diff --git a/.gitattributes b/.gitattributes index 4ceef72dd2fe0f62459636494db6a5172acbf8ae..2e49b0b4d51c4c3b3ed0db7380749dfc0f613afc 100644 --- a/.gitattributes +++ b/.gitattributes @@ -239,3 +239,43 @@ test/tasks/Exam-07-T8/images/Exam-07-T8-4-1-prompt.png filter=lfs diff=lfs merge test/tasks/Exam-07-T8/images/Exam-07-T8-5-1-prompt.png filter=lfs diff=lfs merge=lfs -text test/tasks/Exam-08-T3/images/Exam-08-T3-1-1-prompt.png filter=lfs diff=lfs merge=lfs -text test/tasks/Exam-08-T3/images/Exam-08-T3-2-1-prompt.png filter=lfs diff=lfs merge=lfs -text +test/tasks/Exam-11-T2/images/Exam-11-T2-1-1-prompt.png filter=lfs diff=lfs merge=lfs -text +test/tasks/Exam-14-T7/images/Exam-14-T7-1-1-prompt.png filter=lfs diff=lfs merge=lfs -text +test/tasks/Exam-14-T7/images/Exam-14-T7-2-1-prompt.png filter=lfs diff=lfs merge=lfs -text +test/tasks/Exam-14-T7/images/Exam-14-T7-3-1-prompt.png filter=lfs diff=lfs merge=lfs -text +test/tasks/Exam-17-T2/images/Exam-17-T2-header-1.png filter=lfs diff=lfs merge=lfs -text +test/tasks/Exam-17-T7/images/Exam-17-T7-1-1-prompt.png filter=lfs diff=lfs merge=lfs -text +test/tasks/Exam-17-T7/images/Exam-17-T7-3-1-prompt.png filter=lfs diff=lfs merge=lfs -text +test/tasks/Exam-17-T7/images/Exam-17-T7-7-1-prompt.png filter=lfs diff=lfs merge=lfs -text +test/tasks/Exam-17-T9/images/Exam-17-T9-1-1-prompt.png filter=lfs diff=lfs merge=lfs -text +test/tasks/Exam-17-T9/images/Exam-17-T9-1-2-prompt.png filter=lfs diff=lfs merge=lfs -text +test/tasks/Exam-17-T9/images/Exam-17-T9-2-1-prompt.png filter=lfs diff=lfs merge=lfs -text +test/tasks/Exam-17-T9/images/Exam-17-T9-4-1-prompt.png filter=lfs diff=lfs merge=lfs -text +test/tasks/Exam-17-T9/images/Exam-17-T9-5-1-prompt.png filter=lfs diff=lfs merge=lfs -text +test/tasks/Exam-18-T5/images/Exam-18-T5-header-1.png filter=lfs diff=lfs merge=lfs -text +test/tasks/Exam-18-T9/images/Exam-18-T9-1-1-prompt.png filter=lfs diff=lfs merge=lfs -text +test/tasks/Exam-18-T9/images/Exam-18-T9-2-1-prompt.png filter=lfs diff=lfs merge=lfs -text +test/tasks/Exam-18-T9/images/Exam-18-T9-3-1-prompt.png filter=lfs diff=lfs merge=lfs -text +test/tasks/Exam-18-T9/images/Exam-18-T9-4-1-prompt.png filter=lfs diff=lfs merge=lfs -text +test/tasks/Exam-18-T9/images/Exam-18-T9-5-1-prompt.png filter=lfs diff=lfs merge=lfs -text +test/tasks/Exam-18-T9/images/Exam-18-T9-6-1-prompt.png filter=lfs diff=lfs merge=lfs -text +test/tasks/Exam-18-T9/images/Exam-18-T9-7-1-prompt.png filter=lfs diff=lfs merge=lfs -text +test/tasks/Exam-18-T9/images/Exam-18-T9-8-1-prompt.png filter=lfs diff=lfs merge=lfs -text +test/tasks/Exam-19-T2/images/Exam-19-T2-1-1-prompt.png filter=lfs diff=lfs merge=lfs -text +test/tasks/Exam-19-T2/images/Exam-19-T2-1-2-prompt.png filter=lfs diff=lfs merge=lfs -text +test/tasks/Exam-19-T2/images/Exam-19-T2-3-1-prompt.png filter=lfs diff=lfs merge=lfs -text +test/tasks/Exam-19-T2/images/Exam-19-T2-header-1.png filter=lfs diff=lfs merge=lfs -text +test/tasks/Exam-19-T2/images/Exam-19-T2-header-2.png filter=lfs diff=lfs merge=lfs -text +test/tasks/Exam-19-T2/images/Exam-19-T2-header-3.png filter=lfs diff=lfs merge=lfs -text +test/tasks/Exam-20-T3/images/Exam-20-T3-1-1-prompt.png filter=lfs diff=lfs merge=lfs -text +test/tasks/Exam-20-T3/images/Exam-20-T3-2-1-prompt.png filter=lfs diff=lfs merge=lfs -text +test/tasks/Exam-20-T3/images/Exam-20-T3-4-1-prompt.png filter=lfs diff=lfs merge=lfs -text +test/tasks/Exam-20-T3/images/Exam-20-T3-5-1-prompt.png filter=lfs diff=lfs merge=lfs -text +test/tasks/Exam-21-T10/images/Exam-21-T10-1-1-prompt.png filter=lfs diff=lfs merge=lfs -text +test/tasks/Exam-21-T10/images/Exam-21-T10-3-1-prompt.png filter=lfs diff=lfs merge=lfs -text +test/tasks/Exam-21-T10/images/Exam-21-T10-header-1.png filter=lfs diff=lfs merge=lfs -text +test/tasks/Exam-21-T3/images/Exam-21-T3-3-1-prompt.png filter=lfs diff=lfs merge=lfs -text +test/tasks/Exam-21-T5/images/Exam-21-T5-2-1-prompt.png filter=lfs diff=lfs merge=lfs -text +test/tasks/Exam-21-T6/images/Exam-21-T6-2-1-prompt.png filter=lfs diff=lfs merge=lfs -text +test/tasks/Exam-21-T6/images/Exam-21-T6-3-1-prompt.png filter=lfs diff=lfs merge=lfs -text +test/tasks/Exam-22-T5/images/Exam-22-T5-10-1-prompt.png filter=lfs diff=lfs merge=lfs -text diff --git a/test/tasks/Exam-08-T5/problem.json b/test/tasks/Exam-08-T5/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..f0a3769e966acfb6de2579bd46a2243a2aee03ad --- /dev/null +++ b/test/tasks/Exam-08-T5/problem.json @@ -0,0 +1,34 @@ +{ + "problem_id": "Exam-08-T5", + "header": "Defective Cubic Perovskite\nA polymorph of $\\mathrm{CsFe}_x\\mathrm{Cl}_3$ has a defective cubic-perovskite structure. Relative to cubic perovskite, Cs occupies every Ca site, Cl every O site, and Fe occupies an ordered subset of the Ti sites. Equivalently, Cs and Cl jointly form close-packed layers in a 1:3 atomic ratio, and these layers are cubically stacked. Between every two adjacent Cs-Cl close-packed layers are octahedral voids, some formed only by Cl and others jointly by Cl and Cs. Fe fills some interlayer octahedral voids in the repeating sequence occupied, occupied, vacant, occupied, occupied, vacant, and so on. In each Fe-containing interlayer region, Fe occupies all and only the octahedral voids formed solely by Cl. This defect pattern changes the crystal to the hexagonal crystal family. The cell parameters are $a=736.2$ pm and $c=896.4$ pm. Two Fe atoms are located at $(2/3,1/3,0.18565)$ and $(1/3,2/3,0.81435)$, and their $\\mathrm{FeCl}_6$ octahedra are corner-sharing. In the projection down the crystallographic $c$ axis, the Fe atoms form a graphene-like pattern.", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-08-T5-1", + "tags": [ + "1.3" + ], + "text_open": "5.1 Determine the unit-cell composition using reasonable assumptions and approximations.", + "text_images": [], + "full_mark": 7 + }, + { + "sub_id": "Exam-08-T5-2", + "tags": [ + "1.3" + ], + "text_open": "5.4 Determine the crystal system.", + "text_images": [], + "full_mark": 2 + }, + { + "sub_id": "Exam-08-T5-3", + "tags": [ + "1.3" + ], + "text_open": "5.7 State the number of inversion centers in one primitive cell.", + "text_images": [], + "full_mark": 1 + } + ] +} diff --git a/test/tasks/Exam-11-T2/images/Exam-11-T2-1-1-prompt.png b/test/tasks/Exam-11-T2/images/Exam-11-T2-1-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..d694ca86e1ff7a6b734fd64114351e6d7828398a --- /dev/null +++ b/test/tasks/Exam-11-T2/images/Exam-11-T2-1-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:2d8636e9664746bb57b5519df9ae16b74b6e54a8ab2f09b7e0f30e89fd5d1134 +size 1051769 diff --git a/test/tasks/Exam-11-T2/problem.json b/test/tasks/Exam-11-T2/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..9f9db1d3d731763ae6365397e013663ecd33f73a --- /dev/null +++ b/test/tasks/Exam-11-T2/problem.json @@ -0,0 +1,31 @@ +{ + "problem_id": "Exam-11-T2", + "header": "Problem 2: An Electrochemistry Medley", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-11-T2-1", + "tags": [ + "3.6" + ], + "text_open": "2.3.1 Shown below are the pH-potential diagram for the Cr-H_2O system at 298.15 K (the concentration of each soluble species is 1 mmol/L) and thermodynamic data for selected substances.\n\n| Substance | $\\Delta_f H^\\circ_m /(kJ \\cdot mol^{-1})$ | $\\Delta S^\\circ_m /(J \\cdot mol^{-1} \\cdot K^{-1})$ |\n| :--- | :--- | :--- |\n| H_2O(l) | -285.83 | 69.95 |\n| OH^-(aq) | -229.98 | -10.9 |\n| H_2(g) | - | 130.68 |\n| B | -985.09 | 95.4 |\n| F | -872.95 | 190.4 |\n| G | -890.89 | 5.4 |\n\nNote: the diagram below is a potential diagram for a very dilute aqueous solution (1 mmol/L), and each species contains only one Cr atom. In the pH-potential diagram, the slopes of lines cd, de, and ef are in the ratio 7:4:5. Using the information given and your chemical knowledge, identify the species represented by A-G.", + "text_images": [ + { + "file": "Exam-11-T2-1-1-prompt.png", + "description": "This is a two-dimensional rectangular pH-potential diagram. The unlabeled horizontal axis ranges from 0 to 14, and the vertical axis, labeled “potential/V,” ranges from -1.5 to 1.5 V. The upper and lower boundaries are horizontal lines at 1.5 and -1.5, respectively; the left and right boundaries are vertical lines at 0 and 14, respectively. Thick black lines divide the rectangle into 8 regions: seven labeled A, B, C, D, E, F, and G, and one labeled $Cr^{2+}$. Region E occupies the upper-left corner and is a right trapezoid. Its left boundary is part of the left frame, its right boundary is a vertical line at about 0.4, its upper boundary is part of the top frame, and its lower boundary slopes downward from left to right. The upper-right vertex a is approximately (0.4, 1.5), and the lower-right vertex c is approximately (0.4, 1.3). Region F is pentagonal. Segment ac is its left boundary, segment ab on the top frame is its upper boundary, and segment be at about 6.6 is its right boundary. Its two lower boundaries are segments cd and de, both sloping downward from left to right. Point e is the lower-right vertex; d is approximately (3.2,0.9), and e is approximately (6.6,0.6). The vertical segment ac separates E from F. Segment cd separates A (lower left) from F (upper right), while segment de separates B (below) from F (above). At approximately 3.2, a vertical line extends downward from d to k, with h between them. Segment dh has A on the left and B on the right; segment hk has the $Cr^{2+}$ region on the left and C on the right. Point h is approximately (3.2,-0.4), and k is approximately (3.2,-0.9). At approximately 6.6, a vertical line extends from point b on the top frame down to e. Point b is approximately (6.6,1.5); segment be has F on the left and G on the right, and the vertical line does not continue below e. Region G lies above diagonal ef and below the top frame, extending from about 6.6 to the right boundary at 14; thus G lies above ef and B below it. Region A lies below cd and above horizontal gh, with part of the left frame as its left boundary and vertical dh as its right boundary. Point g is approximately (0.0,-0.4), and h is approximately (3.2,-0.4). Segment gh is horizontal, with A above and the $Cr^{2+}$ region below. Region B lies below the upper broken line def and above the lower diagonal hi. Its left boundary is vertical dh, and fi is part of the right frame. Thus B is to the right of dh and A to the left; B is below ef and G above; B is above hi and C below; B is to the left of vertical fi, with the area to its right outside the frame. Point f, approximately (14.0,-0.1), is the intersection of the upper diagonal with the right boundary. The $Cr^{2+}$ region lies below horizontal gh, above horizontal jk (about -0.9 V), and between 0 and 3.2. Part of the left frame is its left boundary, and hk is its right boundary. Thus $Cr^{2+}$ is below gh and A above; $Cr^{2+}$ is above jk and D below; $Cr^{2+}$ is left of hk and C right of it. Point j is approximately (0.0,-0.9), and k is approximately (3.2,-0.9). Region C lies below diagonal hi and above diagonal kl, extending from 3.2 to 14. Its left boundary is hk and its right boundary is il. Thus C is right of hk and $Cr^{2+}$ left of it; C is below hi and B above; C is above kl and D below; C is left of vertical il, with the area to its right outside the frame. Point i, approximately (14.0,-1.0), is the intersection of hi with the right boundary; point l, approximately (14.0,-1.5), is the intersection of kl with the right boundary. Region D lies below horizontal j-k and diagonal k-l. Its lower boundary is the bottom frame, and it spans 0 to 14. At horizontal coordinate 0 on the left boundary, D lies below j; D is below jk and the $Cr^{2+}$ region above; D is below kl and C above. Reading the coordinates from the axes to one decimal place gives approximately: a≈(0.4,1.5), b≈(6.6,1.5), c≈(0.4,1.3), d≈(3.2,0.9), e≈(6.6,0.6), f≈(14.0,-0.1), g≈(0.0,-0.4), h≈(3.2,-0.4), i≈(14.0,-1.0), j≈(0.0,-0.9), k≈(3.2,-0.9), and l≈(14.0,-1.5). Thus, from top to bottom, the three principal diagonal boundaries are an unlabeled lowercase point-c-d-e-f, h-i, and k-l. Above the first are regions E/F/G in sequence and below it are A/B: specifically, E lies above the unlabeled point-c segment, F above c-d-e, and G above e-f; A lies below the unlabeled point-c-d portion, and B below d-e-f. B is above h-i and C below; C is above k-l and D below. The principal horizontal boundaries are g-h (A above and $Cr^{2+}$ below) and j-k ($Cr^{2+}$ above and D below)." + } + ], + "full_mark": 7 + }, + { + "sub_id": "Exam-11-T2-2", + "tags": [ + "3.4", + "3.6" + ], + "text_open": "2.3.3 Calculate the coordinates of point e from the thermodynamic data.", + "text_images": [], + "full_mark": 12 + } + ] +} diff --git a/test/tasks/Exam-12-T1/problem.json b/test/tasks/Exam-12-T1/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..7d262d3bc7e8f36e829e14f37821d2092f78a35d --- /dev/null +++ b/test/tasks/Exam-12-T1/problem.json @@ -0,0 +1,17 @@ +{ + "problem_id": "Exam-12-T1", + "header": "Problem 1 Metallic Vitamins. Rare-earth elements are core strategic resources in modern materials and chemical industries and are known as \"industrial vitamins.\" Their unique electron configurations give them distinctive physicochemical properties, making them essential materials for emerging industries such as modern semiconductors and new energy. A rare-earth element M has density $7.563 \\ \\mathrm{g \\cdot cm^{-3}}$ and forms an alloy with Co in a fixed ratio for manufacturing strong magnets. It has the following chemical properties. (1) When elemental M is left in air, its metallic luster rapidly disappears and a composite thin layer whose principal component is oxide A forms on its surface. M reacts with hydrochloric acid to form blood-red chloride B. After M has dissolved completely, the blood-red color rapidly fades, producing pale-yellow chloride C and colorless gas D. (2) If M is placed in a tetrahydrofuran solution of an organic compound E containing $90.05\\%$ iodine, reaction at room temperature gives unstable green iodide F and releases gas G. Addition of a small amount of water turns the solution yellow. Gas G is a common plant-ripening hormone.", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-12-T1-1", + "tags": [ + "1.4", + "2.4" + ], + "text_open": "1-4-1 Complexes of M also have interesting structures. Addition of 0.5 equivalents of $\\mathrm{P_2Ph_4}$ to $\\mathrm{MCp*_2(THF)_2}$ gives a chain polymer X. After polymer X is separated, the pentane mother liquor is left for several weeks, yielding a yellow-brown complex Y. Y contains two M atoms and a 14-membered macrocycle. The mass fraction of M is $22.19\\%$ in both X and Y. Write the SMILES structure of the repeat unit of X, using [R] for the site connected to the next repeat unit (for example, polypropylene may be written as C([R])C(C)[R]), and the SMILES structure of Y ($\\mathrm{Cp*}$ = pentamethylcyclopentadienyl anion).", + "text_images": [], + "full_mark": 4 + } + ] +} diff --git a/test/tasks/Exam-12-T4/problem.json b/test/tasks/Exam-12-T4/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..bbd590e6562db33ba8de34ccb6cc425c5ebcd622 --- /dev/null +++ b/test/tasks/Exam-12-T4/problem.json @@ -0,0 +1,36 @@ +{ + "problem_id": "Exam-12-T4", + "header": "Problem 4 Titrimetric Analysis of Etching Acids\nWet chemical etching is a major process in the semiconductor-chip industry. It uses acidic substances to corrode a substrate surface and thereby achieve etching. A mixture of hydrofluoric acid, fluorosilicic acid, and nitric acid is a common etchant. A rapid and accurate laboratory method for determining the contents of these three acids in an etchant is described below.", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-12-T4-1", + "tags": [ + "4.2" + ], + "text_open": "Procedure background: Accurately weigh 3.028 g of the sample. Place 1.502 g of it in a plastic beaker to prepare test solution A, and use the remaining sample to prepare test solution B by the same procedure. During preparation, add 10 mL of saturated $KNO_3$ solution and anhydrous ethanol, mix, dilute with anhydrous ethanol, and let stand for approximately 20 min until the precipitate has completely settled into a separate layer. Label this as test solution A. Prepare test solution B from the remaining sample by the same procedure.\n4.1.2 State the purpose of adding saturated KNO_3 solution.", + "text_images": [], + "full_mark": 2 + }, + { + "sub_id": "Exam-12-T4-2", + "tags": [ + "2.1", + "4.2" + ], + "text_open": "4.2 Titrimetric analysis. Add 30 mL of water and 3 drops of phenolphthalein indicator to test solution A. Titrate with 0.5018 $\\mathrm{mol \\cdot L^{-1}}$ standard NaOH solution until the solution becomes faint pink; 24.39 mL of the standard NaOH solution is consumed. Add 30 mL of water and 2 drops of methyl-orange indicator to test solution B. Adjust the pH with NaOH solution until the solution just turns bright yellow. Using a fluoride-ion combination electrode as the measuring electrode, titrate with 0.08273 $\\mathrm{mol \\cdot L^{-1}}$ standard La(NO_3)_3 solution; precipitate forms continuously during the titration. At the endpoint, 25.11 mL of La(NO_3)_3 solution is consumed. Weigh 0.5003 g of the original sample into a beaker containing 15 mL of 0.5 $\\mathrm{mol \\cdot L^{-1}}$ NaOH solution, add water to 60 mL, and verify with pH paper that the pH is greater than 13. Heat slowly to boiling, maintain a gentle boil for 5 min, and cool to room temperature. Add 2 drops of methyl-red indicator, adjust with nitric acid until red, and then adjust with NaOH solution until the solution just turns bright yellow. Using a fluoride-ion combination electrode as the indicator electrode, titrate with 0.08273 $\\mathrm{mol \\cdot L^{-1}}$ standard La(NO_3)_3 solution. At the endpoint, 15.28 mL is consumed. 4.2.1 Write all reaction equations occurring in Sections 4.1 and 4.2; write repeated reactions only once.", + "text_images": [], + "full_mark": 4 + }, + { + "sub_id": "Exam-12-T4-3", + "tags": [ + "4.7", + "4.5" + ], + "text_open": "4.2.2 Calculate the mass fractions of HF, H_2SiF_6, and HNO_3 in the sample. Note: the fluoride-ion electrode indicates the endpoint by the abrupt change in fluoride-ion concentration.", + "text_images": [], + "full_mark": 6 + } + ] +} diff --git a/test/tasks/Exam-13-T4/problem.json b/test/tasks/Exam-13-T4/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..0f4149d4f1837a41dc20c6695df2e01d95a94cef --- /dev/null +++ b/test/tasks/Exam-13-T4/problem.json @@ -0,0 +1,25 @@ +{ + "problem_id": "Exam-13-T4", + "header": "Question 4 Determination of Iron in Iron Ore. The potassium dichromate method is a classic method for determining iron in iron ore, but its use of $HgCl_2$ creates toxicity and environmental concerns. Selected steps of an improved dichromate method, the $TiCl_3-K_2Cr_2O_7$ method, are as follows. ① Preparation of the $K_2Cr_2O_7$ standard solution: accurately weigh 0.9797 g of $K_2Cr_2O_7$ into a 100 mL beaker, dissolve it in water, quantitatively transfer it to a 250 mL volumetric flask, and dilute to volume to obtain the $K_2Cr_2O_7$ standard solution. ② Titration of the sample: accurately weigh 0.2026 g of powdered magnetite concentrate, add 20 mL of hydrochloric acid, approximately 6 mol/L, cover with a watch glass, and heat gently to near boiling. After the ore decomposes, the mixture is reddish brown; immediately add $SnCl_2$ solution dropwise until it becomes pale yellow. Once most of the ore has dissolved, keep the mixture at a gentle boil for 1-2 min until decomposition is complete and no black particles remain. If the yellow color is too deep, add a little more $SnCl_2$ solution to obtain a pale-yellow solution; if the solution is colorless, add a small amount of $KMnO_4$ until a pale-yellow color appears, and stop heating. Rinse the watch glass and beaker walls with a wash bottle, add 50 mL of water and 8 drops of $Na_2WO_4$ solution, and while swirling add $TiCl_3$ solution to endpoint 1, followed by 2 excess drops. Cool to room temperature and carefully add the $K_2Cr_2O_7$ standard solution to endpoint 2; V1 = 0.02 mL is consumed. Add another 50 mL of water, 10 mL of sulfuric-phosphoric mixed acid with concentrated sulfuric acid:concentrated phosphoric acid:water = 2:3:5 by volume, and 2 drops of sodium diphenylamine sulfonate. Immediately titrate with the $K_2Cr_2O_7$ standard solution to endpoint 3; V2 = 28.12 mL is consumed.", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-13-T4-1", + "tags": [ + "4.7" + ], + "text_open": "4.3 The following operations cause errors in the experiment. Explain the reason for each error. 4.3.1 The ore is heated at too high a temperature or for too long during decomposition with hydrochloric acid.", + "text_images": [], + "full_mark": 2 + }, + { + "sub_id": "Exam-13-T4-2", + "tags": [ + "4.7" + ], + "text_open": "4.3.2 The sulfuric-phosphoric mixed acid is not added before titration.", + "text_images": [], + "full_mark": 2 + } + ] +} diff --git a/test/tasks/Exam-14-T1/problem.json b/test/tasks/Exam-14-T1/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..41cbddf4217d489888993aff9b38ba591f3d5f59 --- /dev/null +++ b/test/tasks/Exam-14-T1/problem.json @@ -0,0 +1,17 @@ +{ + "problem_id": "Exam-14-T1", + "header": "Ancient Metallurgy and Transmutation Chemistry\nBecause natural gold and silver were rare and extremely valuable, people in antiquity devised many ingenious methods for preparing precious metals artificially. Ming-dynasty works such as Tiangong Kaiwu and Shuyuan Zaji recorded methods for converting pure copper into “imitation gold and silver” with certain minerals, while alchemists also performed “transmutation” experiments with arsenic-, mercury-, and lead-bearing minerals. Modern chemistry has shown that these remarkable phenomena involve complex principles of inorganic synthesis and thermodynamics.", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-14-T1-1", + "tags": [ + "1.1", + "2.1" + ], + "text_open": "1.1.2 In the Jin dynasty, Ge Hong recorded in Baopuzi a method for transmuting pure copper into a silvery-white alloy using an arsenic-containing compound. In the actual reaction, heating this compound in a sealed system first causes it to sublime and form gaseous molecule A with an adamantane-type cage; A is then reduced by charcoal to gaseous molecule B with a regular tetrahedral geometry; B penetrates the surface of solid pure copper and ultimately forms a silvery-white copper-arsenic alloy (the atomic ratio of the two elements in the alloy is 1:3). Write the chemical formula of A and the overall equation for formation of the alloy (assume that CO is the only oxidation product of carbon).", + "text_images": [], + "full_mark": 3 + } + ] +} diff --git a/test/tasks/Exam-14-T4/problem.json b/test/tasks/Exam-14-T4/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..d6e7437b821bc0ffe57f1c3778f41e25cf3ef93e --- /dev/null +++ b/test/tasks/Exam-14-T4/problem.json @@ -0,0 +1,27 @@ +{ + "problem_id": "Exam-14-T4", + "header": "Determination of Iron in Tin Concentrate by Cerium(IV) Sulfate Titration\n4.1 Preparation of standard titrant solutions\nIron standard solution: Weigh 2.0000 g of pure iron wire (99.95%) into a beaker, add 50 mL of hydrochloric acid (1+1), cover with a watch glass, and heat on a low-temperature electric hot plate until completely dissolved (do not boil). Remove, cool, transfer to a 1000 mL volumetric flask, dilute to the mark with water, and mix thoroughly.\nCerium(IV) sulfate standard titrant: Weigh 18.1 g of cerium(IV) sulfate [Ce(SO4)2·4H2O] into a beaker, add a small amount of water, and add 80 mL of concentrated sulfuric acid with constant stirring. After complete dissolution, cool, dilute to 1000 mL with water, and mix.\nStandardization: Pipet 30.00 mL of the iron standard solution into a 300 mL conical flask, add 5 mL of concentrated hydrochloric acid, and carry out reduction and standardization according to Section 4.3. Conduct a blank test in parallel.\n4.2 Alkali fusion and leaching of the test portion\nPlace a 0.3126 g portion of tin concentrate in a 10 mL high-alumina crucible, add 2 g of sodium peroxide, and mix carefully with a thin glass rod. Wipe the rod with a small piece of filter paper and place the paper in the crucible. Place the crucible in a furnace preheated to 500 °C, continue heating to 650 °C–700 °C, and fuse until the melt is red and transparent. Remove the crucible and cool.\nPlace the crucible in a 250 mL beaker and add 20 mL of water for leaching. After the vigorous reaction ceases, neutralize to an acidic solution with sulfuric acid (1+1), add 5 mL of concentrated hydrochloric acid, wash the crucible with water, and heat the solution to boiling on a low-temperature electric hot plate.\n4.3 Reduction and titration\nWhile the solution is hot, add hydrochloric-acidified tin(II) chloride solution (100 g/L) dropwise with swirling until the solution is pale yellow. Rinse down the beaker wall, add 40 mL of water, add 5 drops of sodium tungstate solution (50 g/L) as indicator, and add hydrochloric-acidified titanium(III) chloride solution (50 g/L) dropwise until a blue color appears, followed by two excess drops. Cool to room temperature in a cold-water bath.\nAdd 50 mL of copper sulfate-ammonium sulfate solution (containing 0.02 g of copper sulfate and 80 g of ammonium sulfate per liter). After the blue color disappears, allow the solution to stand for 10 min. Add 25 mL of sulfuric acid-phosphoric acid solution (sulfuric acid:phosphoric acid:water = 1:1:4), add 7 drops of sodium diphenylamine sulfonate solution (5 g/L), and titrate with the cerium(IV) sulfate standard titrant to a stable purple-red endpoint.", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-14-T4-1", + "tags": [ + "4.1", + "4.2" + ], + "text_open": "4.3.2 Explain why preliminary reduction with tin(II) chloride solution is required instead of adding sodium tungstate solution directly and reducing with titanium(III) chloride solution.", + "text_images": [], + "full_mark": 2 + }, + { + "sub_id": "Exam-14-T4-2", + "tags": [ + "4.1", + "4.2" + ], + "text_open": "4.3.3 Explain why reduction with tin(II) chloride and titanium(III) chloride must be performed in a hot solution, whereas the subsequent operations require cooling to room temperature.", + "text_images": [], + "full_mark": 2 + } + ] +} diff --git a/test/tasks/Exam-14-T7/images/Exam-14-T7-1-1-prompt.png b/test/tasks/Exam-14-T7/images/Exam-14-T7-1-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..d90321936efc90c9ea288babc8540703d1787f2a --- /dev/null +++ b/test/tasks/Exam-14-T7/images/Exam-14-T7-1-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:7390eafb7a9a4137f120a66f2b350e91bbdddac7718ea67bf5da17dccdc5a67d +size 3938 diff --git a/test/tasks/Exam-14-T7/images/Exam-14-T7-2-1-prompt.png b/test/tasks/Exam-14-T7/images/Exam-14-T7-2-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..2707025a2feff7db7e10084346a92b478e488dfb --- /dev/null +++ b/test/tasks/Exam-14-T7/images/Exam-14-T7-2-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:f1c09c0c7a7fdb915f7751cf31e31588ec5edd70f41bbacfeabaae71beafcee6 +size 5007 diff --git a/test/tasks/Exam-14-T7/images/Exam-14-T7-3-1-prompt.png b/test/tasks/Exam-14-T7/images/Exam-14-T7-3-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..e1bbbc10d7706814479b4f4d28d1c4984f6d4db3 --- /dev/null +++ b/test/tasks/Exam-14-T7/images/Exam-14-T7-3-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:3ace9314ac6c83bb645c2166461cf68dd3be45da5fa83c3d13ffaf52aaea3799 +size 67066 diff --git a/test/tasks/Exam-14-T7/problem.json b/test/tasks/Exam-14-T7/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..0c358303db999c38e23e04ed812dd3e9b3e3854c --- /dev/null +++ b/test/tasks/Exam-14-T7/problem.json @@ -0,0 +1,54 @@ +{ + "problem_id": "Exam-14-T7", + "header": "Bioinorganic Chemistry. 7.1 O2 is in fact highly toxic to many organisms. Oxygen carriers bound to metalloproteins enable the transport and storage of $O_2$ in organisms.", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-14-T7-1", + "tags": [ + "1.4", + "1.6", + "2.4" + ], + "text_open": "7.1.1 Arthropods transport oxygen using hemocyanin. Hemocyanin is an oligomeric protein containing a pair of closely spaced Cu atoms. Colorless deoxyhemocyanin A [Cu(I)] is represented below. A binds one molecule of $\\mathrm{O_2}$ to form bright-blue protein B. Given that Cu is in the +2 oxidation state in B, write the SMILES string of B (for example, A may be represented as '[N][Cu]([N])[N].[N][Cu]([N])[N]').", + "text_images": [ + { + "file": "Exam-14-T7-1-1-prompt.png", + "description": "Simplified representation of the dicopper(I) center in deoxyhemocyanin A. The diagram is labeled with a bold uppercase A at the bottom center and is left-right symmetric. A central $Cu$ atom in the left half is connected by three straight lines to three $N$ atoms at the upper-left, directly left, and lower-left positions. These three $N$ atoms are joined in sequence by two arcs curving outward to the left: the upper-left $N$ is connected to the directly-left $N$, and the directly-left $N$ is connected to the lower-left $N$. A second central $Cu$ atom in the right half is similarly connected by three straight lines to three $N$ atoms at the upper-right, directly right, and lower-right positions. These three $N$ atoms are joined in sequence by two arcs curving outward to the right: the upper-right $N$ is connected to the directly-right $N$, and the directly-right $N$ is connected to the lower-right $N$. A long downward-curving arc at the bottom directly connects the lower-left $N$ and lower-right $N$ atoms." + } + ], + "full_mark": 3 + }, + { + "sub_id": "Exam-14-T7-2", + "tags": [ + "1.4", + "1.8", + "2.4" + ], + "text_open": "7.1.2 Hemerythrin is a member of a special family of proteins containing a diiron center. In addition to coordination by amino-acid side chains and two bridging carboxylates, the two iron atoms are bridged by a common diatomic anion. Hemerythrin C is represented below (the diatomic bridging ligand is not shown), and C binds one molecule of $\\mathrm{O_2}$ to form D. Given that $\\mathrm{O_2}$ coordinates to only one Fe atom and that the diatomic bridging ligand provides some noncovalent stabilization to $\\mathrm{O_2}$, write the SMILES string of D.", + "text_images": [ + { + "file": "Exam-14-T7-2-1-prompt.png", + "description": "Simplified representation of the diiron center in hemerythrin C. C: '[R]/C1=O/[Fe]([*])([*])O/C([*])=O\\[Fe]([*])([*])O1', where * denotes a site at which Fe coordinates to the imidazole group of His and may be omitted here." + } + ], + "full_mark": 4 + }, + { + "sub_id": "Exam-14-T7-3", + "tags": [ + "1.8", + "1.4" + ], + "text_open": "7.1.3 CO is well known to strongly inhibit the binding of $\\mathrm{O_2}$ to myoglobin, so suppressing CO coordination is essential. In fact, the structure of myoglobin weakens the binding of CO to the protein. Using the simplified myoglobin structure above, explain why CO binds less strongly to myoglobin than $\\mathrm{O_2}$, considering the coordination geometries of CO and $\\mathrm{O_2}$ and their noncovalent interactions. (No more than 100 words.)", + "text_images": [ + { + "file": "Exam-14-T7-3-1-prompt.png", + "description": "Simplified structure of the myoglobin coordination site, drawn with black lines on a white background.\n\nA five-membered ring near the top bears the labels HN at the lower left and N at the upper right, with single- and double-bond lines in and around the ring. A downward-curving line from the lower-right side of the ring extends toward the upper-right region of an elliptical structure below.\n\nAn approximately elliptical ring of curved lines occupies the center. Four N labels appear around it at the upper-left, upper-right, lower-left, and lower-right positions, and Fe is labeled at the center.\n\nA vertical line extends downward from Fe through the lower part of the ellipse to the label His below.\n\nThe Chinese caption at the bottom reads “Simplified myoglobin structure” in the English-localized version. SMILES: '[Fe][*].[N].[N].[N].[N].C1=CN=CN1', where * denotes the site at which Fe coordinates to the imidazole group of His and may be omitted here." + } + ], + "full_mark": 4 + } + ] +} diff --git a/test/tasks/Exam-15-T1/problem.json b/test/tasks/Exam-15-T1/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..cbb39d7965a3aa6b5570e21155b1e06e26f5552d --- /dev/null +++ b/test/tasks/Exam-15-T1/problem.json @@ -0,0 +1,35 @@ +{ + "problem_id": "Exam-15-T1", + "header": "Problem 1 The Immortal Mirror. In 1863, two chemists at the Freiberg Mining Academy in Germany unexpectedly observed a previously unknown spectral line while studying locally mined sphalerite. They assigned it to an undiscovered element M and successfully isolated elemental M from the mineral. At the subsequent Paris Exposition, elemental M was publicly displayed as \"the rarest metal,\" causing a sensation. At that time, M was used to make highly reflective mirrors; today it is more commonly an important semiconductor raw material used in conductive films.", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-15-T1-1", + "tags": [ + "2.1" + ], + "text_open": "1.1 Burning M in air gives a pale-yellow amorphous solid A. Heating A under hydrogen at 400~500°C reduces it to a black crystalline solid B (Reaction 1); dissolving B in dilute hydrochloric acid releases a colorless gas C. Above approximately 700°C, B decomposes to A and elemental M (Reaction 2). At high temperature, A reacts with ammonia to form a yellow semiconductor material D and releases a gas (Reaction 3).\n1.1.2 Write the chemical equations for Reactions 1~3.", + "text_images": [], + "full_mark": 3 + }, + { + "sub_id": "Exam-15-T1-2", + "tags": [ + "1.4" + ], + "text_open": "1.2 Chloride E of M is a white flaky solid that is extremely hygroscopic and dissolves in water with substantial heat release. In ether solution, E reacts with LiH to form complex F. F decomposes above 0°C and can be used as an in situ reducing agent in organic synthesis. E forms complex E·3THF with the electron-donor ligand THF. Single-crystal X-ray diffraction shows that the M center in this complex is six-coordinate with octahedral geometry. E has a layered crystal structure consisting of close-packed chloride ions with layers of octahedrally coordinated M centers. By contrast, chloride G of the congener whose period differs by only 1 commonly crystallizes as dimer H. 1.2.1 Write the chemical formulas of E~G, write the SMILES of H, and describe its three-dimensional structure in words.", + "text_images": [], + "full_mark": 5 + }, + { + "sub_id": "Exam-15-T1-3", + "tags": [ + "1.5", + "1.6" + ], + "text_open": "1.3 In $E\\cdot3THF$, the $M$–Cl bond length is approximately 242 pm and the $M$–O bond length is approximately 225 pm. The ionic radius of $M$ is 80 pm, the ionic radius of $Cl^-$ is 181 pm, and the van der Waals radius of O in THF is approximately 152 pm.\n1.3.2 Use the bond-length data to compare the bonding character of the M-Cl and M-O bonds, including their relative ionic/covalent character, and explain.", + "text_images": [], + "full_mark": 3 + } + ] +} diff --git a/test/tasks/Exam-15-T4/problem.json b/test/tasks/Exam-15-T4/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..7c8fbc1b7adf7add2944dd0d69a64e3f7a64b82d --- /dev/null +++ b/test/tasks/Exam-15-T4/problem.json @@ -0,0 +1,48 @@ +{ + "problem_id": "Exam-15-T4", + "header": "Problem 4 Determination of Chloride by the Volhard Method. ① Pipette 20.00 mL of 0.1 mol/L standard silver nitrate titrant into a 250 mL conical flask. Add 30 mL water, 5 mL nitric acid solution, and 2 mL saturated ferric ammonium sulfate solution. While shaking, titrate with standard potassium thiocyanate titrant until a pale brown-red color appears and persists for 1 min. Record the volume of standard potassium thiocyanate consumed: V_{1} = 19.88 mL. ② Weigh 0.10 g sodium chloride that has been ignited at 500 ℃ ~ 600 ℃ to constant mass, accurate to 0.1 mg. Dissolve it in approximately 40 mL water in a beaker and transfer to a 100 mL volumetric flask. Add 5 mL nitric acid solution (1+3), then add 25.00 mL of 0.1 mol/L standard silver nitrate solution while shaking vigorously. Dilute to the mark with water and mix. Let stand in the dark for 5 min, filter through rapid filter paper, and discard the first 10 mL of filtrate. Accurately pipette 50.00 mL filtrate into a 250 mL conical flask, add 2 mL of 1 mol/L ferric ammonium sulfate solution, and titrate with standard potassium thiocyanate while shaking until a pale brown-red color appears and persists for 1 min. Record V_{2} = 4.01 mL. ③ Measure 50 mL test solution into a 100 mL colorimetric tube and otherwise follow the same procedure as in ②. Record V_{3} = 2.98 mL.", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-15-T4-1", + "tags": [ + "4.5", + "2.1" + ], + "text_open": "4.1 Calculate the concentrations of the standard silver nitrate and potassium thiocyanate titrant solutions.", + "text_images": [], + "full_mark": 5 + }, + { + "sub_id": "Exam-15-T4-2", + "tags": [ + "4.1", + "4.7" + ], + "text_open": "4.4.2 Explain why the first 10 mL of filtrate is discarded.", + "text_images": [], + "full_mark": 1 + }, + { + "sub_id": "Exam-15-T4-3", + "tags": [ + "4.2", + "4.5" + ], + "text_open": "4.5 If addition of standard silver nitrate produces a colloidal solution, what additional operation is required before filtration?", + "text_images": [], + "full_mark": 2 + }, + { + "sub_id": "Exam-15-T4-4", + "tags": [ + "4.5", + "2.1", + "4.7" + ], + "text_open": "4.6 Calculate the $Cl^-$ concentration in the sample.", + "text_images": [], + "full_mark": 3 + } + ] +} diff --git a/test/tasks/Exam-17-T2/images/Exam-17-T2-header-1.png b/test/tasks/Exam-17-T2/images/Exam-17-T2-header-1.png new file mode 100644 index 0000000000000000000000000000000000000000..b2d2fe54c4e09cf8b2aea76822e1e241950005d3 --- /dev/null +++ b/test/tasks/Exam-17-T2/images/Exam-17-T2-header-1.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:daf5975087ee04d545688808446e0b7a7839dd9955c786702679b8e18b0c6abc +size 226189 diff --git a/test/tasks/Exam-17-T2/problem.json b/test/tasks/Exam-17-T2/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..c79dccffabeffd71f20138d0d11d67239e0e3797 --- /dev/null +++ b/test/tasks/Exam-17-T2/problem.json @@ -0,0 +1,39 @@ +{ + "problem_id": "Exam-17-T2", + "header": "Problem 2 Determination of Paracetamol by Cyclic Voltammetry and Study of Its Electrochemistry\nCyclic voltammetry (CV) is an important electrochemical technique for investigating electrochemical reaction mechanisms and performing quantitative analysis. It uses a three-electrode system consisting of a working electrode, a reference electrode, and an auxiliary electrode. A triangular-wave potential that varies linearly with time is applied to the working electrode while the response current is recorded. One triangular-wave scan completes both an oxidation scan and a reduction scan, and the resulting current–potential curve is called a cyclic voltammogram. A substance that can undergo oxidation or reduction at an electrode within the scanned potential range is called an electroactive substance. A typical cyclic voltammogram exhibits an oxidation peak (anodic peak) and a reduction peak (cathodic peak). Under suitable conditions, the peak currents ($I_{pa}$ and $I_{pc}$) are proportional to the concentration of the electroactive substance and form the basis of quantitative analysis. The peak potentials ($E_{pa}$ and $E_{pc}$) and the peak-potential separation ($\\Delta E_p = E_{pa} - E_{pc}$) can be used to assess reaction reversibility and the number of electrons transferred. Changing the scan rate ($v$) can distinguish diffusion control from adsorption control and reveal the kinetics of subsequent chemical reactions.\nParacetamol (APAP, also known as acetaminophen) is a commonly used antipyretic analgesic. Studies show that APAP undergoes electrochemical oxidation at an electrode surface, and its oxidation product NAPQI undergoes subsequent chemical reactions in acidic media. A research group used cyclic voltammetry to investigate the electrochemical behavior of APAP systematically.\nAt room temperature, the peak current ($I_p$, in A) for a reversible electrochemical reaction in a quiescent solution at a planar electrode is related to the scan rate ($v$, in V/s), the concentration of the electroactive substance ($C$, in $\\mathrm{mol/cm^3}$), and other parameters by the Randles–Sevcik equation:\n$$I_p = 2.69 \\times 10^5 n^{3/2} A C D^{1/2} v^{1/2}$$\nn: number of electrons transferred in the reaction.\nA: effective area of the working electrode (unit: $\\mathrm{cm^2}$).\nD: diffusion coefficient of the electroactive substance (unit: $\\mathrm{cm^2/s}$), reflecting how rapidly the substance moves through solution.\nThe figure below is a typical cyclic voltammogram. Curve 1 is the background signal, and curve 2 is the normal response signal.", + "header_images": [ + { + "file": "Exam-17-T2-header-1.png", + "description": "Typical cyclic voltammogram. The horizontal axis is potential, with tick marks of approximately 0.2, 0.1, 0, -0.1, and -0.2 from left to right; the vertical axis is current. The peak-current and peak-potential symbols $I_{pc}$, $I_{pa}$, $E_{pa}$, and $E_{pc}$ are labeled. Two closed curves are labeled 1 and 2. As stated in the question, curve 1 is the background signal and curve 2 is the normal response signal." + } + ], + "subquestions": [ + { + "sub_id": "Exam-17-T2-1", + "tags": [ + "4.7" + ], + "text_open": "2.2 At a scan rate of $40\\ \\mathrm{mV\\cdot s^{-1}}$ in a pH = 2.2 buffer, the research group obtained the following cyclic-voltammetry data for standard APAP solutions:\n\n| APAP concentration/(mmol/L) | 0.10 | 0.40 | 1.00 | 3.00 | 5.00 |\n| ----------------- | ----- | ----- | ----- | ----- | ----- |\n| Anodic peak current $I_{pa}$/μA | 2.15 | 5.82 | 13.46 | 38.74 | 63.52 |\n\nUsing concentration as the horizontal coordinate and anodic peak current as the vertical coordinate, obtain the linear regression equation of $I_{pa}$ against APAP concentration and calculate the expected peak current for an APAP solution with a concentration of $3.50\\ \\mathrm{mmol\\cdot L^{-1}}$.", + "text_images": [], + "full_mark": 5 + }, + { + "sub_id": "Exam-17-T2-2", + "tags": [ + "2.3" + ], + "text_open": "2.3 To investigate the oxidation mechanism of APAP, the group performed CV scans at $40\\ \\mathrm{mV\\cdot s^{-1}}$ in media of different acidity (pH = 6.0, pH = 2.2, and $1.8\\ \\mathrm{mol\\cdot L^{-1}}\\ H_2SO_4$) and observed the following:\nAt pH = 6.0, both oxidation and reduction peaks appear, but $I_{pa}$ ≠ $I_{pc}$; at pH = 2.2, only the oxidation peak appears and the reduction peak disappears; under strongly acidic conditions ($1.8\\ \\mathrm{mol\\cdot L^{-1}}\\ H_2SO_4$), the reduction peak reappears.\n2.3.1 Explain why the reduction peak disappears at pH = 2.2.", + "text_images": [], + "full_mark": 2 + }, + { + "sub_id": "Exam-17-T2-3", + "tags": [ + "2.3" + ], + "text_open": "2.3.2 What chemical reaction is indicated by the reappearance of the reduction peak under strongly acidic conditions?", + "text_images": [], + "full_mark": 2 + } + ] +} diff --git a/test/tasks/Exam-17-T4/problem.json b/test/tasks/Exam-17-T4/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..c8327e7dbf062c8cefa4637caf808744644ae4f1 --- /dev/null +++ b/test/tasks/Exam-17-T4/problem.json @@ -0,0 +1,26 @@ +{ + "problem_id": "Exam-17-T4", + "header": "Problem 4 Gravimetric Determination of Tungsten (17 points, 8%)\nTungsten is an important strategic metal that has an irreplaceable strengthening role in cemented carbides, high-speed steel, and specialty steels. Accurate determination of tungsten content is essential for quality control of alloy materials. In the classical gravimetric method, tungsten is converted to a tungstic-acid precipitate, ignited to tungsten trioxide, and weighed. Accurate tungsten content is obtained by correcting for interference from coexisting elements such as molybdenum, chromium, and vanadium. This problem concerns precipitation and separation of tungsten, correction for impurities, and special treatment of niobium-containing samples, and tests understanding of the principles of gravimetric analysis and strategies for eliminating interference.\nThe hydrochloric acid used in this problem is diluted 1:1, the nitric acid is concentrated, and every reported mass has been corrected for the tare mass of the crucible.", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-17-T4-1", + "tags": [ + "2.1" + ], + "text_open": "4.1 Procedure for samples without niobium\nA sample of mass $m$ is weighed into a beaker. Hydrochloric acid and nitric acid are added, and the mixture is heated gently to dissolve the sample and evaporated to a syrup to remove nitrogen oxides. Hydrochloric acid is added to dissolve the salts, the solution is diluted with water and brought to the boil, and cinchonine hydrochloride solution is added slowly to precipitate all tungsten as tungstic acid. The precipitate is aged at 70-80℃ for 4 h and then filtered and washed with a cinchonine hydrochloride wash solution.\nThe precipitate and filter paper are placed in a platinum crucible, carbonized, and ignited at 750-800℃. The residue is moistened with water, hydrofluoric acid and sulfuric acid are added, and the mixture is heated to volatilize silicon. It is then ignited to constant mass, giving a mass $m_1$ of impure $WO_3$.\nSodium carbonate is added to the platinum crucible for fusion. After cooling, the melt is extracted with hot water and filtered, and the residue is ignited and weighed to give $m_2$. The filtrate is analyzed spectrophotometrically for molybdenum, chromium, and vanadium, which are converted to the corresponding oxide masses $m_3$, $m_4$, and $m_5$.\n4.1.1 Write the equation for oxidation of elemental tungsten to tungstic acid, $H_2WO_4$, in hydrochloric acid/nitric acid.", + "text_images": [], + "full_mark": 2 + }, + { + "sub_id": "Exam-17-T4-2", + "tags": [ + "4.7", + "2.1" + ], + "text_open": "4.1.6 During aqueous extraction after sodium carbonate fusion, $WO_3$ is converted to $Na_2WO_4$ and enters solution, whereas $TiO_2$ and $Fe_2O_3$ remain insoluble. Analysis of a tungsten-steel sample gives the following data: $m$ = 0.5123 g, $m_1$ = 0.0842 g, $m_2$ = 0.0018 g, $m_3$ = 0.0023 g, $m_4$ = 0.0015 g, and $m_5$ = 0.0012 g. Calculate the mass fraction of tungsten in the sample.", + "text_images": [], + "full_mark": 2 + } + ] +} diff --git a/test/tasks/Exam-17-T7/images/Exam-17-T7-1-1-prompt.png b/test/tasks/Exam-17-T7/images/Exam-17-T7-1-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..7ddc101c455472d248debc0f71d11ea55e8e27af --- /dev/null +++ b/test/tasks/Exam-17-T7/images/Exam-17-T7-1-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:428534b76acb199c62a5fbb1b02b3f5d6cc1449bad3b72639f158dc91960a5af +size 80032 diff --git a/test/tasks/Exam-17-T7/images/Exam-17-T7-3-1-prompt.png b/test/tasks/Exam-17-T7/images/Exam-17-T7-3-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..190735f830ca0fdc11bc09d8359aaadadfaab63c --- /dev/null +++ b/test/tasks/Exam-17-T7/images/Exam-17-T7-3-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:b6f672b778fe431cb7ec63869662d2f52c62a109825292fb006c371defe87b4c +size 56376 diff --git a/test/tasks/Exam-17-T7/images/Exam-17-T7-7-1-prompt.png b/test/tasks/Exam-17-T7/images/Exam-17-T7-7-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..dfc2da2808f719b939ac9e8d334576c4307bb5ad --- /dev/null +++ b/test/tasks/Exam-17-T7/images/Exam-17-T7-7-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:22c75c472291aebfa4e0195687f123b987e990e5e325ae5e00c018902c7e6552 +size 57525 diff --git a/test/tasks/Exam-17-T7/problem.json b/test/tasks/Exam-17-T7/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..3463440fb19a6f0c80f517b20fcd2c50c72441cd --- /dev/null +++ b/test/tasks/Exam-17-T7/problem.json @@ -0,0 +1,89 @@ +{ + "problem_id": "Exam-17-T7", + "header": "Problem 7 Low-Valent Cluster Compounds of Group 13 Metals\nLow-valent compounds of Group 13 elements such as aluminum and gallium can form metal clusters with diverse structures. Their bonding modes and electronic structures are important for understanding the nature of metal-metal bonds. This problem concerns low-valent gallium and aluminum halides and their assembly reactions with ligands, focusing on structural analysis, oxidation-state calculations, and cluster reactivity.", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-17-T7-1", + "tags": [ + "2.1" + ], + "text_open": "Synthesis and structures of $E_3$ units: Ga-Ga bonds are relatively stable and can support extended compounds.\n7.1 Chain-like $E_3$ structure: reaction of freshly prepared LiM with 1.67 eq GaI affords the trigallane C = $Ga_3M_3I_2$.\n7.1.1 Write the reaction equation for the formation of C from LiM and GaI.", + "text_images": [ + { + "file": "Exam-17-T7-1-1-prompt.png", + "description": "Two-step reaction: '[dipp]N=C=N[dipp]'>t-BuLi (1eq), n-hexane, 0 °C>intermediate Li+M-, where M denotes the ligand and the SMILES for M- is '[dipp]/N=C([N-][dipp])/C(C)(C)C'. The M- anion has an N,N'-bis(dipp) amidinate framework with t-Bu bonded to the central carbon. The second step is '[dipp]/N=C([N-][dipp])/C(C)(C)C'>GaI (1.67eq), Toluene, -78 °C>C, where C = $Ga_3M_3I_2$ and its SMILES is 'I[Ga]1([Ga]2([Ga]3(I)N([dipp])C(C(C)(C)C)=[N]3[dipp])N([dipp])C(C(C)(C)C)=[N]2[dipp])N([dipp])C(C(C)(C)C)=[N]1[dipp]'. Three Ga atoms form a chain through Ga-Ga single bonds, each terminal Ga bears one I, and each Ga is chelated by one M ligand. No wedge or dashed bonds are shown." + } + ], + "full_mark": 2 + }, + { + "sub_id": "Exam-17-T7-2", + "tags": [ + "2.1" + ], + "text_open": "7.1.2 Write the equation for the substitution reaction of trigallane C with $NaBEt_3H$ to form Et(M)Ga-Ga(M)-Ga(M)H.", + "text_images": [], + "full_mark": 2 + }, + { + "sub_id": "Exam-17-T7-3", + "tags": [ + "2.4", + "1.6" + ], + "text_open": "7.2 Cyclic $E_3$ structure: cyclotrigallanes have intriguing electronic properties. Reaction of $(Mes_2C_6H_3)Li$ with $GaCl_3$ gives $(Mes_2C_6H_3)GaCl_2$, which is reduced with potassium metal in diethyl ether to afford the cyclic gallium compound D = $K_2[(Mes_2C_6H_3)Ga]_3$. This compound has a nearly equilateral triangular $Ga_3$ core, with an average Ga-Ga bond length of 2.426 Å that is markedly shorter than a typical single bond, indicating a higher bond order. The two potassium ions lie on opposite sides of the $Ga_3$ plane. Write the SMILES string for D and determine whether its anion is aromatic.", + "text_images": [ + { + "file": "Exam-17-T7-3-1-prompt.png", + "description": "Two-step reaction: $GaCl_3$ reacts with LiAr (1.0 eq) in $Et_2O$ at -78 °C; $K^0$ (4.0 eq) is then added and the mixture is reacted at rt. for 5 days to give $K_2(ArGa)_3·(Et_2O)_2$. The structure defining Ar = $(Mes_2C_6H_3)$ appears at the right: the top carbon of the central benzene ring, drawn in an alternating-double-bond Kekulé form, bears a wavy bond marking the attachment point to Ga, and each of the two ortho carbons bears a 2,4,6-trimethylphenyl group (Mes). The encodable structure of Ar is represented by the SMILES '[*]C1=C(C2=C(C)C=C(C)C=C2C)C=CC=C1C3=C(C)C=C(C)C=C3C', where [*] is the attachment point corresponding to the wavy bond." + } + ], + "full_mark": 3 + }, + { + "sub_id": "Exam-17-T7-4", + "tags": [ + "2.4", + "1.4" + ], + "text_open": "7.4 Eight-membered-ring structure: gaseous GaI is cocondensed with excess $PEt_3$ in toluene at -196 °C and then warmed slowly, giving F = $Ga_8I_8·6PEt_3$. This cluster has $C_{2h}$ symmetry and contains a nearly planar eight-membered $Ga_8$ ring. Two iodine atoms bridge across the ring to form a four-membered $Ga_2I_2$ ring. All eight Ga atoms are four-coordinate, and the average Ga-Ga bond length is 2.466 Å.\n7.4.1 Write the SMILES string for F.", + "text_images": [], + "full_mark": 4 + }, + { + "sub_id": "Exam-17-T7-5", + "tags": [ + "1.7", + "2.4" + ], + "text_open": "7.4.2 The Ga/I/$PEt_3$ system forms the eight-membered-ring cluster F ($Ga_8I_8$), whereas the Al/Br/$NEt_3$ system forms the four-membered-ring cluster E ($Al_4Br_4$) under similar conditions. Which explanations are the most reasonable? Select all that apply.\n(A) Ga has a smaller atomic radius than Al and a shorter Ga-Ga bond; formation of a four-membered ring would therefore cause greater ring strain and bond-angle distortion, so an eight-membered ring is favored to relieve strain.\n(B) I has a larger atomic radius than Br and is well suited for bridging; transannular bridges stabilize the eight-membered ring.\n(C) $PEt_3$ is a stronger electron donor than $NEt_3$, increasing the electron density at Ga and shortening the Ga-Ga bonds, which favors formation of a larger ring system.\n(D) The Ga-Ga bonds in the eight-membered ring have a higher bond order than the Al-Al bonds in the four-membered ring, giving greater thermodynamic stability.", + "text_images": [], + "full_mark": 2 + }, + { + "sub_id": "Exam-17-T7-6", + "tags": [ + "2.4", + "1.2" + ], + "text_open": "7.5 Fused-ring structure: attempts to synthesize amine-stabilized $Ga_8I_8·6NEt_3$ were unsuccessful, but partial oxidation unexpectedly gave G = $Ga_8I_8O_2·6NEt_3$. Its structure has $C_i$ symmetry and consists of two nearly planar $Ga_4O$ rings fused through a shared edge of a central four-membered $Ga_2O_2$ ring. All Ga atoms are four-coordinate. Write the SMILES string for G.", + "text_images": [], + "full_mark": 2 + }, + { + "sub_id": "Exam-17-T7-7", + "tags": [ + "1.1" + ], + "text_open": "7.6 Tetrahedral type: high-temperature gaseous GaBr is cocondensed with a toluene solution of $NHEt_2$. The resulting dark solution is concentrated and allowed to stand at 25 °C, yielding crystals of cluster H = $Ga_5Br_7(NHEt_2)_5$. H contains a $Ga_5$ core in which the central Ga atom is tetrahedrally surrounded by four Ga atoms.\n7.6.2 In the sterically congested H molecule, the 12 heteroatoms—the halogens and the O/N atoms of the ligands—adopt an arrangement that maximizes their mutual separation. What polyhedron is formed by these 12 vertices?", + "text_images": [ + { + "file": "Exam-17-T7-7-1-prompt.png", + "description": "Structure of H = $Ga_5Br_7(NHEt_2)_5$: the central Ga is connected by single bonds to four peripheral Ga atoms in a tetrahedral arrangement. The encodable structure of H is represented by the SMILES '[L][Ga](Br)(Br)[Ga]([Ga]([L])(Br)Br)([Ga]([L])([L])Br)[Ga]([L])(Br)Br', where [L] denotes the $NHEt_2$ ligand labeled in the image. A total of 5 L ligands and 7 Br atoms are distributed among the peripheral Ga atoms. The lower right of the image defines L = $NHEt_2$, and all bonds are solid single bonds." + } + ], + "full_mark": 1 + } + ] +} diff --git a/test/tasks/Exam-17-T9/images/Exam-17-T9-1-1-prompt.png b/test/tasks/Exam-17-T9/images/Exam-17-T9-1-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..4378a6671430a026a482a8896b2e51fc9702f7a9 --- /dev/null +++ b/test/tasks/Exam-17-T9/images/Exam-17-T9-1-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:4de7ab1879e1fec710df4685e95d7be74d374fd8b161ff806f2b90b4abdbccad +size 51169 diff --git a/test/tasks/Exam-17-T9/images/Exam-17-T9-1-2-prompt.png b/test/tasks/Exam-17-T9/images/Exam-17-T9-1-2-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..797469dc3d73515c171cfe55512a7961273a3bf6 --- /dev/null +++ b/test/tasks/Exam-17-T9/images/Exam-17-T9-1-2-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:15c8756f20448077a06a1d1598863078ef1cfe75abd5844e393a107cf142bbe7 +size 43629 diff --git a/test/tasks/Exam-17-T9/images/Exam-17-T9-2-1-prompt.png b/test/tasks/Exam-17-T9/images/Exam-17-T9-2-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..246d1fadb4086cdf9b9714bd8c8907570f6e7b14 --- /dev/null +++ b/test/tasks/Exam-17-T9/images/Exam-17-T9-2-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:65e8da18a04e831d248dc2761061644f695a6873b4152de4029e676c337c30b2 +size 925448 diff --git a/test/tasks/Exam-17-T9/images/Exam-17-T9-4-1-prompt.png b/test/tasks/Exam-17-T9/images/Exam-17-T9-4-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..4058d4217a8ecc9c50632bf80e8e4b1c4ae10b3b --- /dev/null +++ b/test/tasks/Exam-17-T9/images/Exam-17-T9-4-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:d8b885ec10917daccb26b2746a3051483e5fafe27c515e1815ec834700d9095e +size 58032 diff --git a/test/tasks/Exam-17-T9/images/Exam-17-T9-5-1-prompt.png b/test/tasks/Exam-17-T9/images/Exam-17-T9-5-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..b8c7942ddb256f19ddca15cc53747d1a09c267bd --- /dev/null +++ b/test/tasks/Exam-17-T9/images/Exam-17-T9-5-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:64deca9309bd9b50c22bab4ab1eb7d7465356b198435ea8bdfd90ac703368df4 +size 43369 diff --git a/test/tasks/Exam-17-T9/problem.json b/test/tasks/Exam-17-T9/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..950c9e512e757bbfc5e61ff26617a2ada2c97166 --- /dev/null +++ b/test/tasks/Exam-17-T9/problem.json @@ -0,0 +1,81 @@ +{ + "problem_id": "Exam-17-T9", + "header": "Problem 9 Stereochemistry\nCareful design of reactant structures and reaction conditions can provide highly selective stereochemical control.", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-17-T9-1", + "tags": [ + "2.4", + "1.2" + ], + "text_open": "9.1 Garner's aldehyde is frequently used for stereochemical control in synthesis. The route below prepares it from the natural product S-serine.\n9.1.2 Garner's aldehyde can undergo further nucleophilic addition. Write the SMILES strings of the major products under the different reaction conditions below, representing them as anions and specifying stereochemistry.", + "text_images": [ + { + "file": "Exam-17-T9-1-1-prompt.png", + "description": "Two-step reaction: 'OC[C@@H](C(O)=O)N'>>'OC[C@@H]1COC(C)(C)N1C(OC(C)(C)C)=O'; 'OC[C@@H]1COC(C)(C)N1C(OC(C)(C)C)=O'>DMSO, (COCl)2, DIPEA>'[H]C([C@@H]1COC(C)(C)N1C(OC(C)(C)C)=O)=O'." + }, + { + "file": "Exam-17-T9-1-2-prompt.png", + "description": "Two parallel reactions: '[H]C([C@@H]1COC(C)(C)N1C(OC(C)(C)C)=O)=O'>Nu^->[A]; '[H]C([C@@H]1COC(C)(C)N1C(OC(C)(C)C)=O)=O'>Nu^-, [M]>[B]. The products are labeled only [A] and [B]; their structures are not shown." + } + ], + "full_mark": 4 + }, + { + "sub_id": "Exam-17-T9-2", + "tags": [ + "2.7", + "1.2" + ], + "text_open": "9.1.3 In practice, the two reaction pathways compete. Determine which of the following conditions favor formation of product B, and briefly explain why.\n(1) Solvent: a. THF; b. toluene", + "text_images": [ + { + "file": "Exam-17-T9-2-1-prompt.png", + "description": "'[H]C([C@@H]1COC(C)(C)N1C(OC(C)(C)C)=O)=O'>nucleophile, solvent, additive>[A]+[B]. The products are labeled only [A] and [B]; their structures are not shown." + } + ], + "full_mark": 2 + }, + { + "sub_id": "Exam-17-T9-3", + "tags": [ + "2.7", + "1.2" + ], + "text_open": "(3) Additive: a. DMPU; b. $ZnCl_2$", + "text_images": [], + "full_mark": 2 + }, + { + "sub_id": "Exam-17-T9-4", + "tags": [ + "1.2", + "2.7" + ], + "text_open": "9.1.4 Garner's aldehyde can also be synthesized from R-serine. In the reaction below, replacing Boc with Ts reverses the selectivity. Explain why.", + "text_images": [ + { + "file": "Exam-17-T9-4-1-prompt.png", + "description": "'CC1(C)N([R])[C@@H](C=O)CO1'>Et-MgBr, THF>'CC1(C)N([R])[C@@H]([C@@H](O)CC)CO1' + 'CC1(C)N([R])[C@@H]([C@H](O)CC)CO1'. The image also states that the product ratio is 1:9 when R=Boc and 8:1 when R=Ts. R is the placeholder group used in the original diagram." + } + ], + "full_mark": 2 + }, + { + "sub_id": "Exam-17-T9-5", + "tags": [ + "2.4", + "1.2" + ], + "text_open": "9.2 Garner's aldehyde can also participate in multicomponent reactions that construct new stereocenters. Write the SMILES string of the product of the reaction below.", + "text_images": [ + { + "file": "Exam-17-T9-5-1-prompt.png", + "description": "Multicomponent reaction: 'C1(C=CC=C2)=C2NC=C1'.'CC(OC[C@H]1C=O)(C)N1C(OC(C)(C)C)=O'.'O=C1CC(OC(C)(C)O1)=O'>CH3CN, D,L-proline>. No product structure is shown to the right of the arrow." + } + ], + "full_mark": 2 + } + ] +} diff --git a/test/tasks/Exam-18-T3/problem.json b/test/tasks/Exam-18-T3/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..20149e18c8541785e8504a630fd8f3c97a080c70 --- /dev/null +++ b/test/tasks/Exam-18-T3/problem.json @@ -0,0 +1,25 @@ +{ + "problem_id": "Exam-18-T3", + "header": "Problem 3 Latent Monomers and Controlled Metathesis Polymerization\nControlling the temporal and spatial sequence of polymerization is an important topic in conventional polymer synthesis. Chemists have generally developed “latent catalysts” that switch between inactive and active states in response to external stimuli such as light or heat. Recent research, however, has introduced a new “latent monomer” strategy: isomerization of a photoswitchable molecule generates an active monomer in situ and on demand, enabling efficient control of polymerization.\nThis problem concerns a classic class of highly reactive ring-opening metathesis polymerization (ROMP) monomers: norbornadiene (NBD) derivatives. Under ultraviolet irradiation of a specific wavelength, they undergo intramolecular cycloaddition to form highly strained quadricyclane (QC) structures.", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-18-T3-1", + "tags": [ + "2.4" + ], + "text_open": "3.1 Synthesis and Photothermal Isomerization of the Latent Monomer\nMonomer A, an NBD derivative, can be prepared by reacting cyclopentadiene with diethyl acetylenedicarboxylate at room temperature. Under 313 nm ultraviolet irradiation, A is quantitatively converted into latent monomer B, a QC derivative. Adding silica-coated gold nanobipyramids (AuBPs) to B and irradiating at 850 nm with infrared light regenerates A.\n3.1.1 Write the SMILES strings for A and B.", + "text_images": [], + "full_mark": 3 + }, + { + "sub_id": "Exam-18-T3-2", + "tags": [ + "2.4" + ], + "text_open": "3.3 The complete inertness of latent monomers toward the catalyst at room temperature enables the synthesis of complex polymer topologies that are difficult to obtain by conventional metathesis polymerization. The researchers replaced the ethoxycarbonyl groups, $-COOC_2H_5$, in B with pent-4-enyloxycarbonyl groups, $-COO(CH_2)_3CH=CH_2$, to give monomer F. F was mixed with the Grubbs catalyst and reacted at 40℃, releasing gas X and giving a soluble linear polymer G with double bonds in its main chain. G was coated onto a microscope slide placed over a masked LED and then heated to 80℃, rapidly converting the film into an insoluble crosslinked network H. Write the SMILES strings for polymers G and H.", + "text_images": [], + "full_mark": 4 + } + ] +} diff --git a/test/tasks/Exam-18-T4/problem.json b/test/tasks/Exam-18-T4/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..4a477af0b4e0d291f621adde68807c14840ac236 --- /dev/null +++ b/test/tasks/Exam-18-T4/problem.json @@ -0,0 +1,46 @@ +{ + "problem_id": "Exam-18-T4", + "header": "Problem 4 Complexometric Titration\nEthylenediaminetetraacetic acid (EDTA) is an effective chelating ligand widely used in chemical analysis. Approximately 20 g of a standard reagent is roughly weighed, 250mL of water is added, and the mixture is heated to dissolve the solid, cooled, and mixed thoroughly. The measured pH of the solution is 4.2. The stepwise ionization constants of EDTA are shown below.\n\n| | $K_{a1}$ | $K_{a2}$ | $K_{a3}$ | $K_{a4}$ | $K_{a5}$ | $K_{a6}$ |\n| --- | --- | --- | --- | --- | --- | --- |\n| | $10^{-0.9}$ | $10^{-1.6}$ | $10^{-2.0}$ | $10^{-2.67}$ | $10^{-6.16}$ | $10^{-10.26}$ |", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-18-T4-1", + "tags": [ + "4.5", + "3.4" + ], + "text_open": "The standard solution is then standardized. Weigh 0.2500 g of working-reference ZnO that has been ignited to constant mass in a furnace at 800 ℃. Moisten it with a small amount of water, dissolve it in 2 mL of 20% hydrochloric acid solution, and add 100 mL of water. Adjust the solution to pH 8 with hexamethylenetetramine (HMTA) solution, add 10 mL of $(CH_2)_6N_4$-$(CH_2)_6N_4$·HCl buffer to adjust the pH to 10, and then add 5 drops of Eriochrome Black T indicator solution. Titrate with the prepared EDTA sodium-salt solution until the color changes from purple to pure blue. Conduct a blank determination in parallel. After blank correction, the volume of EDTA consumed is $V = 15.35\\ \\mathrm{mL}$.\n4.2.2 The buffering ability of a buffer solution is commonly measured by its buffer capacity, denoted β. It is defined as the amount db in mol of strong base required to raise the pH of 1L of buffer solution by dpH units, with the mathematical expression\n$\\beta = \\frac{db}{dpH} = -2.30[H^+] \\times \\frac{db}{d[H^+]}$\nGiven $pK_b((CH_2)_6N_4) = 4.9$, calculate the buffer capacity of a pH = 10 $(CH_2)_6N_4$-$(CH_2)_6N_4$·HCl buffer in which the analytical concentration of $(CH_2)_6N_4$ is 0.1000 mol/L. Hint: express $b$ as a function of $[H^+]$ and differentiate with respect to $[H^+]$ to obtain $\\frac{db}{d[H^+]}$. B may be used to represent $(CH_2)_6N_4$ in the calculation.", + "text_images": [], + "full_mark": 4 + }, + { + "sub_id": "Exam-18-T4-2", + "tags": [ + "4.5", + "3.4" + ], + "text_open": "4.2.3 Neglecting volume changes during standardization, the change in solution pH over the entire process must not exceed 0.2 pH unit to ensure an accurate endpoint. Determine the minimum value of $[(CH_2)_6N_4]$.", + "text_images": [], + "full_mark": 4 + }, + { + "sub_id": "Exam-18-T4-3", + "tags": [ + "4.1" + ], + "text_open": "4.3 During standardization, why is the $(CH_2)_6N_4$-$(CH_2)_6N_4$·HCl buffer not added directly to adjust the solution to pH=10?", + "text_images": [], + "full_mark": 1 + }, + { + "sub_id": "Exam-18-T4-4", + "tags": [ + "4.5", + "1.4" + ], + "text_open": "4.4 The EDTA standard solution standardized in this problem can be used to determine the concentrations of Cu2+ and Mg2+ in a solution by either of the following methods. Method 1: Masking Cu2+ ions. Adjust the solution to pH 5.5 with an HMTA buffer and add excess thiosulfate to mask Cu2+. In this reaction, thiosulfate acts as both a reducing agent and a ligand. Method 2: Selective determination of $Cu^{2+}$. Adjust the solution to pH 5 with an HMTA buffer. Using 5-Br-PADAT as the indicator, selectively determine $Cu^{2+}$ with an EDTP standard solution whose principal component is disodium ethylenediaminetetrapropionate. 4.4.2 Based on the data below, explain why EDTP shows high selectivity for $Cu^{2+}$.\n\n| | $Cu^{2+}$ | $Zn^{2+}$ | $Cd^{2+}$ | $Mg^{2+}$ |\n| --- | --- | --- | --- | --- |\n| Ionic radius/pm | 73 | 74 | 95 | 72 |\n| $\\lg K_{M-EDTP}$ | 15.4 | 7.8 | 6.0 | 1.8 |\n| $\\lg K_{M-EDTA}$ | 18.8 | 16.50 | 16.46 | 8.7 |", + "text_images": [], + "full_mark": 2 + } + ] +} diff --git a/test/tasks/Exam-18-T5/images/Exam-18-T5-header-1.png b/test/tasks/Exam-18-T5/images/Exam-18-T5-header-1.png new file mode 100644 index 0000000000000000000000000000000000000000..73f3c0b4c311a027a03f151b23fd0064490c5c0a --- /dev/null +++ b/test/tasks/Exam-18-T5/images/Exam-18-T5-header-1.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:986302b14d7f80500ac4611f639c31a47c22d1f92ee366592f6a77b3a7bba163 +size 1339205 diff --git a/test/tasks/Exam-18-T5/problem.json b/test/tasks/Exam-18-T5/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..6d09ad362bb2c3d62f744ba3bd3c782307c91733 --- /dev/null +++ b/test/tasks/Exam-18-T5/problem.json @@ -0,0 +1,30 @@ +{ + "problem_id": "Exam-18-T5", + "header": "Problem 5 Pharmacokinetics and Molecular Design\nPharmacokinetics quantitatively studies the absorption, distribution, metabolism, and excretion of drugs in living organisms and uses mathematical principles and methods to describe how drug concentration in blood varies with time. Pharmacokinetic parameters reveal pharmacological shortcomings of drug molecules and guide molecular modifications that improve specific properties.\nPropranolol (P), metoprolol (M), betaxolol (B), talinolol (T), and atenolol (A) are β-blockers. Propranolol (P) binds both β1 receptors, found mainly in the heart, and β2 receptors, found mainly in bronchial smooth muscle. At therapeutic doses, the other drugs bind β1 receptors with high selectivity. Although their therapeutic mechanisms are similar, their pharmacological properties differ substantially. Their structures are shown below.\nParameters that may be used in this problem are listed below.\n\n| Drug | Abbreviation | lg $P$ | Conjugate-acid $pK_a$ | $F_a$ | $E_{gi}$ | $E_H$ | $V_d$ / (L·kg$^{-1}$) |\n| --- | --- | --- | --- | --- | --- | --- | --- |\n| Propranolol | P | 3.20 | 9.45 | 0.95 | 0.10 | 0.64 | 3.8 |\n| Metoprolol | M | 2.90 | 9.68 | 0.90 | 0.02 | 0.60 | 5.6 |\n| Betaxolol | B | 2.42 | 9.40 | 0.95 | 0.00 | 0.19 | 1.6 |\n| Talinolol | T | 3.20 | 9.40 | 0.95 | 0.36 | 0.09 | 3.3 |\n| Atenolol | A | 0.25 | 9.60 | 0.50 | 0.00 | 0.01 | 0.70 |", + "header_images": [ + { + "file": "Exam-18-T5-header-1.png", + "description": "Propranolol (P) has SMILES 'CC(NCC(COC1=CC=CC2=C1C=CC=C2)O)C'; metoprolol (M) has SMILES 'CC(NCC(COC1=CC=C(CCOC)C=C1)O)C'; betaxolol (B) has SMILES 'CC(NCC(COC1=CC=C(CCOCC2CC2)C=C1)O)C'; talinolol (T) has SMILES 'O=C(NC1=CC=C(OCC(CNC(C)(C)C)O)C=C1)NC2CCCCC2'; and atenolol (A) has SMILES 'NC(CC1=CC=C(OCC(CNC(C)(C)C)O)C=C1)=O'." + } + ], + "subquestions": [ + { + "sub_id": "Exam-18-T5-1", + "tags": [ + "1.7" + ], + "text_open": "5.3 The second pharmacokinetic process is distribution. After a drug molecule enters the systemic circulation, differences in lipophilicity cause some drugs to enter tissues and even the brain rapidly, while others remain in the blood. To compare drug distribution in the human body, the apparent volume of distribution $V_d$ is defined as follows. When drug distribution reaches dynamic equilibrium, assume that the drug concentration in every tissue and organ equals that in plasma; $V_d$ is the total fluid volume that would be required to contain all the drug in the body at that concentration. The formula is:\n$V_d = \\dfrac{\\text{total amount of drug in the body}\\,(\\mathrm{mg/kg})}{\\text{plasma drug concentration}\\,C\\,(\\mathrm{mg/L})}$\n5.3.2 A standard adult weighing 70kg has an actual total body-fluid volume of only approximately 40 L, including approximately 3 L of plasma. Calculate the approximate $V_d$ of a drug confined entirely to the plasma.", + "text_images": [], + "full_mark": 2 + }, + { + "sub_id": "Exam-18-T5-2", + "tags": [ + "2.9" + ], + "text_open": "5.4 The third pharmacokinetic process is clearance, which is divided mainly into hepatic and renal clearance. Highly lipophilic drugs generally tend toward hepatic clearance, and hepatic clearance capacity may be regarded as proportional to $E_h$, whereas highly water-soluble drugs generally tend toward renal clearance.\nThe approximate half-lives of the five drugs are P 4 h, M 5 h, B 18 h, T 12 h, and A 6.5 h. In the absence of other side effects, a drug with a longer half-life is preferred to maintain smooth dosing.\nThe following patients require a β-blocker from the five drugs above. Select the most appropriate drug for each patient and give only its abbreviation.\n5.4.3 Patient C has mild liver disease and cannot use a drug with $E_H$ greater than 0.10.", + "text_images": [], + "full_mark": 1 + } + ] +} diff --git a/test/tasks/Exam-18-T9/images/Exam-18-T9-1-1-prompt.png b/test/tasks/Exam-18-T9/images/Exam-18-T9-1-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..54e271e51a105a8a5013d5735c0571ad01d7688b --- /dev/null +++ b/test/tasks/Exam-18-T9/images/Exam-18-T9-1-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:7a464b22a03bd0e098e3f7ad345b8118fbcb7d751c09820359753613dab4855b +size 78094 diff --git a/test/tasks/Exam-18-T9/images/Exam-18-T9-2-1-prompt.png b/test/tasks/Exam-18-T9/images/Exam-18-T9-2-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..e0df638428925e1535e429546c4210a7a6765604 --- /dev/null +++ b/test/tasks/Exam-18-T9/images/Exam-18-T9-2-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:f949cefce115fff422c2c12e4ed92d8815fbd1fcdfb03ab2a4aa6f57c75df12b +size 80232 diff --git a/test/tasks/Exam-18-T9/images/Exam-18-T9-3-1-prompt.png b/test/tasks/Exam-18-T9/images/Exam-18-T9-3-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..462d0b25eb317041f6a34967493d0260c6253d75 --- /dev/null +++ b/test/tasks/Exam-18-T9/images/Exam-18-T9-3-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:946d84b60387ae95a989a8bdeeb4e0c2be426a2fcb4be0285851f17d1563e8cf +size 105345 diff --git a/test/tasks/Exam-18-T9/images/Exam-18-T9-4-1-prompt.png b/test/tasks/Exam-18-T9/images/Exam-18-T9-4-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..0b58247af816c95d2d4c16eb230c17b7fbdd5f14 --- /dev/null +++ b/test/tasks/Exam-18-T9/images/Exam-18-T9-4-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:fff56cf77d465a4f4a47cf3bf9b24f2beffa295b94e5550bbbbef4cb744a092b +size 61878 diff --git a/test/tasks/Exam-18-T9/images/Exam-18-T9-5-1-prompt.png b/test/tasks/Exam-18-T9/images/Exam-18-T9-5-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..bd2fcdfb576693ea25653621df5255faf127352b --- /dev/null +++ b/test/tasks/Exam-18-T9/images/Exam-18-T9-5-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:07664fe7da79385081d0e90b428127633ad18a0e6b968a3c896a6c44e21f5b02 +size 75371 diff --git a/test/tasks/Exam-18-T9/images/Exam-18-T9-6-1-prompt.png b/test/tasks/Exam-18-T9/images/Exam-18-T9-6-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..285783ceb60b5700e86c9254e17592878228e1d7 --- /dev/null +++ b/test/tasks/Exam-18-T9/images/Exam-18-T9-6-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:438bb20d8e0e82004f40c1f10a51df71e32a16a70fe57cb632c41725aa7474a6 +size 53814 diff --git a/test/tasks/Exam-18-T9/images/Exam-18-T9-7-1-prompt.png b/test/tasks/Exam-18-T9/images/Exam-18-T9-7-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..85ec8505753b04da5bf2ddf2a66378db1843d65f --- /dev/null +++ b/test/tasks/Exam-18-T9/images/Exam-18-T9-7-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:af84d47aaca31fd727d58fd8f71d7537503bd101188276e7dc9eb4a7ff963a3e +size 131099 diff --git a/test/tasks/Exam-18-T9/images/Exam-18-T9-8-1-prompt.png b/test/tasks/Exam-18-T9/images/Exam-18-T9-8-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..cd2cb3408ac8bf282c4a9168473b9607ffa0f72b --- /dev/null +++ b/test/tasks/Exam-18-T9/images/Exam-18-T9-8-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:9ac8815c0aa08a497529c1db935fc5586c5f66bd39e29bba0ef7b1d9738ddc72 +size 83715 diff --git a/test/tasks/Exam-18-T9/problem.json b/test/tasks/Exam-18-T9/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..d267d3c1fc2317eb37e34010c09726eb074d9440 --- /dev/null +++ b/test/tasks/Exam-18-T9/problem.json @@ -0,0 +1,127 @@ +{ + "problem_id": "Exam-18-T9", + "header": "Problem 9 Skeletal Reconstruction of Alkaloids\nLate-stage skeletal modification and precise remodeling of natural products provide an ideal shortcut for rapidly expanding compound libraries and discovering breakthrough drugs. In 2026, Amir H. Hoveyda's team reported an isolable, general-purpose platform generated by Hofmann elimination (HE), followed by a general strategy for divergent construction of new nonnatural frameworks through rearrangement, ring-closing metathesis, or intramolecular hydroamination.", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-18-T9-1", + "tags": [ + "2.2", + "2.4" + ], + "text_open": "9.1.1 Under specific conditions, anhydrovinblastine reacts with a chlorine source to introduce a chlorine atom and form electrically neutral intermediate A1. In the presence of $AgBF_4$ and solvent, intramolecular reorganization occurs through intermediates A2 and A3, ultimately giving vinorelbine, a widely used clinical anticancer drug. Write the SMILES strings for the key intermediates A1, A2, and A3.", + "text_images": [ + { + "file": "Exam-18-T9-1-1-prompt.png", + "description": "Organic reaction scheme: '[H][C@]12CN(CCC3=C([C@]([R])(C(OC)=O)C2)NC4=CC=CC=C43)CC(CC)=C1'>AgBF4, THF/H_2O (1/1)>'[H][C@]1(C[C@@]([R])(C(OC)=O)C2=C(C3)C4=CC=CC=C4N2)CN3CC(CC)=C1'. The chlorine source is also drawn above the arrow and has SMILES 'CN1C2=CC=CC=C2N=N1.[Cl]'. R is the original placeholder in the figure." + } + ], + "full_mark": 3 + }, + { + "sub_id": "Exam-18-T9-2", + "tags": [ + "2.2", + "2.4" + ], + "text_open": "9.1.2 Under further basic conditions, nucleophilic conjugate addition introduces an alkynyl aldehyde fragment into vinorelbine to give A4. C-N bond cleavage then forms A5, followed by cyclization and desilylation to afford a two-carbon ring-expanded analogue. Write the SMILES strings for the two key intermediates A4 and A5.", + "text_images": [ + { + "file": "Exam-18-T9-2-1-prompt.png", + "description": "Organic reaction scheme: '[H][C@]1(C[C@@]([R])(C(OC)=O)C2=C(C3)C4=CC=CC=C4N2)CN3CC(CC)=C1'>'C[Si](C#CC=O)(C)C', Me3Si removal on workup>'[H][C@]1(C[C@]([R])(C(OC)=O)C2=C(C/C(C=O)=C\\3)C4=CC=CC=C4N2)CN3CC(CC)=C1'. R is the original placeholder in the figure." + } + ], + "full_mark": 2 + }, + { + "sub_id": "Exam-18-T9-3", + "tags": [ + "2.2", + "2.3" + ], + "text_open": "9.2 While studying skeletal deconstruction of vinpocetine, the researchers encountered an unexpected product. Reaction of vinpocetine with 5.0 equivalents of allyl iodide efficiently formed a quaternary ammonium salt. Subsequent treatment in potassium ethoxide/ethanol (KOEt/EtOH) at 80°C for 2 h gave not only the expected ring-opened product in 58% yield but also a deallylated secondary-amine byproduct in 32% yield, as shown. Isotopic-labeling experiments demonstrated that formation of the byproduct did not arise from direct $S_N2$ allylic nucleophilic substitution, but proceeded through a rapid and reversible acid-base proton-exchange process.\n9.2.1 Write the SMILES string for the key enammonium cation intermediate B formed after deprotonation of the quaternary ammonium salt and used to explain the deallylated byproduct.", + "text_images": [ + { + "file": "Exam-18-T9-3-1-prompt.png", + "description": "The figure shows a two-step reaction: '[R]C1=C[C@]2(CC)CCCN([C@@H]23)CCC4=C3N1C5=CC=CC=C45'>5.0 equiv. allyl iodide, MeCN, 3 h, 80 °C>'[R]C1=C[C@@]2(CCC[N@@+]3(CCC4=C([C@H]23)N1C5=CC=CC=C45)C(C=C)[H])CC.[I-]'; '[R]C1=C[C@@]2(CCC[N@@+]3(CCC4=C([C@H]23)N1C5=CC=CC=C45)C(C=C)[H])CC.[I-]'>KOEt, EtOH, + aq. workup>'[R]C(N1C2=C(C(C=C)=C1[C@H]3NCCC4)C=CC=C2)=C[C@@]34CC'. R is the original placeholder in the figure." + } + ], + "full_mark": 1 + }, + { + "sub_id": "Exam-18-T9-4", + "tags": [ + "2.3", + "2.2" + ], + "text_open": "9.2.2 If the KOEt/EtOH system above is replaced by the deuterated reagent system KOEt/EtOD and the vincane-derived quaternary ammonium salt is treated at 80°C, partially deuterated products are obtained. Quantitative nuclear magnetic resonance ($^1H$ NMR) analysis shows unequal deuteration at the allylic positions of the recovered diene substrate: 33% at $H^b$ and 16% at $H^c$. At the most acidic benzylic proton $H^a$, no appreciable deuteration is observed because elimination is too rapid.\nWrite the SMILES string for intermediate C in this kinetic proton-exchange process.", + "text_images": [ + { + "file": "Exam-18-T9-4-1-prompt.png", + "description": "Organic reaction scheme: '[R]C1=C[C@@]2(CCC[N@@+]3(CCC4=C([C@H]23)N1C5=CC=CC=C45)[C@@H](/C=C/[Hc])[Hb])CC.[I-]'>KOEt, EtOD>'[R]C1=C[C@@]2(CCC[N@@+]3(CCC4=C([C@H]23)N1C5=CC=CC=C45)[C@@H](/C=C/[Dc])[Db])CC.[I-]'. R is the original placeholder, and [Hb], [Hc], [Db], and [Dc] are the original hydrogen/deuterium labels in the figure." + } + ], + "full_mark": 1 + }, + { + "sub_id": "Exam-18-T9-5", + "tags": [ + "2.7", + "1.2" + ], + "text_open": "9.3 When the proton environment of the reaction is changed, a reactive carbanion intermediate can be directed onto an entirely different rearrangement pathway.\n9.3.1 Vincane is reacted with allyl iodide to form intermediate D, which is then treated with KOEt in anhydrous acetonitrile. Rather than undergoing elimination, it gives a C-allylated product in 65% yield with exceptionally high asymmetric stereoselectivity of >98:2 d.r.\nGive the specific name of the concerted rearrangement in this step, and write the SMILES string for intermediate D with the stereochemistry of its stereogenic centers shown explicitly.", + "text_images": [ + { + "file": "Exam-18-T9-5-1-prompt.png", + "description": "The figure shows a two-step reaction: 'CC[C@]12CCCN3CCC4=C([C@@H]23)N(CC1)C5=CC=CC=C45'>2.0 equiv. allyl iodide, MeCN>[D]; [D]>3.0 equiv. KOEt, (no EtOH), 1 h, 22 °C>'C=CC[C@]12C(N(CC3)C4=CC=CC=C54)=C5CCN1CCC[C@@]32CC'." + } + ], + "full_mark": 2 + }, + { + "sub_id": "Exam-18-T9-6", + "tags": [ + "2.2", + "2.4" + ], + "text_open": "9.3.2 The product from the previous step reacts with MeOTf to form quaternary ammonium salt E, which is again treated with strong base for elimination. DFT calculations show that, to relieve steric compression, a particular C-N bond in E is substantially elongated to 1.61 Å and is therefore highly susceptible to cleavage. Elimination ultimately occurs selectively at the less accessible but more acidic C-H bond, giving the ten-membered-ring exo-bridged diene F. Write the SMILES strings for quaternary ammonium salt E and elimination product F.", + "text_images": [ + { + "file": "Exam-18-T9-6-1-prompt.png", + "description": "The figure shows a two-step reaction: 'C=CC[C@]12C(N(CC3)C4=CC=CC=C54)=C5CCN1CCC[C@@]32CC'>3.0 equiv. MeOTf, DCE, 16 h, 70 °C>[E]; [E]>3.0 equiv. KOEt, EtOH, 3 h, 80 °C>[F]. E and F are shown only as labels; their structures are not drawn." + } + ], + "full_mark": 2 + }, + { + "sub_id": "Exam-18-T9-7", + "tags": [ + "2.2", + "2.4" + ], + "text_open": "9.4 While attempting ring-closing metathesis (RCM) of an oxidized tetrahydropalmatine derivative to construct a medium-sized oxygen-containing ring, the research team encountered an unusual reverse process. When the Lewis acid $BPh_3$ was added to the W catalyst, the reaction did not give the expected ten-membered-ring metathesis product. Instead, it afforded a ring-contracted isomer bearing a stereogenic vinyl substituent in 61% yield.\nA Lewis acid-base adduct forms first, followed by an extremely rare reverse Meisenheimer rearrangement (RMR) to give boron-complex intermediate G. Write the SMILES strings for G and the boron-containing organic byproduct formed with the product.", + "text_images": [ + { + "file": "Exam-18-T9-7-1-prompt.png", + "description": "The figure shows a three-step reaction: 'C=CC(C=C1OC)=C(C=C1OC)[C@@H]2CC3=CC=C(C(OC)=C3CN2OCC=C)OC'>1.2 equiv. BPh_3, Benzene, 15 h, 60 °C, W10%>'COC1=CC2=C(C=C1OC)/C=C\\C[O+]([B-](C3=CC=CC=C3)(C4=CC=CC=C4)C5=CC=CC=C5)N(CC6=C7OC)[C@H]2CC6=CC=C7OC'; 'COC1=CC2=C(C=C1OC)/C=C\\C[O+]([B-](C3=CC=CC=C3)(C4=CC=CC=C4)C5=CC=CC=C5)N(CC6=C7OC)[C@H]2CC6=CC=C7OC'>[G]; [G]>'C=C[C@@H](C1=C2C=C(OC)C(OC)=C1)N3CC4=C(C[C@@H]23)C=CC(OC)=C4OC', with a branch arrow pointing to byproduct ?." + } + ], + "full_mark": 2 + }, + { + "sub_id": "Exam-18-T9-8", + "tags": [ + "2.3", + "2.4" + ], + "text_open": "9.5 While attempting to isomerize the double bonds of the substrate above, the researchers found that treatment in a $KO^tBu$/DMSO system at room temperature (22°C) gave quantitative conversion in only 30 min, with >98% yield, to an aryl-substituted dehydroisoquinoline derivative. In subsequent biological-activity tests, this molecule showed remarkably potent anticancer activity at nanomolar concentrations.\nTheoretical calculations (DFT) and deuterium-control experiments showed that rapid elimination first forms the $(DMSO)_3K^+$-associated π-allyl-type conjugated complex intermediate $H_1$. With assistance from the potassium ion, this intermediate then undergoes a thermodynamically highly favorable anionic [1,6]-hydrogen proton-transfer process, an 8-electron pericyclic reaction with an activation barrier of only 15.9 kcal/mol, to form $H_2$. Write the SMILES strings for intermediates $H_1$ and $H_2$.", + "text_images": [ + { + "file": "Exam-18-T9-8-1-prompt.png", + "description": "The figure shows a three-step reaction: 'C=CC(C=C1OC)=C(C=C1OC)[C@@H]2CC3=CC=C(C(OC)=C3CN2OCC=C)OC'>1.0 equiv. KOt-Bu, DMSO, 0.5 h, 22 °C>[H1]; [H1]>[H2]; [H2]>'[H]CCC(C=C1OC)=C(C=C1OC)C2=CC3=CC=C(C(OC)=C3C=N2)OC'." + } + ], + "full_mark": 2 + } + ] +} diff --git a/test/tasks/Exam-19-T1/problem.json b/test/tasks/Exam-19-T1/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..6378a58b0664480e956fb0c5b2671208c323d6cd --- /dev/null +++ b/test/tasks/Exam-19-T1/problem.json @@ -0,0 +1,34 @@ +{ + "problem_id": "Exam-19-T1", + "header": "Question 1 Chemistry in Language\nMāori is the language of the indigenous Māori people of New Zealand and belongs to the Malayo-Polynesian branch of the Austronesian language family. Whereas element names in many other languages are borrowed from English or Latin, a number of Māori element names are constructed from native morphemes. The Chinese and English names of several elements and their Māori equivalents are given below. Examine these element names, infer the word-formation rules, and use those rules to answer the questions that follow.\n\n| Chinese name (English translation) | English | Māori | Chinese name (English translation) | English | Māori |\n| --- | --- | --- | --- | --- | --- |\n| hydrogen | hydrogen | hauwai | helium | helium | haumāmā |\n| lithium | lithium | konukōhatu | carbon | carbon | waro |\n| sodium | sodium | konutai | oxygen | oxygen | hāora |\n| chlorine | chlorine | haumāota | calcium | calcium | konupūmā |\n| chromium | chromium | konukita | nickel | nickel | konukōreko |\n\nFor reference, English translations of several Māori words are provided below:\n\n| | | | |\n| --- | --- | --- | --- |\n| hā — breath | ora — alive | wai — water | tai — sea |\n| kōhatu — stone | kita — bright | uku — clay | kōreko — shiny |\n| mohe — malleable | pūmā — grayish white | māota — fresh green, vitality | |", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-19-T1-1", + "tags": [ + "1.7" + ], + "text_open": "1.2 One day, a Māori friend tells you about several elements and their compounds. Infer their meanings from the context.\nHygroscopic solid A contains only the elements hautawa and konumohe. It fumes in air and sublimes under vacuum. Thorough heating of A in air produces compound B together with colored fumes (Reaction 1). B is also an important industrial raw material for producing konumohe, but New Zealand relies mainly on imports of this material.\nElemental konumohe has many uses. Dissolve it in a caustic soda solution, adjust the pH until precipitate C forms, and collect the precipitate by filtration. Adding soda ash and C to an aqueous solution of the simple monoprotic acid corresponding to haukōwhai, followed by recrystallization, gives a 3:1 crystal D (Reaction 2). D has a lower melting point than B and can be used in the metallurgical industry.\nThe Māori names haukōwhai and hautawa are both derived from their colors. The two elements belong to the same group of the periodic table, but their elemental substances have different physical states at room temperature.\nElemental haukōwhai reacts with 2% caustic soda solution to form mainly colorless gas E (Reaction 3). The molecule of E has a structure similar to that of water and reacts under acidic conditions with a simple potassium salt of hautawa to form elemental hautawa (Reaction 4).\n1.2.1 From the analysis, directly give the element names corresponding to hautawa, konumohe, and haukōwhai, and the chemical formulas of A–E.", + "text_images": [], + "full_mark": 8 + }, + { + "sub_id": "Exam-19-T1-2", + "tags": [ + "2.1" + ], + "text_open": "1.2.2 Write the equations for Reactions 1–4.", + "text_images": [], + "full_mark": 6 + }, + { + "sub_id": "Exam-19-T1-3", + "tags": [ + "1.7" + ], + "text_open": "1.3.2 Some people argue that the prefix hau in hautawa is inconsistent with the properties of the element. Do you agree? Explain your reasoning.", + "text_images": [], + "full_mark": 2 + } + ] +} diff --git a/test/tasks/Exam-19-T2/images/Exam-19-T2-1-1-prompt.png b/test/tasks/Exam-19-T2/images/Exam-19-T2-1-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..40970b9c454d047ddaa06ce5a8810e1317674cb8 --- /dev/null +++ b/test/tasks/Exam-19-T2/images/Exam-19-T2-1-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:d985b3685526804a252b79f3db46cbd7201627050b759d8e0cab993744d755e6 +size 71881 diff --git a/test/tasks/Exam-19-T2/images/Exam-19-T2-1-2-prompt.png b/test/tasks/Exam-19-T2/images/Exam-19-T2-1-2-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..4c47d76b972a2fe82d291f009b8e0b350fac26e3 --- /dev/null +++ b/test/tasks/Exam-19-T2/images/Exam-19-T2-1-2-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:e137a957af2bcce1d424b2bfc2b0aabc0243cdd43643c87aed21bf064493193c +size 74130 diff --git a/test/tasks/Exam-19-T2/images/Exam-19-T2-3-1-prompt.png b/test/tasks/Exam-19-T2/images/Exam-19-T2-3-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..2ed1cbb5a88ebca66ff4352f15581f0082caff21 --- /dev/null +++ b/test/tasks/Exam-19-T2/images/Exam-19-T2-3-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:013b1009cbbf0802c9c0c81999378a32a419c2421855c1c6c7130c18171d4e4e +size 27744 diff --git a/test/tasks/Exam-19-T2/images/Exam-19-T2-header-1.png b/test/tasks/Exam-19-T2/images/Exam-19-T2-header-1.png new file mode 100644 index 0000000000000000000000000000000000000000..595035509fc56b7338d9f75cab748f282fc0e82b --- /dev/null +++ b/test/tasks/Exam-19-T2/images/Exam-19-T2-header-1.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:e75d53c8612779f2aa64d775dcd91a96b68ab68b099a3e1b4a504adbe0af31d6 +size 147329 diff --git a/test/tasks/Exam-19-T2/images/Exam-19-T2-header-2.png b/test/tasks/Exam-19-T2/images/Exam-19-T2-header-2.png new file mode 100644 index 0000000000000000000000000000000000000000..e7f1d538e756f55e068c2f885786b8637791edbf --- /dev/null +++ b/test/tasks/Exam-19-T2/images/Exam-19-T2-header-2.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:388d5c5fae2cdb823604c0c05c49a430ddc713a54458048b11028c94df60eb98 +size 76883 diff --git a/test/tasks/Exam-19-T2/images/Exam-19-T2-header-3.png b/test/tasks/Exam-19-T2/images/Exam-19-T2-header-3.png new file mode 100644 index 0000000000000000000000000000000000000000..287b7a3839307114af5bdab1bdd22818f67a120f --- /dev/null +++ b/test/tasks/Exam-19-T2/images/Exam-19-T2-header-3.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:09b48af4d8b20c3e573bb2c5c1b2ffc5ca2aba9ca7b2cfdac5796392b5562b98 +size 6312 diff --git a/test/tasks/Exam-19-T2/problem.json b/test/tasks/Exam-19-T2/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..e7c68a723bb6127fa11d39cf9f6e1ff57dee99b5 --- /dev/null +++ b/test/tasks/Exam-19-T2/problem.json @@ -0,0 +1,73 @@ +{ + "problem_id": "Exam-19-T2", + "header": "Question 2 Photochromic Organic Transistor\nOrganic field-effect transistors (OFETs) offer advantages such as flexibility, printability, and low-temperature fabrication, and are widely used in flexible displays, sensors, and large-area electronic circuits. Organic molecules with distinctive photochemical properties can therefore be used to make light-controlled switches or transistors. Researchers fabricated a photochromic organic transistor using a DAE film as the transistor channel layer. The device supports reversible writing and erasing. The molecular structure of DAE, the organic transistor structure, and the structure of a conventional bipolar transistor are shown below.\nHere, Channel means the conduction channel. From a purely functional perspective, the Source (S terminal) of the field-effect transistor is the source electrode and corresponds to emitter E of the bipolar transistor; Gate (G terminal) is the gate electrode and corresponds to base B; Drain (D terminal) is the drain electrode and corresponds to collector C.", + "header_images": [ + { + "file": "Exam-19-T2-header-1.png", + "description": "The SMILES representation of DAE is 'FC1(F)C(F)(F)C(C2=C(C)SC(C3=CC=C(C4=CC=CC=C4)C=C3)=C2)=C(C5=C(C)SC(C6=CC=C(C7=CC=CC=C7)C=C6)=C5)C1(F)F'." + }, + { + "file": "Exam-19-T2-header-2.png", + "description": "The figure is a structural schematic of an organic field-effect transistor (OFET). From bottom to top it shows a p+-Si substrate (with a lead from the bottom connected to a variable DC supply and labeled Gate), a SiO₂ layer, Au electrodes on the left and right, and a 1D channel spanning the two electrodes. The left electrode is grounded by a lead and labeled Source; the right electrode is connected by a lead to a variable DC supply and labeled Drain. Above the device, the tip of a probe (nanomanipulator/AFM probe) points vertically toward the 1D channel. A translucent cover layer at upper right is labeled DAE and PMMA. The lower right is labeled XY-axes stage and includes arrows showing three-dimensional directions of motion." + }, + { + "file": "Exam-19-T2-header-3.png", + "description": "The figure shows the circuit symbol of a conventional bipolar transistor. Inside a circle is an inverted trapezoidal arrangement. A vertical line at the top connects to B (base), a horizontal line at the right connects to C (collector), and a horizontal line at the left connects to E (emitter). The end of the E line carries an arrow pointing toward the lower diagonal line inside the circle." + } + ], + "subquestions": [ + { + "sub_id": "Exam-19-T2-1", + "tags": [ + "3.5", + "1.7" + ], + "text_open": "The channel contains a mixture of DAE and its isomer X. Let α denote the fraction of X in the channel. Experiments show that after sufficiently prolonged irradiation at wavelength λ1, the maximum isomer fraction α_{max} = 0.75. If a further 40s scan at wavelength λ2 is then applied, α decreases by 0.10. After sufficiently prolonged irradiation at wavelength λ2, the minimum isomer fraction α_{max} = 0.10. If a further 40s scan at wavelength λ1 is then applied, α increases by 0.20. For a reversible first-order isomerization, \\alpha(t) = \\alpha_{\\infty} + (\\alpha_0 - \\alpha_{\\infty})e^{-kt}. Here k is the apparent rate constant, \\alpha_{\\infty} is the isomer fraction at the photostationary state, and \\alpha_ is the isomer fraction at t = 0.\n2.3 In the saturation region, the magnitude of the drain current through a single channel satisfies\n$$|I_D|=\\frac{μC_iW}{2L}(V_G-V_T)^2$$\nwhere W is the channel width; L is the channel length (the length of semiconductor traversed by the current from S to D), taken as 40 μm in this part; μ is the carrier mobility in the saturation region, which describes the overall rate of motion of electrons and holes through the material; $C_i$ is the capacitance per unit area of the insulating layer; $V_G$ is the gate voltage; and $V_T$ is the threshold voltage, which may be approximated as a constant independent of $V_G$ and μ. For n identical channels, approximately\n$$|I_D|=\\frac{μnC_iW}{2L}(V_G-V_T)^2$$\nIn this problem, the saturation-region carrier mobility may be taken as proportional to α.\n2.3.1 The source is grounded and the drain voltage $V_D$ is fixed so that the transistor remains in the saturation region. After irradiation with ultraviolet and visible light of specified wavelengths, respectively, the gate voltage $V_G$ is scanned while the drain current is monitored. The results are shown in the left-hand figure below. The region 0 > $V_G$ > -20V is outside the saturation region and is irrelevant to this question. The transistor properties differ markedly after the two irradiations. Determine whether the transistor after ultraviolet irradiation corresponds to curve group A or B, and briefly explain why. (Figure labels: Current; Voltage; Drawing Cycle.)", + "text_images": [ + { + "file": "Exam-19-T2-1-1-prompt.png", + "description": "The figure is a coordinate plot (left-hand figure for 2.3.1). The vertical axis is labeled 'Drain Current (pA)', with tick labels -8, -6, -4, -2, 0, and 2 from top to bottom. The horizontal axis is labeled 'Gate Voltage (V)', with tick labels 0, -20, -40, and -60 from left to right. Several dark solid curves nearly overlap from 0 to -20 V and separate into two groups beyond -20 V. Group A (labeled 1 and 3,5) has currents of approximately -6 pA and -4 pA at -60 V. Group B (labeled 6 and 2,4,0) has currents between 0 and 1 pA at -60 V. The upper left is labeled 'UV: 25 mW/cm²' and 'VIS: 500 mW/cm²'." + }, + { + "file": "Exam-19-T2-1-2-prompt.png", + "description": "The figure is a coordinate plot (right-hand figure for 2.3.1). The vertical axis is labeled 'Drain Current (pA)', with tick labels 2, 0, -2, -4, -6, and -8 from bottom to top. The horizontal axis is labeled 'Drawing Cycle', with ticks at 0, 5, 10, 15, and 20. A black solid polyline with filled black square markers oscillates in a sawtooth pattern between approximately -0.5 and -6 pA. The absolute magnitudes of the peaks and troughs decrease gradually as the cycle number increases. Gray arrows above each peak and below each trough indicate application of light. No legend is present." + } + ], + "full_mark": 2 + }, + { + "sub_id": "Exam-19-T2-2", + "tags": [ + "3.6", + "3.8" + ], + "text_open": "Starting from the $\\lambda_2$ photostationary state, two consecutive $\\lambda_1$ scans under the stated conditions give an isomer-X fraction $\\alpha=0.44$.\nProlonged irradiation at $\\lambda_1$ makes the isomer-X fraction $\\alpha$ reach 0.75, whereas prolonged irradiation at $\\lambda_2$ makes $\\alpha$ reach 0.10.\n2.3.3 For a transistor with n = 1, after prolonged scanning at wavelength $λ_1$, two data points in the saturation region of its $I_D$–$V_G$ curve are measured: (-40 V, -2.0 pA) and (-50 V, -3.9 pA). A transistor with n = 3 is irradiated for a long time at wavelength $λ_2$ and then subjected to two consecutive 40 s scans at wavelength $λ_1$. Neglecting losses, calculate $I_D$ at $V_G$ = -45 V.", + "text_images": [], + "full_mark": 5 + }, + { + "sub_id": "Exam-19-T2-3", + "tags": [ + "3.6" + ], + "text_open": "2.4 This transistor can be used to construct a light-controlled analog adder. Its structure is shown below. The three branches (1, 2, and 3) forming the Y share one gate G, with $V_G$ = -40 V. The distance between S and D is 60 μm. Branch 1 contains 3 channels and branch 2 contains 2 channels. Every channel operates in the saturation region and has the same width. The section connecting the three channels (the section parallel to S and D) lies midway between S and D and may be treated as an ideal conductor. Initially, every channel has been irradiated for a long time at wavelength $λ_1$. (Hint: note the definition of L. Neglect losses.)\n2.4.1 Calculate the initial magnitude of the drain current.", + "text_images": [ + { + "file": "Exam-19-T2-3-1-prompt.png", + "description": "The figure is a schematic of the Y-shaped light-controlled analog adder, shown as a three-dimensional rectangular trough-like device. The left side is labeled S and connected by a bent lead to a ground symbol. The right side is labeled D and connected by a bent lead to a DC-source symbol. The front of the bottom is labeled G and connected by a vertical lead to a ground symbol. Internal stepped layered regions labeled 1, 2, and 3 correspond to the three branches forming the Y; branches 1 and 2 converge into branch 3." + } + ], + "full_mark": 3 + }, + { + "sub_id": "Exam-19-T2-4", + "tags": [ + "3.6", + "3.8" + ], + "text_open": "2.4.2 For a channel in which μ varies, the equivalent mobility of the channel can be obtained by averaging μ as follows:\n$$μ_{eff}=\\frac{L_1+L_2}{\\frac{L_1}{μ_1}+\\frac{L_2}{μ_2}}$$\nEach channel in branch 1 is now subjected to one 40 s scan at wavelength $λ_2$. Calculate the change in the magnitude of the drain current.", + "text_images": [], + "full_mark": 3 + } + ] +} diff --git a/test/tasks/Exam-19-T3/problem.json b/test/tasks/Exam-19-T3/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..7d7747c79b951c164409930daf647088ee4ab835 --- /dev/null +++ b/test/tasks/Exam-19-T3/problem.json @@ -0,0 +1,25 @@ +{ + "problem_id": "Exam-19-T3", + "header": "Question 3 Degradable Thermosetting Resin\nConventional thermosetting polymers form covalently cross-linked networks that provide exceptional mechanical strength and heat resistance, but also prevent melt processing and make recycling and degradation difficult. In recent years, orthogonal polymerization of different functional groups within a single monomer has emerged as a leading strategy for synthesizing polymers that can be degraded at predetermined sites. 2,3-Dihydrofuran (DHF) is a monomer that can be prepared from bio-based feedstocks. Recently, chemists used DHF in a one-pot process involving two successive polymerizations with entirely different mechanisms to produce a thermosetting resin that can be recycled through multiple pathways.", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-19-T3-1", + "tags": [ + "2.4" + ], + "text_open": "3.1 In the presence of a trace amount of second-generation Grubbs catalyst, DHF slowly undergoes ring-opening metathesis polymerization (ROMP) in the dark to give rubbery linear polymer A. In the presence of a photoacid generator (PAG, which produces the superacid HSbF6 upon irradiation), it rapidly undergoes cationic polymerization to give highly rigid linear polymer B.\n3.1.2 Write the SMILES strings for linear polymers A and B.", + "text_images": [], + "full_mark": 2 + }, + { + "sub_id": "Exam-19-T3-2", + "tags": [ + "2.4" + ], + "text_open": "3.2 To prepare a thermosetting resin, the researchers mixed DHF, second-generation Grubbs catalyst, and PAG. The mixture was first allowed to react in the dark for 10 minutes to form an appropriate amount of polymer A, then irradiated with 456 nm blue light for 6 minutes to initiate cationic polymerization. This produced cross-linked thermosetting resin C, which is insoluble and infusible. Write the SMILES string for C when the two polymer chains are fully cross-linked, with no unreacted cross-linking sites. Hint: to simplify the structure, assume that both linear polymers A and B participating in the reaction have degree of polymerization n.", + "text_images": [], + "full_mark": 2 + } + ] +} diff --git a/test/tasks/Exam-20-T3/images/Exam-20-T3-1-1-prompt.png b/test/tasks/Exam-20-T3/images/Exam-20-T3-1-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..5a5ce35aea6f2f58c2aa72da267a3ea800dc294a --- /dev/null +++ b/test/tasks/Exam-20-T3/images/Exam-20-T3-1-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:3ecad2b388ae301b85d9c7c49aca1e231309ba842124a5c5ba8ac5266004fdcd +size 419777 diff --git a/test/tasks/Exam-20-T3/images/Exam-20-T3-2-1-prompt.png b/test/tasks/Exam-20-T3/images/Exam-20-T3-2-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..e1de77dbf7a40cf467cf148fe8696a4eced08222 --- /dev/null +++ b/test/tasks/Exam-20-T3/images/Exam-20-T3-2-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:199dcd0240957126e23443457a3c00867ec79661a1d662d56bc9b86cb7165176 +size 180090 diff --git a/test/tasks/Exam-20-T3/images/Exam-20-T3-4-1-prompt.png b/test/tasks/Exam-20-T3/images/Exam-20-T3-4-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..fe0f2058658eb9e6c57d05c9ab59166855f330a6 --- /dev/null +++ b/test/tasks/Exam-20-T3/images/Exam-20-T3-4-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:bfef5665604e31c7cea834b993e37fffe4a114a4975a6bc780c767107e894832 +size 13317 diff --git a/test/tasks/Exam-20-T3/images/Exam-20-T3-5-1-prompt.png b/test/tasks/Exam-20-T3/images/Exam-20-T3-5-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..5519af0727f894d6008ac273c4c102b6e69b6d83 --- /dev/null +++ b/test/tasks/Exam-20-T3/images/Exam-20-T3-5-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:d7758ef643f5e18c79c79bfeb2f42490f31b00a2475cc81531f8c4f81eceb8a6 +size 126316 diff --git a/test/tasks/Exam-20-T3/problem.json b/test/tasks/Exam-20-T3/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..47cdfa496217caed5a9c28ec493dc1f4869c49fc --- /dev/null +++ b/test/tasks/Exam-20-T3/problem.json @@ -0,0 +1,76 @@ +{ + "problem_id": "Exam-20-T3", + "header": "Halogen-Bond-Donor-Catalyzed Synthesis of Poly(ortho esters)\nPolymers have transformed modern life because of their low cost and practicality. Polyesters are high-performance, versatile engineering plastics and can also be made into polyester fibers and films; the family includes polyester resins and elastomers.", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-20-T3-1", + "tags": [ + "2.4", + "2.3" + ], + "text_open": "3.1 Cyclic ketene acetals undergo radical polymerization in the presence of peroxides to generate poly(regular lactone) chains. Write the SMILES strings of A-D in the reaction scheme below. Polymer end groups need not be shown. Only D is a polymer.", + "text_images": [ + { + "file": "Exam-20-T3-1-1-prompt.png", + "description": "Four-step reaction: 'BrCC1OCCO1'>tBuOK>A; A>ROOR, heat>[B]; [B]>isomerization>[C]; [C]>D. R is the placeholder used in the original diagram, and [B] and [C] denote radical intermediates." + } + ], + "full_mark": 5 + }, + { + "sub_id": "Exam-20-T3-2", + "tags": [ + "2.4", + "2.6" + ], + "text_open": "3.2 Poly(ortho esters) (POEs) are renewable, biodegradable polymers whose degradation kinetics and surface-erosion behavior can be tuned and whose degradation products are harmless. Earlier synthetic methods, however, gave POEs with molecular weights that were too low and properties that could not be well controlled, making them unsuitable as general-purpose materials such as plastics and rubbers. The reaction below uses renewable feedstocks and an innovative initiator to achieve controlled ring-opening polymerization of the monomer and produce POEs with controllable molecular weights.\n3.2.1 In this reaction, two renewable starting materials first form the spiro-ortho-ester monomer E, MSOE, under $BF_3$ catalysis. The target polymer P1 is then synthesized using the initiator MOMBr and the catalyst ion XBD. After monomer E reacts with the active center of the polymer chain, it can isomerize and polymerize to give the acyclic polymer P2. Based on this information, write the SMILES strings of E, P1, and P2. Polymer end groups need not be shown.", + "text_images": [ + { + "file": "Exam-20-T3-2-1-prompt.png", + "description": "Two-step reaction: 'CC1CO1' + 'O=C1CCCCCO1' > $BF_3\\cdot OEt_2$ (5.0 mol%) > E; E + 'BrCOC' (MOMBr) > XBD > P1 (major) + P2 (minor)." + } + ], + "full_mark": 6 + }, + { + "sub_id": "Exam-20-T3-3", + "tags": [ + "2.4" + ], + "text_open": "3.2.2 At pH = 3, P1 can react with water and degrade, allowing recovery of small-molecule ester monomers. Write the SMILES strings of the possible recovered small molecules.", + "text_images": [], + "full_mark": 2 + }, + { + "sub_id": "Exam-20-T3-4", + "tags": [ + "1.6", + "1.1" + ], + "text_open": "3.3 The catalyst ion XBD is a trinuclear halogen-bond-donor catalyst whose mechanism involves a special secondary interaction: the halogen bond. A halogen bond is fundamentally an electrostatic interaction arising from a region of accumulated positive charge at the extension opposite a covalent bond to a halogen atom, especially Cl, Br, or I. This region is called a \"σ hole\" in the literature and is illustrated below.\n3.3.1 Based on this information, write the SMILES strings of the halogen-bonded aggregates $(Cl_2)_2$ and $(RBr)_3$. The latter has a rotational axis of order higher than two; treat R as a spherical ligand. Specify the halogen-bond angles rigorously.", + "text_images": [ + { + "file": "Exam-20-T3-4-1-prompt.png", + "description": "Schematic of a σ hole on a halogen atom: a blue gradient sphere at the center bears the capital letter E. The letter R is connected to the left of the sphere by a short horizontal line. To the right is a concentric region with an outer green ring, a middle yellow ring, and a red central point. A red δ+ at the upper right has a curved arrow pointing to the concentric region, while a blue δ- at the lower left has a curved arrow pointing to the bottom of the sphere." + } + ], + "full_mark": 4 + }, + { + "sub_id": "Exam-20-T3-5", + "tags": [ + "2.3", + "2.1" + ], + "text_open": "3.3.2 The structure of the catalyst ion XBD used in 3.2 is shown below. Write the reaction equations for chain initiation by the initiating system, MOMBr and the catalyst ion XBD. XBD may be abbreviated; there is no need to write a SMILES string containing XBD. On this basis, determine whether the reaction is cationic or anionic polymerization. Hint: chain initiation consists of two elementary steps.", + "text_images": [ + { + "file": "Exam-20-T3-5-1-prompt.png", + "description": "The SMILES for XBD is '[R]N(N=[N+]1C)C(I)=C1C2=CC(C3=C(I)N([R])N=[N+]3C)=CC(C4=C(I)N([R])N=[N+]4C)=C2'. The 3 [R] groups are placeholders used in the original diagram, and the 3 [N+] atoms carry positive-charge labels." + } + ], + "full_mark": 5 + } + ] +} diff --git a/test/tasks/Exam-20-T4/problem.json b/test/tasks/Exam-20-T4/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..0266cbeaacb537fcba9d8150a5ba5a9dad34ce76 --- /dev/null +++ b/test/tasks/Exam-20-T4/problem.json @@ -0,0 +1,44 @@ +{ + "problem_id": "Exam-20-T4", + "header": "Determination of Silicon by the Silicomolybdenum Blue Spectrophotometric Method\nSilicon is an important element in iron ores and steel materials, and its content significantly affects steel properties. In the commonly used silicomolybdenum blue spectrophotometric method, silicic acid reacts with ammonium molybdate in acidic medium to form silicomolybdenum yellow, which is then reduced to silicomolybdenum blue for absorbance measurement. This problem concerns sample fusion, removal of interferences, control of color-development conditions, and quantification by a calibration curve.", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-20-T4-1", + "tags": [ + "4.1", + "4.7" + ], + "text_open": "4.1 Sample pretreatment\nWeigh 0.1284 g of sample into a 50 mL silver crucible, add 2 g of sodium peroxide and mix, then cover with another 0.5 g of sodium peroxide. Cover the crucible, place it in a furnace at 300 °C, slowly raise the temperature to 750 °C-800 °C, and fuse for 20 min. Remove and allow to cool slightly. Add 100 mL of boiling water to leach the melt completely and wash out the crucible. While swirling, slowly add 30 mL of concentrated hydrochloric acid. After cooling, transfer the solution to a 200 mL volumetric flask, dilute to the mark with water, mix, filter, and discard the initial portion of the filtrate.\n4.1.6 The acidified solution must be dry-filtered through medium-speed filter paper, and the first few milliliters of filtrate must be discarded. Explain the purposes of this procedure. If the initial filtrate were not discarded, would the measured result be too high or too low?", + "text_images": [], + "full_mark": 3 + }, + { + "sub_id": "Exam-20-T4-2", + "tags": [ + "4.2" + ], + "text_open": "4.2 Color development for the sample\nTransfer 10.00 mL of filtrate to a 100 mL volumetric flask and add blank-test solution to bring the volume to 20 mL. Add 10 mL of hydrochloric acid (1 + 3), add potassium permanganate solution (20 g/L) dropwise until the solution is pale red, add 10 mL of water, and mix.\nAdd 1 mL of anhydrous ethanol and mix. While continuously swirling, add 5 mL of ammonium molybdate solution (100 g/L), mix, and allow to stand at 15 °C-25 °C for 15 min.\nAdd 15 mL of hydrochloric acid (1 + 3), followed by 5 mL of oxalic acid-tartaric acid mixture containing 50 g/L oxalic acid and 40 g/L tartaric acid. Mix and immediately add 5 mL of ascorbic acid solution (20 g/L), dilute to the mark with water, mix, and allow to stand at 25 °C-35 °C for 15 min.\nTransfer a portion of the colored solution to a 3 cm cuvette. Using the colored blank-test solution carried through with the sample as reference, measure an absorbance of 0.425 at 810 nm.\n4.2.1 Explain why adding potassium permanganate to a pale-red endpoint eliminates interference from lower-valent iron and organic matter, and state the effect of excess potassium permanganate.", + "text_images": [], + "full_mark": 3 + }, + { + "sub_id": "Exam-20-T4-3", + "tags": [ + "4.2" + ], + "text_open": "4.2.4 State which interfering ions are masked by oxalic acid and tartaric acid, respectively, in the color-development system, and explain why ascorbic acid must not be added before the oxalic acid-tartaric acid mixture.", + "text_images": [], + "full_mark": 3 + }, + { + "sub_id": "Exam-20-T4-4", + "tags": [ + "4.7" + ], + "text_open": "4.3 Construction of the calibration curve\nTransfer 0 mL, 1.00 mL, 2.00 mL, 4.00 mL, 6.00 mL, 8.00 mL, and 10.00 mL portions of silicon dioxide standard solution (20.0 μg/mL) to a series of 100 mL volumetric flasks. Add water to 20 mL and develop the color according to the procedure in 4.2. Transfer each colored solution to a 3 cm cuvette and measure the absorbance at 810 nm against a reagent blank. Construct a calibration curve using the amount of silicon dioxide in μg as the horizontal coordinate and absorbance as the vertical coordinate.\n\n| Volume of silicon dioxide standard solution ($V$ / mL) | 0.00 | 1.00 | 2.00 | 4.00 | 6.00 | 8.00 | 10.00 |\n| --- | --- | --- | --- | --- | --- | --- | --- |\n| Amount of silicon dioxide ($m$ / μg) | 0 | 20.00 | 40.00 | 80.00 | 120.0 | 160.0 | 200.0 |\n| Absorbance ($A$) | 0.000 | 0.082 | 0.163 | 0.326 | 0.488 | 0.651 | 0.812 |\n\nCalculate the silicon dioxide content of the sample.", + "text_images": [], + "full_mark": 5 + } + ] +} diff --git a/test/tasks/Exam-21-T10/images/Exam-21-T10-1-1-prompt.png b/test/tasks/Exam-21-T10/images/Exam-21-T10-1-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..0ca2554562547e584d555011636f72f59a4e07f2 --- /dev/null +++ b/test/tasks/Exam-21-T10/images/Exam-21-T10-1-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:439a942469ec91ed4bf26e5bb1be17986f56cd45cdbde8e7bd7b8fff92f64fe1 +size 79629 diff --git a/test/tasks/Exam-21-T10/images/Exam-21-T10-3-1-prompt.png b/test/tasks/Exam-21-T10/images/Exam-21-T10-3-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..eaff1bc45d8b0f0e45a2d0014ff873d91291066f --- /dev/null +++ b/test/tasks/Exam-21-T10/images/Exam-21-T10-3-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:28d0f68052cd21f6c1861279b951623b2c9c11f72ce0fad16a3c94b8caf1c43a +size 79767 diff --git a/test/tasks/Exam-21-T10/images/Exam-21-T10-header-1.png b/test/tasks/Exam-21-T10/images/Exam-21-T10-header-1.png new file mode 100644 index 0000000000000000000000000000000000000000..66f367d9f2138cee1c0850e631e8a5ddb9417894 --- /dev/null +++ b/test/tasks/Exam-21-T10/images/Exam-21-T10-header-1.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:ea60b3b8f101104b9c2495f6572ab483ef3e3c175fd491c893877697a431b974 +size 191664 diff --git a/test/tasks/Exam-21-T10/problem.json b/test/tasks/Exam-21-T10/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..67c41254f3da581289f97ded7d0dc059bc7a9519 --- /dev/null +++ b/test/tasks/Exam-21-T10/problem.json @@ -0,0 +1,52 @@ +{ + "problem_id": "Exam-21-T10", + "header": "Problem 10 Building a Very Familiar Aromatic Ring—the Simple Shikimic Acid Pathway\nAmino acids are essential substances that our bodies require continuously. Some can be synthesized in the body, whereas a few cannot and must be obtained from food, including Phe (phenylalanine), Tyr (tyrosine), and Trp (tryptophan), whose structures are shown below.\nHow do plants, microorganisms, and other organisms synthesize these amino acids? A widespread natural route produces aromatic compounds and many related six-membered cyclic compounds. It is named after a key intermediate: the shikimic acid pathway. Note: stereochemistry is required throughout this problem unless otherwise stated.", + "header_images": [ + { + "file": "Exam-21-T10-header-1.png", + "description": "Phe has the SMILES representation '[H][C@@](N)(C(O)=O)CC1=CC=CC=C1'; Tyr, 'N[C@]([H])(C(O)=O)CC1=CC=C(O)C=C1'; and Trp, 'N[C@]([H])(C(O)=O)CC1=CNC2=C1C=CC=C2'. The three structures are shown side by side, with -NH2 at each stereogenic carbon drawn on a solid wedge." + } + ], + "subquestions": [ + { + "sub_id": "Exam-21-T10-1", + "tags": [ + "2.4", + "1.2" + ], + "text_open": "10.1 The shikimic acid pathway begins with condensation of phosphoenolpyruvate (PEP) and erythrose 4-phosphate (E4P).\n10.1.1 Write the SMILES string for the major organic product A. A carries one negative charge, contains only one phosphorus atom, and the reaction creates one new stereocenter of absolute configuration R.", + "text_images": [ + { + "file": "Exam-21-T10-1-1-prompt.png", + "description": "Single-step reaction: 'O=P(OC(C(O)=O)=C)([O-])O' (PEP, with one negative charge on a phosphate oxygen).'O=P(O)([O-])OC[C@@H](O)[C@@H](O)C([H])=O' (E4P, with one negative charge on a phosphate oxygen)>>[A]. No reaction conditions are written, and the product is shown only as A without an explicit structure." + } + ], + "full_mark": 3 + }, + { + "sub_id": "Exam-21-T10-2", + "tags": [ + "2.4", + "1.2" + ], + "text_open": "10.1.2 A readily isomerizes, so the reaction actually gives a mixture of A and its isomers. Write the SMILES string for the principal component M other than A.", + "text_images": [], + "full_mark": 2 + }, + { + "sub_id": "Exam-21-T10-3", + "tags": [ + "2.2", + "2.3" + ], + "text_open": "10.2 A cyclizes to form 3-dehydroquinate in a process mediated by $NAD^+$ and NADH. Earlier studies proposed the following mechanism. First, A is oxidized by 1 molecule of $NAD^+$ to give Int1, with loss of one stereocenter. Int1 undergoes β-elimination of the phosphate group and tautomerization to give Int2, which contains an α, β-diketone unit. Int2 is then reduced by 1 molecule of NADH to give Int3, restoring the lost stereocenter with its original relative configuration. Int3 undergoes intramolecular condensation to form 3-dehydroquinate ($C_7H_{10}O_6$). Write the SMILES strings for Int1, Int2, Int3, and 3-dehydroquinate.", + "text_images": [ + { + "file": "Exam-21-T10-3-1-prompt.png", + "description": "NAD+ has the SMILES representation 'NC(C1=CC=C[N+]([R])=C1)=O'; NADH has the SMILES representation 'NC(C1=CN([R])C=CC1)=O'. R is the original placeholder in the figure for the substituent on the pyridinium/1,4-dihydropyridine ring nitrogen." + } + ], + "full_mark": 9 + } + ] +} diff --git a/test/tasks/Exam-21-T3/images/Exam-21-T3-3-1-prompt.png b/test/tasks/Exam-21-T3/images/Exam-21-T3-3-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..dd6470088e3c889433dbbcefb0c97b559b5bef63 --- /dev/null +++ b/test/tasks/Exam-21-T3/images/Exam-21-T3-3-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:e860267bb2b12e00750a07d1c5a3ebfb9a3f4d1dbe7a6796d0e03586a56b5290 +size 829850 diff --git a/test/tasks/Exam-21-T3/problem.json b/test/tasks/Exam-21-T3/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..316c256d3f29bc9b1c73c79c60d7c635d9b186c1 --- /dev/null +++ b/test/tasks/Exam-21-T3/problem.json @@ -0,0 +1,50 @@ +{ + "problem_id": "Exam-21-T3", + "header": "Preparation and Properties of a Thermochromic Copper(II) Complex\nMaterials whose compounds exhibit thermochromism through structural phase transitions are widely used in anti-counterfeiting applications. To synthesize the thermochromic complex $[Cu(deen)_2](ClO_4)_2$ (deen denotes N,N-diethylethylenediamine), hydrated copper perchlorate must first be prepared, followed by introduction of the deen ligand.", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-21-T3-1", + "tags": [ + "1.4", + "4.3" + ], + "text_open": "Hydrated copper perchlorate can be prepared by a precipitation–dissolution method. Dissolve $CuSO_4\\cdot 5H_2O$ in water, add $NaOH$ solution with stirring, and heat slowly on a hot plate until the precipitate turns black. Maintain gentle boiling for 5–10 min, allow the mixture to cool slightly, and collect the precipitate by suction filtration. Rinse it with a small amount of deionized water, then use a glass rod to transfer the precipitate into an evaporating dish containing 10 mL of water. Add perchloric acid, gently guide the precipitate into the solution with the glass rod, and stir until it dissolves completely to form a copper perchlorate solution. Concentrate the solution by gentle heating to about 2 mL. On cooling, crystals of hydrated copper perchlorate are obtained. The preparation of $[Cu(deen)_2](ClO_4)_2$ begins with the hydrated copper perchlorate crystals obtained above. Adding absolute ethanol to these crystals gives a clear blue solution. In a dry beaker, dissolve N,N-diethylethylenediamine in ethanol to obtain a clear pale-yellow solution. Slowly add the ethanolic deen solution to the ethanolic copper perchlorate solution with stirring; the mixture immediately turns deep blue-violet. Stir for 2–3min, then allow it to stand for about 10 min. A large amount of pink precipitate forms at the bottom of the vessel while the supernatant remains deep blue-violet. Collect the product by suction filtration and dry it to obtain $[Cu(deen)_2](ClO_4)_2$. 3.2.2 Dissolving the red $[Cu(deen)_2](ClO_4)_2$ in water again gives a deep-blue solution. Explain why. (No more than 50 words)", + "text_images": [], + "full_mark": 4 + }, + { + "sub_id": "Exam-21-T3-2", + "tags": [ + "1.4" + ], + "text_open": "3.3 In a thermochromism experiment on $[Cu(deen)_2](ClO_4)_2$, a small amount of the $[Cu(deen)_2](ClO_4)_2$ complex is held between pieces of adhesive tape and heated with a hair dryer. The sample changes from pink to blue-violet and returns to pink on cooling.\n3.3.1 Based on crystal field theory, describe in words as necessary the structural change of the $[Cu(deen)_2]^{2+}$ unit upon heating and cooling. The cooled unit may be approximated as centrosymmetric, whereas after heating it is no longer centrosymmetric.", + "text_images": [], + "full_mark": 3 + }, + { + "sub_id": "Exam-21-T3-3", + "tags": [ + "4.6" + ], + "text_open": "3.4 To investigate the thermochromism of $[Cu(deen)_2]X_2$ complexes formed with monovalent counteranion $X^-$, differential scanning calorimetry is performed on $[Cu(deen)_2](ClO_4)_2$ and $[Cu(deen)_2](NO_3)_2$. Multiple interactions occur between the cations and anions in both complexes. The results are shown below. For both complexes, the color change is associated with a phase-transition temperature, and the chemical composition remains unchanged across the color change. The temperature at the peak maximum is defined as the transition temperature.\n3.4.1 From the graph, directly give the thermochromic transition temperatures of $[Cu(deen)_2](ClO_4)_2$ and $[Cu(deen)_2](NO_3)_2$.", + "text_images": [ + { + "file": "Exam-21-T3-3-1-prompt.png", + "description": "Untitled coordinate plot. The horizontal axis is labeled \"Temperature/°C\" with major ticks at 50, 100, 150, and 200 and minor ticks at 25, 75, 125, and 175. The vertical axis is labeled \"Thermogravimetry\" vertically and has no numerical ticks. Two solid black curves are shown. The legend at lower left uses short line segments for \"$ClO_4$(a)\" and \"$NO_3$(b),\" and the curves are labeled (a) and (b). Curve (a), corresponding to $[Cu(deen)_2](ClO_4)_2$, has a lower baseline and a sharp upward peak at about 45–50°C, remains flat thereafter, and drops steeply after about 200°C. Curve (b), corresponding to $[Cu(deen)_2](NO_3)_2$, has a higher baseline, a broad upward peak at about 110°C, a sharp upward peak at about 145°C, another smaller upward peak at about 180°C, and a steep drop after about 200°C. No arrows are shown." + } + ], + "full_mark": 4 + }, + { + "sub_id": "Exam-21-T3-4", + "tags": [ + "1.4", + "4.6" + ], + "text_open": "3.4.2 Using the information above, explain the origin of the difference between the transition temperatures of $[Cu(deen)_2](ClO_4)_2$ and $[Cu(deen)_2](NO_3)_2$. (No more than 100 words)", + "text_images": [], + "full_mark": 4 + } + ] +} diff --git a/test/tasks/Exam-21-T4/problem.json b/test/tasks/Exam-21-T4/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..fc2a9923d6d16d8a7dbeda469668e87ffd0710b9 --- /dev/null +++ b/test/tasks/Exam-21-T4/problem.json @@ -0,0 +1,25 @@ +{ + "problem_id": "Exam-21-T4", + "header": "Determination of Manganese by Ammonium Nitrate Oxidation Titrimetry\nManganese is an important element in steels and alloys, and its content directly affects their strength, hardness, and wear resistance. Ammonium nitrate oxidation titrimetry is a classical method for determining manganese: ammonium nitrate oxidizes Mn(II) to Mn(III) in a phosphoric acid medium, and the product is then titrated with a standard ammonium iron(II) sulfate solution. This problem addresses standardization of the titrant, control of the oxidation conditions, and calculation of the result.", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-21-T4-1", + "tags": [ + "1.4" + ], + "text_open": "4.2 Sample treatment and titration\nWeigh 0.2026 g of sample into a 300 mL conical flask. Add 15 mL of concentrated phosphoric acid and 5 mL of concentrated nitric acid, and heat until the sample dissolves completely. Continue heating and evaporating until the surface becomes still and faint fumes just begin to appear (about 220–240 °C), then remove the flask from the heat. Immediately add 3.0 g ± 0.2 g of ammonium nitrate. While swirling the flask, use a rubber-bulb pipette to blow tangentially across the mouth of the flask to remove nitrogen oxides until the yellow fumes disappear completely. Allow the mixture to cool slightly, add 60 mL of distilled water, mix thoroughly, and cool to room temperature. Titrate with the standard ammonium iron(II) sulfate solution until the solution is pale red. Add 2 drops of N-phenylanthranilic acid indicator solution (2 g/L) and continue titrating to a bright-yellow endpoint. The standard ammonium iron(II) sulfate solution consumed is 16.48 mL, and the blank consumes 0.84 mL.\n4.2.2 Give the predominant form of manganese after oxidation by ammonium nitrate. (Hint: the ligand is dihydrogen phosphate.)", + "text_images": [], + "full_mark": 2 + }, + { + "sub_id": "Exam-21-T4-2", + "tags": [ + "2.1" + ], + "text_open": "4.2.4 Write the net ionic equation for the titration. In the products, the metal ions are present as free ions rather than complexes.", + "text_images": [], + "full_mark": 2 + } + ] +} diff --git a/test/tasks/Exam-21-T5/images/Exam-21-T5-2-1-prompt.png b/test/tasks/Exam-21-T5/images/Exam-21-T5-2-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..d194a80721e25a0d2ab44cb6c054d14eefe53cf9 --- /dev/null +++ b/test/tasks/Exam-21-T5/images/Exam-21-T5-2-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:3d7b28e7eeb2d1d381ff4b5b6ff85d5495cf09adf2e5f9321fc64a2e92bfb309 +size 109587 diff --git a/test/tasks/Exam-21-T5/problem.json b/test/tasks/Exam-21-T5/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..d3a3a978e354123a0a35082a3b2b7305e6cb36c7 --- /dev/null +++ b/test/tasks/Exam-21-T5/problem.json @@ -0,0 +1,39 @@ +{ + "problem_id": "Exam-21-T5", + "header": "Structures of Silicides", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-21-T5-1", + "tags": [ + "1.3" + ], + "text_open": "5.1 Consider a hypothetical crystal X. Si atoms form graphite-like layers A stacked along the c direction. The projections of all layers A along c coincide, producing a series of layers of hexagonal-prismatic voids between them. Metal M regularly occupies 2/3 of the hexagonal-prismatic voids in each such layer to form layers B, and the projections of all layers B along c also coincide.\n5.2 Crystal Y is obtained by replacing some Si atoms in layers A of crystal X with M atoms. In crystal Y, no two Si atoms and no two M atoms are adjacent within the same layer A. Likewise, between two neighboring layers A along c, no Si atom is adjacent to another Si atom and no M atom is adjacent to another M atom.\n5.2.1 Give the chemical formula of crystal Y in terms of M and Si, reduced approximately to the simplest ratio, and the number of formula units in one conventional unit cell.", + "text_images": [], + "full_mark": 3 + }, + { + "sub_id": "Exam-21-T5-2", + "tags": [ + "1.3" + ], + "text_open": "5.3 Further distortion of layers A in crystal Y gives hexagonal crystal Z, a real silicide. The projection of crystal Z along c is shown below, where the small dark spheres represent Si atoms and the large light-colored spheres represent M atoms. After distortion, all Si atoms still have identical chemical environments, and all M atoms within the same type of layer likewise have identical chemical environments.\n5.3.1 In crystal Z, one Si atom has coordinates (0.6000, 0, 0.2500), and two M atoms have coordinates (0.2300, 0, 0.2500) and (0.3333, 0.6667, 0.0000), respectively. Give the coordinates of all remaining Si and M atoms.", + "text_images": [ + { + "file": "Exam-21-T5-2-1-prompt.png", + "description": "Projection of crystal Z along the c direction on a white background. Thin gray lines outline a right-leaning parallelogram unit cell, with approximately horizontal top and bottom edges and side edges slanting upward to the right. The diagram contains 22 shaded, highlighted spheres of two types: 12 large light-blue spheres and 10 small dark-blue spheres. There are 16 spheres on the unit-cell boundary, 4 on each edge; every edge has 2 large light-blue spheres and 2 small dark-blue spheres arranged in the order large light-blue, small dark-blue, small dark-blue, large light-blue. The interior contains 6 additional spheres: 4 large light-blue spheres at upper left, upper right, lower left, and lower right, and 2 small dark-blue spheres at center-right and lower-center-left, approximately along the shorter diagonal. At lower left, a red solid arrow points horizontally right and is labeled a; a green solid arrow points upper left and is labeled b. Their common origin is a blue circle containing a white c. As specified in the prompt, the dark spheres are Si atoms and the light-colored spheres are M atoms." + } + ], + "full_mark": 6 + }, + { + "sub_id": "Exam-21-T5-3", + "tags": [ + "1.3" + ], + "text_open": "5.3.2 During the distortion, the coordination polyhedron around an M atom in layer A, formed by Si atoms in the same and adjacent layers, is also distorted. State how many distinct Si–M distances occur in the distorted coordination polyhedron.", + "text_images": [], + "full_mark": 2 + } + ] +} diff --git a/test/tasks/Exam-21-T6/images/Exam-21-T6-2-1-prompt.png b/test/tasks/Exam-21-T6/images/Exam-21-T6-2-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..695f003e67354225370a191c6546fec6f81d3a94 --- /dev/null +++ b/test/tasks/Exam-21-T6/images/Exam-21-T6-2-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:ecdb03a7b10587870a7a4a61a278c75e2779e242b49e18c17b7adef7893c223f +size 670609 diff --git a/test/tasks/Exam-21-T6/images/Exam-21-T6-3-1-prompt.png b/test/tasks/Exam-21-T6/images/Exam-21-T6-3-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..c75e230451c1fb4aa220f7c24ba10f29205e381d --- /dev/null +++ b/test/tasks/Exam-21-T6/images/Exam-21-T6-3-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:1221df2b58f97d026c251ee2500a562053a637de2a1b540f8c4e4c31d79ede3f +size 812829 diff --git a/test/tasks/Exam-21-T6/problem.json b/test/tasks/Exam-21-T6/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..a58141cc97a466e77a6314f3367c8598b28ca95e --- /dev/null +++ b/test/tasks/Exam-21-T6/problem.json @@ -0,0 +1,46 @@ +{ + "problem_id": "Exam-21-T6", + "header": "Pharmacokinetics and Molecular Design (II)\nPharmacokinetics quantitatively studies the absorption, distribution, metabolism, and excretion of drugs in living organisms and uses mathematical principles and methods to describe how plasma drug concentration changes with time. Chemical-kinetic modeling provides a systematic understanding of drug metabolism in vivo. This problem concerns the oxidative metabolism of diclofenac in the body.", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-21-T6-1", + "tags": [ + "3.7" + ], + "text_open": "The Michaelis equation is $r=r_{max}[S]/(K_m+[S])$.\nExperiments show that $[S]$ and $r$ for the 4′-hydroxylation of diclofenac catalyzed by human liver microsomes obey the following relationship:\n| Group | $[S]_1$ / $\\mu\\mathrm{M}$ | $r_1$ / $(\\mathrm{pmol \\cdot min^{-1} \\cdot mg^{-1}})$ | $[S]_2$ / $\\mu\\mathrm{M}$ | $r_2$ / $(\\mathrm{pmol \\cdot min^{-1} \\cdot mg^{-1}})$ |\n| --- | --- | --- | --- | --- |\n| Low concentration | 0.90 | 39.27 | 1.80 | 72.00 |\n| High concentration | 90.0 | 392.73 | 180.0 | 411.43 |\n\n6.2.4 Under the same conditions, the 5-hydroxylation has $K_m = 43\\ \\mu\\mathrm{M}$ and $r_{max} = 15.4\\ \\mathrm{pmol \\cdot min^{-1} \\cdot mg^{-1}}$. Intrinsic clearance $CL_{int}$ is defined by $CL_{int} = r_{max}/K_m$. Calculate $CL_{int}$ for the 4′- and 5-hydroxylation pathways and briefly explain the kinetic meaning of $CL_{int}$. Hint: the drug concentration in the liver is much lower than $K_m$.", + "text_images": [], + "full_mark": 5 + }, + { + "sub_id": "Exam-21-T6-2", + "tags": [ + "2.9", + "3.7" + ], + "text_open": "6.2.5 The kinetic parameters for the individual steps in the pathways above are shown below:\n| Pathway | Substrate → product | Enzyme | $CL_{int}$ / $(\\mu\\mathrm{L}\\cdot\\mathrm{min}^{-1}\\cdot\\mathrm{mg}^{-1})$ |\n| --- | --- | --- | --- |\n| 4′-hydroxylation pathway | DF → 4′-OH-DF | CYP2C9 | / |\n| 4′-hydroxylation pathway | 4′-OH-DF → A | CYP2C9 | 358 |\n| 5-hydroxylation pathway | DF → 5-OH-DF | CYP3A4 | / |\n| 5-hydroxylation pathway | 5-OH-DF → B | CYP2C9 | 103 |\n\nThe researchers also compared detoxification of the quinone products from the two hydroxylation pathways. The measured curves are shown below, where \"conjugate\" denotes a detoxification product.\nExplain why both steps of the 5-hydroxylation pathway have lower $CL_{int}$ values than those of the 4′-hydroxylation pathway, yet the 5-hydroxylation pathway is more toxic. (No more than 100 words)", + "text_images": [ + { + "file": "Exam-21-T6-2-1-prompt.png", + "description": "Coordinate plot containing two side-by-side panels labeled A and B at upper left. Both horizontal axes are labeled 'GSH concentration (mM)' with ticks at 0, 2, 4, and 6; both vertical axes are labeled 'Conjugate formation (%)' with ticks at 0, 50, and 100. Each panel contains one solid black curve with filled circular data markers, some with vertical error bars. In panel A, the curve rises rapidly from near the origin as GSH concentration increases, approaches 100% at about 2 mM, and remains on a plateau. In panel B, the curve rises more gradually from near the origin and approaches 100% at about 5–6 mM, without a clear plateau. There is no legend or title." + } + ], + "full_mark": 2 + }, + { + "sub_id": "Exam-21-T6-3", + "tags": [ + "1.7", + "2.9" + ], + "text_open": "6.3 Diclofenac has a half-life of about 1.0h in humans and is cleared mainly through CYP2C9-catalyzed 4′-hydroxylation of an aromatic ring. The structurally similar fenclorac is more resistant to aromatic hydroxylation and has a half-life exceeding 20h. Its structure is shown below.\nFrom a structural perspective, explain why fenclorac has a longer duration of action. (No more than 50 words)", + "text_images": [ + { + "file": "Exam-21-T6-3-1-prompt.png", + "description": "The SMILES representation of fenclorac is 'OC(CC1=C(OC2=CC=C(Cl)C=C2Cl)C=CC=C1)=O'." + } + ], + "full_mark": 2 + } + ] +} diff --git a/test/tasks/Exam-22-T1/problem.json b/test/tasks/Exam-22-T1/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..9580da63a33c86b1e99c4dbb3f37079a755f156b --- /dev/null +++ b/test/tasks/Exam-22-T1/problem.json @@ -0,0 +1,52 @@ +{ + "problem_id": "Exam-22-T1", + "header": "Problem 1 Chemistry of the Rare-Earth Elements\nRare-earth elements are indispensable strategic resources for modern high technology, new energy, advanced materials, defense, and related fields. Often called the “vitamins of industry” and a “treasure house of new materials,” they comprise scandium, yttrium, and all the lanthanides. For convenience, the symbol Ln is commonly used in coordination and inorganic chemistry to denote the lanthanides or the entire rare-earth series. This problem examines rare-earth chemistry in detail.\nNote: Different letters or alphanumeric combinations in this problem may represent the same element or chemical formula.", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-22-T1-1", + "tags": [ + "1.3" + ], + "text_open": "1.2 Consider a rare-earth element M1 as an example of rare-earth chemistry. M1 has two common oxidation states in its compounds. Elemental M1 is a silvery-white metal that slowly tarnishes in air. M1 dissolves in dilute hydrochloric acid, releasing colorless gas A and giving a pale blue-green solution (Reaction 1). Adding dilute sulfuric acid dropwise to this solution immediately produces white precipitate B, which is insoluble in dilute nitric acid. After B is filtered, washed, and dried, it is ignited to constant mass at 800 °C under Ar; the white precipitate becomes dark-brown powder C (Reaction 2), an oxide of M1. If the mass of B is 2.480 g, 1.760 g of C is obtained after ignition. A separate portion of B is heated with dilute hydrochloric acid and bromine water until it dissolves (Reaction 3), during which the solution changes from pale blue-green to colorless (or very pale pink). The solution is divided into two portions. Excess aqueous ammonia and oxalic acid are added separately to give white gelatinous precipitate D (with aqueous ammonia) and white precipitate E (with oxalic acid), respectively. Filtering and igniting either D or E at 900 °C likewise gives C. C is mixed with Y2O3 in a certain ratio and calcined at high temperature; product F emits bright red fluorescence under ultraviolet irradiation. 1.2.3 Write ionic equations for Reactions 1 and 3 and a chemical equation for Reaction 2.", + "text_images": [], + "full_mark": 6 + }, + { + "sub_id": "Exam-22-T1-2", + "tags": [ + "1.4" + ], + "text_open": "1.4 Coordination Compounds of Rare-Earth Elements\n1 equivalent of elemental rare-earth metal M5 and 2.8 equivalents of ethylene glycol monomethyl ether are added together to a toluene solution to form X. Addition of 2,6-dimethylphenol then precipitates orange crystals Y. Y is filtered and washed; upon slight heating it becomes a different crystal Z. Further characterization provides the following information:\n① The effective magnetic moment of X is 7.9 $\\mu_B$.\n② Crystal diffraction shows that Y belongs to the orthorhombic crystal system, with $a$ = 12.748 Å, $b$ = 12.900 Å, and $c$ = 29.220 Å. The unit cell contains 2 formula units, and the theoretical density of Y is 1.428 $\\mathrm{g/cm^3}$.\n③ Elemental analysis of Z gives the following experimental mass fractions: $\\omega(C)$ = 39.60%, $\\omega(O)$ = 18.86%, and $\\omega(H)$ = 5.74%. Crystal diffraction shows that Z belongs to the tetragonal crystal system, with $a$ = 14.837 Å and $c$ = 15.117 Å. The crystal contains 2 formula units, and the density of Z is 1.1854 times the theoretical density of Y.\n④ The anion of Z contains an $[M5_4O_4]$ cluster framework and exactly two types of ligands. One ligand adopts two different coordination modes. All M5 atoms are 7-coordinate, and a particular conformation has 1 twofold rotation axis.\n⑤ The oxidation state of M5 is the same in X, Y, and Z.\n1.4.2 Determine the chemical formulas of X, Y, and Z, showing your calculations. In your answer, the ethylene glycol monomethyl ether anion may be abbreviated as $mee^-$, and the 2,6-dimethylphenoxide anion as $dmp^-$.\n", + "text_images": [], + "full_mark": 9 + }, + { + "sub_id": "Exam-22-T1-3", + "tags": [ + "1.4" + ], + "text_open": "1.4.3 Write the SMILES strings for the cation and anion of Z. You may treat all $\\sigma$ coordinate bonds as ordinary single bonds and represent the structures using SMILES.", + "text_images": [], + "full_mark": 4 + }, + { + "sub_id": "Exam-22-T1-4", + "tags": [ + "2.1" + ], + "text_open": "1.5 Electrochemistry of the Rare-Earth Elements\nThe following standard electrode potentials are given for acidic medium at 25 °C:\n$O_2(g) + 4H^+ + 4e^- \\rightarrow 2H_2O$ $E° = +1.229\\ \\mathrm{V}$\n$M1^{n+} + e^- \\rightarrow M1^{(n-1)+}$ $E° = -0.35\\ \\mathrm{V}$\n1.5.1 At pH = 0 and 25 °C, calculate the common logarithm $\\lg K$ of the standard equilibrium constant for oxidation of $M1^{(n-1)+}$ to $M1^{n+}$ by $O_2$.", + "text_images": [], + "full_mark": 3 + }, + { + "sub_id": "Exam-22-T1-5", + "tags": [ + "2.1" + ], + "text_open": "1.5.2 The partial pressure of oxygen in air is $p(O_2)$ = 0.21 bar. Use a calculation to determine whether $M1^{(n-1)+}$ will be oxidized by oxygen in air when pH = 4.00 and $[M1^{(n-1)+}]$ = $[M1^{n+}]$ = 1.00 mol·L-1.", + "text_images": [], + "full_mark": 2 + } + ] +} diff --git a/test/tasks/Exam-22-T5/images/Exam-22-T5-10-1-prompt.png b/test/tasks/Exam-22-T5/images/Exam-22-T5-10-1-prompt.png new file mode 100644 index 0000000000000000000000000000000000000000..494cb877903d57d534a2c62a78a6cbc7d8a56631 --- /dev/null +++ b/test/tasks/Exam-22-T5/images/Exam-22-T5-10-1-prompt.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:33c94df44d296f58900fa7567cd30332f506a33ae89282a7b419fce502e73f14 +size 127661 diff --git a/test/tasks/Exam-22-T5/problem.json b/test/tasks/Exam-22-T5/problem.json new file mode 100644 index 0000000000000000000000000000000000000000..801bfdf066f6f086f19b59d245a314890601a64c --- /dev/null +++ b/test/tasks/Exam-22-T5/problem.json @@ -0,0 +1,102 @@ +{ + "problem_id": "Exam-22-T5", + "header": "Unlike most covalent compounds, whose structures can be described using 2c-2e bonds, polyhedral clusters are often electron-deficient systems and require multicenter bonding models.\nWade's rules, more generally known as the polyhedral skeletal electron pair theory (PSEPT), are an important tool for understanding cluster structures. Let $n$ denote the number of atoms or isolobal fragments that form the cluster framework. Subtracting the electrons used in terminal two-center two-electron bonds, nonskeletal lone pairs, and the internal bonding of ligands from the total valence-electron count gives the number of skeletal electrons. Skeletal electron-pair counts of $n+1$, $n+2$, $n+3$, and $n+4$ correspond, respectively, to closo, nido, arachno, and hypho structures. A closo cluster is a complete triangulated polyhedron with $n$ vertices, whereas nido, arachno, and hypho clusters may be regarded as the corresponding closo parent polyhedron with one, two, or three vertices removed, respectively. For example, in $B_5H_9$, 5 BH fragments contribute 10 electrons and 4 additional hydrogen atoms contribute another 4, giving 14 skeletal electrons or 7 skeletal electron pairs. Since the skeletal electron-pair count is $n+2$, the cluster is nido.\nThis problem uses low-valent clusters of Group 13 metals (Al, Ga, In, and Tl) to examine cluster synthesis, structure, bonding, and reactivity.", + "header_images": [], + "subquestions": [ + { + "sub_id": "Exam-22-T5-1", + "tags": [ + "1.6" + ], + "text_open": "5.1 $B_4Cl_4$ and $B_4(t\\text{-}Bu)_4$ have tetrahedral cores, whereas the acetamido-substituted boron compound $B_4(NHAc)_4$ adopts a butterfly-shaped cyclic structure. Explain why.", + "text_images": [], + "full_mark": 2 + }, + { + "sub_id": "Exam-22-T5-2", + "tags": [ + "2.1" + ], + "text_open": "5.2 Write equations for the following reactions.\n5.2.1 $AlCl\\cdot x\\,Et_2O$ ($x>1$) reacts with $MgCp^*_2$ at low temperature in toluene/diethyl ether to form $Al_4Cp^*_4$.", + "text_images": [], + "full_mark": 2 + }, + { + "sub_id": "Exam-22-T5-3", + "tags": [ + "2.1" + ], + "text_open": "5.2.2 $GaI$ reacts with $Li(THF)_3Ge(SiMe_3)_3$ in toluene to form $[GaGe(SiMe_3)_3]_4$.", + "text_images": [], + "full_mark": 2 + }, + { + "sub_id": "Exam-22-T5-4", + "tags": [ + "2.4" + ], + "text_open": "5.4 Tetrahedral clusters not only have unusual structures but also show diverse oxidation chemistry, giving different products under different conditions. Draw the structures of products A, B, D, and E below. 5.4.1 Oxidation of [GaSi(t-Bu)_3]_4 with oxygen gives cubic cluster A = [OGaSi(t-Bu)_3]_4.", + "text_images": [], + "full_mark": 3 + }, + { + "sub_id": "Exam-22-T5-5", + "tags": [ + "2.4" + ], + "text_open": "5.4.2 Further hydrolysis of A gives product B, which contains only one ring and has only one chemical environment for its Ga atoms. Write the SMILES string for B. If necessary, describe its detailed structure in no more than 50 words. Multiple reasonable structures may be possible; writing any one reasonable SMILES string is sufficient.", + "text_images": [], + "full_mark": 2 + }, + { + "sub_id": "Exam-22-T5-6", + "tags": [ + "2.4" + ], + "text_open": "5.4.3 Oxidation of $C = [InC(SiMe_3)_3]_4$ with o-nitrosotoluene gives a cubic cluster and product D, which contains two benzene rings. Write the SMILES string for aromatic product D and, if necessary, describe its detailed structure in no more than 50 words.", + "text_images": [], + "full_mark": 2 + }, + { + "sub_id": "Exam-22-T5-7", + "tags": [ + "2.4" + ], + "text_open": "5.4.4 Oxidation of C with 1,2-thiirane gives a trigonal-bipyramidal cluster E. Write the SMILES string for E and, if necessary, describe its detailed structure in no more than 50 words.", + "text_images": [], + "full_mark": 2 + }, + { + "sub_id": "Exam-22-T5-8", + "tags": [ + "2.2" + ], + "text_open": "5.6 Reaction of $GaI$ with $Li(THF)_3R$ ($R = SiMe(SiMe_3)_2$) in toluene gives the octahedral cluster $G = [GaR]_6$. In contrast, reaction of $GaI$ with $Na(THF)_2R'$ ($R' = Si(t\\text{-}Bu)_3$) in toluene forms the octahedral cluster $H = [Ga_6R'_4Bn_2]^{2-}$, where Bn denotes benzyl.\n5.6.1 State the source of the benzyl groups in H.", + "text_images": [], + "full_mark": 1 + }, + { + "sub_id": "Exam-22-T5-9", + "tags": [ + "1.7" + ], + "text_open": "5.6.3 Explain why G is a highly distorted octahedron whereas H is nearly a regular octahedron.", + "text_images": [], + "full_mark": 3 + }, + { + "sub_id": "Exam-22-T5-10", + "tags": [ + "1.7" + ], + "text_open": "5.8 Electron-rich cluster compounds: When the number of ligands in a cluster is much smaller than the number of metal atoms, its structure approaches that of a bare metal-atom cluster; such species are called electron-rich cluster compounds. A representative example is the metalloid cluster $K = Ga_{84}[N(SiMe_3)_2]_{20}^{4-}$. This cluster contains 84 Ga atoms, 64 of which are “bare.” Its structure has three layers: a central $Ga_2$ dumbbell with a bond length of 235 pm, approaching a $Ga\\equiv Ga$ triple bond; a middle $Ga_{32}$ football-shaped tube; and an outer assembly of three $Ga_{10}$ rings surrounded by 20 $N(SiMe_3)_2$ ligands.\nCompare K with the fullerene $Sc_2@C_{84}$, and identify which structural units in K correspond to the encapsulated $Sc_2$ and to the $C_{84}$ carbon cage, respectively.", + "text_images": [ + { + "file": "Exam-22-T5-10-1-prompt.png", + "description": "Unlabeled three-dimensional ball-and-stick molecular model. The overall structure is a vertically symmetric multilayer assembly. The upper and lower layers each contain a polyhedron made of dark-blue spheres, approximately 12 in total. A larger central polyhedron is formed by approximately 12 purple spheres, and the outermost framework contains approximately 24 orange-yellow spheres. Spheres of the same color are connected by solid lines of the same color, with radial connections between layers. The drawing represents the layered structure of the metalloid cluster $K = Ga_{84}[N(SiMe_3)_2]_{20}^{4-}$ described in the question." + } + ], + "full_mark": 2 + } + ] +}