Cambridge A Level Chemistry 9701 — 2025 Oct/Nov Paper 4 · Variant 3

This document has 28 pages. Any blank pages are indicated. [Turn over Cambridge International AS & A Level DC (CE/JG) 342071/3 © UCLES 2025 CHEMISTRY 9701/43 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. * 4 6 7 0 5 6 2 4 0 0 * ,










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3 9701/43/O/N/25 © UCLES 2025 [Turn over 1 (a) Solutions of Group 2 hydrogencarbonates, M(HCO3)2, decompose on heating to give the corresponding metal carbonate, carbon dioxide and water. (i) Write an equation for the decomposition of strontium hydrogencarbonate, Sr(HCO3)2. … [1] (ii) The thermal stability of Group 2 carbonates increases down the group. Explain this trend. … … … … [2] (b) The hydroxides and fluorides of Group 2 elements show similar trends in solubility. Describe the trend in the solubility of the fluorides of calcium, strontium and barium. Explain your answer. … … … least soluble most soluble explanation … … … … … … … [4] * 0000800000003 * ,










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4 9701/43/O/N/25 © UCLES 2025 (c) (i) Define enthalpy change of hydration, ΔHhyd. … … … [1] (ii) State the main factors that affect the magnitude of enthalpy change of hydration. Explain your answer. … … … … [2] (d) Table 1.1 shows various energy changes. Table 1.1 energy change value / kJ mol–1 lattice energy of MgF2 –2957 enthalpy change of hydration, ΔHhyd, of Mg2+ –1926 enthalpy change of hydration, ΔHhyd, of F– –505 Use data from Table 1.1 to calculate the enthalpy change of solution, ΔHsol, for MgF2(s). It may be helpful to draw a labelled energy cycle. Show your working. ΔHsol of MgF2(s) = … kJ mol–1 [2] * 0000800000004 * ,










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5 9701/43/O/N/25 © UCLES 2025 [Turn over (e) Mercury(I) fluoride, Hg2F2, is sparingly soluble in water. The cation in Hg2F2 exists as the diatomic ion Hg2 2+ with a covalent Hg–Hg bond. (i) Write the expression for the solubility product, Ksp, of Hg2F2. Include the units. Ksp = units … [2] (ii) The solubility of Hg2F2 is 9.20 × 10–3 mol dm–3 at 298 K. Calculate the value of Ksp of Hg2F2 at 298 K. Ksp = … [1] [Total: 15] * 0000800000005 * ,










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6 9701/43/O/N/25 © UCLES 2025 2 (a) Iron can form stable ions in the +2 and +3 oxidation states. Explain why transition elements have variable oxidation states. … … [1] (b) Aqueous solutions of iron(II) salts contain the complex ion [Fe(H2O)6]2+. Define complex ion. … … [1] (c) [Fe(H2O)6]2+ can be converted into [Fe(H2O)4(OH)2]. (i) Suggest a suitable reagent for this conversion. State the type of reaction. reagent … type of reaction … [1] (ii) [Fe(H2O)4(OH)2] is a green precipitate that turns brown on standing in air. Table 2.1 shows electrode potentials for some electrode reactions. Table 2.1 electrode reaction E o / V Fe(H2O)3(OH)3 + H2O + e– Fe(H2O)4(OH)2 + OH– –0.56 O2 + 2H2O + 4e– 4OH– +0.40 Use the information in Table 2.1 to explain why [Fe(H2O)4(OH)2] turns brown on standing in air. Include an equation for this reaction. … … … … … … [3] * 0000800000006 * ,










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7 9701/43/O/N/25 © UCLES 2025 [Turn over (d) The complex [Co(NH3)6]2+ reacts with hydrogen peroxide as shown. reaction 1 2[Co(NH3)6]2+ + H2O2 2[Co(NH3)6]3+ + 2OH– E cell o = +1.67 V Calculate ΔG o , in kJ mol–1, for reaction 1. ΔG o = … kJ mol–1 [2] [Total: 8] * 0000800000007 * ,










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8 9701/43/O/N/25 © UCLES 2025 3 (a) Solid manganese(IV) oxide, MnO2, catalyses the decomposition of hydrogen peroxide. 2H2O2(aq) 2H2O(l) + O2(g) State the type of catalysis for this reaction. Explain your answer. … … [1] (b) Hydrogen peroxide reacts with iodide ions in acidic conditions as shown. H2O2 + 2I– + 2H+ 2H2O + I2 The initial rate of this reaction is investigated with different concentrations of H2O2, I– and H+. The results obtained are shown in Table 3.1. Table 3.1 experiment [H2O2] / mol dm–3 [I–] / mol dm–3 [H+] / mol dm–3 initial rate / mol dm–3 s–1 1 0.0450 0.0300 0.0125 2.42 × 10–3 2 0.0225 0.0600 0.0125 2.42 × 10–3 3 0.0225 0.120 0.0125 4.84 × 10–3 4 0.0450 0.120 0.0500 9.68 × 10–3 (i) Use the information in Table 3.1 to deduce the rate equation for this reaction. Explain your reasoning. … … … … … … … [4] (ii) Use your rate equation from (b)(i) and the data from Experiment 1 to calculate the rate constant, k, for this reaction. Include the units of k. k = … units … [2] * 0000800000008 * ,










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9 9701/43/O/N/25 © UCLES 2025 [Turn over (c) The rate of the thermal decomposition of azomethane, CH3N=NCH3, is investigated. CH3N=NCH3 N2 + C2H6 Fig. 3.1 shows the results obtained. The reaction is first order with respect to CH3N=NCH3. 0.06 0.05 0.04 0.03 0.02 0.01 0 0 50 100 150 200 250 300 350 400 [CH3N=NCH3] / mol dm–3 time / s Fig. 3.1 (i) Use Fig. 3.1 to calculate two half-lives, t 2 1, to show that the reaction is first order. … … … … [2] (ii) Use your answer to (c)(i) to calculate the rate constant, k, for the decomposition of azomethane. k = … s–1 [1] (d) Describe the effect of increasing temperature on the rate constant and on the rate of a reaction. … … … [1] [Total: 11] * 0000800000009 * ,










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10 9701/43/O/N/25 © UCLES 2025 4 (a) Define standard cell potential, E cell o . Include a description of standard conditions. … … … [2] (b) The Daniell cell is an electrochemical cell consisting of a Cu2+(aq) / Cu(s) electrode and a Zn2+(aq) / Zn(s) electrode. (i) Draw a labelled diagram of this electrochemical cell. Include all necessary substances and relevant pieces of apparatus needed to measure the E cell o . It is not necessary to state the conditions used. [3] (ii) State the charge carriers that transfer current through the solutions and through the wire. the solutions … the wire … [1] (iii) The standard electrode potential, E o , for the Zn2+(aq) / Zn(s) electrode is –0.76 V. Water is added to a standard Zn2+(aq) / Zn(s) electrode. The new concentration of Zn2+(aq) is 0.25 mol dm–3. Use the Nernst equation to calculate the electrode potential, E, for this new Zn2+(aq) / Zn(s) electrode. E (Zn2+(aq) / Zn(s)) = … V [2] * 0000800000010 * ,










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11 9701/43/O/N/25 © UCLES 2025 [Turn over (c) An electrochemical cell consists of a ZnO / Zn electrode and a MnO2 / Mn2O3 electrode in an alkaline electrolyte. The standard cell potential, E cell o , for this cell is +1.47 V. The half-equation at each electrode when this cell is discharging is shown. Zn + 2OH– ZnO + H2O + 2e– 2MnO2 + H2O + 2e– Mn2O3 + 2OH– (i) Use this information to determine the change in oxidation state of manganese when this cell is discharging. from … to … [1] (ii) Write the equation for the overall reaction that occurs when this cell is discharging. … [1] (iii) The E o for the ZnO / Zn electrode is –1.28 V. Calculate the standard electrode potential, E o, for the MnO2 / Mn2O3 electrode. E o (MnO2 / Mn2O3) = … V [1] [Total: 11] * 0000800000011 * ,









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12 9701/43/O/N/25 © UCLES 2025 5 (a) Copper shows typical properties of transition elements, including its behaviour as a catalyst. Complete Table 5.1 to show the total number of unpaired electrons in the 3d and 4s orbitals of an isolated gaseous Cu atom and a Cu2+ ion. Table 5.1 species number of unpaired electrons 3d 4s Cu Cu2+ [1] (b) The 3d orbitals in an isolated Cu2+ ion are degenerate. Complete the diagram to show the relative energies of the 3d orbitals in an isolated Cu2+ ion and in Cu2+ in a tetrahedral complex. energy isolated Cu2+ ion Cu2+ in a tetrahedral complex [2] (c) Explain why transition elements behave as catalysts. … … … [2] * 0000800000012 * ,









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13 9701/43/O/N/25 © UCLES 2025 [Turn over (d) CN– is a monodentate ligand. Table 5.2 shows information about two complex ions that contain only CN– ions as ligands. Complete Table 5.2. Table 5.2 metal ion coordination number formula of complex ion charge of complex ion Ag+ 2 Fe2+ 4– [2] (e) The complex ion [Au(CN)2Br2]– displays geometrical (cis / trans) isomerism. Draw the structure of trans-[Au(CN)2Br2]–. State its shape and the Br-Au-Br bond angle. shape … Br-Au-Br bond angle = … [2] * 0000800000013 * ,









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14 9701/43/O/N/25 © UCLES 2025 (f) An impure sample of a vanadium(V) compound of mass 0.250 g is dissolved in aqueous acid. This solution contains VO3 – ions. An excess of zinc is added to this solution. All the VO3 – ions are reduced to V2+ ions and Zn atoms are oxidised to Zn2+ ions. The unreacted zinc is removed and the resulting solution is titrated with acidified MnO4 –. The end-point is reached when 22.5 cm3 of 0.0750 mol dm–3 MnO4 – is added. A redox reaction takes place and all the V2+ reacts forming VO3 –. 3MnO4 – + 5V2+ + 3H2O 3Mn2+ + 5VO3 – + 6H+ (i) Calculate the percentage by mass of vanadium in the 0.250 g of impure sample. Assume the impurities do not contain any vanadium ions. Show your working. percentage of vanadium = … [3] (ii) Complete the equation for the reaction between acidified VO3 – ions and Zn metal. … VO3 – + … Zn + … … V2+ + … Zn2+ + … [2] [Total: 14] * 0000800000014 * ,









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15 9701/43/O/N/25 © UCLES 2025 [Turn over BLANK PAGE * 0000800000015 * ,









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16 9701/43/O/N/25 © UCLES 2025 6 (a) Thin-layer and gas / liquid chromatography can be used to separate mixtures into their individual components. (i) Define the following terms used in chromatography. Rf value … … retention time … … [2] (ii) Each type of chromatography makes use of a stationary phase and a mobile phase. Complete Table 6.1 with a description of each of these. Table 6.1 stationary phase mobile phase thin-layer chromatography gas / liquid chromatography [1] (b) A mixture of two substances A and B is analysed by thin-layer chromatography. The Rf value of substance A is larger than that of substance B. Suggest why substance A has a larger Rf value. … … [1] * 0000800000016 * ,









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17 9701/43/O/N/25 © UCLES 2025 [Turn over (c) The two isomeric compounds Y and Z are analysed by proton (1H) NMR spectroscopy. O O O O Y Z (i) Complete Table 6.2 to predict the number of peaks observed in the proton (1H) NMR spectra for Y and Z. Table 6.2 compound number of peaks observed Y Z [1] (ii) Name all the different splitting patterns observed in the proton (1H) NMR spectra for Y and Z. Y … Z … [2] [Total: 7] * 0000800000017 * ,









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18 9701/43/O/N/25 © UCLES 2025 7 (a) State the relative acidities of bromoethanoic acid, BrCH2COOH, chloroethanoic acid, Cl CH2COOH, ethanoic acid, CH3COOH and ethanol, CH3CH2OH. Explain your answer. … … … … most acidic least acidic … … … … … … … [4] (b) Fig. 7.1 shows the reaction of methylbenzene and ethanedioic acid with KMnO4. Predict the major carbon-containing product for each of these reactions. hot alkaline KMnO4 hot acidified KMnO4 HO O O OH methylbenzene ethanedioic acid Fig. 7.1 [2] * 0000800000018 * ,









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19 9701/43/O/N/25 © UCLES 2025 [Turn over (c) Polyamide X can be synthesised from ethanedioic acid and benzene-1,4-diamine. benzene-1,4-diamine H2N NH2 (i) Draw the repeat unit of polyamide X in the box. The new functional group formed should be shown displayed. polyamide X [2] (ii) Benzene-1,4-diamine can be formed by reduction of 1,4-dinitrobenzene. Complete the equation for this reduction. [H] represents one atom of hydrogen from a reducing agent. H2N + … [H] NH2 + … … [1] * 0000800000019 * ,









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20 9701/43/O/N/25 © UCLES 2025 (d) Fig. 7.2 shows the two-step synthesis of the azo compound W. H2N NH2 step 1 step 2 V C6H4N4Cl2 W C18H14N4O2 OH and NaOH(aq) Fig. 7.2 (i) Suggest structures for compounds V and W and draw them in the boxes in Fig. 7.2. [2] (ii) Give the reagents and conditions for step 1. … [1] [Total: 12] * 0000800000020 * ,










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22 9701/43/O/N/25 © UCLES 2025 8 (a) In the electrophilic substitution of arenes, different substituents can direct to different ring positions. (i) Describe the directing effect of the –NO2 group. Explain your answer. … … [1] (ii) The nitration of arenes uses a mixture of concentrated HNO3 and concentrated H2SO4 to generate the NO2 + electrophile. Write an equation for the formation of the NO2 + electrophile. … [1] (b) Carbon-carbon bond formation is an important reaction in organic synthesis. Fig. 8.1 shows the synthesis of compound Q from benzene in two reaction steps. reaction 1 reaction 2 compound P compound Q O Fig. 8.1 (i) Draw the structure of compound P in the box in Fig. 8.1. [1] (ii) Suggest reagents and conditions for reactions 1 and 2 in Fig. 8.1. reaction 1 … reaction 2 … [2] * 0000800000022 * ,









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23 9701/43/O/N/25 © UCLES 2025 [Turn over (c) Separate samples of C6H5Br and C6H5CH2Br are added to warm AgNO3(aq). State the expected observations, if any. Explain your answer. C6H5Br with AgNO3(aq) … C6H5CH2Br with AgNO3(aq) … explanation … … … … [3] (d) Acyl bromides, RCOBr, react readily with H2O. The mechanism of this reaction is similar to that of the reaction of H2O with acyl chlorides, RCOCl. (i) Name the mechanism of this reaction. … [1] (ii) Complete the mechanism in Fig. 8.2 for the reaction of RCOBr with H2O. Include all relevant lone pairs of electrons, curly arrows, charges and dipoles. Draw the structure of the intermediate. intermediate products O C R Br O H H Fig. 8.2 [4] [Total: 13] * 0000800000023 * ,









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24 9701/43/ O/N/25 © UCLES 2025 9 (a) Explain why amides are much weaker bases than amines. … … … … [2] (b) Fig. 9.1 shows the preparation of 2-phenylethylamine, C6H5CH2CH2NH2, by three different routes. H2/Ni M N O NH2 NH2 reaction 1 2-phenylethylamine NH3 Fig. 9.1 (i) Suggest structures for compounds M and N and draw them in the boxes in Fig. 9.1. [2] (ii) Give the reagents and conditions for reaction 1. … [1] * 0000800000024 * ,









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25 9701/43/O/N/25 © UCLES 2025 [Turn over (c) Fig. 9.2 shows compound H which is a useful starting material in organic synthesis. O N H H Fig. 9.2 H contains an alkene and an amine functional group. Name the other functional group and give the classification of the amine group in H. other functional group in H … classification of amine … [1] (d) Ozonolysis involves the oxidative cleavage of a C=C bond in alkenes using ozone, O3, as shown in Fig. 9.3. R1 R2 R3 O3 R2 R1 R3 R4 R4 O O + Fig. 9.3 Fig. 9.4 shows the first step in this reaction which involves the formation of an ozonide intermediate. R1 R2 R3 R4 R2 R1 R4 R3 O O O O O O + - ozonide Fig. 9.4 (i) On Fig. 9.4, draw three curly arrows to complete the mechanism of this step. [2] * 0000800000025 * ,









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26 9701/43/O/N/25 © UCLES 2025 (ii) L is formed from alkene K, C8H14, by a similar reaction to that shown in Fig. 9.3. O O L Suggest the structure of K. K C8H14 [1] [Total: 9] * 0000800000026 * ,









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27 9701/43/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) * 0000800000027 * ,









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28 9701/43/O/N/25 © UCLES 2025 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 – 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. * 0000800000028 * ,









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This document consists of 19 printed pages. © Cambridge University Press & Assessment 2025 [Turn over Cambridge International AS & A Level CHEMISTRY 9701/43 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.

9701/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 2 of 19 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.

9701/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 3 of 19 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.

9701/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 4 of 19 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 October/November 2025 © Cambridge University Press & Assessment 2025 Page 5 of 19 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 Repetition

9701/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 6 of 19 Annotation Meaning 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 October/November 2025 © Cambridge University Press & Assessment 2025 Page 7 of 19 Question Answer Marks 1(a)(i) Sr(HCO3)2 → SrCO3 + CO2 + H2O 1 1(a)(ii) M1: radius of cation / M2+ increases OR charge density of cation / M2+ decreases M2: less polarisation / less distortion of the anion / carbonate ion 2 1(b) M1: least soluble CaF2 < SrF2 < BaF2 most soluble M2: Hlatt and Hhyd both become less exothermic / less negative M3: Hhyd changes less / becomes less exothermic by a smaller extent OR Hlatt changes more / becomes less exothermic by a larger extent M4: Hsol becomes more exothermic / more negative 4 1(c)(i) enthalpy change when one mole of gaseous ions dissolve in water to form a solution 1 1(c)(ii) M1: ionic radii AND ionic charge M2: (ionic) radii increase / charge density decreases AND ∆Hhyd decreases/less exothermic AND less attraction between water molecules and (gaseous) ions OR as ionic charge increases / charge density increases AND ∆Hhyd increases/more exothermic AND more attraction between water molecules and (gaseous) ions 2 1(d) M1: use of –2957 / –1926 / –505 AND 2  (–505) M2: correct signs and evaluation ∆Hsol of MgF2(s) = –1926 + (2  –505) – (–2957) = (+)21 kJ mol–1 2 1(e)(i) M1: Ksp = [Hg22+] [F–]2 M2: units = mol3 dm–9 2 1(e)(ii) Ksp = 22  (9.20  10–3)3 = 3.11  10–6 min 2sf 1

9701/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 8 of 19 Question Answer Marks 2(a) the (3)d and (4)s sub-shells / orbitals / electrons are close / similar in energy 1 2(b) species or ion formed by a central metal atom/ion AND surrounded by/bonded to (one or more) ligands 1 2(c)(i) NaOH / OH–(aq) AND precipitation / ligand exchange / deprotonation / acid–base 1 2(c)(ii) M1: Fe(H2O)4(OH)2 is oxidised (from + 2 to + 3 on standing) M2: Eo of Fe(H2O)3(OH)3 is more negative than Eo of O2 OR Eocell = 0.40 – (–0.56) = (+)0.96 V M3: 4Fe(H2O)4(OH)2 + O2 → 4Fe(H2O)3(OH)3 + 2H2O 3 2(d) ∆Go = –2  1.67  96 500 = –322 310 J mol–1 M1: ∆Go = –nEocellF AND n = 2 OR –2  1.67  96 500 M2: ∆Go = –322.3 kJ mol–1 2

9701/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 9 of 19 Question Answer Marks 3(a) heterogeneous AND MnO2 is in a different state / phase to the reactants / H2O2 1 3(b)(i) M1: (expt 2 and 3) [I–]  2 initial rate  2 so 1st order wrt [I–] M2: (expt 1 and 2) [I–]  2, [H2O2]  0.5 initial rate  1 so 1st order wrt [H2O2] OR (expt 1 and 4) [I–]  4 initial rate  4 AND [H+]  4 so zero order wrt [H+] OR (expt 1 and 3) [I–]  4, [H2O2]  0.5 initial rate  2 so 1st order wrt [H2O2] M3: (expt 1 and 4) [I–] x 4 , [H+]  4 initial rate  4 so zero order wrt [H+] OR (expt 3 and 4) [H2O2]  2 initial rate  2AND [H+] x 4 so zero order wrt [H+] OR (expt 3 and 4) [H2O2]  2, [H+]  4 initial rate  2 so 1st order wrt [H2O2] M4: rate = k [H2O2] [I–] OR rate = k [H2O2] [I–] [H+]0 4 3(b)(ii) M1: k = 2.42  10–3 ÷ (0.045  0.030) = 1.79 min 2sf M2: units = dm3 mol-1 s-1 2 3(c)(i) M1: calculation of one t1/2 = 60 s ± 5 s M2: two t1/2 calculated that are constant 2 3(c)(ii) k = In2 ÷ 60 = 0.0116 OR k = 0.693 ÷ 60 = 0.0116 min 2sf 1 3(d) increases k and increases the rate of reaction 1 4(a) M1: potential difference / voltage / EMF AND between two half-cells / two electrodes (in a cell) M2: (at concentration of) 1 mol dm–3 AND (pressure of) 1 atm / 101 kPa AND (temperature of) 298 K / 25 °C 2

9701/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 10 of 19 Question Answer Marks 4(b)(i) M1: salt bridge labelled AND voltmeter / V AND wire to electrode and liquid level shown [complete circuit] M2: Cu(s) AND Cu2+(aq) M3: Zn(s) AND Zn2+(aq) 3 4(b)(ii) ions AND electrons 1 4(b)(iii) M1: Nernst equation E = Eo + (0.059/z) log(Zn2+/Zn) M2: E = –0.76 + (0.059 / 2) log(0.25) = –0.778 V OR E = –0.76 + (8.31  298) / (96 500  2) ln(0.25) = –0.778 V min 2sf 2 4(c)(i) (from) +4 (to) +3 1 4(c)(ii) Zn + 2MnO2 → ZnO + Mn2O3 1 4(c)(iii) Eo (MnO2 / Mn2O3) = 1.47 – 1.28 = (+)0.19 V 1

9701/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 11 of 19 Question Answer Marks 5(a) 1 5(b) • five 3d orbitals (lines, boxes) in the isolated Cu2+ ion of the same energy • splitting: three higher and two lower d orbitals • energy of all five d orbitals in complex higher than all d orbitals in isolated ion any two [1], all three [2] 2 5(c) M1: more than one (stable) oxidation state / exist in variable oxidation states M2: vacant / empty (d) orbitals are energetically accessible OR vacant / empty (d) orbitals can form dative bonds with ligands 2 5(d) each row [1] 2

9701/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 12 of 19 Question Answer Marks 5(e) M1: M2: (shape) square planar AND (bond angle) 180o 2 5(f)(i) M1: moles MnO4– = 0.075  0.0225 = 1.6875  10–3 M2: moles V2+ = 5 / 3  1.6875  10–3 = 2.8125  10–3 M3: mass V = 50.9  2.8125  10–3 = 0.143 g % of V = 0.143 / 0.250  100 = 57.3 min 2sf 3 5(f)(ii) 2VO3– + 3Zn + 12H+ → 2V2+ + 3Zn2+ + 6H2O M1: 2 : 3 ratio on both sides M2: rest of the equation correct 2 Question Answer Marks 6(a)(i) M1: (Rf value) is distance moved by a component / spot / solute AND divided by distance moved by solvent OR (Rf value) is the ratio of distance moved by a component / spot / solute AND by distance moved by solvent M2: (retention time) is (the time) between injection and detection (of a component) 2

9701/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 13 of 19 Question Answer Marks 6(a)(ii) thin-layer chromatography: polar solvent OR non-polar solvent OR named solvent gas-liquid chromatography: non-volatile liquid OR high boiling point liquid 1 6(b) A is more soluble in the mobile phase OR A has less adsorption to the stationary phase 1 6(c)(i) 1 6(c)(ii) M1: Y (two) singlet(s) M2: Z triplet AND quartet 2

9701/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 14 of 19 Question Answer Marks 7(a) M1: Cl CH2COOH BrCH2COOH CH3COOH CH3CH2OH most acidic least acidic M2: electron withdrawing groups / electronegative groups / negative inductive effect AND weakens O–H bond / more stable anion OR electron donating groups / positive inductive effect AND strengthens O-H bond / less stable anion M3 M4: • chlorine is more electron-withdrawing / electronegative than bromine (related to XCH2COOH) • electron withdrawing of C=O / negative inductive effect of C=O (related to COOH) • positive inductive effect / electron donating of alkyl / ethyl / R group (related to ROH) any two [1], all three [2] 4 7(b) M1: M2: CO2 2

9701/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 15 of 19 Question Answer Marks 7(c)(i) M1: correct displayed amide bond with C=O, (C6H5)−N and (C–)C=O M2: rest of the structure correct with continuation bonds 2 7(c)(ii) 1 7(d)(i) M1: M2: 2

9701/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 16 of 19 Question Answer Marks 7(d)(ii) HNO2 (and HCl) AND ⩽10 °C OR NaNO2 AND HCl AND ⩽10 °C 1 Question Answer Marks 8(a)(i) NO2 group directs to 3 (and 5) / meta position AND due to being an electron-withdrawing / electronegative group 1 8(a)(ii) HNO3 + H2SO4 → NO2+ + HSO4– + H2O OR HNO3 + 2H2SO4 → NO2+ + 2HSO4– + H3O+ 1 8(b)(i) 1 8(b)(ii) M1: (reaction 1) ethanoyl chloride / CH3COCl AND Al Cl 3 M2: (reaction 2) 2-bromopropane / (CH3)2CHBr AND FeBr3 2 8(c) M1: with C6H5Br no change / no precipitate AND with C6H5CH2Br cream precipitate in C6H5Br M2: lone pair / p-orbital on Br delocalised / overlaps with ring /  system M3: C–Br bond stronger / has partially double bond character 3

9701/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 17 of 19 Question Answer Marks 8(d)(i) (nucleophilic) addition–elimination 1 8(d)(ii) M1 M2: • lone pair on O • curly arrow from (lone pair) O (in H2O) to C (of C=O) • correct dipole on C=O • curly arrow from the C=O bond to O atom any two [1], all four [2] M3: correct intermediate M4: curly arrow from (lone pair on) O(–) to C–O bond AND curly arrow from C–Br to Br 4 Question Answer Marks 9(a) M1: (basicity linked to) ability of lone pair / p-orbital to AND accept / coordinate with a proton / H+ M2: (lone pair of) electrons on N is delocalised into C=O group (and make amides neutral) 2

9701/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 18 of 19 Question Answer Marks 9(b)(i) M1: M M2: N 2 9(b)(ii) LiAl H4 1 9(c) ketone / carbonyl AND secondary / 2o 1 9(d)(i) curly arrow 1: from O lone pair to C atom (in C=C) curly arrow 2: from C=C bond to the right-hand O atom in O=O curly arrow 3: from O=O bond to O+ 2

9701/43 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 19 of 19 Question Answer Marks 9(d)(ii) 1