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Chemistry 9701/42 October/November 2025: question paper and mark scheme

Cambridge AS & A Level Chemistry (9701), Paper 4 A Level Structured Questions, October/November 2025. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.

9701_w25_qp_429701_w25_ms_429701/42/O/N/25

Paper
Paper 4 A Level Structured Questions
Total marks
100
Time
2 hours
Questions
9
Code on the paper
9701/42/O/N/25
9701_w25_ms_42.pdf

9701/42 mark scheme, question by question

Each question as it's printed, and the mark scheme's points for it, as text. Diagrams, tables and graphs are in the PDFs above.

Question 17 marks

Question

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] * 0000800000004 * DFD 4

Mark scheme

1(a) Mg(NO3)2 → MgO + 2NO2 + ½O2 OR 2Mg(NO3)2 → 2MgO + 4NO2 + O2 1 1(b) M1: Mg(NO3)2 / magnesium nitrate 2 AND magnesium ion / Mg2+ AND is smaller / has higher charge density M2: nitrate ion / anion is more distorted / polarised 1(c)(i) magnesium sulfate and water / MgSO4 and H2O 1 1(c)(ii) M1: magnesium sulfate / A has a more exothermic ΔHlatt and ΔHhyd 3 M2: difference in ΔHhyd is greater M3: magnesium sulfate / A has a more exothermic ΔHsol

Question 29 marks

Question

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] * 0000800000005 * DFD 5 (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] * 0000800000006 * DFD 6

Mark scheme

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 2 M2: units = mol–1 dm3 s–1 2(c) not constant half-life because overall second order 1 2(d) 8.17 / 8.2 1 2(e) • reactants are adsorbed onto the catalyst surface 2 • bonds within reactant molecules are weakened • products are desorbed from catalyst surface Any two [1], all three [2]

Question 315 marks

Question

(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, HPO42–. conjugate acid of HPO42– …… conjugate base of HPO42– …… [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] * 0000800000007 * DFD 7 (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 place one tick (✓) / mol dm–3 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] * 0000800000008 * DFD 8 (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] * 0000800000009 * DFD 9

Mark scheme

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 2 M2: [CH3CH2COOH] = 4.67 × 10–3 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 3 M2: alkyl group is electron donating M3: O–H bond strengthened / anion is destabilised 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 3(d)(ii) M1: Ksp = [Mn2+][OH–]2 2 M2: units = mol3 dm–9 3(d)(iii) 2.0 × 10–13 1

Question 46 marks

Question

(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] * 0000800000010 * DFD 10

Mark scheme

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 2 M2: less attraction between nucleus and incoming electron in iodine 4(d) M1: ½ (–486 + (2 × 364) ) 2 M2: +121

Question 515 marks

Question

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] * 0000800000011 * DFD 11 (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 + 14H + 6e– 2– + 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 Cr2O72– / 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] * 0000800000012 * DFD 12

Mark scheme

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 2 • blue • aqueous Any two [1], all three [2] 5(a)(iv) [Co(H2O)6]2+ + 4HCl → [CoCl4]2– + 6H2O + 4H+ OR [Co(H2O)6]2+ + 4Cl– → [CoCl4]2– + 6H2O 1 5(a)(v) ligand exchange 1 5(a)(vi) [Co(H2O)6]2+ + 2OH– → Co(OH) 2(H2O)4 + 2H2O OR 1 [Co(H2O)6]2+ + 2OH– → Co(OH) 2 + 6H2O 5(b)(i) [oxidised] 2 M1: Nernst equation E = Eo + (0.059 / z) × log( ) [reduced] M2: E = –0.28 + (0.059 / 2) × log(0.02) 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 3 M2: coulombs required = 0.00929 × 96500 × 2 = 1792 M3: time required = 1792 / 0.50 = 3584.8 s = 59.7 minutes 3sf

Question 67 marks

Question

(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] * 0000800000013 * DFD 13

Mark scheme

6(a)(i) Ni(CO)4 1 6(a)(ii) • ethanedioate given as C2O4 2 • formula shows one Ni and three ligands • charge = 4– any two [1], all three [2] Ni(C2O4)34– scores [2] 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+] 1 [ Cd 2+] [CH3NH2] 4

Question 711 marks

Question

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] * 0000800000014 * DFD 14 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] * 0000800000015 * DFD 15 (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] * 0000800000016 * DFD 16

Mark scheme

7(a) M1: Q CH3CH2CH(CH3)CHO 2 M2: AND 7(b)(i) T CH3CH2CH2COCH3 AND V (CH3)2CHCOCH3 1 7(b)(ii) • yellow solid 2 • CH3CH2CH2COOH OR (CH3)2CHCOOH • CHI3 Any two [1], all three [2] 7(c)(i) S (CH3)3CCHO 1 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 2 3 5 3 Any two [1], all four [2]

Question 816 marks

Question

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 product of aspartic acid treated with excess of LiAl H4 an excess of LiAl H4 [2] * 0000800000017 * DFD 17 (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] * 0000800000018 * DFD 18 (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] * 0000800000019 * DFD 19 Question 9 starts on page 20. * 0000800000020 * DFD 20

Mark scheme

8(a) the pH at which an amino acid exists as a zwitterion 1 8(b) M1: [HOOCCH(NH3)CH2CONH2]+ 2 M2: [HOOCCH(NH3)CH2COOH]+ 8(c) M1: HOCH2CH(NH2)CH2CH2NH2 2 M2: HOCH2CH(NH2)CH2CH2OH 8(d)(i) HOOCCH2COOH + 2SOCl2 → ClOCCH2COCl + 2SO2 + 2HCl [1] 1 8(d)(ii) HOOCCH(CH2CONH2)NHCOCH2CONHCH(CH2CONH2)COOH 2 M1: three molecules joined, with four correct amide groups only M2: whole structure correct as above, displayed formula accepted, skeletal formula as below accepted 8(e) M1: hydrolysis of amide to carboxylate ion 2 M2: hydrolysis of –COOH to carboxylate ion and rest of molecule –OOCCH(NH2)CH2COO– 8(f) • three monomer residues 3 • 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 8(g)(i) • rotate the plane of the plane polarised light 2 • by the same angle • in opposite directions Any two [1], all three [2] 8(g)(ii) racemic mixture 1

Question 914 marks

Question

Compound X is made from benzene by the route shown in Fig. 9.1. W X benzene NO2 NO2 NH2 step 1 step 2 step 3 step 4 CH3 CH3 COOH COOH 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] * 0000800000021 * DFD 21 (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. intermediate + …… +CH 3 CH3 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] * 0000800000022 * DFD 22 (h) Benzene can also be used as a starting material to make compound Y. compound Y NH2 COOH 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] W X benzene NO2 NO2 NH2 step 1 step 2 step 3 step 4 CH3 CH3 COOH COOH Fig. 9.1 [Total: 14] * 0000800000023 * DFD 23 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) The Periodic Table of Elements Group 1 2 13 14 15 16 17 18 1 2 H He hydrogen helium Key 1.0 4.0 3 4 atomic number 5 6 7 8 9 10 Li Be atomic symbol B C N O F Ne lithium beryllium name boron carbon nitrogen oxygen fluorine neon 6.9 9.0 relative atomic mass 10.8 12.0 14.0 16.0 19.0 20.2 11 12 13 14 15 16 17 18 Na Mg Al Si P S Cl Ar sodium magnesium aluminium silicon phosphorus sulfur chlorine argon DFD 23.0 24.3 3 4 5 6 7 8 9 10 11 12 27.0 28.1 31.0 32.1 35.5 39.9 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 at www.cambridgeinternational.org after the live examination series. K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr potassium calcium scandium titanium vanadium chromium manganese iron cobalt nickel copper zinc gallium germanium arsenic selenium bromine krypton 39.1 40.1 45.0 47.9 50.9 52.0 54.9 55.8 58.9 58.7 63.5 65.4 69.7 72.6 74.9 79.0 79.9 83.8 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 24 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe rubidium strontium yttrium zirconium niobium molybdenum technetium ruthenium rhodium palladium silver cadmium indium tin antimony tellurium iodine xenon 85.5 87.6 88.9 91.2 92.9 95.9 – 101.1 102.9 106.4 107.9 112.4 114.8 118.7 121.8 127.6 126.9 131.3 55 56 57–71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 lanthanoids Cs Ba Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn caesium barium hafnium tantalum tungsten rhenium osmium iridium platinum gold mercury thallium lead bismuth polonium astatine radon 132.9 137.3 178.5 180.9 183.8 186.2 190.2 192.2 195.1 197.0 200.6 204.4 207.2 209.0 – – – 87 88 89–103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 actinoids Fr Ra Rf Db Sg Bh Hs Mt Ds Rg Cn Nh Fl Mc Lv Ts Og francium radium rutherfordium dubnium seaborgium bohrium hassium meitnerium darmstadtium roentgenium copernicium nihonium flerovium moscovium livermorium tennessine oganesson – – – – – – – – – – – – – – – – – 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 lanthanoids La Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu lanthanum cerium praseodymium neodymium promethium samarium europium gadolinium terbium dysprosium holmium erbium thulium ytterbium lutetium 138.9 140.1 140.9 144.2 – 150.4 152.0 157.3 158.9 162.5 164.9 167.3 168.9 173.1 175.0 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 actinoids Ac Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr actinium thorium protactinium uranium neptunium plutonium americium curium berkelium californium einsteinium fermium mendelevium nobelium lawrencium – 232.0 231.0 238.0 – – – – – – – – – – – Assessment International Education Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download To avoid the issue of disclosure of answer-related information to candidates, all copyright acknowledgements are reproduced online in the Cambridge

Mark scheme

9(a) • all C atoms are sp2 hybridised 3 • bonds between C atoms are σ and π • C–C σ bonds are formed by overlap of sp 2 hybrid orbitals • C–C π bonds are formed by overlap of p orbitals • bonds between C and H atoms are σ • C–H bonds are formed by overlap of sp 2 hybrid orbitals and s orbitals Any two [1], any four [2], all six [3] 9(b) CH3Cl and Al Cl 3 1 9(c) 3 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 9(d) M1: HNO3 AND H2SO4 2 M2: concentrated AND 25 °C ⩽ T ⩽ 60 °C 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 2 M2: COOH is 3-directing OR NO2 is 3-directing

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Chemistry 9701/42 for October/November 2025 is out of 100 marks, with 2 hours to answer it, across 9 questions.

Can I get my 9701/42 answers marked?

Yes. Upload your answers to 9701/42/O/N/25 as a PDF or photos, and CIE Buddy marks every question against this paper's official mark scheme, using Cambridge's marking rules. You get your score, each lost mark explained on your own script, and the topics to revise. It's a guide, not an official Cambridge grade.