Cambridge A Level Chemistry 9701 — 2024 Oct/Nov Paper 4 · Variant 1

[Turn over Cambridge International AS & A Level This document has 24 pages. Any blank pages are indicated. * 0 6 7 1 5 2 8 9 0 2 * DC (DE) 346546 © UCLES 2024 CHEMISTRY 9701/41 Paper 4 A Level Structured Questions October/November 2024 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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2 9701/41/O/N/24 © UCLES 2024 1 (a) Disodium phosphate, (Na+)2(HPO4 2–), reacts with an acid to form monosodium phosphate, Na+(H2PO4 –). (i) Identify the ions that are a conjugate acid–base pair in this reaction, using the formulae of the species involved. conjugate acid conjugate base … … [1] (ii) Define buffer solution. … … [2] (iii) Write two equations to show how a mixture of (Na+)2(HPO4 2–) and Na+(H2PO4 –) can act as a buffer solution. equation 1 … equation 2 … [2] (iv) Identify one inorganic ion that acts as a buffer in blood. … [1] (b) Compound E is the hydroxide of a Group 2 element. Compound E is a strong alkali. 2.63 g of E is dissolved in water to make 250 cm3 of solution F. Solution F has a pH of 13.09 at 298 K. (i) Show that the concentration of hydroxide ions in solution F is 0.123 mol dm–3. [2] (ii) Explain why the concentration of compound E in solution F is 0.0615 mol dm–3. … … [1] (iii) Use the concentration given in (ii) to identify compound E. compound E … [1] * 0000800000002 * ,










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3 9701/41/O/N/24 © UCLES 2024 [Turn over (c) Compound E is much more soluble than magnesium hydroxide. A saturated solution of magnesium hydroxide in water has a concentration of 1.40 × 10–4 mol dm–3 at 298 K. Calculate the solubility product, Ksp, of magnesium hydroxide. Include units. Ksp = … units … [3] (d) Explain why compound E is much more soluble than magnesium hydroxide. … … … … … … … … [3] [Total: 16] * 0000800000003 * ,










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4 9701/41/O/N/24 © UCLES 2024 2 (a) Predict and explain the variation in enthalpy change of hydration for the ions F–, Cl –, Br – and I–. … … … [2] (b) Fig. 2.1 shows an incomplete energy cycle involving calcium fluoride, CaF2. process 1: enthalpy change of formation of Ca2+(g) plus twice the enthalpy change of formation of F–(g), ΔHf Ca2+(g) + 2ΔHf F–(g) process 3: enthalpy change of formation of calcium fluoride, ΔHf CaF2(s) process 2: enthalpy change of hydration of calcium ions plus twice the enthalpy change of hydration of fluoride ions, ΔHhyd Ca2+(g) + 2ΔHhyd F–(g) line A: Ca(s) + F2(g) line B: Ca2+(g) + 2F–(g) line C: CaF2(s) line D: process 4: enthalpy change of solution of calcium fluoride, ΔHsol CaF2(s) Fig. 2.1 (i) Complete line D. Include state symbols. [1] (ii) The value of the enthalpy change for process 1 can be calculated using the values of five other enthalpy changes which are not referred to in Fig. 2.1. process 1: Ca(s) + F2(g) Ca2+(g) + 2F–(g) Identify these five other enthalpy changes, using either names or symbols. … … … … … [2] * 0000800000004 * ,










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5 9701/41/O/N/24 © UCLES 2024 [Turn over (iii) Define lattice energy, ΔHlatt. … … … [2] (iv) Complete the expression to give the mathematical relationship between ΔHlatt of calcium fluoride and the enthalpy changes for processes 1 and 3. ΔHlatt = [1] (c) Use data from Table 2.1 to calculate a value for the hydration energy, ΔHhyd, of fluoride ions, F–(g). Table 2.1 value / kJ mol–1 enthalpy change of solution of calcium fluoride, CaF2(s) +13 overall enthalpy change of process 1 in Fig. 2.1 +1395 enthalpy change of formation of calcium fluoride –1214 enthalpy change of hydration of Ca2+(g) –1650 ΔHhyd F–(g) = … kJ mol–1 [2] (d) Define entropy. … … [1] * 0000800000005 * ,










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6 9701/41/O/N/24 © UCLES 2024 (e) At 298 K, the Gibbs free energy change, ΔG, for the solution of compound T is +6.00 kJ mol–1. The enthalpy change of solution, ΔHsol, of compound T is +30.0 kJ mol–1 at 298 K. Calculate the value of the entropy change, ΔS, for the solution of compound T at 298 K. ΔS = … J K–1 mol–1 [2] (f) Predict whether compound T becomes more or less soluble as the water is heated from 298 K to 360 K. Explain your answer. … … [1] [Total: 14] * 0000800000006 * ,










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7 9701/41/O/N/24 © UCLES 2024 [Turn over 3 (a) A and B react together to give product AB. A + B AB When the concentrations of A and B are both 0.0100 mol dm–3, the rate of formation of AB is 7.62 × 10–4 mol dm–3 s–1. When the concentrations of A and B are both 0.0200 mol dm–3, the rate of formation of AB is 3.05 × 10–3 mol dm–3 s–1. (i) Complete the three possible rate equations that are consistent with these data. rate = … rate = … rate = … [2] (ii) Choose one of the rate equations you have written in (i), and calculate the value of the rate constant, k. Include the units of k. k = … units … [2] (iii) Explain why it is not possible to calculate a value for the half-life, t1 2 , of this reaction using the value of the rate constant k calculated in (ii) and the equation k = 0.693 / t1 2 . … … … [1] * 0000800000007 * ,










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8 9701/41/O/N/24 © UCLES 2024 (b) Catalysts may be homogeneous or heterogeneous. (i) Identify two metals that act as heterogeneous catalysts in the removal of NO2 from the exhaust gases of car engines. … and … [1] (ii) Iron acts as a heterogeneous catalyst in the Haber process. Describe the mode of action of this iron catalyst. … … … [2] (iii) Fe2+ ions act as a homogeneous catalyst in the reaction between I–(aq) and S2O8 2–(aq). Write equations for the two reactions that occur when Fe2+(aq) is added to a mixture of I–(aq) and S2O8 2–(aq). equation 1 S2O8 2– + … equation 2 … [2] (iv) Explain the difference between a homogeneous catalyst and a heterogeneous catalyst. … … … [1] * 0000800000008 * ,










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9 9701/41/O/N/24 © UCLES 2024 [Turn over (c) Fe2+ ions can be oxidised to Fe3+ ions under alkaline conditions by suitable oxidising agents. (i) Iron is a transition element. Explain why iron forms stable compounds in both the +2 and the +3 oxidation states. … [1] (ii) The half-equation for the reduction of Fe3+ under alkaline conditions, and its E o value, are shown. Fe(OH)3 + e– Fe(OH)2 + OH– E o = – 0.56 V Four more half-equations for reactions under alkaline conditions, and their E o values, are shown. Al (OH)4 – + 3e– Al + 4OH– E o = –2.35 V Cl O– + H2O + 2e– Cl – + 2OH– E o = +0.89 V O2 + 2H2O + 4e– 4OH– E o = +0.40 V Zn(OH)4 2– + 2e– Zn + 4OH– E o = –1.22 V Select two oxidising agents that can oxidise Fe2+ ions to Fe3+ ions under alkaline conditions. Write an equation, and give the E cell value, for each of the two reactions that occur. oxidising agent 1: … equation: … E cell = … V oxidising agent 2: … equation: … E cell = … V [4] [Total: 16] o o o * 0000800000009 * ,










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10 9701/41/O/N/24 © UCLES 2024 4 Transition metal atoms and transition metal ions form complexes by combining with ligands. (a) Explain why transition elements form complex ions. … … [1] (b) Co2+ ions form complex ion G. Each G ion contains two Co2+ ions, both of which are octahedrally coordinated. Each G ion contains one O2 molecule, which donates one pair of electrons to each Co2+ ion, and one NH2 – ion, which donates one pair of electrons to each Co2+ ion. The remaining ligands are NH3 molecules. (i) Deduce the formula of complex ion G. Include its overall charge. formula of G … [2] (ii) The d-orbitals of the Co2+ ions present in complex ion G are split. State the number of d-orbitals that are at a higher energy level and the number of d-orbitals that are at a lower energy level in each Co2+ ion. number of d-orbitals at a higher energy level number of d-orbitals at a lower energy level [1] (iii) Co2+ ions form a different complex ion, M. Each M ion contains two Co2+ ions, both of which are octahedrally coordinated, but the ligands are different from the ligands in G. Explain why G and M have different colours. … … … … … … [2] * 0000800000010 * ,










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11 9701/41/O/N/24 © UCLES 2024 [Turn over (c) Cadmium forms complex ion X, [Cd(NH3)4]2+. When a solution containing CN– ions is added to an aqueous solution of X, a ligand exchange reaction takes place, forming complex ion Y. Y contains no NH3 ligands and no H2O ligands. Y is in a much higher concentration in the mixture than X. The oxidation state and coordination number of cadmium do not change in this reaction. (i) Write an ionic equation for this reaction, using the formulae of the complex ions. … [2] (ii) Cadmium forms complex ion Z in the same oxidation state and with the same coordination number as in X. All the ligands in Z are Cl – ions. When NaCl (aq) is added to a solution of X, very little Z forms. Write the three cadmium complexes, X, Y and Z, in order of increasing stability constant, Kstab. … … … smallest value of Kstab largest value of Kstab [1] (d) Ethanedioate ions, C2O4 2–, form complexes with transition element ions. The concentration of C2O4 2– ions can be found by reaction with acidified Cr2O7 2– ions. C2O4 2– ions are protonated and form HOOCCOOH molecules which are oxidised by Cr2O7 2–. The half-equations are shown. Cr2O7 2– + 14H+ + 6e– 2Cr3+ + 7H2O 2CO2 + 2H+ + 2e– HOOCCOOH (i) Construct an equation for the reaction between acidified Cr2O7 2– and HOOCCOOH. … [1] (ii) A 25.0 cm3 sample of a solution of Na2C2O4 reacts with exactly 16.20 cm3 of an acidified solution of 0.0500 mol dm–3 K2Cr2O7. Calculate the concentration of the solution of Na2C2O4. [Na2C2O4] = … mol dm–3 [2] [Total: 12] * 0000800000011 * ,









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12 9701/41/O/N/24 © UCLES 2024 5 The shapes of four different complexes, P, Q, R and S, are shown in Table 5.1. The symbol J represents an atom or ion of a transition element. The symbol L is used to represent a monodentate ligand. Table 5.1 L P Q R S L L J L J L L L L J L L L J L L L L L (a) Label one bond angle on each of complexes P, Q, R and S, and identify the size of the angle in degrees. [2] (b) Identify the shapes of complexes P, Q, R and S. P … Q … R … S … [2] (c) Two L ligands are exchanged with two different monodentate ligands X and Y in each of complexes P, Q, R and S. Identify all the complexes which form new complexes that show geometrical isomerism. … [1] (d) Three L ligands are exchanged with three different monodentate ligands X, Y and Z in each of complexes P, Q and R. Identify all the complexes which form new complexes that show optical isomerism. … [1] [Total: 6] * 0000800000012 * ,









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13 9701/41/O/N/24 © UCLES 2024 [Turn over BLANK PAGE * 0000800000013 * ,









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14 9701/41/O/N/24 © UCLES 2024 6 Benzene, C6H6, reacts with chloroethane, C2H5Cl, in the presence of a suitable catalyst to form ethylbenzene, C6H5C2H5. In the presence of the catalyst, the ion C2H5 + is formed. This ion reacts with benzene. (a) Complete the equation for the reaction of C2H5Cl with this catalyst to form C2H5 + as one product. C2H5Cl + … C2H5 + + … [1] (b) Ethylbenzene reacts with more C2H5Cl, forming a mixture containing 1,2-diethylbenzene and 1,4-diethylbenzene. (i) Draw the structures of 1,2-diethylbenzene and 1,4-diethylbenzene. 1,2-diethylbenzene 1,4-diethylbenzene [1] (ii) Explain why there is very little 1,3-diethylbenzene in the product mixture. … … [1] * 0000800000014 * ,









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15 9701/41/O/N/24 © UCLES 2024 [Turn over (c) 1,2-diethylbenzene can be oxidised to benzene-1,2-dioic acid, C6H4(COOH)2. COOH COOH benzene-1,2-dioic acid (i) State the reagent and conditions used for this reaction. … [1] (ii) Complete the overall equation for this reaction. An atom of oxygen from the oxidising agent is represented as [O]. All of the atoms in the two ethyl groups are fully oxidised in this reaction. … + … [O] C6H4(COOH)2 + … + … (1,2-diethylbenzene) [2] (iii) Predict the number of peaks in the carbon-13 NMR spectrum of benzene-1,2-dioic acid. … [1] * 0000800000015 * ,









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16 9701/41/O/N/24 © UCLES 2024 (d) The proton (1H) NMR spectra of ethylbenzene, C6H5C2H5, in CDCl 3 and of benzene-1,2-dioic acid, C6H4(COOH)2, in CDCl 3 are shown. They have not been identified. 11 10 9 8 7 6 5 δ / ppm 4 3 2 1 0 Fig. 6.1 δ / ppm 0 2 4 6 8 10 12 14 16 Fig. 6.2 (i) Explain the use of CDCl 3, instead of CHCl 3, as the solvent when obtaining these spectra. … [1] * 0000800000016 * ,









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17 9701/41/O/N/24 © UCLES 2024 [Turn over (ii) Identify the substance shown by the spectrum in Fig. 6.1, and complete Table 6.1. substance … Table 6.1 peak at δ = 1.2 peak at δ = 2.6 name of splitting pattern group responsible for peak explanation of splitting pattern [3] (iii) Identify the substance shown by the spectrum in Fig. 6.2, and complete Table 6.2. substance … Table 6.2 peak at δ = 7.8 peak at δ = 13.1 group responsible for peak [1] (iv) When D2O is used as a solvent, the spectrum obtained is different from the spectrum in Fig. 6.2. Describe this difference and explain your answer. … … … [1] (e) Benzene-1,2-dioic acid can be used to produce K. COOH COOH C C O O K O benzene-1,2-dioic acid heat Suggest the name of this type of reaction. … [1] [Total: 14] * 0000800000017 * ,









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18 9701/41/O/N/24 © UCLES 2024 7 A reaction scheme is shown in Fig. 7.1. The reagents needed for reaction 2 and reaction 3 are stated. Reaction 5 takes place when C2H5NH2 is mixed with compound V. No special conditions are required. compound U compound W compound V C2H5NH2 C2H5NHC2H5 C2H5NHCOCH3 CH3COOH reaction 1 reaction 6 reaction 2 reagent = LiAl H4 reaction 4 reaction 5 reaction 3 reagent = SOCl2 Fig. 7.1 (a) Identify compound U which contains only three elements. … [1] (b) Describe the reagents and conditions for reaction 1. … [1] (c) Identify compound V. … [1] (d) Complete the equation for reaction 3. CH3COOH + SOCl2 … [1] (e) Identify compound W. … [1] (f) Describe the conditions for reaction 4. … [1] * 0000800000018 * ,









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19 9701/41/O/N/24 © UCLES 2024 [Turn over (g) Suggest the reagent needed for reaction 6. … [1] (h) Complete Table 7.1 by adding the reaction numbers, 1, 2, 3, 4, 5 and 6, to the right-hand column. Use the reaction numbers given in Fig. 7.1. Each of the numbers 1, 2, 3, 4, 5 and 6 should be used once only. Table 7.1 type of reaction reaction number(s) hydrolysis addition reduction substitution [4] (i) Compare the basicities of C2H5NHCOCH3, C2H5NHC2H5 and NH3. Explain your answer. … … … most basic least basic … … … … … … [4] [Total: 15] * 0000800000019 * ,









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20 9701/41/O/N/24 © UCLES 2024 8 (a) An aqueous solution of phenol, C6H5OH, is acidic at 298 K. Explain why phenol is more acidic than water. … … … [2] (b) (i) Name the two products formed when phenol reacts with an excess of Br2(aq). … and … [1] (ii) Draw the structures of the two isomeric organic products, with Mr = 139, that are formed when phenol reacts with HNO3(aq) at room temperature. [1] (iii) Write the equation for the reaction between phenol, C6H5OH, and sodium metal. … [1] (c) Phenol can be produced from phenylamine in a two-step synthesis. phenylamine intermediate compound phenol Describe the reagents and conditions needed in each step. step one: reagents … conditions … step two: reagents … conditions … [2] [Total: 7] step one step two * 0000800000020 * ,










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23 9701/41/O/N/24 © UCLES 2024 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 = 6.022 × 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 9701/41/O/N/24 © UCLES 2024 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 14 printed pages. © Cambridge University Press & Assessment 2024 [Turn over Cambridge International AS & A Level CHEMISTRY 9701/41 Paper 4 A Level Structured Questions October/November 2024 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 2024 series for most Cambridge IGCSE, Cambridge International A and AS Level components, and some Cambridge O Level components.

9701/41 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 2 of 14 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.

9701/41 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 3 of 14 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.

9701/41 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 4 of 14 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.

9701/41 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 5 of 14 Question Answer Marks 1(a)(i) conjugate acid conjugate base H2PO4– HPO42– [1] BOTH 1 1(a)(ii) allows only small changes in pH / resists changes in pH [1] when small amounts of acid or alkali/base / H+ or OH– are added [1] 2 1(a)(iii) Na+2HPO42– + H+ → Na+H2PO4– + Na+ or HPO42– + H+ → H2PO4– [1] Na+H2PO4– + NaOH → Na+2HPO42– + H2O or H2PO4– + OH– → HPO42– + H2O [1] 2 1(a)(iv) HCO3– / hydrogen carbonate [1] 1 1(b)(i) either 10–13.09 or [H+]= 8.1  10–14 seen [1] 8.1x10–14  [OH–] = 1  10–14 seen [1] OR [OH–] = 14 14 1 10 8.1 10 − −   seen ALLOW alternative method pOH = 0.91 [1] [OH–] = 10–0.91 seen [1] 2 1(b)(ii) 2 moles of OH– in one mole of E / X(OH)2 OR there are two hydroxide ions in each formula unit of E / X(OH)2 [1] 1 1(b)(iii) 0.0615  0.25 = 0.0154 moles of E RFM = 2.63 / 0.0154 = 170.8 / 171 OR RAM = 34 – 2.63 / 0.0154 = 137.1 /137 AND barium hydroxide / Ba(OH)2 [1] 1

9701/41 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 6 of 14 Question Answer Marks 1(c) Ksp = [Mg2+][OH–]2 OR Ksp = 1.4  10-4  (2  1.4  10-4)2 [1] Ksp = 1.10  10–11 [1] min 2sf correct answer [2] mol3 dm–9 [1] ecf from M1 3 1(d) M1: Hlatt and Hhyd less exothermic (down the group) OR Hlatt/LE and Hhyd more exothermic/more negative for Mg [1] M2: Hlatt changes more than Hhyd (down the group) OR Hlatt changes more / faster for Mg [1] M3: Hsol becomes more exothermic (down the group) OR Hsol becomes less exothermic / less negative for Mg [1] mark independently 3 Question Answer Marks 2(a) becomes less negative/less exothermic (down the group / from F to I) AND due to increase in (ionic) radius / size [1] decreased attraction to water OR weaker ion-dipole force to water [1] 2 2(b)(i) CaF2(aq) OR Ca2+ (aq) + 2F– (aq) [1] 1 2(b)(ii) • atomisation energy of Ca / Hat(o)m)) • atomisation energy of F(2) / Hat(o)m)) OR F-F bond energy / BE of F-F • first ionisation energy / IE1 of Ca / Hi1 • second ionisation energy / IE2 of Ca / Hi2 • (first) electron affinity /EA of F / Hea any three for [1] ALL five [2] 2

9701/41 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 7 of 14 Question Answer Marks 2(b)(iii) energy change / energy released when one mole of an ionic compound is formed [1] from its gaseous ion(s) [1] 2 2(b)(iv) (Hlatt = ) (change) 3 – (change) 1 [1] OR (Hlatt = ) Hf(CaF2(s)) – Hf(Ca2+(g)) – 2Hf(F–(g)) 1 2(c) expression involves four correct numbers (13, 1395, 1214, 1650) AND 2 times [1] 1395 – 1650 + 2x = –1214+13 Hf(F–(g))  = –473 [1] ecf correct answer [2] 2 2(d) number of possible arrangements of particles / energy in a system [1] 1 2(e) states or clearly uses G = H - TS [1] OR 6000 = 30 000 – (298  S) S = (+)80.5(4) [1] min 3sf ecf correct answer [2] 2 2(f) becomes more soluble AND S is positive / TS is positive / –TS is negative (as T inc) [1] (so G becomes more negative) 1

9701/41 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 8 of 14 Question Answer Marks 3(a)(i) rate = k[A][B] [1] rate = k[A]2 / k[A]2[B]0 AND rate = k[B]2 / k[A]0[B]2 [1] BOTH 2 3(a)(ii) k = 7.62 / 7.625 / 7.63 [1] ecf from 3a(i) min 3sf mol–1 dm3 s–1 [1] ecf from 3a(i) 2 3(a)(iii) reaction is 2nd order OR reaction is not 1st order OR reaction does not have constant half-life [1] 1 3(b)(i) Pt, Pd, Rh any two [1] 1 3(b)(ii) • adsorption (of reactants) / bond forms between the surface/catalyst and reactants • bond weakening (in reactants) • desorption (of reactants / products) / bond breaks between the surface/catalyst and products any two [1] all three [2] 2 3(b)(iii) S2O82– + 2Fe2+ → 2SO42– + 2Fe3+ [1] any multiple 2Fe3+ + 2I– → 2Fe2+ + I2 [1] any multiple 2 3(b)(iv) (heterogeneous) catalyst and reactants in different states / phases AND (homogeneous) catalyst and reactants in same states / phases [1] BOTH 1 3(c)(i) similar energy of the (3)d and (4)s subshells / orbitals [1] 1

9701/41 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 9 of 14 Question Answer Marks 3(c)(ii) M1M2: • Cl O– chosen • O2 chosen • E⦵cell values 1.45 (with Cl O–) • E⦵cell values 0.96 (with O2) any two [1] all four [2] M3: 2Fe(OH)2 + Cl O– + H2O → Cl– + 2Fe(OH)3 [1] M4: 4Fe(OH)2 + O2 + 2H2O → 4Fe(OH)3 [1] 4 Question Answer Marks 4(a) empty (d) orbitals are energetically accessible OR empty (d) orbitals can form dative bonds with ligands OR empty (d) orbitals can accept a lone pair from ligands [1] 1 4(b)(i) [Co2O2NH2(NH3)8]3+ scores [2] must have Co2 • Co2O2NH2 • 8 ammonia so both Co are octahedral (ecf bullet 1) • correct charge based on ligands present (ecf Co and NH2) any two [1] all three [2] 2 4(b)(ii) 2 3 BOTH [1] 1 4(b)(iii) E different OR (d–d) energy gap different [1] different frequency/wavelength/energy from visible light absorbed [1] 2

9701/41 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 10 of 14 Question Answer Marks 4(c)(i) [Cd(NH3)4]2+ + 4CN– → [Cd(CN)4]2– + 4NH3 X Y complex formed is [Cd(CN)4]2– / Na2[Cd(CN)4] [1] equation fully correct [1] ECF 2 4(c)(ii) [CdCl4]2– [Cd(NH3)4]2+ [Cd(CN)4]2– OR Z X Y [1] 1 4(d)(i) Cr2O72– + 8H+ + 3HOOCCOOH → 2Cr3+ + 7H2O + 6CO2 [1] 1 4(d)(ii) • mol Cr2O72– = 0.0500  16.20 / 1000 = 8.10  10–4 • mol HOOCCOOH / C2O42– = 8.10  10–4  3 = 2.43  10–3 ecf • [C2O42–] = 2.43  10–3  40 = 0.0972 mol dm–3 min 2sf ecf  ✓ ✓ [2] 2 Question Answer Marks 5(a) bond angle must go from bond to bond P 109 OR 109.5° Q 90° (or 180° if different angle labelled) R 90° (or 180° if different angle labelled) S 180° any two [1] all four [2] 2 5(b) P tetrahedral Q square planar R octahedral S linear any two [1] all four [2] 2 5(c) Q AND R [1] 1 5(d) P AND R [1] 1

9701/41 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 11 of 14 Question Answer Marks 6(a) C2H5Cl + AlCl3 → C2H5+ + AlCl4– [1] OR C2H5Cl + FeCl3 → C2H5+ + FeCl4– 1 6(b)(i) AND BOTH [1] 1 6(b)(ii) alkyl / ethyl group is 2,4 directing OR ethyl group is electron donating / positive inductive effect [1] 1 6(c)(i) hot alkaline KMnO4 / MnO4– (followed by acid) [1] 1 6(c)(ii) C10H14 / C6H4(C2H5)2 + 12 [O] → C6H4(COOH)2 + 2CO2 + 4H2O all species correct formulae C10H14 + CO2 + H2O [1] correct balancing [1] 2 6(c)(iii) 4 [1] 1 6(d)(i) CDCl3 does not cause a peak OR does not interfere with spectrum / peaks [1] 1 6(d)(ii) • ethylbenzene / C6H5C2H5 • triplet • CH3 • 2H on neighbouring C / next to CH2 – quartet / quadruplet – CH2 – 3H on neighbouring C / next to CH3 / coupling by CH3 any three [1] any five [2] all seven [3] 3

9701/41 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 12 of 14 Question Answer Marks 6(d)(iii) • benzene-1,2-dioic acid / C6H4(COOH)2 • H on the benzene / C6H4 – COOH / carboxylic acid ALL three [1] 1 6(d)(iv) • COOH peak disappears OR remove peak at 13.1 (ecf (d)(iii) if  value used) AND • as proton / hydrogen exchanges with deuterium / D BOTH [1] 1 6(e) dehydration / elimination / (auto)condensation [1] 1 Question Answer Marks 7(a) CH3CN / ethanenitrile [1] 1 7(b) dilute acid / HCl (aq) AND hot / heat [1] three components required (acid / aq / heat) 1 7(c) CH3COCl / ethanoyl chloride [1] 1 7(d) CH3COOH + SOCl2 → CH3COCl + HCl + SO2 [1] 1 7(e) C2H5Br / C2H5Cl / bromoethane / chloroethane [1] 1 7(f) heat in ethanol AND under pressure / in sealed tube [1] 1 7(g) LiAlH4 / lithium aluminium hydride [1] 1

9701/41 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 13 of 14 Question Answer Marks 7(h) Examiner checks first that each of 1–6 is only used once and IGNOREs any repeated unless a CON hydrolysis 1 addition 2 reduction 2, 6 substitution 3, 4, 5 two correct [1] four correct [2] five correct [3] six correct [4] 4 7(i) M1 C2H5NHC2H5 ammonia C2H5NHCOCH3 [1] M2 (basicity linked to) lone pair/ p orbital on N AND being able accept / donate to a proton/H+ [1] M3 (amine stronger because of) +ve inductive effect / electron donating alkyl / R / ethyl group [1] M4 (amide weaker because) lone pair / p orbital on N is delocalised into C=O group [1] 4 Question Answer Marks 8(a) M1: p orbital / lone pair on the oxygen / O AND is delocalised into the ring / overlaps with delocalised ring [1] M2: O-H weakened (in phenol) OR anion / phenoxide ion / conjugate base is stabilised/more stable [1] 2 8(b)(i) 2,4,6-tribromophenol (name) AND hydrogen bromide / HBr [1] 1

9701/41 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 14 of 14 Question Answer Marks 8(b)(ii) AND 1 8(b)(iii) 2C6H5OH + 2Na → 2C6H5ONa + H2 [1] 1 8(c) step one: HNO2 (+ HCl) OR NaNO2 + HCl AND T⩽10 °C step two: H2O AND warm / T> 10 °C any two [1] all four [2] 2