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

This document has 24 pages. [Turn over Cambridge International AS & A Level DC (EV/CT) 359428/3 © UCLES 2025 CHEMISTRY 9701/44 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. * 7 1 2 5 2 8 2 7 6 1 * ,










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2 9701/44/O/N/25 © UCLES 2025 1 (a) Define a transition element. … … … [1] (b) The 3d orbitals in an isolated gaseous Cu2+ ion are degenerate. (i) Define the term degenerate. … … [1] (ii) Complete the electronic configuration of Cu2+. 1s2 … [1] (c) (i) State the colours of the aqueous solutions for the two copper(II) complex ions shown. • [Cu(NH3)4(H2O)2]2+(aq) … • [CuCl 4]2–(aq) … [1] (ii) Explain why aqueous complex ions of transition elements are usually coloured. … … … … … … [3] (d) (i) When an excess of NH3(aq) is added to a solution of [CuCl 4]2–(aq), [Cu(NH3)4(H2O)2]2+(aq) is formed. State the type of reaction. Complete the equation for this reaction. State symbols are not required. type of reaction … equation [CuCl 4]2– + … [Cu(NH3)4(H2O)2]2+ + … [2] * 0000800000002 * ,










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3 9701/44/O/N/25 © UCLES 2025 [Turn over (ii) The [Cu(NH3)4(H2O)2]2+ complex ion shows stereoisomerism. Complete the three‑dimensional diagrams in Fig. 1.1 to show the two different stereoisomers of [Cu(NH3)4(H2O)2]2+. Cu Cu isomer 1 isomer 2 Fig. 1.1 [2] (iii) Deduce which stereoisomer in (d)(ii) is polar. Explain your answer. polar isomer … explanation … … [1] (e) The dianion P can act as a tridentate ligand. P N O O O O H – – (i) Suggest how P can form three dative covalent bonds. … … … [1] (ii) 2 moles of dianion P, C4H5NO4 2–, react with 1 mole of aqueous cobalt(III) ions, [Co(H2O)6]3+ to form 1 mole of complex ion Q. Deduce the formula and charge of Q. … [1] * 0000800000003 * ,










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4 9701/44/O/N/25 © UCLES 2025 (f) Table 1.1 shows values for the stability constants, Kstab, of some silver(I) complexes. Table 1.1 complex value of Kstab [Ag(CN)2]–(aq) 1.1 × 1018 [Ag(NH3)2]+(aq) 1.2 × 107 [Ag(S2O3)2]3–(aq) 2.9 × 1013 (i) Define the stability constant of a complex. … … … [1] (ii) Use the information in Table 1.1 to identify the most stable silver(I) complex. Explain your answer. most stable … explanation … … [1] * 0000800000004 * ,










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5 9701/44/O/N/25 © UCLES 2025 [Turn over (g) Sodium sulfite, Na2SO3, is used as a food preservative. A 3.75 g sample of impure Na2SO3 is dissolved in distilled water and made up to 250 cm3 in a volumetric flask. 10.0 cm3 of this solution requires 18.70 cm3 of acidified 0.0150 mol dm–3 MnO4 –(aq) to reach the end‑point. The equation for the reaction is shown. 2MnO4 – + 5SO3 2– + 6H+ 2Mn2+ + 5SO4 2– + 3H2O Calculate the percentage by mass of Na2SO3 in the sample. percentage by mass of Na2SO3 = … [3] [Total: 19] * 0000800000005 * ,










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6 9701/44/O/N/25 © UCLES 2025 2 (a) The Group 2 sulfates and the Group 2 chromates show similar trends in solubility. Suggest the trend in the solubility of the Group 2 chromates down the group. Explain your answer. … … … … … … [4] (b) Silver(I) chromate, Ag2CrO4, is sparingly soluble in water. (i) Write an ionic equation to show the equilibrium between solid Ag2CrO4 and its aqueous solution. Include state symbols. … [1] (ii) The value of the solubility product, Ksp, of Ag2CrO4 is 1.12 × 10–12 at 298 K. Calculate the equilibrium concentration of Ag+, in mol dm–3, in a saturated solution of Ag2CrO4 at 298 K. equilibrium concentration of Ag+ = … mol dm–3 [3] * 0000800000006 * ,










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7 9701/44/O/N/25 © UCLES 2025 [Turn over (c) The hydrogenchromate ion, HCrO4 –, is a weak acid. The pKa of HCrO4 – is 6.49. (i) Calculate the pH of a 0.0250 mol dm–3 HCrO4 – solution. pH = … [2] (ii) HCrO4 – can show amphoteric behaviour. State the formula of: • the conjugate acid of HCrO4 – … • the conjugate base of HCrO4 –. … [1] * 0000800000007 * ,










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8 9701/44/O/N/25 © UCLES 2025 (d) Table 2.1 shows some energy changes. Table 2.1 energy change value / kJ mol–1 first ionisation energy of silver +731 second ionisation energy of silver +2074 first ionisation energy of sulfur +1000 second ionisation energy of sulfur +2251 first electron affinity of sulfur –200 second electron affinity of sulfur +532 enthalpy change of atomisation of sulfur +279 enthalpy change of formation of silver(I) sulfide, Ag2S(s) –33 lattice energy of silver(I) sulfide, Ag2S(s) –2677 (i) Define the term first electron affinity. … … … [1] (ii) Explain why the value for the second electron affinity of sulfur is positive. … … … [1] (iii) Construct an equation for the lattice energy of Ag2S. Include state symbols. … [1] * 0000800000008 * ,










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9 9701/44/O/N/25 © UCLES 2025 [Turn over (iv) Calculate the enthalpy change of atomisation, ΔHat, in kJ mol–1, of silver using relevant data from Table 2.1. It may be helpful to draw a labelled Born–Haber cycle. Show your working. ΔHat of silver = … kJ mol–1 [3] (e) Suggest how the magnitude for the lattice energy of Ag2S(s) differs from the lattice energy of Cu2S(s). Explain your answer. … … … [1] [Total: 18] * 0000800000009 * ,










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10 9701/44/O/N/25 © UCLES 2025 3 (a) Define the term entropy. … … [1] (b) (i) Place one tick (3) in each row of Table 3.1 to show the sign of the entropy change, ΔS, for each process. Table 3.1 process ΔS is negative ΔS is positive steam condensing into water solid KCl dissolving in water [1] (ii) Chlorine trifluoride, Cl F3, decomposes on heating into its elements, as shown. reaction 1 2Cl F3(g) Cl 2(g) + 3F2(g) Standard entropies are shown in Table 3.2. Table 3.2 substance Cl F3(g) Cl 2(g) F2(g) S ⦵ / J K–1 mol–1 +281.6 +223.1 +203.0 Calculate the standard entropy change, ΔS ⦵, in J K–1 mol–1, for reaction 1. ΔS ⦵ for reaction 1 = … J K–1 mol–1 [2] (c) Group 2 carbonates decompose on heating. The decomposition for one of the Group 2 carbonates, MCO3, is shown in reaction 2. reaction 2 MCO3(s) MO(s) + CO2(g) (i) Predict the sign of the entropy change, ΔS, for reaction 2. Explain your answer. … … [1] * 0000800000010 * ,










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11 9701/44/O/N/25 © UCLES 2025 [Turn over (ii) The Gibbs equation is shown. ΔG ⦵ = ΔH ⦵ – TΔS ⦵ Fig. 3.1 shows values of the Gibbs free energy change, ΔG ⦵, in kJ mol–1, at different temperatures, T, in K, for reaction 2. Assume ΔH ⦵ and ΔS ⦵ values for this reaction remain constant over this temperature range. –100 –50 0 50 100 150 200 0 200 400 600 800 1000 1200 1400 1600 ΔG ⦵ / kJ mol–1 T / K Fig. 3.1 Use the gradient and intercept on the y‑axis in Fig. 3.1 and the Gibbs equation to determine: • ΔS ⦵, in J K–1 mol–1, for reaction 2 • the minimum temperature, T, in K, at which the reaction is feasible • ΔH ⦵, in kJ mol–1, for reaction 2. ΔS ⦵ for reaction 2 = … J K–1 mol–1 minimum temperature, T = … K ΔH ⦵ for reaction 2 = … kJ mol–1 [4] [Total: 9] * 0000800000011 * ,









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12 9701/44/O/N/25 © UCLES 2025 4 (a) Nitrogen monoxide, NO, reacts with hydrogen, as shown in reaction 3. reaction 3 2NO + 2H2 N2 + 2H2O (i) The rate equation for reaction 3 is shown. rate = k[H2][NO]2 Complete Table 4.1. Table 4.1 the order of reaction with respect to [H2] the order of reaction with respect to [NO] the overall order of the reaction [1] (ii) Predict how the initial rate for reaction 3 changes when the concentration of NO is halved. … [1] (iii) Predict how the initial rate for reaction 3 changes when the concentrations of NO and H2 are both increased three times. … [1] (iv) Suggest why reaction 3 is unlikely to proceed by a mechanism involving only a single step. … … [1] (v) Suggest equations for the three steps of the reaction mechanism for reaction 3. Each step involves a reaction between two molecules. step 1 … … step 2 … + … N2O + … step 3 N2O + … … + … [2] (vi) Suggest the role of N2O in this mechanism. Explain your reasoning. … … [1] * 0000800000012 * ,









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13 9701/44/O/N/25 © UCLES 2025 [Turn over (b) Iodine, I2, reacts with thiosulfate ions, S2O3 2–, as shown in reaction 4. reaction 4 2S2O3 2– + I2 S4O6 2– + 2I– Reaction 4 is carried out in the presence of a large excess of I2. Under these conditions, the reaction is first order with respect to [S2O3 2–] and zero order with respect to [I2]. The half‑life, t1 2 , for reaction 4 is 720 s under certain conditions. Calculate the value of the rate constant, k, for reaction 4. Include the units of k. k = … units … [1] (c) The reaction between iodide ions, I–(aq), and peroxydisulfate ions, S2O8 2–(aq), is catalysed by Co3+(aq). The mechanism is similar to the mechanism of this reaction when Fe3+(aq) is used as the catalyst. (i) State the type of catalysis that occurs in this reaction. Explain your reasoning. … … [1] (ii) Write two equations to show how Co3+(aq) catalyses this reaction. equation 1 … equation 2 … [2] (iii) Suggest why this reaction is slow in the absence of Co3+(aq). … … [1] [Total: 12] * 0000800000013 * ,









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14 9701/44/O/N/25 © UCLES 2025 5 (a) Describe and explain the shape of benzene. In your answer, include: • the shape and bond angle in the ring • the hybridisation of the carbon atoms • how orbital overlap forms σ and π bonds between the carbon atoms in the ring. … … … … … … … … [4] (b) Fig. 5.1 shows two reactions of benzoic acid. O COOH COOH COOH reaction 5 reaction 6 Fig. 5.1 (i) Suggest reagents and conditions for reaction 5 and for reaction 6 in Fig. 5.1. reaction 5 … reaction 6 … [2] (ii) State the type of reaction for reaction 5 in Fig. 5.1. … [1] * 0000800000014 * ,









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15 9701/44/O/N/25 © UCLES 2025 [Turn over (c) In the electrophilic substitution of arenes, different substituents can direct to different ring positions. (i) Describe the directing effect of the –CH2CH3 group. Explain your answer. … … [1] (ii) The alkylation of arenes uses a mixture of CH3CH2Br and FeBr3 to generate the CH3CH2 + electrophile. Write an equation for the formation of the CH3CH2 + electrophile. … [1] (iii) Complete the mechanism in Fig. 5.2. Include all relevant curly arrows and charges. Draw the structure of the organic intermediate. COOH COOH +CH2CH3 + … organic intermediate Fig. 5.2 [3] (iv) Write an equation to show how FeBr3 is regenerated after the reaction in Fig. 5.2. … [1] [Total: 13] * 0000800000015 * ,









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16 9701/44/O/N/25 © UCLES 2025 6 (a) Compound Z is used in organic synthesis. Z O NH2 CN H2N Complete Table 6.1 to show the number of sp, sp2 and sp3 hybridised carbon atoms present in one molecule of Z. Table 6.1 type of hybridisation sp sp2 sp3 number of carbon atoms [1] (b) Z can undergo different reactions, as shown in Fig. 6.1. Z O NH2 CN H2N excess HCl (aq) heat CH3COCl excess LiAl H4 reaction 7 reaction 8 reaction 9 Fig. 6.1 (i) Name the two types of reaction occurring in reaction 7 in Fig. 6.1. … and … [1] (ii) Draw the structures of the organic products of reactions 7, 8 and 9 in Fig. 6.1. [4] * 0000800000016 * ,









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17 9701/44/O/N/25 © UCLES 2025 [Turn over (c) Compound Z is dissolved in D2O and analysed by carbon‑13 NMR and proton (1H) NMR spectroscopy. (i) Predict the number of peaks in the carbon‑13 NMR spectrum of Z. … [1] (ii) The proton (1H) NMR spectrum of Z in D2O gives three peaks for the proton environments, labelled a, b and c, as shown on Fig. 6.2. Z O NH2 CN H2N C CH CH2 CH2 a b c Fig. 6.2 Complete Table 6.2 for the proton (1H) NMR spectrum of Z in D2O. Table 6.2 proton environment a b c name of splitting pattern chemical shift range, δ / ppm [2] Table 6.3 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 nitrile –CH2–CN 2.0–3.0 alkyl next to electronegative atom CH3–O, –CH2–O, –CH2–N 3.2– 4.0 attached to alkene =CHR 4.5–6.0 alkyl amine R–NH– 1.0–5.0 amide RCONHR 5.0–12.0 [Total: 9] * 0000800000017 * ,









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18 9701/44/O/N/25 © UCLES 2025 7 Phenylmethanol and 4‑methylphenol are isomers. CH2OH OH phenylmethanol 4-methylphenol (a) Complete Table 7.1 to show the relative acidities of benzoic acid (C6H5COOH), phenylmethanol, 4‑methylphenol and water. Explain your answer. Table 7.1 name of compound most acidic least acidic explanation … … … … … … … … [4] (b) 4‑methylphenol reacts readily with sodium. Complete the equation for this reaction. OH + … [1] * 0000800000018 * ,









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19 9701/44/O/N/25 © UCLES 2025 [Turn over (c) Under certain conditions, ethane‑1,2‑diol, HOCH2CH2OH, reacts with propane‑1,3‑dioic acid, HOOCCH2COOH, to form different organic products, as shown in Fig. 7.1. O O HO HO OH OH + reaction 10 reaction 11 polymer Y X, C5H6O4 Fig. 7.1 (i) X does not react with Na metal. Draw the structure of the organic product X, C5H6O4, shown in Fig. 7.1. [1] (ii) Reactions 10 and 11 in Fig. 7.1 are different types of reaction. Name the type of reaction for reaction 10 and for reaction 11. reaction 10 … reaction 11 … [1] (iii) Draw a section of polymer Y showing only one repeat unit. The new functional group formed should be displayed. [2] [Total: 9] * 0000800000019 * ,









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20 9701/44/O/N/25 © UCLES 2025 8 (a) Describe the difference in reactivity between ethanoyl chloride and chlorobenzene with water. Explain your answer. … … … … … … [2] (b) The structure of compound V is shown. V CN O O O N H O (i) Name all the functional groups in V. … … … [2] (ii) Deduce the number of possible optical isomers for V. … [1] (iii) Suggest one reason, other than better biological activity and lower dosage required, why it is beneficial to synthesise a single optical isomer of V for use as a drug. … … … [1] * 0000800000020 * ,










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21 9701/44/O/N/25 © UCLES 2025 [Turn over (c) A sample of V is hydrolysed with an excess of hot aqueous alkali. The products are isolated from the reaction mixture at pH 12. Draw the structures of the two organic products of the complete alkaline hydrolysis of V in Fig. 8.1. Fig. 8.1 [3] * 0000800000021 * ,









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22 9701/44/O/N/25 © UCLES 2025 (d) A polypeptide formed from four amino acids, A, B, C and D, is completely hydrolysed and then analysed by gas–liquid chromatography. The chromatogram produced is shown in Fig. 8.2. intensity time / minutes A B C D 28 58 13 42 Fig. 8.2 The number above each peak represents the area under the peak. The area under each peak is proportional to the mass of the respective amino acid in the mixture. (i) Calculate the percentage by mass of amino acid A in the original mixture. percentage by mass of amino acid A = … [1] (ii) The retention time for amino acid D is the longest. Explain why D has a longer retention time than the other amino acids A, B and C. … … … … [1] [Total: 11] * 0000800000022 * ,









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23 9701/44/O/N/25 © UCLES 2025 Important values, constants and standards molar gas constant R = 8.31 J K–1 mol–1 Faraday constant F = 9.65 × 104 C mol–1 Avogadro constant L = 6.02 × 1023 mol–1 electronic charge e = –1.60 × 10–19 C molar volume of gas Vm = 22.4 dm3 mol–1 at s.t.p. (101 kPa and 273 K) Vm = 24.0 dm3 mol–1 at room conditions ionic product of water Kw = 1.00 × 10–14 mol2 dm–6 (at 298 K (25 °C)) specific heat capacity of water c = 4.18 kJ kg–1 K–1 (4.18 J g–1 K–1) * 0000800000023 * ,









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24 9701/44/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. * 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 9701/44 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.

9701/44 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 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 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.

9701/44 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 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 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.

9701/44 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 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.

9701/44 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 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 standard isation and monitoring processes and guide the supervising examiners when they are checking the work of examiners within their team. The meaning of annotations and how they are used is specific to each component and is understood by all examiners who mark the component. We publish annotations in our mark schemes to help centres understand the annotations they may see on copies of scripts. Note that there may not be a direct correlation between the number of annotations on a script and the mark awarded. Similarly, the use of an annotation may not be an indication of the quality of the response. The annotations listed below were available to examiners marking this component in this series. Annotations Annotation Meaning Correct point or mark awarded Incorrect point or mark not awarded Unclear Information missing or insufficient for credit Benefit of the doubt given Contradiction in response otherwise markworthy, mark not given Part of the correct answer has been seen. Full credit has not been awarded. Error carried forward applied Incorrect or insufficient point ignored while marking the rest of the response Rounding error

9701/44 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 6 of 16 Annotation Meaning Repetition Blank page or part of script seen Error in number of significant figures Transcription error

9701/44 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 7 of 16 Question Answer Marks 1(a) forms (one or more) stable ions / compounds / oxidation states with incomplete / partially filled (3)d orbital(s) / d shell / d sub-shell 1 1(b)(i) (orbitals) are at the same energy 1 1(b)(ii) (1s2) 2s2 2p6 3s2 3p6 3d9 1 1(c)(i) [Cu(NH3)4(H2O)2]2+(aq) deep / dark / royal blue AND [CuCl4]2–(aq) yellow 1 1(c)(ii) M1: d orbital(s) of different energy / d-d splitting occurs / d sub-shell splits M2: electron(s) promoted / excited M3: light / wavelength / frequency / photon absorbed AND complementary colour seen 3 1(d)(i) M1: ligand exchange / replacement / substitution / displacement M2: [CuCl4]2– + 4NH3 + 2H2O → [Cu(NH3)4(H2O)2]2+ + 4Cl– 2 1(d)(ii) M1: one correct 3D structure of [Cu(H2O)2(NH3)4]2+ M2: correct 3D stereoisomer structure of isomer 1 2 1(d)(iii) cis isomer AND dipoles do not cancel 1 1(e)(i) (two) oxygen and (one) nitrogen (can donate) three lone pair(s) of electrons (to the metal ion) 1

9701/44 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 8 of 16 Question Answer Marks 1(e)(ii) [Co(C4H5NO4)2]– 1 1(f)(i) equilibrium constant for the formation of the complex (ion) in a solvent / water / solution OR from its constituent ions or molecules 1 1(f)(ii) [Ag(CN)2]–(aq) AND largest (value of) Kstab 1 1(g) M1 M2: any two [1], all four [2] • moles MnO4– = 0.015 ×18.70/1000 = 2.805 × 10–4 • moles SO32– = 2.805 × 10–4 × 5/2 = 7.0125 × 10–4 (in 10 cm3) • moles SO32– = 1.753 × 10–2 (in 250 cm3) • mass Na2SO3 = 1.753 × 10–2 × 126.1 = 2.21 g M3: % purity = 100 × 2.21 / 3.75 = 58.9 – 59.0 % min 2sf 3 Question Answer Marks 2(a) M1: (solubility) decreases down the group M2: Hlatt and Hhyd decrease / become less exothermic / less negative M3: Hhyd decreases / changes more / dominant factor / becomes less exothermic by a larger extent OR Hlatt decreases / changes less / changes slower M4: Hsol becomes less exothermic / less negative OR Hsol becomes (more) endothermic / (more) positive 4 2(b)(i) Ag2CrO4(s) ⇌ 2Ag+(aq) + CrO42–(aq) 1 2(b)(ii) M1: Ksp = [Ag+]2 [CrO42-] M2: √(1.12× 10−12 ÷ 4) 3 = 6.54  10–5 M3: [Ag+] = 2  6.54  10–5 = 1.31  10–4 (mol dm–3) min 2sf 3

9701/44 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 9 of 16 Question Answer Marks 2(c)(i) Ka = 10–6.49 = 3.236  10–7 [H+]2 = 8.08984  10–9 M1: [H+] = 8.99  10–5 M2: pH = –log(8.99  10–5) = 4.05 min 2sf 2 2(c)(ii) conjugate acid H2CrO4 AND conjugate base CrO42– 1 2(d)(i) (the energy change / released when) one mole of gaseous atoms become 1– ions / gain one mole of electrons 1 2(d)(ii) due to repulsion between negative ion and incoming / gained electron 1 2(d)(iii) 2Ag+(g) S2–(g) → Ag2S(s) 1 2(d)(iv) M1: selection of correct six numbers only: –33, 731, 279, –200, 532, –2677 M2: use of  2 as only multiplier with Ag M3: correct signs and evaluation of data –33 = 2Hat + (2  731) + 279 + (–200) + 532 + (–2677) 2Hat = 571 (kJ mol –1) Hat = +285.5 (kJ mol –1) 3 2(e) Ag2S smaller / less negative / less exothermic lattice neergy AND larger ionic radius (of cation) 1

9701/44 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 10 of 16 Question Answer Marks 3(a) number of possible arrangements of particles and energy in a system 1 3(b)(i) process S is negative S is positive steam condensing into water ✓ solid KCl dissolving into water ✓ 1 3(b)(ii) M1: three numbers and correct multipliers used M2: correct signs and evaluation So = (3  203.0) + (223.1) – (2  281.6) = (+)268.9 (J K–1 mol–1) 2 3(c)(i) positive AND gas produced / on the right OR less / no gas on left 1 3(c)(ii) M1 and M2: calculation of gradient from the graph for S gradient = (–)140 / 900 = (–)0.1556 So = (–)0.1556  1000 = (+)155.6 ± 5 (J K–1 mol–1) M3: T = 1120 ± 5 (K) M4: Ho = (+)172 ± 10 (kJ mol–1) 4

9701/44 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 11 of 16 Question Answer Marks 4(a)(i) the order of reaction with respect to [H2] 1 the order of reaction with respect to [NO] 2 overall order of the reaction 3 1 4(a)(ii) rate  1 4 1 4(a)(iii) rate  27 1 4(a)(iv) a four-particle collision is unlikely 1 4(a)(v) step 1 2NO → N2O2 step 2 N2O2 + H2 → N2O + H2O step 3 N2O + H2 → N2 + H2O 2 4(a)(vi) intermediate formed (step 2) then used up later (step 3) 1 4(b) k = 0.693 / 720 = 9.625  10–4 min 2sf AND units: s–1 1 4(c)(i) homogeneous AND Co3+ / catalyst / it is in the same phase / same state as the reactants 1 4(c)(ii) M1: (equation 1) 2Co3+ + 2I– → 2Co2+ + I2 M2: (equation 2) 2Co2+ + S2O82– → 2Co3+ + 2SO42– 2 4(c)(iii) repulsion of two negative / same charge ions slows the reaction / raises EA 1

9701/44 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 12 of 16 Question Answer Marks 5(a) M1 and M2: any two [1], all three [2] • (hexagonal ring) planar / (trigonal) planar • 120o • sp2 hybridised M3: p orbitals overlap sideways / laterally (with each other above and below the ring) forming  bonds M4:  bonds form when sp2 / hybridised orbitals overlap end-on-end / head on 4 5(b)(i) M1: (reaction 5) Pt / Ni + H2 (+ heat) M2: (reaction 6) (CH3)2CHCOCl + Al Cl 3 (+ heat) 2 5(b)(ii) reduction / hydrogenation 1 5(c)(i) (CH2CH3 group) directs to 2-, 4- (and 6-) positions AND due to being an electron donating group 1 5(c)(ii) CH3CH2Br + FeBr3 → CH3CH2+ + FeBr4– 1 5(c)(iii) M1: curly arrow from inside / touching the hexagon to (CH2) group / C+ charge M2: structure of the intermediate M3: curly arrow from C-H bond into / onto the ring AND formation / loss of H+ 3 5(c)(iv) H+ + FeBr4– → HBr + FeBr3 1 organic intermediate

9701/44 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 13 of 16 Question Answer Marks 6(a) sp = 1 AND sp2 = 1 AND sp3 = 3 1 6(b)(i) acid-base / neutralisation AND hydrolysis 1 6(b)(ii) any two * [1], any three * [2], any five * [3], all six * [4] 4 6(c)(i) five / 5 1 6(c)(ii) proton environment a b c name of splitting pattern doublet multiplet doublet chemical shift range,  / ppm 2.2–3.0 3.2– 4.0 2.0–3.0 any three [1], all six [2] 2 reaction 7 reaction 8 reaction 9 * * * * * * * *

9701/44 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 14 of 16 Question Answer Marks 7(a) M1: (most acidic) benzoic acid 4-methylphenol water (least acidic) phenylmethanol M2: correct link of acidity to weakens O—H / H+ more easily lost / anion stabilised / conjugate base stabilised M3 and M4: any two [2] • for benzoic acid: negative inductive effect of C=O / electronegative C=O / two electronegative O / –ve charge delocalised (across two O) / resonance in –COO– • for 4-methylphenol: as lone pair on oxygen delocalised into the ring /  system / delocalised system OR p orbital on oxygen overlaps with ring /  system / delocalised system • for phenylmethanol: positive inductive effect of alkyl group / R / CH2 4 7(b) 1 7(c)(i) 1 7(c)(ii) (reaction 10) addition–elimination / condensation / esterification AND (reaction 11) addition–elimination / condensation (polymerisation) / esterification 1

9701/44 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 15 of 16 Question Answer Marks 7(c)(iii) M1: ester linkage shown M2: rest of the structure correct with continuation bonds 2 Question Answer Marks 8(a) • chlorobenzene is less reactive than ethanoyl chloride • p orbital / lone pair on Cl / –Cl will overlap / delocalise into the ring OR C of COCl has most electron deficient OR has an electronegative oxygen atom / two electronegative atoms / electron withdrawing C=O group • due to partial double C–Cl bond OR C–Cl bond strengthened (more) / stronger bond between Cl and benzene / ring OR C–Cl weakened in ROCl any two [1], all three [2] 2 8(b)(i) • nitrile • amide • ketone / carbonyl • ester any two [1], all four [2] 2 8(b)(ii) 8 / eight 1 8(b)(iii) no need to separate optical isomers 1

9701/44 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2025 © Cambridge University Press & Assessment 2025 Page 16 of 16 Question Answer Marks 8(c) M1: amide bond hydrolysed and COO– and NH2 formed M2: nitrile bond hydrolysed and COO– formed M3: ester bond hydrolysed and COO– and OH formed 3 8(d)(i) percentage = 100  28 / (28 + 58 + 13 + 42) = 19.9 (%) min 2sf 1 8(d)(ii) D is attracted more strongly to the stationary phase / forms stronger intermolecular forces with the stationary phase 1