Cambridge A Level Chemistry 9701 — 2025 Oct/Nov Paper 4 · Variant 2

[Turn over Cambridge International AS & A Level DC (WW/SG) 342070/3 © UCLES 2025 This document has 24 pages. Any blank pages are indicated. CHEMISTRY 9701/42 Paper 4 A Level Structured Questions October/November 2025 2 hours You must answer on the question paper. No additional materials are needed. INSTRUCTIONS ● Answer all questions. ● Use a black or dark blue pen. You may use an HB pencil for any diagrams or graphs. ● Write your name, centre number and candidate number in the boxes at the top of the page. ● Write your answer to each question in the space provided. ● Do not use an erasable pen or correction fluid. ● Do not write on any bar codes. ● You may use a calculator. ● You should show all your working and use appropriate units. INFORMATION ● The total mark for this paper is 100. ● The number of marks for each question or part question is shown in brackets [ ]. ● The Periodic Table is printed in the question paper. ● Important values, constants and standards are printed in the question paper. * 0 1 4 9 5 2 0 5 6 0 * ,










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3 9701/42/O/N/25 © UCLES 2025 [Turn over 1 Magnesium nitrate, Mg(NO3)2, and strontium nitrate, Sr(NO3)2, both decompose when heated to form the metal oxide and a mixture of gases. (a) Write an equation for the thermal decomposition of Mg(NO3)2. … [1] (b) State which of Mg(NO3)2 or Sr(NO3)2 decomposes at a lower temperature. Explain your answer. compound that decomposes at a lower temperature … explanation … … … … … [2] (c) Magnesium oxide, MgO, and strontium oxide, SrO, both react with dilute sulfuric acid. MgO forms a soluble salt, A. SrO forms an insoluble salt, B. (i) Identify the products formed when MgO reacts with dilute sulfuric acid. … [1] (ii) Explain why A is more soluble than B. … … … … … … [3] [Total: 7] * 0000800000003 * ,










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4 9701/42/O/N/25 © UCLES 2025 2 Ethanal, CH3CHO, reacts with nitrogen dioxide, NO2. The products of the first step of this reaction are a CH3C• =O radical and a molecule of nitrous acid, HNO2. CH3CHO + NO2 CH3C• =O + HNO2 (a) (i) Use two words to complete the sentence. This reaction involves … … of the single covalent bond between a hydrogen atom and a carbon atom in CH3CHO. [1] (ii) The hydrogen atom mentioned in (a)(i) forms a covalent bond with one of the oxygen atoms of an NO2 molecule. An NO2 molecule has a single, unpaired electron on the nitrogen atom. All electrons are paired in an HNO2 molecule. Draw dot-and-cross diagrams of NO2 and HNO2 in the boxes. Show outer shell electrons only. NO2 HNO2 [1] (iii) Use VSEPR theory to predict the bond angle at the nitrogen atom in an HNO2 molecule. bond angle = … [1] (b) The rate equation for the reaction between CH3CHO and NO2 is shown. rate = k [CH3CHO][NO2] Under certain conditions, when the concentrations of both CH3CHO and NO2 are 0.200 mol dm–3, the rate of the reaction is 1.53 × 10–4 mol dm–3 s–1. Calculate the value of the rate constant, k, under these conditions. Give the units of k. k = … units … [2] * 0000800000004 * ,










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5 9701/42/O/N/25 © UCLES 2025 [Turn over (c) The reaction mixture described in (b) is monitored over a period of time. Predict whether the graph of [NO2] against time shows a constant half-life. Explain your answer. prediction … explanation … … [1] (d) NO2 also reacts with ozone, O3. The rate equation is shown. rate = k1[NO2] Under certain conditions, the value of k1 is 0.0848 s–1. The reaction has a constant half-life under these conditions. Calculate the half-life in seconds. half-life = … s [1] (e) NO2 is present in the exhaust gases of cars. It can react with carbon monoxide, CO, on the surface of a heterogeneous catalyst in the car’s catalytic converter. Describe the mode of action of this heterogeneous catalyst. … … … … … … [2] [Total: 9] * 0000800000005 * ,










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6 9701/42/O/N/25 © UCLES 2025 3 (a) (i) Define conjugate acid–base pair. … … [1] (ii) Give the formulas of the conjugate acid and the conjugate base of the hydrogen phosphate ion, HPO4 2–. conjugate acid of HPO4 2– … conjugate base of HPO4 2– … [1] (b) The Ka of propanoic acid, CH3CH2COOH, is 1.35 × 10–5 mol dm–3 at 298 K. Solution C is a solution of CH3CH2COOH with a pH of 3.60 at 298 K. (i) Calculate the concentration of CH3CH2COOH in solution C. [CH3CH2COOH] = … mol dm–3 [2] (ii) Calculate the concentration of hydroxide ions in solution C. [OH–] = … mol dm–3 [1] (iii) Calculate the concentration of a solution of hydrochloric acid with the same pH as solution C. concentration = … mol dm–3 [1] * 0000800000006 * ,










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7 9701/42/O/N/25 © UCLES 2025 [Turn over (iv) Table 3.1 shows three possible values of the Ka of dimethylpropanoic acid, (CH3)3CCOOH. Place a tick in Table 3.1 to show the correct value. Explain your answer. Table 3.1 value of Ka / mol dm–3 place one tick (✓) in this column 9.33 × 10–6 1.35 × 10–5 3.35 × 10–5 explanation … … … … … [3] (c) Solution D is made by mixing 100 cm3 of 0.100 mol dm–3 CH3CH2COOH and 100 cm3 of 0.100 mol dm–3 NaCl. The pH of solution D is measured as small amounts of H2SO4(aq) are added to it, and when small amounts of NaOH(aq) are added to it. Solution D only acts as a buffer solution when one of these solutions is added to it. (i) Complete the sentence and write an equation for the reaction that occurs. Solution D acts as a buffer when … is added to it. equation … [1] (ii) Complete the sentence and explain why solution D does not act as a buffer when the other solution is added. Solution D does not act as a buffer when … is added to it. explanation … … [1] * 0000800000007 * ,










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8 9701/42/O/N/25 © UCLES 2025 (d) Manganese(II) hydroxide, Mn(OH)2, is only slightly soluble in water. The solubility of Mn(OH)2 in water is 3.28 × 10–3 g dm–3 at 298 K. (i) Calculate the concentration of a saturated solution of Mn(OH)2 at 298 K. [Mn(OH)2] = … mol dm–3 [1] (ii) Write an expression for the Ksp of Mn(OH)2. Give the units of Ksp. Ksp = units = … [2] (iii) Use your answers to (d)(i) and (d)(ii) to calculate the value of Ksp of Mn(OH)2 at 298 K. Ksp = … [1] [Total: 15] * 0000800000008 * ,










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9 9701/42/O/N/25 © UCLES 2025 [Turn over 4 (a) Define enthalpy change of atomisation, ΔHat. … … [1] (b) Define first electron affinity, EA. … … [1] (c) Explain why the first electron affinity of chlorine is more exothermic than the first electron affinity of iodine. … … … … [2] (d) The enthalpy change for the reaction Cl 2(g) + 2e– 2Cl –(g) is – 486 kJ mol–1. The first electron affinity of chlorine is –364 kJ mol–1. Calculate the enthalpy change of atomisation of chlorine. ΔHat of chlorine = … kJ mol–1 [2] [Total: 6] * 0000800000009 * ,










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10 9701/42/O/N/25 © UCLES 2025 5 Cobalt is a transition element which forms compounds containing Co2+ and Co3+ ions. Cobalt(II) sulfate dissolves in water to form a solution containing the [Co(H2O)6]2+ complex ion. (a) (i) Complete the electronic configurations of a Co2+ ion and a Co3+ ion. Co2+ = [Ar] … Co3+ = [Ar] … [1] (ii) Explain why transition elements can form complex ions. … … [1] (iii) An excess of concentrated HCl is added to a solution containing [Co(H2O)6]2+. Describe the colour change observed and the state of the cobalt-containing product. The colour changes from … to … . The state of the cobalt-containing product is … . [2] (iv) Write an equation for the reaction occurring in (a)(iii). … [1] (v) Name the type of reaction occurring in (a)(iii). … [1] (vi) Write an equation for the reaction that occurs when an excess of NaOH(aq) is added to a solution containing [Co(H2O)6]2+. … [1] * 0000800000010 * ,










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11 9701/42/O/N/25 © UCLES 2025 [Turn over (b) Cobalt metal can be oxidised by acidified K2Cr2O7. The relevant half-equations, and their E o– values, are shown. Co2+ + 2e– Co E o– = –0.28 V Cr2O7 2– + 14H+ + 6e– 2Cr3+ + 7H2O E o– = +1.33 V (i) A Co2+/Co electrode is constructed in which [Co2+] is 0.020 mol dm–3 at 298 K. Use the Nernst equation to show that the E value for this Co2+ / Co electrode is –0.33 V. [2] (ii) An electrochemical cell is constructed using the Co2+ / Co electrode described in (b)(i) and a Cr2O7 2– / Cr3+ electrode in which all conditions are standard. Calculate the value of Ecell. Ecell = … [1] (iii) A current is drawn from the electrochemical cell described in (b)(ii). Write an equation for the reaction taking place in the cell. … [1] (iv) Complete the sentences to identify the negative electrode and the direction of electron flow when a current is drawn from the cell described in (b)(ii). The … electrode is the negative electrode. Electrons flow from the … electrode to the … electrode. [1] (c) A molten Co2+ salt is electrolysed using a current of 0.500 A. 0.547 g of cobalt metal forms at the cathode. Under the conditions used no other reduction reaction occurs at the cathode. Calculate the time in minutes for which the current flows to produce this mass of cobalt. Give your answer to three significant figures. time = … min [3] [Total: 15] * 0000800000011 * ,









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12 9701/42/O/N/25 © UCLES 2025 6 (a) Nickel forms complexes. (i) Give the formula and charge of the tetrahedral complex formed by Ni atoms with carbon monoxide molecules. Carbon monoxide is a monodentate ligand. This is complex E. E = … [1] (ii) Give the formula and charge of the octahedral complex formed by Ni2+ ions with ethanedioate ions. This is complex F. F = … [2] (iii) Identify which complex, E or F, exists as a mixture of two stereoisomers and the type of stereoisomerism involved. The complex which exists as a mixture of two stereoisomers is … . The type of stereoisomerism involved is … . [1] (b) Cadmium forms complexes with methylamine, CH3NH2, and 1,2-diaminoethane, en. The values of the stability constants, Kstab, of these complex ions are given in Table 6.1. Table 6.1 complex Kstab [Cd(CH3NH2)4]2+ 3.5 × 106 [Cd(en)2]2+ 4.0 × 1010 (i) Explain, by reference to its structure, why CH3NH2 acts as a monodentate ligand. … [1] (ii) Some Cd2+(aq) is added to a solution containing equal concentrations of CH3NH2 and en. Predict which of the two complexes in Table 6.1 forms at the higher concentration. Explain your answer. complex that forms at the higher concentration … explanation … … [1] (iii) Complete the expression for the Kstab of [Cd(CH3NH2)4]2+. Kstab = [1] [Total: 7] * 0000800000012 * ,









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13 9701/42/O/N/25 © UCLES 2025 [Turn over 7 P, Q, R, S, T, U, and V are the seven structural isomers with molecular formula C5H10O that have a carbonyl group. P CH3(CH2)3CHO Q CH3CH2CH(CH3)CHO R (CH3)2CHCH2CHO S (CH3)3CCHO T CH3CH2CH2COCH3 U CH3CH2COCH2CH3 V (CH3)2CHCOCH3 (a) Only one of these seven compounds has stereoisomers. Draw three-dimensional diagrams of the two stereoisomers of this compound. [2] (b) P, Q, R, S, T, U, and V are treated separately with alkaline I2(aq) and the product mixture is acidified. (i) Identify the two compounds that give a positive result with alkaline I2(aq). … and … [1] (ii) Describe the observations when one of the compounds you have identified in (b)(i) is treated with alkaline I2(aq) and give the structural formulae of the two carbon-containing products of this reaction. observations … two carbon-containing products … and … [2] * 0000800000013 * ,









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14 9701/42/O/N/25 © UCLES 2025 P CH3(CH2)3CHO Q CH3CH2CH(CH3)CHO R (CH3)2CHCH2CHO S (CH3)3CCHO T CH3CH2CH2COCH3 U CH3CH2COCH2CH3 V (CH3)2CHCOCH3 (c) The proton (1H) NMR spectra of P, Q, R, S, T, U, and V are compared. (i) Identify the only compound that gives a spectrum with two singlets and no other peaks. … [1] Fig. 7.1 shows the spectrum obtained from one of the compounds. 11 10 9 8 7 6 5 4 3 2 1 0 δ / ppm Fig. 7.1 (ii) Identify the compound that gives this spectrum. … [1] (iii) Name the splitting pattern of the peak at δ = 1.1 in Fig. 7.1. Give the reason for this splitting. name … reason … [1] (iv) Identify the substance that gives the small peak at δ = 0 in Fig. 7.1. … [1] * 0000800000014 * ,









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15 9701/42/O/N/25 © UCLES 2025 [Turn over (d) The carbon-13 NMR spectra of R, S, T and U are compared. Complete Table 7.1 to state the number of peaks in the spectrum of each compound. Table 7.1 compound number of peaks R (CH3)2CHCH2CHO S (CH3)3CCHO T CH3CH2CH2COCH3 U CH3CH2COCH2CH3 [2] [Total: 11] * 0000800000015 * ,









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16 9701/42/O/N/25 © UCLES 2025 8 Asparagine and aspartic acid are two naturally occurring amino acids. Their structures and isoelectric points are shown in Table 8.1. Table 8.1 amino acid structure isoelectric point asparagine HOOCCH(NH2)CH2CONH2 5.41 aspartic acid HOOCCH(NH2)CH2COOH 2.77 (a) Define isoelectric point. … … [1] (b) Draw the structures of asparagine and aspartic acid at pH 2. asparagine at pH 2 aspartic acid at pH 2 [2] (c) Asparagine and aspartic acid are treated separately with an excess of LiAl H4. Draw the structures of the organic products of these reactions. product of asparagine treated with an excess of LiAl H4 product of aspartic acid treated with an excess of LiAl H4 [2] * 0000800000016 * ,









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17 9701/42/O/N/25 © UCLES 2025 [Turn over (d) Propanedioic acid, HOOCCH2COOH, is treated with an excess of thionyl chloride, SOCl 2. Propanedioyl chloride, Cl OCCH2COCl , is formed. (i) Write an equation for this reaction. … [1] (ii) Propanedioyl chloride reacts with an excess of asparagine to form compound G with molecular formula C11H16N4O8. Each molecule of compound G has four amide groups. Draw the structure of compound G. Compound G, C11H16N4O8 [2] (e) Asparagine is hydrolysed with an excess of hot NaOH(aq). Draw the structure of the organic product of this reaction. [2] (f) A polymer can form from asparagine, HOOCCH(NH2)CH2CONH2, as the only monomer. Draw a length of the polymer chain containing three monomer residues. Clearly label the repeat unit of the polymer on your diagram. [3] * 0000800000017 * ,









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18 9701/42/O/N/25 © UCLES 2025 (g) Aspartic acid exists in two optically active forms. (i) Plane polarised light is passed through pure samples of these two optically active forms in solutions of the same concentration. Describe two similarities and one difference in their effect on the plane polarised light. similarities … … difference … … [2] (ii) Give the term used to describe a mixture of equal amounts of the two optically active forms. … [1] [Total: 16] * 0000800000018 * ,









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19 9701/42/O/N/25 © UCLES 2025 [Turn over Question 9 starts on page 20. * 0000800000019 * ,









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20 9701/42/O/N/25 © UCLES 2025 9 Compound X is made from benzene by the route shown in Fig. 9.1. step 1 CH3 step 2 CH3 NO2 NO2 W NH2 X step 3 COOH step 4 COOH benzene Fig. 9.1 (a) Describe the bonding in benzene, C6H6. Your answer should include: • the hybridisation of the six carbon atoms • the types of bond between the carbon atoms • the orbitals that overlap to produce the bonds between the carbon atoms • the type of bond between the carbon atoms and the hydrogen atoms • the orbitals that overlap to produce the bonds between the carbon atoms and the hydrogen atoms. … … … … … … … [3] (b) Describe the reagents and conditions required for step 1 in Fig. 9.1. … … [1] * 0000800000020 * ,










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21 9701/42/O/N/25 © UCLES 2025 [Turn over (c) In step 1 of Fig. 9.1 benzene reacts with +CH3. Complete Fig. 9.2 to show the mechanism for this reaction, including: • the movement of electron pairs using curly arrows • the structure of the intermediate involved. +CH3 CH3 intermediate + … Fig. 9.2 [3] (d) Describe the reagents and conditions required for step 2 of Fig. 9.1. … … [2] (e) Identify the reagents required for step 3 of Fig. 9.1. Compound W is the product of this step. … [1] (f) Name compound W. … [1] (g) The reagents commonly used for step 4 will not reduce the –COOH group. Identify the reagents and conditions required for step 4 of Fig. 9.1. … [1] * 0000800000021 * ,









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22 9701/42/O/N/25 © UCLES 2025 (h) Benzene can also be used as a starting material to make compound Y. COOH compound Y NH2 Describe how the route described in Fig. 9.1 (repeated below) can be changed to give compound Y instead of compound X. Explain your answer. … … … … [2] step 1 CH3 step 2 CH3 NO2 NO2 W NH2 X step 3 COOH step 4 COOH benzene Fig. 9.1 [Total: 14] * 0000800000022 * ,









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23 9701/42/O/N/25 © UCLES 2025 Important values, constants and standards molar gas constant R = 8.31 J K–1 mol–1 Faraday constant F = 9.65 × 104 C mol–1 Avogadro constant L = 6.02 × 1023 mol–1 electronic charge e = –1.60 × 10–19 C molar volume of gas Vm = 22.4 dm3 mol–1 at s.t.p. (101 kPa and 273 K) Vm = 24.0 dm3 mol–1 at room conditions ionic product of water Kw = 1.00 × 10–14 mol2 dm–6 (at 298 K (25 °C)) specific heat capacity of water c = 4.18 kJ kg–1 K–1 (4.18 J g–1 K–1) * 0000800000023 * ,









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24 9701/42/O/N/25 © UCLES 2025 To avoid the issue of disclosure of answer-related information to candidates, all copyright acknowledgements are reproduced online in the Cambridge Assessment International Education Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download at www.cambridgeinternational.org after the live examination series. Group The Periodic Table of Elements 1 H hydrogen 1.0 2 He helium 4.0 1 2 13 14 15 16 17 18 3 4 5 6 7 8 9 10 11 12 3 Li lithium 6.9 4 Be beryllium 9.0 atomic number atomic symbol Key name relative atomic mass 11 Na sodium 23.0 12 Mg magnesium 24.3 19 K potassium 39.1 20 Ca calcium 40.1 37 Rb rubidium 85.5 38 Sr strontium 87.6 55 Cs caesium 132.9 56 Ba barium 137.3 87 Fr francium – 88 Ra radium – 5 B boron 10.8 13 Al aluminium 27.0 31 Ga gallium 69.7 49 In indium 114.8 81 Tl thallium 204.4 6 C carbon 12.0 14 Si silicon 28.1 32 Ge germanium 72.6 50 Sn tin 118.7 82 Pb lead 207.2 22 Ti titanium 47.9 40 Zr zirconium 91.2 72 Hf hafnium 178.5 104 Rf rutherfordium – 23 V vanadium 50.9 41 Nb niobium 92.9 73 Ta tantalum 180.9 105 Db dubnium – 24 Cr chromium 52.0 42 Mo molybdenum 95.9 74 W tungsten 183.8 106 Sg seaborgium – 25 Mn manganese 54.9 43 Tc technetium – 75 Re rhenium 186.2 107 Bh bohrium – 26 Fe iron 55.8 44 Ru ruthenium 101.1 76 Os osmium 190.2 108 Hs hassium – 27 Co cobalt 58.9 45 Rh rhodium 102.9 77 Ir iridium 192.2 109 Mt meitnerium – 28 Ni nickel 58.7 46 Pd palladium 106.4 78 Pt platinum 195.1 110 Ds darmstadtium – 29 Cu copper 63.5 47 Ag silver 107.9 79 Au gold 197.0 111 Rg roentgenium – 30 Zn zinc 65.4 48 Cd cadmium 112.4 80 Hg mercury 200.6 112 Cn copernicium – 114 Fl flerovium – 116 Lv livermorium – 7 N nitrogen 14.0 15 P phosphorus 31.0 33 As arsenic 74.9 51 Sb antimony 121.8 83 Bi bismuth 209.0 8 O oxygen 16.0 16 S sulfur 32.1 34 Se selenium 79.0 52 Te tellurium 127.6 84 Po polonium – 9 F fluorine 19.0 17 Cl chlorine 35.5 35 Br bromine 79.9 53 I iodine 126.9 85 At astatine – 10 Ne neon 20.2 18 Ar argon 39.9 36 Kr krypton 83.8 54 Xe xenon 131.3 86 Rn radon – 113 Nh nihonium – 115 Mc moscovium – 117 Ts tennessine – 118 Og oganesson – 21 Sc scandium 45.0 39 Y yttrium 88.9 57–71 lanthanoids 89–103 actinoids 57 La lanthanum 138.9 89 Ac lanthanoids actinoids actinium – 58 Ce cerium 140.1 90 Th thorium 232.0 59 Pr praseodymium 140.9 91 Pa protactinium 231.0 60 Nd neodymium 144.2 92 U uranium 238.0 61 Pm promethium – 93 Np neptunium – 62 Sm samarium 150.4 94 Pu plutonium – 63 Eu europium 152.0 95 Am americium – 64 Gd gadolinium 157.3 96 Cm curium – 65 Tb terbium 158.9 97 Bk berkelium – 66 Dy dysprosium 162.5 98 Cf californium – 67 Ho holmium 164.9 99 Es einsteinium – 68 Er erbium 167.3 100 Fm fermium – 69 Tm thulium 168.9 101 Md mendelevium – 70 Yb ytterbium 173.1 102 No nobelium – 71 Lu lutetium 175.0 103 Lr lawrencium – * 0000800000024 * ,









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This document consists of 15 printed pages. © Cambridge University Press & Assessment 2025 [Turn over Cambridge International AS & A Level CHEMISTRY 9701/42 Paper 4 A Level Structured Questions October/November 2025 MARK SCHEME Maximum Mark: 100 Published This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners’ meeting before marking began, which would have considered the acceptability of alternative answers. Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for Teachers. Cambridge International will not enter into discussions about these mark schemes. Cambridge International is publishing the mark schemes for the October/November 2025 series for most Cambridge IGCSE, Cambridge International A and AS Level components, and some Cambridge O Level components.

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 2 of 15 Generic Marking Principles These general marking principles must be applied by all examiners when marking candidate answers. They should be applied alon gside the specific content of the mark scheme or generic level descriptions for a question. Each question paper and mark scheme will also comply with these marking principles. GENERIC MARKING PRINCIPLE 1: Marks must be awarded in line with: • the specific content of the mark scheme or the generic level descriptors for the question • the specific skills defined in the mark scheme or in the generic level descriptors for the question • the standard of response required by a candidate as exemplified by the standardisation scripts. GENERIC MARKING PRINCIPLE 2: Marks awarded are always whole marks (not half marks, or other fractions). GENERIC MARKING PRINCIPLE 3: Marks must be awarded positively: • marks are awarded for correct/valid answers, as defined in the mark scheme. However, credit is given for valid answers which go beyond the scope of the syllabus and mark scheme, referring to your Team Leader as appropriate • marks are awarded when candidates clearly demonstrate what they know and can do • marks are not deducted for errors • marks are not deducted for omissions • answers should only be judged on the quality of spelling, punctuation and grammar when these features are specifically assessed by the question as indicated by the mark scheme. The meaning, however, should be unambiguous. GENERIC MARKING PRINCIPLE 4: Rules must be applied consistently, e.g. in situations where candidates have not followed instructions or in the application of generic level descriptors.

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 3 of 15 GENERIC MARKING PRINCIPLE 5: Marks should be awarded using the full range of marks defined in the mark scheme for the question (however; the use of the full mark range may be limited according to the quality of the candidate responses seen). GENERIC MARKING PRINCIPLE 6: Marks awarded are based solely on the requirements as defined in the mark scheme. Marks should not be awarded with grade thre sholds or grade descriptors in mind. Science-Specific Marking Principles 1 Examiners should consider the context and scientific use of any keywords when awarding marks. Although keywords may be present, marks should not be awarded if the keywords are used incorrectly. 2 The examiner should not choose between contradictory statements given in the same question part, and credit should not be awarded for any correct statement that is contradicted within the same question part. Wrong science that is irrelevant to the question should be ignored. 3 Although spellings do not have to be correct, spellings of syllabus terms must allow for clear and unambiguous separation fro m other syllabus terms with which they may be confused (e.g. ethane / ethene, glucagon / glycogen, refraction / reflection). 4 The error carried forward (ecf) principle should be applied, where appropriate. If an incorrect answer is subsequently used in a scientifically correct way, the candidate should be awarded these subsequent marking points. Further guidance will be included in the mark scheme where necessary and any exceptions to this general principle will be noted. 5 ‘List rule’ guidance For questions that require n responses (e.g. State two reasons …): • The response should be read as continuous prose, even when numbered answer spaces are provided. • Any response marked ignore in the mark scheme should not count towards n. • Incorrect responses should not be awarded credit but will still count towards n. • Read the entire response to check for any responses that contradict those that would otherwise be credited. Credit should not be awarded for any responses that are contradicted within the rest of the response. Where two responses contradict one another, this should be treated as a single incorrect response. • Non-contradictory responses after the first n responses may be ignored even if they include incorrect science.

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 4 of 15 6 Calculation specific guidance Correct answers to calculations should be given full credit even if there is no working or incorrect working, unless the question states ‘show your working’. For questions in which the number of significant figures required is not stated, credit should be awarded for correct answers when rounded by the examiner to the number of significant figures given in the mark scheme. This may not apply to measured values. For answers given in standard form (e.g. a  10n) in which the convention of restricting the value of the coefficient (a) to a value between 1 and 10 is not followed, credit may still be awarded if the answer can be converted to the answer given in the mark scheme. Unless a separate mark is given for a unit, a missing or incorrect unit will normally mean that the final calculation mark is not awarded. Exceptions to this general principle will be noted in the mark scheme. 7 Guidance for chemical equations Multiples / fractions of coefficients used in chemical equations are acceptable unless stated otherwise in the mark scheme. State symbols given in an equation should be ignored unless asked for in the question or stated otherwise in the mark scheme.

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 5 of 15 Annotations guidance for centres Examiners use a system of annotations as a shorthand for communicating their marking decisions to one another. Examiners are trained during the standardisation process on how and when to use annotations. The purpose of annotations is to inform the standard isation and monitoring processes and guide the supervising examiners when they are checking the work of examiners within their team. The meaning of annotations and how they are used is specific to each component and is understood by all examiners who mark the component. We publish annotations in our mark schemes to help centres understand the annotations they may see on copies of scripts. Note that there may not be a direct correlation between the number of annotations on a script and the mark awarded. Similarly, the use of an annotation may not be an indication of the quality of the response. The annotations listed below were available to examiners marking this component in this series. Annotations Annotation Meaning Correct point or mark awarded Incorrect point or mark not awarded Unclear Information missing or insufficient for credit Benefit of the doubt given Contradiction in response otherwise markworthy, mark not given Part of the correct answer has been seen. Full credit has not been awarded. Error carried forward applied Incorrect or insufficient point ignored while marking the rest of the response Rounding error

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 6 of 15 Annotation Meaning Repetition Blank page or part of script seen Error in number of significant figures Transcription error

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 7 of 15 Question Answer Marks 1(a) Mg(NO3)2 → MgO + 2NO2 + ½O2 OR 2Mg(NO3)2 → 2MgO + 4NO2 + O2 1 1(b) M1: Mg(NO3)2 / magnesium nitrate AND magnesium ion / Mg2+ AND is smaller / has higher charge density M2: nitrate ion / anion is more distorted / polarised 2 1(c)(i) magnesium sulfate and water / MgSO4 and H2O 1 1(c)(ii) M1: magnesium sulfate / A has a more exothermic Hlatt and Hhyd M2: difference in Hhyd is greater M3: magnesium sulfate / A has a more exothermic Hsol 3 Question Answer Marks 2(a)(i) homolytic fission 1 2(a)(ii) 1 2(a)(iii) 115–120o 1 2(b) M1: 3.825  10–3 / 0.003825 / 0.00383 / 0.0038 / 3.8  10–3 M2: units = mol–1 dm3 s–1 2 2(c) not constant half-life because overall second order 1

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 8 of 15 Question Answer Marks 2(d) 8.17 / 8.2 1 2(e) • reactants are adsorbed onto the catalyst surface • bonds within reactant molecules are weakened • products are desorbed from catalyst surface Any two [1], all three [2] 2 Question Answer Marks 3(a)(i) two species that differ by one proton / H+ 1 3(a)(ii) H2PO4– AND PO43– 1 3(b)(i) M1: [H+] = 2.51  10–4 M2: [CH3CH2COOH] = 4.67  10–3 2 3(b)(ii) 3.98  10–11 1 3(b)(iii) [HCl ] = 2.51  10–4 1 3(b)(iv) M1: top box ticked AND weaker acid M2: alkyl group is electron donating M3: O–H bond strengthened / anion is destabilised 3 3(c)(i) NaOH AND CH3CH2COOH + OH– → CH3CH2COO– + H2O 1 3(c)(ii) H2SO4 AND conjugate base of CH3CH2COOH is not present 1 3(d)(i) 3.69  10–5 1

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 9 of 15 Question Answer Marks 3(d)(ii) M1: Ksp = [Mn2+][OH–]2 M2: units = mol3 dm–9 2 3(d)(iii) 2.0  10–13 1 Question Answer Marks 4(a) energy required when one mole of gaseous atoms is formed from the element 1 4(b) energy released when one mole of gaseous atoms gains one mole of electrons and becomes one mole of gaseous 1– ions 1 4(c) M1: iodine atom has greater radius M2: less attraction between nucleus and incoming electron in iodine 2 4(d) M1: ½ (–486 + (2  364) ) M2: +121 2 Question Answer Marks 5(a)(i) Co2+ = [Ar] 3d7 AND Co3+ = [Ar] 3d6 1 5(a)(ii) has vacant d orbitals that are energetically accessible 1 5(a)(iii) • pink • blue • aqueous Any two [1], all three [2] 2 5(a)(iv) [Co(H2O)6]2+ + 4HCl → [CoCl4]2– + 6H2O + 4H+ OR [Co(H2O)6]2+ + 4Cl– → [CoCl4]2– + 6H2O 1

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 10 of 15 Question Answer Marks 5(a)(v) ligand exchange 1 5(a)(vi) [Co(H2O)6]2+ + 2OH– → Co(OH)2(H2O)4 + 2H2O OR [Co(H2O)6]2+ + 2OH– → Co(OH)2 + 6H2O 1 5(b)(i) M1: Nernst equation E = Eo + (0.059 / z) × log( [oxidised] [reduced] ) M2: E = –0.28 + (0.059 / 2) × log(0.02) 2 5(b)(ii) 1.66 V 1 5(b)(iii) 3Co + Cr2O72– + 14H+ → 3Co2+ + 2Cr3+ + 7H2O 1 5(b)(iv) Co2+ / Co, Co2+ / Co, Cr2O72– / Cr3+ 1 5(c) M1: moles of cobalt = 0.547 / 58.9 = 0.00929 M2: coulombs required = 0.00929  96500  2 = 1792 M3: time required = 1792 / 0.50 = 3584.8 s = 59.7 minutes 3sf 3 Question Answer Marks 6(a)(i) Ni(CO)4 1 6(a)(ii) • ethanedioate given as C2O4 • formula shows one Ni and three ligands • charge = 4– any two [1], all three [2] Ni(C2O4)34– scores [2] 2

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 11 of 15 Question Answer Marks 6(a)(iii) F AND optical 1 6(b)(i) N atom can donate one lone pair of electrons 1 6(b)(ii) [Cd(en)2]2+ AND has a larger Kstab / is more stable 1 6(b)(iii) [[Cd(CH3NH2)4]2+] [Cd2+][CH3NH2]4 1 Question Answer Marks 7(a) M1: Q CH3CH2CH(CH3)CHO M2: 2 7(b)(i) T CH3CH2CH2COCH3 AND V (CH3)2CHCOCH3 1 7(b)(ii) • yellow solid • CH3CH2CH2COOH OR (CH3)2CHCOOH • CHI3 Any two [1], all three [2] 2 7(c)(i) S (CH3)3CCHO 1 AND

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 12 of 15 Question Answer Marks 7(c)(ii) U CH3CH2COCH2CH3 1 7(c)(iii) triplet AND two hydrogen atoms on neighbouring carbon atoms 1 7(c)(iv) TMS / tetramethylsilane / Si(CH3)4 1 7(d) 4 3 5 3 Any two [1], all four [2] 2 Question Answer Marks 8(a) the pH at which an amino acid exists as a zwitterion 1 8(b) M1: [HOOCCH(NH3)CH2CONH2]+ M2: [HOOCCH(NH3)CH2COOH]+ 2

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 13 of 15 Question Answer Marks 8(c) M1: HOCH2CH(NH2)CH2CH2NH2 M2: HOCH2CH(NH2)CH2CH2OH 2 8(d)(i) HOOCCH2COOH + 2SOCl2 → ClOCCH2COCl + 2SO2 + 2HCl [1] 1 8(d)(ii) HOOCCH(CH2CONH2)NHCOCH2CONHCH(CH2CONH2)COOH M1: three molecules joined, with four correct amide groups only M2: whole structure correct as above, displayed formula accepted, skeletal formula as below accepted 2 8(e) M1: hydrolysis of amide to carboxylate ion M2: hydrolysis of –COOH to carboxylate ion and rest of molecule –OOCCH(NH2)CH2COO– 2

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 14 of 15 Question Answer Marks 8(f) • three monomer residues • all linkages correct • correct trailing bonds • repeat unit marked contains correct atoms Any two [1], any three [2], all four [3] CH2CONH2 CH2CONH2 | | –COCHNH–[–COCHNH–]–COCHNH– | CH2CONH2 3 8(g)(i) • rotate the plane of the plane polarised light • by the same angle • in opposite directions Any two [1], all three [2] 2 8(g)(ii) racemic mixture 1 Question Answer Marks 9(a) • all C atoms are sp2 hybridised • bonds between C atoms are  and  • C–C  bonds are formed by overlap of sp2 hybrid orbitals • C–C  bonds are formed by overlap of p orbitals • bonds between C and H atoms are  • C–H bonds are formed by overlap of sp2 hybrid orbitals and s orbitals Any two [1], any four [2], all six [3] 3

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 15 of 15 Question Answer Marks 9(b) CH3Cl and Al Cl 3 1 9(c) M1: curly arrow from within hexagon towards +C M2: intermediate, usual rules for horseshoe and + charge M3: curly arrow from C–H bond into ring AND H+ product 3 9(d) M1: HNO3 AND H2SO4 M2: concentrated AND 25 °C ⩽ T ⩽ 60 °C 2 9(e) alkaline KMnO4 1 9(f) 4-nitrobenzoic acid 1 9(g) (hot) concentrated HCl and Sn 1 9(h) M1: perform step 3 before step 2 OR perform step 2 before step 1 M2: COOH is 3-directing OR NO2 is 3-directing 2