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

This document has 24 pages. Any blank pages are indicated. [Turn over Cambridge International AS & A Level * 9 2 9 1 9 3 9 0 0 2 * DC (CE/SG) 357257/1 © UCLES 2025 CHEMISTRY 9701/44 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. ,










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3 9701/44/M/J/25 © UCLES 2025 [Turn over 1 (a) (i) Describe the trend in the thermal stabilities of the carbonates of the Group 2 elements. Explain your answer. … … … … … [3] (ii) Copper(II) carbonate decomposes on heating in a similar way to the carbonates of Group 2. Write an equation for the decomposition of copper(II) carbonate. … [1] (b) (i) Complete the electrons in boxes diagram in Fig. 1.1 to show the electronic configuration of a copper(II) ion. 3d [Ar] 4p 4s Fig. 1.1 [1] (ii) There are five different 3d orbitals. Sketch the shape of a 3dz2 orbital in Fig. 1.2. z y x Fig. 1.2 [1] (iii) Copper can form stable complexes in the +1 and +2 oxidation states. Explain why transition elements have variable oxidation states. … … [1] * 0000800000003 * ,










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4 9701/44/M/J/25 © UCLES 2025 (c) (i) 1,2-diaminoethane, H2NCH2CH2NH2, en, can act as a bidentate ligand. Explain what is meant by a bidentate ligand. … … … [2] (ii) The complex [Cu(H2O)2(en)2]2+ exists as stereoisomers. Complete the three-dimensional diagrams in Fig. 1.3 to show the three different stereoisomers of [Cu(H2O)2(en)2]2+. The en ligand can be represented using N N . Cu isomer 1 Cu isomer 2 Cu isomer 3 Fig. 1.3 [3] (iii) State the different types of stereoisomerism shown by [Cu(H2O)2(en)2]2+. … [1] * 0000800000004 * ,










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5 9701/44/M/J/25 © UCLES 2025 [Turn over (iv) Identify one isomer in (c)(ii) that is polar. Explain your answer. isomer … … … [1] (d) (i) The mineral ore cryolite, Na3Al F6, contains a single anion which is a complex ion. Complete Table 1.1 to suggest the formula of the complex ion and to identify the ligand present in Na3Al F6. Table 1.1 complex ion in Na3Al F6 ligand in Na3Al F6 [1] (ii) When a solution of Al 2O3 in molten cryolite is electrolysed, aluminium metal is formed at the cathode. The equation is shown. Al 3+ + 3e– Al Calculate the maximum mass of aluminium produced when a current of 1.5 A is passed through this solution for 30 minutes. Give your answer to two significant figures. mass of aluminium = … g [4] [Total: 19] * 0000800000005 * ,










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6 9701/44/M/J/25 © UCLES 2025 2 (a) Anhydrous barium chloride can be obtained from the hydrated salt, as shown in reaction 1. reaction 1 BaCl 2•2H2O(s) + 2SOCl 2(l) BaCl 2(s) + 2SO2(g) + 4HCl (g) (i) Describe one observation when reaction 1 is carried out. … … [1] (ii) Define the term entropy, S. … … [1] (iii) The entropy change, ΔS o, for reaction 1 at 25 °C is +768 J K–1 mol–1. Explain why ΔS o has a large positive value. … … [1] (iv) Table 2.1 shows the enthalpy changes of formation, ΔH o f , for the compounds in reaction 1. Table 2.1 compound ΔH o f / kJ mol–1 BaCl 2(s) –859 BaCl 2•2H2O(s) –1460 SOCl 2(l) –246 SO2(g) –297 HCl (g) –92 Calculate the standard Gibbs free energy change, ΔGo, in kJ mol–1, for reaction 1 at 25 °C. ΔGo = … kJ mol–1 [3] * 0000800000006 * ,










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7 9701/44/M/J/25 © UCLES 2025 [Turn over (b) When aqueous solutions of BaCl 2 and Na2Cr2O7 are mixed, a yellow precipitate of BaCrO4(s) is produced and an acidic solution remains. (i) Write the ionic equation for this reaction. … [1] (ii) Explain why BaCrO4(s) is coloured. … … … … … … [3] (c) Barium sulfate is the least soluble of the Group 2 sulfates. Explain the trend in the solubilities of the Group 2 sulfates. … … … … … … [3] [Total: 13] * 0000800000007 * ,










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8 9701/44/M/J/25 © UCLES 2025 3 (a) Nickel(II) iodate(V), Ni(IO3)2, is sparingly soluble in water. The concentration of its saturated solution is 2.30 × 10–2 mol dm–3 at 298 K. (i) Complete the expression for the solubility product, Ksp, of Ni(IO3)2. Include the units. Ksp = units = … [2] (ii) Calculate the numerical value for Ksp of Ni(IO3)2 at 298 K. Ksp = … [1] (b) An electrochemical cell is set up as shown in Fig. 3.1. Pt Ni(IO3)2(aq) and I2(aq) Ni(IO3)2(s) CuCl 2(aq) Cu(s) (s) V Fig. 3.1 The relevant standard electrode potentials, E o , for this electrochemical cell are shown. IO3 –(aq) + 6H+(aq) + 5e– 1 2 I2(aq) + 3H2O(l) E o = +1.19 V Cu2+(aq) + 2e– Cu(s) E o = +0.34 V (i) Use this information to calculate the value of E o cell. State which electrode is positive. E o cell = … positive electrode is … [1] (ii) Suggest how the measured Ecell of this cell compares to the E o cell under standard conditions. Explain your answer. … … [1] * 0000800000008 * ,










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9 9701/44/M/J/25 © UCLES 2025 [Turn over (iii) Complete Table 3.1 by placing one tick (3) to indicate how the Ecell of this cell changes when a small amount of NiSO4(aq) is added to the beaker containing Ni(IO3)2(aq) and I2(aq) in Fig. 3.1. Explain your answer. Table 3.1 change in Ecell less positive no change more positive … … … [2] (c) In solution, iodic(V) acid, HIO3, ionises as shown. HIO3(aq) IO3 –(aq) + H+(aq) The pH of a 1.0 mol dm–3 solution of HIO3 is 0.47. (i) Calculate [H+(aq)], in mol dm–3, in a 1.0 mol dm–3 solution of HIO3. [H+(aq)] = … mol dm–3 [1] (ii) Use your answer from (c)(i) to calculate the equilibrium concentrations of HIO3(aq) and IO3 –(aq), in mol dm–3, in a 1.0 mol dm–3 solution of HIO3. [HIO3(aq)] = … mol dm–3 [IO3 –(aq)] = … mol dm–3 [1] (iii) Use your answers from (c)(i) and (c)(ii) to calculate the Ka, in mol dm–3, of HIO3. Ka = … mol dm–3 [1] * 0000800000009 * ,










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10 9701/44/M/J/25 © UCLES 2025 (d) The Dushman reaction is the reaction between iodate(V) ions and iodide ions in acid solution. IO3 –(aq) + 5I–(aq) + 6H+(aq) 3I2(aq) + 3H2O(l) The rate equation for this reaction is shown. rate = k [IO3 –][I–]2[H+]2 The rate of this reaction is investigated in a buffer solution. The initial concentrations are shown. [IO3 –] = 0.500 mol dm–3 [I–] = 1.00 × 10–3 mol dm–3 [H+] = 1.00 × 10–2 mol dm–3 Under these conditions the initial rate of the reaction is 2.10 × 10–2 mol dm–3 s–1. (i) Define buffer solution. … … … [2] (ii) Use the information to calculate the rate constant, k. State its units. k = … units = … [2] (iii) This reaction is repeated at the same temperature and with the same initial values of [IO3 –] and [I–]. The [H+] is increased to 3.00 × 10–2 mol dm–3. Calculate the initial rate of this reaction. rate = … mol dm–3 s–1 [1] [Total: 15] * 0000800000010 * ,










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11 9701/44/M/J/25 © UCLES 2025 [Turn over BLANK PAGE * 0000800000011 * ,









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12 9701/44/M/J/25 © UCLES 2025 4 Table 4.1 shows the structures of sections of three polymers, X, Y and Z. Each polymer is made from only one type of monomer. Table 4.1 polymer structure of section of polymer X H2 C H2 C H2 C C C O O N H H N C H2 C H2 C H2 Y CH COOCH3 CH COOCH3 CH COOCH3 CH COOCH3 CH COOCH3 C H2 C H2 C H2 C H2 Z O O O O C C O C O CH CH CH3 CH3 CH3 CH (a) Complete Table 4.2 to state the type of polymerisation and draw the structure of the monomer for each polymer, X, Y and Z. Table 4.2 polymer type of polymerisation structure of monomer X Y Z [4] * 0000800000012 * ,









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13 9701/44/M/J/25 © UCLES 2025 [Turn over (b) Amino acids can act as monomers. State what is meant by the isoelectric point of an amino acid. … … [1] (c) Electrophoresis can be used to separate and identify amino acids. Table 4.3 shows information about the three amino acids glycine, lysine and glutamic acid. Table 4.3 amino acid structural formula of amino acid isoelectric point glycine (gly) H2NCH2COOH 6.0 lysine (lys) H2NCH(CH2CH2CH2CH2NH2)COOH 9.7 glutamic acid (glu) H2NCH(CH2CH2COOH)COOH 3.2 (i) A mixture containing these three amino acids is analysed in a buffer solution of pH 6.0. Draw and label three spots on Fig. 4.1 to indicate the predicted position of each of these amino acids, gly, lys and glu, after electrophoresis. – + mixture applied here Fig. 4.1 [2] (ii) Electrophoresis is repeated using a buffer solution of pH 11. Predict how the position of glycine will change, if at all, after electrophoresis. … … [1] [Total: 8] * 0000800000013 * ,









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14 9701/44/M/J/25 © UCLES 2025 5 A group of drugs known as statins are used to lower cholesterol in blood. A commonly used statin is atorvastatin. O OH OH NH OH atorvastatin O N F Fig. 5.1 (a) Statins can break down in the acid found in the stomach. (i) Draw a line through the bond in the atorvastatin structure in Fig. 5.1 that could be broken under acid conditions. [1] (ii) By referring to the structure, explain why atorvastatin dissolves in water. … … [1] (iii) Complete the molecular formula of atorvastatin. C H35N O F [1] (iv) Atorvastatin contains chiral carbon atoms. Circle all chiral carbon atoms in Fig. 5.1. [1] (v) The synthetic preparation of atorvastatin requires the production of a single optical isomer. Suggest why. … … [1] * 0000800000014 * ,









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15 9701/44/M/J/25 © UCLES 2025 [Turn over (b) (i) The proton (1H) NMR spectrum of atorvastatin dissolved in CDCl 3 is recorded. Use Table 5.1 to deduce the number of hydrogen atoms that could produce peaks in the region δ = 6.5–13.0 ppm. … [1] (ii) The proton (1H) NMR spectrum of atorvastatin dissolved in D2O is recorded. Predict the number of hydrogen atoms that would not show a peak in this spectrum. Explain your answer. … … … [2] Table 5.1 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 (c) Atorvastatin reacts with an excess of LiAl H4. Name all the functional groups in atorvastatin that react with LiAl H4. Name the new functional group that would be formed in each case. names of functional groups in atorvastatin that react … names of the new functional groups formed … [2] [Total: 10] * 0000800000015 * ,









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17 9701/44/M/J/25 © UCLES 2025 [Turn over 6 (a) A list of tests for different organic groups is given in Table 6.1. Complete Table 6.1 to identify an organic functional group, in aliphatic compounds, that produces a positive result in each test. Table 6.1 test sodium metal Na2CO3(aq) 2,4-DNPH I2(aq) + OH–(aq) warm with Fehling’s reagent Br2(aq) [4] * 0000800000017 * ,









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18 9701/44/M/J/25 © UCLES 2025 (b) Lavandulol is an aliphatic organic compound and the major component of lavender oil. Fig. 6.1 shows a reaction scheme involving lavandulol, A. lavandulol, A (C10H18O) C (C6H8O5) D (C3H6O) E (C6H10O5) F (C6H8O4) + heat with concentrated acidified KMnO4 + + CO2 heat with concentrated H2SO4 heat with concentrated acidified KMnO4 NaBH4 G (C4H4O5) H (C2H4O2) B (C10H16O) Fig. 6.1 * 0000800000018 * ,









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19 9701/44/M/J/25 © UCLES 2025 [Turn over Table 6.2 shows the results obtained when the tests in Table 6.1 are carried out on the eight organic compounds, A–H, in the reaction scheme in Fig. 6.1. Table 6.2 letter of compound test sodium metal Na2CO3(aq) 2,4-DNPH I2(aq) + OH–(aq) warm with Fehling’s reagent Br2(aq) A 3 ✗ ✗ ✗ ✗ 3 B ✗ ✗ 3 ✗ 3 3 C 3 3 3 3 ✗ ✗ D ✗ ✗ 3 3 ✗ ✗ E 3 3 ✗ 3 ✗ ✗ F 3 3 ✗ ✗ ✗ 3 G 3 3 3 ✗ ✗ ✗ H 3 3 ✗ ✗ ✗ ✗ (i) Deduce the functional group present in compound A using both the molecular formulae of A and B and the reaction of B with Fehling’s reagent. … [1] (ii) Name the type of reaction that occurs in each of the following conversions. • A B … • C E … • E F … [3] (iii) Use the information in Table 6.2 and the molecular formulae to deduce structures for G and H. Draw your structures in Fig. 6.1. [2] (iv) Use the information in Table 6.2, the molecular formulae and your answer to (b)(iii) to deduce structures for A, C, D, E and F. Draw your structures in Fig. 6.1. [5] [Total: 15] * 0000800000019 * ,









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20 9701/44/M/J/25 © UCLES 2025 7 (a) State the relative acidities of benzoic acid, C6H5COOH, ethanol, CH3CH2OH, and phenol, C6H5OH, in aqueous solution. Explain your answer. … … … most acidic least acidic … … … … … … [3] (b) Draw the major products from the nitration of benzoic acid and phenol in the boxes in Fig. 7.1. The molecular formula for each major product is given in the boxes. major product from benzoic acid, C7H5NO4 major product from phenol, C6H5NO3 [2] Fig. 7.1 (c) Ethanol reacts with propanoyl chloride, C2H5COCl, to form ester W. O O W Fig. 7.2 (i) Give the systematic name for ester W. … [1] * 0000800000020 * ,










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21 9701/44/M/J/25 © UCLES 2025 [Turn over (ii) Complete the mechanism in Fig. 7.3 for the reaction between C2H5COCl and ethanol. R–OH represents ethanol. Include all relevant lone pairs of electrons, curly arrows, charges and partial charges. C C2H5 Cl O R H O organic intermediate products Fig. 7.3 [4] (iii) Name the mechanism for the reaction shown in Fig. 7.3. … [1] [Total: 11] * 0000800000021 * ,









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22 9701/44/M/J/25 © UCLES 2025 8 (a) (i) Define lattice energy, ΔHlatt. … … [2] (ii) Define enthalpy change of solution, ΔHsol. … … [1] (b) The enthalpy change of hydration can be represented by ΔHhyd. Write the mathematical expression for the ΔHsol of NaCl in terms of ΔHlatt(NaCl ), ΔHhyd(Na+) and ΔHhyd(Cl –). ΔHsol(NaCl ) = … [1] (c) Complete the Born–Haber cycle in Fig. 8.1 for the ionic solid NaCl. Include state symbols of relevant species. Na+(g) + Cl –(g) NaCl (s) ΔHea1 ΔHi1 ΔHi1 ΔHea1 first ionisation energy Key first electron affinity ΔHat ΔHat ΔHf … … … … Fig. 8.1 [3] * 0000800000022 * ,









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23 9701/44/M/J/25 © UCLES 2025 (d) Predict which of the ions, Cl – or NO3 –, has the more negative enthalpy change of hydration. Explain your answer. … … … … … [2] [Total: 9] 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/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 17 printed pages. © Cambridge University Press & Assessment 2025 [Turn over Cambridge International AS & A Level CHEMISTRY 9701/44 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.

9701/44 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 2 of 17 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.

9701/44 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 3 of 17 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.

9701/44 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 4 of 17 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 May/June 2025 © Cambridge University Press & Assessment 2025 Page 5 of 17 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

9701/44 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 6 of 17 Annotation Meaning Rounding error 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 May/June 2025 © Cambridge University Press & Assessment 2025 Page 7 of 17 Question Answer Marks 1(a)(i) M1: (thermal stability) increases (down the group or from Mg to Ba) OR carbonates become more stable (down the group or from Mg to Ba) M2: ion / cation size / radius increases OR charge density on ion / cation decreases M3: less polarised CO3(2)– / anion / carbonate ion OR weakens the C-O / C=O / covalent bonds less OR weakens bonds in the anion less OR lattice enthalpy of MO falls (less exothermic) faster than MCO3 (due to differing sizes of anions) 3 1(a)(ii) CuCO3 ⎯⎯→ CuO + CO2 1 1(b)(i) [Ar] ↿⇂ ↿⇂ ↿⇂ ↿⇂ ↿ 1 1(b)(ii) 1 1(b)(iii) the (3)d and (4)s are close / similar in energy 1 1(c)(i) M1: species with / has two lone pairs of electrons M2: that form dative (covalent) / co-ordinate bond(s) to a (central) transition-element / metal AND atom / ion OR donates electron pair(s) to a (central) transition-element / metal AND atom / ion 2

9701/44 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 8 of 17 Question Answer Marks 1(c)(ii) each correct structure = [1] 3 1(c)(iii) optical AND geometrical / cis-trans 1 1(c)(iv) identification of either cis isomer AND dipoles / polar bonds / partial charges do not cancel OR identification of either cis isomer AND it is asymmetric and has polar bonds 1 1(d)(i) complex ion in Na3Al F6 Al F63– ligand in Na3Al F6 F– 1 1(d)(ii) M1: charge passed = 1.5  30  60 OR 2700 (C) M2: n(e–) = 2700 / 96500 OR 2.80  10–2 (mol) M3: n(Al) = 2.80  10-2 ÷ 3 OR 9.33  10–3 (mol) M4: mass (Al) = (27  9.33  10-3) = 0.25 (g) 4

9701/44 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 9 of 17 Question Answer Marks 2(a)(i) white / steamy fumes 1 2(a)(ii) number of (possible) arrangements of particles AND energy in a system 1 2(a)(iii) due to the large number increase in gas molecules formed (in the reaction) OR there are much more gaseous molecules in the products OR six gas moles are produced 1 2(a)(iv) M1: Hr = (–859) + (2 x –297) + (4 x –92) – (–1460) – (2 x –246) OR +131 kJ mol–1 M2: use of G = H – TS AND use of 298 (or 273+25) for T M3: G = 131 – (298 x 0.768) = –97.9 (kJ mol–1) 3 2(b)(i) 2Ba2+(aq) + Cr2O72-(aq) + H2O(l) ⎯⎯→ 2BaCrO4(s) + 2H+(aq) 1 2(b)(ii) M1: d orbital(s) of two different energies / d-d splitting occurs OR d orbital(s) / d (sub)-shell splits / d-d gap OR (inferred from movement of an electron) from lower d to higher d orbital M2: electron(s) promoted / excited OR electron(s) moves to higher (d–) orbital OR electron(s) jumps up (to d– orbital) / jumps to higher (d–orbital) M3: wavelength / frequency / light / photon / h / hf absorbed OR radiation / energy from visible (region) absorbed AND colour (seen) is complementary OR wavelength / frequency / colour / light not absorbed is transmitted / reflected / seen 3 2(c) M1: Hlatt and Hhyd decrease / both become less exothermic / less negative M2: Hhyd changes more / dominant factor / changes faster OR Hlatt changes less / becomes less exothermic by a smaller extent M3: Hsol becomes less exothermic / less negative OR Hsol becomes (more) endothermic / (more) positive 3

9701/44 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 10 of 17 Question Answer Marks 3(a)(i) M1: Ksp = [Ni2+][IO3–]2 M2: units: mol3 dm–9 2 3(a)(ii) [Ni2+] = 2.3  10–2; [IO3–] = 4.6  10–2 Ksp = (2.3  10–2)  (4.6  10-2)2 = 4.87  10–5 1 3(b)(i) Eocell = 1.19 – (0.34) = (+)0.85 (V) AND (positive electrode is) right hand side / nickel electrode / platinum / iodate 1 3(b)(ii) (Ecell would be) less positive / more negative AND as [IO3–] is less than 1.0 mol dm–3 / has a lower / smaller concentration 1 3(b)(iii) M1: Ecell is less positive ticked M2: [IO3–(aq)] is lowered due to the common ion effect OR Ni(IO3)2 precipitating or shown by equation OR solubility of Ni(IO3)2 decreasing 2 3(c)(i) [H+] = 10–pH = 10–0.47 = 0.34 / 0.339 (mol dm-3) min 2sf 1 3(c)(ii) [HIO3]eqm = 1.0 – 0.34 = 0.66 (mol dm–3) min 2sf AND [H+]eqm = [IO3–]eqm = 0.34 (mol dm–3) min 2sf 1 3(c)(iii) Ka = (0.34)2 / 0.66 = 0.17– 0.18 (mol dm–3) min 2sf 1 3(d)(i) M1: opposes / resists change in pH / controls pH / pH kept within a small range M2: when small amount of acid / H+ or alkali / base / OH– is added 2 3(d)(ii) M1: k = {2.1  10–2}/{0.5  (1  10–3)2  0.012} = 4.20  108 min 2sf M2: mol–4 dm12 s–1 2 3(d)(iii) rate will be (0.03 / 0.01)2 = 9 times as fast, so rate = 9  2.1  10–2 = 0.19 (0.189) (mol dm–3 s–1) min 2sf 1

9701/44 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 11 of 17 Question Answer Marks 4(a) polymer type of polymerisation structure of monomer X condensation H2(CH2)5CO2H OR NH2(CH2)5COCl Y addition CH2=CHCO2CH3 Z condensation CH3CH(OH)CO2H OR CH3CH(OH)COCl Type of polymerisation (all three) = [1] Structure of monomers = [1] for each structure 4 4(b) pH (where) the species is a zwitterion (is the dominant form) OR pH (where) the species is (electrically) neutral OR pH (where) the species has a (net overall) charge of zero 1 4(c)(i) M1: Gly correct M2: Glu and Lys correct 2 4(c)(ii) it would move towards the positive terminal / anode / side / end or to the left 1 Start point + - Glu Gly Lys

9701/44 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 12 of 17 Question Answer Marks 5(a)(i) 1 5(a)(ii) the OH / NH groups / O atoms AND can hydrogen bond with water 1 5(a)(iii) C33H35N2O5F 1 5(a)(iv) 1 5(a)(v) different / better biological activity 1 5(b)(i) 16 / sixteen 1 5(b)(ii) M1: 4 / four M2: proton exchange between NH / OH with D 2

9701/44 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 13 of 17 Question Answer Marks 5(c) groups reacting • carboxylic acid • amide groups forming • (primary) alcohol • amine Any two [1] all four [2] 2 Question Answer Marks 6(a) test sodium metal Na2CO3(aq) 2,4-DNPH I2(aq) + OH–(aq) warm with Fehling’s solution Br2(aq) alcohol / carboxylic acid carboxylic acid carbonyl / ketone / aldehyde (methyl) ketone / ( methyl) alcohol aldehyde alkene Any two [1] any three [2] any five [3] all six [4] 4 6(b)(i) (primary) alcohol 1 6(b)(ii) M1: A → B oxidation/dehydrogenation M2: C → E reduction/(nucleophilic) addition M3: E → F elimination/dehydration 3

9701/44 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 14 of 17 Question Answer Marks 6(b)(iii) each correct structure [1] 2 6(b)(iv) D F E C 5 A any one of the following each correct structure [1]

9701/44 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 15 of 17 Question Answer Marks 7(a) M1: benzoic acid > phenol > ethanol M2 / M3: any two [1] any three [2] • correct link of acidity once AND weakens O—H / OH bond / hydroxyl bond / -O-H OR H+ more easily lost / (carboxylate) anion stabilised u / c • (benzoic acid) due to negative inductive effect / electron withdrawing effect AND of C=O / COOH / carboxyl • (phenol) as lone pair / p-orbital (electrons) on oxygen (on phenol) / AND overlap / delocalised into the ring / π-system • (ethanol) alkyl / ethyl / R group AND is electron donating / positive inductive effect 3 7(b) Each correct structure = [1] 2 7(c)(i) ethyl propanoate 1

9701/44 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 16 of 17 Question Answer Marks 7(c)(ii) M1 / M2: any two [1] all four [2] • lone pair on O • correct arrow from (lone pair) O to C (of C=O) • dipole on C=O • correct arrow on C=O M3: correct intermediate M4: arrow from lone pair or charge on O– to C-O bond AND arrow from C-Cl to Cl 4 7(c)(iii) (nucleophilic) addition – elimination 1 Question Answer Marks 8(a)(i) M1: energy change / released when 1 mole of a (ionic) solid / lattice / crystal / compound is formed M2: from gas (phase) ions / gaseous ions (under standard conditions) 2 8(a)(ii) (enthalpy change when) 1 mole of a substance / solid / solute / molecule AND dissolves in water (to give a solution of infinite dilution) 1 8(b) Hsol(NaCl) = Hhyd(Na+) + Hhyd(Cl –) – Hlatt(NaCl) 1

9701/44 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 17 of 17 Question Answer Marks 8(c) M1 / M2: any two [1] all four [2] M3: all state symbols for the formula are present and correct 3 8(d) M1: (more exothermic because) Cl– OR NO3– because its (ionic) radius / size is smaller M2: (more exothermic because) (ion-dipole) attraction / bond between it and water is stronger OR M1: NO3– because it has lone pairs on O / more lone pairs M2: which can form hydrogen bonds with water 2