mn{O_4}^ - /Mn ^{2+}}^o$$ = 1.51 V
    \nAmong the following, the strongest reducing agent is :", "options": [ { "text": "Mn2+" }, { "text": "Cr3+\n" }, { "text": "Cl–" }, { "text": "Cr" } ], "answer": "Cr", "solution": "**Answer:** Cr\n\n$$E_{C{l_2}/C{l^ - }}^o$$ = 1.36 V, $$E_{C{r^{3 + }}/Cr}^o$$ = - 0.74 V
    \n$$E_{C{r_2}{O_7}^{2 - }/C{r^{3 + }}}^o$$ = 1.33 V, $$E_{Mn{O_4}^ - /Mn ^{2+}}^o$$ = 1.51 V\n

    More negative the E° value of the species, more stronger is the reducing agent. Since Cr3+ is\nhaving least reducing potential, so Cr would be strongest reducing agent.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1420, "subject": "Chemistry", "question": "What is the standard reduction potential (Eo) for Fe3+ $$ \\to $$ Fe ? \n
    Given that :\n
    Fe2+ + 2e$$-$$ $$ \\to $$ Fe; $$E_{F{e^{2 + }}/Fe}^o$$ = $$-$$0.47 V\n
    Fe3+ + e$$-$$ $$ \\to $$ Fe2+; $$E_{F{e^{3 + }}/F{e^{2 + }}}^o$$ = +0.77 V ", "options": [ { "text": "$$-$$ 0.057 V" }, { "text": "+ 0.057 V" }, { "text": "+ 0.30 V" }, { "text": "$$-$$ 0.30 V" } ], "answer": "$$-$$ 0.057 V", "solution": "**Answer:** $$-$$ 0.057 V\n\nFor the given reaction :\n

    Fe3+ + e$$-$$ $$ \\to $$ Fe2+; $$E_{F{e^{3 + }}/F{e^{2 + }}}^o$$ = +0.77 V \n

    $$\\Delta $$Go = -nFEo\n

    $$ \\Rightarrow $$ $$\\Delta G_1^o = - \\left( 1 \\right)F\\left( {0.77} \\right)$$ = -0.77F\n

    For the following reaction :\n

    Fe2+ + 2e$$-$$ $$ \\to $$ Fe; $$E_{F{e^{2 + }}/Fe}^o$$ = $$-$$0.47 V\n

    $$\\Delta G_2^o = - \\left( 2 \\right)F\\left( {-0.47} \\right)$$ = 0.47$$ \\times $$2F\n

    Overall reaction :\n

    Fe3+ + 3e$$-$$ $$ \\to $$ Fe;\n

    $$\\Delta G_3^o = - 3FE_3^o$$\n

    $$ \\Rightarrow $$ $$\\Delta G_1^o$$ + $$\\Delta G_2^o$$ = $$ 3FE_3^o$$\n

    $$ \\Rightarrow $$ -0.77F + 0.47$$ \\times $$2F = $$ 3FE_3^o$$\n

    $$ \\Rightarrow $$ $$E_3^o = - {{0.17} \\over 3}$$ = -0.057 V", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1421, "subject": "Chemistry", "question": "Consider the following standard electrode potentials (Eo in volts) in aqueous solution :\n

    \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    ElementM3+ /MM+ /M
    A1-1.66+ 0.55
    T1+1.26- 0.34
    \n

    Based on these data, which of the following statements is correct ?", "options": [ { "text": "T1+ is more stable than A13+" }, { "text": "A1+ is more stable than A13+ " }, { "text": "T1 + is more stable than A1+" }, { "text": "T13+ is more stable than A13+" } ], "answer": "T1 + is more stable than A1+", "solution": "**Answer:** T1 + is more stable than A1+\n\n

    The standard electrode potential E0 for M+/M become negative for Tl which indicates that Tl+ is more stable than Al+.

    \n

    This can also be be explained by inert pair effect. The atoms of this group have an outer electronic configuration of s2p1. The two s electrons of Tl do not participate in bonding due to poor shielding of d electrons.

    ", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1422, "subject": "Chemistry", "question": "To find the standard potential of M3+/M electrode,the following cell is constituted : Pt/M/M3+(0.001 mol L−1 )/Ag+(0.01 mol L−1 )/Ag \n

    The emf of the cell is found to be 0.421 volt at 298 K. The standard potential of half reaction M3+ + 3e−$$ \\to $$ M at 298 K will be :\n

    (Given $$E_{A{g^ + }\\,/\\,Ag}^ - $$ at 298 K = 0.80 Volt)", "options": [ { "text": "0.38 Volt" }, { "text": "0.32 Volt " }, { "text": "1.28 Volt" }, { "text": "0.66 Volt" } ], "answer": "0.32 Volt ", "solution": "**Answer:** 0.32 Volt \n\n

    According to Nernst equation, for the cell reaction

    \n

    M(s) + 3Ag+(aq) $$\\to$$ M3+(aq) + 3Ag(s)

    \n

    $${E_{cell}} = E_{cell}^o - {{0.059} \\over n}\\log {{[{M^{3 + }}]} \\over {{{[A{g^ + }]}^3}}}$$

    \n

    Substituting the values, we get

    \n

    $$0.421 = E_{cell}^o - {{0.059} \\over 3}\\log {{0.001} \\over {{{(0.01)}^3}}}$$

    \n

    $$E_{cell}^o = 0.48V$$

    \n

    We know that $$E_{cell}^o = E_{A{g^ + }/Ag}^o - E_{{M^{3 + }}/M}^o$$

    \n

    $$0.48 = 0.8 - E_{{M^{3 + }}/M}^o$$

    \n

    $$E_{{M^{3 + }}/M}^o = 0.32V$$

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1423, "subject": "Chemistry", "question": "If the standard electrode potential for a cell is 2 V at 300 K, the equilibrium constant (K) for the reaction \n

    Zn(s) + Cu2+ (aq) $$\\rightleftharpoons$$ Zn2+(aq) + Cu(s) \n

    at 300 K is approximately,\n

    (R = 8 JK$$-$$1mol$$-$$1, F = 96000 C mol$$-$$1)


    ", "options": [ { "text": "e$$-$$80" }, { "text": "e$$-$$160" }, { "text": "e320" }, { "text": "e160" } ], "answer": "e160", "solution": "**Answer:** e160\n\n$$\\Delta $$Go = $$-$$ RT lnk = $$-$$nFEocell\n

    lnk = $${{n \\times F \\times {E^o}} \\over {R \\times T}} = {{2 \\times 96000 \\times 2} \\over {8 \\times 300}}$$\n

    lnk = 160\n

    k = e160", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1424, "subject": "Chemistry", "question": "Given\n

    CO3+ + e– $$ \\to $$ CO2+ ; Eo = + 1.81 V\n

    Pb4+\n + 2e– $$ \\to $$ Pb2+ ; Eo = + 1.67 V\n

    Ce4+\n + e– $$ \\to $$ Ce3+\n ; Eo = + 1.61 V\n

    Bi3+ + 3e– $$ \\to $$ Bi ; Eo = + 0.20 V\n

    Oxidizing power of the species will increase in the order :




    ", "options": [ { "text": "Co3+ < Ce4+\n < Bi3+ < Pb4+" }, { "text": "Co3+ < Pb4+ < Ce4+\n < Bi3+" }, { "text": "Ce4+\n < Pb4+ < Bi3+ < Co3+" }, { "text": "Bi3+ < Ce4+\n < Pb4+ < Co3+" } ], "answer": "Bi3+ < Ce4+\n < Pb4+ < Co3+", "solution": "**Answer:** Bi3+ < Ce4+\n < Pb4+ < Co3+\n\nHigher the value of Standard reduction potential of metal ion electrode, the more strong oxidizing agent is that metal ion electrode and have greater oxidising power.\n

    So, Bi3+ < Ce4+\n < Pb4+ < Co3+", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1425, "subject": "Chemistry", "question": "The standard Gibbs energy for the given cell\nreaction in kJ mol–1 at 298 K is :
    \n
    Zn(s) + Cu2+ (aq) $$ \\to $$ Zn2+ (aq) + Cu (s),
    \n
    E° = 2 V at 298 K
    \n
    (Faraday's constant, F = 96000 C mol–1)


    ", "options": [ { "text": "384" }, { "text": "–192" }, { "text": "–384" }, { "text": "192" } ], "answer": "–384", "solution": "**Answer:** –384\n\nHere Zn is losing two electrons and Cu is gaining two electrons. So only two electrons are involved in the reaction.\n

    $$ \\therefore $$ n = 2\n

    $$\\Delta $$Go = - nFEo\n

    = -2 $$ \\times $$ 96000 $$ \\times $$ 2\n

    = -384 kJ/mol", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1426, "subject": "Chemistry", "question": "Calculate the standard cell potential in (V) of the\ncell in which following reaction takes place :\n

    Fe2+(aq) + Ag+(aq) $$ \\to $$ Fe3+(aq) + Ag (s)\n

    Given that
    \n
    $$E_{A{g^ + }/Ag}^o = xV$$
    \n
    $$E_{Fe^{2+ }/Fe}^o = yV$$
    \n
    $$E_{Fe^{3+ }/Fe}^o = zV$$


    ", "options": [ { "text": "x + 2y - 3z" }, { "text": "x - z" }, { "text": "x - y" }, { "text": "x + y - z" } ], "answer": "x + 2y - 3z", "solution": "**Answer:** x + 2y - 3z\n\nStandard emf,\n

    $${E^0} = E_{A{g^ + }|Ag}^0 - E_{F{e^{3 + }}|F{e^{2 + }}}^0$$ ..............(1)\n

    Given $$E_{Fe^{2+ }/Fe}^o = yV$$\n

    $$ \\therefore $$ Fe2+ + 2e- $$ \\to $$ Fe ..........(2)\n

    $${E^0}$$ = y and $$\\Delta {G^0}$$ = -2Fy\n

    Also given $$E_{Fe^{3+ }/Fe}^o = zV$$\n

    $$ \\therefore $$ Fe3+ + 3e- $$ \\to $$ Fe .........(3)\n

    $${E^0}$$ = z and $$\\Delta {G^0}$$ = -3Fz\n

    Performing (2) - (1), we get\n

    Fe3+ + e- $$ \\to $$ Fe2+\n

    $$ \\therefore $$ $$\\Delta {G^0}$$ = -3Fz + 2Fy\n

    $$ \\Rightarrow $$ -(1)F$${E^0}$$ = -3Fz + 2Fy\n

    $$ \\therefore $$ $$E_{F{e^{3 + }}|F{e^{2 + }}}^0$$ = 3z - 2y\n

    From Equation (1),\n

    $${E^0} = E_{A{g^ + }|Ag}^0 - E_{F{e^{3 + }}|F{e^{2 + }}}^0$$\n

    = x - (3z - 2y) = x + 2y - 3z", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1427, "subject": "Chemistry", "question": "Given that $${E^\\Theta }_{{O_2}/{H_2}O} = 1.23\\,V$$ ;
    \n
    $${E^\\Theta }_{{S_2}O_8^{2 - }/SO_4^{2 - }} = 2.05\\,V$$
    \n
    $${E^\\Theta }_{B{r_2}/B{r^ - }} = 1.09\\,V$$
    \n
    $${E^\\Theta }_{A{u^{3 + }}/Au} = 1.4\\,V$$
    \n
    The strongest oxidizing agent is :


    ", "options": [ { "text": "O2" }, { "text": "Au3+" }, { "text": "Br2" }, { "text": "$${S_2}O_8^{2 - }$$" } ], "answer": "$${S_2}O_8^{2 - }$$", "solution": "**Answer:** $${S_2}O_8^{2 - }$$\n\nWhich electrode have higher value of standard reduction potential (SRP), that electrode will be strongest oxidizing agent.\n

    Tendency to gain electrone is called standard reduction potential. When tendency to gain electron is more then that electrode will have more oxidizing power.\n

    Here given SRP's are :\n

    $${E^\\Theta }_{{O_2}/{H_2}O} = 1.23\\,V$$ ;
    \n
    $${E^\\Theta }_{{S_2}O_8^{2 - }/SO_4^{2 - }} = 2.05\\,V$$
    \n
    $${E^\\Theta }_{B{r_2}/B{r^ - }} = 1.09\\,V$$
    \n
    $${E^\\Theta }_{A{u^{3 + }}/Au} = 1.4\\,V$$
    \n
    You can see $${E^\\Theta }_{{S_2}O_8^{2 - }/SO_4^{2 - }} = 2.05$$,\n

    have highest SRP so $${S_2}O_8^{2 - }$$ is the strongest oxidizing agent.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1428, "subject": "Chemistry", "question": "Given the equilibrium constant: \n

    KC of the reaction : \n

    Cu(s) + 2Ag+ (aq) $$ \\to $$ Cu2+ (aq) + 2Ag(s) is

    10 $$ \\times $$ 1015, calculate the E$$_{cell}^0$$ of this reaciton at 298 K

    [2.303 $${{RT} \\over F}$$ at 298 K = 0.059V]", "options": [ { "text": "0.4736 mV" }, { "text": "0.04736 V" }, { "text": "0.4736 V" }, { "text": "0.04736 mV" } ], "answer": "0.4736 V", "solution": "**Answer:** 0.4736 V\n\nWe know,\n

    $$\\Delta $$Go = -RTln(KC) ....(1)\n

    Also $$\\Delta $$Go = -nF$$E_{cell}^o$$ ....(2)\n

    $$ \\therefore $$ -nF$$E_{cell}^o$$ = -RTln(KC)\n

    $$ \\Rightarrow $$ $$E_{cell}^o$$ = $${{RT} \\over {nF}}\\ln \\left( {{K_C}} \\right)$$\n

    = $$2.303{{RT} \\over {nF}}\\log \\left( {{K_C}} \\right)$$\n

    = $${{0.059} \\over 2}\\log \\left( {10 \\times {{10}^{15}}} \\right)$$\n

    ( n = no of electron transferred = 2 )\n

    = 0.059 $$ \\times $$ $${{16} \\over 2}$$\n

    = 0.059 $$ \\times $$ 8\n

    = 0.472 V", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1429, "subject": "Chemistry", "question": "For the cell Zn(s) |Zn2+ (aq)| |Mx+ (aq)| M(s), different half cells and their standard electrode potentials are given below : \n

    \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    Mx+ (aq)/M(s)Au3+(aq)/Au(s)Ag+(aq)/Ag(s)Fe3+(aq)/Fe2+ (aq)Fe2+(aq)/Fe(s)
    E0Mx+/M/(V)1.400.800.77$$-$$0.44
    \n

    If $$E_{z{n^{2 + }}/zn}^0$$ = $$-$$ 0.76 V, which cathode will give maximum value of Eocell per electron transferred?", "options": [ { "text": "Ag+/Ag" }, { "text": "Fe3+/Fe2+" }, { "text": "Au3+/Au" }, { "text": "Fe2+/Fe" } ], "answer": "Ag+/Ag", "solution": "**Answer:** Ag+/Ag\n\nZn(s) |Zn2+ (aq)| |Mx+ (aq)| M(s)\n
    ---------------------------------------\n
            Anode              Cathode\n

    Eocell = Eocathode – Eoanode\n

    (i) For Ag+/Ag :\n

    Eocell = 0.80 – (– 0.76) = 1.56 V\n

    No of electrons transferred = LCM of valency factor of two electrode\n

    Valency factor of Zn(s) |Zn2+ = 2\n

    Valency factor of Ag+/Ag = 1\n

    LCM of 1 and 2 = 2\n

    $$ \\therefore $$ No of electrons transferred = 2\n

    $$ \\therefore $$ Eocell per electron = $${{1.56} \\over 2}$$ = 0.78\n\n

    (ii) For Fe3+/Fe2+ :\n

    Eocell = 0.77 – (– 0.76) = 1.53 V\n

    No of electrons transferred = LCM of valency factor of two electrode\n

    Valency factor of Zn(s) |Zn2+ = 2\n

    Valency factor of Fe3+/Fe2+ = 1\n

    LCM of 2 and 1 = 2\n

    $$ \\therefore $$ No of electrons transferred = 2\n

    $$ \\therefore $$ Eocell per electron = $${{1.53} \\over 2}$$ = 0.76\n\n

    (iii) For Au3+/Au :\n

    Eocell = 1.40 – (– 0.76) = 2.16 V\n

    No of electrons transferred = LCM of valency factor of two electrode\n

    Valency factor of Zn(s) |Zn2+ = 2\n

    Valency factor of Au3+/Au = 3\n

    LCM of 2 and 3 = 6\n

    $$ \\therefore $$ No of electrons transferred = 6\n

    $$ \\therefore $$ Eocell per electron = $${{2.16} \\over 6}$$ = 0.36\n\n

    (iv) For Fe2+/Fe :\n

    Eocell = –0.44 – (– 0.76) = 0.32 V\n

    No of electrons transferred = LCM of valency factor of two electrode\n

    Valency factor of Zn(s) |Zn2+ = 2\n

    Valency factor of Fe2+/Fe = 2\n

    LCM of 2 and 2 = 2\n

    $$ \\therefore $$ No of electrons transferred = 2\n

    $$ \\therefore $$ Eocell per electron = $${{0.32} \\over 2}$$ = 0.16\n\n

    Eocell is maximum for EoAg+(aq)/Ag(s)\n.", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1430, "subject": "Chemistry", "question": "In the cell \n

    Pt$$\\left| {\\left( s \\right)} \\right|$$H2(g, 1 bar)$$\\left| {HCl\\left( {aq} \\right)} \\right|$$AgCl$$\\left| {\\left( s \\right)} \\right|$$Ag(s)|Pt(s) \n

    the cell potential is 0.92 V when a 10–6 molal HCl solution is used. The standard electrode potential of (AgCl/ AgCl– ) electrode is : \n
    $$\\left\\{ {} \\right.$$Given,  $${{2.303RT} \\over F} = 0.06V$$  at  $$\\left. {298} \\right\\}$$", "options": [ { "text": "0.94 V" }, { "text": "0.40 V" }, { "text": "0.76 V" }, { "text": "0.20 V" } ], "answer": "0.20 V", "solution": "**Answer:** 0.20 V\n\nAnode : H2(g) $$ \\to $$ 2H+(aq) + 2e-\n

    Cathode : AgCl(s) + e- $$ \\to $$ Ag(s) + Cl-(aq)\n
    ---------------------------------------------------------------\n
                H2(g) + 2AgCl(s) $$ \\to $$ 2Ag(s) + 2H+(aq) + 2Cl-(aq)\n

    From Nernst equation we know,\n

    Ecell = E0cell - $${{0.06} \\over n}\\log Q$$\n

    Here,\n

    Ecell = E0cell - $${{0.06} \\over 2}\\log {{{{\\left[ {{H^ + }} \\right]}^2}{{\\left[ {C{l^ - }} \\right]}^2}} \\over {{P_{{H_2}}}}}$$\n

    $$ \\Rightarrow $$ 0.92 = E0AgCl/AgCl - - $$0.03\\log {{{{\\left[ {{{10}^{ - 6}}} \\right]}^2}{{\\left[ {{{10}^{ - 6}}} \\right]}^2}} \\over 1}$$\n

    $$ \\Rightarrow $$ E0AgCl/AgCl = 0.92 - 0.72 = 0.2 V", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1431, "subject": "Chemistry", "question": "Consider the following reduction processes :\n
    Zn2+ + 2e– $$ \\to $$ Zn(s) ; Eo = – 0.76 V\n
    Ca2+ + 2e– $$ \\to $$ Ca(s); Eo = –2.87 V\n
    Mg2+ + 2e– $$ \\to $$ Mg(s) ; Eo = – 2.36 V\n
    Ni2 + 2e– $$ \\to $$ Ni(s) ; Eo = – 0.25 \n
    The reducing power of the metals increases in the order :", "options": [ { "text": "Ca < Mg < Zn < Ni " }, { "text": "Ni < Zn < Mg < Ca" }, { "text": "Zn < Mg < Ni < Ca" }, { "text": "Ca < Zn < Mg < Ni " } ], "answer": "Ni < Zn < Mg < Ca", "solution": "**Answer:** Ni < Zn < Mg < Ca\n\nHigher the oxidation potential better will be reducing power.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1432, "subject": "Chemistry", "question": "The anodic half-cell of lead-acid battery is recharged using electricity of 0.05 Faraday. The amount of PbSO4 electrolyzed in g during the process is : (Molar mass of PbSO4 = 303 g mol$$-$$1)", "options": [ { "text": "22.8" }, { "text": "15.2" }, { "text": "7.6" }, { "text": "11.4" } ], "answer": "7.6", "solution": "**Answer:** 7.6\n\nPb(s) + SO$$_4^{ - 2}$$  $$ \\to $$  PbSO4 + 2e$$-$$\n

    $${{{n_{PbS{O_4}}}} \\over 1} = {{{n_{e - }}} \\over 2}$$\n

    $$ \\Rightarrow $$   $${{{n_{PbS{O_4}}}} \\over 1} = {{0.05} \\over 2}$$\n

    $$ \\therefore $$  Weight of PbSO4 $$=$$ $${{0.05} \\over 2} \\times 303 = 7.6g$$", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 1433, "subject": "Chemistry", "question": "The standard electrode potential $${E^o }$$ and its temperature coefficient $$\\left( {{{d{E^o }} \\over {dT}}} \\right)$$ for a cell are 2V and $$-$$ 5 $$ \\times $$ 10$$-$$4 VK$$-$$1 at 300 K respectively. \n
    The cell reaction is \n
    Zn(s) + Cu2+ (aq) $$\\buildrel \\, \\over\n \\longrightarrow $$ Zn2+ (aq) + Cu(s)\n

    The standard reaction enthalpy ($$\\Delta $$rH$${^o }$$) at 300 K in kJ mol–1 is, [Use R = 8 JK–1 mol–1 and F = 96,000C mol–1] ", "options": [ { "text": "$$-$$ 412.8" }, { "text": "$$-$$ 384.0" }, { "text": "192.0" }, { "text": "206.4" } ], "answer": "$$-$$ 412.8", "solution": "**Answer:** $$-$$ 412.8\n\n$$\\Delta $$G = -nFEcell = -2$$ \\times $$96500$$ \\times $$2 = -386 kJ\n

    $$\\Delta $$S = nF$$\\left( {{{d{E^o }} \\over {dT}}} \\right)$$ = 2$$ \\times $$96500$$ \\times $$ ($$-$$ 5 $$ \\times $$ 10$$-$$4) = -96.5 kJ\n

    At 298 K\n
    T$$\\Delta $$S = 298 $$ \\times $$ (–96.5 J) = – 28.8 kJ\n

    at constant T (=248 K) and pressure\n

    $$\\Delta $$G = $$\\Delta $$H – T$$\\Delta $$S\n

    $$ \\Rightarrow $$ $$\\Delta $$H = $$\\Delta $$G + T$$\\Delta $$S\n

    = -386 - 28.8 = -412.8 kJ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1434, "subject": "Chemistry", "question": "For the disproportionation reaction\n
    2Cu+(aq) ⇌ Cu(s) + Cu2+(aq) at 298 K. ln K\n
    (where K is the equilibrium constant) is\n
    ___________ × 10–1.\n
    Given :\n
    ($$E_{C{u^{2 + }}/C{u^ + }}^0 = 0.16V$$\n
    $$E_{C{u^ + }/Cu}^0 = 0.52V$$\n
    $${{RT} \\over F} = 0.025$$)", "options": [], "answer": "144", "solution": "**Answer:** 144\n\n$$E_{cell}^0$$ = $$E_{C{u^ + }/Cu}^0$$ - $$E_{C{u^{2 + }}/C{u^ + }}^0$$\n

    = 0.52 – 0.16\n

    = 0.36 V\n

    At equilibrium, Ecell = 0\n

    $$E_{cell}^0$$ = $${{RT} \\over {nF}}$$ln K\n

    $$ \\Rightarrow $$ ln K = $${{E_{cell}^0 \\times nF} \\over {RT}}$$\n

    = $${{0.36 \\times 1} \\over {0.025}}$$ = 14.4 = 144 $$ \\times $$ 10-1", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1435, "subject": "Chemistry", "question": "For the given cell :\n

    Cu(s) | Cu2+(C1M) || Cu2+(C2M) | Cu(s)\n

    change in Gibbs energy ($$\\Delta $$G) is negative, if :", "options": [ { "text": "C2 = $$\\sqrt 2 $$C1" }, { "text": "C2 = $${{{C_1}} \\over {\\sqrt 2 }}$$" }, { "text": "C1 = 2C2" }, { "text": "C1 = C2" } ], "answer": "C2 = $$\\sqrt 2 $$C1", "solution": "**Answer:** C2 = $$\\sqrt 2 $$C1\n\nGiven $$\\Delta $$G < 0\n

    $$ \\therefore $$ -nFEcell < 0\n

    $$ \\Rightarrow $$ Ecell > 0\n

    We know, Ecell = $$E_{cell}^0$$ - $${{RT} \\over {2F}}\\ln \\left( {{{{C_1}} \\over {{C_2}}}} \\right)$$\n

    = 0 - $${{RT} \\over {2F}}\\ln \\left( {{{{C_1}} \\over {{C_2}}}} \\right)$$\n

    $$ \\therefore $$ - $${{RT} \\over {2F}}\\ln \\left( {{{{C_1}} \\over {{C_2}}}} \\right)$$ > 0\n

    $$ \\Rightarrow $$ $$\\ln \\left( {{{{C_1}} \\over {{C_2}}}} \\right)$$ < 0\n

    $$ \\Rightarrow $$ C1 < C2\n

    By checking option, we can see\n

    C2 = $$\\sqrt 2 $$C1 satisfy the condition C1 < C2.\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1436, "subject": "Chemistry", "question": "An oxidation-reduction reaction in which
    3 electrons are transferred has a $$\\Delta $$Gº of 17.37 kJ mol–1 at\n
    25 oC. The value of Eo\n
    cell (in V) is ______ × 10–2.\n
    (1 F = 96,500 C mol–1)", "options": [], "answer": "-6", "solution": "**Answer:** -6\n\n$$\\Delta $$Gº = 17.37 kJ ; n = 3\n

    $$\\Delta $$Gº = -nF$$E_{cell}^o$$\n

    $$ \\Rightarrow $$ $$E_{cell}^o$$ = $$ - {{17.37 \\times 1000} \\over {3 \\times 96500}}$$\n

    = -0.06 = -6.00 $$ \\times $$ 10-2", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1437, "subject": "Chemistry", "question": "The photoelectric current from Na (Work function, w0\n = 2.3 eV) is stopped by the output voltage of\nthe cell
    Pt(s) | H2\n(g, 1 Bar) | HCl (aq., pH =1) | AgCl(s) | Ag(s).\n
    The pH of aq. HCl required to stop the photoelectric current form K(w0\n = 2.25 eV), all other\nconditions remaining the same, is _______ $$ \\times $$ 10-2 (to the nearest integer).\n

    Given, 2.303$${{RT} \\over F}$$ = 0.06 V;\n
    $$E_{AgCl|Ag|C{l^ - }}^0$$ = 0.22 V", "options": [], "answer": "142", "solution": "**Answer:** 142\n\n$${1 \\over 2}$$H2 + AgCl $$ \\to $$ H+ + Ag + Cl-\n

    From Nernst Equation,\n

    Ecell = $$E_{cell}^0$$ - $${{0.06} \\over 1}\\log \\left[ {{H^ + }} \\right]\\left[ {C{l^ - }} \\right]$$\n

    = 0.22 – 0.06 log 10–2 = 0.34 V\n

    Work function of Na metal = 2.3 eV\n

    KE of photoelectron = 0.34 eV\n

    Energy of incident radiation = 2.3 + 0.34 = 2.64 eV\n

    For K atom,\n

    Energy of incident radiation for K metal\n= 2.64 eV\n

    Work function of K metal = 2.25 eV\n

    KE of photoelectrons = 2.64 – 2.25 = 0.39 eV\n

    $$ \\therefore $$ Ecell = 0.39 V\n

    Using Nernst Equation,\n

    0.39 = 0.22 - $${{0.06} \\over 1}\\log \\left[ {{H^ + }} \\right]\\left[ {C{l^ - }} \\right]$$\n

    As $$\\left[ {{H^ + }} \\right] = \\left[ {C{l^ - }} \\right]$$\n

    0.39 = 0.22 - $${{0.06} \\over 1}\\log {\\left[ {{H^ + }} \\right]^2}$$\n

    $$ \\Rightarrow $$ 0.39 = 0.22 - $$0.12 \\times \\log \\left[ {{H^ + }} \\right]$$\n

    $$ \\Rightarrow $$ 0.39 = 0.22 + 0.12 $$ \\times $$ pH\n

    $$ \\Rightarrow $$ pH = 1.42 = 142 $$ \\times $$ 10-2", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1438, "subject": "Chemistry", "question": "The Gibbs change (in J) for the given reaction at\n
    [Cu2+] = [Sn2+] = 1 M and 298K is :\n

    Cu(s) + Sn2+(aq.) $$ \\to $$ Cu2+(aq.) + Sn(s);\n

    ($$E_{S{n^{2 + }}|Sn}^0 = - 0.16\\,V$$, \n
    $$E_{C{u^{2 + }}|Cu}^0 = 0.34\\,V$$)\n
    Take F = 96500 C mol–1)", "options": [], "answer": "96500", "solution": "**Answer:** 96500\n\n$$\\Delta $$G = $$\\Delta $$Go + RTln $$\\left[ {{{S{n^{ + 2}}} \\over {C{u^{ + 2}}}}} \\right]$$\n

    = –2 × 96500 [(–0.16) – 0.34] + RT$$\\left[ {{1 \\over 1}} \\right]$$\n

    = 96500 J", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1439, "subject": "Chemistry", "question": "For an electrochemical cell
    \n
    Sn(s) | Sn2+ (aq,1M)||Pb2+ (aq,1M)|Pb(s)
    \n
    the ratio $${{\\left[ {S{n^{2 + }}} \\right]} \\over {\\left[ {P{b^{2 + }}} \\right]}}$$ when this cell attains\nequilibrium is _________.\n

    \n(Given $$E_{S{n^{2 + }}|Sn}^0 = - 0.14V$$,
    \n
    $$E_{P{b^{2 + }}|Pb}^0 = - 0.13V$$, $${{2.303RT} \\over F} = 0.06$$)


    ", "options": [], "answer": "2.13TO2.16", "solution": "**Answer:** 2.13TO2.16\n\nCell reaction is :\n

    Sn(s) + Pb+2(aq) $$ \\to $$ Sn+2(aq) + Pb(s)\n

    Apply Nernst equation :\n

    Ecell = $$E_{cell}^0$$ - $${{0.06} \\over 2}\\log {{\\left[ {S{n^{ + 2}}} \\right]} \\over {\\left[ {P{b^{ + 2}}} \\right]}}$$ ....(1)\n

    $$ \\Rightarrow $$ $$E_{cell}^0$$ = -0.13 + 0.14 = 0.01 V\n

    At equilibrium : Ecell = 0\n

    Substituting in (1), we get\n

    0 = 0.01 - $${{0.06} \\over 2}\\log {{\\left[ {S{n^{ + 2}}} \\right]} \\over {\\left[ {P{b^{ + 2}}} \\right]}}$$\n

    $$ \\Rightarrow $$ $$\\log {{\\left[ {S{n^{ + 2}}} \\right]} \\over {\\left[ {P{b^{ + 2}}} \\right]}}$$ = $${1 \\over 3}$$\n

    $$ \\Rightarrow $$ $${{\\left[ {S{n^{2 + }}} \\right]} \\over {\\left[ {P{b^{2 + }}} \\right]}}$$ = 2.15", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1440, "subject": "Chemistry", "question": "What would be the electrode potential for the given half cell reaction at pH = 5?\n______.\n

    2H2O $$ \\to $$ O2 + 4H$$ \\oplus $$ + 4e– ;\n$$E_{red}^0$$ = 1.23 V\n

    (R = 8.314 J mol–1 K–1 ; Temp = 298 k;

    oxygen under std. atm. pressure of 1 bar)


    ", "options": [], "answer": "1.52TO1.53", "solution": "**Answer:** 1.52TO1.53\n\nE = E0 - $${{0.0591} \\over 4}\\log {\\left[ {{H^ + }} \\right]^4}$$\n

    $$ \\Rightarrow $$ E = 1.23 + 0.0591 × pH\n

    $$ \\Rightarrow $$ E = 1.23 + 0.0591 × (5)\n

    $$ \\Rightarrow $$ E = 1.52", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1441, "subject": "Chemistry", "question": "Given that the standard potentials (Eo) of Cu2+/Cu and Cu+/Cu are 0.34 V and 0.522 V respectively, the Eo of Cu2+/Cu+ :\n", "options": [ { "text": "- 0.182 V" }, { "text": "- 0.158 V" }, { "text": "0.182 V" }, { "text": "+0.158 V" } ], "answer": "+0.158 V", "solution": "**Answer:** +0.158 V\n\nCu+2 + 2e– $$ \\to $$ Cu    Eo = 0.34 V ....(1)\n

    Cu+ + e– $$ \\to $$Cu    Eo = 0.522 V\n

    Cu $$ \\to $$ Cu+ + e–     Eo = -0.522 V ...(2)\n

    By adding (1) and (2) we get,\n

    Cu+2 + e– $$ \\to $$ Cu+ \n

    Eocell = $${{0.34 \\times 2 + 1 \\times \\left( { - 0.522} \\right)} \\over 1}$$ = 0.158 V", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1442, "subject": "Chemistry", "question": "The electrode potential of M2+/M of 3d-series elements shows positive value for :", "options": [ { "text": "Zn" }, { "text": "Fe" }, { "text": "Cu" }, { "text": "Co" } ], "answer": "Cu", "solution": "**Answer:** Cu\n\nIn the electrode potential series, only copper have positive value for electrode potential because copper has lower tendency than hydrogen to form ions. So, if standard hydrogen electrode (ECell = 0) is connected to copper half-cell, the copper with be relatively less negative or less number of electrons.

    $$E_{(C{u^{2 + }}/Cu)}^o$$ = + 0.34 V; $$E_{(F{e^{2 + }}/Fe)}^o$$ = $$-$$ 0.41 V

    $$E_{(C{o^{2 + }}/Co)}^o$$ = $$-$$ 0.28 V; $$E_{(Z{n^{2 + }}/Zn)}^o$$ = $$-$$ 0.76 V

    $$\\therefore$$ Electrode potential of Cu $$E_{(C{u^{2 + }}/Cu)}^o$$ show positive value.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1443, "subject": "Chemistry", "question": "The magnitude of the change in oxidising power of the $$MnO_4^ - /M{n^{2 + }}$$ couple is x $$\\times$$ 10$$-$$4 V, if the H+ concentration is decreased from 1M to 10$$-$$4 M at 25$$^\\circ$$C. (Assume concentration of $$MnO_4^ - $$ and $$M{n^{2 + }}$$ to be same on change in H+ concentration). The value of x is ___________. $$\\left[ {Given\\,:{{2.303RT} \\over F} = 0.059} \\right]$$", "options": [], "answer": "3776", "solution": "**Answer:** 3776\n\nReaction,

    $$MnO_4^ - + {H^ + } + 5{e^ - } \\to M{n^{2 + }} + 4{H_2}O$$

    n = 5

    Applying Nernst equation, $${E_{cell}} = E_{cell}^o - {{0.0591} \\over n}\\log {{[P]} \\over {[R]}}$$

    or $${E_{cell}} = E_{cell}^o - {{0.0591} \\over n}\\log {{[M{n^{2 + }}]} \\over {[MnO_4^ - ]}}{\\left[ {{1 \\over {{H^ + }}}} \\right]^8}$$

    (I) Given, [H+] = 1 M

    $${E_1} = E^\\circ - {{0.0591} \\over 5}\\log {{[M{n^{2 + }}]} \\over {[MnO_4^ - ]}}$$

    (II) Now, [H+] = 10$$-$$4 M

    $${E_2} = E^\\circ - {{0.0591} \\over 5}\\log {{[M{n^{2 + }}]} \\over {[MnO_4^ - ]}} \\times {1 \\over {{{({{10}^{ - 4}})}^8}}}$$

    $$\\therefore$$ $$\\left| {{E_1} - {E_2}} \\right|$$

    $$\\left| {{E_1} - {E_2}} \\right| = {{0.0591} \\over 5} \\times 32 = 0.3776\\,V = 3776 \\times {10^{ - 4}}$$

    x = 3776", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1444, "subject": "Chemistry", "question": "Copper reduces NO$$_3^ - $$ into NO and NO2 depending upon the concentration of HNO3 in solution. (Assuming fixed [Cu2+] and PNO = PNO2), the HNO3 concentration at which the thermodynamic tendency for reduction of NO$$_3^ - $$ into NO and NO2 by copper is same is 10x M. The value of 2x is _______. (Rounded off to the nearest integer)

    [Given, $$E_{C{u^{2 + }}/Cu}^o = 0.34$$ V, $$E_{NO_3^ - /NO}^o = 0.96$$ V, $$E_{NO_3^ - /N{O_2}}^o = 0.79$$ V and at 298 K, $${{RT} \\over F}$$(2.303) = 0.059]", "options": [], "answer": "1", "solution": "**Answer:** 1\n\n

    Cell-I $$(HN{O_3} \\to NO)$$

    \n

    $$3Cu + 2NO_3^ - + 8{H^ + } \\to 3C{u^{2 + }} + 2NO + 4{H_2}O$$

    \n

    $${Q_1} = {{{{[C{u^{2 + }}]}^3} \\times {{({p_{NO}})}^2}} \\over {{{[NO_3^ - ]}^2} \\times {{[{H^ + }]}^8}}}$$

    \n

    $$\\because$$ $$E_1^o = 0.96 - ( - 0.34) = 1.3\\,V$$

    \n

    $${E_1} = 1.3 - {{0.059} \\over 6}\\log {Q_1}$$

    \n

    Cell-II $$(HN{O_3} \\to N{O_2})$$

    \n

    $$Cu + 2NO_3^ - + 4{H^ + } \\to C{u^{2 + }} + 2N{O_2} + 2{H_2}O$$

    \n

    $${Q_2} = {{[C{u^2}] \\times {{({p_{N{O_2}}})}^2}} \\over {{{[NO_3^ - ]}^2} \\times {{[{H^ + }]}^4}}}$$

    \n

    $$\\because$$ $$E_2^o = 0.79 - ( - 0.34)\\,V = 1.13\\,V$$

    \n

    $${E_2} = 1.13 - {{0.059} \\over 2}\\log {Q_2}$$

    \n

    Now, $${E_1} = {E_2}$$

    \n

    $$1.3 - {{0.059} \\over 6}\\log {Q_1} = 1.13 - {{0.059} \\over 2}\\log {Q_2}$$

    \n

    $$0.17 = {{0.059} \\over 6}[\\log {Q_1} - 3\\log {Q_2} - = {{0.059} \\over 6}\\log {{{Q_1}} \\over {{Q_2}}}$$

    \n

    $$ = {{0.059} \\over 6}\\log {{{{[C{u^{2 + }}]}^3} \\times {{({p_{NO}})}^2}} \\over {{{[NO_3^ - ]}^2} \\times {{[{H^ + }]}^8}}} \\times {{{{[NO_3^ - ]}^6} \\times {{[{H^ + }]}^{12}}} \\over {{{[C{u^{2 + }}]}^3} \\times {{({p_{N{O_2}}})}^6}}}$$

    \n

    $$ = {{0.059} \\over 6}\\log {{{{[{H^ + }]}^4} \\times {{[NO_3^ - ]}^4}} \\over {{{({p_{N{O_2}}})}^4}}}$$ [$$\\because$$ $${p_{NO}} = {p_{N{O_2}}}$$]

    \n

    $$ = {{0.059} \\over 6}\\log {{{{[HN{O_3}]}^4}} \\over {{{({p_{N{O_2}}})}^4}}}$$

    \n

    Now, $${p_{N{O_2}}} \\equiv [HN{O_3}]$$

    \n

    So, $$0.17 = {{0.059} \\over 6}\\log {[HN{O_3}]^8}$$

    \n

    $$ = {{0.059} \\over 6} \\times 8\\log [HN{O_3}]$$

    \n

    $$\\log [HN{O_3}] = 2.16$$

    \n

    $$[HN{O_3}] = {10^{2.16}}M = {10^x}M$$

    \n

    $$\\therefore$$ $$x = 2.16$$

    \n

    $$ \\Rightarrow 2x = 2 \\times 2.16 = 4.32 \\simeq 4$$

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1445, "subject": "Chemistry", "question": "Emf of the following cell at 298K in V is x $$\\times$$ 10$$-$$2.

    Zn|Zn2+(0.1 M)||Ag+ (0.01 M)|Ag

    The value of x is _________. (Rounded off to the nearest integer)

    [Given : $$E_{Z{n^{2 + }}/Zn}^\\theta = - 0.76V;E_{A{g^{2 + }}/Ag}^\\theta = + 0.80V;{{2.303RT} \\over F} = 0.059$$]", "options": [], "answer": "147", "solution": "**Answer:** 147\n\nZn | Zn2+(0.1 M) || Ag+ (0.01 M) | Ag\n

    Zn(s) + 2Ag+ $$ \\rightleftharpoons $$ 2Ag(s) + Zn+2\n

    $$E_{cell}^0 = E_{A{g^ + }/Ag}^0 - E_{Z{n^{2 + }}/Zn}^0$$

    $$ = 0.80 - ( - 0.76)$$

    $$ = 1.56V$$

    $${E_{cell}} = 1.56 - {{ 0.059} \\over 2}\\log {{[Z{n^{2 + }}]} \\over {{{[A{g^ + }]}^2}}}$$

    $$ = 1.56 - {{0.059} \\over 2}\\log {{0.1} \\over {{{(0.01)}^2}}}$$

    $$ = 1.56 - {{0.059} \\over 2} \\times 3$$

    $$ = 1.56 - 0.0885$$

    $$ = 1.4715$$

    $$ = 147.15 \\times {10^{ - 2}}$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1446, "subject": "Chemistry", "question": "For the reaction

    2Fe3+(aq) + 2I$$-$$(aq) $$ \\to $$ 2Fe2+(aq) + I2(s)

    the magnitude of the standard molar Gibbs free energy change, $$\\Delta$$rG$$_m^o$$ = $$-$$ ___________ kJ (Round off to the Nearest Integer).

    $$\\left[ {\\matrix{\n {E_{F{e^{2 + }}/Fe(s)}^o = - 0.440V;} & {E_{F{e^{3 + }}/Fe(s)}^o = - 0.036V} \\cr \n {E_{{I_2}/2{I^ - }}^o = 0.539V;} & {F = 96500C} \\cr \n\n } } \\right]$$", "options": [], "answer": "46", "solution": "**Answer:** 46\n\n$$E_{F{e^{3 + }}/Fe}^o = 0.036V$$

    $$F{e^{3 + }} + 3{e^ - } \\to Fe$$

    $$ \\therefore $$ $$\\Delta G_1^o = - nF{E^o}$$

    $$ = - 3F( - 0.036)$$

    $$E_{F{e^{2 + }}/Fe}^o = 0.440V$$

    $$F{e^{2 + }} + 2{e^ - } \\to Fe;{E^o} = - 0.440V$$

    $$Fe \\to F{e^{2 + }} + 2{e^ - };{E^o} = 0.440V$$

    $$ \\therefore $$ $$\\Delta G_2^o = - 2F(0.440)$$

    $$E_{{I_2}/2{I^ - }}^o = 0.539V$$

    $${I_2} + 2{e^ - } \\to 2{I^ - };{E^o} = 0.539V$$

    $$2{I^ - } \\to {I_2} + 2{e^ - };{E^o} = - 0.539V$$

    $$\\Delta G_3^o = - 2F( - 0.539)$$

    $$ \\therefore $$ $$\\Delta {G^o} = 2\\left[ {\\Delta G_1^o + \\Delta G_2^o} \\right] + \\Delta G_3^o$$

    $$ = 2\\left[ {3F(0.036) - 2F(0.440)} \\right] + 2F(0.539)$$

    $$ = - 45934 = - 45.9KJ \\simeq - 46KJ$$", "topic": "Algebra", "subtopic": "Linear Algebra" }, { "id": 1447, "subject": "Chemistry", "question": "The molar conductivities at infinite dilution of barium chloride, sulphuric acid and hydrochloric acid are 280, 860 and 426 S cm2 mol$$-$$1 respectively. The molar conductivity at infinite dilution of barium sulphate is _________ S cm2 mol$$-$$1. (Round off to the Nearest Integer ).", "options": [], "answer": "288", "solution": "**Answer:** 288\n\nFrom Kohlrausch's law

    $$\\Lambda _m^\\infty (BaS{O_4}) = \\lambda _m^\\infty (B{a^{2 + }}) + \\lambda _m^\\infty (SO_4^{2 - })$$

    $$\\Lambda _m^\\infty (BaS{O_4}) = \\Lambda _m^\\infty (BaC{l_2}) + \\Lambda _m^\\infty ({H_2}S{O_4}) - 2\\Lambda _m^\\infty (HCl)$$

    $$ = 280 + 860 - 2(426)$$

    $$ = 288$$ S cm2 mol$$-$$1", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1448, "subject": "Chemistry", "question": "Assume a cell with the following reaction

    $$C{u_{(s)}} + 2A{g^ + }(1 \\times {10^{ - 3}}M) \\to C{u^{2 + }}(0.250M) + 2A{g_{(s)}}$$

    $$E_{cell}^\\Theta = 2.97$$ V

    Ecell for the above reaction is ______________ V. (Nearest integer)

    [Given : log 2.5 = 0.3979, T = 298 K]", "options": [], "answer": "3", "solution": "**Answer:** 3\n\n$$E = {E^o} - {{0.059} \\over 2}\\log {{[C{u^{ + 2}}]} \\over {{{[A{g^ + }]}^2}}}$$

    $$ = 2.97 - {{0.059} \\over 2}\\log {{0.25} \\over {{{({{10}^{ - 3}})}^2}}} = 2.81V$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1449, "subject": "Chemistry", "question": "Consider the cell at 25$$^\\circ$$C

    Zn | Zn2+ (aq), (1M) || Fe3+ (aq), Fe2+ (aq) | Pt(s)

    The fraction of total iron present as Fe3+ ion at the cell potential of 1.500 V is x $$\\times$$ 10$$-$$2. The value of x is ______________. (Nearest integer)

    (Given : $$E_{F{e^{3 + }}/F{e^{2 + }}}^0 = 0.77V$$, $$E_{Z{n^{2 + }}/Zn}^0 = - 0.76V$$)", "options": [], "answer": "24", "solution": "**Answer:** 24\n\n$$Zn\\mathrel{\\mathop{\\kern0pt\\longrightarrow}\n\\limits_{}} Z{n^{2 + }} + 2{e^ - }$$

    $$2F{e^{3 + }}\\mathrel{\\mathop{\\kern0pt\\longrightarrow}\n\\limits_{}} 2{e^ - } + 2{e^{2 + }}$$

    $$Zn + 2F{e^{3 + }}\\buildrel {} \\over\n \\longrightarrow Z{n^{2 + }} + 2F{e^{2 + }}$$

    $$E_{cell}^0 = 0.77 - (0.76)$$

    $$ = 1.53$$ V

    $$1.50 = 1.53 - {{0.06} \\over 2}\\log {\\left( {{{F{e^{2 + }}} \\over {F{e^{3 + }}}}} \\right)^2}$$

    $$\\log \\left( {{{F{e^{2 + }}} \\over {F{e^{3 + }}}}} \\right) = {{0.03} \\over {0.06}} = {1 \\over 2}$$

    $${{[F{e^{2 + }}]} \\over {[F{e^{3 + }}]}} = {10^{1/2}} = \\sqrt {10} $$

    $${{[F{e^{3 + }}]} \\over {[F{e^{2 + }}]}} = {1 \\over {\\sqrt {10} }}$$

    $${{[F{e^{3 + }}]} \\over {[F{e^{2 + }}] + [F{e^{3 + }}]}} = {1 \\over {1 + \\sqrt {10} }} = {1 \\over {4.16}}$$

    = 0.2402

    = 24 $$\\times$$ 10-2", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1450, "subject": "Chemistry", "question": "For the cell

    Cu(s) | Cu2+ (aq) (0.1 M) || Ag+(aq) (0.01 M) | Ag(s)

    the cell potential E1 = 0.3095 V

    For the cell

    Cu(s) | Cu2+ (aq) (0.01 M) || Ag+(aq) (0.001 M) | Ag(s)

    the cell potential = ____________ $$\\times$$ 10$$-$$2 V. (Round off the nearest integer).

    [Use : $${{2.303RT} \\over F}$$ = 0.059]", "options": [], "answer": "28", "solution": "**Answer:** 28\n\nCell reaction is :

    $$Cu(s) + 2A{g^ + }(aq) \\to C{u^{2 + }}(aq) + 2Ag(s)$$

    Now, $${E_{cell}} = E_{cell}^o - {{0.059} \\over 2}\\log {{[C{u^{2 + }}]} \\over {{{[A{g^ + }]}^2}}}$$ .... (1)

    $$\\therefore$$ $${E_1} = 0.3095 = E_{cell}^o - {{0.059} \\over 2}.\\log {{0.01} \\over {{{(0.001)}^2}}}$$ ....(2)

    From (1) and (2), E2 = 0.28 V = 28 $$\\times$$ 10$$-$$2 V", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1451, "subject": "Chemistry", "question": "These are physical properties of an element

    (A) Sublimation enthalpy

    (B) Ionisation enthalpy

    (C) Hydration enthalpy

    (D) Electron gain enthalpy

    The total number of above properties that affect the reduction potential is ____________ (Integer answer)", "options": [], "answer": "3", "solution": "**Answer:** 3\n\nSublimation enthalpy, Ionisation enthalpy and hydration enthalpy affect the reduction potential.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1452, "subject": "Chemistry", "question": "For the galvanic cell,

    Zn(s) + Cu2+ (0.02 M) $$\\to$$ Zn2+ (0.04 M) + Cu(s),

    Ecell = ______________ $$\\times$$ 10$$-$$2 V. (Nearest integer)

    [Use : $$E_{Cu/C{u^{2 + }}}^0$$ = $$-$$ 0.34 V, $$E_{Zn/Z{n^{2 + }}}^0$$ = + 0.76 V, $${{2.303RT} \\over F} = 0.059\\,V$$]", "options": [], "answer": "109", "solution": "**Answer:** 109\n\nGalvanic cell :

    $$Z{n_{(s)}} + \\mathop {Cu_{(aq.)}^{ + 2}}\\limits_{0.02\\,M} \\to \\mathop {Zn_{}^{ + 2}}\\limits_{0.04\\,M} + Cu(s)$$

    Nernst equation = $${F_{cell}} = E_{cell}^o - {{0.059} \\over 2}\\log {{[2{n^{ + 2}}]} \\over {[C{u^{ + 2}}]}}$$

    $$ \\Rightarrow {E_{cell}}\\left[ {E_{cell}^o - E_{Z{n^{ + 2}}/Zn}^o} \\right] - {{0.059} \\over 2}\\log {{0.04} \\over {0.02}}$$

    $$ \\Rightarrow {E_{cell}}[0.34 - ( - 0.76)] - {{0.059} \\over 2}{\\log ^2}$$

    $$ \\Rightarrow {E_{cell}}1 - 1 - {{0.059} \\over 2} \\times 0.3010$$

    = 1.0911 = 109.11 $$\\times$$ 10$$-$$2

    = 109", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1453, "subject": "Chemistry", "question": "Consider the following cell reaction :

    $$C{d_{(s)}} + H{g_2}S{O_{4(s)}} + {9 \\over 5}{H_2}{O_{(l)}}$$ $$\\rightleftharpoons$$ $$CdS{O_4}.{9 \\over 5}{H_2}{O_{(s)}} + 2H{g_{(l)}}$$

    The value of $$E_{cell}^0$$ is 4.315 V at 25$$^\\circ$$C. If $$\\Delta$$H$$^\\circ$$ = $$-$$825.2 kJ mol$$-$$1, the standard entropy change $$\\Delta$$S$$^\\circ$$ in J K$$-$$1 is ___________. (Nearest integer) [Given : Faraday constant = 96487 C mol$$-$$1]", "options": [], "answer": "25", "solution": "**Answer:** 25\n\n$$\\Delta G^\\circ = - nFE^\\circ = \\Delta H^\\circ - T\\Delta S^\\circ $$

    $$ \\therefore $$ $$\\Delta$$S$$^\\circ$$ $$ = {{\\Delta H^\\circ + nFE^\\circ } \\over T}$$

    $$ = {{( - 825.2 \\times {{10}^3}) + (2 \\times 96487 \\times 4.315)} \\over {298}}$$

    $$ = {{ - 825.2 \\times {{10}^3} + 832.682 \\times {{10}^3}} \\over {298}}$$

    $$ = {{7.483 \\times {{10}^3}} \\over {298}} = 25.11$$ JK$$-$$1 mol$$-$$1

    $$\\therefore$$ Nearest integer answer is 25.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1454, "subject": "Chemistry", "question": "

    The cell potential for the given cell at 298 K

    \n

    Pt| H2 (g, 1 bar) | H+ (aq) || Cu2+ (aq) | Cu(s)

    \n

    is 0.31 V. The pH of the acidic solution is found to be 3, whereas the concentration of Cu2+ is 10$$-$$x M. The value of x is ___________.

    \n

    (Given : $$E_{C{u^{2 + }}/Cu}^\\Theta $$ = 0.34 V and $${{2.303\\,RT} \\over F}$$ = 0.06 V)

    ", "options": [], "answer": "7", "solution": "**Answer:** 7\n\n$\\mathrm{Q}=\\frac{\\left[\\mathrm{H}^{+}\\right]^{2}}{\\left[\\mathrm{Cu}^{+2}\\right] \\mathrm{pH}_{2}}=\\frac{10^{-6}}{\\mathrm{C}} \\quad \\mathrm{pH}_{2}=1$\n

    \n$$\n\\begin{aligned}\n&E=E_{\\text {cell }}^{\\circ}-\\frac{0.06}{n} \\log Q \\\\\\\\\n&0.31=0.34-\\frac{0.06}{2} \\log \\frac{10^{-6}}{C} \\\\\\\\\n&\\log \\frac{10^{-6}}{C}=1 \\\\\\\\\n&C=10^{-7} M \\\\\\\\\n&x=7\n\\end{aligned}\n$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1455, "subject": "Chemistry", "question": "

    For the given reactions

    \n

    Sn2+ + 2e$$-$$ $$\\to$$ Sn

    \n

    Sn4+ + 4e$$-$$ $$\\to$$ Sn

    \n

    the electrode potentials are ; $$E_{S{n^{2 + }}/Sn}^o = - 0.140$$ V and $$E_{S{n^{4 + }}/Sn}^o = + 0.010$$ V. The magnitude of standard electrode potential for $$S{n^{4 + }}/S{n^{2 + }}$$ i.e. $$E_{S{n^{4 + }}/S{n^{2 + }}}^o$$ is _____________ $$\\times$$ 10$$-$$2 V. (Nearest integer)

    ", "options": [], "answer": "16", "solution": "**Answer:** 16\n\n$\\mathrm{Sn} \\longrightarrow \\mathrm{Sn}^{2+}+2 \\mathrm{e}^{-} \\quad \\mathrm{E}_{1}^{0}=0.140 \\mathrm{~V}$\n

    \n$$\n\\mathrm{Sn}^{4+}+4 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Sn} \\quad \\mathrm{E}_{2}^{0}=0.010 \\mathrm{~V}\n$$\n

    \n$$\n\\begin{aligned}\n& \\mathrm{Sn}^{4+}+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Sn}^{2+} \\quad \\mathrm{E}_{\\text {cell }}^{0} \\\\\\\\\n& \\mathrm{E}_{\\mathrm{cell}}^{\\mathrm{O}}=\\frac{\\mathrm{n}_{2} \\mathrm{E}_{2}^{\\mathrm{o}}+\\mathrm{n}_{1} \\mathrm{E}_{1}^{0}}{\\mathrm{n}}=\\frac{4(0.010)+2(0.140)}{2} \\\\\\\\\n& \\mathrm{E}_{\\text {cell }}^{0}=0.16 \\mathrm{~V}=16 \\times 10^{-2} \\mathrm{~V}\n\\end{aligned}\n$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1456, "subject": "Chemistry", "question": "

    In 3d series, the metal having the highest M2+/M standard electrode potential is :

    ", "options": [ { "text": "Cr" }, { "text": "Fe" }, { "text": "Cu" }, { "text": "Zn" } ], "answer": "Cu", "solution": "**Answer:** Cu\n\nCr+2/Cr $$ \\to $$ –0.90 V

    \nFe+2/Fe $$ \\to $$ –0.44 V

    \nCu+2/Cu $$ \\to $$ +0.34 V

    \nZn+2/Zn $$ \\to $$ -0.76 V

    \n So Ans. is Cu+2/Cu", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1457, "subject": "Chemistry", "question": "

    For the reaction taking place in the cell :

    \n

    Pt (s)| H2 (g)|H+(aq) || Ag+(aq) |Ag (s)

    \n

    E$$_{cell}^o$$ = + 0.5332 V.

    \n

    The value of $$\\Delta$$fG$$^\\circ$$ is ______________ kJ mol$$-$$1. (in nearest integer)

    ", "options": [], "answer": "51OR103", "solution": "**Answer:** 51OR103\n\n

    \"JEE

    \n

    # At anode, oxidation occur

    \n

    H2 $$\\to$$ 2H+ + 2e$$-$$ ....... (1)

    \n

    # At cathode, reduction occur

    \n

    2Ag+ + 2e$$-$$ $$\\to$$ 2Ag ...... (2)

    \n

    Adding equation (1) and (2), we get n = 2, where n = cancelled out electron

    \n

    Now,

    \n

    $$\\Delta G^\\circ = - nF\\,E_{cell}^o$$

    \n

    $$ = - 2 \\times 96500 \\times 0.5332$$

    \n

    $$ = - 102907.6$$

    \n

    $$ = - 102.9$$ kJ/mol

    \n

    $$ = - 103$$ kJ/mol

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1458, "subject": "Chemistry", "question": "

    The $${\\left( {{{\\partial E} \\over {\\partial T}}} \\right)_P}$$ of different types of half cells are as follows:

    \n

    \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    ABCD
    $$1 \\times {10^{ - 4}}$$$$2 \\times {10^{ - 4}}$$$$0.1 \\times {10^{ - 4}}$$$$0.2 \\times {10^{ - 4}}$$

    \n

    (Where E is the electromotive force)

    \n

    Which of the above half cells would be preferred to be used as reference electrode?

    ", "options": [ { "text": "A" }, { "text": "B" }, { "text": "C" }, { "text": "D" } ], "answer": "C", "solution": "**Answer:** C\n\nA cell with less variation in EMF with temperature\nis preferred as a reference electrode because it can be\nused for a wider range of temperatures without much\nderivation from standard value so a cell with less $${\\left( {{{\\partial E} \\over {\\partial T}}} \\right)_P}$$ is preferred. ", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1459, "subject": "Chemistry", "question": "

    Cu(s) + Sn2+ (0.001M) $$\\to$$ Cu2+ (0.01M) + Sn(s)

    \n

    The Gibbs free energy change for the above reaction at 298 K is x $$\\times$$ 10$$-$$1 kJ mol$$-$$1. The value of x is __________. [nearest integer]

    \n

    [Given : $$E_{C{u^{2 + }}/Cu}^\\Theta = 0.34\\,V$$ ; $$E_{S{n^{2 + }}/Sn}^\\Theta = - 0.14\\,V$$ ; F = 96500 C mol$$-$$1]

    ", "options": [], "answer": "983", "solution": "**Answer:** 983\n\n$$\n\\begin{aligned}\n\\mathrm{Cu} &+\\mathrm{Sn}^{+2} \\longrightarrow \\mathrm{Cu}^{+2}+\\mathrm{Sn}(\\mathrm{s}) \\\\\\\\\n\\mathrm{E}_{\\text {cell }}^{\\circ} &=\\mathrm{E}_{\\mathrm{OX}}^{\\circ}+\\mathrm{E}_{\\text {Red }}^{\\circ} \\\\\\\\\n&=-0.34-0.14 \\\\\\\\\n&=-0.48 \\mathrm{~V} \\\\\\\\\n\\mathrm{E}=& ~\\mathrm{E}^{\\circ}-\\frac{0.0591}{2} \\log \\frac{\\left[\\mathrm{Cu}^{+2}\\right]}{\\left[\\mathrm{Sn}^{+2}\\right]} \\\\\\\\\n=&-0.48-0.0295 \\log 10 \\\\\\\\\n=&-0.5095 \\mathrm{~V} \\\\\\\\\n\\Delta \\mathrm{G}=&-\\mathrm{nFE} \\\\\\\\\n&=-2 \\times 96500 \\times-0.5095 \\mathrm{~J} / \\mathrm{mol} \\\\\\\\\n&=98333.5 \\times 10^{-3} \\mathrm{~kJ} / \\mathrm{mol} \\\\\\\\\n&=983.3 \\times 10^{-1} \\mathrm{~kJ} / \\mathrm{mol} \\\\\\\\\n&=983 \\times 10^{-1} \\mathrm{~kJ} / \\mathrm{mol}\n\\end{aligned}\n$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1460, "subject": "Chemistry", "question": "

    The correct order of reduction potentials of the following pairs is

    \n

    A. Cl2/Cl$$-$$

    \n

    B. I2/I$$-$$

    \n

    C. Ag+/Ag

    \n

    D. Na+/Na

    \n

    E. Li+/Li

    \n

    Choose the correct answer from the options given below.

    ", "options": [ { "text": "A > C > B > D > E" }, { "text": "A > B > C > D > E" }, { "text": "A > C > B > E > D" }, { "text": "A > B > C > E > D" } ], "answer": "A > C > B > D > E", "solution": "**Answer:** A > C > B > D > E\n\n$$\\mathrm{E}_{\\mathrm{C}_{2} / \\mathrm{CI}}^{\\circ}=+1.36 \\mathrm{~V}$$

    \n$$\\mathrm{E}_{\\mathrm{I}_{2} / \\mathrm{I}^{-}}^{0}=+0.54 \\mathrm{~V}$$

    \n$$\\mathrm{E}_{\\mathrm{Ag}^{+} / \\mathrm{Ag}}^{\\circ}=+0.80 \\mathrm{~V}$$

    \n$$\\mathrm{E}_{\\mathrm{Na}^{+} / \\mathrm{Na}}^{\\circ}=-2.71 \\mathrm{~V}$$

    \n$$\\mathrm{E}_{\\mathrm{L}^{+} / \\mathrm{Li}}=-3.05 \\mathrm{~V}$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1461, "subject": "Chemistry", "question": "

    In a cell, the following reactions take place

    \n

    $$\\matrix{\n {F{e^{2 + }} \\to F{e^{3 + }} + {e^ - }} & {E_{F{e^{3 + }}/F{e^{2 + }}}^o = 0.77\\,V} \\cr \n {2{I^ - } \\to {I_2} + 2{e^ - }} & {E_{{I_2}/{I^ - }}^o = 0.54\\,V} \\cr \n\n } $$

    \n

    The standard electrode potential for the spontaneous reaction in the cell is x $$\\times$$ 10$$-$$2 V 298 K. The value of x is ____________. (Nearest Integer)

    ", "options": [], "answer": "23", "solution": "**Answer:** 23\n\n$\\underset{\\text { Cathode }}{\\mathrm{Fe}^{+3}}+\\underset{\\text { anode }}{\\mathrm{I}^{-}} \\longrightarrow \\mathrm{I}_2+\\mathrm{Fe}^{+2}$

    \n$E_{\\text {cell }}^{\\circ}=E_{\\text {cathode }}^{\\circ}-E_{\\text {anode }}^{\\circ}$\n

    \n$$\n\\begin{aligned}\n&=0.77-0.54 \\\\\\\\\n&=0.23 \\mathrm{~V} \\\\\\\\\n&=23 \\times 10^{-2} \\mathrm{~V}\n\\end{aligned}\n$$", "topic": "Algebra", "subtopic": "Linear Algebra" }, { "id": 1462, "subject": "Chemistry", "question": "

    The cell potential for the following cell

    \n

    Pt |H2(g)|H+ (aq)|| Cu2+ (0.01 M)|Cu(s)

    \n

    is 0.576 V at 298 K. The pH of the solution is __________. (Nearest integer)

    \n

    (Given : $$E_{C{u^{2 + }}/Cu}^o = 0.34$$ V and $${{2.303\\,RT} \\over F} = 0.06$$ V)

    ", "options": [], "answer": "5", "solution": "**Answer:** 5\n\n$E_{\\text {cell }}=E_{\\text {cell }}^{0}-\\frac{0.06}{2} \\log \\frac{\\left[\\mathrm{H}^{\\oplus}\\right]^{2}}{\\left[\\mathrm{Cu}^{+2}\\right]}$\n

    \n$0.576=0.34-0.03 \\log \\frac{\\left[\\mathrm{H}^{\\oplus}\\right]^{2}}{[0.01]}$\n

    \n$0.576-0.34=-0.03 \\log \\left[\\mathrm{H}^{\\oplus}\\right]^{2}+0.03 \\log (0.01)$\n

    \n$$\n=0.06 \\,\\mathrm{pH}-0.06\n$$\n

    \n$\\mathrm{pH} \\simeq 4.93 \\simeq 5$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1463, "subject": "Chemistry", "question": "

    In which of the following half cells, electrochemical reaction is pH dependent?

    ", "options": [ { "text": "$$Pt\\,|\\,F{e^{3 + }},\\,F{e^{2 + }}$$" }, { "text": "$$MnO_4^ - \\,|M{n^{2 + }}$$" }, { "text": "$$Ag\\,|\\,AgCl\\,|C{l^{ - 1}}$$" }, { "text": "$${1 \\over 2}{F_2}\\,|{F^ - }$$" } ], "answer": "$$MnO_4^ - \\,|M{n^{2 + }}$$", "solution": "**Answer:** $$MnO_4^ - \\,|M{n^{2 + }}$$\n\nReduction of $\\mathrm{MnO}_{4}^{-}$ is $\\mathrm{pH}$ dependent.\n

    \nIn acidic medium\n

    \n$$\n\\mathrm{MnO}_{4}^{-}+5 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Mn}^{2+}\n$$\n

    \nIn neutral medium\n

    \n$$\n\\mathrm{MnO}_{4}^{-}+3 \\mathrm{e}^{-} \\longrightarrow \\mathrm{Mn}^{4+}\n$$\n

    \nIn basic medium\n

    \n$$\n\\mathrm{MnO}_{4}^{-}+\\mathrm{e} \\longrightarrow \\mathrm{Mn}^{6+}\n$$\n

    \nSo, according to $\\mathrm{pH}$, the reaction and potential of cell changes.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1464, "subject": "Chemistry", "question": "

    The cell potential for $$\\mathrm{Zn}\\left|\\mathrm{Zn}^{2+}(\\mathrm{aq})\\right|\\left|\\mathrm{Sn}^{x+}\\right| \\mathrm{Sn}$$ is $$0.801 \\mathrm{~V}$$ at $$298 \\mathrm{~K}$$. The reaction quotient for the above reaction is $$10^{-2}$$. The number of electrons involved in the given electrochemical cell reaction is ____________.

    \n

    $$\\left(\\right.$$ Given $$: \\mathrm{E}_{\\mathrm{Zn}^{2+} \\mid \\mathrm{Zn}}^{\\mathrm{o}}=-0.763 \\mathrm{~V}, \\mathrm{E}_{\\mathrm{Sn}^{x+} \\mid \\mathrm{Sn}}^{\\mathrm{o}}=+0.008 \\mathrm{~V}$$ and $$\\left.\\frac{2.303 \\mathrm{RT}}{\\mathrm{F}}=0.06 \\mathrm{~V}\\right)$$

    ", "options": [], "answer": "4", "solution": "**Answer:** 4\n\n$$\\mathrm{A}: \\mathrm{Zn} \\rightarrow \\mathrm{Zn}^{2+}+2 e^{-}$$\n

    \n$$\\mathrm{C}: \\mathrm{Sn}^{+\\mathrm{x}}+\\mathrm{xe}^{-} \\rightarrow \\mathrm{Sn}$$\n

    \n$$\\mathrm{E}_{\\mathrm{Cell}}^{\\circ}=\\mathrm{E}_{\\mathrm{Zn} \\mid \\mathrm{Zn}^{2+}}^{\\circ}+\\mathrm{E}_{\\mathrm{Sn}^{+x} \\mid \\mathrm{Sn}}^{\\circ}$$\n

    \n$$\\Rightarrow 0.763+0.008=0.771 \\mathrm{~V}$$\n

    \nFrom the Nernst equation,\n

    \n$$\\mathrm{E}_{\\text {Cell }}=\\mathrm{E}_{\\text {Cell }}^{\\circ} \\frac{-2.303 \\,\\mathrm{RT}}{\\mathrm{nF}} \\log \\mathrm{Q}$$\n

    \n$$0.801=0.771-\\frac{0.06}{\\mathrm{n}} \\log 10^{-2}$$\n

    \n$$0.03=\\frac{0.06}{n} \\times 2$$\n

    \n$$\\mathrm{n}=4$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1465, "subject": "Chemistry", "question": "

    The spin-only magnetic moment value of M3+ ion (in gaseous state) from the pairs Cr3+ / Cr2+, Mn3+ / Mn2+, Fe3+ / Fe2+ and Co3+ / Co2+ that has negative standard electrode potential, is ____________ B.M. [Nearest integer]

    ", "options": [], "answer": "4", "solution": "**Answer:** 4\n\nAmong the pairs given, $$\\mathrm{Cr}^{3+} / \\mathrm{Cr}^{2+}$$ has negative reduction potential which is $$-0.41 \\mathrm{~V}$$.\n

    \n$$\\operatorname{Cr}$$ (III) $$\\Rightarrow d^{3}$$\n

    \nNumber of unpaired electrons $$=3$$\n

    \n$$\\mu=\\sqrt{3(3+2)}=\\sqrt{15} \\simeq 4$$ B.M.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1466, "subject": "Chemistry", "question": "

    For a cell, $$\\mathrm{Cu}(\\mathrm{s})\\left|\\mathrm{Cu}^{2+}(0.001 \\,\\mathrm{M}) \\| \\mathrm{Ag}^{+}(0.01 \\,\\mathrm{M})\\right| \\mathrm{Ag}(\\mathrm{s})$$

    \n

    the cell potential is found to be $$0.43 \\mathrm{~V}$$ at $$298 \\mathrm{~K}$$. The magnitude of standard electrode potential for $$\\mathrm{Cu}^{2+} / \\mathrm{Cu}$$ is _________ $$\\times 10^{-2} \\mathrm{~V}$$.

    \n

    [Given : $$E_{A{g^ + }/Ag}^\\Theta $$ = 0.80 V and $${{2.303RT} \\over F}$$ = 0.06 V]

    ", "options": [], "answer": "34", "solution": "**Answer:** 34\n\nAnode : $$\\mathrm{Cu}(\\mathrm{s}) \\longrightarrow \\mathrm{Cu}^{2+}(\\mathrm{aq})+2 \\mathrm{e}^{-}$$\n

    \nCathode : $$\\left.[\\mathrm{Ag}^{+}+\\mathrm{e}^{-} \\longrightarrow \\mathrm{Ag}(\\mathrm{s})\\right] 2$$

    \n

    \nCus(s) $$+2 \\mathrm{Ag}^{+}(\\mathrm{aq}) \\longrightarrow \\mathrm{Cu}^{2+}(\\mathrm{aq})+2 \\mathrm{Ag}(\\mathrm{s})$$

    \n$$\n\\begin{aligned}\n& E_{\\text {cell }}=E_{\\text {cell }}^{0}-\\frac{0.06}{2} \\log \\frac{\\left[\\mathrm{Cu}^{2+}\\right]}{\\left[\\mathrm{Ag}^{+}\\right]^{2}} \\\\\n& 0.43=\\mathrm{E}_{\\text {cell }}^{0}-\\frac{0.06}{2} \\log \\left(\\frac{10^{-3}}{\\left(10^{-2}\\right)^{2}}\\right) \\\\\n& 0.43=E_{\\text {cell }}^{0}-0.03 \\log 10 \\\\\n& \\mathrm{E}_{\\text {cell }}^{0}=0.46 \\mathrm{~V} \\\\\n& \\mathrm{E}_{\\text {cell }}^{0}=\\mathrm{E}_{\\mathrm{Ag}^{+} / \\mathrm{Ag}}^{0}-\\mathrm{E}_{\\mathrm{Cu}^{2+} / \\mathrm{Cu}}^{0} \\\\\n& \\mathrm{E}_{\\mathrm{Cu}^{2+} / \\mathrm{Cu}}^{0}=(0.80-0.46)=0.34 \\mathrm{~V}=34 \\times 10^{-2}\n\\end{aligned}\n$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1467, "subject": "Chemistry", "question": "

    At what pH, given half cell $$\\mathrm{MnO_{4}^{-}(0.1~M)~|~Mn^{2+}(0.001~M)}$$ will have electrode potential of 1.282 V? ___________ (Nearest Integer)

    \n

    Given $$\\mathrm{E_{MnO_4^ - |M{n^{2 + }}}^o}=1.54~\\mathrm{V},\\frac{2.303\\mathrm{RT}}{\\mathrm{F}}=0.059\\mathrm{V}$$

    ", "options": [], "answer": "3", "solution": "**Answer:** 3\n\n$\\mathrm{MnO}_4^{-}+8 \\mathrm{H}^{+}+5 \\mathrm{e}^{-} \\rightleftharpoons \\mathrm{Mn}^{2+}+4 \\mathrm{H}_2 \\mathrm{O}$\n

    $$\n\\begin{aligned}\n& \\mathrm{E}=\\mathrm{E}^{\\circ}-\\frac{0.059}{5} \\log \\frac{\\left[\\mathrm{Mn}^{2+}\\right]}{\\left[\\mathrm{MnO}_4^{-}\\right]\\left[\\mathrm{H}^{+}\\right]^8} \\\\\\\\\n& \\Rightarrow 1.282=1.54-\\frac{0.059}{5} \\log \\frac{10^{-3}}{10^{-1} \\times\\left[\\mathrm{H}^{+}\\right]^8} \\\\\\\\\n& \\Rightarrow \\frac{0.258 \\times 5}{0.059}=\\log \\frac{10^{-2}}{\\left[\\mathrm{H}^{+}\\right]^8} \\\\\\\\\n& \\Rightarrow 21.86=-2+8 \\mathrm{pH} \\\\\\\\\n& \\therefore \\mathrm{pH}=2.98 \\\\\\\\\n& \\simeq 3 \\\\\\\\\n&\n\\end{aligned}\n$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1468, "subject": "Chemistry", "question": "

    The logarithm of equilibrium constant for the reaction $$\\mathrm{Pd}^{2+}+4 \\mathrm{Cl}^{-} \\rightleftharpoons \\mathrm{PdCl}_{4}^{2-}$$ is ___________ (Nearest integer)

    \n

    Given : $$\\frac{2.303 R \\mathrm{~T}}{\\mathrm{~F}}=0.06 \\mathrm{~V}$$

    \n

    $$\n\\mathrm{Pd}_{(\\mathrm{aq})}^{2+}+2 \\mathrm{e}^{-} \\rightleftharpoons \\mathrm{Pd}(\\mathrm{s}) \\quad \\mathrm{E}^{\\ominus}=0.83 \\mathrm{~V}\n$$

    \n

    $$\n\\begin{aligned}\n& \\mathrm{PdCl}_{4}^{2-}(\\mathrm{aq})+2 \\mathrm{e}^{-} \\rightleftharpoons \\mathrm{Pd}(\\mathrm{s})+4 \\mathrm{Cl}^{-}(\\mathrm{aq}) \\mathrm{E}^{\\ominus}=0.65 \\mathrm{~V}\n\\end{aligned}\n$$

    ", "options": [], "answer": "6", "solution": "**Answer:** 6\n\nSol. $\\Delta G^{\\circ}=-R T \\ell n K$\n

    $$\n\\begin{aligned}\n& -\\mathrm{nFE}_{\\text {cell }}^{\\mathrm{o}}=-\\mathrm{RT} \\times 2.303\\left(\\log _{10} \\mathrm{~K}\\right) \\\\\\\\\n& \\frac{\\mathrm{E}_{\\text {Cell }}^{\\mathrm{o}}}{0.06} \\times \\mathrm{n}=\\log \\mathrm{K} .......(1) \\\\\\\\\n& \\mathrm{Pd}^{+2} \\text { (aq.) }+ \\mathrm{2e}^{-} \\rightleftharpoons \\mathrm{Pd}(\\mathrm{s}), \\mathrm{E}_{\\text {cat, }{ reduction}}^{\\mathrm{o}}=0.83 \\\\\\\\\n& \\mathrm{Pd}(\\mathrm{s})+4 \\mathrm{Cl}^{-} \\text {(aq.) } \\rightleftharpoons \\mathrm{PdCl}_4^{2-}, \\text { (aq) }+2 \\mathrm{e}^{-}, \\mathrm{E}_{\\text {Anode, Oxidation }}^0=0.65\n\\end{aligned}\n$$\n

    Net Reaction $\\rightarrow \\mathrm{Pd}^{2+}$ (aq.) $+4 \\mathrm{Cl}^{-}$(aq.) $\\rightleftharpoons \\mathrm{PdCl}_4^{2-}$ (aq.)\n

    $$\n\\begin{aligned}\n& \\mathrm{E}_{\\text {cell }}^0=\\mathrm{E}_{\\text {cat,red }}^{\\mathrm{o}}-\\mathrm{E}_{\\text {Anded, oxidid }}^0 \\\\\\\\\n& \\mathrm{E}_{\\text {cell }}^{\\mathrm{o}}=0.83-0.65 \\\\\\\\\n& \\mathrm{E}_{\\text {cell }}^0=0.18 .........(2)\n\\end{aligned}\n$$\n

    Also $\\mathrm{n}=2$ .......(3)\n

    Using equation (1), (2) and (3) \n

    $\\log K=6$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1469, "subject": "Chemistry", "question": "The electrode potential of the following half cell at $298 \\mathrm{~K}$\n

    \n$\\mathrm{X}\\left|\\mathrm{X}^{2+}(0.001 \\mathrm{M}) \\| \\mathrm{Y}^{2+}(0.01 \\mathrm{M})\\right| \\mathrm{Y}$ is _______ $\\times 10^{-2} \\mathrm{~V}$ (Nearest integer)\n

    \nGiven: $\\mathrm{E}^{0} _ {\\mathrm{X}^{2+} \\mid \\mathrm{X}}=-2.36 \\mathrm{~V}$\n

    \n$\\mathrm{E}_{\\mathrm{Y}^{2+} \\mid \\mathrm{Y}}^{0}=+0.36 \\mathrm{~V}$\n

    \n$\\frac{2.303 \\mathrm{RT}}{\\mathrm{F}}=0.06 \\mathrm{~V}$", "options": [], "answer": "275", "solution": "**Answer:** 275\n\n$$\n\\begin{aligned}\n& \\text { Cell reaction : } X+Y^{2+}(0.01 \\mathrm{M}) \\longrightarrow X^{2+}(0.001 \\mathrm{M})+Y \\\\\\\\\n& E_{\\text {cell }}^o=0.36-(-2.36)=2.72 \\mathrm{~V} \\\\\\\\\n& E_{\\text {cell }}=E_{\\text {cell }}^{\\circ}-\\frac{2.303 R T}{n F} \\log \\frac{\\left[X^{2+}\\right]}{\\left[Y^{2+}\\right]} \\\\\\\\\n& =2.72-\\frac{0.06}{2} \\log \\frac{(0.001)}{0.01} \\quad[n=2 \\text {, as two electrons are involved in the reaction] } \\\\\\\\\n& =2.72-(0.03) \\times(-1) \\\\\\\\\n& =2.72+0.03 \\\\\\\\\n& =2.75 \\mathrm{~V}=275 \\times 10^{-2} \\mathrm{~V} \\\\\\\\\n&\n\\end{aligned}\n$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1470, "subject": "Chemistry", "question": "

    Consider the cell

    \n

    $$\\mathrm{Pt}_{(\\mathrm{s})}\\left|\\mathrm{H}_{2}(\\mathrm{~g}, 1 \\mathrm{~atm})\\right| \\mathrm{H}^{+}(\\mathrm{aq}, 1 \\mathrm{M})|| \\mathrm{Fe}^{3+}(\\mathrm{aq}), \\mathrm{Fe}^{2+}(\\mathrm{aq}) \\mid \\operatorname{Pt}(\\mathrm{s})$$

    \n

    When the potential of the cell is $$0.712 \\mathrm{~V}$$ at $$298 \\mathrm{~K}$$, the ratio $$\\left[\\mathrm{Fe}^{2+}\\right] /\\left[\\mathrm{Fe}^{3+}\\right]$$ is _____________. (Nearest integer)

    \n

    Given : $$\\mathrm{Fe}^{3+}+\\mathrm{e}^{-}=\\mathrm{Fe}^{2+}, \\mathrm{E}^{\\theta} \\mathrm{Fe}^{3+}, \\mathrm{Fe}^{2+} \\mid \\mathrm{Pt}=0.771$$

    \n

    $$\n\\frac{2.303 \\mathrm{RT}}{\\mathrm{F}}=0.06 \\mathrm{~V}\n$$

    ", "options": [], "answer": "10", "solution": "**Answer:** 10\n\n

    Anode $${H_2} \\to 2{H^ + } + 2{e^ - }$$

    \n

    Cathode $$(F{e^{3 + }} + {e^ - } \\to F{e^{2 + }}) \\times 2$$

    \n

    $${H_2} + 2F{e^{3 + }} \\to 2{H^ + } + 2F{e^{2 + }}$$

    \n

    $${E_{cell}} = E_{cell}^o - {{0.059} \\over 2}\\log {\\left( {{{F{e^{2 + }}} \\over {F{e^{3 + }}}}} \\right)^2}$$

    \n

    $$0.712 = 0.771 - 0.059\\log {{F{e^{2 + }}} \\over {F{e^{3 + }}}}$$

    \n

    $$ - 0.059 = - 0.059\\log {{F{e^{2 + }}} \\over {F{e^{3 + }}}}$$

    \n

    $${{[F{e^{2 + }}]} \\over {[F{e^{3 + }}]}} = 10$$

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1471, "subject": "Chemistry", "question": "

    The equilibrium constant for the reaction

    \n

    $$\\mathrm{Zn(s)+Sn^{2+}(aq)}$$ $$\\rightleftharpoons$$ $$\\mathrm{Zn^{2+}(aq)+Sn(s)}$$ is $$1\\times10^{20}$$ at 298 K. The magnitude of standard electrode potential of $$\\mathrm{Sn/Sn^{2+}}$$ if $$\\mathrm{E_{Z{n^{2 + }}/Zn}^\\Theta = - 0.76~V}$$ is __________ $$\\times 10^{-2}$$ V. (Nearest integer)

    \n

    Given : $$\\mathrm{\\frac{2.303RT}{F}=0.059~V}$$

    ", "options": [], "answer": "17", "solution": "**Answer:** 17\n\n

    $$E_{cell}^o = {{2.303\\,RT} \\over {2F}}\\log k$$

    \n

    $$E_{cell}^o = {{0.059} \\over 2}\\log ({10^{20}})$$

    \n

    $$E_{Z{n^{2 + }}/Zn}^o + 0.76 = 0.59$$

    \n

    $$E_{Z{n^{2 + }}/Zn}^o = 0.59 - 0.76$$

    \n

    $$E_{Zn/Z{n^{2 + }}}^o = 0.17\\,V$$

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1472, "subject": "Chemistry", "question": "

    The standard electrode potential $$\\mathrm{(M^{3+}/M^{2+})}$$ for V, Cr, Mn & Co are $$-$$0.26 V, $$-$$0.41 V, + 1.57 V and + 1.97 V, respectively. The metal ions which can liberate $$\\mathrm{H_2}$$ from a dilute acid are :

    ", "options": [ { "text": "$$\\mathrm{Mn^{2+}}$$ and $$\\mathrm{Co^{2+}}$$" }, { "text": "$$\\mathrm{V^{2+}}$$ and $$\\mathrm{Mn^{2+}}$$" }, { "text": "$$\\mathrm{V^{2+}}$$ and $$\\mathrm{Cr^{2+}}$$" }, { "text": "$$\\mathrm{Cr^{2+}}$$ and $$\\mathrm{Co^{2+}}$$" } ], "answer": "$$\\mathrm{V^{2+}}$$ and $$\\mathrm{Cr^{2+}}$$", "solution": "**Answer:** $$\\mathrm{V^{2+}}$$ and $$\\mathrm{Cr^{2+}}$$\n\nMetal cation with $(-)$ value of reduction potential $\\left(\\mathrm{M}^{+3} / \\mathrm{M}^{+2}\\right)$ or with $(+)$ value of oxidation potential $\\left(\\mathrm{M}^{+2} / \\mathrm{M}^{+3}\\right)$ will liberate $\\mathrm{H}_{2}$

    Therefore they will reduce $\\mathrm{H}^{+}$ i. $\\mathrm{e~V}^{+2}$ and $\\mathrm{Cr}^{+2}$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1473, "subject": "Chemistry", "question": "

    $$Pt(s)|{H_2}(g)(1\\,bar)|{H^ + }(aq)(1\\,M)||{M^{3 + }}(aq),{M^ + }(aq)|Pt(s)$$

    \n

    The $$\\mathrm{E_{cell}}$$ for the given cell is 0.1115 V at 298 K when $${{\\left[ {{M^ + }(aq)} \\right]} \\over {\\left[ {{M^{3 + }}(aq)} \\right]}} = {10^a}$$

    \n

    The value of $$a$$ is ____________

    \n

    Given : $$\\mathrm{E_{{M^{3 + }}/{M^ + }}^\\theta = 0.2}$$ V

    \n

    $${{2.303RT} \\over F} = 0.059V$$

    ", "options": [], "answer": "3", "solution": "**Answer:** 3\n\nOverall reaction :-

    \n$$\n\\begin{aligned}\n& \\mathrm{H}_{2(\\mathrm{~g})}+\\mathrm{M}_{(\\mathrm{aq})}^{3+} \\longrightarrow \\mathrm{M}_{\\text {(aq) }}^{+}+2 \\mathrm{H}_{(\\mathrm{aq})}^{+} \\\\\\\\\n& \\mathrm{E}_{\\text {Cell }}=\\mathrm{E}_{\\text {Cathode }}^{\\mathrm{o}}-\\mathrm{E}_{\\text {anode }}^{\\mathrm{o}}-\\frac{0.059}{2} \\log \\frac{\\left[\\mathrm{M}^{+}\\right] \\times 1^2}{\\left[\\mathrm{M}^{+3}\\right] 1} \\\\\\\\\n& 0.1115=0.2-\\frac{0.059}{2} \\log \\frac{\\left[\\mathrm{M}^{+}\\right]}{\\left[\\mathrm{M}^{+3}\\right]} \\\\\\\\\n& 3=\\log \\frac{\\left[\\mathrm{M}^{+}\\right]}{\\left[\\mathrm{M}^{+3}\\right]} \\\\\\\\\n& \\therefore \\mathrm{a}=3\n\\end{aligned}\n$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1474, "subject": "Chemistry", "question": "

    Consider the cell

    \n

    $$\\mathrm{Pt(s)|{H_2}(g)\\,(1\\,atm)|{H^ + }\\,(aq,[{H^ + }] = 1)||F{e^{3 + }}(aq),F{e^{2 + }}(aq)|Pt(s)}$$

    \n

    Given $$\\mathrm{E_{F{e^{3 + }}/F{e^{2 + }}}^o = 0.771\\,V}$$ and $$\\mathrm{E_{{H^ + }/1/2\\,{H_2}}^o = 0\\,V,\\,T = 298\\,K}$$

    \n

    If the potential of the cell is 0.712 V, the ratio of concentration of Fe$$^{2+}$$ to Fe$$^{3+}$$ is _____________ (Nearest integer)

    ", "options": [], "answer": "10", "solution": "**Answer:** 10\n\n$$\n\\begin{aligned}\n& \\frac{1}{2} \\mathrm{H}_2(\\mathrm{~g})+\\mathrm{Fe}^{3+}(\\mathrm{aq} .) \\longrightarrow \\mathrm{H}^{+}(\\mathrm{aq})+\\mathrm{Fe}^{2+}(\\mathrm{aq} .) \\\\\\\\\n& \\mathrm{E}=\\mathrm{E}^{\\mathrm{o}}-\\frac{0.059}{1} \\log \\frac{\\left[\\mathrm{Fe}^{2+}\\right]}{\\left[\\mathrm{Fe}^{3+}\\right]} \\\\\\\\\n& \\Rightarrow \\quad 0.712=(0.771-0)-\\frac{0.059}{1} \\log \\frac{\\left[\\mathrm{Fe}^{2+}\\right]}{\\left[\\mathrm{Fe}^{3+}\\right]} \\\\\\\\\n& \\Rightarrow \\quad \\log \\frac{\\left[\\mathrm{Fe}^{2+}\\right]}{\\left[\\mathrm{Fe}^{3+}\\right]}=\\frac{(0.771-0712)}{0.059}=1 \\\\\\\\\n& \\Rightarrow \\quad \\frac{\\left[\\mathrm{Fe}^{2+}\\right]}{\\left[\\mathrm{Fe}^{3+}\\right]}=10\n\\end{aligned}\n$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1475, "subject": "Chemistry", "question": "

    At 298 K, a 1 litre solution containing 10 mmol of $$\\mathrm{C{r_2}O_7^{2 - }}$$ and 100 mmol of $$\\mathrm{Cr^{3+}}$$ shows a pH of 3.0.

    \n

    Given : $$\\mathrm{C{r_2}O_7^{2 - } \\to C{r^{3 + }}\\,;\\,E^\\circ = 1.330}$$V

    \n

    and $$\\mathrm{{{2.303\\,RT} \\over F} = 0.059}$$ V

    \n

    The potential for the half cell reaction is $$x\\times10^{-3}$$ V. The value of $$x$$ is __________

    ", "options": [], "answer": "917", "solution": "**Answer:** 917\n\n$$\n\\begin{aligned}\n& \\mathrm{Cr}_2 \\mathrm{O}_7^{2-}+14 \\mathrm{H}^{+}+6 \\mathrm{e}^{-} \\rightarrow 2 \\mathrm{Cr}^{3+}+7 \\mathrm{H}_2 \\mathrm{O} \\\\\\\\\n& \\mathrm{E}=1.33-\\frac{0.059}{6} \\log \\frac{(0.1)^2}{\\left(10^{-2}\\right)\\left(10^{-3}\\right)^{14}} \\\\\\\\\n& \\mathrm{E}=1.33-\\frac{0.059}{6} \\times 42=0.917 \\\\\\\\\n& \\mathrm{E}=917 \\times 10^{-3} \\\\\\\\\n& \\mathrm{x}=917\n\\end{aligned}\n$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1476, "subject": "Chemistry", "question": "

    At $$298 \\mathrm{~K}$$, the standard reduction potential for $$\\mathrm{Cu}^{2+} / \\mathrm{Cu}$$ electrode is $$0.34 \\mathrm{~V}$$.

    \n

    Given : $$\\mathrm{K}_{\\mathrm{sp}} \\mathrm{Cu}(\\mathrm{OH})_{2}=1 \\times 10^{-20}$$

    \n

    Take $$\\frac{2.303 \\mathrm{RT}}{\\mathrm{F}}=0.059 \\mathrm{~V}$$

    \n

    The reduction potential at $$\\mathrm{pH}=14$$ for the above couple is $$(-) x \\times 10^{-2} \\mathrm{~V}$$. The value of $$x$$ is ___________

    ", "options": [], "answer": "25", "solution": "**Answer:** 25\n\nGiven:\n

    \nStandard reduction potential for Cu²⁺/Cu, E° = 0.34 V

    \nKsp of Cu(OH)₂ = 1 × 10⁻²⁰

    \n2.303RT/F = 0.059 V

    \npH = 14\n

    \nFirst, we have the solubility equilibrium for Cu(OH)₂:\n

    \n$$\\mathrm{Cu}(\\mathrm{OH})_2(\\mathrm{~s}) \\rightleftharpoons \\mathrm{Cu}^{2+}(\\mathrm{aq})+2 \\mathrm{OH}^{-}(\\mathrm{aq})$$\n

    \nThe Ksp expression for this reaction is:\n

    \n$$\\mathrm{Ksp}=\\left[\\mathrm{Cu}^{2+}\\right]\\left[\\mathrm{OH}^{-}\\right]^2$$\n

    \nAt pH 14, the concentration of OH⁻ ions is 1 M:\n

    \n$$\\left[\\mathrm{OH}^{-}\\right] = 1 \\mathrm{M}$$\n

    \nNow we can find the concentration of Cu²⁺:\n

    \n$$\\left[\\mathrm{Cu}^{2+}\\right]=\\frac{\\mathrm{Ksp}}{\\left[\\mathrm{OH}^{-}\\right]^2}=\\frac{1 \\times 10^{-20}}{1^2}=10^{-20} \\mathrm{M}$$\n

    \nThe half-cell reaction for the reduction of Cu²⁺ is:\n

    \n$$\\mathrm{Cu}^{2+}(\\mathrm{aq})+2 \\mathrm{e}^{-} \\rightarrow \\mathrm{Cu}(\\mathrm{s})$$\n

    \nNow we can use the Nernst equation to calculate the reduction potential at pH 14:\n

    \n$$E = E° - \\frac{0.059}{n} \\log_{10} \\frac{1}{\\left[\\mathrm{Cu}^{2+}\\right]}$$\n

    \nHere, n = 2 (number of electrons transferred in the Cu²⁺/Cu couple).\n

    \n$$E = 0.34 - \\frac{0.059}{2} \\log_{10} \\frac{1}{10^{-20}}$$

    \n$$E = 0.34 - \\frac{0.059}{2} \\times 20$$

    \n$$E = 0.34 - 0.59$$

    \n$$E = -0.25 \\mathrm{~V}$$\n

    \nThus, the reduction potential at pH 14 for the Cu²⁺/Cu couple is -0.25 V. In terms of x × 10⁻² V:\n

    \n$$(-) x \\times 10^{-2} \\mathrm{~V} = -0.25 \\mathrm{~V}$$\n

    \nThe value of x is 25.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1477, "subject": "Chemistry", "question": "

    The number of correct statements from the following is __________

    \n

    A. $$\\mathrm{E_{\\text {cell }}}$$ is an intensive parameter

    \n

    B. A negative $$\\mathrm{E}^{\\ominus}$$ means that the redox couple is a stronger reducing agent than the $$\\mathrm{H}^{+} / \\mathrm{H}_{2}$$ couple.

    \n

    C. The amount of electricity required for oxidation or reduction depends on the stoichiometry of the electrode reaction.

    \n

    D. The amount of chemical reaction which occurs at any electrode during electrolysis by a current is proportional to the quantity of electricity passed through the electrolyte.

    ", "options": [], "answer": "4", "solution": "**Answer:** 4\n\n

    A. $$\\mathrm{E_{\\text {cell }}}$$ is an intensive parameter.

    \n

    This statement is correct. The standard cell potential ($$\\mathrm{E}^{\\ominus}_{\\text{cell}}$$) is an intensive parameter because it depends only on the nature of the reactants and the products, and not on the size or shape of the electrodes or the amount of material present.

    \n

    B. A negative $$\\mathrm{E}^{\\ominus}$$ means that the redox couple is a stronger reducing agent than the $$\\mathrm{H}^{+} / \\mathrm{H}_{2}$$ couple.

    \n

    This statement is also correct. A negative standard reduction potential ($$\\mathrm{E}^{\\ominus}$$) for a redox couple indicates that the couple is a stronger reducing agent than the standard hydrogen electrode ($$\\mathrm{H}^{+} / \\mathrm{H}_{2}$$), which has a standard reduction potential of 0 V.

    \n

    C. The amount of electricity required for oxidation or reduction depends on the stoichiometry of the electrode reaction.

    \n

    This statement is correct. The amount of electricity required for a particular oxidation or reduction reaction depends on the number of electrons involved in the reaction, which in turn depends on the stoichiometry of the electrode reaction.

    \n

    D. The amount of chemical reaction which occurs at any electrode during electrolysis by a current is proportional to the quantity of electricity passed through the electrolyte.

    \n

    This statement is also correct. Faraday's laws of electrolysis state that the amount of chemical reaction at an electrode during electrolysis is directly proportional to the amount of electricity passed through the electrolyte.

    \n

    Therefore, all of the statements A, B, C, and D are correct.

    \n", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 1478, "subject": "Chemistry", "question": "

    In an electrochemical reaction of lead, at standard temperature, if $$\\mathrm{E}^{0}\\left(\\mathrm{~Pb}^{2+} / \\mathrm{Pb}\\right)=\\mathrm{m}$$ Volt and $$\\mathrm{E}^{0}\\left(\\mathrm{~Pb}^{4+} / \\mathrm{Pb}\\right)=\\mathrm{n}$$ Volt, then the value of $$\\mathrm{E}^{0}\\left(\\mathrm{~Pb}^{2+} / \\mathrm{Pb}^{4+}\\right)$$ is given by $$\\mathrm{m-x n}$$. The value of $$\\mathrm{x}$$ is ___________. (Nearest integer)

    ", "options": [], "answer": "2", "solution": "**Answer:** 2\n\nIn this problem, we're considering three different half-cell reactions involving lead ions.\n\n

    1) The reduction of $$\\mathrm{Pb}^{2+}$$ ions to lead metal :\n

    $$\\mathrm{Pb}^{2+} + 2 \\mathrm{e}^{-} \\rightarrow \\mathrm{Pb}$$\n

    The Gibbs free energy change for this process can be written as $$\\Delta \\mathrm{G}_1^0=-2 \\mathrm{FE}_1^0$$, where $$\\mathrm{E}_1^0$$ is the standard cell potential, F is Faraday's constant, and the factor of 2 is because two electrons are involved in the reaction.\n\n

    2) The reduction of $$\\mathrm{Pb}^{4+}$$ ions to lead metal :\n

    $$\\mathrm{Pb}^{4+} + 4 \\mathrm{e}^{-} \\rightarrow \\mathrm{Pb}$$\n

    This reaction has a Gibbs free energy change of $$\\Delta \\mathrm{G}_2^0=-4 \\mathrm{FE}_2^0$$.\n\n

    3) The oxidation of $$\\mathrm{Pb}^{2+}$$ ions to $$\\mathrm{Pb}^{4+}$$ ions :\n

    $$\\mathrm{Pb}^{2+} \\rightarrow \\mathrm{Pb}^{4+} + 2 \\mathrm{e}^{-}$$\n

    The Gibbs free energy change for this process is $$\\Delta \\mathrm{G}_3^0=-2 \\mathrm{FE}_3^0$$.\n\n

    We can write the third reaction as the difference between the first two reactions (i.e., reaction 2 - reaction 1). This implies that the Gibbs free energy changes for these reactions should add up accordingly :\n

    $$\\Delta \\mathrm{G}_3^0=\\Delta \\mathrm{G}_2^0 - \\Delta \\mathrm{G}_1^0$$\n\n

    Substituting the expressions for the Gibbs free energy changes from the half-cell reactions into this equation, we get :\n

    $$-2 \\mathrm{FE}_3^0 = -4 \\mathrm{FE}_2^0 - (-2 \\mathrm{FE}_1^0) = 2F (2n - m)$$\n\n

    Solving this equation for $$E_3^0$$ gives :\n

    $$E_3^0 = m - 2n$$\n\n

    However, the problem statement tells us that $$E_3^0$$ can also be written as $$m - xn$$. Comparing these two expressions for $$E_3^0$$, we see that $$x$$ must be equal to 2.\n\n

    So, the value of $$x$$ is 2.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1479, "subject": "Chemistry", "question": "

    $$\\mathrm{FeO_4^{2 - }\\buildrel { + 2.2V} \\over\n \\longrightarrow F{e^{3 + }}\\buildrel { + 0.70V} \\over\n \\longrightarrow F{e^{2 + }}\\buildrel { - 0.45V} \\over\n \\longrightarrow F{e^0}}$$

    \n

    $$E_{FeO_4^{2 - }/F{e^{2 + }}}^\\theta $$ is $$x \\times {10^{ - 3}}$$ V. The value of $$x$$ is _________

    ", "options": [], "answer": "1825", "solution": "**Answer:** 1825\n\n$$\n\\begin{aligned}\n& \\mathrm{FeO}_4^{2-} \\stackrel{+2.2 \\mathrm{~V}}{\\longrightarrow} \\mathrm{Fe}^{3+} ~~~\\Delta \\mathrm{G}_1=-6.6 \\mathrm{~F} \\\\\\\\\n& \\mathrm{Fe}^{3+} \\stackrel{+0.70 \\mathrm{~V}}{\\longrightarrow} \\mathrm{Fe}^{2+} ~~~\\Delta \\mathrm{G}_2=-0.7 \\mathrm{~F}\n\\end{aligned}\n$$

    \nHence for

    \n$$\n\\begin{aligned}\n& \\mathrm{FeO}_4^{2-} \\longrightarrow \\mathrm{Fe}^{2+} \\Delta \\mathrm{G} =-7.3 \\mathrm{~F} \\\\\\\\\n& =-\\mathrm{nEF} \\\\\\\\\n& \\mathrm{E}_{\\mathrm{FeO}_4^{2-} / \\mathrm{Fe}^{+2}}^0=\\frac{-7.3 \\mathrm{~F}}{-4 \\mathrm{~F}}= 1.825, \\mathrm{n}=4 \\\\\\\\\n& =1825 \\times 10^{-3} \\mathrm{~V}\n\\end{aligned}\n$$

    \n$n=$ electron exchange of that half cell reaction.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1480, "subject": "Chemistry", "question": "

    The number of incorrect statements from the following is ___________.

    \n

    A. The electrical work that a reaction can perform at constant pressure and temperature is equal to the reaction Gibbs energy.

    \n

    B. $$\\mathrm{E_{cell}^{\\circ}}$$ cell is dependent on the pressure.

    \n

    C. $$\\frac{d E^{\\theta} \\text { cell }}{\\mathrm{dT}}=\\frac{\\Delta_{\\mathrm{r}} \\mathrm{S}^{\\theta}}{\\mathrm{nF}}$$

    \n

    D. A cell is operating reversibly if the cell potential is exactly balanced by an opposing source of potential difference.

    ", "options": [], "answer": "1", "solution": "**Answer:** 1\n\n

    Let's go through the statements one by one:

    \n

    A. The electrical work that a reaction can perform at constant pressure and temperature is equal to the reaction Gibbs energy.

    \n

    This statement is correct. The maximum non-expansion work that a system can perform at constant temperature and pressure is given by the change in Gibbs free energy. This is especially relevant for electrochemical reactions where this non-expansion work appears as electrical work.

    \n

    B. E°cell is dependent on the pressure.

    \n

    This statement is incorrect. The standard cell potential, E°cell, is not dependent on pressure. It is dependent on the nature of the reactants and products (their identities and stoichiometric ratios), as well as temperature, but not pressure. The E°cell is calculated using standard conditions, which includes a standard pressure of 1 bar or approximately 1 atm.

    \n

    C. $$\\frac{d E^{\\theta} \\text { cell }}{\\mathrm{dT}}=\\frac{\\Delta_{\\mathrm{r}} \\mathrm{S}^{\\theta}}{\\mathrm{nF}}$$

    \n

    This statement is correct. This equation is derived from the Gibbs-Helmholtz equation, which describes the temperature dependence of the change in Gibbs free energy. In an electrochemical cell, the change in Gibbs free energy can be related to the cell potential.

    \n

    D. A cell is operating reversibly if the cell potential is exactly balanced by an opposing source of potential difference.

    \n

    This statement is correct. In a reversible electrochemical cell, the cell potential is exactly balanced by an external, opposing potential. This ensures that the reaction proceeds at an infinitesimally slow rate, allowing the system to maintain equilibrium at all times.

    \n

    So, among these four statements, only one (Statement B) is incorrect.

    \n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1481, "subject": "Chemistry", "question": "

    The reaction

    \n

    $$\\frac{1}{2} \\mathrm{H}_{2}(\\mathrm{~g})+\\mathrm{AgCl}(\\mathrm{s}) \\rightleftharpoons \\mathrm{H}^{+}(\\mathrm{aq})+\\mathrm{Cl}^{-}(\\mathrm{aq})+\\mathrm{Ag}(\\mathrm{s})$$

    \n

    occurs in which of the given galvanic cell.

    ", "options": [ { "text": "$$\\mathrm{Ag}|\\mathrm{AgCl}(\\mathrm{s})| \\mathrm{KCl}\\left(\\mathrm{sol}^{\\mathrm{n}}\\right)\\left|\\mathrm{AgNO}_{3}\\right| \\mathrm{Ag}$$" }, { "text": "$$\\mathrm{Pt}\\left|\\mathrm{H}_{2}(\\mathrm{~g})\\right| \\mathrm{HCl}\\left(\\mathrm{sol}^{\\mathrm{n}}\\right)|\\mathrm{AgCl}(\\mathrm{s})| \\mathrm{Ag}$$" }, { "text": "$$\\mathrm{Pt}\\left|\\mathrm{H}_{2}(\\mathrm{~g})\\right| \\mathrm{HCl}\\left(\\mathrm{sol}^{\\mathrm{n}}\\right)\\left|\\mathrm{AgNO}_{3}\\left(\\mathrm{sol}^{\\mathrm{n}}\\right)\\right| \\mathrm{Ag}$$" }, { "text": "$$\\mathrm{Pt}\\left|\\mathrm{H}_{2}(\\mathrm{~g})\\right| \\mathrm{KCl}\\left(\\mathrm{sol}^{\\mathrm{n}}\\right)|\\mathrm{AgCl}(\\mathrm{s})| \\mathrm{Ag}$$" } ], "answer": "$$\\mathrm{Pt}\\left|\\mathrm{H}_{2}(\\mathrm{~g})\\right| \\mathrm{HCl}\\left(\\mathrm{sol}^{\\mathrm{n}}\\right)|\\mathrm{AgCl}(\\mathrm{s})| \\mathrm{Ag}$$", "solution": "**Answer:** $$\\mathrm{Pt}\\left|\\mathrm{H}_{2}(\\mathrm{~g})\\right| \\mathrm{HCl}\\left(\\mathrm{sol}^{\\mathrm{n}}\\right)|\\mathrm{AgCl}(\\mathrm{s})| \\mathrm{Ag}$$\n\n

    The provided reaction is:

    \n

    $$\\frac{1}{2} \\mathrm{H}_{2}(\\mathrm{~g})+\\mathrm{AgCl}(\\mathrm{s}) \\rightleftharpoons \\mathrm{H}^{+}(\\mathrm{aq})+\\mathrm{Cl}^{-}(\\mathrm{aq})+\\mathrm{Ag}(\\mathrm{s})$$

    \n

    This reaction involves the following half-reactions:

    \n
      \n
    1. Oxidation of hydrogen gas to H+ ions: $$\\frac{1}{2} \\mathrm{H}_{2}(\\mathrm{~g}) \\rightleftharpoons \\mathrm{H}^{+}(\\mathrm{aq}) + e^-$$

      2. \n
    2. Reduction of AgCl to Ag: $$\\mathrm{AgCl}(\\mathrm{s}) + e^- \\rightleftharpoons \\mathrm{Ag}(\\mathrm{s}) + \\mathrm{Cl}^{-}(\\mathrm{aq})$$
      3. \n
    \n

    Looking at the options provided:

    \n

    Option A: Doesn't involve H2 gas, so it can't be correct.

    \n

    Option B: This includes the necessary elements - H2, AgCl, and Ag.

    \n

    Option C: Doesn't involve AgCl, so it can't be correct.

    \n

    Option D: Also includes the necessary elements - H2, AgCl, and Ag.

    \n

    However, looking closely, we can see that Option B represents the galvanic cell for this reaction. The reaction requires the oxidation of H2 to H+, which occurs at the anode. The reaction also requires the reduction of AgCl to Ag and Cl-, which occurs at the cathode.

    \n

    In Option B, the anode (on the left) is where H2 is being oxidized to H+. The cathode (on the right) is where AgCl is reduced to Ag and Cl-. The salt bridge or ion exchange component is HCl, which allows for the flow of ions to balance charge in the cell.

    \n

    Therefore, the reaction occurs in the galvanic cell represented by Option B.

    \n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1482, "subject": "Chemistry", "question": "

    The standard electrode potential of $$\\mathrm{M}^{+} / \\mathrm{M}$$ in aqueous solution does not depend on

    ", "options": [ { "text": "Ionisation of a gaseous metal atom" }, { "text": " Sublimation of a solid metal" }, { "text": "Ionisation of a solid metal atom" }, { "text": "Hydration of a gaseous metal ion" } ], "answer": "Ionisation of a solid metal atom", "solution": "**Answer:** Ionisation of a solid metal atom\n\n

    The standard electrode potential (E°) of a metal ion M+/M in aqueous solution is calculated under standard conditions and relates to the tendency of the M+ ion to gain an electron to form the neutral metal atom, M.

    \n

    This is indeed a redox (reduction-oxidation) process, and the standard electrode potential is defined for the reduction half-reaction. E° values are based on the energies involved in the processes of ionization and hydration.

    \n

    However, the standard electrode potential does not depend on the state (solid, liquid, gas) of the atom being ionized. Ionization of a solid metal atom (Option C) would not directly impact the electrode potential as it is not inherently part of the process of reduction at the electrode that the E° values are measuring.

    \n

    The correct answer should be Option C : Ionisation of a solid metal atom.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1483, "subject": "Chemistry", "question": "

    The standard reduction potentials at $$298 \\mathrm{~K}$$ for the following half cells are given below:

    \n

    $$\\mathrm{NO}_{3}^{-}+4 \\mathrm{H}^{+}+3 \\mathrm{e}^{-} \\rightarrow \\mathrm{NO}(\\mathrm{g})+2 \\mathrm{H}_{2} \\mathrm{O} \\quad \\mathrm{E}^{\\theta}=0.97 \\mathrm{~V}$$

    \n

    $$\\mathrm{V}^{2+}(\\mathrm{aq})+2 \\mathrm{e}^{-} \\rightarrow \\mathrm{V} \\quad\\quad\\quad \\mathrm{E}^{\\theta}=-1.19 \\mathrm{~V}$$

    \n

    $$\\mathrm{Fe}^{3+}(\\mathrm{aq})+3 \\mathrm{e}^{-} \\rightarrow \\mathrm{Fe} \\quad\\quad\\quad \\mathrm{E}^{\\theta}=-0.04 \\mathrm{~V}$$

    \n

    $$\\mathrm{Ag}^{+}(\\mathrm{aq})+\\mathrm{e}^{-} \\rightarrow \\mathrm{Ag}(\\mathrm{s}) \\quad\\quad\\quad \\mathrm{E}^{\\theta}=0.80 \\mathrm{~V}$$

    \n

    $$\\mathrm{Au}^{3+}(\\mathrm{aq})+3 \\mathrm{e}^{-} \\rightarrow \\mathrm{Au}(\\mathrm{s}) \\quad\\quad\\quad \\mathrm{E}^{\\theta}=1.40 \\mathrm{~V}$$

    \n

    The number of metal(s) which will be oxidized by $$\\mathrm{NO}_{3}^{-}$$ in aqueous solution is __________.

    ", "options": [], "answer": "3", "solution": "**Answer:** 3\n\n

    To determine the metals that can be oxidized by $NO_3^-$, we need to find the metals that have a standard reduction potential (SRP) lower than 0.97 V (since $NO_3^-$ has an SRP of 0.97 V).

    \n

    From the given data, we can find the metals that satisfy this condition:

    \n
      \n
    1. $V^{2+} (\\text{aq}) + 2e^- \\rightarrow V$, $E^\\theta = -1.19 \\, \\text{V}$ (less than 0.97 V, hence can be oxidized)

      2. \n
    2. $Fe^{3+} (\\text{aq}) + 3e^- \\rightarrow Fe$, $E^\\theta = -0.04 \\, \\text{V}$ (less than 0.97 V, hence can be oxidized)

      3. \n
    3. $Ag^+ (\\text{aq}) + e^- \\rightarrow Ag (\\text{s})$, $E^\\theta = 0.80 \\, \\text{V}$ (less than 0.97 V, hence can be oxidized)

      4. \n
    4. $Au^{3+} (\\text{aq}) + 3e^- \\rightarrow Au (\\text{s})$, $E^\\theta = 1.40 \\, \\text{V}$ (greater than 0.97 V, hence cannot be oxidized)
      5. \n
    \n

    Therefore, $V$, $Fe$, and $Ag$ are the metals that can be oxidized by $NO_3^-$.

    The total number of such metals is 3.

    \n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1484, "subject": "Chemistry", "question": "Consider the following redox reaction :

    \n$$\n\\mathrm{MnO}_4^{-}+\\mathrm{H}^{+}+\\mathrm{H}_2 \\mathrm{C}_2 \\mathrm{O}_4 \\rightleftharpoons \\mathrm{Mn}^{2+}+\\mathrm{H}_2 \\mathrm{O}+\\mathrm{CO}_2\n$$\n

    \nThe standard reduction potentials are given as below $\\left(\\mathrm{E}_{\\text {red }}^0\\right)$ :\n

    $$\n\\begin{aligned}\n& \\mathrm{E}_{\\mathrm{MnO}_4^{-} / \\mathrm{Mn}^{2+}}^{\\circ}=+1.51 \\mathrm{~V} \\\\\\\\\n& \\mathrm{E}_{\\mathrm{CO}_2 / \\mathrm{H}_2 \\mathrm{C}_2 \\mathrm{O}_4}^{\\circ}=-0.49 \\mathrm{~V}\n\\end{aligned}\n$$\n

    \nIf the equilibrium constant of the above reaction is given as $\\mathrm{K}_{\\mathrm{eq}}=10^x$, then the value of $x=$ __________ (nearest integer)", "options": [], "answer": "338", "solution": "**Answer:** 338\n\nCell $\\mathrm{Rx}^{\\mathrm{n}} ; \\mathrm{MnO}_4^{-}+\\mathrm{H}_2 \\mathrm{C}_2 \\mathrm{O}_4 \\rightarrow \\mathrm{Mn}^{2+}+\\mathrm{CO}_2$\n

    $\\mathrm{E}_{\\text {cell }}^{\\circ}=\\mathrm{E}_{\\text {op }}^{\\circ}$ of anode $+\\mathrm{E}_{\\mathrm{RP}}^{\\circ}$ of cathode\n

    $$\n=0.49+1.51=2.00 \\mathrm{~V}\n$$\n\n

    At equilibrium\n

    $$\n\\mathrm{E}_{\\text {cell }}=0 \\text {, }\n$$\n

    $$\n\\mathrm{E}_{\\text {cell }}^{\\circ}=\\frac{0.0591}{\\mathrm{n}} \\log \\mathrm{K}\n$$\n

    $2=\\frac{0.0591}{10} \\log K$\n

    $\\log K=338$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1485, "subject": "Chemistry", "question": "The potential for the given half cell at $298 \\mathrm{~K}$ is (-) __________ $\\times 10^{-2} \\mathrm{~V}$

    \n$$\n\\begin{aligned}\n& 2 \\mathrm{H}_{(\\mathrm{aq})}^{+}+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{H}_2(\\mathrm{~g}) \\\\\\\\\n& {\\left[\\mathrm{H}^{+}\\right]=1 \\mathrm{M}, \\mathrm{P}_{\\mathrm{H}_2}=2 \\mathrm{~atm}}\n\\end{aligned}\n$$

    \n(Given : $2.303 \\mathrm{RT} / \\mathrm{F}=0.06 \\mathrm{~V}, \\log 2=0.3$ )", "options": [], "answer": "1", "solution": "**Answer:** 1\n\n

    To find the potential of the given half cell at 298 K, we can use the Nernst equation, which in this scenario is simplified due to the standard conditions for the hydrogen ion concentration ($[\\mathrm{H}^{+}] = 1 M$). The half-reaction involved is the reduction of hydrogen ions to hydrogen gas:

    \n

    $$2 \\mathrm{H}_{(\\mathrm{aq})}^+ + 2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{H}_2 (g)$$

    \n

    Given:

    \n\n

    The Nernst equation for this reaction under the given conditions becomes:

    \n

    $$\\mathrm{E} = \\mathrm{E}^{\\circ} - \\frac{0.06}{n} \\log \\frac{\\mathrm{P}_{\\mathrm{H}_2}}{[\\mathrm{H}^{+}]^2}$$

    \n

    Where $\\mathrm{E}^{\\circ} = 0.00 V$ is the standard electrode potential for the hydrogen half-cell, $n = 2$ is the number of electrons transferred, and the pressure of $\\mathrm{H}_2$ is given as 2 atm. Inserting the values, we get:

    \n

    $$\\mathrm{E}=0.00 - \\frac{0.06}{2} \\log \\frac{2}{1^{2}} = -\\frac{0.06}{2} \\times 0.3 = -0.03 \\times 0.3 = -0.009 = -0.9 \\times 10^{-2} \\text{V}$$

    \n

    Therefore, the potential of the given half-cell at 298 K is $-0.9 \\times 10^{-2} \\text{V}$.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1486, "subject": "Chemistry", "question": "

    The hydrogen electrode is dipped in a solution of $$\\mathrm{pH}=3$$ at $$25^{\\circ} \\mathrm{C}$$. The potential of the electrode will be _________ $$\\times 10^{-2} \\mathrm{~V}$$.

    \n

    $$\\left(\\frac{2.303 \\mathrm{RT}}{\\mathrm{F}}=0.059 \\mathrm{~V}\\right)$$

    ", "options": [], "answer": "18", "solution": "**Answer:** 18\n\n

    $$\\begin{aligned}\n& 2 \\mathrm{H}_{(\\mathrm{aq} .)}^{+}+2 \\mathrm{e}^{-} \\rightarrow \\mathrm{H}_2(\\mathrm{~g}) \\\\\n& \\mathrm{E}_{\\text {cell }}=\\mathrm{E}_{\\text {cell }}^0-\\frac{0.059}{2} \\log \\frac{\\mathrm{P}_{\\mathrm{H}_2}}{\\left[\\mathrm{H}^{+}\\right]^2} \\\\\n& =0-0.059 \\times 3=-0.177 \\text { volts. }=-17.7 \\times 10^{-2} \\mathrm{~V}\n\\end{aligned}$$

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1487, "subject": "Chemistry", "question": "

    Reduction potential of ions are given below:

    \n

    $$\\begin{array}{ccc}\n\\mathrm{ClO}_4^{-} & \\mathrm{IO}_4^{-} & \\mathrm{BrO}_4^{-} \\\\\n\\mathrm{E}^{\\circ}=1.19 \\mathrm{~V} & \\mathrm{E}^{\\circ}=1.65 \\mathrm{~V} & \\mathrm{E}^{\\circ}=1.74 \\mathrm{~V}\n\\end{array}$$

    \n

    The correct order of their oxidising power is :

    ", "options": [ { "text": "$$\\mathrm{IO}_4^{-}>\\mathrm{BrO}_4^{-}>\\mathrm{ClO}_4^{-}$$\n" }, { "text": "$$\\mathrm{BrO}_4^{-}>\\mathrm{ClO}_4^{-}>\\mathrm{IO}_4^{-}$$\n" }, { "text": "$$\\mathrm{ClO}_4^{-}>\\mathrm{IO}_4^{-}>\\mathrm{BrO}_4^{-}$$\n" }, { "text": "$$\\mathrm{BrO}_4^{-}>\\mathrm{IO}_4^{-}>\\mathrm{ClO}_4^{-}$$" } ], "answer": "$$\\mathrm{BrO}_4^{-}>\\mathrm{IO}_4^{-}>\\mathrm{ClO}_4^{-}$$", "solution": "**Answer:** $$\\mathrm{BrO}_4^{-}>\\mathrm{IO}_4^{-}>\\mathrm{ClO}_4^{-}$$\n\n

    Higher the value of $$\\oplus$$ve SRP (Std. reduction potential) more is tendency to undergo reduction, so better is oxidising power of reactant.

    \n

    Hence, ox. Power:- $$\\mathrm{BrO}_4^{-}>\\mathrm{IO}_4^{-}>\\mathrm{ClO}_4^{-}$$

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1488, "subject": "Chemistry", "question": "

    The standard reduction potentials at $$298 \\mathrm{~K}$$ for the following half cells are given below :

    \n

    $$\\mathrm{Cr}_2 \\mathrm{O}_7^{2-}+14 \\mathrm{H}^{+}+6 \\mathrm{e}^{-} \\rightarrow 2 \\mathrm{Cr}^{3+}+7 \\mathrm{H}_2 \\mathrm{O}, \\quad \\mathrm{E}^{\\circ}=1.33 \\mathrm{~V}$$

    \n

    $$\\begin{array}{ll}\n\\mathrm{Fe}^{3+}(\\mathrm{aq})+3 \\mathrm{e}^{-} \\rightarrow \\mathrm{Fe} & \\mathrm{E}^{\\circ}=-0.04 \\mathrm{~V} \\\\\n\\mathrm{Ni}^{2+}(\\mathrm{aq})+2 \\mathrm{e}^{-} \\rightarrow \\mathrm{Ni} & \\mathrm{E}^{\\circ}=-0.25 \\mathrm{~V} \\\\\n\\mathrm{Ag}^{+}(\\mathrm{aq})+\\mathrm{e}^{-} \\rightarrow \\mathrm{Ag} & \\mathrm{E}^{\\circ}=0.80 \\mathrm{~V} \\\\\n\\mathrm{Au}^{3+}(\\mathrm{aq})+3 \\mathrm{e}^{-} \\rightarrow \\mathrm{Au} & \\mathrm{E}^{\\circ}=1.40 \\mathrm{~V}\n\\end{array}$$

    \n

    Consider the given electrochemical reactions,

    \n

    The number of metal(s) which will be oxidized be $$\\mathrm{Cr}_2 \\mathrm{O}_7^{2-}$$, in aqueous solution is _________.

    ", "options": [], "answer": "3", "solution": "**Answer:** 3\n\n

    To determine the number of metals that will be oxidized by $$\\mathrm{Cr}_2\\mathrm{O}_7^{2-}$$ in aqueous solution, we need to compare their standard reduction potentials with that of the given reduction potential of $$\\mathrm{Cr}_2\\mathrm{O}_7^{2-}$$.

    \n\n

    The reduction reaction for $$\\mathrm{Cr}_2\\mathrm{O}_7^{2-}$$ is:

    \n\n

    $$\\mathrm{Cr}_2\\mathrm{O}_7^{2-} + 14\\mathrm{H}^+ + 6\\mathrm{e}^- \\rightarrow 2\\mathrm{Cr}^{3+}+7\\mathrm{H}_2\\mathrm{O}, \\quad \\mathrm{E}^{\\circ}=1.33 \\mathrm{~V}$$

    \n\n

    This means that $$\\mathrm{Cr}_2\\mathrm{O}_7^{2-}$$ has a strong tendency to get reduced (due to its high reduction potential of 1.33 V). Thus, any metal with a lower standard reduction potential than 1.33 V has the potential to be oxidized by $$\\mathrm{Cr}_2\\mathrm{O}_7^{2-}$$.

    \n\n

    Given the standard reduction potentials:

    \n\n

    $$\\begin{array}{ll} \\mathrm{Fe}^{3+}(\\mathrm{aq})+3 \\mathrm{e}^{-} \\rightarrow \\mathrm{Fe} & \\mathrm{E}^{\\circ}=-0.04 \\mathrm{~V} \\\\ \\mathrm{Ni}^{2+}(\\mathrm{aq})+2 \\mathrm{e}^{-} \\rightarrow \\mathrm{Ni} & \\mathrm{E}^{\\circ}=-0.25 \\mathrm{~V} \\\\ \\mathrm{Ag}^{+}(\\mathrm{aq})+\\mathrm{e}^{-} \\rightarrow \\mathrm{Ag} & \\mathrm{E}^{\\circ}=0.80 \\mathrm{~V} \\\\ \\mathrm{Au}^{3+}(\\mathrm{aq})+3 \\mathrm{e}^{-} \\rightarrow \\mathrm{Au} & \\mathrm{E}^{\\circ}=1.40 \\mathrm{~V} \\end{array}$$

    \n\n

    Now we compare these reduction potentials with that of $$\\mathrm{Cr}_2\\mathrm{O}_7^{2-}$$:

    \n\n\n\n

    From the comparison, we see that $$\\mathrm{Fe}$$, $$\\mathrm{Ni}$$, and $$\\mathrm{Ag}$$ have standard reduction potentials lower than 1.33 V, meaning these metals can be oxidized by $$\\mathrm{Cr}_2\\mathrm{O}_7^{2-}$$, but $$\\mathrm{Au}$$ cannot.

    \n\n

    Therefore, the number of metals that will be oxidized by $$\\mathrm{Cr}_2\\mathrm{O}_7^{2-}$$ in aqueous solution is: 3.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1489, "subject": "Chemistry", "question": "

    One of the commonly used electrode is calomel electrode. Under which of the following categories, calomel electrode comes?

    ", "options": [ { "text": "Metal ion - Metal electrodes\n" }, { "text": "Oxidation - Reduction electrodes\n" }, { "text": "Metal - Insoluble Salt - Anion electrodes\n" }, { "text": "Gas - Ion electrodes" } ], "answer": "Metal - Insoluble Salt - Anion electrodes\n", "solution": "**Answer:** Metal - Insoluble Salt - Anion electrodes\n\n\n

    Calomel electrode is metal-insoluble salt – Anion \nelectrode.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1490, "subject": "Chemistry", "question": "

    What pressure (bar) of $$\\mathrm{H}_2$$ would be required to make emf of hydrogen electrode zero in pure water at $$25^{\\circ} \\mathrm{C}$$ ?

    ", "options": [ { "text": "0.5" }, { "text": "$$10^{-14}$$" }, { "text": "1" }, { "text": "$$10^{-7}$$" } ], "answer": "$$10^{-14}$$", "solution": "**Answer:** $$10^{-14}$$\n\n

    The electromotive force (emf) of a hydrogen electrode can be derived from the Nernst equation, which for the hydrogen half-cell reaction $$\\mathrm{H_2(g)} \\rightarrow 2\\mathrm{H^+(aq)} + 2\\mathrm{e^-}$$ is given by:

    \n\n

    $$ E = E^\\circ + \\frac{RT}{2F} \\ln \\left( \\frac{[\\mathrm{H^+}]^2}{P_{\\mathrm{H_2}}} \\right) $$

    \n\n

    where:

    \n\n\n\n

    In pure water at $$25^{\\circ} \\mathrm{C}$$, $$[\\mathrm{H^+}] = 10^{-7} \\, \\mathrm{M}$$. To make the emf zero, we set E to 0 in the Nernst equation:

    \n\n

    $$ 0 = 0 + \\frac{8.314 \\times 298}{2 \\times 96485} \\ln \\left( \\frac{(10^{-7})^2}{P_{\\mathrm{H_2}}} \\right) $$

    \n\n

    Simplifying the constants:

    \n\n

    $$ \\frac{8.314 \\times 298}{2 \\times 96485} \\approx 0.0128 $$

    \n\n

    Thus, the equation becomes:

    \n\n

    $$ 0 = 0.0128 \\ln \\left( \\frac{(10^{-7})^2}{P_{\\mathrm{H_2}}} \\right) $$

    \n\n

    Since ln term must be zero for this equation to hold true (as 0 divided by any number is still 0), the expression inside the logarithm must equal 1:

    \n\n

    $$ \\frac{(10^{-7})^2}{P_{\\mathrm{H_2}}} = 1 $$

    \n\n

    Simplifying this gives:

    \n\n

    $$ (10^{-7})^2 = P_{\\mathrm{H_2}} $$

    \n\n

    $$ 10^{-14} = P_{\\mathrm{H_2}} $$

    \n\n

    Therefore, the required pressure of $$\\mathrm{H_2}$$ to make the emf of the hydrogen electrode zero in pure water at $$25^{\\circ} \\mathrm{C}$$ is $$10^{-14}$$ bar.

    \n\n

    The correct answer is Option B: $$10^{-14}$$.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1491, "subject": "Chemistry", "question": "

    The emf of cell $$\\mathrm{Tl}\\left|\\underset{(0.001 \\mathrm{M})}{\\mathrm{Tl}^{+}}\\right| \\underset{(0.01 \\mathrm{M})}{\\mathrm{Cu}^{2+}} \\mid \\mathrm{Cu}$$ is $$0.83 \\mathrm{~V}$$ at $$298 \\mathrm{~K}$$. It could be increased by :

    ", "options": [ { "text": "increasing concentration of $$\\mathrm{Tl}^{+}$$ ions\n" }, { "text": "increasing concentration of $$\\mathrm{Cu}^{2+}$$ ions\n" }, { "text": "increasing concentration of both $$\\mathrm{Tl}^{+}$$ and $$\\mathrm{Cu}^{2+}$$ ions\n" }, { "text": "decreasing concentration of both $$\\mathrm{Tl}^{+}$$ and $$\\mathrm{Cu}^{2+}$$ ions" } ], "answer": "increasing concentration of $$\\mathrm{Cu}^{2+}$$ ions\n", "solution": "**Answer:** increasing concentration of $$\\mathrm{Cu}^{2+}$$ ions\n\n\n

    To determine how the emf of the cell, $$\\mathrm{Tl}\\left|\\underset{(0.001 \\mathrm{M})}{\\mathrm{Tl}^{+}}\\right| \\underset{(0.01 \\mathrm{M})}{\\mathrm{Cu}^{2+}} \\mid \\mathrm{Cu}$$, can be increased, we can use the Nernst equation. The Nernst equation for this electrochemical cell is given by:

    \n\n

    $$ E_{\\text{cell}} = E_{\\text{cell}}^\\circ - \\frac{0.0591}{n} \\log \\left( \\frac{[\\mathrm{Tl}^{+}]}{[\\mathrm{Cu}^{2+}]} \\right) $$

    \n\n

    where:

    \n\n\n\n

    Given that the emf of the cell can be expressed in terms of the concentration of the ions involved, we can see that increasing the concentration of $$[\\mathrm{Cu}^{2+}]$$ or decreasing the concentration of $$[\\mathrm{Tl}^{+}]$$ will affect the logarithmic term in the Nernst equation:

    \n\n

    $$ E_{\\text{cell}} = E_{\\text{cell}}^\\circ - \\frac{0.0591}{2} \\log \\left( \\frac{0.001}{0.01} \\right) $$

    \n\n

    To increase the emf ($$ E_{\\text{cell}} $$) of the cell, it is beneficial to have a less negative (or more positive) correction term. This can be achieved by:

    \n\n\n\n

    Therefore, the correct answer is:

    \n\n

    Option B: increasing concentration of $$\\mathrm{Cu}^{2+}$$ ions

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1492, "subject": "Chemistry", "question": "

    The reaction;

    \n

    $$\\frac{1}{2} \\mathrm{H}_{2(\\mathrm{~g})}+\\mathrm{AgCl}_{(\\mathrm{s})} \\rightarrow \\mathrm{H}_{(\\mathrm{aq})}^{+}+\\mathrm{Cl}_{(\\mathrm{aq})}^{-}+\\mathrm{Ag}_{(\\mathrm{s})}$$

    \n

    occurs in which of the following galvanic cell :

    ", "options": [ { "text": "$$\\mathrm{Pt}\\left|\\mathrm{H}_{2(\\mathrm{~g})}\\right| \\mathrm{HCl}_{(\\text {soln.) }}\\left|\\mathrm{AgNO}_{3(\\mathrm{aq})}\\right| \\mathrm{Ag}$$\n" }, { "text": "$$\\mathrm{Ag}\\left|\\mathrm{AgCl}_{(\\mathrm{s})}\\right| \\mathrm{KCl}_{\\text {(soln.) }}\\left|\\mathrm{AgNO}_{3 \\text { (aq.) }}\\right| \\mathrm{Ag}$$\n" }, { "text": "$$\\mathrm{Pt}\\left|\\mathrm{H}_{2(\\mathrm{~g})}\\right| \\mathrm{KCl}_{(\\text {soln.) }}\\left|\\mathrm{AgCl}_{(\\mathrm{s})}\\right| \\mathrm{Ag}$$\n" }, { "text": "$$\\mathrm{Pt}\\left|\\mathrm{H}_{2(\\mathrm{~g})}\\right| \\mathrm{HCl}_{(\\text {soln. })}\\left|\\mathrm{AgCl}_{(\\mathrm{s})}\\right| \\mathrm{Ag}$$" } ], "answer": "$$\\mathrm{Pt}\\left|\\mathrm{H}_{2(\\mathrm{~g})}\\right| \\mathrm{HCl}_{(\\text {soln. })}\\left|\\mathrm{AgCl}_{(\\mathrm{s})}\\right| \\mathrm{Ag}$$", "solution": "**Answer:** $$\\mathrm{Pt}\\left|\\mathrm{H}_{2(\\mathrm{~g})}\\right| \\mathrm{HCl}_{(\\text {soln. })}\\left|\\mathrm{AgCl}_{(\\mathrm{s})}\\right| \\mathrm{Ag}$$\n\n

    To determine which galvanic cell corresponds to the given reaction:

    \n\n

    $$\\frac{1}{2} \\mathrm{H}_{2(\\mathrm{~g})}+\\mathrm{AgCl}_{(\\mathrm{s})} \\rightarrow \\mathrm{H}_{(\\mathrm{aq})}^{+}+\\mathrm{Cl}_{(\\mathrm{aq})}^{-}+\\mathrm{Ag}_{(\\mathrm{s})}$$

    \n\n

    we need to examine the setups of the given cells and check if they produce the same reactants and products as in the above reaction.

    \n\n

    Analyzing the reaction, we observe:

    \n\n\n

    Let's examine each option:

    \n\n

    Option A:

    \n

    $$\\mathrm{Pt}\\left|\\mathrm{H}_{2(\\mathrm{~g})}\\right| \\mathrm{HCl}_{(\\text {soln.) }}\\left|\\mathrm{AgNO}_{3(\\mathrm{aq})}\\right| \\mathrm{Ag}$$

    \n

    This cell setup does not directly involve $$\\mathrm{AgCl}_{(\\mathrm{s})}$$ solid separating into $$\\mathrm{Ag}_{(\\mathrm{s})}$$ and $$\\mathrm{Cl}_{(\\mathrm{aq})}^{-}$$. Hence, it does not match the reaction given.

    \n\n

    Option B:

    \n

    $$\\mathrm{Ag}\\left|\\mathrm{AgCl}_{(\\mathrm{s})}\\right| \\mathrm{KCl}_{(\\text {(soln.) }}\\left|\\mathrm{AgNO}_{3 \\text { (aq.) }}\\right| \\mathrm{Ag}$$

    \n

    This involves silver electrodes and does not have a setup for reaction with hydrogen gas or the splitting of $$\\mathrm{H}_{2(\\mathrm{~g})}$$ into $$\\mathrm{H}_{(\\mathrm{aq})}^{+}$$ ions. So, this does not match either.

    \n\n

    Option C:

    \n

    $$\\mathrm{Pt}\\left|\\mathrm{H}_{2(\\mathrm{~g})}\\right| \\mathrm{KCl}_{(\\text {soln.) }}\\left|\\mathrm{AgCl}_{(\\mathrm{s})}\\right| \\mathrm{Ag}$$

    \n

    While this configuration includes $$\\mathrm{AgCl}_{(\\mathrm{s})}$$ and $$\\mathrm{H}_{2(\\mathrm{~g})}$$, it uses potassium chloride solution instead of hydrochloric acid solution, which alters the chemistry and does not perfectly match the required reaction mechanism.

    \n\n

    Option D:

    \n

    $$\\mathrm{Pt}\\left|\\mathrm{H}_{2(\\mathrm{~g})}\\right| \\mathrm{HCl}_{(\\text {soln. })}\\left|\\mathrm{AgCl}_{(\\mathrm{s})}\\right| \\mathrm{Ag}$$

    \n

    This configuration fits the reaction given. Here hydrogen gas at the platinum electrode splits into $$\\mathrm{H}_{(\\mathrm{aq})}^{+}$$ ions in the hydrochloric acid solution. On the other side, the solid silver chloride separates into $$\\mathrm{Ag}_{(\\mathrm{s})}$$ and $$\\mathrm{Cl}_{(\\mathrm{aq})}^{-}$$ ions. Hence, this accurately matches the reaction.

    \n\n

    Therefore, the correct answer is:

    \n

    Option D

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1493, "subject": "Chemistry", "question": "

    For the electro chemical cell

    \n

    $$\\mathrm{M}\\left|\\mathrm{M}^{2+}\\right||\\mathrm{X}| \\mathrm{X}^{2-}$$

    \n

    If $$\\mathrm{E}_{\\left(\\mathrm{M}^{2+} / \\mathrm{M}\\right)}^0=0.46 \\mathrm{~V}$$ and $$\\mathrm{E}_{\\left(\\mathrm{x} / \\mathrm{x}^{2-}\\right)}^0=0.34 \\mathrm{~V}$$.

    \n

    Which of the following is correct?

    ", "options": [ { "text": "$$\\mathrm{E}_{\\mathrm{cell}}=0.80 \\mathrm{~V}$$\n" }, { "text": "$$\\mathrm{M}+\\mathrm{X} \\rightarrow \\mathrm{M}^{2+}+\\mathrm{X}^{2-}$$ is a spontaneous reaction\n" }, { "text": "$$\\mathrm{E}_{\\text {cell }}=-0.80 \\mathrm{~V}$$\n" }, { "text": "$$\\mathrm{M}^{2+}+\\mathrm{X}^{2-} \\rightarrow \\mathrm{M}+\\mathrm{X}$$ is a spontaneous reaction" } ], "answer": "$$\\mathrm{M}^{2+}+\\mathrm{X}^{2-} \\rightarrow \\mathrm{M}+\\mathrm{X}$$ is a spontaneous reaction", "solution": "**Answer:** $$\\mathrm{M}^{2+}+\\mathrm{X}^{2-} \\rightarrow \\mathrm{M}+\\mathrm{X}$$ is a spontaneous reaction\n\n

    $$\\begin{aligned}\n& \\mathrm{E}_{\\mathrm{cell}}^{\\circ}=\\mathrm{E}_{\\mathrm{X} \\mid \\mathrm{X}^{2-}}^{\\circ}-\\mathrm{E}_{\\mathrm{M}^{2+} \\mid \\mathrm{M}}^{\\circ} \\\\\n& =0.34-0.46 \\\\\n& =-0.12 \\mathrm{~V} \\text { (Non-spontaneous) }\n\\end{aligned}$$

    \n

    So, reverse reaction will be spontaneous.

    \n

    $$\\mathrm{M}^{2+}+\\mathrm{X}^{2-} \\rightarrow \\mathrm{M}+\\mathrm{X}$$

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1494, "subject": "Chemistry", "question": "

    How can an electrochemical cell be converted into an electrolytic cell ?

    ", "options": [ { "text": "Applying an external opposite potential greater than $$\\mathrm{E}_{\\text {cell }}^0$$\n" }, { "text": "Exchanging the electrodes at anode and cathode.\n" }, { "text": "Applying an external opposite potential lower than $$\\mathrm{E}^0{ }_{\\text {cell }}$$\n" }, { "text": "Reversing the flow of ions in salt bridge." } ], "answer": "Applying an external opposite potential greater than $$\\mathrm{E}_{\\text {cell }}^0$$\n", "solution": "**Answer:** Applying an external opposite potential greater than $$\\mathrm{E}_{\\text {cell }}^0$$\n\n\n

    An electrochemical cell can be converted into an electrolytic cell by applying an external potential greater than $$E_{\\text {cell }}^{\\circ}$$. The entire cell reaction will be reversed. The oxidised species will be reduced and the reduced species will be oxidised.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1495, "subject": "Chemistry", "question": "The smog is essentially caused by the presence of :", "options": [ { "text": "O2 and O3c" }, { "text": "O3 and N2" }, { "text": "Oxides of sulphur and nitrogen" }, { "text": "O2 and N2" } ], "answer": "Oxides of sulphur and nitrogen", "solution": "**Answer:** Oxides of sulphur and nitrogen\n\nPhotochemical smog is caused by oxides of sulphur and nitrogen. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1496, "subject": "Chemistry", "question": "Identify the pollutant gases largely responsible for the discoloured and\nlustreless nature of marble of the Taj Mahal. \n ", "options": [ { "text": "O3 and CO2" }, { "text": "CO2 and NO2 " }, { "text": "SO2 and NO2" }, { "text": "SO2 and O3" } ], "answer": "SO2 and NO2", "solution": "**Answer:** SO2 and NO2\n\nSO2 and NO2 are responsible for the discoloured and lustreless nature of marble.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1497, "subject": "Chemistry", "question": "The higher concentration of which gas in air can cause stiffness of flower buds?", "options": [ { "text": "CO2" }, { "text": "SO2" }, { "text": "NO2" }, { "text": "CO" } ], "answer": "SO2", "solution": "**Answer:** SO2\n\nDue to acid rain in plants high concentration of SO2 makes the flower buds stiff and makes them\nfall.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1498, "subject": "Chemistry", "question": "Taj Mahal is being slowly disfigured and discoloured. This is primarily due to : \n", "options": [ { "text": "acid rain" }, { "text": "soil pollution" }, { "text": "global warming" }, { "text": "water pollution " } ], "answer": "acid rain", "solution": "**Answer:** acid rain\n\nTaj mahal is slowely disfigured and discoloured due to acid rain.\n

    The primary acid component of acid rain is sulfuric acid(H2SO4) which reacts with Marble(CaCO3) and gives aqueous ion. This aqueous ions get washed away in the water flow and marble get disfigured and discoloured.\n

    CaCO3(s) + H2SO4 $$ \\to $$ \n

    Ca2+(aq) + SO42-(aq)\n+ H2O + CO2", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1499, "subject": "Chemistry", "question": "The molecule that has minimum /no role in the formation of photochemical smog, is :", "options": [ { "text": "CH2 = O" }, { "text": "O3" }, { "text": "N2" }, { "text": "NO " } ], "answer": "N2", "solution": "**Answer:** N2\n\nPhotochemical smog is produced when ultraviolet light from sun reacts with NO\n, O3\n and hydrocarbon in atmosphere.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1500, "subject": "Chemistry", "question": "Excessive release of CO2 into the atomosphere\nresults in :", "options": [ { "text": "depletion of ozone" }, { "text": "polar vortex" }, { "text": "global warming" }, { "text": "formation of smog" } ], "answer": "global warming", "solution": "**Answer:** global warming\n\nCO2 causes global warming.", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1501, "subject": "Chemistry", "question": "Air pollution that occurs in sunlight is :", "options": [ { "text": "oxidising smog" }, { "text": "reducing smog" }, { "text": "fog" }, { "text": "acid rain" } ], "answer": "oxidising smog", "solution": "**Answer:** oxidising smog\n\nAir pollution caused by sunlight is\nphotochemical smog and it is oxidising.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1502, "subject": "Chemistry", "question": "The correct set of species responsible for the photochemical smog is :", "options": [ { "text": "NO, NO2, O3 and hydrocarbons" }, { "text": "N2, NO2 and hydrocarbons" }, { "text": "CO2, NO2, SO2 and hydrocarbons" }, { "text": "N2, O2, O3 and hydrocarbons" } ], "answer": "NO, NO2, O3 and hydrocarbons", "solution": "**Answer:** NO, NO2, O3 and hydrocarbons\n\nPhotochemical smog contains oxides of\nnitrogen, ozone and hydrocarbons.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1503, "subject": "Chemistry", "question": "The primary pollutant that leads to photochemical smog is :", "options": [ { "text": "acrolein" }, { "text": "ozone" }, { "text": "sulphur dioxide" }, { "text": "nitrogen oxides " } ], "answer": "nitrogen oxides ", "solution": "**Answer:** nitrogen oxides \n\nPhotochemical smog contains oxides of nitrogen.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1504, "subject": "Chemistry", "question": "Peroxyacetyl nitrate (PAN), an eye irritant is produced by : ", "options": [ { "text": "classical smog " }, { "text": "acid rain " }, { "text": "photochemical smog " }, { "text": "organic waste " } ], "answer": "photochemical smog ", "solution": "**Answer:** photochemical smog \n\nPhotochemical smog produce chemicals such as formaldehyde, acrolein and peroxyacetyl nitrate(PAN).", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1505, "subject": "Chemistry", "question": "Thermal power plants can lead to :", "options": [ { "text": "Ozone layer depletion" }, { "text": "Blue baby syndrome" }, { "text": "Eutrophication" }, { "text": "Acid rain" } ], "answer": "Acid rain", "solution": "**Answer:** Acid rain\n\nThermal power plants lead to acid rain.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1506, "subject": "Chemistry", "question": "The incorrect statement(s) among (a) – (d)\nregarding acid rain is (are) :\n

    (a) It can corrode water pipes.\n
    (b) It can damage structures made up of stone.\n
    (c) It cannot cause respiratory ailments in animals.\n
    (d) It is not harmful for trees", "options": [ { "text": "(c) and (d)" }, { "text": "(a), (b) and (d)" }, { "text": "(c) only" }, { "text": "(a), (c) and (d)" } ], "answer": "(c) and (d)", "solution": "**Answer:** (c) and (d)\n\n(1) Acid rain corrodes water pipes resulting in the leaching of heavy of heavy metals such as iron, lead and\ncopper into the drinking water.\n

    (2) Acid rain damages buildings and other structures made of stone or metal.\n

    (3) It causes respiratory aliments in human beings and animals.\n

    (4) It is harmful for agriculture, trees and plants as it wasshes down the nutrients needed for its growth.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1507, "subject": "Chemistry", "question": "The statement that is not true about ozone is :", "options": [ { "text": "In the stratosphere, CFCs release chlorine\nfree radicals (Cl) which reacts with O3 to\ngive chlorine dioxide radicals" }, { "text": "It is a toxic gas and its reaction with NO\ngives NO2." }, { "text": "In the atmosphere, it is depleted by CFCs" }, { "text": "In the stratosphere, it forms a protective\nshield against UV radiation" } ], "answer": "In the stratosphere, CFCs release chlorine\nfree radicals (Cl) which reacts with O3 to\ngive chlorine dioxide radicals", "solution": "**Answer:** In the stratosphere, CFCs release chlorine\nfree radicals (Cl) which reacts with O3 to\ngive chlorine dioxide radicals\n\nIn the stratosphere, CFCs release chlorine free radical ($$\\mathop {Cl}\\limits^ \\bullet $$).\n

    CF2Cl2(g) $$\\buildrel {UV} \\over\n \\longrightarrow $$ $$\\mathop {Cl}\\limits^ \\bullet $$(g) + $$\\mathop {Cl}\\limits^ \\bullet $$F2Cl(g)\n

    Which react with O3 to give chlorine oxide (Cl$$\\mathop O\\limits^ \\bullet $$) radical not chlorine dioxide (Cl$$\\mathop O\\limits^ \\bullet $$2) radical.\n

    $$\\mathop {Cl}\\limits^ \\bullet $$(g) + O3(g) $$ \\to $$ Cl$$\\mathop O\\limits^ \\bullet $$(g) + O2(g)", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1508, "subject": "Chemistry", "question": "Given below are two statements :

    Statement I : An allotrope of oxygen is an important intermediate in the formation of reducing smog.

    Statement II : Gases such as oxides of nitrogen and sulphur present in troposphere contribute to the formation of photochemical smog.

    In the light of the above statements, choose the correct answer from the options given below :", "options": [ { "text": "Statement I is false but Statement II is true" }, { "text": "Both Statement I and Statement II are false" }, { "text": "Both Statement I and Statement II are true" }, { "text": "Statement I is true but Statement II is false" } ], "answer": "Both Statement I and Statement II are false", "solution": "**Answer:** Both Statement I and Statement II are false\n\nReducing smog as is acts as reducing agent, the reducing character is due to presence of\nsulphur dioxide and carbon particles.", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1509, "subject": "Chemistry", "question": "The presence of ozone in troposphere :", "options": [ { "text": "protects us from greenhouse effect " }, { "text": "protects us from the UV radiation" }, { "text": "protects us from the X-ray radiation" }, { "text": "generates photochemical smog" } ], "answer": "generates photochemical smog", "solution": "**Answer:** generates photochemical smog\n\n

    Ozone is a strong oxidising agent like NO2. In troposphere both O3 and NO2 react with the unburnt hydrocarbons to produce the chemical components of photochemical smog, like formaldehyde (HCHO), acrolein (CH2 = CH - CHO) and PAN or peroxyacetyl nitrate.

    \n

    Hence, in troposphere, the presence of ozone generates photochemical smog.

    ", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1510, "subject": "Chemistry", "question": "The type of pollution that gets increased during the day time and in the presence of O3 is :", "options": [ { "text": "Oxidising smog" }, { "text": "Global warming" }, { "text": "Acid rain" }, { "text": "Reducing smog" } ], "answer": "Oxidising smog", "solution": "**Answer:** Oxidising smog\n\nPhotochemical smog occurs in warm, dry, and\nsunny climates. It is also called oxidizing smog. The\nmain components of photochemical smog are\nozone, nitric oxide, acrolein, formaldehyde, and PAN.", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 1511, "subject": "Chemistry", "question": "Reducing smog is a mixture of :", "options": [ { "text": "Smoke, fog and O3" }, { "text": "Smoke, fog and N2O3" }, { "text": "Smoke, fog and CH2 = CH $$-$$ CHO" }, { "text": "Smoke, fog and SO2" } ], "answer": "Smoke, fog and SO2", "solution": "**Answer:** Smoke, fog and SO2\n\nClassical smog occurs in a cool humid climate, it is a mixture of smoke, fog, and SO2. It is known as reducing smog. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1512, "subject": "Chemistry", "question": "Given below are two statements :

    Statement I : Chlorofluoro breakdown by radiation in the visible energy region and release chlorine gas in the atmosphere which then reacts with stratospheric ozone.

    Statement II : Atmospheric ozone reacts with nitric oxide to give nitrogen and oxygen gases, which add to the atmosphere.

    For the above statements choose the correct answer from the options given below :", "options": [ { "text": "Statement I is incorrect but Statement II is true" }, { "text": "Both Statement I and II are false" }, { "text": "Statement I is correct but Statement II is false" }, { "text": "Both Statement I and II are correct" } ], "answer": "Both Statement I and II are false", "solution": "**Answer:** Both Statement I and II are false\n\nStatement (1)

    CFCs are broken down by powerful UV radiation and releases chlorine free radical which reacts with ozone and start chain reaction.

    $$C{F_2}C{l_2}_{(g)}\\buildrel {UV} \\over\n \\longrightarrow \\mathop C\\limits^ \\bullet {l_g} + \\mathop C\\limits^ \\bullet {F_2}C{l_{(g)}}$$

    $$\\mathop C\\limits^ \\bullet {l_{(g)}} + {O_{3(g)}} \\to \\mathop C\\limits^ \\bullet l{O_{(g)}} + {O_{2(g)}}$$

    $$\\mathop C\\limits^ \\bullet l{O_{(g)}} + {O_{(g)}} \\to \\mathop C\\limits^ \\bullet {l_{(g)}} + {O_{2(g)}}$$

    Statement (2)

    Atmosphere ozone reacts with nitric oxide to produce nitrogen dioxide and oxygen.

    $$N{O_{(g)}} + {O_{3(g)}} \\to N{O_{2(g)}} + {O_{2(g)}}$$", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1513, "subject": "Chemistry", "question": "Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).

    Assertion (A) : Photochemical smog causes cracking of rubber.

    Reason (R) : Presence of ozone, nitric oxide, acrolein, formaldehyde and peroxyacetyl nitrate in photochemical smog makes it oxidizing.

    Choose the most appropriate answer from the options given below :", "options": [ { "text": "Both (A) and (R) are true but (R) is not the true explanation of (A)" }, { "text": "(A) is false but (R) is true." }, { "text": "(A) is true but (R) is false." }, { "text": "Both (A) and (R) are true and (R) is the true explanation of (A)" } ], "answer": "Both (A) and (R) are true and (R) is the true explanation of (A)", "solution": "**Answer:** Both (A) and (R) are true and (R) is the true explanation of (A)\n\nPhotochemical smog causes cracking of rubber, the common component of photochemical smog are ozone, nitric oxide, acrolein, formaldehyde and peroxyacetyel nitrate (PAN).", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1514, "subject": "Chemistry", "question": "The gas 'A' is having very low reactivity reaches to stratosphere. It is non-toxic and non-flammable but dissociated by UV-radiations in stratosphere. The intermediates formed initially from the gas 'A' are :", "options": [ { "text": "$$Cl\\mathop O\\limits^ \\bullet + \\mathop C\\limits^ \\bullet {F_2}Cl$$" }, { "text": "$$Cl\\mathop O\\limits^ \\bullet + C\\mathop {{H_3}}\\limits^ \\bullet $$" }, { "text": "$$\\mathop C\\limits^ \\bullet {H_3} + C\\mathop {{F_2}}\\limits^ \\bullet Cl$$" }, { "text": "$$\\mathop C\\limits^ \\bullet l + C\\mathop {{F_2}}\\limits^ \\bullet Cl$$" } ], "answer": "$$\\mathop C\\limits^ \\bullet l + C\\mathop {{F_2}}\\limits^ \\bullet Cl$$", "solution": "**Answer:** $$\\mathop C\\limits^ \\bullet l + C\\mathop {{F_2}}\\limits^ \\bullet Cl$$\n\nIn stratosphere CFCs get broken down by powerful UV radiations releasing $$C{l^ \\bullet }$$

    $$C{F_2}C{l_2}(g)\\buildrel {U.V.} \\over\n \\longrightarrow C{l^ \\bullet }(g) + {}^ \\bullet C{F_2}Cl(g)$$", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1515, "subject": "Chemistry", "question": "In stratosphere most of the ozone formation is assisted by :", "options": [ { "text": "cosmic rays" }, { "text": "$$\\gamma$$-rays" }, { "text": "ultraviolet radiation" }, { "text": "visible radiations" } ], "answer": "ultraviolet radiation", "solution": "**Answer:** ultraviolet radiation\n\nOzone in the stratosphere is a product of UV radiations acting on dioxygen (O2) molecules.

    $${O_2}(g)\\buildrel {UV} \\over\n \\longrightarrow O(g) + O(g)$$

    $$O(g) + {O_2}(g)$$ $$\\buildrel {UV} \\over\n \\rightleftharpoons $$ $${O_3}(g)$$", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1516, "subject": "Chemistry", "question": "The deposition of X and Y on ground surfaces is referred as wet and dry depositions, respectively. X and Y are :", "options": [ { "text": "X = Ammonium salts, Y = CO2" }, { "text": "X = SO2, Y = Ammonium salts" }, { "text": "X = Ammonium salts, Y = SO2" }, { "text": "X = CO2, Y = SO2" } ], "answer": "X = Ammonium salts, Y = SO2", "solution": "**Answer:** X = Ammonium salts, Y = SO2\n\nOxides of nitrogen and sulphur are acidic and settle down on ground as dry deposition.

    Ammonium salts in rain drops result in wet deposition.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1517, "subject": "Chemistry", "question": "

    The acid that is believed to be mainly responsible for the damage of Taj Mahal is :

    ", "options": [ { "text": "sulfuric acid." }, { "text": "hydrofluoric acid." }, { "text": "phosphoric acid." }, { "text": "hydrochloric acid." } ], "answer": "sulfuric acid.", "solution": "**Answer:** sulfuric acid.\n\nCaCO3 + H2SO4 $$ \\to $$ CaSO4 + H2O + CO2

    \nIndustries like oil refinery releases SO2 which causes air pollution. It reacts with water to form acid rain when SO2 mix with water it forms H2SO4 (Sulphuric acid).", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1518, "subject": "Chemistry", "question": "

    Correct statement about photo-chemical smog is :

    ", "options": [ { "text": "It occurs in humid climate." }, { "text": "It is a mixture of smoke, fog and SO2." }, { "text": "It is reducing smog." }, { "text": "It results from reaction of unsaturated hydrocarbons." } ], "answer": "It results from reaction of unsaturated hydrocarbons.", "solution": "**Answer:** It results from reaction of unsaturated hydrocarbons.\n\nPhotochemical smog occurs in warm, dry and\nsunny climate. The main components of\nphotochemical smog result from the action of\nunsaturated hydrocarbons and nitrogen oxides.\nThis is an oxidising smog.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1519, "subject": "Chemistry", "question": "

    On the surface of polar stratospheric clouds, hydrolysis of chlorine nitrite gives A and B while its reaction with its reaction with HCl produces B and C. A, B and C are, respectively :

    ", "options": [ { "text": "HOCl, HNO3, Cl2" }, { "text": "Cl2, HNO3, HOCl" }, { "text": "HClO2, HNO2, HOCl" }, { "text": "HOCl, HNO2, Cl2O" } ], "answer": "HOCl, HNO3, Cl2", "solution": "**Answer:** HOCl, HNO3, Cl2\n\nOn the surface of polar stratospheric clouds,\nhydrolysis of chlorine nitrate as

    \n$$\\mathrm{ClONO}_{2}+\\mathrm{H}_{2} \\mathrm{O} \\longrightarrow \\underset{\\mathrm{A}}{\\mathrm{HOCl}}+\\underset{\\mathrm{B}}{\\mathrm{HNO}_{3}}$$

    \n$$\\mathrm{ClONO}_{2}+\\mathrm{HCl} \\longrightarrow \\underset{\\mathrm{C}}{\\mathrm{Cl}_{2}}+\\underset{\\mathrm{H}}{\\mathrm{HNO}_{3}}$$

    \nHence A, B and C are HOCl, HNO3 and Cl2 respectively.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1520, "subject": "Chemistry", "question": "

    Given below are two Statements:

    \n

    Statement I : Classical smog occurs in cool humid climate. It is a reducing mixture of smoke, fog and sulphur dioxide.

    \n

    Statement II : Photochemical smog has components, ozone, nitric oxide, acrolein, formaldehyde, PAN etc.

    \n

    In the light of the above statements, choose the most appropriate answer from the options given below.

    ", "options": [ { "text": "Both Statement I and Statement II are correct." }, { "text": "Both Statement I and Statement II are incorrect." }, { "text": "Statement I is correct but Statement II is incorrect." }, { "text": "Statement I is incorrect but Statement II is correct." } ], "answer": "Both Statement I and Statement II are correct.", "solution": "**Answer:** Both Statement I and Statement II are correct.\n\n(I) Classical smog occurs in cool humid climate. It is a reducing mixture of smoke, fog and sulphur dioxide. This is a correct statement.\n

    \n(II) This statement is also based on fact and is a correct statement.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1521, "subject": "Chemistry", "question": "

    Polar stratospheric clouds facilitate the formation of :

    ", "options": [ { "text": "ClONO2" }, { "text": "HOCl" }, { "text": "ClO" }, { "text": "CH4" } ], "answer": "HOCl", "solution": "**Answer:** HOCl\n\nPolar stratospheric clouds provide surface on which hydrolysis of $\\mathrm{ClONO}_2$ takes place to form $\\mathrm{HOCl}$ (Hypochlorous acid)\n

    $$\n\\mathrm{ClONO}_2(\\mathrm{~g})+\\mathrm{H}_2 \\mathrm{O}(\\mathrm{g}) \\rightarrow \\mathrm{HOCl}(\\mathrm{g})+\\mathrm{HNO}_3(\\mathrm{~g})\n$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1522, "subject": "Chemistry", "question": "

    The photochemical smog does not generally contain :

    ", "options": [ { "text": "NO" }, { "text": "NO2" }, { "text": "SO2" }, { "text": "HCHO" } ], "answer": "SO2", "solution": "**Answer:** SO2\n\nPhotochemical smog contain:\n

    \nOzone, nitric oxide, organic compounds, nitrogen dioxide, formaldehyde.\n

    \n$$\\therefore \\mathrm{SO}_{2}$$ is not the part of photochemical smog.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1523, "subject": "Chemistry", "question": "

    How can photochemical smog be controlled?

    ", "options": [ { "text": "By using catalytic convertors in the automobiles/industry." }, { "text": "By complete combustion of fuel." }, { "text": "By using tall chimneys." }, { "text": "By using catalyst." } ], "answer": "By using catalytic convertors in the automobiles/industry.", "solution": "**Answer:** By using catalytic convertors in the automobiles/industry.\n\n

    Option A, By using catalytic converters in the automobiles/industry, is the most appropriate answer to control photochemical smog.

    Photochemical smog is a type of air pollution that forms in urban areas when pollutants from automobile and industrial emissions, such as nitrogen oxides (NOx) and volatile organic compounds (VOCs), react in the presence of sunlight.

    Catalytic converters are devices that are installed in the exhaust systems of automobiles and in some industrial processes to remove pollutants from exhaust gases. They work by using a catalyst to convert harmful pollutants, such as NOx and VOCs, into less harmful substances before they are released into the atmosphere. This can significantly reduce the formation of photochemical smog.

    Option B, By complete combustion of fuel, is not a complete or sufficient solution to control photochemical smog. Complete combustion of fuel, which refers to the complete oxidation of the fuel to produce carbon dioxide and water, can reduce the amount of unburned hydrocarbons and carbon monoxide in the exhaust gases.

    However, even with complete combustion, other pollutants, such as nitrogen oxides (NOx), can still be produced and contribute to the formation of photochemical smog. NOx emissions are a result of high temperatures and pressures in the combustion process and are not easily prevented through complete combustion alone.

    Additionally, while complete combustion can reduce the amount of certain pollutants, it does not address other sources of pollution, such as industrial emissions, that can contribute to the formation of photochemical smog.

    Therefore, while complete combustion can be an important step in reducing emissions and controlling photochemical smog, it is not a sufficient solution on its own and must be combined with other methods, such as the use of catalytic converters, to effectively address this issue.

    Option C, using tall chimneys, is not a direct way to control photochemical smog. Tall chimneys can help to disperse pollutants into the atmosphere, but do not necessarily prevent their formation in the first place.

    Option D, By using a catalyst, is not a complete or sufficient solution to control photochemical smog. While catalysts can be used to convert harmful pollutants into less harmful substances, they need to be part of a larger system, such as a catalytic converter, to effectively reduce emissions.

    Using a catalyst alone, without the proper design, engineering, and installation, is unlikely to be effective in controlling photochemical smog. It is important to note that the goal is to reduce the overall emissions of pollutants, such as nitrogen oxides (NOx) and volatile organic compounds (VOCs), which are the precursors to photochemical smog.

    Therefore, while catalysts can be a useful tool in controlling photochemical smog, they must be used in combination with other methods, such as the use of catalytic converters in automobiles and industry, to be most effective.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1524, "subject": "Chemistry", "question": "

    Formation of photochemical smog involves the following reaction in which $$\\mathrm{A}, \\mathrm{B}$$ and $$\\mathrm{C}$$ are respectively.

    \n

    $$\n\\begin{aligned}\n& \\text { i. } \\mathrm{NO}_{2} \\stackrel{\\mathrm{h} v}{\\longrightarrow} \\mathrm{A}+\\mathrm{B} \\\\\n& \\text { ii. } \\mathrm{B}+\\mathrm{O}_{2} \\rightarrow \\mathrm{C} \\\\\n& \\text { iii. } \\mathrm{A}+\\mathrm{C} \\rightarrow \\mathrm{NO}_{2}+\\mathrm{O}_{2}\n\\end{aligned}\n$$

    \n

    Choose the correct answer from the options given below :

    ", "options": [ { "text": "$$\\mathrm{NO}, \\mathrm{O} ~\\& ~\\mathrm{O}_{3}$$" }, { "text": "$$\\mathrm{N}, \\mathrm{O}_{2} ~\\& ~\\mathrm{O}_{3}$$" }, { "text": "$$\\mathrm{O}, \\mathrm{N}_{2} \\mathrm{O} ~\\& ~\\mathrm{NO}$$" }, { "text": "$$\\mathrm{O}, \\mathrm{NO} ~\\mathrm{\\&} ~\\mathrm{NO}_{3}^{-}$$" } ], "answer": "$$\\mathrm{NO}, \\mathrm{O} ~\\& ~\\mathrm{O}_{3}$$", "solution": "**Answer:** $$\\mathrm{NO}, \\mathrm{O} ~\\& ~\\mathrm{O}_{3}$$\n\n

    (i) $$N{O_2}\\buildrel {hv} \\over\n \\longrightarrow {{NO} \\over {(A)}} + {O \\over {(B)}}$$

    \n

    (ii) $${O \\over {(B)}} + {O_2}\\buildrel {} \\over\n \\longrightarrow {{{O_3}} \\over {(C)}}$$

    \n

    (iii) $${{NO} \\over {(A)}} + {{{O_3}} \\over {(C)}}\\buildrel {} \\over\n \\longrightarrow N{O_2} + {O_2}$$

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1525, "subject": "Chemistry", "question": "

    Correct statement about smog is :

    ", "options": [ { "text": "NO$$_2$$ is present in classical smog" }, { "text": "Both NO$$_2$$ and SO$$_2$$ are present in classical smog" }, { "text": "Photochemical smog has high concentration of oxidizing agents" }, { "text": "Classical smog also has high concentration of oxidizing agents" } ], "answer": "Photochemical smog has high concentration of oxidizing agents", "solution": "**Answer:** Photochemical smog has high concentration of oxidizing agents\n\nPhotochemical smog has high concentration of\noxidising agents

    \nNO2 is produced from NO and O3 in the presence\nof sunlight

    \n Classical smog contain smoke, fog and SO2 and it\nis known as reducing smog, as chemically it is\nreducing mixture", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1526, "subject": "Chemistry", "question": "

    Correct statement is :

    ", "options": [ { "text": "An average human being consumes more food than air" }, { "text": "An average human being consumes 100 times more air than food" }, { "text": "An average human being consumes nearly 15 times more air than food" }, { "text": "An average human being consumes equal amount of food and air" } ], "answer": "An average human being consumes nearly 15 times more air than food", "solution": "**Answer:** An average human being consumes nearly 15 times more air than food\n\nThe correct statement is An average human being consumes nearly 15 times more air than food.\n

    \nThis is because humans need to breathe in a certain amount of air to supply oxygen to the body and release carbon dioxide, which is produced by cellular respiration. On the other hand, the amount of food an average human consumes is much less than the amount of air, even though it is also essential for providing energy, nutrients, and other vital substances to the body.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1527, "subject": "Chemistry", "question": "

    Which of the following is true about freons?

    ", "options": [ { "text": "These are radicals of chlorine and chlorine monoxide" }, { "text": "These are chemicals causing skin cancer" }, { "text": "All radicals are called freons" }, { "text": "These are chlorofluorocarbon compounds" } ], "answer": "These are chlorofluorocarbon compounds", "solution": "**Answer:** These are chlorofluorocarbon compounds\n\nFreons are a family of chlorofluorocarbon (CFC) compounds, which are chemicals that contain chlorine, fluorine, and carbon atoms. They are used in a variety of applications, including refrigeration and air conditioning, as well as in aerosol spray cans.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1528, "subject": "Chemistry", "question": "The possibility of photochemical smog formation will be minimum at :", "options": [ { "text": "Mumbai in May" }, { "text": "Kolkata in October" }, { "text": "New-Delhi in August (Summer)" }, { "text": "Srinagar, Jammu and Kashmir in January" } ], "answer": "Srinagar, Jammu and Kashmir in January", "solution": "**Answer:** Srinagar, Jammu and Kashmir in January\n\nThe formation of photochemical smog depends on several factors, including sunlight, temperature, and the presence of pollutants such as nitrogen oxides and volatile organic compounds. \n

    \nOption (A) Mumbai in May may also have some possibility of photochemical smog formation because the temperature and sunlight are relatively high in May.\n

    \nOption (B) Kolkata in October may have some possibility of photochemical smog formation because the temperature and sunlight are still relatively high in October.\n

    \nOption (C) New-Delhi in August (Summer) is more likely to have photochemical smog formation because the temperature and sunlight are at their peak in the summer months, and there is a high level of air pollution in the city due to various anthropogenic activities.\n

    \nOption (D) Srinagar, Jammu and Kashmir in January is less likely to have photochemical smog formation because the winter season typically has less sunlight and lower temperatures, which are not conducive to the formation of photochemical smog.\n

    \nTherefore, the answer is (D) Srinagar, Jammu and Kashmir in January, as it is the option with the least possibility of photochemical smog formation.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1529, "subject": "Chemistry", "question": "

    Which of the following are the Green house gases?

    \n

    A. Water vapour

    \n

    B. Ozone

    \n

    C. I$$_2$$

    \n

    D. Molecular hydrogen

    \n

    Choose the most appropriate answer from the options given below :

    ", "options": [ { "text": "A and D only" }, { "text": "A and B only" }, { "text": "B and C only" }, { "text": "C and D only" } ], "answer": "A and B only", "solution": "**Answer:** A and B only\n\nGreen house gases are $\\mathrm{CO}_2, \\mathrm{CH}_4$, water vapour, nitrous oxide, $\\mathrm{CFC}_8$ and ozone.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1530, "subject": "Chemistry", "question": "

    The radical which mainly causes ozone depletion in the presence of UV radiations is :

    ", "options": [ { "text": "$$\\mathrm{NO}^\\cdot$$" }, { "text": "$$\\mathrm{CH}_{3} ^\\cdot$$" }, { "text": "$$\\dot{\\mathrm{Cl}}$$" }, { "text": "$$\\dot{\\mathrm{O}} \\mathrm{H}$$" } ], "answer": "$$\\dot{\\mathrm{Cl}}$$", "solution": "**Answer:** $$\\dot{\\mathrm{Cl}}$$\n\nThe radical that mainly causes ozone depletion in the presence of UV radiation is $\\dot{\\mathrm{Cl}}$. The chloro radical reacts with ozone molecules to form chlorine monoxide and oxygen:\n

    \n$$\\dot{\\mathrm{Cl}} + \\mathrm{O}_{3} \\rightarrow \\mathrm{ClO} + \\mathrm{O}_{2}$$\n

    \nThis reaction depletes the ozone layer, as the chlorine monoxide can then go on to react with another ozone molecule:\n

    \n$$\\mathrm{ClO} + \\mathrm{O}_{3} \\rightarrow \\mathrm{Cl} + \\mathrm{2O}_{2}$$\n

    \nThus, $\\dot{\\mathrm{Cl}}$, is the correct answer.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1531, "subject": "Chemistry", "question": "

    Which of the following compounds is an example of Freon?

    ", "options": [ { "text": "$$\\mathrm{C}_{2} \\mathrm{H}_{2} \\mathrm{F}_{2}$$" }, { "text": "$$\\mathrm{C}_{2} \\mathrm{HF}_{3}$$" }, { "text": "$$\\mathrm{C}_{2} \\mathrm{F}_{4}$$" }, { "text": "$$\\mathrm{C}_{2} \\mathrm{Cl}_{2} \\mathrm{F}_{2}$$" } ], "answer": "$$\\mathrm{C}_{2} \\mathrm{Cl}_{2} \\mathrm{F}_{2}$$", "solution": "**Answer:** $$\\mathrm{C}_{2} \\mathrm{Cl}_{2} \\mathrm{F}_{2}$$\n\n

    Freons are a class of compounds which are also known as chlorofluorocarbons (CFCs). As the name suggests, these compounds contain carbon, chlorine, and fluorine atoms. So the compound that is an example of a Freon from the options given is :

    \n

    Option D : $$\\mathrm{C}_{2} \\mathrm{Cl}_{2} \\mathrm{F}_{2}$$

    \n

    This compound contains both chlorine and fluorine atoms bonded to carbon, and is therefore a CFC or a Freon.

    \n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1532, "subject": "Chemistry", "question": "

    The delicate balance of $$\\mathrm{CO}_{2}$$ and $$\\mathrm{O}_{2}$$ is NOT disturbed by :

    ", "options": [ { "text": "Burning of Coal" }, { "text": "Burning of petroleum" }, { "text": "Respiration" }, { "text": "Deforestation" } ], "answer": "Respiration", "solution": "**Answer:** Respiration\n\n

    The delicate balance of $\\mathrm{CO}_2$ and $\\mathrm{O}_2$ is disturbed by the burning of coal (Option A), the burning of petroleum (Option B), and deforestation (Option D). These activities contribute to the increase in atmospheric $\\mathrm{CO}_2$ levels, which can lead to global warming and climate change.

    \n

    However, respiration (Option C) does not disturb the delicate balance of $\\mathrm{CO}_2$ and $\\mathrm{O}_2$. During respiration, organisms including humans and animals consume $\\mathrm{O}_2$ and produce $\\mathrm{CO}_2$ as a byproduct. This $\\mathrm{CO}_2$ can be utilized by plants during photosynthesis to produce $\\mathrm{O}_2$, thus maintaining a balance between $\\mathrm{CO}_2$ and $\\mathrm{O}_2$ levels in the atmosphere.

    \n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1533, "subject": "Chemistry", "question": "

    Which of these reactions is not a part of breakdown of ozone in stratosphere?

    ", "options": [ { "text": "$$\\mathrm{C{F_2}C{l_2}(g)\\buildrel {uv} \\over\n \\longrightarrow \\mathop C\\limits^ \\bullet l(g) + \\mathop C\\limits^ \\bullet {F_2}Cl(g)}$$" }, { "text": "$$\\mathrm{C\\mathop l\\limits^ \\bullet (g) + {O_3}(g)\\buildrel {} \\over\n \\longrightarrow Cl\\mathop O\\limits^ \\bullet (g) + {O_2}(g)}$$" }, { "text": "$$\\mathrm{2Cl\\mathop O\\limits^ \\bullet \\buildrel {} \\over\n \\longrightarrow Cl{O_2}(g) + C\\mathop l\\limits^ \\bullet (g)}$$" }, { "text": "$$\\mathrm{Cl\\mathop O\\limits^ \\bullet (g) + O(g)\\buildrel {} \\over\n \\longrightarrow C\\mathop l\\limits^ \\bullet (g) + {O_2}(g)}$$" } ], "answer": "$$\\mathrm{2Cl\\mathop O\\limits^ \\bullet \\buildrel {} \\over\n \\longrightarrow Cl{O_2}(g) + C\\mathop l\\limits^ \\bullet (g)}$$", "solution": "**Answer:** $$\\mathrm{2Cl\\mathop O\\limits^ \\bullet \\buildrel {} \\over\n \\longrightarrow Cl{O_2}(g) + C\\mathop l\\limits^ \\bullet (g)}$$\n\n

    The breakdown of ozone in the stratosphere largely involves a series of reactions often initiated by free radicals, such as chlorine (Cl), bromine (Br), or nitric oxide (NO) radicals. In particular, chlorine atoms, which can come from substances like chlorofluorocarbons (CFCs), can catalyze the destruction of ozone.

    \n

    The first three reactions listed are all part of the ozone breakdown process:

    \n

    Option A: This is the photodissociation of a chlorofluorocarbon, which releases a chlorine atom (a free radical) that can participate in ozone destruction.

    \n

    $$\\mathrm{CF_2Cl_2(g) \\xrightarrow{uv} Cl(g) + CF_2Cl(g)}$$

    \n

    Option B: This reaction shows a chlorine atom catalyzing the destruction of ozone, forming ClO and O2.

    \n

    $$\\mathrm{Cl(g) + O_3(g) \\rightarrow ClO(g) + O_2(g)}$$

    \n

    Option D: This reaction shows a ClO radical reacting with a single oxygen atom to regenerate the Cl radical and form O2. The regenerated Cl radical can then go on to catalyze more ozone destruction, making this a catalytic cycle.

    \n

    $$\\mathrm{ClO(g) + O(g) \\rightarrow Cl(g) + O_2(g)}$$

    \n

    However, option C:

    \n

    $$\\mathrm{2ClO \\rightarrow ClO_2 + Cl}$$

    \n

    This reaction is not typically part of the ozone breakdown process. Chlorine monoxide (ClO) radicals generally react with atomic oxygen (O), not with another ClO radical, to regenerate the chlorine radical and produce oxygen (O2). The product of this reaction, ClO2, does not commonly appear in the standard ozone depletion mechanisms.

    \n", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1534, "subject": "Chemistry", "question": "

    The possibility of photochemical smog formation is more at :

    ", "options": [ { "text": "Industrial areas" }, { "text": "Himalayan villages in winter" }, { "text": "The places with healthy vegetation" }, { "text": "Marshy lands" } ], "answer": "Industrial areas", "solution": "**Answer:** Industrial areas\n\n

    Photochemical smog is a type of air pollution that results from the interaction between sunlight and pollutants such as nitrogen oxides and volatile organic compounds (VOCs). This type of smog is common in urban and industrial areas where there is a high concentration of these pollutants.

    \n

    The conditions favorable for the formation of photochemical smog include strong sunlight, stagnant air, and the presence of pollutants such as those emitted by vehicles and industrial processes.

    \n

    Given the options:

    \n\n

    The correct answer is : Option A : Industrial areas.

    \n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1535, "subject": "Chemistry", "question": "Identify the wrong statements in the following :", "options": [ { "text": "Chlorofluorocarbons are responsible for ozone layer depletion" }, { "text": "Greenhouse effect is responsible for global warming" }, { "text": "Ozone layer does not permit infrared radiation from the sun to reach the earth " }, { "text": "Acid rains is mostly because of oxides of nitrogen and sulphur " } ], "answer": "Ozone layer does not permit infrared radiation from the sun to reach the earth ", "solution": "**Answer:** Ozone layer does not permit infrared radiation from the sun to reach the earth \n\nNOTE : Ozone layer acts as a shield and does not allow ultraviolet radiation from sun to reach earth. It does not prevent infra-red radiation from sun to reach earth. \n

    Thus option $$(c)$$ is wrong statement and so it is the correct answer. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1536, "subject": "Chemistry", "question": "Which one of the following substances used in dry cleaning is a better strategy to control environmental pollution?", "options": [ { "text": "Tetrachloroethylene" }, { "text": "Carbon dioxide" }, { "text": "Sulphur dioxide" }, { "text": "Nitrogen dioxide" } ], "answer": "Carbon dioxide", "solution": "**Answer:** Carbon dioxide\n\n

    Out of the given options, the substance that is a better strategy to control environmental pollution in dry cleaning is carbon dioxide (CO2).

    \n

    Option A: Tetrachloroethylene, also known as perchloroethylene (PERC), is a commonly used solvent in dry cleaning. However, it is a hazardous air pollutant and a potential human carcinogen. Its use has been restricted or banned in many countries due to its harmful effects on the environment and human health.

    \n

    Option B: Carbon dioxide is a more environmentally friendly alternative to traditional dry cleaning solvents like PERC. CO2 is a non-toxic, non-flammable, and readily available gas that can be used as a cleaning solvent when it is in its supercritical state. CO2 dry cleaning is a "green" process that does not produce hazardous waste or emit harmful air pollutants.

    \n

    Option C: Sulphur dioxide (SO2) is a common air pollutant that is emitted from the burning of fossil fuels. It is not used as a dry cleaning solvent, and its use would not be an effective strategy to control pollution in dry cleaning.

    \n

    Option D: Nitrogen dioxide (NO2) is also a common air pollutant that is emitted from the burning of fossil fuels. It is not used as a dry cleaning solvent, and its use would not be an effective strategy to control pollution in dry cleaning.

    \n

    Therefore, the correct option is Carbon dioxide.

    \n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1537, "subject": "Chemistry", "question": "Which of the following is a set of greenhouse gases?", "options": [ { "text": "CH4, O3, N2, SO2 " }, { "text": "O3, N2, CO2, NO2" }, { "text": "O3, NO2, SO2, Cl2 " }, { "text": "CO2, CH4, N2O, O3 " } ], "answer": "CO2, CH4, N2O, O3 ", "solution": "**Answer:** CO2, CH4, N2O, O3 \n\nThe gases that contribute to greenhouse effect are called greenhouse gases. These include CO2 , CH4 , N2O and O3.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1538, "subject": "Chemistry", "question": "The reaction that is NOT involved in the ozone layer depletion mechanism in the stratosphere is : \n", "options": [ { "text": "$$Cl\\mathop O\\limits^ \\bullet \\left( g \\right) + O\\left( g \\right) \\to \\mathop C\\limits^ \\bullet l\\left( g \\right) + {O_2}(g)$$" }, { "text": "$$C{F_2}C{l_2}\\left( g \\right)\\,\\buildrel {uv} \\over\n \\longrightarrow \\mathop {\\,C}\\limits^ \\bullet l\\left( g \\right) + \\mathop C\\limits^ \\bullet {F_2}Cl\\left( g \\right)$$" }, { "text": "$$C{H_4} + 2{O_3}\\, \\to \\,3C{H_2} = O + 3{H_2}O$$" }, { "text": "$$HOCl\\left( g \\right)\\,\\buildrel {hv} \\over\n \\longrightarrow \\,\\mathop O\\limits^ \\bullet H\\left( g \\right) + \\mathop C\\limits^ \\bullet l\\left( g \\right)$$" } ], "answer": "$$C{H_4} + 2{O_3}\\, \\to \\,3C{H_2} = O + 3{H_2}O$$", "solution": "**Answer:** $$C{H_4} + 2{O_3}\\, \\to \\,3C{H_2} = O + 3{H_2}O$$\n\nOzone layer depletion mechanism occurs with Chlorofluorocarbons and involves chlorine radical.\n

    In CH4, there is no Cl so it is not involved in the ozone layer depletion mechanism", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1539, "subject": "Chemistry", "question": "Assertion : Ozone is destroyed by CFCs in the upper stratosphere.

    \nReason : Ozone holes increase the amount of UV radiation the earth.", "options": [ { "text": "Assertion and reason are both correct, and the reason is the correct explanation for the\nassertion." }, { "text": "Assertion and reason are incorrect." }, { "text": "Assertion and reason are correct but, the reason is not the explanation for the assertion" }, { "text": "Assertion is false, but the reason is correct." } ], "answer": "Assertion and reason are correct but, the reason is not the explanation for the assertion", "solution": "**Answer:** Assertion and reason are correct but, the reason is not the explanation for the assertion\n\nCFC’s are responsible for depletion of ozone\nlayer.\n

    When the ozone holes increases more UV rays will enter the earth.\n

    So here both Assertion and reason are correct but reason is not the correct explanation for why CFC is destroying the ozone layer.", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1540, "subject": "Chemistry", "question": "Which is wrong with respect to our responsiblity as a human being to protect our environment?", "options": [ { "text": "Restricting the use of vehicles" }, { "text": "Setting up compost tin in gardens." }, { "text": "Avoiding the use of floodlighted facilities" }, { "text": "Using plastic bags." } ], "answer": "Using plastic bags.", "solution": "**Answer:** Using plastic bags.\n\nPlastic bag is harmful for our eviroment an plastic is non bio-degradable polutant.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1541, "subject": "Chemistry", "question": "The layer of atmosphere between 10 km to\n50 km above the sea level is called as :", "options": [ { "text": "mesosphere" }, { "text": "thermosphere" }, { "text": "troposphere" }, { "text": "stratosphere" } ], "answer": "stratosphere", "solution": "**Answer:** stratosphere\n\nBetween 10-50 km above sea level is the lowest level which is called\nstratosphere.", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1542, "subject": "Chemistry", "question": "The regions of the atmosphere, where clouds form and where we live, respectively, are :", "options": [ { "text": "Stratosphere and Stratosphere" }, { "text": "Stratosphere and Troposphere" }, { "text": "Troposphere and Stratosphere" }, { "text": "Troposphere and Troposphere " } ], "answer": "Troposphere and Troposphere ", "solution": "**Answer:** Troposphere and Troposphere \n\nThe lowest region of atmosphere in which\nhuman beings live is troposphere. It extends up\nto a height of 10 km from sea level. Clouds are\nalso formed in this layer", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1543, "subject": "Chemistry", "question": "Among the gases (a) – (e), the gases that cause greenhouse effect are :\n
    (a) CO2\n
    (b) H2O\n
    (c) CFCs\n
    (d) O2\n
    (e) O3", "options": [ { "text": "(a), (b), (c) and (d)" }, { "text": "(a) and (d)" }, { "text": "(a), (b), (c) and (e)" }, { "text": "(a), (c), (d) and (e)" } ], "answer": "(a), (b), (c) and (e)", "solution": "**Answer:** (a), (b), (c) and (e)\n\nCO2, O3, H2O, CFCs are green house gases.\n

    Here only O2 is not greenhouse gas.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1544, "subject": "Chemistry", "question": "The processes of calcination and roasting in\nmetallurgical industries, respectively, can Iead\nto :", "options": [ { "text": "Global warming and photochemical smog" }, { "text": "Photochemical smog and ozone layer\ndepletion" }, { "text": "Photochemical smog and global warming" }, { "text": "Global warming and acid rain" } ], "answer": "Global warming and acid rain", "solution": "**Answer:** Global warming and acid rain\n\nCalcination Releases $$ \\to $$ CO2 $$ \\to $$ Global warming\n

    Roasting Releases $$ \\to $$ SO2 $$ \\to $$ Acid Rain", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1545, "subject": "Chemistry", "question": "The condition that indicates a polluted environment\nis :", "options": [ { "text": "BOD value of 5 ppm" }, { "text": "eutrophication" }, { "text": "0.03% of CO2 in the atmosphere" }, { "text": "pH of rain water to be 5.6" } ], "answer": "eutrophication", "solution": "**Answer:** eutrophication\n\nIn Eutrophication nutrient enriched water bodies support a dense plant population, which\nkills animal life by depriving it of oxygen and results in subsequent loss of biodiversity. If\nindicates polluted environment.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1546, "subject": "Chemistry", "question": "The gas released during anaerobic degradation of vegetation may lead to :", "options": [ { "text": "Ozone hole" }, { "text": "Corrosion of metals" }, { "text": "Acid rain" }, { "text": "Global warming and cancer" } ], "answer": "Global warming and cancer", "solution": "**Answer:** Global warming and cancer\n\nMethane (CH4)\ngas is evolved due to anaerobic degradation of\nvegetation which causes global warming and cancer. It is a heat\ntrapping gas that forces the planet to warm drastically and quickly.", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 1547, "subject": "Chemistry", "question": "Given below are two statements :

    Statement I : The pH of rain water is normally 5.6.

    Statement II : If the pH of rain water drops below 5.6, it is called acid rain.

    In the light of the above statements, choose the correct answer from the options given below :", "options": [ { "text": "Both Statement I and Statement II are true." }, { "text": "Both Statement I and Statement II are false." }, { "text": "Statement I is true but Statement II is false." }, { "text": "Statement I is false but Statement II is true." } ], "answer": "Both Statement I and Statement II are true.", "solution": "**Answer:** Both Statement I and Statement II are true.\n\n

    Option A

    \n

    Both Statement I and Statement II are true.

    \n

    Explanation :

    \n

    The pH of pure water is 7, which is considered neutral. However, rainwater is naturally slightly acidic due to the presence of dissolved carbon dioxide, which reacts with the water to form carbonic acid. The pH of "normal" rainwater is around 5.6 because of this natural process.

    \n

    Acid rain is a term that refers to any form of precipitation with acidic components, such as sulfuric or nitric acid, that fall to the ground from the atmosphere in wet or dry forms. This can include rain, snow, fog, hail or even dust that is acidic. The increased acidity in rain is caused by high amounts of pollutant components such as sulfur and nitrogen compounds which react in the atmosphere with water, oxygen, and other chemicals to form sulfuric and nitric acid. These then fall to the ground as acid rain. The generally accepted definition of "acid rain" refers to rainfall with a pH of less than 5.6, which is more acidic than normal rainwater.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1548, "subject": "Chemistry", "question": "The green house gas/es is (are) :

    (A) Carbon dioxide

    (B) Oxygen

    (C) Water vapour

    (D) Methane

    Choose the most appropriate answer from the options given below :", "options": [ { "text": "(A) and (C) only" }, { "text": "(A) and (B) only" }, { "text": "(A), (C) and (D) only" }, { "text": "(A) only" } ], "answer": "(A), (C) and (D) only", "solution": "**Answer:** (A), (C) and (D) only\n\nGreenhouse gases are CO2, CH4, Water vapour, Nitrous oxide, CFCs and ozone. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1549, "subject": "Chemistry", "question": "The statements that are TRUE :

    (A) Methane leads to both global warming and photochemical smog

    (B) Methane is generated from paddy fields

    (C) Methane is a stronger global warming gas than CO2

    (D) Methane is a part of reducing smog

    Choose the most appropriate answer from the options given below :", "options": [ { "text": "(A), (B), (C) only" }, { "text": "(A) and (B) only" }, { "text": "(B), (C), (D) only" }, { "text": "(A), (B), (D) only" } ], "answer": "(A), (B), (C) only", "solution": "**Answer:** (A), (B), (C) only\n\nCH4 causes global warming and also a\nconstituent of photochemical smog.
    \nIt is a part of oxidising smog.

    \nCH4 is generated from paddy fields.

    \nDue to high heat capacity, CH4 is a stronger\nglobal warming causing gas than CO2.", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 1550, "subject": "Chemistry", "question": "Given below are two statements :

    Statement I : Non-biodegradable wastes are generated by the thermal power plants.

    Statement II : Bio-degradable detergents leads to eutrophication.

    In the light of the above statements, choose the most appropriate answer from the options given below :", "options": [ { "text": "Statement I is false but statement II is true." }, { "text": "Both statement I and statement II are true." }, { "text": "Both statement I and statement II are false." }, { "text": "Statement I is true but statement is false." } ], "answer": "Both statement I and statement II are true.", "solution": "**Answer:** Both statement I and statement II are true.\n\nNon-biodegradable wastes are generated by thermal power which produce fly ash.

    \nNowadays most of the detergents available are biodegradable. The bacteria responsible for\ndegrading biodegradable detergent feed on it and grow rapidly. While growing, they may use\nup all the oxygen dissolved in water which cause Eutrophication.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1551, "subject": "Chemistry", "question": "Which one of the following gases is reported to retard photosynthesis?", "options": [ { "text": "CO" }, { "text": "CFCs" }, { "text": "CO2" }, { "text": "NO2" } ], "answer": "NO2", "solution": "**Answer:** NO2\n\nAccording to NCERT only NO2 from the given options can retard the photosynthesis process in plants.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1552, "subject": "Chemistry", "question": "Which one of the following is used to remove most of plutonium from spent nuclear fuel?", "options": [ { "text": "ClF3" }, { "text": "O2F2" }, { "text": "I2O5" }, { "text": "BrO3" } ], "answer": "O2F2", "solution": "**Answer:** O2F2\n\nO2F2 oxidizes plutonium to PuF6 and the reaction is used in removing plutonium as PuF6 from spent nuclear fuel.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1553, "subject": "Chemistry", "question": "

    Some gases are responsible for heating of atmosphere (green house effect). Identify from the following the gaseous species which does not cause it.

    ", "options": [ { "text": "CH4" }, { "text": "O3" }, { "text": "H2O" }, { "text": "N2" } ], "answer": "N2", "solution": "**Answer:** N2\n\nAmong the given gases, the green house gases\nwhich are responsible for heating the atmosphere\nare CH4, water vapour and ozone. Nitrogen is not a\ngreen house gas.", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1554, "subject": "Chemistry", "question": "

    Match List I with List II.

    \n

    \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    List I

    Pollutant    		List II

    Source
    (A)Microorganisms(I)Strip mining
    (B)Plant nutrients(II)Domestic sewage
    (C)Toxic heavy metals(III)Chemical fertilizer
    (D)Sediment(IV)Chemical factory

    \n

    Choose the correct answer from the options given below :

    ", "options": [ { "text": "A-II, B-III, C-IV, D-I" }, { "text": "A-II, B-I, C-IV D-III" }, { "text": "A-I, B-IV, C-II, D-III" }, { "text": "A-I, B-IV, C-III, D-II" } ], "answer": "A-II, B-III, C-IV, D-I", "solution": "**Answer:** A-II, B-III, C-IV, D-I\n\n  Pollutant                              Source\n

    \n$$\n\\begin{array}{ll}\n\\text { Microorganisms } & \\rightarrow \\text { Domestic sewage } \\\\\n\\text { Plant nutrients } & \\rightarrow \\text { Chemical fertilizers } \\\\\n\\text { Toxic heavy metals } & \\rightarrow \\text { Chemical factory } \\\\\n\\text { Sediment } & \\rightarrow \\text { Strip mining }\n\\end{array}\n$$", "topic": "Number Theory", "subtopic": "Divisibility / Factorization" }, { "id": 1555, "subject": "Chemistry", "question": "

    Given below are two statements:

    \n

    Statement I : The non bio-degradable fly ash and slag from steel industry can be used by cement industry.

    \n

    Statement II : The fuel obtained from plastic waste is lead free.

    \n

    In the light of the above statements, choose the most appropriate answer from the options given below :

    ", "options": [ { "text": "Both Statement I and Statement II are correct" }, { "text": "Both Statement I and Statement II are incorrect" }, { "text": "Statement I is correct but Statement II is incorrect" }, { "text": "Statement I is incorrect but Statement II is correct" } ], "answer": "Both Statement I and Statement II are correct", "solution": "**Answer:** Both Statement I and Statement II are correct\n\nBoth Statement are correct.\n

    \n- Fuel obtained from plastic waste has high octane rating. It contain no lead and is known as \"green fuel\".\n

    \n- The non bio-degradable fly ash and slag from steel industry can be used by cement industry.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1556, "subject": "Chemistry", "question": "

    The industrial activity held least responsible for global warming is :

    ", "options": [ { "text": "manufacturing of cement" }, { "text": "Electricity generation in thermal power plants" }, { "text": "steel manufacturing" }, { "text": "Industrial production of urea" } ], "answer": "Industrial production of urea", "solution": "**Answer:** Industrial production of urea\n\n

    Option A (manufacturing of cement) is a significant contributor to global greenhouse gas emissions, as the production of cement releases large amounts of carbon dioxide (CO2).\n

    \n

    Option B (electricity generation in thermal power plants) is also a major contributor, as burning fossil fuels such as coal, natural gas, and oil to generate electricity releases large amounts of CO2 and other greenhouse gases into the atmosphere.

    \n\n

    Option C (steel manufacturing) also has a significant impact, as the production of steel requires high temperatures and the use of large amounts of energy, which can come from the burning of fossil fuels.

    \n\n

    Option D (Industrial production of urea) has a lower impact compared to the other options, as it is a less energy-intensive process and does not directly release greenhouse gases into the atmosphere. However, the production of urea does rely on the use of natural gas as a feedstock, which does result in the release of some greenhouse gases.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1557, "subject": "Chemistry", "question": "

    Match List I with List II

    \n

    \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    LIST ILIST II
    A.Nitrogen oxides in airI.Eutrophication
    B.Methane in airII.pH of rain water becomes 5.6
    C.Carbon dioxideIII.Global warming
    D.Phosphate fertilisers in waterIV.Acid rain

    \n

    Choose the correct answer from the options given below :

    ", "options": [ { "text": "A-II, B-III, C-I, D-IV" }, { "text": "A-I, B-II, C-III, D-IV" }, { "text": "A-IV, B-II, C-III, D-I" }, { "text": "A-IV, B-III, C-II, D-I" } ], "answer": "A-IV, B-III, C-II, D-I", "solution": "**Answer:** A-IV, B-III, C-II, D-I\n\nLet's match the items in List I with the items in List II:\n

    \nA. Nitrogen oxides in air - These are responsible for acid rain. So, A-IV.\n

    \nB. Methane in air - Methane is a greenhouse gas, contributing to global warming. So, B-III.\n

    \nC. Carbon dioxide - Normal rain water has a pH of 5.6 (slightly acidic). This is because it is exposed to the carbon dioxide in the atmosphere. So, C-II.\n

    \nD. Phosphate fertilizers in water - These contribute to eutrophication. So, D-I.\n

    \nTherefore, the correct answer is:\n

    \nA-IV, B-III, C-II, D-I", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1558, "subject": "Chemistry", "question": "

    Match List I with List II

    \n

    \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    List I
    Industry    		List II
    Waste Generated
    (A)Steel plants(I)Gypsum
    (B)Thermal power plants(II)Fly ash
    (C)Fertilizer industries(III)Slag
    (D)Paper mills(IV)Bio-degradable wastes

    \n

    Choose the correct answer from the options given below :

    ", "options": [ { "text": "(A)-(III), (B)-(II), (C)-(I), (D)-(IV)" }, { "text": "(A)-(III), (B)-(IV), (C)-(I), (D)-(II)" }, { "text": "(A)-(IV), (B)-(I), (C)-(II), (D)-(III)" }, { "text": "(A)-(II), (B)-(III), (C)-(IV), (D)-(I)" } ], "answer": "(A)-(III), (B)-(II), (C)-(I), (D)-(IV)", "solution": "**Answer:** (A)-(III), (B)-(II), (C)-(I), (D)-(IV)\n\n

    A - Steel plants : They generate waste in the form of slag (III), which is a by-product of the smelting process used in steel production.

    \n

    B - Thermal power plants : These plants generate fly ash (II) as a by-product of coal combustion.

    \n

    C - Fertilizer industries : The manufacture of phosphate fertilizers can produce gypsum (I) as a waste product.

    \n

    D - Paper mills : They generate bio-degradable wastes (IV) as a result of the pulp and papermaking processes.

    \n

    Therefore, the correct option is A : (A)-(III), (B)-(II), (C)-(I), (D)-(IV).

    \n", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1559, "subject": "Chemistry", "question": "

    Which among the following pairs has only herbicides?

    ", "options": [ { "text": "Aldrin and Dieldrin" }, { "text": "Sodium chlorate and Aldrin" }, { "text": "Sodium arsinate and Dieldrin" }, { "text": "Sodium chlorate and sodium arsinite" } ], "answer": "Sodium chlorate and sodium arsinite", "solution": "**Answer:** Sodium chlorate and sodium arsinite\n\nBoth sodium chlorate and sodium arsenate behave as herbicide.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1560, "subject": "Chemistry", "question": "BOD stands for :", "options": [ { "text": "Biological Oxygen Demand" }, { "text": "Bacterial Oxidation Demand" }, { "text": "Biochemical Oxygen Demand" }, { "text": "Biochemical Oxidation Demand" } ], "answer": "Biochemical Oxygen Demand", "solution": "**Answer:** Biochemical Oxygen Demand\n\nBOD stands for Biochemical Oxygen Demand.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1561, "subject": "Chemistry", "question": "The concentration of fluoride, lead, nitrate and iron in a water sample from an underground lake was found\nto be 1000 ppb, 40 ppb, 100 ppm and 0.2 ppm, respectively. This water is unsuitable for drinking due to\nhigh concentration of :", "options": [ { "text": "Lead" }, { "text": "Nitrate" }, { "text": "Iron" }, { "text": "Fluoride" } ], "answer": "Nitrate", "solution": "**Answer:** Nitrate\n\nThe maximum limit of nitrate in drinking water is $$50$$ ppm. Excess nitrate in drinking water can cause disease such as methemoglobinemia ('blue baby' syndrome).", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1562, "subject": "Chemistry", "question": "A water sample has ppm level concentration of following anions
    \nF- = 10; $$SO_4^{2-}$$ = 100; $$NO_3^-$$ = 50
    \nThe anion/anions that make/makes the water sample unsuitable for drinking is/are :", "options": [ { "text": "both $$SO_4^{2-}$$ and $$NO_3^-$$" }, { "text": "only F-" }, { "text": "only $$SO_4^{2-}$$\n" }, { "text": "only $$NO_3^-$$\n" } ], "answer": "only F-", "solution": "**Answer:** only F-\n\nPermissible limit for SO4\n2-\n = 500 ppm\n

    Permissible limit for NO3\n-\n = 50 ppm\n

    Permissible limit for F-\n = 1 ppm", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1563, "subject": "Chemistry", "question": "The recommended concentration of fluoride ion in drinking water is up to 1 ppm as fluoride ion is required\nto make teeth enamel harder by converting
    [3Ca3 (PO4)2.Ca(OH)2] to ", "options": [ { "text": "[3{Ca(OH)2}. CaF2] " }, { "text": "[CaF2] \n" }, { "text": "[3(CaF2). Ca(OH)2] " }, { "text": "[3Ca3(PO4)2. CaF2] " } ], "answer": "[3Ca3(PO4)2. CaF2] ", "solution": "**Answer:** [3Ca3(PO4)2. CaF2] \n\n\"JEE ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1564, "subject": "Chemistry", "question": "Biochemical Oxygen Demand (BOD) value can be a measure of water pollution caused by the organic matter. Which of the following statements is correct ? ", "options": [ { "text": "Aerobic bacteria decrease the BOD value." }, { "text": "Anaerobic bacteria increase the BOD value. " }, { "text": "Clean water has BOD value higher than 10 ppm." }, { "text": "Polluted water has BOD value higher than 10 ppm." } ], "answer": "Polluted water has BOD value higher than 10 ppm.", "solution": "**Answer:** Polluted water has BOD value higher than 10 ppm.\n\nAerobic bacteria decreases the BOD value and aerobic bacteria increase the BOD value. Clear water has BOD value less than 5 ppm. Polluted water has BOD value higher than 10 ppm. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1565, "subject": "Chemistry", "question": "A water sample has ppm level concentration of the following metals : Fe = 0.2; Mn = 5.0; Cu = 3.0; Zn = 5.0. The metal that makes the water sample unsuitable for drinking is : ", "options": [ { "text": "Cu" }, { "text": "Mn" }, { "text": "Fe" }, { "text": "Zn" } ], "answer": "Mn", "solution": "**Answer:** Mn\n\nWhen Mn present in water with concentration 0.05 ppm or more than this then the water is unsuitable for drinking. \n

    Here given concentration for Mn is 5.0, which is grater than 0.05 that is why it is Mn makes water unsuitable for drinking. \n

    For other metal water is suitable for drinking when concentration for Fe = 0.2 ppm and for Cu = 3.0 ppm and for Zn = 5.0 ppm.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1566, "subject": "Chemistry", "question": "Which of the following conditions in drinking water causes methemoglobinemia?", "options": [ { "text": "> 50 ppm of lead" }, { "text": "> 50 ppm of chloride" }, { "text": "> 50 ppm of nitrate" }, { "text": "> 100 ppm of sulphate" } ], "answer": "> 50 ppm of nitrate", "solution": "**Answer:** > 50 ppm of nitrate\n\nConcentration of nitrate > 50 ppm in drinking water causes methemoglobinemia", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1567, "subject": "Chemistry", "question": "Water filled in two glasses A and B have BOD values of 10 and 20, respectively. The correct statement regarding them, is :", "options": [ { "text": "B is more polluted than A. " }, { "text": "A is suitable for drinking, whereas B is not ." }, { "text": "Both A and B are suitable for dirnking " }, { "text": "A is more polluted than B. " } ], "answer": "B is more polluted than A. ", "solution": "**Answer:** B is more polluted than A. \n\nTwo glasses ''A'' and ''B'' have BOD values 10 and ''20'', respectively. \n

    Hence glasses ''B'' is more polluted than glasses ''A''.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1568, "subject": "Chemistry", "question": "The concentration of dissolved oxygen (DO) in cold water can go upto : ", "options": [ { "text": "14 ppm" }, { "text": "16 ppm" }, { "text": "8 ppm" }, { "text": "10 ppm" } ], "answer": "10 ppm", "solution": "**Answer:** 10 ppm\n\nIn cold water, dissolved oxygen (DO) can reach a\nconcentration upto 10 ppm.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1569, "subject": "Chemistry", "question": "Water samples with BOD values of 4 ppm and 18 ppm, respectively are :", "options": [ { "text": "Clean and Highly polluted" }, { "text": "Highly polluted and Clean " }, { "text": "Highly polluted and Highly polluted " }, { "text": "Clean and Clean \n" } ], "answer": "Clean and Highly polluted", "solution": "**Answer:** Clean and Highly polluted\n\nClean water would have BOD value of less than 5 ppm whereas highly polluted water could have a\nBOD value of 17 ppm or more.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1570, "subject": "Chemistry", "question": "The maximum prescribed concentration of\ncopper in drinking water is :", "options": [ { "text": "3 ppm" }, { "text": "5 ppm" }, { "text": "0.5 ppm" }, { "text": "0.05 ppm" } ], "answer": "3 ppm", "solution": "**Answer:** 3 ppm\n\nMaximum prescribed concentration of Cu in\ndrinking water is 3 ppm.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1571, "subject": "Chemistry", "question": "Biochemical Oxygen Demand (BOD) is the\namount of oxygen required (in ppm) :", "options": [ { "text": "for sustaining life in a water body." }, { "text": "by anaerobic bacteria to breakdown\ninorganic waste present in a water body." }, { "text": "for the photochemical breakdown of waste\npresent in 1 m3 volume of a water body." }, { "text": "by bacteria to break-down organic waste in\na certain volume of a water sample." } ], "answer": "by bacteria to break-down organic waste in\na certain volume of a water sample.", "solution": "**Answer:** by bacteria to break-down organic waste in\na certain volume of a water sample.\n\nBiochemical oxygen demand (BOD) is amount\nof oxygen required by bacteria to break down\norganic waste in a certain volume of water\nsample.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1572, "subject": "Chemistry", "question": "The presence of soluble fluoride ion upto\n1 ppm concentration in drinking water, is :", "options": [ { "text": "safe for teeth" }, { "text": "harmful to skin" }, { "text": "harmful for teeth" }, { "text": "harmful to bones" } ], "answer": "safe for teeth", "solution": "**Answer:** safe for teeth\n\nFluoride (F–) ion conc upto 1 ppm makes the\nenamel of teeth much harder by converting.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1573, "subject": "Chemistry", "question": "Given below are two statements :

    Statement I : The value of the parameter \"Biochemical Oxygen Demand (BOD)\" is important for survival of aquatic life.

    Statement II : The optimum value of BOD is 6.5 ppm.

    In the light of the above statements, choose the most appropriate answer from the options given below :", "options": [ { "text": "Statement I is false but Statement II is true" }, { "text": "Statement I is true but Statement II is false" }, { "text": "Both Statement I and Statement II are true" }, { "text": "Both Statement I and Statement II are false" } ], "answer": "Statement I is true but Statement II is false", "solution": "**Answer:** Statement I is true but Statement II is false\n\n

    Clean water would have BOD value of less than 5 ppm whereas highly polluted water could have a BOD value of 17 ppm or more.

    \n\n

    Hence, the value of parameter ‘BOD’ is important for survival of aquatic life but optimum value of BOD is 17 ppm or more. So, statement II is incorrect.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1574, "subject": "Chemistry", "question": "Which of the following statement(s) is (are) incorrect reason for eutrophication?

    (A) excess usage of fertilisers

    (B) excess usage of detergents

    (C) dense plant population in water bodies

    (D) lack of nutrients in water bodies that prevent plant growth

    Choose the most appropriate answer from the options given below :", "options": [ { "text": "(A) Only" }, { "text": "(B) and (D) only" }, { "text": "(C) only" }, { "text": "(D) only" } ], "answer": "(D) only", "solution": "**Answer:** (D) only\n\nNutrient enriched water bodies support dense\nplant populations. This is because of the excess\nusage of fertilizers and detergents. This process is\nknown as eutrophication.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1575, "subject": "Chemistry", "question": "Which one of the following statements is NOT correct?", "options": [ { "text": "Eutrophication indicates that water body is polluted" }, { "text": "The dissolved oxygen concentration below 6 ppm inhibits fish growth" }, { "text": "Eutrophication leads to increase in the oxygen level in water" }, { "text": "Eutrophication leads to anaerobic conditions" } ], "answer": "Eutrophication leads to increase in the oxygen level in water", "solution": "**Answer:** Eutrophication leads to increase in the oxygen level in water\n\nEutrophication leads to decrease in oxygen level of water.

    3rd statement is incorrect.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1576, "subject": "Chemistry", "question": "The conversion of hydroxyapatite occurs due to presence of F$$-$$ ions in water. The correct formula of hydroxyapatite is :", "options": [ { "text": "[3Ca3(PO4)2 . Ca(OH)2]" }, { "text": "[3Ca(OH)2 . CaF2]" }, { "text": "[Ca3(PO4)2 . CaF2]" }, { "text": "[3Ca3(PO4)2 . CaF2]" } ], "answer": "[3Ca3(PO4)2 . Ca(OH)2]", "solution": "**Answer:** [3Ca3(PO4)2 . Ca(OH)2]\n\nThe F$$\\Theta $$ ions make the enamel on teeth much harder by converting hydroxyapatite, [3Ca3(PO4)2] . Ca(OH)2], the enamel on the surface of the teeth into much harder fluroappatite. [3Ca3(PO4)2 . CaF2]", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1577, "subject": "Chemistry", "question": "BOD values (in ppm) for clean water (A) and polluted water (B) are expected respectively :", "options": [ { "text": "A > 50, B < 27" }, { "text": "A > 25, B < 17" }, { "text": "A < 5, B > 17" }, { "text": "A > 15, B > 47" } ], "answer": "A < 5, B > 17", "solution": "**Answer:** A < 5, B > 17\n\nBOD values of clean water (A) is less than 5 ppm

    So, A < 5

    BOD values of polluted water (B) is greater than 17 ppm

    So, B > 17

    So, Ans. is (c)", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1578, "subject": "Chemistry", "question": "Water sample is called cleanest on the basis of which one of the BOD values given below :", "options": [ { "text": "11 ppm" }, { "text": "15 ppm" }, { "text": "3 ppm" }, { "text": "21 ppm" } ], "answer": "3 ppm", "solution": "**Answer:** 3 ppm\n\nBOD is the biological oxygen demand. Cleanest water sample\nwill have BOD value equal to 3 ppm as clean water could\nhave BOD value of less than 5 ppm whereas highly polluted water could have BOD value of 17 ppm or more.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1579, "subject": "Chemistry", "question": "

    The measured BOD values for four different water samples (A-D) are as follows: A = 3 ppm; B = 18 ppm; C = 21 ppm; D = 4 ppm. The water samples which can be called as highly polluted with organic wastes, are :

    ", "options": [ { "text": "A and B" }, { "text": "A and D" }, { "text": "B and C" }, { "text": "B and D" } ], "answer": "B and C", "solution": "**Answer:** B and C\n\nHighly polluted water should have BOD value of 17\nppm or more", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1580, "subject": "Chemistry", "question": "

    Given below are two statements : One is labelled as Assertion A and the other is labelled as Reason R.

    \n

    Assertion A : Polluted water may have a value of BOD of the order of 17 ppm.

    \n

    Reason R : BOD is a measure of oxygen required to oxidise both the bio-degradable an non-biodegradable organic material in water.

    \n

    In the light of the above statements, choose the most appropriate answer from the options given below.

    ", "options": [ { "text": "Both A and R are correct and R is the correct explanation of A." }, { "text": "Both A and R are correct but R is NOT the correct explanation of A." }, { "text": "A is correct but R is not correct." }, { "text": "A is not correct but R is correct." } ], "answer": "A is correct but R is not correct.", "solution": "**Answer:** A is correct but R is not correct.\n\nHighly polluted water could have a BOD value of 17\nppm or more.

    \nThe amount of oxygen required by bacteria to break\ndown the organic matter present in a certain volume\nof a sample of water is called Biochemical Oxygen\ndemand (BOD).

    \nHence A is correct but R is not correct.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1581, "subject": "Chemistry", "question": "

    The eutrophication of water body results in :

    ", "options": [ { "text": "loss of Biodiversity." }, { "text": "breakdown of organic matter." }, { "text": "increase in biodiversity." }, { "text": "decrease in BOD." } ], "answer": "loss of Biodiversity.", "solution": "**Answer:** loss of Biodiversity.\n\nEutrophication is the process in which nutrient\nenriched water bodies support a dense plant\npopulation, which kills animal life by depriving it of\noxygen and results in subsequent loss of\nbiodiversity.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1582, "subject": "Chemistry", "question": "

    Match List I with List II.

    \n

    \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    List I

    Pollutant    		List II

    Disease/sickness
    (A)Sulphate (>500 ppm)(I)Methemoglobinemia
    (B)Nitrate (>50 ppm)(II)Brown mottling of teeth
    (C)Lead (>50 ppb)(III)Laxative effect
    (D)Fluoride (>2 ppm)(IV)Kidney damage

    \n

    Choose the correct answer from the options given below :

    ", "options": [ { "text": "A-IV, B-I, C-II, D-III" }, { "text": "A-III, B-I, C-IV, D-II" }, { "text": "A-II, B-IV, C-I, D-III" }, { "text": "A-II, B-IV, C-III, D-I" } ], "answer": "A-III, B-I, C-IV, D-II", "solution": "**Answer:** A-III, B-I, C-IV, D-II\n\nA. Sulphate (>500 ppm) - Causes Laxative effect\nthat leads to dehydration

    \nB. Nitrate (>50 ppm) - Causes\nMethemoglobinemia, skin appears blue

    \nC. Lead (> 50 ppb) – It damage kidney and RBC

    \nD. Fluoride (>2 ppm) – It Causes Brown mottling\nof teeth", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1583, "subject": "Chemistry", "question": "

    Given below are two statements:

    \n

    Statement I : In polluted water values of both dissolved oxygen and $$\\mathrm{BOD}$$ are very low.

    \n

    Statement II : Eutrophication results in decrease in the amount of dissolved oxygen.

    \n

    In the light of the above statements, choose the most appropriate answer from the options given below :

    ", "options": [ { "text": "Both Statement I and Statement II are true" }, { "text": "Both Statement I and Statement II are false" }, { "text": "Statement I is true but Statement II is false" }, { "text": "Statement I is false but Statement II is true" } ], "answer": "Statement I is false but Statement II is true", "solution": "**Answer:** Statement I is false but Statement II is true\n\nSince eutrophication is result of excessive growth\nof weed in water bodies, which consume dissolved\noxygen of water bodies.

    \n$\\therefore$ Eutrophication decreases amount of dissolved\noxygen in water bodies.

    \n Polluted water has low value of dissolved oxygen,\nbut high valueof BOD (Biological oxygen\ndemand), since chemical and organic matter\nrequires dissolved oxygen to get decompose.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1584, "subject": "Chemistry", "question": "The water quality of a pond was analyzed and its BOD was found to be 4. The pond has :", "options": [ { "text": "Very clean water" }, { "text": "Highly polluted water\n" }, { "text": "Water has high amount of fluoride compounds\n" }, { "text": "Slightly polluted water" } ], "answer": "Very clean water", "solution": "**Answer:** Very clean water\n\n

    A clean water would have BOD value less than 5 ppm.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1585, "subject": "Chemistry", "question": "

    The concentration of dissolved Oxygen in water for growth of fish should be more than $$\\mathrm{\\underline X }$$ ppm and Biochemical Oxygen Demand in clean water should be less than $$\\mathrm{\\underline Y }$$ ppm. X and Y in ppm are, respectively.

    ", "options": [ { "text": "\n\n\n\n\n\n\n \n \n \n \n\n\n \n \n \n \n\n
    $$\\mathrm{X}$$$$\\mathrm{Y}$$
    415
    " }, { "text": "\n\n\n\n\n\n\n \n \n \n \n\n\n \n \n \n \n\n
    $$\\mathrm{X}$$$$\\mathrm{Y}$$
    48
    " }, { "text": "\n\n\n\n\n\n\n \n \n \n \n\n\n \n \n \n \n\n
    $$\\mathrm{X}$$$$\\mathrm{Y}$$
    612
    " }, { "text": "\n\n\n\n\n\n\n \n \n \n \n\n\n \n \n \n \n\n
    $$\\mathrm{X}$$$$\\mathrm{Y}$$
    65
    " } ], "answer": "\n\n\n\n\n\n\n \n \n \n \n\n\n \n \n \n \n\n
    $$\\mathrm{X}$$$$\\mathrm{Y}$$
    65
    ", "solution": "**Answer:** \n\n\n\n\n\n\n \n \n \n \n\n\n \n \n \n \n\n
    $$\\mathrm{X}$$$$\\mathrm{Y}$$
    65
    \n\n

    For high growth, dissolved oxygen should be less than 6 ppm and clean water has dissolved oxygen less than 5 ppm.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1586, "subject": "Chemistry", "question": "

    Given below are two statements:

    \n

    Statement I : If BOD is 4 ppm and dissolved oxygen is 8 ppm, then it is a good quality water.

    \n

    Statement II : If the concentration of zinc and nitrate salts are $$5 ~\\mathrm{ppm}$$ each, then it can be a good quality water.

    \n

    In the light of the above statements, choose the most appropriate answer from the options given below :

    ", "options": [ { "text": "Statement I is incorrect but Statement II is correct" }, { "text": "Both the statements I and II are incorrect" }, { "text": "Both the statements I and II are correct" }, { "text": "Statement I is correct but Statement II is incorrect" } ], "answer": "Both the statements I and II are correct", "solution": "**Answer:** Both the statements I and II are correct\n\n

    Statement I : If BOD is 4 ppm and dissolved oxygen is 8 ppm, then it is a good quality water.

    \n

    Statement II : If the concentration of zinc and nitrate salts are 5 ppm each, then it can be a good quality water.

    \n

    The Biochemical Oxygen Demand (BOD) measures the quantity of oxygen used by microorganisms in the oxidation of organic matter. Lower values of BOD indicate better water quality. A BOD value of less than 5 ppm indicates good quality water. Therefore, in Statement I, the BOD value of 4 ppm indicates good water quality.

    \n

    Dissolved oxygen (DO) refers to the level of free, non-compound oxygen present in water. DO levels above 5-6 ppm are generally good for aquatic life. In Statement I, the DO level of 8 ppm suggests the water is of good quality.

    \n

    The concentration of zinc in clean water should not exceed 5 ppm and the concentration of nitrates should not exceed 50 ppm. In Statement II, both zinc and nitrate concentrations are given as 5 ppm, which are within their respective permissible limits, therefore suggesting the water can be of good quality.

    \n

    Hence, based on this analysis :

    \n

    Option C : Both the statements I and II are correct.

    \n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1587, "subject": "Chemistry", "question": "

    Match List I with List II:

    \n

    \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    LIST I
    (Species)    		LIST II
    (Maximum allowed concentration in ppm in drinking water)
    A.$$\\mathrm{F^-}$$I.< 50 ppm
    B.$$\\mathrm{SO_4^{2-}}$$II.< 5 ppm
    C.$$\\mathrm{NO_3^-}$$III.< 2 ppm
    D.$$\\mathrm{Zn}$$IV.< 500 ppm

    \n

    Choose the correct answer from the options given below :

    ", "options": [ { "text": "A-IV, B-III, C-II, D-I" }, { "text": "A-I, B-II, C-III, D-IV" }, { "text": "A-II, B-I, C-III, D-IV" }, { "text": "A-III, B-IV C-I, D-II" } ], "answer": "A-III, B-IV C-I, D-II", "solution": "**Answer:** A-III, B-IV C-I, D-II\n\n
      \n

    1. Fluoride ($F^-$) : High concentration of fluoride can lead to a condition known as fluorosis, affecting teeth and bones. Hence, the WHO sets the upper limit for fluoride in drinking water to be 1.5 ppm. This can be approximated to less than 2 ppm. Therefore, $F^-$ (List I: A) corresponds to < 2 ppm (List II: III).

      \n

      2. \n

    2. Sulfate ($SO_4^{2-}$) : The standard concentration limit for sulphates in drinking water, according to WHO, is 500 ppm. Hence, $SO_4^{2-}$ (List I: B) matches with < 500 ppm (List II: IV).

      \n

      3. \n

    3. Nitrate ($NO_3^-$) : An excess of nitrate in drinking water can lead to health conditions such as methemoglobinemia. The safe limit, according to the WHO, is 50 ppm. Hence, $NO_3^-$ (List I: C) corresponds to < 50 ppm (List II: I).

      \n

      4. \n

    4. Zinc (Zn) : Zinc is an essential nutrient, but excessive amounts in drinking water can be harmful. The standard limit for zinc in drinking water is 5 ppm. So, Zn (List I: D) corresponds to < 5 ppm (List II: II).

      \n
      5. \n
    \n", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1588, "subject": "Chemistry", "question": "The interaction energy of London forces between two particles is proportional to rx, where r is the distance between the particles. The value of x is :", "options": [ { "text": "3" }, { "text": "$$-$$3" }, { "text": "$$-$$6" }, { "text": "6" } ], "answer": "$$-$$6", "solution": "**Answer:** $$-$$6\n\nFor London dispersion forces.

    $$E \\propto {1 \\over {{r^6}}}$$

    Hence, x = $$-$$6", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1589, "subject": "Chemistry", "question": "Equal masses of methane and oxygen are mixed in an empty container at 25oC. The fraction of the\ntotal pressure exerted by oxygen is ", "options": [ { "text": "2/3" }, { "text": "1/2" }, { "text": "1/3 $$\\times$$ 273/298" }, { "text": "1/3" } ], "answer": "1/3", "solution": "**Answer:** 1/3\n\nLet the mass of methane and oxygen $$=m$$ $$gm.$$\n

    Mole fraction of $${O_2}$$\n

    $$ = {{Moles\\,\\,of\\,\\,{O_2}} \\over {Moles\\,\\,of\\,\\,{O_2}\\, + \\,Moles\\,\\,of\\,\\,C{H_4}}}$$\n

    $$ = {{m/32} \\over {m/32 + m/16}}$$\n

    $$ = {{m/32} \\over {3m/32}} = {1 \\over 3}$$\n

    Partial pressure of $${O_2}$$\n

    $$=$$ Total pressure $$ \\times $$ mole fraction of $${O_2},$$ \n

    $${P_{{O_2}}} = P \\times {1 \\over 3} = {1 \\over 3}P$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1590, "subject": "Chemistry", "question": "A mixture of one mole each of H2\n, He and O2\neach are enclosed in a cylinder of volume V at\ntemperature T. If the partial pressure of H2\n is 2 atm, the total pressure of the gases in the\ncylinder is :", "options": [ { "text": "14 atm" }, { "text": "38 atm" }, { "text": "6 atm" }, { "text": "22 atm" } ], "answer": "6 atm", "solution": "**Answer:** 6 atm\n\nAccording to Dalton’s law of partial pressure, pi\n = xi\n × PT\n

    pi\n = partial pressure of the ith component\n

    xi\n = mole fraction of the ith component\n

    pT = total pressure of mixture\n

    $$ \\Rightarrow $$ 2 atm = $$\\left( {{{{n_{{H_2}}}} \\over {{n_{{H_2}}} + {n_{He}} + {n_{{O_2}}}}}} \\right)$$ $$ \\times $$ pT\n

    $$ \\Rightarrow $$ pT = $$\\left( {{{1 + 1 + 1} \\over 1}} \\right)$$ $$ \\times $$ 2 = 6 atm", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1591, "subject": "Chemistry", "question": "

    A mixture of hydrogen and oxygen contains $$40 \\%$$ hydrogen by mass when the pressure is $$2.2$$ bar. The partial pressure of hydrogen is bar. (Nearest Integer)

    ", "options": [], "answer": "2", "solution": "**Answer:** 2\n\n$$40 \\% \\,\\mathrm{w} / \\mathrm{w}$$ hydrogen gas is given in mixture of $$\\mathrm{H}_{2}$$ and oxygen.\n

    \nWt. of $$\\mathrm{H}_{2}=40 \\mathrm{~g}$$\n

    \nWt. of $$\\mathrm{O}_{2}=60 \\mathrm{~g}$$\n

    \n$$\\chi_{\\mathrm{H}_{2}}=\\frac{\\mathrm{n}_{\\mathrm{H}_{2}}}{\\mathrm{n}_{\\mathrm{H}_{2}}+\\mathrm{n}_{\\mathrm{O}_{2}}}$$\n

    \n$$=\\frac{\\frac{40}{2}}{\\frac{40}{2}+\\frac{60}{32}}$$\n

    \n$$=\\frac{20}{20+1.875}$$\n

    \n$$=\\frac{20}{21.875}=0.914$$\n

    \n$$\\mathrm{P}_{\\mathrm{H}_{2}}=\\chi_{\\mathrm{H}_{2}} \\times \\mathrm{P}_{\\mathrm{T}}$$\n

    \n$$=0.914 \\times 2.2$$\n

    \n$$=2.01 \\simeq 2 \\mathrm{bar}$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1592, "subject": "Chemistry", "question": "

    'x' g of molecular oxygen $$\\left(\\mathrm{O}_{2}\\right)$$ is mixed with $$200 \\mathrm{~g}$$ of neon (Ne). The total pressure of the non-reactive mixture of $$\\mathrm{O}_{2}$$ and Ne in the cylinder is 25 bar. The partial pressure of Ne is 20 bar at the same temperature and volume. The value of 'x' is _________.

    \n

    [Given: Molar mass of $$\\mathrm{O}_{2}=32 \\mathrm{~g} \\mathrm{~mol}^{-1}$$.

    \n

    Molar mass of $$\\mathrm{Ne}=20 \\mathrm{~g} \\mathrm{~mol}^{-1}$$]

    ", "options": [], "answer": "80", "solution": "**Answer:** 80\n\n$$P_{\\mathrm{Ne}}=\\left[P_{\\text {Total }}\\right] X_{\\mathrm{Ne}}$$\n

    $$20=25\\left[X_{N e}\\right]$$\n

    $${\\left[X_{N e}\\right]=\\frac{4}{5} }$$\n

    $$\\Rightarrow\\left[\\frac{\\frac{200}{20}}{\\frac{200}{20}+\\frac{x}{32}}\\right]=\\frac{4}{5}$$\n

    $$ \\Rightarrow $$$$\\frac{10}{10+\\left(\\frac{x}{32}\\right)}=\\frac{4}{5}$$\n

    $$ \\Rightarrow $$$$50=40+\\frac{x}{8}$$\n

    $$ \\Rightarrow $$$$400=320+x$$\n

    $$ \\Rightarrow $$$$x=80$$ gram", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1593, "subject": "Chemistry", "question": "

    The total pressure of a mixture of non-reacting gases $$\\mathrm{X}(0.6 \\mathrm{~g})$$ and $$\\mathrm{Y}(0.45 \\mathrm{~g})$$ in a vessel is $$740 \\mathrm{~mm}$$ of $$\\mathrm{Hg}$$.

    The partial pressure of the gas $$\\mathrm{X}$$ is _______ $$\\mathrm{mm}$$ of $$\\mathrm{Hg}$$. (Nearest Integer)

    \n

    (Given : molar mass $$\\mathrm{X}=20$$ and $$\\mathrm{Y}=45 \\mathrm{~g} \\mathrm{~mol}^{-1}$$)

    ", "options": [], "answer": "555", "solution": "**Answer:** 555\n\n$$\n\\begin{aligned}\n& P_{\\text {Total }}=740 \\mathrm{~mm} \\text { of } \\mathrm{Hg} \\\\\\\\\n& \\mathrm{P}_{\\mathrm{X}}= (\\text { mole fraction of } X) \\mathrm{P}_{\\text {Total }} \\\\\\\\\n& \\mathrm{n}_X=\\frac{0.6}{20}=0.03 \\\\\\\\\n& \\mathrm{n}_Y=\\frac{0.45}{45}=0.01 \\\\\\\\\n& \\text { Mole fraction of } X=\\frac{0.03}{0.01+0.03}=\\frac{3}{4} \\\\\\\\\n& \\text { Partial pressure of } X=\\frac{3}{4} \\times 740 \\\\\\\\\n& =555 \\mathrm{~mm} \\text { of } \\mathrm{Hg}\n\\end{aligned}\n$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1594, "subject": "Chemistry", "question": "A spherical balloon of radius 3 cm containing\nhelium gas has
    a pressure of 48 $$ \\times $$ 10–3 bar. At\nthe same temperature, the pressure, of a\nspherical balloon of radius 12 cm containing\nthe
    same amount of gas will be ________ $$ \\times $$ 10–6\nbar.", "options": [], "answer": "750", "solution": "**Answer:** 750\n\nFor gas when temperature and number of moles are constant then we can apply Boyle's law.\n

    According to Boyle's law,\n

    P1V1 = P2V2\n

    Given, P1 = 48 $$ \\times $$ 10–3 bar\n

    V1 = $${{4 \\over 3}\\pi {{\\left( 3 \\right)}^3}}$$\n

    V2 = $${{4 \\over 3}\\pi {{\\left( 12 \\right)}^3}}$$\n

    P2 = ?\n

    $$ \\therefore $$ P2 = $${{{P_1}{V_1}} \\over {{V_2}}}$$\n

    = $${{48 \\times {{10}^{ - 3}} \\times {{\\left( 3 \\right)}^3}} \\over {{{\\left( {12} \\right)}^3}}}$$\n

    = 7.5 $$ \\times $$ 10-4 = 750 $$ \\times $$ 10-6 bar", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1595, "subject": "Chemistry", "question": "A car tyre is filled with nitrogen gas at 35 psi at 27$$^\\circ$$C. It will burst if pressure exceeds 40 psi. The temperature in $$^\\circ$$C at which the car tyre will burst is _________. (Rounded off to the nearest integer)", "options": [], "answer": "70", "solution": "**Answer:** 70\n\n$${{{P_1}} \\over {{T_1}}} = {{{P_2}} \\over {{T_2}}}$$

    $${{35} \\over {300}} = {{40} \\over {{T_2}}}$$

    $${T_2} = {{40 \\times 300} \\over {35}}$$

    $$ = 342.86$$ K

    $$ = 69.85^\\circ $$ C $$ \\simeq 70^\\circ $$ C", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1596, "subject": "Chemistry", "question": "An LPG cylinder contains gas at a pressure of 300 kPa at 27$$^\\circ$$. The cylinder can withstand the pressure of 1.2 $$\\times$$ 106 Pa. The room in which the cylinder is kept catches fire. The minimum temperature at which the bursting of cylinder will take place is __________$$^\\circ$$ C. (Nearest integer)", "options": [], "answer": "927", "solution": "**Answer:** 927\n\n$${{{P_1}} \\over {{T_1}}} = {{{P_2}} \\over {{T_2}}} \\Rightarrow {{300 \\times {{10}^3}} \\over {300}} = {{1.2 \\times {{10}^6}} \\over {{T_2}}}$$

    $$ \\Rightarrow {T_2} = 1200$$ K

    $$ \\Rightarrow $$ $${T_2} = 927^\\circ $$ C", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1597, "subject": "Chemistry", "question": "

    The pressure of a moist gas at $$27^{\\circ} \\mathrm{C}$$ is $$4 \\mathrm{~atm}$$. The volume of the container is doubled at the same temperature. The new pressure of the moist gas is ________________ $$\\times 10^{-1} \\mathrm{~atm}$$. (Nearest integer)

    \n

    (Given : The vapour pressure of water at $$27^{\\circ} \\mathrm{C}$$ is $$0.4 \\mathrm{~atm} .$$ )

    ", "options": [], "answer": "22", "solution": "**Answer:** 22\n\n

    To solve this problem, we need to determine the new pressure of the moist gas when the volume of the container is doubled while maintaining the same temperature. First, we need to separate the contributions of the gas and the water vapor to the initial pressure and then apply Boyle's Law for the gas component only. The steps are as follows:

    \n\n

    1. Initial pressure of the moist gas (gas + water vapor) is given as $$4 \\, \\text{atm}$$.
    \n\n

    The vapour pressure of water at $$27^{\\circ} \\mathrm{C}$$ is $$0.4 \\, \\text{atm}$$.

    \n\n

    2. The partial pressure of the dry gas component can be calculated by subtracting the vapor pressure of water from the initial pressure:

    \n\n

    $$ P_{\\text{dry gas (initial)}} = 4 \\, \\text{atm} - 0.4 \\, \\text{atm} = 3.6 \\, \\text{atm} $$

    \n\n

    3. When the volume of the container is doubled, the pressure of the dry gas component will change according to Boyle's Law, which states:

    \n\n

    $$ P_1 V_1 = P_2 V_2 $$

    \n\n

    Since the temperature remains constant and the volume is doubled (i.e., $$V_2 = 2V_1$$), we can find the new pressure of the dry gas ($$P_{\\text{dry gas (new)}}$$) with the equation:

    \n\n

    $$ 3.6 \\, \\text{atm} \\times V_1 = P_{\\text{dry gas (new)}} \\times 2V_1 $$

    \n\n

    Solving for $$P_{\\text{dry gas (new)}}$$:

    \n\n

    $$ P_{\\text{dry gas (new)}} = \\frac{3.6 \\, \\text{atm}}{2} = 1.8 \\, \\text{atm} $$

    \n\n

    4. The vapor pressure of water remains unchanged at $$0.4 \\, \\text{atm}$$ because it depends only on the temperature, not on the volume.

    \n\n

    5. The new total pressure of the moist gas is the sum of the new pressure of the dry gas and the unchanged vapor pressure of water:

    \n\n

    $$ P_{\\text{moist gas (new)}} = P_{\\text{dry gas (new)}} + P_{\\text{water vapor}} = 1.8 \\, \\text{atm} + 0.4 \\, \\text{atm} = 2.2 \\, \\text{atm} $$

    \n\n

    6. Finally, as requested, we express the new pressure in the form of $$\\times 10^{-1} \\, \\text{atm}$$ and find the nearest integer:

    \n\n

    $$ 2.2 \\, \\text{atm} = 22 \\times 10^{-1} \\, \\text{atm} $$

    \n\n

    Therefore, the nearest integer is 22.

    ", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1598, "subject": "Chemistry", "question": "

    A $$300 \\mathrm{~mL}$$ bottle of soft drink has $$0.2 ~\\mathrm{M}~ \\mathrm{CO}_{2}$$ dissolved in it. Assuming $$\\mathrm{CO}_{2}$$ behaves as an ideal gas, the volume of the dissolved $$\\mathrm{CO}_{2}$$ at $$\\mathrm{STP}$$ is ____________ $$\\mathrm{mL}$$. (Nearest integer)

    \n

    Given : At STP, molar volume of an ideal gas is $$22.7 \\mathrm{~L} \\mathrm{~mol}^{-1}$$

    ", "options": [], "answer": "1362", "solution": "**Answer:** 1362\n\n

    Moles = 0.3 $$\\times$$ 0.2

    \n

    Volume at STP = $$0.3\\times0.2\\times22.7$$

    \n

    = 1.362 litre

    \n

    = 1362 mL

    ", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1599, "subject": "Chemistry", "question": "

    At constant temperature, a gas is at a pressure of 940.3 mm Hg. The pressure at which its volume decreases by 40% is __________ mm Hg. (Nearest integer)

    ", "options": [], "answer": "1567", "solution": "**Answer:** 1567\n\n

    To find the pressure at which the volume decreases by 40%, we can use Boyle's Law, which states that the pressure and volume of a gas are inversely proportional at constant temperature.

    \n

    Let's assume the initial volume of the gas is V. According to Boyle's Law:

    \n

    P₁V₁ = P₂V₂

    \n

    Where P₁ is the initial pressure (940.3 mm Hg), V₁ is the initial volume, P₂ is the final pressure (which we need to find), and V₂ is the final volume (which is 40% less than the initial volume, or 0.6V).

    \n

    Plugging in the values, we have:

    \n

    940.3 mm Hg × V₁ = P₂ × (0.6V)

    \n

    Now we can solve for P₂:

    \n

    P₂ = (940.3 mm Hg × V₁) / (0.6V)

    \n

    Since the question does not provide the initial volume (V₁), we cannot calculate the exact pressure. However, we can determine the pressure ratio when the volume decreases by 40%.

    \n

    P₂ = (940.3 mm Hg × V₁) / (0.6V)

    \n

    We can simplify this by canceling out V:

    \n

    P₂ = 940.3 mm Hg / 0.6

    \n

    P₂ ≈ 1567 mm Hg (nearest integer)

    \n

    Therefore, the pressure at which the volume decreases by 40% is approximately 1567 mm Hg.

    \n", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1600, "subject": "Chemistry", "question": "Value of gas constant R is", "options": [ { "text": "0.082 litre atm" }, { "text": "0.987 cal mol-1 K-1" }, { "text": "8.3 J mol-1 K-1" }, { "text": "83 erg mol-1 K-1" } ], "answer": "8.3 J mol-1 K-1", "solution": "**Answer:** 8.3 J mol-1 K-1\n\nValue of gas constant\n

    $$\\left( R \\right) = 0.0821L\\,atm\\,{K^{ - 1}}\\,mo{l^{ - 1}}$$\n

    $$ = 8.314 \\times {10^7}\\,\\,ergs\\,{K^{ - 1}}\\,mo{l^{ - 1}}$$\n

    $$ = 8.314J{K^{ - 1}}\\,mo{l^{ - 1}}$$\n

    $$ = 1.987\\,cal\\,{K^{ - 1}}\\,mo{l^{ - 1}}$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1601, "subject": "Chemistry", "question": "For an ideal gas, number of moles per litre in terms of its pressure P, gas constant R and temperature T is", "options": [ { "text": "PT/R" }, { "text": "PRT" }, { "text": "P/RT" }, { "text": "RT/P" } ], "answer": "P/RT", "solution": "**Answer:** P/RT\n\n$$PV = nRT\\,\\,$$ (number of moles $$ = n/V$$) \n

    $$\\therefore$$ $$\\,\\,\\,n/V = P/RT.$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1602, "subject": "Chemistry", "question": "If 10–4 dm3 of water is introduced into a 1.0 dm3 flask to 300 K, how many moles of water are in the vapour phase when equilibrium is established ?
    \n(Given : Vapour pressure of H2O at 300 K is 3170 Pa ; R = 8.314 J K–1 mol–1)", "options": [ { "text": "5.56 x 10–3 mol" }, { "text": "1.53 x 10–2 mol" }, { "text": "4.46 x 10–2 mol" }, { "text": "1.27 x 10–3 mol" } ], "answer": "1.27 x 10–3 mol", "solution": "**Answer:** 1.27 x 10–3 mol\n\nFrom the ideal gas equation : \n

    $$PV = nRT\\,\\,$$ \n

    or $$\\,\\,\\,n = {{PV} \\over {RT}} = {{3170 \\times {{10}^{ - 3}}} \\over {8.314 \\times 300}}$$ \n

    $$ = 1.27 \\times {10^{ - 3}}$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1603, "subject": "Chemistry", "question": "At 300 K, the density of a certain gaseous molecule at 2 bar is double to that of\ndinitrogen (N2 ) at 4 bar. The molar mass of gaseous molecule is :", "options": [ { "text": "28 g mol$$-$$1" }, { "text": "56 g mol$$-$$1" }, { "text": "112 g mol$$-$$1" }, { "text": "224 g mol$$-$$1" } ], "answer": "112 g mol$$-$$1", "solution": "**Answer:** 112 g mol$$-$$1\n\nDensity = $${{Mass} \\over {Volume}}$$\n

    PV = RT $$ \\Rightarrow $$ V = $${{RT} \\over P}$$\n

    So, Density(d) = $${{MP} \\over {RT}}$$\n

    Now, d1\n = x, P1\n = 4, M1\n = 28, d2\n = 2x, P2\n = 2, M2\n = ?\n

    $$ \\therefore $$ $${{{d_1}} \\over {{d_2}}} = {{{M_1}{P_1}} \\over {R{T_1}}} \\times {{R{T_2}} \\over {{M_2}{P_2}}} = {{{M_1}{P_1}} \\over {{M_2}{P_2}}}$$ [As T1 = T2 ]\n

    $$ \\Rightarrow $$ M2 = $${{{M_1}{P_1}{d_2}} \\over {{d_1}{P_2}}}$$\n

    = $${{2x \\times 28 \\times 4} \\over {2 \\times x}}$$ = 112 g mol-1", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1604, "subject": "Chemistry", "question": "Assuming ideal gas behaviour, the ratio of density of ammonia to that of hydrogen chloride at same temperature and pressure is : (Atomic wt. of Cl = 35.5 u) ", "options": [ { "text": "1.46" }, { "text": "0.46" }, { "text": "1.64" }, { "text": "0.64" } ], "answer": "0.46", "solution": "**Answer:** 0.46\n\nWe know, PV = nRT\n

    n = no. of moles = $${m \\over M}$$\n

    So,   PV = $${m \\over M}RT$$\n

    $$ \\Rightarrow $$$$\\,\\,\\,\\,$$ P = $${m \\over V} \\times {{RT} \\over M}$$\n

    $$ \\Rightarrow $$$$\\,\\,\\,\\,$$ P = d $$ \\times $$ $${{RT} \\over M}$$ [ d = density = $${m \\over V}$$ ]\n

    at constant temperature and pressure d $$ \\propto $$ M\n

    Now let d1 and d2 are the density of ammonia and HCl.\n

    $$\\therefore\\,\\,\\,\\,$$ $${{{d_1}} \\over {{d_2}}} = {{{M_{N{H_3}}}} \\over {{M_{HCl}}}}$$\n

    $$ \\Rightarrow $$ $$\\,\\,\\,\\,$$ $${{{d_1}} \\over {{d_2}}}$$ = $${{17} \\over {36.5}}$$ = 0.46", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1605, "subject": "Chemistry", "question": "0.5 moles of gas A and x moles of gas B exert a pressure of 200 Pa in a container of volume 10 m3 at 1000 K. Given R is the gas constant in JK$$-$$1mol$$-$$1, x is : ", "options": [ { "text": "$${{2R} \\over {4 + R}}$$" }, { "text": "$${{2R} \\over {4 - R}}$$" }, { "text": "$${{4 + R} \\over {2R}}$$" }, { "text": "$${{4 - R} \\over {2R}}$$" } ], "answer": "$${{4 - R} \\over {2R}}$$", "solution": "**Answer:** $${{4 - R} \\over {2R}}$$\n\nWe know, \n

    PV = nRT\n

    Given, \n

    P = 200 Pa\n

    V = 10 m3\n

    T = 1000 K\n

    n = 0.5 + x\n

    $$ \\therefore $$   200 $$ \\times $$ 10 = (0.5 + x) R $$ \\times $$ 1000\n

    $$ \\Rightarrow $$    0.5 + x = $${2 \\over R}$$\n

    $$ \\Rightarrow $$    x = $${2 \\over R} - {1 \\over 2}$$\n

    $$ \\Rightarrow $$   x = $${{4 - R} \\over {2R}}$$", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1606, "subject": "Chemistry", "question": "An open vessel at 27oC is heated until two fifth of the air (assumed as an ideal gas) it has escaped from the vessel assuming that the volume of the vessel remains constant, the temperature at which the vessel has been\nheated is - ", "options": [ { "text": "750 K" }, { "text": "500oC" }, { "text": "750oC" }, { "text": "500 K" } ], "answer": "500 K", "solution": "**Answer:** 500 K\n\nWe know,\n

    PV = nRT\n

    As the vessel is open, so the pressure and volume is constant.\n

    $$ \\therefore $$   n1 R T1 = n2 R T2\n

    $$ \\Rightarrow $$   n1 T1 = n2 T1\n

    $${2 \\over 5}$$th of the air escape from the vessel. So the remaining air is $${3 \\over 5}$$ of the total air.\n

    $$ \\therefore $$   n $$ \\times $$ 300 = $${3 \\over 5}$$n $$ \\times $$ T2\n

    $$ \\Rightarrow $$   T2 = 500 K", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1607, "subject": "Chemistry", "question": "NaClO3 is used, even in spacecrafts, to produce\nO2. The daily consumption of pure O2 by a\nperson is 492L at 1 atm, 300K. How much\namount of NaClO3, in grams, is required to\nproduce O2 for the daily consumption of a\nperson at 1 atm, 300 K?
    \nNaClO3(s) + Fe(s) $$ \\to $$ O2(g) + NaCl(s) + FeO(s)
    \nR = 0.082 L atm mol–1 K–1", "options": [], "answer": "2120TO2140", "solution": "**Answer:** 2120TO2140\n\nNaClO3(s) + Fe(s) $$ \\to $$ NaCl(s) + FeO(s) + O2(g)\n

    moles of NaClO3 = moles of O2\n

    moles of O2 = $${{PV} \\over {RT}}$$ = $${{1 \\times 492} \\over {0.082 \\times 300}}$$ = 20 mol\n

    mass of NaClO3 = 20 $$ \\times $$ 106.5 = 2130 g", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1608, "subject": "Chemistry", "question": "The volume occupied by 4.75 g of acetylene gas at 50$$^\\circ$$C and 740 mmHg pressure is __________ L. (Rounded off to the nearest integer)

    [Given R = 0.0826 L atm K$$-$$1 mol$$-$$1]", "options": [], "answer": "5", "solution": "**Answer:** 5\n\nGiven, mass of C2H2(g) = 4.75 g

    Molecular weight = 26 g/mol

    Temperature = 50 + 273 = 323 K

    Pressure = 740 torr/mm of Hg

    Pressure = $${{740} \\over {760}}$$ atm

    R = 0.0821 L atm mol$$-$$1 K$$-$$1

    Hence, no. of mole n = $${{4.75} \\over {26}}$$ mol

    Formula used, pV = nRT (ideal gas)

    $$ \\Rightarrow V = {{nRT} \\over p} = {{4.75} \\over {26}} \\times {{0.0821 \\times 323} \\over {(740/760)}}$$

    $$ = {{96314.078} \\over {19240}} = 5.0059$$ L = 5 L", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1609, "subject": "Chemistry", "question": "3.12 g of oxygen is adsorbed on 1.2 g of platinum metal. The volume of oxygen adsorbed per gram of the adsorbent at 1 atm and 300 K in L is __________.

    [R = 0.0821 L atm K$$-$$1mol$$-$$1]", "options": [], "answer": "2", "solution": "**Answer:** 2\n\nMoles of O2 = $${{3.12} \\over {32}}$$ = 0.0975\n

    Using ideal gas equation : PV = nRT\n

    V = $${{0.0975 \\times 0.082 \\times 300} \\over 1}$$ = 2.4 L\n

    $$ \\therefore $$ Volume of O2(g) adsorbed per gram of the\n

    adsorbent = $${{2.4} \\over {1.2}}$$ = 2", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1610, "subject": "Chemistry", "question": "The pressure exerted by a non-reactive gaseous mixture of 6.4g of methane and 8.8 g of carbon dioxide in a 10L vessel at 27$$^\\circ$$C is __________ kPa. (Round off to the Nearest Integer).

    [Assume gases are ideal, R = 8.314 J mol$$-$$1 K$$-$$1

    Atomic masses : C : 12.0 u, H : 1.0 u, O : 16.0 u]", "options": [], "answer": "150", "solution": "**Answer:** 150\n\nV = 10 L, T = 27$$^\\circ$$ C = 300 K

    (m)methane = 6.4 g, (m)CO2 = 8.8 g

    PV = ntotalRT

    P $$\\times$$ 10 $$\\times$$ 10$$-$$3 = $$\\left( {{{6.4} \\over {16}} + {{8.8} \\over {44}}} \\right)$$ $$\\times$$ 8.314 $$\\times$$ 300

    P $$\\times$$ 10$$-$$2 = (0.4 + 0.2) $$\\times$$ 8.314 $$\\times$$ 300

    P = 149652 Pa

    P = 149.652 KPa $$ \\approx $$ 150 kPa", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1611, "subject": "Chemistry", "question": "The number of chlorine atoms in 20 mL of chlorine gas at STP is _________ 1021. (Round off to the Nearest Integer).

    [Assume chlorine is an ideal gas at STP

    R = 0.083 L bar mol$$-$$1 K$$-$$1, NA = 6.023 $$\\times$$ 1023]", "options": [], "answer": "1", "solution": "**Answer:** 1\n\n$$n = {{PV} \\over {RT}}$$

    $$ = {{1 \\times 20 \\times {{10}^{ - 3}}} \\over {0.083 \\times 273}}$$

    No. of atoms = $$ = {{1 \\times 20 \\times {{10}^{ - 3}}} \\over {0.083 \\times 273}} \\times 2 \\times 6.023 \\times {10^{23}}$$

    $$ = 1.06 \\times {10^{21}}$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1612, "subject": "Chemistry", "question": "A home owner uses 4.00 $$\\times$$ 103 m3 of methane (CH4) gas, (assume CH4 is an ideal gas) in a year to heat his home. Under the pressure of 1.0 atm and 300 K, mass of gas used is x $$\\times$$ 105 g. The value of x is __________. (Nearest integer)

    (Given R = 0.083 L atm K$$-$$1 mol$$-$$1)", "options": [], "answer": "26", "solution": "**Answer:** 26\n\n$$n(C{H_4}) = {{PV} \\over {RT}}$$

    $$ = {{1 \\times 4 \\times {{10}^3} \\times 1000} \\over {0.083 \\times 300}}$$

    Weight of CH4

    $$ = {{40 \\times 16 \\times {{10}^5}} \\over {0.083 \\times 300}}$$ gm

    $$ = 25.7 \\times {10^5}$$ gm", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1613, "subject": "Chemistry", "question": "An empty LPG cylinder weighs 14.8 kg. When full, it weighs 29.0 kg and shows a pressure of 3.47 atm. In the course of use at ambient temperature, the mass of the cylinder is reduced to 23.0 kg. The final pressure inside of the cylinder is _________ atm. (Nearest integer)

    (Assume LPG of be an ideal gas)", "options": [], "answer": "2", "solution": "**Answer:** 2\n\nWeight of empty LPG cylinder = 14.8 kg

    Weight of full LPG cylinder = 29 kg

    $$\\therefore$$ Weight of gas = 29 $$-$$ 14.8 = 14.2 kg

    If weight of full LPG cylinder = 23 kg

    then weight of gas used = 29 $$-$$ 23 = 6 kg at ambient temperature.

    From ideal gas equation, pV = nRT

    or $$pV = {{Weight\\,of\\,solute} \\over {Molecular\\,mass\\,of\\,solute}} \\times RT$$

    or $$pV = {W \\over M} \\times RT$$

    Applying ideal gas to LPG cylinder when gas is full,

    $$pV = nRT$$

    $$3.47\\,atm \\times V = {{14.2kg} \\over M} \\times RT$$ .... (i)

    Applying ideal gas to LPG cylinder when gas is reduced to 23 kg at ambient temperture,

    $$pV = nRT$$

    $$p \\times V = {{8.2kg} \\over M} \\times RT$$ .... (ii)

    Divide Eq. (i) by (ii)

    $${{3.47 \\times V} \\over {p \\times V}} = {{{{14.2kg\\,RT} \\over M}} \\over {{{8.2kg\\, \\times RT} \\over M}}}$$

    $${{3.47} \\over p} = {{14.2} \\over {8.2}}$$

    $$ \\Rightarrow p = {{3.47 \\times 41} \\over {71}} = 2.003$$ atm

    Hence, answer is 2.", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1614, "subject": "Chemistry", "question": "

    Geraniol, a volatile organic compound, is a component of rose oil. The density of the vapour is 0.46 gL$$-$$1 at 257$$^\\circ$$C and 100 mm Hg. The molar mass of geraniol is ____________ g mol$$-$$1. (Nearest Integer)

    \n

    [Given : R = 0.082 L atm K$$-$$1 mol$$-$$1]

    ", "options": [], "answer": "152", "solution": "**Answer:** 152\n\n

    From ideal gas equation we know

    \n

    PV = nRT

    \n

    $$ \\Rightarrow PV = {W \\over M}RT$$

    \n

    $$ \\Rightarrow P = {W \\over V}\\,.\\,{{RT} \\over M}$$

    \n

    $$ \\Rightarrow P = d\\,.\\,{{RT} \\over M}$$ [$$\\because$$ $$d = {W \\over V}$$]

    \n

    We know, 760 mm of Hg = 1 atm

    \n

    $$\\therefore$$ 100 mm of Hg = $${{100} \\over {760}}$$ atm

    \n

    $$\\therefore$$ Pressure (P) = $${{100} \\over {760}}$$ atm

    \n

    Density (d) = 0.46

    \n

    R = 0.082 L atm K$$-$$1 mol$$-$$1

    \n

    T = (257 + 273) K = 530 K

    \n

    Putting the values in above equation, we get

    \n

    $${{100} \\over {760}} = {{0.46 \\times 0.082 \\times 530} \\over M}$$

    \n

    $$\\Rightarrow$$ M = 152

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1615, "subject": "Chemistry", "question": "

    100 g of an ideal gas is kept in a cylinder of 416 L volume at 27$$^\\circ$$C under 1.5 bar pressure. The molar mass of the gas is __________ g mol$$-$$1. (Nearest integer)

    \n

    (Given : R = 0.083 L bar K$$-$$1 mol$$-$$1)

    ", "options": [], "answer": "4", "solution": "**Answer:** 4\n\n

    Given, Mass of ideal gas = 100 gm

    \n

    Let the molar mass of ideal gas = M

    \n

    $$\\therefore$$ Number of moles of gas (n) = $${{100} \\over M}$$

    \n

    Volume of cylinder (V) = 416 L

    \n

    Temperature (T) = (27 + 273)K = 300 K

    \n

    Pressure (P) = 1.5 bar

    \n

    R = 0.083 L bar K$$-$$1 mol$$-$$1

    \n

    Using ideal gas equation,

    \n

    PV = nRT

    \n

    $$ \\Rightarrow 1.5 \\times 416 = {{100} \\over M} \\times 0.083 \\times 300$$

    \n

    $$\\Rightarrow$$ M = 4

    ", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1616, "subject": "Chemistry", "question": "

    An evacuated glass vessel weighs 40.0 g when empty, 135.0 g when filled with a liquid of density 0.95 g mL$$-$$1 and 40.5 g when filled with an ideal gas at 0.82 atm at 250 K. The molar mass of the gas in g mol$$-$$1 is :

    \n

    (Given : R = 0.082 L atm K$$-$$1 mol$$-$$1)

    ", "options": [ { "text": "35" }, { "text": "50" }, { "text": "75" }, { "text": "125" } ], "answer": "125", "solution": "**Answer:** 125\n\n

    Weight of empty glass vessel = 40 gm

    \n

    Weight of glass vessel filled with liquid = 135 gm

    \n

    $$\\therefore$$ Weight of liquid = 135 $$-$$ 40 = 95 gm

    \n

    Given density of liquid = 0.95 gm ml$$-$$1

    \n

    $$\\therefore$$ Volume of liquid $$={{95} \\over {0.95}} = 100$$ ml

    \n

    Weight of glass filled with ideal gas = 40.5 gm

    \n

    $$\\therefore$$ Weight of gas = 40.5 $$-$$ 40 = 0.5 g

    \n

    Let the Molar mass = M

    \n

    $$\\therefore$$ Moles of gas $$ = {{0.5} \\over M}$$

    \n

    $$\\therefore$$ Now applying ideal gas equation,

    \n

    pV = nRT

    \n

    $$ \\Rightarrow 0.82 \\times {{100} \\over {1000}} = {{0.5} \\over M} \\times 0.082 \\times 250$$

    \n

    $$\\Rightarrow$$ M = 0.5 $$\\times$$ 250 = 125 g/mol

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1617, "subject": "Chemistry", "question": "

    2.0 g of H2 gas is adsorbed on 2.5 g of platinum powder at 300 K and 1 bar pressure. The volume of the gas adsorbed per gram of the adsorbent is __________ mL.

    \n

    (Given : R = 0.083 L bar K$$-$$1 mol$$-$$1)

    ", "options": [], "answer": "9960", "solution": "**Answer:** 9960\n\n$\\mathrm{PV}=\\mathrm{nR} T$\n

    \n$$\n\\mathrm{V}=\\frac{2 \\times 0.083 \\times 300}{2 \\times 1}=24.9 \\text { litre }\n$$\n

    \n$\\therefore$ Volume of the gas adsorbed per gram of the adsorbent\n

    \n$$\n\\begin{aligned}\n&=\\frac{24.9}{2.5}=9.96 \\mathrm{~L} \\\\\\\\\n&=9960 ~\\mathrm{ml}\n\\end{aligned}\n$$", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1618, "subject": "Chemistry", "question": "

    A rigid nitrogen tank stored inside a laboratory has a pressure of 30 atm at 06:00 am when the temperature is 27$$^\\circ$$C. At 03:00 pm, when the temperature is 45$$^\\circ$$, the pressure in the tank will be _________ atm. [nearest integer]

    ", "options": [], "answer": "32", "solution": "**Answer:** 32\n\n

    A nitrogen tank of fixed volume used where number of moles of nitrogen is fixed.

    \n

    $$\\therefore$$ V = constant

    \n

    n = constant

    \n

    R = constant

    \n

    From ideal gas equation,

    \n

    PV = nRT

    \n

    $$\\Rightarrow$$ P $$\\propto$$ T [As V, n, R = constant]

    \n

    Here, initially P1 = 30 atm, T1 = 300 K

    \n

    Finally, P2 = ?, T2 = 318 K

    \n

    $$\\therefore$$ $${{{P_1}} \\over {{P_2}}} = {{{T_1}} \\over {{T_2}}}$$

    \n

    $$ \\Rightarrow {{30} \\over {{P_2}}} = {{300} \\over {318}}$$

    \n

    $$ \\Rightarrow {P_2} = 31.8 \\simeq 32$$

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1619, "subject": "Chemistry", "question": "

    At 300 K, a sample of 3.0 g of gas A occupies the same volume as 0.2 g of hydrogen at 200 K at the same pressure. The molar mass of gas A is ____________ g mol$$-$$1. (nearest integer) Assume that the behaviour of gases as ideal.

    \n

    (Given : The molar mass of hydrogen (H2) gas is 2.0 g mol$$-$$1.)

    ", "options": [], "answer": "45", "solution": "**Answer:** 45\n\n

    Both gas A and Hydrogen (H2) gas have same volume at same pressure. Let both 's volume is V and pressure P.

    \n

    For gas A :

    \n

    Pressure = P

    \n

    Temperature (T) = 300 K

    \n

    Volume = V

    \n

    Mass = 3 g

    \n

    Molar mass = M gm/mol

    \n

    using ideal gas equation,

    \n

    PV = nRT

    \n

    $$ \\Rightarrow PV = {3 \\over M} \\times R \\times 300$$ ..... (1)

    \n

    For Hydrogen,

    \n

    Pressure = P

    \n

    Temperature (T) = 200 K

    \n

    Volume = V

    \n

    Mass = 0.2 g

    \n

    Molar mass = 2 gm/mol

    \n

    Using ideal gas equation,

    \n

    $$PV = {{0.2} \\over 2} \\times R \\times 200$$ ...... (2)

    \n

    From (1) and (2), we get

    \n

    $${3 \\over M} \\times R \\times 300 = {{0.2} \\over 2} \\times R \\times 200$$

    \n

    $$\\Rightarrow$$ M = 45

    ", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1620, "subject": "Chemistry", "question": "

    A sealed flask with a capacity of 2 dm3 contains 11 g of propane gas. The flask is so weak that it will burst if the pressure becomes 2 MPa. The minimum temperature at which the flask will burst is ___________ $$^\\circ$$C. [Nearest integer]

    \n

    (Given : R = 8.3 J K$$-$$1 mol$$-$$1 , Atomic masses of C and H are 12u and 1u, respectively.) (Assume that propane behaves as an ideal gas.)

    ", "options": [], "answer": "1655", "solution": "**Answer:** 1655\n\nFrom ideal gas equation,\n

    \n$\\mathrm{PV}=\\mathrm{nRT}$\n

    \n$P=2 \\times 10^{6} \\mathrm{~Pa}$\n

    \n$V=2 \\mathrm{dm}^{3}=2 \\times 10^{-3} \\mathrm{~m}^{3}$\n

    \n$\\mathrm{R}=8.3 \\mathrm{JK}^{-1} \\mathrm{~mol}^{-1}$\n

    \n$\\mathrm{n}=\\frac{11}{44} \\mathrm{~mol}$\n

    \n$2 \\times 10^{6} \\times 2 \\times 10^{-3}=\\frac{11}{44} \\times 8.3 \\times \\mathrm{T}$\n

    \n$\\mathrm{T}=1927.7 \\mathrm{~K}$\n

    \n$\\mathrm{T}\\left(\\right.$ in $\\left.^{\\circ} \\mathrm{C}\\right)=1927.7-273 \\simeq 1655^{\\circ} \\mathrm{C}$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1621, "subject": "Chemistry", "question": "

    A $$10 \\mathrm{~g}$$ mixture of hydrogen and helium is contained in a vessel of capacity $$0.0125 \\mathrm{~m}^{3}$$ at 6 bar and $$27^{\\circ} \\mathrm{C}$$. The mass of helium in the mixture is ____________ g. (nearest integer)

    \n

    Given: $$\\mathrm{R}=8.3 \\mathrm{~J} \\mathrm{~K}^{-1} \\mathrm{~mol}^{-1}$$

    \n

    (Atomic masses of $$\\mathrm{H}$$ and $$\\mathrm{He}$$ are $$1 \\mathrm{u}$$ and $$4 \\mathrm{u}$$, respectively)

    ", "options": [], "answer": "8", "solution": "**Answer:** 8\n\nNumber of moles of mixture of $\\mathrm{H}_{2}$ and $\\mathrm{He}$\n

    \n$$\n\\begin{aligned}\n&=\\frac{P V}{R T} \\\\\n&=\\frac{6 \\times 10^{5} \\times 0.0125}{8.3 \\times 300}=3\n\\end{aligned}\n$$\n

    \nLet the mass of He in $10 \\mathrm{~g}$ mixture be $\\mathrm{xg}$\n

    \n$\\therefore \\frac{x}{4}+\\frac{10-x}{2}=3$\n

    \nOn solving $x=8 \\mathrm{~g}$\n

    \n$\\therefore$ Mass of $\\mathrm{He}$ in the mixture $=8 \\mathrm{~g}$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1622, "subject": "Chemistry", "question": "Kinetic theory of gases proves", "options": [ { "text": "only Boyle's law" }, { "text": "only Charle's law" }, { "text": "only Avogadro's law" }, { "text": "All of these" } ], "answer": "All of these", "solution": "**Answer:** All of these\n\nKinetic theory of gases proves all the given gas laws.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1623, "subject": "Chemistry", "question": "According to the kinetic theory of gases, in an ideal gas, between two successive collisions of a gas molecule travels", "options": [ { "text": "in a wavy path" }, { "text": "in a straight line path" }, { "text": "with an accelerated velocity" }, { "text": "in a circular path" } ], "answer": "in a straight line path", "solution": "**Answer:** in a straight line path\n\nAccording to kinetic theory the gas molecules are in a state of constant rapid motion in all possible directions colloiding in a random manner with one another and with the walls of the container and between two successive collisions molecules travel in a straight line path but show haphazard motion due to collisions.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1624, "subject": "Chemistry", "question": "As the temperature is raised from 20°C to 40°C, the average kinetic energy of neon atoms\nchanges by a factor of which of the following? ", "options": [ { "text": "1/2" }, { "text": "2" }, { "text": "$$313\\over293$$" }, { "text": "$$\\sqrt{313\\over293}$$" } ], "answer": "$$313\\over293$$", "solution": "**Answer:** $$313\\over293$$\n\n$${{K.E\\,\\,of\\,\\,neon\\,\\,at\\,\\,{{40}^ \\circ }C} \\over {K.E\\,\\,of\\,\\,neon\\,\\,at\\,\\,{{20}^ \\circ }C}}$$\n

    $$ = {{{3 \\over 2}K \\times 313} \\over {{3 \\over 2}K \\times 293}}$$\n

    $$ = {{313} \\over {293}}$$", "topic": "Number Theory", "subtopic": "Divisibility / Factorization" }, { "id": 1625, "subject": "Chemistry", "question": "Which one of the following statements is NOT true about the effect of an increase in\ntemperature on the distribution of molecular speeds in a gas? ", "options": [ { "text": "The most probable speed increases" }, { "text": "The fraction of the molecules with the most probable speed increases " }, { "text": "The distribution becomes broader" }, { "text": "The area under the distribution curve remains the same as under the lower temperature " } ], "answer": "The fraction of the molecules with the most probable speed increases ", "solution": "**Answer:** The fraction of the molecules with the most probable speed increases \n\nDistribution of molecular velocities at two different temperature is given shown below. \n

    \"AIEEE\n

    NOTE : At higher temperature more molecules have higher velocities and less molecules have lower velocities. \n

    As evident from fig. thus it is clear that With the increase in temperature the most probable velocity increase but the fraction of such molecules decreases.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1626, "subject": "Chemistry", "question": "For gaseous state, if most probable speed is denoted by C*, average speed by $$\\mathop C\\limits^{\\_\\_} $$ and mean square speed by\nC, then for a large number of molecules the ratios of these speeds are:", "options": [ { "text": "C*: $$\\mathop C\\limits^{\\_\\_} $$ : C = 1.128 : 1.225 : 1" }, { "text": "C*: $$\\mathop C\\limits^{\\_\\_} $$ : C = 1.225 : 1.128 : 1" }, { "text": "C*: $$\\mathop C\\limits^{\\_\\_} $$ : C = 1 : 1.225 : 1.128" }, { "text": "C*: $$\\mathop C\\limits^{\\_\\_} $$ : C = 1 : 1.128 : 1.225 " } ], "answer": "C*: $$\\mathop C\\limits^{\\_\\_} $$ : C = 1 : 1.128 : 1.225 ", "solution": "**Answer:** C*: $$\\mathop C\\limits^{\\_\\_} $$ : C = 1 : 1.128 : 1.225 \n\nMost probable speed $$\\left( {{C^ * }} \\right) = \\sqrt {{{2RT} \\over M}} $$\n

    Average Speed $$\\left( {\\overline C } \\right) = \\sqrt {{{8RT} \\over {\\pi M}}} $$\n

    Root mean square velocity $$\\left( c \\right) = \\sqrt {{{3RT} \\over M}} $$\n

    $${C^ * }:\\overline C :C = \\sqrt {{{2RT} \\over M}} :\\sqrt {{{8RT} \\over {\\pi M}}} :\\sqrt {{{3RT} \\over M}} $$\n

    $$ = 1:\\sqrt {{4 \\over \\pi }} :\\sqrt {{3 \\over 2}} $$\n

    $$ = 1:1.128:1.225$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1627, "subject": "Chemistry", "question": "Initially, the root mean square (rms) velocity of N2 molecules at certain temperature is $$u$$. If this temperature is doubled and all the nitrogen molecules dissociate into nitrogen atoms, then the new rms velocity will be :\n", "options": [ { "text": "u/2" }, { "text": "2u" }, { "text": "4u" }, { "text": "14u" } ], "answer": "2u", "solution": "**Answer:** 2u\n\n

    Root mean square (rms) velocity of N2 molecule (u) = $$\\sqrt {{{3RT} \\over M}} $$.

    \n

    For N2 molecule, molecular mass, M = 28. Therefore,

    \n

    $$u = \\sqrt {{{3RT} \\over {28}}} $$ ..... (1)

    \n

    When the temperature becomes doubled, that is, 2T. Let the new root mean square (rms) velocity of N-atom be u'. For N-atom, the molecular mass, M = 14. Therefore,

    \n

    $$u' = \\sqrt {{{3RT} \\over {14}}} $$....... (2)

    \n

    On dividing Eq. (1) by Eq. (2), we get

    \n

    $${u \\over {u'}} = \\sqrt {{1 \\over 4}} $$

    \n

    $$ = {1 \\over 2} \\Rightarrow u' = 2u$$

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1628, "subject": "Chemistry", "question": "

    At $$600 \\mathrm{~K}$$, the root mean square (rms) speed of gas $$\\mathrm{X}$$ (molar mass $$=40$$ ) is equal to the most probable speed of gas $$\\mathrm{Y}$$ at $$90 \\mathrm{~K}$$. The molar mass of the gas $$\\mathrm{Y}$$ is ___________ $$\\mathrm{g} ~\\mathrm{mol}^{-1}$$. (Nearest integer)

    ", "options": [], "answer": "4", "solution": "**Answer:** 4\n\n

    The root mean square speed ($v_{rms}$) of a gas can be calculated using the following formula:

    \n

    $v_{rms} = \\sqrt{\\frac{3kT}{M}}$

    \n

    where:

    \n\n

    The most probable speed ($v_{mp}$) of a gas can be calculated using the following formula:

    \n

    $v_{mp} = \\sqrt{\\frac{2kT}{M}}$

    \n

    Given that the root mean square speed of gas X at 600 K is equal to the most probable speed of gas Y at 90 K, we can set the two equations equal to each other and solve for the molar mass of gas Y:

    \n

    $\\sqrt{\\frac{3kT_X}{M_X}} = \\sqrt{\\frac{2kT_Y}{M_Y}}$

    \n

    Square both sides to eliminate the square root:

    \n

    $\\frac{3kT_X}{M_X} = \\frac{2kT_Y}{M_Y}$

    \n

    We are asked to find $M_Y$, so let's rearrange the equation to solve for $M_Y$:

    \n

    $M_Y = \\frac{2kT_YM_X}{3kT_X}$

    \n

    We know that $T_X = 600$ K, $M_X = 40$ g/mol $= 40 \\times 10^{-3}$ kg/mol (converted from grams to kilograms), $T_Y = 90$ K, and $k = 1.38 \\times 10^{-23}$ J/K. Substituting these values into the equation gives:

    \n

    $M_Y = \\frac{2 \\times 1.38 \\times 10^{-23} \\text{J/K} \\times 90 \\text{K} \\times 40 \\times 10^{-3} \\text{kg/mol}}{3 \\times 1.38 \\times 10^{-23} \\text{J/K} \\times 600 \\text{K}}$

    \n

    Solving this equation gives:

    \n

    $M_Y = 0.004 \\text{kg/mol}$

    \n

    To convert this value back into grams per mole, multiply by 1000:

    \n

    $M_Y = 4 \\text{g/mol}$

    \n

    So, the molar mass of gas Y is approximately 4 g/mol.

    \n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1629, "subject": "Chemistry", "question": "In Van der Waals equation of state of the gas law, the constant ‘b’ is a measure of ", "options": [ { "text": "intermolecular repulsions" }, { "text": "intermolecular collisions per unit volume " }, { "text": "Volume occupied by the molecules " }, { "text": "intermolecular attraction" } ], "answer": "Volume occupied by the molecules ", "solution": "**Answer:** Volume occupied by the molecules \n\nIn van der Waals equation $$'b'$$ is for volume correction.", "topic": "Algebra", "subtopic": "Equations and Inequalities" }, { "id": 1630, "subject": "Chemistry", "question": "'a’ and `b’ are van der Waals’ constants for gases. Chlorine is more easily liquefied than ethane because", "options": [ { "text": "a and b for Cl2 < a and b for C2H6" }, { "text": "a and b for Cl2 > a and b for C2H6" }, { "text": "a for Cl2 > a for C2H6 and b Cl2 < b for C2H6" }, { "text": "a for Cl2 < a for C2H6 and b Cl2 > b for C2H6" } ], "answer": "a for Cl2 > a for C2H6 and b Cl2 < b for C2H6", "solution": "**Answer:** a for Cl2 > a for C2H6 and b Cl2 < b for C2H6\n\nThe value of $$a$$ is a measure of the magnitude of the attractive forces between the molecules of the gas. Greater the value of $$'a',$$ larger is the attractive inter-molecular force between the gas molecules. \n

    The value of $$b$$ related to the effective size of the gas molecules. It is also termed as excluded volume. \n

    The gases with higher value of $$a$$ and lower value of $$b$$ are more liquefiable, hence for $$C{{\\rm l}_2}$$ $$''a'''$$ should be greater than for $${C_2}{H_6}$$ but for it $$b$$ should be less than for $${C_2}{H_6}.$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1631, "subject": "Chemistry", "question": "The compressibility factor for a real gas at high pressure is :", "options": [ { "text": "1 + RT/pb" }, { "text": "1" }, { "text": "1 + pb/RT" }, { "text": "1–pb/RT" } ], "answer": "1 + pb/RT", "solution": "**Answer:** 1 + pb/RT\n\n$$\\left( {P + {a \\over {{V^2}}}} \\right)\\left( {V - b} \\right) = RT\\,\\,$$ \n

    at high pressure $${a \\over {{V^2}}}$$ can be neglected\n

    $$PV - Pb = RT\\,\\,\\,$$ \n

    and $$\\,\\,\\,PV = RT + Pb$$\n

    $${{PV} \\over {RT}} = 1 + {{Pb} \\over {RT}}$$\n

    $$z = 1 + {{Pb} \\over {RT}};Z > 1\\,\\,\\,$$ at high pressure", "topic": "Number Theory", "subtopic": "Divisibility / Factorization" }, { "id": 1632, "subject": "Chemistry", "question": "If Z is a compressibility factor, van der Waals equation at low pressure can be written as:", "options": [ { "text": "Z = 1 + $$RT \\over Pb$$" }, { "text": "Z = 1 - $$a \\over VRT$$" }, { "text": "Z = 1 - $$Pb \\over RT$$" }, { "text": "Z = 1 + $$Pb \\over RT$$" } ], "answer": "Z = 1 - $$a \\over VRT$$", "solution": "**Answer:** Z = 1 - $$a \\over VRT$$\n\nCompressibility factor $$\\left( Z \\right) = {{PV} \\over {RT}}$$ \n

    (For one mole of real gas)\n

    van der Waals equation\n

    $$\\left( {P + {a \\over {{V^2}}}} \\right)\\left( {V - b} \\right) = RT$$\n

    At low pressure, volume is very large and hence correction term $$b$$ can be neglected in comparison to very large volume of $$V.$$ \n

    i.e.$$\\,\\,\\,V - b \\approx V$$\n

    $$\\left( {P + {a \\over {{V^2}}}} \\right)V = RT;$$ \n

    $$PV + {a \\over V} = RT$$\n

    $$PV = RT - {a \\over V};$$\n

    $${{PV} \\over {RT}} = 1 - {a \\over {VRT}}$$ \n

    Hence, $$\\,\\,\\,z = 1 - {a \\over {VRT}}$$ ", "topic": "Number Theory", "subtopic": "Divisibility / Factorization" }, { "id": 1633, "subject": "Chemistry", "question": "At very high pressures, the compressibility factor of one mole of a gas is given by : ", "options": [ { "text": "$${{pb} \\over {RT}}$$" }, { "text": "1 + $${{pb} \\over {RT}}$$" }, { "text": "1 $$-$$ $${{pb} \\over {RT}}$$" }, { "text": "1 $$-$$ $${b \\over {\\left( {VRT} \\right)}}$$" } ], "answer": "1 + $${{pb} \\over {RT}}$$", "solution": "**Answer:** 1 + $${{pb} \\over {RT}}$$\n\n

    According to van der Waals' equation, for one mole of a gas

    \n

    $$\\left( {p + {a \\over {{V^2}}}} \\right)(V - b) = RT$$ ...... (1)

    \n

    At high pressure, $${a \\over {{V^2}}}$$ can be neglected

    \n

    So, $$p + {a \\over {{V^2}}} \\approx p$$ ...... (2)

    \n

    From Eqs. (1) and (2), we get

    \n

    $$p(V - b) = RT$$

    \n

    $$pV - pb = RT$$

    \n

    $$pV = RT + pb$$

    \n

    Dividing both the sides by RT, we get

    \n

    $${{pV} \\over {RT}} = 1 + {{pb} \\over {RT}}$$

    \n

    $$Z = 1 + {{pb} \\over {RT}}$$

    ", "topic": "Number Theory", "subtopic": "Divisibility / Factorization" }, { "id": 1634, "subject": "Chemistry", "question": "Among the following, the incorrect statement is :", "options": [ { "text": "At low pressure, real gases show ideal behaviour." }, { "text": "At very low temperature, real gases show ideal behaviour." }, { "text": "At very large volume, real gases show ideal behaviour." }, { "text": "At Boyle’s temperature, real gases show ideal behaviour." } ], "answer": "At very low temperature, real gases show ideal behaviour.", "solution": "**Answer:** At very low temperature, real gases show ideal behaviour.\n\nA real gas do not show ideal behaviour at low temperature, high pressure and low volume.\n

    So according to the question, at low temperature real gas show ideal behaviour. This statement is wrong. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1635, "subject": "Chemistry", "question": "Consider the van der Waals constants, a and b,\nfor the following gases.

    \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    GasArNeKrXe
    a/ (atm dm6 mol–2)1.30.25.14.1
    b/ (10–2 dm3 mol–1)3.21.71.05.0
    \n
    Which gas is expected to have the highest\ncritical temperature?", "options": [ { "text": "Ne" }, { "text": "Kr" }, { "text": "Xe" }, { "text": "Ar" } ], "answer": "Kr", "solution": "**Answer:** Kr\n\nWe know,\n

    Critical temperature (Tc) = $${{8a} \\over {27bR}}$$\n

    $$ \\therefore $$ Tc $$ \\propto $$ $${a \\over b}$$\n

    So, species with greatest value of\n$${a \\over b}$$ has\ngreatest value of critical temperature.\n

    Accordig to the given data Kr have highest $${a \\over b}$$ ratio.", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1636, "subject": "Chemistry", "question": "Consider the following table :

    \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    Gasa/(k Pa dm6 mol-1) b/(dm3 mol-1)
    A642.320.05196
    B155.210.04136
    C431.910.05196
    D155.210.4382
    \n
    a and b are vander Waals constants. The correct statement about the gases is :", "options": [ { "text": "Gas C will occupy more volume than gas A; gas B will be more compressible than gas D" }, { "text": "Gas C will occupy lesser volume than gas A; gas B will be more compressible than gas D" }, { "text": "Gas C will occupy lesser volume than gas A; gas B will be lesser compressible than gas D" }, { "text": "Gas C will occupy more volume than gas A; gas B will be lesser compressible than gas D" } ], "answer": "Gas C will occupy more volume than gas A; gas B will be more compressible than gas D", "solution": "**Answer:** Gas C will occupy more volume than gas A; gas B will be more compressible than gas D\n\nvan der Walls equation of state is,\n

    $$z = 1 + {{Pb} \\over {RT}} - {a \\over {{V_m}RT}}$$\n

    b = intermolecule volume of gases.\n

    So when the value of 'b' is higher then the higher value of\n‘b’ will occupy higher volume and according to van der Walls equation of state 'z' will be higher hence gas will be\nless compressible.\n

    $$a$$ = intermolecule force of attraction\n

    The value of 'a' is higher means intermolecule force of attraction is higher then the volume of gas will be lesser and according to van der Walls equation of state 'z' will be smaller hence gas will have higher compressibility.", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1637, "subject": "Chemistry", "question": "At a given temperature T, gases Ne, Ar, Xe and\nKr are found to deviate from ideal gas\nbehaviour. Their equation of state is given as

    \n$$p = {{RT} \\over {V - b}}$$ at T.

    \nHere, b is the van der Waals constant. Which\ngas will exhibit steepest increase in the plot of\nZ (compression factor) vs p?", "options": [ { "text": "Xe" }, { "text": "Ne" }, { "text": "Ar" }, { "text": "Kr" } ], "answer": "Xe", "solution": "**Answer:** Xe\n\nGiven $$p = {{RT} \\over {V - b}}$$\n

    $$ \\Rightarrow $$ $$p\\left( {V - b} \\right) = RT$$\n

    $$ \\Rightarrow $$ $${{pV} \\over {RT}} = {{pb} \\over {RT}} + {{RT} \\over {RT}}$$\n

    $$ \\Rightarrow $$ Z = $${b \\over {RT}}p + 1$$\n

    $$ \\therefore $$ Slope of Z vs p curve (straight line) = $${b \\over {RT}}$$\n

    $$ \\therefore $$ Higher the value of b, more steep will be\nthe curve.\n

    As we know, b = 4vNA = $$4 \\times {4 \\over 3}\\pi {r^3} \\times {N_A}$$\n

    $$ \\therefore $$ b $$ \\propto $$ size of gas molecules\n

    Among Ne, Ar, Xe and\nKr, size of Xe is the most as down the group in the periodic table size increases.", "topic": "Number Theory", "subtopic": "Divisibility / Factorization" }, { "id": 1638, "subject": "Chemistry", "question": "The volume of gas A is twice than that of gas B. The compressibility factor of gas A is thrice than that of gas B at same temperature. The pressures of the gases for equal number of moles are - ", "options": [ { "text": "2PA = 3PB " }, { "text": "3PA = 2PB " }, { "text": "PA = 3PB " }, { "text": "PA = 2PB" } ], "answer": "2PA = 3PB ", "solution": "**Answer:** 2PA = 3PB \n\nZ = PV/nRT\n

    $$ \\Rightarrow $$ P = $${{ZnRT} \\over V}$$\n

    at constant T and mol(n),\n

    P $$ \\propto $$ $${Z \\over V}$$\n

    $$ \\Rightarrow $$ $${{{P_A}} \\over {{P_B}}} = {{{Z_A}} \\over {{Z_A}}} \\times {{{V_B}} \\over {{V_A}}}$$\n

    = $$\\left( {{3 \\over 1}} \\right) \\times \\left( {{1 \\over 2}} \\right)$$ = $${3 \\over 2}$$\n

    $$ \\therefore $$ 2PA = 3PB", "topic": "Number Theory", "subtopic": "Divisibility / Factorization" }, { "id": 1639, "subject": "Chemistry", "question": "A certain gas obeys P(Vm $$-$$ b) = RT. The value of $${\\left( {{{\\partial Z} \\over {\\partial P}}} \\right)_T}$$ is $${{xb} \\over {RT}}$$. The value of x is _________. (Integer answer) (Z : compressibility factor)", "options": [], "answer": "1", "solution": "**Answer:** 1\n\n

    For 1 mole of a real gas, the van der Waals' equation is,

    \n

    $$\\left( {p + {a \\over {V_m^2}}} \\right)({V_m} - b) = RT$$

    \n

    At very high pressure, the equation becomes,

    \n

    $$p({V_m} - b) = RT$$

    \n

    $$ \\Rightarrow p{V_m} = RT + pb \\Rightarrow {{p{V_m}} \\over {RT}} = 1 + {{pb} \\over {RT}}$$

    \n

    $$ \\Rightarrow Z = 1 + {{pb} \\over {RT}}$$ [$$\\because$$ $$Z = {{p{V_m}} \\over {RT}}$$ = compressibility]

    \n

    $$\\therefore$$ $${\\left( {{{\\delta Z} \\over {\\delta p}}} \\right)_T} = 0 + {b \\over {RT}} + {b \\over {RT}} = {{xb} \\over {RT}}$$ (Given)

    \n

    $$ \\Rightarrow x = 1$$

    ", "topic": "Number Theory", "subtopic": "Divisibility / Factorization" }, { "id": 1640, "subject": "Chemistry", "question": "The unit of the van der Waals gas equation parameter 'a' in $$\\left( {P + {{a{n^2}} \\over {{V^2}}}} \\right)(V - nb) = nRT$$ is :", "options": [ { "text": "kg m s$$-$$2" }, { "text": "dm3 mol$$-$$1" }, { "text": "kg m s$$-$$1" }, { "text": "atm dm6 mol$$-$$2" } ], "answer": "atm dm6 mol$$-$$2", "solution": "**Answer:** atm dm6 mol$$-$$2\n\n$${{a{n^2}} \\over {{V^2}}} = atm \\Rightarrow a = atm \\times {{d{m^6}} \\over {mo{l^2}}}$$", "topic": "Algebra", "subtopic": "Equations and Inequalities" }, { "id": 1641, "subject": "Chemistry", "question": "

    For a real gas at $$25^{\\circ} \\mathrm{C}$$ temperature and high pressure (99 bar) the value of compressibility factor is 2, so the value of Vander Waal's constant 'b' should be __________ $$\\times 10^{-2} \\mathrm{~L} \\mathrm{~mol}^{-1}$$ (Nearest integer)

    \n

    (Given $$\\mathrm{R}=0.083 \\mathrm{~L}$$ bar $$\\mathrm{K}^{-1} \\mathrm{~mol}^{-1}$$)

    ", "options": [], "answer": "25", "solution": "**Answer:** 25\n\n For 1 mole at high pressure\n

    \n$P(V-b)=R T$\n

    \n$\\mathrm{PV}-\\mathrm{Pb}=\\mathrm{RT}$\n

    \n$\\frac{\\mathrm{PV}}{\\mathrm{RT}}=1+\\frac{\\mathrm{Pb}}{\\mathrm{RT}}$\n

    \n$Z=1+\\frac{\\mathrm{Pb}}{\\mathrm{RT}}$\n

    \n$1=\\frac{99(\\mathrm{~b})}{0.083 \\times 298}$\n

    \n$b=\\frac{0.083 \\times 298}{99} \\simeq 0.249 \\simeq 25 \\times 10^{-2}$", "topic": "Number Theory", "subtopic": "Divisibility / Factorization" }, { "id": 1642, "subject": "Chemistry", "question": "

    A certain quantity of real gas occupies a volume of $$0.15~ \\mathrm{dm}^{3}$$ at $$100 \\mathrm{~atm}$$ and $$500 \\mathrm{~K}$$ when its compressibility factor is 1.07 . Its volume at 300 atm and $$300 \\mathrm{~K}$$ (When its compressibility factor is 1.4 ) is ___________ $$\\times 10^{-4} ~\\mathrm{dm}^{3}$$ (Nearest integer)

    ", "options": [], "answer": "392", "solution": "**Answer:** 392\n\n$$\n\\begin{aligned}\n& \\mathrm{z}=\\frac{\\mathrm{PV}}{\\mathrm{nRT}} ; \\mathrm{n}=\\frac{\\mathrm{PV}}{\\mathrm{ZRT}} \\\\\\\\\n& \\mathrm{Z}_1=1.07, \\mathrm{P}_1=100 \\mathrm{~atm}, \\mathrm{~V}_1=0.15 \\mathrm{~L}, \\mathrm{~T}_1=500 \\mathrm{~K} \\\\\\\\\n& \\mathrm{Z}_2=1.4, \\mathrm{P}_2=300 \\mathrm{~atm}, \\mathrm{~T}_2=300 \\mathrm{~K}, \\mathrm{~V}_2=? \\\\\\\\\n& \\frac{\\mathrm{P}_1 \\mathrm{~V}_1}{\\mathrm{Z}_1 \\mathrm{RT}_{\\mathrm{I}}}=\\frac{\\mathrm{P}_2 \\mathrm{~V}_2}{\\mathrm{Z}_2 \\mathrm{RT}_2}=\\mathrm{n} \\\\\\\\\n& \\mathrm{V}_2=\\frac{1.4}{1.07} \\times .03 \\\\\\\\\n&=392 \\times 10^{-4} \\mathrm{dm}^3\n\\end{aligned}\n$$", "topic": "Number Theory", "subtopic": "Divisibility / Factorization" }, { "id": 1643, "subject": "Chemistry", "question": "

    Arrange the following gases in increasing order of van der Waals constant 'a'

    \n

    A. Ar

    \n

    B. $$\\mathrm{CH}_{4}$$

    \n

    C. $$\\mathrm{H}_{2} \\mathrm{O}$$

    \n

    D. $$\\mathrm{C}_{6} \\mathrm{H}_{6}$$

    \n

    Choose the correct option from the following.

    ", "options": [ { "text": "D, C, B and A" }, { "text": "A, B, C and D" }, { "text": "B, C, D and A" }, { "text": "C, D, B and A" } ], "answer": "A, B, C and D", "solution": "**Answer:** A, B, C and D\n\n

    The van der Waals constant 'a' is a measure of the strength of the intermolecular forces in a gas. Larger molecules and molecules with stronger intermolecular forces will have larger 'a' constants.

    \n

    Here, we are comparing argon (Ar), methane (CH₄), water (H₂O), and benzene (C₆H₆).

    \n

    Argon (Ar) is a noble gas and is monoatomic. Methane (CH₄) is a small molecule with weak intermolecular forces. Water (H₂O) is a small molecule but has strong hydrogen bonding, making its intermolecular forces stronger than those in methane. Benzene (C₆H₆) is a larger molecule than the others and has relatively strong intermolecular forces due to its larger size and polarizability.

    \n

    So, in the increasing order of van der Waals constant 'a', it should be:

    \n

    Argon (Ar) < Methane (CH₄) < Water (H₂O) < Benzene (C₆H₆)

    \n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1644, "subject": "Chemistry", "question": "The reaction :
    \n$${(CH)_3}C - Br\\buildrel {{H_2}O} \\over\n \\longrightarrow {(CH)_3}C - OH$$", "options": [ { "text": "elimination reaction" }, { "text": "substitution reaction" }, { "text": "free radical reaction" }, { "text": "displacement reaction" } ], "answer": "substitution reaction", "solution": "**Answer:** substitution reaction\n\nThe hydrolysis of $$t$$-butyl bromide is an example of $${S_N}1$$ reaction. The reaction consists of two steps. \n

    \"AIEEE ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1645, "subject": "Chemistry", "question": "Tertiary alkyl halides are practically inert to substitution by SN2 mechanism because of ", "options": [ { "text": "insolubility" }, { "text": "instability" }, { "text": "inductive effect" }, { "text": "steric hindrance" } ], "answer": "steric hindrance", "solution": "**Answer:** steric hindrance\n\nDue to steric hindrance tertiary alkyl halide do not react by $${S_N}2$$ mechanism they react by $${S_N}1$$ mechanism. $${S_N}2$$ mechanism is followed in case of primary and secondary alkyl halides .\n

    The order is \n

    $$C{H_2} - X > C{H_3} - C{H_2}X > {\\left( {C{H_3}} \\right)_2}CH.X > {\\left( {C{H_3}} \\right)_3}C - X$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1646, "subject": "Chemistry", "question": "Alkyl halides react with dialkyl copper reagents to give ", "options": [ { "text": "alkenes" }, { "text": "alkyl copper halides" }, { "text": "alkanes" }, { "text": "alkenyl halides" } ], "answer": "alkanes", "solution": "**Answer:** alkanes\n\nIn Corey House synthesis of alkanes alkyl halide react with lithium dialkyl cuprate\n

    $$R'X + Li{R_2}Cu\\buildrel \\, \\over\n \\longrightarrow R' - R + RCu + LiX$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1647, "subject": "Chemistry", "question": "Elimination of bromine from 2-bromobutane results in the formation of- ", "options": [ { "text": "equimolar mixture of 1 and 2-butene" }, { "text": "predominantly 2-butene" }, { "text": "predominantly 1-butene" }, { "text": "predominantly 2-butyne " } ], "answer": "predominantly 2-butene", "solution": "**Answer:** predominantly 2-butene\n\n\"AIEEE \n

    The formation of $$2$$-butene is in accordance to Saytzeff's rule. The more substituted alkene is formed. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1648, "subject": "Chemistry", "question": "Reaction of trans-2-phenyl-1-bromocyclopentane on reaction with alcoholic KOH produces", "options": [ { "text": "4-phenylcyclopentene" }, { "text": "2-phenylcyclopentene" }, { "text": "1-phenylcyclopentene " }, { "text": "3-phenylcyclopentene" } ], "answer": "3-phenylcyclopentene", "solution": "**Answer:** 3-phenylcyclopentene\n\n\"AIEEE\nHughes and Ingold proposed that bimolecular elimination\nreactions take place when the two groups to be eliminated are\ntrans and lie in one plane with the two carbon atoms to which\nthey are attached i.e. E2 reactions are stereoselectively trans.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1649, "subject": "Chemistry", "question": "Which of the following is the correct order of decreasing SN2 reactivity? ", "options": [ { "text": "RCH2X > R3CX > R2CHX" }, { "text": "RCH2X > R2CHX > \nR3CX" }, { "text": "R3CX > R2CHX > RCH2X" }, { "text": "R2CHX > R3CX > RCH2X" } ], "answer": "RCH2X > R2CHX > \nR3CX", "solution": "**Answer:** RCH2X > R2CHX > \nR3CX\n\nIn $${S_N}2$$ mechanism transition state is pentavelent. For bulky alkyl group it will have sterical hinderance and smaller alkyl group will favour the $${S_N}2$$ mechanism. So the decreasing order of reactivity of alkyl halides is $$RC{H_2}X > {R_2}CHX > {R_3}CX$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1650, "subject": "Chemistry", "question": "The organic chloro compound, which shows complete stereochemical inversion during a SN2 reaction,\nis", "options": [ { "text": "(C2H5)2CHC" }, { "text": "(CH3)3CCl " }, { "text": "(CH3)2CHCl " }, { "text": "CH3Cl" } ], "answer": "CH3Cl", "solution": "**Answer:** CH3Cl\n\n$${S_N}2$$ reaction is favoured by small groups on the carbon atom attached to halogen.\n

    So, the order of reactivity is \n

    $$C{H_3}Cl > {\\left( {C{H_3}} \\right)_2}CHCl > $$ \n

    $$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,{\\left( {C{H_3}} \\right)_3}CCl > {\\left( {C{}_2{H_5}} \\right)_2}CHCl$$ \n

    NOTE : $${S_N}2$$ reaction is shown to maximum extent by primary halides. The only primary halides given is $$C{H_3}Cl,$$ so the correct answer is $$(d). $$", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 1651, "subject": "Chemistry", "question": "How many chiral compounds are possible on monochlorination of 2–methyl butane ?", "options": [ { "text": "8" }, { "text": "2" }, { "text": "4" }, { "text": "6" } ], "answer": "2", "solution": "**Answer:** 2\n\nThe reaction involved is

    \n\"AIEEE
    \nOut of all the four isomers formed only two compounds are optically active.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1652, "subject": "Chemistry", "question": "In SN2 reactions, the correct order of reactivity for the following compounds:
    \nCH3Cl, CH3CH\n2Cl, (CH3)2CHCl and (CH3)3CCl is: ", "options": [ { "text": "CH3CH2Cl > CH3Cl > (CH3)2CHCl > (CH3)3CCl" }, { "text": "(CH3)2CHCl > CH3CH2Cl > CH3Cl > (CH3)3CCl" }, { "text": "CH3Cl > (CH3)2CHCl > CH3CH2Cl > (CH3)3CCl" }, { "text": "CH3Cl > CH3CH2Cl > (CH3)2CHCl > (CH3)3CCl" } ], "answer": "CH3Cl > CH3CH2Cl > (CH3)2CHCl > (CH3)3CCl", "solution": "**Answer:** CH3Cl > CH3CH2Cl > (CH3)2CHCl > (CH3)3CCl\n\nSteric congestion around the carbon atom undergoing the inversion process will slow down the $${S_N}2$$ reaction, hence less congestion faster will the reaction. So, the order is \n

    $$C{H_3}Cl > \\left( {C{H_3}} \\right)C{H_2} - Cl > $$\n

    $$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,{\\left( {C{H_3}} \\right)_2}CH - Cl > {\\left( {C{H_3}} \\right)_3}CCl$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1653, "subject": "Chemistry", "question": "The synthesis of alkyl fluorides is best accomplished by:", "options": [ { "text": "Sandmeyer’s reaction" }, { "text": "Finkelstein reaction" }, { "text": "Swarts reaction" }, { "text": "Free radical fluorination" } ], "answer": "Swarts reaction", "solution": "**Answer:** Swarts reaction\n\nAlkyl fluorides are more conveniently prepared by heating suitable chloro $$-$$ or bromo-alkanes with organic fluorides such as $$As{F_3},\\,\\,Sb{F_3},\\,\\,Co{F_2},\\,\\,AgF,\\,H{g_2}{F_2}\\,\\,$$ etc.\n

    This reaction is called Swarts reaction. \n

    $$C{H_3}Br + AgF\\buildrel \\, \\over\n \\longrightarrow C{H_3}F + AgBr$$\n

    $$2C{H_3}C{H_2}Cl + H{g_2}{F_2}\\buildrel \\, \\over\n \\longrightarrow $$ \n

    $$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,2C{H_3}C{H_2}F + H{g_2}Cl{}_2$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1654, "subject": "Chemistry", "question": "3-Methyl-pent-2-ene on reaction with HBr in presence of peroxide forms an addition product. The number\nof possible stereoisomers for the product is :", "options": [ { "text": "Zero" }, { "text": "Two" }, { "text": "Four" }, { "text": "Six" } ], "answer": "Four", "solution": "**Answer:** Four\n\n\"JEE\n

    A compound having two chiral centres can exist in 4 stereoisomeric forms (2n).\n

    \"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1655, "subject": "Chemistry", "question": "Which one of the following is likely to give a precipitate with AgNO3 solution ?", "options": [ { "text": "CHCl3" }, { "text": "(CH3)3CCl" }, { "text": "CCl4" }, { "text": "CH2=CH–Cl" } ], "answer": "(CH3)3CCl", "solution": "**Answer:** (CH3)3CCl\n\nThe molecule which will produce stable carbocation will react with AgNO3 solution.\n

    (CH3)3CCl $$\\buildrel {AgN{O_3}} \\over\n \\longrightarrow $$ (CH3)3C+NO3- + AgCl($$ \\downarrow $$)", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1656, "subject": "Chemistry", "question": "An 'Assertion' and a 'Reason' are given below. Choose the correct answer from the following options :\n
    Assertion (A) : Vinyl halides do not undergo nucleophilic substitution easily.\n
    Reason (R) : Even though the intermediate carbocation is stabilized by loosely held p-electrons, the cleavage\nis difficult because of strong bonding.", "options": [ { "text": "Both (A) and (R) are correct statements but (R) is not the correct explanation of (A)" }, { "text": "Both (A) and (R) are correct statements and (R) is the correct explanation of (A)" }, { "text": "(A) is correct statement but (R) is a wrong statement" }, { "text": "Both (A) and (R) are wrong statements." } ], "answer": "(A) is correct statement but (R) is a wrong statement", "solution": "**Answer:** (A) is correct statement but (R) is a wrong statement\n\n\"JEE\n
    Due to resonance C - Cl bond strength increses, that is why C - Cl bond do not break easily and don't show nucleophilic substitution reaction easily.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1657, "subject": "Chemistry", "question": "Which one of the following alkenes when\ntreated with HCl yields majorly an anti\nMarkovnikov product?", "options": [ { "text": "Cl – CH = CH2" }, { "text": "F3C – CH = CH2" }, { "text": "CH3O – CH = CH2" }, { "text": "H2N – CH = CH2" } ], "answer": "F3C – CH = CH2", "solution": "**Answer:** F3C – CH = CH2\n\nMarkovnikov rule says negative part of the reagent attacks on the carbon of the double bond which have less number of hydrogen atom.\n

    So in Anti-Markovnikov rule negative part of the reagent attacks on the carbon of the double bond which have more number of hydrogen atom.\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1658, "subject": "Chemistry", "question": "The major product formed in the following reaction is :\n
    CH3CH = CHCH(CH3)2 $$\\buildrel {HBr} \\over\n \\longrightarrow $$", "options": [ { "text": "Br(CH2)3CH(CH3)2" }, { "text": "CH3CH(Br)CH2CH(CH3)2" }, { "text": "CH3CH2CH(Br)CH(CH3)2" }, { "text": "CH3CH2CH2C(Br)(CH3)2" } ], "answer": "CH3CH2CH2C(Br)(CH3)2", "solution": "**Answer:** CH3CH2CH2C(Br)(CH3)2\n\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1659, "subject": "Chemistry", "question": "The correct pair(s) of the ambident nucleophiles is(are)

    (A) AgCN/KCN

    (B) RCOOAg/RCOOK

    (C) AgNO2/KNO2

    (D) AgI/KI", "options": [ { "text": "(B) and (C) only" }, { "text": "(B) only" }, { "text": "(A) and (C) only" }, { "text": "(A) only" } ], "answer": "(A) and (C) only", "solution": "**Answer:** (A) and (C) only\n\nAmbident ligands are those which have two donor sites but at once it has the ability to donate either site.
    \nHence in AgCN/KCN, AgNO2/KNO2, CN-, NO2- have two donar sites act as ambident Nu-.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1660, "subject": "Chemistry", "question": "A chloro compound \"A\".

    (i) forms aldehydes on ozonolysis followed by the hydrolysis.

    (ii) when vaporized completely 1.53 g of A, gives 448 mL of vapour at STP.

    The number of carbon atoms in a molecule of compound A is ___________.", "options": [], "answer": "3", "solution": "**Answer:** 3\n\n448 ml of A $$\\Rightarrow$$ 1.53 gm A

    22400 ml of A $$\\Rightarrow$$ $${{1.53} \\over {448}}$$ $$\\times$$ 22400 gm A = 76.5 Agm

    $$\\mathop {{H_3}CHC - CH - Cl}\\limits_{It\\,has\\,3\\,carbon\\,atoms} \\mathrel{\\mathop{\\kern0pt\\longrightarrow}\n\\limits_{Zn/{H_2}O}^{{O_3}}} \\mathop {C{H_3} - CH = O}\\limits_{Aldehyde} $$

    & MM is 36 + 5 + 35.5 = 76.5", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1661, "subject": "Chemistry", "question": "

    The isomeric deuterated bromide with molecular formula $$\\mathrm{C_4H_8DBr}$$ having two chiral carbon atoms is

    ", "options": [ { "text": "2 - Bromo - 1 - deuterobutane" }, { "text": "2 - Bromo - 1 - deutero - 2 - methylpropane" }, { "text": "2 - Bromo - 3 - deuterobutane" }, { "text": "2 - Bromo - 2 - deuterobutane" } ], "answer": "2 - Bromo - 3 - deuterobutane", "solution": "**Answer:** 2 - Bromo - 3 - deuterobutane\n\n

    The isomeric deuterated bromide with molecular formula $\\mathrm{C}_{4} \\mathrm{H}_{8} \\mathrm{DBr}$ having two chiral carbon atoms is

    \n

    \"JEE

    \n

    2-Bromo-3-deuterobutane

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1662, "subject": "Chemistry", "question": "

    Assertion A : Hydrolysis of an alkyl chloride is a slow reaction but in the presence of NaI, the rate of the hydrolysis increases.

    \n

    Reason R : I$$^{-}$$ is a good nucleophile as well as a good leaving group.

    \n

    In the light of the above statements, choose the correct answer from the options given below

    ", "options": [ { "text": "A is true but R is false" }, { "text": "A is false but R is true" }, { "text": "Both A and R are true and R is the correct explanation of A" }, { "text": "Both A and R are true but R is NOT the correct explanation of A" } ], "answer": "Both A and R are true and R is the correct explanation of A", "solution": "**Answer:** Both A and R are true and R is the correct explanation of A\n\nThe rate of hydrolysis of alkyl chloride improves because of better Nucleophilicity of I–.", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 1663, "subject": "Chemistry", "question": "

    The pair from the following pairs having both compounds with net non-zero dipole moment is

    ", "options": [ { "text": "cis-butene, trans-butene" }, { "text": "CH$$_2$$Cl$$_2$$, CHCl$$_3$$" }, { "text": "Benzene, anisidine" }, { "text": "1,4-Dichlorobenzene, 1,3-Dichlorobenzene" } ], "answer": "CH$$_2$$Cl$$_2$$, CHCl$$_3$$", "solution": "**Answer:** CH$$_2$$Cl$$_2$$, CHCl$$_3$$\n\n

    The correct option from the pairs having both compounds with a net non-zero dipole moment is:

    \n

    Option A: $\\text{cis-butene, trans-butene}$\ncis-butene has a net dipole moment because the two alkyl groups are on the same side of the molecule, causing an imbalance in the distribution of electrons. However, trans-butene is symmetric, so the dipole moments of the two alkyl groups cancel each other out, giving it a net dipole moment of zero. Therefore, this pair does not have both compounds with a net non-zero dipole moment.

    \n

    Option B: $\\text{CH}_2\\text{Cl}_2, \\text{CHCl}_3$\nBoth $\\text{CH}_2\\text{Cl}_2$ (dichloromethane) and $\\text{CHCl}_3$ (chloroform) have a net non-zero dipole moment due to the difference in electronegativity between the hydrogen and chlorine atoms. Hence, this pair has both compounds with a net non-zero dipole moment.

    \n

    Option C: $\\text{Benzene, Anisidine}$\nBenzene has a symmetrical structure, leading to a net zero dipole moment. Anisidine, however, does have a net dipole moment due to the asymmetry in its structure. Therefore, this pair does not have both compounds with a net non-zero dipole moment.

    \n

    Option D: $\\text{1,4-Dichlorobenzene, 1,3-Dichlorobenzene}$\n$\\text{1,4-Dichlorobenzene}$ has a symmetrical structure, leading to a net zero dipole moment. However, $\\text{1,3-Dichlorobenzene}$ has a non-zero dipole moment due to the asymmetry in its structure. Therefore, this pair does not have both compounds with a net non-zero dipole moment.

    \n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1664, "subject": "Chemistry", "question": "Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).

    \nAssertion (A) : Haloalkanes react with KCN to form alkyl cyanides as a main product while with $\\mathrm{AgCN}$ form isocyanide as the main product.\n

    \nReason (R) : $\\mathrm{KCN}$ and $\\mathrm{AgCN}$ both are highly ionic compounds.

    \nIn the light of the above statements, choose the most appropriate answer from the options given below :\n", "options": [ { "text": "(A) is correct but $(\\mathbf{R})$ is not correct" }, { "text": "Both (A) and (R) are correct and (R) is the correct explanation of (A)" }, { "text": "(A) is not correct but (R) is correct" }, { "text": "Both (A) and (R) are correct but (R) is not the correct explanation of (A)" } ], "answer": "(A) is correct but $(\\mathbf{R})$ is not correct", "solution": "**Answer:** (A) is correct but $(\\mathbf{R})$ is not correct\n\n$\\begin{aligned} & \\mathrm{R}-\\mathrm{X}+\\mathrm{KCN} \\longrightarrow \\mathrm{R}-\\mathrm{CN}+\\mathrm{KX} \\\\\\\\ & \\mathrm{R}-\\mathrm{X}+\\mathrm{AgCN} \\longrightarrow \\mathrm{R}-\\mathrm{NC}+\\mathrm{AgX}\\end{aligned}$\n

    $\\mathrm{KCN}$ is ionic so it provides cyanide ions in solution and attacks from carbon side on alkyl halide.\n\n

    But $\\mathrm{AgCN}$ is covalent it cannot form cyanide ion so it attacks from nitrogen side and isocyanide is formed predominantly.\n

    $\\mathrm{A}$ is correct $\\mathrm{R}$ is not correct.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1665, "subject": "Chemistry", "question": "

    The correct statement regarding nucleophilic substitution reaction in a chiral alkyl halide is ;

    ", "options": [ { "text": "Retention occurs in $$S_N 1$$ reaction and inversion occurs in $$\\mathrm{S}_{\\mathrm{N}} 2$$ reaction.\n" }, { "text": "Racemisation occurs in $$\\mathrm{S}_{\\mathrm{N}} 1$$ reaction and retention occurs in $$\\mathrm{S}_{\\mathrm{N}} 2$$ reaction.\n" }, { "text": "Racemisation occurs in both $$\\mathrm{S}_{\\mathrm{N}} 1$$ and $$\\mathrm{S}_{\\mathrm{N}} 2$$ reactions.\n" }, { "text": "Racemisation occurs in $$S_N 1$$ reaction and inversion occurs in $$\\mathrm{S}_{\\mathrm{N}} 2$$ reaction." } ], "answer": "Racemisation occurs in $$S_N 1$$ reaction and inversion occurs in $$\\mathrm{S}_{\\mathrm{N}} 2$$ reaction.", "solution": "**Answer:** Racemisation occurs in $$S_N 1$$ reaction and inversion occurs in $$\\mathrm{S}_{\\mathrm{N}} 2$$ reaction.\n\n

    The correct statement regarding nucleophilic substitution reaction in a chiral alkyl halide is Option D. Let's break down why this is the case by understanding both the $$S_N 1$$ and $$S_N 2$$ reactions and their effects on the chirality of the alkyl halide.

    \n\n

    $$S_N 2$$ Reaction:

    \n\n

    The $$S_N 2$$ (Substitution Nucleophilic Bimolecular) reaction is a one-step process where the bond formation (nucleophile attacking the substrate) and the bond-breaking (leaving group leaving the substrate) processes occur simultaneously. This reaction type is associated with an inversion of configuration at the chiral center. This means that if the reactant molecule is chiral and the nucleophile attacks from the side opposite to the leaving group, the product will have the opposite configuration to the reactant. This phenomenon is known as Walden inversion. Therefore, it does not lead to racemisation but rather to inversion of the stereochemistry at the chiral center.

    \n\n

    $$S_N 1$$ Reaction:

    \n\n

    The $$S_N 1$$ (Substitution Nucleophilic Unimolecular) reaction is a two-step process. The first step involves the formation of a carbocation intermediate by the departure of the leaving group. This intermediate is planar, and hence, the nucleophile can attack from either side of the plane with equal probability. This results in the formation of products with both configurations - one retaining the original configuration and the other being its mirror image (inverted configuration). As both enantiomers are formed, the process leads to racemisation, especially in situations where the carbocation can freely rotate and is attacked with equal probability from both sides. Over time, if this reaction goes to completion and there's no bias in nucleophile attack direction, you'd expect a racemic mixture of products.

    \n\n

    Thus, Option D is correct: Racemisation occurs in $$S_N 1$$ reaction, and inversion occurs in the $$S_N 2$$ reaction.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1666, "subject": "Chemistry", "question": "

    The product (C) in the below mentioned reaction is :

    \n

    $$\\mathrm{CH}_3-\\mathrm{CH}_2-\\mathrm{CH}_2-\\mathrm{Br} \\xrightarrow[\\Delta]{\\mathrm{KOH}_{(\\text {alc) }}} \\mathrm{A} \\xrightarrow{\\mathrm{HBr}} \\mathrm{B} \\xrightarrow[\\mathrm{KOH}_{(\\mathrm{aq})}]{\\Delta} \\mathrm{C}$$

    ", "options": [ { "text": "Propan-1-ol" }, { "text": "Propyne" }, { "text": "Propan-2-ol" }, { "text": "Propene" } ], "answer": "Propan-2-ol", "solution": "**Answer:** Propan-2-ol\n\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1667, "subject": "Chemistry", "question": "

    Alkyl halide is converted into alkyl isocyanide by reaction with

    ", "options": [ { "text": "$$\\mathrm{KCN}$$\n" }, { "text": "$$\\mathrm{NH}_4 \\mathrm{CN}$$\n" }, { "text": "$$\\mathrm{NaCN}$$\n" }, { "text": "$$\\mathrm{AgCN}$$" } ], "answer": "$$\\mathrm{AgCN}$$", "solution": "**Answer:** $$\\mathrm{AgCN}$$\n\n

    Covalent character of $$\\mathrm{AgCN}$$.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1668, "subject": "Chemistry", "question": "

    Given below are two statements:

    \n

    Statement - I: High concentration of strong nucleophilic reagent with secondary alkyl halides which do not have bulky substituents will follow $$\\mathrm{S}_{\\mathrm{N}}{ }^2$$ mechanism.

    \n

    Statement - II: A secondary alkyl halide when treated with a large excess of ethanol follows $$\\mathrm{S}_{\\mathrm{N}}{ }^1$$ mechanism.

    \n

    In the light of the above statements, choose the most appropriate from the options given below:

    ", "options": [ { "text": "Both Statement I and Statement II are true.\n" }, { "text": "Statement I is true but Statement II is false.\n" }, { "text": "Both Statement I and Statement II are false.\n" }, { "text": "Statement I is false but Statement II is true." } ], "answer": "Both Statement I and Statement II are true.\n", "solution": "**Answer:** Both Statement I and Statement II are true.\n\n\n

    Statement - I: Rate of $$\\mathrm{S}_{\\mathrm{N}} 2 \\propto[\\mathrm{R}-\\mathrm{X}]\\left[\\mathrm{Nu}^{-}\\right]$$

    \n

    $$\\mathrm{S}_{\\mathrm{N}} 2$$ reaction is favoured by high concentration of nucleophile $$(\\mathrm{Nu}^{-})$$ & less crowding in the substrate molecule.

    \n

    Statement - II: Solvolysis follows $$\\mathrm{S}_{\\mathrm{N}} 1$$ path.

    \n

    Both are correct Statements.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1669, "subject": "Chemistry", "question": "

    2-chlorobutane $$+\\mathrm{Cl}_2 \\rightarrow \\mathrm{C}_4 \\mathrm{H}_8 \\mathrm{Cl}_2$$ (isomers)

    \n

    Total number of optically active isomers shown by $$\\mathrm{C}_4 \\mathrm{H}_8 \\mathrm{Cl}_2$$, obtained in the above reaction is _________.

    ", "options": [], "answer": "6", "solution": "**Answer:** 6\n\n

    \"JEE

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1670, "subject": "Chemistry", "question": "

    Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).

    \n

    Assertion (A) : $$\\mathrm{CH}_2=\\mathrm{CH}-\\mathrm{CH}_2-\\mathrm{Cl}$$ is an example of allyl halide.

    \n

    Reason (R) : Allyl halides are the compounds in which the halogen atom is attached to $$\\mathrm{sp}^2$$ hybridised carbon atom.

    \n

    In the light of the above statements, choose the most appropriate answer from the options given below :

    ", "options": [ { "text": "(A) is true but (R) is false" }, { "text": "(A) is false but (R) is true" }, { "text": "Both (A) and (R) are true but (R) is not the correct explanation of (A)" }, { "text": "Both (A) and (R) are true and (R) is the correct explanation of (A)" } ], "answer": "(A) is true but (R) is false", "solution": "**Answer:** (A) is true but (R) is false\n\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1671, "subject": "Chemistry", "question": "

    Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).

    \n

    Assertion (A) : $$\\mathrm{S}_{\\mathrm{N}} 2$$ reaction of $$\\mathrm{C}_6 \\mathrm{H}_5 \\mathrm{CH}_2 \\mathrm{Br}$$ occurs more readily than the $$\\mathrm{S}_{\\mathrm{N}} 2$$ reaction of $$\\mathrm{CH}_3 \\mathrm{CH}_2 \\mathrm{Br}$$.

    \n

    Reason (R) : The partially bonded unhybridized p-orbital that develops in the trigonal bipyramidal transition state is stabilized by conjugation with the phenyl ring.

    \n

    In the light of the above statements, choose the most appropriate answer from the options given below :

    ", "options": [ { "text": "Both (A) and (R) are correct but (R) is not the correct explanation of (A)" }, { "text": "Both (A) and (R) are correct and (R) is the correct explanation of (A)" }, { "text": "(A) is not correct but (R) is correct" }, { "text": "(A) is correct but (R) is not correct" } ], "answer": "Both (A) and (R) are correct and (R) is the correct explanation of (A)", "solution": "**Answer:** Both (A) and (R) are correct and (R) is the correct explanation of (A)\n\n

    To determine the correct answer, we need to analyze both the Assertion (A) and the Reason (R) independently, as well as see if the Reason (R) provides the correct explanation for the Assertion (A).

    \n\n

    Assertion (A) Analysis:
    \n\n

    Assertion (A) states that $$\\mathrm{S}_{\\mathrm{N}} 2$$ reaction of $$\\mathrm{C}_6 \\mathrm{H}_5 \\mathrm{CH}_2 \\mathrm{Br}$$ occurs more readily than the $$\\mathrm{S}_{\\mathrm{N}} 2$$ reaction of $$\\mathrm{CH}_3 \\mathrm{CH}_2 \\mathrm{Br}$$. In an $$\\mathrm{S}_{\\mathrm{N}} 2$$ reaction, the rate depends on the nucleophile as well as the substrate. Specifically, $$\\mathrm{S}_{\\mathrm{N}} 2$$ reactions proceed through a backside attack mechanism, leading to the formation of a transition state where the nucleophile and the leaving group are simultaneously bonded to the carbon atom. The benzyl group $$\\mathrm{C}_6 \\mathrm{H}_5 \\mathrm{CH}_2-$$ is more stable for $$\\mathrm{S}_{\\mathrm{N}} 2$$ reactions compared to the ethyl group $$\\mathrm{CH}_3 \\mathrm{CH}_2-$$ because the benzyl carbon's partial positive charge in the transition state is stabilized by the resonance of the phenyl ring.

    \n\n

    Reason (R) Analysis:
    \n\n

    Reason (R) explains that the partially bonded unhybridized p-orbital that develops in the trigonal bipyramidal transition state is stabilized by conjugation with the phenyl ring. This means that the presence of the phenyl ring can delocalize and stabilize the charge in the transition state, facilitating the $$\\mathrm{S}_{\\mathrm{N}} 2$$ reaction.

    \n\n

    Combining Assertion (A) and Reason (R):
    \n\n

    Given that the phenyl ring in $$\\mathrm{C}_6 \\mathrm{H}_5 \\mathrm{CH}_2 \\mathrm{Br}$$ can indeed stabilize the transition state through conjugation more effectively than an ethyl group, both the Assertion (A) and Reason (R) are correct. Furthermore, Reason (R) provides a valid explanation for Assertion (A), reinforcing why the $$\\mathrm{S}_{\\mathrm{N}} 2$$ reaction occurs more readily for $$\\mathrm{C}_6 \\mathrm{H}_5 \\mathrm{CH}_2 \\mathrm{Br}$$.

    \n\n

    Thus, the most appropriate answer is:

    \n\n

    Option B: Both (A) and (R) are correct and (R) is the correct explanation of (A)

    ", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 1672, "subject": "Chemistry", "question": "Bottles containing C6H5l and C6H5CH2I lost their original labels. They were labelled A and B for testing A and B were separately taken in test tubes and boiled with NaOH solution. The end solution in each tube was made acidic with dilute HNO3 and then some AgNO3 solution was added. Substance B gave a yellow precipitate. Which one of the following statements is true for this experiment?", "options": [ { "text": "A and C6H5CH2I" }, { "text": "B and C6H5I" }, { "text": "Addition of HNO3 was unnecessary" }, { "text": "A was C6H5I" } ], "answer": "A was C6H5I", "solution": "**Answer:** A was C6H5I\n\n$${C_6}{H_5}I\\,\\,\\buildrel {NaOH} \\over\n \\longrightarrow \\,\\,{C_6}{H_5}ONa\\,\\,\\buildrel {HN{O_3}/{H^ + }} \\over\n \\longrightarrow $$ \n

    $${C_6}{H_5}OH\\,\\,\\buildrel {AgN{O_3}} \\over\n \\longrightarrow \\,\\,$$ No yellow ppt.\n

    $${C_6}{H_5}C{H_2}I\\,\\,\\buildrel {NaOH} \\over\n \\longrightarrow \\,\\,{C_6}{H_5}C{H_2}ONa\\,\\,\\buildrel {HN{O_3}/{H^ + }} \\over\n \\longrightarrow $$\n

    $${C_6}{H_5}C{H_2}OH\\,\\,\\buildrel {AgN{O_3}} \\over\n \\longrightarrow .$$ $$\\,\\,\\,$$ yellow ppt.\n

    Since benzyl iodide gives yellow ppt. hence this is compound $$B$$ and $$A$$ was phenyl iodide $$\\left( {{C_6}{H_5}I} \\right).$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1673, "subject": "Chemistry", "question": "Fluorobenzene (C6H5F) can be synthesized in the laboratory ", "options": [ { "text": "by heating phenol with HF and KF " }, { "text": "from aniline by diazotisation followed by heating the diazonium salt with HBF4" }, { "text": "by direct fluorination of benzene with F2 gas" }, { "text": "by reacting bromobenzene with NaF solution" } ], "answer": "from aniline by diazotisation followed by heating the diazonium salt with HBF4", "solution": "**Answer:** from aniline by diazotisation followed by heating the diazonium salt with HBF4\n\n\"AIEEE ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1674, "subject": "Chemistry", "question": "A solution of (–$$l$$) – chloro –1 – phenylethane in toluene racemises slowly in the presence of a small\namount of SbCl5, due to the formation of :", "options": [ { "text": "carbene" }, { "text": "carbocation" }, { "text": "free radical" }, { "text": "carbanion" } ], "answer": "carbocation", "solution": "**Answer:** carbocation\n\n\"JEE ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1675, "subject": "Chemistry", "question": "Which of the following reaction is an example of ammonolysis?", "options": [ { "text": "C6H5COCl + C6H5NH2 $$ \\to $$ C6H5CONHC6H5" }, { "text": "C6H5CH2CN $$\\buildrel {[H]} \\over\n \\longrightarrow $$ C6H5CH2CH2NH2" }, { "text": "C6H5CH2CN $$\\buildrel {HCl} \\over\n \\longrightarrow $$ C6H5$$\\mathop N\\limits^ + $$H3Cl$$-$$" }, { "text": "C6H5CH2Cl + NH3 $$ \\to $$ C6H5CH2NH2" } ], "answer": "C6H5CH2Cl + NH3 $$ \\to $$ C6H5CH2NH2", "solution": "**Answer:** C6H5CH2Cl + NH3 $$ \\to $$ C6H5CH2NH2\n\nAmmonolysis of alkyl halides is the reaction of an alkyl\nhalide with NH3 which leads to the preparation of\namines.

    \nC6H5CH2Cl + NH3 $$ \\to $$ C6H5CH2NH2", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1676, "subject": "Chemistry", "question": "Given below are two statements :

    Statement I : C2H5OH and AgCN both can generate nucleophile.

    Statement II : KCN and AgCN both will generate nitrile nucleophile with all reaction conditions.

    Choose the most appropriate option :", "options": [ { "text": "Statement I is false but statement II is true." }, { "text": "Both statement I and statement II are true." }, { "text": "Both statement I and statement II are false." }, { "text": "Statement I is true but statement II is false." } ], "answer": "Statement I is true but statement II is false.", "solution": "**Answer:** Statement I is true but statement II is false.\n\nBoth C2H5OH and AgCN can generate nucleophile.

    \nKCN generates nitriles on substitution reactions\nwith haloalkanes where AgCN generates isonitriles\non substitution reactions with haloalkanes. Because\nKCN is ionic and has ‘C’ nucleophilic centre\nwhereas AgCN is covalent and has ‘N’ nucleophilic\ncentre.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1677, "subject": "Chemistry", "question": "

    In the presence of sunlight, benzene reacts with Cl2 to give product, X. The number of hydrogens in X is _____________.

    ", "options": [], "answer": "6", "solution": "**Answer:** 6\n\n

    \"JEE

    \n

    Total number of hydrogens are 6.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1678, "subject": "Chemistry", "question": "In the following halogenated organic compounds, the one with the maximum number of chlorine atoms in its structure is :", "options": [ { "text": "Gammaxene" }, { "text": "Chloropicrin" }, { "text": "Chloral" }, { "text": "Freon-12" } ], "answer": "Gammaxene", "solution": "**Answer:** Gammaxene\n\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1679, "subject": "Chemistry", "question": "

    Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R :

    \n

    Assertion A : Aryl halides cannot be prepared by replacement of hydroxyl group of phenol by halogen atom.

    \n

    Reason R : Phenols react with halogen acids violently.

    \n

    In the light of the above statements, choose the most appropriate from the options given below :

    ", "options": [ { "text": "A is true but R is false" }, { "text": "Both A and R are true but R is NOT the correct explanation of A" }, { "text": "A is false but R is true" }, { "text": "Both A and R are true and R is the correct explanation of A" } ], "answer": "A is true but R is false", "solution": "**Answer:** A is true but R is false\n\n

    Assertion (A): Given statement is correct because in phenol hydroxyl group cannot be replaced by halogen atom.

    \n

    Reason (R) :

    \n

    \"JEE

    \n

    Given reason is false.

    \n

    Hence Assertion (A) is correct but Reason (R) is false.

    ", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 1680, "subject": "Chemistry", "question": "The compound formed on heating chlorobenzene with chloral in the presence concentrated\nsulphuric acid is ", "options": [ { "text": "gammexene" }, { "text": "hexachloroethane " }, { "text": "Freon" }, { "text": "DDT" } ], "answer": "DDT", "solution": "**Answer:** DDT\n\n$$DDT$$ is prepared by heating chlorbenzene and chloral with concentrated sulphuric acid\n

    \"AIEEE ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1681, "subject": "Chemistry", "question": "Which of the following on heating with aqueous KOH, produces acetaldehyde ? ", "options": [ { "text": "CH3COCl" }, { "text": "CH3CH2Cl" }, { "text": "CH2ClCH2Cl" }, { "text": "CH3CHCl2" } ], "answer": "CH3CHCl2", "solution": "**Answer:** CH3CHCl2\n\n\"AIEEE ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1682, "subject": "Chemistry", "question": "What is DDT among the following?", "options": [ { "text": "Greenhouse gas" }, { "text": "A fertilizer" }, { "text": "Biodegradable pollutant" }, { "text": "Non–biodegradable pollutant\n" } ], "answer": "Non–biodegradable pollutant\n", "solution": "**Answer:** Non–biodegradable pollutant\n\n\n$$DDT$$ is a non-biodegradable pollutant.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1683, "subject": "Chemistry", "question": "Iodoform can be prepared from all except :", "options": [ { "text": "Ethyl methyl ketone" }, { "text": "Isopropyl alcohol" }, { "text": "3–Methyl – 2– butanone " }, { "text": "Isobutyl alcohol" } ], "answer": "Isobutyl alcohol", "solution": "**Answer:** Isobutyl alcohol\n\nIodoform test is given by methyl ketones, acetaldehyde and methyl secondary alcohols. \n
    \"AIEEE \n

    isobutyl alcohol is a primary alcohol hence does'nt give positive iodoform test. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1684, "subject": "Chemistry", "question": "The major organic compound formed by the reaction of 1, 1, 1– trichloroethane with silver powder is:\n", "options": [ { "text": "2- Butyne" }, { "text": "2- Butene" }, { "text": "Acetylene" }, { "text": "Ethene" } ], "answer": "2- Butyne", "solution": "**Answer:** 2- Butyne\n\n\"JEE ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1685, "subject": "Chemistry", "question": "The reaction of toluene with Cl2 in presence of FeCl3 gives predominantly", "options": [ { "text": "benzoyl chloride" }, { "text": "benzyl chloride" }, { "text": "o-and p-chlorotoluene" }, { "text": "m-chlorotoluene " } ], "answer": "o-and p-chlorotoluene", "solution": "**Answer:** o-and p-chlorotoluene\n\n$$FeC{l_3}\\,\\,$$ is Lewis acid. In presence of $$FeC{l_3}\\,\\,$$ side chain hydrogen atoms of toluene are substituted. \n

    \"AIEEE ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1686, "subject": "Chemistry", "question": "Toluene is nitrated and the resulting product is reduced with tin and hydrochloric acid. The product so\nobtained is diazotised and then heated with cuprous bromide. The reaction mixture so formed\ncontains ", "options": [ { "text": "mixture of o− and p−bromotoluenes " }, { "text": "mixture of o− and p−dibromobenzenes " }, { "text": "mixture of o− and p−bromoanilines " }, { "text": "mixture of o− and m−bromotoluenes" } ], "answer": "mixture of o− and p−bromotoluenes ", "solution": "**Answer:** mixture of o− and p−bromotoluenes \n\n\"AIEEE \n

    \"AIEEE \n

    on reduction with $$Sn/HCl$$ they will form corresponding anilines in which $$ - N{O_2}$$ group changes to $$ - N{H_2}.$$ \n

    \"AIEEE \n

    These anilines when diazotized and then treated with $$CuBr$$ forms $$o-, p-$$ bromotoluenes. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1687, "subject": "Chemistry", "question": "Polysubstitution is a major drawback in:", "options": [ { "text": "Reimer Tiemann reaction " }, { "text": "Friedel Craft's alkylation" }, { "text": "Acetylation of aniline" }, { "text": "Friedel Craft's acylation" } ], "answer": "Friedel Craft's alkylation", "solution": "**Answer:** Friedel Craft's alkylation\n\nPolysubstitution is a major drawback of Friedal–Craft alkylation.\n

    –CH3 group in highly activating group\n due to +H effect, so after first time addition of -CH3 group in benzene ring the reactivity of the benzene ring increases.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1688, "subject": "Chemistry", "question": "The correct sequence of reagents used in the preparation of 4-bromo-2-nitroethyl benzene from benzene is :", "options": [ { "text": "$$C{H_3}COCl/AlC{l_3},B{r_2}/AlB{r_3},HN{O_3}/{H_2}S{O_4},Zn/HCl$$" }, { "text": "$$C{H_3}COCl/AlC{l_3},Zn - Hg/HCl,B{r_2}/AlB{r_3},HN{O_3}/{H_2}S{O_4}$$" }, { "text": "$$B{r_2}/AlB{r_3},C{H_3}COCl/AlC{l_3},HN{O_3}/{H_2}S{O_4},Zn/HCl$$" }, { "text": "$$HN{O_3}/{H_2}S{O_4},B{r_2}/AlC{l_3},C{H_3}COCl/AlC{l_3},Zn - Hg/HCl$$" } ], "answer": "$$C{H_3}COCl/AlC{l_3},Zn - Hg/HCl,B{r_2}/AlB{r_3},HN{O_3}/{H_2}S{O_4}$$", "solution": "**Answer:** $$C{H_3}COCl/AlC{l_3},Zn - Hg/HCl,B{r_2}/AlB{r_3},HN{O_3}/{H_2}S{O_4}$$\n\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1689, "subject": "Chemistry", "question": "Benzene on nitration gives nitrobenzene in presence of HNO3 and H2SO4 mixture, where :", "options": [ { "text": "both H2SO4 and HNO3 act as a bases" }, { "text": "HNO3 acts as an acid and H2SO4 acts as a base" }, { "text": "both H2SO4 and HNO3 act as an acids" }, { "text": "HNO3 acts as a base and H2SO4 acts as an acid" } ], "answer": "HNO3 acts as a base and H2SO4 acts as an acid", "solution": "**Answer:** HNO3 acts as a base and H2SO4 acts as an acid\n\nReagent for nitration of Benzene

    $$\\mathop {{H_2}S{O_4}}\\limits_{(Acid)} + \\mathop {HN{O_3}}\\limits_{(Base)} \\mathbin{\\lower.3ex\\hbox{$\\buildrel\\textstyle\\rightarrow\\over\n {\\smash{\\leftarrow}\\vphantom{_{\\vbox to.5ex{\\vss}}}}$}} HSO_4^\\Theta + {H_2}\\mathop N\\limits^ \\oplus {O_3}$$

    $${H_2}\\mathop N\\limits^ \\oplus {O_3} \\mathbin{\\lower.3ex\\hbox{$\\buildrel\\textstyle\\rightarrow\\over\n {\\smash{\\leftarrow}\\vphantom{_{\\vbox to.5ex{\\vss}}}}$}} {H_2}O + \\mathop N\\limits^ \\oplus {O_2}$$

    \"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1690, "subject": "Chemistry", "question": "Which one of the following has the minimum boiling point?", "options": [ { "text": "n-butane" }, { "text": "isobutane" }, { "text": "1- butene " }, { "text": "1- butyne " } ], "answer": "isobutane", "solution": "**Answer:** isobutane\n\nNOTE : Among isomeric alkanes, the straight chain isomer has higher boiling point than the branched chain isomer. The greater the branching of the chain, the lower is the boiling point. Further due to the presence of $$\\pi $$ electrons, these molecules are slightly polar and hence have higher boiling points than the corresponding alkanes. \n

    Thus $$B.pt.$$ follows the order\n

    alkynes $$>$$ alkene $$>$$ alkanes (straight chain) > branched chain alkanes. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1691, "subject": "Chemistry", "question": "Which branched chain isomer of the hydrocarbon with molecular mass 72u gives only one isomer of\nmono substituted alkyl halide ?", "options": [ { "text": "Tertiary butyl chloride" }, { "text": "Neopentane" }, { "text": "Isohexane" }, { "text": "Neohexane" } ], "answer": "Neopentane", "solution": "**Answer:** Neopentane\n\n\"AIEEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1692, "subject": "Chemistry", "question": "

    Choose the correct set of reagents for the following conversion.

    \n

    trans $$\\left(\\mathrm{Ph}-\\mathrm{CH}=\\mathrm{CH}-\\mathrm{CH}_{3}\\right) \\rightarrow \\operatorname{cis}\\left(\\mathrm{Ph}-\\mathrm{CH}=\\mathrm{CH}-\\mathrm{CH}_{3}\\right)$$

    ", "options": [ { "text": "$$\\mathrm{Br_2,alc\\cdot KOH,NaNH_2,H_2}$$ Lindlar Catalyst" }, { "text": "$$\\mathrm{Br_2,aq\\cdot KOH,NaNH_2,Na}$$ (Liq NH$$_3$$)" }, { "text": "$$\\mathrm{Br_2,alc\\cdot KOH,NaNH_2,Na}$$ (Liq NH$$_3$$)" }, { "text": "$$\\mathrm{Br_2,aq\\cdot KOH,NaNH_2,H_2}$$ Lindlar Catalyst" } ], "answer": "$$\\mathrm{Br_2,alc\\cdot KOH,NaNH_2,H_2}$$ Lindlar Catalyst", "solution": "**Answer:** $$\\mathrm{Br_2,alc\\cdot KOH,NaNH_2,H_2}$$ Lindlar Catalyst\n\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1693, "subject": "Chemistry", "question": "At 300 K and 1 atm, 15 mL of a gaseous hydrocarbon requires 375 mL air containing 20% O2 by volume\nfor complete combustion. After combustion the gases occupy 345 mL. Assuming that the water formed is\nin liquid form and the volumes were measured at the same temperature and pressure, the formula of the hydrocarbon is:", "options": [ { "text": "C3H8" }, { "text": "C4H8" }, { "text": "C4H10 " }, { "text": "C3H6" } ], "answer": "C3H8", "solution": "**Answer:** C3H8\n\n$${C_x}{H_{y\\left( g \\right)}} + \\left( {{{4x + y} \\over 4}} \\right){O_{2\\left( g \\right)}}$$\n$$ \\to xC{O_{2\\left( g \\right)}} + {y \\over 2}{H_2}O\\left( l \\right)$$\n

    Volume of $${O_2}$$ used \n

    $$ = 375 \\times {{20} \\over {100}} = 75ml$$ \n

    $$\\therefore$$ From the reaction of combination\n

    $$1\\,\\,ml\\,{C_x}{H_y}\\,\\,$$ requires $$ = {{4x + y} \\over 4}ml\\,{O_2}$$ \n

    $$15\\left( {{{4x + y} \\over 4}} \\right) = 75$$ \n

    So, $$4x+y=20$$ \n

    $$x=3$$ \n

    $$y=8$$ \n

    $$\\therefore$$ $${C_3}{H_8}$$ ", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1694, "subject": "Chemistry", "question": "The trans-alkenes are formed by the reduction of alkynes with", "options": [ { "text": "Sn - HCl" }, { "text": "H2 – Pd/C, BaSO4" }, { "text": "NaBH4" }, { "text": "Na/liq. NH3" } ], "answer": "Na/liq. NH3", "solution": "**Answer:** Na/liq. NH3\n\n\"JEE ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1695, "subject": "Chemistry", "question": "When 2-butyne is treate with H2/Lindlar's catalyst, compound X is produced as the major product and when treated with Na/liq. NH3 it produces Y as the major product Which of the following statements is correct ? ", "options": [ { "text": "X will have higher dipole moment and higher boiling point than Y. " }, { "text": "Y will have higher dipole moment and higher boiling point than X. " }, { "text": "X will have lower dipole moment and lower boiling point than Y. " }, { "text": "Y will have higher dipole momet and lower boiling point than X. " } ], "answer": "X will have higher dipole moment and higher boiling point than Y. ", "solution": "**Answer:** X will have higher dipole moment and higher boiling point than Y. \n\n\"JEE\n

    Here dipole moment of Y = 0\n

    cis-isomer(X) will have higher dipole moment and higher boiling point than trans (Y). ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1696, "subject": "Chemistry", "question": "Consider the following reaction :

    \n$${\\left( {C{H_3}} \\right)_3}CCH\\left( {OH} \\right)C{H_3}\\buildrel {conc.\\,\\,\\,{H_2}S{O_4}} \\over\n \\longrightarrow $$

    \n$${\\left( {C{H_3}} \\right)_2}CHCH\\left( {Br} \\right)C{H_3}\\buildrel {alc.\\,\\,KOH} \\over\n \\longrightarrow $$

    \n$${\\left( {C{H_3}} \\right)_2}{\\rm{ }}CHCH\\left( {Br} \\right)C{H_3}\\buildrel {{{(C{H_3})}_3}{O^\\Theta }{K^ \\oplus }} \\over\n \\longrightarrow $$

    \n$${\\left( {C{H_3}} \\right)_2}\\mathop C\\limits_{\\mathop |\\limits_{OH} } - C{H_2} - CHO\\buildrel \\Delta \\over\n \\longrightarrow $$

    \nWhich of these reaction(s) will not produce Saytzeff product ?", "options": [ { "text": "(d) only" }, { "text": "(b) and (d)" }, { "text": "(a), (c) and (d) " }, { "text": "(c) only" } ], "answer": "(c) only", "solution": "**Answer:** (c) only\n\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1697, "subject": "Chemistry", "question": "In $$\\mathop C\\limits^1 {H_2} = \\mathop C\\limits^2 = \\mathop C\\limits^3 H - \\mathop C\\limits^4 {H_3}$$ molecule, the hybridization of carbon 1, 2, 3 and 4 respectively, are :", "options": [ { "text": "sp2, sp, sp2, sp3" }, { "text": "sp2, sp2, sp2, sp3" }, { "text": "sp3, sp, sp3, sp3" }, { "text": "sp2, sp3, sp2, sp3" } ], "answer": "sp2, sp, sp2, sp3", "solution": "**Answer:** sp2, sp, sp2, sp3\n\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1698, "subject": "Chemistry", "question": "

    Given below are two statements :

    \n

    Statement I : The presence of weaker $$\\pi$$-bonds make alkenes less stable than alkanes.

    \n

    Statement II : The strength of the double bond is greater than that of carbon-carbon single bond.

    \n

    In the light of the above statements, choose the correct answer from the options given below :

    ", "options": [ { "text": "Both Statement I and Statement II are correct." }, { "text": "Both Statement I and Statement II are incorrect." }, { "text": "Statement I is correct but Statement II is incorrect." }, { "text": "Statement I is incorrect but Statement II is correct." } ], "answer": "Both Statement I and Statement II are correct.", "solution": "**Answer:** Both Statement I and Statement II are correct.\n\nThe $\\pi$-bond present is alkenes is weaker than $\\sigma$ bond present in alkanes. That makes alkenes less stable than alkanes. Therefore, statement-I is correct.\n

    \nCarbon-carbon double bond is stronger than Carbon-carbon single bond because more energy is required to break 1 sigma and 1 pi bond than to break 1 sigma bond only. Therefore, statement-II is also correct.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1699, "subject": "Chemistry", "question": "The hydrocarbon which can react with sodium in liquid ammonia is ", "options": [ { "text": "CH3CH2CH2C≡CCH2CH2CH3 " }, { "text": "CH3CH2C≡CH" }, { "text": "CH3CH=CHCH3 " }, { "text": "CH3CH2C≡CCH2CH3" } ], "answer": "CH3CH2C≡CH", "solution": "**Answer:** CH3CH2C≡CH\n\nAlkynes having terminal $$ - C \\equiv H$$ react with Na in liquid ammonia to yield $${H_2}$$ gas of the given compounds $$C{H_3}C{H_2}C \\equiv CH$$ can react with Na in liquid $$N{H_3}$$ so the correct answer is $$(b).$$ \n

    \"AIEEE ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1700, "subject": "Chemistry", "question": "The correct order for acid strength of compounds :

    CH $$ \\equiv $$ CH, CH3–C $$ \\equiv $$ CH and CH2 = CH2 is as follows ", "options": [ { "text": "CH $$ \\equiv $$ CH > CH2 = CH2 > CH3 – C $$ \\equiv $$ CH " }, { "text": "CH3 – C $$ \\equiv $$ CH > CH2 = CH2 > HC $$ \\equiv $$ CH" }, { "text": "HC $$ \\equiv $$ CH > CH3 – C $$ \\equiv $$ CH > CH2 = CH2 " }, { "text": "CH3 – CH $$ \\equiv $$ CH > CH $$ \\equiv $$ CH > CH2 = CH2 " } ], "answer": "HC $$ \\equiv $$ CH > CH3 – C $$ \\equiv $$ CH > CH2 = CH2 ", "solution": "**Answer:** HC $$ \\equiv $$ CH > CH3 – C $$ \\equiv $$ CH > CH2 = CH2 \n\nAmong CH $$ \\equiv $$ CH and CH2 = CH2, in CH $$ \\equiv $$ CH compound H atom is attached with sp hybridised C atom and in CH2 = CH2 compound H atom is attached with sp2 hybridised C atom.\n

    As we know, electronegitivity of sp carbon is more compare to sp2 carbon atom and that H is more acidic which is attached with more electronegative carbon atom. So CH $$ \\equiv $$ CH is more acidic than CH2 = CH2.\n

    Among CH $$ \\equiv $$ CH and CH3–C $$ \\equiv $$ CH, in CH $$ \\equiv $$ CH both the H atom is connected to sp carbon atom but in CH3–C $$ \\equiv $$ CH, with one sp carbon atom H atom attached and with other sp carbon atom -CH3 group attached which have +I effect and we know +I effect decreases acidic strength.\n
    So, CH $$ \\equiv $$ CH is more acidic than CH3–C $$ \\equiv $$ CH.\n

    So correct order is\n

    HC $$ \\equiv $$ CH > CH3 – C $$ \\equiv $$ CH > CH2 = CH2 \n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1701, "subject": "Chemistry", "question": "Match List - I with List - II :

    List - I (Chemicals)

    (a) Alcoholic potassium hydroxide

    (b) Pd/BaSO4

    (c) BHC (Benzene hexachloride)

    (d) Polyacetylene

    List - II (Use / Preparation / Constituent)

    (i) Electrodes in batteries

    (ii) Obtained by addition reaction

    (iii) Used for $$\\beta$$ - elimination reaction

    (iv) Lindlar's catalyst

    Choose the most appropriate match :", "options": [ { "text": "(a) - (ii), (b) - (i), (c) - (iv), (d) - (iii)" }, { "text": "(a) - (iii), (b) - (iv), (c) - (ii), (d) - (i)" }, { "text": "(a) - (iii), (b) - (i), (c) - (iv), (d) - (ii)" }, { "text": "(a) - (ii), (b) - (iv), (c) - (i), (d) - (iii)" } ], "answer": "(a) - (iii), (b) - (iv), (c) - (ii), (d) - (i)", "solution": "**Answer:** (a) - (iii), (b) - (iv), (c) - (ii), (d) - (i)\n\nAlc. KOH causes elimination

    \nPd / BaSO4 – Lindlar’s catalyst

    \nBHC is obtained by the addition reaction of Cl2\nwith benzene in presence of U.V.

    \nThin film of polyacetylene can be used as\nelectrode in batteries.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1702, "subject": "Chemistry", "question": "

    The incorrect statement regarding ethyne is

    ", "options": [ { "text": "The carbon - carbon bonds in ethyne is weaker than that in ethene\n" }, { "text": "Both carbons are sp hybridised\n" }, { "text": "The $$\\mathrm{C}-\\mathrm{C}$$ bonds in ethyne is shorter than that in ethene\n" }, { "text": "Ethyne is linear" } ], "answer": "The carbon - carbon bonds in ethyne is weaker than that in ethene\n", "solution": "**Answer:** The carbon - carbon bonds in ethyne is weaker than that in ethene\n\n\n

    To evaluate which statement regarding ethyne is incorrect, it's necessary to review the characteristics of the chemical bonds in ethyne (acetylene, $$\\mathrm{C_2H_2}$$) compared to ethene (ethylene, $$\\mathrm{C_2H_4}$$).

    \n\n

    Option A: The statement that the carbon-carbon bonds in ethyne are weaker than that in ethene is incorrect. In ethyne, the carbon-carbon triple bond is composed of one sigma ($\\sigma$) bond and two pi ($\\pi$) bonds. This configuration makes the triple bond in ethyne much stronger and shorter compared to the double bond in ethene, which consists of one $\\sigma$ bond and one $\\pi$ bond. Thus, the carbon-carbon bond in ethyne is actually stronger than that in ethene.

    \n\n

    Option B: The statement that both carbons are sp hybridised in ethyne is correct. In ethyne, each carbon atom is bonded to the other carbon atom and a single hydrogen atom, necessitating the sp hybridisation to form a linear molecule with $180^\\circ$ angles between the bonds.

    \n\n

    Option C: The statement that the $$\\mathrm{C-C}$$ bonds in ethyne are shorter than that in ethene is correct. The presence of a triple bond between the carbon atoms in ethyne results in a shorter bond length compared to the double-bonded carbons in ethene. This is because the additional pi bonds in a triple bond pull the carbon atoms closer together.

    \n\n

    Option D: The statement that ethyne is linear is correct. Due to the sp hybridisation of carbon atoms in ethyne, the molecule adopts a linear geometry with bond angles of $180^\\circ$.

    \n\n

    Therefore, the incorrect statement regarding ethyne is Option A: The carbon-carbon bonds in ethyne are weaker than that in ethene.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1703, "subject": "Chemistry", "question": "On mixing a certain alkane with chlorine and irradiating it with ultravioletlight, it forms only one\nmonochloroalkane. This alkane could be", "options": [ { "text": "pentane" }, { "text": "isopentane" }, { "text": "neopentane" }, { "text": "propane" } ], "answer": "neopentane", "solution": "**Answer:** neopentane\n\nIn neopentane all the $$H$$ atoms are same $$\\left( {{1^ \\circ }} \\right).$$ \n

    \"AIEEE ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1704, "subject": "Chemistry", "question": "Of the five isomeric hexanes, the isomer which can give two monochlorinated\ncompounds is ", "options": [ { "text": "n-hexane" }, { "text": "2, 3-dimethylbutane " }, { "text": "2,2-dimethylbutane" }, { "text": "2-methylpentane" } ], "answer": "2, 3-dimethylbutane ", "solution": "**Answer:** 2, 3-dimethylbutane \n\n\"AIEEE \n

    Since it contains only two types of $$H$$-atoms hence it will give only two mono chlorinated compounds viz. \n

    \"AIEEE \n

    and \n

    \"AIEEE ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1705, "subject": "Chemistry", "question": "2 methylbutane on reacting with bromine in the presence of sunlight gives mainly", "options": [ { "text": "1 – bromo -2 - methylbutane" }, { "text": "2 – bromo -2 - methylbutane " }, { "text": "2 – bromo -3 - methylbutane" }, { "text": "1 – bromo -3 – methylbutane " } ], "answer": "2 – bromo -2 - methylbutane ", "solution": "**Answer:** 2 – bromo -2 - methylbutane \n\n\"AIEEE \n

    Ease of replacement of $$H$$-atom $${3^ \\circ } > {2^ \\circ } > {1^ \\circ }.$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1706, "subject": "Chemistry", "question": "5 L of an alkane requires 25 L of oxygen for its complete combustion. If all volumes are measured at constant temperature and pressure, the alkane is :", "options": [ { "text": "Ethane " }, { "text": "Propane " }, { "text": "Butane" }, { "text": "Isobutane" } ], "answer": "Propane ", "solution": "**Answer:** Propane \n\nWe know, conbustion of Hydrocarbon \n

    Cx Hy + (x + $${y \\over 4}$$)O2  $$ \\to $$  x CO2 + $${y \\over 2}$$ H2O\n

    So, to consume 1 mole of Cx Hy we need \n

    (x + $${y \\over 4}$$) mole of O2 gas.\n

    As both Cx Hy and O2 are gas,\n

    So avogadro's law is applicable on them.\n

    At constant temperature and pressure according to avogadro's law, volume $$ \\propto $$ mole\n

    So, for 5L of Cx Hy the \n

    volume of O2 = 5(x + $${y \\over 4}$$)\n

    According to the question,\n

    5(x + $${y \\over 4}$$) = 25\n

    $$\\therefore\\,\\,\\,$$ (x + $${y \\over 4}$$) = 5 . . . . . (1)\n

    For Ethane (C2 H6), x = 2 and y = 6\n

    $$\\therefore\\,\\,\\,$$ x + $${y \\over 4}$$ = 2 + $${6 \\over 4}$$ $$ \\ne $$ 5\n

    $$\\therefore\\,\\,\\,$$ C2 H6 can't be the required alkane. \n

    For Propane (C3 H8), x = 3. and y = 8\n

    $$\\therefore\\,\\,\\,$$ x + $${y \\over 4}$$ = 3 + $${8 \\over 4}$$ = 5\n

    $$\\therefore\\,\\,\\,$$ Propane (C3 H8) is the right alkane.\n

    Similarly for Butane (C4 H10) and Isobutane (C4 H10) you can check \n
    x + $${y \\over 4}$$ $$ \\ne $$ 5.", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1707, "subject": "Chemistry", "question": "The minimum number of moles of O2\n required for complete combustion of 1 mole of propane and 2\nmoles of butane is _____.", "options": [], "answer": "18", "solution": "**Answer:** 18\n\nC3H8(g) + 5O2(g) $$ \\to $$ 3CO2(g) + 4H2O(l)\n

    C4H10(g) + $${{13} \\over 2}$$O2(g) $$ \\to $$ 4CO2(g) + 5H2O(l)\n

    $$ \\therefore $$ No. of moles of O2 required to oxidise 1 mole of propane = 5\n

    $$ \\therefore $$ No. of moles of O2 required to oxidise 1 mole of butane = $${{13} \\over 2}$$\n

    So, No. of moles of O2 required to oxidise 1 mole of\n

    propane and 2 moles of butane = 5 + 2 $$ \\times $$ $${{13} \\over 2}$$ = 18", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1708, "subject": "Chemistry", "question": "Given below are two statements :

    Statement I : 2-methylbutane on oxidation with KMnO4 gives 2-methylbutan-2-ol.

    Statement II : n-alkanes can be easily oxidised to corresponding alcohols with KMnO4.

    Choose the correct option :", "options": [ { "text": "Both statement I and statement II are incorrect" }, { "text": "Both statement I and statement II are correct" }, { "text": "Statement I is correct but statement II is incorrect" }, { "text": "Statement I is incorrect but statement II is correct" } ], "answer": "Statement I is correct but statement II is incorrect", "solution": "**Answer:** Statement I is correct but statement II is incorrect\n\nAlkanes having tertiary H can be oxidized to\ncorresponding alcohols by KMnO4.

    \n\"JEE
    whereas ordinary alkanes resist oxidation.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1709, "subject": "Chemistry", "question": "Presence of which reagent will affect the reversibility of the following reaction, and change it to a irreversible reaction :

    CH4 + I2 $$\\mathrel{\\mathop{\\kern0pt\\rightleftharpoons}\n\\limits_{{\\mathop{\\rm Re}\\nolimits} versible}^{hv}} $$ CH3 $$-$$ I + HI", "options": [ { "text": "HOCl" }, { "text": "dilute HNO2" }, { "text": "Liquid NH3" }, { "text": "Concentrated HIO3" } ], "answer": "Concentrated HIO3", "solution": "**Answer:** Concentrated HIO3\n\nIodination of alkane is reversible reaction. It can be irreversible in the presence of strong oxidising agent like conc. HNO3 or conc. HIO3.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1710, "subject": "Chemistry", "question": "

    $$\n\\stackrel{\\bullet}{\\mathrm{Cl}}+\\mathrm{CH}_{4} \\rightarrow \\mathrm{A}+\\mathrm{B}\n$$

    \n

    A and B in the above atmospheric reaction step are :

    ", "options": [ { "text": "$$\\mathrm{C}_{2} \\mathrm{H}_{6}$$ and $$\\mathrm{Cl}_{2}$$" }, { "text": "$$\\stackrel{\\bullet}{\\mathrm{C}} \\mathrm{HCl}_{2}$$ and $$\\mathrm{H}_{2}$$" }, { "text": "$$\\stackrel{\\bullet}{\\mathrm{C}}\\mathrm{H}_{3}$$ and $$\\mathrm{HCl}$$" }, { "text": "$$\\mathrm{C}_{2} \\mathrm{H}_{6}$$ and $$\\mathrm{HCl}$$" } ], "answer": "$$\\stackrel{\\bullet}{\\mathrm{C}}\\mathrm{H}_{3}$$ and $$\\mathrm{HCl}$$", "solution": "**Answer:** $$\\stackrel{\\bullet}{\\mathrm{C}}\\mathrm{H}_{3}$$ and $$\\mathrm{HCl}$$\n\n$$\n\\stackrel{\\bullet}{\\mathrm{Cl}}+\\mathrm{CH}_4 \\longrightarrow {\\stackrel{\\bullet}{\\mathrm{C}}}\\mathrm{H}_3+\\mathrm{HCl}\n$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1711, "subject": "Chemistry", "question": "

    The one giving maximum number of isomeric alkenes on dehydrohalogenation reaction is (excluding rearrangement)

    ", "options": [ { "text": "2-Bromo-3,3-dimethylpentane" }, { "text": "2-Bromopropane" }, { "text": "1-Bromo-2-methylbutane" }, { "text": "2-Bromopentane" } ], "answer": "2-Bromopentane", "solution": "**Answer:** 2-Bromopentane\n\n

    \"JEE\n\"JEE\n\"JEE\n\n

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1712, "subject": "Chemistry", "question": "

    Maximum number of isomeric monochloro derivatives which can be obtained from 2,2,5,5-tetramethylhexane by chlorination is _____________

    ", "options": [], "answer": "3", "solution": "**Answer:** 3\n\n

    \"JEE

    \n

    Total numbers of isomer = 03

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1713, "subject": "Chemistry", "question": "

    Number of bromo derivatives obtained on treating ethane with excess of $$\\mathrm{Br}_{2}$$ in diffused sunlight is ___________

    ", "options": [], "answer": "9", "solution": "**Answer:** 9\n\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1714, "subject": "Chemistry", "question": "

    Given below are two statements :

    \n

    Statement (I) : $$\\mathrm{S}_{\\mathrm{N}} 2$$ reactions are 'stereospecific', indicating that they result in the formation of only one stereo-isomer as the product.

    \n

    Statement (II) : $$\\mathrm{S}_{\\mathrm{N}} 1$$ reactions generally result in formation of product as racemic mixtures.

    \n

    In the light of the above statements, choose the correct answer from the options given below :

    ", "options": [ { "text": "Both Statement I and Statement II are false\n" }, { "text": "Statement I is false but Statement II is true\n" }, { "text": "Statement I is true but Statement II is false\n" }, { "text": "Both Statement I and Statement II are true" } ], "answer": "Both Statement I and Statement II are true", "solution": "**Answer:** Both Statement I and Statement II are true\n\n

    Statement (I) explains a characteristic of $$\\mathrm{S}_{\\mathrm{N}}2$$ (Substitution Nucleophilic Bimolecular) reactions. In $$\\mathrm{S}_{\\mathrm{N}}2$$ reactions, the attack of the nucleophile and the departure of the leaving group occur simultaneously in a single step. This reaction has a back-side attack mechanism, where the nucleophile attacks the electrophilic carbon from the opposite side of the leaving group. This back-side attack leads to the inversion of configuration at the carbon atom undergoing substitution, a phenomenon often referred to as 'Walden inversion'. Thus, $$\\mathrm{S}_{\\mathrm{N}}2$$ reactions are stereospecific, leading to the formation of a product with a specific configuration (either inversion or retention of configuration, but typically inversion in the case of $$\\mathrm{S}_{\\mathrm{N}}2$$). Therefore, Statement I is true.

    \n

    Statement (II) concerns $$\\mathrm{S}_{\\mathrm{N}}1$$ (Substitution Nucleophilic Unimolecular) reactions, which proceed in two steps: first, the leaving group departs, forming a carbocation intermediate, and then the nucleophile attacks. The attack of the nucleophile can occur from either side of the planar carbocation, resulting in the formation of both enantiomers of the product. When the reactant is chiral, the product is usually a racemic mixture, consisting of equal amounts of both enantiomers, assuming that there is no steric or electronic preference for the attack on the carbocation. This makes $$\\mathrm{S}_{\\mathrm{N}}1$$ reactions non-stereospecific. Therefore, Statement II is true as well.

    \n

    Based on the explanations provided, the correct answer is Option D: Both Statement I and Statement II are true.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1715, "subject": "Chemistry", "question": "Butene-1 may be converted to butane by reaction with", "options": [ { "text": "Sn - HCI" }, { "text": "Zn - Hg" }, { "text": "Pd/H2" }, { "text": "Zn - HCI" } ], "answer": "Pd/H2", "solution": "**Answer:** Pd/H2\n\nAlkenes combine with hydrogen under pressure and in presence of a catalyst $$(Ni, Pt$$ or $$Pd)$$ and form alkanes. \n

    \"AIEEE ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1716, "subject": "Chemistry", "question": "Reaction of one molecule of HBr with one molecule of 1,3-butadiene at 40oC gives\npredominantly ", "options": [ { "text": "3-bromobutene under kinetically controlled conditions" }, { "text": "1-bromo-2-butene under thermodymically controlled conditions" }, { "text": "3-bromobutene under thermodynamically controlled conditions" }, { "text": "1-bromo-2-butene under kinetically controlled conditions " } ], "answer": "1-bromo-2-butene under thermodymically controlled conditions", "solution": "**Answer:** 1-bromo-2-butene under thermodymically controlled conditions\n\n\"AIEEE ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1717, "subject": "Chemistry", "question": "Which types of isomerism is shown by 2,3-dichlorobutane? ", "options": [ { "text": "Diastereo " }, { "text": "Optical " }, { "text": "Geometric " }, { "text": "Structural " } ], "answer": "Optical ", "solution": "**Answer:** Optical \n\n\"AIEEE \n

    $$2,3$$-dichloro butane will exhibit optical isomerism due to the presence of two asymmetric carbon atom. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1718, "subject": "Chemistry", "question": "The reaction of propene with HOCl (Cl2 + H2O) proceeds through the intermediate:", "options": [ { "text": "$$\\mathrm{CH}_{3}-\\mathrm{CH}^{+}-\\mathrm{CH}_{2}-\\mathrm{OH}$$" }, { "text": "$$\\mathrm{CH}_{3}-\\mathrm{CHCl}-\\mathrm{CH}_{2}^{+}$$" }, { "text": "CH3 - CH+ - CH2 - OH" }, { "text": "$$\\mathrm{CH}_{3}-\\mathrm{CH}^{+}-\\mathrm{CH}_{2}-\\mathrm{Cl}$$" } ], "answer": "$$\\mathrm{CH}_{3}-\\mathrm{CH}^{+}-\\mathrm{CH}_{2}-\\mathrm{Cl}$$", "solution": "**Answer:** $$\\mathrm{CH}_{3}-\\mathrm{CH}^{+}-\\mathrm{CH}_{2}-\\mathrm{Cl}$$\n\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1719, "subject": "Chemistry", "question": "An organic compound 'A' C4H8 on treatment with KMnO4/H+ yields compound 'B' C3H6O.

    Compound 'A' also yields compound 'B' an ozonolysis. Compound 'A' is :", "options": [ { "text": "2-Methylpropene" }, { "text": "1-Methylcyclopropane" }, { "text": "But-2-ene" }, { "text": "Cyclobutane" } ], "answer": "2-Methylpropene", "solution": "**Answer:** 2-Methylpropene\n\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1720, "subject": "Chemistry", "question": "The stereoisomers that are formed by electrophilic addition of bromine to trans-but-2-ene is/are :", "options": [ { "text": "2 enantiomers and 2 mesomers" }, { "text": "2 identical mesomers" }, { "text": "2 enantiomers" }, { "text": "1 racemic and 2 enantiomers" } ], "answer": "2 identical mesomers", "solution": "**Answer:** 2 identical mesomers\n\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1721, "subject": "Chemistry", "question": "

    Two isomers 'A' and 'B' with molecular formula C4H8 give different products on oxidation with KMnO4 in acidic medium. Isomer 'A' on reaction with KMnO4/H+ results in effervescence of a gas and gives ketone. The compound 'A' is

    ", "options": [ { "text": "But-1-ene." }, { "text": "cis-But-2-ene." }, { "text": "trans-But-2-ene." }, { "text": "2-methyl propene." } ], "answer": "2-methyl propene.", "solution": "**Answer:** 2-methyl propene.\n\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1722, "subject": "Chemistry", "question": "A hydrocarbon ' $\\mathrm{X}$ ' with formula $\\mathrm{C}_{6} \\mathrm{H}_{8}$ uses two moles of $\\mathrm{H}_{2}$ on catalytic hydrogenation of its one mole. On ozonolysis, ' $\\mathrm{X}$ ' yields two moles of methane dicarbaldehyde. The hydrocarbon ' $\\mathrm{X}$ ' is:", "options": [ { "text": "hexa-1, 3, 5-triene" }, { "text": "cyclohexa-1, 4-diene" }, { "text": "cyclohexa - 1,3 - diene" }, { "text": "1-methylcyclopenta-1, 4-diene" } ], "answer": "cyclohexa-1, 4-diene", "solution": "**Answer:** cyclohexa-1, 4-diene\n\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1723, "subject": "Chemistry", "question": "

    The number of possible isomeric products formed when 3-chloro-1-butene reacts with $$\\mathrm{HCl}$$ through carbocation formation is __________.

    ", "options": [], "answer": "4", "solution": "**Answer:** 4\n\n\"JEE\n

    Total Possible Isomeric product = 1+3 = 4", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1724, "subject": "Chemistry", "question": "

    Molar mass of the hydrocarbon (X) which on ozonolysis consumes one mole of $$\\mathrm{O}_{3}$$ per mole of $$(\\mathrm{X})$$ and gives one mole each of ethanal and propanone is _________ $$\\mathrm{g}~ \\mathrm{mol}^{-1}$$

    (Molar mass of $$\\mathrm{C}: 12 \\mathrm{~g} \\mathrm{~mol}^{-1}, \\mathrm{H}: 1 \\mathrm{~g} \\mathrm{~mol}^{-1}$$ )

    ", "options": [], "answer": "70", "solution": "**Answer:** 70\n\n\"JEE
    Hydrocarbon $(\\mathrm{X})$ is 2-methyl-but-2-ene $\\left(\\mathrm{C}_5 \\mathrm{H}_{10}\\right)$

    And the molecular mass is $5(12)+10(1)=70$.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1725, "subject": "Chemistry", "question": "Which of these will not react with acetylene ?", "options": [ { "text": "NaOH" }, { "text": "ammonical AgNO3\n " }, { "text": "Na" }, { "text": "HCl" } ], "answer": "NaOH", "solution": "**Answer:** NaOH\n\nAcetylene reacts with the other three as : \n

    \"AIEEE ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1726, "subject": "Chemistry", "question": "What is the product when acetylene reacts with hypochlorous acid ?", "options": [ { "text": "CH3COCI" }, { "text": "ClCH2CHO" }, { "text": "Cl2CHCHO" }, { "text": "ClCHCOOH" } ], "answer": "Cl2CHCHO", "solution": "**Answer:** Cl2CHCHO\n\n\"AIEEE ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1727, "subject": "Chemistry", "question": "Acetylene does not react with :", "options": [ { "text": "$\\mathrm{Na}$" }, { "text": "ammoniacal $\\mathrm{AgNO}_3$" }, { "text": "$\\mathrm{HCl}$" }, { "text": "$\\mathrm{NaOH}$" } ], "answer": "$\\mathrm{NaOH}$", "solution": "**Answer:** $\\mathrm{NaOH}$\n\n1. Acetylene reacts with sodium :\n

    $$\\mathrm{2CH} \\equiv \\mathrm{CH} + 2\\mathrm{Na} \\longrightarrow 2\\mathrm{CH} \\equiv \\mathrm{C}^{-}\\mathrm{Na}^{+} + \\mathrm{H}_2$$\n

    Here, acetylene reacts with sodium to produce sodium acetylide and hydrogen gas.\n\n

    2. Acetylene reacts with ammoniacal silver nitrate :\n

    $$\\mathrm{CH} \\equiv \\mathrm{CH} + 2\\mathrm{AgNO}_3 \\longrightarrow \\mathrm{Ag}_2\\mathrm{C}_2 + 2\\mathrm{HNO}_3$$\n

    In this reaction, acetylene reacts with ammoniacal silver nitrate to form a white precipitate of silver acetylide and nitric acid.\n\n

    3. Acetylene reaction with hydrochloric acid in the presence of a $\\mathrm{Hg}^{2+}$ catalyst (oxymercuration-demercuration) :\n

    $$\\mathrm{CH} \\equiv \\mathrm{CH} + \\mathrm{Hg(OAc)_2} + \\mathrm{H}_2\\mathrm{O} \\longrightarrow \\mathrm{CH}_2=\\mathrm{CH}-\\mathrm{OH} + \\mathrm{HgOAc}$$\n

    $$\\mathrm{CH}_2=\\mathrm{CH}-\\mathrm{OH} + \\mathrm{NaBH}_4 \\longrightarrow \\mathrm{CH}_2=\\mathrm{CH}-\\mathrm{H} + \\mathrm{B(OH)_3} + \\mathrm{NaOAc}$$\n

    Acetylene does not readily react with $\\mathrm{HCl}$. However, in the presence of a catalyst like $\\mathrm{Hg}^{2+}$, a specific type of electrophilic addition known as oxymercuration can occur. The result is vinyl chloride. Note that $\\mathrm{NaBH}_4$ is used to reduce the mercurinium ion intermediate and to replace the $\\mathrm{Hg}$ group with a hydrogen atom.\n\n

    4. Acetylene does not react with sodium hydroxide :\n

    $$\\mathrm{CH} \\equiv \\mathrm{CH} + \\mathrm{NaOH} \\longrightarrow \\text { no reaction }$$\n

    According to the principles of the Hard and Soft Acids and Bases (HSAB) theory, soft acids prefer to bind with soft bases and hard acids prefer to bind with hard bases. In this case, $\\mathrm{OH}^-$ is a hard base, while the $\\mathrm{C}$ in acetylene is a soft acid, so they are not particularly reactive with each other.\n\n

    5. Acetylene reacts with sodium amide :\n

    $$\\mathrm{CH} \\equiv \\mathrm{CH} + \\mathrm{NaNH}_2 \\longrightarrow \\mathrm{CH} \\equiv \\mathrm{C}^{-}\\mathrm{Na}^{+} + \\mathrm{NH}_3$$\n

    Here, sodium amide acts as a strong base and nucleophile, deprotonating the acetylene to form sodium acetylide and ammonia.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1728, "subject": "Chemistry", "question": "Which of the following reactions will yield 2, 2-dibromopropane? ", "options": [ { "text": "CH3 - CH = CH2 + HBr $$\\to$$" }, { "text": "CH3 - C $$\\equiv$$ CH + 2HBr $$\\to$$" }, { "text": "CH3CH = CHBr + HBr $$\\to$$" }, { "text": "CH $$\\equiv$$ CH + 2HBr $$\\to$$" } ], "answer": "CH3 - C $$\\equiv$$ CH + 2HBr $$\\to$$", "solution": "**Answer:** CH3 - C $$\\equiv$$ CH + 2HBr $$\\to$$\n\nThe reaction follows Markownikoff rule which states that when unsymmetrical reagent adds across unsymmetrical double or triple bond the negative part adds to carbon atom having lesser number of hydrogen atoms. \n

    \"AIEEE ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1729, "subject": "Chemistry", "question": "In the following sequence of reactions, the alkene affords the compound ‘B’
    \nCH3 - CH = CH - CH3 $$\\buildrel {{O_3}} \\over\n \\longrightarrow $$ A $$\\mathrel{\\mathop{\\kern0pt\\longrightarrow}\n\\limits_{Zn}^{H{}_2O}} $$ B
    \nThe compound B is", "options": [ { "text": "CH3CH2CHO " }, { "text": "CH3COCH3" }, { "text": "CH3CH2COCH3" }, { "text": "CH3CHO" } ], "answer": "CH3CHO", "solution": "**Answer:** CH3CHO\n\nCompleting the sequence of given reactions, \n

    \"AIEEE \n

    Thus $$'B'\\,\\,$$ is $$\\,\\,C{H_3}CHO$$ \n

    Hence $$(d)$$ is correct answer.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1730, "subject": "Chemistry", "question": "Ozonolysis of an organic compound gives formaldehyde as one of the products. This confirms the\npresence of :", "options": [ { "text": "a vinyl group" }, { "text": "an isopropyl group " }, { "text": "an acetylenic triple bond" }, { "text": "two ethylenic double bonds" } ], "answer": "a vinyl group", "solution": "**Answer:** a vinyl group\n\n\"AIEEE \n
    \"AIEEE ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1731, "subject": "Chemistry", "question": "2–Hexyne gives trans–2–Hexene on treatment with :", "options": [ { "text": "Pt/H2" }, { "text": "Li/NH3" }, { "text": "Pd/BaSO4" }, { "text": "LiAlH4" } ], "answer": "Li/NH3", "solution": "**Answer:** Li/NH3\n\nAnti addition of hydrogen atoms to the the triple bond occurs when alkynes are reduced with sodium (or lithium) metal in ammonia, ethylamine, or alcohol at low temperatures. This reaction called, a dissolving metal reduction, produces an $$(E)$$- or $$trans$$-alkene. \n

    Sodium in liq. $$N{H_3}$$ is used as a source of electrons in the reduction of an alkyne to a $$trans$$ alkene. \n

    \"AIEEE ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1732, "subject": "Chemistry", "question": "Which of the following is Lindlar catalyst?", "options": [ { "text": "Partially deactivated palladised charcoal" }, { "text": "Zinc chloride and HCl" }, { "text": "Sodium and Liquid NH3" }, { "text": "Cold dilute solution of KMnO4" } ], "answer": "Partially deactivated palladised charcoal", "solution": "**Answer:** Partially deactivated palladised charcoal\n\nLindlar's catalyst $$ \\Rightarrow $$ Pd/CaCO3 + (CH3COO)2 Pb + quinolene", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1733, "subject": "Chemistry", "question": "Metallic sodium does not react normally with :", "options": [ { "text": "gaseous ammonia" }, { "text": "But-2-yne" }, { "text": "Ethyne" }, { "text": "tert-butyl alcohol" } ], "answer": "But-2-yne", "solution": "**Answer:** But-2-yne\n\nMetallic sodium does not react with 2-butyne because 2-butyne does not have acidic hydrogen.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1734, "subject": "Chemistry", "question": "

    The major product of the following reaction is P.

    \n

    $$\\mathrm{CH}_3 \\mathrm{C}\\equiv\\mathrm{C}-\\mathrm{CH}_3$$ $$\\xrightarrow[\\substack{\\text { (ii) dil. } \\mathrm{KMnO}_4 \\\\ 273 \\mathrm{~K}}]{\\text { (i) } \\mathrm{Na} \\text { /liq. } \\mathrm{NH}_3}$$ 'P'

    \n

    Number of oxygen atoms present in product '$$\\mathrm{P}$$' is _______. (nearest integer)

    ", "options": [], "answer": "2", "solution": "**Answer:** 2\n\n

    Product is

    \n

    \"JEE

    \n

    Number of oxygen atoms = 2

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1735, "subject": "Chemistry", "question": "Which one of the following process will produce hard water?", "options": [ { "text": "Saturation of water with MgCO3" }, { "text": "Saturation of water with CaSO4" }, { "text": "Addition of Na2SO4 to water" }, { "text": "Saturation of water with CaCO3" } ], "answer": "Saturation of water with CaSO4", "solution": "**Answer:** Saturation of water with CaSO4\n\nPermanent hardness of water is due to chlorides and sulphates of calcium and magnesium i.e. $$CaC{l_2},CaS{O_4},MgC{l_2}\\,$$ and $$\\,\\,MgS{O_4}.$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1736, "subject": "Chemistry", "question": "The temporary hardness of water is due to : ", "options": [ { "text": "Na2SO4" }, { "text": "NaCl" }, { "text": "Ca(HCO3)2" }, { "text": "CaCl2" } ], "answer": "Ca(HCO3)2", "solution": "**Answer:** Ca(HCO3)2\n\nCa(HCO3)2 is responsible for temporary hardness of water.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1737, "subject": "Chemistry", "question": "100 mL of a water sample contains 0.81 g of calcium bicrabonate and 0.73 g of magnesium\nbicarbonate. The hardness of this water sample expressed in terms of equivalents of CaCO3 is :\n(Molar mass of calcium bicarbonate is 162 g mol–1 and magnesium bicarbonate is 146 g mol–1)", "options": [ { "text": "1,000 ppm" }, { "text": "100 ppm" }, { "text": "10,000 ppm" }, { "text": "5,000 ppm" } ], "answer": "10,000 ppm", "solution": "**Answer:** 10,000 ppm\n\nHere hardness of water is expressed in terms of CaCO3.\n

    $$ \\therefore $$ Equivalent of CaCO3\n

    = Equivalent of Ca(HCO3)2 + Equivalent of Mg(HCO3)2\n

    $$ \\Rightarrow $$ 2 $$ \\times $$ $${W \\over {100}}$$ = $${{0.81} \\over {162}} \\times 2 + {{0.73} \\over {146}} \\times 2$$\n

    $$ \\Rightarrow $$ W = 1 gm\n

    Volume of water = 100 mL\n

    $$ \\therefore $$ Mass of water = 100 g\n

    $$ \\therefore $$ Hardness = $${{1.0} \\over {100}} \\times {10^6}$$ = 10000 ppm", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1738, "subject": "Chemistry", "question": "The temporary hardness of a water sample is due to compound X. Boiling this sample converts X to\ncompound Y. X and Y, respectively, are : ", "options": [ { "text": "Mg(HCO3)2 and MgCO3" }, { "text": "Ca(HO3)2 and CaO" }, { "text": "Ca(HCO3)2 and Ca(OH)2" }, { "text": "Mg(HCO3)2 and Mg(OH)2" } ], "answer": "Mg(HCO3)2 and Mg(OH)2", "solution": "**Answer:** Mg(HCO3)2 and Mg(OH)2\n\nAccoring to NCERT this is the right reaction\n
    Mg(HCO3)2 $$\\buildrel \\Delta \\over\n \\longrightarrow $$ Mg(OH)2 $$ \\downarrow $$ + CO2 $$ \\uparrow $$ + H2O\n

    But right reaction should be,\n
    Mg(HCO3)2 $$\\buildrel \\Delta \\over\n \\longrightarrow $$ MgCO3 $$ \\downarrow $$ + CO2 $$ \\uparrow $$ + H2O", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1739, "subject": "Chemistry", "question": "In comparison to the zeolite process for the removal of permanent hardness, the synthetic resins\nmethod is :", "options": [ { "text": "less efficient as the resins cannot be regenerated.\n" }, { "text": "more efficient as it can exchange only cations." }, { "text": "more efficient as it can exchange both cations as well as anions." }, { "text": "less efficient as it exchanges only anions." } ], "answer": "more efficient as it can exchange both cations as well as anions.", "solution": "**Answer:** more efficient as it can exchange both cations as well as anions.\n\nSynthetic resin method is more efficient because it can exchange both cation & anion.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1740, "subject": "Chemistry", "question": "Amongst the following, the form of water with\nthe lowest ionic conductance at 298 K is :", "options": [ { "text": "saline water used for intravenous injection" }, { "text": "sea water" }, { "text": "distilled water" }, { "text": "water from a well" } ], "answer": "distilled water", "solution": "**Answer:** distilled water\n\nThe form of H2O with the lowest ionic conductance at 298 K is distilled water.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1741, "subject": "Chemistry", "question": "The one that is NOT suitable for the removal of permanent hardness of water is :", "options": [ { "text": "Ion-exchange method\n" }, { "text": "Calgon's method" }, { "text": "Treatment with sodium carbonate" }, { "text": "Clark's method" } ], "answer": "Clark's method", "solution": "**Answer:** Clark's method\n\nFor Removal of Temporary hardness two methods are present \n
    (1) Clark’s method \n
    (2) Boiling\n

    While permanent hardness of water can be\nremoved 4 methods\n
    (1) Washong soda method (Na2CO3 method), \n
    (2) Calgon method\n
    (3) Permutit Process\n
    (4) Ion-exchange Resin method.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1742, "subject": "Chemistry", "question": "The functional groups that are responsible for the ion-exchange property of cation and anion exchange resins, respectively, are :", "options": [ { "text": "$$-$$SO3H and $$-$$NH2" }, { "text": "$$-$$NH2 and COOH" }, { "text": "$$-$$NH2 and $$-$$SO3H" }, { "text": "$$-$$SO3H and $$-$$COOH" } ], "answer": "$$-$$SO3H and $$-$$NH2", "solution": "**Answer:** $$-$$SO3H and $$-$$NH2\n\nCation exchange resins contain large organic\na molecule with –SO3H group.

    \nIn the cation exchange process H+ exchanges for Na+,\nCa2+, Mg2+ and other cations present in water.

    \nWhile anion exchange resins contain –NH2 in form\nof –NH3+ OH– where OH– exchanges for anions like\nCl–, HCO3–, SO4\n2–, etc.", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 1743, "subject": "Chemistry", "question": "Which one of the following methods is most suitable for preparing deionized water?", "options": [ { "text": "Synthetic resin method" }, { "text": "Clark's method" }, { "text": "Calgon's method" }, { "text": "Permutit method" } ], "answer": "Synthetic resin method", "solution": "**Answer:** Synthetic resin method\n\nPure demineralised (de-ionized) water free from all soluble mineral salts is obtained by passing water successively through a cation exchange (in the H+ form) and an anion exchange (in the OH$$-$$ form) resins.", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 1744, "subject": "Chemistry", "question": "

    Boiling of hard water is helpful in removing the temporary hardness by converting calcium hydrogen carbonate and magnesium hydrogen carbonate to :

    ", "options": [ { "text": "CaCO3 and Mg(OH)2" }, { "text": "CaCO3 and MgCO3" }, { "text": "Ca(OH)2 and MgCO3" }, { "text": "Ca(OH)2 and Mg(OH)2" } ], "answer": "CaCO3 and Mg(OH)2", "solution": "**Answer:** CaCO3 and Mg(OH)2\n\n$$\\mathrm{Ca}\\left(\\mathrm{HCO}_{3}\\right)_{2} \\stackrel{\\Delta}{\\longrightarrow} \\mathrm{CaCO}_{3} \\downarrow+\\mathrm{H}_{2} \\mathrm{O}+\\mathrm{CO}_{2} \\uparrow$$

    \n$$\\mathrm{Mg}\\left(\\mathrm{HCO}_{3}\\right)_{2} \\stackrel{\\mathrm{A}}{\\longrightarrow} \\mathrm{Mg}(\\mathrm{OH})_{2}+2 \\mathrm{CO}_{2} \\uparrow$$\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1745, "subject": "Chemistry", "question": "

    The metal salts formed during softening of hardwater using Clark's method are :

    ", "options": [ { "text": "$$\\mathrm{Ca}(\\mathrm{OH})_{2}$$ and $$\\mathrm{Mg}(\\mathrm{OH})_{2}$$" }, { "text": "$$\\mathrm{CaCO}_{3}$$ and $$\\mathrm{Mg}(\\mathrm{OH})_{2}$$" }, { "text": "$$\\mathrm{Ca}(\\mathrm{OH})_{2}$$ and $$\\mathrm{MgCO}_{3}$$" }, { "text": "$$\\mathrm{CaCO}_{3}$$ and $$\\mathrm{MgCO}_{3}$$" } ], "answer": "$$\\mathrm{CaCO}_{3}$$ and $$\\mathrm{Mg}(\\mathrm{OH})_{2}$$", "solution": "**Answer:** $$\\mathrm{CaCO}_{3}$$ and $$\\mathrm{Mg}(\\mathrm{OH})_{2}$$\n\nIn Clark's method, calculated amount of lime is added to hard water. It precipitates out calcium carbonate and magnesium hydroxide which can filtered off.\n

    \n$$\n\\begin{aligned}\n&\\mathrm{Ca}\\left(\\mathrm{HCO}_{3}\\right)_{2}+\\mathrm{Ca}(\\mathrm{OH})_{2} \\rightarrow 2 \\mathrm{CaCO}_{3} \\downarrow+2 \\mathrm{H}_{2} \\mathrm{O} \\\\\\\\\n&\\mathrm{Mg}\\left(\\mathrm{HCO}_{3}\\right)_{2}+2 \\mathrm{Ca}(\\mathrm{OH})_{2} \\rightarrow 2 \\mathrm{CaCO}_{3} \\downarrow+\\mathrm{Mg}(\\mathrm{OH})_{2}+2 \\mathrm{H}_{2} \\mathrm{O}\n\\end{aligned}\n$$", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 1746, "subject": "Chemistry", "question": "

    The products obtained during treatment of hard water using Clark's method are :

    ", "options": [ { "text": "$$\\mathrm{CaCO}_{3}$$ and $$\\mathrm{MgCO}_{3}$$" }, { "text": "$$\\mathrm{Ca}(\\mathrm{OH})_{2}$$ and $$\\mathrm{Mg}(\\mathrm{OH})_{2}$$" }, { "text": "$$\\mathrm{CaCO}_{3}$$ and $$\\mathrm{Mg}(\\mathrm{OH})_{2}$$" }, { "text": "$$\\mathrm{Ca}(\\mathrm{OH})_{2}$$ and $$\\mathrm{MgCO}_{3}$$" } ], "answer": "$$\\mathrm{CaCO}_{3}$$ and $$\\mathrm{Mg}(\\mathrm{OH})_{2}$$", "solution": "**Answer:** $$\\mathrm{CaCO}_{3}$$ and $$\\mathrm{Mg}(\\mathrm{OH})_{2}$$\n\nIn Clark's method lime water is used

    \n$$\n\\begin{aligned}\n&\\mathrm{Ca}\\left(\\mathrm{HCO}_3\\right)_2+2 \\mathrm{Ca}(\\mathrm{OH})_2 \\rightarrow 2 \\mathrm{CaCO}_3+2 \\mathrm{H}_2 \\mathrm{O} \\\\\\\\\n&\\mathrm{Mg}\\left(\\mathrm{HCO}_3\\right)_2+2 \\mathrm{Ca}(\\mathrm{OH})_2 \\rightarrow 2 \\mathrm{CaCO}_3+\\mathrm{Mg}(\\mathrm{OH})_2+2 \\mathrm{H}_2 \\mathrm{O}\n\\end{aligned}\n$$", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 1747, "subject": "Chemistry", "question": "

    Given below are two statements :

    \n

    Statement I : Permutit process is more efficient compared to the synthetic resin method for the softening of water.

    \n

    Statement II : Synthetic resin method results in the formation of soluble sodium salts.

    \n

    In the light of the above statements, choose the most appropriate answer from the options given below:

    ", "options": [ { "text": "Statement I is correct but Statement II is incorrect" }, { "text": "Both the Statements I and II are correct" }, { "text": "Both the Statements I and II are incorrect" }, { "text": "Statement I is incorrect but Statement II is correct" } ], "answer": "Both the Statements I and II are incorrect", "solution": "**Answer:** Both the Statements I and II are incorrect\n\nI. Synthetic resin is more efficient as cations as well as anions responsible for removal of permanent hardness.

    \nII. In synthetic resin method, cations and anions are removed in the form of ppt.

    \nHence, both statements are incorrect.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1748, "subject": "Chemistry", "question": "NaH is an example of : ", "options": [ { "text": "metallic hydride " }, { "text": "saline hydride " }, { "text": "electron-rich hydrid" }, { "text": "molecular hydride " } ], "answer": "saline hydride ", "solution": "**Answer:** saline hydride \n\nNaH is an example of ionic hydride which is also known as saline hydride.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1749, "subject": "Chemistry", "question": "The correct statements among (a) to (d) are :\n
    (a) saline hydrides produce H2 gas when reacted with H2O.\n
    (b) reaction of LiAlH4 with BF3 leads to B2H6.\n
    (c) PH3 and CH4 are electron - rich and electron - precise hydrides, respectively.\n
    (d) HF and CH4 are called as molecular ", "options": [ { "text": "(a), (b), (c) and (d)" }, { "text": "(a), (c) and (d) only" }, { "text": "(c) and (d) only" }, { "text": "(a), (b) and (c) only" } ], "answer": "(a), (b), (c) and (d)", "solution": "**Answer:** (a), (b), (c) and (d)\n\n(a) NaH + H2O $$ \\to $$ NaOH + H2\n

    (b) 3LiAlH4 + 4BF3 $$ \\to $$ 2B2H6 + 3LiF + 3AlF3\n

    (c) PH3 is electron rich due to one lone pair while CH4 is electron\nprecise hydride.\n

    (d) HF and CH4 are molecular hydrides.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1750, "subject": "Chemistry", "question": "If the boiling point of H2O is 373 K, the boiling point of H2S will be :", "options": [ { "text": "greater than 300 K but less than 373 K" }, { "text": "equal to 373 K" }, { "text": "more than 373 K" }, { "text": "less than 300 K" } ], "answer": "less than 300 K", "solution": "**Answer:** less than 300 K\n\nBoiling point of H2S < Boiling point of H2O\n

    Boiling point of H2S = 213 K", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1751, "subject": "Chemistry", "question": "

    Match List I with List II

    \n

    \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    LIST I
    Type of Hydride    		LIST II
    Example
    A.Electron deficient hydrideI.$$\\mathrm{MgH_2}$$
    B.Electron rich hydrideII.$$\\mathrm{HF}$$
    C.Electron precise hydrideIII.$$\\mathrm{B_2H_6}$$
    D.Saline hydrideIV.$$\\mathrm{CH_4}$$

    \n

    Choose the correct answer from the options given below:

    ", "options": [ { "text": "A-III, B-II, C-I, D-IV" }, { "text": "A-II, B-III, C-I, D-IV" }, { "text": "A-III, B-II, C-IV, D-I" }, { "text": "A-II, B-III, C-IV, D-I" } ], "answer": "A-III, B-II, C-IV, D-I", "solution": "**Answer:** A-III, B-II, C-IV, D-I\n\nA Electron deficient hydride -> III $\\mathrm{B}_2 \\mathrm{H}_6$

    \nB Electron rich hydride -> II $\\mathrm{HF}$

    \nC Electron precise hydride -> IV $\\mathrm{CH}_4$

    \nD Saline hydride -> I $\\mathrm{MgH}_2$", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1752, "subject": "Chemistry", "question": "

    Which hydride among the following is less stable?

    ", "options": [ { "text": "NH$$_3$$" }, { "text": "HF" }, { "text": "LiH" }, { "text": "BeH$$_2$$" } ], "answer": "BeH$$_2$$", "solution": "**Answer:** BeH$$_2$$\n\n

    The stability of hydrides depends on their bonding and the electronegativity difference between the elements. The greater the electronegativity difference, the more stable the hydride. Based on this, the order of stability of the given hydrides is:

    \n

    HF > NH$$_3$$ > LiH > BeH$$_2$$

    \n

    Therefore, the less stable hydride among the given options is BeH$$_2$$.

    \n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1753, "subject": "Chemistry", "question": "From the following statements regarding H2O2, choose the incorrect statement :", "options": [ { "text": "It decomposes on exposure to light" }, { "text": "It has to be stored in plastic or wax lined glass bottles in dark" }, { "text": "It has to be kept away from dust" }, { "text": "It can act only as an oxidizing agent" } ], "answer": "It can act only as an oxidizing agent", "solution": "**Answer:** It can act only as an oxidizing agent\n\n$${H_2}{O_2}$$ has oxidizing and reducing properties both. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1754, "subject": "Chemistry", "question": "In which of the following reactions, hydrogen peroxide acts as an oxidizing\nagent ?\n", "options": [ { "text": "HOCl + H2O2 $$ \\to $$ H3O+ + Cl$$-$$ + O2 " }, { "text": "I2 + H2O2 + 2OH$$-$$ $$ \\to $$ 2I$$-$$ + 2H2O + O2 " }, { "text": "2Mn$$O_4^ - $$ + 3H2O2 $$ \\to $$ 2 MnO2 + 3O2 + 2H2O + 2OH$$-$$" }, { "text": "PbS + 4H2O2 $$ \\to $$ PbSO4 + 4H2O" } ], "answer": "PbS + 4H2O2 $$ \\to $$ PbSO4 + 4H2O", "solution": "**Answer:** PbS + 4H2O2 $$ \\to $$ PbSO4 + 4H2O\n\n\"JEE\n

    Since, H2O2 gains electron to convert into H2O. Hence, H2O2 acts as an oxidizing agent. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1755, "subject": "Chemistry", "question": "Hydrogen peroxide oxidises [Fe(CN)6]4- to [Fe(CN)6]3- in acidic medium but reduces [Fe(CN)6]3- to [Fe(CN)6]4- in alkaline medium. The other products formed are, respectively :", "options": [ { "text": "H2O and (H2O + OH-)" }, { "text": "(H2O + O2) and H2O " }, { "text": "(H2O + O2) and (H2O + OH-) " }, { "text": "H2O and (H2O + O2)" } ], "answer": "H2O and (H2O + O2)", "solution": "**Answer:** H2O and (H2O + O2)\n\nIn acidic medium, [Fe(CN)6]-3 is oxidise to [Fe(CN)6]-3 by H2O2. So H2O2 will perform reduction reaction.\n

    Reduction reaction of H2O2.\n

    $${H_2}O_2^{ - 1}\\mathrel{\\mathop{\\kern0pt\\longrightarrow}\n\\limits_{{\\mathop{\\rm Re}\\nolimits} duction}} {H_2}\\mathop O\\limits^{ - 2} $$\n

    Complete reaction : \n

    [ Fe (CN)6 ]$$-$$4 + H2O2 + 2H+ $$\\buildrel \\, \\over\n \\longrightarrow $$ 2 [ Fe(CN)6 ] $$-$$3 + 2H2O\n

    In alkaline medium, [Fe (CN)6]$$-$$3 reduce to [ Fe(CN)6 ]$$-$$4 \n

    So H2O2 will oxidize. \n

    Oxidization of H2O2 :\n

    H2$$O_2^{ - 1}$$ $$\\buildrel \\, \\over\n \\longrightarrow $$ $$O_2^0$$\n

    Complete reaction : \n

    [Fe (CN)6 ]$$-$$3 + H2O2 + 2OH$$-$$ $$\\buildrel \\, \\over\n \\longrightarrow $$ 2 [Fe(CN)6 ] $$-$$4 + 2H2O + O2\n

    So, ion acidic medium H2O and in alkaline medium O2 + H2O are produced as by product.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1756, "subject": "Chemistry", "question": "The chemical nature of hydrogen peroxide is : ", "options": [ { "text": "Oxidising agent in acidic medium, but not in basic medium \n" }, { "text": "Oxidising and reducing agent in both acidic and basic medium " }, { "text": "Reducing agent in basic medium, but not in acidic medium \n" }, { "text": "Oxidising and reducing agent in acidic medium, but not in basic medium." } ], "answer": "Oxidising and reducing agent in both acidic and basic medium ", "solution": "**Answer:** Oxidising and reducing agent in both acidic and basic medium \n\nH2O2 act as oxidising agent and reducing agent in acidic medium as well as basic medium.\n

    H2O2 Act as oxidant : - \n

    H2O2 + 2H$$^ \\oplus $$ + 2e$$-$$ $$ \\to $$  2H2O (In acidic medium)\n

    H2O2 + 2e$$-$$ $$ \\to $$ 2OH$$-$$ (In basic medium)\n

    H2O2 Act as reductant : -\n

    H2O2 $$ \\to $$ 2H+ + O2 + 2e$$-$$ (In acidic medium)\n

    H2O2 + 2OH$$-$$ $$ \\to $$ 2H2O + O2 + 2e$$-$$ (In basic medium)", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1757, "subject": "Chemistry", "question": "The volume strength of 1M H2O2 is : (Molar mass of H2O2 = 34 g mol–1)", "options": [ { "text": "5.6" }, { "text": "22.4" }, { "text": "11.35" }, { "text": "16.8" } ], "answer": "11.35", "solution": "**Answer:** 11.35\n\nAt   STP, \n

    Volume strength = 11.35 $$ \\times $$ Molarity\n

    $$ \\therefore $$  Volume strength of H2O2 = 11.35 $$ \\times $$ 1 = 11.35", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1758, "subject": "Chemistry", "question": "The strength of 11.2 volume solution of H2O2\nis : [Given that molar mass of H = 1 g mol–1\nand O = 16 g mol–1]", "options": [ { "text": "3.4%" }, { "text": "34%" }, { "text": "13.6%" }, { "text": "1.7%" } ], "answer": "3.4%", "solution": "**Answer:** 3.4%\n\nStrength means $${{Weight} \\over {Volume}}$$ %.\n

    Molarity of H2O2 = $${{volume\\,strength} \\over {11.2}}$$ = $${{11.2} \\over {11.2}}$$ = 1 M = 34 gm/lit\n

    In 1000 ml solution 34 gm H2O2 present\n

    $$ \\therefore $$ In 100 ml solution 3.4 gm H2O2 present.", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1759, "subject": "Chemistry", "question": "Among the statements (a) - (d), the correct ones are :
    \n(a) Decomposition of hydrogen peroxide gives dioxygen.
    \n(b) Like hydrogen peroxide, compounds, such as KCIO3, Pb(NO3)2 and NaNO3 when heated liberate dioxygen.
    \n(c) 2-Ethylanthraquinone is useful for the industrial preparation of hydrogen peroxide.
    \n(d) Hydrogen peroxide is used for the manufacture of sodium perborate.\n", "options": [ { "text": "(a), (b), (c) and (d)" }, { "text": "(a) and (c) only" }, { "text": "(a), (b) and (c) only" }, { "text": "(a), (c) and (d) only" } ], "answer": "(a), (b), (c) and (d)", "solution": "**Answer:** (a), (b), (c) and (d)\n\nH2O2 $$\\buildrel \\Delta \\over\n \\longrightarrow $$ 2H2O +O2\n

    KClO3, Pb(NO3)2, NaNO3 on heating will release\nO2 gas.\n

    2H3BO3 + 2NaOH + 2H2O2\n$$ \\to $$ Na2[B2(O2)(OH)4] + 4H2O", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1760, "subject": "Chemistry", "question": "Hydrogen peroxide, in the pure state, is :", "options": [ { "text": "Linear and blue in color" }, { "text": "Linear and almost colorless" }, { "text": "Non-planar and almost colorless" }, { "text": "Planar and bluein color" } ], "answer": "Non-planar and almost colorless", "solution": "**Answer:** Non-planar and almost colorless\n\n\"JEE\n

    Hydrogen peroxide, in the pure state, has openbook like structure which is non- planar and almost colourless (very pale blue) liquid.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1761, "subject": "Chemistry", "question": "(A) HOCI + H2O2 → H3O+ + Cl- + O2

    \n(B) I2 + H2O2 + 2OH- → 2I- + 2H2O + O2

    \nChoose the correct option.", "options": [ { "text": "H2O2 acts as reducing and oxidising agent respectively in equations (A) and (B)." }, { "text": "H2O2 act as oxidizing and reducing agent respectively in equations (A) and (B)." }, { "text": "H2O2 acts as oxidising agent in equations (A) and (B)." }, { "text": "H2O2 acts as reducing agent in equations (A) and (B)." } ], "answer": "H2O2 acts as reducing agent in equations (A) and (B).", "solution": "**Answer:** H2O2 acts as reducing agent in equations (A) and (B).\n\nIn equation (A), HOCl undergoes reduction in presence of H2O2.\n\"JEE\n
    Here, oxidation state of Cl changes from +1 to –1\n(i.e. reduces)\n

    $$ \\therefore $$ H2O2 act as reducing agent I2\nreduces to I-\nin presence of H2O2.\n

    In equation (B),\n\"JEE\n
    Here oxidation state of iodine decreases (from 0 to –1)\n

    $$ \\therefore $$ H2O2 act as reducing agent in both the equations.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1762, "subject": "Chemistry", "question": "Which of the following equation depicts the oxidizing nature of H2O2?", "options": [ { "text": "$$C{l_2} + {H_2}{O_2} \\to 2HCl + {O_2}$$" }, { "text": "$$KI{O_4} + {H_2}{O_2} \\to KI{O_3} + {H_2}O + {O_2}$$" }, { "text": "$$2{I^ - } + {H_2}{O_2} + 2{H^ + } \\to {I_2} + 2{H_2}O$$" }, { "text": "$${I_2} + {H_2}{O_2} + 2O{H^ - } \\to 2{I^ - } + 2{H_2}O + {O_2}$$" } ], "answer": "$$2{I^ - } + {H_2}{O_2} + 2{H^ + } \\to {I_2} + 2{H_2}O$$", "solution": "**Answer:** $$2{I^ - } + {H_2}{O_2} + 2{H^ + } \\to {I_2} + 2{H_2}O$$\n\n$$2{I^ - } + {H_2}{O_2} + 2{H^ + } \\to {I_2} + 2{H_2}O$$\n

    Oxygen reduces from –1 to –2.\n

    So, its reduction will takes place. Hence it will behave as oxidising agent or it shows oxidising\nnature.\n

    While in other option it change from (–1) to 0.", "topic": "Algebra", "subtopic": "Equations and Inequalities" }, { "id": 1763, "subject": "Chemistry", "question": "Given below are two statements :

    Statement I : H2O2 can act as both oxidising and reducing agent in basic medium.

    Statement II : In the hydrogen economy, the energy is transmitted in the form of dihydrogen.

    In the light of the above statements, choose the correct answer from the options given below :", "options": [ { "text": "Both statement I and statement II are false" }, { "text": "Statement I is false but statement II is true" }, { "text": "Statement I is true but statement II is false" }, { "text": "Both statement I and statement II are true" } ], "answer": "Both statement I and statement II are true", "solution": "**Answer:** Both statement I and statement II are true\n\nOxidising action in basic medium

    \n2Fe2+ + H2O2 $$ \\to $$ 2Fe3+ + 2OH–

    \nReducing action in basic medium

    \nI2 + H2O2 + 2OH– $$ \\to $$ 2I– + 2H2O + O2

    \nAdvantage of hydrogen economy is that energy is\ntransmitted in the form of dihydrogen and not as\nelectric power", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1764, "subject": "Chemistry", "question": "The correct statements about H2O2 are :

    (A) used in the treatment of effluents.

    (B) used as both oxidising and reducing agents.

    (C) the two hydroxyl groups lie in the same plane.

    (D) miscible with water.

    Choose the correct answer from the options given below :", "options": [ { "text": "(A), (C) and (D) only" }, { "text": "(A), (B), (C) and (D)" }, { "text": "(A), (B) and (D) only" }, { "text": "(B), (C) and (D) only" } ], "answer": "(A), (B) and (D) only", "solution": "**Answer:** (A), (B) and (D) only\n\nH2O2 act as both oxidising and reducing agent.

    \nH2O2 is miscible with water

    \ndue to open book like structure both -OH group are not in the same plane

    \nH2O2 used in the treatment of effluents.", "topic": "Geometry", "subtopic": "Plane Geometry" }, { "id": 1765, "subject": "Chemistry", "question": "In basic medium, H2O2 exhibits which of the following reactions?

    (A) Mn2+ $$ \\to $$ Mn4+

    (B) I2 $$ \\to $$ I$$-$$

    (C) PbS $$ \\to $$ PbSO4

    Choose the most appropriate answer from the options given below :", "options": [ { "text": "(B) only" }, { "text": "(A), (B) only" }, { "text": "(A) only" }, { "text": "(A), (C) only" } ], "answer": "(A), (B) only", "solution": "**Answer:** (A), (B) only\n\nIn basic medium, oxidising action of H2O2.

    \nMn2+ + H2O2 $$ \\to $$ Mn+4 + 2OH–

    \nIn basic medium, reducing action of H2O2.

    \nI2 + H2O2 + 2OH– $$ \\to $$ 2I– + 2H2O + O2

    \nIn acidic medium, oxidising action of H2O2.

    \nPbS(s) + 4H2O2(aq) $$ \\to $$ PbSO4(s) + 4H2O(l)", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1766, "subject": "Chemistry", "question": "Given below are two statements : One is labelled as Assertion A and other is labelled as Reason R.

    Assertion A : The dihedral angles in H2O2 in gaseous phase is 90.2$$^\\circ$$ and in solid phase is 111.5$$^\\circ$$.

    Reason R : The change in dihedral angle in solid and gaseous phase is due to the difference in the intermolecular forces.

    Choose the most appropriate answer from the options given below for A and R.", "options": [ { "text": "A is correct but R is not correct." }, { "text": "Both A and R are correct but R is not the correct explanation of A." }, { "text": "Both A and R are correct and R is the correct explanation of A." }, { "text": "A is not correct but R is correct." } ], "answer": "A is not correct but R is correct.", "solution": "**Answer:** A is not correct but R is correct.\n\n\"JEE\n

    (a) H2O2 structure in gas phase, dihedral angle is 111.5°.\n

    (b) H2O2 structure in solid phase at 110K, dihedral angle is 90.2°\n

    Hence given statement (A) is not correct.\n

    But statement (B) is correct.", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1767, "subject": "Chemistry", "question": "The oxide that gives H2O2 most readily on treatment with H2O is :", "options": [ { "text": "PbO2" }, { "text": "Na2O2" }, { "text": "SnO2" }, { "text": "BaO2 . 8H2O" } ], "answer": "Na2O2", "solution": "**Answer:** Na2O2\n\n(a) PbO2 + 2H2O $$\\to$$ Pb(OH)4

    (b) Na2O2 + 2H2O $$\\to$$ 2NaOH + H2O2

    this reaction is possible at room temperature

    (c) SnO2 + 2H2O $$\\to$$ Sn(OH)4

    (d) Acidified BaO2 . 8H2O gives H2O2 after evaporation.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1768, "subject": "Chemistry", "question": "Hydrogen peroxide reacts with iodine in basic medium to give :", "options": [ { "text": "IO$$_4^ - $$" }, { "text": "IO$$-$$" }, { "text": "I$$-$$" }, { "text": "IO$$_3^ - $$" } ], "answer": "I$$-$$", "solution": "**Answer:** I$$-$$\n\nI2 + H2O2 + 2OH$$-$$ $$\\to$$ 2I$$-$$ + 2H2O + O2", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1769, "subject": "Chemistry", "question": "

    Which one of the following reactions indicates the reducing ability of hydrogen peroxide in basic medium?

    ", "options": [ { "text": "HOCl + H2O2 $$\\to$$ H3O+ + Cl$$-$$ + O2" }, { "text": "PbS + 4H2O2 $$\\to$$ PbSO4 + 4H2O" }, { "text": "2MnO$$_4^ - $$ + 3H2O2 $$\\to$$ 2MnO2 + 3O2 + 2H2O + 2OH$$-$$" }, { "text": "Mn2+ + H2O2 $$\\to$$ Mn4+ + 2OH$$-$$" } ], "answer": "2MnO$$_4^ - $$ + 3H2O2 $$\\to$$ 2MnO2 + 3O2 + 2H2O + 2OH$$-$$", "solution": "**Answer:** 2MnO$$_4^ - $$ + 3H2O2 $$\\to$$ 2MnO2 + 3O2 + 2H2O + 2OH$$-$$\n\n$2 \\mathrm{Mn}\\overset{+7}{\\mathrm{O}_{4}^{-}}+3 \\mathrm{H}_{2} \\mathrm{O}_{2} \\rightarrow$\n

    \n$2 \\mathrm{Mn}\\overset{+4}{\\mathrm{O}_{2}}+3 \\mathrm{O}_{2}+2 \\mathrm{H}_{2} \\mathrm{O}+2 \\mathrm{O} \\overline{\\mathrm{H}}$\n

    \nIn basic medium $\\mathrm{MnO}_{4}^{-}$is reduced to $\\mathrm{MnO}_{2}$, whereas in acidic medium it is reduced to $\\mathrm{Mn}^{+2}$.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1770, "subject": "Chemistry", "question": "

    Consider the following reaction:

    \n

    $$2HSO_4^ - (aq)\\mathrel{\\mathop{\\kern0pt\\longrightarrow}\n\\limits_{(2)\\,Hydrolysis}^{(1)\\,Electrolysis}} 2HSO_4^ - + 2{H^ + } + A$$

    \n

    The dihedral angle in product A in its solid phase at 110 K is :

    ", "options": [ { "text": "104$$^\\circ$$" }, { "text": "111.5$$^\\circ$$" }, { "text": "90.2$$^\\circ$$" }, { "text": "111.0$$^\\circ$$" } ], "answer": "90.2$$^\\circ$$", "solution": "**Answer:** 90.2$$^\\circ$$\n\nA should be $\\mathrm{H}_{2} \\mathrm{O}_{2}$\n

    \nStructure of $\\mathrm{H}_{2} \\mathrm{O}_{2}$ is solid phase

    \n\"JEE
    \nDihedral angle $=90.2^{\\circ}$", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1771, "subject": "Chemistry", "question": "

    The reaction of $$\\mathrm{H}_{2} \\mathrm{O}_{2}$$ with potassium permanganate in acidic medium leads to the formation of mainly :

    ", "options": [ { "text": "Mn2+" }, { "text": "Mn4+" }, { "text": "Mn3+" }, { "text": "Mn6+" } ], "answer": "Mn2+", "solution": "**Answer:** Mn2+\n\nThe reaction of $$\\mathrm{KMnO}_{4}$$ with $$\\mathrm{H}_{2} \\mathrm{O}_{2}$$ in acidic medium is as\n

    \n$$\n\\begin{aligned}\n2 \\mathrm{KMnO}_{4}+3 \\mathrm{H}_{2} \\mathrm{SO}_{4}+5 \\mathrm{H}_{2} \\mathrm{O}_{2} \\\\\n& \\rightarrow \\mathrm{K}_{2} \\mathrm{SO}_{4}+2 \\mathrm{MnSO}_{4}+8 \\mathrm{H}_{2} \\mathrm{O}+5 \\mathrm{O}_{2}\n\\end{aligned}\n$$

    \n$$\\therefore \\mathrm{Mn}^{2+}$$ will be formed as the product.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1772, "subject": "Chemistry", "question": "

    Which of the following can be used to prevent the decomposition of $$\\mathrm{H}_{2} \\mathrm{O}_{2}$$ ?

    ", "options": [ { "text": "Urea" }, { "text": "Formaldehyde" }, { "text": "Formic acid" }, { "text": "Ethanol" } ], "answer": "Urea", "solution": "**Answer:** Urea\n\nUrea is used as a stabilizer for the storage of $$\\mathrm{H}_{2} \\mathrm{O}_{2}$$.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1773, "subject": "Chemistry", "question": "The products obtained from a reaction of hydrogen peroxide and acidified potassium permanganate are :", "options": [ { "text": "$$\\mathrm{Mn}^{4+}, \\mathrm{H}_{2} \\mathrm{O}$$ only" }, { "text": "$$\\mathrm{Mn}^{2+}, \\mathrm{H}_{2} \\mathrm{O}$$ only" }, { "text": "$$\\mathrm{Mn}^{4+}, \\mathrm{H}_{2} \\mathrm{O}, \\mathrm{O}_{2}$$ only" }, { "text": "$$\\mathrm{Mn}^{2+}, \\mathrm{H}_{2} \\mathrm{O}, \\mathrm{O}_{2}$$ only" } ], "answer": "$$\\mathrm{Mn}^{2+}, \\mathrm{H}_{2} \\mathrm{O}, \\mathrm{O}_{2}$$ only", "solution": "**Answer:** $$\\mathrm{Mn}^{2+}, \\mathrm{H}_{2} \\mathrm{O}, \\mathrm{O}_{2}$$ only\n\n$$2 \\mathrm{MnO}_{4}^{-}+6 \\mathrm{H}^{+}+5 \\mathrm{H}_{2} \\mathrm{O}_{2} \\longrightarrow 2 \\mathrm{Mn}^{2+}+8 \\mathrm{H}_{2} \\mathrm{O}+5 \\mathrm{O}_{2}$$\n

    \nThis reaction shows reducing action of $$\\mathrm{H}_{2} \\mathrm{O}_{2}$$ in acidic medium.\n

    \nThe products formed are $$\\mathrm{Mn}^{2+}, \\mathrm{H}_{2} \\mathrm{O}$$ and $$\\mathrm{O}_{2}$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1774, "subject": "Chemistry", "question": "

    Given below are two statements.

    \n

    Statement I : Hydrogen peroxide can act as an oxidizing agent in both acidic and basic conditions.

    \n

    Statement II : Density of hydrogen peroxide at $$298 \\mathrm{~K}$$ is lower than that of $$\\mathrm{D}_{2} \\mathrm{O}$$.

    \n

    In the light of the above statements, choose the correct answer from the options given below.

    ", "options": [ { "text": "Both Statement I and Statement II are true." }, { "text": "Both Statement I and Statement II are false." }, { "text": "Statement I is true but Statement II is false." }, { "text": "Statement I is false but Statement II is true." } ], "answer": "Statement I is true but Statement II is false.", "solution": "**Answer:** Statement I is true but Statement II is false.\n\nDensity of $\\mathrm{H}_{2} \\mathrm{O}_{2}$ is more as compared to $\\mathrm{D}_{2} \\mathrm{O}$\n

    \n$$\n\\begin{aligned}\n&\\mathrm{d}_{\\mathrm{H}_{2} \\mathrm{O}_{2}}=1.44 \\mathrm{~g} / \\mathrm{cc} \\\\\\\\\n&\\mathrm{d}_{\\mathrm{D}_{2} \\mathrm{O}}=1.106 \\mathrm{~g} / \\mathrm{cc}\n\\end{aligned}\n$$\n

    \nAnd hydrogen peroxide acts as an oxidising as well as reducing agent in both acidic and basic medium.\n

    \n$\\therefore $ Statement $\\mathrm{I}$ is correct.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1775, "subject": "Chemistry", "question": "

    The oxidation state of manganese in the product obtained in a reaction of potassium permanganate and hydrogen peroxide in basic medium is ____________.

    ", "options": [], "answer": "4", "solution": "**Answer:** 4\n\n$2 \\mathrm{KMnO}_4+3 \\mathrm{H}_2 \\mathrm{O}_2 \\stackrel{\\text { basic medium }}{\\longrightarrow} \\overset{+4}{2 \\mathrm{MnO}_2}+3 \\mathrm{O}_2+2 \\mathrm{H}_2 \\mathrm{O}+2 \\mathrm{KOH}$

    \nOxidation state of Mn in MnO2 = +4", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1776, "subject": "Chemistry", "question": "

    On reaction with stronger oxidizing agent like $$\\mathrm{KIO}_{4}$$, hydrogen peroxide oxidizes with the evolution of $$\\mathrm{O}_{2}$$. The oxidation number of $$\\mathrm{I}$$ in $$\\mathrm{KIO}_{4}$$ changes to __________.

    ", "options": [], "answer": "5", "solution": "**Answer:** 5\n\n$\\overset{+7}{\\mathrm{KIO}_{4}}+\\mathrm{H}_{2} \\mathrm{O}_{2} \\longrightarrow \\overset{+5}{\\mathrm{KIO}_{3}}+\\mathrm{H}_{2} \\mathrm{O}+\\mathrm{O}_{2}$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1777, "subject": "Chemistry", "question": "

    The starting material for convenient preparation of deuterated hydrogen peroxide ($$\\mathrm{D_{2}O_{2}}$$) in :

    ", "options": [ { "text": "$$\\mathrm{BaO_{2}}$$" }, { "text": "2-ethylanthraquinol" }, { "text": "$$\\mathrm{BaO}$$" }, { "text": "$$\\mathrm{K_{2}S_{2}O_{8}}$$" } ], "answer": "$$\\mathrm{K_{2}S_{2}O_{8}}$$", "solution": "**Answer:** $$\\mathrm{K_{2}S_{2}O_{8}}$$\n\n$\\mathrm{K}_2 \\mathrm{~S}_2 \\mathrm{O}_8$ is used in the laboratory preparation of $\\mathrm{D}_2 \\mathrm{O}_2$\n

    $$\n\\mathrm{K}_2 \\mathrm{~S}_2 \\mathrm{O}_8(\\mathrm{~s})+2 \\mathrm{D}_2 \\mathrm{O}(\\mathrm{l}) \\rightarrow 2 \\mathrm{KDSO}_4(\\mathrm{aq})+\\mathrm{D}_2 \\mathrm{O}_2(\\mathrm{l})\n$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1778, "subject": "Chemistry", "question": "

    Given below are two statements :

    \n

    Statement I : $ \\mathrm{H}_{2} \\mathrm{O}_{2}$ is used in the synthesis of Cephalosporin

    \n

    Statement II : $ \\mathrm{H}_{2} \\mathrm{O}_{2}$ is used for the restoration of aerobic conditions to sewage wastes.

    \n

    In the light of the above statements, choose the most appropriate answer from the options given below :

    ", "options": [ { "text": "Statement $\\mathbf{I}$ is incorrect but Statement $\\mathbf{II}$ is correct" }, { "text": "Statement $\\mathbf{I}$ is correct but Statement $\\mathbf{II}$ is incorrect" }, { "text": "Both Statement $\\mathbf{I}$ and Statement $\\mathbf{II}$ are incorrect" }, { "text": "Both Statement $\\mathbf{I}$ and Statement $\\mathbf{II}$ are correct" } ], "answer": "Both Statement $\\mathbf{I}$ and Statement $\\mathbf{II}$ are correct", "solution": "**Answer:** Both Statement $\\mathbf{I}$ and Statement $\\mathbf{II}$ are correct\n\nH2O2 is used in the synthesis of hydroquinone,\ntartaric acid and certain food products and\npharmaceuticals (cephalosporin).\n

    Nowadays it is also used in environmental (green)\nchemistry for example in pollution control\ntreatment of domestic and industrial effluents,\noxidation of cyanides restoration of aerobic\ncondition to sewage waste.\n

    Hence both statements are correct.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1779, "subject": "Chemistry", "question": "

    $$\\mathrm{H}_{2} \\mathrm{O}_{2}$$ acts as a reducing agent in :

    ", "options": [ { "text": "$$\\mathrm{Mn}^{2+}+2 \\mathrm{H}_{2} \\mathrm{O}_{2} \\rightarrow \\mathrm{MnO}_{2}+2 \\mathrm{H}_{2} \\mathrm{O}$$" }, { "text": "$$2 \\mathrm{Fe}^{2+}+2 \\mathrm{H}^{+}+\\mathrm{H}_{2} \\mathrm{O}_{2} \\rightarrow 2 \\mathrm{Fe}^{3+}+2 \\mathrm{H}_{2} \\mathrm{O}$$" }, { "text": "$$\\mathrm{Na}_{2} \\mathrm{S}+4 \\mathrm{H}_{2} \\mathrm{O}_{2} \\rightarrow \\mathrm{Na}_{2} \\mathrm{SO}_{4}+4 \\mathrm{H}_{2} \\mathrm{O}$$" }, { "text": "$$2 \\mathrm{NaOCl}+\\mathrm{H}_{2} \\mathrm{O}_{2} \\rightarrow 2 \\mathrm{NaCl}+\\mathrm{H}_{2} \\mathrm{O}+\\mathrm{O}_{2}$$" } ], "answer": "$$2 \\mathrm{NaOCl}+\\mathrm{H}_{2} \\mathrm{O}_{2} \\rightarrow 2 \\mathrm{NaCl}+\\mathrm{H}_{2} \\mathrm{O}+\\mathrm{O}_{2}$$", "solution": "**Answer:** $$2 \\mathrm{NaOCl}+\\mathrm{H}_{2} \\mathrm{O}_{2} \\rightarrow 2 \\mathrm{NaCl}+\\mathrm{H}_{2} \\mathrm{O}+\\mathrm{O}_{2}$$\n\n$\\mathrm{H}_{2} \\mathrm{O}_{2}$ act as a reducing agent.\n\n

    $2 \\stackrel{+1}{\\mathrm{Na}}\\stackrel{-2}{\\mathrm{O}} \\stackrel{+1}{\\mathrm{Cl}}+\\mathrm{H}_{2} \\mathrm{O}_{2} \\longrightarrow 2 \\mathrm{NaCl}+\\mathrm{H}_{2} \\mathrm{O}+\\mathrm{O}_{2}$\n\n

    $\\mathrm{Cl}$ from (+1) state changes to $\\mathrm{Cl}^{-1}$.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1780, "subject": "Chemistry", "question": "

    In which of the following reactions the hydrogen peroxide acts as a reducing agent?

    ", "options": [ { "text": "$$\\mathrm{M{n^{2 + }} + {H_2}{O_2} \\to M{n^{4 + }} + 2O{H^ - }}$$" }, { "text": "$$\\mathrm{HOCl + {H_2}{O_2} \\to {H_3}{O^ + } + C{l^ - } + {O_2}}$$" }, { "text": "$$\\mathrm{PbS + 4{H_2}{O_2} \\to PbS{O_4} + 4{H_2}O}$$" }, { "text": "$$\\mathrm{2F{e^{2 + }} + {H_2}{O_2} \\to 2F{e^{3 + }} + 2O{H^ - }}$$" } ], "answer": "$$\\mathrm{HOCl + {H_2}{O_2} \\to {H_3}{O^ + } + C{l^ - } + {O_2}}$$", "solution": "**Answer:** $$\\mathrm{HOCl + {H_2}{O_2} \\to {H_3}{O^ + } + C{l^ - } + {O_2}}$$\n\n$\\mathrm{HOCl}+\\mathrm{H}_{2} \\mathrm{O}_{2} \\longrightarrow \\mathrm{H}_{3} \\mathrm{O}^{\\oplus}+\\mathrm{Cl}^{\\ominus}+\\mathrm{O}_{2}$\n

    \nIn this reaction $\\mathrm{H}_{2} \\mathrm{O}_{2}$ is acting as a reducing agent as $\\mathrm{Cl}$ is undergoing a change in oxidation state from $+1$ to $-1$.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1781, "subject": "Chemistry", "question": "

    Which of the following can reduce decomposition of $$\\mathrm{H}_{2} \\mathrm{O}_{2}$$ on exposure to light :

    ", "options": [ { "text": "Urea" }, { "text": "Glass containers" }, { "text": "Alkali" }, { "text": "Dust" } ], "answer": "Urea", "solution": "**Answer:** Urea\n\nUrea is a negative catalyst, which means that it slows down the rate of a chemical reaction. In the case of hydrogen peroxide, urea absorbs the light that would otherwise cause it to decompose. This prevents the reaction from happening, or at least slows it down significantly.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1782, "subject": "Chemistry", "question": "Which of the following could act as a propellant for rockets?", "options": [ { "text": "Liquid oxygen + liquid argon" }, { "text": "Liquid hydrogen + liquid oxygen" }, { "text": "Liquid nitrogen + liquid oxygen" }, { "text": "Liquid hydrogen + liquid nitrogen" } ], "answer": "Liquid hydrogen + liquid oxygen", "solution": "**Answer:** Liquid hydrogen + liquid oxygen\n\nLiquid hydrogen and liquid oxygen are used as excellent fuel for rockets. $${H_2}\\left( \\ell \\right)$$ has low mass and high enthalpy of combustion whereas oxygen is a strong supporter of combustion.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1783, "subject": "Chemistry", "question": "Which of the following statements in relation to the hydrogen atom is correct? ", "options": [ { "text": "3s orbital is lower in energy than 3p orbital " }, { "text": "3p orbital is lower in energy than 3d orbital " }, { "text": "3s and 3p orbitals are of lower energy than 3d orbital " }, { "text": "3s, 3p and 3d orbitals all have the same energy" } ], "answer": "3s, 3p and 3d orbitals all have the same energy", "solution": "**Answer:** 3s, 3p and 3d orbitals all have the same energy\n\nNOTE : In one electron species, such as $$H$$-atom, the energy of orbital depends only on the principal quantum number, $$n.$$ \n

    Hence answer $$(d)$$ \n

    $$i.e.$$ is $$ 2s = 2p < 3s = 3p = 3d < 4s = 4p = 4d = 4f$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1784, "subject": "Chemistry", "question": "In context with the industrial preparation of hydrogen from water gas (CO + H2), which of the following\nis the correct statement? ", "options": [ { "text": "CO and H2 are fractionally separated using differences in their densities " }, { "text": "CO is removed by absorption in aqueous Cu2Cl2 solution" }, { "text": "H2 is removed through occlusion with Pd " }, { "text": "CO is oxidised to CO2 with steam in the presence of a catalyst followed by absorption of CO2 in\nalkali " } ], "answer": "CO is oxidised to CO2 with steam in the presence of a catalyst followed by absorption of CO2 in\nalkali ", "solution": "**Answer:** CO is oxidised to CO2 with steam in the presence of a catalyst followed by absorption of CO2 in\nalkali \n\nOn the industrial scale hydrogen is prepared from water gas according to following reaction sequence \n

    $$\\underbrace {CO + {H_2}}_{water\\,\\,\\,gas} + \\underbrace {{H_2}O}_{\\left( {steam} \\right)}\\,\\,\\buildrel {catalyst} \\over\n \\longrightarrow C{O_2} + 2{H_2}$$\n

    $$\\,\\,\\,\\,\\,\\,\\,\\,\\mathrel{\\mathop{\\kern0pt\\longrightarrow}\n\\limits_{\\left( {alkali} \\right)}^{2NaOH}} N{a_2}C{O_3} + {H_2}O$$ \n

    From the above it is clear that $$CO$$ is first oxidised to $$C{O_2}$$ which is then ansorbed in $$NaOH.$$ \n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1785, "subject": "Chemistry", "question": "Very pure hydrogen (99.9%) can be made by which of the following processes?", "options": [ { "text": "Reaction of methane with steam" }, { "text": "Mixing natural hydrocarbons of high molecular weight" }, { "text": "Electrolysis of water" }, { "text": "Reaction of salt like hydrides with water" } ], "answer": "Reaction of salt like hydrides with water", "solution": "**Answer:** Reaction of salt like hydrides with water\n\nVery pure hydrogen (99.9%) can be prepared by the action of water on any hydrides i.e, sodium hydride(NaH), Calcium Hydride(CaH2) etc.\n

    $$NaH + {H_2}O \\to NaOH + {H_2}$$ (very pure Hydrogen) \n

    By electrolysis of water we get 99.5% pure hydrogen.\n

    2H2O $$\\buildrel {Electrolysis} \\over\n \\longrightarrow $$ 2H2 + O2\n

    Note :\n

    By electrolysis of water we can't get 99.9% pure hydrogen we only get around 99.5% pure hydrogen. That is why option (C) is wrong.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1786, "subject": "Chemistry", "question": "Identify the reaction which does not liberate hydrogen :", "options": [ { "text": "Reaction of zinc with aqueous alkali. " }, { "text": "Electrolysis of acidified water using Pt electrodes." }, { "text": "Allowing a solution of sodium in liquid ammonia to stand." }, { "text": "Reaction of lithium hydride with B2H6." } ], "answer": "Reaction of lithium hydride with B2H6.", "solution": "**Answer:** Reaction of lithium hydride with B2H6.\n\n2LiH + B2H6 $$\\buildrel {Ether} \\over\n \\longrightarrow $$ 2LiBH4", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1787, "subject": "Chemistry", "question": "The isotopes of hydrogen are : ", "options": [ { "text": "Tritium and protium only " }, { "text": "Protium and Deuterium only" }, { "text": "Protium, deuterium and tritium" }, { "text": "Deuterium and tritium only" } ], "answer": "Protium, deuterium and tritium", "solution": "**Answer:** Protium, deuterium and tritium\n\nIsotopes of hydrogen are\n
    (1)  Protium (1H1)\n
    (2)  Deuterium (1H2)\n
    (3)  Tritium (1H3)", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1788, "subject": "Chemistry", "question": "The total number of isotopes of hydrogen and number of radioactive isotopes among them, respectively, are : ", "options": [ { "text": "3 and 2" }, { "text": "2 and 1" }, { "text": "2 and 0" }, { "text": "3 and 1" } ], "answer": "3 and 1", "solution": "**Answer:** 3 and 1\n\nTotal number of isotopes of hydrogen is 3\n

    $$ \\Rightarrow \\,\\,{}_1^1H,{}_1^2H$$  or  $${}_1^2D,\\,{}_1^3H\\,\\,$$or  $${}_1^3T$$\n

    and only $${}_1^3H$$ or $${}_1^3T$$ is an Radioactive element. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1789, "subject": "Chemistry", "question": "The correct statements among (a) to (d) regarding H2 as a fuel are :\n

    (a)   It produces less pollutants than petrol.\n

    (b)   A cylinder of compressed dihydrogen weighs $$ \\sim $$ 30 times more than a petrol tank producing the same amount of energy. \n

    (c)   Dihydrogen is stored in tanks of metal alloys like NaNi5.\n\n

    (d)   On combustion, values of energy released per gram of liquid dihydrogen and LPG are 50 and 142 KJ, respectively.", "options": [ { "text": "(a) and (c) only" }, { "text": "(b) and (d) only" }, { "text": "(a), (b) and (c) only" }, { "text": "(b), (c) and (d) only" } ], "answer": "(a), (b) and (c) only", "solution": "**Answer:** (a), (b) and (c) only\n\n

    (I) H2 is a 100% pollution free fuel. So, option (I) is correct.

    \n

    (II) Molecular weight of H2(2u).

    \n

    $$ = {1 \\over {29}} \\times $$ molecular weight of butane,

    \n

    C4H10 (LPG) [58u].

    \n

    So, compressed H2 weighs $$\\sim$$ 30 times more than a petrol tank and option (b) is correct.

    \n

    (III) NaNi5, Ti - TiH2 etc. are used for storage of H2 in small quantities. Thus, option (c) is correct.

    \n

    (IV) On combustion values of energy released per gram of liquid dihydrogen (H2) : 142 kJ g$$-$$1, and for LPG : 50 kJ g$$-$$1. So, option (d) is incorrect.

    ", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1790, "subject": "Chemistry", "question": "The metal that gives hydrogen gas upon treatment with both acid as well as base is :", "options": [ { "text": "mercury" }, { "text": "zinc" }, { "text": "iron " }, { "text": "magnesium" } ], "answer": "zinc", "solution": "**Answer:** zinc\n\nZn + 2HCl $$ \\to $$ ZnCl2 + H2\n

    Zn + NaOH $$ \\to $$ Na2ZnO2 + H2\n

    Here zinc(Zn) is amphoteric metal, that is why it reacts with both acid and base.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1791, "subject": "Chemistry", "question": "Dihydrogen of high purity (> 99.95%) is\nobtained through :", "options": [ { "text": "the electrolysis of acidified water using Pt\nelectrodes" }, { "text": "the electrolysis of warm Ba(OH)2 solution\nusing Ni electrodes" }, { "text": "the electrolysis of brine solution" }, { "text": "the reaction of Zn with dilute HCl" } ], "answer": "the electrolysis of warm Ba(OH)2 solution\nusing Ni electrodes", "solution": "**Answer:** the electrolysis of warm Ba(OH)2 solution\nusing Ni electrodes\n\nTo obtain H2 of high purity (> 99.95 %)\nelectrolysis of Ba(OH)2 solution is done using\nNi electrodes.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1792, "subject": "Chemistry", "question": "Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R.

    Assertion A : Hydrogen is the most abundant element in the Universe, but it is not the most abundant gas in the troposphere.

    Reason R : Hydrogen is the lightest element.

    In the light of the above statements, choose the correct answer from the options given below :", "options": [ { "text": "Both A and R are true but R is NOT the correct explanation of A" }, { "text": "Both A and R are true and R is the correct explanation of A" }, { "text": "A is true but R is false" }, { "text": "A is false but R is true." } ], "answer": "Both A and R are true and R is the correct explanation of A", "solution": "**Answer:** Both A and R are true and R is the correct explanation of A\n\n

    As we know, hydrogen is most abundant element in the universe but the most abundant gas in the troposphere is nitrogen, because hydrogen is lightest element.

    \n\n

    Troposphere contains 3 quarters of mass of the entire atmosphere. The air here contain 78% nitrogen and 21% oxygen. The last 1% is made of argon, water vapour and carbon dioxide.

    ", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1793, "subject": "Chemistry", "question": "The single largest industrial application of dihydrogen is :", "options": [ { "text": "Manufacture of metal hybrides" }, { "text": "Rocket fuel in space research" }, { "text": "In the synthesis of ammonia" }, { "text": "In the synthesis of nitric acid" } ], "answer": "In the synthesis of ammonia", "solution": "**Answer:** In the synthesis of ammonia\n\nThe single largest industrial application of dihydrogen is in the synthesis of ammonia which is mainly used in the manufacture of fertiliser.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1794, "subject": "Chemistry", "question": "At 298.2 K the relationship between enthalpy of bond dissociation (in kJ mol$$-$$1) for hydrogen (EH) and its isotope, deuterium (ED), is best described by :", "options": [ { "text": "$${E_H} = {1 \\over 2}{E_D}$$" }, { "text": "$${E_H} = {E_D}$$" }, { "text": "$${E_H} \\simeq {E_D} - 7.5$$" }, { "text": "$${E_H} = 2{E_D}$$" } ], "answer": "$${E_H} \\simeq {E_D} - 7.5$$", "solution": "**Answer:** $${E_H} \\simeq {E_D} - 7.5$$\n\nEnthalpy of bond dissociation (kJ/mole) at 298.2 K

    For, Hydrogen = 435.88

    For, Deuterium = 443.35

    $$\\therefore$$ $${E_H} \\simeq {E_D} - 7.5$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1795, "subject": "Chemistry", "question": "The number of neutrons and electrons, respectively, present in the radioactive isotope of hydrogen is :", "options": [ { "text": "1 and 1" }, { "text": "3 and 1" }, { "text": "2 and 1" }, { "text": "2 and 2" } ], "answer": "2 and 1", "solution": "**Answer:** 2 and 1\n\nRadioactive isotope of hydrogen is Tritium ($$_1^3$$T)

    No. of neutrons (A $$-$$ Z) = 3 $$-$$ 1 = 2

    No. of electrons = 1", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1796, "subject": "Chemistry", "question": "Deuterium resembles hydrogen in properties but :", "options": [ { "text": "reacts slower than hydrogen" }, { "text": "reacts vigorously than hydrogen" }, { "text": "reacts just as hydrogen" }, { "text": "emits $$\\beta$$+ particles" } ], "answer": "reacts slower than hydrogen", "solution": "**Answer:** reacts slower than hydrogen\n\nThe bond dissociation energy of D2 is greater than H2 and therefore D2 reacts slower than H2.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1797, "subject": "Chemistry", "question": "Given below are two statements :

    Statement I : The process of producing syn-gas is called gasification of coal.

    Statement II : The composition of syn-gas is CO + CO2 + H2 (1 : 1 : 1)

    In the light of the above statements, choose the most appropriate answer from the options given below :", "options": [ { "text": "Statement I is false but Statement II is true" }, { "text": "Statement I is true but Statement II is false" }, { "text": "Both Statement I and Statement II are false" }, { "text": "Both Statement I and Statement II are true" } ], "answer": "Statement I is true but Statement II is false", "solution": "**Answer:** Statement I is true but Statement II is false\n\nThe process of producing syn-gas from coal is called gasification of coal.

    Syn-gas having composition of CO & H2 in 1 : 1", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1798, "subject": "Chemistry", "question": "Which one of the following statements is incorrect?", "options": [ { "text": "Atomic hydrogen is produced when H2 molecules at a high temperature are irradiated with UV radiation." }, { "text": "At around 2000 K, the dissociation of dihydrogen into its atoms is nearly 8.1%" }, { "text": "Bond dissociation enthalpy of H2 is highest among diatomic gaseous molecules which contain a single bond." }, { "text": "Dihydrogen is produced on reacting zinc with HCl as well as NaOH(aq)." } ], "answer": "At around 2000 K, the dissociation of dihydrogen into its atoms is nearly 8.1%", "solution": "**Answer:** At around 2000 K, the dissociation of dihydrogen into its atoms is nearly 8.1%\n\nAtomic hydrogen is produced at high temperature in an electric are or under ultraviolet radiations.

    The dissociation of dihydrogen at 2000 K is only 0.081%.

    H-H bond dissociation enthalpy is highest for a single bond for any diatomic molecule.

    Dihydrogen can be produced on reacting Zn with dil. HCl as well as NaOH (aq.)", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1799, "subject": "Chemistry", "question": "

    Hydrogen has three isotopes : protium (1H), deuterium (2H or D) and tritium (3H or T). They have nearly same chemical properties but different physical properties. They differ in :

    ", "options": [ { "text": "number of protons." }, { "text": "atomic number." }, { "text": "electronic configuration." }, { "text": "atomic mass." } ], "answer": "atomic mass.", "solution": "**Answer:** atomic mass.\n\n1H, 2H and 3H have same atomic number but their\natomic masses are different.

    \nIsotopes have same atomic number (i.e. same number of protons)", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1800, "subject": "Chemistry", "question": "

    Dihydrogen reacts with CuO to give :

    ", "options": [ { "text": "CuH2" }, { "text": "Cu" }, { "text": "Cu2O" }, { "text": "Cu(OH)2" } ], "answer": "Cu", "solution": "**Answer:** Cu\n\nCuO + H2 $$ \\to $$ Cu + H2O (under hot conditions) ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1801, "subject": "Chemistry", "question": "

    The highest industrial consumption of molecular hydrogen is to produce compounds of element :

    ", "options": [ { "text": "Carbon" }, { "text": "Nitrogen" }, { "text": "Oxygen" }, { "text": "Chlorine" } ], "answer": "Nitrogen", "solution": "**Answer:** Nitrogen\n\nHydrogen combines with nitrogen to produce Ammonia in Haber's process.\n

    \n$\\mathrm{N}_{2}(\\mathrm{g})+3 \\mathrm{H}_{2}(\\mathrm{g}) \\rightleftharpoons 2 \\mathrm{NH}_{3}(\\mathrm{g})$\n

    \nIn this process, iron oxide is used with small amounts of $\\mathrm{K}_{2} \\mathrm{O}$ and $\\mathrm{Al}_{2} \\mathrm{O}_{3}$ to increase the rate of attainment of equilibrium.\n

    \nOptimum conditions for the production of ammonia are a pressure of $200 \\mathrm{~atm}$ and a temperature of $700 \\mathrm{~K}$.\n

    \nEarlier, iron was used as a catalyst with molybdenum as a promoter in this reaction.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1802, "subject": "Chemistry", "question": "

    High purity (> 99.95%) dihydrogen is obtained by :

    ", "options": [ { "text": "reaction of zinc with aqueous alkali." }, { "text": "electrolysis of acidified water using platinum electrodes." }, { "text": "electrolysis of warm aqueous barium hydroxide solution between nickel electrodes." }, { "text": "reaction of zinc with dilute acid." } ], "answer": "electrolysis of warm aqueous barium hydroxide solution between nickel electrodes.", "solution": "**Answer:** electrolysis of warm aqueous barium hydroxide solution between nickel electrodes.\n\nHigh purity (>99.95 %) $$\\mathrm{H}_{2}$$ is obtained by electrolysing warm aqueous $$\\mathrm{Ba}(\\mathrm{OH})_{2}$$ solution between nickel electrodes.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1803, "subject": "Chemistry", "question": "

    Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R

    \n

    Assertion A : Hydrogen is an environment friendly fuel.

    \n

    Reason R : Atomic number of hydrogen is 1 and it is a very light element.

    \n

    In the light of the above statements, choose the correct answer from the options given below :

    ", "options": [ { "text": "Both A and R are true but R is NOT the correct explanation of A" }, { "text": "A is false but R is true" }, { "text": "A is true but R is false" }, { "text": "Both A and R are true and R is the correct explanation of A" } ], "answer": "Both A and R are true but R is NOT the correct explanation of A", "solution": "**Answer:** Both A and R are true but R is NOT the correct explanation of A\n\n

    The correct answer is (A) Both A and R are true but R is NOT the correct explanation of A.

    \n\n

    Assertion A is true that hydrogen is an environmentally friendly fuel because when hydrogen is combusted it produces only water vapor and no harmful pollutants or greenhouse gases.

    \n\n

    Reason R is also true that the atomic number of hydrogen is 1 and it is a very light element. However, it is not a correct explanation for Assertion A as the environmental friendliness of hydrogen as a fuel is not solely determined by its atomic number or weight, but rather by how it is produced, stored, and transported. The environmental impact of hydrogen as a fuel source depends on various factors, such as the method used for its production and the emissions associated with its transportation and storage.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1804, "subject": "Chemistry", "question": "

    Which of the given compounds can enhance the efficiency of hydrogen storage tank?

    ", "options": [ { "text": "SiH$$_4$$" }, { "text": "Li/P$$_4$$" }, { "text": "NaNi$$_5$$" }, { "text": "Di-isobutylaluminium hydride" } ], "answer": "NaNi$$_5$$", "solution": "**Answer:** NaNi$$_5$$\n\nTanks of metal alloy like $\\mathrm{NaNi}_{5}, \\mathrm{Ti}{-\\mathrm{TiH}_{2},} \\mathrm{Mg}-{\\mathrm{MgH}}_{2}$ are used for the storage of dihydrogen.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1805, "subject": "Chemistry", "question": "

    Match List I with List II

    \n

    \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    List IList II
    A.Cobalt catalystI.$$\\mathrm{(H_2+Cl_2)}$$ production
    B.SyngasII.Water gas production
    C.Nickel catalystIII.Coal gasification
    D.Brine solutionIV.Methanol production

    \n

    Choose the correct answer from the options given below :

    ", "options": [ { "text": "A-IV, B-III, C-I, D-II" }, { "text": "A-IV, B-III, C-II, D-I" }, { "text": "A-II, B-III, C-IV, D-I" }, { "text": "A-IV, B-I, C-II, D-III" } ], "answer": "A-IV, B-III, C-II, D-I", "solution": "**Answer:** A-IV, B-III, C-II, D-I\n\n(A) Hydrogen reacts with carbon monoxide in presence of cobalt catalyst to give methanol\n

    \n$$\n2 \\mathrm{H}_{2}(\\mathrm{~g})+\\mathrm{CO}(\\mathrm{g}) \\stackrel{\\text { cobalt catalyst }}{\\longrightarrow} \\mathrm{CH}_{3} \\mathrm{OH}(\\mathrm{I})\n$$\n

    \n(B) Syn gas is produced from coal and the process is called coal gasification.\n

    \n$$\n\\mathrm{C}(\\mathrm{s})+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g}) \\stackrel{1270 \\mathrm{~K}}{\\longrightarrow} \\underset{(\\text { syn gas) }}{\\mathrm{CO}(\\mathrm{g})+\\mathrm{H}_{2}(\\mathrm{~g})}\n$$\n

    \n(C) Reaction of steam with hydrocarbons or coke at high temperature in presence of nickel catalyst gives a mixture of $\\mathrm{CO}$ and $\\mathrm{H}_{2}$, called water gas\n

    \n$$\n\\mathrm{CH}_{4}(\\mathrm{~g})+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g}) \\underset{{\\mathrm{Ni}}}{\\stackrel{1270 \\mathrm{~K}}{\\longrightarrow}}\\underset{(\\text { water gas })}{\\mathrm{CO}(\\mathrm{g})+3 \\mathrm{H}_{2}(\\mathrm{~g})}\n$$

    \n\n(D) Electrolysis of brine solution produces $\\mathrm{H}_{2}$ gas at cathode and $\\mathrm{Cl}_{2}$ gas at anode\n

    \n$\\mathrm{NaCl}(\\mathrm{aq}) \\longrightarrow \\mathrm{Na}^{+}(\\mathrm{aq})+\\mathrm{Cl}^{-}(\\mathrm{aq})$\n

    \nCathode : $2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l})+2 \\mathrm{e}^{-} \\longrightarrow \\mathrm{H}_{2}(\\mathrm{~g})+2 \\mathrm{OH}^{-}(\\mathrm{aq})$\n

    \nAnode : $2 \\mathrm{Cl}^{-}(\\mathrm{aq}) \\longrightarrow \\mathrm{Cl}_{2}(\\mathrm{~g})+2 \\mathrm{e}^{-}$", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1806, "subject": "Chemistry", "question": "

    Given below are two statements, one is labelled as Assertion A and the other is labelled as Reason R.

    \n

    Assertion A : Isotopes of hydrogen have almost same chemical properties, but difference in their rates of reaction.

    \n

    Reason R : Isotopes of hydrogen have different enthalpy of bond dissociation.

    \n

    In the light of the above statements, choose the most appropriate answer from the options given below:

    ", "options": [ { "text": "A is not correct but R is correct" }, { "text": "Both A and R are correct but R is NOT the correct explanation of A" }, { "text": "A is correct but R is not correct" }, { "text": "Both A and R are correct and R is the correct explanation of A" } ], "answer": "Both A and R are correct and R is the correct explanation of A", "solution": "**Answer:** Both A and R are correct and R is the correct explanation of A\n\nAssertion A : Isotopes of hydrogen (protium, deuterium, and tritium) have almost the same chemical properties because they have the same number of protons and electrons, which dictate chemical behavior. However, they differ in their rates of reaction due to differences in their atomic masses, which affect reaction kinetics.\n

    \nReason R : The isotopes of hydrogen have different enthalpy of bond dissociation due to their differences in atomic mass. The stronger bond in deuterium compared to protium, for example, leads to higher bond dissociation enthalpy. This, in turn, influences the rate of reactions involving these isotopes, as breaking the bonds requires different amounts of energy.\n

    \nTherefore, the correct answer is Both A and R are correct, and R is the correct explanation of A.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1807, "subject": "Chemistry", "question": "

    Given below are two statements:

    \n

    Statement-I : Methane and steam passed over a heated $$\\mathrm{Ni}$$ catalyst produces hydrogen gas.

    \n

    Statement-II : Sodium nitrite reacts with $$\\mathrm{NH}_{4} \\mathrm{Cl}$$ to give $$\\mathrm{H}_{2} \\mathrm{O}, \\mathrm{N}_{2}$$ and $$\\mathrm{NaCl}$$.

    \n

    In the light of the above statements, choose the most appropriate answer from the options given below :

    ", "options": [ { "text": "Statement I is correct but Statement II is incorrect" }, { "text": "Both the statements I and II are incorrect" }, { "text": "Both the statements I and II are correct" }, { "text": "Statement I is incorrect but Statement II is correct" } ], "answer": "Both the statements I and II are correct", "solution": "**Answer:** Both the statements I and II are correct\n\n

    Let's consider each statement:

    \n

    Statement-I : Methane (CH₄) and steam (H₂O) can indeed react in the presence of a heated nickel catalyst to produce carbon monoxide (CO) and hydrogen gas (H₂). This is known as the steam reforming or steam methane reforming reaction and is industrially significant for hydrogen production. Hence, Statement-I is correct.

    \n

    Statement-II : Sodium nitrite (NaNO₂) reacts with ammonium chloride (NH₄Cl) to produce nitrogen gas (N₂), water (H₂O), and sodium chloride (NaCl). This is a standard reaction that is often used in inorganic chemistry to produce nitrogen gas. Hence, Statement-II is also correct.

    \n

    Therefore, the correct answer is Option C: Both the statements I and II are correct.

    \n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1808, "subject": "Chemistry", "question": "

    Given below are two statements: one is labelled as Assertion $$\\mathbf{A}$$ and the other is labelled as Reason $$\\mathbf{R}$$

    \n

    Assertion A : Physical properties of isotopes of hydrogen are different.

    \n

    Reason R : Mass difference between isotopes of hydrogen is very large.

    \n

    In the light of the above statements, choose the correct answer from the options given below:

    ", "options": [ { "text": "$$\\mathbf{A}$$ is true but $$\\mathbf{R}$$ is false" }, { "text": "Both $$\\mathbf{A}$$ and $$\\mathbf{R}$$ are true but $$\\mathbf{R}$$ is NOT the correct explanation of $$\\mathbf{A}$$" }, { "text": "$$\\mathbf{A}$$ is false but $$\\mathbf{R}$$ is true" }, { "text": "Both $$\\mathbf{A}$$ and $$\\mathbf{R}$$ are true and $$\\mathrm{R}$$ is the correct explanation of $$\\mathbf{A}$$" } ], "answer": "Both $$\\mathbf{A}$$ and $$\\mathbf{R}$$ are true and $$\\mathrm{R}$$ is the correct explanation of $$\\mathbf{A}$$", "solution": "**Answer:** Both $$\\mathbf{A}$$ and $$\\mathbf{R}$$ are true and $$\\mathrm{R}$$ is the correct explanation of $$\\mathbf{A}$$\n\n\n

    This is true because the different isotopes of hydrogen have different masses. The mass of an atom determines its physical properties, such as its melting point, boiling point, and density.

    \n\n

    This is also true. The three isotopes of hydrogen are protium (1H), deuterium (2H), and tritium (3H). Protium has a mass of 1, deuterium has a mass of 2, and tritium has a mass of 3. The difference in mass between these isotopes is relatively large, which is why they have different physical properties.

    \n

    The reason is the correct explanation of the assertion because it explains why the physical properties of the isotopes of hydrogen are different. The different masses of the isotopes cause them to have different physical properties.

    \n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1809, "subject": "Chemistry", "question": "

    Given below are two reactions, involved in the commercial production of dihydrogen (H$$_2$$). The two reactions are carried out at temperature \"T$$_1$$\" and \"T$$_2$$\", respectively

    \n

    $$C(s) + {H_2}O(g)\\buildrel {{T_1}} \\over\n \\longrightarrow CO(g) + {H_2}(g)$$

    \n

    $$CO(g) + {H_2}O(g)\\mathrel{\\mathop{\\kern0pt\\longrightarrow}\n\\limits_{Catalyst}^{{T_2}}} C{O_2}(g) + {H_2}(g)$$

    \n

    The temperatures T$$_1$$ and T$$_2$$ are correctly related as :

    ", "options": [ { "text": "T$$_1$$ < T$$_2$$" }, { "text": "T$$_1$$ = T$$_2$$" }, { "text": "T$$_1$$ = 100 K, T$$_2$$ = 1270 K" }, { "text": "T$$_1$$ > T$$_2$$" } ], "answer": "T$$_1$$ > T$$_2$$", "solution": "**Answer:** T$$_1$$ > T$$_2$$\n\n

    These reactions are part of the water-gas shift reaction used in the industrial production of hydrogen. The first reaction is the endothermic steam reforming of coal (carbon), while the second reaction is the exothermic water-gas shift reaction.

    \n
      \n

    1. Carbon reacts with steam at high temperatures (around 1000°C or 1273 K) to form carbon monoxide and hydrogen:

      \nC(s) + H$_2$O(g) $\\xrightarrow[T_1]{}$ CO(g) + H$_2$(g)

      \n

      2. \n

    2. Then carbon monoxide reacts with steam in the presence of a catalyst to produce carbon dioxide and more hydrogen:

      \nCO(g) + H$_2$O(g) $\\xrightarrow[Catalyst]{[T_2]}$ CO$_2$(g) + H$_2$(g)

      \n
      3. \n
    \n

    Since the first reaction is endothermic (absorbs heat), it requires a higher temperature to proceed than the second reaction, which is exothermic (releases heat). Therefore, the temperature T$_1$ should be higher than T$_2$.

    \n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1810, "subject": "Chemistry", "question": "

    Given below are two statements, one is labelled as Assertion A and the other is labelled as Reason R.

    \n

    Assertion A : Loss of electron from hydrogen atom results in nucleus of $$\\sim 1.5 \\times 10^{-3} \\mathrm{pm}$$ size.

    \n

    Reason R : Proton $$\\left(\\mathrm{H}^{+}\\right)$$ always exists in combined form.

    \n

    In the light of the above statements, choose the most appropriate answer from the options given below:

    ", "options": [ { "text": "Both $$\\mathbf{A}$$ and $$\\mathbf{R}$$ are correct and $$\\mathbf{R}$$ is the correct explanation of $$\\mathbf{A}$$" }, { "text": "A is correct but R is not correct" }, { "text": "A is not correct but R is correct" }, { "text": "Both $$\\mathbf{A}$$ and $$\\mathbf{R}$$ are correct but $$\\mathbf{R}$$ is NOT the correct explanation of $$\\mathbf{A}$$" } ], "answer": "Both $$\\mathbf{A}$$ and $$\\mathbf{R}$$ are correct but $$\\mathbf{R}$$ is NOT the correct explanation of $$\\mathbf{A}$$", "solution": "**Answer:** Both $$\\mathbf{A}$$ and $$\\mathbf{R}$$ are correct but $$\\mathbf{R}$$ is NOT the correct explanation of $$\\mathbf{A}$$\n\nSize of nucleus is of order

    \n$1.5 \\times 10^{-15} \\mathrm{~m}$ or $1.5 \\times 10^{-3} \\mathrm{pm}$

    \n$\\mathrm{H}^{+}$always exists in combined form.

    There is no relation between two statements and hence option (D) is the answer.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1811, "subject": "Chemistry", "question": "Which one of the following statements about water is FALSE?", "options": [ { "text": "Water is oxidized to oxygen during photosynthesis" }, { "text": "Water can act both as an acid and as a base." }, { "text": "There is extensive intramolecular hydrogen bonding in the condensed phase. " }, { "text": "Ice formed by heavy water sinks in normal water" } ], "answer": "There is extensive intramolecular hydrogen bonding in the condensed phase. ", "solution": "**Answer:** There is extensive intramolecular hydrogen bonding in the condensed phase. \n\nThere is extensive inter-molecular hydrogen bonding in the condensed phase instead of intra-molecular $$H$$-bonding. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1812, "subject": "Chemistry", "question": "Identify the incorrect statement regarding heavy water :", "options": [ { "text": "It reacts with Al4C3 to produce CD4 and Al(OD)3\n." }, { "text": "It is used as a coolant in nuclear reactors." }, { "text": "It reacts with CaC2 to produce C2D2 and Ca(OD)2\n." }, { "text": "It reacts with SO3 to form deuterated sulphuric acid (D2SO4)." } ], "answer": "It is used as a coolant in nuclear reactors.", "solution": "**Answer:** It is used as a coolant in nuclear reactors.\n\nHeavy water is used as moderator in nuclear reactors\nto control the speed of neutrons.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1813, "subject": "Chemistry", "question": "The number of water molecule(s) not\ncoordinated to copper ion directly in\nCuSO4.5H2O, is :", "options": [ { "text": "2" }, { "text": "4" }, { "text": "3" }, { "text": "1" } ], "answer": "1", "solution": "**Answer:** 1\n\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1814, "subject": "Chemistry", "question": "The synonym for water gas when used in the production of methanol is :", "options": [ { "text": "fuel gas\n" }, { "text": "laughing gas" }, { "text": "syn gas" }, { "text": "natural gas " } ], "answer": "syn gas", "solution": "**Answer:** syn gas\n\nH2O(g) + C $$ \\to $$ CO + H2\n

    The gaseous mixture CO + H2 which is produced when steam is passed over red hot coke and this mixture called water gas or synthesis gas or syn gas.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1815, "subject": "Chemistry", "question": "The equation that represents the water-gas shift\nreaction is :", "options": [ { "text": "CO(g) + H2O(g) $$\\mathrel{\\mathop{\\kern0pt\\longrightarrow}\n\\limits_{Catalyst}^{673K}} $$ CO2(g) + H2(g)" }, { "text": "CH4(g) + H2O(g) $$\\mathrel{\\mathop{\\kern0pt\\longrightarrow}\n\\limits_{Ni}^{1270K}} $$ CO(g) + 3H2(g)" }, { "text": "C(s) + H2O(g) $$\\buildrel {1270K} \\over\n \\longrightarrow $$ CO(g) + H2(g)" }, { "text": "2C(s) + O2(g) + 4N2(g) $$\\buildrel {1273K} \\over\n \\longrightarrow $$ 2CO(g) + 4N2(g)" } ], "answer": "CO(g) + H2O(g) $$\\mathrel{\\mathop{\\kern0pt\\longrightarrow}\n\\limits_{Catalyst}^{673K}} $$ CO2(g) + H2(g)", "solution": "**Answer:** CO(g) + H2O(g) $$\\mathrel{\\mathop{\\kern0pt\\longrightarrow}\n\\limits_{Catalyst}^{673K}} $$ CO2(g) + H2(g)\n\n(A) Water gas shift reaction\n
    CO(g) + H2O(g) $$\\mathrel{\\mathop{\\kern0pt\\longrightarrow}\n\\limits_{Catalyst}^{673K}} $$ CO2(g) + H2(g)\n

    (B) Water gas is produced by this reaction\n
    CH4(g) + H2O(g) $$\\mathrel{\\mathop{\\kern0pt\\longrightarrow}\n\\limits_{Ni}^{1270K}} $$ CO(g) + 3H2(g)\n

    (C) Water gas is produced by this reaction\n
    C(s) + H2O(g) $$\\buildrel {1270K} \\over\n \\longrightarrow $$ CO(g) + H2(g)\n

    (D) Producer gas is produced by this reaction\n
    2C(s) + O2(g) + 4N2(g) $$\\buildrel {1273K} \\over\n \\longrightarrow $$ 2CO(g) + 4N2(g)", "topic": "Algebra", "subtopic": "Equations and Inequalities" }, { "id": 1816, "subject": "Chemistry", "question": "Water does not produce CO on reacting with :", "options": [ { "text": "C" }, { "text": "CH4" }, { "text": "C3H8" }, { "text": "CO2" } ], "answer": "CO2", "solution": "**Answer:** CO2\n\nCO2 + H2O $$ \\to $$ H2CO3

    \nC + H2O(steam) $$ \\to $$ CO + H2

    \nCH4 + H2O $$ \\to $$ CO + 3H2

    \nC3H8 + H2O $$ \\to $$ 3CO + H2", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1817, "subject": "Chemistry", "question": "Statements about heavy water are given below :

    A. Heavy water is used in exchange reactions for the study of reaction mechanisms.

    B. Heavy water is prepared by exhaustive electrolysis of water.

    (C) Heavy water has higher boiling point than ordinary water.

    (D) Viscosity of H2O is greater than D2O.

    Choose the most appropriate answer from the options given below :", "options": [ { "text": "A, B, and C only" }, { "text": "A and C only" }, { "text": "A and D only" }, { "text": "A and B only" } ], "answer": "A, B, and C only", "solution": "**Answer:** A, B, and C only\n\nViscosity of D2O is greater than H2O.\n

    B.P. of D2O is greater than H2O.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1818, "subject": "Chemistry", "question": "The INCORRECT statement(s) about heavy water is (are)

    (A) used as a moderator in nuclear reactor

    (B) obtained as a by-product in fertilizer industry

    (C) used for the study of reaction mechanism

    (D) has a higher dielectric constant than water

    Choose the correct answer from the options given below :", "options": [ { "text": "(D) only" }, { "text": "(B) and (D) only" }, { "text": "(B) only" }, { "text": "(C) only" } ], "answer": "(D) only", "solution": "**Answer:** (D) only\n\nHeavy water (D2O) is obtained as a by-product in\nthe fertilizer industry. It is used as a moderator in\nnuclear reactors and for the study of the reaction\nmechanisms. Its dielectric constant is lower than that\nof H2O.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1819, "subject": "Chemistry", "question": "The water having more dissolved O2 is :", "options": [ { "text": "boiling water" }, { "text": "water at 80$$^\\circ$$C" }, { "text": "polluted water" }, { "text": "water at 4$$^\\circ$$C" } ], "answer": "water at 4$$^\\circ$$C", "solution": "**Answer:** water at 4$$^\\circ$$C\n\nOn heating concentration of O2 in water decreases. So boiling water and water at 80$$^\\circ$$C having less O2 concentration. Polluted water also having less O2 concentration. So water at 4$$^\\circ$$C having maximum O2 concentration.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1820, "subject": "Chemistry", "question": "Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).

    Assertion (A) : Heavy water is used for the study of reaction mechanism.

    Reason (R) : The rate of reaction for the cleavage of O - H bond is slower than that of O - D bond.

    Choose the most appropriate answer from the options given below :", "options": [ { "text": "Both (A) and (R) are true but (R) is not the true explanation of (A)." }, { "text": "Both (A) and (R) are true and (R) is the true explanation of (A)." }, { "text": "(A) is false but (R) is true." }, { "text": "(A) is true but (R) is false." } ], "answer": "(A) is true but (R) is false.", "solution": "**Answer:** (A) is true but (R) is false.\n\nD2O in used for the study of reaction mechanism. Rate of reaction for the cleavage of O-H bond > O-D bond.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1821, "subject": "Chemistry", "question": "The normal rain water is slightly acidic and its $\\mathrm{pH}$ value is $5.6$ because of which one of the following?", "options": [ { "text": "$\\mathrm{N}_{2} \\mathrm{O}_{5}+\\mathrm{H}_{2} \\mathrm{O} \\rightarrow 2 \\mathrm{HNO}_{3}$" }, { "text": "$2 \\mathrm{SO}_{2}+\\mathrm{O}_{2}+2 \\mathrm{H}_{2} \\mathrm{O} \\rightarrow 2 \\mathrm{H}_{2} \\mathrm{SO}_{4}$" }, { "text": "$4 \\mathrm{NO}_{2}+\\mathrm{O}_{2}+2 \\mathrm{H}_{2} \\mathrm{O} \\rightarrow 4 \\mathrm{HNO}_{3}$" }, { "text": "$\\mathrm{CO}_{2}+\\mathrm{H}_{2} \\mathrm{O} \\rightarrow \\mathrm{H}_{2} \\mathrm{CO}_{3}$" } ], "answer": "$\\mathrm{CO}_{2}+\\mathrm{H}_{2} \\mathrm{O} \\rightarrow \\mathrm{H}_{2} \\mathrm{CO}_{3}$", "solution": "**Answer:** $\\mathrm{CO}_{2}+\\mathrm{H}_{2} \\mathrm{O} \\rightarrow \\mathrm{H}_{2} \\mathrm{CO}_{3}$\n\nNormally rain water has a pH of 5.6 due to\npresence of H+ ions formed by the reaction of rain\nwater with carbon dioxide present in the\natmosphere.\n

    $$\n\\begin{aligned}\n& \\mathrm{H}_2 \\mathrm{O}(\\mathrm{l})+\\mathrm{CO}_2(\\mathrm{~g}) \\leftrightharpoons \\mathrm{H}_2 \\mathrm{CO}_3(\\mathrm{aq}) \\\\\\\\\n& \\mathrm{H}_2 \\mathrm{CO}_3(\\mathrm{aq}) \\leftrightharpoons \\mathrm{H}^{+}(\\mathrm{aq})+\\mathrm{HCO}_3^{-}(\\mathrm{aq})\n\\end{aligned}\n$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1822, "subject": "Chemistry", "question": "Species acting as both Bronsted acid and base is :", "options": [ { "text": "(HSO4)-1" }, { "text": "Na2CO3" }, { "text": "NH3" }, { "text": "OH-1" } ], "answer": "(HSO4)-1", "solution": "**Answer:** (HSO4)-1\n\n$${\\left( {HS{O_4}} \\right)^ - }\\,\\,\\,$$ can accept and donate a proton\n

    $${\\left( {HS{O_4}} \\right)^ - } + {H^ + } \\to {H_2}S{O_4}\\,\\,\\,$$ (acting as base)\n

    $${\\left( {HS{O_4}} \\right)^ - } - {H^ + } \\to {H_2}S{O_4}^{2 - }.\\,\\,\\,$$ (acting as acid)", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1823, "subject": "Chemistry", "question": "The conjugate base of H2PO4- is :", "options": [ { "text": "$$PO_4^{3-}$$" }, { "text": "$$HPO_4^{2-}$$" }, { "text": "H3PO4 " }, { "text": "P2O5" } ], "answer": "$$HPO_4^{2-}$$", "solution": "**Answer:** $$HPO_4^{2-}$$\n\nNOTE : Conjugate acid-base differ by $${H^ + }$$\n

    $$\\mathop {{H_2}PO_4^ - }\\limits_{Acid} \\mathop \\to \\limits^{ - H} \\,\\mathop {\\,HPO{{_4^ - }^ - }}\\limits_{conjugate\\,\\,base} $$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1824, "subject": "Chemistry", "question": "What is the conjugate base of OH-?", "options": [ { "text": "O2" }, { "text": "H2O " }, { "text": "O-" }, { "text": "O-2" } ], "answer": "O-2", "solution": "**Answer:** O-2\n\nConjugate acid-base pair differ by only one proton.\n

    $$O{H^ - } \\to {H^ + } + {O^2}^ - \\,\\,$$ Conjugate base of $$O{H^ - }$$ is $${O^{2 - }}$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1825, "subject": "Chemistry", "question": "The first and second dissociation constants of an acid H2A are 1.0 $$\\times$$ 10−5\n and 5.0 $$\\times$$ 10−10 respectively. The overall dissociation constant of the acid will be :", "options": [ { "text": "5.0 $$\\times$$ 10−5" }, { "text": "5.0 $$\\times$$ 1015" }, { "text": "5.0 $$\\times$$ 10−15" }, { "text": "5.0 $$\\times$$ 105" } ], "answer": "5.0 $$\\times$$ 10−15", "solution": "**Answer:** 5.0 $$\\times$$ 10−15\n\n$${H_2}A\\,\\rightleftharpoons\\,{H^ + }\\,\\, + \\,\\,H{A^ - }$$\n

    $$\\therefore$$ $$\\,\\,\\,{K_1} = 1.0 \\times {10^{ - 5}} = {{\\left[ {{H^ + }} \\right]\\left[ {H{A^ - }} \\right]} \\over {\\left[ {{H_2}A} \\right]}}$$ (Given)\n

    $$H{A^ - } \\to {H^ + } + {A^ - }$$\n

    $$\\therefore$$ $$\\,\\,\\,{K_2} = 5.0 \\times {10^{ - 10}}$$\n

    $$ = {{\\left[ {{H^ + }} \\right]\\left[ {{A^{ - - }}} \\right]} \\over {\\left[ {H{A^ - }} \\right]}}$$ (Given)\n

    $$K = {{{{\\left[ {{H^ + }} \\right]}^2}\\left[ {{A^{2 - }}} \\right]} \\over {\\left[ {{H_2}A} \\right]}}$$\n

    $$ = {K_1} \\times {K_2}$$\n

    $$ = \\left( {1.0 \\times {{10}^{ - 5}}} \\right) \\times \\left( {5 \\times {{10}^{ - 10}}} \\right)$$\n

    $$ = 5 \\times {10^{ - 15}}$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1826, "subject": "Chemistry", "question": "Four species are listed below
    \ni. $$HCO_3^−$$
    \nii. $$H_3O^+$$
    \niii. $$HSO_4^−$$
    \niv. $$HSO_3F$$
    \nWhich one of the following is the correct sequence of their acid strength?", "options": [ { "text": "iv < ii < iii < I " }, { "text": "ii < iii < i < iv " }, { "text": "i < iii < ii < iv " }, { "text": "iii < i < iv < ii " } ], "answer": "i < iii < ii < iv ", "solution": "**Answer:** i < iii < ii < iv \n\nThe correct order of acidic strength of the given species in\n

    $$\\mathop {HS{O_3}F}\\limits_{\\left( {iv} \\right)} > \\mathop {{H_3}{O^ + }}\\limits_{\\left( {ii} \\right)} > \\mathop {HS{O_4}^ - }\\limits_{\\left( {iii} \\right)} > \\mathop {HC{O_3}^ - }\\limits_{\\left( i \\right)} $$\n

    or$$\\,\\,\\,\\left( i \\right) < \\left( {iii} \\right) < \\left( {ii} \\right) < \\left( {iv} \\right)$$ \n

    It corresponds to choice $$(c)$$ which is correct answer. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1827, "subject": "Chemistry", "question": "Three reactions involving $$H_2PO_4^−$$ are given below :

    \n(i) H3PO4 + H2O $$\\to$$ H3O+ + $$H_2PO_4^−$$

    \n(ii) $$H_2PO_4^−$$ + H2O $$\\to$$ $$HPO_4^{2−}$$ + H3O+

    \n(iii) $$H_2PO_4^−$$ + OH- $$\\to$$H3PO4 + O2-

    \nIn which of the above does $$H_2PO_4^−$$ act as an acid?



    ", "options": [ { "text": "(ii) only" }, { "text": "(i) and (ii)" }, { "text": "(iii) only" }, { "text": "(i) only" } ], "answer": "(ii) only", "solution": "**Answer:** (ii) only\n\n

    To determine in which reactions $ H_2PO_4^- $ acts as an acid, we need to understand the Bronsted-Lowry concept of acids and bases. According to this concept, an acid is a substance that donates a proton (H⁺) to another substance, while a base is a substance that accepts a proton.

    \n

    Let's analyze each reaction :

    \n
      \n

    1. 1. $ H_3PO_4 + H_2O \\to H_3O^+ + H_2PO_4^- $

      \n\n
      2. \n

    2. 2. $ H_2PO_4^- + H_2O \\to HPO_4^{2−} + H_3O^+ $

      \n\n
      3. \n

    3. 3. $ H_2PO_4^- + OH^- \\to H_3PO_4 + O^{2-} $

      \n\n
      4. \n
    \n

    Given the information, the correct option is :

    \n

    Option A : (ii) only

    \n

    This is because in reaction (ii), $ H_2PO_4^- $ is clearly acting as an acid by donating a proton to water. In reaction (i), $ H_2PO_4^- $ is not acting as an acid but rather is formed as a product. Reaction (iii) as written is chemically incorrect, but even in a corrected form, it would show $ H_2PO_4^- $ acting as an acid.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1828, "subject": "Chemistry", "question": "Which of the following are Lewis acids?\n", "options": [ { "text": "BCl3 and AlCl3" }, { "text": "PH3 and BCl3" }, { "text": "AlCl3 and SiCl4 " }, { "text": "PH3 and SiCl4\n" } ], "answer": "BCl3 and AlCl3", "solution": "**Answer:** BCl3 and AlCl3\n\nThe compound which have the ability to accepted at least are lone pair electron. \n

    Structure of BCl3 is \n

    \"JEE\n

    Here B is electron deficient atom, so it can accepted lone pair. So it is a lewis acid. \n

    Structure of AlCl3 \n

    \"JEE\n

    Here, Al also a electron deficient atom so it has vacant orbital and in that vacant orbital it can take lone pair. So it is also lewis acid. \n

    \"JEE\n

    Here in PH3 there is vacant 3d orbital but it Can't take. Lone pair in 3d orbital because P is more electro-negative than H so around P atom negative charge density is created and tendency of accepting electron decreases. So PH3 is not lewis acid. \n

    \"JEE\n

    Here octet of Si full ut in has a tendency of accepting lone pair in vacant 3d orbital. \n

    This can be shown by following reaction. \n

    \"JEE\n

    So, here option (A) and (C) both are correct. But as SiCl4 is not as strong lewis acid as BCl3 and AlCl3, So we can say option (A) is more correct.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1829, "subject": "Chemistry", "question": "Which of the following is a Lewis acid?", "options": [ { "text": "PH3" }, { "text": "B(CH3)3" }, { "text": "NaH" }, { "text": "NF3" } ], "answer": "B(CH3)3", "solution": "**Answer:** B(CH3)3\n\n\"JEE\n

    As B can accept electron pair, so it is Lewis Acid.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1830, "subject": "Chemistry", "question": "The decreasing order of electrical conductivity of the following aqueous solutions is :\n

    0.1 M Formic acid (A),\n

    0.1 M Acetic acid (B),\n

    0.1 M Benzoic acid (C)


    ", "options": [ { "text": "A > B > C" }, { "text": "C > B > A" }, { "text": "A > C > B" }, { "text": "C > A > B" } ], "answer": "A > C > B", "solution": "**Answer:** A > C > B\n\n0.1 M Formic acid(HCOOH) (A),\n
    0.1 M Acetic acid(CH3COOH) (B),\n
    0.1 M Benzoic acid(C6H5COOH) (C)\n

    Conductivity (K) depends on no of ions present in unit volume of solution. When no of ions increases conductivity also increases.\n
    Also no of ions depends on degree of dissociation($$\\alpha $$).\n
    We know, $$\\alpha $$ = $$\\sqrt {{{{K_a}} \\over C}} $$\n

    For all solutions C = 0.1 M\n
    So, $$\\alpha $$ depends on only K$$a$$ here.\n
    K$$a$$ of HCOOH = 10-4\n
    K$$a$$ of CH3COOH = 1.8 $$ \\times $$ 10-5\n
    and K$$a$$ of C6H5COOH = 6 $$ \\times $$ 10-5\n

    $$ \\therefore $$ $$\\alpha $$(HCOOH) > $$\\alpha $$(C6H5COOH) > $$\\alpha $$(CH3COOH)\n

    Therefore conductivity order = A > C > B", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1831, "subject": "Chemistry", "question": "Which of the following compound CANNOT act as a Lewis base?", "options": [ { "text": "SF4" }, { "text": "NF3" }, { "text": "ClF3" }, { "text": "PCl5" } ], "answer": "PCl5", "solution": "**Answer:** PCl5\n\nNF3 has no vacant orbital neither in nitrogen nor in fluorine so it cannot\naccept the electron & hence cannot acts as lewis acid and but for PCl5 P has\nno L.P & hence it cannot acts as base but ClF3 (3 B.P + 2 L.P) & SF4 (4 B.P +\n1 L.P)", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1832, "subject": "Chemistry", "question": "

    Given below are two statements one is labelled as Assertion A and the other is labelled as Reason R :

    \n

    Assertion A : The amphoteric nature of water is explained by using Lewis acid/base concept.

    \n

    Reason R : Water acts as an acid with NH3 and as a base with H2S.

    \n

    In the light of the above statements choose the correct answer from the options given below :

    ", "options": [ { "text": "Both A and R are true and R is the correct explanation of A." }, { "text": "Both A and R are true but R is NOT the correct explanation of A." }, { "text": "A is true but R is false." }, { "text": "A is false but R is true." } ], "answer": "A is false but R is true.", "solution": "**Answer:** A is false but R is true.\n\nThe amphoteric nature of water is explained by\nusing Bronsted-Lowry acid base concept

    \n$$\\underset{\\text { (acid) }}{\\mathrm{H}_{2} \\mathrm{O}}+\\mathrm{NH}_{3} \\rightleftharpoons \\mathrm{OH}^{-}+\\mathrm{NH}_{4}^{+}$$

    \n$$\\underset{\\text { (base) }}{\\mathrm{H}_{2} \\mathrm{O}}+\\mathrm{H}_{2} \\mathrm{~S} \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}+\\mathrm{HS}^{-}$$

    \nHence, A is false but R is true", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1833, "subject": "Chemistry", "question": "

    $${K_{{a_1}}}$$, $${K_{{a_2}}}$$ and $${K_{{a_3}}}$$ are the respective ionization constants for the following reactions (a), (b) and (c).

    \n

    (a) $${H_2}{C_2}{O_4} \\mathbin{\\lower.3ex\\hbox{$\\buildrel\\textstyle\\rightarrow\\over\n {\\smash{\\leftarrow}\\vphantom{_{\\vbox to.5ex{\\vss}}}}$}} {H^ + } + H{C_2}O_4^ - $$

    \n

    (b) $$H{C_2}O_4^ - \\mathbin{\\lower.3ex\\hbox{$\\buildrel\\textstyle\\rightarrow\\over\n {\\smash{\\leftarrow}\\vphantom{_{\\vbox to.5ex{\\vss}}}}$}} {H^ + } + {C_2}O_4^{2 - }$$

    \n

    (c) $${H_2}{C_2}O_4^{} \\mathbin{\\lower.3ex\\hbox{$\\buildrel\\textstyle\\rightarrow\\over\n {\\smash{\\leftarrow}\\vphantom{_{\\vbox to.5ex{\\vss}}}}$}} 2{H^ + } + {C_2}O_4^{2 - }$$

    \n

    The relationship between $${K_{{a_1}}}$$, $${K_{{a_2}}}$$ and $${K_{{a_3}}}$$ is given as :

    ", "options": [ { "text": "$${K_{{a_3}}}$$ $$=$$ $${K_{{a_1}}}$$ $$+$$ $${K_{{a_2}}}$$" }, { "text": "$${K_{{a_3}}}$$ $$=$$ $${K_{{a_1}}}$$ $$-$$ $${K_{{a_2}}}$$" }, { "text": "$${K_{{a_3}}}$$ $$=$$ $${K_{{a_1}}}$$ $$/$$ $${K_{{a_2}}}$$" }, { "text": "$${K_{{a_3}}}$$ $$=$$ $${K_{{a_1}}}$$ $$\\times$$ $${K_{{a_2}}}$$" } ], "answer": "$${K_{{a_3}}}$$ $$=$$ $${K_{{a_1}}}$$ $$\\times$$ $${K_{{a_2}}}$$", "solution": "**Answer:** $${K_{{a_3}}}$$ $$=$$ $${K_{{a_1}}}$$ $$\\times$$ $${K_{{a_2}}}$$\n\n$$\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4} \\rightleftharpoons 2 \\mathrm{H}^{+}+\\mathrm{C}_{2} \\mathrm{O}_{4}^{2-} \\quad \\mathrm{K}_{\\mathrm{a}_{3}}$$\n

    \n$$\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4} \\rightleftharpoons \\mathrm{H}^{+}+\\mathrm{HC}_{2} \\mathrm{O}_{4}^{-} \\quad \\mathrm{K}_{\\mathrm{a}_{1}}$$\n

    \n$$\\mathrm{HC}_{2} \\mathrm{O}_{4}^{-} \\rightleftharpoons \\mathrm{H}^{+}+\\mathrm{C}_{2} \\mathrm{O}_{4}^{2-} \\quad \\mathrm{K}_{\\mathrm{a}_{2}}$$\n

    \n$$\\mathrm{K}_{\\mathrm{a}_{3}}=\\frac{\\left[\\mathrm{H}^{+}\\right]^{2}\\left[\\mathrm{C}_{2} \\mathrm{O}_{4}^{2-}\\right]}{\\left[\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right]}$$\n

    \n$$\\mathrm{K}_{\\mathrm{a}_{1}}=\\frac{\\left[\\mathrm{H}^{+}\\right]\\left[\\mathrm{HC}_{2} \\mathrm{O}_{4}^{-}\\right]}{\\left[\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right]}, \\mathrm{K}_{\\mathrm{a}_{2}}=\\frac{\\left[\\mathrm{H}^{+}\\right]\\left[\\mathrm{C}_{2} \\mathrm{O}_{4}^{-}\\right]}{\\left[\\mathrm{HC}_{2} \\mathrm{O}_{4}^{-}\\right]}$$\n

    \n$$\\mathrm{K}_{\\mathrm{a}_{3}}=\\mathrm{K}_{\\mathrm{a}_{1}} \\times \\mathrm{K}_{\\mathrm{a}_{2}}$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1834, "subject": "Chemistry", "question": "

    In the titration of $$\\mathrm{KMnO}_{4}$$ and oxalic acid in acidic medium, the change in oxidation number of carbon at the end point is ___________.

    ", "options": [], "answer": "1", "solution": "**Answer:** 1\n\n$16 \\mathrm{H}^{+}+2 \\mathrm{MnO}_{4}^{-}+5 \\mathrm{C}_{2} \\mathrm{O}_{4}^{2-} \\rightarrow 10 \\mathrm{CO}_{2}+2 \\mathrm{Mn}^{2+}+8 \\mathrm{H}_{2} \\mathrm{O}$\n

    \nDuring titration of oxalic acid by $\\mathrm{KMnO}_{4}$, oxalic acid converts into $\\mathrm{CO}_{2}$.\n

    \n$\\therefore$ Change in oxidation state of carbon $=1$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1835, "subject": "Chemistry", "question": "

    Given below are two statements : One is labelled as Assertion A and the other is labelled as Reason R

    \n

    Assertion A : Permanganate titrations are not performed in presence of hydrochloric acid.

    \n

    Reason R : Chlorine is formed as a consequence of oxidation of hydrochloric acid.

    \n

    In the light of the above statements, choose the correct answer from the options given below

    ", "options": [ { "text": "Both $$\\mathrm{A}$$ and $$\\mathrm{R}$$ are true and $$\\mathrm{R}$$ is the correct explanation of $$\\mathrm{A}$$" }, { "text": "Both $$\\mathrm{A}$$ and $$\\mathrm{R}$$ are true but $$\\mathrm{R}$$ is NOT the correct explanation of $$\\mathrm{A}$$" }, { "text": "$$\\mathrm{A}$$ is true but $$\\mathrm{R}$$ is false" }, { "text": "$$\\mathrm{A}$$ is false but $$\\mathrm{R}$$ is true" } ], "answer": "Both $$\\mathrm{A}$$ and $$\\mathrm{R}$$ are true and $$\\mathrm{R}$$ is the correct explanation of $$\\mathrm{A}$$", "solution": "**Answer:** Both $$\\mathrm{A}$$ and $$\\mathrm{R}$$ are true and $$\\mathrm{R}$$ is the correct explanation of $$\\mathrm{A}$$\n\n$2 \\mathrm{KMnO}_4+16 \\mathrm{HCl} \\rightarrow 2 \\mathrm{MnCl}_2+2 \\mathrm{KCl}+8 \\mathrm{H}_2 \\mathrm{O}+\\mathrm{Cl}_2$

    $\\mathrm{HCl}$ is not used in the process of titration because it reacts with the $\\left(\\mathrm{KMnO}_{4}\\right)$ that is used in the process and gets oxidized.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1836, "subject": "Chemistry", "question": "

    Consider the dissociation of the weak acid HX as given below

    \n

    $$\\mathrm{HX}(\\mathrm{aq}) \\rightleftharpoons \\mathrm{H}^{+}(\\mathrm{aq})+\\mathrm{X}^{-}(\\mathrm{aq}), \\mathrm{Ka}=1.2 \\times 10^{-5}$$

    \n

    [$$\\mathrm{K}_{\\mathrm{a}}$$ : dissociation constant]

    \n

    The osmotic pressure of $$0.03 \\mathrm{M}$$ aqueous solution of $$\\mathrm{HX}$$ at $$300 \\mathrm{~K}$$ is _________ $$\\times 10^{-2}$$ bar (nearest integer).

    \n

    [Given : $$\\mathrm{R}=0.083 \\mathrm{~L} \\mathrm{~bar} \\mathrm{~mol}^{-1} \\mathrm{~K}^{-1}$$]

    ", "options": [], "answer": "76", "solution": "**Answer:** 76\n\n

    $$\\begin{aligned}\n& \\mathrm{Ka}=\\frac{\\mathrm{C} \\alpha^2}{1-\\alpha} \\\\\n& 1.2 \\times 10^{-5}=(0.03)\\left(\\alpha^2\\right) \\\\\n& \\alpha=0.02 \\\\\n& \\pi=\\mathrm{iCRT} \\\\\n&=(1.02)(0.03)(0.083)(300) \\\\\n&=0.76194 \\\\\n&=76.194 \\times 10^{-2} \\mathrm{bar} \\\\\n& \\text { Nearest integer }=76\n\\end{aligned}$$

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1837, "subject": "Chemistry", "question": "1 M NaCL and 1 M HCL are present in an aqueous solution. The solution is", "options": [ { "text": "not a buffer solution with pH < 7" }, { "text": "not a buffer solution with pH > 7" }, { "text": "a buffer solution with pH < 7" }, { "text": "a buffer solution with pH > 7" } ], "answer": "not a buffer solution with pH < 7", "solution": "**Answer:** not a buffer solution with pH < 7\n\nNOTE : A buffer is a solution of weak acid and its salt with strong base and vice versa. \n

    $$HCL$$ is strong acid and $$NaCL$$ is its salt with strong base. \n

    $$pH$$ is less than $$7$$ due to $$HCL$$.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1838, "subject": "Chemistry", "question": "Which one of the following statements is not true?", "options": [ { "text": "pH + pOH = 14 for all aqueous solutions" }, { "text": "The pH of 1 $$\\times$$ 10-8 M HCl is 8" }, { "text": "96,500 coulombs of electricity when passed through a CuSO4 solution deposits 1 gram equivalent of copper at the cathode" }, { "text": "The conjugate base of $$H_2PO_4^-$$ is $$HPO_4^{2-}$$" } ], "answer": "The pH of 1 $$\\times$$ 10-8 M HCl is 8", "solution": "**Answer:** The pH of 1 $$\\times$$ 10-8 M HCl is 8\n\n$$pH$$ of an acidic solution should be less than $$7.$$ The reason is that from $${H_2}O.\\left[ {{H^ + }} \\right] = {10^{ - 7}}M$$ which cannot be neglected in comparison to $${10^{ - 8}}M.$$ The $$pH$$ can be calculated as. \n

    from acid, $$\\,\\,\\,\\left[ {{H^ + }} \\right] = {10^{ - 8}}M.$$\n

    from $${H_2}O,\\,\\,\\left[ {{H^ + }} \\right] = {10^{ - 7}}M$$\n

    $$\\therefore$$ $$\\,\\,\\,$$ Total $$\\left[ {{H^ + }} \\right] = {10^{ - 8}} + {10^{ - 7}}$$\n

    $$ = {10^{ - 8}}\\left( {1 + 10} \\right) = 11 \\times {10^{ - 8}}$$\n

    $$\\therefore$$ $$\\,\\,\\,pH = - \\log \\left[ {{H^ + }} \\right]$$\n

    $$ = - \\log 11 \\times {10^{ - 8}}$$\n

    $$ = - \\left[ {\\log 11 + 8\\log \\,10} \\right]$$\n

    $$ = - \\left[ {1.0414 - 8} \\right] = 6.9586$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1839, "subject": "Chemistry", "question": "When rain is accompanied by a thunderstorm, the collected rain water will have a pH value :", "options": [ { "text": "slightly higher than that when the thunderstorm is not there" }, { "text": "uninfluenced by occurence of thunderstorm" }, { "text": "which depends on the amount of dust in air" }, { "text": "slightly lower than that of rain water without thunderstorm" } ], "answer": "slightly lower than that of rain water without thunderstorm", "solution": "**Answer:** slightly lower than that of rain water without thunderstorm\n\nThe rain water after thunderstorm contains dissolved acid and therefore the $$pH$$ is less than rain water without thunderstorm. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1840, "subject": "Chemistry", "question": "Hydrogen ion concentration in mol / L in a solution of pH = 5.4 will be :", "options": [ { "text": "3.98 $$\\times$$ 10-6" }, { "text": "3.68 $$\\times$$ 10-6" }, { "text": "3.88 $$\\times$$ 106" }, { "text": "3.98 $$\\times$$ 108" } ], "answer": "3.98 $$\\times$$ 10-6", "solution": "**Answer:** 3.98 $$\\times$$ 10-6\n\n$$pH = - \\log \\left[ {{H^ + }} \\right] = \\log {1 \\over {\\left[ {{H^ + }} \\right]}};$$ \n

    $$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,5.4 = \\log {1 \\over {\\left[ {{H^ + }} \\right]}}$$\n

    On solving, $$\\left[ {{H^ + }} \\right] = 3.98 \\times {10^{ - 6}}$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1841, "subject": "Chemistry", "question": "The pKa of a weak acid (HA) is 4.5. The pOH of an aqueous buffered solution of HA in which 50% of\nthe acid is ionized is :", "options": [ { "text": "7.0" }, { "text": "4.5" }, { "text": "2.5" }, { "text": "9.5" } ], "answer": "9.5", "solution": "**Answer:** 9.5\n\nFor acidic buffer $$pH = p{K_a} + \\log \\left[ {{{salt} \\over {acid}}} \\right]$$\n

    or $$\\,\\,\\,pH = p{K_a} + \\log {{\\left[ {{A^ - }} \\right]} \\over {\\left[ {HA} \\right]}}$$\n

    Given $$p{K_a} = 4.5\\,\\,$$ and acid is $$50\\% $$ ionised.\n

    $$\\left[ {HA} \\right] = \\left[ {{A^ - }} \\right]\\,\\,\\,$$ (when acid is $$50\\% $$ ionised)\n

    $$\\therefore$$ $$\\,\\,\\,pH = p{K_a} + \\log \\,1$$\n

    $$\\therefore$$ $$\\,\\,\\,pH = p{K_a} = 4.5$$ \n

    $$pOH = 14 - pH$$\n

    $$ = 14 - 4.5$$\n

    $$ = 9.5$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1842, "subject": "Chemistry", "question": "The pKa of a weak acid, HA, is 4.80. The pKb of a weak base, BOH, is 4.78. The pH of an aqueous\nsolution of the corresponding salt, BA, will be", "options": [ { "text": "9.58" }, { "text": "4.79" }, { "text": "7.01" }, { "text": "9.22" } ], "answer": "7.01", "solution": "**Answer:** 7.01\n\nIn aqueous solution $$BA$$ (salt) hydrolyses to give \n

    $$BA\\,\\, + \\,\\,{H_2}O\\,\\rightleftharpoons\\,BOH\\,\\, + \\,\\,HA$$ \n

    $$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,$$ $$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,$$ Base acid\n

    Now $$pH$$ is given by\n

    $$pH = {1 \\over 2}p{K_w} + {1 \\over 2}pKa - {1 \\over 2}p{K_b}$$\n

    Substituting given values, we get\n

    $$pH = {1 \\over 2}\\left( {14 + 4.80 - 4.78} \\right) = 7.01$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1843, "subject": "Chemistry", "question": "At 25°C, the solubility product of Mg(OH)2 is 1.0 $$\\times$$ 10–11. At which pH, will Mg2+ ions start precipitating in the form of Mg(OH)2 from a solution of 0.001 M Mg2+ ions?", "options": [ { "text": "9" }, { "text": "10" }, { "text": "11" }, { "text": "8" } ], "answer": "10", "solution": "**Answer:** 10\n\n$$Mg{\\left( {OH} \\right)_2}\\,\\rightleftharpoons\\,M{g^{ + + }} + 2O{H^ - }$$\n

    $${K_{sp}} = \\left[ {M{g^{ + + }}} \\right]{\\left[ {O{H^ - }} \\right]^2}$$\n

    $$1.0 \\times {10^{ - 11}} = {10^{ - 3}} \\times {\\left[ {O{H^ - }} \\right]^2}$$\n

    $$\\left[ {O{H^ - }} \\right] = \\sqrt {{{{{10}^{ - 11}}} \\over {{{10}^{ - 3}}}}} = {10^{ - 4}}$$\n

    $$\\therefore$$ $$\\,\\,\\,pOH = 4$$\n

    $$\\therefore$$ $$\\,\\,\\,pH + pOH = 14$$\n

    $$\\therefore$$ $$\\,\\,\\,pH = 10$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1844, "subject": "Chemistry", "question": "The pH of a 0.1 molar solution of the acid HQ is 3. The value of the ionization constant, Ka of this acid is :", "options": [ { "text": "3 $$\\times$$ 10–1" }, { "text": "1 $$\\times$$ 10–3" }, { "text": "1 $$\\times$$ 10–5" }, { "text": "1 $$\\times$$ 10–7" } ], "answer": "1 $$\\times$$ 10–5", "solution": "**Answer:** 1 $$\\times$$ 10–5\n\n$${H^ + } = C\\alpha ;\\alpha = {{\\left[ {{H^ + }} \\right]} \\over C}$$\n

    or $$\\,\\,\\,\\alpha = {{{{10}^{ - 3}}} \\over {0.1}} = {10^{ - 2}}$$\n

    $$Ka = C{\\alpha ^2} = 0.1 \\times {10^{ - 2}} \\times {10^{ - 2}}$$ \n

    $$ = {10^{ - 5}}$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1845, "subject": "Chemistry", "question": "How many litres of water must be added to 1 litre of an aqueous solution of HCl with a pH of 1 to create an\naqueous solution with pH of 2? ", "options": [ { "text": "0.1 L" }, { "text": "0.9 L" }, { "text": "2.0 L" }, { "text": "9.0 L" } ], "answer": "9.0 L", "solution": "**Answer:** 9.0 L\n\nAs $$\\,\\,\\,pH = 1;\\,\\,{H^ + } = {10^{ - 1}} = 0.1\\,M$$ \n

    $$\\,\\,\\,pH = 2;\\,\\,{H^ + } = {10^{ - 2}} = 0.01\\,M$$\n

    $$\\therefore$$ $$\\,\\,\\,{M_1} = 0.1\\,\\,{V_1} = 1;\\,\\,$$ \n

    $${M_2} = 0.01\\,\\,{V_2} = ?$$\n

    From\n

    $${M_1}{V_1} = {M_2}{V_2};0.1 \\times 1 = 0.01 \\times {V_2}$$\n

    $${V_2} = 10\\,$$ litre\n

    $$\\therefore$$ volume of water added $$ = 10 - 1 = 9\\,$$ litre.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1846, "subject": "Chemistry", "question": "pKa of a weak acid (HA) and pKb of a weak base (BOH) are 3.2 and 3.4, respectively. The pH of their salt (AB) solution is :", "options": [ { "text": "6.9" }, { "text": "7.0" }, { "text": "1.0" }, { "text": "7.2" } ], "answer": "6.9", "solution": "**Answer:** 6.9\n\nThe salt (AB) given is a salt is of weak acid and weak base. Hence its pH can be calculated by\nthe formula\n

    $$ \\therefore $$ pH = 7 + $${1 \\over 2}p{K_a} - {1 \\over 2}p{K_b}$$\n

    = 7 + $${1 \\over 2}\\left( {3.2} \\right) - {1 \\over 2}\\left( {3.4} \\right)$$\n

    = 6.9", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1847, "subject": "Chemistry", "question": "Additin of sodium hydroxide solution to a weak acid (HA)results in a buffer of pH 6. If ionition constant of HA is 10$$-$$5, the ratio of salt to acid concentration in the buffer solution will be :", "options": [ { "text": "4 : 5" }, { "text": "1 : 10" }, { "text": "10 : 1" }, { "text": "5 : 4" } ], "answer": "10 : 1", "solution": "**Answer:** 10 : 1\n\nHA   $$\\rightleftharpoons$$    H+ + A$$-$$\n

    Ka = $${{\\left[ {{H^ + }} \\right]\\left[ {{A^ - }} \\right]} \\over {\\left[ {HA} \\right]}}$$ = 10$$-$$5\n

    pH = pKa + log $${{\\left[ {Salt} \\right]} \\over {\\left[ {Acid} \\right]}}$$\n

    $$ \\Rightarrow $$$$\\,\\,\\,$$ 6 = $$-$$ log [10$$-$$5] + log $${{\\left[ {Salt} \\right]} \\over {\\left[ {Acid} \\right]}}$$\n

    $$ \\Rightarrow $$$$\\,\\,\\,$$ 6 = 5 + log $${{\\left[ {Salt} \\right]} \\over {\\left[ {Acid} \\right]}}$$\n

    $$ \\Rightarrow $$ log $${{\\left[ {Salt} \\right]} \\over {\\left[ {Acid} \\right]}}$$ = 1\n

    $$ \\Rightarrow $$$$\\,\\,\\,$$ $${{\\left[ {Salt} \\right]} \\over {\\left[ {Acid} \\right]}}$$ = 10\n

    $$\\therefore\\,\\,\\,$$ Salt : Acid = 10 : 1", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1848, "subject": "Chemistry", "question": "50 mL of 0.2 M ammonia solution is treated with 25 mL of 0.2 M HCl. If pKb of ammonia solution is 4.75, the pH of the mixture will be :\n", "options": [ { "text": "3.75" }, { "text": "4.75" }, { "text": "8.25" }, { "text": "9.25" } ], "answer": "9.25", "solution": "**Answer:** 9.25\n\nNH3\n + HCl $$ \\to $$ NH4Cl\n

    moles of HCl = 0.2 M × 25 × 10–3 L = 0.005 moles HCl (total consumed)\n
    moles of NH3\n = 0.2 M × 50 × 10–3 L = 0.01 moles HCl\n
    excess NH3\n = 0.01 – 0.005 = 0.005 moles\n

    From reaction,\n1 mole ammonia = 1 mole NH4Cl\n

    $$ \\therefore $$ 0.005 NH3\n = 0.005 NH4Cl\n

    Total Volume = VHCl + VNH3 = 25 + 50 = 75 mL\n

    [NH3] = [NH4Cl] = $${{0.005} \\over {75 \\times {{10}^{ - 3}}}}$$ = 0.066 M\n

    pOH = pKb + log $${{\\left[ {salt} \\right]} \\over {\\left[ {base} \\right]}}$$\n

    = pKb + log$${{\\left[ {N{H_4}Cl} \\right]} \\over {\\left[ {N{H_3}} \\right]}}$$\n

    = 4.75 + log$${{\\left[ {0.066} \\right]} \\over {\\left[ {0.066} \\right]}}$$\n

    $$ \\Rightarrow $$ pOH = 4.75\n

    $$ \\therefore $$ pH = 14 – 4.75 = 9.25", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1849, "subject": "Chemistry", "question": "An alkali is titrated against an acid with methyl orange as indicator, which of the following is a correct combination?", "options": [ { "text": "\n\n\n \n \n \n \n \n\n\n \n \n \n \n \n\n
    BaseAcidEnd point
    WeakStrongColourless to pink
    " }, { "text": "\n \n \n \n \n \n \n \n \n \n \n
    BaseAcidEnd point
    StrongStrongPinkish red to yellow
    " }, { "text": "\n \n \n \n \n \n \n \n \n \n \n
    BaseAcidEnd point
    WeakStrongYellow to pinkish red
    " }, { "text": "\n \n \n \n \n \n \n \n \n \n \n
    BaseAcidEnd point
    StrongStrongPink to colourless
    " } ], "answer": "\n \n \n \n \n \n \n \n \n \n \n
    BaseAcidEnd point
    WeakStrongYellow to pinkish red
    ", "solution": "**Answer:** \n \n \n \n \n \n \n \n \n \n \n
    BaseAcidEnd point
    WeakStrongYellow to pinkish red
    \n\n

    Methyl orange is a pH indicator frequently used in titrations because of its clear and distinct colour change. Under acidic conditions, it is red or pinkish red and under alkaline conditions, it turns yellow.

    \n

    Let's consider each of the options :

    \n

    Option A : When a weak base is titrated against a strong acid, the solution at the end point is slightly acidic. However, the color change from colorless to pink is not associated with methyl orange.

    \n

    Option B : When a strong base is titrated against a strong acid, the solution is neutral at the end point. Methyl orange changes color from pinkish-red in acidic conditions to yellow in basic conditions, so this color change doesn't match the description.

    \n

    Option C : When a weak base is titrated against a strong acid, the solution is slightly acidic at the end point. Methyl orange would change from yellow in the basic solution to red in the acidic solution at the end point. This corresponds to a "yellow to pinkish-red" color change.

    \n

    Option D : When a strong base is titrated against a strong acid, the solution is neutral at the end point. The color change described here is "pink to colorless" which does not correspond with the characteristic changes of methyl orange.

    \n

    Therefore, the correct answer is Option C: Weak base, Strong acid, Yellow to pinkish-red.

    ", "topic": "Discrete Mathematics", "subtopic": "Combinatorics" }, { "id": 1850, "subject": "Chemistry", "question": "Following four solutions are prepared by mixing different volumes of NaOH and HCl of different concentrations, pH of which one of them will be equal to 1 ? ", "options": [ { "text": "100 mL $${M \\over {10}}$$ HCl + 100 mL $${M \\over {10}}$$ NaOH" }, { "text": "75 mL $${M \\over {5}}$$ HCl + 25 mL $${M \\over {5}}$$ NaOH" }, { "text": "60 mL $${M \\over {10}}$$ HCl + 40 mL $${M \\over {10}}$$ NaOH" }, { "text": "55 mL $${M \\over {10}}$$ HCl + 45 mL $${M \\over {10}}$$ NaOH" } ], "answer": "75 mL $${M \\over {5}}$$ HCl + 25 mL $${M \\over {5}}$$ NaOH", "solution": "**Answer:** 75 mL $${M \\over {5}}$$ HCl + 25 mL $${M \\over {5}}$$ NaOH\n\n(a)    100 mL  $${M \\over {10}}$$ NaOH will nutalise\n

    100 mL   $${M \\over {10}}$$ HCl, so number extra\n

    HCl   will remain.\n

    This will be neutral solution\n

    $$\\therefore\\,\\,\\,$$ pH = 7\n

    (b)    Here 25 mL  $${M \\over 5}$$ NaOH   nutralise \n

    25 mL  $${M \\over 5}$$ HCl\n

    $$\\therefore\\,\\,\\,$$ Extra HCl   =   75 $$-$$ 25   =   50 mL\n

    Total volume   =   75 + 25   =   100 mL\n

    $$\\therefore\\,\\,\\,$$ Milimole of HCl   =   $${{50} \\over 5}$$ = 10\n

    $$\\therefore\\,\\,\\,$$ Concentration of HCl  =  $${{10} \\over {100}}$$  = 0.1\n

    $$\\therefore\\,\\,\\,$$ pH  =  $$-$$ log[H+]  =  $$-$$ log(0.1)  =  1\n

    (c)    HCl left  =  60 $$-$$ 40  =  20 mL\n

    $$\\therefore\\,\\,\\,$$ milimole of HCl   =   $${{20} \\over {10}}$$   =  2\n

    $$\\therefore\\,\\,\\,$$ Concentration of HCl   =   $${2 \\over {100}}$$   =  0.02 M\n

    $$\\therefore\\,\\,\\,$$ pH  =  $$-$$ log (0.02)   =   1.69\n

    (d)    HCl left  = 55 $$-$$ 45  =  10 mL\n

    $$\\therefore\\,\\,\\,$$ milimole of HCl   =  $${{10} \\over {10}}$$  = 1\n

    $$\\therefore\\,\\,\\,$$ Concentration of HCl  = $${{1} \\over {100}}$$   =  0.01 M\n

    $$\\therefore\\,\\,\\,$$ pH = $$-$$ log (0.01) = 2", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1851, "subject": "Chemistry", "question": "20 mL of 0.1 M H2SO4 solution is added to 30 mL of of 0.2 M NH4OH solution. The pH of the resultant mixture is : [pkb of NH4OH = 4.7].", "options": [ { "text": "5.2" }, { "text": "9.0" }, { "text": "5.0" }, { "text": "9.4" } ], "answer": "9.0", "solution": "**Answer:** 9.0\n\nH2SO4 + 2NH4OH $$ \\to $$ (NH4)2SO4 + H2O\n

    Initially, \n

    H2SO4 present = 20 $$ \\times $$ 0.1 $$ \\times $$ 2 = 4 miliequivalent\n

    NH4OH present = 30 $$ \\times $$ 0.2 = 6 miliequivalent\n

    Here H2SO4 is the limiting reagent,\n

    So, finally. H2SO4 present = 0\n

    and NH4OH present = (6 $$-$$ 4) = 2\n

    and (NH4)2SO4 produced = 4 miliequivalent.\n

    As in the solution there is (NH4)2 SO4 present so it a basic buffer. \n

    $$ \\therefore $$   POH = PKb + log $${{\\left[ {Salt} \\right]} \\over {\\left[ {base} \\right]}}$$\n

    = 4.7 + log $${4 \\over 2}$$\n

    = 4.7 + log2\n

    = 4.7 + 0.3\n

    = 5\n

    $$ \\therefore $$   PH = 14 $$-$$ POH\n

    = 14 $$-$$ 5\n

    = 9", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1852, "subject": "Chemistry", "question": "Consider the following statements\n
    (a) The pH of a mixture containing 400 mL of 0.1 M H2SO4 and 400 mL of 0.1 M NaOH will be approximately 1.3\n
    (b) Ionic product of water is temperature dependent.\n
    (c) A monobasic acid with Ka = 10–5\n has pH = 5. The degree of dissociation of this acid is 50 %.\n
    (d) The Le Chatelier's principle is not applicable to common-ion effect. \n

    The correct statements are :", "options": [ { "text": "(a) and (b)" }, { "text": "(a), (b) and (c)" }, { "text": "(a), (b) and (d)" }, { "text": "(b) and (c)" } ], "answer": "(a), (b) and (c)", "solution": "**Answer:** (a), (b) and (c)\n\n(a) Equivalance of strong acid = 0.1 $$ \\times $$ 2 $$ \\times $$ 400 = 80\n

    Equivalance of strong base = 0.1 $$ \\times $$ 400 = 40\n

    $$ \\therefore $$ [H+] of mixture = $${{80 - 40} \\over {800}}$$ = $${1 \\over {20}}$$\n

    $$ \\therefore $$ pH = $$ - \\log \\left[ {{H^ + }} \\right]$$ = $$ - \\log \\left( {{1 \\over {20}}} \\right)$$ = 1.3\n

    (b) Ionic product of water increases with\nincrease of temperature because ionisation\nof water is endothermic.\n

    (c) \"JEE\n

    ka = $${{{{10}^{ - 5}} \\times c\\alpha } \\over {c\\left( {1 - \\alpha } \\right)}}$$\n

    $$ \\Rightarrow $$ 10-5 = $${{{{10}^{ - 5}} \\times \\alpha } \\over {\\left( {1 - \\alpha } \\right)}}$$\n

    $$ \\Rightarrow $$ $${\\alpha \\over {\\left( {1 - \\alpha } \\right)}}$$ = 1\n

    $$ \\Rightarrow $$ $$\\alpha $$ = 0.5\n

    $$ \\therefore $$ Degree of dissociation($$\\alpha $$) = 50%\n

    (d) The Le Chatelier's principle is always applicable to common-ion effect.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1853, "subject": "Chemistry", "question": "The pH of rain water, is approximately : ", "options": [ { "text": "5.6 " }, { "text": "7.5" }, { "text": "7.0" }, { "text": "6.5" } ], "answer": "5.6 ", "solution": "**Answer:** 5.6 \n\npH of rain water is approximate 5.6", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1854, "subject": "Chemistry", "question": "Two solutions, A and B, each of 100L was made by dissolving 4g of NaOH and 9.8 g of H2SO4 in water, respectively. The pH of the resultant solution obtained from mixing 40L of solution A and 10L of solution B is :", "options": [], "answer": "10.6", "solution": "**Answer:** 10.6\n\nIn 100 L solution H2SO4 present = 9.8 gm\n

    $$ \\therefore $$ In 10 L solution H2SO4 present = $$9.8 \\times {{10} \\over {100}}$$ gm\n

    $$ \\Rightarrow $$ In 10 L solution moles of H2SO4 present = $${{9.8} \\over {98}} \\times {{10} \\over {100}}$$\n

    In one molecule of H2SO4 two H+ ion present.\n

    $$ \\therefore $$ In 10 L solution moles of H+ present = 2$$ \\times $$$${{9.8} \\over {98}} \\times {{10} \\over {100}}$$ = 0.02 moles\n

    Also In 100 L solution NaOH present = 4 gm\n

    $$ \\therefore $$ In 40 L solution NaOH present = $$4 \\times {{40} \\over {100}}$$\n

    $$ \\Rightarrow $$ In 40 L solution moles of NaOH present = $${4 \\over {40}} \\times {{40} \\over {100}}$$\n

    In one molecule of NaOH one OH- ion present.\n

    $$ \\therefore $$ In 40 L solution moles of OH- ion present = $${4 \\over {40}} \\times {{40} \\over {100}}$$ = 0.04 moles\n

    As moles of OH- ion is more than H+ ion, so solution is basic.\n

    $$ \\therefore $$ Final Conc. of OH– = $${{0.04 - 0.02} \\over {40 + 10}}$$ = 4 $$ \\times $$ 10-4\n

    $$ \\therefore $$ pOH = – log (4 ×10–4) = 3.4\n

    pH = 14 – 3.4 = 10.6", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1855, "subject": "Chemistry", "question": "3 g of acetic acid is added to 250 mL of 0.1 M HCL and the solution made up to 500 mL. To 20 mL\nof this solutions $${1 \\over 2}$$ mL of 5 M NaOH is added. The pH of the solution is __________.

    \n[Given : pKa of acetic acid = 4.75, molar mass of acetic of acid = 60 g/mol, log 3 = 0.4771]\nNeglect any changes in volume.\n", "options": [], "answer": "5.22TO5.24", "solution": "**Answer:** 5.22TO5.24\n\nmilimole of acetic acid in 20 ml \n
    = $${3 \\over {60}} \\times 1000 \\times {{20} \\over {500}}$$ = 2 \n

    milimole of HCl in 20 ml = 25 $$ \\times $$ $${{20} \\over {500}}$$ = 1\n

    milimole of NaOH 20 ml = $${1 \\over 2} \\times 5$$ = 2.5\n

    \n\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n
    NaOH+CH3COOH$$ \\to $$CH3COONa+H2O
    1.5200
    00.51.5
    \n

    pH = pKa + log$${{1.5} \\over {0.5}}$$\n

    = 4.74 + log 3 = 5.22", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1856, "subject": "Chemistry", "question": "The strength of an aqueous NaOH solution is most accurately determined by titrating :
    (Note : consider that\nan appropriate indicator is used)", "options": [ { "text": "Aq. NaOH in a pipette and aqueous oxalic acid in a burette" }, { "text": "Aq. NaOH in a burette and aqueous oxalic acid in a conical flask" }, { "text": "Aq. NaOH in a volumetric flask and concentrated H2SO4 in a conical flask" }, { "text": "Aq. NaOH in a burette and concentrated H2SO4 in a conical flask" } ], "answer": "Aq. NaOH in a burette and aqueous oxalic acid in a conical flask", "solution": "**Answer:** Aq. NaOH in a burette and aqueous oxalic acid in a conical flask\n\nAq. NaOH in a burette and aqueous oxalic acid in a conical flask.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1857, "subject": "Chemistry", "question": "For the following Assertion and Reason, the\ncorrect option is :\n

    Assertion : The pH of water increases with\nincrease in temperature.
    \n
    Reason : The dissociation of water into H+ and\nOH– is an exothermic reaction.", "options": [ { "text": "Both assertion and reason are false." }, { "text": "Both assertion and reason are true, but the\nreason is not the correct explanation for the\nassertion." }, { "text": "Both assertion and reason are true, and the\nreason is the correct explanation for the\nassertion." }, { "text": "Assertion is not true, but reason is true." } ], "answer": "Both assertion and reason are false.", "solution": "**Answer:** Both assertion and reason are false.\n\nH2O ⇌ H+\n + OH-\n

    As $$\\Delta $$H $$>$$ 0 for this reaction. So Dissociation of H2O is endothermic.\n

    pH = $$ - \\ln \\left[ {{H^ + }} \\right]$$\n

    kw = $$\\left[ {{H^ + }} \\right]\\left[ {O{H^ - }} \\right]$$\n

    $$\\ln \\left\\{ {{{{k_{{T_2}}}} \\over {{k_{{T_1}}}}}} \\right\\} = {{\\Delta H^\\circ } \\over R}\\left\\{ {{1 \\over {{T_1}}} - {1 \\over {{T_2}}}} \\right\\}$$\n

    When temperature increases then

    $${{\\Delta H^\\circ } \\over R}\\left\\{ {{1 \\over {{T_1}}} - {1 \\over {{T_2}}}} \\right\\} > 0$$, as reaction is endothermic.\n

    $$ \\therefore $$ kT2 $$>$$ kT1 \n

    $$ \\Rightarrow $$ $${\\left[ {{H^ + }} \\right]_{{T_2}}}$$ $$>$$ $${\\left[ {{H^ + }} \\right]_{{T_1}}}$$\n

    $$ \\therefore $$ [pH]T2 $$<$$ [pH]T1\n\n\n\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1858, "subject": "Chemistry", "question": "An acidic buffer is obtained on mixing :", "options": [ { "text": "100 mL of 0.1 M HCl and 200 mL of 0.1 M CH3COONa" }, { "text": "100 mL of 0.1 M HCl and 200 mL of 0.1 M NaCl" }, { "text": "100 mL of 0.1 M CH3COOH and 100 mL of 0.1 M NaOH" }, { "text": "100 mL of 0.1 M CH3COOH and 200 mL of 0.1 M NaOH" } ], "answer": "100 mL of 0.1 M HCl and 200 mL of 0.1 M CH3COONa", "solution": "**Answer:** 100 mL of 0.1 M HCl and 200 mL of 0.1 M CH3COONa\n\n\"JEE\n

    So finally we get mixture of\n

    CH3COOH + CH3COONa that will work like acidic buffer solution.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1859, "subject": "Chemistry", "question": "A soft drink was bottled with a partial pressure of CO2\n of 3 bar over the liquid at room temperature.\nThe partial pressure of CO2\n over the solution approaches a value of 30 bar when 44 g of CO2\n is\ndissolved in 1 kg of water at room temperature. The approximate pH of the soft drink is ______ $$ \\times $$ 10–1.\n
    (First dissociation constant of
    H2CO3\n = 4.0 $$ \\times $$ 10–7; log 2 = 0.3; density
    of the soft drink = 1 g mL–1)\n.\n", "options": [], "answer": "37", "solution": "**Answer:** 37\n\nCO2\n + H2O $$ \\to $$ H2CO3\n

    At 30 bar pressure mass of CO2 in 1 kg water\n= 44 gm\n

    At 3 bar pressure mass of CO2 in 1 kg water\n= 4.4 gm\n

    $$ \\therefore $$ Moles of CO2 in 1 kg water = $${{4.4} \\over {44}}$$ = 0.1\n

    \n\n\n \n \n \n \n \n \n \n \n\n\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\n
    H2CO3⇌H++HCO3-
    t = 00.100
    t = teq0.1(1 - $$\\alpha $$)0.1$$\\alpha $$0.1$$\\alpha $$
    \n

    4.0 $$ \\times $$ 10–7 = $${{0.1{\\alpha ^2}} \\over {1 - \\alpha }}$$\n

    $${1 - \\alpha }$$ $$ \\simeq $$ 1\n

    $$ \\Rightarrow $$ 0.1$${{\\alpha ^2}}$$ = 4 $$ \\times $$ 10-7\n

    $$ \\Rightarrow $$ $$\\alpha $$ = 2 $$ \\times $$ 10-3\n

    [H+] = 0.1$$\\alpha $$ = 2 $$ \\times $$ 10-4\n

    $$ \\therefore $$ pH = –[– 4 × log(2)] = 3.7 = 37 × 10–1", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1860, "subject": "Chemistry", "question": "Arrange the following solutions in the\ndecreasing order of pOH :\n

    (A) 0.01 M HCl\n
    (B) 0.01 M NaOH\n
    (C) 0.01 M CH3COONa\n
    (D) 0.01 M NaCl\n", "options": [ { "text": "(B) > (C) > (D) > (A)" }, { "text": "(A) > (D) > (C) > (B)" }, { "text": "(A) > (C) > (D) > (B)" }, { "text": "(B) > (D) > (C) > (A)" } ], "answer": "(A) > (D) > (C) > (B)", "solution": "**Answer:** (A) > (D) > (C) > (B)\n\n(A) 10–2 M HCl $$ \\Rightarrow $$ [H+] = 10–2 M $$ \\Rightarrow $$ pH = 2 $$ \\Rightarrow $$ pOH = 14 - 2 = 12\n
    (B) 10–2 M NaOH $$ \\Rightarrow $$ [OH–] = 10–2 M $$ \\Rightarrow $$ pOH = 2\n
    (C) 10–2 M CH3COO–Na+ $$ \\Rightarrow $$ [OH+] > 10–7 $$ \\Rightarrow $$ pOH < 7\n
    (D) 10–2 M NaCl $$ \\Rightarrow $$ Neutral pOH = 7\n

    Order of pOH value A > D > C > B", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1861, "subject": "Chemistry", "question": "The pH of ammonium phosphate solution, if pka of phosphoric acid and pkb of ammonium hydroxide are 5.23 and 4.75 respectively, is ___________.", "options": [], "answer": "7", "solution": "**Answer:** 7\n\nSince (NH4)3PO4 is salt of weak acid (H3PO4) & weak base (NH4OH).

    pH = 7 + $${1 \\over 2}$$(pka $$-$$ pkb)

    = 7 + $${1 \\over 2}$$ (5.23 $$-$$ 4.75)

    = 7.24 $$ \\approx $$ 7.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1862, "subject": "Chemistry", "question": "Sulphurous acid (H2SO3) has Ka1 = 1.7 $$\\times$$ 10$$-$$2 and Ka2 = 6.4 $$\\times$$ 10$$-$$8. The pH of 0.588 M H2SO3 is __________. (Round off to the Nearest Integer).", "options": [], "answer": "1", "solution": "**Answer:** 1\n\n$$K{a_1}$$ of $${H_2}S{O_3} > > K{a_2}$$ of $${H_2}S{O_3}$$

    $$ \\therefore $$ The contribution of H+ from 2nd dissociation of H2SO3 can be neglected.

    \"JEE

    $$ \\Rightarrow $$ $${{c{\\alpha ^2}} \\over {1 - \\alpha }} = 1.7 \\times {10^{ - 2}}$$

    $$ \\Rightarrow {{0.588{\\alpha ^2}} \\over {1 - \\alpha }} = 1.7 \\times {10^{ - 2}}$$

    $$ \\Rightarrow 58.8{\\alpha ^2} = 1.7 - 1.7\\alpha $$

    $$ \\Rightarrow 58.8{\\alpha ^2} + 1.7\\alpha - 1.7 = 0$$

    $$\\alpha = {{ - 1.7 + \\sqrt {{{1.7}^2} + 4 \\times 1.7 \\times 58.8} } \\over {2 \\times 58.8}} = 0.156$$

    $$[{H^ + }] = c\\alpha = 0.092$$

    $$pH = - \\log [{H^ + }]$$

    $$ = 1.036$$

    $$ \\approx 1$$

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1863, "subject": "Chemistry", "question": "In order to prepare a buffer solution of pH 5.74, sodium acetate is added to acetic acid. If the concentration of acetic acid in the buffer is 1.0 M, the concentration of sodium acetate in the buffer is ___________ M. (Round off to the Nearest Integer). [Given : pKa (acetic acid) = 4.74]", "options": [], "answer": "10", "solution": "**Answer:** 10\n\n$$C{H_3}COOH + C{H_3}COONa$$ [Acidic Buffer]

    $${pH} = {p{ka}} + \\log {{[C{H_3}COONa]} \\over {[C{H_3}COOH]}}$$

    $$ \\Rightarrow 5.74 = 4.74 + \\log {{[C{H_3}COONa]} \\over {[C{H_3}COOH]}}$$

    $$ \\Rightarrow 1 = \\log {{[C{H_3}COONa]} \\over {[C{H_3}COOH]}}$$

    $$ \\Rightarrow {{[C{H_3}COONa]} \\over {[C{H_3}COOH]}} = 10$$

    $$ \\Rightarrow [C{H_3}COONa] = 10 \\times 1 = 10$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1864, "subject": "Chemistry", "question": "Given below are two statements.

    Statement I : In the titration between strong acid and weak base methyl orange is suitable as an indicator.

    Statement II : For titration of acetic acid with NaOH phenolphthalein is not a suitable indicator.

    In the light of the above statements, choose the most appropriate answer from the options given below :", "options": [ { "text": "Statement I is false but Statement II is true" }, { "text": "Statement I is true but Statement II is false" }, { "text": "Both Statement I and Statement II are true" }, { "text": "Both Statement I and Statement II are false" } ], "answer": "Statement I is true but Statement II is false", "solution": "**Answer:** Statement I is true but Statement II is false\n\nTitration curve for strong acid and weak base initially a buffer of weak base and conjugate acid is :

    \"JEE

    Formed, thus pH falls slowly and after equivalence point, so the pH falls sharply so methyl orange, having pH range of 3.2 to 4.4 will weak as indicator. So, statement I is correct.

    \"JEE

    Titration curve for weak acid and strong base (NaOH)

    Initially weak acid will form a buffer so pH increases slowly but after equivalence point. It rises sharply covering range of phenolphthalein so it will be suitable indicator so statement II is false.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1865, "subject": "Chemistry", "question": "The pH of a solution obtained by mixing 50 mL of 1 M HCl and 30 mL of 1 M NaOH is x $$\\times$$ 10$$-$$4. The value of x is ____________. (Nearest integer) [log 2.5 = 0.3979]", "options": [], "answer": "6021", "solution": "**Answer:** 6021\n\n\"JEE

    $$[HCl] = {{20} \\over {80}} = {1 \\over 4}M = 2.5 \\times {10^{ - 1}}M$$

    pH = $$-$$log 2.15 $$\\times$$ 10-1 = 1 $$-$$ 0.3979 = 0.6021

    pH = 6021 $$\\times$$ 10-4", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1866, "subject": "Chemistry", "question": "

    A student needs to prepare a buffer solution of propanoic acid and its sodium salt with pH 4. The ratio of $${{[C{H_3}C{H_2}CO{O^ - }]} \\over {[C{H_3}C{H_2}COOH]}}$$ required to make buffer is ___________.

    \n

    Given : $${K_a}(C{H_3}C{H_2}COOH) = 1.3 \\times {10^{ - 5}}$$

    ", "options": [ { "text": "0.03" }, { "text": "0.13" }, { "text": "0.23" }, { "text": "0.33" } ], "answer": "0.13", "solution": "**Answer:** 0.13\n\n$\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{COOH} \\rightleftharpoons \\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{COO}^{-}+\\mathrm{H}^{+}$\n

    \nFrom Henderson equation\n

    \n$\\mathrm{pH}=\\mathrm{pK}_{\\mathrm{a}}+\\log \\frac{\\left[\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{COO}^{-}\\right]}{\\left[\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{COOH}\\right]}$\n

    \n$4=-\\log 1.3 \\times 10^{-5}+\\log \\frac{\\left[\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{COO}^{-}\\right]}{\\left[\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{COOH}\\right]}$\n

    \n$-\\log 10^{-4}=-\\log 1.3 \\times 10^{-5}+\\log \\frac{\\left[\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{COO}^{-}\\right]}{\\left[\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{COOH}\\right]}$\n

    \n$-\\log 10^{-4}=-\\log 1.3 \\times 10^{-5} \\frac{\\left[\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{COOH}\\right]}{\\left[\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{COO}^{-}\\right]}$\n

    \n$10^{-4}=1.3 \\times 10^{-5} \\frac{\\left[\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{COOH}\\right]}{\\left[\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{COO}^{-}\\right]}$\n

    \n$\\frac{\\left[\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{COO}^{-}\\right]}{\\left[\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{COOH}\\right]}=0.13$\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1867, "subject": "Chemistry", "question": "

    pH value of 0.001 M NaOH solution is ____________.

    ", "options": [], "answer": "11", "solution": "**Answer:** 11\n\n$$\n\\begin{aligned}\n&{\\left[\\mathrm{OH}^{-}\\right]=0.001=10^{-3} \\mathrm{M}} \\\\\\\\\n&{\\left[\\mathrm{H}^{+}\\right]\\left[\\mathrm{OH}^{-}\\right]=10^{-14}} \\\\\\\\\n&\\quad\\left[\\mathrm{H}^{+}\\right]=10^{-11} \\\\\\\\\n&\\begin{aligned}\n\\mathrm{pH} &=-\\log \\left[\\mathrm{H}^{+}\\right] \\\\\\\\\n=&-\\log \\left(10^{-11}\\right) \\\\\\\\\n\\mathrm{pH} &=11\n\\end{aligned}\n\\end{aligned}\n$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1868, "subject": "Chemistry", "question": "

    50 mL of 0.1 M CH3COOH is being titrated against 0.1 M NaOH. When 25 mL of NaOH has been added, the pH of the solution will be _____________ $$\\times$$ 10$$-$$2. (Nearest integer)

    \n

    (Given : pKa (CH3COOH) = 4.76)

    \n

    log 2 = 0.30

    \n

    log 3 = 0.48

    \n

    log 5 = 0.69

    \n

    log 7 = 0.84

    \n

    log 11 = 1.04

    ", "options": [], "answer": "476", "solution": "**Answer:** 476\n\n

    CH3COOH + NaOH $$\\to$$ CH3COONa + H2O

    \n

    After adding 25 ml of NaOH volume of mixture = 50 + 25 = 75 ml

    \n

    Initially,

    \n

    Number of millimole of NaOH = 25 $$\\times$$ 0.1 = 2.5 mm

    \n

    Number of millimole of CH3COOH = 50 $$\\times$$ 0.1 = 5 mm

    \n

    After nutrilisation,

    \n

    Millimole of NaOH = 0

    \n

    Millimole of CH3COOH = 5 $$-$$ 2.5 = 2.5 mm

    \n

    Millimole of CH3COONa = 2.5

    \n

    After nutrilisation,

    \n

    Concentration of CH3COOH = $$[C{H_3}COOH] = {{5 - 2.5} \\over {75}} = {1 \\over {30}}$$

    \n

    Concentration of CH3COONa = $$[C{H_3}COONa] = {{ 2.5} \\over {75}} = {1 \\over {30}}$$

    \n

    $${P^H} = {P^{Ka}} + \\log {{[C{H_3}COONa]} \\over {[C{H_3}COOH]}}$$

    \n

    $$ = 4.76 + \\log {{{1 \\over {30}}} \\over {{1 \\over {30}}}}$$

    \n

    $$ = 4.76 + \\log (1)$$

    \n

    $$ = 4.76 + 0$$

    \n

    $$ = 4.76$$

    \n

    $$ = 4.76 \\times {10^{ - 2}}$$

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1869, "subject": "Chemistry", "question": "

    Class XII students were asked to prepare one litre of buffer solution of $$\\mathrm{pH} \\,8.26$$ by their Chemistry teacher: The amount of ammonium chloride to be dissolved by the student in $$0.2\\, \\mathrm{M}$$ ammonia solution to make one litre of the buffer is :

    \n

    (Given: $$\\mathrm{pK}_{\\mathrm{b}}\\left(\\mathrm{NH}_{3}\\right)=4.74$$

    \n

    Molar mass of $$\\mathrm{NH}_{3}=17 \\mathrm{~g} \\mathrm{~mol}^{-1}$$

    \n

    Molar mass of $$\\mathrm{NH}_{4} \\mathrm{Cl}=53.5 \\mathrm{~g} \\mathrm{~mol}^{-1}$$ )

    ", "options": [ { "text": "53.5 g" }, { "text": "72.3 g" }, { "text": "107.0 g" }, { "text": "126.0 g" } ], "answer": "107.0 g", "solution": "**Answer:** 107.0 g\n\nFor basic Buffer, $$\\mathrm{pOH}=\\mathrm{pK}_{\\mathrm{b}}+\\log \\frac{[\\text { salt }]}{[\\text { Base }]}$$\n

    \n$$\\mathrm{pOH}=14-8.26=5.74$$\n

    \n$$5.74=4.74+\\log \\frac{\\left[\\mathrm{NH}_{4} \\mathrm{Cl}\\right]}{0.2}$$\n

    \n$$\\left[\\mathrm{NH}_{4} \\mathrm{Cl}\\right]=2 \\mathrm{M}$$\n

    \nMoles of $$\\mathrm{NH}_{4} \\mathrm{Cl}=2 \\times 1=2$$ moles\n

    \nWeight of $$\\mathrm{NH}_{4} \\mathrm{Cl}=2 \\times 53.5=107 \\mathrm{~g}$$\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1870, "subject": "Chemistry", "question": "

    $$\\mathrm{K}_{\\mathrm{a}}$$ for butyric acid $$\\left(\\mathrm{C}_{3} \\mathrm{H}_{7} \\mathrm{COOH}\\right)$$ is $$2 \\times 10^{-5}$$. The $$\\mathrm{pH}$$ of $$0.2 \\,\\mathrm{M}$$ solution of butyric acid is __________ $$\\times 10^{-1}$$. (Nearest integer)

    \n

    [Given $$\\log 2=0.30$$]

    ", "options": [], "answer": "27", "solution": "**Answer:** 27\n\n$\\mathrm{K}_{\\mathrm{a}}$ of Butyric acid $\\Rightarrow 2 \\times 10^{-5} \\,\\mathrm{PKa}=4.7$

    $\\mathrm{pH}$ of $0.2 \\mathrm{M}$ solution,

    \n$$\n\\mathrm{pH}=\\frac{1}{2} \\mathrm{pK}_{\\mathrm{a}}-\\frac{1}{2} \\log \\mathrm{C}\n$$

    \n$$\n\\begin{aligned}\n&=\\frac{1}{2}(4 \\cdot 7) - \\frac{1}{2} \\log (0.2) \\\\\\\\\n&=2.35+0.35=2.7\n\\end{aligned}\n$$

    \n$$\n\\mathrm{pH}=27 \\times 10^{-1}\n$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1871, "subject": "Chemistry", "question": "

    $$200 \\mathrm{~mL}$$ of $$0.01 \\,\\mathrm{M} \\,\\mathrm{HCl}$$ is mixed with $$400 \\mathrm{~mL}$$ of $$0.01 \\,\\mathrm{M} \\,\\mathrm{H}_{2} \\mathrm{SO}_{4}$$. The $$\\mathrm{pH}$$ of the mixture is _________.

    \n

    Given: $$\\log {2}=0.30, \\log 3=0.48, \\log 5=0.70, \\log 7=0.84, \\log 11=1.04$$

    ", "options": [ { "text": "1.14" }, { "text": "1.78" }, { "text": "2.34" }, { "text": "3.02" } ], "answer": "1.78", "solution": "**Answer:** 1.78\n\n$$\\begin{aligned} {\\left[\\mathrm{H}^{+}\\right] } &=\\frac{0.01 \\times 200+2 \\times 0.01 \\times 400}{600} \\\\ &=\\frac{0.01+2 \\times 0.01 \\times 2}{3} \\\\ &=\\frac{0.01+0.04}{3} \\\\ &=\\frac{5}{3} \\times 10^{-2} \\\\ \\mathrm{pH} &=-\\log \\left[\\mathrm{H}^{+}\\right] \\\\ &=-\\log \\left(\\frac{5}{3} \\times 10^{-2}\\right) \\\\ &=-\\left[\\log \\frac{5}{3}+\\log 10^{-2}\\right] \\\\ &=-[\\log 5-\\log 3-2] \\\\ &=-0.7+0.48+2 \\\\ &=2.48-0.7 \\\\ &=1.78 \\end{aligned}$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1872, "subject": "Chemistry", "question": "The incorrect statement for the use of indicators in acid-base titration is :", "options": [ { "text": "Phenolphthalein is a suitable indicator for a weak acid vs strong base titration." }, { "text": "Methyl orange may be used for a weak acid vs weak base titration." }, { "text": "Methyl orange is a suitable indicator for a strong acid vs weak base titration." }, { "text": "Phenolphthalein may be used for a strong acid vs strong base titration." } ], "answer": "Methyl orange may be used for a weak acid vs weak base titration.", "solution": "**Answer:** Methyl orange may be used for a weak acid vs weak base titration.\n\n\n\n\n \n \n \n \n\n\n \n \n \n \n \n \n \n \n\n
    IndicatorpH range
    Methyl orange3.2 - 4.5
    Phenolpthalein8.3 - 10.5
    \n
    Option B is the incorrect statement for the use of indicators in acid-base titration.\n\n

    Methyl orange is a suitable indicator for a strong acid vs strong base titration, not for a weak acid vs weak base titration. In a weak acid vs weak base titration, the pH at the equivalence point is typically around 7, which is within the pH range where methyl orange undergoes a color change. Therefore, methyl orange is not suitable for use in weak acid vs weak base titrations.\n\n

    The other options are correct:\n\n

    Phenolphthalein is a suitable indicator for a weak acid vs strong base titration because the pH at the equivalence point is basic, which causes phenolphthalein to undergo a color change.\n

    Methyl orange is a suitable indicator for a strong acid vs weak base titration because the pH at the equivalence point is acidic, which causes methyl orange to undergo a color change.\n

    Phenolphthalein may be used for a strong acid vs strong base titration because the pH at the equivalence point is basic, which causes phenolphthalein to undergo a color change.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1873, "subject": "Chemistry", "question": "

    $$600 \\mathrm{~mL}$$ of $$0.01~\\mathrm{M} ~\\mathrm{HCl}$$ is mixed with $$400 \\mathrm{~mL}$$ of $$0.01~\\mathrm{M} ~\\mathrm{H}_{2} \\mathrm{SO}_{4}$$. The $$\\mathrm{pH}$$ of the mixture is ___________ $$\\times 10^{-2}$$. (Nearest integer)

    \n

    [Given $$\\log 2=0.30$$

    \n

    $$\\log 3=0.48$$

    \n

    $$\\log 5=0.69$$

    \n

    $$\\log 7=0.84$$

    \n

    $$\\log 11=1.04]$$

    ", "options": [], "answer": "186", "solution": "**Answer:** 186\n\n

    $$\\mathrm{[H^+]=\\frac{6+8}{1000}=14\\times10^{-3}}$$

    \n

    $$\\mathrm{pH=3-\\log14}$$

    \n

    $$=3-.3-.84$$

    \n

    $$=1.86=186\\times10^{-2}$$

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1874, "subject": "Chemistry", "question": "

    Millimoles of calcium hydroxide required to produce 100 mL of the aqueous solution of pH 12 is $$x\\times10^{-1}$$. The value of $$x$$ is ___________ (Nearest integer).

    \n

    Assume complete dissociation.

    ", "options": [], "answer": "5", "solution": "**Answer:** 5\n\n$$\n\\begin{aligned}\n& \\mathrm{pH}=12, \\\\\\\\\n& \\mathrm{pH}+\\mathrm{pOH}=14 \\\\\\\\\n& \\mathrm{pOH}=14-12=2 \\\\\\\\\n& {\\left[\\mathrm{OH}^{-}\\right]=10^{-2} \\mathrm{~mol} \\mathrm{~L}} \\\\\\\\\n& \\underset{5 \\times 10^{-3}}{\\mathrm{Ca}(\\mathrm{OH})_2} \\longrightarrow \\underset{5 \\times 10^{-3}}{\\mathrm{Ca}^{2+}}+\\underset{10^{-2}}{2 \\mathrm{OH}^{-}}\n\\end{aligned}\n$$\n

    Moles of $\\mathrm{Ca}(\\mathrm{OH})_2$ in $1000 \\mathrm{~mL}=5 \\times 10^{-3}$

    Millimoles in $100 \\mathrm{~mL}=5 \\times 10^{-1}$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1875, "subject": "Chemistry", "question": "

    When the hydrogen ion concentration [H$$^+$$] changes by a factor of 1000, the value of pH of the solution __________

    ", "options": [ { "text": "decreases by 2 units" }, { "text": "increases by 2 units" }, { "text": "decreases by 3 units" }, { "text": "increases by 1000 units" } ], "answer": "decreases by 3 units", "solution": "**Answer:** decreases by 3 units\n\nLet the initial concentration of $\\mathrm{H}^{+}$ be 1\n

    \n$\\therefore \\left[\\mathrm{H}^{+}\\right]_{\\mathrm{i}}=1 \\Rightarrow \\mathrm{pH}=0$\n

    \nIt changes by 1000 units\n

    \n$\\therefore \\left[\\mathrm{H}^{+}\\right]_{\\mathrm{f}}=10^{3} \\Rightarrow \\mathrm{pH}=-3$\n

    \n$\\therefore \\mathrm{pH}$ decreases by 3 units", "topic": "Number Theory", "subtopic": "Divisibility / Factorization" }, { "id": 1876, "subject": "Chemistry", "question": "

    A litre of buffer solution contains 0.1 mole of each of NH$$_3$$ and NH$$_4$$Cl. On the addition of 0.02 mole of HCl by dissolving gaseous HCl, the pH of the solution is found to be _____________ $$\\times$$ 10$$^{-3}$$ (Nearest integer)

    \n

    [Given : $$\\mathrm{pK_b(NH_3)=4.745}$$

    \n

    $$\\mathrm{\\log2=0.301}$$

    \n

    $$\\mathrm{\\log3=0.477}$$

    \n

    $$\\mathrm{T=298~K]}$$

    ", "options": [], "answer": "9079", "solution": "**Answer:** 9079\n\nIn resultant solution

    \n$$\n\\begin{aligned}\n& \\mathrm{n}_{\\mathrm{NH}_3}= 0.1-0.02=0.08 \\\\\\\\\n& \\mathrm{n}_{\\mathrm{NH}_4 \\mathrm{Cl}}= \\mathrm{n}_{\\mathrm{NH}_4^{+}}=0.1+0.02=0.12 \\\\\\\\\n& \\mathrm{pOH}= \\mathrm{pK}_{\\mathrm{b}}+\\log \\frac{\\left[\\mathrm{NH}_4^{+}\\right]}{\\left[\\mathrm{NH}_3\\right]} \\\\\\\\\n&=4.745+\\log \\frac{0.12}{0.08} \\\\\\\\\n&=4.745+\\log \\frac{3}{2} \\\\\\\\\n&=4.745+0.477-0.301 \\\\\\\\\n& \\mathrm{pOH}=4.921 \\\\\\\\\n& \\mathrm{pH}=14-\\mathrm{pH} \\\\\\\\\n&=9.079 = 9079\\times 10^{-3}\n\\end{aligned}\n$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1877, "subject": "Chemistry", "question": "

    The dissociation constant of acetic acid is $$x\\times10^{-5}$$. When 25 mL of 0.2 $$\\mathrm{M~CH_3COONa}$$ solution is mixed with 25 mL of 0.02 $$\\mathrm{M~CH_3COOH}$$ solution, the pH of the resultant solution is found to be equal to 5. The value of $$x$$ is ____________

    ", "options": [], "answer": "10", "solution": "**Answer:** 10\n\n$\\mathrm{pH}=\\mathrm{pK}_{\\mathrm{a}}+\\log \\left(\\frac{25 \\times 0.2}{25 \\times 0.02}\\right)$\n

    \n$5=\\mathrm{pK}_{\\mathrm{a}}+\\log 10$\n

    \n$\\mathrm{pK}_{\\mathrm{a}}=4 \\Rightarrow \\mathrm{K}_{\\mathrm{a}}=10^{-4}=10 \\times 10^{-5}$\n

    \n$x=10$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1878, "subject": "Chemistry", "question": "Which of the following statement(s) is/are correct?

    \n(A) The $\\mathrm{pH}$ of $1 \\times 10^{-8}~ \\mathrm{M} ~\\mathrm{HCl}$ solution is 8 .

    \n(B) The conjugate base of $\\mathrm{H}_{2} \\mathrm{PO}_{4}^{-}$ is $\\mathrm{HPO}_{4}^{2-}$.

    \n(C) $\\mathrm{K}_{\\mathrm{w}}$ increases with increase in temperature.

    \n(D) When a solution of a weak monoprotic acid is titrated against a strong base at half neutralisation point, $\\mathrm{pH}=\\frac{1}{2} \\mathrm{pK}_{\\mathrm{a}}$

    \n\nChoose the correct answer from the options given below:", "options": [ { "text": "$(\\mathrm{A}),(\\mathrm{B}),(\\mathrm{C})$" }, { "text": "(B), (C)" }, { "text": "(B), (C), (D)" }, { "text": "(A), (D)" } ], "answer": "(B), (C)", "solution": "**Answer:** (B), (C)\n\n(A) The $\\mathrm{pH}$ of $1 \\times 10^{-8}~ \\mathrm{M} ~\\mathrm{HCl}$ solution is 8.\n

    \nThis statement is incorrect. For a strong acid like HCl, the concentration of H+ ions will be the same as the concentration of the acid, i.e., $1 \\times 10^{-8}~\\mathrm{M}$. The pH can be calculated using the formula:\n

    \n$\\mathrm{pH} = -\\log [\\mathrm{H}^+] = -\\log (1 \\times 10^{-8}) = 8$\n

    \nHowever, because the concentration is so low, it approaches the range where water auto-ionization becomes significant. In this case, the solution pH will be slightly higher than 7, but not exactly 8.\n

    \n(B) The conjugate base of $\\mathrm{H}_{2} \\mathrm{PO}_{4}^{-}$ is $\\mathrm{HPO}_{4}^{2-}$.\n

    \nThis statement is correct. The conjugate base of an acid is formed when it loses one H+ ion:\n

    \n$\\mathrm{H}_{2} \\mathrm{PO}_{4}^{-} \\rightarrow \\mathrm{HPO}_{4}^{2-} + \\mathrm{H}^{+}$\n

    \n(C) $\\mathrm{K}_{\\mathrm{w}}$ increases with an increase in temperature.\n

    \nThis statement is correct. The ion product of water, $\\mathrm{K}_{\\mathrm{w}}$, increases with increasing temperature. This is because the auto-ionization of water is an endothermic process, meaning it absorbs heat:\n

    \n$\\mathrm{H}_{2} \\mathrm{O} \\rightleftharpoons \\mathrm{H}^{+} + \\mathrm{OH}^{-}$\n

    \nAs the temperature increases, the equilibrium shifts towards the formation of more $\\mathrm{H}^{+}$ and $\\mathrm{OH}^{-}$ ions, leading to an increase in $\\mathrm{K}_{\\mathrm{w}}$.\n

    \n(D) When a solution of a weak monoprotic acid is titrated against a strong base at the half-neutralization point, $\\mathrm{pH}=\\frac{1}{2} \\mathrm{pK}_{\\mathrm{a}}$\n

    \nThis statement is incorrect. At the half-neutralization point, the concentration of the weak acid ([HA]) is equal to the concentration of its conjugate base ([A-]). According to the Henderson-Hasselbalch equation:\n

    \n$\\mathrm{pH} = \\mathrm{pK}_{\\mathrm{a}} + \\log \\frac{[\\mathrm{A}^{-}]}{[\\mathrm{HA}]}$\n

    \nAt the half-neutralization point, the ratio of [A-] to [HA] is 1, so the equation becomes:\n

    \n$\\mathrm{pH} = \\mathrm{pK}_{\\mathrm{a}} + \\log (1) = \\mathrm{pK}_{\\mathrm{a}}$\n

    \nTherefore, the correct answer is:\n

    \n(B) and (C) are correct.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1879, "subject": "Chemistry", "question": "

    20 mL of $$0.1 ~\\mathrm{M} ~\\mathrm{NaOH}$$ is added to $$50 \\mathrm{~mL}$$ of $$0.1 ~\\mathrm{M}$$ acetic acid solution. The $$\\mathrm{pH}$$ of the resulting solution is ___________ $$\\times 10^{-2}$$ (Nearest integer)

    \n

    Given : $$\\mathrm{pKa}\\left(\\mathrm{CH}_{3} \\mathrm{COOH}\\right)=4.76$$

    \n

    $$\\log 2=0.30$$

    \n

    $$\\log 3=0.48$$

    ", "options": [], "answer": "458", "solution": "**Answer:** 458\n\nFirst, we need to find the moles of NaOH and acetic acid (CH₃COOH) in the solution:\n

    \nMoles of NaOH = Volume × Molarity = 20 mL × 0.1 M = 2 mmol

    \nMoles of acetic acid = Volume × Molarity = 50 mL × 0.1 M = 5 mmol\n

    \nSince NaOH is a strong base, it will react with acetic acid to form acetate ions (CH₃COO⁻) and water:\n

    \nCH₃COOH + OH⁻ → CH₃COO⁻ + H₂O\n

    \n2 mmol of NaOH will react with 2 mmol of acetic acid, resulting in 2 mmol of acetate ions and leaving 3 mmol of acetic acid unreacted.\n

    \nNext, we need to find the concentrations of acetic acid and acetate ions in the resulting 70 mL solution:\n

    \nConcentration of acetic acid = Moles / Total volume = 3 mmol / 70 mL = 0.04286 M

    \nConcentration of acetate ions = Moles / Total volume = 2 mmol / 70 mL = 0.02857 M\n

    \nNow we can use the Henderson-Hasselbalch equation to find the pH of the resulting solution:\n

    \npH = pKa + log ([A⁻] / [HA])\n

    \nGiven the pKa of acetic acid is 4.76, we can substitute the values:\n

    \npH = 4.76 + log (0.02857 / 0.04286)\n

    \nUsing the given log values, we can approximate the log value:\n

    \nlog (0.02857 / 0.04286) ≈ log (2/3) ≈ log 2 - log 3 ≈ 0.30 - 0.48 = -0.18\n

    \nNow substitute this value back into the Henderson-Hasselbalch equation:\n

    \npH = 4.76 - 0.18 = 4.58\n

    \nTo express the pH as the nearest integer multiplied by 10⁻², multiply the pH value by 100:\n

    \n4.58 × 100 = 458\n

    \nSo, the pH of the resulting solution is approximately 458 × 10⁻².", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1880, "subject": "Chemistry", "question": "

    An analyst wants to convert $$1 \\mathrm{~L} \\mathrm{~HCl}$$ of $$\\mathrm{pH}=1$$ to a solution of $$\\mathrm{HCl}$$ of $$\\mathrm{pH} ~2$$. The volume of water needed to do this dilution is __________ $$\\mathrm{mL}$$. (Nearest integer)

    ", "options": [], "answer": "9000", "solution": "**Answer:** 9000\n\n

    We can use the formula for pH to calculate the concentration of hydrogen ions in each solution:

    \n

    $$\\mathrm{pH} = -\\log_{10}[\\mathrm{H}^+]$$

    \n

    For the first solution, we have:

    \n

    $$1 = -\\log_{10}[\\mathrm{H}^+]$$

    \n

    Solving for $[\\mathrm{H}^+]$, we get:

    \n

    $$[\\mathrm{H}^+] = 0.1 \\mathrm{~M}$$

    \n

    For the second solution, we want:

    \n

    $$2 = -\\log_{10}[\\mathrm{H}^+]$$

    \n

    Solving for $[\\mathrm{H}^+]$, we get:

    \n

    $$[\\mathrm{H}^+] = 0.01 \\mathrm{~M}$$

    \n

    To dilute the first solution to the desired concentration, we can use the dilution equation:

    \n

    $$C_1V_1 = C_2V_2$$

    \n

    where $C_1$ and $V_1$ are the initial concentration and volume, and $C_2$ and $V_2$ are the final concentration and volume.

    \n

    We know $C_1 = 0.1 \\mathrm{~M}$, $C_2 = 0.01 \\mathrm{~M}$, and $V_1 = 1 \\mathrm{~L}$. Solving for $V_2$, we get:

    \n

    $$V_2 = \\frac{C_1V_1}{C_2} = \\frac{(0.1 \\mathrm{~M})(1 \\mathrm{~L})}{0.01 \\mathrm{~M}} = 10 \\mathrm{~L}$$

    \n

    Therefore, we need to add $10-1=9$ liters of water to the initial solution to obtain the desired pH. Converting liters to milliliters, we get:

    \n

    $$9 \\mathrm{~L} \\times \\frac{1000 \\mathrm{~mL}}{1 \\mathrm{~L}} = 9000 \\mathrm{~mL}$$

    \n

    So the volume of water needed is 9000 mL or 9,000 mL (nearest integer).

    \n", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1881, "subject": "Chemistry", "question": "$\\mathrm{K}_{\\mathrm{a}}$ for $\\mathrm{CH}_3 \\mathrm{COOH}$ is $1.8 \\times 10^{-5}$ and $\\mathrm{K}_{\\mathrm{b}}$ for $\\mathrm{NH}_4 \\mathrm{OH}$ is $1.8 \\times 10^{-5}$. The $\\mathrm{pH}$ of ammonium acetate solution will be _________.", "options": [], "answer": "7", "solution": "**Answer:** 7\n\n

    To find the pH of an ammonium acetate solution, we use the fact that ammonium acetate is a salt resulting from the neutralization of a weak acid (acetic acid, $CH_3COOH$) by a weak base (ammonium hydroxide, $NH_4OH$). The respective ionization constant values for the weak acid $K_a$ and the weak base $K_b$ are given as $1.8 \\times 10^{-5}$.

    Firstly, calculate the $pK_a$ and $pK_b$ values.

    Given that the values of $K_a$ and $K_b$ are the same, their $pK_a$ and $pK_b$ values will also be the same, establishing a neutral condition where the effects of the acid and base neutralize each other.

    The formula used to determine the pH of a solution of such a salt is:

    $$pH = \\frac{1}{2}(pK_w + pK_a - pK_b)$$

    Where $pK_w$ is the ionic product of water, which is 14 at 25°C. In this specific case, since $pK_a = pK_b$, the formula simplifies to:

    $$pH = \\frac{1}{2}(14 + pK_a - pK_a)$$

    $$pH = \\frac{1}{2}(14)$$

    $$pH = 7$$

    The pH of an ammonium acetate solution in this scenario is 7, indicating a neutral solution.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1882, "subject": "Chemistry", "question": "

    The $$\\mathrm{pH}$$ of an aqueous solution containing $$1 \\mathrm{M}$$ benzoic acid $$\\left(\\mathrm{pK}_{\\mathrm{a}}=4.20\\right)$$ and $$1 \\mathrm{M}$$ sodium benzoate is 4.5. The volume of benzoic acid solution in $$300 \\mathrm{~mL}$$ of this buffer solution is _________ $$\\mathrm{mL}$$. (given : $$\\log 2=0.3$$)

    ", "options": [], "answer": "100", "solution": "**Answer:** 100\n\n

    $$\\begin{array}{ccc} \n& \\text { 1M Benzoic acid } & +1 \\mathrm{M} \\text { Sodium Benzoate } \\\\\n& \\left(\\mathrm{V}_{\\mathrm{a}} \\mathrm{ml}\\right) & \\left(\\mathrm{V}_{\\mathrm{s}} \\mathrm{ml}\\right) \\\\\n\\text { Millimole } & \\mathrm{V}_{\\mathrm{a}} \\times 1 & \\mathrm{~V}_{\\mathrm{s}} \\times 1\n\\end{array}$$

    \n

    $$\\begin{aligned}\n& \\mathrm{pH}=4.5 \\\\\n& \\mathrm{pH}=\\mathrm{pka}+\\log \\frac{[\\text { salt }]}{[\\text { acid }]} \\\\\n& 4.5=4.2+\\log \\left(\\frac{\\mathrm{V}_{\\mathrm{s}}}{\\mathrm{V}_{\\mathrm{a}}}\\right) \\\\\n& \\frac{\\mathrm{V}_{\\mathrm{s}}}{\\mathrm{V}_{\\mathrm{a}}}=2 \\quad \\text{..... (1)}\\\\\n& \\mathrm{~V}_{\\mathrm{s}}+\\mathrm{V}_{\\mathrm{a}}=300 \\quad \\text{..... (2)}\\\\\n& \\mathrm{~V}_{\\mathrm{a}}=100 \\mathrm{~ml}\n\\end{aligned}$$

    ", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1883, "subject": "Chemistry", "question": "

    Given below are two statements :

    \n

    Statement (I) : A Buffer solution is the mixture of a salt and an acid or a base mixed in any particular quantities

    \n

    Statement (II) : Blood is naturally occurring buffer solution whose $$\\mathrm{pH}$$ is maintained by $$\\mathrm{H}_2 \\mathrm{CO}_3 / \\mathrm{HCO}_3{ }^{\\ominus}$$ concentrations.

    \n

    In the light of the above statements, choose the correct answer from the options given below :

    ", "options": [ { "text": "Both Statement I and Statement II are false\n" }, { "text": "Both Statement I and Statement II are true\n" }, { "text": "Statement I is false but Statement II is true\n" }, { "text": "Statement I is true but Statement II is false" } ], "answer": "Statement I is false but Statement II is true\n", "solution": "**Answer:** Statement I is false but Statement II is true\n\n\n

    Let's analyze both statements given to determine the correct answer.

    \n\n

    Statement (I) describes a buffer solution as a mixture of a salt and an acid or a base mixed in any particular quantities. However, this definition is partially incorrect. A buffer solution is more accurately defined as a mixture of a weak acid and its conjugate base or a weak base and its conjugate acid. The purpose of a buffer solution is to maintain a stable pH when small amounts of acid (H⁺ ions) or alkali (OH⁻ ions) are added. It is not just any mixture of a salt and an acid or base but must involve components that can react with added acid or base to minimize changes in pH.

    \n\n

    Therefore, Statement (I) is misleading or incomplete as it omits the necessity for a weak acid and its conjugate base or a weak base and its conjugate acid to form a true buffer solution.

    \n\n

    Statement (II) mentions that blood is a naturally occurring buffer solution whose pH is maintained by the $$\\mathrm{H}_2\\mathrm{CO}_3/\\mathrm{HCO}_3^{-}$$ (carbonic acid/bicarbonate) system. This is accurate. The carbonic acid-bicarbonate buffering system is one of the main buffering systems in human blood, helping to maintain the pH within a narrow range (typically around 7.35 to 7.45). This buffer system works by the reversible reaction between carbon dioxide (CO₂) and water to form carbonic acid ($$\\mathrm{H}_2\\mathrm{CO}_3$$), which can then dissociate into bicarbonate ion ($$\\mathrm{HCO}_3^{-}$$) and a hydrogen ion (H⁺). This system effectively moderates changes in pH by either consuming or releasing H⁺ ions.

    \n\n

    Given this analysis:

    \n\n

    Statement I is false because it inaccurately or incompletely describes a buffer solution.

    \n\n

    Statement II is true as it correctly identifies blood as a naturally occurring buffer solution that uses the $$\\mathrm{H}_2\\mathrm{CO}_3/\\mathrm{HCO}_3^{-}$$ system to maintain pH.

    \n\n

    Therefore, the correct answer is:

    \n\n

    Option C: Statement I is false but Statement II is true.

    ", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 1884, "subject": "Chemistry", "question": "

    The equilibrium $$\\mathrm{Cr}_2 \\mathrm{O}_7^{2-} \\rightleftharpoons 2 \\mathrm{CrO}_4^{2-}$$ is shifted to the right in :

    ", "options": [ { "text": "a weakly acidic medium\n" }, { "text": "a basic medium\n" }, { "text": "a neutral medium\n" }, { "text": "an acidic medium" } ], "answer": "a basic medium\n", "solution": "**Answer:** a basic medium\n\n\n

    The equilibrium $$\\mathrm{Cr}_2 \\mathrm{O}_7^{2-} \\rightleftharpoons 2 \\mathrm{CrO}_4^{2-}$$ can be influenced by changes in the pH of the medium, according to Le Chatelier's principle. This principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change.

    \n\n

    In this case:

    \n\n

    $$\\mathrm{Cr}_2 \\mathrm{O}_7^{2-}$$ is dichromate, which exists more prominently in acidic conditions. On the other hand, $$\\mathrm{CrO}_4^{2-}$$ is chromate, which is favored in basic conditions. When the medium is basic, hydrogen ion concentration decreases, leading to the formation of more chromate ions ($$\\mathrm{CrO}_4^{2-}$$).

    \n\n

    Thus, in a basic medium, the reaction shifts to the right, resulting in the formation of $$2 \\mathrm{CrO}_4^{2-}$$ from $$\\mathrm{Cr}_2 \\mathrm{O}_7^{2-}$$.

    \n\n

    Therefore, the correct option is:

    \n\n

    Option B: a basic medium

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1885, "subject": "Chemistry", "question": "An aqueous solution contains 0.10 M H2S and 0.20 M HCl. If the equilibrium constants for the formation of HS– from H2S is 1.0 $$\\times$$ 10–7 and that of S2- from HS– ions is 1.2 $$\\times$$ 10–13 then the concentration of S2- ions in aqueous solution is :", "options": [ { "text": "5 $$\\times$$ 10–19" }, { "text": "5 $$\\times$$ 10–8" }, { "text": "3 $$\\times$$ 10–20" }, { "text": "6 $$\\times$$ 10–21" } ], "answer": "3 $$\\times$$ 10–20", "solution": "**Answer:** 3 $$\\times$$ 10–20\n\nHCl $$ \\to $$ H+ + Cl$$-$$ \n

    H+ concentration is = 0.2 M. \n

    H2S $$\\rightleftharpoons$$ H+ + HS$$-$$; K1 = 1.0 $$ \\times $$ 10$$-$$7 \n

    HS$$-$$ $$\\rightleftharpoons$$ H+ + S2$$-$$; K2 = 1.2 $$ \\times $$ 10$$-$$13 \n

    H2S $$\\rightleftharpoons$$ S2$$-$$ + 2H+ \n

    K = K1 $$ \\times $$ K2 = 1.0 $$ \\times $$ 10$$-$$7 $$ \\times $$ 1.2 $$ \\times $$ 1.0$$-$$13 = 1.2 $$ \\times $$ 10$$-$$20 \n

    as K1 and K2 both are very low for this reaction so dissociation of H2S and HS$$-$$ will be very low so, the produced H+ from this reaction will also be very low. \n

    So, we can say the concentration of H+ will be almost same as H+ in HCl. \n

    $$\\therefore\\,\\,\\,$$ [ H+ ] = 0.2 M.\n

    From the reaction, H2S $$\\rightleftharpoons\\,$$ 2H+ + S2$$-$$ \n

    We get [ H+ ]2 [ S2$$-$$] = K $$ \\times $$ [ H2 S ]\n

    $$ \\Rightarrow \\,\\,\\,$$ [ S2$$-$$ ] = $$ {{1.2 \\times {{10}^{ - 20}} \\times 0.1} \\over {{{\\left( {0.2} \\right)}^2}}}$$\n

    $$ \\Rightarrow \\,\\,\\,$$ [ S2$$-$$ ] = 3 $$ \\times $$ 10$$-$$20 M ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1886, "subject": "Chemistry", "question": "Which of the following salts is the most basic in aqueous solution?", "options": [ { "text": "Pb(CH3COO)2" }, { "text": "Al(CN)3" }, { "text": "CH3COOK" }, { "text": "FeCl3" } ], "answer": "CH3COOK", "solution": "**Answer:** CH3COOK\n\nNote :\n
    When a salt is made with strong base and weak acid then that salt will be basic nature in aqueous solution. \n

    (a)   Pb(CH3COO)2  is a salt of weak base Pb(OH)2 and weak acid CH3COOH.\n

    (b)   Al(CN)3  is a salt of weak base Al(OH)3 and weak acid HCN.\n

    (c)   CH3COOK is a salt of weak acid CH3COOH and strong base KOH.\n

    (d)   FeCl3 is a salt of weak base Fe(OH)3 and strong acid HCl.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1887, "subject": "Chemistry", "question": "The pH of a 0.02 M NH4Cl solution will be :\n
    [given Kb (NH4OH) = 10–5\n and log 2 = 0.301]", "options": [ { "text": "2.56" }, { "text": "5.35" }, { "text": "4.35" }, { "text": "4.65" } ], "answer": "5.35", "solution": "**Answer:** 5.35\n\nNH4+ + H2O ⇋ NH4OH + H+\n

    [H+] = c$$\\alpha $$\n

    = $$\\sqrt {{k_a}\\left( {NH_4^ + } \\right) \\times c} $$\n

    = $$\\sqrt {{{{k_w}} \\over {{k_b}\\left( {N{H_4}OH} \\right)}} \\times c} $$\n

    = $$\\sqrt {{{{{10}^{ - 14}}} \\over {{{10}^{ - 5}}}} \\times 0.02} $$\n

    = $$\\sqrt {20} \\times {10^{ - 6}}$$\n

    $$ \\therefore $$ pH = $$ - \\log \\left( {\\sqrt {20} \\times {{10}^{ - 6}}} \\right)$$ = 5.35\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1888, "subject": "Chemistry", "question": "Given below are two statements : One is labelled as Assertion A and the other labelled as reason R

    Assertion A : During the boiling of water having temporary hardness, Mg(HCO3)2 is converted to MgCO3.

    Reason R : The solubility product of Mg(OH)2 is greater than that of MgCO3.

    In the light of the above statements, choose the most appropriate answer from the options given below :", "options": [ { "text": "Both A and R are true but R is not the correct explanation of A" }, { "text": "A is true but R is false" }, { "text": "Both A and R are true and R is the correct explanation of A" }, { "text": "A is false but R is true" } ], "answer": "A is false but R is true", "solution": "**Answer:** A is false but R is true\n\nA : For temporary hardness,
    Mg(HCO3)2 $$\\buildrel {heating} \\over\n \\longrightarrow $$ Mg(OH)2 + 2CO2 Assertion is false. MgCO3 has\nhigh solubility product than Mg(OH)2.

    \nR : MgCO3 is more water soluble than Mg(OH)2.

    \nKSP(Mg(OH)2) = 1.8 × 10–11

    \nKSP(MgCO3) = 3.5 × 10–8", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 1889, "subject": "Chemistry", "question": "

    $$20 \\mathrm{~mL}$$ of $$0.1\\, \\mathrm{M} \\,\\mathrm{NH}_{4} \\mathrm{OH}$$ is mixed with $$40 \\mathrm{~mL}$$ of $$0.05 \\mathrm{M} \\mathrm{HCl}$$. The $$\\mathrm{pH}$$ of the mixture is nearest to :

    \n

    (Given : $$\\mathrm{K}_{\\mathrm{b}}\\left(\\mathrm{NH}_{4} \\mathrm{OH}\\right)=1 \\times 10^{-5}, \\log 2=0.30, \\log 3=0.48, \\log 5=0.69, \\log 7=0.84, \\log 11= 1.04)$$

    ", "options": [ { "text": "3.2" }, { "text": "4.2" }, { "text": "5.2" }, { "text": "6.2" } ], "answer": "5.2", "solution": "**Answer:** 5.2\n\n

    \"JEE

    \n

    $\\left[\\mathrm{NH}_4^{+}\\right]=\\frac{2 \\mathrm{mmole}}{60 \\mathrm{ml}}=\\frac{1}{30} \\mathrm{M}$

    \n

    $\\mathrm{pH}=\\frac{\\mathrm{pK}_{\\mathrm{w}}-\\mathrm{pK}_{\\mathrm{b}}-\\log \\mathrm{C}}{2}=\\frac{14-5+1.48}{2}=5.24$

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1890, "subject": "Chemistry", "question": "

    Given below are two statements :

    \n

    Statement (I) : Aqueous solution of ammonium carbonate is basic.

    \n

    Statement (II) : Acidic/basic nature of salt solution of a salt of weak acid and weak base depends on $$K_a$$ and $$K_b$$ value of acid and the base forming it.

    \n

    In the light of the above statements, choose the most appropriate answer from the options given below :

    ", "options": [ { "text": "Both Statement I and Statement II are correct\n" }, { "text": "Statement I is correct but Statement II is incorrect\n" }, { "text": "Both Statement I and Statement II are incorrect\n" }, { "text": "Statement I is incorrect but Statement II is correct" } ], "answer": "Both Statement I and Statement II are correct\n", "solution": "**Answer:** Both Statement I and Statement II are correct\n\n\n

    The correct answer is Option A: Both Statement I and Statement II are correct.

    \n\n

    Explanation:

    \n\n

    Statement I explains that an aqueous solution of ammonium carbonate is basic. This is because ammonium carbonate ($$\\text{(NH}_4)_2\\text{CO}_3$$) dissociates in water to form ammonium ions ($$\\text{NH}_4^+$$) and carbonate ions ($$\\text{CO}_3^{2-}$$). The ammonium ion ($$\\text{NH}_4^+$$) is a weak acid and can donate a proton to form ammonia ($$\\text{NH}_3$$) and a hydronium ion ($$\\text{H}_3\\text{O}^+$$), but this happens to a minimal extent due to its weak acidic nature. On the other hand, the carbonate ion ($$\\text{CO}_3^{2-}$$) can accept a hydrogen ion ($$\\text{H}^+$$) from water to form bicarbonate ($$\\text{HCO}_3^-$$) and hydroxide ions ($${OH}^-$$), which makes the solution basic. The reaction of carbonate ions removing hydrogen ions from water is more pronounced than the reaction of ammonium ions donating hydrogen ions to water, hence the overall solution becomes basic.

    \n\n

    Statement II is about the general principle that determines the acidic or basic nature of a salt solution formed by the salt of a weak acid and a weak base. The acidic or basic nature of such a salt solution depends on the acid dissociation constant ($$K_a$$) of the acid and the base dissociation constant ($$K_b$$) of the base. If $$K_a > K_b$$, the solution tends to be acidic because the weak acid is stronger (more willing to donate protons) than the weak base is at accepting protons. Conversely, if $$K_b > K_a$$, the solution tends to be basic because the weak base is stronger (more willing to accept protons) than the weak acid is at donating protons. In the case of ammonium carbonate, carbonic acid ($$H_2CO_3$$) is a weak acid, and ammonia ($$NH_3$$) is a weak base. The basic nature of ammonium carbonate solution suggests that the $$K_b$$ of ammonia exceeds the $$K_a$$ of carbonic acid for the reactions in an aqueous solution, making the solution basic.

    \n\n

    Therefore, both statements I and II are correct, thereby making Option A the right choice.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1891, "subject": "Chemistry", "question": "Let the solubility of an aqueous solution of Mg(OH)2 be x then its Ksp is :", "options": [ { "text": "4x3" }, { "text": "108x5" }, { "text": "27x4" }, { "text": "9x" } ], "answer": "4x3", "solution": "**Answer:** 4x3\n\n$$Mg{\\left( {OH} \\right)_2} \\to \\mathop {\\left[ {M{g^{2 + }}} \\right]}\\limits_x + 2\\mathop {\\left[ {O{H^ - }} \\right]}\\limits_{2x} $$\n

    $${K_{sp}} = \\left[ {Mg} \\right]{\\left[ {OH} \\right]^2}$$\n

    $$ = \\left[ x \\right]{\\left[ {2x} \\right]^2}$$\n

    $$ = x.4{x^2} = 4{x^3}.$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1892, "subject": "Chemistry", "question": "The solubility in water of a sparingly soluble salt AB2 is 1.0 $$\\times$$ 10-5 mol L-1. Its solubility product number will be :", "options": [ { "text": "4 $$\\times$$ 10-10" }, { "text": "1 $$\\times$$ 10-15" }, { "text": "1 $$\\times$$ 10-10" }, { "text": "4 $$\\times$$ 10-15" } ], "answer": "4 $$\\times$$ 10-15", "solution": "**Answer:** 4 $$\\times$$ 10-15\n\n$$A{B_2}\\rightleftharpoons\\,{A^{ + 2}} + 2{B^ - }$$\n

    $$\\left[ A \\right] = 1.0 \\times {10^{ - 5}},\\,\\,$$ \n

    $$\\left[ B \\right] = \\left[ {2.0 \\times {{10}^{ - 5}}} \\right],$$\n

    $${K_{sp}} = {\\left[ B \\right]^2}\\left[ A \\right]$$\n

    $$ = {\\left[ {2 \\times {{10}^{ - 5}}} \\right]^2}\\left[ {1.0 \\times {{10}^{ - 5}}} \\right]$$\n

    $$ = 4 \\times {10^{ - 15}}$$ ", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 1893, "subject": "Chemistry", "question": "The molar solubility (in ol L-1) of a sparingly soluble salt MX4 is \"s\". The corresponding solubility product is Ksp. 's' is given in term of Ksp by the relation :", "options": [ { "text": "s = (256 Ksp)1/5" }, { "text": "s = (128 Ksp)1/4" }, { "text": "s = ( Ksp / 128)1/4" }, { "text": "s = (Ksp / 256)1/5" } ], "answer": "s = (Ksp / 256)1/5", "solution": "**Answer:** s = (Ksp / 256)1/5\n\n$$M{X_4}\\rightleftharpoons\\,\\mathop {{M^{4 + }}}\\limits_{S\\,\\,\\,\\,\\,\\,\\,} + \\mathop {4{X^ - }}\\limits_{4S} $$ \n

    $${K_{sp}} = \\left[ s \\right]{\\left[ {4s} \\right]^4} = 256\\,{s^5}$$\n

    $$\\therefore$$ $$\\,\\,\\,\\,\\,s = {\\left( {{{{K_{sp}}} \\over {256}}} \\right)^{1/5}}$$ ", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 1894, "subject": "Chemistry", "question": "The solubility product of a salt having general formula MX2, in water is: 4 $$\\times$$ 10-12 . The\nconcentration of M2+ ions in the aqueous solution of the salt is :", "options": [ { "text": "2.0 $$\\times$$ 10-6 M" }, { "text": "4.0 $$\\times$$ 10-10 M" }, { "text": "1.0 $$\\times$$ 10-4 M" }, { "text": "1.6 $$\\times$$ 10-4 M" } ], "answer": "1.0 $$\\times$$ 10-4 M", "solution": "**Answer:** 1.0 $$\\times$$ 10-4 M\n\n$$M{X_2}\\,\\rightleftharpoons\\,\\,\\mathop {{M^{ 2 + }}}\\limits_{s\\,\\,\\,\\,\\,\\,\\,\\,} \\,\\, + \\,\\,\\mathop {2{X^ - }}\\limits_{2s} $$ \n

    Where $$s$$ is the solubility of $$M{X_2}$$ \n

    then $${K_{sp}} = $$ $$s \\times {\\left( {2s} \\right)^2}$$ = $$4{s^3}$$;\n

    $$ \\Rightarrow $$$$ 4 \\times {10^{ - 12}} = 4{s^3};$$

    $$ \\Rightarrow $$ $$s = 1 \\times {10^{ - 4}}$$\n

    $$\\left[ {{M^{ 2+ }}} \\right] = s $$\n$$ = 1 \\times 10^ {- 4}$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1895, "subject": "Chemistry", "question": "In a sautrated solution of the sparingly soluble strong electrolyte AgIO3 (Molecular mass = 283) the\nequilibrium which sets in is
    \nAgIO3(s) $$\\leftrightharpoons$$ Ag+(aq) + $$IO_3^-$$
    \nIf the solubility product constant Ksp of AgIO3 at a given temperature is 1.0 $$\\times$$10−8, what is the mass of AgIO3 contained in 100 ml of its saturated solution?", "options": [ { "text": "28.3 × 10−2 g" }, { "text": "2.83 × 10−3 g" }, { "text": "1.0 × 10−7 g" }, { "text": "1.0 × 10−4 g" } ], "answer": "2.83 × 10−3 g", "solution": "**Answer:** 2.83 × 10−3 g\n\nLet $$\\,\\,\\,s = \\,\\,\\,$$ solubility\n

    $$Ag{\\rm I}{O_3}\\,\\rightleftharpoons\\,\\mathop {A{g^ + }}\\limits_s \\,\\,\\mathop {{\\rm I}{O_3}^ - }\\limits_s $$\n

    $${K_{sp}} = \\left[ {A{g^ + }} \\right]\\left[ {{\\rm I}{O_3}^ - } \\right]$$\n

    $$ = s \\times s = {s^2}$$\n

    Given $$\\,\\,\\,{K_{sp}} = 1 \\times {10^{ - 8}}$$\n

    $$\\therefore$$ $$\\,\\,\\,s = \\sqrt {{K_{sp}}} = \\sqrt {1 \\times {{10}^{ - 8}}} $$\n

    $$ = 1.0 \\times {10^4}\\,\\,mol/lit$$\n

    $$ = 1.0 \\times {10^{ - 4}} \\times 283\\,g/lit$$\n

    (as Molecular mass of $$Ag\\,{\\rm I}{O_3} = 283$$ )\n

    $$ = {{1.0 \\times {{10}^{ - 4}} \\times 283 \\times 100} \\over {1000}}gm/100ml$$\n

    $$ = 2.83 \\times {10^{ - 3}}\\,gm/100\\,ml$$", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 1896, "subject": "Chemistry", "question": "Solid Ba(NO3)2 is gradually dissolved in a 1.0 $$\\times$$ 10-4 M Na2CO3 solution. At what concentration of Ba2+ will a precipitate begin to form ?
    \n(Ksp for BaCO3 = 5.1 $$\\times$$ 10−9 )", "options": [ { "text": "5.1 $$\\times$$ 10-5 M" }, { "text": "8.1 $$\\times$$ 10-8 M" }, { "text": "8.1 $$\\times$$ 10-7 M" }, { "text": "4.1 $$\\times$$ 10-5 M" } ], "answer": "5.1 $$\\times$$ 10-5 M", "solution": "**Answer:** 5.1 $$\\times$$ 10-5 M\n\n$$\\mathop {N{a_2}C{O_3}}\\limits_{1 \\times {{10}^{ - 4}}M} \\to \\mathop {2N{a^ + }}\\limits_{1 \\times {{10}^{ - 4}}M} \\,\\, + \\,\\,\\mathop {C{O_3}^{2 - }}\\limits_{1 \\times {{10}^{ - 4}}M} $$ \n

    $${K_{SP\\left( {BaC{O_3}} \\right)}} = \\left[ {B{a^{2 + }}} \\right]\\left[ {CO_3^{2 - }} \\right]$$ \n

    $$\\left[ {B{a^{2 + }}} \\right] = {{5.1 \\times {{10}^{ - 9}}} \\over {1 \\times {{10}^{ - 4}}}}$$\n

    $$ = 5.1 \\times {10^{ - 5}}M$$", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1897, "subject": "Chemistry", "question": "Solubility product of silver bromide is 5.0 $$\\times$$ 10–13. The quantity of potassium bromide (molar mass taken as 120g of mol–1) to be added to 1 litre of 0.05 M solution of silver nitrate to start the\nprecipitation of AgBr is :", "options": [ { "text": "1.2 $$\\times$$ 10–10 g" }, { "text": "1.2 $$\\times$$ 10–9 g" }, { "text": "6.2 $$\\times$$ 10–5 g" }, { "text": "5.0 $$\\times$$ 10–8 g" } ], "answer": "1.2 $$\\times$$ 10–9 g", "solution": "**Answer:** 1.2 $$\\times$$ 10–9 g\n\n$$AgBr\\,\\rightleftharpoons\\,A{g^ + } + B{r^ - }$$\n

    $${K_{sp}} = \\left[ {A{g^ + }} \\right]\\left[ {B{r^ - }} \\right]$$\n

    For precipitation to occur\n

    Ionic product $$>$$ Solubility product\n

    $$\\left[ {B{r^ - }} \\right] = {{{K_{sp}}} \\over {\\left[ {A{g^ + }} \\right]}} = {{5 \\times {{10}^{ - 13}}} \\over {0.05}} = {10^{ - 11}}$$ \n

    i.e., precipitation just starts when $${10^{ - 11}}\\,$$ \n

    moles of $$KBr$$ is added to $$1\\ell \\,AgN{O_3}\\,$$ solution\n

    $$\\therefore$$$$\\,\\,\\,$$ Number of moles of $$\\,\\,\\,B{r^ - }\\,\\,$$ needed \n

    from $$KBr = {10^{ - 11}}$$\n

    $$\\therefore$$ $$\\,\\,\\,$$ Mass of $$KBr = {10^{ - 11}} \\times 120$$\n

    $$ = 1.2 \\times {10^{ - 9}}g$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1898, "subject": "Chemistry", "question": "An aqueous solution contains an unknown concentration of Ba2+. When 50 mL of a 1 M solution of Na2SO4 is added, BaSO4 just begins to precipitate. The final volume is 500 mL. The solubility product of BaSO4 is 1 $$\\times$$ 10–10. What is the original concentration of Ba2+? ", "options": [ { "text": "1.0 $$\\times$$ 10–10 M " }, { "text": "5 $$\\times$$ 10–9 M " }, { "text": "2 $$\\times$$ 10–9 M " }, { "text": "1.1 $$\\times$$ 10–9 M " } ], "answer": "1.1 $$\\times$$ 10–9 M ", "solution": "**Answer:** 1.1 $$\\times$$ 10–9 M \n\nLet initially concentration of Ba+2 = x m. \n

    After adding 50 ml Na2SO4 in Ba+2 solution final volume becomes 500 ml.\n

    $$\\therefore\\,\\,\\,$$ Initial volume of Ba+2 solution \n

    = (500 $$-$$ 50) ml = 450 ml\n

    As at the begining of precipitation, ionic product = solubility product.\n

    $$ \\Rightarrow \\,\\,\\,$$ [Ba2+] [SO$${_4^{ - 2}}$$] = Ksp of BaSO4 \n

    $$ \\Rightarrow \\,\\,\\,$$ [Ba2+] $$\\left( {{{50 \\times 1} \\over {500}}} \\right)$$ = 1 $$ \\times $$ 10$$-$$10 \n

    $$ \\Rightarrow \\,\\,\\,$$ [Ba2+] = 10$$-$$9 M. \n

    So, the concentration of Ba+2 in final solution is 10$$-$$9 M. \n

    $$\\therefore\\,\\,\\,$$ concentration of Ba+2 in original solution, \n

    M1 V1 = M2 V2\n

    $$ \\Rightarrow \\,\\,\\,$$ x $$ \\times $$ 450 = 10$$-$$9 $$ \\times $$ 500\n

    $$ \\Rightarrow \\,\\,\\,$$ x = 1.1 $$ \\times $$ 10$$-$$9 M", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1899, "subject": "Chemistry", "question": "The minimum volume of water required to dissolve 0.1 g lead (II) chloride to get a saturated solution (Ksp of PbCl2 = 3.2 $$ \\times $$ 10-8 atomic mass of Pb = 207 u ) is :", "options": [ { "text": "0.36 L" }, { "text": "17.98 L" }, { "text": "0.18 L" }, { "text": "1.798 L" } ], "answer": "0.18 L", "solution": "**Answer:** 0.18 L\n\nGiven, \n

    Ksp of PbCl2 = 3.2 $$ \\times $$10$$-$$8 \n

    \n\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n
    PbCl2 $$\\rightleftharpoons$$Pb2+2Cl
    Initially100
    At equilibrium1-ss2s
    \n

    Ksp = [s] [2s]2\n

    $$\\therefore\\,\\,\\,\\,$$ Ksp = 4s3\n

    $$\\therefore\\,\\,\\,\\,$$ 4s3 = 3.2 $$ \\times $$ 10$$-$$8 \n

    $$ \\Rightarrow $$ $$\\,\\,\\,\\,$$ s3 = 8 $$ \\times $$ 10$$-$$9 \n

    $$ \\Rightarrow $$ $$\\,\\,\\,\\,$$ s = 2 $$ \\times $$ 10 $$-$$3 M\n

    Solubility = $${{no.\\,\\,of\\,moles\\,\\,of\\,\\,pbc{l_2}\\,} \\over {volume\\,(in\\,\\,litre)}}$$\n

    $$\\therefore\\,\\,\\,\\,$$ 2 $$ \\times $$ 10$$-$$3 = $${{0.1} \\over {278}}$$ $$ \\times $$ $${1 \\over V}$$\n

    $$ \\Rightarrow $$ $$\\,\\,\\,\\,$$ V = $${{0.1} \\over {278}}$$ $$ \\times $$ $${{{{10}^3}} \\over 2}$$ = 0.18 L", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1900, "subject": "Chemistry", "question": "If solublity product of Zr3(PO4)4 is denoted by Ksp and its molar solubility is denoted by S, then\nwhich of the following relation between S and Ksp is correct ?", "options": [ { "text": "$$S = {\\left( {{{{K_{sp}}} \\over {929}}} \\right)^{1/9}}$$" }, { "text": "$$S = {\\left( {{{{K_{sp}}} \\over {6912}}} \\right)^{1/7}}$$" }, { "text": "$$S = {\\left( {{{{K_{sp}}} \\over {144}}} \\right)^{1/6}}$$" }, { "text": "$$S = {\\left( {{{{K_{sp}}} \\over {216}}} \\right)^{1/7}}$$" } ], "answer": "$$S = {\\left( {{{{K_{sp}}} \\over {6912}}} \\right)^{1/7}}$$", "solution": "**Answer:** $$S = {\\left( {{{{K_{sp}}} \\over {6912}}} \\right)^{1/7}}$$\n\n\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n
    Zr3(PO4)4⇌3Zr+4+4PO4-3
    S3S4S
    \n

    Ksp = [Zr+4]3 [PO4-3]4\n

    = [3S]3 [4S]4\n

    = 27S3 $$ \\times $$ 256S4\n

    = 6912S7\n

    $$ \\therefore $$ $$S = {\\left( {{{{K_{sp}}} \\over {6912}}} \\right)^{1/7}}$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1901, "subject": "Chemistry", "question": "The molar solubility of Cd(OH)2 is 1.84 × 10–5\n M in water. The expected solubility of Cd(OH)2 in a buffer\nsolution of pH = 12 is : ", "options": [ { "text": "2.49 × 10–10 M" }, { "text": "1.84 × 10–9\n M" }, { "text": "6.23 × 10–11 M" }, { "text": "$${{2.49} \\over {1.84}} \\times {10^{ - 9}}M$$" } ], "answer": "2.49 × 10–10 M", "solution": "**Answer:** 2.49 × 10–10 M\n\n\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n
    Cd(OH)2(s)⇌Cd2+(aq)+2OH$$-$$(aq)
    S2S
    \n
    $$ \\therefore $$ Ksp[Cd(OH)2] = 4(S)3 = 4(1.84 × 10–5)3 [Given S = 1.84 × 10–5]\n

    For buffer solution of pH = 12 :\n
    As pH = 12 so pOH = 2\n
    $$ \\Rightarrow $$ [OH$$-$$] = 10-2\n
    Let the solubilty = S1\n

    \n\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n
    Cd(OH)2(s)⇌Cd2+(aq)+2OH$$-$$(aq)
    S12S1 + 10-2
    \n
    $$ \\therefore $$ Ksp[Cd(OH)2] = S1(10-2)2 = 4(1.84 × 10–5)3\n
    $$ \\Rightarrow $$ S1 = 2.49 × 10–10 M", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1902, "subject": "Chemistry", "question": "What is the molar solubility of Al(OH)3 in 0.2 M NaOH solution ? Given that, solubility product of Al(OH)3 = 2.4 × 10–24\n :", "options": [ { "text": "3 × 10–22" }, { "text": "3 × 10–19" }, { "text": "12 × 10–21" }, { "text": "12 × 10–22" } ], "answer": "3 × 10–22", "solution": "**Answer:** 3 × 10–22\n\n\"JEE\n

    $${K_{sp}} = $$ [Al3+][OH-]3\n

    $$ \\Rightarrow $$ 2.4 × 10–24 = s(0.2)3\n

    $$ \\Rightarrow $$ s = $${{2.4 \\times {{10}^{ - 24}}} \\over {8 \\times {{10}^{ - 3}}}}$$\n

    $$ \\Rightarrow $$ s = 3 × 10–22", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1903, "subject": "Chemistry", "question": "If Ksp of Ag2CO3 is 8 $$ \\times $$ 10–12, the molar solubility of Ag2CO3 in 0.1 M AgNO3 is - ", "options": [ { "text": "8 $$ \\times $$ 10–12 M " }, { "text": "8 $$ \\times $$ 10–10 M " }, { "text": "8 $$ \\times $$ 10–13 M " }, { "text": "8 $$ \\times $$ 10–11 M " } ], "answer": "8 $$ \\times $$ 10–10 M ", "solution": "**Answer:** 8 $$ \\times $$ 10–10 M \n\n\"JEE\n
    $$ \\therefore $$   8 $$ \\times $$ 10$$-$$12 = (2s + 0.1)2 s\n

    $$ \\Rightarrow $$   s $$ \\times $$ 10$$-$$2 = 8 $$ \\times $$ 10$$-$$12\n

    $$ \\Rightarrow $$   s = 8 $$ \\times $$ 10$$-$$10", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1904, "subject": "Chemistry", "question": "A mixture of 100 m mol of Ca(OH)2 and 2 g of sodium sulphate was dissolved in water and the volume was made up to 100 mL. The mass of calcium sulphate formed and the concentration of OH– in resulting solution, respectively, are : (Molar mass of Ca (OH)2, Na2SO4 and CaSO4 are 74, 143 and 136 g mol–1\n , respectively; Ksp of Ca(OH)2 is 5.5 × 10–6\n) \n", "options": [ { "text": "13.6g, 0.28 mol L$$-$$1" }, { "text": "13.6g, 0.14 mol L$$-$$1" }, { "text": "1.9g, 0.28 mol L$$-$$1" }, { "text": "1.9g, 0.14 mol L$$-$$1" } ], "answer": "1.9g, 0.28 mol L$$-$$1", "solution": "**Answer:** 1.9g, 0.28 mol L$$-$$1\n\nCa(OH)2 + Na2SO4 $$ \\to $$ CaSO4 + 2NaOH \n

    100 m mol 14 m mol      $$-$$     $$-$$      $$-$$\n

    $$-$$    $$-$$      $$-$$  14 m mol 28 m mol\n

    wCasO4 = 14 $$ \\times $$ 10$$-$$3 $$ \\times $$ 136 = 1.9 gm\n

    [OH$$-$$] = $${{28} \\over {100}}$$ = 0.28 M", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1905, "subject": "Chemistry", "question": "The Ksp for the following dissociation is\n1.6 × 10–5

    \n$$PbC{l_{2(s)}} \\leftrightharpoons Pb_{(aq)}^{2 + } + 2Cl_{(aq)}^ - $$

    \nWhich of the following choices is correct for\na mixture of 300 mL 0.134 M Pb(NO3)2 and\n100 mL 0.4 M NaCl ?", "options": [ { "text": "Q > Ksp" }, { "text": "Not enough data provided" }, { "text": "Q < Ksp" }, { "text": "Q = Ksp" } ], "answer": "Q > Ksp", "solution": "**Answer:** Q > Ksp\n\n[Pb2+] = $${{300 \\times 0.134} \\over {400}}$$\n

    = 1.005 × 10–1 M\n

    [Cl-] = $${{100 \\times 0.4} \\over {400}}$$\n

    = 10–1 M\n

    $$PbC{l_{2(s)}} \\leftrightharpoons Pb_{(aq)}^{2 + } + 2Cl_{(aq)}^ - $$\n

    Q = [Pb2+] × [Cl–]2\n

    = 0.1005 × (0.1)2\n

    = 1.005 × 10–3\n

    Given Ksp\n= 1.6 × 10–5\n

    $$ \\therefore $$ Q > Ksp", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1906, "subject": "Chemistry", "question": "The solubility product of Cr(OH)3 at 298 K is\n6.0 × 10–31. The concentration of hydroxide ions\nin a saturated solution of Cr(OH)3 will be :", "options": [ { "text": "(2.22 × 10–31)1/4" }, { "text": "(4.86 × 10–29)1/4" }, { "text": "(18 × 10–31)1/4" }, { "text": "(18 × 10–31)1/2" } ], "answer": "(18 × 10–31)1/4", "solution": "**Answer:** (18 × 10–31)1/4\n\n\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n
    Cr(OH)3⇌Cr+3+3OH-
    S3S
    \n

    Ksp = [Cr3+] [OH–\n]3\n

    $$ \\Rightarrow $$ 6 × 10–31 = S × (3S)3\n

    $$ \\Rightarrow $$ 6 × 10–31 = 27 S4\n

    S = $${\\left( {{6 \\over {27}} \\times {{10}^{31}}} \\right)^{{1 \\over 4}}}$$\n

    As [OH–] = 3S\n

    = $$3{\\left( {{6 \\over {27}} \\times {{10}^{31}}} \\right)^{{1 \\over 4}}}$$\n

    = (18 × 10–31)1/4 M", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1907, "subject": "Chemistry", "question": "For the following Assertion and Reason, the\ncorrect option is\n

    Assertion (A): When Cu (II) and sulphide ions\nare mixed, they react together\nextremely quickly to give a\nsolid.\n

    Reason (R): The equilibrium constant of\n
    Cu2+(aq) + S2–(aq) ⇌ CuS(s) is\nhigh because the solubility\nproduct is low.", "options": [ { "text": "(A) is false and (R) is true." }, { "text": "Both (A) and (R) are true but (R) is not the\nexplanation for (A)." }, { "text": "Both (A) and (R) are true and (R) is the\nexplanation for (A)." }, { "text": "Both (A) and (R) are false." } ], "answer": "Both (A) and (R) are true and (R) is the\nexplanation for (A).", "solution": "**Answer:** Both (A) and (R) are true and (R) is the\nexplanation for (A).\n\nKsp value of CuS is very low 10–36 (3.6 × 10–36)\ndue to low Ksp value Cu+2 ion gets precipitated\nvery quickly even with very low concentration\nof S–2 ion.", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1908, "subject": "Chemistry", "question": "If the solubility product of AB2 is 3.20 $$ \\times $$ 10–11 M3,\nthen the solubility of AB2 in pure water is\n_____ $$ \\times $$ 10–4 mol L–1. \n
    [Assuming that neither kind\nof ion reacts with water]", "options": [], "answer": "2", "solution": "**Answer:** 2\n\n\n\n \n \n \n \n \n \n \n\n\n \n \n \n \n \n \n \n\n
    AB2⇌A2+(aq)+2B-(aq)
    s2s
    \n

    Ksp = 4s3 = 3.2 × 10–11\n

    $$ \\Rightarrow $$ s3 = 8 × 10–12\n

    $$ \\Rightarrow $$ s = 2 × 10–4", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1909, "subject": "Chemistry", "question": "The solubility product of PbI2 is 8.0 $$\\times$$ 10$$-$$9. The solubility of lead iodide in 0.1 molar solution of lead nitrate is x $$\\times$$ 10$$-$$6. mol/L. The value of x is __________. (Rounded off to the nearest integer) [Given $$\\sqrt 2 $$ = 1.41]", "options": [], "answer": "141", "solution": "**Answer:** 141\n\nGiven, $${[{K_{sp}}]_{Pb{l_2}}} = 8 \\times {10^{ - 9}}$$

    To calculate solubility of Pbl2 in 0.1 M solution of Pb(NO3)2,

    (I) $$\\mathop {Pb{{(N{O_3})}_2}}\\limits_{0. 1 M} \\to \\mathop {P{b^{2 + }}(aq)}\\limits_{0.1 M} + \\mathop {2NO_3^ - (aq)}\\limits_{0. 2 M} $$

    (II) $$Pb{I_2}(s)$$ $$\\rightleftharpoons$$ $$\\mathop {P{b^{2 + }}(aq)}\\limits_S + \\mathop {2{I^ - }(aq)}\\limits_{2S} $$

    $$\\therefore$$ [Pb2+] = S + 0.1 $$\\approx$$ 0.1

    $$\\because$$ S < < 0.1

    Now, Ksp = 8 $$\\times$$ 10$$-$$9

    [Pb2] [I$$-$$]2 = 8 $$\\times$$ 10$$-$$9

    0.1 $$\\times$$ (2S)2 = 8 $$\\times$$ 10$$-$$9

    4S2 = 8 $$\\times$$ 10$$-$$8 $$\\Rightarrow$$ S = 141 $$\\times$$ 10$$-$$6 M

    $$ \\therefore $$ x = 141", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1910, "subject": "Chemistry", "question": "The solubility of AgCN in a buffer solution of pH = 3 is x. The value of x is : [Assume : No cyano complex is formed; Ksp(AgCN) = 2.2 $$\\times$$ 10$$-$$16 and Ka(HCN) = 6.2 $$\\times$$ 10$$-$$10]", "options": [ { "text": "1.9 $$\\times$$ 10$$-$$5" }, { "text": "1.6 $$\\times$$ 10$$-$$6" }, { "text": "2.2 $$\\times$$ 10$$-$$16" }, { "text": "0.625 $$\\times$$ 10$$-$$6" } ], "answer": "1.9 $$\\times$$ 10$$-$$5", "solution": "**Answer:** 1.9 $$\\times$$ 10$$-$$5\n\nLet solubility is x

    \"JEE

    $${K_{sp}} \\times {1 \\over {{K_a}}} = [A{g^ + }][C{N^ - }] \\times {{[HCN]} \\over {[{H^ + }][C{N^ - }]}}$$

    $$2.2 \\times {10^{ - 16}} \\times {1 \\over {6.2 \\times {{10}^{ - 10}}}} = {{[S][S]} \\over {{{10}^{ - 3}}}}$$

    $${S^2} = {{2.2} \\over {6.2}} \\times {10^{ - 9}}$$

    $${S^2} = 3.55 \\times {10^{ - 10}}$$

    $$S = \\sqrt {3.55 \\times {{10}^{ - 10}}} $$

    $$S = 1.88 \\times {10^{ - 5}} = 1.9 \\times {10^{ - 5}}$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1911, "subject": "Chemistry", "question": "The solubility of Ca(OH)2 in water is :

    [Given : The solubility product of Ca(OH)2 in water = 5.5 $$\\times$$ 10$$-$$6]", "options": [ { "text": "1.11 $$\\times$$ 10$$-$$6" }, { "text": "1.11 $$\\times$$ 10$$-$$2" }, { "text": "1.77 $$\\times$$ 10$$-$$6" }, { "text": "1.77 $$\\times$$ 10$$-$$2" } ], "answer": "1.11 $$\\times$$ 10$$-$$2", "solution": "**Answer:** 1.11 $$\\times$$ 10$$-$$2\n\n

    Let, solubility of Ca(OH)2 in pure water = S mol/L

    \n

    $$Ca{(OH)_2}$$ $$\\rightleftharpoons$$ $$\\mathop {C{a^{2 + }}}\\limits_{S\\,mol/L} + \\mathop {2O{H^ - }}\\limits_{2 \\times S\\,(mol/L)} $$

    \n

    Ksp = [Ca2+] [OH$$-$$]2 = S $$\\times$$ (2S)2 = 4 S3 (mol/L)

    \n

    The expression of Ksp can also be written as,

    \n

    Ksp = xx . yy . Sx + y

    \n

    = 11 . 22 . S1 + 2

    \n

    = 4 S3 [$$\\because$$ For Ca(OH)2 : x = 1, y = 2]

    \n

    x and y are the coefficients of cations and anions respectively

    \n

    $$S = {\\left( {{{{K_{sp}}} \\over 4}} \\right)^{1/3}} = {\\left( {{{5.5 \\times {{10}^{ - 6}}} \\over 4}} \\right)^{1/3}}$$

    \n

    $$ = 1.11 \\times {10^{ - 2}}$$ ml/L

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1912, "subject": "Chemistry", "question": "Two salts A2X and MX have the same value of solubility product of 4.0 $$\\times$$ 10$$-$$12. The ratio of their molar solubilities i.e. $${{S({A_2}X)} \\over {S(MX)}}$$ = __________. (Round off to the Nearest Integer)", "options": [], "answer": "50", "solution": "**Answer:** 50\n\nFor A2X

    $$\\matrix{\n {{A_2}X} & \\to & {2{A^ + }} & {{X^{2 - }}} \\cr \n {} & {} & {2{S_1}} & {{S_1}} \\cr \n\n } $$

    $${K_{sp}} = 4S_1^3 = 4 \\times {10^{ - 12}}$$

    S1 = 10$$-$$4

    for MX

    $$\\matrix{\n {MX} & \\to & {{M^ + }} & {{X^ - }} \\cr \n {} & {} & {{S_2}} & {{S_2}} \\cr \n\n } $$

    $${K_{sp}} = S_2^2 = 4 \\times {10^{ - 12}}$$

    S2 = 2 $$\\times$$ 10$$-$$6

    so, $${{{S_{{A_2}X}}} \\over {{S_{MX}}}} = {{{{10}^{ - 4}}} \\over {2 \\times {{10}^{ - 6}}}} = 50$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1913, "subject": "Chemistry", "question": "The solubility of CdSO4 in water is 8.0 $$\\times$$ 10$$-$$4 mol L$$-$$1. Its solubility in 0.01 M H2SO4 solution is __________ $$\\times$$ 10$$-$$6 mol L$$-$$1. (Round off to the Nearest Integer). (Assume that solubility is much less than 0.01 M)", "options": [], "answer": "64", "solution": "**Answer:** 64\n\nIn pure water

    \"JEE

    $$ \\therefore $$ Ksp = (s)2 = (8 $$\\times$$ 10-4)2 = 64 $$\\times$$ 10-8

    In H2SO4 solution,

    \"JEE

    As S1 < < 0.01 so, S1 + 0.01 $$ \\simeq $$ 0.01

    $$ \\therefore $$ Ksp = [Cd+2] [So$$_4^{ - 2}$$]

    $$ \\Rightarrow $$ 64 $$\\times$$ 10-8 = S1 $$\\times$$ 0.01

    $$ \\Rightarrow $$ S1 = 64 $$\\times$$ 10-6", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1914, "subject": "Chemistry", "question": "A solution is 0.1 M in Cl$$-$$ and 0.001 M in CrO$$_4^{2 - }$$. Solid AgNO3 is gradually added to it. Assuming that the addition does not change in volume and Ksp(AgCl) = 1.7 $$\\times$$ 10$$-$$10 M2 and Ksp(Ag2CrO4) = 1.9 $$\\times$$ 10$$-$$12 M3.

    Select correct statement from the following :\n", "options": [ { "text": "AgCl precipitates first because its Ksp is high." }, { "text": "Ag2CrO4 precipitates first as its Ksp is low." }, { "text": "Ag2CrO4 precipitates first because the amount of Ag+ needed is low." }, { "text": "AgCl will precipitate first as the amount of Ag+ needed to precipitate is low." } ], "answer": "AgCl will precipitate first as the amount of Ag+ needed to precipitate is low.", "solution": "**Answer:** AgCl will precipitate first as the amount of Ag+ needed to precipitate is low.\n\nConc. of Cl$$-$$ = 0.1 M = 10$$-$$1 M

    Conc. of CrO$$_4^{2 - }$$ = 0.001 M = 10$$-$$3 M

    Ksp(AgCl) = [Ag+][Cl$$-$$]

    [Ag+]AgCl = $${{1.7 \\times {{10}^{ - 10}}} \\over {{{10}^{ - 1}}}} = 1.7 \\times {10^{ - 9}}$$

    $${K_{sp}}(A{g_2}Cr{O_4}) = {[A{g^ + }]^2}[CrO_4^{2 - }]$$

    $$[A{g^ + }] = \\sqrt {{{1.9 \\times {{10}^{ - 12}}} \\over {{{10}^{ - 3}}}}} = \\sqrt {19} \\times {10^{ - 4}}$$

    $$\\therefore$$ AgCl will be precipitate first", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1915, "subject": "Chemistry", "question": "A3B2 is a sparingly soluble salt of molar mass M (g mol$$-$$1) and solubility x g L$$-$$1. The solubility product satisfies $${K_{sp}} = a{\\left( {{x \\over M}} \\right)^5}$$. The value of a is _____________. (Integer answer)", "options": [], "answer": "108", "solution": "**Answer:** 108\n\n\"JEE

    KSP = (3s)3 (2s)2

    KSP = 108 s5 & s = (x/M)

    KSP = 108$${\\left( {{x \\over M}} \\right)^5}$$

    given $${K_{sp}} = a{\\left( {{x \\over M}} \\right)^5}$$

    comparing a = 108", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 1916, "subject": "Chemistry", "question": "The molar solubility of Zn(OH)2 in 0.1 M NaOH solution is x $$\\times$$ 10$$-$$18 M. The value of x is _________ (Nearest integer)

    (Given : The solubility product of Zn(OH)2 is 2 $$\\times$$ 10$$-$$20)", "options": [], "answer": "2", "solution": "**Answer:** 2\n\n\"JEE

    Due to common-ion effect (presence of NaOH) the concentration of OH$$-$$ will be (2S + 0.1) $$\\approx$$ 0.1

    ($$\\because$$ 0.1 > > 2 S)

    $$\\therefore$$ Solubility of product,

    $${K_{sp}} = {(0.1)^2} \\times S$$

    $$2 \\times {10^{ - 20}} = 0.01 \\times S$$

    $$ \\Rightarrow S = {{2 \\times {{10}^{ - 20}}} \\over {0.01}} = 2 \\times {10^{ - 18}}$$

    $$\\therefore$$ x = 2", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1917, "subject": "Chemistry", "question": "

    The solubility of AgCl will be maximum in which of the following?

    ", "options": [ { "text": "0.01 M KCl" }, { "text": "0.01 M HCl" }, { "text": "0.01 M AgNO3" }, { "text": "Deionised water" } ], "answer": "Deionised water", "solution": "**Answer:** Deionised water\n\n

    In deionized water no common ion effect will take\nplace so maximum solubility.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1918, "subject": "Chemistry", "question": "

    The solubility product of a sparingly soluble salt A2X3 is 1.1 $$\\times$$ 10$$-$$23. If specific conductance of the solution is 3 $$\\times$$ 10$$-$$5 S m$$-$$1, the limiting molar conductivity of the solution is $$x \\,\\times$$ 10$$-$$3 S m2 mol$$-$$1. The value of x is ___________.

    ", "options": [], "answer": "3", "solution": "**Answer:** 3\n\n$A_{2} X_{3} \\rightleftharpoons \\underset{2S}{2 \\mathrm{~A}}+\\underset{3S}{3 \\mathrm{X}}$\n

    \n$$\n\\begin{aligned}\n&\\mathrm{K}_{\\mathrm{sp}}=(2 \\mathrm{~s})^{2}(3 s)^{3}=1.1 \\times 10^{-23} \\\\\\\\\n&\\mathrm{~S} \\approx 10^{-5}\n\\end{aligned}\n$$\n

    \nFor sparingly soluble salts\n

    \n$$\n\\begin{aligned}\n\\wedge_{m} &=\\wedge_{m}^{0} \\\\\\\\\n\\wedge_{m} &=\\frac{\\mathrm{k}}{\\mathrm{S} \\times 10^{3}} \\\\\\\\\n&=\\frac{3 \\times 10^{-5}}{10^{-5}} \\times 10^{-3} \\\\\\\\\n&=3 \\times 10^{-3} ~ \\mathrm{Sm}^{2} \\mathrm{~mol}^{-1}\n\\end{aligned}\n$$", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 1919, "subject": "Chemistry", "question": "

    The Ksp for bismuth sulphide (Bi2S3) is 1.08 $$\\times$$ 10$$-$$73. The solubility of Bi2S3 in mol L$$-$$1 at 298 K is :

    ", "options": [ { "text": "1.0 $$\\times$$ 10$$-$$15" }, { "text": "2.7 $$\\times$$ 10$$-$$12" }, { "text": "3.2 $$\\times$$ 10$$-$$10" }, { "text": "4.2 $$\\times$$ 10$$-$$8" } ], "answer": "1.0 $$\\times$$ 10$$-$$15", "solution": "**Answer:** 1.0 $$\\times$$ 10$$-$$15\n\n\"JEE

    \nKsp = (2s)2(3s)3 = 108s5

    \n108s5 = 108 × 10–75

    \ns = 1.0 × 10–15 mol/L", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1920, "subject": "Chemistry", "question": "

    At $$310 \\mathrm{~K}$$, the solubility of $$\\mathrm{CaF}_{2}$$ in water is $$2.34 \\times 10^{-3} \\mathrm{~g} / 100 \\mathrm{~mL}$$. The solubility product of $$\\mathrm{CaF}_{2}$$ is ____________ $$\\times 10^{-8}(\\mathrm{~mol} / \\mathrm{L})^{3}$$. (Give molar mass : $$\\mathrm{CaF}_{2}=78 \\mathrm{~g} \\mathrm{~mol}^{-1}$$)

    ", "options": [], "answer": "0", "solution": "**Answer:** 0\n\n$\\mathrm{CaF}_{2} \\stackrel{\\mathrm{s}}{\\rightleftharpoons} \\underset{ \\mathrm{s}}{\\mathrm{Ca}^{2+}}+\\underset{2 \\mathrm{~s}}{2 \\mathrm{~F}^{-}}$\n

    \n$$\n\\begin{aligned}\n\\mathrm{K}_{\\mathrm{sp}} &=\\mathrm{s}(2 \\mathrm{~s})^{2} \\\\\\\\\n&=4 \\mathrm{~s}^{3}\n\\end{aligned}\n$$\n

    \nSolubility $(\\mathrm{s})=2.34 \\times 10^{-3} \\mathrm{~g} / 100 \\mathrm{~mL}$

    $=\\frac{2 \\cdot 34 \\times 10^{-3} \\times 10}{78}$ mole $/$ lit

    $=3 \\times 10^{-4} \\mathrm{~mole} / \\mathrm{lit}$\n

    \n$\\therefore \\mathrm{K}_{\\mathrm{sp}}=4 \\times\\left(3 \\times 10^{-4}\\right)^{3}$\n

    \n$$\n\\begin{aligned}\n&=108 \\times 10^{-12} \\\\\\\\\n&=0.0108 \\times 10^{-8}(\\mathrm{~mole} / \\mathrm{lit})^{3}\n\\end{aligned}\n$$

    \n$$\n\\begin{aligned}\n& \\therefore x \\approx 0\n\\end{aligned}\n$$\n

    \n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1921, "subject": "Chemistry", "question": "

    If the solubility product of PbS is 8 $$\\times$$ 10$$-$$28, then the solubility of PbS in pure water at 298 K is x $$\\times$$ 10$$-$$16 mol L$$-$$1. The value of x is __________. (Nearest Integer)

    \n

    [Given : $$\\sqrt2$$ = 1.41]

    ", "options": [], "answer": "282", "solution": "**Answer:** 282\n\n$$\\mathrm{K}_{\\mathrm{sp}}=\\mathrm{S}^{2}$$\n

    \n$$\\mathrm{S}=\\sqrt{K_{s p}}=\\sqrt{8 \\times 10^{-28}}=2 \\sqrt{2} \\times 10^{-14}$$\n

    \n$$=2.82 \\times 10^{-14}$$\n

    \n$$=282 \\times 10^{-16}$$\n

    $$ \\therefore $$ Ans. 282", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1922, "subject": "Chemistry", "question": "At $298 \\mathrm{~K}$, the solubility of silver chloride in water is $1.434 \\times 10^{-3} \\mathrm{~g} \\mathrm{~L}^{-1}$. The value of $-\\log \\mathrm{K}_{\\mathrm{sp}}$ for silver chloride is _________.\n

    \n(Given mass of $\\mathrm{Ag}$ is $107.9 \\mathrm{~g} \\mathrm{~mol}^{-1}$ and mass of $\\mathrm{Cl}$ is $35.5 \\mathrm{~g} \\mathrm{~mol}^{-1}$ )", "options": [], "answer": "10", "solution": "**Answer:** 10\n\n\"JEE\n

    $$\n\\begin{aligned}\n& \\mathrm{K}_{\\mathrm{sp}}=\\left(\\mathrm{Ag}^{+}\\right)\\left(\\mathrm{Cl}^{-}\\right)=\\mathrm{S} \\times \\mathrm{S}=\\mathrm{S}^2 \\\\\\\\\n& \\mathrm{~S}=\\sqrt{\\mathrm{K}_{\\mathrm{sp}}} \\\\\\\\\n& \\mathrm{S}=\\frac{1.434 \\times 10^{-3}}{143.4}=10^{-5} \\\\\\\\\n& \\mathrm{~K}_{\\mathrm{sp}}=\\mathrm{S}^2=10^{-10} \\\\\\\\\n& \\Rightarrow-\\log \\left(\\mathrm{K}_{\\mathrm{sp}}\\right)=-\\log \\left(10^{-10}\\right)=10\n\\end{aligned}\n$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1923, "subject": "Chemistry", "question": "

    $$25.0 \\mathrm{~mL}$$ of $$0.050 ~\\mathrm{M} ~\\mathrm{Ba}\\left(\\mathrm{NO}_{3}\\right)_{2}$$ is mixed with $$25.0 \\mathrm{~mL}$$ of $$0.020 ~\\mathrm{M} ~\\mathrm{NaF} . \\mathrm{K}_{\\mathrm{Sp}}$$ of $$\\mathrm{BaF}_{2}$$ is $$0.5 \\times 10^{-6}$$ at $$298 \\mathrm{~K}$$. The ratio of $$\\left[\\mathrm{Ba}^{2+}\\right]\\left[\\mathrm{F}^{-}\\right]^{2}$$ and $$\\mathrm{K}_{\\mathrm{sp}}$$ is ___________.

    \n

    (Nearest integer)

    ", "options": [], "answer": "5", "solution": "**Answer:** 5\n\nInitial concentrations before mixing:

    \n$$[\\mathrm{Ba(NO_3)_2}] = 0.050\\, \\mathrm{M}$$

    \n$$[\\mathrm{NaF}] = 0.020\\, \\mathrm{M}$$\n

    \nVolumes of the solutions:

    \n$$V_{\\mathrm{Ba(NO_3)_2}} = V_{\\mathrm{NaF}} = 25.0\\, \\mathrm{mL}$$\n

    \nAfter mixing, the total volume becomes:

    \n$$V_{\\mathrm{total}} = 25.0\\, \\mathrm{mL} + 25.0\\, \\mathrm{mL} = 50.0\\, \\mathrm{mL}$$\n

    \nNow, we calculate the initial concentrations after mixing:\n

    \n$$[\\mathrm{Ba}^{2+}] = \\frac{25.0\\, \\mathrm{mL} \\times 0.050\\, \\mathrm{M}}{50.0\\, \\mathrm{mL}} = 0.025\\, \\mathrm{M}$$\n

    \n$$[\\mathrm{F}^{-}] = \\frac{25.0\\, \\mathrm{mL} \\times 0.020\\, \\mathrm{M}}{50.0\\, \\mathrm{mL}} = 0.010\\, \\mathrm{M}$$\n

    \nNext, we calculate the reaction quotient (Q) for the precipitation of BaF₂:\n

    \n$$Q = [\\mathrm{Ba}^{2+}][\\mathrm{F}^{-}]^2$$\n

    \n$$Q = (0.025\\, \\mathrm{M})(0.010\\, \\mathrm{M})^2 = 2.5 \\times 10^{-6}$$\n

    \nKsp of BaF₂ is given as $$5 \\times 10^{-7}$$.\n

    \nNow, we find the ratio of $$[\\mathrm{Ba}^{2+}][\\mathrm{F}^{-}]^2$$ to Ksp:\n

    \n$$\\text{Ratio} = \\frac{(2.5 \\times 10^{-6})}{(5 \\times 10^{-7})} = 5$$\n

    \nSo, the correct ratio of $$[\\mathrm{Ba}^{2+}][\\mathrm{F}^{-}]^2$$ to Ksp is 5.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1924, "subject": "Chemistry", "question": "

    $$25 \\mathrm{~mL}$$ of silver nitrate solution (1M) is added dropwise to $$25 \\mathrm{~mL}$$ of potassium iodide $$(1.05 \\mathrm{M})$$ solution. The ion(s) present in very small quantity in the solution is/are :

    ", "options": [ { "text": "$$\\mathrm{I^-}$$ only" }, { "text": "$$\\mathrm{K^+}$$ only" }, { "text": "$$\\mathrm{Ag^+}$$ and $$\\mathrm{I^-}$$ both" }, { "text": "$$\\mathrm{NO_3^-}$$ only" } ], "answer": "$$\\mathrm{Ag^+}$$ and $$\\mathrm{I^-}$$ both", "solution": "**Answer:** $$\\mathrm{Ag^+}$$ and $$\\mathrm{I^-}$$ both\n\nThe reaction between silver nitrate (AgNO3) and potassium iodide (KI) forms silver iodide (AgI), which is practically insoluble in water. The reaction is as follows :\n\n

    AgNO3(aq) + KI(aq) → AgI(s) + KNO3(aq)\n\n

    Although the reaction stoichiometry indicates that iodide ions (I-) and silver ions (Ag+) are produced, silver iodide (AgI) precipitates out of the solution due to its low solubility, and very little Ag+ and I- ions remain in the solution. Hence, their concentration in the solution will be negligible.\n\n

    So, the correct answer should be :\n\n

    Option C : Ag+ and I- both", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1925, "subject": "Chemistry", "question": "

    The solubility product of $$\\mathrm{BaSO}_{4}$$ is $$1 \\times 10^{-10}$$ at $$298 \\mathrm{~K}$$. The solubility of $$\\mathrm{BaSO}_{4}$$ in $$0.1 ~\\mathrm{M} ~\\mathrm{K}_{2} \\mathrm{SO}_{4}(\\mathrm{aq})$$ solution is ___________ $$\\times 10^{-9} \\mathrm{~g} \\mathrm{~L}^{-1}$$ (nearest integer).

    \n

    Given: Molar mass of $$\\mathrm{BaSO}_{4}$$ is $$233 \\mathrm{~g} \\mathrm{~mol}^{-1}$$

    ", "options": [], "answer": "233", "solution": "**Answer:** 233\n\n

    Barium sulfate, BaSO4, is a sparingly soluble salt. Its dissolution can be represented by the following equilibrium reaction:

    \n

    $$\\mathrm{BaSO}_4 (\\mathrm{s}) \\rightleftharpoons \\mathrm{Ba}^{2+}(\\mathrm{aq}) + \\mathrm{SO}_4^{2-}(\\mathrm{aq})$$

    \n

    The solubility product constant, Ksp, is given by:

    \n

    $$\\mathrm{K}_{sp} = [\\mathrm{Ba}^{2+}][\\mathrm{SO}_4^{2-}]$$

    \n

    In this case, the salt is being dissolved in a solution that already contains sulfate ions, $\\mathrm{SO}_4^{2-}$, from the $\\mathrm{K}_2\\mathrm{SO}_4$.

    \n

    When $\\mathrm{K}_2\\mathrm{SO}_4$ dissolves completely in water, it forms $\\mathrm{K}^+$ and $\\mathrm{SO}_4^{2-}$ ions. Because its concentration is 0.1 M, the $\\mathrm{SO}_4^{2-}$ concentration due to $\\mathrm{K}_2\\mathrm{SO}_4$ is 0.1 M.

    \n

    Therefore, the concentration of $\\mathrm{SO}_4^{2-}$ ions is now not just due to the BaSO4 dissolving, but also the added $\\mathrm{K}_2\\mathrm{SO}_4$. Hence, the total concentration of $\\mathrm{SO}_4^{2-}$ ions is $\\mathrm{S} + 0.1$ where S is the solubility of $\\mathrm{BaSO}_4$.

    \n

    We then substitute into the Ksp expression:

    \n

    $$1 \\times 10^{-10} = [\\mathrm{S}][\\mathrm{S} + 0.1]$$

    \n

    However, because BaSO4 is sparingly soluble, S is very small compared to 0.1. Therefore, we can make the approximation that $\\mathrm{S} + 0.1$ is approximately 0.1. This simplifies the equation to:

    \n

    $$1 \\times 10^{-10} = 0.1[\\mathrm{S}]$$

    \n

    Solving for S (the molar solubility of BaSO4) gives:

    \n

    $$S = \\frac{1 \\times 10^{-10}}{0.1} = 1 \\times 10^{-9} \\, \\mathrm{mol/L}$$

    \n

    To convert this molarity to a mass per volume concentration, we multiply by the molar mass of BaSO4, which is 233 g/mol:

    \n

    $$\\mathrm{Solubility} = S \\times \\text{Molar mass of BaSO4} = 1 \\times 10^{-9} \\, \\mathrm{mol/L} \\times 233 \\, \\mathrm{g/mol} = 233 \\times 10^{-9} \\, \\mathrm{g/L}$$

    \n

    So, the solubility of BaSO4 in a 0.1 M K2SO4 solution is 233 x $10^{-9}$ g/L, or 233 ng/L.

    \n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1926, "subject": "Chemistry", "question": "Solubility of calcium phosphate (molecular mass, M) in water is $\\mathrm{W_{g}}$ per $100 \\mathrm{~mL}$ at $25^{\\circ} \\mathrm{C}$. Its solubility product at $25^{\\circ} \\mathrm{C}$ will be approximately.", "options": [ { "text": "$10^7\\left(\\frac{W}{M}\\right)^3$" }, { "text": "$10^3\\left(\\frac{\\mathrm{W}}{\\mathrm{M}}\\right)^5$" }, { "text": "$10^7\\left(\\frac{W}{M}\\right)^5$" }, { "text": "$10^5\\left(\\frac{\\mathrm{W}}{\\mathrm{M}}\\right)^5$" } ], "answer": "$10^7\\left(\\frac{W}{M}\\right)^5$", "solution": "**Answer:** $10^7\\left(\\frac{W}{M}\\right)^5$\n\n

    To determine the solubility product (Ksp) of calcium phosphate, we need to consider its chemical formula and how it dissociates in water. The formula for calcium phosphate is $\\text{Ca}_3(\\text{PO}_4)_2$. When dissolved in water, it dissociates according to the following equation:

    \n

    $$\n\\text{Ca}_3(\\text{PO}_4)_2 (s) \\rightleftharpoons 3 \\text{Ca}^{2+} (aq) + 2 \\text{PO}_4^{3-} (aq)\n$$

    \n

    Let $S$ be the molar solubility of calcium phosphate in $\\text{mol}/L$. According to the stoichiometry of the dissociation, if $S$ moles per liter of calcium phosphate dissolve, $3S$ moles per liter of calcium ions and $2S$ moles per liter of phosphate ions are produced.

    \n

    The expression for the solubility product constant (Ksp) for calcium phosphate in water is the product of the concentrations of the ions raised to the power of their respective coefficients in the balanced equation:

    \n

    $$\nK_{sp} = [\\text{Ca}^{2+}]^3 [\\text{PO}_4^{3-}]^2\n$$

    \n

    Substituting the concentrations with $3S$ for $\\text{Ca}^{2+}$ and $2S$ for $\\text{PO}_4^{3-}$, we get:

    \n

    $$\nK_{sp} = (3S)^3 (2S)^2 = 27S^3 \\cdot 4S^2 = 108S^5\n$$

    \n

    If the solubility of calcium phosphate is $\\text{W}_{g}$ per $100 \\text{ mL}$ of water at $25^\\circ\\text{C}$, we need to convert grams to moles to find $S$. Solubility in moles per liter (which is the same as moles per $1000 \\text{ mL}$) is:

    \n

    $$\nS = \\frac{\\text{W}}{\\text{M}} \\times \\frac{1000 \\text{ mL}}{100 \\text{ mL}} = 10\\frac{\\text{W}}{\\text{M}}\n$$

    \n

    Now, substituting $S$ with $10\\frac{\\text{W}}{\\text{M}}$ into the expression for $K_{sp}$:

    \n

    $$\nK_{sp} = 108 \\left(10\\frac{\\text{W}}{\\text{M}}\\right)^5\n$$

    \n

    This simplifies to:

    \n

    $$\nK_{sp} = 108 \\times 10^5 \\left(\\frac{\\text{W}}{\\text{M}}\\right)^5\n$$

    \n

    Since $108$ is on the order of $10^2$, we can approximate this to:

    \n

    $$\nK_{sp} \\approx 10^7 \\left(\\frac{\\text{W}}{\\text{M}}\\right)^5\n$$

    \n

    Therefore, the correct answer based on the options provided is:

    \n

    Option C: $10^7 \\left(\\frac{\\text{W}}{\\text{M}}\\right)^5$

    \n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1927, "subject": "Chemistry", "question": "

    The $$\\mathrm{pH}$$ at which $$\\mathrm{Mg}(\\mathrm{OH})_2\\left[\\mathrm{~K}_{\\mathrm{sp}}=1 \\times 10^{-11}\\right]$$ begins to precipitate from a solution containing $$0.10 \\mathrm{~M} \\mathrm{~Mg}^{2+}$$ ions is __________.

    ", "options": [], "answer": "9", "solution": "**Answer:** 9\n\n

    $$\\begin{aligned}\n& \\text { Precipitation when } Q_{s p}=K_{s p} \\\\\n& {\\left[\\mathrm{Mg}^{2+}\\right]\\left[\\mathrm{OH}^{-}\\right]^2=10^{-11}} \\\\\n& 0.1 \\times\\left[\\mathrm{OH}^{-}\\right]^2=10^{-11} \\Rightarrow\\left[\\mathrm{OH}^{-}\\right]=10^{-5} \\\\\n& \\Rightarrow \\mathrm{pOH}=5 \\quad \\Rightarrow \\mathrm{pH}=9\n\\end{aligned}$$

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1928, "subject": "Chemistry", "question": "

    For a sparingly soluble salt $$\\mathrm{AB}_2$$, the equilibrium concentrations of $$\\mathrm{A}^{2+}$$ ions and $$B^{-}$$ ions are $$1.2 \\times 10^{-4} \\mathrm{M}$$ and $$0.24 \\times 10^{-3} \\mathrm{M}$$, respectively. The solubility product of $$\\mathrm{AB}_2$$ is :

    ", "options": [ { "text": "$$0.069 \\times 10^{-12}$$\n" }, { "text": "$$0.276 \\times 10^{-12}$$\n" }, { "text": "$$6.91 \\times 10^{-12}$$\n" }, { "text": "$$27.65 \\times 10^{-12}$$" } ], "answer": "$$6.91 \\times 10^{-12}$$\n", "solution": "**Answer:** $$6.91 \\times 10^{-12}$$\n\n\n

    For a sparingly soluble salt $\\mathrm{AB}_2$, the dissolution in water can be represented by the following equilibrium equation:

    \n\n

    $ \\mathrm{AB}_2 \\rightleftharpoons \\mathrm{A}^{2+} + 2\\mathrm{B}^{-} $

    \n\n

    When $\\mathrm{AB}_2$ dissolves in water, it generates one $\\mathrm{A}^{2+}$ ion and two $\\mathrm{B}^{-}$ ions. The solubility product constant, $K_{sp}$, for this reaction can be expressed as:

    \n\n

    $ K_{sp} = [\\mathrm{A}^{2+}][\\mathrm{B}^{-}]^2 $

    \n\n

    Given in the problem, the equilibrium concentrations are:

    \n\n

    $ [\\mathrm{A}^{2+}] = 1.2 \\times 10^{-4} \\mathrm{M} $

    \n\n

    $ [\\mathrm{B}^{-}] = 0.24 \\times 10^{-3} \\mathrm{M} = 2.4 \\times 10^{-4} \\mathrm{M} $

    \n\n

    Substituting these values into the $K_{sp}$ expression:

    \n\n

    $ K_{sp} = (1.2 \\times 10^{-4}) \\times (2.4 \\times 10^{-4})^2 $

    \n\n

    Calculating $K_{sp}$:

    \n\n

    $ K_{sp} = 1.2 \\times 10^{-4} \\times (5.76 \\times 10^{-8}) $

    \n\n

    $ K_{sp} = 6.912 \\times 10^{-12} $

    \n\n

    Therefore, the correct option is:

    \n\n

    Option C: $6.91 \\times 10^{-12}$

    ", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 1929, "subject": "Chemistry", "question": "Which one of the following ores is best concentrated by froth – floatation method?", "options": [ { "text": "Magnetite" }, { "text": "Malachite " }, { "text": "Galena" }, { "text": "Cassiterite" } ], "answer": "Galena", "solution": "**Answer:** Galena\n\nNOTE : Galena is PbS and thus purified by froth floatation method.\n

    Froath floatation method is used to concentrate sulphide ores. This method is based on the preferential wetting properties with the froathing agent and water.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1930, "subject": "Chemistry", "question": "Which of the following factors is of no significance for roasting sulphide ores to the oxides and not\nsubjecting the sulphide ores to carbon reduction directly? ", "options": [ { "text": "Metal sulphides are thermodynamically more stable than CS2" }, { "text": "CO2 is thermodynamically more stable than CS2" }, { "text": "Metal sulphides are less stable than the corresponding oxides" }, { "text": "CO2 is more volatile than CS2" } ], "answer": "Metal sulphides are less stable than the corresponding oxides", "solution": "**Answer:** Metal sulphides are less stable than the corresponding oxides\n\nNOTE : The reduction of metal sulphides by carbon reduction process is not spontaneous because $$\\Delta G$$ for such a process is positive. The reduction of metal oxide by carbon reduction process is spontaneous as $$\\Delta G$$ for such a process is negative.\n

    From this we find that on thermodynamic considerations $$C{O_2}$$ is more stable than $$C{S_2}$$ and the metal sulphides are more stable than corresponding oxides. \n

    In view of above the factor listed in choice (c) is incorrect and so is of no significance.\n

    Hence the correct answer is $$(c)$$ ", "topic": "Number Theory", "subtopic": "Divisibility / Factorization" }, { "id": 1931, "subject": "Chemistry", "question": "Which one of the following ores is best concentrated by froth floatation method?", "options": [ { "text": "Siderite" }, { "text": "Galena" }, { "text": "Malachite" }, { "text": "Magnetite" } ], "answer": "Galena", "solution": "**Answer:** Galena\n\nFroth floatation method is mainly applicable for sulphide ores.\n

    $$(1)$$ Malachite ore : $$Cu{\\left( {OH} \\right)_2},\\,\\,CuC{O_3}$$ \n

    $$(2)$$ Magnetite ore : $$F{e_3}{O_4}$$ \n

    $$(3)$$ Siderite ore : $$FeC{O_3}$$\n

    $$(4)$$ Galena ore : $$PbS$$ (Sulphide Ore) ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1932, "subject": "Chemistry", "question": "In the leaching method, bauxite ore is digested with a concentrated solution of NaOH that produces 'X'. When CO2 gas is passed through the aqueous solution of 'X', a hydrated compound 'Y' is precipitated. 'X' and 'Y' respectively are : ", "options": [ { "text": "NaAlO2   and   Al2(CO3)3.x H2O" }, { "text": "A(OH)3   and   Al2O3.x H2O" }, { "text": "Na[Al(OH)4]   and   Al2O3.x H2O" }, { "text": "Na[Al(OH)4]   and   Al2(CO3)3.x H2O" } ], "answer": "Na[Al(OH)4]   and   Al2O3.x H2O", "solution": "**Answer:** Na[Al(OH)4]   and   Al2O3.x H2O\n\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1933, "subject": "Chemistry", "question": "The idea of froth floatation method came from a person X and this method is related to the process Y of ores. X and Y, respectively, are :", "options": [ { "text": "washer woman and concentration" }, { "text": "fisher woman and concentration" }, { "text": "wisher man and reduction" }, { "text": "fisher man and reduction" } ], "answer": "washer woman and concentration", "solution": "**Answer:** washer woman and concentration\n\nFroth floatation is a method of concentration\nand it was discovered by a washer women.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1934, "subject": "Chemistry", "question": "Consider two chemical reactions (A) and (B) that take place during metallurgical process:

    (A) $$ZnC{O_{3(s)}}\\buildrel \\Delta \\over\n \\longrightarrow Zn{O_{(s)}} + C{O_{2(g)}}$$

    (B) $$2Zn{S_{(s)}} + 3{O_{2(g)}}\\buildrel \\Delta \\over\n \\longrightarrow 2Zn{O_{(s)}} + 2S{O_{2(g)}}$$

    The correct option of names given to them respectively is :", "options": [ { "text": "(A) is calcination and (B) is roasting" }, { "text": "Both (A) and (B) are producing same product so both are roasting" }, { "text": "Both (A) and (B) are producing same product so both are calcination" }, { "text": "(A) is roasting and (B) is calcination" } ], "answer": "(A) is calcination and (B) is roasting", "solution": "**Answer:** (A) is calcination and (B) is roasting\n\n(A) $$ZnC{O_{3(s)}}\\buildrel \\Delta \\over\n \\longrightarrow Zn{O_{(s)}} + C{O_{2(g)}}$$

    Heating in absence of oxygen in calcination.

    (B) $$2Zn{S_{(s)}} + 3{O_{2(g)}}\\buildrel \\Delta \\over\n \\longrightarrow 2Zn{O_{(s)}} + 2S{O_{2(g)}}$$

    Heating in presence of oxygen in roasting.

    Hence, (A) is calcination while (B) in roasting.", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 1935, "subject": "Chemistry", "question": "In the leaching of alumina from bauxite, the ore expected to leach out in the process by reacting with NaOH is :", "options": [ { "text": "TiO2" }, { "text": "Fe2O3" }, { "text": "ZnO" }, { "text": "SiO2" } ], "answer": "SiO2", "solution": "**Answer:** SiO2\n\nIn bauxite impurities of Fe2O3, TiO2 and SiO2 are present, Fe2O3 and TiO2 are basic oxides therefore does not reacts with or dissolve in NaOH whereas SiO2 is acidic oxide it gets dissolve in NaOH, hence leach out

    $$Si{O_2} + 2NaOH \\to N{a_2}Si{O_3}(aq.) + {H_2}O$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1936, "subject": "Chemistry", "question": "Match List I with List II : (Both having metallurgical terms)

    \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    List - IList - II
    (a)Concentration of Ag ore(i)Reverberatory furnace
    (b)Blast furnace(ii)Pig iron
    (c)Blister copper(iii)Leaching with dilute NaCN solution
    (d)Froth floatation method(iv)Sulfide ores

    Choose the correct answer from the options given below :", "options": [ { "text": "(a)-(iii), (b)-(ii), (c)-(i), (d)-(iv)" }, { "text": "(a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)" }, { "text": "(a)-(iv), (b)-(i), (c)-(iii), (d)-(ii)" }, { "text": "(a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)" } ], "answer": "(a)-(iii), (b)-(ii), (c)-(i), (d)-(iv)", "solution": "**Answer:** (a)-(iii), (b)-(ii), (c)-(i), (d)-(iv)\n\n(a) Concentration of Ag is performed by leaching with dilute NaCN solution

    (b) Pig iron is formed in blast furnace

    (c) Blister Cu is produced in Bessemer converter

    (d) Froth floatation method is used for sulphide ores.

    Note : During extraction of Cu reverberatory furnace is involved.", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1937, "subject": "Chemistry", "question": "

    Statement I : Leaching of gold with cyanide ion in absence of air / O2 leads to cyano complex of Au(III).

    \n

    Statement II : Zinc is oxidized during the displacement reaction carried out for gold extraction.

    \n

    In the light of the above statements, choose the correct answer from the options given below.

    ", "options": [ { "text": "Both Statement I and Statement II are correct." }, { "text": "Both Statement I and Statement II are incorrect." }, { "text": "Statement I is correct but Statement II is incorrect." }, { "text": "Statement I is incorrect but Statement II is correct." } ], "answer": "Statement I is incorrect but Statement II is correct.", "solution": "**Answer:** Statement I is incorrect but Statement II is correct.\n\nLeaching of gold with cyanide ion is done in\npresence of air/O2 leading to cyano complex [Au(CN)2]– where Au is in +1 oxidation state.

    \n4Au(s) + 8CN–(aq) + 2H2O(aq) + O2(g) $$\\longrightarrow 4\\left[\\mathrm{Au}(\\mathrm{CN})_{2}\\right]_{\\mathrm{aq}}^{-}+4 \\mathrm{OH}^{-}$$

    \n$$\\begin{aligned} 2\\left[\\mathrm{Au}(\\mathrm{CN})_{2}\\right]_{(\\mathrm{aq})}^{-} &+\\mathrm{Zn}(\\mathrm{s}) \\\\ & \\longrightarrow\\left[\\mathrm{Zn}(\\mathrm{CN})_{4}\\right]_{(\\mathrm{aq})}^{2-}+2 \\mathrm{Au}(\\mathrm{s}) \\end{aligned}$$

    \nZinc is oxidised from (0) to +2 oxidation state during\ndisplacement reaction carried out for gold\nextraction. \n", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 1938, "subject": "Chemistry", "question": "

    Match List-I with List-II

    \n

    \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    List - IList - II
    (A)Concentration of gold ore(I)Aniline
    (B)Leaching of alumina(II)NaOH
    (C)Froth stabiliser(III)SO$$_2$$
    (D)Blister copper(IV)NaCN

    \n

    Choose the correct answer from the options given below :

    ", "options": [ { "text": "(A) - (IV), (B) - (III), (C) - (II), (D) - (I)" }, { "text": "(A) - (IV), (B) - (II), (C) - (I), (D) - (III)" }, { "text": "(A) - (III), (B) - (II), (C) - (I), (D) - (IV)" }, { "text": "(A) - (II), (B) - (IV), (C) - (III), (D) - (I)" } ], "answer": "(A) - (IV), (B) - (II), (C) - (I), (D) - (III)", "solution": "**Answer:** (A) - (IV), (B) - (II), (C) - (I), (D) - (III)\n\nGold is concentrated by cyanidation

    \nLeaching of alumina is done by NaOH

    \n Froth stabiliser is aniline

    \n Blister copper has condensed SO2 on the surface ", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1939, "subject": "Chemistry", "question": "

    The role of depressants in 'Froth Flotation method' is to

    ", "options": [ { "text": "selectively prevent one component of the ore from coming to the froth." }, { "text": "reduce the consumption of oil for froth formation." }, { "text": "stabilize the froth." }, { "text": "enhance non-wettability of the mineral particles." } ], "answer": "selectively prevent one component of the ore from coming to the froth.", "solution": "**Answer:** selectively prevent one component of the ore from coming to the froth.\n\nDepressant prevent one component from coming to\nthe froth.

    \nFor eg., in Galena ore, the depressant (NaCN)\nprevents impurity (ZnS) from coming to the froth. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1940, "subject": "Chemistry", "question": "

    Leaching of gold with dilute aqueous solution of NaCN in presence of oxygen gives complex [A], which on reaction with zinc forms the elemental gold and another complex [B]. [A] and [B], respectively are :

    ", "options": [ { "text": "$${[Au{(CN)_4}]^ - }$$ and $${[Zn{(CN)_2}{(OH)_2}]^{2 - }}$$" }, { "text": "$${[Au{(CN)_2}]^ - }$$ and $${[Zn{(OH)_4}]^{2 - }}$$" }, { "text": "$${[Au{(CN)_2}]^{ - }}$$ and $${[Zn{(CN)_4}]^{2 - }}$$" }, { "text": "$${[Au{(CN)_4}]^{2 - }}$$ and $${[Zn{(CN)_6}]^{4 - }}$$" } ], "answer": "$${[Au{(CN)_2}]^{ - }}$$ and $${[Zn{(CN)_4}]^{2 - }}$$", "solution": "**Answer:** $${[Au{(CN)_2}]^{ - }}$$ and $${[Zn{(CN)_4}]^{2 - }}$$\n\nIn the metallurgy of gold

    \n$$4 \\mathrm{Au}+8 \\mathrm{CN}^{-}+2 \\mathrm{H}_{2} \\mathrm{O}+\\mathrm{O}_{2} \\rightarrow 4\\left[\\mathrm{Au}(\\mathrm{CN})_{2}\\right]^{-}+4 \\mathrm{OH}^{-}$$

    \n$$2\\left[\\mathrm{Au}(\\mathrm{CN})_{2}\\right]^{-}+\\mathrm{Zn} \\rightarrow\\left[\\mathrm{Zn}(\\mathrm{CN})_{4}\\right]^{2-}+2 \\mathrm{Au}$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1941, "subject": "Chemistry", "question": "

    Which of the reaction is suitable for concentrating ore by leaching process ?

    ", "options": [ { "text": "$$\n2 \\mathrm{Cu}_{2} \\mathrm{~S}+3 \\mathrm{O}_{2} \\rightarrow 2 \\mathrm{Cu}_{2} \\mathrm{O}+2 \\mathrm{SO}_{2}\n$$" }, { "text": "$$\\mathrm{Fe}_{3} \\mathrm{O}_{4}+\\mathrm{CO} \\rightarrow 3 \\mathrm{FeO}+\\mathrm{CO}_{2}$$" }, { "text": "$$\\mathrm{Al}_{2} \\mathrm{O}_{3}+2 \\mathrm{NaOH}+3 \\mathrm{H}_{2} \\mathrm{O} \\rightarrow 2 \\mathrm{Na}\\left[\\mathrm{Al}(\\mathrm{OH})_{4}\\right]$$" }, { "text": "$$\\mathrm{Al}_{2} \\mathrm{O}_{3}+6 \\mathrm{Mg} \\rightarrow 6 \\mathrm{MgO}+4 \\mathrm{Al}$$" } ], "answer": "$$\\mathrm{Al}_{2} \\mathrm{O}_{3}+2 \\mathrm{NaOH}+3 \\mathrm{H}_{2} \\mathrm{O} \\rightarrow 2 \\mathrm{Na}\\left[\\mathrm{Al}(\\mathrm{OH})_{4}\\right]$$", "solution": "**Answer:** $$\\mathrm{Al}_{2} \\mathrm{O}_{3}+2 \\mathrm{NaOH}+3 \\mathrm{H}_{2} \\mathrm{O} \\rightarrow 2 \\mathrm{Na}\\left[\\mathrm{Al}(\\mathrm{OH})_{4}\\right]$$\n\nLeaching involves the treatment of ore with a suitable reagent so as it make it soluble while impurities remain insoluble.\n

    \n$\\mathrm{Al}_{2} \\mathrm{O}_{3}+2 \\mathrm{NaOH}+3 \\mathrm{H}_{2} \\mathrm{O} \\rightarrow \\underset{\\text{Soluble complex}}{2 \\mathrm{Na}\\left[\\mathrm{Al}(\\mathrm{OH})_{4}\\right]}$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1942, "subject": "Chemistry", "question": "

    The methods NOT involved in concentration of ore are

    \n

    A. Liquation

    \n

    B. Leaching

    \n

    C. Electrolysis

    \n

    D. Hydraulic washing

    \n

    E. Froth floatation

    \n

    Choose the correct answer from the options given below :

    ", "options": [ { "text": "A and C only" }, { "text": "C, D and E only" }, { "text": "B, D and C only" }, { "text": "B, D and E only" } ], "answer": "A and C only", "solution": "**Answer:** A and C only\n\nMethods involved in concentration of one are\n

    (i) Hydraulic Washing\n

    (ii) Froth Floatation\n

    (iii) Magnetic Separation\n

    (iv) Leaching", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1943, "subject": "Chemistry", "question": "

    Given below are two statements :

    \n

    Statement I : In froth floatation method of rotating paddle agitates the mixture to drive air out of it.

    \n

    Statement II : Iron pyrites and generally avoided for extraction of iron due to environmental reasons.

    \n

    In the light of the above statements, choose the correct answer from the options given below :

    ", "options": [ { "text": "Statement I is true but Statement II is false" }, { "text": "Statement I is false but Statement II is true" }, { "text": "Both Statement I and Statement II are true" }, { "text": "Both Statement I and Statement II are false" } ], "answer": "Statement I is false but Statement II is true", "solution": "**Answer:** Statement I is false but Statement II is true\n\n

    Statement I is false because the rotating paddle in\nfroth floatation method agitates the mixture to\ngenerate froth and not to drive air out of it.

    \n

    Statement II is true because iron is commercially\nextracted from haematite ore and not from iron\npyrites to minimize environmental pollution.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1944, "subject": "Chemistry", "question": "Which one of the following is not an example of calcination?\n", "options": [ { "text": "$\\mathrm{CaCO}_{3} \\cdot \\mathrm{MgCO}_{3} \\stackrel{\\Delta}{\\longrightarrow} \\mathrm{CaO}+\\mathrm{MgO}+2 \\mathrm{CO}_{2}$" }, { "text": "$2 \\mathrm{PbS}+3 \\mathrm{O}_{2} \\stackrel{\\Delta}{\\longrightarrow} 2 \\mathrm{PbO}+2 \\mathrm{SO}_{2}$" }, { "text": "$\\mathrm{CaCO}_{3} \\stackrel{\\Delta}{\\longrightarrow} \\mathrm{CaO}+\\mathrm{CO}_{2}$" }, { "text": "$\\mathrm{Fe}_{2} \\mathrm{O}_{3} \\cdot \\mathrm{xH}_{2} \\mathrm{O} \\stackrel{\\Delta}{\\longrightarrow} \\mathrm{Fe}_{2} \\mathrm{O}_{3}+\\mathrm{xH}_{2} \\mathrm{O}$" } ], "answer": "$2 \\mathrm{PbS}+3 \\mathrm{O}_{2} \\stackrel{\\Delta}{\\longrightarrow} 2 \\mathrm{PbO}+2 \\mathrm{SO}_{2}$", "solution": "**Answer:** $2 \\mathrm{PbS}+3 \\mathrm{O}_{2} \\stackrel{\\Delta}{\\longrightarrow} 2 \\mathrm{PbO}+2 \\mathrm{SO}_{2}$\n\nCalcination is the process of heating a solid material in the absence of air or with limited air to remove volatile components or to cause thermal decomposition. Among the given reactions:\n

    \n\n(A) $\\mathrm{CaCO}_3 \\cdot \\mathrm{MgCO}_3 \\stackrel{\\Delta}{\\longrightarrow} \\mathrm{CaO}+\\mathrm{MgO}+2 \\mathrm{CO}_2$ is an example of calcination, where a mixture of calcium carbonate and magnesium carbonate is heated to produce calcium oxide, magnesium oxide, and carbon dioxide.\n\n

    \n(C) $\\mathrm{CaCO}_3 \\stackrel{\\Delta}{\\longrightarrow} \\mathrm{CaO}+\\mathrm{CO}_2$ is an example of calcination, where calcium carbonate is heated to produce calcium oxide and carbon dioxide.

    \n(D) $\\mathrm{Fe}_2 \\mathrm{O}_3 \\cdot \\mathrm{xH}_2 \\mathrm{O} \\stackrel{\\Delta}{\\longrightarrow} \\mathrm{Fe}_2 \\mathrm{O}_3+\\mathrm{xH}_2 \\mathrm{O}$ is an example of calcination, where hydrated iron(III) oxide loses water molecules to form anhydrous iron(III) oxide.

    \nHowever, (B) $2 \\mathrm{PbS}+3 \\mathrm{O}_2 \\stackrel{\\Delta}{\\longrightarrow} 2 \\mathrm{PbO}+2 \\mathrm{SO}_2$ is not an example of calcination. This reaction is an example of roasting, a process in which a sulfide ore is heated in the presence of air to convert it into its oxide and release sulfur dioxide gas. Roasting is typically used for metal extraction from sulfide ores.\n

    \nSo the correct answer is:\n

    \n(B) $2 \\mathrm{PbS}+3 \\mathrm{O}_2 \\stackrel{\\Delta}{\\longrightarrow} 2 \\mathrm{PbO}+2 \\mathrm{SO}_2$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1945, "subject": "Chemistry", "question": "

    Given below are two statements :

    \n

    Statement I : In the metallurgy process, sulphide ore is converted to oxide before reduction.

    \n

    Statement II : Oxide ores in general are easier to reduce.

    \n

    In the light of the above statements, choose the most appropriate answer from the options given below:

    ", "options": [ { "text": "Both Statement I and Statement II are correct" }, { "text": "Both Statement I and Statement II are incorrect" }, { "text": "Statement I is correct but Statement II is incorrect" }, { "text": "Statement I is incorrect but Statement II is correct" } ], "answer": "Both Statement I and Statement II are correct", "solution": "**Answer:** Both Statement I and Statement II are correct\n\nOption A is the correct answer. \n\n

    Statement I : In the metallurgy process, sulphide ore is converted to oxide before reduction.\n\n

    This is correct. In metallurgy, it is common to first convert sulphide ores to oxides because it's typically easier to reduce oxides to extract the metal. This process is known as roasting, where the sulphide ore is heated in the presence of air to convert it into an oxide.\n\n

    $$\n2 \\mathrm{ZnS}+3 \\mathrm{O}_2 \\rightarrow 2 \\mathrm{ZnO}+2 \\mathrm{SO}_2\n$$\n\n

    Statement II : Oxide ores in general are easier to reduce.\n\n

    This is also correct. Oxides are generally easier to reduce than most other types of compounds. This is because the bond between metal and oxygen in metal oxides is often weaker than the bond between metal and other elements in various types of ores. For example, the C-O bond in CO2 (a product of the reduction of a metal oxide) is weaker and more easily broken than the C-S bond in CS2 (a product of the reduction of a metal sulphide), making CO2 more volatile and easier to remove from the system.\n\n

    So, both Statement I and Statement II are correct, making Option A the right choice.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1946, "subject": "Chemistry", "question": "

    Which of the following is used as a stabilizer during the concentration of sulphide ores?

    ", "options": [ { "text": "Cresols" }, { "text": "Xanthates" }, { "text": "Fatty acids" }, { "text": "Pine oils" } ], "answer": "Cresols", "solution": "**Answer:** Cresols\n\nIn the froth floatation process pine oils, fatty acids, and xanthates are collectors, and froth stabilizers cresols, and aniline are used.", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 1947, "subject": "Chemistry", "question": "With respect to an ore, Ellingham diagram helps to predict the feasibility of its.", "options": [ { "text": "Electrolysis" }, { "text": "Thermal reducation" }, { "text": "Vapour phase refining" }, { "text": "Zone refining" } ], "answer": "Thermal reducation", "solution": "**Answer:** Thermal reducation\n\nEllingham diagram gives idea that which sustance or metal is suitable for thermal reduction of oxide.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1948, "subject": "Chemistry", "question": "An Ellingham diagram provides information\nabout :", "options": [ { "text": "the temperature dependence of the\nstandard Gibbs energies of formation of\nsome metal oxides" }, { "text": "the pressure dependence of the standard\nelectrode potentials of reduction reactions\ninvolved in the extraction of metals" }, { "text": "the conditions of pH and potential under\nwhich a species is thermodynamically\nstable" }, { "text": "the kinetics of the reduction process" } ], "answer": "the temperature dependence of the\nstandard Gibbs energies of formation of\nsome metal oxides", "solution": "**Answer:** the temperature dependence of the\nstandard Gibbs energies of formation of\nsome metal oxides\n\nAn Ellingham diagram provides information\nabout, the temperature dependence of the\nstandard gibbs energies of formation of some\nmetal oxides.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1949, "subject": "Chemistry", "question": "Ellingham diagram is a graphical representation of :", "options": [ { "text": "$$\\Delta$$G vs P" }, { "text": "$$\\Delta$$G vs T" }, { "text": "($$\\Delta$$G $$-$$ T$$\\Delta$$S) vs T" }, { "text": "$$\\Delta$$H vs T" } ], "answer": "$$\\Delta$$G vs T", "solution": "**Answer:** $$\\Delta$$G vs T\n\nEllingham diagram is a graphical\nrepresentation of $$\\Delta $$G vs T when metal heated\nwith oxygen to form metal oxide.", "topic": "Discrete Mathematics", "subtopic": "Other" }, { "id": 1950, "subject": "Chemistry", "question": "The statement is incorrect about Ellingham diagram is ", "options": [ { "text": "provides idea about the reaction rate." }, { "text": "provides idea about free energy change." }, { "text": "provides idea about changes in the phases during the reaction." }, { "text": "provides idea about reduction of metal oxide." } ], "answer": "provides idea about the reaction rate.", "solution": "**Answer:** provides idea about the reaction rate.\n\nEllingham diagram is a plot between $$\\Delta$$G$$^\\circ$$ and T and does not give any information regarding rate of reaction.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1951, "subject": "Chemistry", "question": "Given below are two statements :

    Statement I : The choice of reducing agents for metals extraction can be made by using Ellingham diagram, a plot of $$\\Delta$$G vs temperature.

    Statement II : The value of $$\\Delta$$S increases from left to right in Ellingham diagram.

    In the light of the above statements, choose the most appropriate answer from the options given below :", "options": [ { "text": "Both Statement I and Statement II are true" }, { "text": "Statement I is false but Statement II is true" }, { "text": "Both Statement I and Statement II are false" }, { "text": "Statement I is true but Statement II is false" } ], "answer": "Statement I is true but Statement II is false", "solution": "**Answer:** Statement I is true but Statement II is false\n\nGiven Statement I is true as in a number of processes, one element is used to reduce the oxide of another metal. Any element will reduce the oxide of other metal which lie above it in the Ellingham diagram because the free energy change will become more negative.

    Given Statement II is false as the value of $$\\Delta$$S is decreases from left to right in Ellingham diagram.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1952, "subject": "Chemistry", "question": "

    Given below are two statements :

    \n

    Statement I : According to Ellingham diagram, any metal oxide with higher $$\\Delta$$G$$^\\circ$$ is more stable than the one with lower $$\\Delta$$G$$^\\circ$$.

    \n

    Statement II : The metal involved in the formation of oxide placed lower in the Ellingham diagram can reduce the oxide of a metal placed higher in the diagram.

    \n

    In the light of the above statements, choose the most appropriate answer from the options given below :

    ", "options": [ { "text": "Both Statement I and Statement II are correct." }, { "text": "Both Statement I and Statement II are incorrect." }, { "text": "Statement I is correct but Statement II is incorrect." }, { "text": "Statement I is incorrect but Statement II is correct." } ], "answer": "Statement I is incorrect but Statement II is correct.", "solution": "**Answer:** Statement I is incorrect but Statement II is correct.\n\nEllingham diagram is plot of $$\\Delta $$G vs T.\nThe criterion for the feasibility of a thermal\nreduction is that at a given temperature Gibbs\nenergy change of a reaction must be negative. The\nchange in Gibbs energy, $$\\Delta $$G for any process at any\nspecified temperature is given by the equation

    \n$$\\Delta $$G = $$\\Delta $$H - T$$\\Delta $$S

    \nwhere $$\\Delta $$H = enthalpy change and $$\\Delta $$S = entropy change

    \nAccording to the ellingham diagram, any metal\noxide with higher $$\\Delta $$G° has a tendency of getting\nreduced by the metal whose metal oxide has lower\nvalue of $$\\Delta $$G°.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1953, "subject": "Chemistry", "question": "

    Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R

    \n

    Assertion A: In an Ellingham diagram, the oxidation of carbon to carbon monoxide shows a negative slope with respect to temperature.

    \n

    Reason R: CO tends to get decomposed at higher temperature.

    \n

    In the light of the above statements, choose the correct answer from the options given below

    ", "options": [ { "text": "A is correct but R is not correct" }, { "text": "Both A and R are correct but R is NOT the correct explanation of A" }, { "text": "A is not correct but R is correct" }, { "text": "Both A and R are correct and R is the correct explanation of A" } ], "answer": "A is correct but R is not correct", "solution": "**Answer:** A is correct but R is not correct\n\n$$\n2 \\mathrm{C}(\\mathrm{s})+\\mathrm{O}_2(\\mathrm{~g}) \\rightarrow 2 \\mathrm{CO}(\\mathrm{g})\n$$\n

    $\\Delta_{\\mathrm{r}} \\mathrm{S}^0$ is $+v e, \\Delta_{\\mathrm{r}} \\mathrm{G}^0=\\Delta_{\\mathrm{r}} \\mathrm{H}^0-\\mathrm{T} \\Delta_{\\mathrm{r}} \\mathrm{S}^0$;\n

    Thus slope is negative.\n\n

    As temperature increases $\\Delta_{\\mathrm{r}} G^{\\mathrm{o}}$ becomes more negative thus it has lower tendency to get decomposed.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1954, "subject": "Chemistry", "question": "

    Given below are two statements related to Ellingham diagram:

    \n

    Statement I : Ellingham diagrams can be constructed for formation of oxides, sulfides and halides of metals.

    \n

    Statement II : It consists of plots of $$\\Delta_{\\mathrm{f}} \\mathrm{H}^{0}$$ vs $$\\mathrm{T}$$ for formation of oxides of elements.

    \n

    In the light of the above statements, choose the most appropriate answer from the options given below:

    \n

    ", "options": [ { "text": "Both Statement I and Statement II are correct" }, { "text": "Both Statement I and Statement II are incorrect" }, { "text": "Statement I is correct but Statement II is incorrect" }, { "text": "Statement I is incorrect but Statement II is correct" } ], "answer": "Statement I is correct but Statement II is incorrect", "solution": "**Answer:** Statement I is correct but Statement II is incorrect\n\n

    Statement I is correct, Ellingham diagram can be constructed for formation of oxides, sulphides and halides of metals. (As per NCERT)

    \n\n

    Statement II is incorrect because Ellingham diagram consists of $\\Delta_t G^0$ vs $T$ for formation of oxides of elements.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1955, "subject": "Chemistry", "question": "

    Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R

    \n

    Assertion A : In the Ellingham diagram, a sharp change in slope of the line is observed for $$\\mathrm{Mg} \\rightarrow \\mathrm{MgO}$$ at $$\\sim 1120^{\\circ} \\mathrm{C}$$

    \n

    Reason R: There is a large change of entropy associated with the change of state

    \n

    In the light of the above statements, choose the correct answer from the options given below

    ", "options": [ { "text": "Both $$\\mathbf{A}$$ and $$\\mathbf{R}$$ are true and $$\\mathbf{R}$$ is the correct explanation of $$\\mathbf{A}$$" }, { "text": "$$\\mathbf{A}$$ is false but $$\\mathbf{R}$$ is true" }, { "text": "Both $$\\mathbf{A}$$ and $$\\mathbf{R}$$ are true but $$\\mathbf{R}$$ is NOT the correct explanation of $$\\mathbf{A}$$" }, { "text": "$$\\mathbf{A}$$ is true but $$\\mathbf{R}$$ is false" } ], "answer": "Both $$\\mathbf{A}$$ and $$\\mathbf{R}$$ are true and $$\\mathbf{R}$$ is the correct explanation of $$\\mathbf{A}$$", "solution": "**Answer:** Both $$\\mathbf{A}$$ and $$\\mathbf{R}$$ are true and $$\\mathbf{R}$$ is the correct explanation of $$\\mathbf{A}$$\n\nIn the Ellingham diagram, a sharp change in slope of the line is observed for $\\mathrm{Mg}-\\mathrm{MgO}$ at $\\sim 1120^{\\circ} \\mathrm{C}$ because that is the boiling point of magnesium.

    \nThere is a large increase in entropy associated with the change of state of magnesium. So, both Assertion (A) and Reason (R) are true and (R) is the correct explanation of $(A)$.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1956, "subject": "Chemistry", "question": "Aluminium is extracted by the electrolysis of", "options": [ { "text": "bauxite" }, { "text": "alumina" }, { "text": "alumina mixed with molten cryolite" }, { "text": "molten cryolite" } ], "answer": "alumina mixed with molten cryolite", "solution": "**Answer:** alumina mixed with molten cryolite\n\nPure aluminium can be obtained by electrolysis of a mixture containing alumina, crayolite and fluorspar in the ratio $$20:24:20.$$ The fusion temperature of this mixture is $${900^ \\circ }C$$ and it is a good conductor of electricity. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1957, "subject": "Chemistry", "question": "The metal extracted by leaching with a cyanide is", "options": [ { "text": "Mg" }, { "text": "Ag" }, { "text": "Cu" }, { "text": "Na" } ], "answer": "Ag", "solution": "**Answer:** Ag\n\nSilver ore forms a soluble complex with $$NaCN$$ from which silver is precipitated using scrap zinc.\n

    $$A{g_2}S + 2NaCN \\to $$\n

    $$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,Na\\left[ {Ag{{\\left( {CN} \\right)}_2}} \\right]\\buildrel {Zn} \\over\n \\longrightarrow $$\n

    $$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,N{a_2}\\left[ {Zn{{\\left( {CN} \\right)}_4}} \\right] + Ag \\downarrow $$\n

    $$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,$$sodargento-cynanide(soluble) ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1958, "subject": "Chemistry", "question": "Extraction of copper by smelting uses silica as an additive to remove :", "options": [ { "text": "Cu2S" }, { "text": "FeO" }, { "text": "FeS" }, { "text": "Cu2O" } ], "answer": "FeO", "solution": "**Answer:** FeO\n\n

    The gangue material in the ore is removed by adding\nanother chemical substance called flux. If ore has basic\nimpurities, such as FeO, CaO, MgO, etc. suitable acid\nflux such as SiO2\n, P2O5\n, etc. are used. For example, in the\nextraction of copper, ferrous oxide (FeO), a basic impurity, is removed by adding silica. FeO (gangue) and SiO2\n(flux) together form slag FeSiO3.

    \n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1959, "subject": "Chemistry", "question": "In the extraction of copper from its sulphate ore, metal is finally obtained by the oxidation of cuprous sulphide with : ", "options": [ { "text": "Fe2O3 " }, { "text": "Cu2O" }, { "text": "SO2 " }, { "text": "CO" } ], "answer": "Cu2O", "solution": "**Answer:** Cu2O\n\n2Cu2S  +  3O2   $$ \\to $$   2Cu2O  +   2SO2\n

    2Cu2O  +  Cu2S  $$ \\to $$  6Cu  +  SO2\n

    It is a example of auto reduction.\n

    First a part of sulphide ore is converted into oxide which then reacts with remaining sulphide to give the metal.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1960, "subject": "Chemistry", "question": "Hall- Heroult's process is given by :", "options": [ { "text": "Cu2+ (aq) + H2(g) $$ \\to $$ Cu(s) + 2H+ (aq)" }, { "text": "Cr2O3 + 2AI  $$ \\to $$ Al2O3 + 2Cr" }, { "text": "2Al2O3 + 3C  $$ \\to $$ 4Al + 3CO2" }, { "text": "ZnO + C   $$\\buildrel {Coke,\\,1673\\,K} \\over\n \\longrightarrow $$ Zn + CO" } ], "answer": "2Al2O3 + 3C  $$ \\to $$ 4Al + 3CO2", "solution": "**Answer:** 2Al2O3 + 3C  $$ \\to $$ 4Al + 3CO2\n\nIn Hall-Heroult's process is given by\n
    \"JEE\n
    At cathode :- $$4Al_{\\left( \\ell \\right)}^{3 + }{\\mkern 1mu} {\\mkern 1mu} + {\\mkern 1mu} {\\mkern 1mu} 12{e^ - }{\\mkern 1mu} {\\mkern 1mu} \\,\\mathrel{\\mathop{\\kern0pt\\longrightarrow}\n\\limits_\\,} \\,\\,4Al\\left( \\ell \\right)$$\n

    At Anode :- $$6O_{\\left( \\ell \\right)}^{2 - }\\,\\,\\mathrel{\\mathop{\\kern0pt\\longrightarrow}\n\\limits_\\,} \\,\\,3{O_2}(g) + 12{e^ - }$$\n

    $$3C + 3{O_2}\\,\\,\\mathrel{\\mathop{\\kern0pt\\longrightarrow}\n\\limits_\\,} \\,\\,3C{O_2}\\left( \\uparrow \\right)$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1961, "subject": "Chemistry", "question": "In the Hall-Heroult process, aluminium is formed at the cathode. The cathode is made out of - ", "options": [ { "text": "Copper " }, { "text": "Platinum " }, { "text": "Carbon" }, { "text": "Pure aluminium " } ], "answer": "Carbon", "solution": "**Answer:** Carbon\n\nIn the Hall-Heroult process the cathode is made of carbon.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1962, "subject": "Chemistry", "question": "The pair that does NOT require calcination is - ", "options": [ { "text": "Fe2O3 and CaCO3⋅MgCO3" }, { "text": "ZnCO3 and CaO " }, { "text": "ZnO and Fe2O3⋅xH2O" }, { "text": "ZnO and MgO " } ], "answer": "ZnO and MgO ", "solution": "**Answer:** ZnO and MgO \n\nIn calcination, we convert a compound or ore into oxide.\n

    In option (D) you can see there is ZnO and MgO, as they are already in oxide form so no calcination required for then.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1963, "subject": "Chemistry", "question": "Assertion: For the extraction of iron, haematite\nore is used.

    \nReason: Haematite is a carbonate ore of iron.", "options": [ { "text": "Both the assertion and reason are correct\nand the reason is the correct explanation for\nthe assertion." }, { "text": "Only the reason is correct." }, { "text": "Both the assertion and reason are correct,\nbut the reason is not the correct explanation\nfor the assertion." }, { "text": "Only the assertion is correct." } ], "answer": "Only the assertion is correct.", "solution": "**Answer:** Only the assertion is correct.\n\nFor the extraction of iron, haematite ore in\nused.\n

    Haematite = Fe2O3 which is a oxide ore not carbonate ore.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1964, "subject": "Chemistry", "question": "The correct statement is :", "options": [ { "text": "leaching of bauxite using concentrated NaOH solution gives sodium aluminate and sodium silicate" }, { "text": "the Hall-Heroult process is used for the production of aluminium and iron" }, { "text": "pig iron is obtained from cast iron" }, { "text": "the blistered appearance of copper during the metallurgical process is due to the evolution of CO2" } ], "answer": "leaching of bauxite using concentrated NaOH solution gives sodium aluminate and sodium silicate", "solution": "**Answer:** leaching of bauxite using concentrated NaOH solution gives sodium aluminate and sodium silicate\n\n(A) Leaching of bauxite :\n
    Al2O3.H2O + NaOH $$ \\to $$ Na[Al(OH)4] + Na2SiO3\n
    So, option (A) is correct.\n

    (B) Hall Heroult’s process is used for aluminium only not for iron.\n
    So, option (B) is wrong.\n

    (C) From Blast Furnace, at the end we get pig iron and in pig iron percentage of carbon is high and after reducing the carbon percantage we get cast iron. So we can say cast iron is obtained from pig iron.\n
    So, option (C) is wrong.\n

    (D) In Bessemer converter following reaction happens :\n
    Cu2S + Air $$ \\to $$ Cu2O + SO2\n
    Cu2O + Cu2S $$ \\to $$ Cu + SO2\n

    So we can say, Appearance of Blister copper is due to evolution of SO2\n gas.\n

    $$ \\therefore $$ Option (D) is wrong.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1965, "subject": "Chemistry", "question": "The purest from of commercial iron is :\n", "options": [ { "text": "pig iron" }, { "text": "wrought iron " }, { "text": "cast iron" }, { "text": "scrap iron and pig iron" } ], "answer": "wrought iron ", "solution": "**Answer:** wrought iron \n\nWrought iron is purest from of commercial\niron.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1966, "subject": "Chemistry", "question": "Among the reactions (a) - (d), the reaction(s)\nthat does/do not occur in the blast furnace\nduring the extraction of iron is/are :
    \n(a) CaO + SiO2 $$ \\to $$ CaSiO3
    \n(b) 3Fe2O3 + CO $$ \\to $$ 2Fe3O4 + CO2
    \n(c) FeO + SiO2 $$ \\to $$ FeSiO3
    \n(d) FeO $$ \\to $$ Fe + $${{1 \\over 2}}$$O2", "options": [ { "text": "(a) and (d)" }, { "text": "(c) and (d)" }, { "text": "(a)" }, { "text": "(d)" } ], "answer": "(c) and (d)", "solution": "**Answer:** (c) and (d)\n\nIn blast furnace (metallugy of iron) involved\nreactions are\n

    (a) CaO + SiO2 $$ \\to $$ CaSiO3\n

    (b) 3Fe2O3 + CO $$ \\to $$ 2Fe3O4 + CO2", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 1967, "subject": "Chemistry", "question": "Cast iron is used for the manufacture of :", "options": [ { "text": "wrought iron, pig iron and steel" }, { "text": "wrought iron and pig iron" }, { "text": "wrought iron and steel" }, { "text": "pig iron, scrap iron and steel" } ], "answer": "wrought iron and steel", "solution": "**Answer:** wrought iron and steel\n\nCast iron is used for manufacturing of wrought\niron and steel.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1968, "subject": "Chemistry", "question": "Match List - I with List - II.

    \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    List - I List - II
    (a)Sodium Carbonate(i)Deacon
    (b)Titanium(ii)Castner-Kellner
    (c)Chlorine(iii)van-Arkel
    (d)Sodium hydroxide(iv)Solvay


    Choose the correct answer from the options given below :", "options": [ { "text": "(a) $$ \\to $$ (iv), (b) $$ \\to $$ (iii), (c) $$ \\to $$ (i), (d) $$ \\to $$ (ii)" }, { "text": "(a) $$ \\to $$ (iv), (b) $$ \\to $$ (i), (c) $$ \\to $$ (ii), (d) $$ \\to $$ (iii)" }, { "text": "(a) $$ \\to $$ (iii), (b) $$ \\to $$ (ii), (c) $$ \\to $$ (i), (d) $$ \\to $$ (iv)" }, { "text": "(a) $$ \\to $$ (i), (b) $$ \\to $$ (iii), (c) $$ \\to $$ (iv), (d) $$ \\to $$ (ii)" } ], "answer": "(a) $$ \\to $$ (iv), (b) $$ \\to $$ (iii), (c) $$ \\to $$ (i), (d) $$ \\to $$ (ii)", "solution": "**Answer:** (a) $$ \\to $$ (iv), (b) $$ \\to $$ (iii), (c) $$ \\to $$ (i), (d) $$ \\to $$ (ii)\n\n(a) Sodium carbonate is prepared by Solvay\nprocess

    \n(b) Titanium is refined by Van-Arkel process

    \n(c) Chlorine is prepared by Deacon process

    \n(d) Sodium hydroxide is prepared by Castner-Kellner process\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1969, "subject": "Chemistry", "question": "Calgon is used for water treatment. Which of the following statement is NOT true about Calgon?", "options": [ { "text": "It is also known as Graham's salt. " }, { "text": "It is polymeric compound and is water soluble." }, { "text": "It does not remove Ca2+ ion by precipitation." }, { "text": "Calgon contains the 2nd most abundant element by weight in the Earth's crust." } ], "answer": "Calgon contains the 2nd most abundant element by weight in the Earth's crust.", "solution": "**Answer:** Calgon contains the 2nd most abundant element by weight in the Earth's crust.\n\nCalgon is sodium hexametaphosphate, a\npolymeric compound also called as Graham’s\nsalt.

    \nOrder of abundance of element in earth crust is

    \nO > Si > Al > Fe > Ca > Na > Mg > K

    \nSo second most abundant element in earth crust is Si not Ca.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1970, "subject": "Chemistry", "question": "The process that involves the removal of sulphur from the ores is :", "options": [ { "text": "Roasting" }, { "text": "Smelting" }, { "text": "Leaching" }, { "text": "Refining" } ], "answer": "Roasting", "solution": "**Answer:** Roasting\n\nRemoval of sulphur from the ore is done by\nRoasting.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1971, "subject": "Chemistry", "question": "Which of the following reduction reaction CANNOT be carried out with coke?", "options": [ { "text": "ZnO $$ \\to $$ Zn" }, { "text": "Fe2O3 $$ \\to $$ Fe" }, { "text": "Cu2O $$ \\to $$ Cu" }, { "text": "Al2O3 $$ \\to $$ Al" } ], "answer": "Al2O3 $$ \\to $$ Al", "solution": "**Answer:** Al2O3 $$ \\to $$ Al\n\nAl2O3 is reduced by the electrolytic reduction method.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1972, "subject": "Chemistry", "question": "The chemical that is added to reduce the melting point of the reaction mixture during the extraction of aluminium is :", "options": [ { "text": "Cryolite" }, { "text": "Bauxite" }, { "text": "Calamine" }, { "text": "Kaolite" } ], "answer": "Cryolite", "solution": "**Answer:** Cryolite\n\nAlumina is mixed with Na3AlF6 (Cryolite) and CaF2 to lower the melting point of mixture.", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 1973, "subject": "Chemistry", "question": "Which one of the following set of elements can be detected using sodium fusion extract?", "options": [ { "text": "Sulfur, Nitrogen, Phosphorous, Halogens" }, { "text": "Phosphorous, Oxygen, Nitrogen, Halogens" }, { "text": "Nitrogen, Phosphorous, Carbon, Sulfur" }, { "text": "Halogens, Nitrogen, Oxygen, Sulfur" } ], "answer": "Sulfur, Nitrogen, Phosphorous, Halogens", "solution": "**Answer:** Sulfur, Nitrogen, Phosphorous, Halogens\n\nBy sodium fusion extract we can detect Sulphur, Nitrogen, Phosphorous and Halogens, because they are converted in to their ionic form with sodium metal. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1974, "subject": "Chemistry", "question": "The addition of silica during the extraction of copper from its sulphide ore :-", "options": [ { "text": "converts copper sulphide into copper silicate" }, { "text": "converts iron oxide into iron silicate" }, { "text": "reduces copper sulphide into metallic copper" }, { "text": "reduces the melting point of the reaction mixture" } ], "answer": "converts iron oxide into iron silicate", "solution": "**Answer:** converts iron oxide into iron silicate\n\nSilica is used to remove FeO impurity from the ore of copper

    $$FeO + Si{O_2} \\to \\mathop {FeSi{O_3}}\\limits_{iron\\,silicate\\,(Slag)} $$", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 1975, "subject": "Chemistry", "question": "Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).

    Assertion (A) : Aluminium is extracted from bauxite by the electrolysis of molten mixture of Al2O3 with cryolite.

    Reason (R) : The oxidation state of Al in cryolite is +3.

    In the light of the above statements, choose the most appropriate answer from the options given below :", "options": [ { "text": "(A) is true but (R) is false." }, { "text": "(A) is false but (R) is true." }, { "text": "Both (A) and (R) are correct and (R) is the correct explanation of (A)" }, { "text": "Both (A) and (R) are correct but (R) is not the correct explanation of (A)" } ], "answer": "Both (A) and (R) are correct but (R) is not the correct explanation of (A)", "solution": "**Answer:** Both (A) and (R) are correct but (R) is not the correct explanation of (A)\n\n(A) Aluminium is reactive metal so Aluminium is extracted by electrolysis of Alumina with molten mixture of Cryolite

    (B) Cryolite, Na3AlF6

    Here, Al is in +3 O.S.

    So, answer is (d).", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1976, "subject": "Chemistry", "question": "

    In isolation of which one of the following metals from their ores, the use of cyanide salt is not commonly involved?

    ", "options": [ { "text": "Zinc" }, { "text": "Gold" }, { "text": "Silver" }, { "text": "Copper" } ], "answer": "Copper", "solution": "**Answer:** Copper\n\nIn the extraction of Silver and Gold, NaCN is used\nto leach the metal.

    \n$$4\\text{Au(s)} +8C\\overline{N} (aq)+2H_{2}O+O_{2}(g) $$

    \n     $$ \\to 4\\left[ Au(CN)_{2}\\right]^{-} (aq)+4OH^{-}(aq)$$

    \n$$\n2\\left[\\mathrm{Au}(\\mathrm{CN})_{2}\\right]^{-}(\\mathrm{aq})+\\mathrm{Zn} \\rightarrow 2 \\mathrm{Au}(\\mathrm{s})+\\left[\\left(\\mathrm{Zn}(\\mathrm{CN})_{4}\\right)\\right]^{2-}(\\mathrm{aq})\n$$

    \nIn case of ore containing ZnS and PbS, the depressant used is NaCN.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1977, "subject": "Chemistry", "question": "

    In the metallurgical extraction of copper, following reaction is used :

    \n

    FeO + SiO2 $$\\to$$ FeSiO3

    \n

    FeO and FeSiO3 respectively are.

    ", "options": [ { "text": "gangue and flux." }, { "text": "flux and slag." }, { "text": "slag and flux." }, { "text": "gangue and slag." } ], "answer": "gangue and slag.", "solution": "**Answer:** gangue and slag.\n\nFeO + SiO2 → FeSiO3

    \nFeO = Gangue and FeSiO3 = Slag", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1978, "subject": "Chemistry", "question": "

    Given are two statements one is labelled as Assertion A and other is labelled as Reason R.

    \n

    Assertion A : Magnesium can reduce Al2O3 at a temperature below 1350$$^\\circ$$C, while above 1350$$^\\circ$$C, while above 1350$$^\\circ$$C aluminium can reduce MgO.

    \n

    Reason R : The melting and boiling points of magnesium are lower than those of aluminium.

    \n

    n light of the above statements, choose most appropriate answer from the options given below :

    ", "options": [ { "text": "Both A and R are correct, and R is correct explanation of A." }, { "text": "Both A and R are correct, but R is NOT the correct explanation of A." }, { "text": "A is correct R is not correct." }, { "text": "A is not correct, R is correct." } ], "answer": "Both A and R are correct, but R is NOT the correct explanation of A.", "solution": "**Answer:** Both A and R are correct, but R is NOT the correct explanation of A.\n\nMagnesium can reduce Al2O3 at a temperature\nbelow 1350°C while above 1350°C aluminium can\nreduce MgO because below 1350°C $$\\Delta $$G of MgO\n(formation) is more negative and above 1350°C $$\\Delta $$G\nof Al2O3 (formation) is more negative.

    \nThe melting and boiling point of magnesium are\nlower than those of aluminium but it is not the\ncorrect reason.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1979, "subject": "Chemistry", "question": "

    Which of the following chemical reactions represents Hall-Heroult Process?

    ", "options": [ { "text": "Cr2O3 + 2Al $$\\to$$ Al2O3 + 2Cr" }, { "text": "2Al2O3 + 3C $$\\to$$ 4Al + 3CO2" }, { "text": "FeO + CO $$\\to$$ Fe + CO2" }, { "text": "2[Au(CN)2]$$_{(aq)}^ - $$ + Zn(s) $$\\to$$ 2Au(s) + [Zn(CN4)]2$$-$$" } ], "answer": "2Al2O3 + 3C $$\\to$$ 4Al + 3CO2", "solution": "**Answer:** 2Al2O3 + 3C $$\\to$$ 4Al + 3CO2\n\nHall-Herault process is used for the extraction of aluminium by electrolysis molten $$\\mathrm{Al}_{2} \\mathrm{O}_{3}$$\n

    \n$$\n2 \\mathrm{Al}_{2} \\mathrm{O}_{3}+3 \\mathrm{C} \\rightarrow 4 \\mathrm{Al}+3 \\mathrm{CO}_{2}\n$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1980, "subject": "Chemistry", "question": "

    The compound(s) that is (are) removed as slag during the extraction of copper is :

    \n

    (A) CaO

    \n

    (B) FeO

    \n

    (C) Al2O3

    \n

    (D) ZnO

    \n

    (E) NiO

    \n

    Choose the correct answer from the options given below:

    ", "options": [ { "text": "(C), (D) only" }, { "text": "(A), (B), (E) only" }, { "text": "(A), (B) only" }, { "text": "(B) only" } ], "answer": "(B) only", "solution": "**Answer:** (B) only\n\nThe compound(s) that are removed as a slag during the extraction of copper is :\n

    \n$$\n\\mathrm{FeS} \\stackrel{\\mathrm{O}_2 / \\mathrm{SiO}_2}{\\longrightarrow} \\underset{\\text { slag }}{\\mathrm{FeSiO}_3}+\\mathrm{SO}_2\n$$\n

    \n$\\therefore$ Only iron oxide (FeO) formed slag during extraction of copper.", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 1981, "subject": "Chemistry", "question": "

    Given below are two statements.

    \n

    Statement I : Pig iron is obtained by heating cast iron with scrap iron.

    \n

    Statement II : Pig iron has a relatively lower carbon content than that of cast iron.

    \n

    In the light of the above statements, choose the correct answer from the options given below.

    ", "options": [ { "text": "Both Statement I and Statement II are correct." }, { "text": "Both Statement I and Statement II are not correct." }, { "text": "Statement I is correct but Statement II is not correct." }, { "text": "Statement I is not correct but Statement II is correct." } ], "answer": "Both Statement I and Statement II are not correct.", "solution": "**Answer:** Both Statement I and Statement II are not correct.\n\nCast iron is made by melting pig iron with scrap iron\nand coke using hot air blast.

    \nHence Statement-I is incorrect

    \nBut Pig iron has relatively more carbon content

    \nHence statement-II is incorrect", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1982, "subject": "Chemistry", "question": "

    Refining using liquation method is the most suitable for metals with :

    ", "options": [ { "text": "Low melting point" }, { "text": "High boiling point" }, { "text": "High electrical conductivity" }, { "text": "Less tendency to be soluble in melts than impurities" } ], "answer": "Low melting point", "solution": "**Answer:** Low melting point\n\n Refining using liquation method is the most suitable\nfor metals with low melting point", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1983, "subject": "Chemistry", "question": "

    The metal that is not extracted from its sulfide ore is :

    ", "options": [ { "text": "Aluminium" }, { "text": "Iron" }, { "text": "Lead" }, { "text": "Zinc" } ], "answer": "Aluminium", "solution": "**Answer:** Aluminium\n\nAluminium is not extracted from sulphide ore. It is\nusually extracted from bauxite ore, leaching of\nbauxite ore is done followed by electrolytic\nreduction.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1984, "subject": "Chemistry", "question": "Which one of the following statements is incorrect?", "options": [ { "text": "Cast iron is obtained by melting pig iron with scrap iron and coke using hot air blast." }, { "text": "The malleable iron is prepared from cast iron by oxidising impurities in a reverberatory furnace." }, { "text": "van Arkel method is used to purify tungsten." }, { "text": "Boron and Indium can be purified by zone refining method." } ], "answer": "van Arkel method is used to purify tungsten.", "solution": "**Answer:** van Arkel method is used to purify tungsten.\n\n

    Option A: Cast iron is obtained by melting pig iron (crude iron) with scrap iron and coke (a fuel made from coal) using a hot air blast. The resulting molten metal is poured into molds to form cast iron products.

    Option B: Wrought Iron or Malleable iron is produced from cast iron by heating it in a reverberatory furnace lined with Haematite, which removes impurities through oxidation. The resulting product is malleable iron, which is a type of iron that can be shaped and molded.

    \n

    Option C: van Arkel Method for Refining Zirconium or Titanium: This method is very useful for removing all the oxygen and nitrogen present in the form of impurity in certain metals like $\\mathrm{Zr}$ and Ti. The crude metal is heated in an evacuated vessel with iodine. The metal iodide being more covalent, volatilises:

    \n

    $$\n\\mathrm{Zr}+2 \\mathrm{I}_2 \\rightarrow \\mathrm{ZrI}_4\n$$

    \n

    The metal iodide is decomposed on a tungsten filament, electrically heated to about $1800 \\mathrm{~K}$. The pure metal deposits on the filament.

    \n

    $$\n\\mathrm{ZrI}_4 \\rightarrow \\mathrm{Zr}+2 \\mathrm{I}_2\n$$

    \n

    Option D: Zone refining is a method used to purify certain metals such as boron, indium, germanium, silicon, and galium. It involves heating the metal to a liquid state and then slowly moving it along a column to allow impurities to migrate to one end of the column, where they can be removed.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1985, "subject": "Chemistry", "question": "

    In the extraction of copper, its sulphide ore is heated in a reverberatory furnace after mixing with silica to:

    ", "options": [ { "text": "remove $$\\mathrm{FeO}$$ as $$\\mathrm{FeSiO}_{3}$$" }, { "text": "remove calcium as $$\\mathrm{CaSiO}_{3}$$" }, { "text": "decrease the temperature needed for roasting of $$\\mathrm{Cu}_{2} \\mathrm{~S}$$" }, { "text": "separate $$\\mathrm{CuO}$$ as $$\\mathrm{CuSiO}_{3}$$" } ], "answer": "remove $$\\mathrm{FeO}$$ as $$\\mathrm{FeSiO}_{3}$$", "solution": "**Answer:** remove $$\\mathrm{FeO}$$ as $$\\mathrm{FeSiO}_{3}$$\n\n

    $$\\mathrm{\\mathop {FeO}\\limits_{Basic} + \\mathop {Si{O_2}}\\limits_{Acidic} \\buildrel {} \\over\n \\longrightarrow \\mathop {FeSi{O_3}}\\limits_{(Slag)}}$$

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1986, "subject": "Chemistry", "question": "

    Which one of the following reactions does not occur during extraction of copper?

    ", "options": [ { "text": "$$\\mathrm{2FeS + 3{O_2} \\to 2FeO + 2S{O_2}}$$" }, { "text": "$$\\mathrm{FeO + Si{O_2} \\to FeSi{O_3}}$$" }, { "text": "$$\\mathrm{CaO + Si{O_2} \\to CaSi{O_3}}$$" }, { "text": "$$\\mathrm{2C{u_2}s + 3{O_2} \\to 2C{u_2}O + 2S{O_2}}$$" } ], "answer": "$$\\mathrm{CaO + Si{O_2} \\to CaSi{O_3}}$$", "solution": "**Answer:** $$\\mathrm{CaO + Si{O_2} \\to CaSi{O_3}}$$\n\n$$\n\\begin{aligned}\n& \\mathrm{CuFeS}_2+\\mathrm{O}_2 \\stackrel{\\text { Partial roasting }}{\\longrightarrow} \\\\\\\\\n& \\mathrm{Cu}_2 \\mathrm{~S}+\\mathrm{FeO}+\\mathrm{SO}_2+\\underset{\\text { very small }}{\\mathrm{FeS}}+\\underset{\\text { very small }}{\\mathrm{Cu}_2 \\mathrm{O}} \\\\\\\\\n& \\mathrm{Cu}_2 \\mathrm{~S}+\\mathrm{O}_2 \\rightarrow \\mathrm{Cu}_2 \\mathrm{O}+\\mathrm{SO}_2 \\\\\\\\\n& \\mathrm{FeS}+\\mathrm{O}_2 \\rightarrow \\mathrm{FeO}+\\mathrm{SO}_2 \\\\\\\\\n& \\mathrm{FeO}+\\mathrm{SiO}_2 \\rightarrow \\mathrm{FeSiO}_3\n\\end{aligned}\n$$

    \nNo formation of calcium silicate $\\left(\\mathrm{CaSiO}_3\\right)$ in extraction of $\\mathrm{Cu}$.", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 1987, "subject": "Chemistry", "question": "

    The metal which is extracted by oxidation and subsequent reduction from its ore is :

    ", "options": [ { "text": "Al" }, { "text": "Cu" }, { "text": "Ag" }, { "text": "Fe" } ], "answer": "Ag", "solution": "**Answer:** Ag\n\nAg is first extracted by oxidation and then the subsequent reduction is carried out to obtain\n

    \n$$\n\\begin{aligned}\n& \\begin{aligned}\n4 \\mathrm{Ag}+8 \\mathrm{NaCN}+2 \\mathrm{H}_{2} \\mathrm{O}+\\mathrm{O}_{2}\n\\end{aligned} \\stackrel{\\text { Oxidation }}{\\longrightarrow} 4 \\mathrm{Na}\\left[\\mathrm{Ag}(\\mathrm{CN})_{2}\\right]+4 \\mathrm{NaOH} \\\\\\\\\n& 2 \\mathrm{Na}\\left[\\mathrm{Ag}(\\mathrm{CN})_{2}\\right]+\\mathrm{Zn} \\stackrel{\\text { reduction }}{\\longrightarrow} \\mathrm{Na}_{2}\\left[\\mathrm{Zn}(\\mathrm{CN})_{4}\\right]+2 \\mathrm{Ag}\n\\end{aligned}\n$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1988, "subject": "Chemistry", "question": "

    Match List I with List II

    \n

    \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    List-I
    		List-II
    A.Reverberatory furnaceI.Pig Iron
    B.Electrolytic cellII.Aluminium
    C.Blast furnaceIII.Silicon
    D.Zone Refining furnaceIV.Copper

    \n

    Choose the correct answer from the options given below :

    ", "options": [ { "text": "A-III, B-IV, C-I, D-II" }, { "text": "A-IV, B-II, C-I, D-III" }, { "text": "A-I, B-IV, C-II, D-III" }, { "text": "A-I, B-III, C-II, D-IV" } ], "answer": "A-IV, B-II, C-I, D-III", "solution": "**Answer:** A-IV, B-II, C-I, D-III\n\n(A) Reverberatory furnance is used for extraction of copper.\n

    \n(B) Electrolytic cell is used for obtaining highly reactive metals like aluminium.\n

    \n(C) Blast furnace is used for pig iron.\n

    \n(D) Zone refining is used for purification of silicon.\n", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1989, "subject": "Chemistry", "question": "

    In the extraction process of copper, the product obtained after carrying out the reactions

    \n

    (i) $$2 \\mathrm{Cu}_{2} \\mathrm{~S}+3 \\mathrm{O}_{2} \\rightarrow 2 \\mathrm{Cu}_{2} \\mathrm{O}+2 \\mathrm{SO}_{2}$$

    \n

    (ii) $$2 \\mathrm{Cu}_{2} \\mathrm{O}+\\mathrm{Cu}_{2} \\mathrm{~S} \\rightarrow 6 \\mathrm{Cu}+\\mathrm{SO}_{2}$$ is called

    ", "options": [ { "text": "Copper matte" }, { "text": "Blister copper" }, { "text": "Reduced copper" }, { "text": "Copper scrap" } ], "answer": "Blister copper", "solution": "**Answer:** Blister copper\n\nThe sequence of reactions stated here is part of the copper extraction process, specifically the smelting and conversion stages. Initially, copper sulfide (Cu$_2$S) reacts with oxygen (O$_2$) to form copper(I) oxide (Cu$_2$O) and sulfur dioxide (SO$_2$), according to the equation :\n\n

    $$2 \\mathrm{Cu}_2 \\mathrm{S} + 3 \\mathrm{O}_2 \\rightarrow 2 \\mathrm{Cu}_2 \\mathrm{O} + 2 \\mathrm{SO}_2.$$\n\n

    Subsequently, the resulting copper(I) oxide then reacts with remaining copper sulfide to yield copper (Cu) and additional sulfur dioxide :\n\n

    $$2 \\mathrm{Cu}_2 \\mathrm{O} + \\mathrm{Cu}_2 \\mathrm{S} \\rightarrow 6 \\mathrm{Cu} + \\mathrm{SO}_2.$$\n\n

    The output of this process, which contains about 98-99% copper, is known as \"blister copper\". This name derives from the blister-like appearance of the copper surface, which results from the release of sulfur dioxide gas during the reaction.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1990, "subject": "Chemistry", "question": "

    In Hall - Heroult process, the following is used for reducing $$\\mathrm{Al}_{2} \\mathrm{O}_{3}$$ :-

    ", "options": [ { "text": "Magnesium" }, { "text": "$$\\mathrm{CaF}_{2}$$" }, { "text": "Graphite" }, { "text": "$$\\mathrm{Na}_{3} \\mathrm{AlF}_{6}$$" } ], "answer": "Graphite", "solution": "**Answer:** Graphite\n\n

    In the Hall-Héroult process, graphite anodes are used to reduce alumina (Al₂O₃). During the electrolytic reduction, Al³⁺ ions are reduced at the cathode to form aluminum, while at the anode, oxide ions are oxidized to form oxygen gas. The oxygen gas then reacts with the carbon in the graphite anodes to produce carbon dioxide. This means that graphite is consumed during the process, and the anodes have to be replaced periodically.

    \n

    So, in the context of the question, graphite (Option C) is the correct answer, as it is the material of the anode used for the electrolytic reduction of Al₂O₃ in the Hall-Héroult process.

    \n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1991, "subject": "Chemistry", "question": "

    Which of the following metals can be extracted through alkali leaching technique?

    ", "options": [ { "text": "$$\\mathrm{Sn}$$" }, { "text": "Cu" }, { "text": "$$\\mathrm{Au}$$" }, { "text": "$$\\mathrm{Pb}$$" } ], "answer": "$$\\mathrm{Au}$$", "solution": "**Answer:** $$\\mathrm{Au}$$\n\n

    The alkali leaching technique is a method to extract metals from their ores by using an alkaline solution, typically a sodium hydroxide (NaOH) solution, sometimes in the presence of other substances like a cyanide for gold and silver ores. Among the options given, the metal that can be extracted through alkali leaching technique is:

    \n\n

    Option C: $$\\mathrm{Au}$$ (Gold)

    \n\n

    Gold can be extracted using a process known as cyanidation or cyanide leaching. In this process, finely ground gold-bearing ore is treated with a dilute solution of sodium cyanide (NaCN) or potassium cyanide (KCN) in the presence of oxygen to dissolve the gold. The chemical reaction for the dissolution of gold in the cyanide solution is given by:

    \n\n$$\n\\mathrm{4Au + 8NaCN + O_2 + 2H_2O \\rightarrow 4Na[Au(CN)_2] + 4NaOH}\n$$\n\n

    The gold is then recovered from the solution by various methods, such as adsorption onto activated carbon, followed by elution and eventually electro-winning or by precipitation with zinc dust (the Merrill-Crowe process). Because of environmental and safety concerns, alternatives to cyanide leaching have been researched, yet cyanidation remains the primary method for the recovery of gold from its ores.

    \n\n

    As for the other options:

    \n\n

    Option A: $$\\mathrm{Sn}$$ (Tin) is generally extracted through pyrometallurgical processes such as smelting, rather than alkali leaching.

    \n\n

    Option B: Copper (Cu) is often extracted using hydrometallurgical processes like sulfuric acid leaching rather than alkali leaching. Copper can also be extracted using the process called solvent extraction and electrowinning (SX-EW) after the leaching step.

    \n\n

    Option D: $$\\mathrm{Pb}$$ (Lead) is typically extracted using pyrometallurgical methods, including smelting and refining, and not alkali leaching.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1992, "subject": "Chemistry", "question": "Match the ores (column A) with the metals (column B) : \n

    \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    (Column A) ores(Column B) Metals
    (I)Siderite(a)Zinc
    (II)Kaolinite(b)Copper
    (III)Malachite(c)Iron
    (IV)Calamine(d)Aluminium
    ", "options": [ { "text": "(I)-(c); (II)-(d); (III)-(b); (IV)-(a)" }, { "text": "(I)-(a); (II)-(b); (III)-(c); (IV)-(d)" }, { "text": "(I)-(b); (II)-(c); (III)-(d); (IV)-(a)" }, { "text": "(I)-(c); (II)-(d); (III)-(a); (IV)-(b)" } ], "answer": "(I)-(c); (II)-(d); (III)-(b); (IV)-(a)", "solution": "**Answer:** (I)-(c); (II)-(d); (III)-(b); (IV)-(a)\n\nSiderite : FeCO3\n

    Kaolinite : Al2(OH)4Si2O5\n

    Malachite : CuCO3\n.Cu(OH)2\n

    Calamine : ZnCO3\n

    Note : ( Remember all those ores names. Any one of those can be asked in the exam.)\n

    Oxides Ores :\n

    (1) ZnO - Zincite

    (2) Fe2O3 - Haematite

    (3) Fe3O4 - Magnetite (FeO + Fe2O3 mixture)

    (4) Fe2O3 . 3H2O - Limonite

    (5) MnO2 - Pyrolusite

    (6) Cu2O - Cuprite or Ruby Copper

    (7) TiO2 - Rutile

    (8) FeCr2O4 - Chromite (FeO + Cr2O3)

    (9) FeTiO3 - Illmenite (FeO + TiO2)

    (10) Na2B4O7 . 10H2O - Borax or Tincal

    (11) U3O8 - Pitch Blende

    (12) SnO2 - Tin Stone or Cassiterite

    (13) Ca2B6O11 . 5H2O - Colemanite (2 Cao + 3 B2O3)

    (14) Al2O3 . 2H2O - Bauxite

    (15) Al2O3 . H2O - Diaspore

    (16) Al2O3 - Corundum\n

    Sulphides Ores :\n

    (1) ZnS - Zinc Blende or Sphalerite

    (2) PbS - Galena

    (3) Ag2S - Argentite or Silver Glance

    (4) HgS - Cinnabar

    (5) Cu2S - Chalcocite or Copper glance

    (6) CuFeS2 - Copper pyrites or Chalco pyrites (Cu2S + Fe2S3 mixture)

    (7) FeS2 - Iron pyrites or Fool's Gold

    (8) 3Ag2S . Sb2S2 - Pyrargyrite or ruby silver\n

    Halides Ores :

    (1) NaCl - Rock Salt

    (2) KCl - Sylvine

    (3) Na3AlF6 - Cryolite [3NaF + AlF6]

    (4) CaF2 - Fluorspar

    (5) KCl . MgCl2 . 6H2O - Carnalite

    (6) AgCl - Horn Silver\n

    Carbonates Ores :

    (1) CaCO3 - Limestone

    (2) MgCO3 - Magnesite

    (3) CaCO3 . MgCO3 - Dolomite

    (4) ZnCO3 - Calamine

    (5) PbCO3 - Cerrusite

    (6) FeCO3 - Siderite

    (7) CuCO3 . Cu(OH)2 or Cu2CO3(OH)2 - Malachite or Basic Copper Carbonates

    (8) 2 CuCO3 . Cu(OH)2 - Azurite\n

    Sulphates Ores :

    (1) CuSO4 . 2H2O - Gypsum

    (2) MgSO4 . 7H2O - Epson Salt

    (3) Na2SO4 . 10 H2O - Glauber's Salt

    (4) PbSO4 - Anglesite

    (5) ZnSO4 . 7H2O - White Vitriol

    (6) FeSO4 . 7H2O - Green Vitriol

    (7) CuSO4 . 5H2O - Blue Vitriol or Chalcanthite

    Nitrate Ores :

    (1) KNO3 - Indian Saltpetre

    (2) NaNO3 - Chile Saltpetre

    Arsenides Ores :

    (1) NiAs - Kupfernickel

    (2) NiAsS - Nickel glance", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 1993, "subject": "Chemistry", "question": "The ore that contains the metal in the form of\nfluoride is :", "options": [ { "text": "sphalerite" }, { "text": "magnetite" }, { "text": "malachite" }, { "text": "cryolite" } ], "answer": "cryolite", "solution": "**Answer:** cryolite\n\nMagnetite $$ \\to $$ Fe3O4\n

    Sphalerite $$ \\to $$ ZnS\n

    Cryolite $$ \\to $$ Na3AlF6\n

    Malachite $$ \\to $$ CuCO3. Cu(OH)2", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1994, "subject": "Chemistry", "question": "The one that is not a carbonate ore is :", "options": [ { "text": "siderite" }, { "text": "bauxite" }, { "text": "malachite" }, { "text": "calamine" } ], "answer": "bauxite", "solution": "**Answer:** bauxite\n\nBauxite = AlOx. (OH)3–2x\n

    Calamine = ZnCO3\n

    Siderite = FeCO3\n

    Malachite = CuCO3. Cu(OH)2", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1995, "subject": "Chemistry", "question": "The ore that contains both iron and copper is : ", "options": [ { "text": "Copper Pyrites " }, { "text": "Malachite " }, { "text": "Dolomite " }, { "text": "Azurite" } ], "answer": "Copper Pyrites ", "solution": "**Answer:** Copper Pyrites \n\n\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n
    Copper pyrites:CuFeS2
    Malachite:CuCO3 . Cu(OH)2
    Azurite:2CuCO3 . Cu(OH)2
    Dolomite:CaCO3 . MgCO3
    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1996, "subject": "Chemistry", "question": "The incorrect statement is :", "options": [ { "text": "bronze is an alloy of copper and tin" }, { "text": "cast iron is used to manufacture wrought\niron" }, { "text": "german silver is an alloy of zinc, copper\nand nickel" }, { "text": "brass is an alloy of copper and nickel" } ], "answer": "brass is an alloy of copper and nickel", "solution": "**Answer:** brass is an alloy of copper and nickel\n\nBronze contains 75 – 90% Cu and 10 – 25% Sn\nCast iron is used to make wraugut iron.\n

    German silver contains 56% Cu, 24% Zn and\n20% Ni.\n

    Brass contains 60% Cu and 40% Zn.\n

    Note :

    \n\n\n \n \n \n \n\n\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\n
    AlloysComposition
    Brass60% Cu + 40% Zn
    Bronze75-90% Cu + 10-25% Sn
    Gun-metal90% Cu + 8% Sn + 2% Zn
    Aluminium bronze90% Cu + 10% Al
    Monel metal30% Cu + 67% Ni + 3% (Fe + Mn)
    German Silver56% Cu + 24% Zn + 20% Ni
    Manganin78-86% Cu + 13-18% Mn + 1-4% Ni
    Bell metal80% Cu + 20% Sn
    Constantan60% Cu + 40% Ni
    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1997, "subject": "Chemistry", "question": "The major components in \"Gun Metal” are :", "options": [ { "text": "Al, Cu, Mg and Mn" }, { "text": "Cu, Ni and Fe" }, { "text": "Cu, Sn and Zn" }, { "text": "Cu, Zn and Ni" } ], "answer": "Cu, Sn and Zn", "solution": "**Answer:** Cu, Sn and Zn\n\nComposition of Gun metal $$ \\to $$ 90% Cu + 8% Sn + 2% Zn

    Note :

    \n\n\n \n \n \n \n\n\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\n
    AlloysComposition
    Brass60% Cu + 40% Zn
    Bronze75-90% Cu + 10-25% Sn
    Gun-metal90% Cu + 8% Sn + 2% Zn
    Aluminium bronze90% Cu + 10% Al
    Monel metal30% Cu + 67% Ni + 3% (Fe + Mn)
    German Silver56% Cu + 24% Zn + 20% Ni
    Manganin78-86% Cu + 13-18% Mn + 1-4% Ni
    Bell metal80% Cu + 20% Sn
    Constantan60% Cu + 40% Ni
    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 1998, "subject": "Chemistry", "question": "Which of the following ore is concentrated using group 1 cyanide salt ?", "options": [ { "text": "Sphalerite" }, { "text": "Malachite" }, { "text": "Siderite" }, { "text": "Calamine" } ], "answer": "Sphalerite", "solution": "**Answer:** Sphalerite\n\n

    The chemical formulas of given ore in options are as follows

    \n

    Sphalerite ore : ZnS

    \n

    Calamine ore : ZnCO3

    \n

    Siderite ore : FeCO3

    \n

    Malachite ore : Cu(OH)2.CuCO3

    \n

    Sphalerite ore containing ZnS is concentrated using group is cyanide\nsalt (NaCN) as a depressant by froth flotation method.

    \n
    Note : ( Remember all those ores names. Any one of those can be asked in the exam.)\n

    Oxides Ores :\n

    (1) ZnO - Zincite

    (2) Fe2O3 - Haematite

    (3) Fe3O4 - Magnetite (FeO + Fe2O3 mixture)

    (4) Fe2O3 . 3H2O - Limonite

    (5) MnO2 - Pyrolusite

    (6) Cu2O - Cuprite or Ruby Copper

    (7) TiO2 - Rutile

    (8) FeCr2O4 - Chromite (FeO + Cr2O3)

    (9) FeTiO3 - Illmenite (FeO + TiO2)

    (10) Na2B4O7 . 10H2O - Borax or Tincal

    (11) U3O8 - Pitch Blende

    (12) SnO2 - Tin Stone or Cassiterite

    (13) Ca2B6O11 . 5H2O - Colemanite (2 Cao + 3 B2O3)

    (14) Al2O3 . 2H2O - Bauxite

    (15) Al2O3 . H2O - Diaspore

    (16) Al2O3 - Corundum\n

    Sulphides Ores :\n

    (1) ZnS - Zinc Blende or Sphalerite

    (2) PbS - Galena

    (3) Ag2S - Argentite or Silver Glance

    (4) HgS - Cinnabar

    (5) Cu2S - Chalcocite or Copper glance

    (6) CuFeS2 - Copper pyrites or Chalco pyrites (Cu2S + Fe2S3 mixture)

    (7) FeS2 - Iron pyrites or Fool's Gold

    (8) 3Ag2S . Sb2S2 - Pyrargyrite or ruby silver\n

    Halides Ores :

    (1) NaCl - Rock Salt

    (2) KCl - Sylvine

    (3) Na3AlF6 - Cryolite [3NaF + AlF6]

    (4) CaF2 - Fluorspar

    (5) KCl . MgCl2 . 6H2O - Carnalite

    (6) AgCl - Horn Silver\n

    Carbonates Ores :

    (1) CaCO3 - Limestone

    (2) MgCO3 - Magnesite

    (3) CaCO3 . MgCO3 - Dolomite

    (4) ZnCO3 - Calamine

    (5) PbCO3 - Cerrusite

    (6) FeCO3 - Siderite

    (7) CuCO3 . Cu(OH)2 or Cu2CO3(OH)2 - Malachite or Basic Copper Carbonates

    (8) 2 CuCO3 . Cu(OH)2 - Azurite\n

    Sulphates Ores :

    (1) CuSO4 . 2H2O - Gypsum

    (2) MgSO4 . 7H2O - Epson Salt

    (3) Na2SO4 . 10 H2O - Glauber's Salt

    (4) PbSO4 - Anglesite

    (5) ZnSO4 . 7H2O - White Vitriol

    (6) FeSO4 . 7H2O - Green Vitriol

    (7) CuSO4 . 5H2O - Blue Vitriol or Chalcanthite

    Nitrate Ores :

    (1) KNO3 - Indian Saltpetre

    (2) NaNO3 - Chile Saltpetre

    Arsenides Ores :

    (1) NiAs - Kupfernickel

    (2) NiAsS - Nickel glance", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 1999, "subject": "Chemistry", "question": "Math List - I with List - II.

    \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    List I (Metal)List II (Ores)
    (a)Aluminium(i)Siderite
    (b)Iron(ii)Calamine
    (c)Copper(iii)Kaolinite
    (d)Zinc(iv)Malachite


    Choose the correct answer from the options given below :", "options": [ { "text": "(a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)" }, { "text": "(a)-(i), (b)-(ii), (c)-(iii), (d)-(iv)" }, { "text": "(a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)" }, { "text": "(a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)" } ], "answer": "(a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)", "solution": "**Answer:** (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)\n\nSiderite $$ \\to $$ FeCO3\n

    Calamine $$ \\to $$ ZnCO3\n

    Kaolinite $$ \\to $$ Si2Al2O5(OH)4 or Al2O3.2SiO2.2H2O\n

    Malachite $$ \\to $$ CuCO3.Cu(OH)2\n

    Note : ( Remember all those ores names. Any one of those can be asked in the exam.)\n

    Oxides Ores :\n

    (1) ZnO - Zincite

    (2) Fe2O3 - Haematite

    (3) Fe3O4 - Magnetite (FeO + Fe2O3 mixture)

    (4) Fe2O3 . 3H2O - Limonite

    (5) MnO2 - Pyrolusite

    (6) Cu2O - Cuprite or Ruby Copper

    (7) TiO2 - Rutile

    (8) FeCr2O4 - Chromite (FeO + Cr2O3)

    (9) FeTiO3 - Illmenite (FeO + TiO2)

    (10) Na2B4O7 . 10H2O - Borax or Tincal

    (11) U3O8 - Pitch Blende

    (12) SnO2 - Tin Stone or Cassiterite

    (13) Ca2B6O11 . 5H2O - Colemanite (2 Cao + 3 B2O3)

    (14) Al2O3 . 2H2O - Bauxite

    (15) Al2O3 . H2O - Diaspore

    (16) Al2O3 - Corundum\n

    Sulphides Ores :\n

    (1) ZnS - Zinc Blende or Sphalerite

    (2) PbS - Galena

    (3) Ag2S - Argentite or Silver Glance

    (4) HgS - Cinnabar

    (5) Cu2S - Chalcocite or Copper glance

    (6) CuFeS2 - Copper pyrites or Chalco pyrites (Cu2S + Fe2S3 mixture)

    (7) FeS2 - Iron pyrites or Fool's Gold

    (8) 3Ag2S . Sb2S2 - Pyrargyrite or ruby silver\n

    Halides Ores :

    (1) NaCl - Rock Salt

    (2) KCl - Sylvine

    (3) Na3AlF6 - Cryolite [3NaF + AlF6]

    (4) CaF2 - Fluorspar

    (5) KCl . MgCl2 . 6H2O - Carnalite

    (6) AgCl - Horn Silver\n

    Carbonates Ores :

    (1) CaCO3 - Limestone

    (2) MgCO3 - Magnesite

    (3) CaCO3 . MgCO3 - Dolomite

    (4) ZnCO3 - Calamine

    (5) PbCO3 - Cerrusite

    (6) FeCO3 - Siderite

    (7) CuCO3 . Cu(OH)2 or Cu2CO3(OH)2 - Malachite or Basic Copper Carbonates

    (8) 2 CuCO3 . Cu(OH)2 - Azurite\n

    Sulphates Ores :

    (1) CuSO4 . 2H2O - Gypsum

    (2) MgSO4 . 7H2O - Epson Salt

    (3) Na2SO4 . 10 H2O - Glauber's Salt

    (4) PbSO4 - Anglesite

    (5) ZnSO4 . 7H2O - White Vitriol

    (6) FeSO4 . 7H2O - Green Vitriol

    (7) CuSO4 . 5H2O - Blue Vitriol or Chalcanthite

    Nitrate Ores :

    (1) KNO3 - Indian Saltpetre

    (2) NaNO3 - Chile Saltpetre

    Arsenides Ores :

    (1) NiAs - Kupfernickel

    (2) NiAsS - Nickel glance", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2000, "subject": "Chemistry", "question": "The major components of German Silver are :", "options": [ { "text": "Zn, Ni and Ag" }, { "text": "Cu, Zn and Ni" }, { "text": "Cu, Zn and Ag" }, { "text": "Ge, Cu and Ag" } ], "answer": "Cu, Zn and Ni", "solution": "**Answer:** Cu, Zn and Ni\n\nComposition of German Silver $$ \\to $$ 56% Cu + 24% Zn + 20% Ni

    Note :

    \n\n\n \n \n \n \n\n\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\n
    AlloysComposition
    Brass60% Cu + 40% Zn
    Bronze75-90% Cu + 10-25% Sn
    Gun-metal90% Cu + 8% Sn + 2% Zn
    Aluminium bronze90% Cu + 10% Al
    Monel metal30% Cu + 67% Ni + 3% (Fe + Mn)
    German Silver56% Cu + 24% Zn + 20% Ni
    Manganin78-86% Cu + 13-18% Mn + 1-4% Ni
    Bell metal80% Cu + 20% Sn
    Constantan60% Cu + 40% Ni
    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2001, "subject": "Chemistry", "question": "Match List - I with List - II.

    \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    List - I (Ore)List - II (Element Present)
    (a)Kernite(i)Tin
    (b)Cassiterite(ii)Boron
    (c)Calamine(iii)Fluorine
    (d)Cryolite(iv)Zinc


    Choose the most appropriate answer from the options given below :", "options": [ { "text": "(a) $$ \\to $$ (i), (b) $$ \\to $$ (iii), (c) $$ \\to $$ (iv), (d) $$ \\to $$ (ii)" }, { "text": "(a) $$ \\to $$ (iii), (b) $$ \\to $$ (i), (c) $$ \\to $$ (ii), (d) $$ \\to $$ (iv)" }, { "text": "(a) $$ \\to $$ (ii), (b) $$ \\to $$ (i), (c) $$ \\to $$ (iv), (d) $$ \\to $$ (iii)" }, { "text": "(a) $$ \\to $$ (ii), (b) $$ \\to $$ (iv), (c) $$ \\to $$ (i), (d) $$ \\to $$ (iii)" } ], "answer": "(a) $$ \\to $$ (ii), (b) $$ \\to $$ (i), (c) $$ \\to $$ (iv), (d) $$ \\to $$ (iii)", "solution": "**Answer:** (a) $$ \\to $$ (ii), (b) $$ \\to $$ (i), (c) $$ \\to $$ (iv), (d) $$ \\to $$ (iii)\n\nKernite : Na2B4O7.4H2O (Boron)\n

    Cassiterite : SnO2 (Tin)\n

    Calamine : ZnCO3 (Zinc)\n

    Cryolite : Na3AlF6 (Fluorine)\n

    Note : ( Remember all those ores names. Any one of those can be asked in the exam.)\n

    Oxides Ores :\n

    (1) ZnO - Zincite

    (2) Fe2O3 - Haematite

    (3) Fe3O4 - Magnetite (FeO + Fe2O3 mixture)

    (4) Fe2O3 . 3H2O - Limonite

    (5) MnO2 - Pyrolusite

    (6) Cu2O - Cuprite or Ruby Copper

    (7) TiO2 - Rutile

    (8) FeCr2O4 - Chromite (FeO + Cr2O3)

    (9) FeTiO3 - Illmenite (FeO + TiO2)

    (10) Na2B4O7 . 10H2O - Borax or Tincal

    (11) U3O8 - Pitch Blende

    (12) SnO2 - Tin Stone or Cassiterite

    (13) Ca2B6O11 . 5H2O - Colemanite (2 Cao + 3 B2O3)

    (14) Al2O3 . 2H2O - Bauxite

    (15) Al2O3 . H2O - Diaspore

    (16) Al2O3 - Corundum\n

    Sulphides Ores :\n

    (1) ZnS - Zinc Blende or Sphalerite

    (2) PbS - Galena

    (3) Ag2S - Argentite or Silver Glance

    (4) HgS - Cinnabar

    (5) Cu2S - Chalcocite or Copper glance

    (6) CuFeS2 - Copper pyrites or Chalco pyrites (Cu2S + Fe2S3 mixture)

    (7) FeS2 - Iron pyrites or Fool's Gold

    (8) 3Ag2S . Sb2S2 - Pyrargyrite or ruby silver\n

    Halides Ores :

    (1) NaCl - Rock Salt

    (2) KCl - Sylvine

    (3) Na3AlF6 - Cryolite [3NaF + AlF6]

    (4) CaF2 - Fluorspar

    (5) KCl . MgCl2 . 6H2O - Carnalite

    (6) AgCl - Horn Silver\n

    Carbonates Ores :

    (1) CaCO3 - Limestone

    (2) MgCO3 - Magnesite

    (3) CaCO3 . MgCO3 - Dolomite

    (4) ZnCO3 - Calamine

    (5) PbCO3 - Cerrusite

    (6) FeCO3 - Siderite

    (7) CuCO3 . Cu(OH)2 or Cu2CO3(OH)2 - Malachite or Basic Copper Carbonates

    (8) 2 CuCO3 . Cu(OH)2 - Azurite\n

    Sulphates Ores :

    (1) CuSO4 . 2H2O - Gypsum

    (2) MgSO4 . 7H2O - Epson Salt

    (3) Na2SO4 . 10 H2O - Glauber's Salt

    (4) PbSO4 - Anglesite

    (5) ZnSO4 . 7H2O - White Vitriol

    (6) FeSO4 . 7H2O - Green Vitriol

    (7) CuSO4 . 5H2O - Blue Vitriol or Chalcanthite

    Nitrate Ores :

    (1) KNO3 - Indian Saltpetre

    (2) NaNO3 - Chile Saltpetre

    Arsenides Ores :

    (1) NiAs - Kupfernickel

    (2) NiAsS - Nickel glance", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2002, "subject": "Chemistry", "question": "Match List - I with List - II.

    \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    List - I List - II
    (a)Siderite(i)Cu
    (b)Calamine(ii)Ca
    (c)Malachite(iii)Fe
    (d)Cryolite(iv)Al
    (v)Zn


    Choose the correct answer from the options given below :", "options": [ { "text": "(a) $$ \\to $$ (i), (b) $$ \\to $$ (ii), (c) $$ \\to $$ (iii), (d) $$ \\to $$ (iv)" }, { "text": "(a) $$ \\to $$ (iii), (b) $$ \\to $$ (i), (c) $$ \\to $$ (v), (d) $$ \\to $$ (ii)" }, { "text": "(a) $$ \\to $$ (i), (b) $$ \\to $$ (ii), (c) $$ \\to $$ (v), (d) $$ \\to $$ (iii)" }, { "text": "(a) $$ \\to $$ (iii), (b) $$ \\to $$ (v), (c) $$ \\to $$ (i), (d) $$ \\to $$ (iv)" } ], "answer": "(a) $$ \\to $$ (iii), (b) $$ \\to $$ (v), (c) $$ \\to $$ (i), (d) $$ \\to $$ (iv)", "solution": "**Answer:** (a) $$ \\to $$ (iii), (b) $$ \\to $$ (v), (c) $$ \\to $$ (i), (d) $$ \\to $$ (iv)\n\nSiderite $$ \\to $$ FeCO3\n

    Calamine $$ \\to $$ ZnCO3

    Malachite $$ \\to $$ CuCO3.Cu(OH)2

    Cryolite $$ \\to $$ Na3AlF6\n

    Note : ( Remember all those ores names. Any one of those can be asked in the exam.)\n

    Oxides Ores :\n

    (1) ZnO - Zincite

    (2) Fe2O3 - Haematite

    (3) Fe3O4 - Magnetite (FeO + Fe2O3 mixture)

    (4) Fe2O3 . 3H2O - Limonite

    (5) MnO2 - Pyrolusite

    (6) Cu2O - Cuprite or Ruby Copper

    (7) TiO2 - Rutile

    (8) FeCr2O4 - Chromite (FeO + Cr2O3)

    (9) FeTiO3 - Illmenite (FeO + TiO2)

    (10) Na2B4O7 . 10H2O - Borax or Tincal

    (11) U3O8 - Pitch Blende

    (12) SnO2 - Tin Stone or Cassiterite

    (13) Ca2B6O11 . 5H2O - Colemanite (2 Cao + 3 B2O3)

    (14) Al2O3 . 2H2O - Bauxite

    (15) Al2O3 . H2O - Diaspore

    (16) Al2O3 - Corundum\n

    Sulphides Ores :\n

    (1) ZnS - Zinc Blende or Sphalerite

    (2) PbS - Galena

    (3) Ag2S - Argentite or Silver Glance

    (4) HgS - Cinnabar

    (5) Cu2S - Chalcocite or Copper glance

    (6) CuFeS2 - Copper pyrites or Chalco pyrites (Cu2S + Fe2S3 mixture)

    (7) FeS2 - Iron pyrites or Fool's Gold

    (8) 3Ag2S . Sb2S2 - Pyrargyrite or ruby silver\n

    Halides Ores :

    (1) NaCl - Rock Salt

    (2) KCl - Sylvine

    (3) Na3AlF6 - Cryolite [3NaF + AlF6]

    (4) CaF2 - Fluorspar

    (5) KCl . MgCl2 . 6H2O - Carnalite

    (6) AgCl - Horn Silver\n

    Carbonates Ores :

    (1) CaCO3 - Limestone

    (2) MgCO3 - Magnesite

    (3) CaCO3 . MgCO3 - Dolomite

    (4) ZnCO3 - Calamine

    (5) PbCO3 - Cerrusite

    (6) FeCO3 - Siderite

    (7) CuCO3 . Cu(OH)2 or Cu2CO3(OH)2 - Malachite or Basic Copper Carbonates

    (8) 2 CuCO3 . Cu(OH)2 - Azurite\n

    Sulphates Ores :

    (1) CuSO4 . 2H2O - Gypsum

    (2) MgSO4 . 7H2O - Epson Salt

    (3) Na2SO4 . 10 H2O - Glauber's Salt

    (4) PbSO4 - Anglesite

    (5) ZnSO4 . 7H2O - White Vitriol

    (6) FeSO4 . 7H2O - Green Vitriol

    (7) CuSO4 . 5H2O - Blue Vitriol or Chalcanthite

    Nitrate Ores :

    (1) KNO3 - Indian Saltpetre

    (2) NaNO3 - Chile Saltpetre

    Arsenides Ores :

    (1) NiAs - Kupfernickel

    (2) NiAsS - Nickel glance", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2003, "subject": "Chemistry", "question": "Match List - I with List - II :

    \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    List - IList - II
    (a)Haematite(i)$$A{l_2}{O_3}\\,.\\,x{H_2}O$$
    (b)Bauxite(ii)$$F{e_2}{O_3}$$
    (c)Magnetite(iii)$$CuC{O_3}\\,.\\,Cu{(OH)_2}$$
    (d)Malachite(iv)$$F{e_3}{O_4}$$


    Choose the correct answer from the options given below :", "options": [ { "text": "(a)-(i), (b)-(iii), (c)-(ii), (d)-(iv)" }, { "text": "(a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)" }, { "text": "(a)-(ii), (b)-(iii), (c)-(i), (d)-(iv)" }, { "text": "(a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)" } ], "answer": "(a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)", "solution": "**Answer:** (a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)\n\n(a) Haematite $$ \\to $$ (ii) Fe2O3

    \n(b) Bauxite $$ \\to $$ (i) Al2O3.xH2O

    \n(c) Magnetite $$ \\to $$ (iv) Fe3O4

    \n(d) Malachite $$ \\to $$ (iii) CuCO3.Cu(OH)2", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2004, "subject": "Chemistry", "question": "Sulphide ion is soft base and its ores are common for metals.

    (a) Pb (b) Al (c) Ag (d) Mg", "options": [ { "text": "(a) and (c) only" }, { "text": "(a) and (d) only" }, { "text": "(a) and (b) only" }, { "text": "(c) and (d) only" } ], "answer": "(a) and (c) only", "solution": "**Answer:** (a) and (c) only\n\nPb and Ag commonly exist in the form of sulphide ore like Pbs (galena) and Ag2S (Argentite) 'Al' is mainly found in the form of oxide ore whereas 'Mg' is found in the form of halide ore.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2005, "subject": "Chemistry", "question": "Given below are two statements :

    Statement I : Sphalerite is a sulphide or of zinc and copper glance is a sulphide ore of copper.

    Statement II : It is possible to separate two sulphide ores by adjusting proportion of oil to water or by using 'depressants' in a forth flotation method.

    Choose the most appropriate answer from the options given below :", "options": [ { "text": "Statement I is true but Statement II is false." }, { "text": "Both Statement I and Statement II are true." }, { "text": "Statement I is false but Statement II is true." }, { "text": "Both Statement I and Statement II are false." } ], "answer": "Both Statement I and Statement II are true.", "solution": "**Answer:** Both Statement I and Statement II are true.\n\nsphalerite-ZnS, copper glance - Cu2S two sulphide ores can be separated by adjusting proportions of oil to water or by using 'Depressants'.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2006, "subject": "Chemistry", "question": "Match List-I with List-II :

    \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    List - I
    (Name of ore/mineral)    		List - II
    (Chemical formula)
    (a)Calmine(i)$$Zns$$
    (b)Malachite(ii)$$FeC{O_3}$$
    (c)Siderite(iii)$$ZnC{O_3}$$
    (d)Sphalerite(iv)$$CuC{O_3}.Cu{(OH)_2}$$


    Choose the most appropriate answer from the options given below :", "options": [ { "text": "(a)-(iii), (b)-(iv), (c)-(ii), (d)-(i)" }, { "text": "(a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)" }, { "text": "(a)-(iv), (b)-(iii), (c)-(i), (d)-(ii)" }, { "text": "(a)-(iii), (b)-(ii), (c)-(iv), (d)-(i)" } ], "answer": "(a)-(iii), (b)-(iv), (c)-(ii), (d)-(i)", "solution": "**Answer:** (a)-(iii), (b)-(iv), (c)-(ii), (d)-(i)\n\n(Name of ore/mineral)

    (a) Calamine - ZnCO3

    (b) Malachite - CuCO3 . Cu(OH)2

    (c) Siderite - FeCO3

    (d) Sphalerite - ZnS\n

    Note : ( Remember all those ores names. Any one of those can be asked in the exam.)\n

    Oxides Ores :\n

    (1) ZnO - Zincite

    (2) Fe2O3 - Haematite

    (3) Fe3O4 - Magnetite (FeO + Fe2O3 mixture)

    (4) Fe2O3 . 3H2O - Limonite

    (5) MnO2 - Pyrolusite

    (6) Cu2O - Cuprite or Ruby Copper

    (7) TiO2 - Rutile

    (8) FeCr2O4 - Chromite (FeO + Cr2O3)

    (9) FeTiO3 - Illmenite (FeO + TiO2)

    (10) Na2B4O7 . 10H2O - Borax or Tincal

    (11) U3O8 - Pitch Blende

    (12) SnO2 - Tin Stone or Cassiterite

    (13) Ca2B6O11 . 5H2O - Colemanite (2 Cao + 3 B2O3)

    (14) Al2O3 . 2H2O - Bauxite

    (15) Al2O3 . H2O - Diaspore

    (16) Al2O3 - Corundum\n

    Sulphides Ores :\n

    (1) ZnS - Zinc Blende or Sphalerite

    (2) PbS - Galena

    (3) Ag2S - Argentite or Silver Glance

    (4) HgS - Cinnabar

    (5) Cu2S - Chalcocite or Copper glance

    (6) CuFeS2 - Copper pyrites or Chalco pyrites (Cu2S + Fe2S3 mixture)

    (7) FeS2 - Iron pyrites or Fool's Gold

    (8) 3Ag2S . Sb2S2 - Pyrargyrite or ruby silver\n

    Halides Ores :

    (1) NaCl - Rock Salt

    (2) KCl - Sylvine

    (3) Na3AlF6 - Cryolite [3NaF + AlF6]

    (4) CaF2 - Fluorspar

    (5) KCl . MgCl2 . 6H2O - Carnalite

    (6) AgCl - Horn Silver\n

    Carbonates Ores :

    (1) CaCO3 - Limestone

    (2) MgCO3 - Magnesite

    (3) CaCO3 . MgCO3 - Dolomite

    (4) ZnCO3 - Calamine

    (5) PbCO3 - Cerrusite

    (6) FeCO3 - Siderite

    (7) CuCO3 . Cu(OH)2 or Cu2CO3(OH)2 - Malachite or Basic Copper Carbonates

    (8) 2 CuCO3 . Cu(OH)2 - Azurite\n

    Sulphates Ores :

    (1) CuSO4 . 2H2O - Gypsum

    (2) MgSO4 . 7H2O - Epson Salt

    (3) Na2SO4 . 10 H2O - Glauber's Salt

    (4) PbSO4 - Anglesite

    (5) ZnSO4 . 7H2O - White Vitriol

    (6) FeSO4 . 7H2O - Green Vitriol

    (7) CuSO4 . 5H2O - Blue Vitriol or Chalcanthite

    Nitrate Ores :

    (1) KNO3 - Indian Saltpetre

    (2) NaNO3 - Chile Saltpetre

    Arsenides Ores :

    (1) NiAs - Kupfernickel

    (2) NiAsS - Nickel glance", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2007, "subject": "Chemistry", "question": "Calamine and Malachite, respectively, are the ores of : ", "options": [ { "text": "Nickel and Aluminium" }, { "text": "Zinc and Copper" }, { "text": "Copper and Iron" }, { "text": "Aluminium and Zinc" } ], "answer": "Zinc and Copper", "solution": "**Answer:** Zinc and Copper\n\nCalamine $$\\Rightarrow$$ ZnCO3

    Malachite $$\\Rightarrow$$ Cu(OH)2 . CuCO3\n

    Note : ( Remember all those ores names. Any one of those can be asked in the exam.)\n

    Oxides Ores :\n

    (1) ZnO - Zincite

    (2) Fe2O3 - Haematite

    (3) Fe3O4 - Magnetite (FeO + Fe2O3 mixture)

    (4) Fe2O3 . 3H2O - Limonite

    (5) MnO2 - Pyrolusite

    (6) Cu2O - Cuprite or Ruby Copper

    (7) TiO2 - Rutile

    (8) FeCr2O4 - Chromite (FeO + Cr2O3)

    (9) FeTiO3 - Illmenite (FeO + TiO2)

    (10) Na2B4O7 . 10H2O - Borax or Tincal

    (11) U3O8 - Pitch Blende

    (12) SnO2 - Tin Stone or Cassiterite

    (13) Ca2B6O11 . 5H2O - Colemanite (2 Cao + 3 B2O3)

    (14) Al2O3 . 2H2O - Bauxite

    (15) Al2O3 . H2O - Diaspore

    (16) Al2O3 - Corundum\n

    Sulphides Ores :\n

    (1) ZnS - Zinc Blende or Sphalerite

    (2) PbS - Galena

    (3) Ag2S - Argentite or Silver Glance

    (4) HgS - Cinnabar

    (5) Cu2S - Chalcocite or Copper glance

    (6) CuFeS2 - Copper pyrites or Chalco pyrites (Cu2S + Fe2S3 mixture)

    (7) FeS2 - Iron pyrites or Fool's Gold

    (8) 3Ag2S . Sb2S2 - Pyrargyrite or ruby silver\n

    Halides Ores :

    (1) NaCl - Rock Salt

    (2) KCl - Sylvine

    (3) Na3AlF6 - Cryolite [3NaF + AlF6]

    (4) CaF2 - Fluorspar

    (5) KCl . MgCl2 . 6H2O - Carnalite

    (6) AgCl - Horn Silver\n

    Carbonates Ores :

    (1) CaCO3 - Limestone

    (2) MgCO3 - Magnesite

    (3) CaCO3 . MgCO3 - Dolomite

    (4) ZnCO3 - Calamine

    (5) PbCO3 - Cerrusite

    (6) FeCO3 - Siderite

    (7) CuCO3 . Cu(OH)2 or Cu2CO3(OH)2 - Malachite or Basic Copper Carbonates

    (8) 2 CuCO3 . Cu(OH)2 - Azurite\n

    Sulphates Ores :

    (1) CuSO4 . 2H2O - Gypsum

    (2) MgSO4 . 7H2O - Epson Salt

    (3) Na2SO4 . 10 H2O - Glauber's Salt

    (4) PbSO4 - Anglesite

    (5) ZnSO4 . 7H2O - White Vitriol

    (6) FeSO4 . 7H2O - Green Vitriol

    (7) CuSO4 . 5H2O - Blue Vitriol or Chalcanthite

    Nitrate Ores :

    (1) KNO3 - Indian Saltpetre

    (2) NaNO3 - Chile Saltpetre

    Arsenides Ores :

    (1) NiAs - Kupfernickel

    (2) NiAsS - Nickel glance", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2008, "subject": "Chemistry", "question": "

    Match List I with List II.

    \n

    \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    List-I
    Ore    		List-II
    Composition
    (A)Siderite(I)FeCO$$_3$$
    (B)Malachite(II)CuCO$$_3$$ . Cu(OH)$$_2$$
    (C)Sphalerite(III)ZnS
    (D)Calamine(IV)ZnCO$$_3$$

    \n

    Choose the correct answer from the options given below :

    ", "options": [ { "text": "A-I, B-II, C-III, D-IV" }, { "text": "A-III, B-IV, C-II, D-I" }, { "text": "A-IV, B-III, C-I, D-II" }, { "text": "A-I, B-II, C-IV, D-III" } ], "answer": "A-I, B-II, C-III, D-IV", "solution": "**Answer:** A-I, B-II, C-III, D-IV\n\n

    Siderite → FeCO3

    \n

    Malachite → CuCO3. Cu(OH)2

    \n

    Sphalerite → ZnS

    \n

    Calamine → ZnCO3

    \nNote : ( Remember all those ores names. Any one of those can be asked in the exam.)\n

    Oxides Ores :\n

    (1) ZnO - Zincite

    (2) Fe2O3 - Haematite

    (3) Fe3O4 - Magnetite (FeO + Fe2O3 mixture)

    (4) Fe2O3 . 3H2O - Limonite

    (5) MnO2 - Pyrolusite

    (6) Cu2O - Cuprite or Ruby Copper

    (7) TiO2 - Rutile

    (8) FeCr2O4 - Chromite (FeO + Cr2O3)

    (9) FeTiO3 - Illmenite (FeO + TiO2)

    (10) Na2B4O7 . 10H2O - Borax or Tincal

    (11) U3O8 - Pitch Blende

    (12) SnO2 - Tin Stone or Cassiterite

    (13) Ca2B6O11 . 5H2O - Colemanite (2 Cao + 3 B2O3)

    (14) Al2O3 . 2H2O - Bauxite

    (15) Al2O3 . H2O - Diaspore

    (16) Al2O3 - Corundum\n

    Sulphides Ores :\n

    (1) ZnS - Zinc Blende or Sphalerite

    (2) PbS - Galena

    (3) Ag2S - Argentite or Silver Glance

    (4) HgS - Cinnabar

    (5) Cu2S - Chalcocite or Copper glance

    (6) CuFeS2 - Copper pyrites or Chalco pyrites (Cu2S + Fe2S3 mixture)

    (7) FeS2 - Iron pyrites or Fool's Gold

    (8) 3Ag2S . Sb2S2 - Pyrargyrite or ruby silver\n

    Halides Ores :

    (1) NaCl - Rock Salt

    (2) KCl - Sylvine

    (3) Na3AlF6 - Cryolite [3NaF + AlF6]

    (4) CaF2 - Fluorspar

    (5) KCl . MgCl2 . 6H2O - Carnalite

    (6) AgCl - Horn Silver\n

    Carbonates Ores :

    (1) CaCO3 - Limestone

    (2) MgCO3 - Magnesite

    (3) CaCO3 . MgCO3 - Dolomite

    (4) ZnCO3 - Calamine

    (5) PbCO3 - Cerrusite

    (6) FeCO3 - Siderite

    (7) CuCO3 . Cu(OH)2 or Cu2CO3(OH)2 - Malachite or Basic Copper Carbonates

    (8) 2 CuCO3 . Cu(OH)2 - Azurite\n

    Sulphates Ores :

    (1) CuSO4 . 2H2O - Gypsum

    (2) MgSO4 . 7H2O - Epson Salt

    (3) Na2SO4 . 10 H2O - Glauber's Salt

    (4) PbSO4 - Anglesite

    (5) ZnSO4 . 7H2O - White Vitriol

    (6) FeSO4 . 7H2O - Green Vitriol

    (7) CuSO4 . 5H2O - Blue Vitriol or Chalcanthite

    Nitrate Ores :

    (1) KNO3 - Indian Saltpetre

    (2) NaNO3 - Chile Saltpetre

    Arsenides Ores :

    (1) NiAs - Kupfernickel

    (2) NiAsS - Nickel glance", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 2009, "subject": "Chemistry", "question": "

    (a) Baryte, (b) Galena, (c) Zinc blende and (d) Copper pyrites. How many of these minerals are sulphide based?

    ", "options": [], "answer": "3", "solution": "**Answer:** 3\n\nBaryte : $ \\mathrm{BaSO}_{4}$\n

    \nGalena :$\\mathrm{PbS}$\n

    \nZinc blende: $\\mathrm{ZnS}$\n

    \nCopper pyrites : $\\mathrm{CuFeS}_{2}$\n

    \nOf the given minerals, only 3 are sulphide based.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2010, "subject": "Chemistry", "question": "

    Match List - I with List - II :

    \n

    \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    List - IList -II
    (A)Sphalerite(I)FeCO$$_3$$
    (B)Calamine(II)PbS
    (C)Galena(III)ZnCO$$_3$$
    (D)Siderite(IV)ZnS

    \n

    Choose the most appropriate answer from the options given below :

    ", "options": [ { "text": "(A)-(IV), (B)-(III), (C)-(II), (D)-(I)" }, { "text": "(A)-(IV), (B)-(I), (C)-(II), (D)-(III)" }, { "text": "(A)-(II), (B)-(III), (C)-(I), (D)-(IV)" }, { "text": "(A)-(III), (B)-(IV), (C)-(II), (D)-(I)" } ], "answer": "(A)-(IV), (B)-(III), (C)-(II), (D)-(I)", "solution": "**Answer:** (A)-(IV), (B)-(III), (C)-(II), (D)-(I)\n\n\n\n\n\n\n\n\n\n\n \n \n \n \n \n \n\n\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\n
    List - IList -II
    (A)Sphalerite(IV)ZnS
    (B)Calamine(III)ZnCO$$_3$$
    (C)Galena(II)PbS
    (D)Siderite(I)FeCO$$_3$$

    \nHence, the most appropriate option is (A).", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 2011, "subject": "Chemistry", "question": "

    Among the following ores Bauxite, Siderite, Cuprite, Calamine, Haematite, Kaolinite, Malachite, Magnetite, Sphalerite, Limonite, Cryolite, the number of principal ores if iron is _____________.

    ", "options": [], "answer": "4", "solution": "**Answer:** 4\n\nThe principal ores of iron are :

    \nSiderite, Haematite, Magnetite, Limonite.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2012, "subject": "Chemistry", "question": "

    In metallurgy the term \"gangue\" is used for :

    ", "options": [ { "text": "Contamination of undesired earthy materials." }, { "text": "Contamination of metals, other than desired metal." }, { "text": "Minerals which are naturally occurring in pure form" }, { "text": "Magnetic impurities in an ore." } ], "answer": "Contamination of undesired earthy materials.", "solution": "**Answer:** Contamination of undesired earthy materials.\n\nEarthy and undesired materials present in the ore other then the desired metal is known as gangue. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2013, "subject": "Chemistry", "question": "

    The major component of which of the following ore is sulphide based mineral?

    ", "options": [ { "text": "Sphalerite" }, { "text": "Siderite" }, { "text": "Malachite" }, { "text": "Calamine" } ], "answer": "Sphalerite", "solution": "**Answer:** Sphalerite\n\n

    Sphalerite $$\\to$$ $$\\mathrm{ZnS}$$

    \n

    Calamine $$\\to$$ $$\\mathrm{ZnCO_3}$$

    \n

    Malachite $$\\to$$ $$\\mathrm{CuCO_3.Cu(H)_2}$$

    \n

    Siderite $$\\to$$ $$\\mathrm{FeCO_3}$$

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2014, "subject": "Chemistry", "question": "

    The setting time of Cement is increased by adding :

    ", "options": [ { "text": "Gypsum" }, { "text": "Limestone" }, { "text": "Clay" }, { "text": "Silica" } ], "answer": "Gypsum", "solution": "**Answer:** Gypsum\n\nThe setting time of cement is increased by adding gypsum.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2015, "subject": "Chemistry", "question": "During the process of electrolytic refining of copper, some metals present as impurity\nsettle as ‘anode mud’ These are ", "options": [ { "text": "Sn and Ag" }, { "text": "Pb and Zn" }, { "text": "Ag and Au" }, { "text": "Fe and Ni " } ], "answer": "Ag and Au", "solution": "**Answer:** Ag and Au\n\nDuring the process of electrolytic refining Ag and Au are obtained as anode mud. ", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 2016, "subject": "Chemistry", "question": "Which method of purification is represented by the following equation :
    \nTi (s) + 2I2 (g) $$\\buildrel {523K} \\over\n \\longrightarrow $$ TiI4 (g) $$\\buildrel {1700K} \\over\n \\longrightarrow $$ Ti (s) + 2I2 (g)", "options": [ { "text": "zone refining" }, { "text": "cupellation" }, { "text": "Poling" }, { "text": "Van Arkel" } ], "answer": "Van Arkel", "solution": "**Answer:** Van Arkel\n\nVan Arkel is a method in which heat treatment it used to purify metal in this process metals are converted into other metal compound for loosly coupled like as iodine to make metal iodide which are easily decomposed and give pure metal. \n

    The process is known as Van Arkel method. ", "topic": "Algebra", "subtopic": "Equations and Inequalities" }, { "id": 2017, "subject": "Chemistry", "question": "In the context of the Hall-Heroult process for the extraction of Al, which of the following statements is\nfalse? ", "options": [ { "text": "Al2O3 is mixed with CaF2 which lowers the melting point of the mixture and brings conductivity" }, { "text": "Al3+ is reduced at the cathode to form Al" }, { "text": "Na3AlF6 serves as the electrolyte" }, { "text": "CO and CO2 are produced in this process" } ], "answer": "Na3AlF6 serves as the electrolyte", "solution": "**Answer:** Na3AlF6 serves as the electrolyte\n\nIn the metallurgy of aluminium purified $$A{l_2}{O_3}$$ is mixed with $$N{a_3}Al{F_6}\\,\\,$$ $$\\,\\,\\,$$ or $$\\,\\,\\,$$ $$Ca{F_2}$$ which lowers the melting point of the mix and brings conductivity. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2018, "subject": "Chemistry", "question": "The electrolytes usually used in the electroplating of gold and silver, respectively are :", "options": [ { "text": "[Au(CN)2]– and [Ag(CN)2]–" }, { "text": "[Au(CN)2]– and [Ag(Cl)2]–" }, { "text": "[Au(OH)4]– and [Ag(OH)2]–" }, { "text": "[Au(NH3)2]– and [Ag(CN)2]–" } ], "answer": "[Au(CN)2]– and [Ag(CN)2]–", "solution": "**Answer:** [Au(CN)2]– and [Ag(CN)2]–\n\n[Au(CN)2]– and [Ag(CN)2]– both are soluble complexes.\n

    [Au(CN)2]– ⇌ Au+ + 2CN-\n

    [Ag(CN)2]– ⇌ Ag+ + 2CN-\n

    The concentration of produced Au+ ion and Ag+ ion due to ionisation is very less which ensure slow deposition and this slow deposition will generate uniform layer on the substance to be electroplated.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2019, "subject": "Chemistry", "question": "The Mond process is used for the", "options": [ { "text": "Extraction of Zn" }, { "text": "extraction of Mo" }, { "text": "Purification of Zr and Ti" }, { "text": "Purification of Ni" } ], "answer": "Purification of Ni", "solution": "**Answer:** Purification of Ni\n\nMond process is used for purification of Ni.\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2020, "subject": "Chemistry", "question": "Match the refining methods (Column I) with metals (Column II).\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    Column I
    (Refining Methods)    		Column II
    (Metals)
    (I) Liquation(a) Zr
    (II) Zone Refining(b) Ni
    (III) Mond Process(c) Sn
    (IV) Van Arkel Method(d) Ga
    ", "options": [ { "text": "(I)-(c) ; (II)-(a) ; (III)-(b) ; (IV)-(d)" }, { "text": "(I)-(c) ; (II)-(d) ; (III)-(b) ; (IV)-(a)" }, { "text": "(I)-(b) ; (II)-(d) ; (III)-(a) ; (IV)-(c)" }, { "text": "(I)-(b) ; (II)-(c) ; (III)-(d) ; (IV)-(a)" } ], "answer": "(I)-(c) ; (II)-(d) ; (III)-(b) ; (IV)-(a)", "solution": "**Answer:** (I)-(c) ; (II)-(d) ; (III)-(b) ; (IV)-(a)\n\nLiquation is used\nfor Sn.\n

    Zone refining is used for Ga.\n

    Van\nArkel method is used for Zr.\n

    Mond’s process is used for refining of Ni.", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 2021, "subject": "Chemistry", "question": "The correct statement is :", "options": [ { "text": "zincite is a carbonate ore." }, { "text": "Sodium cyanide cannot be used in the metallurgy of silver." }, { "text": "aniline is a froth stabilizer." }, { "text": "zone refining process is used for the refining of titanium. " } ], "answer": "aniline is a froth stabilizer.", "solution": "**Answer:** aniline is a froth stabilizer.\n\nZincite is ZnO.\n

    Sodium cyanide(AgCN) is used in Macarthur-forest Cyanide Process whre Ag is\nleached by dilute solution of NaCN.\n

    Aniline is used as froth stabilizer in Froth flotation process.\n

    Titanium is refined by Van Arkel method.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2022, "subject": "Chemistry", "question": "The refining method used when the metal and the impurities have low and high melting temperatures, respectively, is :", "options": [ { "text": "liquation\n" }, { "text": "vapour phase refining " }, { "text": "distillation" }, { "text": "zone refining " } ], "answer": "liquation\n", "solution": "**Answer:** liquation\n\n\nLiquation method is used when the melting point\nof metal is less compare to the melting point of\nthe associated impurity.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2023, "subject": "Chemistry", "question": "Among statements (a) - (d), the correct ones are\n
    (a) Lime stone is decomposed to CaO during\nthe extraction of iron from its oxides.\n
    (b) In the extraction of silver, silver is extracted\nas an anionic complex.\n
    (c) Nickel is purified by Mond’s process.\n
    (d) Zr and Ti are purified by Van Arkel method.", "options": [ { "text": "(a), (c) and (d) only" }, { "text": "(c) and (d) only" }, { "text": "(b), (c) and (d) only" }, { "text": "(a), (b), (c) and (d)" } ], "answer": "(a), (b), (c) and (d)", "solution": "**Answer:** (a), (b), (c) and (d)\n\n(a) CaCO3 $$\\buildrel \\Delta \\over\n \\longrightarrow $$\n CaO + CO2 {In Blast furnace}\n

    (b) In extraction of silver, silver is extracted as an\nanionic complex [Ag(CN)2]–.\n

    Ag/Ag2S + CN- $$ \\to $$ [Ag(CN)2]\n–\n

    (c) Ni is purified by mond's process\n

    Ni + 4CO $$\\buildrel {330 - 350K} \\over\n \\longrightarrow $$ Ni(CO)4\n

    Ni(CO)4 $$\\buildrel {450 - 470K} \\over\n \\longrightarrow $$ Ni + 4CO\n

    (d) Zr and Ti are purified by van arkel method\n\n", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 2024, "subject": "Chemistry", "question": "The element that can be refined by distillation\nis :", "options": [ { "text": "gallium" }, { "text": "nickel" }, { "text": "zinc" }, { "text": "tin" } ], "answer": "zinc", "solution": "**Answer:** zinc\n\nImpure zinc is refined by distillation.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2025, "subject": "Chemistry", "question": "The method used for the purification of Indium is :", "options": [ { "text": "zone refining" }, { "text": "vapour phase refining" }, { "text": "liquation" }, { "text": "van Arkel method" } ], "answer": "zone refining", "solution": "**Answer:** zone refining\n\nZone refining is used for the purification of\nindium.\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2026, "subject": "Chemistry", "question": "Match List - I with List - II :

    \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    List - IList - II
    (a)Mercury(i)Vapour phase refining
    (b)Copper(ii)Distillation refining
    (c)Silicon(iii)Electrolytic refining
    (d)Nickel(iv)Zone refining


    Choose the most appropriate answer from the option given below :", "options": [ { "text": "(a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)" }, { "text": "(a)-(ii), (b)-(iv), (c)-(iii), (d)-(i)" }, { "text": "(a)-(ii), (b)-(iii), (c)-(i), (d)-(iv)" }, { "text": "(a)-(i), (b)-(iv), (c)-(ii), (d)-(iii)" } ], "answer": "(a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)", "solution": "**Answer:** (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)\n\n(a) Mercury $$ \\to $$ Distillation refining

    \n(b) Copper $$ \\to $$ Electrolytic refining

    \n(c) Silicon $$ \\to $$ Zone refining

    \n(d) Nickel $$ \\to $$ Vapour phase refining", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2027, "subject": "Chemistry", "question": "The metal that can be purified economically by fractional distillation method is :", "options": [ { "text": "Fe" }, { "text": "Zn" }, { "text": "Cu" }, { "text": "Ni" } ], "answer": "Zn", "solution": "**Answer:** Zn\n\nZinc can be purified economically by fractional distillation.\nFractional distillation process utilises the boiling point difference\nbetween metal and that of impurity. Using this process, crude zinc\ncontaining Cd, Fe and Pb as impurities can be refined.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2028, "subject": "Chemistry", "question": "Which refining process is generally used in the purification of low melting metals?", "options": [ { "text": "Chromatographic method" }, { "text": "Liquation" }, { "text": "Electrolysis" }, { "text": "Zone refining" } ], "answer": "Liquation", "solution": "**Answer:** Liquation\n\nLiquation method is used to purify those impure metals which has lower melting point than the melting point of impurities associated.

    $$\\therefore$$ This method is used for metal having low melting point.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2029, "subject": "Chemistry", "question": "In the electrolytic refining of blister copper, the total number of main impurities, from the following, removed as anode mud is ______________

    Pb, Sb, Se, Te, Ru, Ag, Au and Pt", "options": [], "answer": "6", "solution": "**Answer:** 6\n\nAnode mud contains Sb, Se, Te, Ag, Au and Pt", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2030, "subject": "Chemistry", "question": "

    Given below are two statements.

    \n

    Statement I : During electrolytic refining, blister copper deposits precious metals.

    \n

    Statement II : In the process of obtaining pure copper by electrolysis method, copper blister is used to make the anode.

    \n

    In the light of the above statements, choose the correct from the options given below.

    ", "options": [ { "text": "Both Statement I and Statement II are true." }, { "text": "Both Statement I and Statement II are false." }, { "text": "Statement I is true but Statement II is false." }, { "text": "Statement I is false but Statement II is true." } ], "answer": "Both Statement I and Statement II are true.", "solution": "**Answer:** Both Statement I and Statement II are true.\n\nCopper is refined using an electrolytic method.\nAnodes are of impure copper and pure copper\nstrips are taken as cathode.

    \nImpurities from the blister copper deposit as anode\nmud which contains antimony, selenium, tellurium,\nsilver, gold and platinum.

    \nHence both statements are true", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 2031, "subject": "Chemistry", "question": "

    Which of the following methods are not used to refine any metal?

    \n

    A. Liquation

    \n

    B. Calcination

    \n

    C. Electrolysis

    \n

    D. Leaching

    \n

    E. Distillation

    \n

    Choose the correct answer from the options given below:

    ", "options": [ { "text": "B and D only" }, { "text": "A, B, D and E only" }, { "text": "B, D and E only" }, { "text": "A, C and E only" } ], "answer": "B and D only", "solution": "**Answer:** B and D only\n\nLeaching and calcination are the processes which are involved in the extraction of the metals.\n

    \nLiquation, Electrolytic refining, Distillation are used in the refining or purification of metal.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2032, "subject": "Chemistry", "question": "Given below are two statements:\n

    \nStatement I: During Electrolytic refining, the pure metal is made to act as anode and its impure metallic form is used as cathode.\n

    \nStatement II: During the Hall-Heroult electrolysis process, purified $\\mathrm{Al}_{2} \\mathrm{O}_{3}$ is mixed with $\\mathrm{Na}_{3} \\mathrm{AlF}_{6}$ to lower the melting point of the mixture.\n

    \nIn the light of the above statements, choose the most appropriate answer from the options given below:", "options": [ { "text": "Statement $\\mathrm{I}$ is correct but Statement II is incorrect" }, { "text": "Statement I is incorrect but Statement II is correct" }, { "text": "Both Statement I and Statement II are incorrect" }, { "text": "Both Statement I and Statement II are correct" } ], "answer": "Statement I is incorrect but Statement II is correct", "solution": "**Answer:** Statement I is incorrect but Statement II is correct\n\n

    During electrolytic refining, the pure metal is made to act as cathode and impure metal used as anode.

    ", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 2033, "subject": "Chemistry", "question": "

    The reaction representing the Mond process for metal refining is _________.

    ", "options": [ { "text": "$$\\mathrm{ZnO+C\\buildrel \\Delta \\over\n \\longrightarrow Zn+CO}$$" }, { "text": "$$\\mathrm{2K[Au(CN)_2]+Zn\\buildrel \\Delta \\over\n \\longrightarrow K_2[Zn(CN)_4]+2Au}$$" }, { "text": "$$\\mathrm{Zr+2I_2 \\buildrel \\Delta \\over\n \\longrightarrow ZrI_4}$$" }, { "text": "$$\\mathrm{Ni+4CO\\buildrel \\Delta \\over\n \\longrightarrow Ni(CO)_4}$$" } ], "answer": "$$\\mathrm{Ni+4CO\\buildrel \\Delta \\over\n \\longrightarrow Ni(CO)_4}$$", "solution": "**Answer:** $$\\mathrm{Ni+4CO\\buildrel \\Delta \\over\n \\longrightarrow Ni(CO)_4}$$\n\nMond process is need for the purification of $\\mathrm{Ni}$ metal\n

    \n$$\n\\begin{aligned}\n& \\therefore \\underset{\\text { (impure) }}{\\mathrm{Ni}}+4 \\mathrm{CO} \\stackrel{330-350 \\mathrm{k}}{\\longrightarrow} \\mathrm{Ni}(\\mathrm{CO})_{4} \\\\\\\\\n& \\mathrm{Ni}(\\mathrm{CO})_{4} \\stackrel{450-470 \\mathrm{k}}{\\longrightarrow} \\underset{\\text { (pure) }}{\\mathrm{Ni}}+4 \\mathrm{CO}\n\\end{aligned}\n$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2034, "subject": "Chemistry", "question": "

    Which one of the following is most likely a mismatch ?

    ", "options": [ { "text": "Titanium - van Arkel Method" }, { "text": "Zinc - Liquation" }, { "text": "Nickel - Mond process" }, { "text": "Copper - Electrolysis" } ], "answer": "Zinc - Liquation", "solution": "**Answer:** Zinc - Liquation\n\nThe most likely mismatch among the given options is : Zinc - Liquation.\n

    \nLiquation is a process of refining metals that involves melting the metal at a temperature slightly below its melting point and allowing it to cool slowly. This causes impurities to separate out as they solidify at different temperatures from the main metal. Liquation is typically used for metals with low melting points, such as tin and lead.\n

    \nZinc, on the other hand, has a relatively high melting point of 419.53 °C. Therefore, it is unlikely that zinc would be refined using the liquation process. Instead, zinc is commonly produced through the electrolysis of its ores or by distillation in the retort process.\n

    \nOn the other hand, the other options are more likely to be matched with their respective refining processes:

    \n- Titanium is commonly refined using the van Arkel method, which involves the reaction of titanium tetrachloride with a metal like sodium or magnesium to form a volatile titanium compound that can be purified through fractional distillation.

    \n- Nickel is commonly refined using the Mond process, which involves the reaction of nickel with carbon monoxide to form a volatile nickel carbonyl compound that can be purified through fractional distillation.

    \n- Copper is commonly refined using electrolysis, which involves the use of an electric current to separate copper from other metals in an electrolyte solution.\n

    \nTherefore, the most likely mismatch among the given options is Option B - Zinc - Liquation.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2035, "subject": "Chemistry", "question": "Beryllium and aluminium exhibit many properties which are similar. But the two elements\ndiffer in :", "options": [ { "text": "exhibiting maximum covalency in compound " }, { "text": "exhibiting amphoteric nature in their oxides " }, { "text": "forming covalent halides " }, { "text": "forming polymeric hydrides " } ], "answer": "exhibiting maximum covalency in compound ", "solution": "**Answer:** exhibiting maximum covalency in compound \n\nThe maximum valency of beryllium is $$+2$$ white that of aluminium is $$+3.$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2036, "subject": "Chemistry", "question": "Which one of the following is the correct statement?", "options": [ { "text": "Boric acid is a protonic acid " }, { "text": "Beryllium exhibits coordination number of six " }, { "text": "Chlorides of both beryllium and aluminium have bridged chloride structures in solid phase" }, { "text": "B2H6.2NH3 is known as ‘inorganic benzene’" } ], "answer": "Chlorides of both beryllium and aluminium have bridged chloride structures in solid phase", "solution": "**Answer:** Chlorides of both beryllium and aluminium have bridged chloride structures in solid phase\n\n\"AIEEE \n

    The coordination number exhibited by beryllium is $$4$$ and not $$6$$ so statement $$(b)$$ is incorrect.\n

    Both $$BeC{l_2}$$ and $$AlC{l_3}$$ exhibit bridged structures is solid state so $$(c)$$ is correct statement. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2037, "subject": "Chemistry", "question": "The correct statements among I to III regarding\ngroup 13 element oxides are,
    \n(I) Boron trioxide is acidic.
    \n(II) Oxides of aluminium and gallium are\namphoteric.
    \n(III) Oxides of indium and thalliumare basic.", "options": [ { "text": "(II) and (III) only" }, { "text": "(I), (II) and (III)" }, { "text": "(I) and (III) only" }, { "text": "(I) and (II) only" } ], "answer": "(I), (II) and (III)", "solution": "**Answer:** (I), (II) and (III)\n\nB2O3 is an acidic oxide.\n

    Al2O3 and Ga2O3 are amphoteric oxide.\n

    In2O3 and Tl2O are basic oxide.", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 2038, "subject": "Chemistry", "question": "The reaction of H3N3B3Cl3 (A) with LiBH4 in\ntetrahydrofuran gives inorganic benzene (B).\nFurther, the reaction of (A) with (C) leads to\nH3N3B3(Me)3. Compounds (B) and (C)\nrespectively, are :", "options": [ { "text": "Borazine and MeBr" }, { "text": "Diborane and MeMgBr" }, { "text": "Boron nitride and MeBr" }, { "text": "Borazine and MeMgBr" } ], "answer": "Borazine and MeMgBr", "solution": "**Answer:** Borazine and MeMgBr\n\nB3N3H3Cl3(A) + LiBH4 $$ \\to $$ B3N3H6 (B)\n

    B = inorganic\n benzene or Borazine\n

    B3N3H3Cl +\nMeMgBr(C) $$ \\to $$ B3N3H3(CH3)3 + 3MgBrCl", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2039, "subject": "Chemistry", "question": "Boron and silicon of very high purity can be obtained through :", "options": [ { "text": "Liquation" }, { "text": "Electrolytic refining" }, { "text": "Zone refining" }, { "text": "Vapour phase refining" } ], "answer": "Zone refining", "solution": "**Answer:** Zone refining\n\nFor boron and silicon zone refining is used.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2040, "subject": "Chemistry", "question": "Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R.

    Assertion A : In TlI3, isomorphous to CsI3, the metal is present in +1 oxidation state.

    Reason R : Tl metal has fourteen f electrons in its electronic configuration.

    In the light of the above statements, choose the most appropriate answer from the options given below :", "options": [ { "text": "A is correct but R is not correct" }, { "text": "A is not correct but R is correct" }, { "text": "Both A and R are correct but R is NOT the correct explanation of A" }, { "text": "Both A and R are correct and R is the correct explanation of A" } ], "answer": "Both A and R are correct but R is NOT the correct explanation of A", "solution": "**Answer:** Both A and R are correct but R is NOT the correct explanation of A\n\nA : Due to inert pair effect, Tl is more stable in\n+1 oxidation state
    \n     Hence TlI3 and CSI3 are isomorphous

    \nR : Electronic configuration of Tl (81) =\n[Xe] 4f14 5d10 6s2 6p1

    \nBoth A and R are correct but R is not the\ncorrect explanation of A.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2041, "subject": "Chemistry", "question": "

    Choose the correct stability order of group 13 elements in their + 1 oxidation state :

    ", "options": [ { "text": "Al < Ga < In < Tl" }, { "text": "Tl < In < Ga < Al" }, { "text": "Al < Ga < Tl < In" }, { "text": "Al < Tl < Ga < In" } ], "answer": "Al < Ga < In < Tl", "solution": "**Answer:** Al < Ga < In < Tl\n\nDue to inert pair effect, stability of +3 oxidation state\ndecreases and that of +1 oxidation state increases\nfor (down the group) group 13 elements.

    \nSo, the correct order of stability of group 13\nelements in their +1 oxidation state is Al < Ga < In\n< Tl.", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 2042, "subject": "Chemistry", "question": "Given below are two statements :

    \nStatement I: Upon heating a borax bead dipped in cupric sulphate in a luminous flame, the colour of the bead becomes green

    \nStatement II: The green colour observed is due to the formation of copper(I) metaborate

    \nIn light of the above statements, choose the most appropriate answer from the options given below :", "options": [ { "text": "Both Statement I and Statement II are false" }, { "text": "Both Statement I and Statement II are true" }, { "text": "Statement I is false but Statement II is true" }, { "text": "Statement I is true but Statement II is false" } ], "answer": "Both Statement I and Statement II are false", "solution": "**Answer:** Both Statement I and Statement II are false\n\nOn treatment with metal salt, boric anhydride\nforms metaborate of the metal which gives\ndifferent colours in oxidising and reducing flame.\nFor example, in the case of copper sulphate,\nfollowing reactions occur.\n

    $$\n\\mathrm{CuSO}_4+\\mathrm{B}_2 \\mathrm{O}_3 \\mathrel{\\mathop{\\kern0pt\\longrightarrow}\n\\limits_{\\left( {Oxidation} \\right)}^{Non - lu\\min ous\\,flame}} \\mathrm{Cu}\\left(\\mathrm{BO}_2\\right)_2+\\mathrm{SO}_3\n$$\n

    Two reactions may take place in reducing flame (Luminous flame)\n

    (i) The blue-green $\\mathrm{Cu}\\left(\\mathrm{BO}_2\\right)_2$ is reduced to colourless cuprous metaborate as :\n

    $$\n\\begin{aligned}\n& 2 \\mathrm{Cu}\\left(\\mathrm{BO}_2\\right)_2+2 \\mathrm{NaBO}_2+\\mathrm{C} \\stackrel{\\text { Luminous }}{\\longrightarrow} \\\\\\\\\n& 2 \\mathrm{CuBO}_2+\\mathrm{Na}_2 \\mathrm{~B}_4 \\mathrm{O}_7+\\mathrm{CO}\n\\end{aligned}\n$$\n

    (ii) Cupric metaborate may be reduced to metallic copper and bead appears red opaque.\n

    $$\n\\begin{aligned}\n& 2 \\mathrm{Cu}\\left(\\mathrm{BO}_2\\right)_2+4 \\mathrm{NaBO}_2+2 \\mathrm{C} \\underset{\\text { flame }}{\\stackrel{\\text { Luminous }}{\\longrightarrow}} \\\\\\\\\n& 2 \\mathrm{Cu}+2 \\mathrm{Na}_2 \\mathrm{~B}_4 \\mathrm{O}_7+2 \\mathrm{CO}\n\\end{aligned}\n$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2043, "subject": "Chemistry", "question": "

    Potassium dichromate acts as a strong oxidizing agent in acidic solution. During this process, the oxidation state changes from :

    ", "options": [ { "text": "+ 2 to + 1" }, { "text": "+ 6 to + 2" }, { "text": "+ 6 to + 3" }, { "text": "+ 3 to + 1" } ], "answer": "+ 6 to + 3", "solution": "**Answer:** + 6 to + 3\n\n

    $\\mathrm{K}_{2} \\mathrm{Cr}_{2} \\mathrm{O}_{7}$ acts as a strong oxidising agent in acidic medium. During this process, oxidation state of $\\mathrm{Cr}$ changes from $+6$ to $+3$.

    \n

    $$\n\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-}+14 \\mathrm{H}^{+}+6 \\mathrm{e}^{-} \\rightarrow 2 \\mathrm{Cr}^{3+}+7 \\mathrm{H}_{2} \\mathrm{O}\n$$

    ", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 2044, "subject": "Chemistry", "question": "

    The covalency and oxidation state respectively of boron in $$\\left[\\mathrm{BF}_{4}\\right]^{-}$$, are :

    ", "options": [ { "text": "3 and 4" }, { "text": "4 and 3" }, { "text": "4 and 4" }, { "text": "3 and 5" } ], "answer": "4 and 3", "solution": "**Answer:** 4 and 3\n\nIn the tetrafluoroborate anion ($$\\left[\\mathrm{BF}_{4}\\right]^{-}$$), boron is bonded to four fluorine atoms through covalent bonds. Therefore, the covalency of boron in this ion is 4.\n

    \nThe oxidation state of boron can be calculated by considering the charges on the atoms involved in the anion. Fluorine has an oxidation state of -1, and there are four fluorine atoms in the anion, which contributes a total of -4. Since the overall charge on the tetrafluoroborate anion is -1, the oxidation state of boron must be +3 in order to balance the charges.\n

    \nSo, the covalency and oxidation state of boron in $$\\left[\\mathrm{BF}_{4}\\right]^{-}$$ are 4 and 3, respectively.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2045, "subject": "Chemistry", "question": "

    Given below are two statements :

    \n

    Statement I : Boron is extremely hard indicating its high lattice energy.

    \n

    Statement II : Boron has highest melting and boiling point compared to its other group members.

    \n

    In the light of the above statements, choose the most appropriate answer from the options given below :

    ", "options": [ { "text": "Both Statement I and Statement II are incorrect" }, { "text": "Statement I is incorrect but Statement II is correct" }, { "text": "Statement I is correct but Statement II is incorrect" }, { "text": "Both statement I and Statement II are correct" } ], "answer": "Both statement I and Statement II are correct", "solution": "**Answer:** Both statement I and Statement II are correct\n\n

    Boron is indeed a very hard substance, which does indicate high lattice energy. This is because the lattice energy is a measure of the strength of the forces between the ions in an ionic solid; the greater the lattice energy, the stronger the forces and the harder and more stable the compound. This makes Statement I correct.

    \n

    Statement II is also correct. Boron, as the first member of group 13 elements, does indeed have the highest melting and boiling points among its group members (Al, Ga, In, Tl). This can be attributed to the strong covalent bonding in boron due to its small size, which requires a high amount of energy to break.

    \n

    Therefore, Both statement I and Statement II are correct is the correct answer.

    \n", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 2046, "subject": "Chemistry", "question": "

    For compound having the formula $$\\mathrm{GaAlCl}_{4}$$, the correct option from the following is :

    ", "options": [ { "text": "$$\\mathrm{Cl}$$ forms bond with both $$\\mathrm{Al}$$ and $$\\mathrm{Ga}$$ in $$\\mathrm{GaAlCl}_{4}$$" }, { "text": "Oxidation state of $$\\mathrm{Ga}$$ in the salt $$\\mathrm{GaAlCl}_{4}$$ is +3 ." }, { "text": "$$\\mathrm{Ga}$$ is coordinated with $$\\mathrm{Cl}$$ in $$\\mathrm{GaAlCl}_{4}$$" }, { "text": "$$\\mathrm{Ga}$$ is more electronegative than $$\\mathrm{Al}$$ and is present as a cationic part of the salt $$\\mathrm{GaAlCl}_{4}$$" } ], "answer": "$$\\mathrm{Ga}$$ is more electronegative than $$\\mathrm{Al}$$ and is present as a cationic part of the salt $$\\mathrm{GaAlCl}_{4}$$", "solution": "**Answer:** $$\\mathrm{Ga}$$ is more electronegative than $$\\mathrm{Al}$$ and is present as a cationic part of the salt $$\\mathrm{GaAlCl}_{4}$$\n\nOption A : Cl forms bond with both Al and Ga in GaAlCl4\n

    - This is incorrect. In GaAlCl4, chloride ions (Cl-) form bonds with Al to create the AlCl4- ion. There is an ionic bond between the Ga+ cation and the AlCl4- anion.\n

    \"JEE\n\n
    Option B : The oxidation state of Ga in the salt GaAlCl4 is +3\n

    - This is incorrect. The oxidation state of Ga in GaAlCl4 is actually +1.\n\n

    Option C : Ga is coordinated with Cl in GaAlCl4\n

    - This is incorrect. Ga is not coordinated with Cl. The Ga+ ion forms an ionic bond with the AlCl4- ion.\n\n

    Option D : Ga is more electronegative than Al and is present as a cationic part of the salt GaAlCl4\n

    - This is correct. According to the Pauling scale, aluminum (Al) has an electronegativity of 1.61 and gallium (Ga) has an electronegativity of 1.81, which makes Ga more electronegative than Al, and Ga exists as a cationic part (Ga+) in the compound.\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2047, "subject": "Chemistry", "question": "

    Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).

    \n

    Assertion (A) : Melting point of Boron (2453 K) is unusually high in group 13 elements.

    \n

    Reason (R) : Solid Boron has very strong crystalline lattice.

    \n

    In the light of the above statements, choose the most appropriate answer from the options given below ;

    ", "options": [ { "text": "Both (A) and (R) are correct but (R) Is not the correct explanation of (A)\n" }, { "text": "Both (A) and (R) are correct and (R) is the correct explanation of (A)\n" }, { "text": "(A) is true but (R) is false\n" }, { "text": "(A) is false but (R) is true\n" } ], "answer": "Both (A) and (R) are correct and (R) is the correct explanation of (A)\n", "solution": "**Answer:** Both (A) and (R) are correct and (R) is the correct explanation of (A)\n\n\n

    Solid Boron has very strong crystalline lattice so its melting point unusually high in group 13 elements

    ", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 2048, "subject": "Chemistry", "question": "

    Given below are two statements:

    \n

    Statement I : Group 13 trivalent halides get easily hydrolyzed by water due to their covalent nature.

    \n

    Statement II : $$\\mathrm{AlCl}_3$$ upon hydrolysis in acidified aqueous solution forms octahedral $$\\left[\\mathrm{Al}\\left(\\mathrm{H}_2 \\mathrm{O}\\right)_6\\right]^{3+}$$ ion.

    \n

    In the light of the above statements, choose the correct answer from the options given below :

    ", "options": [ { "text": "Statement I is false but statement II is true\n" }, { "text": "Both statement I and statement II are true\n" }, { "text": "Both statement I and statement II are false\n" }, { "text": "Statement I is true but statement II is false" } ], "answer": "Both statement I and statement II are true\n", "solution": "**Answer:** Both statement I and statement II are true\n\n\n

    In trivalent state most of the compounds being covalent are hydrolysed in water. Trichlorides on hydrolysis in water form tetrahedral $$\\left[\\mathrm{M}(\\mathrm{OH})_4\\right]^{-}$$ species, the hybridisation state of element $$\\mathrm{M}$$ is $$\\mathrm{sp}^3$$.

    \n

    In case of aluminium, acidified aqueous solution forms octahedral $$\\left[\\mathrm{Al}\\left(\\mathrm{H}_2 \\mathrm{O}\\right)_6\\right]^{3+}$$ ion.

    ", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 2049, "subject": "Chemistry", "question": "

    Give below are two statements: One is labelled as Assertion A and the other is labelled as Reason R:

    \n

    Assertion A: The stability order of +1 oxidation state of $$\\mathrm{Ga}$$, In and $$\\mathrm{Tl}$$ is Ga < In < Tl.

    \n

    Reason R: The inert pair effect stabilizes the lower oxidation state down the group.

    \n

    In the light of the above statements, choose the correct answer from the options given below:

    ", "options": [ { "text": "A is true but R is false." }, { "text": "Both A and R are true but R is NOT the correct explanation of A." }, { "text": "A is false but R is true." }, { "text": "Both A and R are true and R is the correct explanation of A." } ], "answer": "Both A and R are true and R is the correct explanation of A.", "solution": "**Answer:** Both A and R are true and R is the correct explanation of A.\n\n

    To evaluate the statements provided, let's first understand the concepts involved:

    \n\n

    Assertion A: The stability order of the +1 oxidation state of $$\\mathrm{Ga}$$, In, and $$\\mathrm{Tl}$$ is Ga < In < Tl.

    \n\n

    Reason R: The inert pair effect stabilizes the lower oxidation state down the group.

    \n\n

    The elements $$\\mathrm{Ga}$$ (Gallium), $$\\mathrm{In}$$ (Indium), and $$\\mathrm{Tl}$$ (Thallium) belong to Group 13 of the periodic table. As we move down this group, the +3 oxidation state becomes less stable while the +1 oxidation state becomes more stable. This trend is attributed to the inert pair effect.

    \n\n

    The inert pair effect refers to the reluctance of the s-electrons to participate in bonding as we move down the group in the periodic table. For Group 13 elements, this effect causes the lower +1 oxidation state to be more stable compared to the +3 oxidation state down the group.

    \n\n

    Given this, the Reason R correctly explains why the +1 oxidation state is more stable for $$\\mathrm{Tl}$$ compared to $$\\mathrm{Ga}$$ and $$\\mathrm{In}$$. Therefore, the stability order of the +1 oxidation state should indeed be Ga < In < Tl.

    \n\n

    Conclusion: Both Assertion A and Reason R are true, and Reason R is the correct explanation for Assertion A.

    \n\n

    Therefore, the correct answer is:

    \n\n

    Option D: Both A and R are true and R is the correct explanation of A.

    ", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 2050, "subject": "Chemistry", "question": "

    The number of neutrons present in the more abundant isotope of boron is '$$x$$'. Amorphous boron upon heating with air forms a product, in which the oxidation state of boron is '$$y$$'. The value of $$x+y$$ is _________ .

    ", "options": [ { "text": "9" }, { "text": "6" }, { "text": "4" }, { "text": "3" } ], "answer": "9", "solution": "**Answer:** 9\n\n

    The most abundant isotope of Boron is $${ }_5 \\mathrm{~B}^{11}$$.

    \n

    No. of neutrons in it $$=x=6$$

    \n

    $$2 \\mathrm{B}(\\mathrm{s})+\\mathrm{N}_2(\\mathrm{g}) \\xrightarrow{\\Delta} 2 \\mathrm{BN}(\\mathrm{s})$$

    \n

    Oxidation state of boron in $$\\mathrm{B N=y=+3}$$

    \n

    So, $$x+y=6+3$$

    \n

    $$=9$$

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2051, "subject": "Chemistry", "question": "

    The correct statements from the following are :

    \n

    (A) The decreasing order of atomic radii of group 13 elements is $$\\mathrm{Tl}>\\mathrm{In}>\\mathrm{Ga}>\\mathrm{Al}>\\mathrm{B}$$.

    \n

    (B) Down the group 13 electronegativity decreases from top to bottom.

    \n

    (C) $$\\mathrm{Al}$$ dissolves in dil. $$\\mathrm{HCl}$$ and liberates $$\\mathrm{H}_2$$ but conc. $$\\mathrm{HNO}_3$$ renders $$\\mathrm{Al}$$ passive by forming a protective oxide layer on the surface.

    \n

    (D) All elements of group 13 exhibits highly stable +1 oxidation state.

    \n

    (E) Hybridisation of $$\\mathrm{Al}$$ in $$[\\mathrm{Al}(\\mathrm{H}_2 \\mathrm{O})_6]^{3+}$$ ion is $$\\mathrm{sp}^3 \\mathrm{d}^2$$.

    \n

    Choose the correct answer from the options given below :

    ", "options": [ { "text": "(A), (B), (C) and (E) only\n" }, { "text": "(C) and (E) only\n" }, { "text": "(A), (C) and (E) only\n" }, { "text": "(A) and (C) only" } ], "answer": "(C) and (E) only\n", "solution": "**Answer:** (C) and (E) only\n\n\n

    (A) Group-13 size : $$\\mathrm{B}<\\mathrm{Ga}<\\mathrm{Al}<\\mathrm{In} \\simeq$$ $$\\mathrm{TI}$$

    \n

    (B) In Group-13, Al has exceptional lower EN

    \n

    $$\\mathrm{EN}: \\mathrm{B}>\\mathrm{TI}>\\mathrm{In}>\\mathrm{Ga}>\\mathrm{Al}$$

    \n

    $$\\begin{aligned}\n\\text{(C)} \\quad & 2 \\mathrm{Al}+6 \\mathrm{HCl} \\rightarrow 2 \\mathrm{AlCl}_3+3 \\mathrm{H}_2 \\\\\n& \\mathrm{Al}+\\mathrm{HNO}_3 \\rightarrow \\text { No reaction (Due to protective oxide layer)}\n\\end{aligned}$$

    \n

    (D) +1 oxidation state is observed mainly for heavier Group-13 element like $$\\mathrm{TI}$$.

    \n

    $$\\mathrm{B}$$ and $$\\mathrm{Al}$$ do not show +1 oxidation state

    \n

    (E) $$\\left[\\mathrm{Al}\\left(\\mathrm{H}_2 \\mathrm{O}\\right)_6\\right]^{3+} \\rightarrow$$ Hybridisation : $$s p^3 d^2$$

    ", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 2052, "subject": "Chemistry", "question": "

    Given below are two statements :

    \n

    Statement I : Gallium is used in the manufacturing of thermometers.

    \n

    Statement II : A thermometer containing gallium is useful for measuring the freezing point $$(256 \\mathrm{~K})$$ of brine solution.

    \n

    In the light of the above statements, choose the correct answer from the options given below :

    ", "options": [ { "text": "Both Statement I and Statement II are true\n" }, { "text": "Statement I is true but Statement II is false\n" }, { "text": "Statement I is false but Statement II is true\n" }, { "text": "Both Statement I and Statement II are false" } ], "answer": "Statement I is true but Statement II is false\n", "solution": "**Answer:** Statement I is true but Statement II is false\n\n\n

    S-1 is correct as gallium is used in manufacturing of thermometers.

    \n

    S-2 is incorrect

    ", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 2053, "subject": "Chemistry", "question": "Glass is a :", "options": [ { "text": "super-cooled liquid" }, { "text": "gel" }, { "text": "polymeric mixture" }, { "text": "micro-crystalline solid" } ], "answer": "super-cooled liquid", "solution": "**Answer:** super-cooled liquid\n\nGlass is a translucent or transparent amorphous super-cooled solid solution or we can say super cooled liquid of silicates and borats having a general formula $${R_2}O.\\,MO.\\,6Si{O_2}.$$ where $$R=Na$$ or $$K$$ and $$M$$ $$=Ca, Ba, Zn$$ or $$Pb.$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2054, "subject": "Chemistry", "question": "The stability of dihalides of $$Si,Ge,Sn$$ and $$Pb$$ increases steadily in the sequence :", "options": [ { "text": "$$Pb{X_2} < < Sn{X_2} < < Ge{X_2} < < Si{X_2}$$ " }, { "text": "$$Ge{X_2} < < Si{X_2} < < Sn{X_2} < < Pb{X_2}$$ " }, { "text": "$$Si{X_2} < < Ge{X_2} < < Pb{X_2} < < Sn{X_2}$$ " }, { "text": "$$Si{X_2} < < Ge{X_2} < < Sn{X_2} < < Pb{X_2}$$ " } ], "answer": "$$Si{X_2} < < Ge{X_2} < < Sn{X_2} < < Pb{X_2}$$ ", "solution": "**Answer:** $$Si{X_2} < < Ge{X_2} < < Sn{X_2} < < Pb{X_2}$$ \n\nReductance of valence shell electrons to participate in bonding is called inert pair effect. The stability of lower oxidation state ($${ + 2\\,\\,}$$ for group $$14$$ element) increases on going down the group. So the correct order is \n

    $$Si{X_2} < Ge{X_2} < Sn{X_2} < Pb{X_2}$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2055, "subject": "Chemistry", "question": "In graphite and diamond, the percentage of p-characters of the hybrid orbitals in hybridisation are respectively : ", "options": [ { "text": "33 and 25 " }, { "text": "33 and 75 " }, { "text": "50 and 75" }, { "text": "67 and 75 " } ], "answer": "67 and 75 ", "solution": "**Answer:** 67 and 75 \n\nHybridization of carbon in graphite is sp2.\n

    So,   %   of p - in graphite = $${2 \\over 3} \\times 100$$ = 67%\n

    Hybridization of carbon in diamond is sp3 .\n

    So,   %   of p - in diamond = $${3 \\over 4} \\times 100$$ = 75%", "topic": "Discrete Mathematics", "subtopic": "Other" }, { "id": 2056, "subject": "Chemistry", "question": "The one that is extensively used as a piezoelectric material is : ", "options": [ { "text": "tridymite" }, { "text": "amorphous silica" }, { "text": "quartz" }, { "text": "mica" } ], "answer": "quartz", "solution": "**Answer:** quartz\n\nThose materials which produce electricity under mechanical stress are called piezoelectric material.\n

    Quartz is an example of piezoelectric material.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2057, "subject": "Chemistry", "question": "The element that does NOT show catenation is :", "options": [ { "text": "Ge" }, { "text": "Si" }, { "text": "Sn" }, { "text": "Pb" } ], "answer": "Pb", "solution": "**Answer:** Pb\n\n\"JEE\n
    All the four given element belongs to the carbon family.\n

    Pb will show last catenation.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2058, "subject": "Chemistry", "question": "The element that shows greater ability to form p$$\\pi $$-p$$\\pi $$ multiple bonds, is :", "options": [ { "text": "Si" }, { "text": "Ge" }, { "text": "C" }, { "text": "Sn" } ], "answer": "C", "solution": "**Answer:** C\n\n\"JEE\n
    In carbon family size of carbon is smallest. \n

    A size of carbon atom is small, so between 2p orbital of two carbon atom effective overlapping happens. \n

    But when size of atom increases then effective overlapping does not happens between atoms. \n

    That is why in carbon family, C will have strongest p$$\\pi $$ $$-$$ p$$\\pi $$ bond.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2059, "subject": "Chemistry", "question": "Which one of the following compounds of Group-14 elements is not known?", "options": [ { "text": "[GeCl6]2$$-$$" }, { "text": "[Sn(OH)6]2$$-$$" }, { "text": "[SiCl6]2$$-$$" }, { "text": "[SiF6]2$$-$$" } ], "answer": "[SiCl6]2$$-$$", "solution": "**Answer:** [SiCl6]2$$-$$\n\n[SiCl6]2$$-$$ does not exist due to steric crowding of surrounding atoms.", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 2060, "subject": "Chemistry", "question": "

    In liquation process used for tin (Sn), the metal :

    ", "options": [ { "text": "is reacted with acid." }, { "text": "is dissolved in water." }, { "text": "is brought to molten form which is made to flow on a slope." }, { "text": "is fused with NaOH" } ], "answer": "is brought to molten form which is made to flow on a slope.", "solution": "**Answer:** is brought to molten form which is made to flow on a slope.\n\nLiquation

    \nIn this method a low melting metal like tin can be made to flow on a sloping surface. In this way it is\nseparated from higher melting impurities", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2061, "subject": "Chemistry", "question": "

    Given below are two statements.

    \n

    Statement I : Stannane is an example of a molecular hydride.

    \n

    Statement II : Stannane is a planar molecule.

    \n

    In the light of the above statement, choose the most appropriate answer from the options given below.

    ", "options": [ { "text": "Both Statement I and Statement II are true." }, { "text": "Both Statement I and Statement II are false." }, { "text": "Statement I is true but Statement II is false." }, { "text": "Statement I is false but Statement II is true." } ], "answer": "Statement I is true but Statement II is false.", "solution": "**Answer:** Statement I is true but Statement II is false.\n\nStannane or tin hydride is an inorganic compound with formula $\\mathrm{SnH}_{4}$\n

    \"JEE\n

    $$ \\therefore $$ It is a non-planar molecule.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2062, "subject": "Chemistry", "question": " During water-gas shift reaction", "options": [ { "text": "water is evaporated in presence of catalyst." }, { "text": "carbon monoxide is oxidized to carbon dioxide." }, { "text": "carbon is oxidized to carbon monoxide." }, { "text": "carbon dioxide is reduced to carbon monoxide." } ], "answer": "carbon monoxide is oxidized to carbon dioxide.", "solution": "**Answer:** carbon monoxide is oxidized to carbon dioxide.\n\n

    During the water-gas shift reaction, carbon monoxide (CO) + H2 reacts with steam (H2O) to produce carbon dioxide (CO2) and hydrogen gas (H2). The chemical equation for the reaction is :

    \n\n

    \"JEE

    \n\n

    Therefore, option B is correct, which states that carbon monoxide is oxidized to carbon dioxide. This reaction is typically catalyzed by metal catalysts such as iron, nickel, or copper, which promote the reaction rate by providing a surface for the reactants to adsorb and react.

    \n

    Option A is not correct as water is not evaporated during the reaction, but rather reacts with carbon monoxide to produce hydrogen gas and carbon dioxide.\n

    \n

    Option C is not correct as carbon is not directly involved in the reaction.

    \n\n

    Option D is also not correct as carbon dioxide is not reduced to carbon monoxide during the water-gas shift reaction.

    ", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 2063, "subject": "Chemistry", "question": "Given below are two statements :\n

    Statement (I) : $\\mathrm{SiO}_2$ and $\\mathrm{GeO}_2$ are acidic while $\\mathrm{SnO}$ and $\\mathrm{PbO}$ are amphoteric in nature.\n

    Statement (II) : Allotropic forms of carbon are due to property of catenation and $\\mathrm{p} \\pi-\\mathrm{d} \\pi$ bond formation.\n

    \nIn the light of the above statements, choose the most appropriate answer from the options given below :", "options": [ { "text": "Statement I is false but Statement II is true" }, { "text": "Both Statement I and Statement II are false" }, { "text": "Both Statement I and Statement II are true" }, { "text": "Statement I is true but Statement II is false" } ], "answer": "Statement I is true but Statement II is false", "solution": "**Answer:** Statement I is true but Statement II is false\n\n

    \nBoth statements need to be evaluated to determine which (if any) are true:\n

    \nStatement (I) : $\\mathrm{SiO}_2$ and $\\mathrm{GeO}_2$ are acidic while $\\mathrm{SnO}$ and $\\mathrm{PbO}$ are amphoteric in nature.\n

    \nThis statement is generally true. The oxides of silicon ($\\mathrm{SiO}_2$) and germanium ($\\mathrm{GeO}_2$) are indeed acidic. These elements belong to group 14 of the periodic table and as you move down the group, the acidic character of the oxides decreases while the basic character increases. Tin oxide ($\\mathrm{SnO}$) and lead oxide ($\\mathrm{PbO}$) are lower in the group and they are amphoteric, which means they show both acidic and basic properties depending on the reacting substance.\n

    \nStatement (II) : Allotropic forms of carbon are due to the property of catenation and $\\mathrm{p} \\pi-\\mathrm{d} \\pi$ bond formation.\n

    \nThis statement is false. Catenation is indeed the property that carbon has to form long chains and rings with other carbon atoms, which is responsible for the vast number of organic compounds. However, in the context of pure carbon allotropes—such as diamond, graphite, and fullerene—the different physical structures and properties are mainly due to the $\\mathrm{sp}^3$, $\\mathrm{sp}^2$, and $\\mathrm{sp}$ hybridizations of carbon, not to $\\mathrm{p} \\pi-\\mathrm{d} \\pi$ bonding. Carbon does not use d-orbitals in forming the allotropes. The $\\mathrm{p} \\pi-\\mathrm{d} \\pi$ bonding is typically discussed in the context of transition metal complexes, not allotropic forms of carbon.\n

    \nThus, the correct answer is Option D : Statement I is true but Statement II is false.\n

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2064, "subject": "Chemistry", "question": "

    Consider the oxides of group 14 elements\n$$\\mathrm{SiO}_2, \\mathrm{GeO}_2, \\mathrm{SnO}_2, \\mathrm{PbO}_2, \\mathrm{CO}$$ and $$\\mathrm{GeO}$$. The amphoteric oxides are

    ", "options": [ { "text": "$$\\mathrm{SnO}_2, \\mathrm{CO}$$\n" }, { "text": "$$\\mathrm{SiO}_2, \\mathrm{GeO}_2$$\n" }, { "text": "$$\\mathrm{SnO}_2, \\mathrm{PbO}_2$$\n" }, { "text": "$$\\mathrm{GeO}, \\mathrm{GeO}_2$$" } ], "answer": "$$\\mathrm{SnO}_2, \\mathrm{PbO}_2$$\n", "solution": "**Answer:** $$\\mathrm{SnO}_2, \\mathrm{PbO}_2$$\n\n\n

    SnO$$_2$$ and PbO$$_2$$ are amphoteric.

    ", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 2065, "subject": "Chemistry", "question": "

    The compound that is white in color is

    ", "options": [ { "text": "ammonium sulphide\n" }, { "text": "ammonium arsinomolybdate\n" }, { "text": "lead iodide\n" }, { "text": "lead sulphate" } ], "answer": "lead sulphate", "solution": "**Answer:** lead sulphate\n\n

    Lead sulphate-white

    \n

    Ammonium sulphide-soluble

    \n

    Lead iodide-Bright yellow

    \n

    Ammonium arsinomolybdate-yellow

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2066, "subject": "Chemistry", "question": "

    Evaluate the following statements related to group 14 elements for their correctness.

    \n

    (A) Covalent radius decreases down the group from $$\\mathrm{C}$$ to $$\\mathrm{Pb}$$ in a regular manner.

    \n

    (B) Electronegativity decreases from $$\\mathrm{C}$$ to $$\\mathrm{Pb}$$ down the group gradually.

    \n

    (C) Maximum covalance of $$\\mathrm{C}$$ is 4 whereas other elements can expand their covalance due to presence of d orbitals.

    \n

    (D) Heavier elements do not form $$\\mathrm{p} \\pi-\\mathrm{p} \\pi$$ bonds.

    \n

    (E) Carbon can exhibit negative oxidation states.

    \n

    Choose the correct answer from the options given below :

    ", "options": [ { "text": "(A), (B) and (C) Only\n" }, { "text": "(A) and (B) Only\n" }, { "text": "(C), (D) and (E) Only\n" }, { "text": "(C) and (D) Only" } ], "answer": "(C), (D) and (E) Only\n", "solution": "**Answer:** (C), (D) and (E) Only\n\n\n

    (A) Covalent radius of group 14 elements increases from $$\\mathrm{C}$$ to $$\\mathrm{Pb}$$.

    \n

    (B) Electronegativity decreases from $$\\mathrm{C}$$ to $$\\mathrm{Si}$$ and thereafter it remains more or less constant.

    \n

    (C) Maximum covalency of $$\\mathrm{C}$$ is 4 as the outermost shell of $$\\mathrm{C}$$ is $$2^{\\text {nd }}$$ shell and it cannot expand its octet. But other elements can expand their covalency due to the availability of vacant d-orbitals.

    \n

    (D) Heavier elements do not form $$\\mathrm{p} \\pi-\\mathrm{p} \\pi$$ bonds due to their larger atomic size.

    \n

    (E) Carbon can show negative oxidation states when it is covalently bonded to less electronegative elements like $$\\mathrm{CH}_4$$.

    ", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 2067, "subject": "Chemistry", "question": "

    Which of the following material is not a semiconductor.

    ", "options": [ { "text": "Copper oxide\n" }, { "text": "Graphite\n" }, { "text": "Silicon\n" }, { "text": "Germanium" } ], "answer": "Graphite\n", "solution": "**Answer:** Graphite\n\n\n

    To determine which of the following materials is not a semiconductor, we must understand what a semiconductor is and then look at the properties of each material listed.

    \n\n

    Semiconductors are materials that have a conductivity level between conductors (such as metals) and insulators (such as most ceramic materials). They can conduct electricity under certain conditions or when doped with impurities. The most common semiconductors are silicon and germanium, but some compound materials like gallium arsenide and silicone carbide also behave as semiconductors.

    \n\n

    Now, analyzing each option given:

    \n\n

    Option A: Copper oxide - Copper oxide can behave as a semiconductor, particularly in some specific structures or when used in thin-film solar cells. Thus, it can exhibit semiconductor properties.

    \n\n

    Option B: Graphite - Graphite is a form of carbon where the carbon atoms are bonded together in a sheet-like form. It is known for its excellent electrical conductivity due to the delocalization of pi electrons across the carbon layers. Its behavior is more characteristic of a conductor rather than a semiconductor.

    \n\n

    Option C: Silicon - Silicon is the most well-known and widely used semiconductor material in electronic devices, including transistors, diodes, and integrated circuits. Its semiconducting properties are due to its band structure, which can be altered by doping with other elements.

    \n\n

    Option D: Germanium - Germanium is another classic semiconductor material used in the early development of electronic devices. Though not as commonly used as Silicon in modern electronics, due to its similar crystal structure and semiconducting properties, Germanium remains an important semiconductor material.

    \n\n

    Based on the provided information, Option B: Graphite is not a semiconductor. Its properties align more with those of a conductor rather than a semiconductor, distinguishing it from the other options listed which all have semiconducting capabilities under certain conditions.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2068, "subject": "Chemistry", "question": "Which of the following statements is true? ", "options": [ { "text": "H3PO3 is a stronger acid than H2SO3" }, { "text": "In aqueous medium HF is a stronger acid than HCl " }, { "text": "HClO4 is a weaker acid than HClO3" }, { "text": "HNO3 is a stronger acid than HNO2" } ], "answer": "HNO3 is a stronger acid than HNO2", "solution": "**Answer:** HNO3 is a stronger acid than HNO2\n\nThe $$\\mathop {HN{O_3}}\\limits^{ + 5} $$ is stronger than $$\\mathop {HN{O_2}}\\limits^{ + 3} .$$ The more the oxidation state of $$N,$$ the more is the acid character.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2069, "subject": "Chemistry", "question": "Which of the following statement is wrong?", "options": [ { "text": "Nitrogen cannot form $$d\\pi - p\\pi$$ bond. " }, { "text": "Single N – N bond is weaker than the single P – P bond" }, { "text": "N2O4 has two resonance structures" }, { "text": "The stability of hydrides increases from NH3 to BiH3 in group 15 of the periodic table" } ], "answer": "The stability of hydrides increases from NH3 to BiH3 in group 15 of the periodic table", "solution": "**Answer:** The stability of hydrides increases from NH3 to BiH3 in group 15 of the periodic table\n\nThe case of formation and stability of hydrides decreases rapidly from $$N{H_3}$$ to $$Bi{H_3}.$$. This is evident from their dissociation temperature which decreases from $$N{H_3}$$ to $$B{H_3}.$$ As we go down the group the size of central atom increases and thus metal-hydrogen bond becomes weaker due to decreased overlap between the large central atom and small hydrogen atom. \n

    $$\\mathop {N{H_3} > }\\limits_{(most\\,\\,stable)} P{H_3} > As{H_3} > Sb{H_3}\\mathop { > Bi{H_3}}\\limits_{(least\\,stable)} $$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2070, "subject": "Chemistry", "question": "The molecule having smallest bond angle is :", "options": [ { "text": "NCl3" }, { "text": "AsCl3" }, { "text": "SbCl3" }, { "text": "PCl3" } ], "answer": "SbCl3", "solution": "**Answer:** SbCl3\n\nAll the members form volatile halides of the type $$A{X_3}.$$ All halides are pyramidal in shape. The bond angle decreases on moving down the group due to decrease in bond pair-bond pair repulsion.\n

    $$\\eqalign{\n & NC{L_3}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,PC{l_3}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,AsC{l^3} \\cr \n & {107^ \\circ }\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,{94^ \\circ }\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,{92^ \\circ } \\cr} $$ ", "topic": "Geometry", "subtopic": "Plane Geometry" }, { "id": 2071, "subject": "Chemistry", "question": "Which one of the following properties is not shown by NO?", "options": [ { "text": "It combines with oxygen to form nitrogen dioxide" }, { "text": "It’s bond order is 2.5" }, { "text": "It is diamagnetic in gaseous state" }, { "text": "It is a neutral oxide" } ], "answer": "It is diamagnetic in gaseous state", "solution": "**Answer:** It is diamagnetic in gaseous state\n\nNitric oxide is paramagnetic in the gaseous state because of the presence of one unpaired electron in its outermost shell.\n

    The electronic configuration of $$NO$$ is \n

    $$\\sigma _{1s}^2\\,\\sigma _{1s}^{ * 2}\\,\\sigma _{2s}^2\\,\\sigma _{2s}^{ \\circ 2}\\,\\sigma _{{2_{{p_z}}}}^2\\,\\pi _{{2_{{p_x}}}}^2$$\n

    $$ = \\pi _{{2_{Py}}}^2\\,\\pi _{{2_{Px}}}^{ * 1}$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2072, "subject": "Chemistry", "question": "A metal ‘M’ reacts with nitrogen gas to afford ‘M3N’. ‘M3N’ on heating at high temperature gives back ‘M’ and on\nreaction with water produces a gas ‘B’. Gas ‘B’ reacts with aqueous \n solution of CuSO4 to form a deep blue compound. ‘M’ and\n‘B’ respectively are :", "options": [ { "text": "Li and NH3" }, { "text": "Ba and N2" }, { "text": "Na and NH3" }, { "text": "Al and N2" } ], "answer": "Li and NH3", "solution": "**Answer:** Li and NH3\n\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2073, "subject": "Chemistry", "question": "The compound that does not produce nitrogen gas by the thermal decomposition is :", "options": [ { "text": "(NH4)2SO4" }, { "text": "Ba(N3)2" }, { "text": "(NH4)2Cr2O7" }, { "text": "NH4NO2" } ], "answer": "(NH4)2SO4", "solution": "**Answer:** (NH4)2SO4\n\nThermal decomposition reaction of the given compounds are . . .\n

    (NH4)2SO4 $$\\buildrel \\Delta \\over\n \\longrightarrow $$ 2 NH3(g) + H2SO4\n

    Ba (N3)2 $$\\buildrel \\Delta \\over\n \\longrightarrow $$ Ba + 3N2(g)\n

    NH4NO2 $$\\buildrel \\Delta \\over\n \\longrightarrow $$ N2(g) + 2H2O\n

    (NH4)2Cr2O7 $$\\buildrel \\Delta \\over\n \\longrightarrow $$ N2(g) + 4H2O + Cr2O3 \n

    So, here you can see only (NH4)2SO4 does not give N2 by thermal decomposition it gives NH3 while other give N2 gas ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2074, "subject": "Chemistry", "question": "A metal (A) on heating in nitrogen gas gives\ncompound B. B on treatment with H2O gives\na colourless gas which when passed through\nCuSO4 solution gives a dark blue-violet\ncoloured solution. A and B respectively, are :", "options": [ { "text": "Na and NaNO3" }, { "text": "Na and Na3N" }, { "text": "Mg and Mg3N2" }, { "text": "Mg and Mg(NO3)2" } ], "answer": "Mg and Mg3N2", "solution": "**Answer:** Mg and Mg3N2\n\n3Mg(A) + N2 $$\\buildrel \\Delta \\over\n \\longrightarrow $$\nMg3N2(B) $$\\buildrel {6{H_2}O} \\over\n \\longrightarrow $$
    3Mg(OH)2 + 2NH3 (colourless gas)\n

    CuSO + 4NH3 $$ \\to $$ [Cu(NH3)4]SO4 (deep blue solution)", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2075, "subject": "Chemistry", "question": "The correct statement with respect to\ndinitrogen is :", "options": [ { "text": "N2 is paramagnetic in nature" }, { "text": "it can be used as an inert diluent for\nreactive chemicals" }, { "text": "it can combine with dioxygen at 25oC" }, { "text": "liquid dinitrogen is not used in cryosurgery" } ], "answer": "it can be used as an inert diluent for\nreactive chemicals", "solution": "**Answer:** it can be used as an inert diluent for\nreactive chemicals\n\n(1) N2\n is diamagnetic, with no unpaired elctrons.\n

    (2) In iron and chemical Industry because of its inertness it is used where an\ninert atmosphere is required.\n

    (3) N2\n does not combine with oxygen, hydrogen or most other elements. Nitrogen will combine with\noxygen, however in the presence of lightining or a spark.\n

    (4) Liquid nitrogen is used as a refrigerant to preserve biological material food items and in cryosurgery.\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2076, "subject": "Chemistry", "question": "Find A, B and C in the following reactions :

    $$N{H_3} + A + C{O_2} \\to {(N{H_4})_2}C{O_3}$$

    $${(N{H_4})_2}C{O_3} + {H_2}O + B \\to N{H_4}HC{O_3}$$

    $$N{H_4}HC{O_3} + NaCl \\to N{H_4}Cl + C$$", "options": [ { "text": "$$A - {O_2};B - C{O_2};C - N{a_2}C{O_3}$$" }, { "text": "$$A - {H_2}O;B - {O_2};C - NaHC{O_3}$$" }, { "text": "$$A - {H_2}O;B - {O_2};C - N{a_2}C{O_3}$$" }, { "text": "$$A - {H_2}O;B - C{O_2};C - NaHC{O_3}$$" } ], "answer": "$$A - {H_2}O;B - C{O_2};C - NaHC{O_3}$$", "solution": "**Answer:** $$A - {H_2}O;B - C{O_2};C - NaHC{O_3}$$\n\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2077, "subject": "Chemistry", "question": "A group of 15 element, which is a metal and forms a hydride with strongest reducing power among group 15 hydrides. The element is :", "options": [ { "text": "As" }, { "text": "Sb" }, { "text": "Bi" }, { "text": "P" } ], "answer": "Bi", "solution": "**Answer:** Bi\n\nBiH3 is the strongest reducing agent among the hydrides of 15 group elements as Bi – H bond dissociation energy is very less.", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 2078, "subject": "Chemistry", "question": "Which one of the following group-15 hydride is the strongest reducing agent?", "options": [ { "text": "AsH3" }, { "text": "BiH3" }, { "text": "PH3" }, { "text": "SbH3" } ], "answer": "BiH3", "solution": "**Answer:** BiH3\n\nAmong 15th group hydrides, BiH3 is strongest reducing agent.", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 2079, "subject": "Chemistry", "question": "

    Nitrogen gas is obtained by thermal decomposition of :

    ", "options": [ { "text": "Ba(NO3)2" }, { "text": "Ba(N3)2" }, { "text": "NaNO2" }, { "text": "NaNO3" } ], "answer": "Ba(N3)2", "solution": "**Answer:** Ba(N3)2\n\n$$\\mathrm{Ba}\\left(\\mathrm{N}_{3}\\right)_{2} \\stackrel{\\Delta}{\\longrightarrow} \\mathrm{Ba}+3 \\mathrm{~N}_{2}$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2080, "subject": "Chemistry", "question": "

    The oxide which contains an odd electron at the nitrogen atom is :

    ", "options": [ { "text": "N2O" }, { "text": "NO2" }, { "text": "N2O3" }, { "text": "N2O5" } ], "answer": "NO2", "solution": "**Answer:** NO2\n\nThe oxide of nitrogen which contains odd electron\nis NO2

    \n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2081, "subject": "Chemistry", "question": "PCl5 is well known, but NCl5 is not. because,", "options": [ { "text": "nitrogen is less reactive than phosphorus." }, { "text": "nitrogen doesn't have d-orbitals in its valence shell." }, { "text": "catenation tendency is weaker in nitrogen than phosphorus." }, { "text": "size of phosphorus is larger than nitrogen." } ], "answer": "nitrogen doesn't have d-orbitals in its valence shell.", "solution": "**Answer:** nitrogen doesn't have d-orbitals in its valence shell.\n\n$$\\mathrm{PCl}_{5}$$ is well known but $$\\mathrm{NCl}_{5}$$ is not because nitrogen does not have vacant d-orbitals in its valence shell.

    So, nitrogen cannot expand its octet. On the other hand, phosphorus has vacant $$\\mathrm{d}$$-orbitals in its valence shell which enables it to expand its octet.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2082, "subject": "Chemistry", "question": "

    The most stable trihalide of nitrogen is :

    ", "options": [ { "text": "NF3" }, { "text": "NCl3" }, { "text": "NBr3" }, { "text": "NI3" } ], "answer": "NF3", "solution": "**Answer:** NF3\n\n

    The\nstability of trihalides decreases down the group due\nto weakening of N – X bond and inability of N to\naccommodate large sized halogen atoms (Cl, Br, I)\naround it.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2083, "subject": "Chemistry", "question": "

    A $$\\mathrel{\\mathop{\\kern0pt\\longrightarrow}\n\\limits_{}^{573\\,K}} $$ Red phosphorus $$\\mathrel{\\mathop{\\kern0pt\\longrightarrow}\n\\limits_{under\\,pressure}^{heat\\,;\\,803\\,K}} $$ B

    \n

    Red phosphorus is obtained by heating \"A\" at 573 K, and can be converted to \"B\" by heating at 803 K under pressure.

    \n

    A and B, respectively, are

    ", "options": [ { "text": "$$\\beta$$-black phosphorus and white phosphorus." }, { "text": "white phosphorus and $$\\beta$$-black phosphorus." }, { "text": "$$\\alpha$$-black phosphorus and white phosphorus." }, { "text": "white phosphorus and $$\\alpha$$-black phosphorus." } ], "answer": "white phosphorus and $$\\alpha$$-black phosphorus.", "solution": "**Answer:** white phosphorus and $$\\alpha$$-black phosphorus.\n\nRed phosphrous is obtained by heating white phosphorous at $573 \\mathrm{~K}$ in inert atmosphere. When red phosphorous is heated in a sealed tube at $803 \\mathrm{~K}$, we get $\\alpha$-black phosphorous.\n

    \n$\\beta$-Black phosphorous is prepared by heating white phosphorous at $473 \\mathrm{~K}$ under high pressure.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2084, "subject": "Chemistry", "question": "

    Match List - I with List - II.

    \n

    \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    List I
    (Processes / Reactions)    		List II
    (Catalyst)
    (A)2SO$$_2$$(g) + O$$_2$$(g) $$ \\to $$ 2SO$$_3$$(g)(I)Fe(s)
    (B)4NH$$_3$$(g) + 5O$$_2$$(g) $$ \\to $$ 4NO(g) + 6H$$_2$$O(g)(II)Pt(s) $$ - $$ Rh(s)
    (C)N$$_2$$(g) + 3H$$_2$$(g) $$ \\to $$ 2NH$$_3$$(g)(III)V$$_2$$O$$_5$$
    (D)Vegetable oil(l) + H$$_2$$ $$ \\to $$ Vegetable ghee(s)(IV)Ni(s)

    \n

    Choose the correct answer from the options given below :

    ", "options": [ { "text": "$$(\\mathrm{A})-(\\mathrm{III}),(\\mathrm{B})-(\\mathrm{I}),(\\mathrm{C})-(\\mathrm{II}),(\\mathrm{D})-(\\mathrm{IV})$$" }, { "text": "$$(\\mathrm{A})-(\\mathrm{III}),(\\mathrm{B})-(\\mathrm{II}),(\\mathrm{C})-(\\mathrm{I}),(\\mathrm{D})-(\\mathrm{IV})$$" }, { "text": "$$(\\mathrm{A})-(\\mathrm{IV}),(\\mathrm{B})-(\\mathrm{III}),(\\mathrm{C})-(\\mathrm{I}),(\\mathrm{D})-(\\mathrm{II})$$" }, { "text": "$$(\\mathrm{A})-(\\mathrm{IV}),(\\mathrm{B})-(\\mathrm{II}),(\\mathrm{C})-(\\mathrm{III}),(\\mathrm{D})-(\\mathrm{I})$$" } ], "answer": "$$(\\mathrm{A})-(\\mathrm{III}),(\\mathrm{B})-(\\mathrm{II}),(\\mathrm{C})-(\\mathrm{I}),(\\mathrm{D})-(\\mathrm{IV})$$", "solution": "**Answer:** $$(\\mathrm{A})-(\\mathrm{III}),(\\mathrm{B})-(\\mathrm{II}),(\\mathrm{C})-(\\mathrm{I}),(\\mathrm{D})-(\\mathrm{IV})$$\n\n(A) $$2 \\mathrm{SO}_{2}(\\mathrm{g})+\\mathrm{O}_{2}(\\mathrm{g}) \\stackrel{\\mathrm{V}_{2} \\mathrm{O}_{5}}{\\longrightarrow} 2 \\mathrm{SO}_{3}$$\n

    \n(B) $$4 \\mathrm{NH}_{3}(\\mathrm{g})+5 \\mathrm{O}_{2}(\\mathrm{g}) \\stackrel{\\mathrm{Pt}(\\mathrm{s})-\\mathrm{Rh}(\\mathrm{s})}{\\longrightarrow}\n4 \\mathrm{NO}(\\mathrm{g})+6 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})$$\n

    \n(C) $$\\mathrm{N}_{2}(\\mathrm{g})+3 \\mathrm{H}_{2}(\\mathrm{g}) \\stackrel{\\mathrm{Fe}(\\mathrm{s})}{\\longrightarrow} 2 \\mathrm{NH}_{3}(\\mathrm{g})$$\n

    \n(D) Vegetable oil(I) $$+\\mathrm{H}_{2} \\stackrel{\\mathrm{Ni(s})}{\\longrightarrow}$$ Vegetable ghee(s)", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 2085, "subject": "Chemistry", "question": "

    Dinitrogen and dioxygen, the main constituents of air do not react with each other in atmosphere to form oxides of nitrogen because :

    ", "options": [ { "text": "$$\\mathrm{N}_{2}$$ is unreactive in the condition of atmosphere." }, { "text": "Oxides of nitrogen are unstable." }, { "text": "Reaction between them can occur in the presence of a catalyst." }, { "text": "The reaction is endothermic and require very high temperature." } ], "answer": "The reaction is endothermic and require very high temperature.", "solution": "**Answer:** The reaction is endothermic and require very high temperature.\n\n$\\mathrm{N}_{2}$ is unreactive, its reaction with oxides is endothermic and require very high temperature.

    \n$$\n\\mathrm{N}_2+\\mathrm{O}_2 \\overset{{(1483-2000 \\mathrm{~K})}}\\leftrightharpoons 2 \\mathrm{NO}\n$$

    \n(Endothermic and feasible at high temperature)", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 2086, "subject": "Chemistry", "question": "

    Given below are two statements :

    \n

    Statement I : $$\\mathrm{SbCl}_{5}$$ is more covalent than $$\\mathrm{SbCl}_{3}$$

    \n

    Statement II: The higher oxides of halogens also tend to be more stable than the lower ones.

    \n

    In the light of the above statements, choose the most appropriate answer from the options given below :

    ", "options": [ { "text": "Both statement I and Statement II are correct" }, { "text": "Both Statement I and Statement II are incorrect" }, { "text": "Statement I is correct but Statement II is incorrect" }, { "text": "Statement I is incorrect but Statement II is correct" } ], "answer": "Both statement I and Statement II are correct", "solution": "**Answer:** Both statement I and Statement II are correct\n\n

    Statement I : SbCl₅ is more covalent than SbCl₃

    \n

    In general, the covalent character of a compound is directly proportional to the polarization, and the polarization depends upon the size of the cation and the charge on the cation. If the charge on the cation increases or if the size of the cation decreases, the covalent character increases.

    \n

    In the case of SbCl₅ and SbCl₃, the antimony (Sb) cation in SbCl₅ has a higher charge due to the greater number of chlorine (Cl) atoms, thus increasing the polarization and making SbCl₅ more covalent than SbCl₃.

    \n

    Statement II : The higher oxides of halogens also tend to be more stable than the lower ones.

    \n

    This statement is also correct. Halogens can form a variety of oxides, and generally, the stability of these oxides increases with increasing oxidation state. This is because the halogens, being highly electronegative, stabilize the high positive oxidation state. For example, in the case of chlorine, Cl₂O₇ (where chlorine is in the +7 oxidation state) is more stable than Cl₂O (where chlorine is in the +1 oxidation state).

    \n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2087, "subject": "Chemistry", "question": "The strongest reducing agent among the following is :", "options": [ { "text": "$\\mathrm{SbH}_3$" }, { "text": "$\\mathrm{NH}_3$" }, { "text": "$\\mathrm{BiH}_3$" }, { "text": "$\\mathrm{PH}_3$" } ], "answer": "$\\mathrm{BiH}_3$", "solution": "**Answer:** $\\mathrm{BiH}_3$\n\n\n

    The reducing strength of a substance is generally associated with its ability to donate electrons to other substances. In the context of hydrides of Group 15 elements, such as $\\mathrm{SbH}_3$, $\\mathrm{NH}_3$, $\\mathrm{BiH}_3$, and $\\mathrm{PH}_3$, the reducing strength can be considered based on their chemical reactivity, stability, and tendency to donate electrons.

    \n\n

    In general, the stability of hydrides decreases as we move down the group in the periodic table, which means that heavier hydrides are less stable and hence act as stronger reducing agents. This trend is primarily due to the increase in the size of the central atom as we move down the group, leading to weaker bonds between the central atom and hydrogen in the hydride. As the bond strength decreases, the hydrides can more readily release hydrogen in the form of protons (H+) or hydride ions (H−), making them stronger reducing agents.

    \n\n

    The order of reducing strength for Group 15 hydrides is typically as follows:

    \n\n

    $$\\mathrm{BiH}_3 > \\mathrm{SbH}_3 > \\mathrm{PH}_3 > \\mathrm{NH}_3$$

    \n\n

    Among the options given, $\\mathrm{BiH}_3$ (bismuthine) is the strongest reducing agent. It is the least stable of the listed hydrides, due to the large size and low electronegativity of bismuth compared to antimony, phosphorus, and nitrogen. This makes $\\mathrm{BiH}_3$ more prone to releasing electrons and acting as a reducing agent.

    \n\n

    In contrast, $\\mathrm{NH}_3$ (ammonia) is the most stable among the given hydrides and, therefore, the weakest reducing agent. Its bond with hydrogen is comparatively stronger, and nitrogen's higher electronegativity makes the molecule more resistant to donating electrons.

    \n\n

    Thus, the correct answer is Option C, $\\mathrm{BiH}_3$.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2088, "subject": "Chemistry", "question": "The number of white coloured salts, among the following is

    \n(a) $\\mathrm{SrSO}_4$

    \n(b) $\\mathrm{Mg}\\left(\\mathrm{NH}_4\\right) \\mathrm{PO}_4$

    \n(c) $\\mathrm{BaCrO}_4$

    \n(d) $\\mathrm{Mn}(\\mathrm{OH})_2$

    \n(e) $\\mathrm{PbSO}_4$

    \n(f) $\\mathrm{PbCrO}_4$

    \n(g) $\\mathrm{AgBr}$

    \n(h) $\\mathrm{PbI}_2$

    \n(i) $\\mathrm{CaC}_2 \\mathrm{O}_4$

    \n(j) $\\left[\\mathrm{Fe}(\\mathrm{OH})_2\\left(\\mathrm{CH}_3 \\mathrm{COO}\\right)\\right]$", "options": [], "answer": "5", "solution": "**Answer:** 5\n\n

    To determine the number of white coloured salts among the listed compounds, we look at their common color characteristics:

    (a) $\\mathrm{SrSO}_4$ - Generally, most sulfate salts are white unless the cation imparts a color. Strontium sulfate is indeed white.

    (b) $\\mathrm{Mg}\\left(\\mathrm{NH}_4\\right)\\mathrm{PO}_4$ - Magnesium ammonium phosphate is also known to be white. This is typical for many phosphates with colorless cations.

    (c) $\\mathrm{BaCrO}_4$ - Barium chromate is well-known for its yellow color, a characteristic of many chromates.

    (d) $\\mathrm{Mn}(\\mathrm{OH})_2$ - Manganese hydroxide can have a faint pink to white color, but in this context, we consider it white for counting purposes.

    (e) $\\mathrm{PbSO}_4$ - Lead sulfate is white, consistent with many lead salts not influenced by strong coloring anions.

    (f) $\\mathrm{PbCrO}_4$ - Lead chromate is distinguished by its yellow color, again a feature of chromate salts.

    (g) $\\mathrm{AgBr}$ - Silver bromide is known for its pale yellow to virtually white appearance, but it is more accurately described as pale yellow.

    (h) $\\mathrm{PbI}_2$ - Lead iodide exhibits a bright yellow color, typical for many iodides with heavy metal cations.

    (i) $\\mathrm{CaC}_2\\mathrm{O}_4$ - Calcium oxalate is white, as most insoluble oxalates are.

    (j) $\\left[\\mathrm{Fe}(\\mathrm{OH})_2\\left(\\mathrm{CH}_3\\mathrm{COO}\\right)\\right]$ - This complex containing iron is brown red due to the presence of iron ions.

    Summarizing, the white salts among the given compounds are $\\mathrm{SrSO}_4$, $\\mathrm{Mg}\\left(\\mathrm{NH}_4\\right)\\mathrm{PO}_4$, $\\mathrm{Mn}(\\mathrm{OH})_2$, $\\mathrm{PbSO}_4$, and $\\mathrm{CaC}_2\\mathrm{O}_4$, making a total of 5 white colored salts.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2089, "subject": "Chemistry", "question": "

    On passing a gas, '$$\\mathrm{X}$$', through Nessler's regent, a brown precipitate is obtained. The gas '$$\\mathrm{X}$$' is

    ", "options": [ { "text": "$$\\mathrm{Cl}_2$$\n" }, { "text": "$$\\mathrm{CO}_2$$\n" }, { "text": "$$\\mathrm{NH}_3$$\n" }, { "text": "$$\\mathrm{H}_2 \\mathrm{S}$$" } ], "answer": "$$\\mathrm{NH}_3$$\n", "solution": "**Answer:** $$\\mathrm{NH}_3$$\n\n\n

    Nessler's Reagent Reaction :

    \n

    $$\\underset{\\text { (Nessler's Reagent) }}{2 \\mathrm{~K}_2 \\mathrm{HgI}_4}+\\mathrm{NH}_3+3 \\mathrm{KOH} \\rightarrow \\underset{\\substack{\\text { (lodine of Millon's base } \\\\ \\text { (Brown precipitate }}}{\\mathrm{HgO} . \\mathrm{Hg}\\left(\\mathrm{NH}_2\\right) \\mathrm{I}}+7 \\mathrm{KI}+2 \\mathrm{H}_2 \\mathrm{O}$$

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2090, "subject": "Chemistry", "question": "

    Choose the correct statements about the hydrides of group 15 elements.

    \n

    A. The stability of the hydrides decreases in the order $$\\mathrm{NH}_3>\\mathrm{PH}_3>\\mathrm{AsH}_3> \\mathrm{SbH}_3>\\mathrm{BiH}_3$$.

    \n

    B. The reducing ability of the hydride increases in the order $$\\mathrm{NH}_3<\\mathrm{PH}_3<\\mathrm{AsH}_3 <\\mathrm{SbH}_3<\\mathrm{BiH}_3$$.

    \n

    C. Among the hydrides, $$\\mathrm{NH}_3$$ is strong reducing agent while $$\\mathrm{BiH}_3$$ is mild reducing agent.

    \n

    D. The basicity of the hydrides increases in the order $$\\mathrm{NH}_3<\\mathrm{PH}_3<\\mathrm{AsH}_3< \\mathrm{SbH}_3<\\mathrm{BiH}_3$$.

    \n

    Choose the most appropriate from the options given below :

    ", "options": [ { "text": "C and D only" }, { "text": "A and D only" }, { "text": "A and B only" }, { "text": "B and C only" } ], "answer": "A and B only", "solution": "**Answer:** A and B only\n\n

    On moving down the group, bond strength of $$\\mathrm{M}-\\mathrm{H}$$ bond decreases, which reduces the thermal stability but increases reducing nature of hydrides, hence A and B are correct statements.

    ", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 2091, "subject": "Chemistry", "question": "

    Identify the incorrect statements about group 15 elements :

    \n

    (A) Dinitrogen is a diatomic gas which acts like an inert gas at room temperature.

    \n

    (B) The common oxidation states of these elements are $$-3,+3$$ and +5.

    \n

    (C) Nitrogen has unique ability to form $$\\mathrm{p} \\pi-\\mathrm{p} \\pi$$ multiple bonds.

    \n

    (D) The stability of +5 oxidation states increases down the group.

    \n

    (E) Nitrogen shows a maximum covalency of 6.

    \n

    Choose the correct answer from the options given below :

    ", "options": [ { "text": "(A), (C), (E) only\n" }, { "text": "(D) and (E) only\n" }, { "text": "(A), (B), (D) only\n" }, { "text": "(B), (D), (E) only" } ], "answer": "(D) and (E) only\n", "solution": "**Answer:** (D) and (E) only\n\n\n

    Let's analyze each statement step by step to identify the incorrect ones:

    \n\n

    (A) Dinitrogen is a diatomic gas which acts like an inert gas at room temperature.

    \n\n

    This statement is generally correct. Dinitrogen ($$N_2$$) is indeed a diatomic gas and is very inert at room temperature due to the strong triple bond between the nitrogen atoms.

    \n\n

    (B) The common oxidation states of these elements are $$-3,+3$$ and +5.

    \n\n

    This statement is also correct. The group 15 elements commonly exhibit oxidation states of $$-3$$ (in nitrides), $$+3$$ (as in $$\\mathrm{NH}_3$$), and +5 (as in $$\\mathrm{HNO_3}$$).

    \n\n

    (C) Nitrogen has unique ability to form $$\\mathrm{p}\\pi-\\mathrm{p}\\pi$$ multiple bonds.

    \n\n

    This statement is correct. Nitrogen has a unique ability to form $$p\\pi-p\\pi$$ multiple bonds, which is why it can form a triple bond ($$N_2$$).

    \n\n

    (D) The stability of +5 oxidation states increases down the group.

    \n\n

    This statement is incorrect. The stability of +5 oxidation states actually decreases down the group due to the effect of inert pair effect. For example, bismuth in +5 state is less stable than nitrogen in +5 state.

    \n\n

    (E) Nitrogen shows a maximum covalency of 6.

    \n\n

    This statement is incorrect. Nitrogen can exhibit a maximum covalency of 4 (as in $$\\mathrm{NH_4^+}$$). It cannot expand its octet to show a covalency of 6.

    \n\n

    Based on the above analysis, the incorrect statements are (D) and (E). Therefore, the correct answer is:

    \n\n

    Option B: (D) and (E) only

    ", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 2092, "subject": "Chemistry", "question": "

    In qualitative test for identification of presence of phosphorous, the compound is heated with an oxidising agent. Which is further treated with nitric acid and ammonium molybdate respectively. The yellow coloured precipitate obtained is :

    ", "options": [ { "text": "$$\\mathrm{Na}_3 \\mathrm{PO}_4 \\cdot 12 \\mathrm{MoO}_3$$\n" }, { "text": "$$\\left(\\mathrm{NH}_4\\right)_3 \\mathrm{PO}_4 \\cdot 12\\left(\\mathrm{NH}_4\\right)_2 \\mathrm{MoO}_4$$\n" }, { "text": "$$\\mathrm{MoPO}_4 \\cdot 21 \\mathrm{NH}_4 \\mathrm{NO}_3$$\n" }, { "text": "$$\\left(\\mathrm{NH}_4\\right)_3 \\mathrm{PO}_4 \\cdot 12 \\mathrm{MoO}_3$$" } ], "answer": "$$\\left(\\mathrm{NH}_4\\right)_3 \\mathrm{PO}_4 \\cdot 12 \\mathrm{MoO}_3$$", "solution": "**Answer:** $$\\left(\\mathrm{NH}_4\\right)_3 \\mathrm{PO}_4 \\cdot 12 \\mathrm{MoO}_3$$\n\n

    For test of phosphorus:

    \n

    $$\\mathrm{{H_3}P{O_4} + 12{(N{H_4})_2}Mo{O_4} + 21HN{O_3} \\to \\mathop {{{(N{H_4})}_3}\\,.\\,12Mo{O_3}}\\limits_{(Yellow\\,ppt)} + 21N{H_4}N{O_3} + 12{H_2}O}$$

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2093, "subject": "Chemistry", "question": "Which one of the following substances has the highest proton affinity?", "options": [ { "text": "H2S" }, { "text": "NH3" }, { "text": "PH3" }, { "text": "H2O" } ], "answer": "NH3", "solution": "**Answer:** NH3\n\nAmong the given compounds, the $$N{H_3}$$ is most basic. Hence has highest proton affinity.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2094, "subject": "Chemistry", "question": "Regular use of which of the following fertilizer increases the acidity of soil? ", "options": [ { "text": "Potassium nitrate" }, { "text": "Urea " }, { "text": "Superphosphate of lime" }, { "text": "Ammonium sulphate " } ], "answer": "Ammonium sulphate ", "solution": "**Answer:** Ammonium sulphate \n\n$$\\,{\\left( {N{H_4}} \\right)_2}S{O_4} + 2{H_2}O\\buildrel \\, \\over\n \\longrightarrow 2{H_2}S{O_4} + N{H_4}OH$$\n

    $${H_2}S{O_4}\\,\\,$$ is strong acid and increases the acidity of soil.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2095, "subject": "Chemistry", "question": "Which of the following statements regarding sulphur is incorrect?", "options": [ { "text": "The vapour at 200oC consists mostly of S8 rings " }, { "text": "At 600oC the gas mainly consists of S2 molecules " }, { "text": "The oxidation state of sulphur is never less than +4 in its compounds " }, { "text": "S2 molecule is paramagnetic. " } ], "answer": "The oxidation state of sulphur is never less than +4 in its compounds ", "solution": "**Answer:** The oxidation state of sulphur is never less than +4 in its compounds \n\nOxidation of sulphur varies from $$-2$$ to $$+6$$ in its various compounds. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2096, "subject": "Chemistry", "question": "Assertion : Nitrogen and Oxygen are the main components in the atmosphere but these do not react to\nform oxides of nitrogen.

    \nReason : The reaction between nitrogen and oxygen requires high temperature.
    ", "options": [ { "text": "Both assertion and reason are correct, and the reason is the correct explanation for the assertion" }, { "text": "Both assertion and reason are correct, but the reason is not the correct explanation for the assertion" }, { "text": "The assertion is incorrect, but the reason is correct" }, { "text": "Both the assertion and reason are incorrect" } ], "answer": "Both assertion and reason are correct, and the reason is the correct explanation for the assertion", "solution": "**Answer:** Both assertion and reason are correct, and the reason is the correct explanation for the assertion\n\nNitrogen and oxygen in air do not react to form oxides of nitrogen in atmosphere because the reaction between nitrogen and oxygen requires high temperature.", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 2097, "subject": "Chemistry", "question": "In KO2, the nature of oxygen species and the oxydation state of oxygen atom are, respectively : ", "options": [ { "text": "Oxide and $$-$$ 2" }, { "text": "Superoxide and $$-$$ 1/2" }, { "text": "Peroxide and $$-$$ 1/2" }, { "text": "Superoxide and $$-$$ 1" } ], "answer": "Superoxide and $$-$$ 1/2", "solution": "**Answer:** Superoxide and $$-$$ 1/2\n\nKO2 is called potassium superoxide. Here O$$_2^ - $$ is the superoxide ion.\n

    Oxidation state of K is +1 and let oxidation state of oxygent atom is = x \n

    $$\\therefore\\,\\,\\,$$ 1 + 2(x) = 0\n

    $$ \\Rightarrow $$$$\\,\\,\\,$$ x = $$-$$ $${1 \\over 2}$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2098, "subject": "Chemistry", "question": "Match List - I with List - II

    \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    List - I
    (process)    		List - II
    (catalyst)
    (a)Deacron's process(i)ZSM-5
    (b)Contact process(ii)$$CuC{l_2}$$
    (c)Cracking of hydrocarbons(iii)Particles 'Ni'
    (d)Hydrogenation of vegetable oils(iv)$${V_2}{O_5}$$


    Choose the most appropriate answer from the options given below :", "options": [ { "text": "(a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)" }, { "text": "(a)-(i), (b)-(iii), (c)-(ii), (d)-(iv)" }, { "text": "(a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)" }, { "text": "(a)-(iv), (b)-(ii), (c)-(i), (d)-(iii)" } ], "answer": "(a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)", "solution": "**Answer:** (a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)\n\nIn manufacture of H2SO4 (contact process), V2O5 is used as a catalyst.

    \nNi catalysts enables the hydrogenation of fats.\nCuCl2 is used as catalyst in Deacon's process.\nZSM-5 used as catalyst in cracking of\nhydrocarbons.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2099, "subject": "Chemistry", "question": "Given below are two statement : one is labelled as Assertion A and the other is labelled as Reason R.

    Assertion A : SO2(g) is absorbed to a large extent than H2(g) on activated charcoal.

    Reason R : SO2(g) has a higher critical temperature than H2(g).

    In the light of the above statements, choose the most appropriate answer from the options given below.", "options": [ { "text": "Both A and R are correct but R is not the correct explanation of A." }, { "text": "Both A and R are correct and R is the correct explanation of A." }, { "text": "A is not correct but R is correct." }, { "text": "A is correct but R is not correct." } ], "answer": "Both A and R are correct and R is the correct explanation of A.", "solution": "**Answer:** Both A and R are correct and R is the correct explanation of A.\n\nGases having higher critical temperature absorb to a greater extent.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2100, "subject": "Chemistry", "question": "Match List - I with List - II :

    \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    List - I
    (compound)    		List - II
    (effect/affected species)
    (a)Carbon monoxide(i)Carcinogenic
    (b)Sulphur dioxide(ii)Metabolized by pyrus plants
    (c)Polychlorinated biphenyls(iii)haemoglobin
    (d)Oxides of Nitrogen(iv)Stiffness of flower buds


    Choose the correct answer from the options given below :", "options": [ { "text": "(a) - (iii), (b) - (iv), (c) - (i), (d) - (ii)" }, { "text": "(a) - (iv), (b) - (i), (c) - (iii), (d) - (ii)" }, { "text": "(a) - (i), (b) - (ii), (c) - (iii), (d) - (iv)" }, { "text": "(a) - (iii), (b) - (iv), (c) - (ii), (d) - (i)" } ], "answer": "(a) - (iii), (b) - (iv), (c) - (i), (d) - (ii)", "solution": "**Answer:** (a) - (iii), (b) - (iv), (c) - (i), (d) - (ii)\n\n

    A. If haemoglobin binds CO, it prevents the binding of O2\nto\nhaemoglobin due to the competition for same binding sites.

    \n

    B. Higher conc. SO2\nin air can cause stiffness of flower buds.

    \n

    C. Polychlorinated biphenyls are carcinogenic (cancer causing) in\nhumans as well as animals.

    \n

    D. Certain plants, e.g. pinus, Juniparus, quercus, pyrus and vitis\ncan metabolise nitrogen oxides. Hence, the correct match is

    \n(a) - (iii), (b) - (iv), (c) - (i), (d) - (ii)", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 2101, "subject": "Chemistry", "question": "In polythionic acid, H2SxO6 (x = 3 to 5) the oxidation state(s) of sulphur is/are :", "options": [ { "text": "+ 5 only" }, { "text": "+ 6 only" }, { "text": "+ 3 and + 5 only" }, { "text": "0 and + 5 only" } ], "answer": "0 and + 5 only", "solution": "**Answer:** 0 and + 5 only\n\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2102, "subject": "Chemistry", "question": "

    Sulphur dioxide is one of the components of polluted air. SO2 is also a major contributor to acid rain. The correct and complete reaction to represent acid rain caused by SO2 is :

    ", "options": [ { "text": "2SO2 + O2 $$\\to$$ 2SO3" }, { "text": "SO2 + O3 $$\\to$$ SO3 + O2" }, { "text": "SO2 + H2O2 $$\\to$$ H2SO4" }, { "text": "2SO2 + O2 + 2H2O $$\\to$$ 2H2SO4" } ], "answer": "2SO2 + O2 + 2H2O $$\\to$$ 2H2SO4", "solution": "**Answer:** 2SO2 + O2 + 2H2O $$\\to$$ 2H2SO4\n\n

    $$2\\text{SO}_{2} +\\text{O}_{2} +\\text{2H}_{2} \\text{O} \\longrightarrow \\text{2H}_{2} \\text{SO}_{4} $$

    \n

    Acid rain occurs due to increased concentration of\noxides of sulphur and Nitrogen.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2103, "subject": "Chemistry", "question": "

    Match List I with List II.

    \n

    \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    List-I
    (Anion)    		List-II
    (gas evolved on reaction with dil.H$$_2$$SO$$_2$$)
    (A)CO$${_3^{2 - }}$$I.Colourless gas which turns lead acetate paper black.
    (B)S$$^{2 - }$$II.Colourless gas which turns acidified potassium dichromate solution green.
    (C)SO$${_3^{2 - }}$$III.Brown fumes which turns acidified KI solution containing starch blue.
    (D)NO$${_2^ - }$$IV.Colourless gas evolved with brisk effervescence, which turns lime water milky.

    \n

    Choose the correct answer from the options given below :

    ", "options": [ { "text": "A-III, B-I, C-II, D-IV" }, { "text": "A-II, B-I, C-IV, D-III" }, { "text": "A-IV, B-I, C-III, D-II" }, { "text": "A-IV, B-I, C-II, D-III" } ], "answer": "A-IV, B-I, C-II, D-III", "solution": "**Answer:** A-IV, B-I, C-II, D-III\n\n$\\mathrm{CO}_{3}^{2-}$ : On action of dil sulphuric acid, $\\mathrm{CO}_{2}$ gas is released which turns lime water milky.\n

    \n$\\mathrm{S}^{2-} $ : On action of dil sulphuric acid, $\\mathrm{H}_{2} \\mathrm{~S}$ gas is released which turns lead acetate paper black.\n

    \n$\\mathrm{SO}_{3}^{2-}$ : On action of dil $\\mathrm{H}_{2} \\mathrm{SO}_{4}, \\mathrm{SO}_{2}$ gas is evolved which turns acidified potassium dichromate solution green.\n

    \n$\\mathrm{NO}_{2}^{-}$: On action of dil $\\mathrm{H}_{2} \\mathrm{SO}_{4}, \\mathrm{NO}_{2}$ gas is evolved which turns $\\mathrm{KI}$ solution containg starch blue.", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 2104, "subject": "Chemistry", "question": "

    Which of the following elements is considered as a metalloid?

    ", "options": [ { "text": "Sc" }, { "text": "Pb" }, { "text": "Bi" }, { "text": "Te" } ], "answer": "Te", "solution": "**Answer:** Te\n\nSc, Pb, Bi are metals

    \nTe is a metalloid", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2105, "subject": "Chemistry", "question": "

    Match List I with List II:

    \n

    \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    List IList II
    (A)Sulphate(I)Pesticide
    (B)Fluoride(II)Bending of bones
    (C)Nicotine(III)Laxative effect
    (D)Sodium arsinite(IV)Herbicide

    \n

    Choose the correct answer from the options given below :

    ", "options": [ { "text": "A-II, B-III, C-IV, D-I" }, { "text": "A-IV, B-III, C-II, D-I" }, { "text": "A-III, B-II, C-I, D-IV" }, { "text": "A-III, B-II, C-IV, D-I" } ], "answer": "A-III, B-II, C-I, D-IV", "solution": "**Answer:** A-III, B-II, C-I, D-IV\n\nSodium arsinite - Herbicide\n

    \nNicotine - Pesticide\n

    \nSulphate - Laxative effect\n

    \nFluoride - Bending of bones", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 2106, "subject": "Chemistry", "question": "

    Given below are two statements. One is labelled as Assertion A and the other is labelled as Reason R.

    \n

    Assertion A : Activated charcoal adsorbs SO2 more efficiently than CH4.

    \n

    Reason R : Gases with lower critical temperatures are readily adsorbed by activated charcoal.

    \n

    In the light of the above statements, choose the correct answer from the options given below.

    ", "options": [ { "text": "Both A and R are correct and R is the correct explanation of A." }, { "text": "Both A and R are correct but R is NOT the correct explanation of A." }, { "text": "A is correct but R is not correct." }, { "text": "A is not correct but R is correct." } ], "answer": "A is correct but R is not correct.", "solution": "**Answer:** A is correct but R is not correct.\n\nMore polar gases easily adsorbs on activated charcoal.\n

    \nAnd more polar gases has more (higher) critical temperature as compared to non-polar or less polar gases.\n

    \n$\\therefore$ Gases with higher critical temperature adsorbed more.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2107, "subject": "Chemistry", "question": "

    $$\\mathrm{K_2Cr_2O_7}$$ paper acidified with dilute $$\\mathrm{H_2SO_4}$$ turns green when exposed to :

    ", "options": [ { "text": "Sulphur dioxide" }, { "text": "Sulphur trioxide" }, { "text": "Hydrogen sulphide" }, { "text": "Carbon dioxide" } ], "answer": "Sulphur dioxide", "solution": "**Answer:** Sulphur dioxide\n\n$\\mathrm{SO}_{2}$ gets oxidised in presence of $\\mathrm{K}_{2} \\mathrm{Cr}_{2} \\mathrm{O}_{7}$ and it converts to $\\mathrm{Cr}^{+3}$ in presence of dil. $\\mathrm{H}_{2} \\mathrm{SO}_{4}$.\n

    \nSimilarly, $\\mathrm{H}_{2} \\mathrm{~S}$ can also get oxidized to sulphur.\n

    \nHowever, most appropriate is (A) Sulphur dioxide.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2108, "subject": "Chemistry", "question": "

    Sum of $$\\pi$$-bonds present in peroxodisulphuric acid and pyrosulphuric acid is ___________

    ", "options": [], "answer": "8", "solution": "**Answer:** 8\n\n

    Peroxodisulphuric acid -

    \n

    \"JEE

    \n

    No. of $$\\pi $$ – bonds = 4

    \n

    Pyro sulphuric acid -

    \n

    \"JEE

    \n

    No. of $$\\pi $$ – bonds = 4

    \n

    Total $$\\pi $$ – bonds = 4 + 4 = 8

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2109, "subject": "Chemistry", "question": "

    Which one of the following pairs is an example of polar molecular solids?

    ", "options": [ { "text": "$$\\mathrm{HCl}(\\mathrm{s}), \\mathrm{AlN}(\\mathrm{s})$$" }, { "text": "$$\\mathrm{MgO}(\\mathrm{s}), \\mathrm{SO}_{2}(\\mathrm{s})$$" }, { "text": "$$\\mathrm{SO}_{2}(\\mathrm{s}), \\mathrm{NH}_{3}(\\mathrm{s})$$" }, { "text": "$$\\mathrm{SO}_{2}(\\mathrm{s}), \\mathrm{CO}_{2}(\\mathrm{s})$$" } ], "answer": "$$\\mathrm{SO}_{2}(\\mathrm{s}), \\mathrm{NH}_{3}(\\mathrm{s})$$", "solution": "**Answer:** $$\\mathrm{SO}_{2}(\\mathrm{s}), \\mathrm{NH}_{3}(\\mathrm{s})$$\n\n

    Polar molecular solids are formed by molecules that have polar covalent bonds and possess a net dipole moment. A net dipole moment arises when there is an asymmetric distribution of charge in the molecule, resulting in a separation of positive and negative charges.

    \n

    Out of the given options, only option C consists of two polar molecular solids - $\\mathrm{SO}_2(\\mathrm{s})$ and $\\mathrm{NH}_3(\\mathrm{s})$. $\\mathrm{SO}_2$ has polar covalent bonds due to the difference in electronegativity between sulfur and oxygen, and the molecule has a net dipole moment due to its bent molecular geometry. Similarly, $\\mathrm{NH}_3$ has polar covalent bonds due to the difference in electronegativity between nitrogen and hydrogen, and the molecule has a net dipole moment due to its trigonal pyramidal molecular geometry.

    \n

    Therefore, the correct answer is - $\\mathrm{SO}_{2}(\\mathrm{s}), \\mathrm{NH}_{3}(\\mathrm{s})$.

    \n", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 2110, "subject": "Chemistry", "question": "Among the following oxides of p-block elements, number of oxides having amphoteric nature is ________.

    $\\mathrm{Cl}_2 \\mathrm{O}_7, \\mathrm{CO}, \\mathrm{PbO}_2, \\mathrm{~N}_2 \\mathrm{O}, \\mathrm{NO}, \\mathrm{Al}_2 \\mathrm{O}_3, \\mathrm{SiO}_2, \\mathrm{~N}_2 \\mathrm{O}_5, \\mathrm{SnO}_2$", "options": [], "answer": "3", "solution": "**Answer:** 3\n\n

    An amphoteric oxide is one that displays both acidic and basic properties; such substances react with both acids and bases. In the p-block, several elements can form amphoteric oxides, particularly those elements that are metallic or metalloid in nature. It's essential to recognize which oxides among the provided list have an amphoteric character:

    \n\n\n\n

    From the list, the amphoteric oxides are:

    \n\n\n\n

    So, the number of amphoteric oxides in the list provided is 3.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2111, "subject": "Chemistry", "question": "

    From the given list, the number of compounds with +4 oxidation state of Sulphur ________.

    \n

    $$\\mathrm{SO}_3, \\mathrm{H}_2 \\mathrm{SO}_3, \\mathrm{SOCl}_2, \\mathrm{SF}_4, \\mathrm{BaSO}_4, \\mathrm{H}_2 \\mathrm{S}_2 \\mathrm{O}_7\n$$

    ", "options": [], "answer": "3", "solution": "**Answer:** 3\n\n

    \n\n\n\n\n\n\n\n\n\n\n\n \n \n \n \n \n \n \n \n \n\n\n \n \n \n \n \n \n \n \n \n\n
    Compounds$$\\mathrm{SO_3}$$$$\\mathrm{H_2SO_3}$$$$\\mathrm{SOCl_2}$$$$\\mathrm{SF_4}$$$$\\mathrm{BaSO_4}$$$$\\mathrm{H_2S_2O_7}$$
    O.S. of Sulphur:+6+4+4+4+6+6

    \n

    To determine the number of compounds with a +4 oxidation state of sulfur, we need to examine the oxidation states of sulfur in each of the listed compounds. The +4 oxidation state means that sulfur has lost 4 electrons compared to its elemental state.

    \n\n

    Let's go through each compound:

    \n\n\n\n

    So, from the given list, only three compounds have sulfur in the +4 oxidation state: $$\\mathrm{H}_2 \\mathrm{SO}_3$$, $$\\mathrm{SOCl}_2$$, and $$\\mathrm{SF}_4$$. Therefore, the number of compounds with a +4 oxidation state of sulfur is three.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2112, "subject": "Chemistry", "question": "

    Given below are two statements :

    \n

    Statement (I) : Oxygen being the first member of group 16 exhibits only -2 oxidation state.

    \n

    Statement (II) : Down the group 16 stability of +4 oxidation state decreases and +6 oxidation state increases.

    \n

    In the light of the above statements, choose the most appropriate answer from the options given below :

    ", "options": [ { "text": "Statement I is correct but Statement II is incorrect\n" }, { "text": "Both Statement I and Statement II are incorrect\n" }, { "text": "Statement I is incorrect but Statement II is correct\n" }, { "text": "Both Statement I and Statement II are correct" } ], "answer": "Both Statement I and Statement II are incorrect\n", "solution": "**Answer:** Both Statement I and Statement II are incorrect\n\n\n

    Statement-I: Oxygen can have oxidation state from -2 to +2 , so statement I is incorrect

    \n

    Statement- II: On moving down the group stability of +4 oxidation state increases whereas stability of +6 oxidation state decreases down the group, according to inert pair effect.

    \n

    So both statements are wrong.

    ", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 2113, "subject": "Chemistry", "question": "

    1 mole of $$\\mathrm{PbS}$$ is oxidised by \"$$\\mathrm{X}$$\" moles of $$\\mathrm{O}_3$$ to get \"$$\\mathrm{Y}$$\" moles of $$\\mathrm{O}_2$$. $$\\mathrm{X}+\\mathrm{Y}=$$ _________.

    ", "options": [], "answer": "8", "solution": "**Answer:** 8\n\n

    $$\\begin{aligned}\n& \\mathrm{PbS}+4 \\mathrm{O}_3 \\rightarrow \\mathrm{PbSO}_4+4 \\mathrm{O}_2 \\\\\n& \\mathrm{x}=4, \\mathrm{y}=4\n\\end{aligned}$$

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2114, "subject": "Chemistry", "question": "

    Given below are two statements:

    \n

    Statement I : $$\\mathrm{S}_8$$ solid undergoes disproportionation reaction under alkaline conditions to form $$\\mathrm{S}^{2-}$$ and $$\\mathrm{S}_2 \\mathrm{O}_3{ }^{2-}$$.

    \n

    Statement II : $$\\mathrm{ClO}_4^{-}$$ can undergo disproportionation reaction under acidic condition.

    \n

    In the light of the above statements, choose the most appropriate answer from the options given below :

    ", "options": [ { "text": "Statement I is correct but statement II is incorrect\n" }, { "text": "Both statement I and statement II are incorrect\n" }, { "text": "Statement I is incorrect but statement II is correct\n" }, { "text": "Both statement I and statement II are correct" } ], "answer": "Statement I is correct but statement II is incorrect\n", "solution": "**Answer:** Statement I is correct but statement II is incorrect\n\n\n

    $$\\mathrm{S}_1: \\mathrm{S}_8+12 \\mathrm{OH}^{\\ominus} \\rightarrow 4 \\mathrm{~S}^{2-}+2 \\mathrm{~S}_2 \\mathrm{O}_3^{2-}+6 \\mathrm{H}_2 \\mathrm{O}$$

    \n

    $$\\mathrm{S}_2: \\mathrm{ClO}_4^{\\ominus}$$ cannot undergo disproportionation reaction as chlorine is present in it's highest oxidation state.

    ", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 2115, "subject": "Chemistry", "question": "

    Choose the correct statements from the following

    \n

    A. All group 16 elements form oxides of general formula $$\\mathrm{EO}_2$$ and $$\\mathrm{EO}_3$$, where $$\\mathrm{E}=\\mathrm{S}, \\mathrm{Se}, \\mathrm{Te}$$ and $$\\mathrm{Po}$$. Both the types of oxides are acidic in nature.

    \n

    B. $$\\mathrm{TeO}_2$$ is an oxidising agent while $$\\mathrm{SO}_2$$ is reducing in nature.

    \n

    C. The reducing property decreases from $$\\mathrm{H}_2 \\mathrm{~S}$$ to $$\\mathrm{H}_2$$ Te down the group.

    \n

    D. The ozone molecule contains five lone pairs of electrons.

    \n

    Choose the correct answer from the options given below:

    ", "options": [ { "text": "A and B only" }, { "text": "C and D only" }, { "text": "A and D only" }, { "text": "B and C only" } ], "answer": "A and B only", "solution": "**Answer:** A and B only\n\n

    (A) All group 16 elements form oxides of the $$\\mathrm{EO}_2$$ and $$\\mathrm{EO}_3$$ type where $$\\mathrm{E}=\\mathrm{S}, \\mathrm{Se}, \\mathrm{Te}$$ or $$\\mathrm{Po}$$.

    \n

    (B) $$\\mathrm{SO}_2$$ is reducing while $$\\mathrm{TeO}_2$$ is an oxidising agent.

    \n

    (C) The reducing property increases from $$\\mathrm{H}_2 \\mathrm{S}$$ to $$\\mathrm{H}_2$$ Te down the group.

    \n

    (D) \"JEE

    ", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 2116, "subject": "Chemistry", "question": "

    A and B formed in the following reactions are:

    \n

    $$\\begin{aligned}\n& \\mathrm{CrO}_2 \\mathrm{Cl}_2+4 \\mathrm{NaOH} \\rightarrow \\mathrm{A}+2 \\mathrm{NaCl}+2 \\mathrm{H}_2 \\mathrm{O}, \\\\\n& \\mathrm{A}+2 \\mathrm{HCl}+2 \\mathrm{H}_2 \\mathrm{O}_2 \\rightarrow \\mathrm{B}+3 \\mathrm{H}_2 \\mathrm{O}\n\\end{aligned}$$

    ", "options": [ { "text": "$$\\mathrm{A}=\\mathrm{Na}_2 \\mathrm{Cr}_2 \\mathrm{O}_7, \\mathrm{~B}=\\mathrm{CrO}_5$$\n" }, { "text": "$$\\mathrm{A}=\\mathrm{Na}_2 \\mathrm{CrO}_4, \\mathrm{~B}=\\mathrm{CrO}_5$$\n" }, { "text": "$$\\mathrm{A}=\\mathrm{Na}_2 \\mathrm{Cr}_2 \\mathrm{O}_4, \\mathrm{~B}=\\mathrm{CrO}_4$$\n" }, { "text": "$$\\mathrm{A}=\\mathrm{Na}_2 \\mathrm{Cr}_2 \\mathrm{O}_7, \\mathrm{~B}=\\mathrm{CrO}_3$$" } ], "answer": "$$\\mathrm{A}=\\mathrm{Na}_2 \\mathrm{CrO}_4, \\mathrm{~B}=\\mathrm{CrO}_5$$\n", "solution": "**Answer:** $$\\mathrm{A}=\\mathrm{Na}_2 \\mathrm{CrO}_4, \\mathrm{~B}=\\mathrm{CrO}_5$$\n\n\n

    $$\\mathrm{Cr{O_2} + 4NaOH \\to \\mathop {N{a_2}Cr{O_4}}\\limits_{(A)} + 2NaCl + 2{H_2}O}$$

    \n

    $$\\mathrm{N{a_2}Cr{O_4} + 2{H_2}{O_2} + 2HCl \\to \\mathop {Cr{O_5}}\\limits_{(B)} + \\mathop {2NaCl}\\limits_{(Mis\\sin g\\,from\\,balanced\\,equation)} + 3{H_2}O}$$

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2117, "subject": "Chemistry", "question": "

    Given below are two statements :

    \n

    Statement (I) : The gas liberated on warming a salt with dil $$\\mathrm{H}_2 \\mathrm{SO}_4$$, turns a piece of paper dipped in lead acetate into black, it is a confirmatory test for sulphide ion.

    \n

    Statement (II) : In statement-I the colour of paper turns black because of formation of lead sulphite. In the light of the above statements, choose the most appropriate answer from the options given below :

    ", "options": [ { "text": "Both Statement I and Statement II are false\n" }, { "text": "Statement I is true but Statement II is false\n" }, { "text": "Both Statement I and Statement II are true\n" }, { "text": "Statement I is false but Statement II is true" } ], "answer": "Statement I is true but Statement II is false\n", "solution": "**Answer:** Statement I is true but Statement II is false\n\n\n

    $$\\mathrm{Na}_2 \\mathrm{S}+\\mathrm{H}_2 \\mathrm{SO}_4 \\rightarrow \\mathrm{Na}_2 \\mathrm{SO}_4+\\mathrm{H}_2 \\mathrm{S}$$

    \n

    $$\\mathrm{{(C{H_3}COO)_2}Pb + {H_2}S \\to \\mathop {PbS}\\limits_{Black\\,lead\\,sulphide} + 2C{H_3}COOH}$$

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2118, "subject": "Chemistry", "question": "

    Number of oxygen atoms present in chemical formula of fuming sulphuric acid is ___________.

    ", "options": [], "answer": "7", "solution": "**Answer:** 7\n\n

    Fuming sulfuric acid, often known as oleum, can be represented chemically using the formula $ H_2S_2O_7 $. This compound is formed by adding excess sulfur trioxide ($ SO_3 $) to sulfuric acid ($ H_2SO_4 $). The chemical reaction can be represented as follows:

    \n\n

    $ H_2SO_4 + SO_3 \\rightarrow H_2S_2O_7 $

    \n\n

    From the formula $ H_2S_2O_7 $, to determine the number of oxygen atoms, we look directly at the subscript corresponding to oxygen in the chemical formula. In this case, the subscript is 7, which indicates that there are 7 oxygen atoms present in the molecule of fuming sulfuric acid.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2119, "subject": "Chemistry", "question": "

    Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).

    \n

    Assertion (A) : Both rhombic and monoclinic sulphur exist as $$\\mathrm{S}_8$$ while oxygen exists as $$\\mathrm{O}_2$$.

    \n

    Reason (R) : Oxygen forms $$p \\pi-p \\pi$$ multiple bonds with itself and other elements having small size and high electronegativity like $$\\mathrm{C}, \\mathrm{N}$$, which is not possible for sulphur.

    \n

    In the light of the above statements, choose the most appropriate answer from the options given below :

    ", "options": [ { "text": "Both (A) and (R) are correct but (R) is not the correct explanation of (A)\n" }, { "text": "(A) is correct but (R) is not correct\n" }, { "text": "(A) is not correct but (R) is correct\n" }, { "text": "Both $$(\\mathbf{A})$$ and $$(\\mathbf{R})$$ are correct and $$(\\mathbf{R})$$ is the correct explanation of $$(\\mathbf{A})$$" } ], "answer": "Both $$(\\mathbf{A})$$ and $$(\\mathbf{R})$$ are correct and $$(\\mathbf{R})$$ is the correct explanation of $$(\\mathbf{A})$$", "solution": "**Answer:** Both $$(\\mathbf{A})$$ and $$(\\mathbf{R})$$ are correct and $$(\\mathbf{R})$$ is the correct explanation of $$(\\mathbf{A})$$\n\n

    Oxygen exists as $$\\mathrm{O}_2$$

    \n

    Sulphur exists as $$\\mathrm{S}_8$$

    \n

    Because of small size of oxygen, oxygen can form $$2 \\mathrm{p}_\\pi-2 \\mathrm{p}_\\pi$$ bond.

    \n

    $$\\Rightarrow$$ Assertion (A) is correct.

    \n

    $$\\Rightarrow$$ Reason (R) is correct.

    \n

    Reason (R) is correct explanation of Assertion (A).

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2120, "subject": "Chemistry", "question": "

    On reaction of Lead Sulphide with dilute nitric acid which of the following is not formed?

    ", "options": [ { "text": "Sulphur\n" }, { "text": "Nitric oxide\n" }, { "text": "Lead nitrate\n" }, { "text": "Nitrous oxide" } ], "answer": "Nitrous oxide", "solution": "**Answer:** Nitrous oxide\n\n

    When lead sulphide (PbS) reacts with dilute nitric acid, the reaction is a redox process where lead sulphide is oxidized and nitric acid is reduced. The overall reaction involves the transformation of PbS and HNO3 to form lead nitrate (Pb(NO3)2), sulphur, and nitrogen oxides, along with water. The specific nitrogen oxide formed depends on the reaction conditions, including the concentration of nitric acid and the temperature. Typically, with dilute nitric acid, nitric oxide (NO) and nitrogen dioxide (NO2) are more common products than nitrous oxide (N2O).

    \n\n

    The general reaction can be represented as:

    \n\n$$\\text{PbS} + 4\\text{HNO}_3 \\rightarrow \\text{Pb(NO}_3\\text{)}_2 + \\text{SO}_2 + 2\\text{H}_2\\text{O} + 2\\text{NO}$$\n\n

    Or, under some conditions, sulphur (S) may also be one of the products instead of or in addition to SO2:

    \n\n$$\\text{PbS} + 2\\text{HNO}_3 \\rightarrow \\text{Pb(NO}_3\\text{)}_2 + \\text{S} + 2\\text{H}_2\\text{O}$$\n\n

    From the reaction above, you can see:

    \n\n\n\n

    Therefore, the correct answer to the question is Option D: Nitrous oxide. It is not a typical product of the reaction of lead sulphide with dilute nitric acid under normal conditions.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2121, "subject": "Chemistry", "question": "

    Identify the correct statements about p-block elements and their compounds.

    \n

    (A) Non metals have higher electronegativity than metals.

    \n

    (B) Non metals have lower ionisation enthalpy than metals.

    \n

    (C) Compounds formed between highly reactive nonmetals and highly reactive metals are generally ionic.

    \n

    (D) The non-metal oxides are generally basic in nature.

    \n

    (E) The metal oxides are generally acidic or neutral in nature.

    \n

    Choose the correct answer from the options given below :

    ", "options": [ { "text": "(B) and (D) only\n" }, { "text": "(B) and (E) only\n" }, { "text": "(D) and (E) only\n" }, { "text": "(A) and (C) only" } ], "answer": "(A) and (C) only", "solution": "**Answer:** (A) and (C) only\n\n

    Let's analyze each statement provided about p-block elements and determine their correctness:

    \n\n

    (A) Non metals have higher electronegativity than metals.

    \n

    This statement is correct. Non-metals, especially those in the upper right of the periodic table like fluorine, oxygen, and nitrogen, have higher electronegativity values compared to metals.

    \n\n

    (B) Non metals have lower ionisation enthalpy than metals.

    \n

    This statement is incorrect. Non-metals generally have higher ionization enthalpy compared to metals because they hold onto their electrons more tightly due to higher nuclear charge and smaller atomic radii.

    \n\n

    (C) Compounds formed between highly reactive nonmetals and highly reactive metals are generally ionic.

    \n

    This statement is correct. Highly reactive nonmetals (like halogens) tend to gain electrons while highly reactive metals (like alkali metals and alkaline earth metals) tend to lose electrons, resulting in the formation of ionic compounds.

    \n\n

    (D) The non-metal oxides are generally basic in nature.

    \n

    This statement is incorrect. Non-metal oxides are generally acidic in nature. They react with bases to form salts and water.

    \n\n

    (E) The metal oxides are generally acidic or neutral in nature.

    \n

    This statement is incorrect. Metal oxides are usually basic in nature, although some metal oxides can be amphoteric (both acidic and basic). Most metal oxides react with acids to form salts and water.

    \n\n

    After evaluating all the statements, the correct statements are (A) and (C).

    \n\n

    Therefore, the correct answer is:

    \n

    Option D: (A) and (C) only

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2122, "subject": "Chemistry", "question": "Oxidation number of Cl in CaOCl2 (bleaching powder) is :", "options": [ { "text": "zero, since it contains Cl2" }, { "text": "-1, since it contains Cl-" }, { "text": "+1, since it contains ClO-" }, { "text": "+1 and -1 since it contains ClO- and Cl-" } ], "answer": "+1 and -1 since it contains ClO- and Cl-", "solution": "**Answer:** +1 and -1 since it contains ClO- and Cl-\n\n$$CaOC{l_2} - $$or it can also be written as $$Ca\\mathop {\\left( {OCl} \\right)}\\limits_{{x_1}} \\mathop {Cl}\\limits_{{x_2}} $$\n

    hence oxidation no of $$Cl$$ in $$OC{l^ - }$$ is \n

    $$\\eqalign{\n & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, - 2 + {x_2} = - 1 \\cr \n & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,{x^2} = 2 - 1 = + 1 \\cr} $$\n

    now oxidation no. of another $$Cl$$ is $$-1$$ as it is present as $$C{l^ - }.$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2123, "subject": "Chemistry", "question": "The correct statement for the molecule, CsI3 is :", "options": [ { "text": "it contains Cs3+ and I- ions." }, { "text": "it contains Cs+, I- and lattice I2 molecule." }, { "text": "it is a covalent molecule" }, { "text": "it contains Cs+ and $$I_3^-$$ ions" } ], "answer": "it contains Cs+ and $$I_3^-$$ ions", "solution": "**Answer:** it contains Cs+ and $$I_3^-$$ ions\n\n$$Cs{l_3}\\,\\,$$ dissociates as \n

    $$Cs{l_3} \\to C{s^ + } + I_3^ - $$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2124, "subject": "Chemistry", "question": "The non-metal that does not exhibit positive oxidation state is :", "options": [ { "text": "Oxygen" }, { "text": "Iodine" }, { "text": "Chlorine" }, { "text": "Fluorine" } ], "answer": "Fluorine", "solution": "**Answer:** Fluorine\n\nFluorine is the most electronegative element and it\nshows only –1 oxidation state.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2125, "subject": "Chemistry", "question": "Aqueous solution of which salt will not contain ions with the electronic\nconfiguration 1s22s22p63s23p6 ?", "options": [ { "text": "NaF" }, { "text": "NaCl" }, { "text": "KBr" }, { "text": "CaI2" } ], "answer": "NaF", "solution": "**Answer:** NaF\n\nNaF is composed of Na+\n and F–.\n

    Na+ : 1s22s22p6\n

    F- : 1s22s22p6\n

    Hence do not match with the configuration given in the question.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2126, "subject": "Chemistry", "question": "The species given below that does NOT show disproportionation reaction is :", "options": [ { "text": "$$BrO_4^ - $$" }, { "text": "$$Br{O^ - }$$" }, { "text": "$$BrO_2^ - $$" }, { "text": "$$BrO_3^ - $$" } ], "answer": "$$BrO_4^ - $$", "solution": "**Answer:** $$BrO_4^ - $$\n\nIn BrO4-\n, Br is in maximum oxidation state (+7), Soit cannot oxidise further it only reduced hence\nit cannot show disproportionation reaction.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2127, "subject": "Chemistry", "question": "

    Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R.

    \n

    Assertion A : Fluorine forms one oxoacid.

    \n

    Reason R : Fluorine has smallest size amongst all halogens and is highly electronegative.

    \n

    In the light of the above statements, choose the most appropriate answer from the options given below :

    ", "options": [ { "text": "Both A and R are correct and R is the correct explanation of A." }, { "text": "Both A and R are correct but R is NOT the correct explanation of A." }, { "text": "A is correct but R is not correct." }, { "text": "A is not correct but R is correct." } ], "answer": "Both A and R are correct and R is the correct explanation of A.", "solution": "**Answer:** Both A and R are correct and R is the correct explanation of A.\n\nDue to its smaller size, fluorine forms only one oxoacid.

    \nBoth the Assertion and Reason are correct and The reason is the correct explanation.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2128, "subject": "Chemistry", "question": "

    White precipitate of AgCl dissolves in aqueous ammonia solution due to formation of :

    ", "options": [ { "text": "[Ag(NH3)4]Cl2" }, { "text": "[Ag(Cl)2(NH3)2]" }, { "text": "[Ag(NH3)2]Cl" }, { "text": "[Ag(NH3)Cl]Cl" } ], "answer": "[Ag(NH3)2]Cl", "solution": "**Answer:** [Ag(NH3)2]Cl\n\n$$\n\\mathrm{AgCl} +2\\mathrm{NH}_{3} \\rightarrow \\begin{gathered}\\left[ \\mathrm{Ag} \\left( \\mathrm{NH}_{3} \\right)_{2} \\right]^{+} \\mathrm{Cl}^{-} \\\\ \\text{Soluble} \\end{gathered} \n$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2129, "subject": "Chemistry", "question": "

    For electron gain enthalpies of the elements denoted as $$\\Delta_{\\mathrm{eg}} \\mathrm{H}$$, the incorrect option is :

    ", "options": [ { "text": "$$\\Delta_{\\mathrm{eg}} \\mathrm{H}(\\mathrm{I})<\\Delta_{\\mathrm{eg}} \\mathrm{H}(\\mathrm{At})$$" }, { "text": "$$\\Delta_{\\mathrm{eg}} \\mathrm{H}(\\mathrm{Te})<\\Delta_{\\mathrm{eg}} \\mathrm{H}(\\mathrm{Po})$$" }, { "text": "$$\\Delta_{\\mathrm{eg}} \\mathrm{H}(\\mathrm{Cl})<\\Delta_{\\mathrm{eg}} \\mathrm{H}(\\mathrm{F})$$" }, { "text": "$$\\Delta_{\\mathrm{eg}} \\mathrm{H}(\\mathrm{Se})<\\Delta_{\\mathrm{eg}} \\mathrm{H}(\\mathrm{S})$$" } ], "answer": "$$\\Delta_{\\mathrm{eg}} \\mathrm{H}(\\mathrm{Cl})<\\Delta_{\\mathrm{eg}} \\mathrm{H}(\\mathrm{F})$$", "solution": "**Answer:** $$\\Delta_{\\mathrm{eg}} \\mathrm{H}(\\mathrm{Cl})<\\Delta_{\\mathrm{eg}} \\mathrm{H}(\\mathrm{F})$$\n\n$\n\n \\Delta \\mathrm{H}_{\\mathrm{eg}}(\\mathrm{Cl})=-349 \\mathrm{~kJ} / \\text { mole } $\n

    $\\Delta \\mathrm{H}_{\\mathrm{eg}}(\\mathrm{F})=-333 \\mathrm{~kJ} / \\mathrm{mole} $

    $\n \\Delta \\mathrm{H}_{\\mathrm{eg}}(\\mathrm{I})=-296 \\mathrm{~kJ} / \\mathrm{mole} $

    $\n \\Delta \\mathrm{H}_{\\mathrm{eg}}(\\mathrm{Se})=-195 \\mathrm{~kJ} / \\mathrm{mole} $

    $\n \\Delta \\mathrm{H}_{\\mathrm{eg}}(\\mathrm{S})=-200 \\mathrm{~kJ} / \\text { mole } $

    $\n \\Delta \\mathrm{H}_{\\mathrm{eg}}(\\mathrm{Te})=-190 \\mathrm{~kJ} / \\mathrm{mole} $

    $\n \\Delta \\mathrm{H}_{\\mathrm{eg}}(\\mathrm{Po})=-174 \\mathrm{~kJ} / \\mathrm{mole}\n\n$\n

    Electron gain enthalpy of $\\mathrm{Se}$ is less negative than that of sulphur.\n

    Option A: ΔegH(I) < ΔegH(At)\n

    Iodine (I) is above Astatine (At) in Group 17 (Halogens), so ΔegH(I) should be more negative than ΔegH(At). This statement is correct.\n\n

    Option B: ΔegH(Te) < ΔegH(Po)\n

    Tellurium (Te) is above Polonium (Po) in Group 16 (Chalcogens), so ΔegH(Te) should be more negative than ΔegH(Po). This statement is correct.\n\n

    Option C: ΔegH(Cl) < ΔegH(F)\n

    Chlorine (Cl) is below Fluorine (F) in Group 17 (Halogens). However, the electron gain enthalpy of Chlorine is more negative than that of Fluorine. So, this statement is incorrect.\n\n

    Option D: ΔegH(Se) < ΔegH(S)\n

    Selenium (Se) is below Sulfur (S) in Group 16 (Chalcogens), so ΔegH(Se) should be less negative (higher) than ΔegH(S). This statement is correct.\n

    Note :\n

    Although the general trend across a period in the periodic table is for the electron gain enthalpy to become more negative, there is an exception between Fluorine (F) and Chlorine (Cl) in the halogen group (Group 17).

    \n

    The electron gain enthalpy of Chlorine is more negative than that of Fluorine for the following reasons:

    \n
      \n

    1. Small atomic size of Fluorine: Fluorine has the smallest atomic size among all the elements in the halogen group. As a result, the electron-electron repulsion in the 2p orbital of Fluorine is significant when an extra electron is added. This repulsion reduces the energy released during the electron gain process, leading to a less negative electron gain enthalpy value for Fluorine compared to Chlorine.

      \n
      2. \n

    2. High effective nuclear charge in Fluorine: Fluorine has a higher effective nuclear charge than Chlorine, which means the positively charged nucleus attracts the negatively charged electrons more strongly in Fluorine. The high effective nuclear charge in Fluorine also contributes to the electron-electron repulsion in the 2p orbital, further reducing the energy released when gaining an electron.

      \n
      3. \n
    \n

    Despite the general trend of electron gain enthalpy becoming more negative across a period, the combination of Fluorine's small atomic size and high effective nuclear charge results in Chlorine having a more negative electron gain enthalpy than Fluorine.

    \n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2130, "subject": "Chemistry", "question": "

    A. Ammonium salts produce haze in atmosphere.

    \n

    B. Ozone gets produced when atmospheric oxygen reacts with chlorine radicals.

    \n

    C. Polychlorinated biphenyls act as cleansing solvents.

    \n

    D. 'Blue baby' syndrome occurs due to the presence of excess of sulphate ions in water.

    \n

    Choose the correct answer from the options given below :

    ", "options": [ { "text": "B and C only" }, { "text": "A and D only" }, { "text": "A, B and C only" }, { "text": "A and C only" } ], "answer": "A and C only", "solution": "**Answer:** A and C only\n\n

    Ammonium salts produce haze in atmosphere.\nOzone is produced when atmospheric oxygen\nreacts with oxygen atoms and not chlorine atoms.\nPolychlorinated biphenyls have number of\napplications including their use as cleansing\nsolvents.

    \n

    ‘Blue baby’ syndrome occurs due to the presence\nof excess of nitrate ions and not sulphate ions in\nwater.

    ", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 2131, "subject": "Chemistry", "question": "

    In the following reactions, the total number of oxygen atoms in X and Y is ___________.

    \n

    Na$$_2$$O + H$$_2$$O $$\\to$$ 2X

    \n

    Cl$$_2$$O$$_7$$ + H$$_2$$O $$\\to$$ 2Y

    ", "options": [], "answer": "5", "solution": "**Answer:** 5\n\n$$\n\\begin{aligned}\n& \\mathrm{Na}_2 \\mathrm{O}+\\mathrm{H}_2 \\mathrm{O} \\longrightarrow 2 \\mathrm{NaOH}(\\mathrm{X}) \\\\\\\\\n& \\mathrm{Cl}_2 \\mathrm{O}_7+\\mathrm{H}_2 \\mathrm{O} \\longrightarrow 2 \\mathrm{HClO}_4(\\mathrm{Y})\n\\end{aligned}\n$$

    \n$X$ has one $O$ and $Y$ has four $O$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2132, "subject": "Chemistry", "question": "

    Element not showing variable oxidation state is :

    ", "options": [ { "text": "Bromine\n" }, { "text": "Iodine\n" }, { "text": "Chlorine\n" }, { "text": "Fluorine" } ], "answer": "Fluorine", "solution": "**Answer:** Fluorine\n\n

    Fluorine is the element among the options that does not exhibit a variable oxidation state, unlike its fellow halogens. This is because fluorine is the most electronegative element, and it always forms compounds in the oxidation state of -1 as it has a strong tendency to gain one electron to achieve a stable electronic configuration.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2133, "subject": "Chemistry", "question": "

    Identify the incorrect pair from the following :

    ", "options": [ { "text": "Fluoroapatite $$-3 \\mathrm{Ca}_3\\left(\\mathrm{PO}_4\\right)_2 \\cdot \\mathrm{CaF}_2$$\n" }, { "text": "Carnallite $$-\\mathrm{KCl} \\cdot \\mathrm{MgCl}_2 \\cdot 6 \\mathrm{H}_2 \\mathrm{O}$$\n" }, { "text": "Cryolite $$-\\mathrm{Na}_3 \\mathrm{AlF}_6$$\n" }, { "text": "Fluorspar $$-\\mathrm{BF}_3$$" } ], "answer": "Fluorspar $$-\\mathrm{BF}_3$$", "solution": "**Answer:** Fluorspar $$-\\mathrm{BF}_3$$\n\n

    (1) Fluorspar is $$\\mathrm{CaF}_2$$

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2134, "subject": "Chemistry", "question": "

    The Lassiagne's extract is boiled with dil $$\\mathrm{HNO}_3$$ before testing for halogens because,

    ", "options": [ { "text": "Silver halides are soluble in $$\\mathrm{HNO}_3$$.\n" }, { "text": "$$\\mathrm{AgCN}$$ is soluble in $$\\mathrm{HNO}_3$$.\n" }, { "text": "$$\\mathrm{Na}_2 \\mathrm{S}$$ and $$\\mathrm{NaCN}$$ are decomposed by $$\\mathrm{HNO}_3$$.\n" }, { "text": "$$\\mathrm{Ag}_2 \\mathrm{S}$$ is soluble in $$\\mathrm{HNO}_3$$." } ], "answer": "$$\\mathrm{Na}_2 \\mathrm{S}$$ and $$\\mathrm{NaCN}$$ are decomposed by $$\\mathrm{HNO}_3$$.\n", "solution": "**Answer:** $$\\mathrm{Na}_2 \\mathrm{S}$$ and $$\\mathrm{NaCN}$$ are decomposed by $$\\mathrm{HNO}_3$$.\n\n\n

    If nitrogen or sulphur is also present in the compound, the sodium fusion extract is first boiled with concentrated nitric acid to decompose cyanide or sulphide of sodium during Lassaigne's test

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2135, "subject": "Chemistry", "question": "

    Among the following halogens

    \n

    $$\\mathrm{F}_2, \\mathrm{Cl}_2, \\mathrm{Br}_2 \\text { and } \\mathrm{I}_2$$

    \n

    Which can undergo disproportionation reactions?

    ", "options": [ { "text": "$$\\mathrm{F}_2$$ and $$\\mathrm{Cl}_2$$\n" }, { "text": "$$\\mathrm{Cl}_2, \\mathrm{Br}_2$$ and $$\\mathrm{I}_2$$\n" }, { "text": "Only $$\\mathrm{I}_2$$\n" }, { "text": "$$\\mathrm{F}_2, \\mathrm{Cl}_2$$ and $$\\mathrm{Br}_2$$" } ], "answer": "$$\\mathrm{Cl}_2, \\mathrm{Br}_2$$ and $$\\mathrm{I}_2$$\n", "solution": "**Answer:** $$\\mathrm{Cl}_2, \\mathrm{Br}_2$$ and $$\\mathrm{I}_2$$\n\n\n

    To determine which halogens among $$\\mathrm{F}_2, \\mathrm{Cl}_2, \\mathrm{Br}_2$$, and $$\\mathrm{I}_2$$ can undergo disproportionation reactions, we need to understand what a disproportionation reaction is.

    \n\n

    In a disproportionation reaction, a single substance is simultaneously oxidized and reduced, forming two different products. The halogen X2 disproportionates in water as follows:

    \n\n

    \n\n

    $$ \\mathrm{X}_2 + \\mathrm{H}_2\\mathrm{O} \\rightarrow \\mathrm{HX} + \\mathrm{HXO} $$

    \n\n

    \n\n

    Here, X represents a halogen.

    \n\n

    Let's analyze each halogen to see if disproportionation is possible:

    \n\n

    Fluorine ($$\\mathrm{F}_2$$): Fluorine is the most electronegative element and has a very high oxidation potential. It does not undergo disproportionation because it prefers to remain in the $$-1$$ oxidation state and does not form higher oxidation states easily. Therefore, $$\\mathrm{F}_2$$ cannot undergo a disproportionation reaction.

    \n\n

    Chlorine ($$\\mathrm{Cl}_2$$): Chlorine can undergo disproportionation. In water, chlorine disproportionates as follows:

    \n\n

    \n\n

    $$ \\mathrm{Cl}_2 + \\mathrm{H}_2\\mathrm{O} \\rightarrow \\mathrm{HCl} + \\mathrm{HOCl} $$

    \n\n

    \n\n

    Here, chlorine is both reduced to $$\\mathrm{HCl}$$ (chloride, -1 oxidation state) and oxidized to $$\\mathrm{HOCl}$$ (hypochlorite, +1 oxidation state).

    \n\n

    Bromine ($$\\mathrm{Br}_2$$): Bromine can also undergo disproportionation. In water, bromine disproportionates as follows:

    \n\n

    \n\n

    $$ \\mathrm{Br}_2 + \\mathrm{H}_2\\mathrm{O} \\rightarrow \\mathrm{HBr} + \\mathrm{HOBr} $$

    \n\n

    \n\n

    Here, bromine is both reduced to $$\\mathrm{HBr}$$ (bromide, -1 oxidation state) and oxidized to $$\\mathrm{HOBr}$$ (hypobromite, +1 oxidation state).

    \n\n

    Iodine ($$\\mathrm{I}_2$$): Iodine can also undergo disproportionation. In alkaline solutions, iodine disproportionates as follows:

    \n\n

    \n\n

    $$ \\mathrm{I}_2 + \\mathrm{OH}^- \\rightarrow \\mathrm{I}^- + \\mathrm{IO}^- $$

    \n\n

    \n\n

    Here, iodine is both reduced to $$\\mathrm{I}^-$$ (iodide, -1 oxidation state) and oxidized to $$\\mathrm{IO}^-$$ (hypoiodite, +1 oxidation state).

    \n\n

    Based on the above analysis, all halogens except $$\\mathrm{F}_2$$ can undergo disproportionation. Therefore, the correct answer is:

    \n\n

    Option B

    \n\n

    $$\\mathrm{Cl}_2, \\mathrm{Br}_2$$ and $$\\mathrm{I}_2$$

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2136, "subject": "Chemistry", "question": "Which one the following statement regarding helium is incorrect? ", "options": [ { "text": "It is used to fill gas balloons instead of hydrogen because it is lighter and non–inflammable " }, { "text": "It is used in gas – cooled nuclear reactors" }, { "text": "It is used to produce and sustain powerful superconducting reagents " }, { "text": "It is used as cryogenic agent for carrying out experiments at low temperatures " } ], "answer": "It is used to fill gas balloons instead of hydrogen because it is lighter and non–inflammable ", "solution": "**Answer:** It is used to fill gas balloons instead of hydrogen because it is lighter and non–inflammable \n\nHelium is heavier than hydrogen although it is non-inflammable", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2137, "subject": "Chemistry", "question": "Which one has the highest boiling point?", "options": [ { "text": "Ne" }, { "text": "Kr" }, { "text": "Xe" }, { "text": "He" } ], "answer": "Xe", "solution": "**Answer:** Xe\n\n$$Xe.$$ As we move down the group, the melting and boiling points show a regular increase due to corresponding increase in the magnitude of their van der waal forces of attraction as the size of the atom increases. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2138, "subject": "Chemistry", "question": "Which intermolecular force is most responsible in allowing xenon gas to liquefy?", "options": [ { "text": "Dipole - dipole" }, { "text": "Ion - dipole" }, { "text": "Instantaneous dipole - induced dipole" }, { "text": "Ionic" } ], "answer": "Instantaneous dipole - induced dipole", "solution": "**Answer:** Instantaneous dipole - induced dipole\n\nInstantaneous dipole-induced dipole forces or van der Waals’ forces are most responsible in allowing xenon gas to\nliquify.\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2139, "subject": "Chemistry", "question": "The noble gas that does not occur in the atmosphere is : ", "options": [ { "text": "Ne" }, { "text": "He" }, { "text": "Kr" }, { "text": "Ra" } ], "answer": "Ra", "solution": "**Answer:** Ra\n\nInert gas Radon(Ra) is not present in atmosphere. Remaining all the inert gas can be found in the atmosphere.", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 2140, "subject": "Chemistry", "question": "

    Give below are two statements:

    \n

    Statement - I: Noble gases have very high boiling points.

    \n

    Statement - II: Noble gases are monoatomic gases. They are held together by strong dispersion forces. Because of this they are liquefied at very low temperature. Hence, they have very high boiling points.

    \n

    In the light of the above statements, choose the correct answer from the options given below:

    ", "options": [ { "text": "Both Statement I and Statement II are false.\n" }, { "text": "Statement I is true but Statement II is false.\n" }, { "text": "Statement I is false but Statement II is true.\n" }, { "text": "Both Statement I and Statement II are true." } ], "answer": "Both Statement I and Statement II are false.\n", "solution": "**Answer:** Both Statement I and Statement II are false.\n\n\n

    Option A is the correct answer. Both Statement I and Statement II are false.

    \n

    Noble gases actually have low boiling points, not high. This is because noble gases are indeed monoatomic gases that exist as single atoms. They have a completely filled valence shell and do not easily form chemical bonds with other atoms. As a result, the intermolecular forces present in noble gases are only temporary dipole-induced dipole attractions, also known as dispersion forces (or London forces). However, these dispersion forces are very weak compared to other types of intermolecular forces like hydrogen bonding or dipole-dipole interactions that occur in other substances.

    \n

    The dispersion forces in noble gases are weak because these gases are non-polar and since they are monoatomic, the electron cloud around the atom is symmetrical, meaning that there are no permanent dipoles within the molecules. As the atomic size of noble gases increases down the group, the dispersion forces do increase in strength due to the larger electron cloud that can be more easily distorted, resulting in higher boiling points for the heavier noble gases. Still, compared to other substances, the overall strength of the dispersion forces in noble gases is low, which means they can be liquefied at low temperatures and hence their boiling points are low, not high.

    \n

    Therefore, the correct statement would be: \"Noble gases are monoatomic gases. They are held together by weak dispersion forces. Because of this they are liquefied at very low temperatures. Hence, they have very low boiling points.\"

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2141, "subject": "Chemistry", "question": "The correct order of ionic radius is ", "options": [ { "text": "$$Ce > Sm > Tb > Lu$$ " }, { "text": "$$Lu > Tb > Sm > Ce$$ " }, { "text": "$$Tb > Lu > Sm > Ce$$ " }, { "text": "$$Sm > Tb > Lu > Ce$$ " } ], "answer": "$$Ce > Sm > Tb > Lu$$ ", "solution": "**Answer:** $$Ce > Sm > Tb > Lu$$ \n\nThose are the elements of lanthanides series. \n

    \n\"AIEEE\n
    In lanthanide series the extra electron is added in $$4f$$ subshell and we know in periodic table from left to right the effective nuclear charge increases as electron is added so the size of elements will decreases. \n

    So, the correct order is $$Ce > Sm > Tb > Lu.$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2142, "subject": "Chemistry", "question": "$$C{e^{ + 3}},\\,\\,L{a^{ + 3}},\\,\\,P{m^{ + 3}}\\,\\,$$ and $$Y{b^{ + 3}}\\,\\,$$ have ionic radial in the increasing order as ", "options": [ { "text": "$$L{a^{ + 3}}\\,\\, < C{e^{ + 3}}\\,\\, < P{m^{ + 3}}\\,\\, < Y{b^{ + 3}}$$ " }, { "text": "$$Y{b^{ + 3}}\\,\\, < P{m^{ + 3}}\\,\\, < C{e^{ + 3}}\\,<\\,L{a^{ + 3}}$$ " }, { "text": "$$L{a^{ + 3}}\\,\\, = C{e^{ + 3}}\\,\\, < P{m^{ + 3}}\\,\\, < Y{b^{ + 3}}$$ " }, { "text": "$$Y{b^{ + 3}}\\,\\, < P{m^{ + 3}}\\,\\, < L{a^{ + 3}}\\,\\, < C{e^{ + 3}}$$ " } ], "answer": "$$Y{b^{ + 3}}\\,\\, < P{m^{ + 3}}\\,\\, < C{e^{ + 3}}\\,<\\,L{a^{ + 3}}$$ ", "solution": "**Answer:** $$Y{b^{ + 3}}\\,\\, < P{m^{ + 3}}\\,\\, < C{e^{ + 3}}\\,<\\,L{a^{ + 3}}$$ \n\n\"AIEEE\n
    Here $$C{e^{ + 3}},\\,\\,P{m^{ + 3}},\\,\\,Y{b^{ + 3}}$$ are from lanthanide series.\n

    $$L{a^{ + 3}}$$ atomic number is $$57.$$ This is a $$d$$ block element but all the property of $$La$$ is similar to lanthanides. It is present before $$Ce.$$ \n

    As we know in lanthanides the radius of elements decreases from left to right of periodic table. \n

    So, order will be $$L{a^{ + 3}} > C{e^{ + 3}} > P{m^{ + 3}} > Y{b^{ + 3}}$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2143, "subject": "Chemistry", "question": "The reduction in atomic size with increase in atomic number is a characteristic of elements of ", "options": [ { "text": "$$d$$ - block" }, { "text": "$$f$$ - block" }, { "text": "Radioactive series " }, { "text": "High atomic masses." } ], "answer": "High atomic masses.", "solution": "**Answer:** High atomic masses.\n\nThis can't be $$f$$ - block elements as $$f$$ block means only lanthanides and actinides but we know lanthanide contraction happens on the elements before lantanidine elements also. So, radius also decrease for those element when increase the atomic number. \n

    Radioactive Series starts from atomic number $$84$$ ( Polonium ) and affer that all are radioactive. But we know before $$84$$ atomic number also the atomic size decrease with increase in atomic number. So, only radioactive series don't show this characteristic. \n

    So, correct answer should be high atomic mass. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2144, "subject": "Chemistry", "question": "Which one of the following ions has the highest value of ionic radius?", "options": [ { "text": "O2-" }, { "text": "B3+" }, { "text": "Li+" }, { "text": "F-" } ], "answer": "O2-", "solution": "**Answer:** O2-\n\nFrom a neutral atom where a electron is removed then it become cation and when a electron is added to the atom then it becomes anion.\n

    For a atom $$x$$ when it loose a electron then it becomes cation $$x - {e^ - } \\to {x^ + }$$\n

    Let atomic no. of $$x$$ is $$z$$\n

    then no. of proton in $${x^ + } = z$$\n

    and no. of electron in $${x^ + } = z - 1$$\n

    So the ratio $${z \\over e}$$ as no. of electrons decreases. And when no of electron decreases then per electron attraction increases from nucleus and radius of cation deceases. \n

    Similarly when atom $$x$$ becomes $${x^ - }$$ by adding a electron $$\\left( {x + {e^ - } \\to {x^ - }} \\right)$$ then no. of proton in $${x^ - } = z$$ and no. of electron in $${x^ - } = z + 1.$$\n

    Now the ratio of $${z \\over e}\\,\\,$$ in $${x^ - }$$ decreases as no. of electron increases. As electron increases in $${x^ - }$$ then attraction from nucleus decreases on per electron and the distance between electron and nucleus increases so the radius also increase in anion.\n

    All $$4$$ ions are from second period and for $${O^{2 - }}\\,\\,$$ and $${F^ - }$$ both have $$1s,$$ $$2s$$ and $$2p$$ shell and for $$L{i^ + }\\,\\,$$ and $${B^{3 + }}$$ both have $$1s$$ shell.\n

    So, size of $${O^{2 - }}\\,\\,\\,$$ and $$\\,\\,\\,{F^ - }\\,\\,\\,$$ are more then $$L{i^ + }\\,\\,\\,\\,$$ \n

    and $$\\,\\,\\,\\,{B^{3 + }}$$ and among $${O^{2 - }}\\,\\,\\,$$ and $$\\,\\,\\,\\,{F^ - }$$ \n

    For $$\\,\\,\\,$$ $${O^{2 - }},\\,\\,{z \\over e} = {8 \\over {10}} = 0.8$$\n

    For $$\\,\\,\\,$$ $${F^ - },\\,\\,\\,{z \\over e} = {9 \\over {10}} = 0.9$$ \n

    as for $${O^{2 - }},\\,\\,{z \\over e}\\,\\,\\,$$ is less so per electron attraction from nucleus decreases and radius increases more than $${F^ - }.$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2145, "subject": "Chemistry", "question": "The set representing the correct order of ionic radius is : ", "options": [ { "text": "$$N{a^ + } > L{i^ + } > M{g^{2 + }} > B{e^{2 + }}$$ " }, { "text": "$$L{i^ + } > N{a^ + } > M{g^{2 + }} > B{e^{2 + }}$$ " }, { "text": "$$M{g^{2 + }} > B{e^{2 + }} > L{i^ + } > N{a^ + }$$ " }, { "text": "$$L{i^ + } > B{e^{2 + }} > N{a^ + } > M{g^{2 + }}$$ " } ], "answer": "$$N{a^ + } > L{i^ + } > M{g^{2 + }} > B{e^{2 + }}$$ ", "solution": "**Answer:** $$N{a^ + } > L{i^ + } > M{g^{2 + }} > B{e^{2 + }}$$ \n\nIn periodic table $$Li, Na, Mg, Be$$ are present in following ways :-\n

    \n\n \n \n \n \n \n \n \n \n \n \n \n \n
    Group  1Group  2
    LiBe
    NaMg
    \n

    Note :\n

    $$(1)\\,\\,\\,$$ In periodic table from left to right the size of element decreases. \n

    $$(2)\\,\\,\\,$$ In periodic table, from top to bottom size of element increases.\n

    So, $$B{e^{ + 2}}$$ will be the smallest in size. \n

    Among $$L{i^ + }$$ and $$N{a^ + },$$ size of $$N{a^ + } > L{i^ + }$$ and among $$B{e^{ + 2}}$$ and $$M{g^{ + 2}},$$ size of $$M{g^{ + 2}} > B{e^{ + 2}}$$.\n

    Note : Size of $$L{i^ + } > M{g^{ + 2}},$$ this is a exception case.\n

    So, the correct order will be, $$N{a^ + } > L{i^ + } > M{g^{ + 2}} > B{e^{ + 2}}$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2146, "subject": "Chemistry", "question": "The correct sequence which shows decreasing order of the ionic radii of the elements is", "options": [ { "text": "Al3+ > Mg2+ > Na+ > F- > O2-" }, { "text": "Na+ > Mg2+ > Al3+ > O2– > F–" }, { "text": "Na+ > F– > Mg2+ > O2– > Al3+" }, { "text": "O2– > F– > Na+ > Mg2+ > Al3+" } ], "answer": "O2– > F– > Na+ > Mg2+ > Al3+", "solution": "**Answer:** O2– > F– > Na+ > Mg2+ > Al3+\n\nAs you can see all of them are isoelectronic and for isoelectric elements which ion has more positive charge will have lesser size. And which ion has more negative charge that ion's size will be more. \n

    So, the correct order is \n

    $${O^{2 - }} > {F^ - } > N{a^ + } > m{g^{2 + }} > A{l^{3 + }}$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2147, "subject": "Chemistry", "question": "The increasing order of the ionic radii of the given isoelectronic species is :", "options": [ { "text": "Cl–, Ca2+ , K+ , S2–" }, { "text": "S2–, Cl–, Ca2+ , K+" }, { "text": "Ca2+ , K+ , Cl– , S2–" }, { "text": "K+, S2–, Ca2+, Cl–" } ], "answer": "Ca2+ , K+ , Cl– , S2–", "solution": "**Answer:** Ca2+ , K+ , Cl– , S2–\n\n

    Iso-electronic ions have same number of electrons. So, for iso-electronic ions, number of electrons = constant. \n
    $$ \\therefore $$ $$\\sigma $$(Slaten's Constant) = Constant. As $$\\sigma $$ depends on number of electrons. If a element's number of electrons increases then that element's $$\\sigma $$ increases.

    \n

    Also we know, Zeffective = Z - $$\\sigma $$

    \n

    As for iso-electronic ions, $$\\sigma $$(Slaten's Constant) = Constant. So Zeffective depend on only value of Z. If Z of an ion increases then Zeffective also increases and if Z of an ion decreases then Zeffective also decreases. And when Zeffective increases then nuclear attraction towards outermost electrons increase and size of ion decreases. Similarly when Zeffective decrease then nuclear attraction towards outermost electrons decreases and size of ion increases.

    \n

    $$ \\therefore $$ Size $$ \\propto $$ $${1 \\over {{Z_{eff}}\\,or\\,Z}}$$

    \n

    \n\n\n \n \n \n \n \n \n \n\n\n \n \n \n \n \n \n \n\n
    Ca2+K+Cl-S2-
    Z20191716

    \n

    $$ \\therefore $$ Ionic radius order\n

    Ca2+ < K+ < Cl– < S2–

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2148, "subject": "Chemistry", "question": "The ionic radii (in Å) of N3–, O2– and F– are respectively:", "options": [ { "text": "1.36, 1.71 and 1.40" }, { "text": "1.71, 1.40 and 1.36" }, { "text": "1.71, 1.36 and 1.40" }, { "text": "1.36, 1.40 and 1.71" } ], "answer": "1.71, 1.40 and 1.36", "solution": "**Answer:** 1.71, 1.40 and 1.36\n\nHere all of them are isoelectric and for isoelectric species size of anion increases as negative charge increases.\n

    So, correct order is :\n

    N3– > O2– > F–\n

    $$ \\therefore $$ Radius of N3– = 1.71 Å\n

    and Radius of \nO2– = 1.40 Å\n

    and Radius of \nF– = 1.36 Å", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2149, "subject": "Chemistry", "question": "For Na+, Mg2+, F- and O2-; the correct order of increasing ionic radii is :", "options": [ { "text": "O2- < F- < Na+ < Mg2+" }, { "text": "Na+ < Mg2+ < F- < O2-" }, { "text": "Mg2+ < Na+ < F- < O2-" }, { "text": "Mg2+ < O2- < Na+ < F-" } ], "answer": "Mg2+ < Na+ < F- < O2-", "solution": "**Answer:** Mg2+ < Na+ < F- < O2-\n\nHere all of them are isoelectric. For isoelectric anion size is more than cation. \nAmong two anions which anion has more negative charge will have more radius and among two cation which cation has less positive charge will have more radius.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2150, "subject": "Chemistry", "question": "The correct order of the atomic radii of C, Cs, Al, and S is : ", "options": [ { "text": "S < C < Al < Cs " }, { "text": "S < C < Cs < Al" }, { "text": "C < S < Al < Cs " }, { "text": "C < S < Cs < Al \n" } ], "answer": "C < S < Al < Cs ", "solution": "**Answer:** C < S < Al < Cs \n\nAtomic radii increase by moving down the group and\ndecrease across a period.

    Hence, the correct order of atomic\nradii is : C < S < Al < Cs.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2151, "subject": "Chemistry", "question": "The correct order of the ionic radii of
    O2–, N3–, F–\n, Mg2+, Na+ and Al3+ is :", "options": [ { "text": "N3– < O2– < F–\n < Na+ < Mg2+ < Al3+" }, { "text": "N3– < F–\n < O2– < Mg2+ < Na+ < Al3+" }, { "text": "Al3+ < Na+ < Mg2+ < O2– < F–\n < N3–" }, { "text": "Al3+ < Mg2+ < Na+ < F–\n < O2– < N3–" } ], "answer": "Al3+ < Mg2+ < Na+ < F–\n < O2– < N3–", "solution": "**Answer:** Al3+ < Mg2+ < Na+ < F–\n < O2– < N3–\n\nFor isoelectronic species, as the no. of protons\nincreases, size of ions decreases.\n

    $$ \\therefore $$ Correct order of size for isoelectronic species\n

    Al3+ < Mg2+ < Na+ < F–\n < O2– < N3–", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2152, "subject": "Chemistry", "question": "The ionic radii of O2–, F–, Na+ and Mg2+ are in\nthe order :", "options": [ { "text": "F– > O2– > Na+ > Mg2+" }, { "text": "Mg2+ > Na+ > F– > O2–" }, { "text": "O2– > F– > Mg2+ > Na+\n" }, { "text": "O2– > F– > Na+ > Mg2+" } ], "answer": "O2– > F– > Na+ > Mg2+", "solution": "**Answer:** O2– > F– > Na+ > Mg2+\n\nAmong isoelectronic species, greater the Zeff\n smaller will be the radius.\n

    Order of Zeff : Mg2+ > Na+ > F– > O2–\n

    Order of Ionic Radii : O2– > F– > Na+ > Mg2+", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2153, "subject": "Chemistry", "question": "The increasing order of the atomic radii of the\nfollowing elements is :-
    \n(a) C (b) O
    (c) F (d) Cl
    \n(e) Br", "options": [ { "text": "(a) < (b) < (c) < (d) < (e)" }, { "text": "(c) < (b) < (a) < (d) < (e)" }, { "text": "(b) < (c) < (d) < (a) < (e)" }, { "text": "(d) < (c) < (b) < (a) < (e)" } ], "answer": "(c) < (b) < (a) < (d) < (e)", "solution": "**Answer:** (c) < (b) < (a) < (d) < (e)\n\nGenerally in a period Left to Right Atomic radius decrease.\n

    In a Group Top to Bottom Atomic radius increase.\n

    Atomic radius order : Br > Cl > C > O > F", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2154, "subject": "Chemistry", "question": "The ionic radius of Na+ ions is 1.02 $$\\mathop A\\limits^o $$. The ionic radii (in $$\\mathop A\\limits^o $$) of Mg2+ and Al3+, respectively, are ", "options": [ { "text": "1.05 and 0.99" }, { "text": "0.72 and 0.54" }, { "text": "0.85 and 0.99" }, { "text": "0.68 and 0.72" } ], "answer": "0.72 and 0.54", "solution": "**Answer:** 0.72 and 0.54\n\n

    The ionic radii (in $$\\mathop A\\limits^o $$) of $$\\text{Mg}^{2+}$$ and $$\\text{Al}^{3+}$$, respectively, are:

    \n

    Option B:

    \n\n

    $$0.72 \\mathop A\\limits^o$$ for $$\\text{Mg}^{2+}$$

    \n

    $$0.54 \\mathop A\\limits^o$$ for $$\\text{Al}^{3+}$$

    \n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2155, "subject": "Chemistry", "question": "The ionic radii of K+, Na+, Al3+ and Mg2+ are in the order:", "options": [ { "text": "Na+ < K+ < Mg2+ < Al3+" }, { "text": "Al3+ < Mg2+ < K+ < Na+" }, { "text": "Al3+ < Mg2+ < Na+ < K+" }, { "text": "K+ < Al3+ < Mg2+ < Na+" } ], "answer": "Al3+ < Mg2+ < Na+ < K+", "solution": "**Answer:** Al3+ < Mg2+ < Na+ < K+\n\nAl3+, Mg2+ and Na+ are isoelectronic ionic species. For monoatomic ionic isoelectronic species as positive charge increases ionic size decreases.

    The order of size of Na+ & K+ is Na+ < K+

    $$\\therefore$$ Order of ionic radii is : Al3+ < Mg2+ < Na+ < K+", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2156, "subject": "Chemistry", "question": "The ionic radii of F$$-$$ and O2$$-$$ respectively are 1.33$$\\mathop A\\limits^o $$ and 1.4$$\\mathop A\\limits^o $$, while the covalent radius of N is 0.74$$\\mathop A\\limits^o $$.

    The correct statement for the ionic radius of N3$$-$$ from the following is :", "options": [ { "text": "It is smaller than F$$-$$ and N" }, { "text": "It is bigger than O2$$-$$ and F$$-$$" }, { "text": "It is bigger than F$$-$$ and N, but smaller than of O2$$-$$" }, { "text": "It is smaller than O2$$-$$ and F$$-$$, but bigger than of N" } ], "answer": "It is bigger than O2$$-$$ and F$$-$$", "solution": "**Answer:** It is bigger than O2$$-$$ and F$$-$$\n\nF$$-$$, O2$$-$$ and N3$$-$$ all are isoelectronic species in which N3$$-$$ have least number of protons due to which it's size increases as least nuclear attraction is experienced by the outer shell electrons. Size order

    N3$$-$$ > O2$$-$$ > F$$-$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2157, "subject": "Chemistry", "question": "Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).

    Assertion (A) : Barium carbonate is insoluble in water and is highly stable.

    Reason (R) : The thermal stability of the carbonates increases with increasing cationic size.", "options": [ { "text": "Both (A) and (R) are true and (R) is the true explanation of (A)" }, { "text": "(A) is true but (R) is false" }, { "text": "Both (A) and (R) are true but (R) is not the true explanation of (A)" }, { "text": "(A) is false but (R) is true." } ], "answer": "Both (A) and (R) are true and (R) is the true explanation of (A)", "solution": "**Answer:** Both (A) and (R) are true and (R) is the true explanation of (A)\n\nIn IIA group on moving down the group size of cation increases and show thermal stability of carbonate increases.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2158, "subject": "Chemistry", "question": "The correct order of ionic radii for the ions, P3$$-$$, S2$$-$$, Ca2+, K+, Cl$$-$$ is :", "options": [ { "text": "P3$$-$$ > S2$$-$$ > Cl$$-$$ > K+ > Ca2+" }, { "text": "Cl$$-$$ > S2$$-$$ > P3$$-$$ > Ca2+ > K+" }, { "text": "P3$$-$$ > S2$$-$$ > Cl$$-$$ > Ca2+ > K+" }, { "text": "K+ > Ca2+ > P3$$-$$ > S2$$-$$ > Cl$$-$$" } ], "answer": "P3$$-$$ > S2$$-$$ > Cl$$-$$ > K+ > Ca2+", "solution": "**Answer:** P3$$-$$ > S2$$-$$ > Cl$$-$$ > K+ > Ca2+\n\nP3$$-$$ > S2$$-$$ > Cl$$-$$ > K+ > Ca2+ (Correct order of ionic radii)

    all the given species are isoelectronic species. In isoelectronic species size increases with increase of negative charge and size decreases with increase in positive charge.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2159, "subject": "Chemistry", "question": "

    The correct order of increasing ionic radii is

    ", "options": [ { "text": "Mg2+ < Na+ < F$$-$$ < O2$$-$$ < N3$$-$$" }, { "text": "N3$$-$$ < O2$$-$$ < F$$-$$ < Na+ < Mg2+" }, { "text": "F$$-$$ < Na+ < O2$$-$$ < Mg2+ < N3$$-$$" }, { "text": "Na+ < F$$-$$ < Mg2+ < O2$$-$$ < N3$$-$$" } ], "answer": "Mg2+ < Na+ < F$$-$$ < O2$$-$$ < N3$$-$$", "solution": "**Answer:** Mg2+ < Na+ < F$$-$$ < O2$$-$$ < N3$$-$$\n\nFor isoelectronic species

    \nIonic radii $$\\propto \\frac{(-) \\text { ve charge }}{(+) \\text { ve charge }}$$

    \nHence, correct order of ionic radii is

    \nMg2+ < Na+ < F– < O2– < N3–", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2160, "subject": "Chemistry", "question": "

    Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).

    \n

    Assertion (A) : The ionic radii of O2$$-$$ and Mg2+ are same.

    \n

    Reason (R) : Both O2$$-$$ and Mg2+ are isoelectronic species.

    \n

    In the light of the above statements, choose the correct answer from the options given below.

    ", "options": [ { "text": "Both (A) and (R) are true and (R) is the correct explanation of (A)." }, { "text": "Both (A) and (R) are true but (R) is not the correct explanation of (A)." }, { "text": "(A) is true but (R) is false." }, { "text": "(A) is false but (R) is true." } ], "answer": "(A) is false but (R) is true.", "solution": "**Answer:** (A) is false but (R) is true.\n\nCorrect order of ionic radii:

    \nO–2 > Mg+2

    \nThis is because among isoelectronic species, the size of anions are greater than the size of cations. Statement (II) is correct as both O–2 and Mg+2 are isoelectronic.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2161, "subject": "Chemistry", "question": "

    Given two statements below :

    \n

    Statement I : In $$\\mathrm{Cl}_{2}$$ molecule the covalent radius is double of the atomic radius of chlorine.

    \n

    Statement II : Radius of anionic species is always greater than their parent atomic radius.

    \n

    Choose the most appropriate answer from options given below :

    ", "options": [ { "text": "Both Statement I and Statement II are correct." }, { "text": "Both Statement I and Statement II are incorrect." }, { "text": "Statement I is correct but Statement II is incorrect." }, { "text": "Statement I is incorrect but Statement II is correct." } ], "answer": "Statement I is incorrect but Statement II is correct.", "solution": "**Answer:** Statement I is incorrect but Statement II is correct.\n\n

    The given statements pertain to the concepts of atomic and ionic radii in chemical bonding and periodic properties. Let's analyze each statement individually for clarity.

    \n\n

    Statement I: In $$\\mathrm{Cl}_{2}$$ molecule the covalent radius is double of the atomic radius of chlorine.

    \n

    The atomic radius of an element like chlorine refers to the size of its atoms, typically measured when the element is in its gas phase and not bonded to anything else. The covalent radius, on the other hand, is a measure of the size of an atom that forms part of a single covalent bond - it's essentially half the distance between two atoms bonded together.

    \n

    In the case of a $$\\mathrm{Cl}_{2}$$ molecule, the covalent bond is formed between two chlorine atoms. Therefore, the distance from the nucleus of one chlorine atom to the nucleus of the other (the bond length) is essentially twice the covalent radius of chlorine. This makes the statement true only if correctly interpreted: the covalent bond length is double the covalent radius, but it tends to be conflated with the concept of the atomic radius. However, it is important to clarify that the atomic radius and covalent radius are different measures. In a diatomic molecule like $$\\mathrm{Cl}_2$$, saying the covalent radius is \"double of the atomic radius\" is inaccurate. The precise statement should be that the bond length (twice the covalent radius) is not directly twice the atomic radius but rather the distance between the nuclei of the bonding atoms. Therefore, this statement as phrased is misleading or incorrect.

    \n\n

    Statement II: Radius of anionic species is always greater than their parent atomic radius.

    \n

    This statement is based on the idea that when an atom gains electrons and becomes an anion, the increased electron-electron repulsions in the electron cloud cause it to expand. This is generally true across the periodic table. For example, when chlorine gains an electron to become $$\\mathrm{Cl}^{-}$$, its radius increases due to the addition of an extra electron which increases repulsion among electrons and expands the electron cloud. This principle holds for virtually all anions compared to their parent atoms, so Statement II is correct.

    \n\n

    Given the analysis, the correct option is:

    \n

    Option D: Statement I is incorrect but Statement II is correct.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2162, "subject": "Chemistry", "question": "

    The correct increasing order of the ionic radii is

    ", "options": [ { "text": "$$\\mathrm{Cl^- < Ca^{2+} < K^+ < S^{2-}}$$" }, { "text": "$$\\mathrm{K^+ < S^{2-} < Ca^{2+} < Cl^-}$$" }, { "text": "$$\\mathrm{Ca^{2+} < K^+ < Cl^- < S^{2-}}$$" }, { "text": "$$\\mathrm{S^{2-} < Cl^- < Ca^{2+} < K^+}$$" } ], "answer": "$$\\mathrm{Ca^{2+} < K^+ < Cl^- < S^{2-}}$$", "solution": "**Answer:** $$\\mathrm{Ca^{2+} < K^+ < Cl^- < S^{2-}}$$\n\nIn isoelectronic species size $\\propto \\frac{1}{Z}$\n

    $$\n\\mathrm{Ca}^{2+}<\\mathrm{K}^{+}<\\mathrm{Cl}^{-}<\\mathrm{S}^{2-} \\text { : Size }\n$$\n

    $\\begin{array}{lllll}Z: & 20 & 19 & 17 & 18\\end{array}$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2163, "subject": "Chemistry", "question": "

    The statement(s) that are correct about the species $$\\mathrm{O}^{2-}, \\mathrm{F}^{-}, \\mathrm{Na}^{+}$$ and $$\\mathrm{Mg}^{2+}$$.

    \n

    (A) All are isoelectronic

    \n

    (B) All have the same nuclear charge

    \n

    (C) $$\\mathrm{O}^{2-}$$ has the largest ionic radii

    \n

    (D) $$\\mathrm{Mg}^{2+}$$ has the smallest ionic radii

    \n

    Choose the most appropriate answer from the options given below :

    ", "options": [ { "text": "(C) and (D) only\n" }, { "text": "(A), (C) and (D) only\n" }, { "text": "(B), (C) and (D) only\n" }, { "text": "(A), (B), (C) and (D)" } ], "answer": "(A), (C) and (D) only\n", "solution": "**Answer:** (A), (C) and (D) only\n\n\n

    To answer this question, let's address each statement individually.

    \n\n

    (A) All are isoelectronic

    \n\n

    An isoelectronic species is a group of ions or atoms which have the same number of electrons. The electron configuration for each species is as follows:

    \n\n\n\n

    All of these ions have the same number of electrons (10), making them isoelectronic. Therefore, statement (A) is correct.

    \n\n

    (B) All have the same nuclear charge

    \n\n

    The nuclear charge refers to the total charge within the nucleus, which is determined by the number of protons. The nuclear charges are:

    \n\n\n\n

    Since the number of protons varies among these species, they do not have the same nuclear charge. Therefore, statement (B) is incorrect.

    \n\n

    (C) $$\\mathrm{O}^{2-}$$ has the largest ionic radii

    \n\n

    In a series of isoelectronic ions, the ionic radius decreases with increasing nuclear charge because the greater the nuclear charge, the more strongly the electrons are pulled towards the nucleus, reducing the size of the ion. As $$\\mathrm{O}^{2-}$$ has the lowest nuclear charge among the given ions, it will have the largest ionic radius. Therefore, statement (C) is correct.

    \n\n

    (D) $$\\mathrm{Mg}^{2+}$$ has the smallest ionic radii

    \n\n

    Following the same logic as above, $$\\mathrm{Mg}^{2+}$$, having the highest nuclear charge among the given isoelectronic species, will have the smallest ionic radius because its electrons are held most tightly by the nucleus. Thus, statement (D) is correct.

    \n\n

    Given the evaluations above, the most appropriate answer is:

    \n\n

    Option B (A), (C) and (D) only.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2164, "subject": "Chemistry", "question": "

    Given below are two statements :

    \n

    Statement I : The metallic radius of $$\\mathrm{Na}$$ is $$1.86 \\mathrm{~A}^{\\circ}$$ and the ionic radius of $$\\mathrm{Na}^{+}$$ is lesser than $$1.86 \\mathrm{~A}^{\\circ}$$

    \n

    Statement II : Ions are always smaller in size than the corresponding elements.

    \n

    In the light of the above statements, choose the correct answer from the options given below :

    ", "options": [ { "text": "Both Statement I and Statement II are true\n" }, { "text": "Statement I is incorrect but Statement II is true\n" }, { "text": "Both Statement I and Statement II are false\n" }, { "text": "Statement I is correct but Statement II is false" } ], "answer": "Statement I is correct but Statement II is false", "solution": "**Answer:** Statement I is correct but Statement II is false\n\n

    To determine the correctness of the given statements, let's analyze each one individually based on atomic and ionic radii concepts.

    \n
    \n

    Statement I:

    \n

    The metallic radius of $\\mathrm{Na}$ is $1.86\\, \\text{Å}$ and the ionic radius of $\\mathrm{Na}^+$ is lesser than $1.86\\, \\text{Å}$.

    \n

    Analysis:

    \n\n

    Metallic Radius of Sodium ($\\mathrm{Na}$):

    \n

    Sodium is a metal, and its metallic radius is indeed approximately $1.86\\, \\text{Å}$.

    \n

    The metallic radius refers to half the distance between the nuclei of two adjacent atoms in a metallic lattice.

    \n

    Ionic Radius of Sodium Ion ($\\mathrm{Na}^+$):

    \n

    When sodium loses an electron to form $\\mathrm{Na}^+$, it loses its outermost electron shell (the 3s orbital).

    \n

    This results in a significant decrease in size due to:

    \n\n

    Decrease in Electron-Electron Repulsion: Fewer electrons mean less repulsion among them.

    \n

    Unchanged Nuclear Charge: The number of protons remains the same, so the effective nuclear charge per electron increases, pulling the remaining electrons closer to the nucleus.

    \n

    The ionic radius of $\\mathrm{Na}^+$ is approximately $0.95\\, \\text{Å}$, which is significantly smaller than $1.86\\, \\text{Å}$.

    \n\n

    Conclusion:

    \n\n

    Statement I is correct.

    \n\n
    \n

    Statement II:

    \n

    Ions are always smaller in size than the corresponding elements.

    \n

    Analysis:

    \n\n

    General Trends:

    \n

    Cations ($\\text{Positive Ions}$):

    \n\n

    Formed by the loss of one or more electrons.

    \n

    Result: Cations are smaller than their parent atoms due to loss of electron(s) and decreased electron-electron repulsion.

    \n

    Example: $\\mathrm{Na} \\rightarrow \\mathrm{Na}^+$ (size decreases).

    \n

    Anions ($\\text{Negative Ions}$):

    \n\n

    Formed by the gain of one or more electrons.

    \n

    Result: Anions are larger than their parent atoms due to added electron(s) and increased electron-electron repulsion.

    \n

    Example: $\\mathrm{Cl} \\rightarrow \\mathrm{Cl}^-$ (size increases).

    \n

    Exceptions to Statement II:

    \n

    Since anions are larger than their corresponding neutral atoms, the statement that \"ions are always smaller in size than the corresponding elements\" is incorrect.

    \n\n

    Conclusion:

    \n\n

    Statement II is false.

    \n\n
    \n

    Final Answer:

    \n\n

    Statement I is correct but Statement II is false.

    \n\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2165, "subject": "Chemistry", "question": "The size of the iso-electronic species Cl–,  Ar and Ca2+ is affected by :", "options": [ { "text": "electron-electron interaction in the outer orbitals" }, { "text": "Principal quantum number of valence shell" }, { "text": "nuclear charge" }, { "text": "azimuthal quantum number of valence shell" } ], "answer": "nuclear charge", "solution": "**Answer:** nuclear charge\n\n\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n
    Cl-ArCa+2
    Protons171820
    Electrons181818
    \n
    Nuclear charge means number of protons present in the nucleus. Here maximum number of protons present in the Ca+2 ion, so the attraction towards those 18 electrons wll be most by the nucleus of the Ca+2 ion. That is why size of the Ca+2 ion wll be least because of highest nuclear charge.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2166, "subject": "Chemistry", "question": "Among the following, the energy of 2s orbital is lowest in :", "options": [ { "text": "Li" }, { "text": "H" }, { "text": "Na" }, { "text": "K" } ], "answer": "K", "solution": "**Answer:** K\n\nHere Z of K = 15, Na = 11, H = 1, Li = 3.\n

    In K, because of more number of protons (high atomic\nnumber) the 2s electron experiences a higher effective nuclear\ncharge and is closer to nucleus, thus having less energy", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2167, "subject": "Chemistry", "question": "In which of the following arrangements the order is NOT according to the property\nindicated against it? ", "options": [ { "text": "Li < Na < K < Rb Increasing metallic radius" }, { "text": "I < Br < F < Cl Increasing electron gain enthalpy (with negative sign) " }, { "text": "B < C < N < O Increasing first ionization enthalpy" }, { "text": "Al3+ < Mg2+ < Na+ < F- Increasing ionic size " } ], "answer": "B < C < N < O Increasing first ionization enthalpy", "solution": "**Answer:** B < C < N < O Increasing first ionization enthalpy\n\n$$\\left( a \\right)\\,\\,\\,\\,\\,$$It is True. \n

    Here $$Li, Na, K$$ and $$Rb$$ all are belongs to the same group, so their effective nuclear charge is same but in a group from top to bottom the no. of shells increase in a atom. So the radius of atom increases. \n

    $$\\left( b \\right)\\,\\,\\,\\,\\,$$ It is True.\n

    In entire periodic table $$Cl$$ (chlorine) has the highest, and in a group it decreases from top to bottom. \n

    So the correct order is $${\\rm I} < Br < F < Cl.$$ \n

    Note : Among F and Cl, when an electron is added to the F atom, electron comes to the 2p orbital and for Cl atom electron is added in 3p orbital. As 2p orbital is closer to the nucleus than 3p orbital as 3p orbital is larger in size, so when a new electron comes to 2p orbital then it will face a strong repulsion force by the nucleus than if electron comes to 3p orbital.\n

    So, F have lesser tendency of gaining electron than Cl.\n

    $$\\left( c \\right)\\,\\,\\,\\,\\,$$ It is False.\n

    Electronic configuration of \n

    $$B\\left( 5 \\right)\\,\\,\\, = \\,\\,\\,1{s^2}2{s^2}2{p^1}$$ \n

    $$C\\left( 6 \\right)\\,\\,\\, = \\,\\,\\,1{s^2}2{s^2}2{p^2}$$ \n

    $$N\\left( 7 \\right)\\,\\,\\, = \\,\\,\\,1{s^2}2{s^2}2{p^3}$$\n

    $$O\\left( 8 \\right)\\,\\,\\, = \\,\\,\\,1{s^2}2{s^2}2{p^4}$$ \n

    In general in periodic table from left to right effective nuclear change increase and it becomes more difficult to remove electron from an atom. That is why ionization enthalpy increases. \n

    But for those atoms which has half filled or full filled outer most shell, those atoms will be more stable and more energy is needed to remove electron from those atoms. Here $$N\\left( {1{s^2}\\,2{s^2}\\,2{p^3}} \\right)$$ has stable half filled $$2p$$ subshell so it is more stable than $$O.$$ \n

    So, correct order is $$B < C < O < N$$ \n

    $$\\left( d \\right)\\,\\,\\,\\,\\,$$ It is True.\n

    $$A{l^{3 + }},M{g^{2 + }},N{a^ + }\\,\\,$$ and \n

    $$\\,\\,{F^ - }$$ are isoelectric. \n

    So all have $$10$$ electrons. So, \n

    $$\\,\\,{Z \\over e}\\,\\,$$ of $$A{l^{3 + }} = {{13} \\over {10}} = 1.3$$ \n

    $$\\,\\,{Z \\over e}\\,\\,$$ of $$M{g^{2 + }} = {{12} \\over {10}} = 1.2$$ \n

    $$\\,\\,{Z \\over e}\\,\\,$$ of $$N{a^ + } = {{11} \\over {10}} = 1.1$$\n

    $$\\,\\,{Z \\over e}\\,\\,$$ of $${F^ - } = {9 \\over {10}} = 0.9$$ \n

    We know $${z \\over e}$$ means $$z$$ no of protons in nucleus is pulling $$e$$ no. of electrons present in the orbital. \n

    So for $$A{l^{3 + }},\\,13$$ protons is pulling $$10$$ electron so the size of $$A{l^{3 + }}$$ will decrease most. That is why more $${z \\over e}$$ means less size of ion. \n

    So, the correct order is \n

    $$A{l^{3 + }} < M{g^{2 + }} < N{a^ + } < {F^ - }$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2168, "subject": "Chemistry", "question": "The correct order of electron affinity is : ", "options": [ { "text": "F > Cl > O" }, { "text": "F > O > Cl" }, { "text": "Cl > F > O" }, { "text": "O > F > Cl" } ], "answer": "Cl > F > O", "solution": "**Answer:** Cl > F > O\n\nElectron affinity means tendency of gaining an electron by an atom.\n

    In a period from left ot right the electron affinity increases and in a group it decreases from top to bottom.\n

    So according to this theory Fluorine(F) should have most electron affinity. But when an electron is added to the F atom, electron comes to the 2p orbital and for Cl atom electron is added in 3p orbital. As 2p orbital is closer to the nucleus than 3p orbital as 3p orbital is larger in size, so when a new electron comes to 2p orbital then it will face a strong repulsion force by the nucleus than if electron comes to 3p orbital.\n

    So, F have lesser tendency of gaining electron than Cl.\n\n

    Note : In entire periodic table Cl have highest electron affinity.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2169, "subject": "Chemistry", "question": "When the first electron gain enthalpy $$\\left( {{\\Delta _{eg}}H} \\right)$$ of oxygen is $$-$$ 141 kJ/mol, its second electron gain enthalpy is : ", "options": [ { "text": "a more negative value than the first" }, { "text": "almost the same as that of the first" }, { "text": "negative, but less negative than the first" }, { "text": "a positive value" } ], "answer": "a positive value", "solution": "**Answer:** a positive value\n\nSecond electron gain enthalpy is always positive for every element. \n
    O$$-$$(g) + e$$-$$ $$ \\to $$ O$$-$$2(g) ; $$\\Delta $$H = positive", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2170, "subject": "Chemistry", "question": "The electron gain enthalpy (in KJ/mol) of fluorine, chlorine, bromine and iodine, respectively are:", "options": [ { "text": "-296, -325, -333 and -349" }, { "text": "349, -333, -325 and -296 " }, { "text": "-333, -349, -325 and -296" }, { "text": "-333, -325, -349 and -296\n" } ], "answer": "-333, -349, -325 and -296", "solution": "**Answer:** -333, -349, -325 and -296\n\nOrder of electron gain enthalpy (magnitude) is\nCl > F > Br > I\n

    Note: Electron gain enthalpy increases with electro negativity but chlorine has higher electron gain enthalpy than\nfluorine (exception).", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2171, "subject": "Chemistry", "question": "Within each pair of elements F & Cl, S & Se, and Li & Na, respectively, the elements that release\nmore energy upon and electron gain are :", "options": [ { "text": "F, S and Li" }, { "text": "Cl, Se and Na" }, { "text": "Cl, S and Li " }, { "text": "F, Se and Na" } ], "answer": "Cl, S and Li ", "solution": "**Answer:** Cl, S and Li \n\nElectron affinity of second period p-block\nelement is less than third period p-block element\ndue to small size of second period p-block element.\n

    $$ \\therefore $$ Electron affinity order : F $$<$$ Cl\n

    Down the group electron affinity decreases due\nto size increases.\n

    $$ \\therefore $$ Electron affinity E.A. order : S $$>$$ Se\n

    Li $$>$$ Na", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2172, "subject": "Chemistry", "question": "The correct order of electron gain enthalpy is :", "options": [ { "text": "Te > Se > S > O" }, { "text": "S > Se > Te > O" }, { "text": "S > O > Se > Te" }, { "text": "O > S > Se > Te" } ], "answer": "S > Se > Te > O", "solution": "**Answer:** S > Se > Te > O\n\nOxygen is the second most electronegative element in comparison to fluorine. In group - 16 family $(\\mathrm{O}, \\mathrm{S}, \\mathrm{Se}, \\mathrm{Te})$, O-atom is smallest in size. So, electron density on O-atom is very high in group -16\nDuring addition of a free electron to gaseous $\\mathrm{O}$-atom,

    \n$$\n\\mathrm{O}(\\mathrm{g})+\\mathrm{e}^{-} \\longrightarrow \\mathrm{O}^{-}(\\mathrm{g})\n$$

    \nWe have to supply a significant amount of energy (endothermic) to overcome the electrostatic repulsion between the approaching electron and O-atom of very high electron density. So, the net value of electron affinity (EA) or (negative) electron gain enthalpy $\\left[\\Delta_{\\mathrm{eg}} H\\right.$ or $\\left.\\left|\\Delta_{\\mathrm{eg}} H\\right|\\right]$ of oxygen decreases to a higher extent in comparison to other elements of group -16 who have larger size and lower electronegativity.\n

    \nSo, the correct order of EA or $\\left|\\Delta_{\\mathrm{eg}} \\mathrm{H}\\right|$ of group $-16$ elements will be $\\mathrm{S}$ $>\\mathrm{Se}>\\mathrm{Te}>\\mathrm{O}$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2173, "subject": "Chemistry", "question": "The absolute value of the electron gain enthalpy of halogens satisfies :", "options": [ { "text": "Cl > Br > F > I" }, { "text": "Cl > F > Br > I" }, { "text": "I > Br > Cl > F" }, { "text": "F > Cl > Br > I" } ], "answer": "Cl > F > Br > I", "solution": "**Answer:** Cl > F > Br > I\n\nThe magnitude of electron gain enthalpy of halogen\natoms down the group show abnormal behavior.\nThe |ΔHeg| of F is lower than that of Cl due to its\nsmaller size. The incoming electron experiences\nhigher repulsive force due to valence electrons of\nF than Cl. The correct order is Cl > F > Br > I.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2174, "subject": "Chemistry", "question": "Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).

    Assertion (A) : Metallic character decreases and non-metallic character increases on moving from left to right in a period.

    Reason (R) : It is due to increase in ionisation enthalpy and decrease in electron gain enthalpy, when one moves from left to right in a period.

    In the light of the above statements, choose the most appropriate answer from the options given below :", "options": [ { "text": "(A) is false but (R) is true." }, { "text": "(A) is true but (R) is false" }, { "text": "Both (A) and (R) are correct and (R) is the correct explanation of (A)" }, { "text": "Both (A) and (R) are correct but (R) is not the correct explanation of (A)" } ], "answer": "(A) is true but (R) is false", "solution": "**Answer:** (A) is true but (R) is false\n\nFrom left to right in periodic table :-

    Metallic character decreases

    Non-metallic character increases

    $$\\Rightarrow$$ It is due to increase in ionization enthalpy and increase in electron gain enthalpy.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2175, "subject": "Chemistry", "question": "

    The correct order of electron gain enthalpies of Cl, F, Te and Po is

    ", "options": [ { "text": "F < Cl < Te < Po" }, { "text": "Po < Te < F < Cl" }, { "text": "Te < Po < Cl < F" }, { "text": "Cl < F < Te < Po" } ], "answer": "Po < Te < F < Cl", "solution": "**Answer:** Po < Te < F < Cl\n\nTe → –190 kJ mol–1

    \nPo → –174 kJ mol–1

    \nF → –333 kJ mol–1

    \nCl → –349 kJ mol–1

    \nHence, correct order is Cl > F > Te > Po", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2176, "subject": "Chemistry", "question": "

    The correct order of electron gain enthalpy ($$-$$ ve value) is :

    ", "options": [ { "text": "O > S > Se > Te" }, { "text": "O < S < Se < Te" }, { "text": "O < S > Se > Te" }, { "text": "O < S > Se < Te" } ], "answer": "O < S > Se > Te", "solution": "**Answer:** O < S > Se > Te\n\nDown the group, the size of atom increases, so electron gain enthalpy decreases.\n

    \nIn case of Oxygen, due to its small size, which leads to electronelectron repulsion, results in less electron gain enthalpy than sulphur.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2177, "subject": "Chemistry", "question": "

    In which of the following pairs, electron gain enthalpies of constituent elements are nearly the same or identical?

    \n

    (A) Rb and Cs

    \n

    (B) Na and K

    \n

    (C) Ar and Kr

    \n

    (D) I and At

    \n

    Choose the correct answer from the options given below :

    ", "options": [ { "text": "(A) and (B) only" }, { "text": "(B) and (C) only" }, { "text": "(A) and (C) only" }, { "text": "(C) and (D) only" } ], "answer": "(A) and (C) only", "solution": "**Answer:** (A) and (C) only\n\n$\\mathrm{Rb} \\,\\& \\,\\mathrm{Cs}$ have nearly same electron gain enthalpy electron gain enthalpy $=-46\\, \\mathrm{kj} / \\mathrm{ml}$

    \nAr & Kr have same $\\Delta \\mathrm{H}_{\\mathrm{eq}} .$ Value is $+96 \\,\\mathrm{kj} / \\mathrm{ml}$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2178, "subject": "Chemistry", "question": "

    Inert gases have positive electron gain enthalpy. Its correct order is :

    ", "options": [ { "text": "He < Ne < Kr < Xe" }, { "text": "He < Xe < Kr < Ne" }, { "text": "Xe < Kr < Ne < He" }, { "text": "He < Kr < Xe < Ne" } ], "answer": "He < Xe < Kr < Ne", "solution": "**Answer:** He < Xe < Kr < Ne\n\n\n\n\n \n \n \n \n \n \n \n \n\n\n \n \n \n \n \n \n \n \n\n
    $\\begin{gathered}\\text { Electron } \\\\\\text { gain }\\end{gathered}$$\\mathrm{He}$$\\mathrm{Ne}$$\\mathrm{Ar}$$\\mathrm{Kr}$$\\mathrm{Xe}$
    $\\begin{gathered}\\text { Enthalpyl } / \\\\\\mathrm{kJ} ~\\mathrm{mol}^{-1}\\end{gathered}$48116969677
    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2179, "subject": "Chemistry", "question": "

    The difference between electron gain enthalpies will be maximum between :

    ", "options": [ { "text": "Ar and Cl" }, { "text": "Ne and Cl" }, { "text": "Ne and F" }, { "text": "Ar and F" } ], "answer": "Ne and Cl", "solution": "**Answer:** Ne and Cl\n\n$\\Delta \\mathrm{H}_{\\mathrm{eg}}$ for chlorine $=-349 \\mathrm{~kJ} \\mathrm{~mole}^{-1}$

    \n$\\Delta \\mathrm{H}_{\\mathrm{eg}}$ for Neon $=+116 \\mathrm{~kJ} \\mathrm{~mole}^{-1}$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2180, "subject": "Chemistry", "question": "

    The correct sequence of electron gain enthalpy of the elements listed below is

    \n

    A. Ar

    \n

    B. Br

    \n

    C. F

    \n

    D. S

    \n

    Choose the most appropriate from the options given below:

    ", "options": [ { "text": "$$\\mathrm{A>D>C>B}$$\n" }, { "text": "$$\\mathrm{A}>\\mathrm{D}>\\mathrm{B}>\\mathrm{C}$$\n" }, { "text": "$$\\mathrm{D}>\\mathrm{C}>\\mathrm{B}>\\mathrm{A}$$\n" }, { "text": "$$\\mathrm{C}>\\mathrm{B}>\\mathrm{D}>\\mathrm{A}$$" } ], "answer": "$$\\mathrm{A}>\\mathrm{D}>\\mathrm{B}>\\mathrm{C}$$\n", "solution": "**Answer:** $$\\mathrm{A}>\\mathrm{D}>\\mathrm{B}>\\mathrm{C}$$\n\n\n

    $$\\begin{array}{ll}\n\\text { Element } & \\Delta_{\\mathrm{eg}} \\mathrm{H}(\\mathrm{kJ} / \\mathrm{mol}) \\\\\n\\mathrm{F} & -333 \\\\\n\\mathrm{~S} & -200 \\\\\n\\mathrm{Br} & -325 \\\\\n\\mathrm{Ar} & +96\n\\end{array}$$

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2181, "subject": "Chemistry", "question": "

    Given below are two statements:

    \n

    Statement I : Fluorine has most negative electron gain enthalpy in its group.

    \n

    Statement II : Oxygen has least negative electron gain enthalpy in its group.

    \n

    In the light of the above statements, choose the most appropriate from the options given below

    ", "options": [ { "text": "Both Statement I and Statement II are true\n" }, { "text": "Both Statement I and Statement II are false\n" }, { "text": "Statement I is false but Statement II is true\n" }, { "text": "Statement I is true but Statement II is false" } ], "answer": "Statement I is false but Statement II is true\n", "solution": "**Answer:** Statement I is false but Statement II is true\n\n\n

    Statement- 1 is false because chlorine has most negative electron gain enthalpy in its group.

    ", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 2182, "subject": "Chemistry", "question": "

    Given below are two statements :

    \n

    Statement (I) : The oxidation state of an element in a particular compound is the charge acquired by its atom on the basis of electron gain enthalpy consideration from other atoms in the molecule.

    \n

    Statement (II) : $$\\mathrm{p} \\pi-\\mathrm{p} \\pi$$ bond formation is more prevalent in second period elements over other periods.

    \n

    In the light of the above statements, choose the most appropriate answer from the options given below :

    ", "options": [ { "text": "Statement I is correct but Statement II is incorrect\n" }, { "text": "Both Statement I and Statement II are correct\n" }, { "text": "Statement I is incorrect but Statement II is correct\n" }, { "text": "Both Statement I and Statement II are incorrect" } ], "answer": "Statement I is incorrect but Statement II is correct\n", "solution": "**Answer:** Statement I is incorrect but Statement II is correct\n\n\n

    To evaluate the given statements, we need to consider the concepts of oxidation states and bonding tendencies across different periods in the periodic table.

    \n\n

    Statement (I) describes the oxidation state of an element in a compound as the hypothetical charge an atom would have if all bonds to atoms of different elements were 100% ionic. This consideration does involve looking at how electrons are shared or transferred between atoms, but it's not solely based on electron gain enthalpy. Rather, the oxidation state is determined by a set of rules which include, but are not limited to, electronegativity differences and known oxidation states of elements. Therefore, the statement is somewhat misleading in its specificity about electron gain enthalpy being the sole consideration. The electron gain enthalpy is a measure of the energy change when an electron is added to a neutral atom in the gaseous state to form a negative ion, which indirectly influences the discussion of oxidation states but is not the direct basis for their determination.

    \n\n

    Statement (II) points out that $$\\mathrm{p}\\pi-\\mathrm{p}\\pi$$ bond formation is more prevalent among second period elements compared to elements in other periods. This is correct and can be attributed to the smaller size of the second period elements (such as carbon, nitrogen, and oxygen), which allows their p orbitals to overlap more effectively for $$\\mathrm{p}\\pi-\\mathrm{p}\\pi$$ bonding. As we move down the periods, the atomic size increases, which leads to less effective overlapping of p orbitals for $$\\mathrm{p}\\pi-\\mathrm{p}\\pi$$ bonding due to increased distance between the valence electrons and nucleus. This makes $$\\mathrm{p}\\pi-\\mathrm{p}\\pi$$ bonds less common and less strong in elements of higher periods compared to those in the second period.

    \n\n

    Therefore, the most accurate assessment of the statements given the explanations above is:

    \n\n\n

    Hence, the correct option is:

    \n

    Option C: Statement I is incorrect but Statement II is correct.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2183, "subject": "Chemistry", "question": "

    The electron affinity value are negative for

    \n

    A. $$\\mathrm{Be} \\rightarrow \\mathrm{Be}^{-}$$

    \n

    B. $$\\mathrm{N} \\rightarrow \\mathrm{N}^{-}$$

    \n

    C. $$\\mathrm{O} \\rightarrow \\mathrm{O}^{2-}$$

    \n

    D. $$\\mathrm{Na} \\rightarrow \\mathrm{Na}^{-}$$

    \n

    E. $$\\mathrm{Al} \\rightarrow \\mathrm{Al}^{-}$$

    \n

    Choose the most appropriate answer from the options given below :

    ", "options": [ { "text": "A, B, D and E only\n" }, { "text": "D and E only\n" }, { "text": "A and D only\n" }, { "text": "A, B and C only" } ], "answer": "A, B and C only", "solution": "**Answer:** A, B and C only\n\n

    To determine which elements have negative electron affinity values, we'll analyze each option individually.

    \n

    Understanding Electron Affinity

    \n\n

    Electron Affinity (EA): The energy change when an electron is added to a neutral atom in the gas phase to form a negative ion.

    \n

    Convention:

    \n

    If energy is released (exothermic process), the electron affinity is considered positive.

    \n

    If energy is absorbed (endothermic process), the electron affinity is considered negative.

    \n\n
    \n

    Option A: $\\mathrm{Be} \\rightarrow \\mathrm{Be}^{-}$

    \n\n

    Electronic Configuration of Be: $1s^2\\,2s^2$

    \n

    Analysis:

    \n

    Adding an electron to beryllium means placing it into the higher energy $2p$ orbital.

    \n

    Beryllium has a filled $2s$ subshell, so the added electron experiences higher energy and electron-electron repulsion.

    \n

    Result: Energy is absorbed; the process is endothermic.

    \n

    Conclusion: Electron affinity of Be is negative.

    \n\n
    \n

    Option B: $\\mathrm{N} \\rightarrow \\mathrm{N}^{-}$

    \n\n

    Electronic Configuration of N: $1s^2\\,2s^2\\,2p^3$

    \n

    Analysis:

    \n

    Nitrogen has a half-filled $2p$ subshell.

    \n

    Adding an electron introduces repulsion in the half-filled orbital.

    \n

    Result: Energy is absorbed; the process is endothermic.

    \n

    Conclusion: Electron affinity of N is negative.

    \n\n
    \n

    Option C: $\\mathrm{O} \\rightarrow \\mathrm{O}^{2-}$

    \n\n

    Electronic Configuration of O: $1s^2\\,2s^2\\,2p^4$

    \n

    Analysis:

    \n

    First Electron Affinity (O to O⁻): Exothermic (energy released).

    \n

    Second Electron Affinity (O⁻ to O²⁻): Endothermic because adding an electron to a negative ion requires energy due to electron-electron repulsion.

    \n

    Overall Process (O to O²⁻): The second step dominates, making the overall process endothermic.

    \n

    Conclusion: Electron affinity for forming $\\mathrm{O}^{2-}$ is negative.

    \n\n
    \n

    Option D: $\\mathrm{Na} \\rightarrow \\mathrm{Na}^{-}$

    \n\n

    Electronic Configuration of Na: $1s^2\\,2s^2\\,2p^6\\,3s^1$

    \n

    Analysis:

    \n

    Adding an electron fills the $3s$ orbital.

    \n

    Sodium tends to lose an electron to form $\\mathrm{Na}^+$, not gain one.

    \n

    However, adding an electron is still an exothermic process due to the low energy of the $3s$ orbital.

    \n

    Result: Energy is released; the process is exothermic.

    \n

    Conclusion: Electron affinity of Na is positive.

    \n\n
    \n

    Option E: $\\mathrm{Al} \\rightarrow \\mathrm{Al}^{-}$

    \n\n

    Electronic Configuration of Al: $1s^2\\,2s^2\\,2p^6\\,3s^2\\,3p^1$

    \n

    Analysis:

    \n

    Adding an electron to the $3p$ orbital.

    \n

    The process releases energy due to the addition of an electron to a partially filled orbital.

    \n

    Result: Energy is released; the process is exothermic.

    \n

    Conclusion: Electron affinity of Al is positive.

    \n\n
    \n

    Final Conclusion

    \n\n

    Negative Electron Affinity Values (Endothermic Processes):

    \n

    Option A: Beryllium ($\\mathrm{Be}$)

    \n

    Option B: Nitrogen ($\\mathrm{N}$)

    \n

    Option C: Oxygen forming $\\mathrm{O}^{2-}$

    \n\n

    Therefore, the correct answer is:

    \n

    Option D: A, B, and C only

    \n
    \n

    Answer: Option D

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2184, "subject": "Chemistry", "question": "The formation of the oxide ion O2-(g) requires first an exothermic and then an endothermic\nstep as shown below

    \nO(g) + e- = $$O_{(g)}^{-}$$ $$\\Delta $$Ho = -142 kJmol-1

    \n$$O_{(g)}^{-}$$ + e- = $$O_{(g)}^{2-}$$ $$\\Delta $$Ho = 844 kJmol-1

    \nThis because ", "options": [ { "text": "O- ion will tend to resist the addition of another electron " }, { "text": "Oxygen has high electron affinity" }, { "text": "Oxygen is more electronegative " }, { "text": "O- ion has comparatively larger size than oxygen atom " } ], "answer": "O- ion will tend to resist the addition of another electron ", "solution": "**Answer:** O- ion will tend to resist the addition of another electron \n\n$$O$$ atom is highly electronegative so will add first electron easily by releasing energy. So it is an exothermic. \n

    After adding first electron $$O$$ becomes $${O^ - }\\,\\,$$ and size of $${O^ - }\\,\\,$$ becomes slightly more than $$O$$ atom. Now when a new electron is trying to add into $$\\,\\,{O^ - }$$ ion, two forces will be act on new electron $$ \\to $$ \n

    $$\\left( 1 \\right)\\,\\,\\,\\,$$ attraction between nucleus and new electron\n

    $$\\left( 2 \\right)\\,\\,\\,\\,$$ Repulsion between outer most shell electrons and electron. \n

    But repulsion between electrons are more compare to the attraction between nucleus and electron as distance between nucleus and electron is more compare to electrons of outer most shell and new electron. \n

    So, to overcome the repulsion energy should be added. \n

    That is why $${O^ - }\\left( g \\right) + {e^ - }\\buildrel \\, \\over\n \\longrightarrow {O^{2 - }}\\left( g \\right)\\,\\,\\,$$ is an endothermic reaction. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2185, "subject": "Chemistry", "question": "In general, the properties that decrease and increase down a group in the periodic table, respectively, are : ", "options": [ { "text": "Atomic Radius and Electronegativity" }, { "text": "Electron Gain Enthalpy and Electronegativity. " }, { "text": "Electronegativity and Atomic Radius." }, { "text": "Electronegativity and Electron Gain Enthalpy." } ], "answer": "Electronegativity and Atomic Radius.", "solution": "**Answer:** Electronegativity and Atomic Radius.\n\nElectronegativity decreases down the group because the increased number of energy levels puts the outer electrons very far away from the pull of nucleus. \n

    Atomic radius increases down the group because the number of energy levels increases when you move down the group. Each subsequent energy level is further from the molecules than the last. That is why atomic radius increases down the group. ", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 2186, "subject": "Chemistry", "question": "The electronegativity of aluminium is similar to : ", "options": [ { "text": "Beryllium" }, { "text": "Carbon" }, { "text": "Boron" }, { "text": "Lithium" } ], "answer": "Beryllium", "solution": "**Answer:** Beryllium\n\nE.N. of Al = (1.5) $$ \\cong $$ Be (1.5)", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2187, "subject": "Chemistry", "question": "The correct option with respect to the Pauling electronegativity values of the element is : ", "options": [ { "text": "Ga < Ge" }, { "text": "Si < Al" }, { "text": "Te > Se" }, { "text": "P > S" } ], "answer": "Ga < Ge", "solution": "**Answer:** Ga < Ge\n\nElectronegativity increases from left to right in a period and decreases down the group. The correct orders are\n

    Si > Al, Ga < Ge, Te < Se and P < S.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2188, "subject": "Chemistry", "question": "In general the property (magnitudes only) that\nshow an opposite trend in comparison to other\nproperties across a period is", "options": [ { "text": "Electron gain enthalpy" }, { "text": "Electronegativity" }, { "text": "Ionization enthalpy" }, { "text": "Atomic radius" } ], "answer": "Atomic radius", "solution": "**Answer:** Atomic radius\n\nAtomic radius decreases on moving left to right\nin periodic table, while other three properties\ngiven increases (in magnitude) on moving left\nto right across a period.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2189, "subject": "Chemistry", "question": "

    The correct order of electronegativity for given elements is:

    ", "options": [ { "text": "$$ \\mathrm{C}>\\mathrm{P}>\\mathrm{At}>\\mathrm{Br}$$" }, { "text": "$$\\mathrm{Br}>\\mathrm{C}>\\mathrm{At}>\\mathrm{P}$$" }, { "text": "$$\\mathrm{Br}>\\mathrm{P}>\\mathrm{At}>\\mathrm{C}$$" }, { "text": "$$\\mathrm{P}>\\mathrm{Br}>\\mathrm{C}>\\mathrm{At}$$" } ], "answer": "$$\\mathrm{Br}>\\mathrm{C}>\\mathrm{At}>\\mathrm{P}$$", "solution": "**Answer:** $$\\mathrm{Br}>\\mathrm{C}>\\mathrm{At}>\\mathrm{P}$$\n\n

    Electronegativity is a measure of an atom's ability to attract shared electrons to itself. On the periodic table, electronegativity generally increases as we move from left to right across a period due to increase in effective nuclear charge.

    \n

    Here are the Pauling electronegativities for the elements:

    \n\n

    Based on these values, the correct order of electronegativity from highest to lowest is Br > C > At > P.

    \n

    Therefore, Option B is the correct answer:

    \n

    Br > C > At > P

    \n\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2190, "subject": "Chemistry", "question": "

    Given below are two statements :

    \n

    Statement I : The electronegativity of group 14 elements from $$\\mathrm{Si}$$ to $$\\mathrm{Pb}$$, gradually decreases.

    \n

    Statement II : Group 14 contains non-metallic, metallic, as well as metalloid elements.

    \n

    In the light of the above statements, choose the most appropriate from the options given below :

    ", "options": [ { "text": "Statement I is false but Statement II is true\n" }, { "text": "Both Statement I and Statement II are true\n" }, { "text": "Both Statement I and Statement II are false\n" }, { "text": "Statement I is true but Statement II is false" } ], "answer": "Statement I is false but Statement II is true\n", "solution": "**Answer:** Statement I is false but Statement II is true\n\n\n

    $$\\begin{array}{ll}\n\\text { Gr-14 } & \\text { EN } \\\\\n\\mathrm{C} & 2.5 \\\\\n\\mathrm{Si} & 1.8 \\\\\n\\mathrm{Ge} & 1.8 \\\\\n\\mathrm{Sn} & 1.8 \\\\\n\\mathrm{~Pb} & 1.9\n\\end{array}$$

    \n

    The electronegativity values for elements from $$\\mathrm{Si}$$ to $$\\mathrm{Pb}$$ are almost same. So Statement $$\\mathrm{I}$$ is false.

    ", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 2191, "subject": "Chemistry", "question": "The atomic numbers of Vanadium (V), Chromium (cr), Manganese (Mn) and Iron (Fe), respectively, $$23,24,25$$ and $$26$$. Which one of these may be expected to have the higher second ionization enthalpy? ", "options": [ { "text": "Cr" }, { "text": "Mn" }, { "text": "Fe" }, { "text": "V" } ], "answer": "Cr", "solution": "**Answer:** Cr\n\nIonization enthalpy is the energy required to remove the electron from the outer most orbit. \n

    Electronic configuration : \n

    $$\\eqalign{\n & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,V\\left( {23} \\right) = \\left[ {Ar} \\right]\\,\\,\\,3{d^3}\\,\\,\\,4{S^2} \\cr \n & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,Cr\\left( {24} \\right) = \\left[ {Ar} \\right]\\,\\,\\,3{d^5}\\,\\,\\,4{S^1} \\cr \n & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,Mn\\left( {25} \\right) = \\left[ {Ar} \\right]\\,\\,\\,3{d^5}\\,\\,\\,4{S^2} \\cr \n & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,Fe\\left( {26} \\right) = \\left[ {Ar} \\right]\\,\\,\\,3{d^6}\\,\\,\\,4{S^2} \\cr} $$ \n

    After first ionization enthalpy the electronic configuration will be, \n
    $$\\eqalign{\n & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,V\\left( {23} \\right) = \\left[ {Ar} \\right]\\,\\,\\,3{d^3}\\,\\,\\,4{S^1} \\cr \n & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,Cr\\left( {24} \\right) = \\left[ {Ar} \\right]\\,\\,\\,3{d^5} \\cr \n & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,Mn\\left( {25} \\right) = \\left[ {Ar} \\right]\\,\\,\\,3{d^5}\\,\\,\\,4{S^1} \\cr \n & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,Fe\\left( {26} \\right) = \\left[ {Ar} \\right]\\,\\,\\,3{d^6}\\,\\,\\,4{S^1} \\cr} $$ \n

    Now in 2nd ionization enthalpy one more electron will be removed. And after second ionization enthalpy electronic configuration will be, \n
    $$\\eqalign{\n & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,V\\left( {23} \\right) = \\left[ {Ar} \\right]\\,\\,\\,3{d^3} \\cr \n & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,Cr\\left( {24} \\right) = \\left[ {Ar} \\right]\\,\\,\\,3{d^4} \\cr \n & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,Mn\\left( {25} \\right) = \\left[ {Ar} \\right]\\,\\,\\,3{d^5} \\cr \n & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,Fe\\left( {26} \\right) = \\left[ {Ar} \\right]\\,\\,\\,3{d^6} \\cr} $$\n

    For $$V(23), Mn (25)$$ and $$Fe (26)$$ electron is removed from $$4S$$ orbital but for $$Cr (24)$$ electron is removed from $$3d$$ orbital.\n

    We know distance of $$4S\\,\\, > \\,\\,3d$$ from the nucleus. So, the attraction on the electrons of $$45$$ shell is less compared to $$3d$$ shell by the nucleus. So, the removed of electron will be easier for the electrons of $$4S$$ shell than $$3d$$ shell.\n

    So, for $$Cr(24)$$ second ionization is more than other as electron is removed from $$3d$$ shell.\n

    And for $$Cr(24)$$ outer most shell $$''3d''$$ was half filled after $$1st$$ ionization enthalpy, so $$3d$$ shell was stable. So removal of electron will be difficult from stable shell. That is why also $$2nd$$ ionization enthalpy of $$Cr(24)$$ is higher compare to other given atoms. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2192, "subject": "Chemistry", "question": "The increasing order of the first ionization enthalpies of the elements B, P, S and F (lowest first) is :", "options": [ { "text": "F < S < P < B " }, { "text": "P < S < B < F" }, { "text": "B < P < S < F " }, { "text": "B < S < P < F " } ], "answer": "B < S < P < F ", "solution": "**Answer:** B < S < P < F \n\nThe correct order of ionisation enthalpies is $$F > P > S > B$$\n

    NOTE : On moving along a period ionization enthalpy increases from left to right and decreases from top to bottom in a group. But this trend breaks up in case of atom having fully or half filled stable orbitals. \n

    In this case $$P$$ has a stable half filled electronic configuration hence its ionisation enthalpy is greater in comparison to $$S.$$ For B(5) electronic configuration is = 1s22s22p1 and P(15) electronic configuration is = 1s22s22p63s23p3. As P has half filled valence shell so removing electron will be harder compare to B. Hence by checking all options the correct possible order is\n

    B < S < P < F", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2193, "subject": "Chemistry", "question": "The first ionization potential of Na is 5.1 eV. The value of electron gain enthalpy of Na+\n will be: ", "options": [ { "text": "– 5.1 eV " }, { "text": "–10.2 eV" }, { "text": "+ 2.55 eV" }, { "text": "– 2.55 eV" } ], "answer": "– 5.1 eV ", "solution": "**Answer:** – 5.1 eV \n\nIonization potential of Na means energy required to convert of Na to Na+ ion.\n

    Na $$ \\to $$ Na+ + e-          IE = 5.1 ev\n

    Electron gain enthalpy of Na+ means energy required to convert Na+ ion to Na.\n

    Na $$ \\to $$ Na+ + e-          $$\\Delta {H_{eq}}$$ = - 5.1 ev\n

    According to Lavoisier and Laplace law of thermochemistry, when we change the direction of reaction then the sign of energy also changes.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2194, "subject": "Chemistry", "question": "Which of the following represents the correct order of increasing first ionization enthalpy for Ca, Ba, S, Se\nand Ar?", "options": [ { "text": "S < Se < Ca < Ba < Ar " }, { "text": "Ba < Ca < Se < S < Ar " }, { "text": "Ca < Ba < S < Se < Ar " }, { "text": "Ca < S < Ba < Se < Ar " } ], "answer": "Ba < Ca < Se < S < Ar ", "solution": "**Answer:** Ba < Ca < Se < S < Ar \n\nThose elements are present in periodic table like this -\n

    \n\n\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n
    Group 2Group 16Group 18
    Period 3Ca..........S.........Ar
    Period 4Sc
    Period 5Ba
    \n

    In periodic table,\n

    1. From left to right ionization energy increases.\n

    2. From top to bottom ionization energy decreases.\n

    $$ \\therefore $$ Ba < Ca and Se < S and S < Ar\n

    $$ \\therefore $$ Correct order is -\n
    Ba < Ca < Sc < S < Ar", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2195, "subject": "Chemistry", "question": "Consider the following ionization enthalpies of two elements 'A' and 'B' .\n

    \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    ElementIonization enthalpy (kJ/mol)
    1st2nd3rd
    A899175714847
    B73714507731
    \n

    Which of the following statements is correct ?\n", "options": [ { "text": "Both ‘A’ and ‘B’ belong to group-1where ‘B’ comes below ‘A’.\n" }, { "text": "Both ‘A’ and ‘B’ belong to group-1 where ‘A’ comes below ‘B’." }, { "text": "Both ‘A’ and ‘B’ belong to group-2 where ‘B’ comes below ‘A’." }, { "text": "Both ‘A’ and ‘B’ belong to group-2 where ‘A’ comes below ‘B’." } ], "answer": "Both ‘A’ and ‘B’ belong to group-2 where ‘B’ comes below ‘A’.", "solution": "**Answer:** Both ‘A’ and ‘B’ belong to group-2 where ‘B’ comes below ‘A’.\n\nFrom the table you can see ionization enthalpy of A is greater than B in all the cases.\n

    So, B comes below A in the group as in a group from top to bottom ionization enthalpy decreases.\n

    After 1st ionization enthalpy each element become cation by removing a electron. In A+ and B+, after removing one electron from each element, effective nuclear charge increases as per-electron attraction increase by the neuclers. So, the removal of next electron will be more difficult, therefore more energy is required for 2nd ionization energy. From the table you can see 2nd ionization energy is more than first ionization energy for both elements. \n

    But you can see 3rd ionization enthalpy is so much higher than 2nd ionization enthalpy. It means after 2nd ionization enthalpy outermost shell is empty and in 3rd ionization enthalpy from a new shell electron is removed, as the new shell is closer to the neucleus so the attraction by the nucleus to the electrons of this shell is more and to remove a electron from this shell you have to provide very high energy, that is why 3rd ionization enthalpy is so high.\n

    If A and B are from group 1 then after 1st ionization enthalpy outermost shell will be empty and 2nd electron will be removed from inner shell, so 2nd ionization will be so high. But from table you can see 2nd ionization energy is around double of 1st ionization energy, it is not so much high than 1st onization energy. So A and B can't be from group 1.\n

    A and B are from group 2 as in group 2 element, outermost shell has 2 electron and after 2nd ionization enthalpy outermost shell will be empty and in 3rd ionization enethalpy, third electron will removed from innershell so 3rd ionization enthalpy will be very high. ", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 2196, "subject": "Chemistry", "question": "The electronic configuration with the highest ionization enthalpy is :", "options": [ { "text": "[Ne] 3s2 3p1" }, { "text": "[Ne] 3s2 3p2 " }, { "text": "[Ne] 3s2 3p3" }, { "text": "[Ar] 3d10 4s2 4p3" } ], "answer": "[Ne] 3s2 3p3", "solution": "**Answer:** [Ne] 3s2 3p3\n\nIn option (D) electron is removed from 4p subshell for 1st ionization enthalpy, but in all other options electron is removed from 3p subshell. As 3p subshell is closer to nucleus than 4p, then attraction to the electrons in 3p subshell is more compared to the electrons in 4p subhell. That is why removal of electrons from 4p subshell is easy compared to 3p subshell. So among all the options, option (D) will have least ionization enthalpy.\n

    Among (A), (B), (C) options, (C) has half filled 3p subshell so it is more stable compare to the others and you have to provide more energy to remove electrons from stable 3p subshell. That is why electron configuration [Ne] 3s2 3p3 has highest ionization enthalpy.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2197, "subject": "Chemistry", "question": "The element having greatest difference between\nits first and second ionization energies, is :", "options": [ { "text": "Ba" }, { "text": "Ca" }, { "text": "Sc" }, { "text": "K" } ], "answer": "K", "solution": "**Answer:** K\n\nElectronic configuration of Potassium (K) :\n

    1s22s22p63s23p64s1\n

    After first ionisation enthalpy its configuration becomes :\n

    1s22s22p63s23p6\n

    which is a inert gas configuration. To remove an electron from inert gas configuration requires very high energy.\n

    So Potassium (K) have high difference in the first\nionisation and the second ionisation energy as\nit achieve stable noble gas configuration\nafter first ionisation.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2198, "subject": "Chemistry", "question": "In comparison to boron, berylium has :", "options": [ { "text": "lesser nuclear charge and greater first ionisation enthalpy" }, { "text": "greater nuclear charge and greater first ionisation enthalpy" }, { "text": "greater nuclear charge and lesser first ionisation enthalpy" }, { "text": "lesser nuclear charge and lesser first ionisation ethalpy" } ], "answer": "lesser nuclear charge and greater first ionisation enthalpy", "solution": "**Answer:** lesser nuclear charge and greater first ionisation enthalpy\n\nElectronic configuration of Be(4) = 1s22s2\n
    Electronic configuration of B(5) = 1s22s22p1\n

    Removing electron from outer most shell of Be is harder compare to B, as for Be 2s orbital is full filled so it is stable therefore required more ionization enthalpy to remove an electron.\n
    We know in a period from left to righ effective nuclear charge increases, so B will have more nuclear charge.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2199, "subject": "Chemistry", "question": "The third ionization enthalpy is minimum for :", "options": [ { "text": "Ni" }, { "text": "Co" }, { "text": "Mn" }, { "text": "Fe" } ], "answer": "Fe", "solution": "**Answer:** Fe\n\nElectronic configuration of\n

    25Mn = [Ar]3d54s2\n

    25Mn2+ = [Ar]3d54s0\n

    26Fe = [Ar]3d64s2\n

    26Fe2+ = [Ar]3d64s0\n

    27Co = [Ar]3d74s2\n

    27Co2+ = [Ar]3d74s0\n

    28Ni = [Ar]3d84s2\n

    28Ni2+ = [Ar]3d84s0\n

    So third ionisation energy is minimum for Fe.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2200, "subject": "Chemistry", "question": "The first ionization energy (in kJ/mol) of Na, Mg, Al and Si respectively, are :", "options": [ { "text": "496, 577, 786, 737" }, { "text": "496, 737, 577, 786" }, { "text": "786, 737, 577, 496" }, { "text": "496, 577, 737, 786" } ], "answer": "496, 737, 577, 786", "solution": "**Answer:** 496, 737, 577, 786\n\nElecronic configuration of Na = [Ne] 3s1\n
    Mg = [Ne] 3s2\n
    Al = [Ne] 3s23p1\n
    Si = [Ne] 3s23p2\n

    Correct order is :\n

    Na $$<$$ Al $$<$$ Mg $$<$$ Si", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2201, "subject": "Chemistry", "question": "B has a smaller first ionization enthalpy than\nBe. Consider the following statements :

    \n(I) It is easier to remove 2p electron than 2s\nelectron

    \n(II) 2p electron of B is more shielded from the\nnucleus by the inner core of electrons than\nthe 2s electrons of Be.

    \n(III) 2s electron has more penetration power\nthan 2p electron.

    \n(IV) atomic radius of B is more than Be\n(Atomic number B = 5, Be = 4)

    \nThe correct statements are :", "options": [ { "text": "(I), (III) and (IV)" }, { "text": "(II), (III) and (IV)" }, { "text": "(I), (II) and (IV)" }, { "text": "(I), (II) and (III)" } ], "answer": "(I), (II) and (III)", "solution": "**Answer:** (I), (II) and (III)\n\n1st I.E. of Be > B\n

    In case of Be, electron is removed from 2s\norbital which has more penetration power,\nwhile in case of B electron is removed from 2p\norbital which has less penetration power.\n

    2p electron of B is more shielded from nucleus\nby the inner electrons than 2s electrons of Be\n

    $$ \\therefore $$ It is easier to remove 2p electron than 2s\nelectron.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2202, "subject": "Chemistry", "question": "The first and second ionisation enthalpies of a\nmetal are 496 and 4560 kJ mol–1, respectively.\nHow many moles of HCl and H2SO4,\nrespectively, will be needed to react completely\nwith 1 mole of the metal hydroxide ?", "options": [ { "text": "1 and 2" }, { "text": "1 and 0.5" }, { "text": "1 and 1" }, { "text": "2 and 0.5" } ], "answer": "1 and 0.5", "solution": "**Answer:** 1 and 0.5\n\nFirst ionization enthalpies = 496 kJ/mole\n

    Second ionization enthalpies = 4560 kJ/mol\n

    According to the given information, the difference between first and second ionization enthalpy is very high so Metal belong to 1st group i.e. Monovalent cation.\n

    Metal hydroxide will be of type, MOH.\n

    MOH + HCl $$ \\to $$ MCl + H2O\n

    MOH + $${1 \\over 2}$$H2SO4\n$$ \\to $$\n$${1 \\over 2}$$M2SO4 + H2O\n

    So one mole of HCl required to react with one\nmole MOH.\n

    And $${1 \\over 2}$$ mole of H2SO4 required to react with one\nmole MOH.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2203, "subject": "Chemistry", "question": "Among the statements (I – IV), the correct ones\nare:\n
    (I) Be has smaller atomic radius compared to\nMg.\n
    (II) Be has higher ionization enthalpy than Al.\n
    (III) Charge/radius ratio of Be is greater than\nthat of Al.\n
    (IV) Both Be and Al form mainly covalent\ncompounds.", "options": [ { "text": "(I), (II) and (IV)" }, { "text": "(II), (III) and (IV)" }, { "text": "(I), (III) and (IV)" }, { "text": "(I), (II) and (III)" } ], "answer": "(I), (II) and (IV)", "solution": "**Answer:** (I), (II) and (IV)\n\nBe < Mg (atomic radius)\n

    Be > Al (I.E1 greater than Al because of fully filled valence shell of Be)\n

    Charge/radius ratio of Be is less than that of Al.\n

    Both Be and Al form mainly covalent\ncompound.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2204, "subject": "Chemistry", "question": "The five successive ionization enthalpies of an\nelement are 800, 2427, 3658, 25024 and 32824\nkJ mol–1. The number of valence electrons in\nthe element is :", "options": [ { "text": "2" }, { "text": "3" }, { "text": "4" }, { "text": "5" } ], "answer": "3", "solution": "**Answer:** 3\n\nThere is a sudden jump after 3rd I.E. due to\nattainment of noble gas configuration.\n

    So, the number of valence electrons in this\nelement are 3.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2205, "subject": "Chemistry", "question": "Consider the elements Mg, Al, S, P and Si, the correct increasing order of their first ionization\nenthalpy is :", "options": [ { "text": "Al < Mg < Si < S < P" }, { "text": "Al < Mg < S < Si < P" }, { "text": "Mg < Al < Si < S < P" }, { "text": "Mg < Al < Si < P < S" } ], "answer": "Al < Mg < Si < S < P", "solution": "**Answer:** Al < Mg < Si < S < P\n\n

    On moving left to right in a period of the periodic table,\nIonisation energy (I.E.) increases due to increase in effective nuclear\ncharge i.e. Zeff.

    \n

    But due to extra stability of fully filled and half-filled electronic\nconfiguration of Mg and P required ionisation enthalpy is more from\nneighbouring elements.

    \n

    i.e. First ionisation enthalpy order is Al < Mg < Si < S < P.

    \n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2206, "subject": "Chemistry", "question": "Match List - I with List - II.

    \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    List I
    Electronic configuration    		List II
    $${\\Delta _i}H$$ in kJ $$mo{l^{ - 1}}$$
    (a)$$1{s^2}2{s^2}$$(i)801
    (b)$$1{s^2}2{s^2}2{p^4}$$(ii)899
    (c)$$1{s^2}2{s^2}2{p^3}$$(iii)1314
    (d)$$1{s^2}2{s^2}2{p^1}$$(iv)1402

    Choose the most appropriate answer from the options given below :", "options": [ { "text": "(a) $$ \\to $$ (i), (b) $$ \\to $$ (iii), (c) $$ \\to $$ (iv), (d) $$ \\to $$ (iii)" }, { "text": "(a) $$ \\to $$ (iv), (b) $$ \\to $$ (i), (c) $$ \\to $$ (ii), (d) $$ \\to $$ (iii)" }, { "text": "(a) $$ \\to $$ (ii), (b) $$ \\to $$ (iii), (c) $$ \\to $$ (iv), (d) $$ \\to $$ (i)" }, { "text": "(a) $$ \\to $$ (i), (b) $$ \\to $$ (iv), (c) $$ \\to $$ (iii), (d) $$ \\to $$ (ii)" } ], "answer": "(a) $$ \\to $$ (ii), (b) $$ \\to $$ (iii), (c) $$ \\to $$ (iv), (d) $$ \\to $$ (i)", "solution": "**Answer:** (a) $$ \\to $$ (ii), (b) $$ \\to $$ (iii), (c) $$ \\to $$ (iv), (d) $$ \\to $$ (i)\n\n(a) 1s2 2s2 $$ \\to $$ Be

    \n(b) 1s2 2s2 2p4 $$ \\to $$ O

    \n(c) 1s2 2s2 2p3 $$ \\to $$ N

    \n(d) 1s2 2s2 2p1 $$ \\to $$ B

    \nThe ionization enthalpy order is

    \nB < Be < O < N

    \nBe has more IE compared to B due to extra\nstability & N has more IE compared to oxygen\ndue to extra stability.

    \nHence, N $$ \\to $$ 1402 kJ/mol

    \nO $$ \\to $$ 1314 kJ/mol

    \nB $$ \\to $$ 801 kJ/mol

    \nBe $$ \\to $$ 899 kJ/mol", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2207, "subject": "Chemistry", "question": "Identify the elements X and Y using the ionisation energy values given below :

    \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    Ionization energy (kJ/mol)
    1$${st}$$2$${nd}$$
    X4954563
    Y7311450
    ", "options": [ { "text": "X = F; Y = Mg" }, { "text": "X = Mg; Y = F" }, { "text": "X = Na; Y = Mg" }, { "text": "X = Mg; Y = Na" } ], "answer": "X = Na; Y = Mg", "solution": "**Answer:** X = Na; Y = Mg\n\nDue to 2p6, noble gas electronic configuration, the\nsecond ionisation enthalpy of Na is very high. That’s\nwhy has large difference between IE1, and IE2

    \nMg+ is 2p6, 3s1.

    \nAfter the loss of one electron, Mg+ will be formed\nwith noble gas electronic configuration. That’s why\nhas less difference between I.E1 and I.E2, but it's I.E3 is very high.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2208, "subject": "Chemistry", "question": "The first ionization energy of magnesium is smaller as compared to that of elements X and Y, but higher than that of Z. The elements X, Y and Z, respectively, are :", "options": [ { "text": "neon, sodium and chlorine" }, { "text": "argon, chlorine and sodium" }, { "text": "chlorine, lithium and sodium" }, { "text": "argon, lithium and sodium" } ], "answer": "argon, chlorine and sodium", "solution": "**Answer:** argon, chlorine and sodium\n\nOrder of I.E.

    \n3rd period $$ \\to $$ Na < Al < Mg < Si < S < P < Cl < Ar", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2209, "subject": "Chemistry", "question": "The correct order of first ionisation enthalpy is :", "options": [ { "text": "Mg < S < Al < P" }, { "text": "Mg < Al < S < P" }, { "text": "Al < Mg < S < P" }, { "text": "Mg < Al < P < S" } ], "answer": "Al < Mg < S < P", "solution": "**Answer:** Al < Mg < S < P\n\n$$\\matrix{\n {Mg} & {Al} & P & {S \\to IE} \\cr \n\n } $$ order $$\\Rightarrow$$ Al < Mg < S < P

    Valence [Ne] : $$\\matrix{\n {\\mathop {3{s^2}}\\limits^{Mg} } & {\\mathop {3{s^2}3{P^1}}\\limits^{Al} } & {\\mathop {3{s^2}3{P^3}}\\limits^P } & {\\mathop {3{s^2}3{P^4}}\\limits^S } \\cr \n\n } $$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2210, "subject": "Chemistry", "question": "

    Given below are two statements. One is labelled as Assertion A and the other is labelled as Reason R.

    \n

    Assertion A : The first ionization enthalpy for oxygen is lower than that of nitrogen.

    \n

    Reason R : The four electrons in 2p orbitals of oxygen experience more electron-electron repulsion.

    \n

    In the light of the above statements, choose the correct answer from the options given below.

    ", "options": [ { "text": "Both A and R are are correct and R is the correct explanation of A." }, { "text": "Both A and R are are correct but R is NOT the correct explanation of A." }, { "text": "A is correct but R is not correct." }, { "text": "A is not correct but R is correct." } ], "answer": "Both A and R are are correct but R is NOT the correct explanation of A.", "solution": "**Answer:** Both A and R are are correct but R is NOT the correct explanation of A.\n\nNitrogen has half-filled p-orbitals which is stable.\nDue to this, its 1st ionization energy is more than\noxygen.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2211, "subject": "Chemistry", "question": "

    The first ionization enthalpies of Be, B, N and O follow the order

    ", "options": [ { "text": "O < N < B < Be" }, { "text": "Be < B < N < O" }, { "text": "B < Be < N < O" }, { "text": "B < Be < O < N" } ], "answer": "B < Be < O < N", "solution": "**Answer:** B < Be < O < N\n\nThe first ionization energy increase from left to right along $2^{\\text {nd }}$ period with the following exceptions\n

    \n$$\\mathrm{IE}_{1}: \\mathrm{Be}>\\mathrm{B}$$ and $$\\mathrm{N}>\\mathrm{O}$$\n

    \nThis is due to stable configuration of $$\\mathrm{Be}$$ in comparison to $$\\mathrm{B}$$ and that of $$\\mathrm{N}$$ in comparison to $$\\mathrm{O}$$.\n

    \nHence the correct order is $$\\mathrm{N}>\\mathrm{O}>\\mathrm{Be}>\\mathrm{B}$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2212, "subject": "Chemistry", "question": "

    The incorrect statement is

    ", "options": [ { "text": "The first ionization enthalpy of K is less than that of Na and Li." }, { "text": "Xe does not have the lowest first ionization enthalpy in its group." }, { "text": "The first ionization enthalpy of element with atomic number 37 is lower than that of the element with atomic number 38." }, { "text": "The first ionization enthalpy of Ga is higher than that of the d-block element with atomic number 30." } ], "answer": "The first ionization enthalpy of Ga is higher than that of the d-block element with atomic number 30.", "solution": "**Answer:** The first ionization enthalpy of Ga is higher than that of the d-block element with atomic number 30.\n\nOn moving down in a group ionisation energy decrease\n

    \n$\\therefore 1^{\\text {st }}$ ionisation enthalpy order is $\\mathrm{Li}>\\mathrm{Na}>\\mathrm{K}$\n

    \n$\\mathrm{Zn}$ has more ionisation energy as compared to Ga because of their pseudo inert gas configuration.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2213, "subject": "Chemistry", "question": "

    The first ionization enthalpy of Na, Mg and Si, respectively, are : 496, 737 and $$786 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$$. The first ionization enthalpy ($$\\mathrm{kJ} \\,\\mathrm{mol}^{-1}$$) of $$\\mathrm{Al}$$ is :

    ", "options": [ { "text": "487" }, { "text": "768" }, { "text": "577" }, { "text": "856" } ], "answer": "577", "solution": "**Answer:** 577\n\nI.E. $\\mathrm{Na}<\\mathrm{Al}<\\mathrm{Mg}<\\mathrm{Si}$\n

    \n$\\because 496<$ I.E. (Al) $<737$\n

    \nOption (C), matches the condition

    i.e. I.E. $(\\mathrm{Al})=577 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2214, "subject": "Chemistry", "question": "Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R)\n

    \nAssertion (A): The first ionization enthalpy of $3 \\mathrm{~d}$ series elements is more than that of group 2 metals\n

    \nReason (R): In 3d series of elements successive filling of d-orbitals takes place.\n

    \nIn the light of the above statements, choose the correct answer from the options given below : ", "options": [ { "text": "Both (A) and (R) are true and (R) is the correct explanation of (A)" }, { "text": "(A) is true but (R) is false" }, { "text": "Both (A) and (R) are true but $(\\mathbf{R})$ is not the correct explanation of (A)" }, { "text": "(A) is false but (R) is true" } ], "answer": "(A) is false but (R) is true", "solution": "**Answer:** (A) is false but (R) is true\n\nFrom Sc to Mn ionization energy is less than that of Mg.\n

    \"JEE\n

    Reason (R) : The statement \"In $3d$ series of elements successive filling of $d$-orbitals takes place\" is indeed true. It describes the characteristic of $3d$ series elements where there is successive filling of $d$ orbitals as we move across the period.", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 2215, "subject": "Chemistry", "question": "

    Given below are two statements :

    \n

    Statement I : The decrease in first ionization enthalpy from B to Al is much larger than that from Al to Ga.

    \n

    Statement II : The d orbitals in Ga are completely filled.

    \n

    In the light of the above statements, choose the most appropriate answer from the options given below

    ", "options": [ { "text": "Statement I is correct but statement II is incorrect" }, { "text": "Statement I is incorrect but statement II is correct" }, { "text": "Both the statements I and II are correct" }, { "text": "Both the statements I and II are incorrect" } ], "answer": "Both the statements I and II are correct", "solution": "**Answer:** Both the statements I and II are correct\n\n

    Statement I is correct as the decrease in first ionization enthalpy from B to Al is indeed much larger than from Al to Ga. This is due to the shielding effect of the electron in the outermost shell and the increased effective nuclear charge as one moves from B to Al and from Al to Ga.

    \n

    Statement II is also correct, as the d orbitals in Ga are indeed completely filled. This results in the shielding effect being maximum for Ga and thus, the decrease in first ionization enthalpy from Al to Ga is smaller compared to that from B to Al.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2216, "subject": "Chemistry", "question": "

    For elements $$\\mathrm{B}, \\mathrm{C}, \\mathrm{N}, \\mathrm{Li}, \\mathrm{Be}, \\mathrm{O}$$ and $$\\mathrm{F}$$, the correct order of first ionization enthalpy is

    ", "options": [ { "text": "$$\\mathrm{Li}<\\mathrm{Be}<\\mathrm{B}<\\mathrm{C}<\\mathrm{O}<\\mathrm{N}<\\mathrm{F}$$" }, { "text": "$$\\mathrm{B}>\\mathrm{Li}>\\mathrm{Be}>\\mathrm{C}>\\mathrm{N}>\\mathrm{O}>\\mathrm{F}$$" }, { "text": "$$\\mathrm{Li}<\\mathrm{Be}<\\mathrm{B}<\\mathrm{C}<\\mathrm{N}<\\mathrm{O}<\\mathrm{F}$$" }, { "text": "$$\\mathrm{Li}<\\mathrm{B}<\\mathrm{Be}<\\mathrm{C}<\\mathrm{O}<\\mathrm{N}<\\mathrm{F}$$" } ], "answer": "$$\\mathrm{Li}<\\mathrm{B}<\\mathrm{Be}<\\mathrm{C}<\\mathrm{O}<\\mathrm{N}<\\mathrm{F}$$", "solution": "**Answer:** $$\\mathrm{Li}<\\mathrm{B}<\\mathrm{Be}<\\mathrm{C}<\\mathrm{O}<\\mathrm{N}<\\mathrm{F}$$\n\n

    The ionization energy of an atom or molecule describes the minimum amount of energy required to remove an electron from the atom or molecule in the gaseous state. It generally increases across a period (from left to right) on the periodic table because the number of protons is increasing, making the nucleus more positively charged and therefore more strongly attracting the negatively charged electrons.

    \n

    However, there are exceptions to this trend when half-filled or fully-filled sub-shells are present, which are particularly stable configurations. The increased stability of these configurations makes it harder to remove an electron, increasing the ionization energy.

    \n

    The order of first ionization enthalpy for these elements, considering the above factors, should be:

    \n

    Li < B < Be < C < O < N < F

    \n

    This is because:

    \n\n

    So the correct answer is: $$\\mathrm{Li}<\\mathrm{B}<\\mathrm{Be}<\\mathrm{C}<\\mathrm{O}<\\mathrm{N}<\\mathrm{F}$$

    \n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2217, "subject": "Chemistry", "question": "

    Given below are two statements: one is labelled as Assertion $$\\mathbf{A}$$ and the other is labelled as Reason $$\\mathbf{R}$$

    \n

    Assertion A : The energy required to form $$\\mathrm{Mg}^{2+}$$ from $$\\mathrm{Mg}$$ is much higher than that required to produce $$\\mathrm{Mg}^+$$

    \n

    Reason $$\\mathbf{R}: \\mathrm{Mg}^{2+}$$ is small ion and carry more charge than $$\\mathrm{Mg}^{+}$$

    \n

    In the light of the above statements, choose the correct answer from the options given below.

    ", "options": [ { "text": "A is false but R is true" }, { "text": "Both A and R are correct but R is NOT the correct explanation of A" }, { "text": "Both A and R are correct and R is the correct explanation of A" }, { "text": "A is true but R is false" } ], "answer": "Both A and R are correct and R is the correct explanation of A", "solution": "**Answer:** Both A and R are correct and R is the correct explanation of A\n\n

    The correct answer is Option C: Both A and R are correct and R is the correct explanation of A.

    \n\n

    This is true because it takes more energy to remove two electrons from an atom than it does to remove one electron. The second electron is more tightly bound to the nucleus than the first electron, so it requires more energy to remove it.

    \n\n

    This is also true. Mg2+ has a smaller radius than Mg+ because it has lost two electrons. The smaller radius means that the protons in the nucleus are more tightly packed together, which makes the charge of the nucleus more effective at attracting the electrons. This makes it more difficult to remove an electron from Mg2+ than it is to remove an electron from Mg+.

    \n

    The reason is the correct explanation of the assertion because it explains why it takes more energy to form Mg2+ than it does to form Mg+. The smaller radius of Mg2+ and the higher charge of the nucleus make it more difficult to remove electrons from Mg2+, which requires more energy.

    \n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2218, "subject": "Chemistry", "question": "The transition metal having highest $3^{\\text {rd }}$ ionisation enthalpy is :", "options": [ { "text": "$\\mathrm{Mn}$" }, { "text": "$\\mathrm{Fe}$" }, { "text": "$\\mathrm{Cr}$" }, { "text": "$V$" } ], "answer": "$\\mathrm{Mn}$", "solution": "**Answer:** $\\mathrm{Mn}$\n\n

    The third ionization enthalpy refers to the energy required to remove the third electron from a di-positive ion ($M^{2+}$) to form a tri-positive ion ($M^{3+}$). In the case of transition metals, the energies involved in removing the third electron are typically higher than for the first and second electrons due to an increasing effective nuclear charge and the electrons being removed from closer energy levels to the nucleus or from a half-filled or full-filled d-subshell which is relatively more stable.

    \n\n

    Let us look at the electronic configurations of the ions of each metal listed to determine which would require the highest third ionisation enthalpy. Since we are talking about $3^{\\text{rd}}$ ionization, we would be considering the removal of three electrons, ending up with a $+3$ oxidation state for each metal.

    \n\n\n\n

    After removing two electrons (assuming they come from the 4s orbital and the next available d orbital), the configurations would be:

    \n\n\n\n

    Now we look at the energy required to remove the third electron from the $M^{2+}$ ions:

    \n\n\n\n

    In terms of stability, the half-filled $d^5$ subshell in manganese ($\\mathrm{Mn^{2+}}$) provide extra stability. So, the $3^{\\text{rd}}$ ionisation enthalpy of $\\mathrm{Mn^{2+}}$ is the highest because it involves breaking a half-filled configuration which is more stable.

    \n\n

    Thus, out of the given options, manganese ($\\mathrm{Mn}$) would have the highest third ionization enthalpy. Therefore, the correct answer is:

    \n\n

    Option A : $\\mathrm{Mn}$

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2219, "subject": "Chemistry", "question": "

    Given below are two statements : one is labelled as Assertion $$\\mathbf{A}$$ and the other is labelled as Reason $$\\mathbf{R}$$ :

    \n

    Assertion A : The first ionisation enthalpy decreases across a period.

    \n

    Reason $$\\mathbf{R}$$ : The increasing nuclear charge outweighs the shielding across the period.

    \n

    In the light of the above statements, choose the most appropriate from the options given below :

    ", "options": [ { "text": "$$\\mathbf{A}$$ is false but $$\\mathbf{R}$$ is true\n" }, { "text": "Both $$\\mathbf{A}$$ and $$\\mathbf{R}$$ are true but $$\\mathbf{R}$$ is NOT the correct explanation of $$\\mathbf{A}$$\n" }, { "text": "Both $$\\mathbf{A}$$ and $$\\mathbf{R}$$ are true and $$\\mathbf{R}$$ is the correct explanation of $$\\mathbf{A}$$\n" }, { "text": "$$\\mathbf{A}$$ is true but $$\\mathbf{R}$$ is false" } ], "answer": "$$\\mathbf{A}$$ is false but $$\\mathbf{R}$$ is true\n", "solution": "**Answer:** $$\\mathbf{A}$$ is false but $$\\mathbf{R}$$ is true\n\n\n

    First ionisation energy increases along the period. Along the period $$\\mathrm{Z}$$ increases which outweighs the shielding effect.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2220, "subject": "Chemistry", "question": "

    The element having the highest first ionization enthalpy is

    ", "options": [ { "text": "$$\\mathrm{C}$$\n" }, { "text": "$$\\mathrm{Al}$$\n" }, { "text": "$$\\mathrm{Si}$$\n" }, { "text": "$$\\mathrm{N}$$" } ], "answer": "$$\\mathrm{N}$$", "solution": "**Answer:** $$\\mathrm{N}$$\n\n

    $$\\mathrm{Al}<\\mathrm{Si}<\\mathrm{C}<\\mathrm{N} ; \\mathrm{IE}_1 \\text { order. }$$

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2221, "subject": "Chemistry", "question": "

    Match List I with List II

    \n

    \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    LIST ILIST II
    A.Melting Point $$[\\mathrm{K}]$$I.$$\\mathrm{T} 1>\\mathrm{In}>\\mathrm{Ga}>\\mathrm{A} 1>\\mathrm{B}$$
    B.\tIonic Radius $$[\\mathrm{M}^{+3} / \\mathrm{pm}]$$II.$$\\mathrm{B}>\\mathrm{T} 1>\\mathrm{Al} \\approx \\mathrm{Ga}>\\mathrm{In}$$
    C.$$\\Delta_{\\mathrm{i}} \\mathrm{H}_1[\\mathrm{~kJ} \\mathrm{~mol}^{-1}]$$III.$$\\mathrm{T} 1>\\mathrm{In}>\\mathrm{Al}>\\mathrm{Ga}>\\mathrm{B}$$
    D.Atomic Radius [pm]IV.$$\\mathrm{B}>\\mathrm{A} 1>\\mathrm{T} 1>\\mathrm{In}>\\mathrm{Ga}$$

    \n

    Choose the correct answer from the options given below:

    ", "options": [ { "text": "A-IV, B-I, C-II, D-III\n" }, { "text": "A-I, B-II, C-III, D-IV\n" }, { "text": "A-III, B-IV, C-I, D-II\n" }, { "text": "A-II, B-III, C-IV, D-I" } ], "answer": "A-IV, B-I, C-II, D-III\n", "solution": "**Answer:** A-IV, B-I, C-II, D-III\n\n\n

    Melting point : $$\\mathrm{B}>\\mathrm{A} \\ell>\\mathrm{T} \\ell>\\mathrm{In}>\\mathrm{Ga}$$

    \n

    Ionic radius $$(\\mathrm{M}^{+3} / \\mathrm{pm}): \\mathrm{T} \\ell>\\mathrm{In}>\\mathrm{Ga}>\\mathrm{A} \\ell>\\mathrm{B}$$

    \n

    $$\\left(\\Delta_{\\mathrm{IE}} \\mathrm{H}\\right)_1\\left[\\frac{\\mathrm{kJ}}{\\mathrm{mol}}\\right]: \\mathrm{B}>\\mathrm{T} \\ell>\\mathrm{A} \\ell \\approx \\mathrm{Ga}>\\mathrm{In}$$

    \n

    Atomic radius (in $$\\mathrm{pm}$$) : $$\\mathrm{T} \\ell>\\mathrm{In}>\\mathrm{A} \\ell>\\mathrm{Ga}>\\mathrm{B}$$

    ", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 2222, "subject": "Chemistry", "question": "

    The correct order of first ionization enthalpy values of the following elements is :

    \n

    (A) O

    \n

    (B) N

    \n

    (C) Be

    \n

    (D) F

    \n

    (E) B

    \n

    Choose the correct answer from the options given below :

    ", "options": [ { "text": "$$\\mathrm{A}<\\mathrm{B}<\\mathrm{D}<\\mathrm{C}<\\mathrm{E}$$\n" }, { "text": "$$\\mathrm{C}<\\mathrm{E}<\\mathrm{A}<\\mathrm{B}<\\mathrm{D}$$\n" }, { "text": "$$\\mathrm{E}<\\mathrm{C}<\\mathrm{A}<\\mathrm{B}<\\mathrm{D}$$\n" }, { "text": "$$\\mathrm{B}<\\mathrm{D}<\\mathrm{C}<\\mathrm{E}<\\mathrm{A}$$" } ], "answer": "$$\\mathrm{E}<\\mathrm{C}<\\mathrm{A}<\\mathrm{B}<\\mathrm{D}$$\n", "solution": "**Answer:** $$\\mathrm{E}<\\mathrm{C}<\\mathrm{A}<\\mathrm{B}<\\mathrm{D}$$\n\n\n

    Correct ionization enthalpy order :

    \n

    $$\\quad$$B < Be < O < N < F

    \n

    or E < C < A < B < D

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2223, "subject": "Chemistry", "question": "

    The correct order of the first ionization enthalpy is

    ", "options": [ { "text": "$$\\mathrm{Ga}>\\mathrm{Al}>\\mathrm{B}$$\n" }, { "text": "$$\\mathrm{Tl}>\\mathrm{Ga}>\\mathrm{Al}$$\n" }, { "text": "$$\\mathrm{Al}>\\mathrm{Ga}>\\mathrm{Tl}$$\n" }, { "text": "$$\\mathrm{B}>\\mathrm{Al}>\\mathrm{Ga}$$" } ], "answer": "$$\\mathrm{Tl}>\\mathrm{Ga}>\\mathrm{Al}$$\n", "solution": "**Answer:** $$\\mathrm{Tl}>\\mathrm{Ga}>\\mathrm{Al}$$\n\n\n

    I. $$\\mathrm{E}_1$$ of $$\\mathrm{TI}=589 \\mathrm{~kJ} / \\mathrm{mol}$$

    \n

    I. $$\\mathrm{E}_1$$ of $$\\mathrm{Ga}=579 \\mathrm{~kJ} / \\mathrm{mol}$$

    \n

    I. $$\\mathrm{E_1}$$ of $$\\mathrm{Al}=577 \\mathrm{~kJ} / \\mathrm{mol}$$

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2224, "subject": "Chemistry", "question": "

    Given below are two statements :

    \n

    Statement I : The correct order of first ionization enthalpy values of $$\\mathrm{Li}, \\mathrm{Na}, \\mathrm{F}$$ and $$\\mathrm{Cl}$$ is $$\\mathrm{Na}<\\mathrm{Li}<\\mathrm{Cl}<\\mathrm{F}$$.

    \n

    Statement II : The correct order of negative electron gain enthalpy values of $$\\mathrm{Li}, \\mathrm{Na}, \\mathrm{F}$$ and $$\\mathrm{Cl}$$ is $$\\mathrm{Na}<\\mathrm{Li}<\\mathrm{F}<\\mathrm{Cl}$$

    \n

    In the light of the above statements, choose the correct answer from the options given below :

    ", "options": [ { "text": "Statement I is true but Statement II is false\n" }, { "text": "Both Statement I and Statement II are false\n" }, { "text": "Statement I is false but Statement II is true\n" }, { "text": "Both Statement I and Statement II are true" } ], "answer": "Both Statement I and Statement II are true", "solution": "**Answer:** Both Statement I and Statement II are true\n\n

    First ionization enthalpy of $$\\mathrm{F}>\\mathrm{Cl}$$ as first IE decreases down the group.

    \n

    Similarly, first IE of $$\\mathrm{Li}>\\mathrm{Na}$$.

    \n

    So, statement-I is correct.

    \n

    Electron gain enthalpy of given elements are negative. Considering the magnitude the given order is correct.

    \n

    Thus, statement-II is correct

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2225, "subject": "Chemistry", "question": "Which one of the following is an amphoteric oxide?", "options": [ { "text": "Na2O" }, { "text": "SO2" }, { "text": "B2O3" }, { "text": "ZnO" } ], "answer": "ZnO", "solution": "**Answer:** ZnO\n\nThese elements are called amphoteric which reacts with both acid and base. \n

    Following elements are amphoteric \n

    $$\\left( 1 \\right)\\,\\,\\,\\,Zn$$\n

    $$\\left( 2 \\right)\\,\\,\\,\\,Be$$\n

    $$\\left( 3 \\right)\\,\\,\\,\\,Al$$\n

    $$\\left( 4 \\right)\\,\\,\\,\\,B$$\n

    $$\\left( 5 \\right)\\,\\,\\,\\,Cr$$\n

    $$\\left( 6 \\right)\\,\\,\\,\\,Ga$$\n

    $$\\left( 7 \\right)\\,\\,\\,\\,Pb$$\n

    $$\\left( 8 \\right)\\,\\,\\,\\,Sn$$\n

    Here $$Zn,Be,Al,Pb\\,\\,\\,$$ and $$\\,$$ $$Sn$$ are mostly amphoteric.\n

    $$B, Cr$$ and $$Ga$$ are less amphoteric. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2226, "subject": "Chemistry", "question": "Among Al2O3, SiO2, P2O3 and SO2 the correct order of acid strength is", "options": [ { "text": "SO2 < P2O3 < SiO2 < Al2O3" }, { "text": "Al2O3 < SiO2 < P2O3 < SO2" }, { "text": "Al2O3 < SiO2 < SO2 < P2O3" }, { "text": "SiO2 < SO2 < Al2O3 < P2O3" } ], "answer": "Al2O3 < SiO2 < P2O3 < SO2", "solution": "**Answer:** Al2O3 < SiO2 < P2O3 < SO2\n\nIn periodic table from left to right (group $$1$$ to group $$18$$ ) acidic strength increases. \n

    $$Al$$ belongs to group $$13,$$ $$Si$$ belongs to group $$14,$$ $$P$$ belongs to group $$15$$ and $$S$$ belongs to group $$16$$. \n

    So, $$\\,\\,\\,A{l_2}{O_3}$$ will be least acidic and $$S{o_2}$$ will most acidic among then. \n

    So, right order will be \n

    $$A{l_2}{O_3} < Si{O_2} < {P_2}{O_3} < S{O_2}$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2227, "subject": "Chemistry", "question": "Which of the following oxides is amphoteric in character? ", "options": [ { "text": "CaO" }, { "text": "CO2" }, { "text": "SiO2 " }, { "text": "SnO2" } ], "answer": "SnO2", "solution": "**Answer:** SnO2\n\nThese elements are called amphoteric which reacts with both acid and base. \n

    Following elements are amphoteric \n

    $$\\left( 1 \\right)\\,\\,\\,\\,Zn$$\n

    $$\\left( 2 \\right)\\,\\,\\,\\,Be$$\n

    $$\\left( 3 \\right)\\,\\,\\,\\,Al$$\n

    $$\\left( 4 \\right)\\,\\,\\,\\,B$$\n

    $$\\left( 5 \\right)\\,\\,\\,\\,Cr$$\n

    $$\\left( 6 \\right)\\,\\,\\,\\,Ga$$\n

    $$\\left( 7 \\right)\\,\\,\\,\\,Pb$$\n

    $$\\left( 8 \\right)\\,\\,\\,\\,Sn$$\n

    Here $$Zn,Be,Al,Pb\\,\\,\\,$$ and $$\\,$$ $$Sn$$ are mostly amphoteric.\n

    $$B, Cr$$ and $$Ga$$ are less amphoteric. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2228, "subject": "Chemistry", "question": "In which of the following arrangements, the sequence is not strictly according to the property written against it", "options": [ { "text": "CO2 < SiO2 < SnO2 < PbO2 : increasing oxidising power" }, { "text": "HF < HCl < HBr, HI : increasing acid strength" }, { "text": "NH3 < PH3 < AsH3 < SbH3 : increasing basic strength" }, { "text": "B < C < O < N : increasing first ionization enthalpy" } ], "answer": "NH3 < PH3 < AsH3 < SbH3 : increasing basic strength", "solution": "**Answer:** NH3 < PH3 < AsH3 < SbH3 : increasing basic strength\n\n$$(A)\\,\\,\\,$$ Option A is true.\n

    Here in $$Pb{O_2}$$ oxidation number of $$Pb$$ is $$+4,$$ due to inert pair effect the oxidation number $$+4$$ is not stable for $$Pb$$, So, to become stable oxide $$+2$$ $$Pb$$ will oxidize other and reduced to $$P{b^{ + 2.}}$$ So, $$Pb{O_2}$$ has the highest oxidizing power.\n

    $$(B)\\,\\,\\,\\,$$ Option B is true.\n

    Bond length from top to bottom in periodic table increases, So the bond length order is $$HI > HBr > HCl > HF$$ \n

    The compound which can give $${H^ + }$$ ion easily that compound has higher acidic strength. \n

    As the bond length between $$H$$ and $$I$$ in $$HI$$ is longest then it will be carier to break the bond between $$H$$ and $$I$$. So $$HI$$ will have highest acidic strength. \n

    $$(C)\\,\\,\\,\\,$$ Option C is false. \n

    The structure of those $$4$$ molecules are \n

    \"AIEEE \n
    Due to the lone pair electron all of them are basic nature. The molecule which can easily donate the lone pair electron will have more basic strength. \n

    Here, Nitrogen (N) is a element of $$2$$nd period so in $$N$$ new electron is added in the second period and as size of $$N$$ is small so electron density is more on the outer shell, so electron become unstable due to repulsion with each other. That is why electron pair is easily removable from $$N{H_3}.$$ \n

    On the other hand $$p$$ is a $$3$$rd block, As is a $$4$$th block and $$Sb$$ is a $$5$$th block elements. So, their size is bigger. So, electron density on the outer shell is less and their repulsion also decreases. So, outer shell electrons are more stable and $$P{H_3},As{H_3}$$ and $$Sb{H_3}$$ will have less ability to donate the electron. \n

    So the correct order is $$Sb{H_3} < As{H_3} < P{H_3} < N{H_3}$$\n

    (D) Option (D) is true\n

    Electronic configuration of \n

    $$B\\left( 5 \\right)\\,\\,\\, = \\,\\,\\,1{s^2}2{s^2}2{p^1}$$ \n

    $$C\\left( 6 \\right)\\,\\,\\, = \\,\\,\\,1{s^2}2{s^2}2{p^2}$$ \n

    $$N\\left( 7 \\right)\\,\\,\\, = \\,\\,\\,1{s^2}2{s^2}2{p^3}$$\n

    $$O\\left( 8 \\right)\\,\\,\\, = \\,\\,\\,1{s^2}2{s^2}2{p^4}$$ \n

    In general in periodic table from left to right effective nuclear change increase and it becomes more difficult to remove electron from an atom. That is why ionization enthalpy increases. \n

    But for those atoms which has half filled or full filled outer most shell, those atoms will be more stable and more energy is needed to remove electron from those atoms. Here $$N\\left( {1{s^2}\\,2{s^2}\\,2{p^3}} \\right)$$ has stable half filled $$2p$$ subshell so it is more stable than $$O.$$ \n

    So, correct order is $$B < C < O < N$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2229, "subject": "Chemistry", "question": "Which one of the following order represents the correct sequence of the increasing basic nature of the\ngiven oxides ?", "options": [ { "text": "MgO < K2O < Al2O3 < Na2O" }, { "text": "Al2O3 < MgO < Na2O < K2O" }, { "text": "Na2O < K2O < MgO < Al2O3" }, { "text": "K2O < Na2O < Al2O3 < MgO" } ], "answer": "Al2O3 < MgO < Na2O < K2O", "solution": "**Answer:** Al2O3 < MgO < Na2O < K2O\n\nYou should know that, \n

    (1)$$\\,\\,\\,$$ In a period, from left to right the acidic strength increases that mean basic nature decreases.\n

    (2)$$\\,\\,\\,$$ In a group, from top to bottom the basic nature increases that means acidic nature decreases. \n

    Na, Mg, Al and K are present in periodic table like this : -\n

    \n\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n
    Group 1Group 2Group 13
    Period 3NaMg.....Al
    Period 4K
    \n

    So, you can see Na, Mg and Al are present in same period so the order of basic nature among Na, Mg and Al is Na > Mg > Al.\n

    And Na and K are from same group so basic nature of K > Na.\n

    So the correct order of oxides \n

    Al2O3 < MgO < Na2O < K2O ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2230, "subject": "Chemistry", "question": "Which one of the following is an oxide?", "options": [ { "text": "KO2 " }, { "text": "BaO2 " }, { "text": "SiO2 " }, { "text": "CsO2" } ], "answer": "SiO2 ", "solution": "**Answer:** SiO2 \n\nNa, H, Ba and Sr produce Peroxide.\n

    K, Rb and Cs produces Superoxide.\n

    So, SiO2 will be oxide. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2231, "subject": "Chemistry", "question": "The acidic, basic and amphoteric oxides,\nrespectively, are :", "options": [ { "text": "Cl2O, CaO, P4O10" }, { "text": "N2O3, Li2O, Al2O3" }, { "text": "MgO, Cl2O, Al2O3" }, { "text": "Na2O, SO3, Al2O3" } ], "answer": "N2O3, Li2O, Al2O3", "solution": "**Answer:** N2O3, Li2O, Al2O3\n\nNon metal oxide $$ \\to $$ acidic.\n

    Metal oxide $$ \\to $$ basic.\n

    Al2O3 amphoteric.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2232, "subject": "Chemistry", "question": "Three elements X, Y and Z are in the 3rd period\nof the periodic table. The oxides of X, Y and Z,\nrespectively, are basic, amphoteric and acidic.\nThe correct order of the atomic numbers of X,\nY and Z is :", "options": [ { "text": "X < Z < Y" }, { "text": "Y < X < Z" }, { "text": "Z < Y < X" }, { "text": "X < Y < Z" } ], "answer": "X < Y < Z", "solution": "**Answer:** X < Y < Z\n\nWhen we are moving from left to right in a\nperiodic table along with atomic number of atom acidic character of oxides also increases.\n

    $$ \\therefore $$ atomic number : X < Y < Z\n

    then acidic character : X < Y < Z", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2233, "subject": "Chemistry", "question": "Number of amphoteric compounds among the following is ________.
    \n(A) BeO
    \n(B) BaO
    \n(C) Be(OH)2
    \n(D) Sr(OH)2", "options": [], "answer": "2", "solution": "**Answer:** 2\n\n\nAn amphoteric compound is a molecule or ion that can react with both as an acid or as a base.

    BeO = Amphoteric

    BaO = Basic

    Be(OH)2 = Amphoteric

    Sr(OH)2 = Basic

    Both beryllium compound BeO and Be(OH)2 are amphoteric in nature while compound BaO and Sr(OH)2 are basic in nature, they form alkaline solution in H2O.\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2234, "subject": "Chemistry", "question": "Which pair of oxides is acidic in nature?", "options": [ { "text": "CaO, SiO2" }, { "text": "B2O3, CaO" }, { "text": "B2O3, SiO2" }, { "text": "N2O, BaO" } ], "answer": "B2O3, SiO2", "solution": "**Answer:** B2O3, SiO2\n\nB2O3 and SiO2 both are oxides of non-metal and hence are acidic in nature.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2235, "subject": "Chemistry", "question": "Match List - I with List - II :

    \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    List - IList - II
    (a)NaOH(i)Acidic
    (b)$$Be{(OH)_2}$$(ii)Basic
    (c)$$Ca{(OH)_2}$$(iii)Amphoteric
    (d)$$B{(OH)_3}$$
    (e)$$Al{(OH)_3}$$


    Choose the most appropriate answer from the options given below ", "options": [ { "text": "(a)-(ii), (b)-(ii), (c)-(iii), (d)-(ii), (e)-(iii)" }, { "text": "(a)-(ii), (b)-(iii), (c)-(ii), (d)-(i), (e)-(iii)" }, { "text": "(a)-(ii), (b)-(ii), (c)-(iii), (d)-(i), (e)-(iii)" }, { "text": "(a)-(ii), (b)-(i), (c)-(ii), (d)-(iii), (e)-(iii)" } ], "answer": "(a)-(ii), (b)-(iii), (c)-(ii), (d)-(i), (e)-(iii)", "solution": "**Answer:** (a)-(ii), (b)-(iii), (c)-(ii), (d)-(i), (e)-(iii)\n\nNaOH $$\\to$$ Basic

    Be(OH)2 $$\\to$$ Amphoteric

    Ca(OH)2 $$\\to$$ Basic

    B(OH)3 $$\\to$$ Acidic

    Al(OH)3 $$\\to$$ Amphoteric", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2236, "subject": "Chemistry", "question": "

    Match List - I with List - II :

    \n

    \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    List-I (Oxide)List-II (Nature)
    (A)$$C{l_2}{O_7}$$(I)Amphoteric
    (B)$$N{a_2}O$$(II)Basic
    (C)$$A{l_2}{O_3}$$(III)Neutral
    (D)$${N_2}O$$(IV)Acidic

    \n

    Choose the correct answer from the options given below :

    ", "options": [ { "text": "(A) - (IV), (B) - (III), (C) - (I), (D) - (II)" }, { "text": "(A) - (IV), (B) - (II), (C) - (I), (D) - (III)" }, { "text": "(A) - (II), (B) - (IV), (C) - (III), (D) - (I)" }, { "text": "(A) - (I), (B) - (II), (C) - (III), (D) - (IV)" } ], "answer": "(A) - (IV), (B) - (II), (C) - (I), (D) - (III)", "solution": "**Answer:** (A) - (IV), (B) - (II), (C) - (I), (D) - (III)\n\n(A) Cl2O7 → Acidic

    \n(B) Na2O → Basic

    \n(C) Al2O3 → Amphoteric

    \n(D) N2O → Neutral

    \nOxides of metals are basic in nature whereas oxides of non-metals are acidic in nature. N2O is a neutral oxide.", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 2237, "subject": "Chemistry", "question": "

    Among the following, basic oxide is :

    ", "options": [ { "text": "SO3" }, { "text": "SiO2" }, { "text": "CaO" }, { "text": "Al2O3" } ], "answer": "CaO", "solution": "**Answer:** CaO\n\nSince, oxides of metals are basic in nature. Hence\nCaO is a basic oxide.

    \nSO3 and SiO2 are acidic oxides and Al2O3 is a amphoteric oxide.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2238, "subject": "Chemistry", "question": "

    Given below are the oxides:

    \n

    Na2O, As2O3, N2O, NO and Cl2O7

    \n

    Number of amphoteric oxides is :

    ", "options": [ { "text": "0" }, { "text": "1" }, { "text": "2" }, { "text": "3" } ], "answer": "1", "solution": "**Answer:** 1\n\nOxides

    $\\mathrm{Na}_2 \\mathrm{O} \\longrightarrow$ Basic

    \n$\\mathrm{As}_2 \\mathrm{O}_3 \\longrightarrow$ Amphoteric

    \n$\\mathrm{N}_2 \\mathrm{O} \\longrightarrow$ Neutral

    \n$\\mathrm{NO} \\longrightarrow$ Neutral

    \n$\\mathrm{Cl}_2 \\mathrm{O}_7 \\longrightarrow$ Acidic

    \nHence, only one amphoteric oxide is present.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2239, "subject": "Chemistry", "question": "

    Total number of acidic oxides among

    \n

    $$\\mathrm{N_2O_3,NO_2,N_2O,Cl_2O_7,SO_2,CO,CaO,Na_2O}$$ and $$\\mathrm{NO}$$ is ____________.

    ", "options": [], "answer": "4", "solution": "**Answer:** 4\n\n

    $$\\mathrm{N_2O_3,NO_2,Cl_2O_7,SO_2}$$ are acidic in nature.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2240, "subject": "Chemistry", "question": "

    Which of the following statements are not correct?

    \n

    A. The electron gain enthalpy of $$\\mathrm{F}$$ is more negative than that of $$\\mathrm{Cl}$$.

    \n

    B. Ionization enthalpy decreases in a group of periodic table.

    \n

    C. The electronegativity of an atom depends upon the atoms bonded to it.

    \n

    D. $$\\mathrm{Al}_{2} \\mathrm{O}_{3}$$ and $$\\mathrm{NO}$$ are examples of amphoteric oxides.

    \n

    Choose the most appropriate answer from the options given below :

    ", "options": [ { "text": "A, B, C and D" }, { "text": "A, C and D Only" }, { "text": "A, B and D Only" }, { "text": "B and D Only" } ], "answer": "A, C and D Only", "solution": "**Answer:** A, C and D Only\n\nLet's analyze the statements:\n

    \nA. The electron gain enthalpy of $\\mathrm{F}$ is more negative than that of $\\mathrm{Cl}$.

    \nThis statement is incorrect. The electron gain enthalpy of $\\mathrm{F}$ is less negative than that of $\\mathrm{Cl}$ due to its small size, which leads to increased electron-electron repulsion when an electron is added.\n

    \nB. Ionization enthalpy decreases in a group of the periodic table.

    \nThis statement is correct. Ionization enthalpy generally decreases down a group of the periodic table due to the increase in atomic size, which results in a weaker attraction between the nucleus and the outermost electrons.\n

    \nC. The electronegativity of an atom depends upon the atoms bonded to it.

    \nThis statement is incorrect. Electronegativity is a property of an atom that depends on the effective nuclear charge and the distance of the outermost electrons from the nucleus. While the difference in electronegativity between atoms in a bond can affect the bond's polarity, the electronegativity itself does not depend on the atoms bonded to it.\n

    \nD. $\\mathrm{Al}_{2} \\mathrm{O}_{3}$ and $\\mathrm{NO}$ are examples of amphoteric oxides.

    \nThis statement is incorrect. $\\mathrm{Al}_{2} \\mathrm{O}_{3}$ is an example of an amphoteric oxide, meaning it can react with both acids and bases. However, $\\mathrm{NO}$ is not an amphoteric oxide; it is a neutral oxide.\n

    \nHence, the incorrect statements are A, C, and D.", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 2241, "subject": "Chemistry", "question": "

    Group-13 elements react with $$\\mathrm{O}_{2}$$ in amorphous form to form oxides of type $$\\mathrm{M}_{2} \\mathrm{O}_{3}~(\\mathrm{M}=$$ element). Which among the following is the most basic oxide?

    ", "options": [ { "text": "Al$$_2$$O$$_3$$" }, { "text": "TI$$_2$$O$$_3$$" }, { "text": "B$$_2$$O$$_3$$" }, { "text": "Ga$$_2$$O$$_3$$" } ], "answer": "TI$$_2$$O$$_3$$", "solution": "**Answer:** TI$$_2$$O$$_3$$\n\nAs electropositive character increases basic character of oxide increases.

    \n$$\n\\underbrace{\\mathrm{B}_2 \\mathrm{O}_3}_{\\text {acidic }}<\\underbrace{\\mathrm{Al}_2 \\mathrm{O}_3<\\mathrm{Ga}_2 \\mathrm{O}_3}_{\\text {amphoteric }}<\\underbrace{\\mathrm{In}_2 \\mathrm{O}_3<\\mathrm{Tl}_2 \\mathrm{O}_3}_{\\text {basic }}\n$$", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 2242, "subject": "Chemistry", "question": "

    Given below are two statements:

    \n

    Statement - I: Along the period, the chemical reactivity of the elements gradually increases from group 1 to group 18 .

    \n

    Statement - II: The nature of oxides formed by group 1 elements is basic while that of group 17 elements is acidic.

    \n

    In the light of the above statements, choose the most appropriate from the options given below:

    ", "options": [ { "text": "Statement I is False but Statement II is true\n" }, { "text": "Both Statement I and Statement II are False\n" }, { "text": "Statement I is True But Statement II is False\n" }, { "text": "Both Statement I and Statement II are True" } ], "answer": "Statement I is False but Statement II is true\n", "solution": "**Answer:** Statement I is False but Statement II is true\n\n\n

    Chemical reactivity of elements decreases along the period therefore statement - I is false.

    \n

    Group - 1 elements from basic nature oxides while group - 17 elements form acidic oxides therefore statement - II is true.

    ", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 2243, "subject": "Chemistry", "question": "According to the Periodic Law of elements, the variation in properties of elements is related to their", "options": [ { "text": "nuclear masses" }, { "text": "atomic numbers" }, { "text": "nuclear neutron-proton number ratios" }, { "text": "atomic masses" } ], "answer": "atomic numbers", "solution": "**Answer:** atomic numbers\n\nAccording to modified modern periodic law, physical and chemical properties of the elements are periodic functions of their atomic numbers. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2244, "subject": "Chemistry", "question": "Following statements regarding the periodic trends of chemical reactivity of the alkali metals and the\nhalogens are given. Which of these statements gives the correct picture? ", "options": [ { "text": "The reactivity decreases in the alkali metals but increases in the halogens with increase in atomic\nnumber down the group " }, { "text": "In both the alkali metals and the halogens the chemical reactivity decreases with increase in\natomic number down the group" }, { "text": "Chemical reactivity increases with increase in atomic number down the group in both the alkali\nmetals and halogens " }, { "text": "In alkali metals the reactivity increases but in the halogens it decreases with increase in atomic\nnumber down the group " } ], "answer": "In alkali metals the reactivity increases but in the halogens it decreases with increase in atomic\nnumber down the group ", "solution": "**Answer:** In alkali metals the reactivity increases but in the halogens it decreases with increase in atomic\nnumber down the group \n\nFor alkali metals on going down the group, number of electron shells increase, so forces between the oppositely charged electron and nucleus is less so electron can be removed easily, therefore, reactivity increases. But for halogens they react by abstracting electron from less electronegative atoms and the attraction for additional electron is stronger when there are fewer filled electron shells between the valence electrons and atomic nucleus.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2245, "subject": "Chemistry", "question": "Outer electronic configuration of Gd (Atomic no : 64) is -", "options": [ { "text": "4f85d06s2" }, { "text": "4f45d46s2" }, { "text": "4f75d16s2" }, { "text": "4f35d56s2" } ], "answer": "4f75d16s2", "solution": "**Answer:** 4f75d16s2\n\nElectronic configuration of Gd(64) : [Xe]4f75s25p65d16s2\n

    So the outer electronic configuration will be : 4f75d16s2\n

    Shortcut Method : Gd belongs to 4f series which is also called lanthanide series. In this lanthanide series for Cerium (Ce), Gadolinium (Gd) and Lutetium (Lu) outer d orbital has 1 electron and for all other lanthanide series elements outer d orbital has 0 electron. From this you can easily say option (C) is correct.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2246, "subject": "Chemistry", "question": "The following statements concern elements in the periodic table. Which of the following is true ?", "options": [ { "text": "All the elements in Group 17 are gases." }, { "text": "The Group 13 elements are all metals." }, { "text": "Elements of Group 16 have lower ionization enthalpy values compared\nto those of Group 15 in the corresponding periods." }, { "text": "For Group 15 elements, the stability of + 5 oxidation state increases down\nthe group." } ], "answer": "Elements of Group 16 have lower ionization enthalpy values compared\nto those of Group 15 in the corresponding periods.", "solution": "**Answer:** Elements of Group 16 have lower ionization enthalpy values compared\nto those of Group 15 in the corresponding periods.\n\n

    (A) Group 17 elements are halogens; they exist in all three states of matter at room temperature.

    \n\n

    (B) Except for the lightest element (boron), all other Group 13 elements are relatively electropositive and are thus metals.

    \n\n

    (C) Group 15 elements have half-filled electronic configuration, thus have high values of ionization enthalpy, while Group 16 elements achieve halffilled configuration on loss of one electron and therefore have lower ionisation enthalpy.

    \n\n(i) In Group 15 elements, due to the inert pair affect the stability of +5 oxidation state decreases down the group.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2247, "subject": "Chemistry", "question": "Aluminium is usually found in +3 oxidation state. In contrast, thallium exists in + 1 and + 3 oxidation states. This is due to : ", "options": [ { "text": "inert pair effect " }, { "text": "diagonal relationship " }, { "text": "lattice structure" }, { "text": "lanthanoid contraction" } ], "answer": "inert pair effect ", "solution": "**Answer:** inert pair effect \n\nElectronic configuration of Thallium (Tl) is \n
    = [Xe]4f14 5d10 6s2 6p1\n

    Here because of poor sheilding of 4f and 5d orbital on 6s orbital, electrons of 6s orbital is more tightly held by the nucleus and perticipate less in bond formation. This is called as innert pair effect. And that is why Tl become stable in +1 oxidation state. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2248, "subject": "Chemistry", "question": "The element with Z = 120 (not yet discovered) will be an/a - ", "options": [ { "text": "Alkaline earth metal" }, { "text": "Alkali metal " }, { "text": "Transition metal " }, { "text": "Inner transition metal " } ], "answer": "Alkaline earth metal", "solution": "**Answer:** Alkaline earth metal\n\nElectronic configuration of element Og (118) is\n
    [Rn] 5f146d\n107s\n27p6,\nwhere Og is oganesson.\n

    [Og118] 8s2\nis configuration for Z = 120,\n

    As per the configuration it is in IInd group. Thus,\nelement with Z = 120 will be an alkaline earth metal.\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2249, "subject": "Chemistry", "question": "The lathanide ion that would show colour is :", "options": [ { "text": "Sm3+" }, { "text": "Gd3+" }, { "text": "Lu3+" }, { "text": "La3+" } ], "answer": "Sm3+", "solution": "**Answer:** Sm3+\n\nElectronic configuration of \n

    Sm = [Xe] 4f6 6s2\n

    $$ \\therefore $$ Electronic configuration of \n

    Sm+3 = [Xe] 4f5\n

    Sm+3 shows yellow colour due to partially filled f orbital and f-f transition.\n

    Electronic configuration of \n

    La+3 = [Xe] 4f0\n

    Lu+3 = [Xe] 4f14\n

    Gd+3 = [Xe] 4f7\n

    La+3, Lu+3 and Gd+3 are colourless due to absence of f-f transition.\n\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2250, "subject": "Chemistry", "question": "The IUPAC symbol for the element with atomic\nnumber 119 would be :", "options": [ { "text": "uun" }, { "text": "une" }, { "text": "uue" }, { "text": "unh" } ], "answer": "uue", "solution": "**Answer:** uue\n\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2251, "subject": "Chemistry", "question": "The electronic configurations of bivalent\neuropium and trivalent cerium are
    \n(atomic number : Xe = 54, Ce = 58, Eu = 63)", "options": [ { "text": "[Xe] 4f7 6s2 and [Xe] 4f2 6s2" }, { "text": "[Xe] 4f2 and [Xe] 4f7" }, { "text": "[Xe] 4f4 and [Xe] 4f9" }, { "text": "[Xe] 4f7 and [Xe] 4f1" } ], "answer": "[Xe] 4f7 and [Xe] 4f1", "solution": "**Answer:** [Xe] 4f7 and [Xe] 4f1\n\nEu2+ : [xe] 4f7\n

    Ce3+ : [xe] 4f1", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2252, "subject": "Chemistry", "question": "The atomic number of the element unnilennium is :", "options": [ { "text": "109" }, { "text": "102" }, { "text": "119" }, { "text": "108" } ], "answer": "109", "solution": "**Answer:** 109\n\nFor Unnilennium\n

    IUPAC symbol – Une\n

    Atomic No. – 109", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2253, "subject": "Chemistry", "question": "The atomic number of Unnilunium is _______.", "options": [], "answer": "101", "solution": "**Answer:** 101\n\nIUPAC symbol = Unu\n

    Atomic no. (Z) = 101", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2254, "subject": "Chemistry", "question": "The correct order of bond dissociation enthalpy of halogens is :", "options": [ { "text": "F2 > Cl2 > Br2 > I2" }, { "text": "Cl2 > Br2 > F2 > I2" }, { "text": "I2 > Br2 > Cl2 > F2" }, { "text": "Cl2 > F2 > Br2 > I2" } ], "answer": "Cl2 > Br2 > F2 > I2", "solution": "**Answer:** Cl2 > Br2 > F2 > I2\n\n

    Among halogens (F2, Cl2, Br2 and I2), bond dissociation enthalpy ($$\\Delta$$dissH$$^\\circ$$) of I2, is minimum because of larger size of I-atom there is a steric repulsion between bonded I-atoms, which makes I-I bond weakest.

    \n

    Whereas, smaller size and highest electronegativity of F-atom cause highest electron density on F-atom of F2 molecule. As a result, F-F bond becomes weaker due to electrostatic repulsion between bonded F-atoms.

    \n

    Thus, the order of $$\\Delta$$dissH$$^\\circ$$ (in kJ mol$$-$$1) is

    \n

    $$\\mathop {Cl - Cl}\\limits_{(242.6)} > \\mathop {Br - Br}\\limits_{(192.3)} > \\mathop {F - F}\\limits_{(158.8)\\,Electrostatic\\,repulsion} > \\mathop {I - I}\\limits_{(151.1)\\,Steric\\,repulsion} $$

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2255, "subject": "Chemistry", "question": "The characteristics of elements X, Y and Z with atomic numbers, respectively, 33, 53 and 83 are :", "options": [ { "text": "X and Y are metalloids and Z is a metal." }, { "text": "X and Z are non-metals and Y is a metalloid." }, { "text": "X is a metalloid, Y is a non-metal and Z is a metal." }, { "text": "X, Y and Z are metals." } ], "answer": "X is a metalloid, Y is a non-metal and Z is a metal.", "solution": "**Answer:** X is a metalloid, Y is a non-metal and Z is a metal.\n\nX(Z = 33) = As (metalloid)

    \nY(Z = 53) = I (non-metal)

    \nZ(Z = 83) = Bi (metal)", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2256, "subject": "Chemistry", "question": "Which one of the following statements for D.I. Mendeleeff, is incorrect?", "options": [ { "text": "He authored the textbook - Principles of Chemistry" }, { "text": "At the time, he proposed Periodic Table of elements structure of atom was known." }, { "text": "Element with atomic number 101 is named after him." }, { "text": "He invented accurate barometer." } ], "answer": "At the time, he proposed Periodic Table of elements structure of atom was known.", "solution": "**Answer:** At the time, he proposed Periodic Table of elements structure of atom was known.\n\nIn 1869, the Russian chemist Dmitri Mendeleev came to prominence with his tabular diagram of known elements, but basic atomic structure the idea that everything is made of atoms was pioneered by John Dalton (1766-1844) in a book he published in 1808.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2257, "subject": "Chemistry", "question": "Match List - I with List - II :

    \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    List - I
    (Metal Ion)    		List - II
    (Group of Qualitative analysis)
    (a)$$M{n^{2 + }}$$(i)Group - III
    (b)$$A{s^{3 + }}$$(ii)Group - IIA
    (c)$$C{u^{2 + }}$$(iii)Group - IV
    (d)$$A{l^{3 + }}$$(iv)Group - IIB


    Choose the most appropriate answer from the options given below :", "options": [ { "text": "(a)-(i), (b)-(ii), (c)-(iii), (d)-(iv)" }, { "text": "(a)-(iii), (b)-(iv), (c)-(ii), (d)-(i)" }, { "text": "(a)-(i), (b)-(iv), (c)-(ii), (d)-(iii)" }, { "text": "(a)-(iv), (b)-(ii), (c)-(iii), (d)-(i)" } ], "answer": "(a)-(iii), (b)-(iv), (c)-(ii), (d)-(i)", "solution": "**Answer:** (a)-(iii), (b)-(iv), (c)-(ii), (d)-(i)\n\nMn2+ $$\\to$$ IV group

    As3+ $$\\to$$ II B group

    Cu2+ $$\\to$$ II A group

    Al3+ $$\\to$$ III group\n

    Note :\n

    IVth group cations – Zn2+\n,Mn2+\n,Ni2+\n, Co2+\n

    II B group cations – As3+\n, Sb3+\n, Sn2+\n

    II A group cations – Hg2+\n, Pb2+\n, Bi3+\n, Cu2+\n, Cd2+\n

    IIIrd group cations – Al3+\n, Cr3+\n, Fe3+\n", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 2258, "subject": "Chemistry", "question": "

    Element \"E\" belongs to the period 4 and group 16 of the periodic table. The valence shell electron configuration of the element, which is just above \"E\" in the group is

    ", "options": [ { "text": "3s2, 3p4" }, { "text": "3d10, 4s2, 4p4" }, { "text": "4d10, 5s2, 5p4" }, { "text": "2s2, 2p4" } ], "answer": "3s2, 3p4", "solution": "**Answer:** 3s2, 3p4\n\nElement E is Selenium (Se)

    \nElectronic configuration of E is [Ar] 3d10 4s2 4p4

    \nThe element which is just above ‘E’ in periodic table\nis sulphur, its electronic configuration is [Ne] 3s2 3p4", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 2259, "subject": "Chemistry", "question": "

    The IUPAC nomenclature of an element with electronic configuration [Rn] $$5 \\mathrm{f}^{14} 6 \\mathrm{d}^{1} 7 \\mathrm{s}^{2}$$ is :

    ", "options": [ { "text": "Unnilbium" }, { "text": "Unnilunium" }, { "text": "Unnilquadium" }, { "text": "Unniltrium" } ], "answer": "Unniltrium", "solution": "**Answer:** Unniltrium\n\nThe element with electronic configuration [Rn] $$5 \\mathrm{f}^{14} 6 \\mathrm{~d}^{1} 7 \\mathrm{~s}^{2}$$ has atomic number $\\rightarrow 103$\n

    \n$$\\therefore$$ Its IUPAC name is: Unniltrium", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2260, "subject": "Chemistry", "question": "Which of the following elements have half-filled f-orbitals in their ground state?\n

    \n(Given : atomic number $\\mathrm{Sm}=62 ; \\mathrm{Eu}=63 ; \\mathrm{Tb}=65 ; \\mathrm{Gd}=64, \\mathrm{Pm}=61$ )

    \nA. $\\mathrm{Sm}$

    \nB. Eu

    \nC. $\\mathrm{Tb}$

    \nD. Gd

    \nE. $\\mathrm{Pm}$\n

    \nChoose the correct answer from the options given below :", "options": [ { "text": " B and D only" }, { "text": "A and $B$ only" }, { "text": "$C$ and $\\mathrm{D}$ only" }, { "text": "A and E only" } ], "answer": " B and D only", "solution": "**Answer:** B and D only\n\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2261, "subject": "Chemistry", "question": "Bond dissociation energy of \"E-H\" bond of the \"$\\mathrm{H}_{2} \\mathrm{E}$ \" hydrides of group 16 elements (given below), follows order.

    \nA. $\\mathrm{O}$

    \nB. $\\mathrm{S}$

    \nC. Se

    \nD. $\\mathrm{Te}$\n

    \nChoose the correct from the options given below:", "options": [ { "text": "$\\mathrm{A}>\\mathrm{B}>\\mathrm{C}>\\mathrm{D}$" }, { "text": "$\\mathrm{D}>\\mathrm{C}>\\mathrm{B}>\\mathrm{A}$\n" }, { "text": "$\\mathrm{B}>\\mathrm{A}>\\mathrm{C}>\\mathrm{D}$" }, { "text": "$\\mathrm{A}>\\mathrm{B}>\\mathrm{D}>\\mathrm{C}$" } ], "answer": "$\\mathrm{A}>\\mathrm{B}>\\mathrm{C}>\\mathrm{D}$", "solution": "**Answer:** $\\mathrm{A}>\\mathrm{B}>\\mathrm{C}>\\mathrm{D}$\n\n

    The correct order of bond strength is

    \n

    $$\\mathrm{H_2O > H_2S > H_2Se > H_2Te}$$

    ", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 2262, "subject": "Chemistry", "question": "

    Match List I with List II

    \n

    \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    List I
    (Atomic number)    		List II
    (Block of periodic table)
    A.37I.p-block
    B.78II.d-block
    C.52III.f-block
    D.65IV.s-block

    \n

    Choose the correct answer from the options given below :

    ", "options": [ { "text": "A - II, B - IV, C - I, D - III" }, { "text": "A - IV, B - III, C - II, D - I" }, { "text": "A - IV, B - II, C - I, D - III" }, { "text": "A - I, B - III, C - IV, D - II" } ], "answer": "A - IV, B - II, C - I, D - III", "solution": "**Answer:** A - IV, B - II, C - I, D - III\n\n

    \n\n\n\n\n\n\n\n\n \n \n \n \n \n \n\n\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\n
    List I
    (Atomic number)    		List II
    (Block of periodic table)
    A.37 (Rb - Rubidium)IV.s-block
    B.78 (Pt - Platinum)II.d-block
    C.52 (Te - Tellurium)I.p-block
    D.65 (Tb - Terbium)III.f-block

    ", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 2263, "subject": "Chemistry", "question": "

    Which one of the following elements will remain as liquid inside pure boiling water?

    ", "options": [ { "text": "Li" }, { "text": "Br" }, { "text": "Cs" }, { "text": "Ga" } ], "answer": "Ga", "solution": "**Answer:** Ga\n\n

    The melting point of gallium is very low (about 29.76°C or 85.57°F). Therefore, gallium can melt just from the heat of one's hand. The boiling point of gallium, on the other hand, is quite high (2400°C or 4352°F).

    \n

    On the contrary, water boils at 100°C (212°F) under normal atmospheric pressure. Therefore, inside boiling water, gallium would melt into a liquid but would not boil or evaporate, unlike the other elements listed. Lithium (Li), Bromine (Br), and Cesium (Cs) have boiling points below 100°C, so they would turn into a gas in boiling water.

    \n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2264, "subject": "Chemistry", "question": "Given below are two statements :

    \nStatement (I) : Both metals and non-metals exist in p and d-block elements.

    \nStatement (II) : Non-metals have higher ionisation enthalpy and higher electronegativity than the metals.\n

    \nIn the light of the above statements, choose the most appropriate answer from the options given below :", "options": [ { "text": "Both Statement I and Statement II are false" }, { "text": "Both Statement I and Statement II are true" }, { "text": "Statement I is false but Statement II is true" }, { "text": "Statement I is true but Statement II is false" } ], "answer": "Statement I is false but Statement II is true", "solution": "**Answer:** Statement I is false but Statement II is true\n\n

    To answer this question, let's investigate both statements in detail.

    \n

    Statement (I): Both metals and non-metals exist in p and d-block elements.

    \n

    The periodic table is divided into blocks based on the electron configuration of the elements. These blocks are labeled s, p, d, and f. The p-block contains a mix of metals, non-metals, and metalloids. For example, the p-block contains nonmetals such as oxygen (O) and nitrogen (N), as well as metals like aluminum (Al) and lead (Pb).

    \n

    The d-block, also known as the transition metals block, primarily consists of metals. However, the d-block does not typically contain elements that are traditionally classified as non-metals. The elements in the d-block have a wide range of properties but are generally characterized by their metallic traits such as conductivity and malleability.

    \n

    Therefore, Statement I is not entirely true because while both metals and nonmetals exist in the p-block of the periodic table, the d-block is principally composed of metals.

    \n

    Statement (II): Non-metals have higher ionisation enthalpy and higher electronegativity than the metals.

    \n

    Ionization enthalpy (or ionization energy) is the energy required to remove an electron from a gaseous atom or ion. Non-metals generally have higher ionization energies compared to metals because non-metals have more tightly bound electrons to their nucleus. This is partly due to the fact that non-metals tend to have higher electronegativities and smaller atomic radii, meaning that their outer electrons are closer to the nucleus and more strongly attracted to it.

    \n

    Electronegativity is a measure of an atom's ability to attract and bond with electrons. Non-metals have higher electronegativities typically because they are more eager to gain electrons to achieve a full valence shell, reflecting their position on the right side of the periodic table. Metals, on the other hand, are more inclined to lose electrons and form positive ions, indicating their lower electronegativities.

    \n

    Hence, Statement II is true as non-metals indeed have higher ionisation enthalpy and higher electronegativity than metals.

    \n

    Given the analysis:

    \n\n

    Thus, the most appropriate answer is:

    \n

    Option C - Statement I is false but Statement II is true.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2265, "subject": "Chemistry", "question": "

    Given below are two statements :

    \n

    Statement (I) : The $$4 \\mathrm{f}$$ and $$5 \\mathrm{f}$$ - series of elements are placed separately in the Periodic table to preserve the principle of classification.

    \n

    Statement (II) : S-block elements can be found in pure form in nature. \n

    In the light of the above statements, choose the most appropriate answer from the options given below :

    ", "options": [ { "text": "Statement I is false but Statement II is true\n" }, { "text": "Both Statement I and Statement II are true\n" }, { "text": "Statement I is true but Statement II is false\n" }, { "text": "Both Statement I and Statement II are false" } ], "answer": "Statement I is true but Statement II is false\n", "solution": "**Answer:** Statement I is true but Statement II is false\n\n\n

    Statement I is correct because the 4f and 5f series elements, commonly referred to as the lanthanides and actinides, respectively, have unique properties and configurations that distinguish them from other elements. Placing them separately in the Periodic table helps to maintain the orderly classification based on atomic numbers and avoids disrupting the pattern established by the periodic law.

    Statement II is incorrect because s-block elements, which include Group 1 (alkali metals) and Group 2 (alkaline earth metals), are indeed highly reactive. They typically do not exist in their pure form in nature due to their reactivity. Instead, they are mostly found in compound forms, combined with other elements. For instance, sodium (Na) and potassium (K) from Group 1 react with water and are usually found as salts, not in their metallic, pure state.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2266, "subject": "Chemistry", "question": "

    Anomalous behavior of oxygen is due to its

    ", "options": [ { "text": "large size and low electronegativity\n" }, { "text": "small size and high electronegativity\n" }, { "text": "small size and low electronegativity\n" }, { "text": "large size and high electronegativity" } ], "answer": "small size and high electronegativity\n", "solution": "**Answer:** small size and high electronegativity\n\n\n

    Fact Based.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2267, "subject": "Chemistry", "question": "

    Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R:

    \n

    Assertion A: $$\\mathrm{H}_2 \\mathrm{Te}$$ is more acidic than $$\\mathrm{H}_2 \\mathrm{~S}$$.

    \n

    Reason R: Bond dissociation enthalpy of $$\\mathrm{H}_2 \\mathrm{Te}$$ is lower than $$\\mathrm{H}_2 \\mathrm{~S}$$.

    \n

    In the light of the above statements, choose the most appropriate from the options given below:

    ", "options": [ { "text": "Both $$A$$ and $$R$$ are true and $$R$$ is the correct explanation of $$A$$.\n" }, { "text": "Both $$A$$ and $$R$$ are true but $$R$$ is NOT the correct explanation of $$A$$.\n" }, { "text": "$$A$$ is false but $$R$$ is true.\n" }, { "text": "$$A$$ is true but $$R$$ is false." } ], "answer": "Both $$A$$ and $$R$$ are true and $$R$$ is the correct explanation of $$A$$.\n", "solution": "**Answer:** Both $$A$$ and $$R$$ are true and $$R$$ is the correct explanation of $$A$$.\n\n\n

    Due to lower Bond dissociation enthalpy of $$\\mathrm{H}_2 \\mathrm{Te}$$ it ionizes to give $$\\mathrm{H}^{+}$$ more easily than $$\\mathrm{H}_2 \\mathrm{S}$$.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2268, "subject": "Chemistry", "question": "

    Match List I with List II.

    \n

    \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    List I
    Species    		List II
    Electronic distribution
    (A)$$\\mathrm{Cr^{+2}}$$(I)$$\\mathrm{3d^8}$$
    (B)$$\\mathrm{Mn^+}$$(II)$$\\mathrm{3d^34s^1}$$
    (C)$$\\mathrm{Ni^{+2}}$$(III)$$\\mathrm{3d^4}$$
    (D)$$\\mathrm{V^+}$$(IV)$$\\mathrm{3d^54s^1}$$

    \n

    Choose the correct answer from the options given below :

    ", "options": [ { "text": "(A)-(III), (B)-(IV), (C)-(I), (D)-(II)" }, { "text": "(A)-(I), (B)-(II), (C)-(III), (D)-(IV)" }, { "text": "(A)-(IV), (B)-(III), (C)-(I), (D)-(II)" }, { "text": "(A)-(II), (B)-(I), (C)-(IV), (D)-(III)" } ], "answer": "(A)-(III), (B)-(IV), (C)-(I), (D)-(II)", "solution": "**Answer:** (A)-(III), (B)-(IV), (C)-(I), (D)-(II)\n\n

    $$\\begin{gathered}\n{ }_{24} \\mathrm{Cr} \\rightarrow[\\mathrm{Ar}] 3 \\mathrm{~d}^5 4 \\mathrm{~s}^1 ; \\mathrm{Cr}^{2+} \\rightarrow[\\mathrm{Ar}] 3 \\mathrm{~d}^4 \\\\\n{ }_{25} \\mathrm{Mn} \\rightarrow[\\mathrm{Ar}] 3 \\mathrm{~d}^5 4 \\mathrm{s}^2 ; \\mathrm{Mn}^{+} \\rightarrow[\\mathrm{Ar}] 3 \\mathrm{~d}^5 4 \\mathrm{s}^1 \\\\\n{ }_{28} \\mathrm{Ni} \\rightarrow[\\mathrm{Ar}] 3 \\mathrm{~d}^8 4 \\mathrm{s}^2 ; \\mathrm{Ni}^{2+} \\rightarrow[\\mathrm{Ar}] 3 \\mathrm{~d}^8 \\\\\n{ }_{23} \\mathrm{V} \\rightarrow[\\mathrm{Ar}] 3 \\mathrm{~d}^3 4 \\mathrm{s}^2 ; \\mathrm{V}^{+} \\rightarrow[\\mathrm{Ar}] 3 \\mathrm{~d}^3 4 \\mathrm{s}^1\n\\end{gathered}$$

    ", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 2269, "subject": "Chemistry", "question": "

    If IUPAC name of an element is \"Unununnium\" then the element belongs to nth group of Periodic table. The value of n is ________.

    ", "options": [], "answer": "11", "solution": "**Answer:** 11\n\n

    The IUPAC name \"Unununnium\" is derived from a systematic element naming method based on the atomic number of elements, primarily used for elements that have not yet been discovered or were not yet officially named at the time. The name \"Unununnium\" is composed of Latin and Greek numerical roots and can be broken down into \"Un-un-un-nium\" where \"un\" represents one and \"nnium\" is the suffix used for elements. The full name \"Unununnium\" hence corresponds to the temporary elemental name for an element with the atomic number 111 (one-one-one).

    \n\n

    The elements in the periodic table are arranged by their atomic number and grouped by their electron configurations, particularly the valence electrons which determine their chemical properties. Each column of the periodic table is known as a group. The atomic number 111 element, which was named \"Unununnium\" before its official name \"Roentgenium (Rg)\" was adopted, is in the same group as copper (Cu), silver (Ag), and gold (Au). This group is group 11 of the periodic table. Therefore:

    \n\n$$ n = 11 $$", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 2270, "subject": "Chemistry", "question": "

    Match List I with List II

    \n

    \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    LIST I
    (Element)    		LIST II
    (Electronic Configuration)
    A.$$\\mathrm{N}$$I.$$[\\mathrm{Ar}] 3 \\mathrm{~d}^{10} 4 \\mathrm{~s}^2 4 \\mathrm{p}^5$$
    B.$$\\mathrm{S}$$II.$$[\\mathrm{Ne}] 3 \\mathrm{~s}^2 3 \\mathrm{p}^4$$
    C.$$\\mathrm{Br}$$III.$$[\\mathrm{He}] 2 \\mathrm{~s}^2 2 \\mathrm{p}^3$$
    D.$$\\mathrm{Kr}$$IV.$$[\\mathrm{Ar}] 3 \\mathrm{~d}^{10} 4 \\mathrm{~s}^2 4 \\mathrm{p}^6$$

    \n

    Choose the correct answer from the options given below:

    ", "options": [ { "text": "A-IV, B-III, C-II, D-I" }, { "text": "A-I, B-IV, C-III, D-II\n" }, { "text": "A-III, B-II, C-I, D-IV\n" }, { "text": "A-II, B-I, C-IV, D-III" } ], "answer": "A-III, B-II, C-I, D-IV\n", "solution": "**Answer:** A-III, B-II, C-I, D-IV\n\n\n

    To match each element with its correct electronic configuration, let's consider the electronic configurations of each element based on their positions in the periodic table:

    \n\n

    Therefore, the correct match is:

    \n
      \n
    1. A-III (Nitrogen matches with $$[\\mathrm{He}] 2 \\mathrm{s}^2 2 \\mathrm{p}^3$$)
      2. \n
    2. B-II (Sulfur matches with $$[\\mathrm{Ne}] 3 \\mathrm{s}^2 3 \\mathrm{p}^4$$)
      3. \n
    3. C-I (Bromine matches with $$[\\mathrm{Ar}] 3 \\mathrm{d}^{10} 4 \\mathrm{s}^2 4 \\mathrm{p}^5$$)
      4. \n
    4. D-IV (Krypton matches with $$[\\mathrm{Ar}] 3 \\mathrm{d}^{10} 4 \\mathrm{s}^2 4 \\mathrm{p}^6$$)
      5. \n
    \n

    Thus, the correct answer is Option C: A-III, B-II, C-I, D-IV.

    ", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 2271, "subject": "Chemistry", "question": "

    Number of elements from the following that CANNOT form compounds with valencies which match with their respective group valencies is ________. B, C, N, S, O, F, P, Al, Si

    ", "options": [ { "text": "7" }, { "text": "3" }, { "text": "5" }, { "text": "6" } ], "answer": "3", "solution": "**Answer:** 3\n\n

    N, F and O will not satisfy the given condition.

    ", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 2272, "subject": "Chemistry", "question": "Match List I with List II\n

    \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    LIST I
    (Elements)    		LIST II
    (Properties in their respective groups)
    A.$$\\mathrm{Cl,S}$$I.Elements with highest electronegativity
    B.\t$$\\mathrm{Ge,As}$$II.Elements with largest atomic size
    C.$$\\mathrm{Fr,Ra}$$III.Elements which show properties of both metals and non-metal
    D.$$\\mathrm{F,O}$$IV.Elements with highest negative electron gain enthalpy

    \n

    Choose the correct answer from the options given below:

    ", "options": [ { "text": "A-III, B-II, C-I, D-IV\n" }, { "text": "A-II, B-I, C-IV, D-III\n" }, { "text": "A-II, B-III, C-IV, D-I\n" }, { "text": "A-IV, B-III, C-II, D-I" } ], "answer": "A-IV, B-III, C-II, D-I", "solution": "**Answer:** A-IV, B-III, C-II, D-I\n\n

    To match the elements from List I with their corresponding properties in List II, let's analyze each pair:

    \n\n

    A. $$\\mathrm{Cl,S}$$
    \n\n

    These elements are known for having high negative electron gain enthalpy (energy released when an electron is added to an atom). Chlorine ($$\\mathrm{Cl}$$) and sulfur ($$\\mathrm{S}$$) are both elements that release a significant amount of energy when they gain an electron.

    \n\n

    Hence, they match with IV: Elements with highest negative electron gain enthalpy.

    \n\n

    B. $$\\mathrm{Ge,As}$$
    \n\n

    Germanium ($$\\mathrm{Ge}$$) and arsenic ($$\\mathrm{As}$$) are metalloids, meaning they show properties of both metals and non-metals.

    \n\n

    Hence, they match with III: Elements which show properties of both metals and non-metal.

    \n\n

    C. $$\\mathrm{Fr,Ra}$$
    \n\n

    Francium ($$\\mathrm{Fr}$$) and radium ($$\\mathrm{Ra}$$) are elements that have very large atomic sizes, being at the bottom of their respective groups on the periodic table.

    \n\n

    Hence, they match with II: Elements with largest atomic size.

    \n\n

    D. $$\\mathrm{F,O}$$
    \n\n

    Fluorine ($$\\mathrm{F}$$) and oxygen ($$\\mathrm{O}$$) are elements with the highest electronegativity, meaning they strongly attract electrons toward themselves.

    \n\n

    Hence, they match with I: Elements with highest electronegativity.

    \n\n

    Summarizing, we have:

    \n\n

    A-IV, B-III, C-II, D-I

    \n\n

    The correct option is:

    \n\n

    Option D
    \n\n

    A-IV, B-III, C-II, D-I

    ", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 2273, "subject": "Chemistry", "question": "

    Given below are two statements:

    \n

    Statement I: In group 13, the stability of +1 oxidation state increases down the group.

    \n

    Statement II : The atomic size of gallium is greater than that of aluminium.

    \n

    In the light of the above statements, choose the most appropriate answer from the options given below :

    ", "options": [ { "text": "Statement I is incorrect but Statement II is correct\n" }, { "text": "Statement I is correct but Statement II is incorrect\n" }, { "text": "Both Statement I and Statement II are correct\n" }, { "text": "Both Statement I and Statement II are incorrect" } ], "answer": "Statement I is correct but Statement II is incorrect\n", "solution": "**Answer:** Statement I is correct but Statement II is incorrect\n\n\n

    Statement I:

    \n

    \"In group 13, the stability of +1 oxidation state increases down the group.\"

    \n

    Analysis:

    \n\n

    Group 13 Elements:

    \n

    Boron (B)

    \n

    Aluminum (Al)

    \n

    Gallium (Ga)

    \n

    Indium (In)

    \n

    Thallium (Tl)

    \n

    Common Oxidation States:

    \n

    The typical oxidation state for group 13 elements is +3.

    \n

    However, due to the inert pair effect, the +1 oxidation state becomes more stable as we move down the group.

    \n

    Inert Pair Effect:

    \n

    The reluctance of the s-electrons (ns²) in the valence shell to participate in bonding.

    \n

    This effect becomes more significant in heavier elements due to poor shielding by inner d and f orbitals.

    \n

    Stability Trend:

    \n

    Boron (B): Exhibits only the +3 oxidation state.

    \n

    Aluminum (Al): Predominantly +3 oxidation state.

    \n

    Gallium (Ga): +3 is more stable, but +1 oxidation state starts appearing.

    \n

    Indium (In): +1 oxidation state becomes significant.

    \n

    Thallium (Tl): The +1 oxidation state is more stable than the +3 oxidation state.

    \n\n

    Conclusion for Statement I:

    \n\n

    Correct.

    \n

    The stability of the +1 oxidation state increases as we move down group 13 due to the inert pair effect.

    \n\n
    \n

    Statement II:

    \n

    \"The atomic size of gallium is greater than that of aluminium.\"

    \n

    Analysis:

    \n\n

    Expected Trend:

    \n

    Generally, atomic size increases down a group because each successive element has an additional electron shell.

    \n

    Actual Atomic Radii:

    \n

    Aluminum (Al): Approximately 143 picometers (pm)

    \n

    Gallium (Ga): Approximately 135 pm

    \n

    Anomalous Behavior:

    \n

    **Gallium's atomic radius is *slightly smaller* than that of aluminum.**

    \n

    This anomaly is due to the presence of 10 d-electrons in gallium's electron configuration (Ga: [Ar] 3d¹⁰ 4s² 4p¹).

    \n

    Poor Shielding Effect:

    \n\n

    3d electrons do not shield the nuclear charge effectively.

    \n

    As a result, the effective nuclear charge increases, pulling the outer electrons closer to the nucleus.

    \n

    This causes gallium to have a smaller atomic radius compared to aluminum.

    \n\n

    Conclusion for Statement II:

    \n\n

    Incorrect.

    \n

    The atomic size of gallium is less than that of aluminum due to poor shielding by d-electrons.

    \n\n
    \n

    Final Answer:

    \n\n

    Statement I is correct, as the stability of the +1 oxidation state increases down group 13.

    \n

    Statement II is incorrect, because gallium has a smaller atomic radius than aluminum.

    \n\n
    \n

    Answer: Option B

    ", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 2274, "subject": "Chemistry", "question": "Which of the following is fully fluorinated polymer? ", "options": [ { "text": "Neoprene " }, { "text": "Teflon " }, { "text": "Thiokol" }, { "text": "PVC " } ], "answer": "Teflon ", "solution": "**Answer:** Teflon \n\nTeflon is polymer of $$C{F_2} = C{F_2}.$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2275, "subject": "Chemistry", "question": "Bakelite is obtained from phenol by reacting with ", "options": [ { "text": "(CH2OH)2" }, { "text": "CH3CHO " }, { "text": "CH3COCH3" }, { "text": "HCHO " } ], "answer": "HCHO ", "solution": "**Answer:** HCHO \n\nBakelite is formed by the reaction of formaldehyde $$(HCHO)$$ and phenol so the correct answer is $$(d).$$\n

    \"AIEEE ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2276, "subject": "Chemistry", "question": "The polymer containing strong intermolecular forces e.g. hydrogen bonding, is", "options": [ { "text": "teflon" }, { "text": "nylon 6,6" }, { "text": "polystyrene" }, { "text": "natural rubber" } ], "answer": "nylon 6,6", "solution": "**Answer:** nylon 6,6\n\nNylon $$6-6$$ has amide linkage capable of forming hydrogen bonding. \n

    \"AIEEE ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2277, "subject": "Chemistry", "question": "The species which can best serve as an initiator for the cationic polymerization is :", "options": [ { "text": "LiAlH4" }, { "text": "HNO3" }, { "text": "AlCl3" }, { "text": "BuLi" } ], "answer": "AlCl3", "solution": "**Answer:** AlCl3\n\nLewis acids are the most common compounds used for initiation of cationic polymerisation. The more popular Lewis acids are $$SnC{l_4},\\,\\,AlC{l_3},\\,\\,B{F_3}$$ and $$TiC{l_4}.$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2278, "subject": "Chemistry", "question": "Which one is classified as a condensation polymer?", "options": [ { "text": "Teflon" }, { "text": "Acrylonitrile" }, { "text": "Dacron" }, { "text": "Neoprene" } ], "answer": "Dacron", "solution": "**Answer:** Dacron\n\nExcept Dacron all are additive polymers. Terephthalic acid condenses with ethylene glycol to give Dacron.\n

    \"JEE ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2279, "subject": "Chemistry", "question": "Assertion : \n
    Rayon is a semisynthetic polymer whose properties are better than natural cotton.\n

    Reason :\n
    Mechanical and aesthetic properties of cellulose can be improved by acetylation.", "options": [ { "text": "Both assertion and reason are correct, and the reason is the correct explanation for the assertion." }, { "text": "Both assertion and reason are correct, but the reason is not the correct explanation for the assertion.\n" }, { "text": "Assertion is incorrect statement, but the reason is correct." }, { "text": "Both assertion and reason are incorrect." } ], "answer": "Both assertion and reason are correct, but the reason is not the correct explanation for the assertion.\n", "solution": "**Answer:** Both assertion and reason are correct, but the reason is not the correct explanation for the assertion.\n\n\nThe strength of cellulose is improved by acetylation to form rayon, a semisynthetic polymer which is better than natural cotton. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2280, "subject": "Chemistry", "question": "Which of the following polymers is synthesized using a free radical\npolymerization technique ?", "options": [ { "text": "Teflon" }, { "text": "Terylene" }, { "text": "Melamine polymer" }, { "text": "Nylon 6, 6" } ], "answer": "Teflon", "solution": "**Answer:** Teflon\n\nTerylene, Melamine and Nylon-6, 6 are condensation polymers. Teflon is an addition polymer which is formed by free radical polymerization of its monomer, (CF2 = CF2). ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2281, "subject": "Chemistry", "question": "The formation of which of the following polymers involves hydrolysis reaction?", "options": [ { "text": "Bakelite\n" }, { "text": "Nylon 6, 6" }, { "text": "Terylene" }, { "text": "Nylon 6" } ], "answer": "Nylon 6", "solution": "**Answer:** Nylon 6\n\nFormation of nylon-6 involves hydrolysis of caprolactum, (its monomer) in initial state.\n
    \"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2282, "subject": "Chemistry", "question": "Which of the following statements is not true ? ", "options": [ { "text": "Step growth polymerisation requires a bifunctional monomer." }, { "text": "Nylon 6 is an example of step-growth polymerisation. " }, { "text": "Chain growth polymerisation includes both homopolymerisation and copylymerisation." }, { "text": "Chain growth polymerisation involves homopolymerisation only. " } ], "answer": "Chain growth polymerisation involves homopolymerisation only. ", "solution": "**Answer:** Chain growth polymerisation involves homopolymerisation only. \n\nChain growth polymerisation is an addition polymerisation which involves homopolymerisation and copolymerisation both.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2283, "subject": "Chemistry", "question": "Which of the following is a thermosetting polymer?", "options": [ { "text": "Buna-N" }, { "text": "PVC" }, { "text": "NyIon6" }, { "text": "Bakelite" } ], "answer": "Bakelite", "solution": "**Answer:** Bakelite\n\nBakelite is an example of thermosetting\npolymer.\n

    When thermosetting\npolymer once heated they change their shape which cannot be brought back to their initial shape.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2284, "subject": "Chemistry", "question": "The correct match between Item-I and Item-II is : \n

    \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    Item - IItem - II
    (a)High density polythene(I)Peroxide catalyst
    (b)Polyacrylonitrile(II)Condensation at high
    temperature & pressure
    (c)Novolac(III)Ziegler-Natta Catalyst
    (d)Nylon 6(IV)Acid or base catalyst
    ", "options": [ { "text": "(a) $$ \\to $$ (IV), (b) $$ \\to $$ (II), (c) $$ \\to $$ (I), (d) $$ \\to $$ (III)" }, { "text": "(a) $$ \\to $$ (III), (b) $$ \\to $$ (I), (c) $$ \\to $$ (IV), (d) $$ \\to $$ (II)" }, { "text": "(a) $$ \\to $$ (II), (b) $$ \\to $$ (IV), (c) $$ \\to $$ (I), (d) $$ \\to $$ (III)" }, { "text": "(a) $$ \\to $$ (III), (b) $$ \\to $$ (I), (c) $$ \\to $$ (II), (d) $$ \\to $$ (IV)" } ], "answer": "(a) $$ \\to $$ (III), (b) $$ \\to $$ (I), (c) $$ \\to $$ (IV), (d) $$ \\to $$ (II)", "solution": "**Answer:** (a) $$ \\to $$ (III), (b) $$ \\to $$ (I), (c) $$ \\to $$ (IV), (d) $$ \\to $$ (II)\n\nHigh density polythene is obtained by using\nZiegler-Natta Catalyst.\n

    Polyacrylonitrile is obtained by acrylonitrile using peroxide catalyst.\n

    Novalac is obtained by acid\nor base catalyzed polymerization of phenol.\n

    Nylon–6 is condensation polymer which is made at higher temperature and pressure. ", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 2285, "subject": "Chemistry", "question": "Which of the following is a condensation polymer?", "options": [ { "text": "Neoprene" }, { "text": "Buna-S " }, { "text": "Nylon 6, 6" }, { "text": "Teflon" } ], "answer": "Nylon 6, 6", "solution": "**Answer:** Nylon 6, 6\n\nNylon 6, 6 is obtained by condensation\npolymerisation of hexamethylenediamine and\nadipic acid.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2286, "subject": "Chemistry", "question": "Poly $$\\beta $$ –hydroxybutyrate-co-$$\\beta $$-hydroxyvalerate (PHBV) is a copolymer of ", "options": [ { "text": "2-hydroxybutanoic acid and 3-hydroxypentanoic acid" }, { "text": "3-hydroxybutanoic acid and 2-hydroxypentanoic acid" }, { "text": "3-hydroxybutanoic acid and 3-hydroxypentanoic acid " }, { "text": "3-hydroxybutanoic acid and 4-hydroxypentanoic acid" } ], "answer": "3-hydroxybutanoic acid and 3-hydroxypentanoic acid ", "solution": "**Answer:** 3-hydroxybutanoic acid and 3-hydroxypentanoic acid \n\nPHBV is obtained by copolymerization of 3-Hydroxybutanioic acid and 3-Hydroxypentanoic\nacid.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2287, "subject": "Chemistry", "question": "The two monomers for the synthesis of Nylon 6, 6 are", "options": [ { "text": "HOOC(CH2)6COOH, H2N(CH2)4NH2 " }, { "text": "HOOC(CH2)4COOH, H2N(CH2)6NH2" }, { "text": "HOOC(CH2)4COOH, H2N(CH2)4NH2" }, { "text": "HOOC(CH2)6COOH, H2N(CH2)6NH2" } ], "answer": "HOOC(CH2)4COOH, H2N(CH2)6NH2", "solution": "**Answer:** HOOC(CH2)4COOH, H2N(CH2)6NH2\n\nNylon $$-$$6, 6 is an amide polymer.\n

    Amide polymer is created using the monomer of amine and acid.\n

    Nylon $$-$$ 6, 6 is created using 6 carbon amine and 6 carbon acid. \n

    So, the amine is,\n

          H2N $$-$$ (CH2)6 $$-$$ NH2\n

    and the acid is,\n

          HOOC(CH2)4COOH", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2288, "subject": "Chemistry", "question": "Which one of the following polymers is not obtained by condensation polymerisation?", "options": [ { "text": "Bakelite" }, { "text": "Nylon 6" }, { "text": "Buna-N" }, { "text": "Nylon 6, 6" } ], "answer": "Buna-N", "solution": "**Answer:** Buna-N\n\nBuna-N : Obtained by addition polymerisation\n

    Nylon 6, Bakelite, Nylon 6, 6 : Obtained by\ncondensation polymerisation.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2289, "subject": "Chemistry", "question": "The correct match between Item - I and Item - II\nis :\n

    \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    Item - IItem - II
    (a) Natural rubber(I) 1, 3-butadiene\n+ styrene
    (b) Neoprene(II) 1, 3-butadiene\n+ acrylonitrile
    (c) Buna-N(III) Chloroprene
    (d) Buna-S(IV) Isoprene
    ", "options": [ { "text": "(a) - (IV), (b) - (III), (c) - (II), (d) - (I)" }, { "text": "(a) - (III), (b) - (IV), (c) - (I), (d) - (II)" }, { "text": "(a) - (III), (b) - (IV), (c) - (II), (d) - (I)" }, { "text": "(a) - (IV), (b) - (III), (c) - (I), (d) - (II)" } ], "answer": "(a) - (IV), (b) - (III), (c) - (II), (d) - (I)", "solution": "**Answer:** (a) - (IV), (b) - (III), (c) - (II), (d) - (I)\n\nNatural rubber – Polymer of isoprene\n

    Neoprene – Polymer of chloroprene\n

    Buna N – Polymer of 1,3-Butadiene and\nacrylonitrile\n

    Buna S – Polymer of 1,3-Butadiene and\nstyrene", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 2290, "subject": "Chemistry", "question": "Consider the Assertion and Reason given\nbelow.\n

    Assertion (A) : Ethene polymerized in the\npresence of Ziegler Natta Catalyst at high\ntemperature and pressure is used to make\nbuckets and dustbins.\n

    Reason (R) : High density polymers are closely\npacked and are chemically inert.\n

    Choose the correct answer from the following:", "options": [ { "text": "Both (A) and (R) are correct and (R) is the\ncorrect explanation of (A)." }, { "text": "Both (A) and (R) are correct but (R) is not\nthe correct explanation of (A)." }, { "text": "(A) is correct but (R) is wrong." }, { "text": "(A) and (R) both are wrong." } ], "answer": "Both (A) and (R) are correct and (R) is the\ncorrect explanation of (A).", "solution": "**Answer:** Both (A) and (R) are correct and (R) is the\ncorrect explanation of (A).\n\nHigh density polyethene is hard and chemically\ninert thats why used to make buckets and\ndustbins.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2291, "subject": "Chemistry", "question": "Which polymer has 'chiral' monomer(s) ?", "options": [ { "text": "Nylon 6,6" }, { "text": "Buna-N" }, { "text": "PHBV" }, { "text": "Neoprene" } ], "answer": "PHBV", "solution": "**Answer:** PHBV\n\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2292, "subject": "Chemistry", "question": "Preparation of Bakelite proceeds via reactions.", "options": [ { "text": "Electrophilic substitution and dehydration" }, { "text": "Condensation and elimination" }, { "text": "Nucleophilic addition and dehydration" }, { "text": "Electrophilic addition and dehydration" } ], "answer": "Electrophilic substitution and dehydration", "solution": "**Answer:** Electrophilic substitution and dehydration\n\nFormation of Bakelite follows electrophillic substitution of phenol and formaldehyde followed by\ndehydration.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2293, "subject": "Chemistry", "question": "Match List I with List II.

    \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    List I
    (Monomer Unit)    		List II
    (Polymer)
    (a) Caprolactum(i) Natural rubber
    (b) 2-Chloro-1, 3-butadiene(ii) Buna-N
    (c) Isoprene(iii) Nylon 6
    (d) Acrylonitrile(iv) Neoprene

    \nChoose the correct answer from the options given below:", "options": [ { "text": "(a) → (iv), (b) → (iii), (c) → (ii), (d) → (i)" }, { "text": "(a) → (iii), (b) → (iv), (c) → (i), (d) → (ii)" }, { "text": "(a) → (ii), (b) → (i), (c) → (iv), (d) → (iii)" }, { "text": "(a) → (i), (b) → (ii), (c) → (iii), (d) → (iv)" } ], "answer": "(a) → (iii), (b) → (iv), (c) → (i), (d) → (ii)", "solution": "**Answer:** (a) → (iii), (b) → (iv), (c) → (i), (d) → (ii)\n\n\"JEE\n\"JEE\n\"JEE\n\"JEE", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 2294, "subject": "Chemistry", "question": "In polymer Buna-S : 'S' stands for :", "options": [ { "text": "Sulphur" }, { "text": "Styrene" }, { "text": "Strength" }, { "text": "Sulphonation" } ], "answer": "Styrene", "solution": "**Answer:** Styrene\n\n

    Buna-S is the co-polymer of but 1, 3-diene and styrene as follows.

    \n\"JEE\n

    Styrene butadiene rubber is also called buna-S, in which ‘Bu’ stands for butadiene, ‘na’ stands for sodium and ‘S’ stands for styrene.

    \n

    Since, butadiene and styrene is one of the constituent monomer of given polymer.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2295, "subject": "Chemistry", "question": "Which statement is correct?", "options": [ { "text": "Buna-N is a natural polymer." }, { "text": "Synthesis of Buna-S needs nascent oxygen." }, { "text": "Buna-S is a synthetic and linear thermosetting polymer." }, { "text": "Neoprene is an addition copolymer used in plastic bucket manufacturing." } ], "answer": "Synthesis of Buna-S needs nascent oxygen.", "solution": "**Answer:** Synthesis of Buna-S needs nascent oxygen.\n\nBuna-S is an elastomer\n

    Buna-N is a synthetic polymer\n

    Buna-S is polymerised by addition\npolymerisation method which needs\nradical initiator for chain propagation step.\nNascent oxygen can be used as an Radical\ninitiator.\n

    Neoprene is a synthetic rubber.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2296, "subject": "Chemistry", "question": "Orlon fibres are made up of :", "options": [ { "text": "Polyacrylonitrile" }, { "text": "Polyesters" }, { "text": "Polyamide" }, { "text": "Cellulose" } ], "answer": "Polyacrylonitrile", "solution": "**Answer:** Polyacrylonitrile\n\nOrlon fibers are made up of Polyacrylonitrile.\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2297, "subject": "Chemistry", "question": "Bakelite is a cross-linked polymer of formaldehyde and :", "options": [ { "text": "PHBV" }, { "text": "Buna-S" }, { "text": "Novolac" }, { "text": "Dacron" } ], "answer": "Novolac", "solution": "**Answer:** Novolac\n\nNovolac (phenol formaldehyde Resin) $$\\to$$ Bakelite", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2298, "subject": "Chemistry", "question": "A biodegradable polyamide can be made from :", "options": [ { "text": "Glycine and isoprene" }, { "text": "Hexamethylene diamine and adipic acid" }, { "text": "Glycine and aminocaproic acid" }, { "text": "Styrene and caproic acid" } ], "answer": "Glycine and aminocaproic acid", "solution": "**Answer:** Glycine and aminocaproic acid\n\nA biodegradable polyamide nylon-2-Nylon-6 is made from glycine and amino caproic acid", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2299, "subject": "Chemistry", "question": "The polymer formed on heating Novolac with formaldehyde is :", "options": [ { "text": "Bakelite" }, { "text": "Polyester" }, { "text": "Melamine" }, { "text": "Nylon 6, 6" } ], "answer": "Bakelite", "solution": "**Answer:** Bakelite\n\nNovolac + formaldehyde $$\\to$$ Bakelite", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2300, "subject": "Chemistry", "question": "Monomer of Novolac is :", "options": [ { "text": "3-Hydroxybutanoic acid" }, { "text": "phenol and melamine" }, { "text": "o-Hydroxymethylphenol" }, { "text": "1, 3-Butadiene and styrene" } ], "answer": "o-Hydroxymethylphenol", "solution": "**Answer:** o-Hydroxymethylphenol\n\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2301, "subject": "Chemistry", "question": "Which among the following is not a polyester?", "options": [ { "text": "Novolac" }, { "text": "PHBV" }, { "text": "Dacron" }, { "text": "Glyptal" } ], "answer": "Novolac", "solution": "**Answer:** Novolac\n\nNovolac is a linear polymer of [Ph-Oh + HCHO].

    So ester linkage not present.

    So novolac is not a polyester.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2302, "subject": "Chemistry", "question": "Monomer units of Dacron polymer are :", "options": [ { "text": "ethylene glycol and phthalic acid" }, { "text": "ethylene glycol and terephthalic acid" }, { "text": "glycerol and terephthalic acid" }, { "text": "glycerol and phthalic acid" } ], "answer": "ethylene glycol and terephthalic acid", "solution": "**Answer:** ethylene glycol and terephthalic acid\n\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2303, "subject": "Chemistry", "question": "

    The polymer, which can be stretched and remains its original status on releasing the force is

    ", "options": [ { "text": "Bakelite." }, { "text": "Nylon 6, 6." }, { "text": "Buna-N." }, { "text": "Terylene." } ], "answer": "Buna-N.", "solution": "**Answer:** Buna-N.\n\n

    Buna – N is synthetic rubber which can be\nstretched and retains its original status on releasing\nthe force.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2304, "subject": "Chemistry", "question": "

    Given below are two statements : One is labelled as Assertion A and the other is labelled as Reason R.

    \n

    Assertion A : Dacron is an example of polyester polymer.

    \n

    Reason R : Dacron is made up of ethylene glycol and terephthalic acid monomers.

    \n

    In the light of the above statements, choose the most appropriate answer from the options given below.

    ", "options": [ { "text": "Both A and R are correct and R is the correct explanation of A." }, { "text": "Both A and R are correct but R is NOT the correct explanation of A." }, { "text": "A is correct but R is not correct." }, { "text": "A is not correct but R is correct." } ], "answer": "Both A and R are correct and R is the correct explanation of A.", "solution": "**Answer:** Both A and R are correct and R is the correct explanation of A.\n\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2305, "subject": "Chemistry", "question": "

    Given below are two statements, one is Assertion (A) and other is Reason (R).

    \n

    Assertion (A) : Natural rubber is a linear polymer of isoprene called cis-polyisoprene with elastic properties.

    \n

    Reason (R) : The cis-polyisoprene molecules consist of various chains held together by strong polar interactions with coiled structure.

    \n

    In the light of the above statements, choose the correct one from the options given below :

    ", "options": [ { "text": "Both (A) and (R) are true and (R) is the correct explanation of (A)." }, { "text": "Both (A) and (R) are true but (R) is not the correct explanation of (A)." }, { "text": "(A) is true but (R) is false." }, { "text": "(A) is false but (R) is true." } ], "answer": "(A) is true but (R) is false.", "solution": "**Answer:** (A) is true but (R) is false.\n\nNatural rubber is linear polymer of isoprene (2-\nmethyl-1,3-butadiene) and is also called cis-1,4-\npolyisoprene. The cis-polyisoprene molecules\nconsists of various chains held together by weak\nVander Waal‟s interactions and has a coiled\nstructure.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2306, "subject": "Chemistry", "question": "

    Which one of the following is NOT a copolymer?

    ", "options": [ { "text": "Buna-S" }, { "text": "Neoprene" }, { "text": "PHBV" }, { "text": "Butadiene-styrene" } ], "answer": "Neoprene", "solution": "**Answer:** Neoprene\n\n

    A copolymer is a polymer that consists of two or more different types of monomers.

    \n

    Here are the options examined:

    \n\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2307, "subject": "Chemistry", "question": "

    Which is true about Buna-N?

    ", "options": [ { "text": "It is a linear polymer of 1, 3-butadiene." }, { "text": "It is obtained by copolymerization of 1, 3-butadiene and styrene." }, { "text": "It is obtained by copolymerization of 1, 3-butadiene and acrylonitrile." }, { "text": "The suffix N in Buna-N stands for its natural occurrence." } ], "answer": "It is obtained by copolymerization of 1, 3-butadiene and acrylonitrile.", "solution": "**Answer:** It is obtained by copolymerization of 1, 3-butadiene and acrylonitrile.\n\nIt is copolymerization of 1, 3-butadiene and acrylonitrile.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2308, "subject": "Chemistry", "question": "

    Which of the following sets are correct regarding polymer?

    \n

    (A) Copolymer : Buna-S

    \n

    (B) Condensation polymer : Nylon-6,6

    \n

    (C) Fibres : Nylon-6,6

    \n

    (D) Thermosetting polymer : Terylene

    \n

    (E) Homopolymer : Buna-N

    \n

    Choose the correct answer from given options below :

    ", "options": [ { "text": "(A), (B) and (C) are correct" }, { "text": "(B), (C) and (D) are correct" }, { "text": "(A), (C) and (E) are correct" }, { "text": "(A), (B) and (D) are correct" } ], "answer": "(A), (B) and (C) are correct", "solution": "**Answer:** (A), (B) and (C) are correct\n\n(A) Buna-S $\\quad-\\quad$ Copolymer\n

    \n(B) Nylon-6,6 $\\quad-\\quad$ Condensation polymer\n

    \n(C) Nylon-6,6 - Fibre\n

    \n(D) Terylene $\\quad-\\quad$ Thermoplastic\n

    \n(E) Buna-N $\\quad-\\quad$ Copolymer\n

    \n$$ \\therefore $$ A, B and C are correct.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2309, "subject": "Chemistry", "question": "

    Which of the following is not an example of a condensation polymer?

    ", "options": [ { "text": "Nylon 6,6" }, { "text": "Dacron" }, { "text": "Buna-N" }, { "text": "Silicone" } ], "answer": "Buna-N", "solution": "**Answer:** Buna-N\n\nNylon 6,6 is a condensation polymer of hexamethylene diamine and adipic acid\n

    \nDacron is a condensation polymer of terephthalic acid and ethylene glycol.\n

    \nBuna- $\\mathrm{N}$ is an addition polymer of 1, 3-butadiene and acrylonitrile\n

    \nSilicone is a condensation polymer of dialkyl silanediol.\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2310, "subject": "Chemistry", "question": "

    Which of the following is an example of polyester?

    ", "options": [ { "text": "Butadiene-styrene copolymer" }, { "text": "Melamine polymer" }, { "text": "Neoprene" }, { "text": "Poly-$$\\beta$$-hydroxybutyrate-co-$$\\beta$$-hydroxyvalerate" } ], "answer": "Poly-$$\\beta$$-hydroxybutyrate-co-$$\\beta$$-hydroxyvalerate", "solution": "**Answer:** Poly-$$\\beta$$-hydroxybutyrate-co-$$\\beta$$-hydroxyvalerate\n\nPolyesters are formed by condensation reaction between alcohols and carboxylic acid.\n

    \nPoly- $\\beta$-hydroxybutyrate-co- $\\beta$-hydroxy valerate (PHBV) is a polymer obtained by condensation reaction of 3-hydroxybutanoic acid with 3-hydroxypentanoic acid.

    \n\"JEE
    \nHence, PHBV is a polyester.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2311, "subject": "Chemistry", "question": "

    Vulcanization of rubber is carried out by heating a mixture of

    ", "options": [ { "text": "isoprene and styrene" }, { "text": "neoprene and sulphur" }, { "text": "isoprene and sulphur" }, { "text": "neoprene and styrene" } ], "answer": "isoprene and sulphur", "solution": "**Answer:** isoprene and sulphur\n\nWhen a mixture of isoprene and sulphur is heated,\nisoprene gets polymerised to natural rubber and\nthen vulcanization of natural rubber with sulphur\ntakes place.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2312, "subject": "Chemistry", "question": "

    Match List I with List II.

    \n

    \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    List I

    Polymers    		List II

    Commercial names
    (A)Phenol-formaldehyde resin(I)Glyptal
    (B)Copolymer of 1,3-butadiene and sterene(II)Novolac
    (C)Polyester of glycol and phthalic acid(III)Buna-S
    (D)Polyester of glycol and terephthalic acid(IV)Dacron

    \n

    Choose the correct answer from the options given below:

    ", "options": [ { "text": "A-II, B-III, C-IV, D-I" }, { "text": "A-II, B-III, C-I, D-IV" }, { "text": "A-II, B-I, C-III, D-IV" }, { "text": "A-III, B-II, C-IV, D-I" } ], "answer": "A-II, B-III, C-I, D-IV", "solution": "**Answer:** A-II, B-III, C-I, D-IV\n\n\"JEE

    \n\"JEE

    \n\"JEE

    \n\"JEE", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 2313, "subject": "Chemistry", "question": "

    Match List - I with List - II.

    \n

    \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    List - I
    (Polymer)    		List - II
    (Monomer)
    (A)Neoprene(I)Acrylonitrile
    (B)Teflon(II)Chloroprene
    (C)Acrilan(III)Tetrafluoroethene
    (D)Natural rubber(IV)Isoprene

    \n

    Choose the correct answer from the options given below :

    ", "options": [ { "text": "$$(\\mathrm{A})-(\\mathrm{II}),(\\mathrm{B})-(\\mathrm{III}),(\\mathrm{C})-(\\mathrm{I}),(\\mathrm{D})-(\\mathrm{IV})$$" }, { "text": "$$(\\mathrm{A})-(\\mathrm{II}),(\\mathrm{B})-(\\mathrm{I}),(\\mathrm{C})-(\\mathrm{III}),(\\mathrm{D})-(\\mathrm{IV})$$" }, { "text": "$$(\\mathrm{A})-(\\mathrm{II}),(\\mathrm{B})-(\\mathrm{I}),(\\mathrm{C})-(\\mathrm{IV}),(\\mathrm{D})-(\\mathrm{III})$$" }, { "text": "$$(\\mathrm{A})-(\\mathrm{I}),(\\mathrm{B})-(\\mathrm{II}),(\\mathrm{C})-(\\mathrm{III}),(\\mathrm{D})-(\\mathrm{IV})$$" } ], "answer": "$$(\\mathrm{A})-(\\mathrm{II}),(\\mathrm{B})-(\\mathrm{III}),(\\mathrm{C})-(\\mathrm{I}),(\\mathrm{D})-(\\mathrm{IV})$$", "solution": "**Answer:** $$(\\mathrm{A})-(\\mathrm{II}),(\\mathrm{B})-(\\mathrm{III}),(\\mathrm{C})-(\\mathrm{I}),(\\mathrm{D})-(\\mathrm{IV})$$\n\n\"JEE", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 2314, "subject": "Chemistry", "question": "

    Terylene polymer is obtained by condensation of :

    ", "options": [ { "text": "Ethane-1,2-diol and Benzene-1,3 dicarboxylic acid" }, { "text": "Propane-1, 2-diol and Benzene-1, 4 dicarboxylic acid" }, { "text": "Ethane-1,2-diol and Benzene-1, 4 dicarboxylic acid" }, { "text": "Ethane-1,2-diol and Benzene-1, 2 dicarboxylic acid" } ], "answer": "Ethane-1,2-diol and Benzene-1, 4 dicarboxylic acid", "solution": "**Answer:** Ethane-1,2-diol and Benzene-1, 4 dicarboxylic acid\n\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2315, "subject": "Chemistry", "question": "

    Which of the following is NOT a natural polymer?

    ", "options": [ { "text": "Protein" }, { "text": "Starch" }, { "text": "Rubber" }, { "text": "Rayon" } ], "answer": "Rayon", "solution": "**Answer:** Rayon\n\n

    Starch, Protein, Rubber are natural polymer.

    \n

    Rayon is a semi-synthetic fiber, made from natural sources of regenerated cellulose, such as wood and\nrelated agricultural products.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2316, "subject": "Chemistry", "question": "

    Caprolactam when heated at high temperature in presence of water, gives

    ", "options": [ { "text": "Dacron" }, { "text": "Nylon 6" }, { "text": "Teflon" }, { "text": "Nylon 6,6" } ], "answer": "Nylon 6", "solution": "**Answer:** Nylon 6\n\n

    Caprolactum $$\\buildrel \\Delta \\over\n \\longrightarrow $$ Nylon $$-$$ 6

    \n

    \"JEE

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2317, "subject": "Chemistry", "question": "

    Match List I with List II

    \n

    \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    List IList II
    A.Elastomeric polymerI.Urea formaldehyde resin
    B.Fibre PolymerII.Polystyrene
    C.Thermosetting PolymerIII.Polyester
    D.Thermoplastic PolymerIV.Neoprene

    \n

    Choose the correct answer from the options given below :

    ", "options": [ { "text": "A-II, B-I, C-IV, D-III" }, { "text": "A-IV, B-I, C-III, D-II" }, { "text": "A-IV, B-III, C-I, D-II" }, { "text": "A-II, B-III, C-I, D-IV" } ], "answer": "A-IV, B-III, C-I, D-II", "solution": "**Answer:** A-IV, B-III, C-I, D-II\n\n

    \n\n\n\n\n\n\n\n\n \n \n \n \n \n \n\n\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\n
    A.Elastomeric fibre(IV)Neoprene
    B.Fibre polymer(III)Polyester
    C.Thermosetting polymer(I)Urea Formaldehyde Resin
    D.Thermoplastic polymer(II)Polystyrene

    ", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 2318, "subject": "Chemistry", "question": "Match List I with List II:

    \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    List I - (Monomer)List II - (Polymer)
    (A) Tetrafluoroethene(I) Orlon
    (B) Acrylonitrile(II) Natural rubber
    (C) Caprolactam(III) Teflon
    (D) Isoprene(IV) Nylon-6

    \nChoose the correct answer from the options given below :", "options": [ { "text": "(A)-(III), (B)-(I), (C)-(IV), (D)-(II)" }, { "text": "(A)-(II), (B)-(III), (C)-(IV), (D)-(I)" }, { "text": "(A)-(IV), (B)-(I), (C)-(II), (D)-(III)" }, { "text": "(A)-(III), (B)-(IV), (C)-(II), (D)-(I) " } ], "answer": "(A)-(III), (B)-(I), (C)-(IV), (D)-(II)", "solution": "**Answer:** (A)-(III), (B)-(I), (C)-(IV), (D)-(II)\n\nThe correct match between List I (Monomer) and List II (Polymer) is as follows:\n

    \n(A) Tetrafluoroethene → (III) Teflon

    \n(B) Acrylonitrile → (I) Orlon

    \n(C) Caprolactam → (IV) Nylon-6

    \n(D) Isoprene → (II) Natural rubber\n

    \nSo, the correct answer is:\n

    \nA-III, B-I, C-IV, D-II", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 2319, "subject": "Chemistry", "question": "

    Match List I with List II

    \n

    \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    LIST ILIST II
    A.Weak intermolecular forces of attractionI.Hexamethylenediamine + adipic acid
    B.Hydrogen bondingII.$$\\mathrm{AlEt_3+TiCl_4}$$
    C.Heavily branched polymerIII.2 - chloro - 1, 3 - butadiene
    D.High density polymerIV.Phenol + formaldehyde

    \n

    Choose the correct answer from the options given below:

    ", "options": [ { "text": "A-III, B-I, C-IV, D-II" }, { "text": "A-IV, B-I, C-III, D-II" }, { "text": "A-II, B-IV, C-I, D-III" }, { "text": "A-IV, B-II, C-III, D-I" } ], "answer": "A-III, B-I, C-IV, D-II", "solution": "**Answer:** A-III, B-I, C-IV, D-II\n\n(A) Weak Intermolecular Forces are present in polymer of 2-chloro, 1, 3-butadiene

    \n(B) Hydrogen Bonding is present in NYLON-66 which is a polymer of Hexamethylenediamine and adipic acid

    \n(C) Heavily branched polymer is Bakelite which is polymer of phenol and formaldehyde

    \n(D) High density polymer prepration requires $\\mathrm{AL}(\\mathrm{Et})_3$ and $\\mathrm{TiCl}_4$ as a catalyst (Ziegler Natta Catalyst)", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 2320, "subject": "Chemistry", "question": "

    Match the following

    \n

    \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    Column AColumn B
    (a)Nylon 6(I)Natural Rubber
    (b)Vulcanized Rubber(II)Cross Linked
    (c)cis-1, 4-polyisoprene(III)Caprolactam
    (d)Polychloroprene(IV)Neoprene

    \n

    Choose the correct answer from options given below:

    ", "options": [ { "text": "$$\\mathrm{a} \\rightarrow \\mathrm{III}, \\mathrm{b} \\rightarrow \\mathrm{II}, \\mathrm{c} \\rightarrow \\mathrm{I}, \\mathrm{d} \\rightarrow \\mathrm{IV}$$" }, { "text": "$$\\mathrm{a} \\rightarrow \\mathrm{II}, \\mathrm{b} \\rightarrow \\mathrm{III}, \\mathrm{c} \\rightarrow \\mathrm{IV}, \\mathrm{d} \\rightarrow \\mathrm{I}$$" }, { "text": "$$\\mathrm{a} \\rightarrow \\mathrm{IV}, \\mathrm{b} \\rightarrow \\mathrm{III}, \\mathrm{c} \\rightarrow \\mathrm{I}, \\mathrm{d} \\rightarrow \\mathrm{II}$$" }, { "text": "$$\\mathrm{a} \\rightarrow \\mathrm{III}, \\mathrm{b} \\rightarrow \\mathrm{IV}, \\mathrm{c} \\rightarrow \\mathrm{I}, \\mathrm{d} \\rightarrow \\mathrm{II}$$" } ], "answer": "$$\\mathrm{a} \\rightarrow \\mathrm{III}, \\mathrm{b} \\rightarrow \\mathrm{II}, \\mathrm{c} \\rightarrow \\mathrm{I}, \\mathrm{d} \\rightarrow \\mathrm{IV}$$", "solution": "**Answer:** $$\\mathrm{a} \\rightarrow \\mathrm{III}, \\mathrm{b} \\rightarrow \\mathrm{II}, \\mathrm{c} \\rightarrow \\mathrm{I}, \\mathrm{d} \\rightarrow \\mathrm{IV}$$\n\nNylon 6 is a synthetic polymer made from caprolactam, so it corresponds to Column A, option (c), which matches with Column B, option (I).\n

    \nVulcanized rubber is rubber that has been treated with sulfur or other additives to make it more durable, so it corresponds to Column A, option (b), which matches with Column B, option (II).\n

    \nCis-1,4-polyisoprene is a naturally occurring polymer found in natural rubber, so it corresponds to Column A, option (a), which matches with Column B, option (III).\n

    \nPolychloroprene is a synthetic polymer used to make neoprene, a type of synthetic rubber, so it corresponds to Column A, option (d), which matches with Column B, option (IV).\n

    \nTherefore, the correct matching is:\n

    \n$$\\mathrm{a} \\rightarrow \\mathrm{III}, \\mathrm{b} \\rightarrow \\mathrm{II}, \\mathrm{c} \\rightarrow \\mathrm{I}, \\mathrm{d} \\rightarrow \\mathrm{IV}$$\n

    \nOption A is the correct answer.", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 2321, "subject": "Chemistry", "question": "

    Match List I with List II

    \n

    \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    LIST I
    (Examples)    		LIST II
    (Type)
    A.2-chloro-1,3- butadieneI.Biodegradable polymer
    B.Nylon 2-nylon 6II.Synthetic Rubber
    C.PolyacrylonitrileIII.Polyester
    D.DacronIV.Addition Polymer

    \n

    Choose the correct answer from the options given below:

    ", "options": [ { "text": "A-II, B-IV, C-I, D-III" }, { "text": "A-IV, B-I, C-III, D-II" }, { "text": "A-II, B-I, C-IV, D-III" }, { "text": "A-IV, B-III, C-I, D-II" } ], "answer": "A-II, B-I, C-IV, D-III", "solution": "**Answer:** A-II, B-I, C-IV, D-III\n\n

    Let's look at each example in List I and try to identify its type:

    \n

    A. 2-chloro-1,3-butadiene: This is an example of a synthetic rubber, which is a type of polymer.

    \n

    B. Nylon 2-nylon 6: This is a type of nylon polymer, which is a type of polyester.

    \n

    C. Polyacrylonitrile: This is an example of a biodegradable polymer, which can be broken down by microorganisms.

    \n

    D. Dacron: This is an example of an addition polymer, which is a polymer formed by the addition of monomers.

    \n

    Now, let's look at the types in List II and match them with their corresponding examples:

    \n

    I. Biodegradable polymer: This matches with example C, which is Polyacrylonitrile.

    \n

    II. Synthetic Rubber: This matches with example A, which is 2-chloro-1,3-butadiene.

    \n

    III. Polyester: This matches with example B, which is Nylon 2-nylon 6.

    \n

    IV. Addition polymer: This matches with example D, which is Dacron.

    \n

    Therefore, the correct matching of List I with List II is:

    \n

    A - II (Synthetic rubber)

    \nB - III (Polyester)

    \nC - I (Biodegradable polymer)

    \nD - IV (Addition polymer)

    \n

    Hence, the correct answer is A-II, B-I, C-IV, D-III.

    \n", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 2322, "subject": "Chemistry", "question": "

    Given below are two statements :

    \n

    Statement I : Ethene at 333 to $$343 \\mathrm{~K}$$ and $$6$$-$$7$$ atm pressure in the presence of $$\\mathrm{AlEt}_{3}$$ and $$\\mathrm{TiCl}_{4}$$ undergoes addition polymerization to give LDP.

    \n

    Statement II : Caprolactam at $$533$$-$$543 \\mathrm{~K}$$ in $$\\mathrm{H}_{2} \\mathrm{O}$$ through step growth polymerizes to give Nylon 6.

    \n

    In the light of the above statements, choose the correct answer from the options given below:

    ", "options": [ { "text": "Both Statement I and Statement II are true" }, { "text": "Statement I is true but Statement II is false" }, { "text": "Statement I is false but Statement II is true" }, { "text": "Both Statement I and Statement II are false" } ], "answer": "Statement I is false but Statement II is true", "solution": "**Answer:** Statement I is false but Statement II is true\n\n

    Statement I is false. Ethene (ethylene) under certain conditions (high pressure, high temperature) in the presence of catalysts does undergo addition polymerization, but it forms polyethylene, not LDP (Low-Density Polyethylene) specifically. The formation of low-density polyethylene (LDPE) vs high-density polyethylene (HDPE) depends on the specific conditions and catalysts used. The catalysts mentioned (AlEt3 and TiCl4) are associated with the production of HDPE through the Ziegler-Natta process, not LDPE.

    \n

    Statement II is true. Caprolactam polymerizes to form Nylon 6 via a ring-opening polymerization mechanism, which is a type of step-growth polymerization. The process is often carried out at high temperatures, and water is a byproduct of the reaction, so the description in Statement II is accurate.

    \n

    Therefore, Option C (Statement I is false but Statement II is true) is the correct answer.

    \n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2323, "subject": "Chemistry", "question": "

    The polymer $$\\mathrm{X}$$ - consists of linear molecules and is closely packed. It is prepared in the presence of triethylaluminium and titanium tetrachloride under low pressure. The polymer $$\\mathrm{X}$$ is-

    ", "options": [ { "text": "High density polythene" }, { "text": "Polytetrafluoroethane" }, { "text": "Low density polythene" }, { "text": "Polyacrylonitrile" } ], "answer": "High density polythene", "solution": "**Answer:** High density polythene\n\n

    The polymer X described in the question is synthesized using triethylaluminium and titanium tetrachloride under low pressure. This is a clear indication that the Ziegler-Natta catalyst is used, which is typically employed in the synthesis of high-density polyethylene (HDPE).

    \n

    High-density polyethylene is made up of linear molecules that pack closely together, giving the material its high density. This is in contrast to low-density polyethylene (LDPE), which has a branched structure that prevents the chains from packing closely together.

    \n

    So, the polymer X is :

    \n

    Option A: High-density polyethylene.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2324, "subject": "Chemistry", "question": "

    Match List I with List II

    \n

    \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    List I
    Polymer    		List II
    Type/Class
    (A)Nylon-2-Nylon-6(I)Thermosetting polymer
    (B)Buna-N(II)Biodegradable polymer
    (C)Urea-formaldehyde resin(III)Synthetic rubber
    (D)Dacron(IV)Polyester

    \n

    Choose the correct answer from the options given below:

    ", "options": [ { "text": "(A)-(IV), (B)-(I), (C)-(III), (D)-(II)" }, { "text": "(A)-(II), (B)-(I), (C)-(IV), (D)-(III)" }, { "text": "(A)-(II), (B)-(III), (C)-(I), (D)-(IV)" }, { "text": "(A)-(IV), (B)-(III), (C)-(I), (D)-(II)" } ], "answer": "(A)-(II), (B)-(III), (C)-(I), (D)-(IV)", "solution": "**Answer:** (A)-(II), (B)-(III), (C)-(I), (D)-(IV)\n\n\n", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 2325, "subject": "Chemistry", "question": "Polymer formation from monomers starts by", "options": [ { "text": "condensation reaction between monomers" }, { "text": "coordinate reaction between monomers" }, { "text": "conversion of monomer to monomer ions by protons" }, { "text": "hydrolysis of monomers" } ], "answer": "condensation reaction between monomers", "solution": "**Answer:** condensation reaction between monomers\n\nPolymerisation starts either by condensation or addition reactions between monomers. Condensation polymers are formed by the combination of monomers with the elimination of simple molecules. Where as the addition polymers are formed by the addition together of the molecules of the monomer or monomers to form a large molecule without elimination of anything.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2326, "subject": "Chemistry", "question": "Nylon threads are made of", "options": [ { "text": "polyester polymer" }, { "text": "polyamide polymer" }, { "text": "polyethylene polymer " }, { "text": "polyvinyl polymer" } ], "answer": "polyamide polymer", "solution": "**Answer:** polyamide polymer\n\nNylon is a polyamide polymer.\n\"AIEEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2327, "subject": "Chemistry", "question": "Which of the following is a polyamide? ", "options": [ { "text": "Teflon" }, { "text": "Nylon – 6, 6" }, { "text": "Terylene" }, { "text": "Bakelite " } ], "answer": "Nylon – 6, 6", "solution": "**Answer:** Nylon – 6, 6\n\nNylon is a general name for all synthetic fibres forming polyamides.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2328, "subject": "Chemistry", "question": "Which polymer is used in the manufacture of paints and lacquers?", "options": [ { "text": "Glyptal" }, { "text": "Polypropene" }, { "text": "Poly vinyl chloride" }, { "text": "Bakelite" } ], "answer": "Glyptal", "solution": "**Answer:** Glyptal\n\nGlyptal is used in the manufacture of paints and lacquers.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2329, "subject": "Chemistry", "question": "Which of the following statements about low density polythene is FALSE ?", "options": [ { "text": "Its synthesis requires high pressure." }, { "text": "It is a poor conductor of electricity." }, { "text": "Its synthesis requires dioxygen or a peroxide initiator as a catalyst." }, { "text": "It is used in the manufacture of buckets, dust-bins etc." } ], "answer": "It is used in the manufacture of buckets, dust-bins etc.", "solution": "**Answer:** It is used in the manufacture of buckets, dust-bins etc.\n\nHigh density polythene is used in the manufacture of housewares like buckets, dustbins, bottles, pipes etc. Low density polythene is used for insulating electric wires and in the manufacture of flexible pipes, toys. coats, bottles etc. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2330, "subject": "Chemistry", "question": "Which of the following polymer is used in the manufacture of wood laminates?", "options": [ { "text": "Melamine formaldehyde resin" }, { "text": "Urea formaldehyde resin" }, { "text": "cis-poly isoprene" }, { "text": "Phenol and formaldehyde resin" } ], "answer": "Urea formaldehyde resin", "solution": "**Answer:** Urea formaldehyde resin\n\nUrea-formaldehyde resin is used in wood laminates.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2331, "subject": "Chemistry", "question": "

    Match List - I with List - II

    \n

    \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    List - I
    (Polymer)    		List - II
    (Used in)
    (A)Bakelite(I)Radio and television cabinets
    (B)Glyptal(II)Electrical switches
    (C)PVC(III)Paints and Lacquers
    (D)Polystyrene(IV)Water pipes

    \n

    Choose the correct answer from the options given below :

    ", "options": [ { "text": "(A) - (II), (B) - (III), (C) - (IV), (D) - (I)" }, { "text": "(A) - (I), (B) - (II), (C) - (III), (D) - (IV)" }, { "text": "(A) - (IV), (B) - (III), (C) - (II), (D) - (I)" }, { "text": "(A) - (II), (B) - (III), (C) - (I), (D) - (IV)" } ], "answer": "(A) - (II), (B) - (III), (C) - (IV), (D) - (I)", "solution": "**Answer:** (A) - (II), (B) - (III), (C) - (IV), (D) - (I)\n\n

    \n\n\n\n\n\n\n\n\n \n \n \n \n \n \n\n\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\n
    List - I
    (Polymer)    		List - II
    (Used in)
    (A)Bakelite(II)Electrical switches
    (B)Glyptal(III)Paints and Lacqures
    (C)PVC(IV)Water pipes
    (D)Polystyrene(I)Radio and television Cabinets

    \n

    Therefore, the correct option is (A).

    ", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 2332, "subject": "Chemistry", "question": "

    The Novolac polymer has mass of 963 g. The number of monomer units present in it are

    ", "options": [], "answer": "9", "solution": "**Answer:** 9\n\nNovolac is

    \n\"JEE
    \nMolar mass of monomer is $107 \\mathrm{~g} / \\mathrm{mol}$\n

    \n$$\n\\mathrm{n}=\\frac{963}{107}=9\n$$\n

    \nThe number of monomer units present in it is 9.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2333, "subject": "Chemistry", "question": "

    Match List I with List II :

    \n

    \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    List IList II
    (A)Nylon 6, 6(I)Buckets
    (B)Low density polythene(II)Non-stick utensils
    (C)High density polythene(III)Bristles of brushes
    (D)Teflon(IV)Toys

    \n

    Choose the correct answer from the options given below :

    ", "options": [ { "text": "A-III, B-I, C-IV, D-II" }, { "text": "A-III, B-IV, C-I, D-II" }, { "text": "A-II, B-I, C-IV, D-III" }, { "text": "A-II, B-IV, C-I, D-III" } ], "answer": "A-III, B-IV, C-I, D-II", "solution": "**Answer:** A-III, B-IV, C-I, D-II\n\nNylon $$6,6 \\rightarrow$$ used in making bristles of brushes

    Low density polythene $$\\rightarrow$$ used in making Toys

    High density polythene $$\\rightarrow$$ used in making Buckets

    Teflon $$\\rightarrow$$ used in making non-stick utensils", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 2334, "subject": "Chemistry", "question": "

    Match List I with List II

    \n

    \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    List I (Name of polymer)List II (Uses)
    A.GlyptalI.Flexible pipes
    B.NeopreneII.Synthetic wool
    C.AcrilanIII.Paints and Lacquers
    D.LDPIV.Gaskets

    \n

    Choose the correct answer from the options given below :

    ", "options": [ { "text": "A-III, B-I, C-IV, D-II" }, { "text": "A-III, B-IV, C-I, D-II" }, { "text": "A-III, B-II, C-IV, D-I" }, { "text": "A-III, B-IV, C-II, D-I" } ], "answer": "A-III, B-IV, C-II, D-I", "solution": "**Answer:** A-III, B-IV, C-II, D-I\n\n(A) Glyptal - (III) Paints and Lacquers\n

    \n(B) Neoprene - (IV) Gaskets\n

    \n(C) Acrilan - (II) Synthetic wool\n

    \n(D) LDP - (I) Flexible pipes\n", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 2335, "subject": "Chemistry", "question": "

    Polymer used in orlon is :

    ", "options": [ { "text": "Polyethene" }, { "text": "Polyacrylonitrile" }, { "text": "Polycarbonate" }, { "text": "Polyamide" } ], "answer": "Polyacrylonitrile", "solution": "**Answer:** Polyacrylonitrile\n\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2336, "subject": "Chemistry", "question": "The compound formed in the positive test for nitrogen with the Lassaigne solution of an\norganic compound is", "options": [ { "text": "Fe4[Fe(CN)6]3 " }, { "text": "Na4[Fe(CN)5NOS] " }, { "text": "Fe(CN)3" }, { "text": "Na3[Fe(CN)6] " } ], "answer": "Fe4[Fe(CN)6]3 ", "solution": "**Answer:** Fe4[Fe(CN)6]3 \n\nIf nitrogen is present in organic compound, then sodium extract contains NaCN.\n

    Na + C + N $$\\buildrel {Fuse} \\over\n \\longrightarrow $$ NaCN\n

    FeSO4 + 6NaCN $$ \\to $$ Na4[Fe(CN)6] + Na2SO4\n

    Na4[Fe(CN)6] changes to prussian blue Fe4[Fe(CN)6]3 on reaction with FeCl3.\n

    FeCl3 + Na4[Fe(CN)6] $$ \\to $$ Fe4[Fe(CN)6]3 + 12NaCl", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2337, "subject": "Chemistry", "question": "29.5 mg of an organic compound containing nitrogen was digested according to Kjeldahl’s method\nand the evolved ammonia was absorbed in 20 mL of 0.1 M HCl solution. The excess of the acid\nrequired 15 mL of 0.1 M NaOH solution for complete neutralization. The percentage of nitrogen in\nthe compound is", "options": [ { "text": "59.0" }, { "text": "47.4" }, { "text": "23.7" }, { "text": "29.5" } ], "answer": "23.7", "solution": "**Answer:** 23.7\n\nMoles of $$HCl$$ taken $$ = 20 \\times 0.1 \\times {10^{ - 3}} = 2 \\times {10^{ - 3}}$$\n

    Moles of $$HCl$$ neutralised by $$NaOH$$ solution\n

    $$ = 15 \\times 0.1 \\times {10^{ - 3}} = 1.5 \\times {10^{ - 3}}$$ \n

    Moles of $$HCl$$ neutralised by ammonia\n

    $$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, = 2 \\times {10^{ - 3}} - 1.5 \\times {10^{ - 3}}$$\n

    $$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, = 0.5 \\times {10^{ - 3}}$$ \n

    $$\\% $$ of nitrogen $$ = {{1.4 \\times N \\times V} \\over {w.t.\\,\\,of\\,\\,Subs\\tan ce}} \\times 100$$ \n

    $$ = {{1.4 \\times 0.5 \\times {{10}^{ - 3}}} \\over {29.5 \\times {{10}^{ - 3}}}} \\times 100 = 23.7\\% $$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2338, "subject": "Chemistry", "question": "The correct match between Item-I and Item-II is : \n

    \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    Item-I (drug)Item-II (test)
    AChloroxylenolPCarbylamine test
    BNorethindroneQSodium hydrogen carbonet test
    CSulphapyridineRFerric chloride test
    DPenicillinSBayer's test
    ", "options": [ { "text": "A $$ \\to $$ R;  B $$ \\to $$ P;  C $$ \\to $$ S;  D $$ \\to $$ Q" }, { "text": "A $$ \\to $$ Q;  B $$ \\to $$ S;  C $$ \\to $$ P;  D $$ \\to $$ R" }, { "text": "A $$ \\to $$ R;  B $$ \\to $$ S;  C $$ \\to $$ P;  D $$ \\to $$ Q" }, { "text": "A $$ \\to $$ Q;  B $$ \\to $$ P;  C $$ \\to $$ S;  D $$ \\to $$ R" } ], "answer": "A $$ \\to $$ R;  B $$ \\to $$ S;  C $$ \\to $$ P;  D $$ \\to $$ Q", "solution": "**Answer:** A $$ \\to $$ R;  B $$ \\to $$ S;  C $$ \\to $$ P;  D $$ \\to $$ Q\n\n(1)  To react with FeCl3, with benzene ring there should be a $$-$$OH group attached. \n

    (2)  For carbylamine test there should be $$-$$NH2 group in a compound. \n

    (3)  NaHCO3 reacts when there is $$-$$COOH group in a compound. \n

    (4)  For Bayer's test there should be double bond or tripple bond between two carbon. \n

    \"JEE", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 2339, "subject": "Chemistry", "question": "An organic compound 'A' is oxidizod with Na2O2 followed by boiling with HNO3. The resultant solution is\nthen treated with ammonium molybdate to yield a yellow precipitate. \n
    Based on above observation, the\nelement present in the given compound is:", "options": [ { "text": "Fluorine" }, { "text": "Phosphorus" }, { "text": "Nitrogen" }, { "text": "Sulphur" } ], "answer": "Phosphorus", "solution": "**Answer:** Phosphorus\n\nPhosphorus is detected in the form of canary\nyellow ppt on reaction with ammonium\nmolybdate.\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2340, "subject": "Chemistry", "question": "If you spill a chemical toilet cleaning liquid on\nyour hand, your first aid would be", "options": [ { "text": "aqueous NaOH" }, { "text": "aqueous NaHCO3" }, { "text": "aqueous NH3" }, { "text": "vinegar" } ], "answer": "aqueous NaHCO3", "solution": "**Answer:** aqueous NaHCO3\n\nToilet cleaning liquid has about 10.5% w/v HCl ; to neutralise its affect aqueous NaHCO3 is used while\nNaOH is avoided for this purpose because of its highly corosive in nature and can burn body.\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2341, "subject": "Chemistry", "question": "Which of the following compound is added to the sodium extract before addition of silver nitrate for testing of halogens?", "options": [ { "text": "Nitric acid" }, { "text": "Sodium hydroxide" }, { "text": "Ammonia" }, { "text": "Hydrochloric acid" } ], "answer": "Nitric acid", "solution": "**Answer:** Nitric acid\n\nNaCN + HNO3 $$ \\to $$ NaNO3 + HCN $$ \\uparrow $$

    \nNa2S + HNO3 $$ \\to $$ NaNO3 + H2S $$ \\uparrow $$

    \nNilnic acid decomposed NaCN & Na2S, else they precipitate in test & misquite the resolve\nFor testing of halogens, Nitric acid is added\nto the sodium extract because if CN–\n or S2–\nare present then they will be oxidised and\nremoved before the test of halides.\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2342, "subject": "Chemistry", "question": "Which of the following is 'a' FALSE statement ?", "options": [ { "text": "Carius method is used for the estimation of nitrogen in an organic compound." }, { "text": "Kjeldahl's method is used for the estimation of nitrogen in an organic compound." }, { "text": "Carius tube is used in the estimation of sulphur in an organic compound." }, { "text": "Phosphoric acid produced on oxidation of phosphorus present in an organic compound is precipitated as Mg2P2O7 by adding magnesia mixture." } ], "answer": "Carius method is used for the estimation of nitrogen in an organic compound.", "solution": "**Answer:** Carius method is used for the estimation of nitrogen in an organic compound.\n\nCarius method is used in the estimation of\nhalogen and sulphur in organic compounds.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2343, "subject": "Chemistry", "question": "Seliwanoff test and Xanthoproteic test are used for the identification of ___________ and ________ respectively.", "options": [ { "text": "ketoses, proteins" }, { "text": "proteins, ketoses" }, { "text": "aldoses, ketoses" }, { "text": "ketoses, aldoses" } ], "answer": "ketoses, proteins", "solution": "**Answer:** ketoses, proteins\n\nSeliwanoff test and Xanthaproteic test are used for identification of 'Ketoses' and proteins\nrespectively.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2344, "subject": "Chemistry", "question": "Match List - I with List - II :

    \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    List - I
    Test/Reagents/Observation(s)    		List - II
    Species detected
    (a)Lassaigne's Test(i)Carbon
    (b)Cu(II) oxide(ii)Sulphur
    (c)Silver nitrate(iii)N, S, P, and halogen
    (d)The sodium fusion extract
    gives black precipitate with
    acetic acid and lead acetate    		(iv)Halogen Specifically


    The correct match is :", "options": [ { "text": "(a)-(iii), (b)-(i), (c)-(ii), (d)-(iv)" }, { "text": "(a)-(i), (b)-(iv), (c)-(iii), (d)-(ii)" }, { "text": "(a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)" }, { "text": "(a)-(i), (b)-(ii), (c)-(iv), (d)-(iii)" } ], "answer": "(a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)", "solution": "**Answer:** (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)\n\n(i) Lassaigne test is used to detect N, S, P, X element

    \n(ii) Carbon and hydrogen are detected by heating the compound with copper (ii) oxide.

    \n(iii) Halides are detected by silver nitrate

    \n(iv) Sodium fusion extract gives black ppt. with acetic acid and lead acetate to confirm the presence of\nsulphur. ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2345, "subject": "Chemistry", "question": "Methylation of 10 g of benzene gave 9.2 g of toluene. Calculate the percentage yield of toluene __________. (Nearest integer)", "options": [], "answer": "78", "solution": "**Answer:** 78\n\n$${C_6}{H_6} + C{H_3}Cl\\mathrel{\\mathop{\\kern0pt\\longrightarrow}\n\\limits_{}} {C_6}{H_5}C{H_3} + HCl$$

    10 gm of C6H6 = $${{10} \\over {78}}$$ mole\n

    Moles of methylbenzene should be obtained = $${{10} \\over {78}}$$ mole\n

    = $$\\left( {{{10} \\over {78}} \\times 92} \\right)gm $$

    $$ \\therefore $$ $${{{A_y}} \\over {{T_y}}} = \\% $$ yield $$ = {{9.2} \\over {920}} \\times 78 \\times 100 \\Rightarrow 78\\% $$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2346, "subject": "Chemistry", "question": "An organic compound is subjected to chlorination to get compound A using 5.0 g of chlorine. When 0.5 g of compound A is reacted with AgNO3 [Carius Method], the percentage of chlorine in compound A is ____________ when it forms 0.3849 g of AgCl. (Round off to the Nearest Integer)

    (Atomic masses of Ag and Cl are 107.87 and 35.5 respectively)", "options": [], "answer": "19", "solution": "**Answer:** 19\n\nMass of organic compound = 0.5 gm.

    Mass of formed AgCl = 0.3849 gm

    % of Cl = $${{atomic\\,mass\\,of\\,Cl \\times mass\\,formed\\,AgCl} \\over {molecular\\,mass\\,of\\,AgCl \\times mass\\,of\\,organic\\,compound}} \\times 100$$

    $$ = {{35.5 \\times 0.3849} \\over {143.37 \\times 0.5}} \\times 100$$

    = 19.06

    $$ \\approx $$ 19", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2347, "subject": "Chemistry", "question": "Which one of the following complexes is violet in colour?", "options": [ { "text": "[Fe(CN)6]4$$-$$" }, { "text": "[Fe(SCN)6]4$$-$$" }, { "text": "Fe4[Fe(CN6)]3 . H2O" }, { "text": "[Fe(CN)5NOS]4$$-$$" } ], "answer": "[Fe(CN)5NOS]4$$-$$", "solution": "**Answer:** [Fe(CN)5NOS]4$$-$$\n\n(a) [Fe(CN)6]4$$-$$ $$\\to$$ Pale yellow solution

    (b) [Fe(SCN)6]4$$-$$ $$\\to$$ Blood red colour

    (c) Fe4[Fe(CN6)]3 . H2O $$\\to $$ Prussian blue

    (d) [Fe(CN)5NOS]4$$-$$ $$\\to$$ Violet colour", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2348, "subject": "Chemistry", "question": "In the sulphur estimation, 0.471 g of an organic compound gave 1.44 g of barium sulphate. The percentage of sulphur in the compound is ____________%. (Nearest integer)

    (Atomic Mass of Ba = 137 u)", "options": [], "answer": "42", "solution": "**Answer:** 42\n\nMolecular mass of BaSO4 = 233 g

    $$\\because$$ 233 BaSO4 contain $$\\to$$ 32 g sulphur

    $$\\therefore$$ 1.44 g BaSO4 contain $$\\to$$ $${{32} \\over {233}}$$ $$\\times$$ 1.44 g sulphur given : 0.471 g of organic compound

    % of S = $${{32 \\times 1.44} \\over {233 \\times 0.471}} \\times 100 = 41.98\\% \\approx 42\\% $$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2349, "subject": "Chemistry", "question": "Acidic ferric chloride solution on treatment with excess of potassium ferrocyanide gives a Prussian blue coloured colloidal species. It is :", "options": [ { "text": "Fe4[Fe(CN)6]3" }, { "text": "K5Fe[Fe(CN)6]2" }, { "text": "HFe[Fe(CN)6]" }, { "text": "KFe[Fe(CN)6]" } ], "answer": "Fe4[Fe(CN)6]3", "solution": "**Answer:** Fe4[Fe(CN)6]3\n\n

    Prussian blue is a complex compound that has the formula $ \\text{Fe}_{4}[\\text{Fe(CN)}_{6}]_{3} $. It is formed by the reaction of ferric ions with ferrocyanide ions.

    \n

    The reaction can be explained as follows:

    \n
      \n

    1. Potassium ferrocyanide, $ \\text{K}_{4}[\\text{Fe(CN)}_{6}] $, dissociates in water to release ferrocyanide ions, $ [\\text{Fe(CN)}_{6}]^{4-} $.

      \n

      2. \n

    2. The ferric ions, $ \\text{Fe}^{3+} $, present in the acidic ferric chloride solution will react with the ferrocyanide ions.

      \n

      3. \n

    3. In an acidic medium, the ferric ions hydrolyze, forming a complex with ferrocyanide ions, leading to the formation of Prussian blue.

      \n
      4. \n
    \n

    The overall reaction can be represented as:

    \n

    $ \\text{Fe}^{3+} + 4[\\text{Fe(CN)}_{6}]^{4-} \\rightarrow \\text{Fe}_{4}[\\text{Fe(CN)}_{6}]_{3}$

    \n

    The Prussian blue complex is a colloidal species and exhibits an intense blue color. It has been historically used as a pigment in paints and also has applications in medical treatments as an antidote for certain types of heavy metal poisoning.

    \n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2350, "subject": "Chemistry", "question": "

    In the estimation of bromine, 0.5 g of an organic compound gave 0.40 g of silver bromide. The percentage of bromine in the given compound is _________ % (nearest integer)

    \n

    (Relative atomic masses of Ag and Br are 108u and 80u, respectively).

    ", "options": [], "answer": "34", "solution": "**Answer:** 34\n\n$188 \\mathrm{~g} ~ \\mathrm{AgBr}$ has $80 \\mathrm{~g}$ of $\\mathrm{Br}$\n

    \n$$\n\\therefore 0.4 \\mathrm{~g} \\mathrm{AgBr}=\\frac{80}{188} \\times 0.4\n$$\n

    \n$\\%$ of $\\mathrm{Br}$ in given organic compound\n

    \n$$\n\\begin{aligned}\n&=\\frac{80 \\times 0.4}{188 \\times 0.5} \\times 100 \\\\\\\\\n&\\approx 34 \\%\n\\end{aligned}\n$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2351, "subject": "Chemistry", "question": "

    0.25 g of an organic compound containing chlorine gave 0.40 g of silver chloride in Carius estimation. The percentage of chlorine present in the compound is ___________. [in nearest integer]

    \n

    (Given : Molar mass of Ag is 108 g mol$$-$$1 and that of Cl is 35.5 g mol$$-$$1)

    ", "options": [], "answer": "40", "solution": "**Answer:** 40\n\n

    Given, weight of organic compound = 0.25 g

    \n

    Moles of AgCl = $${{0.4} \\over M}$$

    \n

    Molecular mass of AgCl (M) = 143.5 gm

    \n

    $$\\therefore$$ Moles of AgCl = $${{0.4} \\over 143.5}$$

    \n

    $$\\therefore$$ Mass of Cl = $${{0.4} \\over 143.5}$$ $$\\times$$ 35.5

    \n

    Mass % of Cl in the organic compound

    \n

    $$ = {{{{35.5 \\times 0.4} \\over {143.5}}} \\over {0.25}} \\times 35.5$$

    \n

    $$ = 39.58$$

    \n

    $$ \\simeq 40$$

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2352, "subject": "Chemistry", "question": "

    0.2 g of an organic compound was subjected to estimation of nitrogen by Dumas method in which volume of N2 evolved (at STP) was found to be 22.400 mL. The percentage of nitrogen in the compound is _________. [nearest integer]

    \n

    (Given : Molar mass of N2 is 28 g mol$$-$$1. Molar volume of N2 at STP : 22.4 L)

    ", "options": [], "answer": "14", "solution": "**Answer:** 14\n\n

    Given volume of N2 = 22.400 mL

    \n

    $$\\therefore$$ Moles of N2 = $${{22.400} \\over {22400}} = {10^{ - 3}}$$ mole

    \n

    $$\\therefore$$ Moles of N atoms = 2 $$\\times$$ 10$$-$$3 mole

    \n

    $$\\therefore$$ Weigh of N atoms = 14 $$\\times$$ 2 $$\\times$$ 10$$-$$3 mole

    \n

    = 28 $$\\times$$ 10$$-$$3 mole

    \n

    $$\\therefore$$ % of N atom in the compound

    \n

    $$ = {{28 \\times {{10}^{ - 3}}} \\over {0.2}} \\times 100$$

    \n

    = 14

    ", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 2353, "subject": "Chemistry", "question": "

    An organic compound with 51.6% sulfur is heated in a Carius tube. The amount of this compound which will form 0.752 g of barium sulphate is ___________ $$\\times$$ 10$$-$$1 g.

    \n

    (Given molar mass of barium sulphate 233 g mol$$-$$1) (Nearest integer).

    ", "options": [], "answer": "2", "solution": "**Answer:** 2\n\n$233 \\mathrm{~g}\\, \\mathrm{BaSO}_{4}$ contains $=32 \\mathrm{~g}$ sulphur\n

    \n$0.752 \\mathrm{~g}\\, \\mathrm{BaSO}_{4}$ contains $=\\frac{32}{233} \\times 0.752=0.1005 \\mathrm{~g}$\n

    \nSulfur $\\%=\\frac{\\text { Weight of Sulphur }}{\\text { Weight of Compound }} \\times 100$\n

    \n$51.6=\\frac{0.1005}{\\text { Weight of compound }} \\times 100$

    \n$\\Rightarrow$ Weight of compound $=\\frac{0.1005}{51.6} \\times 100=1.9 \\times 10^{-1} \\mathrm{~g}$\n

    \n$$\n\\approx 2 \\times 10^{-1} \\mathrm{~g}\n$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2354, "subject": "Chemistry", "question": "

    In bromination of Propyne, with Bromine, 1, 1, 2, 2-tetrabromopropane is obtained in 27% yield. The amount of 1, 1, 2, 2-tetrabromopropane obtained from 1 g of Bromine in this reaction is ___________ $$\\times$$ 10$$-$$1 g. (Nearest integer)

    \n

    (Molar Mass : Bromine = 80 g/mol)

    ", "options": [], "answer": "3", "solution": "**Answer:** 3\n\n\"JEE
    \n2 moles $\\mathrm{Br}_{2} \\equiv 1$ mole 1,1,2,2-tetrabromopropane $\\frac{1}{160}$ mole $\\mathrm{Br}_{2}$\n

    \n$\\equiv \\frac{1}{2} \\times \\frac{1}{160}$ mole 1,1,2,2-tetrabromopropane\n

    \nBut yield of reaction is only $27 \\%$\n

    \nMoles of 1,1,2,2-tetrabromopropane\n

    \n$$\n=\\frac{1}{2} \\times \\frac{1}{160} \\times \\frac{27}{100}\n$$\n

    \nMolar mass of 1,1,2,2-tetrabromopropane $=360 \\mathrm{~g}$ Mass of 1,1,2,2-tetrabromopropane\n

    \n$$\n\\begin{aligned}\n&=\\frac{1}{2} \\times \\frac{1}{160} \\times \\frac{27}{100} \\times 360 \\mathrm{~g} \\\\\\\\\n&\\approx 3 \\times 10^{-1} \\mathrm{~g}\n\\end{aligned}\n$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2355, "subject": "Chemistry", "question": "In Kjeldahl's method for estimation of nitrogen, $\\mathrm{CuSO}_4$ acts as :", "options": [ { "text": "catalytic agent" }, { "text": "hydrolysis agent" }, { "text": "reducing agent" }, { "text": "oxidising agent" } ], "answer": "catalytic agent", "solution": "**Answer:** catalytic agent\n\n

    The correct option here is :

    \n

    Option A : catalytic agent

    \n\n

    In the Kjeldahl method for the estimation of nitrogen, $\\mathrm{CuSO}_4$ or copper sulfate acts as a catalytic agent. The method consists of three main steps: digestion, neutralization, and titration. During the digestion phase, the organic substance containing nitrogen is heated with concentrated sulfuric acid ($\\mathrm{H_2SO_4}$), which digests or breaks down the organic matrix, releasing the nitrogen in the form of ammonium sulfate ($\\mathrm{(NH_4)_2SO_4}$).

    \n\n

    Organic compound $+\\mathrm{H}_2 \\mathrm{SO}_4 \\xrightarrow{\\mathrm{CuSO}_4}\\left(\\mathrm{NH}_4\\right)_2 \\mathrm{SO}_4$

    \n\n

    Catalysts such as $\\mathrm{CuSO}_4$, along with other salts like potassium sulfate ($\\mathrm{K_2SO_4}$), are added to the reaction mixture to raise the boiling point of the sulfuric acid and to accelerate the oxidation of the organic material. This increases the rate of the digestion process, ensuring that the nitrogen within the sample is fully converted to ammonium ions ($\\mathrm{NH_4}^+$). The catalyst itself is not consumed in the reaction and does not become a part of the product; its role is to increase the reaction rate. After digestion, the mixture is neutralized with a base, and then the liberated ammonia is distilled off and quantitatively measured, typically by titration with a standard acid solution.

    \n\n

    Therefore, $\\mathrm{CuSO}_4$ serves as a catalyst in this chemical analysis technique for quantitatively determining the nitrogen content of organic substances.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2356, "subject": "Chemistry", "question": "

    The technique used for purification of steam volatile water immiscible substances is :

    ", "options": [ { "text": "Fractional distillation\n" }, { "text": "Distillation\n" }, { "text": "Fractional distillation under reduced pressure\n" }, { "text": "Steam distillation" } ], "answer": "Steam distillation", "solution": "**Answer:** Steam distillation\n\n

    The correct option for the purification of steam volatile water-immiscible substances is Option D, Steam distillation.

    \n\n

    Steam distillation is a separation technique used to purify or isolate temperature-sensitive materials, such as natural aromatic compounds, that are volatile in steam at a temperature that is lower than their decomposition temperatures. It exploits the fact that immiscible liquids can be distilled at a lower temperature than the boiling points of the individual components.

    \n\n

    Here is how steam distillation works:

    \n\n\n\n

    Steam distillation is different from fractional distillation (Option A), which is used for the separation of a mixture into its component parts, or fractions, of compounds that are miscible and have different boiling points. Fractional distillation often involves the use of a fractionating column, which brings more efficiency in separating substances with close boiling points.

    \n\n

    Simple distillation (Option B) is suitable for separating a liquid from impurities or for separating liquids with significantly different boiling points (usually the difference in boiling points should be at least $25^\\circ C$).

    \n\n

    On the other hand, fractional distillation under reduced pressure (Option C) is used when a liquid has a very high boiling point or might decompose at high temperatures. By reducing the pressure, the boiling point of the liquid is lowered and it can be distilled at a temperature much lower than its normal boiling point, thus avoiding decomposition.

    \n\n

    Therefore, for volatile substances that are not miscible with water, steam distillation remains the ideal choice.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2357, "subject": "Chemistry", "question": "

    The fragrance of flowers is due to the presence of some steam volatile organic compounds called essential oils. These are generally insoluble in water at room temperature but are miscible with water vapour in vapour phase. A suitable method for the extraction of these oils from the flowers is -

    ", "options": [ { "text": "distillation\n" }, { "text": "steam distillation\n" }, { "text": "distillation under reduced pressure\n" }, { "text": "crystallisation" } ], "answer": "steam distillation\n", "solution": "**Answer:** steam distillation\n\n\n

    Steam distillation technique is applied to separate substances which are steam volatile and are immiscible with water.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2358, "subject": "Chemistry", "question": "

    Match List I with List II

    \n

    \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    LIST I
    (Test)    		LIST II
    (Observation)
    A.$$\\mathrm{Br_2}$$ water testI.Yellow orange or orange red precipitate formed
    B.\tCeric ammonium nitrate testII.Reddish orange colour disappears
    C.Ferric chloride testIII.Red colour appears
    D.2, 4 - DNP testIV.Blue, Green, Violet or Red colour appear

    \n

    Choose the correct answer from the options given below:

    ", "options": [ { "text": "A-II, B-III, C-IV, D-I\n" }, { "text": "A-III, B-IV, C-I, D-II\n" }, { "text": "A-I, B-II, C-III, D-IV\n" }, { "text": "A-IV, B-I, C-II, D-III" } ], "answer": "A-II, B-III, C-IV, D-I\n", "solution": "**Answer:** A-II, B-III, C-IV, D-I\n\n\n

    Let's match the given tests in List I with the correct observations in List II based on known chemical reactions and their typical outcomes:

    \n\n

    List I with List II

    \n\nA. $\\mathrm{Br_2}$ water test\n\n

    \n
    B. Ceric ammonium nitrate test\n\n

    \n
    C. Ferric chloride test\n\n

    \n
    D. 2,4-DNP test\n\n

    \n

    Conclusion:

    \n\n

    Based on these observations, the correct matching of tests with outcomes should be:

    \n\n\n

    This makes the correct choice:

    \n\nOption A: A-II, B-III, C-IV, D-I", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 2359, "subject": "Chemistry", "question": "

    Which of the following statements are correct?

    \n

    A. Glycerol is purified by vacuum distillation because it decomposes at its normal boiling point.

    \n

    B. Aniline can be purified by steam distillation as aniline is miscible in water.

    \n

    C. Ethanol can be separated from ethanol water mixture by azeotropic distillation because it forms azeotrope.

    \n

    D. An organic compound is pure, if mixed M.P. is remained same.

    \n

    Choose the most appropriate answer from the options given below :

    ", "options": [ { "text": "A, C, D only\n" }, { "text": "B, C, D only\n" }, { "text": "A, B, D only\n" }, { "text": "A, B, C only" } ], "answer": "A, C, D only\n", "solution": "**Answer:** A, C, D only\n\n\n

    Statements A, C and D are correct.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2360, "subject": "Chemistry", "question": "For standardizing NaOH solution, which of the following is used as a primary standard ? ", "options": [ { "text": "Ferrous Ammonium Sulfate" }, { "text": "dil. HCl" }, { "text": "Oxalic acid " }, { "text": "Sodium tetraborate " } ], "answer": "Oxalic acid ", "solution": "**Answer:** Oxalic acid \n\n

    The solution that can be prepared directly by dissolving an accurately weighed amount of the substance in water and making up the volume to a known amount by water is known as primary standard solution. Oxalic acid is an example of primary standard while NaOH is a secondary standard solution. Hence, strength of NaOH solution can be determined by titrating against oxalic acid. This process is known as standardisation.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2361, "subject": "Chemistry", "question": "Consider titration of NaOH solution versus 1.25 M oxalic acid solution. At the end point following burette readings were obtained.

    (i) 4.5 mL

    (ii) 4.5 mL

    (iii) 4.4 mL

    (iv) 4.4 mL

    (v) 4.4 mL

    If the volume of oxalic acid taken was 10.0 mL then the molarity of the NaOH solution is ________ M. (Rounded off to the nearest integer)", "options": [], "answer": "6", "solution": "**Answer:** 6\n\n

    Average burette reading = Volume of NaOH solution (V1)

    \n

    $$ = {{4.5 + 4.5 + 4.4 + 4.4 + 4.4} \\over 5}$$

    \n

    = 4.44 mL

    \n

    Strength of NaOH solution = S1(M) (say) = S1(N)

    \n

    Volume of oxalic acid solution (V2) = 10 mL

    \n

    Strength of oxalic acid solution (S2) = 1.25 M = 1.25 $$\\times$$ 2 N

    \n

    So, $${V_1}{S_1} = {V_2}{S_2}$$ ($$\\because$$ Law of equivalence)

    \n

    $$ \\Rightarrow {S_1} = {{{V_2}{S_2}} \\over {{V_1}}} = {{10 \\times (1.25 \\times 2)} \\over {4.44}} = 5.63$$ N

    \n

    $$ \\simeq 6M = 6M$$

    ", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 2362, "subject": "Chemistry", "question": "When 10 mL of an aqueous solution of KMnO4 was titrated in acidic medium, equal volume of 0.1 M of an aqueous solution of ferrous sulphate was required for complete discharge of colour. The strength of KMnO4 in grams per litre is __________ $$\\times$$ 10$$-$$2. (Nearest integer)

    [Atomic mass of K = 39, Mn = 55, O = 16]", "options": [], "answer": "316", "solution": "**Answer:** 316\n\nLet molarity of KMnO4 = x

    \"JEE
    (Equivalents of KMnO4 reacted) = (Equivalents of FeSO4 reacted)

    $$\\Rightarrow$$ (5 $$\\times$$ x $$\\times$$ 10 ml) = 1 $$\\times$$ 0.1 $$\\times$$ 10 ml

    $$\\Rightarrow$$ x = 0.02 M

    Molar mass of KMnO4 = 158 gm/mol

    $$\\Rightarrow$$ Strength = (x $$\\times$$ 158) = 3.16 g/l", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 2363, "subject": "Chemistry", "question": "

    A 2.0 g sample containing MnO2 is treated with HCl liberating Cl2. The Cl2 gas is passed into a solution of KI and 60.0 mL of 0.1 M Na2S2O3 is required to titrate the liberated iodine. The percentage of MnO2 in the sample is _____________. (Nearest integer)

    \n

    [Atomic masses (in u) Mn = 55; Cl = 35.5; O = 16, I = 127, Na = 23, K = 39, S = 32]

    ", "options": [], "answer": "13", "solution": "**Answer:** 13\n\n

    First Step :

    \n

    MnO2 + 4HCl $$\\to$$ MnCl2 + Cl2 + 2H2O

    \n

    Here 1 mol of MnO2 produce 1 mol of Cl2

    \n

    $$\\therefore$$ Mole ratio of $${n_{Mn{O_2}}}:{n_{C{l_2}}} = 1:1$$

    \n

    Second Step :

    \n

    Cl2 + 2KI $$\\to$$ 2KCl + I2

    \n

    Here, 1 mol of Cl2 produce 1 mol of I2

    \n

    Mole ratio of $${n_{C{l_2}}}:{n_{{I_2}}} = 1:1$$

    \n

    Third Step :

    \n

    I2 + 2Na2S2O3 $$\\to$$ 2NaI + Na2S2O3

    \n

    1 mol of I2 react with 2 mol of Na2S2O3

    \n

    Mole ratio of $${n_{{I_2}}}:{n_{{N_2}{S_2}{O_3}}} = 1:2$$

    \n

    Given Na2S2O3 is 60 mL of 0.1 M

    \n

    $$\\therefore$$ Number of moles of Na2S2O3

    \n

    = V (in L) $$\\times$$ M (Molarity)

    \n

    = $${{60} \\over {1000}} \\times 0.1$$

    \n

    = 0.006 mol

    \n

    $$\\therefore$$ Number of moles of I2

    \n

    $$ = {1 \\over 2}(0.006)$$

    \n

    = 0.003

    \n

    $$\\therefore$$ Moles of MnO2 = 0.003 (as mole ratio of MnO2 and Cl2 = 1 : 1)

    \n

    Molar mass of MnO2 = 55 + 32 = 87

    \n

    $$\\therefore$$ Mass of MnO2 = 0.003 $$\\times$$ 87 = 0.261 gm

    \n

    Given MnO2 = 2g

    \n

    $$\\therefore$$ % of MnO2 = $${{0.261} \\over 2} \\times 100 = 13\\% $$

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2364, "subject": "Chemistry", "question": "

    In base vs. acid titration, at the end point methyl orange is present as

    ", "options": [ { "text": "quinonoid form" }, { "text": "heterocyclic form" }, { "text": "phenolic form" }, { "text": "benzenoid form" } ], "answer": "quinonoid form", "solution": "**Answer:** quinonoid form\n\n

    \"JEE

    \n

    Hence at the end point methyl orange is present as\nquinonoid form.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2365, "subject": "Chemistry", "question": "

    $$\\mathrm{KMnO}_{4}$$ is titrated with ferrous ammonium sulphate hexahydrate in presence of dilute $$\\mathrm{H}_{2} \\mathrm{SO}_{4}$$. Number of water molecules produced for 2 molecules of $$\\mathrm{KMnO}_{4}$$ is ___________.

    ", "options": [], "answer": "68", "solution": "**Answer:** 68\n\nThe balanced redox equation considering the ferrous ammonium sulfate hexahydrate is as follows:\n

    \n$$\n10\\left[\\mathrm{FeSO}_4 \\cdot\\left(\\mathrm{NH}_4\\right)_2 \\mathrm{SO}_4 \\cdot 6 \\mathrm{H}_2 \\mathrm{O}\\right]+2 \\mathrm{KMnO}_4+8 \\mathrm{H}_2 \\mathrm{SO}_4 \\rightarrow \n$$

    \n     $5 \\mathrm{Fe}_2\\left(\\mathrm{SO}_4\\right)_3+2 \\mathrm{MnSO}_4+10\\left(\\mathrm{NH}_4\\right)_2 \\mathrm{SO}_4+\\mathrm{K}_2 \\mathrm{SO}_4+68 \\mathrm{H}_2 \\mathrm{O}$\n

    \nNow, we are asked to find the number of water molecules produced for 2 molecules of KMnO₄.\n

    \nFrom the balanced equation, we can observe that 2 moles of KMnO₄ produce 68 moles of H₂O. Therefore, for 2 molecules of KMnO₄, the number of water molecules produced will be:\n

    \n2 molecules × 34 water molecules per molecule of KMnO₄ = 68 water molecules\n

    \nSo, 68 water molecules are produced for 2 molecules of KMnO₄ in this titration.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2366, "subject": "Chemistry", "question": "

    Given below are two statements:

    \n

    Statement I : Aqueous solution of K$$_2$$Cr$$_2$$O$$_7$$ is preferred as a primary standard in volumetric analysis over Na$$_2$$Cr$$_2$$O$$_7$$ aqueous solution.

    \n

    Statement II : K$$_2$$Cr$$_2$$O$$_7$$ has a higher solubility in water than Na$$_2$$Cr$$_2$$O$$_7$$.

    \n

    In the light of the above statements, choose the correct answer from the options given below:

    ", "options": [ { "text": "Statement I is true but Statement II is false" }, { "text": "Both Statement I and Statement II are false" }, { "text": "Statement I is false but Statement II is true" }, { "text": "Both Statement I and Statement II are true" } ], "answer": "Statement I is true but Statement II is false", "solution": "**Answer:** Statement I is true but Statement II is false\n\n

    Statement I: Potassium dichromate (K$_2$Cr$_2$O$_7$) is indeed commonly used as a primary standard in volumetric (titrimetric) analysis. It is a powerful oxidizing agent and has a high equivalent weight. Furthermore, it is readily available, it can be obtained in a high state of purity, it is not hygroscopic, and its solutions are stable. However, the reason it is preferred over sodium dichromate (Na$_2$Cr$_2$O$_7$) is not due to its solubility, but rather its stability and molar mass.

    \n

    Statement II: Typically, the solubility of salts increases with the size of the cation. Sodium ions (Na$^+$) are smaller than potassium ions (K$^+$), so sodium salts tend to be more soluble in water than corresponding potassium salts. Thus, we would expect sodium dichromate (Na$_2$Cr$_2$O$_7$) to be more soluble in water than potassium dichromate (K$_2$Cr$_2$O$_7$).

    \n

    Therefore, the correct answer is Option A: Statement I is true but Statement II is false.

    \n", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 2367, "subject": "Chemistry", "question": "

    Given below are two statements:

    \n

    Statement I : Methyl orange is a weak acid.

    \n

    Statement II : The benzenoid form of methyl orange is more intense/deeply coloured than the quinonoid form.

    \n

    In the light of the above statement, choose the most appropriate answer from the options given below:

    ", "options": [ { "text": "Both Statement I and Statement II are incorrect" }, { "text": "Statement $$\\mathrm{I}$$ is incorrect but Statement II is correct" }, { "text": "Both statement I and Statement II are correct" }, { "text": "Statement I is correct but Statement II is incorrect" } ], "answer": "Both Statement I and Statement II are incorrect", "solution": "**Answer:** Both Statement I and Statement II are incorrect\n\n

    Statement I: Methyl orange is a weak base, not a weak acid.

    \n

    This is because Methyl orange is actually a pH indicator commonly used in titrations. In its deprotonated form, it appears yellow, and in its protonated form, it appears red. Methyl orange has the ability to accept a proton ($H^+$) from water (a characteristic of a base) and is therefore considered a weak base.

    \n

    Chemically, this can be represented as:

    \n

    $$\\text{MO}^- + \\text{H}_2\\text{O} \\leftrightharpoons \\text{MOH} + \\text{OH}^-$$

    \n

    Here, $\\text{MO}^-$ is the methyl orange anion, $\\text{MOH}$ is the protonated form of methyl orange, and $\\text{OH}^-$ is the hydroxide ion.

    \n

    Statement II: The quinonoid form of methyl orange is more intensely colored than the benzenoid form.

    \n

    Methyl orange exists in two resonance forms: a benzenoid form (yellow) and a quinonoid form (red). In acidic solution, the equilibrium shifts towards the quinonoid form, which is red and more intensely colored than the yellow benzenoid form.

    \n

    The equilibrium can be represented as follows:

    \n

    Yellow Benzenoid form $\\leftrightharpoons$ Red Quinonoid form

    \n

    In conclusion, both statements are incorrect. Methyl orange is a weak base and the quinonoid form is more intensely colored. So, the correct answer is option A. Both Statement I and Statement II are incorrect.

    \n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2368, "subject": "Chemistry", "question": "Following Kjeldahl's method, $1 \\mathrm{~g}$ of organic compound released ammonia, that neutralised $10 \\mathrm{~mL}$ of $2 \\mathrm{M} ~\\mathrm{H}_2 \\mathrm{SO}_4$. The percentage of nitrogen in the compound is ___________ $\\%$.", "options": [], "answer": "56", "solution": "**Answer:** 56\n\n

    To determine the percentage of nitrogen using the Kjeldahl's method, we first need to find out how much ammonia (NH3) was produced and then calculate the equivalent amount of nitrogen in the sample.

    \n\n

    The ammonia released reacts with sulfuric acid (H2SO4) in the following stoichiometry :

    \n\n$$ 2 \\mathrm{NH}_3 + \\mathrm{H}_2 \\mathrm{SO}_4 \\rightarrow (\\mathrm{NH}_4)_2 \\mathrm{SO}_4 $$\n\n

    Each mole of H2SO4 reacts with 2 moles of NH3. Since 10 mL of 2 M sulfuric acid is neutralized by the ammonia, we can calculate the amount (in moles) of ammonia :

    \n\n$$ n(\\mathrm{NH}_3) = 2 \\times n(\\mathrm{H}_2 \\mathrm{SO}_4) $$\n\n

    Now calculate the moles of H2SO4 used :

    \n\n

    $$ n(\\mathrm{H}_2 \\mathrm{SO}_4) = M \\times V $$

    \n

    $$ n(\\mathrm{H}_2 \\mathrm{SO}_4) = 2 \\, \\mathrm{M} \\times 10 \\, \\mathrm{mL} \\times \\frac{1 \\, \\mathrm{L}}{1000\\,\\mathrm{mL}} $$

    \n

    $$ n(\\mathrm{H}_2 \\mathrm{SO}_4) = 0.02 \\, \\mathrm{mol} $$

    \n\n

    Therefore, the number of moles of NH3 released will be twice that of the moles of H2SO4 neutralized :

    \n\n

    $$ n(\\mathrm{NH}_3) = 2 \\times 0.02 \\, \\mathrm{mol} $$

    \n

    $$ n(\\mathrm{NH}_3) = 0.04 \\, \\mathrm{mol} $$\n

    \n

    Next, we use the molar mass of nitrogen (14 g/mol) to find the mass of nitrogen :

    \n\n

    $$ m(N) = n(N) \\times M(N) $$

    \n

    $$ m(N) = 0.04 \\, \\mathrm{mol} \\times 14 \\, \\mathrm{g/mol} $$

    \n

    $$ m(N) = 0.56 \\, \\mathrm{g} $$

    \n\n

    To find the percentage of nitrogen in the compound, we take the mass of nitrogen divided by the mass of the original sample and multiply by 100% :

    \n\n

    $$ \\mathrm{Percent \\, nitrogen} = \\left( \\frac{m(N)}{m(\\mathrm{sample})} \\right) \\times 100\\% $$

    \n

    $$ \\mathrm{Percent \\, nitrogen} = \\left( \\frac{0.56\\, \\mathrm{g}}{1\\, \\mathrm{g}} \\right) \\times 100\\% $$

    \n

    $$ \\mathrm{Percent \\, nitrogen} = 56\\% $$\n

    \n

    The percentage of nitrogen in the compound is 56%.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2369, "subject": "Chemistry", "question": "Given below are two statements :

    \nStatement (I) : Potassium hydrogen phthalate is a primary standard for standardisation of sodium hydroxide solution.\n

    \nStatement (II) : In this titration phenolphthalein can be used as indicator.

    \nIn the light of the above statements, choose the most appropriate answer from the options given below :\n", "options": [ { "text": "Both Statement I and Statement II are correct" }, { "text": "Statement I is incorrect but Statement II is correct" }, { "text": "Both Statement I and Statement II are incorrect" }, { "text": "Statement I is correct but Statement II is incorrect" } ], "answer": "Both Statement I and Statement II are correct", "solution": "**Answer:** Both Statement I and Statement II are correct\n\n

    Let's examine each statement one by one to choose the most appropriate answer :

    \n\n

    Statement (I) : Potassium hydrogen phthalate is a primary standard for standardisation of sodium hydroxide solution.

    \n

    This statement is correct. Potassium hydrogen phthalate (often abbreviated as KHP, with the chemical formula $$ C_8H_5KO_4 $$) is commonly used as a primary standard in titrations because it is pure, stable, non-hygroscopic, and has a high molar mass. It reacts with sodium hydroxide (NaOH) in a stoichiometric reaction, where one mole of KHP reacts with one mole of NaOH. The equation for the reaction is as follows :

    \n

    $$ C_8H_5KO_4 + NaOH \\rightarrow C_8H_4KO_4Na + H_2O $$

    \n\n

    Statement (II) : In this titration phenolphthalein can be used as indicator.

    \n

    This statement is also correct. Phenolphthalein is a suitable indicator for the titration between KHP and NaOH. It changes color in the pH range of about 8.2 to 10, which is appropriate for the endpoint of a titration involving a weak acid (KHP) and a strong base (NaOH). In an acidic solution, phenolphthalein is colorless, but it turns pink in a basic solution, signaling the end of the titration when slight excess NaOH is present.

    \n\n

    Given that both statements are correct, the most appropriate answer is :

    \n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2370, "subject": "Chemistry", "question": "

    Identify the incorrect statements regarding primary standard of titrimetric analysis.

    \n

    (A) It should be purely available in dry form.

    \n

    (B) It should not undergo chemical change in air.

    \n

    (C) It should be hygroscopic and should react with another chemical instantaneously and stoichiometrically.

    \n

    (D) It should be readily soluble in water.

    \n

    (E) $$\\mathrm{KMnO}_4$$ & $$\\mathrm{NaOH}$$ can be used as primary standard.

    \n

    Choose the correct answer from the options given below :

    ", "options": [ { "text": "(C) and (E) only" }, { "text": "(A) and (B) only" }, { "text": "(B) and (E) only" }, { "text": "(C) and (D) only" } ], "answer": "(C) and (E) only", "solution": "**Answer:** (C) and (E) only\n\n

    Let's evaluate each statement to identify the incorrect ones regarding the primary standard in titrimetric analysis:

    \n\n

    (A) It should be purely available in dry form.

    \n\n

    This statement is generally correct. A primary standard should be pure and available in a stable, dry form to ensure accurate weight measurements.

    \n\n

    (B) It should not undergo chemical change in air.

    \n\n

    This statement is also correct. A primary standard should be stable in air and not react with components of the atmosphere such as oxygen or carbon dioxide.

    \n\n

    (C) It should be hygroscopic and should react with another chemical instantaneously and stoichiometrically.

    \n\n

    This statement is incorrect. A primary standard should not be hygroscopic (it should not absorb moisture from the air) because hygroscopic substances cannot maintain a constant weight. The part about reacting stoichiometrically and instantaneously is generally correct, but being hygroscopic disqualifies a substance from being a primary standard.

    \n\n

    (D) It should be readily soluble in water.

    \n\n

    This statement is correct. A primary standard should be soluble in water to ensure that it can be dissolved to make a standard solution for titration.

    \n\n

    (E) $$\\mathrm{KMnO}_4$$ & $$\\mathrm{NaOH}$$ can be used as primary standard.

    \n\n

    This statement is incorrect. Both $$\\mathrm{KMnO}_4$$ (potassium permanganate) and $$\\mathrm{NaOH}$$ (sodium hydroxide) are not suitable as primary standards. $$\\mathrm{KMnO}_4$$ is unstable and can decompose, while $$\\mathrm{NaOH}$$ is hygroscopic, which makes precise weight measurements difficult.

    \n\n

    Hence, the correct answer is:

    \n\n

    Option A

    \n\n

    (C) and (E) only

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2371, "subject": "Chemistry", "question": "Which of the following reagents may be used to distinguish between phenol and benzoic acid ?", "options": [ { "text": "Tollen’s reagent" }, { "text": "Molisch reagent" }, { "text": "Neutral FeCl3" }, { "text": "Aqueous NaOH " } ], "answer": "Neutral FeCl3", "solution": "**Answer:** Neutral FeCl3\n\nPhenol gives a violet colour with neutral ferric chloride solution whereas benzoic acid does not give this test.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2372, "subject": "Chemistry", "question": "Sodium extract is heated with concentrated HNO3 before testing for halogens because :\n", "options": [ { "text": "Silver halides are totally insoluble in nitric acid." }, { "text": "Ag2S and AgCN are soluble in acidic medium." }, { "text": "S2− and CN−, if present, are\ndecomposed by conc. HNO3 and hence do not interfere in the test.\n" }, { "text": "Ag reacts faster with halides in acidic medium." } ], "answer": "S2− and CN−, if present, are\ndecomposed by conc. HNO3 and hence do not interfere in the test.\n", "solution": "**Answer:** S2− and CN−, if present, are\ndecomposed by conc. HNO3 and hence do not interfere in the test.\n\n\n

    Sodium fusion extract is heated with dilute HNO3 to decompose NaCN or Na2S (if present) to HCN and H2S gases, respectively.

    \n\"JEE\n\n

    The presence of N or S interferes with the test of halogens by forming precipitate with AgNO3.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2373, "subject": "Chemistry", "question": "The correct match between item 'I' and item 'II' is : \n

    \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    Item 'I' (compound)Item 'II'
    (reagent)
    (A)Lysine(P)1-naphthol
    (B)Furfural(Q)ninhydrin
    (C)Benzylalcohol(R)KMnO4
    (D)Styrene(S)Ceric ammonium
    nitrate
    ", "options": [ { "text": "(A)  $$ \\to $$  (Q);  (B)  $$ \\to $$  (R);  (C)  $$ \\to $$  (S);  \n(D)  $$ \\to $$  (P)" }, { "text": "(A)  $$ \\to $$  (Q);  (B)  $$ \\to $$  (P);  (C)  $$ \\to $$  (S);  \n(D)  $$ \\to $$  (R)" }, { "text": "(A)  $$ \\to $$  (R);  (B)  $$ \\to $$  (P);  (C)  $$ \\to $$  (Q);  \n(D)  $$ \\to $$  (S)" }, { "text": "(A)  $$ \\to $$  (Q);  (B)  $$ \\to $$  (P);  (C)  $$ \\to $$  (R);  \n(D)  $$ \\to $$  (S)" } ], "answer": "(A)  $$ \\to $$  (Q);  (B)  $$ \\to $$  (P);  (C)  $$ \\to $$  (S);  \n(D)  $$ \\to $$  (R)", "solution": "**Answer:** (A)  $$ \\to $$  (Q);  (B)  $$ \\to $$  (P);  (C)  $$ \\to $$  (S);  \n(D)  $$ \\to $$  (R)\n\nLysine is an amino acid which gives nindydrin test.\n

    Furfural has aldehyde group that is why it reacts with 1-napthol to give purple colouration.\n

    Benzyl alcohol undergoes reaction with ceric ammonium nitrate to give pink colouration.\n

    Styrene reacts with KMnO4 and gives benzoic acid and brown MnO2.", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 2374, "subject": "Chemistry", "question": "A, B and C are three biomolecules. The results\nof the tests performed on them are given\nbelow:

    \n$$\n\\begin{array}{|l|l|l|l|}\n\\hline & \\begin{array}{l}\n\\text { Molisch's } \\\\\n\\text { Test }\n\\end{array} & \\begin{array}{l}\n\\text { Barfoed } \\\\\n\\text { Test }\n\\end{array} & \\begin{array}{l}\n\\text { Biuret } \\\\\n\\text { Test }\n\\end{array} \\\\\n\\hline \\text { A } & \\text { Positive } & \\text { Negative } & \\text { Negative } \\\\\n\\hline \\text { B } & \\text { Positive } & \\text { Positive } & \\text { Negative } \\\\\n\\hline \\text { C } & \\text { Negative } & \\text { Negative } & \\text { Positive } \\\\\n\\hline\n\\end{array}\n$$\n

    \nA, B and C are respectively :", "options": [ { "text": "A = Lactose, B = Glucose, C = Albumin" }, { "text": "A = Lactose, B = Fructose, C = Alanine" }, { "text": "A = Lactose, B = Glucose, C = Alanine" }, { "text": "A = Glucose, B = Fructose, C = Albumin" } ], "answer": "A = Lactose, B = Glucose, C = Albumin", "solution": "**Answer:** A = Lactose, B = Glucose, C = Albumin\n\n

    The given table provides the results of three tests: Molisch's test, Barfoed's test, and Biuret test for three different biomolecules A, B, and C.

    \n
      \n

    1. Molisch's Test: Detects carbohydrates. Positive results for A and B indicate that they are carbohydrates.

      \n

      2. \n

    2. Barfoed's Test: Specifically for detecting monosaccharides. A positive result for B indicates that B is a monosaccharide, while A's negative result indicates it's not a monosaccharide, implying it might be a disaccharide.

      \n

      3. \n

    3. Biuret Test: Used to detect the presence of proteins. Positive result for C indicates that it is a protein.

      \n
      4. \n
    \n

    Given the options, we can analyze them as follows:

    \n\n

    Therefore, the correct option is:\nOption A: A = Lactose, B = Glucose, C = Albumin.

    \n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2375, "subject": "Chemistry", "question": "In an estimation of bromine by Carius method,\n1.6 g of an organic compound gave 1.88 g of\nAgBr. The mass percentage of bromine in the\ncompound is _______.\n
    (Atomic mass, Ag = 108, Br = 80 g mol–1)", "options": [], "answer": "50", "solution": "**Answer:** 50\n\nMass of organic compound = 1.6 gm\n

    Mass of AgBr = 1.88 gm\n

    Moles of Br = Moles of AgBr = $${{1.88} \\over {188}}$$ = 0.01\n

    Mass of Br = 0.01 × 80 = 0.80 gm\n

    % of Br = $${{0.80 \\times 100} \\over {1.60}}$$ = 50 %", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2376, "subject": "Chemistry", "question": "Match the following :\n

    \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    Test/MethodReagent
    (i) Lucas Test(a) C6H5SO2Cl/\naq. KOH
    (ii) Dumas method(b) HNO3/\nAgNO3
    (iii) Kjeldahl’s method(c) CuO/CO2
    (iv) Hinsberg test(d) Conc. HCl\nand ZnCl2
    \n(e) H2SO4
    ", "options": [ { "text": "(i)-(b), (ii)-(d), (iii)-(e), (iv)-(a)" }, { "text": "(i)-(d), (ii)-(c), (iii)-(e), (iv)-(a)" }, { "text": "(i)-(b), (ii)-(a), (iii)-(c), (iv)-(d)" }, { "text": "(i)-(d), (ii)-(c), (iii)-(b), (iv)-(e)" } ], "answer": "(i)-(d), (ii)-(c), (iii)-(e), (iv)-(a)", "solution": "**Answer:** (i)-(d), (ii)-(c), (iii)-(e), (iv)-(a)\n\nLucas reagent – Conc. HCl/ZnCl2\n

    Dumas method – CuO/CO2\n

    Kjeldahl’s method – H2SO4\n

    Hinsberg test – C6H5SO2Cl/aq. KOH", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 2377, "subject": "Chemistry", "question": "In Duma's method of estimation of nitrogen, 0.1840 g of an organic compound gave 30 mL of nitrogen collected at 287 K and 758 mm of Hg pressure. The percentage composition of nitrogen in the compound is __________. (Round off to the Nearest Integer). [Given : Aqueous tension at 287 K = 14 mm of Hg]", "options": [], "answer": "19", "solution": "**Answer:** 19\n\nAqueous tension at 287 K = 14 mm of Hg.\n

    Hence actual pressure = (758 – 14)\n
    = 744 mm of Hg.\n

    Moles of $${N_2} = {{\\left( {758 - 14} \\right)} \\over {760}} \\times {{30 \\times {{10}^{ - 3}}} \\over {0.0821 \\times 287}}$$

    $$ = 1.246 \\times {10^{ - 3}}$$ mol

    mass of $${N_2} = 1.246 \\times {10^{ - 3}} \\times 28$$

    mass % of N2 $$ = {{mass\\,of\\,'N'} \\over {total\\,mass}} \\times 100$$

    $$ = {{1.246 \\times 28 \\times {{10}^{ - 3}}} \\over {0.184}} \\times 100$$

    $$ = {{124.6 \\times 28} \\over {0.184}}\\% = 18.96\\% $$

    $$ \\simeq 19\\% $$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2378, "subject": "Chemistry", "question": "Reagent, 1-napthylamine and sulphanilic acid in acetic acid is used for the detection of ", "options": [ { "text": "N2O" }, { "text": "NO$$_3^ - $$" }, { "text": "NO" }, { "text": "NO$$_2^ - $$" } ], "answer": "NO$$_2^ - $$", "solution": "**Answer:** NO$$_2^ - $$\n\n1-naphthyl amine and sulphanilic acid in acetic acid\nis used for the detection of $$NO_2^ - $$.

    \"JEE\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2379, "subject": "Chemistry", "question": "In Carius method, halogen containing organic compound is heated with fuming nitric acid in the presence of :", "options": [ { "text": "HNO3" }, { "text": "AgNO3" }, { "text": "CuSO4" }, { "text": "BaSO4" } ], "answer": "AgNO3", "solution": "**Answer:** AgNO3\n\nOrganic compound is heated with fuming nitric acid in the presence of silver nitrate in carius method.

    Lunar caustic (AgNO3) is used as reagent hare to distinguish Cl$$-$$, Br- and I$$-$$ respectively as follows.

    $$C{l^ - }(aq)\\buildrel {AgN{O_3}} \\over\n \\longrightarrow AgCl{ \\downarrow _{ppt}}$$ white

    $$B{r^ - }(aq)\\buildrel {AgN{O_3}} \\over\n \\longrightarrow AgBr{ \\downarrow _{ppt}}$$ pale yellow

    $${I^ - }(aq)\\buildrel {AgN{O_3}} \\over\n \\longrightarrow AgI{ \\downarrow _{ppt}}$$ Dark yellow", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2380, "subject": "Chemistry", "question": "0.8 g of an organic compound was analyzed by Kjeldahl's method for the estimation of nitrogen. If the percentage of nitrogen in the compound was found to be 42%, then ______________ mL of 1 M H2SO4 would have been neutralized by the ammonia evolved during the analysis.", "options": [], "answer": "12", "solution": "**Answer:** 12\n\nOrganic compound : 0.8 gm

    wt. of N = $$\\left( {{{42} \\over {100}} \\times 0.8} \\right)$$ gm

    mole of N = $${{{42 \\times 0.8} \\over {100 \\times 14}} = {{2.4} \\over {100}}}$$ mol

    moles of NH3 = $${{{2.4} \\over {100}}}$$

    \"JEE

    $${{{1.2} \\over {100}}}$$ = 1 $$\\times$$ V(l)

    $${ \\Rightarrow {V_{{H_2}S{O_4}}} = {{1.2} \\over {100}}l = 12}$$ ml", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 2381, "subject": "Chemistry", "question": "The number of moles of CuO, that will be utilized in Dumas method for estimation nitrogen in a sample of 57.5 g of N, N-dimethylaminopentane is _____________ $$\\times$$ 10$$-$$2. (Nearest integer)", "options": [], "answer": "1125", "solution": "**Answer:** 1125\n\nMoles of N in N, N - dimethylaminopentane = $$\\left( {{{57.5} \\over {115}}} \\right)$$ = 0.5 mol

    $$ \\Rightarrow {C_7}{H_{17}}N + {{45} \\over 2}CuO \\to 7C{O_2} + {{17} \\over 2}{H_2}O + {1 \\over 2}{N_2} + {{45} \\over 2}Cu$$

    $${{{n_{CuO}}\\,reacted} \\over {\\left( {{{45} \\over 2}} \\right)}} = {{{n_{{C_7}{H_{17}}N}}\\,reacted} \\over 1}$$

    $$\\Rightarrow$$ nCuO reacted = $$\\left( {{{45} \\over 2}} \\right) \\times 0.5 = 11.25$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2382, "subject": "Chemistry", "question": "Which one of the following tests used for the identification of functional groups in organic compounds does not use copper reagent?", "options": [ { "text": "Barfoed's test" }, { "text": "Seliwanoff's test" }, { "text": "Benedict's test" }, { "text": "Biuret test for peptide bond" } ], "answer": "Seliwanoff's test", "solution": "**Answer:** Seliwanoff's test\n\nIn Seliwanoff's reagent, Cu is not present.

    In Barfoed, Biuret and in Benediet reagent Cu is present.", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 2383, "subject": "Chemistry", "question": "The transformation occurring in Duma's method is given below :

    $${C_2}{H_7}N + \\left( {2x + {y \\over 2}} \\right)CuO \\to xC{O_2} + {y \\over 2}{H_2}O + {z \\over 2}{N_2} + \\left( {2x + {y \\over 2}} \\right)Cu$$

    The value of y is ______________. (Integer answer)", "options": [], "answer": "7", "solution": "**Answer:** 7\n\n$${C_2}{H_7}N + \\left( {2x + {y \\over 2}} \\right)CuO \\to xC{O_2} + {y \\over 2}{H_2}O + {z \\over 2}{N_2} + \\left( {2x + {y \\over 2}} \\right)Cu$$

    On balancing

    $${C_2}{H_7}N + {{15} \\over 2}CuO \\to 2C{O_2} + {7 \\over 2}{H_2}O + {1 \\over 2}{N_2} + {{15} \\over 2}Cu$$

    On comparing

    y = 7", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 2384, "subject": "Chemistry", "question": "

    Kjeldahl's method was used for the estimation of nitrogen in an organic compound. The ammonia evolved from 0.55 g of the compound neutralised 12.5 mL of 1 M H2SO4 solution. The percentage of nitrogen in the compound is _____________. (Nearest integer)

    ", "options": [], "answer": "64", "solution": "**Answer:** 64\n\nMeq. of $\\mathrm{H}_2 \\mathrm{SO}_4=$ Meq. of $\\mathrm{NH}_3$

    \n12. $5 \\times 1 \\times 2=25$ meq. of $\\mathrm{NH}_3$ $=25$ millimoles of $\\mathrm{NH}_3$

    \nSo Millimoles of ${ }^{\\prime} \\mathrm{N}^{\\prime}=25$

    \nMoles of $\\mathrm{N}^{\\prime}=25 \\times 10^{-3}$

    \nwt. of $N=14 \\times 25 \\times 10^{-3}$

    \n$$\n\\begin{aligned}\n& \\% \\mathrm{~N}=\\frac{14 \\times 25 \\times 10^{-3}}{0.55} \\times 100 \\\\\\\\\n& =63.66 \\\\\\\\\n& \\approx 64 \\%\n\\end{aligned}\n$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2385, "subject": "Chemistry", "question": "

    The formula of the purple colour formed in Laissaigne's test for sulphur using sodium nitroprusside is :

    ", "options": [ { "text": "NaFe[Fe(CN)6]" }, { "text": "Na[Cr(NH3)2(NCS)4]" }, { "text": "Na2[Fe(CN)5(NO)]" }, { "text": "Na4[Fe(CN)5(NOS)]" } ], "answer": "Na4[Fe(CN)5(NOS)]", "solution": "**Answer:** Na4[Fe(CN)5(NOS)]\n\n$\\left.\\mathrm{S}^{2-}+\\left[\\mathrm{Fe}(\\mathrm{CN})_{5} \\mathrm{NO}\\right)\\right]^{2-} \\rightarrow\\left[\\mathrm{Fe}(\\mathrm{CN})_{5}(\\mathrm{NOS})\\right]^{4-}$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2386, "subject": "Chemistry", "question": "

    Given below are two statements :

    \n

    Statement I : In 'Lassaigne's Test', when both nitrogen and sulphur are present in an organic compound, sodium thiocyanate is formed.

    \n

    Statement II : If both nitrogen and sulphur are present in an organic compound, then the excess of sodium used in sodium fusion will decompose the sodium thiocyanate formed to give NaCN and Na2S.

    \n

    In the light of the above statements, choose the most appropriate answer from the options given below :

    ", "options": [ { "text": "Both Statement I and Statement II are correct." }, { "text": "Both Statement I and Statement II are incorrect." }, { "text": "Statement I is correct but Statement II is incorrect." }, { "text": "Statement I is incorrect but Statement II is correct." } ], "answer": "Both Statement I and Statement II are correct.", "solution": "**Answer:** Both Statement I and Statement II are correct.\n\nBoth statement I & statement II are correct

    \n$\\mathrm{NaSCN}+2 \\mathrm{Na} \\rightarrow \\mathrm{NaCN}+\\mathrm{Na}_{2} \\mathrm{~S}$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2387, "subject": "Chemistry", "question": "

    The reagent neutral ferric chloride is used to detect the presence of ______________

    ", "options": [ { "text": "sulphide ion and alcoholic $$-$$OH group." }, { "text": "acetate ion and phenolic $$-$$OH group." }, { "text": "sulphide ion and phenolic $$-$$OH group." }, { "text": "acetate ion and alcoholic $$-$$OH group." } ], "answer": "acetate ion and phenolic $$-$$OH group.", "solution": "**Answer:** acetate ion and phenolic $$-$$OH group.\n\nAcetate ions gives deep red colour ppt on reaction with netural ferric chloride solution.\n

    \n$6 \\mathrm{CH}_{3} \\mathrm{COO}^{-}+3 \\mathrm{Fe}^{3+}+2 \\mathrm{H}_{2} \\mathrm{O} \\longrightarrow {\\left[\\mathrm{Fe}_{3}(\\mathrm{OH})_{2}\\left(\\mathrm{CH}_{3} \\mathrm{COO}\\right)_{6}\\right]^{+} + 2 \\mathrm{H}^{+}}$\n

    \n${\\left[\\mathrm{Fe}_{3}(\\mathrm{OH})_{2}\\left(\\mathrm{CH}_{3} \\mathrm{COO}\\right)_{6}\\right]^{+} 4 \\mathrm{H}_{2} \\mathrm{O} \\longrightarrow} \\left.\\left.\\underset{\\text{Brown red ppt}}{3\\left[\\mathrm{Fe}(\\mathrm{OH})_{2}\\left(\\mathrm{CH}_{3} \\mathrm{COO}\\right)\\right]}+3 \\mathrm{CH}_{3} \\mathrm{COOH}\\right)_{2}\\left(\\mathrm{CH}_{3} \\mathrm{COO}\\right)_{6}\\right]^{+}+2 \\mathrm{H}^{+}$\n

    \nPhenol reacts with freshly prepared ferric chloride solution and gives violet colour complex.\n

    \n$$\n6 \\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{OH}+\\mathrm{FeCl}_{3} \\longrightarrow\\underset{\\text{Violet colour}}{\\left[\\mathrm{Fe}\\left(\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{O}\\right)_{6}\\right]^{3-}}+3 \\mathrm{HCl}+3 \\mathrm{H}^{+}\n$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2388, "subject": "Chemistry", "question": "

    While estimating the nitrogen present in an organic compound by Kjeldahl's method, the ammonia evolved from $$0.25 \\mathrm{~g}$$ of the compound neutralized $$2.5 \\mathrm{~mL}$$ of $$2 \\,\\mathrm{M} \\,\\mathrm{H}_{2} \\mathrm{SO}_{4}$$. The percentage of nitrogen present in organic compound is ______________.

    ", "options": [], "answer": "56", "solution": "**Answer:** 56\n\n$$\\mathrm{NH}_{3}$$ gas is neutralized by $$2.5 \\mathrm{~mL}$$ of $$2 \\mathrm{M} \\,\\,\\mathrm{H}_{2} \\mathrm{SO}_{4}$$\n

    \n$$\\therefore$$ Moles of $$\\mathrm{NH}_{3}$$ neutralized $$=2.5 \\times 2 \\times 2$$ millimole\n

    \n$$\n=10 \\times 10^{-3} \\text { moles }\n$$\n

    \n$$\\therefore$$ Weight of $$\\mathrm{N}$$ present in the compound will be\n

    \n$$\n\\begin{aligned}\n&=10 \\times 10^{-3} \\times 14 \\\\\n&=0.14 \\mathrm{~g}\n\\end{aligned}\n$$\n

    \n$$\\therefore \\%$$ of ' $$\\mathrm{N}$$ ' in compound\n

    \n$$\n\\begin{aligned}\n&=\\frac{0.14}{0.25} \\times 100 \\\\\n&=56 \\%\n\\end{aligned}\n$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2389, "subject": "Chemistry", "question": "

    A sample of $$0.125 \\mathrm{~g}$$ of an organic compound when analyzed by Duma's method yields $$22.78 \\mathrm{~mL}$$ of nitrogen gas collected over $$\\mathrm{KOH}$$ solution at $$280 \\mathrm{~K}$$ and $$759 \\mathrm{~mm}\\, \\mathrm{Hg}$$. The percentage of nitrogen in the given organic compound is __________. (Nearest integer)

    \n

    Given :

    \n

    (a) The vapour pressure of water of $$280 \\mathrm{~K}$$ is $$14.2 \\mathrm{~mm} \\,\\mathrm{Hg}$$.

    \n

    (b) $$\\mathrm{R}=0.082 \\mathrm{~L}$$ atm $$\\mathrm{K}^{-1} \\mathrm{~mol}^{-1}$$

    ", "options": [], "answer": "22", "solution": "**Answer:** 22\n\n$P_{\\text {actual }}=759-14.2=744.8 \\,\\mathrm{mmHg}$\n

    \n$$\n\\begin{aligned}\n\\mathrm{n}_{\\mathrm{N}_{2}} &=\\frac{744.8 \\times 22.78}{760 \\times 0.0821 \\times 280 \\times 1000} \\\\\\\\\n&=0.000971 \\mathrm{~mol}\n\\end{aligned}\n$$\n

    \nMass of $\\mathrm{N}_{2}=0.02719 \\,\\mathrm{gm}$\n

    \nPercentage of nitrogen\n

    \n$$\n=\\frac{0.0271}{0.125} \\times 100=21.75 \\simeq 22\n$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2390, "subject": "Chemistry", "question": "

    Given below are two statements :

    \n

    Statement I : Sulphanilic acid gives esterification test for carboxyl group.

    \n

    Statement II : Sulphanilic acid gives red colour in Lassigne's test for extra element detection.

    \n

    In the light of the above statements, choose the most appropriate answer from the options given below :

    ", "options": [ { "text": "Both Statement I and Statement II are correct" }, { "text": "Statement I is correct but Statement II is incorrect" }, { "text": "Both Statement I and Statement II are incorrect" }, { "text": "Statement I is incorrect but Statement II is correct" } ], "answer": "Statement I is incorrect but Statement II is correct", "solution": "**Answer:** Statement I is incorrect but Statement II is correct\n\n

    Sulphanilic acid is p-amino benzene sulphonic acid

    \n

    \"JEE

    \n

    Since it contain both N and S so it give red colour in Lassaigne's test.

    ", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 2391, "subject": "Chemistry", "question": "In Dumas method for the estimation of $\\mathrm{N}_{2}$, the sample is heated with copper oxide and the gas evolved is passed over :", "options": [ { "text": "Copper oxide" }, { "text": "Copper gauze" }, { "text": "$\\mathrm{Pd}$" }, { "text": "$\\mathrm{Ni}$" } ], "answer": "Copper gauze", "solution": "**Answer:** Copper gauze\n\nIn dumas method for the estimation of N2, the\nsample is heated with copper oxide and the gas\nevolved is passed over copper gauze.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2392, "subject": "Chemistry", "question": "

    Compound that will give positive Lassaigne's test for both nitrogen and halogen is :

    ", "options": [ { "text": "$$\\mathrm{NH_2OH.HCl}$$" }, { "text": "$$\\mathrm{N_2H_4.HCl}$$" }, { "text": "$$\\mathrm{CH_3NH_2.HCl}$$" }, { "text": "$$\\mathrm{NH_4Cl}$$" } ], "answer": "$$\\mathrm{CH_3NH_2.HCl}$$", "solution": "**Answer:** $$\\mathrm{CH_3NH_2.HCl}$$\n\n$\\mathrm{CH}_{3} \\mathrm{NH}_{2} \\cdot \\mathrm{HCl}$ will give positive Lassaigne's test for both nitrogen and halogen.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2393, "subject": "Chemistry", "question": "

    $$0.400 \\mathrm{~g}$$ of an organic compound $$(\\mathrm{X})$$ gave $$0.376 \\mathrm{~g}$$ of $$\\mathrm{AgBr}$$ in Carius method for estimation of bromine. $$\\%$$ of bromine in the compound $$(\\mathrm{X})$$ is ___________.

    \n

    (Given: Molar mass $$\\mathrm{AgBr=188~g~mol^{-1}}$$

    \n

    $$\\mathrm{Br}=80 \\mathrm{~g} \\mathrm{~mol}^{-1}$$)

    ", "options": [], "answer": "40", "solution": "**Answer:** 40\n\nTo calculate the percentage of bromine in the compound X, we first need to determine the moles of AgBr formed and then the moles of bromine present in the compound. Finally, we can find the percentage of bromine by weight.\n

    \n1. Calculate the moles of AgBr formed:\n

    \nMoles of AgBr = mass / molar mass = 0.376 g / 188 g/mol = 0.002 mol\n

    \nSince 1 mol of AgBr contains 1 mol of Br, the moles of Br in the compound X are also 0.002 mol.\n

    \n2. Calculate the mass of bromine in the compound:\n

    \nMass of Br = moles of Br × molar mass of Br = 0.002 mol × 80 g/mol = 0.16 g\n

    \n3. Calculate the percentage of bromine in the compound X:\n

    \nPercentage of Br = (mass of Br / mass of compound X) × 100 = (0.16 g / 0.400 g) × 100 = 40%\n

    \nSo, the percentage of bromine in the compound X is 40%.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2394, "subject": "Chemistry", "question": "

    When a solution of mixture having two inorganic salts was treated with freshly prepared ferrous sulphate in acidic medium, a dark brown ring was formed whereas on treatment with neutral $$\\mathrm{FeCl}_{3}$$, it gave deep red colour which disppeared on boiling and a brown red ppt was formed. The mixture contains :

    ", "options": [ { "text": "SO$$_3^{2-}$$ & C$$_2$$O$$_4^{2-}$$" }, { "text": "$$\\mathrm{CH}_{3} \\mathrm{COO}^{-} \\& ~\\mathrm{NO}_{3}^{-}$$" }, { "text": "$$\\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-} \\& ~\\mathrm{NO}_{3}^{-}$$" }, { "text": "$$\\mathrm{SO}_{3}^{2-} \\& ~\\mathrm{CH}_{3} \\mathrm{COO}^{-}$$" } ], "answer": "$$\\mathrm{CH}_{3} \\mathrm{COO}^{-} \\& ~\\mathrm{NO}_{3}^{-}$$", "solution": "**Answer:** $$\\mathrm{CH}_{3} \\mathrm{COO}^{-} \\& ~\\mathrm{NO}_{3}^{-}$$\n\nThe mixture of salts gives a dark brown ring when treated with ferrous sulphate in acidic medium. This is a characteristic reaction of nitrate ions (NO$_3^-$). The brown ring is due to the formation of a complex between NO and Fe$^{2+}$.\n\n

    Additionally, the mixture gives a deep red color when treated with neutral ferric chloride (FeCl$_3$), which then disappears upon boiling to form a red-brown precipitate. This is a classic reaction of acetate ions (CH$_3$COO$^-$) with Fe$^{3+}$. Initially, the acetate ions form a complex with ferric ions, leading to the deep red color. Upon boiling, hydroxide ions displace some acetate ions in the complex, resulting in the red-brown precipitate.\n\n

    Therefore, the mixture contains acetate (CH$_3$COO$^-$) and nitrate (NO$_3^-$) ions.", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 2395, "subject": "Chemistry", "question": "

    In Carius tube, an organic compound '$$\\mathrm{X}$$' is treated with sodium peroxide to form a mineral acid 'Y'.

    \n

    The solution of $$\\mathrm{BaCl}_{2}$$ is added to '$$\\mathrm{Y}$$' to form a precipitate 'Z'. 'Z' is used for the quantitative estimation of an extra element. '$$\\mathrm{X}$$' could be

    ", "options": [ { "text": "Cytosine" }, { "text": "Methionine" }, { "text": "Chloroxylenol" }, { "text": "A nucleotide" } ], "answer": "Methionine", "solution": "**Answer:** Methionine\n\n$\\mathrm{X} \\stackrel{\\mathrm{Na}_2 \\mathrm{O}_2}{\\longrightarrow} \\mathrm{Y} \\stackrel{\\mathrm{BaCl}_2}{\\longrightarrow} \\underset{\\left[\\mathrm{BaSO}_4\\right]}{\\mathrm{Z}}$

    \nMethionine: $\\mathrm{C}_5 \\mathrm{H}_{11} \\mathrm{NO}_2 \\mathrm{S}$

    \n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2396, "subject": "Chemistry", "question": "

    A compound '$$\\mathrm{X}$$' when treated with phthalic anhydride in presence of concentrated $$\\mathrm{H}_{2} \\mathrm{SO}_{4}$$ yields '$$\\mathrm{Y}$$'. '$$\\mathrm{Y}$$' is used as an acid/base indicator. '$$\\mathrm{X}$$' and '$$\\mathrm{Y}$$' are respectively

    ", "options": [ { "text": "Toludine, Phenolphthalein" }, { "text": "Carbolic acid, Phenolphthalein" }, { "text": "Salicylaldehyde, Phenolphthalein" }, { "text": "Anisole, methyl orange" } ], "answer": "Carbolic acid, Phenolphthalein", "solution": "**Answer:** Carbolic acid, Phenolphthalein\n\nPhenolphthalein is a commonly used acid/base indicator that can be synthesized from phthalic anhydride and phenol (carbolic acid) in the presence of concentrated sulfuric acid (H₂SO₄). Therefore, compound 'X' is phenol and compound 'Y' is phenolphthalein.

    \n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2397, "subject": "Chemistry", "question": "

    Match List I with List II

    \n

    \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    LIST I
    Element detected    		LIST II
    Reagent used / Product formed
    A.NitrogenI.$$\\mathrm{Na_2[Fe(CN)_5NO]}$$
    B.SulphurII.$$\\mathrm{AgNO_3}$$
    C.PhosphorousIII.$$\\mathrm{Fe_4[Fe(CN)_6]_3}$$
    D.HalogenIV.$$\\mathrm{(NH_4)_2MoO_4}$$

    \n

    Choose the correct answer from the options given below:

    ", "options": [ { "text": "A-III, B-I, C-IV, D-II" }, { "text": "A-II, B-I, C-IV, D-III" }, { "text": "A-IV, B-II, C-I, D-III" }, { "text": "A-II, B-IV, C-I, D-III" } ], "answer": "A-III, B-I, C-IV, D-II", "solution": "**Answer:** A-III, B-I, C-IV, D-II\n\n

    Nitrogen Detection (Lassaigne's Test)

    \n

    Nitrogen in organic compounds is detected by converting it into sodium cyanide, which further reacts with iron(II) sulfate to form sodium hexacyanoferrate(II). By reacting this with iron(III) ions, Prussian blue is formed:

    \n

    $\n\\begin{aligned}\n&\\mathrm{Na}+\\mathrm{C}+\\mathrm{N} \\rightarrow \\mathrm{NaCN} \\\\\\\\\n&6 \\mathrm{NaCN}+\\mathrm{FeSO}_4 \\rightarrow \\mathrm{Na}_4\\left[\\mathrm{Fe}(\\mathrm{CN})_6\\right]+\\mathrm{Na}_2 \\mathrm{SO}_4 \\\\\\\\\n&\\mathrm{Na}_4\\left[\\mathrm{Fe}(\\mathrm{CN})_6\\right]+\\mathrm{Fe}^{3+} \\rightarrow \\mathrm{Fe}_4\\left[\\mathrm{Fe}(\\mathrm{CN})_6\\right]_3 \\text{ (Prussian blue)}\n\\end{aligned}\n$

    \n

    Sulphur Detection

    \n

    Sulphur can be detected using Sodium nitroprusside:

    \n

    $\n\\mathrm{Na}_2\\left[\\mathrm{Fe}(\\mathrm{CN})_5 \\mathrm{NO}\\right]+\\mathrm{Na}_2 \\mathrm{S} \\rightarrow \\mathrm{Na}_4\\left[\\mathrm{Fe}(\\mathrm{CN})_5 \\mathrm{NOS}\\right] \\text{ [Purple]}\n$

    \n

    Phosphorous Detection

    \n

    Phosphorus detection involves reaction with ammonium molybdate to form a canary yellow compound:

    \n

    $\n\\begin{aligned}\n&\\mathrm{Na}_3 \\mathrm{PO}_4+3 \\mathrm{HNO}_3 \\rightarrow \\mathrm{H}_3 \\mathrm{PO}_4+3 \\mathrm{NaNO}_3 \\\\\\\\\n&\\mathrm{H}_3 \\mathrm{PO}_4+12\\left(\\mathrm{NH}_4\\right)_2 \\mathrm{MoO}_4+21 \\mathrm{HNO}_3 \\rightarrow\\\\\\\\& \\left(\\mathrm{NH}_4\\right)_3 \\mathrm{PO}_4 \\cdot 12 \\mathrm{MoO}_3+21 \\mathrm{NH}_4 \\mathrm{NO}_3+12 \\mathrm{H}_2 \\mathrm{O} \\text{ (canary yellow)}\n\\end{aligned}\n$

    \n

    Halogen Detection

    \n

    Halogens react with silver nitrate to form specific colored precipitates:

    \n

    $\n\\begin{aligned}\n&\\mathrm{NaCl}+\\mathrm{AgNO}_3(\\mathrm{aq}) \\rightarrow \\mathrm{AgCl}+\\mathrm{NaNO}_3 \\text{ (White)} \\\\\\\\\n&\\mathrm{NaBr}+\\mathrm{AgNO}_3(\\mathrm{aq}) \\rightarrow \\mathrm{AgBr}+\\mathrm{NaNO}_3 \\text{ (Pale yellow)} \\\\\\\\\n&\\mathrm{NaI}+\\mathrm{AgNO}_3(\\mathrm{aq}) \\rightarrow \\mathrm{AgI}+\\mathrm{NaNO}_3 \\text{ (Yellow)}\n\\end{aligned}\n$

    \n

    Based on the above reactions, the correct matching is:

    \nOption A

    \nA-III (Nitrogen), B-I (Sulphur), C-IV (Phosphorous), D-II (Halogen).

    \n", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 2398, "subject": "Chemistry", "question": "

    Formation of which complex, among the following, is not a confirmatory test of $$\\mathrm{Pb}^{2+}$$ ions :

    ", "options": [ { "text": "lead chromate" }, { "text": "lead iodide" }, { "text": " lead sulphate" }, { "text": "lead nitrate" } ], "answer": "lead nitrate", "solution": "**Answer:** lead nitrate\n\n

    A confirmatory test for $Pb^{2+}$ ions involves the formation of a specific compound that indicates the presence of lead ions in a solution.

    \n

    Let's go through the options:

    \n

    Option A: Lead chromate ($\\mathrm{PbCrO}_4$) forms a yellow precipitate, which is a confirmatory test for $Pb^{2+}$ ions.

    \n

    Option B: Lead iodide ($\\mathrm{PbI}_2$) forms a yellow precipitate, which is a confirmatory test for $Pb^{2+}$ ions.

    \n

    Option C: Lead sulfate ($\\mathrm{PbSO}_4$) forms a white precipitate, which is a confirmatory test for $Pb^{2+}$ ions.

    \n

    Option D: Lead nitrate ($\\mathrm{Pb(NO_3)_2}$) is soluble in water and does not form a precipitate under normal conditions. Thus, its formation is not used as a confirmatory test for $Pb^{2+}$ ions.

    \n

    So, the correct answer is Option D: lead nitrate.

    \n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2399, "subject": "Chemistry", "question": "

    Match List I with List II

    \n

    \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    LIST I
    (Test)    		LIST II
    (Identification)
    A.Bayer's testI.Phenol
    B.\tCeric ammonium nitrate testII.Aldehyde
    C.Phthalein dye testIII.Alcoholic-OH group
    D.Schiff's testIV.Unsaturation

    \n

    Choose the correct answer from the options given below :

    ", "options": [ { "text": "(A)-(IV), (B)-(I), (C)-(II), (D)-(III)\n" }, { "text": "(A)-(III), (B)-(I), (C)-(IV), (D)-(II)\n" }, { "text": "(A)-(II), (B)-(III), (C)-(IV), (D)-(I)\n" }, { "text": "(A)-(IV), (B)-(III), (C)-(I), (D)-(II)" } ], "answer": "(A)-(IV), (B)-(III), (C)-(I), (D)-(II)", "solution": "**Answer:** (A)-(IV), (B)-(III), (C)-(I), (D)-(II)\n\n

    To match the tests in List I with the corresponding identifications in List II, let's analyze each test:

    \n\n

    \nBayer's test: This test is used to test for unsaturation in compounds, such as alkenes and alkynes. The presence of unsaturation is indicated by the decolorization of potassium permanganate solution.\n

    \n\n

    \nCeric ammonium nitrate test: This test is used to identify alcohols. A positive result is indicated by a color change.\n

    \n\n

    \nPhthalein dye test: This test is used to identify phenols. Phenols react with phthalic anhydride to form a colored compound.\n

    \n\n

    \nSchiff's test: This test is used for the detection of aldehydes. A positive result indicates the formation of a magenta or pink colored compound.\n

    \n\n

    Based on the information above, the correct matching is:

    \n\n\n\n

    Thus, the correct answer is:

    \n\n

    Option D: (A)-(IV), (B)-(III), (C)-(I), (D)-(II)

    ", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 2400, "subject": "Chemistry", "question": "

    Chromatographic technique/s based on the principle of differential adsorption is / are

    \n

    A. Column chromatography

    \n

    B. Thin layer chromatography

    \n

    C. Paper chromatography

    \n

    Choose the most appropriate answer from the options given below:

    ", "options": [ { "text": "B only" }, { "text": "A only" }, { "text": "A & B only" }, { "text": "C only" } ], "answer": "A & B only", "solution": "**Answer:** A & B only\n\n

    Memory Based

    ", "topic": "Discrete Mathematics", "subtopic": "Combinatorics" }, { "id": 2401, "subject": "Chemistry", "question": "

    On a thin layer chromatographic plate, an organic compound moved by $$3.5 \\mathrm{~cm}$$, while the solvent moved by $$5 \\mathrm{~cm}$$. The retardation factor of the organic compound is ________ $$\\times 10^{-1}$$.

    ", "options": [], "answer": "7", "solution": "**Answer:** 7\n\n

    Retardation factor $$=\\frac{\\text { Distance travelled by sample/organic compound }}{\\text { Distance travelled by solvent }}$$

    \n

    $$=\\frac{3.5}{5}=7 \\times 10^{-1}$$

    ", "topic": "Number Theory", "subtopic": "Divisibility / Factorization" }, { "id": 2402, "subject": "Chemistry", "question": "

    The correct statement among the following, for a \"chromatography\" purification method is :

    ", "options": [ { "text": "$$\\mathrm{R}_f$$ of a polar compound is smaller than that of a non-polar compound.\n" }, { "text": "Non-polar compounds are retained at top and polar compounds come down in column chromatography.\n" }, { "text": "Organic compounds run faster than solvent in the thin layer chromatographic plate.\n" }, { "text": "$$\\mathrm{R}_f$$ is an integral value." } ], "answer": "$$\\mathrm{R}_f$$ of a polar compound is smaller than that of a non-polar compound.\n", "solution": "**Answer:** $$\\mathrm{R}_f$$ of a polar compound is smaller than that of a non-polar compound.\n\n\n

    Organic compounds run slower than solvent in thin layer chromatography. $$\\mathrm{R}_{\\mathrm{f}}$$ is not an integral value. It is the ratio of distance travelled by the organic compound to that of solvent. $$\\mathrm{R}_{\\mathrm{f}}$$ value of a polar compound is smaller than that of non-polar compound as the polar compound is retained more by the adsorbent than non-polar compound.

    ", "topic": "Discrete Mathematics", "subtopic": "Other" }, { "id": 2403, "subject": "Chemistry", "question": "A solution containing a group-IV cation gives a precipitate on passing H2S. A solution of this precipitate in dil.HCl produces a white precipitate with NaOH solution and bluish-white precipitate with basic potassium ferrocyanide. The cation is :\n ", "options": [ { "text": "Co2+" }, { "text": "Ni2+" }, { "text": "Mn2+ " }, { "text": "Zn2+" } ], "answer": "Zn2+", "solution": "**Answer:** Zn2+\n\n

    Solution containing Zn2+ ions forms white precipitate on reaction with H2S gas.

    \n\"JEE", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 2404, "subject": "Chemistry", "question": "Sodium salt of an organic acid ‘X’ produces effervescence with conc. H2SO4. ‘X’ reacts with the acidified\naqueous CaCl2 solution to give a white precipitate which decolourises acidic solution of KMnO4. ‘X’ is :", "options": [ { "text": "CH3COONa" }, { "text": "Na2C2O4" }, { "text": "C6H5COONa" }, { "text": "HCOONa" } ], "answer": "Na2C2O4", "solution": "**Answer:** Na2C2O4\n\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2405, "subject": "Chemistry", "question": "A solution of m-chloronailimne, m-chlorophenol and m-chlorobenzoic acid in ethyal acetate was extracted initially with a saturated solution of NaHCO3\n to give fraction A. The left over organic phase\nwas extracted with dilute NaoH solution tio give fraction B. The final organiic layer was labelled as\nfraction C. Fractions A,B and C, contain respectively :", "options": [ { "text": "m-chlorobenzoic acid, m-chloroaniline and m-chlorophenol." }, { "text": "m-chlorobenzoic acid, m-chlorophenol and m-chloroaniline." }, { "text": "m-chlorophenol, m-chlorobenzoic acid and m-chloronailine." }, { "text": "m-chloroaniline m-chlorobenzoic acid and m-chlorophenol." } ], "answer": "m-chlorobenzoic acid, m-chlorophenol and m-chloroaniline.", "solution": "**Answer:** m-chlorobenzoic acid, m-chlorophenol and m-chloroaniline.\n\nm-chlorobenzoic acid dissolves in aq NaHCO3\nwhile m-chlorophenol dissolves in aq NaOH.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2406, "subject": "Chemistry", "question": "Given below are two statements :

    Statement I : The identification of Ni2+ is carried out by dimethyl glyoxime in the presence of NH4OH.

    Statement II : The dimethyl glyoxime is a bidentate neutral ligand.

    In the light of the above statements, choose the correct answer from the options given below :", "options": [ { "text": "Statement I is false but Statement II is true." }, { "text": "Statement I is true but Statement II is false." }, { "text": "Both Statement I and Statement II are false." }, { "text": "Both Statement I and Statement II are true." } ], "answer": "Both Statement I and Statement II are true.", "solution": "**Answer:** Both Statement I and Statement II are true.\n\n

    Statement I is true. Dimethyl glyoxime (DMG) is a chelating reagent commonly used for the identification of Ni2+ ions in aqueous solution. In the presence of ammonium hydroxide (NH4OH), Ni2+ reacts with dimethyl glyoxime to form a deep red-colored complex, which can be used to confirm the presence of Ni2+ ions.

    \n\n

    Statement II is also true. Dimethyl glyoxime is a neutral bidentate ligand, which means it has two donor atoms that can coordinate with a metal ion, and its overall charge is neutral.

    \n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2407, "subject": "Chemistry", "question": "An inorganic Compound 'X' on treatment with concentrated H2SO4 produces brown fumes and gives dark brown ring with FeSO4 in presence of concentrated H2SO4. Also Compound 'X' gives precipitate 'Y', when its solution in dilute HCl is treated with H2S gas. The precipitate 'Y' on treatment with concentrated HNO3 followed by excess of NH4OH further gives deep blue coloured solution, Compound 'X' is :", "options": [ { "text": "Co(NO3)2" }, { "text": "Pb(NO2)2" }, { "text": "Cu(NO3)2" }, { "text": "Pb(NO3)2" } ], "answer": "Cu(NO3)2", "solution": "**Answer:** Cu(NO3)2\n\nCompound ‘X’ is copper nitrate, i.e. Cu(NO3)2 is an inorganic\ncompound. On treatment with concentrated H2SO4 produces brown\nfumes and gives dark brown ring with FeSO4\nin presence of\nconcentrated H2SO4. Chemical reaction is as follows\n

    \"JEE\n

    Cu2+\nis a group II cation with group II reagents\n(HCl/H2S)\n, it gives black coloured precipitates. These precipitates\ngives blue colour solution on treatment with HNO3\nfollowed by\nexcess of NH4OH.\n

    \"JEE", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 2408, "subject": "Chemistry", "question": "Cu2+ salt reacts with potassium iodide to give", "options": [ { "text": "Cu2I2" }, { "text": "Cu2I3" }, { "text": "Cul" }, { "text": "Cu(I3)2" } ], "answer": "Cu2I2", "solution": "**Answer:** Cu2I2\n\n$$2C{u^{ + 2}} + 4{I^ - }\\buildrel {} \\over\n \\longrightarrow C{u_2}{I_2}(s) + {I_2}$$

    $$2C{u^{ + 2}} + 3{I^ - }\\buildrel {} \\over\n \\longrightarrow 2CuI + {I_2}$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2409, "subject": "Chemistry", "question": "When silver nitrate solution is added to potassium iodide solution then the sol produced is :", "options": [ { "text": "AgI / I$$-$$" }, { "text": "AgI / Ag+" }, { "text": "KI / NO$$_3^ - $$" }, { "text": "AgNO3 / NO$$_3^ - $$" } ], "answer": "AgI / I$$-$$", "solution": "**Answer:** AgI / I$$-$$\n\n$$\\mathop {AgN{O_3}(aq.)}\\limits_{(drop\\,by\\,drop)} + \\mathop {KI(aq.)}\\limits_{excess} \\buildrel {} \\over\n \\longrightarrow \\mathop {AgI/{I^ - }}\\limits_{Sol} $$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2410, "subject": "Chemistry", "question": "To an aqueous solution containing ions such as Al3+, Zn2+, Ca2+, Fe3+, Ni2+, Ba2+ and Cu2+ was added conc. HCl, followed by H2S.

    The total number of cations precipitated during this reaction is/are :", "options": [ { "text": "1" }, { "text": "3" }, { "text": "4" }, { "text": "2" } ], "answer": "1", "solution": "**Answer:** 1\n\nAl3+ and Fe3+ sulphides hydrolyse in water.

    Ni2+ and Zn2+ require basic medium with H2S to form ppt

    Ca2+ and Ba2+ sulphides are soluble

    hence, we will receive only CuS ppt.", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 2411, "subject": "Chemistry", "question": "

    A white precipitate was formed when BaCl2 was added to water extract of an inorganic salt. Further, a gas 'X' with characteristic odour was released when the formed white precipitate was dissolved in dilute HCl. The anion present in the inorganic salt is

    ", "options": [ { "text": "I$$-$$" }, { "text": "SO32$$-$$" }, { "text": "S2$$-$$" }, { "text": "NO2$$-$$" } ], "answer": "SO32$$-$$", "solution": "**Answer:** SO32$$-$$\n\nAnion is $\\mathrm{SO}_{3}^{-2}$\n

    \n$$\n\\mathrm{BaSO}_{3} \\stackrel{\\text { dil } \\mathrm{HCl}}{\\longrightarrow} \\underset{x(\\text { gas })}{\\mathrm{SO}_{2} \\uparrow}\n$$\n

    \nGas is released with smell of burning sulphur.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2412, "subject": "Chemistry", "question": "

    During halogen test, sodium fusion extract is boiled with concentrated HNO3 to

    ", "options": [ { "text": "remove unreacted sodium" }, { "text": "decompose cyanide or sulphide of sodium" }, { "text": "extract halogen from organic compound" }, { "text": "maintain the pH of extract." } ], "answer": "decompose cyanide or sulphide of sodium", "solution": "**Answer:** decompose cyanide or sulphide of sodium\n\nDuring test for halogen, if nitrogen or sulphur is also\npresent in the compound, then sodium fusion\nextract is first boiled with concentrated nitric acid to\ndecompose cyanide or sulphide of sodium formed\nduring Lassaigne's test.", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 2413, "subject": "Chemistry", "question": "

    In the flame test of a mixture of salts, a green flame with blue centre was observed. Which one of the following cations may be present?

    ", "options": [ { "text": "Cu2+" }, { "text": "Sr2+" }, { "text": "Ba2+" }, { "text": "Ca2+" } ], "answer": "Cu2+", "solution": "**Answer:** Cu2+\n\nCupric salts give green flame with blue centre. The colour of other salts are\n

    \n$\\begin{array}{ll}\\mathrm{Sr}^{2+} & \\text { Crimson red } \\\\ \\mathrm{Ca}^{2+} & \\text { Brick red } \\\\ \\mathrm{Ba}^{2+} & \\text { Green }\\end{array}$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2414, "subject": "Chemistry", "question": "

    During the qualitative analysis of salt with cation y2+, addition of a reagent (X) to alkaline solution of the salt gives a bright red precipitate. The reagent (X) and the cation (y2+) present respectively are :

    ", "options": [ { "text": "Dimethylglyoxime and Ni2+" }, { "text": "Dimethylglyoxime and Co2+" }, { "text": "Nessler's reagent and Hg2+" }, { "text": "Nessler's reagent and Ni2+" } ], "answer": "Dimethylglyoxime and Ni2+", "solution": "**Answer:** Dimethylglyoxime and Ni2+\n\nOn addition of dimethylglyoxime to alkaline solution of $\\mathrm{Ni}^{+2}$, a bright red ppt. is obtained.\n

    \n$\\mathrm{Ni}^{+2}+2 \\mathrm{dmg} \\rightarrow\\left[\\mathrm{Ni}(\\mathrm{dmg})_{2}\\right]^{+2}$ (Bright red ppt)", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 2415, "subject": "Chemistry", "question": "

    In the wet tests for identification of various cations by precipitation, which transition element cation doesn't belong to group IV in qualitative inorganic analysis?

    ", "options": [ { "text": "Co2+" }, { "text": "Ni2+" }, { "text": "Zn2+" }, { "text": "Fe3+" } ], "answer": "Fe3+", "solution": "**Answer:** Fe3+\n\n

    Fe$$^{3+}$$ belongs to III$$^{rd}$$ group

    ", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 2416, "subject": "Chemistry", "question": "

    Match List I with List II

    \n

    \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    List I
    Elements    		List II
    Colour imparted to the flame
    A.KI.Brick Red
    B.CaII.Violet
    C.SrIII.Apple Green
    D.BaIV.Crimson Red

    \n

    Choose the correct answer from the options given below :

    ", "options": [ { "text": "A-II, B-IV, C-I, D-III" }, { "text": "A-II, B-I, C-IV, D-III" }, { "text": "A-II, B-I, C-III, D-IV" }, { "text": "A-IV, B-III, C-II, D-I" } ], "answer": "A-II, B-I, C-IV, D-III", "solution": "**Answer:** A-II, B-I, C-IV, D-III\n\n\n\n\n \n \n \n \n \n\n\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\n
    Element$\\begin{aligned}  &  \\text { Colour imparted to } \\\\ &  \\text { the flame }\\end{aligned}$
    (A)$\\mathrm{K}$Violet
    (B)$\\mathrm{Ca}$Brick red
    (C)$\\mathrm{Sr}$Crimson red
    (D)$\\mathrm{Ba}$Apple green
    ", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 2417, "subject": "Chemistry", "question": "

    Match the List-I with List-II :

    \n

    \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    List-I
    Cations    		List-II
    Group reagents
    A.$$\\mathrm{Pb^{2+},Cu^{2+}}$$i)$$\\mathrm{H_2S}$$ gas in presence of dilute HCl
    B.$$\\mathrm{Al^{3+},Fe^{3+}}$$ii)$$\\mathrm{(NH_4)_2CO_3}$$ in presence of $$\\mathrm{NH_4OH}$$
    C.$$\\mathrm{Co^{2+},Ni^{2+}}$$iii)$$\\mathrm{NH_4OH}$$ in presence of $$\\mathrm{NH_4Cl}$$
    D.$$\\mathrm{Ba^{2+},Ca^{2+}}$$iv)H2S in presence of NH4OH

    \n

    Correct match is -

    ", "options": [ { "text": "A $$\\to$$ i, B $$\\to$$ iii, C $$\\to$$ iv, D $$\\to$$ ii" }, { "text": "A $$\\to$$ i, B $$\\to$$ iii, C $$\\to$$ ii, D $$\\to$$ iv" }, { "text": "A $$\\to$$ iii, B $$\\to$$ i, C $$\\to$$ iv, D $$\\to$$ ii" }, { "text": "A $$\\to$$ iv, B $$\\to$$ ii, C $$\\to$$ iii, D $$\\to$$ i" } ], "answer": "A $$\\to$$ i, B $$\\to$$ iii, C $$\\to$$ iv, D $$\\to$$ ii", "solution": "**Answer:** A $$\\to$$ i, B $$\\to$$ iii, C $$\\to$$ iv, D $$\\to$$ ii\n\n\n\n\n \n \n \n \n \n\n\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\n
    Cations Group No.Group reagent
    $\\mathrm{Pb}^{+2}, \\mathrm{Cu}^{+2}$II$\\mathrm{H}_2 \\mathrm{S}(\\mathrm{g}) \\text { in presence of } \\text { dil HCl }$
    $\\mathrm{Al}^{+3}, \\mathrm{Fe}^{+3}$III$\\begin{aligned}  &  \\mathrm{NH}_4 \\mathrm{OH} \\text { in presence of } \\\\ &  \\mathrm{NH}_4 \\mathrm{Cl}\\end{aligned}$
    $\\mathrm{CO}^{+2}, \\mathrm{Ni}^{+2}$$\\mathrm{IV}$$\\begin{aligned}  &  \\mathrm{H}_2 \\mathrm{~S} \\text { in presence of } \\\\ &  \\mathrm{NH}_4 \\mathrm{OH}\\end{aligned}$
    $\\mathrm{Ba}^{+2}, \\mathrm{Ca}^{+2}$$\\mathrm{~V}$$\\begin{aligned}  &  \\left(\\mathrm{NH}_4\\right)_2 \\mathrm{CO}_3 \\text { in presence } \\\\ &  \\text { of } \\mathrm{NH}_4 \\mathrm{OH}\\end{aligned}$
    ", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 2418, "subject": "Chemistry", "question": "

    In the wet tests for detection of various cations by precipitation, $$\\mathrm{Ba}^{2+}$$ cations are detected by obtaining precipitate of

    ", "options": [ { "text": "$$\\mathrm{BaSO}_{4}$$" }, { "text": "$$\\mathrm{Ba}(\\mathrm{OAc})_{2}$$" }, { "text": "$$\\mathrm{BaCO}_{3}$$" }, { "text": "$$\\mathrm{Ba}(\\mathrm{ox})$$ : Barium oxalate" } ], "answer": "$$\\mathrm{BaCO}_{3}$$", "solution": "**Answer:** $$\\mathrm{BaCO}_{3}$$\n\n

    In the wet test for detection of Group 5 cations, including Ba²⁺, Ca²⁺, and Sr²⁺, ammonium carbonate ((NH₄)₂CO₃) is used as a group reagent. The reaction with barium ions (Ba²⁺) can be written as :

    \n\n

    $$\\mathrm{Ba}^{2+} + \\left(\\mathrm{NH}_4\\right)_2\\mathrm{CO}_3 \\rightarrow \\underset{\\text{(white precipitate)}}{\\mathrm{BaCO}_3 \\downarrow} + 2\\mathrm{NH}_4^+$$

    \n\n

    Upon adding ammonium carbonate to the solution containing Ba²⁺ ions, a white precipitate of barium carbonate (BaCO₃) is formed, which can be used to identify the presence of barium ions in the sample.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2419, "subject": "Chemistry", "question": "

    In chromyl chloride test for confirmation of $$\\mathrm{Cl}^{-}$$ ion, a yellow solution is obtained. Acidification of the solution and addition of amyl alcohol and $$10 \\% \\mathrm{~H}_2 \\mathrm{O}_2$$ turns organic layer blue indicating formation of chromium pentoxide. The oxidation state of chromium in that is

    ", "options": [ { "text": "+6" }, { "text": "+5" }, { "text": "+3" }, { "text": "+10" } ], "answer": "+6", "solution": "**Answer:** +6\n\n

    \"JEE

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2420, "subject": "Chemistry", "question": "

    Appearance of blood red colour, on treatment of the sodium fusion extract of an organic compound with $$\\mathrm{FeSO}_4$$ in presence of concentrated $$\\mathrm{H}_2 \\mathrm{SO}_4$$ indicates the presence of element/s

    ", "options": [ { "text": "Br" }, { "text": "S" }, { "text": "N and S" }, { "text": "N" } ], "answer": "N and S", "solution": "**Answer:** N and S\n\n

    $$\\begin{aligned}\n& \\mathrm{Fe}^{2+} \\xrightarrow[\\text { Conc. } \\mathrm{H}_2 \\mathrm{SO}_4]{\\mathrm{H}^{+}} \\mathrm{Fe}^{+3} \\\\\n& \\mathrm{Fe}^{+3} \\xrightarrow{- \\text { SCN }} \\mathrm{Fe}(\\mathrm{SCN})_3 \\text { (blood red colour) }\n\\end{aligned}$$

    \n

    Appearance of blood red colour indicates presence of both nitrogen and sulphur.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2421, "subject": "Chemistry", "question": "

    In the precipitation of the iron group (III) in qualitative analysis, ammonium chloride is added before adding ammonium hydroxide to :

    ", "options": [ { "text": "decrease concentration of $${ }^{-} \\mathrm{OH}$$ ions\n" }, { "text": "increase concentration of $$\\mathrm{Cl}^{-}$$ ions\n" }, { "text": "prevent interference by phosphate ions\n" }, { "text": "increase concentration of $$\\mathrm{NH}_4{ }^{+}$$ ions" } ], "answer": "decrease concentration of $${ }^{-} \\mathrm{OH}$$ ions\n", "solution": "**Answer:** decrease concentration of $${ }^{-} \\mathrm{OH}$$ ions\n\n\n

    In qualitative analysis, the precipitation of the iron group (III) requires careful control of conditions to ensure selective precipitation of the desired ions. Ammonium chloride is added before ammonium hydroxide to achieve this control effectively.

    \n\n

    Let's analyze the options to understand why ammonium chloride is used:

    \n\n

    Option A: decrease concentration of $$\\mathrm{OH}^{-}$$ ions

    \n\n

    This option is correct. Adding ammonium chloride ($$\\mathrm{NH}_4\\mathrm{Cl}$$) introduces a common ion effect, where the concentration of $$\\mathrm{OH}^{-}$$ ions is reduced due to the formation of $$\\mathrm{NH}_4\\mathrm{OH}$$. This reaction can be represented as:

    \n\n

    $$\\mathrm{NH}_4^+ + \\mathrm{OH}^- \\leftrightarrow \\mathrm{NH}_4\\mathrm{OH}$$

    \n\n

    By reducing the concentration of $$\\mathrm{OH}^{-}$$ ions, it prevents the premature precipitation of hydroxides of metals that might otherwise not selectively precipitate in the desired group.

    \n\n

    Option B: increase concentration of $$\\mathrm{Cl}^{-}$$ ions

    \n\n

    While it is true that ammonium chloride increases the concentration of $$\\mathrm{Cl}^{-}$$ ions, this is not the primary reason for its addition in this context. The focus is more on regulating $$\\mathrm{OH}^{-}$$ ion concentration.

    \n\n

    Option C: prevent interference by phosphate ions

    \n\n

    This is not relevant in the context of adding ammonium chloride, as the main idea behind adding ammonium chloride is to control the $$\\mathrm{OH}^{-}$$ ion concentration rather than dealing directly with phosphate ions.

    \n\n

    Option D: increase concentration of $$\\mathrm{NH}_4^+$$ ions

    \n\n

    While ammonium chloride does increase the concentration of $$\\mathrm{NH}_4^+$$ ions, which in turn helps in maintaining the $$\\mathrm{OH}^{-}$$ concentration at a lower level, this is again a secondary consideration to the primary goal of controlling the $$\\mathrm{OH}^{-}$$ ion concentration to ensure selective precipitation.

    \n\n

    Therefore, the correct answer is:

    \n\n

    Option A: decrease concentration of $$\\mathrm{OH}^{-}$$ ions

    ", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 2422, "subject": "Chemistry", "question": "

    Match List I with List II

    \n

    \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    LIST I
    (Name of the test)    		LIST II
    (Reaction sequence involved) [M is metal]
    A.Borax bead testI.$$\\mathrm{MCO}_3 \\rightarrow \\mathrm{MO} \\xrightarrow[+\\Delta]{\\mathrm{Co}\\left(\\mathrm{NO}_3\\right)_2} \\mathrm{CoO} \\cdot \\mathrm{MO}$$
    B.\tCharcoal cavity testII.$$\\mathrm{MCO}_3 \\rightarrow \\mathrm{MCl}_2 \\rightarrow \\mathrm{M}^{2+}$$
    C.Cobalt nitrate testIII.$$\\mathrm{MSO}_4 \\xrightarrow[\\Delta]{\\mathrm{Na}_2 \\mathrm{~B}_4 \\mathrm{O}_7} \\mathrm{M}\\left(\\mathrm{BO}_2\\right)_2 \\rightarrow \\mathrm{MBO}_2 \\rightarrow \\mathrm{M}$$
    D.Flame testIV.$$\\mathrm{MSO}_4 \\xrightarrow[\\Delta]{\\mathrm{Na}_2 \\mathrm{CO}_3} \\mathrm{MCO}_3 \\rightarrow \\mathrm{MO} \\rightarrow \\mathrm{M}$$

    \n

    Choose the correct answer from the options given below:

    ", "options": [ { "text": "A-III, B-I, C-IV, D-II\n" }, { "text": "A-III, B-IV, C-I, D-II\n" }, { "text": "A-III, B-II, C-IV, D-I\n" }, { "text": "A-III, B-I, C-II, D-IV" } ], "answer": "A-III, B-IV, C-I, D-II\n", "solution": "**Answer:** A-III, B-IV, C-I, D-II\n\n\n

    (A) Borax bead test

    \n

    $$\\mathrm{MSO}_4 \\xrightarrow[\\Delta]{\\mathrm{Na}_2 \\mathrm{~B}_4 \\mathrm{O}_7} \\mathrm{M}\\left(\\mathrm{BO}_2\\right)_2 \\rightarrow \\mathrm{MBO}_2 \\rightarrow \\mathrm{M}$$

    \n

    (B) Charcoal cavity test

    \n

    $$\\mathrm{MSO}_4 \\xrightarrow[\\Delta]{\\mathrm{Na}_2 \\mathrm{CO}_3} \\mathrm{MCO}_3 \\rightarrow \\mathrm{MO} \\rightarrow \\mathrm{M}$$

    \n

    (C) Cobalt nitrate test

    \n

    $$\\mathrm{MCO}_3 \\rightarrow \\mathrm{MO} \\xrightarrow[+\\Delta]{\\mathrm{CO}\\left(\\mathrm{NO}_3\\right)_2} \\mathrm{CoO} \\cdot \\mathrm{MO}$$

    \n

    (D) Flame test

    \n

    $$\\mathrm{MCO}_3 \\rightarrow \\mathrm{MCl}_2 \\rightarrow \\mathrm{M}^{2+}$$

    \n

    So, $$\\mathrm{A} \\rightarrow$$ (III), $$\\mathrm{B} \\rightarrow$$ (IV), $$\\mathrm{C} \\rightarrow$$ (I), $$\\mathrm{D} \\rightarrow$$ (II).

    ", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 2423, "subject": "Chemistry", "question": "

    During the detection of acidic radical present in a salt, a student gets a pale yellow precipitate soluble with difficulty in $$\\mathrm{NH}_4 \\mathrm{OH}$$ solution when sodium carbonate extract was first acidified with dil. $$\\mathrm{HNO}_3$$ and then $$\\mathrm{AgNO}_3$$ solution was added. This indicates presence of :

    ", "options": [ { "text": "$$\\mathrm{I}^{-}$$\n" }, { "text": "$$\\mathrm{CO}_3{ }^{2-}$$\n" }, { "text": "$$\\mathrm{Cl}^{-}$$\n" }, { "text": "$$\\mathrm{Br}^{-}$$" } ], "answer": "$$\\mathrm{Br}^{-}$$", "solution": "**Answer:** $$\\mathrm{Br}^{-}$$\n\n

    The $$\\mathrm{Br}^{-}$$ ion present in the salt gives pale yellow ppt. of $$\\mathrm{AgBr}$$ with $$\\mathrm{AgNO}_3$$, insoluble in dil $$\\mathrm{HNO}_3$$ but partially soluble in aq. $$\\mathrm{NH}_4 \\mathrm{OH}$$.

    \n

    $$\\mathrm{Br}^{-}+\\mathrm{AgNO}_3 \\rightarrow \\underset{\\text { Pale yellow }}{\\mathrm{AgBr} \\downarrow}+\\mathrm{NO}_3^{-}$$

    \n

    $$\\mathrm{AgBr}+2 \\mathrm{NH}_4 \\mathrm{OH} \\rightarrow \\underset{\\text { Partially soluble }}{\\left[\\mathrm{Ag}\\left(\\mathrm{NH}_3\\right)_2\\right] \\mathrm{Br}}+2 \\mathrm{H}_2 \\mathrm{O}$$

    ", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 2424, "subject": "Chemistry", "question": "

    Match List I with List II

    \n

    \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    LIST I
    (Precipitating reagent and conditions)    		LIST II
    (Cation)
    A.$$
    \\mathrm{NH}_4 \\mathrm{Cl}+\\mathrm{NH}_4 \\mathrm{OH}
    $$    		I.$$\\mathrm{Mn^{2+}}$$
    B.\t$$
    \\mathrm{NH}_4 \\mathrm{OH}+\\mathrm{Na}_2 \\mathrm{CO}_3
    $$    		II.$$\\mathrm{Pb^{2+}}$$
    C.$$
    \\mathrm{NH}_4 \\mathrm{OH}+\\mathrm{NH}_4 \\mathrm{Cl}+\\mathrm{H}_2 \\mathrm{~S} \\text { gas }
    $$    		III.$$\\mathrm{Al^{3+}}$$
    D.dilute $$\\mathrm{HCl}$$IV.$$\\mathrm{Sr^{2+}}$$

    \n

    Choose the correct answer from the options given below:

    ", "options": [ { "text": "A-III, B-IV, C-I, D-II\n" }, { "text": "A-III, B-IV, C-II, D-I\n" }, { "text": "A-IV, B-III, C-II, D-I\n" }, { "text": "A-IV, B-III, C-I, D-II" } ], "answer": "A-III, B-IV, C-I, D-II\n", "solution": "**Answer:** A-III, B-IV, C-I, D-II\n\n\n

    $$\\begin{array}{lll}\n\\mathrm{A} & \\rightarrow \\mathrm{Al}^{3+}(\\mathrm{III}) \\\\\n\\mathrm{B} & \\rightarrow \\mathrm{Sr}^{2+}(\\mathrm{IV}) \\\\\n\\mathrm{C} & \\rightarrow \\mathrm{Mn}^{2+}(\\mathrm{I}) \\\\\n\\mathrm{D} & \\rightarrow \\mathrm{Pb}^{2+}(\\mathrm{II})\n\\end{array}$$

    ", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 2425, "subject": "Chemistry", "question": "Consider the following reaction:

    \n$$xMnO_4^- + yC_2O_4^{2-}$$ + zH+ $$\\to$$ xMn2+ + 2yCO2 + $${z \\over 2}{H_2}O$$

    \nThe value's of x, y and z in the reaction are, respectively :", "options": [ { "text": "5, 2 and 16" }, { "text": "2, 5 and 8" }, { "text": "2, 5 and 16" }, { "text": "5, 2 and 8" } ], "answer": "2, 5 and 16", "solution": "**Answer:** 2, 5 and 16\n\nAfter balancing the reaction we get\n

    $$2MnO_4^- + 5C_2O_4^{2-}$$ + 16H+ $$\\to$$ 2Mn2+ + 10CO2 + $$8{H_2}O$$\n

    $$\\therefore$$ $$x$$ = 2, $$y$$ = 5 and z = 16", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2426, "subject": "Chemistry", "question": "

    See the following chemical reaction:

    \n

    $$\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-}+\\mathrm{XH}^{+}+6 \\mathrm{~F}_{e}^{2+} \\rightarrow \\mathrm{YCr}^{3+}+6 \\mathrm{~F}_{e}^{3+}+\\mathrm{Z} \\mathrm{H}_{2} \\mathrm{O}$$

    \n

    The sum of $$\\mathrm{X}, \\mathrm{Y}$$ and $$\\mathrm{Z}$$ is ___________

    ", "options": [], "answer": "23", "solution": "**Answer:** 23\n\nTo find the sum of X, Y, and Z, we need to balance the given chemical reaction:\n

    \n$$\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-} + \\mathrm{XH}^{+} + 6 \\mathrm{Fe}^{2+} \\rightarrow \\mathrm{YCr}^{3+} + 6 \\mathrm{Fe}^{3+} + \\mathrm{ZH}_{2} \\mathrm{O}$$\n

    \nLet's balance the reaction step by step:\n

    \n1. Balance the chromium atoms:

    \nThere are 2 chromium atoms on the left side of the equation, so we need 2 chromium atoms on the right side. Thus, Y = 2.\n

    \n$$\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-} + \\mathrm{XH}^{+} + 6 \\mathrm{Fe}^{2+} \\rightarrow 2 \\mathrm{Cr}^{3+} + 6 \\mathrm{Fe}^{3+} + \\mathrm{ZH}_{2} \\mathrm{O}$$\n

    \n2. Balance the oxygen atoms:

    \nThere are 7 oxygen atoms on the left side of the equation, so we need 7 oxygen atoms on the right side. Since each water molecule (H₂O) contains one oxygen atom, Z = 7.\n

    \n$$\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-} + \\mathrm{XH}^{+} + 6 \\mathrm{Fe}^{2+} \\rightarrow 2 \\mathrm{Cr}^{3+} + 6 \\mathrm{Fe}^{3+} + 7 \\mathrm{H}_{2} \\mathrm{O}$$\n

    \n3. Balance the hydrogen atoms and charges:

    \nThere are 14 hydrogen atoms on the right side of the equation, so we need 14 hydrogen atoms on the left side. Thus, X = 14.\n

    \n$$\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-} + 14 \\mathrm{H}^{+} + 6 \\mathrm{Fe}^{2+} \\rightarrow 2 \\mathrm{Cr}^{3+} + 6 \\mathrm{Fe}^{3+} + 7 \\mathrm{H}_{2} \\mathrm{O}$$\n

    \nNow the chemical reaction is balanced:\n

    \n$$\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-} + 14 \\mathrm{H}^{+} + 6 \\mathrm{Fe}^{2+} \\rightarrow 2 \\mathrm{Cr}^{3+} + 6 \\mathrm{Fe}^{3+} + 7 \\mathrm{H}_{2} \\mathrm{O}$$\n

    \nThe sum of X, Y, and Z is:\n

    \nX + Y + Z = 14 + 2 + 7 = 23\n

    \nSo, the sum of X, Y, and Z is 23.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2427, "subject": "Chemistry", "question": "

    $$2 \\mathrm{IO}_{3}^{-}+x \\mathrm{I}^{-}+12 \\mathrm{H}^{+} \\rightarrow 6 \\mathrm{I}_{2}+6 \\mathrm{H}_{2} \\mathrm{O}$$

    \n

    What is the value of $$x$$ ?

    ", "options": [ { "text": "10" }, { "text": "2" }, { "text": "12" }, { "text": "6" } ], "answer": "10", "solution": "**Answer:** 10\n\n

    The reaction given is a disproportionation reaction where iodine ($\\mathrm{I}_2$) undergoes both oxidation and reduction. Disproportionation reactions are a type of redox reaction where an element is simultaneously oxidized and reduced.

    \n

    In the process, iodine gets oxidized from 0 oxidation state (in $\\mathrm{I}_2$) to +5 oxidation state (in $\\mathrm{IO}_3^{-}$), and reduced from 0 oxidation state (in $\\mathrm{I}_2$) to -1 oxidation state (in $\\mathrm{I}^{-}$).

    \n

    The n-factor is the total change in oxidation state per molecule that undergoes the redox reaction. In this case, the n-factor for $\\mathrm{IO}_3^{-}$ is 5 (as iodine goes from 0 to +5) and for $\\mathrm{I}^{-}$, it's 1 (as iodine goes from 0 to -1).

    \n

    Now, to balance the redox reaction, the total increase in oxidation state (total oxidation) must equal the total decrease in oxidation state (total reduction). Hence, the molar ratio of $\\mathrm{IO}_3^{-}$ to $\\mathrm{I}^{-}$ must be 1:5.

    \n

    So, the balanced reaction would be:

    \n

    $\\mathrm{IO}_3^{-}+6 \\mathrm{H}^{+}+5 \\mathrm{I}^{-} \\rightarrow 3 \\mathrm{I}_2+3 \\mathrm{H}_2 \\mathrm{O}$

    \n

    This equation yields 3 moles of $\\mathrm{I}_2$, but the original equation needs to produce 6 moles of $\\mathrm{I}_2$, so the entire equation is multiplied by 2:

    \n

    $2 \\mathrm{IO}_3^{-}+12 \\mathrm{H}^{+}+10 \\mathrm{I}^{-} \\rightarrow 6 \\mathrm{I}_2+6 \\mathrm{H}_2 \\mathrm{O}$

    \n

    This tells us that to get 6 moles of $\\mathrm{I}_2$, you need 10 moles of $\\mathrm{I}^{-}$. So, $x = 10$.

    \n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2428, "subject": "Chemistry", "question": "In acidic medium, $\\mathrm{K}_2 \\mathrm{Cr}_2 \\mathrm{O}_7$ shows oxidising action as represented in the half reaction:

    \n$$\n\\mathrm{Cr}_2 \\mathrm{O}_7{ }^{2-}+\\mathrm{XH}^{+}+\\mathrm{Ye}^{\\ominus} \\rightarrow 2 \\mathrm{~A}+\\mathrm{ZH}_2 \\mathrm{O}\n$$

    \n$\\mathrm{X}, \\mathrm{Y}, \\mathrm{Z}$ and $\\mathrm{A}$ are respectively are :", "options": [ { "text": "$14,7,6$ and $\\mathrm{Cr}^{3+}$" }, { "text": "$14,6,7$ and $\\mathrm{Cr}^{3+}$" }, { "text": "$8,4,6$ and $\\mathrm{Cr}_2 \\mathrm{O}_3$" }, { "text": "$8,6,4$ and $\\mathrm{Cr}_2 \\mathrm{O}_3$" } ], "answer": "$14,6,7$ and $\\mathrm{Cr}^{3+}$", "solution": "**Answer:** $14,6,7$ and $\\mathrm{Cr}^{3+}$\n\n\n
      \n
    1. Chromium (Cr) : Already balanced with 2 Cr on each side.

      2. \n
    2. Oxygen (O): Add 7 H₂O to the right:
      3. \n
    \n

    $$\\mathrm{Cr}_2 \\mathrm{O}_7{ }^{2-} + \\mathrm{XH}^{+} + \\mathrm{Ye}^{\\ominus} \\rightarrow 2 \\mathrm{~A} + 7\\mathrm{H}_2 \\mathrm{O}$$

    \n\n
      \n
    1. Hydrogen (H) : Add 14 H⁺ to the left:
      2. \n
    \n

    $$\\mathrm{Cr}_2 \\mathrm{O}_7{ }^{2-} + 14\\mathrm{H}^{+} + \\mathrm{Ye}^{\\ominus} \\rightarrow 2 \\mathrm{~A} + 7\\mathrm{H}_2 \\mathrm{O}$$

    \n\n
      \n
    1. Charge : Add 6e⁻ to the left:
      2. \n
    \n

    $$\\mathrm{Cr}_2 \\mathrm{O}_7{ }^{2-} + 14\\mathrm{H}^{+} + 6\\mathrm{e}^{\\ominus} \\rightarrow 2 \\mathrm{~A} + 7\\mathrm{H}_2 \\mathrm{O}$$

    \n\n\n

    The reduction of Cr₂O₇²⁻ in acidic medium forms Cr³⁺ ions.

    \n\n\n

    $$\\mathrm{Cr}_2 \\mathrm{O}_7{ }^{2-} + 14\\mathrm{H}^{+} + 6\\mathrm{e}^{\\ominus} \\rightarrow 2 \\mathrm{Cr}^{3+} + 7\\mathrm{H}_2 \\mathrm{O}$$

    \n\nAnswer\n\n

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2429, "subject": "Chemistry", "question": "

    Number of moles of $$\\mathrm{H}^{+}$$ ions required by $$1 \\mathrm{~mole}$$ of $$\\mathrm{MnO}_4^{-}$$ to oxidise oxalate ion to $$\\mathrm{CO}_2$$ is _________.

    ", "options": [], "answer": "8", "solution": "**Answer:** 8\n\n

    $$2 \\mathrm{MnO}_4^{-}+5 \\mathrm{C}_2 \\mathrm{O}_4^{2-}+16 \\mathrm{H}^{+} \\longrightarrow 2 \\mathrm{Mn}^{2+}+10 \\mathrm{CO}_2+8 \\mathrm{H}_2 \\mathrm{O}$$

    \n

    $$\\therefore$$ Number of moles of $$\\mathrm{H}^{+}$$ ions required by 1 mole of $$\\mathrm{MnO}_4^{-}$$ to oxidise oxalate ion to $$\\mathrm{CO}_2$$ is 8

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2430, "subject": "Chemistry", "question": "

    Chlorine undergoes disproportionation in alkaline medium as shown below :

    \n

    $$\\mathrm{aCl}_{2(\\mathrm{~g})}+\\mathrm{b} \\mathrm{OH}_{(\\mathrm{aq})}^{-} \\rightarrow \\mathrm{c} \\mathrm{ClO}_{(\\mathrm{aq)}}^{-}+\\mathrm{d} \\mathrm{Cl}_{(\\mathrm{aq})}^{-}+\\mathrm{e} \\mathrm{H}_2 \\mathrm{O}_{(\\mathrm{l})}$$

    \n

    The values of $$a, b, c$$ and $$d$$ in a balanced redox reaction are respectively :

    ", "options": [ { "text": "3, 4, 4 and 2" }, { "text": "1, 2, 1 and 1" }, { "text": "2, 4, 1 and 3" }, { "text": "2, 2, 1 and 3" } ], "answer": "1, 2, 1 and 1", "solution": "**Answer:** 1, 2, 1 and 1\n\n

    \"JEE

    \n

    $$$\\Rightarrow \\mathrm{Cl}_2+2 \\overline{\\mathrm{O}} \\mathrm{H} \\longrightarrow \\mathrm{Cl}^{-}+\\mathrm{ClO}^{-}+\\mathrm{H}_2 \\mathrm{O}$$$

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2431, "subject": "Chemistry", "question": "

    $$2 \\mathrm{MnO}_4^{-}+\\mathrm{bI}^{-}+\\mathrm{cH}_2 \\mathrm{O} \\rightarrow x \\mathrm{I}_2+y \\mathrm{MnO}_2+z \\overline{\\mathrm{O}} \\mathrm{H}$$

    \n

    If the above equation is balanced with integer coefficients, the value of $$z$$ is ________.

    ", "options": [], "answer": "8", "solution": "**Answer:** 8\n\n

    \n\n\n\n\n\n\n \n \n \n \n\n\n \n \n \n \n \n \n \n \n\n
    Reduction HalfOxidation Half
    $$2 \\mathrm{MnO}_4^{-} \\rightarrow 2 \\mathrm{MnO}_2$$$$2 \\mathrm{I}^{-} \\rightarrow \\mathrm{I}_2+2 \\mathrm{e}^{-}$$
    $$2 \\mathrm{MnO}_4^{-}+4 \\mathrm{H}_2 \\mathrm{O}+6 \\mathrm{e}^{-} \\rightarrow 2 \\mathrm{MnO}_2+8 \\mathrm{OH}^{-}$$$$6 \\mathrm{I}^{-} \\rightarrow 3 \\mathrm{I}_2+6 \\mathrm{e}^{-}$$

    \n

    Adding oxidation half and reduction half, net reaction is

    \n

    $$\\begin{aligned}\n& 2 \\mathrm{MnO}_4^{-}+6 \\mathrm{I}^{-}+4 \\mathrm{H}_2 \\mathrm{O} \\rightarrow 3 \\mathrm{I}_2+2 \\mathrm{MnO}_2+8 \\mathrm{OH}^{-} \\\\\n\\Rightarrow \\quad & \\mathrm{z}=8 \\\\\n\\Rightarrow \\quad & \\text { Ans } 8\n\\end{aligned}$$

    ", "topic": "Algebra", "subtopic": "Equations and Inequalities" }, { "id": 2432, "subject": "Chemistry", "question": "The volume of 0.1N dibasic acid sufficient to neutralize 1 g of a base that furnishes 0.04 mole of OH− in aqueous solution is :", "options": [ { "text": "200 mL " }, { "text": "400 mL" }, { "text": "600 mL" }, { "text": "800 mL" } ], "answer": "400 mL", "solution": "**Answer:** 400 mL\n\nAccording to law of equivalence, \n

    Equivalence of acid = Equivalence of base.\n

    Equivalence of acid = Normality x volume = 0.1 $$ \\times $$ v\n

    As we know base produce OH$$-$$ ion, so moles of base is same as moles of OH$$-$$ ion = 0.04\n

    Another formula of equivalence = n factor $$ \\times $$ number of moles\n

    $$\\therefore\\,\\,\\,$$ Equivalance of base = n factor of OH$$-$$ $$ \\times $$ moles of OH$$-$$ = 1 $$ \\times $$ 0.04\n

    As for any ion, the charge of that ion is the n factor of that ion. Here OH$$-$$ has 1 negative charge so it's n factor = 1\n

    $$\\therefore\\,\\,\\,$$ 0.1 $$ \\times $$ v = 1 $$ \\times $$ 0.04\n

    $$ \\Rightarrow $$$$\\,\\,\\,$$ v = 0.4 L\n

    =   0.4 $$ \\times $$ 1000\n

    =    400 ml.", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 2433, "subject": "Chemistry", "question": "The hardness of a water sample (in terms of equivalents of CaCO3) containing 10–3 M CaSO4 is (Molar mass of CaSO4 = 136 g mol–1) ", "options": [ { "text": "90 ppm" }, { "text": "100 ppm" }, { "text": "50 ppm" }, { "text": "10 ppm" } ], "answer": "100 ppm", "solution": "**Answer:** 100 ppm\n\nnCaSO4 $$ \\times $$ Van't hoff factor = nCaCO3 $$ \\times $$ Van't hoff factor\n

    $$ \\Rightarrow $$ 10-3 $$ \\times $$ 2 = nCaCO3 $$ \\times $$ 2\n

    $$ \\Rightarrow $$ nCaCO3 = 10-3 mol in 1 L\n

    $$ \\therefore $$ wCaCO3 = 100 $$ \\times $$ 10-3 g CaCO3 in 1 L solution\n

    $$ \\therefore $$ hardness in terms of CaCO3\n

    = $${{{w_{CaC{O_3}}}} \\over {{w_{Total}}}} \\times {10^6}$$\n

    = $${{100 \\times {{10}^{ - 3}}} \\over {1000}} \\times {10^6}$$\n

    = 100 ppm", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2434, "subject": "Chemistry", "question": "In order to oxidise a mixture of one mole of each of FeC2O4, Fe2(C2O4)3, FeSO4 and Fe2(SO4)3 in\nacidic medium, the number of moles of KMnO4 required is :", "options": [ { "text": "1.5" }, { "text": "3" }, { "text": "2" }, { "text": "1" } ], "answer": "2", "solution": "**Answer:** 2\n\n\"JEE\n

    $$ \\therefore $$ Equivalent of KMnO4 = Eq of FeC2O4 + Eq of Fe2(C2O4)3 + FeSO4\n

    $$ \\Rightarrow $$ moles of KMnO4 $$ \\times $$ 5 = 3 $$ \\times $$ 1 + 6 $$ \\times $$ 1 + 1 $$ \\times $$ 1\n

    $$ \\Rightarrow $$ moles of KMnO4 = $${{10} \\over 5}$$ = 2", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2435, "subject": "Chemistry", "question": "A 100 mL solution was made by adding 1.43 g\nof Na2CO3.xH2O. The normality of the solution\nis 0.1 N. The value of x is _____.\n

    (The atomic mass of Na is 23 g/mol)\n", "options": [], "answer": "10", "solution": "**Answer:** 10\n\nMolar mass of Na2CO3.xH2O\n

    = 23 × 2 + 12 + 48 + 18x\n

    = 46 + 12 + 48 + 18x\n

    = (106 + 18x)\n

    As nfactor in dissolution will be determined from net cationic or anionic charge; which is 2\n

    Eq wt = $${M \\over 2}$$ = (53 + 9x)\n

    volume = 100 ml\n= 0.1 Litre\n

    Normality = \n
    No. of equivalents of solute
    \n
    Volume of solution (in L)
    \n    \n

    $$ \\Rightarrow $$ 0.1= $${{{{1.43} \\over {53 + 9x}}} \\over {0.1}}$$\n

    $$ \\Rightarrow $$ 53 + 9x = 143\n

    $$ \\Rightarrow $$ 9x = 90\n

    $$ \\Rightarrow $$ x = 10", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2436, "subject": "Chemistry", "question": "

    0.01 M KMnO4 solution was added to 20.0 mL of 0.05 M Mohr's salt solution through a burette. The initial reading of 50 mL burette is zero. The volume of KMnO4 solution left in the burette after the end point is _____________ mL. (nearest integer)

    ", "options": [], "answer": "30", "solution": "**Answer:** 30\n\nMeq of oxidizing agent $=$ Meq of reducing agent\n

    \n$\\left(\\mathrm{M} \\times \\mathrm{V} \\times \\mathrm{n}_{\\text{F}}\\right)_{\\mathrm{KMnO}_{4}}=\\left(\\mathrm{M} \\times \\mathrm{V} \\times \\mathrm{n}_{\\text{F}}\\right)_{\\text {Mohr's salt }}$\n

    \n$0.01 \\times 20 \\times 5=0.05 \\times V \\times 1$\n

    \nVolume required $=20 ~ \\mathrm{ml}$\n

    \nSince initial volume of $\\mathrm{KMnO}_{4}$ in burette is $50 ~\\mathrm{ml}$.\n

    \nHence volume of $\\mathrm{KMnO}_{4}$ left in the burette after end point is $30 ~\\mathrm{ml}$.", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 2437, "subject": "Chemistry", "question": "

    The neutralization occurs when 10 mL of 0.1M acid 'A' is allowed to react with 30 mL of 0.05 M base M(OH)2. The basicity of the acid 'A' is __________.

    \n

    [M is a metal]

    ", "options": [], "answer": "3", "solution": "**Answer:** 3\n\nMilieq of acid $A=$ Milieq of base $M(O H)_{2}$\n

    \n$$\n\\begin{aligned}\n& \\left(\\mathrm{M} \\times \\mathrm{V} \\times \\mathrm{n}-\\text { Factor }_{A}\\right.=(\\mathrm{M} \\times \\mathrm{V} \\times \\mathrm{n}-\\text { Factor })_{\\mathrm{M}(\\mathrm{OH})_{2}} \\\\\\\\\n& \\left[\\mathrm{n}-\\text { Factor of } \\mathrm{M}(\\mathrm{OH})_{2}=2\\right] \\\\\\\\\n&0.1 \\times 10 \\times \\mathrm{n} \\text {-Factor }= 0.05 \\times 30 \\times 2 \\\\\\\\\n& (\\mathrm{n}-\\text { Factor })_{\\mathrm{A}} =3\n\\end{aligned}\n$$\n

    \nHence basicity of acid $A$ is 3.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2438, "subject": "Chemistry", "question": "

    The normality of $$\\mathrm{H}_{2} \\mathrm{SO}_{4}$$ in the solution obtained on mixing $$100 \\mathrm{~mL}$$ of $$0.1 \\,\\mathrm{M} \\,\\mathrm{H}_{2} \\mathrm{SO}_{4}$$ with $$50 \\mathrm{~mL}$$ of $$0.1 \\,\\mathrm{M}\\, \\mathrm{NaOH}$$ is _______________ $$\\times 10^{-1} \\mathrm{~N}$$. (Nearest Integer)

    ", "options": [], "answer": "1", "solution": "**Answer:** 1\n\nNo. of equivalents of $\\mathrm{H}_2 \\mathrm{SO}_4=100 \\times 0.1 \\times 2=20$

    \nNo. of equivalents of $\\mathrm{NaOH}=50 \\times 0.1=5$

    \nNo. of equivalents of $\\mathrm{H}_2 \\mathrm{SO}_4$ left $=20-5=15$

    \n$$\n\\begin{aligned}\n&\\Rightarrow 150 \\times \\mathrm{x}=15 \\\\\\\\\n&\\mathrm{x}=\\frac{1}{10}=0.1 \\mathrm{~N}=1 \\times 10^{-1} \\mathrm{~N}\n\\end{aligned}\n$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2439, "subject": "Chemistry", "question": "Which of the following is a redox reaction ?", "options": [ { "text": "NaCl + KNO3 $$\\to$$ NaNO3 + KCl" }, { "text": "CaC2O4 + 2HCl $$\\to$$ CaCl2 + H2C2O4" }, { "text": "Mg(OH)2 + 2NH4Cl $$\\to$$ MgCl2 + 2NH4OH" }, { "text": "Zn + 2AgCN $$\\to$$ 2Ag+ Zn(CN)2" } ], "answer": "Zn + 2AgCN $$\\to$$ 2Ag+ Zn(CN)2", "solution": "**Answer:** Zn + 2AgCN $$\\to$$ 2Ag+ Zn(CN)2\n\n\"AIEEE \n

    The oxidation state shows a change only in (d)", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2440, "subject": "Chemistry", "question": "Which of the following chemical reactions depicts the oxidizing behaviour of H2SO4?", "options": [ { "text": "NaCl + H2SO4 $$\\to$$ NaHSO4 + HCl" }, { "text": "2PCl5 + H2SO4 $$\\to$$ 2POCl3 + 2HCl + SO2Cl2" }, { "text": "2HI + H2SO4 $$\\to$$ I2 + SO2 + 2H2O" }, { "text": "Ca(OH)2 + H2SO4 $$\\to$$ CaSO4 + 2H2O" } ], "answer": "2HI + H2SO4 $$\\to$$ I2 + SO2 + 2H2O", "solution": "**Answer:** 2HI + H2SO4 $$\\to$$ I2 + SO2 + 2H2O\n\n$$2H{I^{ - 1}} + {H_2}\\mathop {S{O_4}}\\limits^{ + 6} \\to I_2^0 + \\mathop {S{O_2}}\\limits^{ + 4} + 2{H_2}O$$ \n

    in this reaction oxidation number of $$S$$ is decreasing from $$+6$$ to $$+4$$ hence undergoing reduction and for $$HI$$ oxidation Number of $$I$$ is increasing from $$-1$$ to $$0$$ hence undergoing oxidation therefore $${H_2}S{O_4}$$ is acting as oxidising agent.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2441, "subject": "Chemistry", "question": "In which of the following reactions H2O2 acts as a reducing agent?\n
    1. H2O2 + 2H+ + 2e- $$\\to$$ 2H2O\n
    2. H2O2 - 2e- $$\\to$$ O2 + 2H+\n
    3. H2O2 + 2e- $$\\to$$ 2OH-\n
    4. H2O2 + 2OH- - 2e- $$\\to$$ O2 + 2H2O", "options": [ { "text": "1, 2" }, { "text": "3, 4" }, { "text": "1, 3" }, { "text": "2, 4" } ], "answer": "2, 4", "solution": "**Answer:** 2, 4\n\nThe reducing agent loses electron during redox reaction i.e. oxidises itself. \n

    $$\\left( 1 \\right)\\,\\,\\,\\,\\,\\,{H_2}\\mathop {{O_2}}\\limits^{ - 1} + 2{H^ + } + 2{e^ - }\\,\\,\\buildrel \\, \\over\n \\longrightarrow \\,\\,2{H_2}\\mathop O\\limits^{ - 2} \\,\\,({\\mathop{\\rm Re}\\nolimits} d.)$$\n

    $$\\left( 2 \\right)\\,\\,\\,\\,\\,\\,{H_2}\\mathop {{O_2}}\\limits^{ - 1} \\,\\,\\buildrel \\, \\over\n \\longrightarrow \\mathop {{O_2}}\\limits^0 + 2{H^ + } + 2{e^ - }\\,\\,(Ox.)$$\n

    $$\\left( 3 \\right)\\,\\,\\,\\,\\,\\,{H_2}\\mathop {{O_2}}\\limits^{ - 1} + 2{e^ - }\\,\\,\\buildrel \\, \\over\n \\longrightarrow \\,\\,2\\mathop O\\limits^{ - 2} {H^ - }\\,\\,(Red.)$$\n

    $$\\left( 4 \\right)\\,\\,\\,\\,\\,\\,{H_2}O_2^{ - 1} + 2O{H^ - }\\,\\buildrel \\, \\over\n \\longrightarrow \\mathop {{O_2}}\\limits^0 + {H_2}O + 2{e^ - }\\,\\,\\left( {Ox.} \\right)$$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2442, "subject": "Chemistry", "question": "Which of the following reactions is an example of a redox reaction?", "options": [ { "text": "$$Xe{F_6}$$ + H2O $$\\to$$ $$XeO{F_4}$$ + 2HF" }, { "text": "$$Xe{F_6}$$ + 2H2O $$\\to$$ $$XeO_2{F_2}$$ + 4HF" }, { "text": "$$Xe{F_4}$$ + O2F2 $$\\to$$ $$Xe{F_6}$$ + O2 " }, { "text": "$$Xe{F_2}$$ + PF5 $$\\to$$ $${\\left[ {XeF} \\right]^ + }PF_6^ - $$" } ], "answer": "$$Xe{F_4}$$ + O2F2 $$\\to$$ $$Xe{F_6}$$ + O2 ", "solution": "**Answer:** $$Xe{F_4}$$ + O2F2 $$\\to$$ $$Xe{F_6}$$ + O2 \n\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2443, "subject": "Chemistry", "question": "In the reaction of oxalate with permanganate in acidic medium, the number of electrons involved in producing one molecule of CO2 is : ", "options": [ { "text": "10" }, { "text": "2" }, { "text": "1" }, { "text": "5" } ], "answer": "1", "solution": "**Answer:** 1\n\n

    The balanced reaction of oxalate with permanganate in acidic medium is :

    \n

    $$5 \\, \\text{C}_2\\text{O}_4^{2-} + 2 \\, \\text{MnO}_4^- + 16 \\, \\text{H}^+ \\rightarrow 10 \\, \\text{CO}_2 + 2 \\, \\text{Mn}^{2+} + 8 \\, \\text{H}_2\\text{O}$$

    \n

    In this redox reaction, the oxalate ion $\\text{C}_2\\text{O}_4^{2-}$ is oxidized to carbon dioxide $\\text{CO}_2$, and the permanganate ion $\\text{MnO}_4^-$ is reduced to manganese(II) ion $\\text{Mn}^{2+}$.

    \n

    The oxidation half-reaction, representing the change for the oxalate ion, is :

    \n

    $$\\text{C}_2\\text{O}_4^{2-} \\rightarrow 2 \\, \\text{CO}_2 + 2 \\, e^-$$

    \n

    From this half-reaction, it's clear that each oxalate ion produces two CO2 molecules and releases two electrons in the process.

    \n

    So, the number of electrons involved in producing one molecule of CO2 is :

    \n

    $$\\frac{2 \\, e^-}{2 \\, \\text{CO}_2} = 1 \\, e^-/\\text{CO}_2$$

    \n

    Therefore, the correct answer is Option C: One electron is involved in the production of one molecule of CO2.

    \n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2444, "subject": "Chemistry", "question": "The redox reaction among the following is :", "options": [ { "text": "reaction of H2SO4 with NaOH." }, { "text": "formation of ozone form atmosphere oxygen in the presence of sunlight." }, { "text": "combination of dinitrogen with dioxygen at 2000 K" }, { "text": "reaction of [Co(H2O)6]Cl3 With AgNO3" } ], "answer": "combination of dinitrogen with dioxygen at 2000 K", "solution": "**Answer:** combination of dinitrogen with dioxygen at 2000 K\n\nN2 + O2 $$ \\to $$ 2NO\n

    during the reaction, oxidation of nitrogen take place\nfrom 0 to 2 and reduction of oxygen take place\nfrom 0 to –2. It means this reaction is redox\nreaction.\n

    3O2 $$ \\to $$ 2O3 (Non - redox reaction)\n

    2NaOH + H2SO4 $$ \\to $$ Na2SO4 + 2H2O (neutralization reaction)\n

    [Co(H2O)6]Cl3 + 3AgNO3\n$$ \\to $$ 3AgCl $$ \\downarrow $$ + [Co(H2O)6](NO3)3\n

    Reaction of [CO(H2O)6]Cl3 with AgNO3 is not\nredox reaction. It is a precipitation reaction.\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2445, "subject": "Chemistry", "question": "Statement I : Sodium hydride can be used as an oxidising agent.

    Statement II : The lone pair of electrons on nitrogen in pyridine makes it basic.

    Choose the CORRECT answer from the options given below :", "options": [ { "text": "Statement I is false but statement II is true" }, { "text": "Both statement I and statement II are true" }, { "text": "Statement I is true but statement II is false" }, { "text": "Both statement I and statement II are false" } ], "answer": "Statement I is false but statement II is true", "solution": "**Answer:** Statement I is false but statement II is true\n\nNaH is a strong H– (hydride) donor. Hence\ncannot be used as an oxidising agent.

    \n\"JEE\n
    \nIn Pyridine, the lone pair of ‘N’ is localised, makes it\nbasic.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2446, "subject": "Chemistry", "question": "

    Which of the following cannot function as an oxidising agent?

    ", "options": [ { "text": "$$\\mathrm{SO}_4^{2-}$$\n" }, { "text": "$$\\mathrm{MnO}_4^{-}$$\n" }, { "text": "$$\\mathrm{N}^{3-}$$\n" }, { "text": "$$\\mathrm{BrO}_3^{-}$$" } ], "answer": "$$\\mathrm{N}^{3-}$$\n", "solution": "**Answer:** $$\\mathrm{N}^{3-}$$\n\n\n

    The ability of a species to function as an oxidizing agent depends on its ability to gain electrons (be reduced). For a species to be an oxidizing agent, it must contain an element that has a positive oxidation state and is capable of being reduced (gain electrons), thereby oxidizing another species.

    \n\n

    Let's look at each of the options :

    \n\n

    Option A: $$\\mathrm{SO}_4^{2-}$$ (sulfate ion) has sulfur in a +6 oxidation state. Sulfate is the fully oxidized form of sulfur in aqueous solution, and while it can undergo certain reductions under specific conditions (to form, for example, $\\mathrm{SO}_3^{2-}$ or sulfite), it is generally stable and not commonly considered a strong oxidizing agent.

    \n\n

    Option B: $$\\mathrm{MnO}_4^{-}$$ (permanganate ion) has manganese in a +7 oxidation state. This ion is a very strong oxidizing agent and is well known for its ability to oxidize a wide range of organic and inorganic substances, with manganese being reduced to a lower oxidation state (often to +2 as $\\mathrm{Mn}^{2+}$ in acidic solution).

    \n\n

    Option C: $$\\mathrm{N}^{3-}$$ is the nitride ion, with nitrogen in a -3 oxidation state. Since nitrogen is in its lowest oxidation state, it cannot be further reduced (gain more electrons), and hence cannot act as an oxidizing agent. In fact, it would more likely act as a reducing agent because it has room to be oxidized (lose electrons).

    \n\n

    Option D: $$\\mathrm{BrO}_3^{-}$$ (bromate ion) has bromine in a +5 oxidation state. This ion can act as an oxidizing agent because the bromine can be reduced to a lower oxidation state (often to -1 as $\\mathrm{Br}^{-}$).

    \n\n

    Therefore, the species that cannot function as an oxidizing agent due to its inability to undergo reduction is Option C, the $$\\mathrm{N}^{3-}$$ ion.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2447, "subject": "Chemistry", "question": "

    In alkaline medium, $$\\mathrm{MnO}_4^{-}$$ oxidises $$\\mathrm{I}^{-}$$ to

    ", "options": [ { "text": "$$I_2$$\n" }, { "text": "$$\\mathrm{IO}_3^{-}$$\n" }, { "text": "$$\\mathrm{IO}^{-}$$\n" }, { "text": "$$\\mathrm{IO}_4^{-}$$" } ], "answer": "$$\\mathrm{IO}_3^{-}$$\n", "solution": "**Answer:** $$\\mathrm{IO}_3^{-}$$\n\n\n

    $$$2 \\mathrm{MnO}_4^{-}+\\mathrm{H}_2 \\mathrm{O}+\\mathrm{I}^{-} \\xrightarrow{\\text { alkaline medium }} 2 \\mathrm{MnO}_2+2 \\mathrm{OH}^{-}+\\mathrm{IO}_3^{-}$$$

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2448, "subject": "Chemistry", "question": "

    When $$\\mathrm{MnO}_2$$ and $$\\mathrm{H}_2 \\mathrm{SO}_4$$ is added to a salt $$(\\mathrm{A})$$, the greenish yellow gas liberated as salt (A) is :

    ", "options": [ { "text": "$$\\mathrm{NH}_4 \\mathrm{Cl}$$\n" }, { "text": "$$\\mathrm{CaI}_2$$\n" }, { "text": "$$\\mathrm{KNO}_3$$\n" }, { "text": "$$\\mathrm{NaBr}$$" } ], "answer": "$$\\mathrm{NH}_4 \\mathrm{Cl}$$\n", "solution": "**Answer:** $$\\mathrm{NH}_4 \\mathrm{Cl}$$\n\n\n

    When a salt reacts with $$\\mathrm{MnO}_2$$ and concentrated $$\\mathrm{H}_2 \\mathrm{SO}_4$$, the type of gas evolved depends on the anion present in the salt. In this case, a greenish yellow gas is evolved, which indicates the formation of chlorine gas ($$\\mathrm{Cl}_2$$). $$\\mathrm{Cl}_2$$ gas is greenish-yellow in color and is typically produced from halide salts (namely chlorides, bromides, or iodides) when they are oxidized. Among the given options, we need to identify a salt that contains a halide which can produce $$\\mathrm{Cl}_2$$ upon reaction with $$\\mathrm{MnO}_2$$ and concentrated $$\\mathrm{H}_2 \\mathrm{SO}_4$$.

    \n\n

    Option A ($$\\mathrm{NH}_4 \\mathrm{Cl}$$) contains chloride ions, and when heated with $$\\mathrm{MnO}_2$$ and concentrated $$\\mathrm{H}_2 \\mathrm{SO}_4$$, it can undergo a redox reaction where the chloride ions are oxidized to $$\\mathrm{Cl}_2$$ gas. The corresponding reaction can be represented as follows:

    \n\n

    $$\\mathrm{MnO}_2 + 4\\mathrm{HCl} \\longrightarrow \\mathrm{MnCl}_2 + \\mathrm{Cl}_2 + 2\\mathrm{H}_2\\mathrm{O}$$

    \n\n

    Option B ($$\\mathrm{CaI}_2$$) contains iodide ions, which would lead to the liberation of iodine or $$\\mathrm{I}_2$$, not a greenish yellow gas.

    \n\n

    Option C ($$\\mathrm{KNO}_3$$) contains nitrate ions and does not produce a halogen gas upon reaction with $$\\mathrm{MnO}_2$$ and $$\\mathrm{H}_2 \\mathrm{SO}_4$$.

    \n\n

    Option D ($$\\mathrm{NaBr}$$) contains bromide ions which would lead to the formation of bromine ($$\\mathrm{Br}_2$$), a reddish-brown gas, not greenish yellow.

    \n\n

    Therefore, the correct answer is Option A ($$\\mathrm{NH}_4 \\mathrm{Cl}$$), since it is the salt that can produce a greenish yellow gas ($$\\mathrm{Cl}_2$$) when reacted with $$\\mathrm{MnO}_2$$ and concentrated $$\\mathrm{H}_2 \\mathrm{SO}_4$$.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2449, "subject": "Chemistry", "question": "

    Thiosulphate reacts differently with iodine and bromine in the reactions given below:

    \n

    $$\\begin{aligned}\n& 2 \\mathrm{~S}_2 \\mathrm{O}_3^{2-}+\\mathrm{I}_2 \\rightarrow \\mathrm{S}_4 \\mathrm{O}_6^{2-}+2 \\mathrm{I}^{-} \\\\\n& \\mathrm{S}_2 \\mathrm{O}_3^{2-}+5 \\mathrm{Br}_2+5 \\mathrm{H}_2 \\mathrm{O} \\rightarrow 2 \\mathrm{SO}_4^{2-}+4 \\mathrm{Br}^{-}+10 \\mathrm{H}^{+}\n\\end{aligned}$$

    \n

    Which of the following statement justifies the above dual behaviour of thiosulphate?

    ", "options": [ { "text": "Thiosulphate undergoes oxidation by bromine and reduction by iodine in these reactions\n" }, { "text": "Bromine is a weaker oxidant than iodine\n" }, { "text": "Bromine is a stronger oxidant than iodine\n" }, { "text": "Bromine undergoes oxidation and iodine undergoes reduction in these reactions" } ], "answer": "Bromine is a stronger oxidant than iodine\n", "solution": "**Answer:** Bromine is a stronger oxidant than iodine\n\n\n

    In the given reactions:

    \n

    $$\\begin{aligned}\n& 2 \\mathrm{S}_2\\mathrm{O}_3^{2-} + \\mathrm{I}_2 \\rightarrow \\mathrm{S}_4\\mathrm{O}_6^{2-} + 2 \\mathrm{I}^{-} \\\\\\\\\n& \\mathrm{S}_2\\mathrm{O}_3^{2-} + 5 \\mathrm{Br}_2 + 5 \\mathrm{H}_2\\mathrm{O} \\rightarrow 2 \\mathrm{SO}_4^{2-} + 4 \\mathrm{Br}^{-} + 10 \\mathrm{H}^{+}\n\\end{aligned}$$

    \n

    In the first reaction with iodine ($I_2$), thiosulfate ($S_2O_3^{2-}$) acts as a reducing agent, converting $I_2$ to $I^{-}$, while it is itself oxidized to $S_4O_6^{2-}$. This indicates that iodine is acting as an oxidant (oxidizing agent), getting reduced in the process.

    \n\n

    In the second reaction with bromine ($Br_2$), thiosulfate is oxidized more extensively, being converted all the way to sulfate ($SO_4^{2-}$), with hydrogen ions ($H^+$) and bromide ions ($Br^-$) also produced. This shows that bromine acts as a stronger oxidant compared to iodine, as it promotes a complete oxidation of $S_2O_3^{2-}$ to $SO_4^{2-}$.

    \n\n

    Given these observations, the correct option that justifies the dual behavior of thiosulfate reacting with iodine and bromine differently is:

    \n\n

    Option C: Bromine is a stronger oxidant than iodine

    \n\n

    This statement correctly explains why thiosulfate is oxidized to $S_4O_6^{2-}$ in the presence of iodine, and to $SO_4^{2-}$ in the presence of bromine. Bromine's stronger oxidizing ability leads to a more complete oxidation of thiosulfate compared to the oxidation by iodine.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2450, "subject": "Chemistry", "question": "

    The number of ions from the following that are expected to behave as oxidising agent is :

    \n

    $$\\mathrm{Sn}^{4+}, \\mathrm{Sn}^{2+}, \\mathrm{Pb}^{2+}, \\mathrm{Tl}^{3+}, \\mathrm{Pb}^{4+}, \\mathrm{Tl}^{+}$$

    ", "options": [ { "text": "3" }, { "text": "4" }, { "text": "2" }, { "text": "1" } ], "answer": "2", "solution": "**Answer:** 2\n\n

    Due to inert pair effect $$\\mathrm{Pb}^{2+}$$ is more stable than $$\\mathrm{Pb}^{4+}$$ and $$\\mathrm{Tl}^{+}$$ is more stable than $$\\mathrm{Tl}^{3+}$$. Therefore, $$\\mathrm{Pb}^{4+}$$ and $$\\mathrm{T}^{3+}$$ will function as oxidising agents and easily get reduced to $$\\mathrm{Pb}^{2+}$$ and $$\\mathrm{TI}^{+}$$ respectively.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2451, "subject": "Chemistry", "question": "The correct order of the oxidation states of\nnitrogen in NO, N2O, NO2 and N2O3 is :", "options": [ { "text": "NO2 < NO < N2O3 < N2O" }, { "text": "NO2 < N2O3 < NO < N2O" }, { "text": "N2O < NO < N2O3 < NO2" }, { "text": "N2O < N2O3 < NO < NO2" } ], "answer": "N2O < NO < N2O3 < NO2", "solution": "**Answer:** N2O < NO < N2O3 < NO2\n\nIn N2O oxidation states of\nnitrogen = +1\n

    In NO oxidation states of\nnitrogen = +2\n

    In N2O3 oxidation states of\nnitrogen = +3\n

    In NO2 oxidation states of\nnitrogen = +4", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2452, "subject": "Chemistry", "question": "Oxidation number of potassium in K2O. K2O2 and KO2 respectively is :", "options": [ { "text": "+1, +2 and + 4" }, { "text": "+1, +1 and + 1" }, { "text": "+2, +1 and $$ + {1 \\over 2}$$" }, { "text": "+1, +4, and +2" } ], "answer": "+1, +1 and + 1", "solution": "**Answer:** +1, +1 and + 1\n\nPotasisum has an oxidation of +1 (only) in\ncombined state.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2453, "subject": "Chemistry", "question": "The compound that cannot act both as oxidising\nand reducing agent is :", "options": [ { "text": "H3PO4" }, { "text": "H2SO3" }, { "text": "H2O2" }, { "text": "HNO2" } ], "answer": "H3PO4", "solution": "**Answer:** H3PO4\n\nIn H3PO4, P is present in +5 oxidation state and\nit can act as reducing agent only.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2454, "subject": "Chemistry", "question": "The oxidation states of transition metal atoms\nin K2Cr2O7, KMnO4 and K2FeO4, respectively,\nare x, y and z. The sum of x, y and z is _______.", "options": [], "answer": "19", "solution": "**Answer:** 19\n\nK2Cr2O7\n

    Let oxidation state of Ce\n

    2(+1) + 2x + 7(–2) = 0\n

    $$ \\Rightarrow $$ x = +6\n

    In K2Cr2O7, Transition metal (Cr) present in +6 oxidation state.\n

    KMnO4\n

    (+1) + y + 4(–2) = 0\n
    $$ \\Rightarrow $$ x = +7\n

    In KMnO4, transition metal (Mn) present in +7\noxidation state.\n

    K2FeO4\n

    2(+1) + z + 4(–2) = 0\n
    $$ \\Rightarrow $$x = +6\n

    In K2FeO4, transition metal (Fe) present in +6\noxidation state.\n

    $$ \\therefore $$ x + y + z = 6 + 7 + 6 = 19", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2455, "subject": "Chemistry", "question": "Dichromate ion is treated with base, the oxidation number of Cr in the product formed is ___________.", "options": [], "answer": "6", "solution": "**Answer:** 6\n\n$$C{r_2}O_7^{2 - } + 2O{H^ - }$$ ⇌ $$2CrO_4^{2 - } + {H_2}O$$\n

    Let Oxidation state of Cr in $$CrO_4^{2 - }$$ is = x.\n

    $$ \\therefore $$ x + (–2 × 4) = –2\n

    $$ \\Rightarrow $$ x = 6", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2456, "subject": "Chemistry", "question": "The oxidation states of nitrogen in NO, NO2, N2O and NO$$_3^ - $$ are in the order of :", "options": [ { "text": "N2O > NO2 > NO > NO$$_3^ - $$" }, { "text": "NO > NO2 > N2O > NO$$_3^ - $$" }, { "text": "NO2 > NO$$_3^ - $$ > NO > N2O" }, { "text": "NO$$_3^ - $$ > NO2 > NO > N2O" } ], "answer": "NO$$_3^ - $$ > NO2 > NO > N2O", "solution": "**Answer:** NO$$_3^ - $$ > NO2 > NO > N2O\n\nThe oxidation states of Nitrogen in following\nmolecules are as follows

    \n$$NO_3^ - \\to $$ +5

    \nNO2 $$ \\to $$ +4

    \nNO $$ \\to $$ +2

    \nN2O $$ \\to $$ +1", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2457, "subject": "Chemistry", "question": "Identify the process in which change in the oxidation state is five :", "options": [ { "text": "$$C{r_2}O_7^{2 - } \\to 2C{r^{3 + }}$$" }, { "text": "$$MnO_4^ - \\to M{n^{2 + }}$$" }, { "text": "$$CrO_4^{2 - } \\to C{r^{3 + }}$$" }, { "text": "$${C_2}O_4^{2 - } \\to 2C{O_2}$$" } ], "answer": "$$MnO_4^ - \\to M{n^{2 + }}$$", "solution": "**Answer:** $$MnO_4^ - \\to M{n^{2 + }}$$\n\n$$MnO_4^ - + 5e \\to M{n^{ + 2}}$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2458, "subject": "Chemistry", "question": "The oxidation states of 'P' in H4P2O7, H4P2O5 and H4P2O6, respectively, are :", "options": [ { "text": "7, 5 and 6" }, { "text": "5, 4 and 3" }, { "text": "5, 3 and 4" }, { "text": "6, 4 and 5" } ], "answer": "5, 3 and 4", "solution": "**Answer:** 5, 3 and 4\n\nOxidation state of P in H4P2O7, H4P2O5 and H4P2O6 is 5, 3 & 4 respectively

    H4P2O7

    2x + 4(+ 1) + 7 ($$-$$2) = 0

    x = + 5

    $${H_4}\\underline {{P_2}} {O_5}$$

    2x + 4(+ 1) + 5 ($$-$$2) = 0

    x = + 3

    $${H_4}\\underline {{P_2}} {O_6}$$

    2x + 4 (+ 1) + 6($$-$$2) = 0

    x = +4", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2459, "subject": "Chemistry", "question": "

    Sum of oxidation states of bromine in bromic acid and perbromic acid is ___________.

    ", "options": [], "answer": "12", "solution": "**Answer:** 12\n\n

    The sum of the oxidation states of bromine in bromic acid (HBrO3) and perbromic acid (HBrO4) is +12.

    \n\n

    In bromic acid, the oxidation state of bromine is +5, while in perbromic acid, it is +7. The sum of the oxidation states of bromine in both compounds is (+5) + (+7) = +12.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2460, "subject": "Chemistry", "question": "

    The oxidation state of phosphorus in hypophosphoric acid is + _____________.

    ", "options": [], "answer": "4", "solution": "**Answer:** 4\n\nHypophosphoric acid H4P2O6\n

    Oxidation state is +4\n

    \"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2461, "subject": "Chemistry", "question": "

    The sum of oxidation state of the metals in $$\\mathrm{Fe}(\\mathrm{CO})_{5}, \\mathrm{VO}^{2+}$$ and $$\\mathrm{WO}_{3}$$ is ___________.

    ", "options": [], "answer": "10", "solution": "**Answer:** 10\n\n

    To find the sum of the oxidation states of the metals in $\\mathrm{Fe(CO)}_{5}$, $\\mathrm{VO}^{2+}$, and $\\mathrm{WO}_{3}$, we first need to find the oxidation states of the metals in each compound.

    \n
      \n

    1. $\\mathrm{Fe(CO)}_{5}$: The compound $\\mathrm{CO}$ is a neutral ligand, so it does not contribute to the oxidation state of the metal. Hence, the oxidation state of Fe in $\\mathrm{Fe(CO)}_{5}$ is 0.

      \n

      2. \n

    2. $\\mathrm{VO}^{2+}$: The compound $\\mathrm{O}$ has an oxidation state of -2, and since the overall charge of the compound is +2, the oxidation state of V must be +4 to balance this out.

      \n

      3. \n

    3. $\\mathrm{WO}_{3}$: The compound $\\mathrm{O}$ has an oxidation state of -2, and since there are three oxygen atoms, the total oxidation state contributed by oxygen is -6. To balance this out, the oxidation state of W must be +6.

      \n
      4. \n
    \n

    Therefore, the sum of the oxidation states of the metals in these three compounds is $0 + 4 + 6 = 10$.

    \n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2462, "subject": "Chemistry", "question": "

    In ammonium - phosphomolybdate, the oxidation state of Mo is + ___________

    ", "options": [], "answer": "6", "solution": "**Answer:** 6\n\nIn ammonium phosphomolybdate, the compound typically found in the reagent used for the quantitative determination of phosphorus, the molybdenum (Mo) atom is in the +6 oxidation state.\n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2463, "subject": "Chemistry", "question": "

    In the reaction of potassium dichromate, potassium chloride and sulfuric acid (conc.), the oxidation state of the chromium in the product is $$(+)$$ _________.

    ", "options": [], "answer": "6", "solution": "**Answer:** 6\n\n

    $$\\begin{aligned}\n& \\mathrm{K}_2 \\mathrm{Cr}_2 \\mathrm{O}_7(\\mathrm{~s})+4 \\mathrm{KCl}(\\mathrm{s})+6 \\mathrm{H}_2 \\mathrm{SO}_4 \\text { (conc.) } \\\\\n& \\rightarrow 2 \\mathrm{CrO}_2 \\mathrm{Cl}_2(\\mathrm{~g})+6 \\mathrm{KHSO}_4+3 \\mathrm{H}_2 \\mathrm{O}\n\\end{aligned}$$

    \n

    This reaction is called chromyl chloride test.

    \n

    Here oxidation state of $$\\mathrm{Cr}$$ is +6.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2464, "subject": "Chemistry", "question": "

    The oxidation number of iron in the compound formed during brown ring test for NO$$_3^-$$ iron is ________.

    ", "options": [], "answer": "1", "solution": "**Answer:** 1\n\n

    $$\\left[\\mathrm{Fe}\\left(\\mathrm{H}_2 \\mathrm{O}\\right)_5(\\mathrm{NO})\\right]^{2+} \\text {, }$$

    \n

    Oxidation no. of $$\\mathrm{Fe}=+1$$

    ", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 2465, "subject": "Chemistry", "question": "50 mL of 0.5 M oxalic acid is needed to neutralize 25 mL of sodium hydroxide solution. The amount of NaOH in 50 mL of the given sodium hydroxide solution is -", "options": [ { "text": "20 g " }, { "text": "4 g" }, { "text": "80 g" }, { "text": "10 g" } ], "answer": "4 g", "solution": "**Answer:** 4 g\n\n

    The reaction between sodium hydroxide (NaOH) and oxalic acid (H2C2O4) is as follows :

    \n

    $$ \\text{H}_2\\text{C}_2\\text{O}_4 + 2\\text{NaOH} \\rightarrow \\text{Na}_2\\text{C}_2\\text{O}_4 + 2\\text{H}_2\\text{O} $$

    \n

    We can see that 1 mole of oxalic acid ($\\text{H}_2\\text{C}_2\\text{O}_4$) reacts with 2 moles of sodium hydroxide (NaOH).

    \n

    Given that 50 mL of 0.5 M oxalic acid is needed to neutralize the sodium hydroxide, we can find the number of moles of oxalic acid that reacted :

    \n

    $$ \\text{Moles of H}_2\\text{C}_2\\text{O}_4 = \\text{Molarity} \\times \\text{Volume (L)} = 0.5 \\, \\text{mol/L} \\times 0.05 \\, \\text{L} = 0.025 \\, \\text{mol} $$

    \n

    Since 1 mole of $\\text{H}_2\\text{C}_2\\text{O}_4$ reacts with 2 moles of $\\text{NaOH}$, the number of moles of $\\text{NaOH}$ in 25 mL solution would be twice that of $\\text{H}_2\\text{C}_2\\text{O}_4$ :

    \n

    $$ \\text{Moles of NaOH} = 0.025 \\, \\text{mol} \\times 2 = 0.05 \\, \\text{mol} $$

    \n

    Now, to find the mass of NaOH, we multiply the number of moles by the molar mass of NaOH (40 g/mol) :

    \n

    $$ \\text{Mass of NaOH} = \\text{Number of moles} \\times \\text{Molar mass} = 0.05 \\, \\text{mol} \\times 40 \\, \\text{g/mol} = 2 \\, \\text{g} $$

    \n

    However, the question asks for the amount of $\\text{NaOH}$ in 50 mL of the given solution. Since we've found the amount in 25 mL, we just need to double our result to find the amount in 50 mL :

    \n

    $$ 2 \\, \\text{g} \\times 2 = 4 \\, \\text{g} $$

    \n

    So, the correct answer is Option B : 4 g.

    \n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2466, "subject": "Chemistry", "question": "While titrating dilute HCl solution with aqueous\nNaOH, which of the following will not be\nrequired?", "options": [ { "text": "Pipette and distilled water\n" }, { "text": "Clamp and phenolphthalein" }, { "text": "Burette and porcelain tile" }, { "text": "Bunsen burner and measuring cylinder" } ], "answer": "Bunsen burner and measuring cylinder", "solution": "**Answer:** Bunsen burner and measuring cylinder\n\nBunsen Burner and measuring cylinder is not\nrequired for titration.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2467, "subject": "Chemistry", "question": "The volume, in mL, of 0.02 M K2Cr2O7 solution\nrequired to react with 0.288 g of ferrous\noxalate in acidic medium is _______.\n
    (Molar mass of Fe = 56 g mol–1)", "options": [], "answer": "50", "solution": "**Answer:** 50\n\nK2Cr2O7 + FeC2O4 $$ \\to $$ Cr+3 + Fe+3 + CO2\n

    nfactor of K2Cr2O7 = 3 $$ \\times $$ 2 = 6\n

    nfactor of FeC2O4 = 1 + 2 = 3\n

    m. eq. of K2Cr2O7 = m. eq. of FeC2O4\n

    $$ \\Rightarrow $$ $${{6 \\times 0.02 \\times vol} \\over {1000}}$$ = $${{0.288} \\over {144}} \\times 3$$\n

    $$ \\Rightarrow $$ vol = 50 ml", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 2468, "subject": "Chemistry", "question": "0.4 g mixture of NaOH, Na2CO3 and some inert impurities was first titrated with $${N \\over {10}}$$ HCl using phenolphthalein as an indicator, 17.5 mL of HCl was required at the end point. After this methyl orange was added and titrated. 1.5 mL of same HCl was required for the next end point. The weight percentage of Na2CO3 in the mixture is ________. (Rounded off to the nearest integer)", "options": [], "answer": "4", "solution": "**Answer:** 4\n\n1st end point reaction :

    $$NaOH + HCl\\buildrel {} \\over\n \\longrightarrow NaCl + {H_2}O$$

    $$nf = 1$$

    $$NaC{O_3} + HCl\\buildrel {} \\over\n \\longrightarrow NaHC{O_3}$$

    $$nf = 1$$

    Eq of HCl used $$ = {n_{NaOH}} \\times 1 + {n_{N{a_2}C{O_3}}} \\times 1$$

    $$17.5 \\times {1 \\over {10}} \\times {10^{ - 3}} = {n_{NaOH}} + {n_{N{a_2}C{O_3}}}$$

    2nd end point :

    $$NaHC{O_3} + HCl\\buildrel {} \\over\n \\longrightarrow {H_2}C{O_3}$$

    $$1.5 \\times {1 \\over {10}} \\times {10^{ - 3}} = {n_{NaHC{O_3}}} \\times 1 = {n_{NaHC{O_3}}}$$

    0.15 mmol = $${n_{N{a_2}C{O_3}}}$$

    $$0.15 = {n_{N{a_2}C{O_3}}}$$

    $${w_{N{a_2}C{O_3}}} = {{0.15 \\times 106 \\times {{10}^{ - 3}}} \\over {0.4}} \\times 100 \\times 10$$

    = 3.975%\n

    $$ \\simeq $$ 4%", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2469, "subject": "Chemistry", "question": "10.0 mL of 0.05 M KMnO4 solution was consumed in a titration with 10.0 mL of given oxalic acid dihydrate solution. The strength of given oxalic acid solution is _________ $$\\times$$ 10$$-$$2 g/L.

    (Round off to the nearest integer)", "options": [], "answer": "1575", "solution": "**Answer:** 1575\n\nneq KMnO4 = neq H2C2O4 . 2H2O

    or, $${{10 \\times 0.05} \\over {1000}} \\times 5 = {{10 \\times M} \\over {1000}} \\times 2$$

    $$\\therefore$$ Conc. of oxalic acid solution = 0.125 M

    = 0.125 $$\\times$$ 125 g/L = 15.75 g/L

    = 1575 $$\\times$$ 10$$-$$2 g/L", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2470, "subject": "Chemistry", "question": "

    Given below are two statements: one is labelled as Assertion $$\\mathbf{A}$$ and the other is labelled as Reason $$\\mathbf{R}$$.

    \n

    Assertion A: Phenolphthalein is a $$\\mathrm{pH}$$ dependent indicator, remains colourless in acidic solution and gives pink colour in basic medium.

    \n

    Reason R: Phenolphthalein is a weak acid. It doesn't dissociate in basic medium.

    \n

    In the light of the above statements, choose the most appropriate answer from the options given below.

    ", "options": [ { "text": "Both $$\\mathbf{A}$$ and $$\\mathbf{R}$$ are true and $$\\mathbf{R}$$ is the correct explanation of $$\\mathbf{A}$$." }, { "text": "Both $$\\mathbf{A}$$ and $$\\mathbf{R}$$ are true but $$\\mathbf{R}$$ is NOT the correct explanation of $$\\mathbf{A}$$." }, { "text": "A is true but R is false." }, { "text": "A is false but R is true." } ], "answer": "A is true but R is false.", "solution": "**Answer:** A is true but R is false.\n\nPhenolphthalein is a pH dependent indicator. It is a\nweak acid which is colourless in the acidic solution\nbut gives pink colour in basic medium. The pink\ncolour is due to its conjugate form. Therefore,\nassertion (A) is true but Reason (R) is false.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2471, "subject": "Chemistry", "question": "

    $$20 \\mathrm{~mL}$$ of $$0.02\\, \\mathrm{M}$$ hypo solution is used for the titration of $$10 \\mathrm{~mL}$$ of copper sulphate solution, in the presence of excess of KI using starch as an indicator. The molarity of $$\\mathrm{Cu}^{2+}$$ is found to be ____________ $$\\times 10^{-2} \\,\\mathrm{M}$$. [nearest integer]

    \n

    Given : $$2 \\,\\mathrm{Cu}^{2+}+4 \\,\\mathrm{I}^{-} \\rightarrow \\mathrm{Cu}_{2} \\mathrm{I}_{2}+\\mathrm{I}_{2}$$

    \n

    $$\n\\mathrm{I}_{2}+2 \\mathrm{~S}_{2} \\mathrm{O}_{3}^{2-} \\rightarrow 2 \\mathrm{I}^{-}+\\mathrm{S}_{4} \\mathrm{O}_{6}^{2-}\n$$

    ", "options": [], "answer": "4", "solution": "**Answer:** 4\n\n$2 \\mathrm{Cu}^{2+}+4 \\mathrm{I}^{-} \\rightarrow \\mathrm{Cu}_{2} \\mathrm{l}_{2}+\\mathrm{I}_{2}$\n

    \n$\\mathrm{I}_{2}+\\mathrm{S}_{2} \\mathrm{O}_{3}^{2-} \\rightarrow 2 \\mathrm{I}^{-}+\\mathrm{S}_{4} \\mathrm{O}_{6}^{2-}$\n

    \nMilliequivalents of hypo solution $=0.02 \\times 20=0.4$

    Milliequivalents of $\\mathrm{Cu}^{2+}$ in $10 \\mathrm{~mL}$ solution =\n

    \nMilliequivalents of $\\mathrm{I}_{2}=$ Milliequivalents of hypo = 0.4\n

    \n\nMillimoles of $\\mathrm{Cu}^{2+}$ ions in $10 \\mathrm{~mL}=0.4$\n

    \nMolarity of $\\mathrm{Cu}^{2+}$ ions $=\\frac{0.4}{10}=0.04 \\,\\mathrm{M}$\n

    \n$$\n=4 \\times 10^{-2} \\,\\mathrm{M}\n$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2472, "subject": "Chemistry", "question": "

    A compound 'X' is a weak acid and it exhibits colour change at pH close to the equivalence point during neutralization of NaOH with $$\\mathrm{CH}_{3} \\mathrm{COOH}$$. Compound 'X' exists in ionized form in basic medium. The compound 'X' is

    ", "options": [ { "text": "methyl orange" }, { "text": "methyl red" }, { "text": "phenolphthalein" }, { "text": "erichrome Black T" } ], "answer": "phenolphthalein", "solution": "**Answer:** phenolphthalein\n\nPhenolphthalein is weak acid give colour in basic\nmedium.", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 2473, "subject": "Chemistry", "question": "

    The density of a monobasic strong acid (Molar mass 24.2 g/mol) is 1.21 kg/L. The volume of its solution required for the complete neutralization of 25 mL of 0.24 M NaOH is __________ $$\\times$$ 10$$^{-2}$$ mL (Nearest integer)

    ", "options": [], "answer": "12", "solution": "**Answer:** 12\n\n$$\n\\begin{aligned}\n& \\text { millimole of } \\mathrm{NaOH}=0.24 \\times 25 \\\\\\\\\n& \\therefore \\text { millimole of acid }=0.24 \\times 25 \\\\\\\\\n& \\Rightarrow \\text { mass of acid }=0.24 \\times 25 \\times 24.2 \\mathrm{mg} \\\\\\\\\n& \\text { for pure acid, } \\\\\\\\\n& \\mathrm{V}=\\frac{\\mathrm{w}}{\\mathrm{d}} ;(\\mathrm{d}=1.21 \\mathrm{~kg} / \\mathrm{L}=1.21 \\mathrm{~g} / \\mathrm{ml}) \\\\\\\\\n& \\therefore \\mathrm{V}=\\frac{0.24 \\times 25 \\times 24.2}{1.12} \\times 10^{-3} \\\\\\\\\n& =120 \\times 10^{-3} ~\\mathrm{ml} \\\\\\\\\n& =12 \\times 10^{-2} ~\\mathrm{ml}\n\\end{aligned}\n$$", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 2474, "subject": "Chemistry", "question": "

    In alkaline medium, the reduction of permanganate anion involves a gain of __________ electrons.

    ", "options": [], "answer": "3", "solution": "**Answer:** 3\n\n

    In alkaline medium, the reduction of permanganate anion involves a gain of 3 electrons.

    \n

    The balanced chemical equation for the reduction of permanganate anion in alkaline medium is:

    \n$\n\\begin{equation}\n\\ce{MnO4^{-}(aq) + 3 e^{-} + 2 H2O(l) -> MnO2(s) + 4 OH^{-}(aq)}\n\\end{equation}$

    In this reaction, the permanganate anion (MnO4-) is reduced to manganese dioxide (MnO2) by gaining 3 electrons. The water molecules (H2O) are oxidized by losing 2 electrons. The hydroxide ions (OH-) are spectator ions that do not participate in the redox reaction.

    \n\n

    The reduction of permanganate anion in alkaline medium is a useful reaction for a variety of analytical and synthetic purposes. For example, it can be used to quantitatively determine the concentration of an analyte in a solution, or to synthesize a variety of organic compounds.

    \n", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2475, "subject": "Chemistry", "question": "

    Given below are two statements:

    \n

    Statement I : In redox titration, the indicators used are sensitive to change in $$\\mathrm{pH}$$ of the solution.

    \n

    Statement II : In acid-base titration, the indicators used are sensitive to change in oxidation potential.

    \n

    In the light of the above statements, choose the most appropriate answer from the options given below

    ", "options": [ { "text": "Both Statement I and Statement II are incorrect" }, { "text": "Statement I is correct but Statement II is incorrect" }, { "text": "Statement I is incorrect but Statement II is correct" }, { "text": "Both Statement I and Statement II are correct" } ], "answer": "Both Statement I and Statement II are incorrect", "solution": "**Answer:** Both Statement I and Statement II are incorrect\n\n

    The correct answer is Option A: Both Statement I and Statement II are incorrect.

    \n

    Explanation :

    \n

    In redox titrations, the indicators used are sensitive to a change in oxidation potential, not pH. A redox titration (also called an oxidation-reduction titration) can accurately determine the concentration of an unknown analyte by measuring the amount of an oxidizing or reducing agent that it consumes. These indicators work by undergoing a definite color change at a particular electrode potential.

    \n

    In contrast, in acid-base titrations, the indicators used are sensitive to a change in pH, not oxidation potential. Acid-base titrations are based on the neutralization reaction between the acid and the base. The point at which all the acid or base has reacted with the other component is called the equivalence point, and it often results in a sudden change in pH. This sudden change in pH can be detected using a pH-sensitive indicator.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2476, "subject": "Chemistry", "question": "

    Only $$2 \\mathrm{~mL}$$ of $$\\mathrm{KMnO}_4$$ solution of unknown molarity is required to reach the end point of a titration of $$20 \\mathrm{~mL}$$ of oxalic acid $$(2 \\mathrm{M})$$ in acidic medium. The molarity of $$\\mathrm{KMnO}_4$$ solution should be ________ M.

    ", "options": [], "answer": "8", "solution": "**Answer:** 8\n\n

    $$\\begin{aligned}\n& \\mathrm{(M) \\times(2) \\times(5)=2 \\times 20 \\times 2} \\\\\n& \\mathrm{M=8}\n\\end{aligned}\n$$

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2477, "subject": "Chemistry", "question": "An example of a disproportionation reaction is :", "options": [ { "text": "2NaBr + Cl2 $$ \\to $$ 2NaCl + Br2" }, { "text": "2KMnO4 $$ \\to $$ 2KMnO4 + MnO2 + O2\n(3) (4)\n" }, { "text": "2CuBr $$ \\to $$ CuBr2 + Cu" }, { "text": "2MnO4\n + 10I–\n + 16H+ $$ \\to $$ 2Mn2+ + 5I2 + 8H2O" } ], "answer": "2CuBr $$ \\to $$ CuBr2 + Cu", "solution": "**Answer:** 2CuBr $$ \\to $$ CuBr2 + Cu\n\nIn disproportionation reaction one element undergoes both oxidation and reduction.\n\"JEE\n

    Here disproportionation reaction is :\n\"JEE", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2478, "subject": "Chemistry", "question": "

    Which one of the following is an example of disproportionation reaction ?

    ", "options": [ { "text": "$$3MnO_4^{2 - } + 4{H^ + } \\to 2MnO_4^ - + Mn{O_2} + 2{H_2}O$$" }, { "text": "$$MnO_4^ - + 4{H^ + } + 4{e^ - } \\to Mn{O_2} + 2{H_2}O$$" }, { "text": "$$10{I^ - } + 2MnO_4^ - + 16{H^ + } \\to 2M{n^{2 + }} + 8{H_2}O + 5{I_2}$$" }, { "text": "$$8MnO_4^ - + 3{S_2}O_3^{2 - } + {H_2}O \\to 8Mn{O_2} + 6SO_4^{2 - } + 2O{H^ - }$$" } ], "answer": "$$3MnO_4^{2 - } + 4{H^ + } \\to 2MnO_4^ - + Mn{O_2} + 2{H_2}O$$", "solution": "**Answer:** $$3MnO_4^{2 - } + 4{H^ + } \\to 2MnO_4^ - + Mn{O_2} + 2{H_2}O$$\n\n\"JEE

    \n$$MnO_4^{2 - }$$ is an intermediate oxidation state and is\nconverted into compounds having higher and lower\noxidation states.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2479, "subject": "Chemistry", "question": "

    Which of the given reactions is not an example of disproportionation reaction?

    ", "options": [ { "text": "$$2 \\mathrm{H}_{2} \\mathrm{O}_{2} \\rightarrow 2 \\mathrm{H}_{2} \\mathrm{O}+\\mathrm{O}_{2}$$" }, { "text": "$$2 \\mathrm{NO}_{2}+\\mathrm{H}_{2} \\mathrm{O} \\rightarrow \\mathrm{HNO}_{3}+\\mathrm{HNO}_{2}$$" }, { "text": "$$\\mathrm{MnO}_{4}^{-}+4 \\mathrm{H}^{+}+3 \\mathrm{e}^{-} \\rightarrow \\mathrm{MnO}_{2}+2 \\mathrm{H}_{2} \\mathrm{O}$$" }, { "text": "$$3 \\mathrm{MnO}_{4}^{2-}+4 \\mathrm{H}^{+} \\rightarrow 2 \\mathrm{MnO}_{4}^{-}+\\mathrm{MnO}_{2}+2 \\mathrm{H}_{2} \\mathrm{O}$$" } ], "answer": "$$\\mathrm{MnO}_{4}^{-}+4 \\mathrm{H}^{+}+3 \\mathrm{e}^{-} \\rightarrow \\mathrm{MnO}_{2}+2 \\mathrm{H}_{2} \\mathrm{O}$$", "solution": "**Answer:** $$\\mathrm{MnO}_{4}^{-}+4 \\mathrm{H}^{+}+3 \\mathrm{e}^{-} \\rightarrow \\mathrm{MnO}_{2}+2 \\mathrm{H}_{2} \\mathrm{O}$$\n\n$$\n\\begin{aligned}\n& 2 \\mathrm{H}_2 \\overset{-1}{\\mathrm{O}_2} \\longrightarrow 2 \\mathrm{H}_2 \\overset{2-}{\\mathrm{O}}+\\overset{0}{\\mathrm{O}_2}: \\text { Disproportionation } \\\\\\\\\n& 2 \\overset{+4}{\\mathrm{NO}_2}+\\mathrm{H}_2 \\mathrm{O} \\rightarrow \\overset{+5}{\\mathrm{HNO}_3}+\\overset{+3}{\\mathrm{HNO}_2} \\text { : Disproportionation } \\\\\\\\\n& \\mathrm{MnO}_4^{-}+4 \\mathrm{H}^{+}+3 \\mathrm{e}^{-} \\rightarrow \\mathrm{MnO}_2+2 \\mathrm{H}_2 \\mathrm{O}: \\text { reduction } \\\\\\\\\n& 3 \\overset{+6}{\\mathrm{MnO}_4^{2-}}+4 \\mathrm{H}^{+} \\rightarrow 2 \\overset{+7}{\\mathrm{MnO}_4^{-}}+\\overset{+4}{\\mathrm{MnO}_2}+2 \\mathrm{H}_2 \\mathrm{O}: \\text { Disproportionation }\n\\end{aligned}\n$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2480, "subject": "Chemistry", "question": "

    Which of the following options are correct for the reaction

    \n

    $$2\\left[\\mathrm{Au}(\\mathrm{CN})_{2}\\right]^{-}(\\mathrm{aq})+\\mathrm{Zn}(\\mathrm{s}) \\rightarrow 2 \\mathrm{Au}(\\mathrm{s})+\\left[\\mathrm{Zn}(\\mathrm{CN})_{4}\\right]^{2-}(\\mathrm{aq})$$

    \n

    A. Redox reaction

    \n

    B. Displacement reaction

    \n

    C. Decomposition reaction

    \n

    D. Combination reaction

    \n

    Choose the correct answer from the options given below:

    ", "options": [ { "text": "A and B only" }, { "text": "C and D only" }, { "text": "A only" }, { "text": "A and D only" } ], "answer": "A and B only", "solution": "**Answer:** A and B only\n\n$2\\left[\\stackrel{+1}{\\mathrm{Au}}(\\mathrm{CN})_2\\right]^{-}+\\stackrel{0}{\\mathrm{Z}} \\mathrm{n}(\\mathrm{s}) \\longrightarrow 2 \\stackrel{0}{\\mathrm{Au}}+\\left[\\stackrel{+2}{\\mathrm{Zn}}(\\mathrm{CN})_4\\right]^{-2}$

    $\\mathrm{Zn}$ displaced $\\mathrm{Au}^{+}$

    \nReduction and Oxidation both are taking place.", "topic": "Discrete Mathematics", "subtopic": "Combinatorics" }, { "id": 2481, "subject": "Chemistry", "question": "Which of the following reactions are disproportionation reactions?

    \n(A) $\\mathrm{Cu}^{+} \\rightarrow \\mathrm{Cu}^{2+}+\\mathrm{Cu}$

    \n(B) $3 \\mathrm{MnO}_4^{2-}+4 \\mathrm{H}^{+} \\longrightarrow 2

    \\mathrm{MnO}_4^{-}+\\mathrm{MnO}_2+2 \\mathrm{H}_2 \\mathrm{O}$

    \n(C) $2 \\mathrm{KMnO}_4 \\longrightarrow \\mathrm{K}_2 \\mathrm{MnO}_4+\\mathrm{MnO}_2+\\mathrm{O}_2$

    \n(D) $2 \\mathrm{MnO}_4^{-}+3 \\mathrm{Mn}^{2+}+2 \\mathrm{H}_2 \\mathrm{O} \\longrightarrow 5 \\mathrm{MnO}_2+4 \\mathrm{H}^{+}$\n

    \nChoose the correct answer from the options given below :", "options": [ { "text": "(A), (B)" }, { "text": "(A), (D)" }, { "text": "(B), (C), (D)" }, { "text": "(A), (B), (C)" } ], "answer": "(A), (B)", "solution": "**Answer:** (A), (B)\n\n

    A disproportionation reaction is a chemical reaction in which a single substance is simultaneously reduced and oxidized, forming two different products. To identify disproportionation reactions, one should look for a specified element in the reactant that gains electrons (reduction) and for the same element that lose electrons (oxidation).\n\n

    Here's an analysis of each reaction :

    \n\n

    (A) $\\mathrm{Cu}^{+} \\rightarrow \\mathrm{Cu}^{2+}+\\mathrm{Cu}$

    \n\n

    Here, copper (Cu) is in the +1 oxidation state and forms two products, where it is oxidized to the +2 state in $\\mathrm{Cu}^{2+}$ and reduced to the 0 state in Cu. This is a classic example of disproportionation. So, (A) is true for disproportionation.

    \n\n

    (B) $3 \\mathrm{MnO}_4^{2-}+4 \\mathrm{H}^{+} \\longrightarrow 2 \\mathrm{MnO}_4^{-}+\\mathrm{MnO}_2+2 \\mathrm{H}_2 \\mathrm{O}$

    \n\n

    In this reaction, $\\mathrm{MnO}_4^{2-}$ (where Mn is in +6 oxidation state) is converted into $\\mathrm{MnO}_4^{-}$ (where Mn is in +7 oxidation state, oxidation has occurred) and $\\mathrm{MnO}_2$ (where Mn is in +4 oxidation state, reduction has occurred). Here, again, we have the same element, manganese (Mn), undergoing both reduction and oxidation. Hence, (B) is also a disproportionation reaction.

    \n\n

    (C) $2 \\mathrm{KMnO}_4 \\longrightarrow \\mathrm{K}_2 \\mathrm{MnO}_4+\\mathrm{MnO}_2+\\mathrm{O}_2$

    \n\n

    For the reaction with potassium permanganate, $\\mathrm{KMnO}_4$, manganese starts in the +7 oxidation state. It forms $\\mathrm{K}_2 \\mathrm{MnO}_4$ where Mn is in +6 state (reduction) and $\\mathrm{MnO}_2$ where Mn is in +4 state (further reduction), but no increase in oxidation state of manganese is observed. Instead, oxygen is released, so the manganese is not being oxidized in any of its products; it is only reduced twice. This is not a disproportionation reaction because the same element should be undergoing both oxidation and reduction, which is not the case here. Hence, (C) is not a disproportionation reaction.

    \n\n

    (D) $2 \\mathrm{MnO}_4^{-}+3 \\mathrm{Mn}^{2+}+2 \\mathrm{H}_2 \\mathrm{O} \\longrightarrow 5 \\mathrm{MnO}_2+4 \\mathrm{H}^{+}$

    \n\n

    In this reaction, the $\\mathrm{MnO}_4^{-}$ ion (where Mn is at +7 oxidation state) and $\\mathrm{Mn}^{2+}$ ion (where Mn is at +2 oxidation state) react to form $\\mathrm{MnO}_2$ (where Mn is at +4 oxidation state). This reaction is not a case of disproportionation; it is a comproportionation or synproportionation reaction, where two species with the same element in different oxidation states produce a product with the element at an intermediate oxidation state.

    \n\n

    Therefore, the disproportionation reactions among the given choices are:

    \n\n\n

    Option A would be the correct choice: (A), (B).

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2482, "subject": "Chemistry", "question": "

    Total number of species from the following which can undergo disproportionation reaction is ________.

    \n

    $$\\mathrm{H}_2 \\mathrm{O}_2, \\mathrm{ClO}_3^{-}, \\mathrm{P}_4, \\mathrm{Cl}_2, \\mathrm{Ag}, \\mathrm{Cu}^{+1}, \\mathrm{~F}_2, \\mathrm{NO}_2, \\mathrm{K}^{+}$$

    ", "options": [], "answer": "6", "solution": "**Answer:** 6\n\n

    Intermediate oxidation state of element can undergo disproportionation.

    \n

    $$\\mathrm{H}_2 \\mathrm{O}_2, \\mathrm{ClO}_3^{-}, \\mathrm{P}_4, \\mathrm{Cl}_2, \\mathrm{Cu}^{+1}, \\mathrm{NO}_2$$

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2483, "subject": "Chemistry", "question": "

    Match List I with List II.

    \n

    \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    LIST I
    Reaction    		LIST II
    Type of redox reaction
    A.$$
    \\mathrm{N}_{2(\\mathrm{~g})}+\\mathrm{O}_{2(\\mathrm{~g})} \\rightarrow 2 \\mathrm{NO}_{(\\mathrm{g})}
    $$
    		I.Decomposition
    B.$$
    2 \\mathrm{~Pb}\\left(\\mathrm{NO}_3\\right)_{2(\\mathrm{~s})} \\rightarrow 2 \\mathrm{PbO}_{(\\mathrm{s})}+4 \\mathrm{NO}_{2(\\mathrm{~g})}+\\mathrm{O}_{2(\\mathrm{~g})}
    $$
    		II.Displacement
    C.$$
    2 \\mathrm{Na}_{(\\mathrm{s})}+2 \\mathrm{H}_2 \\mathrm{O}_{(\\mathrm{l})} \\rightarrow 2 \\mathrm{NaOH}_{(\\mathrm{aq} .)}+\\mathrm{H}_{2(\\mathrm{~g})}
    $$
    		III.Disproportionation
    D.$$
    2 \\mathrm{NO}_{2(\\mathrm{~g})}+2^{-} \\mathrm{OH}(\\text { aq. }) \\rightarrow \\mathrm{NO}_{2(\\mathrm{aq} .)}^{-}+\\mathrm{NO}_{3(\\text { aq. })}^{-}+\\mathrm{H}_2 \\mathrm{O}_{(\\mathrm{l})}
    $$
    		IV.Combination

    \n

    Choose the correct answer from the options given below :

    ", "options": [ { "text": "(A)-(IV), (B)-(I), (C)-(II), (D)-(III)\n" }, { "text": "(A)-(I), (B)-(II), (C)-(III), (D)-(IV)\n" }, { "text": "(A)-(II), (B)-(III), (C)-(IV), (D)-(I)\n" }, { "text": "(A)-(III), (B)-(II), (C)-(I), (D)-(IV)" } ], "answer": "(A)-(IV), (B)-(I), (C)-(II), (D)-(III)\n", "solution": "**Answer:** (A)-(IV), (B)-(I), (C)-(II), (D)-(III)\n\n\n

    (A) $$\\stackrel{\\circ}{\\mathrm{N}}_2+\\stackrel{\\circ}{\\mathrm{O}}_2 \\longrightarrow 2 \\mathrm{NO}_{(\\mathrm{g})}$$

    \n

    Combination reaction

    \n

    (B) $$2 \\mathrm{~Pb}\\left(\\mathrm{NO}_3\\right)_{2(\\mathrm{~s})} \\longrightarrow 2 \\mathrm{PbO}+4 \\mathrm{NO}_2+\\mathrm{O}_2$$

    \n

    Decomposition reaction

    \n

    (C) $$2 \\mathrm{Na}_{(\\mathrm{s})}+2 \\mathrm{H}_2 \\mathrm{O}(\\mathrm{I}) \\longrightarrow 2 \\mathrm{NaOH}_{(\\mathrm{aq})}+\\mathrm{H}_{2(\\mathrm{~g})}$$

    \n

    Displacement reaction

    \n

    (D) $$2 \\mathrm{NO}_{2(\\mathrm{~g})}+2 \\mathrm{OH}_{(\\mathrm{g})}^{-} \\longrightarrow \\mathrm{NO}_{2 \\text { (aq) }}^{-}+\\mathrm{NO}_{3(\\text { (qq) }}^{-}+\\mathrm{H}_2 \\mathrm{O}$$

    \n

    Nitrogen oxidises and reduces both. So it is a disproportionation reaction.

    \n

    $$\\mathrm{A} \\rightarrow(\\mathrm{IV}), \\mathrm{B} \\rightarrow(\\mathrm{I}), \\mathrm{C} \\rightarrow(\\mathrm{II}), \\mathrm{D} \\rightarrow(\\mathrm{III})$$

    ", "topic": "Discrete Mathematics", "subtopic": "Combinatorics" }, { "id": 2484, "subject": "Chemistry", "question": "KO2 (potassium super oxide) is used in oxygen cylinders in space and submarines because it :", "options": [ { "text": "absorbs CO2 and increases O2 content" }, { "text": "eliminates moisture" }, { "text": "absorbs CO2" }, { "text": "produces ozone" } ], "answer": "absorbs CO2 and increases O2 content", "solution": "**Answer:** absorbs CO2 and increases O2 content\n\n$$2K{O_2} + 2{H_2}O$$\n

    $$ \\to 2KOH + {H_2}{O_2} + {O_2}$$\n

    $$K{O_2}$$ is used as an oxidising agent. It is used as air purifier in space capsules. Submarines and breathing masks as it produces oxygen and removes carbon dioxide. ", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 2485, "subject": "Chemistry", "question": "Which of the following on thermal decomposition yields a basic as well as an acidic oxide?", "options": [ { "text": "NaNO3" }, { "text": "KClO3" }, { "text": "CaCO3" }, { "text": "NH4NO3" } ], "answer": "CaCO3", "solution": "**Answer:** CaCO3\n\nCalcium carbonate on thermal decomposition gives $$CaO$$ (Basic oxide) and $$C{O_2}$$ (Acidic oxide)\n

    $$CaC{O_3}\\buildrel \\Delta \\over\n \\longrightarrow \\mathop {\\,\\,\\,\\,\\,CaO\\,\\,\\,\\,\\,}\\limits_{Basic\\,\\,oxide} + \\,\\,\\mathop {\\,\\,\\,\\,C{O_2} \\uparrow }\\limits_{Acidic\\,\\,oxide} $$ ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2486, "subject": "Chemistry", "question": "A metal on combustion in excess air forms X. X upon hydrolysis with water yields H2O2 and O2 along with another product. The metal is :", "options": [ { "text": "Rb" }, { "text": "Mg" }, { "text": "Na" }, { "text": "Li" } ], "answer": "Rb", "solution": "**Answer:** Rb\n\nK, Rb and Cs form super oxides on reaction with excess\nair.\n

    Rb + O2 $$ \\to $$ RbO2\n

    2RbO2\n + 2H2O $$ \\to $$ 2RbOH + H2O2\n + O2$$ \\uparrow $$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2487, "subject": "Chemistry", "question": "A hydrated solid X on heating initially gives a monohydrated compound Y. Y upon heating above 373 K leads\nto an anhydrous white powder Z. X and Z, respectively, are :", "options": [ { "text": "Baking soda and dead burnt plaster.\n" }, { "text": "Washing soda and soda ash" }, { "text": "Washing soda and dead burnt plaster." }, { "text": "Baking soda and soda as. " } ], "answer": "Washing soda and soda ash", "solution": "**Answer:** Washing soda and soda ash\n\n\"JEE", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 2488, "subject": "Chemistry", "question": "One of the by-products formed during the recovery of NH3 from Solvay process is :", "options": [ { "text": "Ca(OH)2" }, { "text": "NH4Cl" }, { "text": "NaHCO3" }, { "text": "CaCl2" } ], "answer": "CaCl2", "solution": "**Answer:** CaCl2\n\nRecovery of NH3 :

    \n$$2N{H_4}Cl + Ca{\\left( {OH} \\right)_2} \\to 2N{H_3} + CaC{l_2} + {H_2}O$$

    \nBy product : CaCl2", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 2489, "subject": "Chemistry", "question": "A s-block element (M) reacts with oxygen to form an oxide of the formula MO2. The oxide is pale yellow in colour and paramagnetic. The element (M) is :", "options": [ { "text": "Mg" }, { "text": "Na" }, { "text": "Ca" }, { "text": "K" } ], "answer": "K", "solution": "**Answer:** K\n\n

    The element (M) is potassium (K). It reacts with O2 to form KO2, which is paramagnetic in nature. All other elements form oxides or peroxides which are diamagnetic in nature.

    \n

    \"JEE

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2490, "subject": "Chemistry", "question": "Match List I with List II :

    \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    List-I
    Elements    		List-II
    Properties
    (a)Li(i)Poor water solubility of $${I^ - }$$ salt
    (b)Na(ii)Most abundant element in cell fluid
    (c)K(iii)Bicarbonate salt used in fire extinguisher
    (d)Cs(iv)Carbonate salt decomposes easily on heating


    Choose the correct answer from the options given below :", "options": [ { "text": "(a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)" }, { "text": "(a)-(i), (b)-(iii), (c)-(ii), (d)-(iv)" }, { "text": "(a)-(iv), (b)-(ii), (c)-(iii), (d)-(i)" }, { "text": "(a)-(i), (b)-(ii), (c)-(iii), (d)-(iv)" } ], "answer": "(a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)", "solution": "**Answer:** (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)\n\n(a) CsI salt is poor water soluble due to it's low hydration energy

    (b) NaHCO3 is used in fire extinguisher

    (c) K is most abundant element in cell fluid

    (d) Li2CO3 decomposes easily due to high covalent character caused by small size Li+ cation.", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 2491, "subject": "Chemistry", "question": "Match List - I with List - II :

    \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    List - IList - II
    (a)Li(i)photoelectric cell
    (b)Na(ii)absorbent of $$C{O_2}$$
    (c)K(iii)coolant in fast breeder nuclear reactor
    (d)Cs(iv)treatment of cancer
    (v)bearings for motor engines


    Choose the correct answer from the options given below :", "options": [ { "text": "(a) - (v), (b) - (i), (c) - (ii), (d) - (iv)" }, { "text": "(a) - (v), (b) - (ii), (c) - (iv), (d) - (i)" }, { "text": "(a) - (iv), (b) - (iii), (c) - (i), (d) - (ii)" }, { "text": "(a) - (v), (b) - (iii), (c) - (ii), (d) - (i)" } ], "answer": "(a) - (v), (b) - (iii), (c) - (ii), (d) - (i)", "solution": "**Answer:** (a) - (v), (b) - (iii), (c) - (ii), (d) - (i)\n\nLi makes alloy with Lead to make white metal bearings for motor engines

    Liquid Na metal is used s coolant in fast breeder nuclear reactor

    K is a very absorbent of CO2

    Cs is used in making photoelectric cell", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 2492, "subject": "Chemistry", "question": "Choose the correct statement from the following :", "options": [ { "text": "The standard enthalpy of formation for alkali metal bromides becomes less negative on descending the group." }, { "text": "The low solubility of CsI in water is due to its high lattice enthalpy." }, { "text": "Among the alkali metal halides, LiF is least soluble in water." }, { "text": "LiF has least negative standard enthalpy of formation among alkali metal fluorides." } ], "answer": "Among the alkali metal halides, LiF is least soluble in water.", "solution": "**Answer:** Among the alkali metal halides, LiF is least soluble in water.\n\na. Standard enthalpy of formation for alkali metal bromides becomes more negative on descending down the group.

    b. In case of CsI, lattice energy is less, but Cs+ is having less hydration enthalpy due to which it is less soluble in water.

    c. For alkali metal fluorides, the solubility in water increases from lithium to caesium. LiF is least soluble in water.

    d. Standard enthalpy of formation for LiF is most negative among alkali metal fluorides.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2493, "subject": "Chemistry", "question": "

    Amongst baking soda, caustic soda and washing soda, carbonate anion is present in :

    ", "options": [ { "text": "washing soda only." }, { "text": "washing soda and caustic soda only." }, { "text": "washing soda and baking soda only." }, { "text": "baking soda, caustic soda and washing soda. " } ], "answer": "washing soda only.", "solution": "**Answer:** washing soda only.\n\n

    Baking soda $$ \\to $$ NaHCO3

    \n

    Washing soda $$ \\to $$ Na2CO3.10H2O

    \n

    Caustic soda $$ \\to $$ NaOH

    \n

    $$\\text{CO}^{-2}_{3} $$ ion is present only in washing soda.

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2494, "subject": "Chemistry", "question": "

    Which one of the following compounds is used as a chemical in certain type of fire extinguishers?

    ", "options": [ { "text": "Baking soda" }, { "text": "Soda ash" }, { "text": "Washing soda" }, { "text": "Caustic soda" } ], "answer": "Baking soda", "solution": "**Answer:** Baking soda\n\nBaking soda $$\\left(\\mathrm{NaHCO}_{3}\\right)$$ is used in certain type of fire extinguishers because it decomposes at high temperature to produce $$\\mathrm{CO}_{2}$$ which extinguishes fire

    \n$$\n2 \\mathrm{NaHCO}_{3}(\\mathrm{~s}) \\stackrel{\\Delta}{\\longrightarrow} \\mathrm{Na}_{2} \\mathrm{CO}_{3}(\\mathrm{~s})+\\mathrm{H}_{2} \\mathrm{O} \\uparrow + \\,\\mathrm{CO}_{2} \\uparrow\n$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2495, "subject": "Chemistry", "question": "

    Correct statement about alkali metal oxides is :

    ", "options": [ { "text": "peroxides are colored." }, { "text": "superoxides are paramagnetic." }, { "text": "oxides are paramagnetic." }, { "text": "peroxides are both colored and paramagnetic." } ], "answer": "superoxides are paramagnetic.", "solution": "**Answer:** superoxides are paramagnetic.\n\nThe peroxide and oxides of alkali metals are colourless when pure.\n

    \nSuperoxides are paramagnetic while peroxides are diamagnetic. Electronic configuration of $\\mathrm{O}_{2}^{2-}$ (peroxide)\n

    \n$\\left(\\sigma 1 s^{2}\\right)\\left(\\sigma^{*} 1 s^{2}\\right)\\left(\\sigma 2 s^{2}\\right)\\left(\\sigma^{*} 2 s^{2}\\right)\\left(\\sigma 2 p_{z}^{2}\\right)\\left(\\pi 2 p_{x}^{2} \\pi 2 p_{y}^{2}\\right)$ $\\left(\\pi^{*} 2 p_{x}^{2} \\pi^{*} 2 p_{y}^{2}\\right)$\n

    \nElectronic configuration of $\\mathrm{O}_{2}^{-}$

    $\\left(\\sigma 1 s^{2}\\right)\\left(\\sigma^{*} 1 s^{2}\\right)\\left(\\sigma 2 s^{2}\\right)\\left(\\sigma^{*} 2 s^{2}\\right)\\left(\\sigma 2 p_{z}^{2}\\right)\\left(\\pi 2 p_{x}^{2} \\quad \\pi 2 p_{y}^{2}\\right)$ $\\left(\\pi^{*} 2 p_{x}^{2} \\pi^{*} 2 p_{y}^{1}\\right)$\n

    \nsuperoxide is paramagnetic due 1 unpaired electron.", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2496, "subject": "Chemistry", "question": "

    Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.

    \n

    Assertion A: $$\\mathrm{LiF}$$ is sparingly soluble in water.

    \n

    Reason R: The ionic radius of $$\\mathrm{Li}^{+}$$ ion is smallest among its group members, hence has least hydration enthalpy.

    \n

    In the light of the above statements, choose the most appropriate answer from the options given below.

    ", "options": [ { "text": "Both A and R are true and R is the correct explanation of A." }, { "text": "Both A and R are true but R is NOT the correct explanation of A." }, { "text": "A is true but R is false." }, { "text": "A is false but R is true." } ], "answer": "A is true but R is false.", "solution": "**Answer:** A is true but R is false.\n\nLiF is sparingly soluble in water.\n

    \nThe low solubility of LiF in water is due to its high lattice enthalpy (Since $$\\mathrm{Li}^{+}$$ and $$\\mathrm{F}^{-}$$ are small in size). Also, due to small size of $$\\mathrm{Li}^{+}$$, its hydration enthalpy is high.\n

    \nHence, Assertion is true but Reason is false", "topic": "Algebra", "subtopic": "Abstract Algebra" }, { "id": 2497, "subject": "Chemistry", "question": "

    Lithium nitrate and sodium nitrate, when heated separately, respectively, give :

    ", "options": [ { "text": "$$\\mathrm{LiNO}_{2}$$ and $$\\mathrm{NaNO}_{2}$$" }, { "text": "$$\\mathrm{Li}_{2} \\mathrm{O}$$ and $$\\mathrm{Na}_{2} \\mathrm{O}$$" }, { "text": "$$\\mathrm{Li}_{2} \\mathrm{O}$$ and $$\\mathrm{NaNO}_{2}$$" }, { "text": "$$\\mathrm{LiNO}_{2}$$ and $$\\mathrm{Na}_{2} \\mathrm{O}$$" } ], "answer": "$$\\mathrm{Li}_{2} \\mathrm{O}$$ and $$\\mathrm{NaNO}_{2}$$", "solution": "**Answer:** $$\\mathrm{Li}_{2} \\mathrm{O}$$ and $$\\mathrm{NaNO}_{2}$$\n\n$$\\mathrm{Li}_{2} \\mathrm{O}, \\mathrm{NaNO}_{2}$$\n

    \nAs per NCERT lithium nitrate when heated gives lithium oxide, $$\\mathrm{Li}_{2} \\mathrm{O}$$. Whereas other alkali metal nitrates decompose to give the corresponding nitrite.\n

    \n$$4 \\mathrm{LiNO}_{3} \\longrightarrow 2 \\mathrm{Li}_{2} \\mathrm{O}+4 \\mathrm{NO}_{2}+\\mathrm{O}_{2}$$\n

    \n$$2 \\mathrm{NaNO}_{3} \\longrightarrow 2 \\mathrm{NaNO}_{2}+\\mathrm{O}_{2}$$\n

    \nHowever, the decomposition product of $$\\mathrm{NaNO}_{3}$$ is temperature dependent process as shown in the below reaction.

    \n\n$$\\mathrm{NaNO}_{3} \\underset{500^{\\circ} \\mathrm{C}}{\\stackrel{\\Delta}{\\longrightarrow}} \\mathrm{NaNO}_{2}(\\mathrm{~s})+\\frac{1}{2} \\mathrm{O}_{2}(\\mathrm{~g})$$\n
    \n                      $$\\Delta \\downarrow, 800^{\\circ} \\mathrm{C}$$\n
    \n                      $$\\mathrm{Na}_{2} \\mathrm{O}(\\mathrm{s})+\\mathrm{N}_{2}(\\mathrm{~g})+\\mathrm{O}_{2}(\\mathrm{~g})$$\n

    \nAs the temperature is not mentioned, we can go by the answer. (C)", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2498, "subject": "Chemistry", "question": "

    Match List I with List II :

    \n

    \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
    List IList II
    A.Slaked limeI.$$\\mathrm{NaOH}$$
    B.Dead burnt plasterII.$$\\mathrm{Ca(OH)_2}$$
    C.Caustic sodaIII.$$\\mathrm{Na_2CO_3\\cdot 10H_2O}$$
    D.Washing sodaIV.$$\\mathrm{CaSO_4}$$

    \n

    Choose the correct answer from the options given below :

    ", "options": [ { "text": "(A) - III, (B) - IV, (C) - II, (D) - I" }, { "text": "(A) - III, (B) - II, (C) - IV, (D) - I" }, { "text": "(A) - II, (B) - IV, (C) - I, (D) - III" }, { "text": "(A) - I, (B) - IV, (C) - II, (D) - III" } ], "answer": "(A) - II, (B) - IV, (C) - I, (D) - III", "solution": "**Answer:** (A) - II, (B) - IV, (C) - I, (D) - III\n\nA : Slaked lime $\\quad: \\mathrm{Ca}(\\mathrm{OH})_{2}$\n\n

    B : Dead burnt plaster $\\quad$: $\\mathrm{CaSO}_{4}$\n\n

    C : Caustic Soda $\\quad$: $\\mathrm{NaOH}$\n\n

    D : Washing Soda $\\quad$: $\\mathrm{Na}_{2} \\mathrm{CO}_{3} \\cdot 10 \\mathrm{H}_{2} \\mathrm{O}$", "topic": "Geometry", "subtopic": "Solid Geometry" }, { "id": 2499, "subject": "Chemistry", "question": "Which of the following reaction is correct?", "options": [ { "text": "$4 \\mathrm{LiNO}_{3} \\stackrel{\\Delta}{\\longrightarrow} 2 \\mathrm{Li}_{2} \\mathrm{O}+4 \\mathrm{NO}_{2}+\\mathrm{O}_{2}$" }, { "text": "$2 \\mathrm{LiNO}_{3} \\longrightarrow 2 \\mathrm{Li}+2 \\mathrm{NO}_{2}+\\mathrm{O}_{2}$" }, { "text": "$4 \\mathrm{LiNO}_{3} \\stackrel{\\Delta}{\\longrightarrow} 2 \\mathrm{Li}_{2} \\mathrm{O}+2 \\mathrm{~N}_{2} \\mathrm{O}_{4}+\\mathrm{O}_{2}$" }, { "text": "$2 \\mathrm{LiNO}_{3} \\stackrel{\\Delta}{\\longrightarrow} 2 \\mathrm{NaNO}_{2}+\\mathrm{O}_{2}$" } ], "answer": "$4 \\mathrm{LiNO}_{3} \\stackrel{\\Delta}{\\longrightarrow} 2 \\mathrm{Li}_{2} \\mathrm{O}+4 \\mathrm{NO}_{2}+\\mathrm{O}_{2}$", "solution": "**Answer:** $4 \\mathrm{LiNO}_{3} \\stackrel{\\Delta}{\\longrightarrow} 2 \\mathrm{Li}_{2} \\mathrm{O}+4 \\mathrm{NO}_{2}+\\mathrm{O}_{2}$\n\n

    $$\\mathrm{4LiNO_3}$$ $$\\buildrel {\\mathrm{Heat}} \\over\n \\longrightarrow $$ $$\\mathrm{2Li_2O+4NO_2+O_2}$$

    \n

    $$\\mathrm{LiNO_3}$$ on heating produces $$\\mathrm{Li_2O,NO_2}$$ and $$\\mathrm{O_2}$$

    ", "topic": "Algebra", "subtopic": "Intermediate Algebra" }, { "id": 2500, "subject": "Chemistry", "question": "

    Lithium aluminium hydride can be prepared from the reaction of :

    ", "options": [ { "text": "$$\\mathrm{LiCl}$$ and $$\\mathrm{Al_2H_6}$$" }, { "text": "$$\\mathrm{LiH}$$ and $$\\mathrm{Al}(\\mathrm{OH})_{3}$$" }, { "text": "$$\\mathrm{LiH}$$ and $$\\mathrm{Al}_{2} \\mathrm{Cl}_{6}$$" }, { "text": "$$\\mathrm{LiCl}, \\mathrm{Al}$$ and $$\\mathrm{H}_{2}$$" } ], "answer": "$$\\mathrm{LiH}$$ and $$\\mathrm{Al}_{2} \\mathrm{Cl}_{6}$$", "solution": "**Answer:** $$\\mathrm{LiH}$$ and $$\\mathrm{Al}_{2} \\mathrm{Cl}_{6}$$\n\n$$\\mathrm{8LiH+Al_2Cl_6\\to2LiAlH_4+6LiCl}$$", "topic": "Algebra", "subtopic": "Intermediate Algebra" } ]