Cambridge A Level Chemistry 9701 — 2025 May/June Paper 4 · Variant 3

[Turn over Cambridge International AS & A Level This document has 24 pages. Any blank pages are indicated. DC (WW) 354973 © UCLES 2025 CHEMISTRY 9701/43 Paper 4 A Level Structured Questions May/June 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. * 6 0 5 3 0 1 8 7 6 4 * ,










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3 9701/43/M/J/25 © UCLES 2025 [Turn over 1 Both calcium carbonate, CaCO3, and barium carbonate, BaCO3, decompose when heated to form the metal oxide and a gas. (a) Write an equation for the thermal decomposition of CaCO3. … [1] (b) State which of CaCO3 and BaCO3 decomposes at a lower temperature. Explain your answer. The compound that decomposes at a lower temperature is … . explanation … … … … … [2] (c) Calcium oxide, CaO, reacts with water to form compound A. Barium oxide, BaO, reacts with water to form compound B. (i) Identify A. … [1] (ii) Explain why A is less soluble than B. … … … … … … [3] [Total: 7] * 0000800000003 * ,










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4 9701/43/M/J/25 © UCLES 2025 2 Three experiments are carried out to investigate the reaction of nitrogen oxide, NO, with chlorine. 2NO + Cl 2 2NOCl The rate equation for this reaction is shown. rate = k [NO]2[Cl 2] (a) Under the conditions used in experiments 1 and 2, the value of k is 26.4. (i) The rate of the reaction is measured in mol dm–3 s–1. State the units of k. units of k = … [1] (ii) In experiment 1, the initial concentrations of NO and Cl 2 are equal. The initial rate of the reaction in experiment 1 is 2.57 × 10–6 mol dm–3 s–1. Calculate the initial concentration of NO. Show your working. initial concentration of NO = … mol dm–3 [2] (iii) In experiment 2, the initial concentrations of NO and Cl 2 are both ten times greater than the initial concentrations used in experiment 1. Calculate the initial rate of the reaction in experiment 2. initial rate of reaction in experiment 2 = … mol dm–3 s–1 [1] * 0000800000004 * ,










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5 9701/43/M/J/25 © UCLES 2025 [Turn over (b) Experiment 3 uses a large excess of NO. The initial concentration of Cl 2 is 2.00 × 10– 4 mol dm–3. (i) The graph of [Cl 2] against time shows that the reaction has a constant half-life, t ½. Explain this observation. … … [1] (ii) Under the conditions used in experiment 3, the value of the rate constant is 105.6. Show that t ½ of [Cl 2] is 6.56 × 10–3 s under these conditions. [1] (iii) Calculate the time taken, in s, for [Cl 2] to fall to 1.25 × 10–5 mol dm–3 in experiment 3. time = … s [1] (c) Sulfur dioxide, SO2, reacts very slowly with oxygen in the atmosphere, forming sulfur trioxide, SO3. This reaction is much faster in the presence of NO. Explain the role of NO in this process. Include chemical equations in your answer. … … … … [3] [Total: 10] * 0000800000005 * ,










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6 9701/43/M/J/25 © UCLES 2025 3 (a) Chromium(III) hydroxide, Cr(OH)3, is only slightly soluble in water. The value of the solubility product, Ksp, of Cr(OH)3 is 1.0 × 10–33 at 298 K. (i) Complete the expression for Ksp of Cr(OH)3. Include the units. Ksp = units = … [2] (ii) Calculate the solubility, in g dm–3, of Cr(OH)3 in pure water at 298 K. Show your working. solubility = … g dm–3 [3] (iii) Cr(OH)3 is less soluble in 0.100 mol dm–3 NaOH than it is in pure water. Explain this observation. … … [1] (b) The value of the acid dissociation constant, Ka, of butanoic acid, CH3CH2CH2COOH, is 1.51 × 10–5 at 298 K. (i) Calculate the pH of 0.100 mol dm–3 CH3CH2CH2COOH at 298 K. Show your working. pH = … [2] (ii) Calculate the pH of 0.100 mol dm–3 NaOH at 298 K. pH = … [1] * 0000800000006 * ,










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7 9701/43/M/J/25 © UCLES 2025 [Turn over (iii) 5.00 cm3 of 0.100 mol dm–3 NaOH is added to 10.00 cm3 of 0.100 mol dm–3 CH3CH2CH2COOH. Calculate the pH of the resulting solution. Show your working. pH = … [2] (c) 80.0 cm3 of an aqueous solution containing 0.704 g of CH3CH2CH2COOH is shaken with 100 cm3 of benzene, C6H6. There is 0.556 g of CH3CH2CH2COOH in the 100 cm3 of C6H6 at equilibrium. Calculate the partition coefficient, Kpc, of CH3CH2CH2COOH between C6H6 and water. Show your working. Kpc = … [2] [Total: 13] * 0000800000007 * ,










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8 9701/43/M/J/25 © UCLES 2025 4 Table 4.1 gives the enthalpy changes of hydration, ΔHhyd, of three ions, F–, K+ and Ca2+. Table 4.1 ion ΔHhyd / kJ mol–1 F– –506 K+ –322 Ca2+ –1650 (a) (i) Define enthalpy change of hydration. … … … [1] (ii) Explain the relative magnitudes of the enthalpy changes of hydration of K+ and Ca2+. … … … … [2] (iii) Define lattice energy. … … [1] (iv) The lattice energy, ΔHlatt, of calcium fluoride, CaF2, is –2602 kJ mol–1. Calculate the enthalpy change of solution, ΔHsol, in kJ mol–1, of CaF2. ΔHsol of CaF2 = … kJ mol–1 [2] * 0000800000008 * ,










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9 9701/43/M/J/25 © UCLES 2025 [Turn over (b) The formation of CaF2 at 298 K is shown. Ca(s) + F2(g) CaF2(s) ΔHo = –1214 kJ mol–1 ΔGo = –1162 kJ mol–1 Calculate the entropy change, ΔSo, in J K–1 mol–1, for this reaction. ΔSo = … J K–1 mol–1 [2] [Total: 8] * 0000800000009 * ,










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10 9701/43/M/J/25 © UCLES 2025 5 Copper is a transition element. (a) (i) Complete the electronic configurations of a Cu+ ion and a Cu2+ ion. Cu+ ion: [Ar] … Cu2+ ion: [Ar] … [1] (ii) Explain why transition elements have variable oxidation states. … … [1] (b) Aqueous copper(II) sulfate, CuSO4, contains the [Cu(H2O)6]2+ complex ion. (i) A few drops of NH3(aq) are added to CuSO4(aq). Describe any observations made. … [1] (ii) Write an equation for the reaction taking place. … [1] (c) (i) An excess of NH3(aq) is added to CuSO4(aq). Describe any further observations made. … [1] (ii) Write an equation for the reaction taking place. … [1] (iii) State the name for the type of reaction taking place. … [1] * 0000800000010 * ,










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11 9701/43/M/J/25 © UCLES 2025 [Turn over (d) Copper metal can be oxidised by acidified KMnO4. The relevant half-equations and their standard electrode potentials, Eo, are shown. Cu2+ + 2e– Cu Eo = +0.34 V MnO4 – + 8H+ + 5e– Mn2+ + 4H2O Eo = +1.52 V (i) A MnO4 –/Mn2+ electrode is constructed using 0.0020 mol dm–3 MnO4 –, 1.0 mol dm–3 Mn2+ and 1.0 mol dm–3 H+. The temperature used is 298 K. Use the Nernst equation to show that the E value for this MnO4 –/Mn2+ electrode is +1.49 V. [2] (ii) An electrochemical cell is constructed using a standard Cu2+/Cu electrode and the MnO4 –/Mn2+ electrode described in (d)(i). Calculate the value of Ecell. Ecell = … V [1] (iii) Write an equation for the reaction taking place in the electrochemical cell described in (d)(ii). … [1] (iv) Complete the sentences for the electrochemical cell described in (d)(ii). The … electrode is the negative electrode. Electrons flow from the … electrode to the … electrode when the cell is in use. [1] (e) A solution containing [Cu(H2O)6]2+ is electrolysed for 5.00 hours using a constant electric current. 0.764 g of copper metal is formed at the cathode. No other reduction reaction takes place. Calculate the electric current, in A, used. Give your answer to three significant figures. current = … A [3] [Total: 15] * 0000800000011 * ,









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12 9701/43/M/J/25 © UCLES 2025 6 (a) Iron forms complex ions with the monodentate ligand, CN–. (i) Complex ion A contains one Fe3+ ion and six CN– ligands. Complex ion B contains one Fe2+ ion and six CN– ligands. State the formulae of these two complex ions. Include the overall charge of each complex ion. complex ion A … complex ion B … [1] (ii) Explain why a solution containing complex ion A and a solution containing complex ion B are different colours. … … … [2] (iii) In complex ion A, the carbon atom of each CN– ligand bonds to the Fe3+ ion. State the type of bonding involved. … [1] * 0000800000012 * ,









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13 9701/43/M/J/25 © UCLES 2025 [Turn over (b) Complex ions have different geometries. Complex ion A is octahedral. Ag+ ions form a linear complex with ammonia. Ni atoms form a tetrahedral complex with carbon monoxide molecules. The carbon atom in the monodentate carbon monoxide ligand bonds to the nickel atom. Pd2+ ions form a square planar complex with chloride ions. Complete Fig. 6.1 to show the geometry of each of these four ions, using three-dimensional bonds where necessary. Label one bond angle on each complex ion. complex ion A Fe Ag+ with ammonia Ag Ni with carbon monoxide Ni Pd2+ with chloride Pd Fig. 6.1 [4] [Total: 8] * 0000800000013 * ,









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14 9701/43/M/J/25 © UCLES 2025 7 A hydrocarbon is known to be either compound D or compound E. CH2CH3 CH2CH3 D CH2CH3 CH2CH3 E Fig. 7.1 (a) Give the systematic name of E. … [1] (b) The proton (1H) NMR spectra of D and E are compared. They are very similar. The proton (1H) NMR spectrum of D is shown in Fig. 7.2. 5 6 7 4 3 chemical shift δ / ppm 2 1 0 8 9 10 Fig. 7.2 (i) Suggest a suitable solvent for obtaining the spectrum in Fig. 7.2. … [1] (ii) The proton (1H) NMR spectrum of E is obtained twice, once before and once after shaking with D2O. Describe any differences between these two spectra. Explain your answer. … … [1] * 0000800000014 * ,









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15 9701/43/M/J/25 © UCLES 2025 [Turn over (iii) Complete Table 7.1 for the proton (1H) NMR spectrum of D. Table 7.1 chemical shift δ / ppm number of 1H atoms responsible for the peak group responsible for the peak splitting pattern 1.3 2.7 7.1 [3] Table 7.2 environment of proton example chemical shift range δ / ppm alkane –CH3, –CH2–, >CH– 0.9–1.7 alkyl next to C=O CH3–C=O, –CH2–C=O, >CH–C=O 2.2–3.0 alkyl next to aromatic ring CH3–Ar, –CH2–Ar, >CH–Ar 2.3–3.0 alkyl next to electronegative atom CH3–O, –CH2–O, –CH2–Cl 3.2–4.0 attached to alkene =CHR 4.5–6.0 attached to aromatic ring H–Ar 6.0–9.0 aldehyde HCOR 9.3–10.5 alcohol ROH 0.5–6.0 phenol Ar–OH 4.5–7.0 carboxylic acid RCOOH 9.0–13.0 alkyl amine R–NH– 1.0–5.0 aryl amine Ar–NH2 3.0–6.0 amide RCONHR 5.0–12.0 (iv) Compounds D and E can be distinguished by carbon-13 NMR spectroscopy. State the number of peaks in each spectrum. The carbon-13 NMR spectrum of D has … peaks. The carbon-13 NMR spectrum of E has … peaks. [1] * 0000800000015 * ,









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16 9701/43/M/J/25 © UCLES 2025 (c) Compound D can be oxidised to compound F by alkaline KMnO4 followed by dilute acid. COOH COOH F CH2CH3 CH2CH3 D Fig. 7.3 (i) Write an equation for this reaction using molecular formulae for D and F. The products of this reaction are F, water and carbon dioxide. Use [O] to represent one atom of oxygen from the oxidising agent. … [1] * 0000800000016 * ,









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17 9701/43/M/J/25 © UCLES 2025 [Turn over (ii) F reacts with an excess of SOCl 2 to form compound G. The molecular formula of G is C8H4Cl 2O2. G reacts with ethane-1,2-diol, HOCH2CH2OH, to form a mixture of products that includes compounds J, molecular formula C10H8O4, and K, molecular formula C18H12Cl 2O6. Draw the structures of compounds G, J and K in Fig. 7.4. G, C8H4Cl 2O2 J, C10H8O4 K, C18H12Cl 2O6 Fig. 7.4 [3] [Total: 11] * 0000800000017 * ,









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18 9701/43/M/J/25 © UCLES 2025 8 Bromine reacts with methylbenzene in the dark in the presence of a suitable catalyst to form HBr and compound L, C7H7Br. L is one of three isomers that can form in this reaction. (a) (i) The mechanism for the reaction involves methylbenzene reacting with a Br+ ion. This ion is produced when bromine reacts with the catalyst. Complete the equation for the reaction of Br2 with the catalyst. Br2 + … Br+ + … [1] (ii) One of the isomers of L forms in much smaller amounts than the other two isomers. Draw the structure of this isomer and explain why it forms in the smallest amount. … … [2] (iii) Complete the mechanism in Fig. 8.1 for the reaction between methylbenzene and the Br+ ion. Include all relevant curly arrows and charges. CH3 Br+ intermediate + … Fig. 8.1 [3] * 0000800000018 * ,









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19 9701/43/M/J/25 © UCLES 2025 [Turn over (b) Chlorobutane and chlorobenzene are added separately to samples of warm aqueous AgNO3. One of the chloro-compounds reacts slowly and the other does not react. (i) Identify the chloro-compound that reacts and describe any observations. … … [1] (ii) Write two equations to explain any observations in (b)(i). … … [2] (iii) Explain the difference in reactivity of chlorobutane and chlorobenzene with warm aqueous AgNO3. … … … … [2] [Total: 11] * 0000800000019 * ,









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20 9701/43/M/J/25 © UCLES 2025 9 (a) Phenylamine, C6H5NH2, and propylamine, CH3CH2CH2NH2, can be produced by different reduction reactions. (i) Identify an organic compound that can be converted into C6H5NH2 by a reduction reaction. State the reagents and conditions for this reaction. organic compound … reagents … conditions … [2] (ii) Identify an organic compound that can be converted into CH3CH2CH2NH2 by a reduction reaction. State the reagent for this reaction. organic compound … reagent … [2] (iii) Identify a single test that will distinguish between C6H5NH2 and CH3CH2CH2NH2 by producing a white precipitate with only one of these amines. Draw the structure of the compound that is precipitated. testing reagent … amine that gives a precipitate … structure of the compound that is precipitated: [2] * 0000800000020 * ,










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21 9701/43/M/J/25 © UCLES 2025 [Turn over (b) Describe the relative basicities of C6H5NH2, CH3CH2CH2NH2 and NH3. Explain your answer. … < … < … least basic most basic … … … … … [4] [Total: 10] * 0000800000021 * ,









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22 9701/43/M/J/25 © UCLES 2025 10 Propanoic acid, methanoic acid and ethanedioic acid are all weak acids. (a) Draw the displayed formula of ethanedioic acid. [1] (b) The three acids, propanoic acid, methanoic acid and ethanedioic acid, can be distinguished using a combination of two chemical tests. Neither testing reagent is an acid–base indicator. Identify two suitable testing reagents and complete Table 10.1 to show the observations from each test. reagent 1 … reagent 2 … Table 10.1 observation when treated with reagent 1 observation when treated with reagent 2 propanoic acid methanoic acid ethanedioic acid [4] (c) When propanoic acid is treated with chlorine gas in the presence of ultraviolet light, a mixture of products is formed. One of these products is 2,2-dichloropropanoic acid. Explain why 2,2-dichloropropanoic acid is stronger than propanoic acid. Refer to the structure of each compound in your answer. … … … … [2] [Total: 7] * 0000800000022 * ,









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23 9701/43/M/J/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 K w = 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 9701/43/M/J/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 16 printed pages. © Cambridge University Press & Assessment 2025 [Turn over Cambridge International AS & A Level CHEMISTRY 9701/43 Paper 4 A Level Structured Questions May/June 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 May/June 2025 series for most Cambridge IGCSE, Cambridge International A and AS Level components, and some Cambridge O Level components.

9701/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 2 of 16 Generic Marking Principles These general marking principles must be applied by all examiners when marking candidate answers. They should be applied alongside 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.

9701/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 3 of 16 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 thresholds 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 from 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.

9701/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 4 of 16 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.

9701/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 5 of 16 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 standardisation 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 Benefit of the doubt not applied in this instance

9701/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 6 of 16 Annotation Meaning Rounding error Repetition Blank page or part of script seen Error in number of significant figures Transcription error

9701/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 7 of 16 Question Answer Marks 1(a) CaCO3 → CaO + CO2 [1] 1 1(b) M1: calcium carbonate AND calcium ion / Ca2+ smaller / ionic radius bigger down group M2: carbonate ion / anion / CO32– is more distorted/polarised by Ca2+ 2 1(c)(i) calcium hydroxide / Ca(OH)2 1 1(c)(ii) M1: A / calcium hydroxide has more exothermic Hlatt AND Hhyd than B / barium hydroxide M2: difference in Hlatt is greater for A / calcium hydroxide than for B / barium hydroxide M3: Hsol is less exothermic for A / calcium hydroxide than for B / barium hydroxide 3 Question Answer Marks 2(a)(i) mol–2 dm6 s–1 1 2(a)(ii) M1: 2.57  10–6 = 26.4  [concentration]3 M2: [concentration] = 4.60  10–3 2 2(a)(iii) 2.57  10–3 1 2(b)(i) overall 1st order / pseudo 1st order OR 1st order with respect to Cl2 AND concentration of NO doesn’t change 1 2(b)(ii) ln2 / 105.6 or 0.693 / 105.6 1 2(b)(iii) 0.0262 / 2.62  10–2 1

9701/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 8 of 16 Question Answer Marks 2(c) M1: homogeneous catalyst OR catalyst because it is regenerated M2: 2NO + O2 → 2NO2 M3: NO2 + SO2 → NO + SO3 3 Question Answer Marks 3(a)(i) M1: Ksp = [Cr3+] [OH–]3 M2: mol4 dm–12 2 3(a)(ii) M1: [OH–] = 3[Cr3+] M2: 2.47  10–9 mol dm–3 M3: 2.54  10–7 g dm–3 3 3(a)(iii) common ion effect 1 3(b)(i) M1: [H+] = 1.23  10–3 M2: pH = 2.91 2 3(b)(ii) 13.0 1 3(b)(iii) M1: [CH3CH2CH2COOH] = [CH3CH2CH2COO–] M2: pH = 4.82 2 3(c) M1: 0.148 g is left in water M2: Kpc = 3.01 OR 0.333 2

9701/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 9 of 16 Question Answer Marks 4(a)(i) energy change when one mole of gaseous ions dissolves in water 1 4(a)(ii) M1: K+ has larger radius AND K+ has smaller charge M2: Ca2+ has larger attraction for H2O 2 4(a)(iii) energy change when one mole of ionic solid forms from gaseous ions 1 4(a)(iv) M1: –1650–(2  506) + 2602 M2: –60 2 4(b) M1: G = H –TS AND T = 298 M2: –174.5 [1] 2 Question Answer Marks 5(a)(i) (Ar) 3d10 AND (Ar) 3d9 1 5(a)(ii) similar energy of 3d and 4s subshells 1 5(b)(i) pale blue precipitate 1 5(b)(ii) [Cu(H2O)6]2+ + 2OH– → Cu(OH)2(H2O)4 + 2H2O 1 5(c)(i) deep blue solution 1 5(c)(ii) [Cu(H2O)6]2+ + 4NH3 → [Cu(NH3)4(H2O)2]2+ + 4H2O 1 5(c)(iii) ligand exchange 1 5(d)(i) M1: E = Eo +(0.059 / z) log(ox/red) M2: E = 1.52 + (0.059 / 5)  log (0.002) 2

9701/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 10 of 16 Question Answer Marks 5(d)(ii) 1.15 1 5(d)(iii) 5Cu + 2MnO4– + 16H+ → 5Cu2+ + 2Mn2+ + 8H2O 1 5(d)(iv) Cu2+ / Cu, Cu2+ / Cu, MnO4– / Mn2+ / platinum 1 5(e) M1: moles of copper = 0.0120 mol M2: coulombs required = 2322 C M3: current required = 0.129 A 3 Question Answer Marks 6(a)(i) [Fe(CN)6]3– AND [Fe(CN)6]4– 1 6(a)(ii) M1: different E M2: absorption of different frequency of visible light OR wavelength of visible light 2 6(a)(iii) dative covalent / coordinate 1

9701/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 11 of 16 Question Answer Marks 6(b) Each correct structure = [1] 4

9701/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 12 of 16 Question Answer Marks 7(a) 1,3-diethylbenzene 1 7(b)(i) CDCl3 1 7(b)(ii) no difference AND no H that can be exchanged with D 1 7(b)(iii) Each correct column = [1]  / ppm number of protons group responsible name of splitting pattern 1.3 6 CH3 triplet 2.7 4 CH2 quartet / quadruplet 7.1 4 3 7(b)(iv) 5 AND 6 1 7(c)(i) C10H14 + 12[O] → C8H6O4 + 2CO2 + 4H2O 1

9701/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 13 of 16 Question Answer Marks 7(c)(ii) Each correct structure = [1] compound G compound J compound K 3

9701/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 14 of 16 Question Answer Marks 8(a)(i) Br2 + Al Br3 → Br+ + Al Br4– 1 8(a)(ii) • reference to methyl group having an effect • methyl group is 2,4,6 directing Correct structure and one correct point [1] correct structure and two correct points [2] 2 8(a)(iii) → → M1: first curly arrow from inside hexagon towards Br+ M2: correct intermediate M3: second curly arrow from bond into hexagon AND organic product, H+ 3 8(b)(i) chlorobutane AND white precipitate 1 8(b)(ii) M1: C4H9Cl + H2O → C4H9OH + H+ + Cl – M2: Cl – + Ag+ → AgCl [1] 2

9701/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 15 of 16 Question Answer Marks 8(b)(iii) M1: delocalisation of lone pair of Cl into benzene ring /  system M2: strengthens C–Cl bond / causes C–Cl bond to have partial double bond nature 2 9(a)(i) • nitrobenzene / C6H5NO2 • Sn and HCl • concentrated HCl and heat / boil / reflux Two correct point [1] all three correct points [2] 2 9(a)(ii) M1: propylamide / propenamide / C2H5CONH2 OR propanenitrile / C2H5CN M2: LiAl H4 2 9(a)(iii) M1: bromine (aq) AND phenylamine / C6H5NH2 M2: correct structure 2 9(b) M1: phenylamine ammonia propylamine M2: a base has a lone pair that can accept a proton / H+ M3: the lone pair on the nitrogen atom of phenylamine is delocalised into the benzene ring / -system M4: propylamine has an electron donating alkyl group 4

9701/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 16 of 16 Question Answer Marks 10(a) 1 10(b) M1: reagent 1:Fehling’s OR Tollens’ reagent M2: acidified MnO4– observation when treated with reagent 1 observation when treated with reagent 2 propanoic acid no change no change methanoic acid red ppt OR silver mirror colourless ethanedioic acid no change colourless M3 / M4: One completely correct column [1] both completely correct columns [2] 4 10(c) M1: chlorine atoms are electron withdrawing / electronegative M2: weakening O–H bond OR stabilising conjugate base 2