Cambridge A Level Chemistry 9701 — 2024 Oct/Nov Paper 4 · Variant 2

This document has 24 pages. Any blank pages are indicated. [Turn over * 5 9 6 1 1 9 7 9 3 9 * Cambridge International AS & A Level CHEMISTRY 9701/42 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. DC (DE/JG) 336276/3 © UCLES 2024 ,










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2 9701/42/O/N/24 © UCLES 2024 1 (a) The equation for reaction 1 is shown. reaction 1 X 2Y Reaction 1 is first order with respect to the concentration of X. The half-life of the reaction, t1 2 , is 900 s at 20 °C. (i) A solution of X with a concentration of 0.180 mol dm–3 is prepared at 20 °C. Calculate the average rate of reaction 1 over the first 1800 s. average rate of reaction 1 = … [2] (ii) Complete the rate equation for reaction 1. rate = … [1] (iii) Show that the rate constant, k, is 7.70 × 10–4 s–1 at 20 °C. [1] (iv) Calculate the initial rate of reaction 1 when the concentration of X is 0.150 mol dm–3. Include units. rate = … units … [2] (b) Catalysts may be homogeneous or heterogeneous. (i) Platinum is a transition element. Explain why transition elements behave as catalysts. … … … [1] (ii) Name the metal catalyst in the Haber process and explain why it is a heterogeneous catalyst. metal … … [1] * 0000800000002 * ,










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3 9701/42/O/N/24 © UCLES 2024 [Turn over (iii) Platinum acts as a heterogeneous catalyst in the removal of nitrogen dioxide, NO2, from the exhaust gases of car engines. Describe the mode of action of a platinum catalyst in this process. … … … … [2] (iv) NO2 acts as a homogeneous catalyst in the oxidation of atmospheric sulfur dioxide, SO2. Write equations for the two reactions that occur. equation 1 … equation 2 … [1] (c) SO2 dissolves in water, forming H2SO3. H2SO3 can be oxidised under acidic conditions. The relevant electrode reaction and its E o– value are shown. SO4 2– + 4H+ + 2e– H2SO3 + H2O E o– = +0.17 V Four more half-equations for reactions occurring under acidic conditions, and their E o– values, are shown. H3BO3 + 3H+ + 3e– B + 3H2O E o– = –0.73 V BiO+ + 2H+ + 3e– Bi + H2O E o– = +0.28 V S + 2H+ + 2e– H2S E o– = +0.14 V Sb + 3H+ + 3e– SbH3 E o– = –0.51 V Select the oxidising agent that could oxidise H2SO3 to SO4 2– ions under acidic conditions. Write an equation, and give the E o– cell value, for the reaction that occurs. oxidising agent … equation … E o– cell = … V [3] [Total: 14] * 0000800000003 * ,










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4 9701/42/O/N/24 © UCLES 2024 2 (a) Predict and explain the variation in enthalpy change of hydration for the ions Na+, Mg2+ and Al 3+. … … … … … [3] (b) Fig. 2.1 shows an incomplete energy cycle. change 1: lattice energy of magnesium chloride, ΔHlatt MgCl 2(s) change 3: enthalpy change of solution of magnesium chloride, ΔHsol MgCl 2(s) change 2: line A: MgCl 2(s) line C: line B: Mg2+(g) + 2Cl –(g) Fig. 2.1 (i) Complete line C on Fig. 2.1. Include state symbols. [1] (ii) Use both words and symbols to identify change 2 on Fig. 2.1. Use changes 1 and 3 as examples of how this should be done. [2] * 0000800000004 * ,










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5 9701/42/O/N/24 © UCLES 2024 [Turn over (iii) Calculate a value for the lattice energy of magnesium chloride, ΔHlatt MgCl 2(s), by selecting and using appropriate data from Table 2.1. Table 2.1 energy change value / kJ mol–1 enthalpy change of solution of magnesium chloride –155 enthalpy change of formation of magnesium chloride –642 first ionisation energy of magnesium +736 second ionisation energy of magnesium +1450 electron affinity of chlorine –349 enthalpy change of hydration of Mg2+ –1920 enthalpy change of hydration of Cl – –364 ΔHlatt MgCl 2(s) = … kJ mol–1 [3] (c) Define entropy. … … [1] (d) At 25 °C the enthalpy change of solution of compound Z is +26 kJ mol–1. The entropy change of solution of Z at the same temperature is +52 J K–1 mol–1. Calculate the value of the Gibbs free energy change, ΔG, for the solution of Z at 25 °C. ΔG = … kJ mol–1 [2] * 0000800000005 * ,










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6 9701/42/O/N/24 © UCLES 2024 (e) (i) Use your answer to (d) to predict whether or not Z is soluble in water at 25 °C. Explain your answer. … … [1] (ii) Predict whether Z becomes more or less soluble as the water is heated from 25 °C to 95 °C. Explain your answer. … … [1] [Total: 14] * 0000800000006 * ,










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7 9701/42/O/N/24 © UCLES 2024 [Turn over BLANK PAGE * 0000800000007 * ,










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8 9701/42/O/N/24 © UCLES 2024 3 (a) The pH of a saturated solution of calcium hydroxide is 12.35 at 298 K. (i) Show that the concentration of hydroxide ions in a saturated solution of calcium hydroxide is 0.0224 mol dm–3 at 298 K. [2] (ii) Use data given in (i) to calculate the solubility product, Ksp, of calcium hydroxide at 298 K. Include the units of Ksp in your answer. Ksp = … units … [3] (iii) A spatula measure of solid calcium chloride is stirred into a sample of saturated calcium hydroxide solution. All of the calcium chloride dissolves. Describe one other observation that would be made and give an estimated value of the pH of the solution obtained. Explain both your answers. observation … pH of solution … explanation … … [3] (iv) Calcium hydroxide reacts with dilute sulfuric acid to form calcium sulfate. Barium hydroxide behaves in a similar way, forming barium sulfate. Explain why calcium sulfate is more soluble in water than barium sulfate. … … … … … [3] * 0000800000008 * ,










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9 9701/42/O/N/24 © UCLES 2024 [Turn over (b) Some solid calcium is added to an excess of aqueous ethanoic acid, CH3COOH, and left until all the calcium has reacted. The resulting mixture, mixture D, contains no undissolved solids. (i) Write an equation for the reaction of calcium with CH3COOH. … [1] (ii) Use formulae of molecules and ions to identify two conjugate acid–base pairs present in mixture D. Pair 1 should consist of organic species. pair 1: … … conjugate acid conjugate base pair 2: … … conjugate acid conjugate base [2] (iii) Write the expression for the Ka of CH3COOH. Ka = [1] (iv) The concentration of calcium ethanoate, (CH3COO)2Ca, in mixture D is 0.394 mol dm–3. The concentration of CH3COOH in mixture D is 0.270 mol dm–3. The Ka of CH3COOH is 1.74 × 10–5 mol dm–3 at 298 K. Calculate the pH of mixture D. pH = … [2] (v) Write two equations to show how mixture D can act as a buffer solution. equation 1 … equation 2 … [2] [Total: 19] * 0000800000009 * ,










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10 9701/42/O/N/24 © UCLES 2024 4 Transition metal atoms and transition metal ions form complexes by combining with species called ligands. (a) When NaOH(aq) is added to an aqueous solution containing [Co(H2O)6]2+ a precipitation reaction occurs accompanied by a colour change. In this reaction two of the water ligands each lose one H+ ion. The H+ ions are gained by OH– ions from the NaOH(aq). (i) State the colour change seen in this precipitation reaction. from … to … [1] (ii) Complete the ionic equation for this precipitation reaction. [Co(H2O)6]2+ + … … + … [1] (iii) This precipitation reaction can also be described as a different type of reaction. Name this type of reaction. … [1] (b) L is an uncharged tridentate ligand. L donates three lone pairs to a metal atom or ion. Cobalt forms an octahedral complex ion, E, with L. Complex ion E has a 2+ charge. (i) Give the formula of E. … [1] (ii) Identify the oxidation state of cobalt in E. … [1] (iii) The d-orbitals of the cobalt atom or ion present in E are split in energy. 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. number of d-orbitals at a higher energy level number of d-orbitals at a lower energy level [1] (iv) Define the term non-degenerate d-orbitals. … … [1] * 0000800000010 * ,










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11 9701/42/O/N/24 © UCLES 2024 [Turn over (c) The mineral chromite contains a compound which has the formula FeCrnO4. The oxidation state of iron in FeCrnO4 is +2. A sample of 4.18 g of FeCrnO4 is dissolved in an excess of sulfuric acid. The resulting solution is made up to 250 cm3. This is solution F. All the Fe2+ ions in 25.0 cm3 of solution F are oxidised to Fe3+ ions by exactly 18.7 cm3 of 0.0200 mol dm–3 KMnO4. One MnO4 – ion reacts with five Fe2+ ions. Assume no other oxidation reaction occurs. (i) Write an equation for the reaction of Fe2+ ions with MnO4 – ions in acid solution. … [1] (ii) Calculate the number of moles of Fe2+ ions in 25.0 cm3 of solution F. number of moles of Fe2+ ions = … [2] (iii) Calculate the Mr of FeCrnO4 and use your answer to deduce the value of n. Mr of FeCrnO4 = … value of n = … [2] [Total: 12] * 0000800000011 * ,









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12 9701/42/O/N/24 © UCLES 2024 5 Ni2+ ions form a number of different complex ions, including [Ni(H2O)6]2+, [Ni(NH3)6]2+ and [Ni(en)3]2+. The abbreviation en represents 1,2-diaminoethane. The numerical values of two stability constants, Kstab, are given in Table 5.1. Table 5.1 complex Kstab [Ni(NH3)6 ]2+ 4.8 × 107 [Ni(en)3]2+ 2.0 × 1018 (a) Complete the expression for the Kstab of [Ni(en)3]2+. Kstab = [1] (b) A solution of [Ni(H2O)6]2+ is added to a solution that contains 0.10 mol dm–3 NH3 and 0.10 mol dm–3 en. (i) Predict which complex ion, [Ni(NH3)6 ]2+ or [Ni(en)3]2+, is present in the resulting mixture in the highest concentration. Explain your answer. complex ion present in largest concentration = … explanation … [1] (ii) Complete the equation for the ligand exchange reaction occurring in (i). [Ni(H2O)6]2+ + … … + … [1] * 0000800000012 * ,









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13 9701/42/O/N/24 © UCLES 2024 [Turn over (c) Complete Fig. 5.1 to show the three-dimensional structures of the two isomers of [Ni(en)3]2+. Use N N to represent the en ligand. Name the type of isomerism shown. Ni Ni Fig. 5.1 type of isomerism shown … [3] [Total: 6] * 0000800000013 * ,









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14 9701/42/O/N/24 © UCLES 2024 6 Fig. 6.1 shows two reactions of ethanedioic acid, HOOCCOOH. an excess of SOCl 2 G 2CO2 + H2O HOOCCOOH Fig. 6.1 (a) (i) Draw the organic product G in the box in Fig. 6.1. [1] (ii) In Fig. 6.1, SOCl 2 is given as the reagent that reacts with HOOCCOOH to produce G. Identify a different reagent that also reacts with HOOCCOOH to produce G. … [1] (b) Identify two different reagents that oxidise HOOCCOOH to form carbon dioxide and water. … … [2] (c) HOOCCOOH ionises as shown. HOOCCOOH HOOCCOO– + H+ HOOCCOOH is a much stronger acid than methanoic acid, HCOOH. Suggest an explanation for this difference in acidity. … … … [2] * 0000800000014 * ,









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15 9701/42/O/N/24 © UCLES 2024 [Turn over (d) Benzene-1,4-dicarboxylic acid, HOOCC6H4COOH, can be made from benzene, C6H6, in two steps as shown in Fig. 6.2. J step 1 step 2 HOOC COOH benzene-1,4-dicarboxylic acid Fig. 6.2 (i) Suggest the identity of J by drawing its structure in the box in Fig. 6.2. [1] (ii) Identify the reagents and conditions for step 1 and step 2. step 1 … step 2 … [2] (iii) Draw the structure of exactly one repeat unit of the polymer formed when benzene-1,4-dicarboxylic acid reacts with ethane-1,2-diol, HOCH2CH2OH. The linkage formed between the monomers should be shown fully displayed. [2] (iv) State the type of polymerisation that occurs when benzene-1,4-dicarboxylic acid reacts with ethane-1,2-diol and name the linkage formed between the monomers. type of polymerisation … linkage … [1] [Total: 12] * 0000800000015 * ,









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16 9701/42/O/N/24 © UCLES 2024 7 Benzene reacts with chlorine gas to form chlorobenzene. This reaction can be described as the reaction between benzene molecules and Cl + ions. The Cl + ions are formed by adding a suitable catalyst to the chlorine gas. (a) Give the name or formula of a catalyst that can be used for this reaction. … [1] (b) The mechanism for this reaction is shown. Cl + Cl Cl x y step 1 step 2 diagram 1 diagram 2 diagram 3 H + (i) The movement of a pair of electrons is represented by x in diagram 1. • State where this pair of electrons is before step 1 takes place. … • State where this pair of electrons is after step 1 has taken place. … [2] (ii) The movement of another pair of electrons is represented by y in diagram 2. • State where this pair of electrons is before step 2 takes place. … • State where this pair of electrons is after step 2 has taken place. … [2] (c) There are six carbon atoms in diagram 2. State how many of these carbon atoms are sp hybridised, sp2 hybridised, and sp3 hybridised. sp hybridised … sp2 hybridised … sp3 hybridised … [1] * 0000800000016 * ,









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17 9701/42/O/N/24 © UCLES 2024 [Turn over (d) Complete the equation for this reaction between benzene and chlorine. C6H6 + … … + … [1] (e) The mechanism for this reaction is electrophilic substitution. Complete the following sentence. Write formulae in the gaps provided. During this reaction, the electrophile is … and a … atom in benzene is substituted by a … atom. [1] (f) Chloroethane reacts with NaOH(aq). Chlorobenzene does not. (i) Name the mechanism of the reaction that chloroethane undergoes with NaOH(aq), and identify the major organic product that is formed. mechanism … major organic product … [1] (ii) Explain the difference in reactivity of chloroethane and chlorobenzene when treated with NaOH(aq). … … … [2] [Total: 11] * 0000800000017 * ,









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18 9701/42/O/N/24 © UCLES 2024 8 The amino acid serine, HOCH2CH(NH2)COOH, exists in two optically active forms. These optical isomers, isomer P and isomer Q, are shown in Fig. 8.1. isomer P isomer Q HOOC COOH H H NH2 H2N CH2OH CH2OH C C Fig. 8.1 (a) Isomer P and isomer Q have identical physical and chemical properties, with the exception of two specific properties. One of these two properties is their differing effect on plane polarised light. State the other property by which they differ. … … [1] (b) A solution of pure isomer P of a particular concentration rotates plane polarised light by 5.0° in a clockwise direction. Describe how a solution of pure isomer Q of the same concentration affects plane polarised light. … … [1] (c) State another term, in addition to stereoisomers, optical isomers and non-superimposable mirror images, which can be used to describe this pair of chiral compounds, isomer P and isomer Q. … [1] (d) Give the term used to describe a mixture containing equal amounts of isomer P and isomer Q. … [1] (e) Describe one way in which a single pure optical isomer of serine can be produced, instead of making a mixture of isomer P and isomer Q. … [1] * 0000800000018 * ,









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19 9701/42/O/N/24 © UCLES 2024 [Turn over (f) Complete Table 8.1 to describe the peaks seen in the proton (1H) NMR spectrum of HOCH2CH(NH2)COOH dissolved in D2O. Use as many rows in Table 8.1 as you need to, leaving the other rows blank. Table 8.1 group responsible for peak name of splitting pattern shown by peak explanation for splitting pattern [3] (g) Proline is a naturally occurring amino acid. The skeletal formula of proline is shown. O proline OH NH State the number of peaks in the carbon-13 (13C) NMR spectrum of proline. … [1] * 0000800000019 * ,









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20 9701/42/O/N/24 © UCLES 2024 (h) Glutamic acid is a naturally occurring amino acid. The skeletal formula of glutamic acid is shown. HO O NH2 O OH glutamic acid The isoelectric point of glutamic acid is pH 3. A sample of glutamic acid is dissolved in a solution of pH 1. A strong alkali is then added until the pH of the mixture reaches pH 14. During this process all possible ionised forms of glutamic acid are present at different times, depending on the pH of the solution. Complete the boxes below to show four different ionised forms of glutamic acid that are present at the stated pH values. at pH 1 at pH 3 at pH 9 at pH 14 [3] [Total: 12] * 0000800000020 * ,










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23 9701/42/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 9701/42/O/N/24 © UCLES 2024 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 12 printed pages. © Cambridge University Press & Assessment 2024 [Turn over Cambridge International AS & A Level CHEMISTRY 9701/42 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.

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 2 of 12 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.

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 3 of 12 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.

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 4 of 12 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.

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 5 of 12 Question Answer Marks 1(a)(i) Change in concentration = 0.135 [1] 0.135 / 1800 = 7.5  10–5 [1] 2 1(a)(ii) Rate = k[X] [1] 1 1(a)(iii) k = 0.693 / t½ / k = 0.693 / 900 / k = ln2 / t½ / k = ln2 / 900 / t½ = ln2 / k [1] 1 1(a)(iv) 1.16  10–4 [1] mol dm–3 s–1 [1] 2 1(b)(i) Any two from: • variable oxidation state • vacant / empty / unfilled d orbitals [1] • can form dative bonds / can accept electrons 1 1(b)(ii) Iron AND iron is solid, reactants are gases OR catalyst and reactants are in different phases / states [1] 1 1(b)(iii) Two for one mark, three for two marks: • adsorption of reactants by Pt • bonds of reactants weaken [1] • desorption of products [1] 2 1(b)(iv) NO2 + SO2 → NO + SO3 AND 2NO + O2 → 2NO2 [1] 1 1(c) BiO+ [1] 3H2SO3 + 2BiO+ + H2O → 3SO42- + 8H+ + 2Bi [1] 0.11 V [1] 3

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 6 of 12 Question Answer Marks 2(a) Hhyd increases from left to right due to increase in charge [1] ionic radius decreases from left to right [1] causing increased attractive force to water molecules [1] 3 2(b)(i) Mg2+ (aq) + 2Cl– (aq) [1] 1 2(b)(ii) Enthalpy change of hydration of magnesium ions and chloride ions [1] HhydMg2+ + 2HhydCl– [1] 2 2(b)(iii) Selects 155, 1920 and 364 only [1] 2  364 [1] answer –2493 [1] 3 2(c) The number of arrangements of the particles and of the energy in the system [1] 1 2(d) G = H – TS [1] answer +10.5 [1] 2 2(e)(i) No, G is positive [1] 1 2(e)(ii) Becomes more / soluble because G becomes more negative / less positive / smaller / closer to zero [1] 1 Question Answer Marks 3(a)(i) [H+] = 10–12.35 OR [H+] = 4.47  10–13 [1] Kw / 4.47  10–13 [1] OR pOH = 1.65 [1] [OH–] = 10–1.65 [1] 2 3(a)(ii) Ksp = (0.0112)(0.0224)2 OR Ksp = [Ca2+][OH–]2 [1] answer 5.62  10–6 [1] mol3 dm–9 [1] 3

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 7 of 12 Question Answer Marks 3(a)(iii) white solid / white ppt AND pH between 7.01 and 12.34 [1] common ion effect [1] hydroxide removed by precipitation OR ppt is Ca(OH)2 [1] 3 3(a)(iv) lattice energy and hydration energy greater for CaSO4 OR lattice energy and hydration energy decrease down group [1] hydration energy decreases more / is dominant factor [1] enthalpy of solution is more endothermic for BaSO4 OR is more endothermic down grp [1] 3 3(b)(i) Ca + 2CH3COOH → Ca(CH3COO)2 + H2 [1] 1 3(b)(ii) Pair 1: CH3COOH and CH3COO– [1] Pair 2: H3O+ and H2O OR H2O and OH– [1] 2 3(b)(iii) + 3 3 [H ][CH COO ] [CH COOH] − [1] 1 3(b)(iv) [CH3COO–] = 0.788, [CH3COOH] = 0.270 [H+] = 5.96  10–6 [1] pH = 5.22 [1] 2 3(b)(v) CH3COO– + H+ → CH3COOH OR Ca(CH3COO)2 + 2H+ → 2CH3COOH + Ca2+ [1] CH3COOH + OH– → CH3COO– + H2O [1] 2 Question Answer Marks 4(a)(i) red/pink blue [1] 1 4(a)(ii) [Co(H2O)6]2+ + 2OH– → Co(H2O)4(OH)2 + 2H2O [1] 1 4(a)(iii) acid-base [1] 1 4(b)(i) [CoL2]2+ [1] 1

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 8 of 12 Question Answer Marks 4(b)(ii) +2 [1] 1 4(b)(iii) 2 3 [1] 1 4(b)(iv) Not of the same energy / have different energy [1] 1 4(c)(i) 5Fe2+ + MnO4– + 8H+ → 5Fe3+ + Mn2+ + 4H2O [1] 1 4(c)(ii) 0.02  18.7/1000 = 3.74  10–4 [1] 3.74  10–4  5 = 1.87  10–3 [1] 2 4(c)(iii) 1.87  10–3  10 = 1.87  10–2 4.18/1.87  10–2 = 224 / 223.5(294) [1] n = 2 [1] 2 Question Answer Marks 5(a) 2 3 stab 2 3 [[Ni( ) ] ] [Ni ][en] en K + + = OR 2 3 2 3 2 6 [[Ni( ) ] ] [Ni(H O) ][en] en + + [1] 1 5(b)(i) [Ni(en)3]2+ AND larger Kstab OR more stable [1] 1 5(b)(ii) [Ni(H2O)6]2+ + 3en → [Nien3]2+ + 6H2O [1] 1

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 9 of 12 Question Answer Marks 5(c) octahedral with correct 3D for one [Ni(en)3]2+ [1] second optical isomer [1] optical isomerism [1] 3 Question Answer Marks 6(a)(i) ClOCCOCl [1] 1 6(a)(ii) PCl3 / PCl5 [1] 1 6(b) oxygen [1] acidified KMnO4 [1] 2 6(c) C=O are electron withdrawing / electronegative [1] weakens O–H bond OR stabilises anion [1] 2 6(d)(i) [1] 1 6(d)(ii) step 1: CH3Cl + AlCl3 [1] step 2: hot alkaline KMnO4 [1] 2

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 10 of 12 Question Answer Marks 6(d)(iii) correctly displayed ester linkage [1] rest of structure [1] 2 6(d)(iv) condensation AND ester [1] 1 Question Answer Marks 7(a) aluminium chloride OR AlCl3 [1] 1 7(b)(i) delocalised system / delocalised ring / pi system / pi ring [1] C–Cl bond [1] 2 7(b)(ii) C–H bond [1] delocalised system / delocalised ring / pi system / pi ring [1] 2 7(c) 0, 5, 1 [1] 1 7(d) C6H6 + Cl2 → C6H5Cl + HCl [1] 1 7(e) Cl + hydrogen / H chlorine / Cl [1] 1 7(f)(i) nucleophilic substitution AND ethanol [1] 1 7(f)(ii) delocalisation of LP of Cl with  system [1] C–Cl bond is stronger in chlorobenzene / partly double [1] 2

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 11 of 12 Question Answer Marks 8(a) biological activity [1] 1 8(b) rotates plane polarised light 5.0° in anticlockwise direction [1] 1 8(c) enantiomers [1] 1 8(d) racemic [1] 1 8(e) use of chiral catalyst [1] 1 8(f) CH2 and CH only in column one [1] CH2 gives a doublet, CH gives a triplet [1] doublet due to 1 proton on neighbouring carbons triplet due to 2 protons on neighbouring carbons [1] 3 8(g) 5 [1] 1

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 12 of 12 Question Answer Marks 8(h) at pH 1 [1] at pH 3 AND or at pH 9 [1] at pH 14 [1] 3