Cambridge A Level Chemistry 9701 — 2023 May/June Paper 4 · Variant 2

This document has 24 pages. [Turn over Cambridge International AS & A Level * 8 9 3 6 5 2 3 2 3 2 * DC (CE/SG) 311841/5 © UCLES 2023 CHEMISTRY 9701/42 Paper 4 A Level Structured Questions May/June 2023 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.

2 9701/42/M/J/23 © UCLES 2023 1 (a) Group 2 carbonates decompose when heated to form the metal oxide and carbon dioxide. (i) Suggest a mechanism for the decomposition of the carbonate ion by adding two curly arrows in Fig. 1.1. O O– –O C CO2 O2– + Fig. 1.1 [1] (ii) Describe the variation in the thermal stability of Group 2 carbonates. Explain your answer. … … … … … [3] (b) (i) Define lattice energy. … … … [2] (ii) The lattice energy of the Group 2 carbonates, ΔH o latt(MCO3), becomes less exothermic down the group. The lattice energy of the Group 2 oxides, ΔH o latt(MO), also becomes less exothermic down the group. ΔH o latt(MCO3) and ΔH o latt(MO) change by different amounts going down the group. Suggest how the standard enthalpy change of the decomposition reaction for Group 2 carbonates changes down the group. Explain your reasoning in terms of the relative sizes of the anions and the relative changes in lattice energy down the group. … … … … … [2]

3 9701/42/M/J/23 © UCLES 2023 [Turn over (c) Potassium sulfite, K2SO3, is used as a food additive. The concentration of sulfite ions, SO3 2–, can be determined by titration using aqueous acidified manganate(VII) ions, MnO4 –. • A 250 cm3 solution contains 3.40 g of impure K2SO3. • 25.0 cm3 of this solution requires 22.40 cm3 of 0.0250 mol dm–3 acidified MnO4 – to reach the end-point. All the SO3 2– ions are oxidised. None of the other species in the impure K2SO3 are oxidised. The reaction occurs as shown by the two half-equations. H2O + SO3 2– SO4 2– + 2H+ + 2e– MnO4 – + 8H+ + 5e– Mn2+ + 4H2O (i) Give the ionic equation for the reaction between SO3 2– and acidified MnO4 –. … [1] (ii) Calculate the percentage purity of the sample of K2SO3. Show your working. percentage purity of K2SO3 = … [3] (d) Potassium disulfite, K2S2O5, is another food additive. The disulfite ion, S2O5 2–, has the displayed formula shown in Fig. 1.2. O O O O– O– S S α Fig. 1.2 Deduce the geometry (shape) around the S(α) atom in S2O5 2–. geometry around S(α) … [1] [Total: 13]

4 9701/42/M/J/23 © UCLES 2023 2 (a) State two typical chemical properties of a transition element. 1 … 2 … [1] (b) Aqueous solutions of cobalt(II) salts contain the complex ion [Co(H2O)6]2+. (i) Define complex ion. … … [1] (ii) Samples of [Co(H2O)6]2+ are reacted separately with an excess of aqueous ammonia, with an excess of concentrated HCl and with an excess of aqueous sodium hydroxide, as shown in Fig. 2.1. excess NH3(aq) excess NaOH(aq) excess concentrated HCl [Co(H2O)6]2+(aq) A C B Fig. 2.1 Complete Table 2.1 about the reactions shown by [Co(H2O)6]2+. Table 2.1 reagent added to [Co(H2O)6]2+(aq) formula of cobalt species formed colour and state of cobalt species formed type of reaction an excess of NH3(aq) A = an excess of concentrated HCl B = an excess of NaOH(aq) C = [4]

5 9701/42/M/J/23 © UCLES 2023 [Turn over (c) The ethanedioate ion, C2O4 2–, can act as a bidentate ligand. (i) Explain what is meant by a bidentate ligand. … … … [2] (ii) The complex [Co(H2O)2(C2O4)BrCl ]− exists as stereoisomers. Complete the three-dimensional diagrams in Fig. 2.2 to show four stereoisomers of [Co(H2O)2(C2O4)BrCl ]−. The C2O4 2– ligand is represented using O O . O Co O O Co O O Co O O Co O Fig. 2.2 [3] (iii) State the oxidation state of cobalt in this complex and a type of stereoisomerism shown. oxidation state of cobalt … type of stereoisomerism … [1] [Total: 12]

6 9701/42/M/J/23 © UCLES 2023 3 (a) Complete Table 3.1 by placing one tick (3) in each row to indicate the sign of each type of energy change under standard conditions. Table 3.1 energy change always positive always negative can be either negative or positive bond energy enthalpy change of atomisation enthalpy change of formation [1] (b) Define standard enthalpy change of atomisation. … … … [2] (c) Table 3.2 shows some energy changes. Table 3.2 energy change value / kJ mol–1 standard enthalpy change of atomisation of silver +285 first ionisation energy of silver +731 second ionisation energy of silver +2074 bond energy of O=O +496 bond energy of O–O +150 first electron affinity of oxygen –141 second electron affinity of oxygen +798 first ionisation energy of oxygen +1314 standard enthalpy change of formation of silver oxide, Ag2O(s) –31

7 9701/42/M/J/23 © UCLES 2023 [Turn over Calculate the lattice energy, ΔH o latt, of Ag2O(s) using relevant data from Table 3.2. It may be helpful to draw a labelled energy cycle. Show your working. ΔH o latt of Ag2O(s) = … kJ mol–1 [3] (d) Suggest the trend in the magnitude of the lattice energies of the silver compounds Ag2S, Ag2O and Ag2Se. Explain your answer. … … … least exothermic most exothermic … … … [2]

8 9701/42/M/J/23 © UCLES 2023 (e) Silver sulfite, Ag2SO3(s), is sparingly soluble in water. (i) Give an expression for the solubility product, Ksp, of Ag2SO3. Ksp = [1] (ii) Calculate the equilibrium concentration of Ag+ in a saturated solution of Ag2SO3 at 298 K. [Ksp: Ag2SO3, 1.50 × 10–14 mol3 dm–9 at 298 K] [Ag+] = … mol dm–3 [1] (f) The standard enthalpy change of solution, ΔH o sol, of AgNO3(s) in water is +22.6 kJ mol–1. Suggest how the feasibility of dissolving AgNO3(s) in water changes with temperature. Explain your answer. … … … … [2] [Total: 12]

9 9701/42/M/J/23 © UCLES 2023 [Turn over 4 (a) In aqueous solution, iron(III) ions react with iodide ions, as shown. 2Fe3+ + 2I– 2Fe2+ + I2 A series of experiments is carried out using different concentrations of Fe3+ and I–, as shown in Table 4.1. Table 4.1 experiment [Fe3+] / mol dm–3 [I–] / mol dm–3 initial rate / mol dm–3 s–1 1 0.0400 0.0200 2.64 × 10–4 2 0.1200 0.0200 7.92 × 10–4 3 0.0800 0.0400 2.11 × 10–3 (i) Explain what is meant by overall order of reaction. … … … [1] (ii) Use the data in Table 4.1 to deduce the order of reaction with respect to Fe3+ and with respect to I–. Explain your reasoning. … … … … [2] (iii) Use your answer to (a)(ii) to construct the rate equation for this reaction. rate = [1] (iv) Use your answer to (a)(iii) and the data from experiment 1 to calculate the rate constant, k, for this reaction. Include the units of k. k = … units … [2] (v) Describe qualitatively the effect of an increase in temperature on the rate constant and on the rate of this reaction. … … [1]

10 9701/42/M/J/23 © UCLES 2023 (b) In aqueous solution, iodide ions react with acidified hydrogen peroxide, as shown. 2I– + H2O2 + 2H+ I2 + 2H2O The initial rate of reaction is found to be first order with respect to I–, first order with respect to H2O2 and zero order with respect to H+. Fig. 4.1 shows a possible four-step mechanism for this reaction. step 1 H2O2 + I– IO– + H2O step 2 H+ + IO– HIO step 3 HIO + I– I2 + OH– step 4 OH– + H+ H2O Fig. 4.1 (i) Suggest which of the steps, 1, 2, 3 or 4, in this mechanism is the rate-determining step. Explain your answer. … … … [1] (ii) Identify a step in Fig. 4.1 that involves a redox reaction. Explain your answer in terms of oxidation numbers. … … … [1] (iii) Suggest the role of HIO in this mechanism. Explain your reasoning. … … … [1] [Total: 10]

11 9701/42/M/J/23 © UCLES 2023 [Turn over 5 (a) Methylbenzene can undergo different reactions, as shown in Fig. 5.1. reaction 1 H2 / Pt Br2 / UV light Br2 / FeBr3 + reaction 2 reaction 3 Fig. 5.1 (i) Draw structures in Fig. 5.1 for the possible organic products of the three reactions shown. [3] (ii) Complete Table 5.1. Table 5.1 type of reaction mechanism reaction 1 reaction 2 [2] (b) When methylbenzene reacts with an electrophile, a substitution reaction occurs. No addition reaction takes place under these conditions. Explain why no addition reaction takes place. … … [1]

12 9701/42/M/J/23 © UCLES 2023 (c) The reaction of methylbenzene with thionyl bromide, SOBr2, in the presence of an iron(III) bromide catalyst, FeBr3, is shown in Fig. 5.2. FeBr3 + SOBr2 + HBr S O Br Fig. 5.2 The mechanism of this reaction is similar to that of the bromination of benzene. The first step of the mechanism generates the SOBr+ electrophile, as shown. SOBr2 + FeBr3 SOBr+ + FeBr4 – (i) The reaction of methylbenzene with SOBr+ ions is shown in Fig. 5.3. Complete the mechanism in Fig. 5.3. Include all relevant curly arrows and charges. Draw the structure of the organic intermediate. S+ O Br intermediate + … S O Br Fig. 5.3 [3]

13 9701/42/M/J/23 © UCLES 2023 [Turn over (ii) The reaction shown in Fig. 5.2 produces a small amount of a by-product, P, with the molecular formula C14H14OS. Suggest a structure for by-product P. [1] (d) Acyl bromides, RCOBr, can be synthesised by the reaction of a carboxylic acid and SOBr2. This is a similar reaction to the synthesis of acyl chlorides using SOCl 2. (i) Give an equation for the reaction between ethanoic acid and SOBr2. … [1] (ii) Suggest the relative ease of hydrolysis of acyl bromides, RCOBr, acyl chlorides, RCOCl, and alkyl chlorides, RCl. Explain your answer. … > … > … easiest to hydrolyse hardest to hydrolyse … … … … … [3] [Total: 14]

14 9701/42/M/J/23 © UCLES 2023 6 (a) Perindopril is a drug used to treat heart disease. O N N O O O OH H perindopril Fig. 6.1 (i) State the number of chiral carbon atoms present in one molecule of perindopril. … [1] (ii) Suggest one benefit and one disadvantage of producing a drug such as perindopril as a single pure optical isomer. benefit … … disadvantage … … [2] (b) (i) Name all the functional groups in perindopril. … … [2]

15 9701/42/M/J/23 © UCLES 2023 [Turn over (ii) A sample of perindopril is hydrolysed with hot aqueous acid. Draw the structures of the three organic products of the complete acid hydrolysis of perindopril. [3] [Total: 8]

16 9701/42/M/J/23 © UCLES 2023 7 (a) Explain why phenol is brominated much more easily than benzene is brominated. … … … … … [3] (b) Iodine monobromide, I–Br, reacts with benzene in the presence of an Al Br3 catalyst. Predict whether the organic product will be bromobenzene or iodobenzene. Explain your answer. … … … [1] (c) Fig. 7.1 shows some reactions of phenol. OH OH OH N N O2N Q R reaction 1 reaction 2 excess Br2(aq) Na(s) Fig. 7.1 (i) Give an equation for the reaction of phenol with Na(s). … [1]

17 9701/42/M/J/23 © UCLES 2023 [Turn over (ii) Draw the structure of the organic product, R, formed when phenol reacts with an excess of Br2(aq). [1] (iii) State the reagents and conditions for reaction 1 and reaction 2 in Fig. 7.1. reaction 1 … reaction 2 … [2] [Total: 8]

18 9701/42/M/J/23 © UCLES 2023 8 (a) Describe the difference in reactivity between chloroethane and chlorobenzene with OH–(aq). Explain your answer. … … … … [2] (b) Compound T, C5H9O2Cl, is a useful synthetic intermediate. Fig. 8.1 shows some reactions of T. T Z, C6H13NO2 O O Cl O O NC step 1 step 2 W X excess NaOH(aq) OH heat + Y + excess HCl (aq) heat Fig. 8.1

19 9701/42/M/J/23 © UCLES 2023 [Turn over (i) Give the systematic name for T. … [1] (ii) Draw the structures of W, X, Y and Z in Fig. 8.1. [4] (iii) State the reagents and conditions for steps 1 and 2 in Fig. 8.1. step 1 … step 2 … [2]

20 9701/42/M/J/23 © UCLES 2023 (c) The proton (1H) NMR spectrum of compound T, C5H9O2Cl, in CDCl 3 is shown in Fig. 8.2. T O O 5 4 3 chemical shift δ / ppm δ 3.9 δ 3.7 δ 2.8 δ 1.2 2 1 0 Cl Fig. 8.2 Table 8.1 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

21 9701/42/M/J/23 © UCLES 2023 [Turn over (i) Suggest why CDCl 3 is used as a solvent instead of CHCl 3 for the proton (1H) NMR spectrum. … … [1] (ii) Complete Table 8.2 for the proton (1H) NMR spectrum of T. Table 8.2 chemical shift δ / ppm environment of proton splitting pattern number of 1H atoms responsible for the peak 1.2 2.8 3.7 3.9 [4] (iii) Explain the splitting pattern of the peak at δ 3.9 ppm. … … [1] [Total: 15]

22 9701/42/M/J/23 © UCLES 2023 9 (a) Define standard cell potential, E o cell. … … [1] (b) An electrochemical cell is set up to measure E o cell of a cell consisting of an Fe3+ / Fe2+ half-cell and a Cl 2 / Cl – half-cell. Draw a labelled diagram of this electrochemical cell. Include all necessary substances. It is not necessary to state conditions used. [3] (c) The cell reaction for the electrochemical cell in (b) is shown. Cl 2 + 2Fe2+ 2Fe3+ + 2Cl – E o cell = +0.59 V Calculate ΔG o , in kJ mol–1, for this cell reaction. ΔG o = … kJ mol–1 [2]

23 9701/42/M/J/23 © UCLES 2023 (d) Another experiment is set up using the same electrochemical cell. In this experiment the Fe2+ concentration is 0.15 mol dm–3. All other concentrations remain at their standard values. The Nernst equation is shown. E = E o + (0.059 / z) log [oxidised species] [reduced species] (i) Use the Nernst equation to calculate the electrode potential, E, for the Fe3+ / Fe2+ half-cell in this experiment. [E o : Fe3+ / Fe2+ = +0.77 V] E = … V [1] (ii) Use your answer to (d)(i) to calculate Ecell for this electrochemical cell. Ecell = … V [1] [Total: 8] 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)

24 9701/42/M/J/23 © UCLES 2023 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 144.4 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 –

This document consists of 19 printed pages. © UCLES 2023 [Turn over Cambridge International AS & A Level CHEMISTRY 9701/42 Paper 4 A Level Structured Questions May/June 2023 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 2023 series for most Cambridge IGCSE, Cambridge International A and AS Level and Cambridge Pre-U components, and some Cambridge O Level components.

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2023 © UCLES 2023 Page 2 of 19 Generic Marking Principles These general marking principles must be applied by all examiners when marking candidate answers. They should be applied alongside the specific content of the mark scheme or generic level descriptors 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 May/June 2023 © UCLES 2023 Page 3 of 19 GENERIC MARKING PRINCIPLE 5: Marks should be awarded using the full range of marks defined in the mark scheme for the question (however; the use of the full mark range may be limited according to the quality of the candidate responses seen). GENERIC MARKING PRINCIPLE 6: Marks awarded are based solely on the requirements as defined in the mark scheme. Marks should not be awarded with grade thresholds or grade descriptors in mind.

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2023 © UCLES 2023 Page 4 of 19 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 May/June 2023 © UCLES 2023 Page 5 of 19 6 Calculation specific guidance Correct answers to calculations should be given full credit even if there is no working or incorrect working, unless the question states ‘show your working’. For questions in which the number of significant figures required is not stated, credit should be awarded for correct answers when rounded by the examiner to the number of significant figures given in the mark scheme. This may not apply to measured values. For answers given in standard form (e.g. a  10n) in which the convention of restricting the value of the coefficient (a) to a value between 1 and 10 is not followed, credit may still be awarded if the answer can be converted to the answer given in the mark scheme. Unless a separate mark is given for a unit, a missing or incorrect unit will normally mean that the final calculation mark is not awarded. Exceptions to this general principle will be noted in the mark scheme. 7 Guidance for chemical equations Multiples / fractions of coefficients used in chemical equations are acceptable unless stated otherwise in the mark scheme. State symbols given in an equation should be ignored unless asked for in the question or stated otherwise in the mark scheme.

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2023 © UCLES 2023 Page 6 of 19 Question Answer Marks 1(a)(i) ARROW 1 starts from O– to the C–O bond AND ARROW 2 starts at the C-O bond to other O– ion 1 1(a)(ii) M1 increases (down the group) M2 (cat)ionic radius / ion size increases (down the group) OR charge density of M2+ decreases M3 less polarisation / distortion of anion / carbonate ion / CO3(2)– 3 1(b)(i) M1 energy released when one mole of a ionic solid / compound is formed M2 from gas (phase) ion(s) / gaseous ion(s) (under standard conditions) 2 1(b)(ii)  (Hdecomp / it) becomes more positive/less negative (down the group)  size /(ionic) radii of oxide ion is smaller (than carbonate ion) ORA  so Hlatt of oxides becomes ORA less exothermic faster OR less negative faster OR changes more OR changes faster Any two [1], all three [2] 2 1(c)(i) 2MnO4– + 6H+ + 5SO32–  2Mn2+ + 3H2O + 5SO42– 1

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2023 © UCLES 2023 Page 7 of 19 Question Answer Marks 1(c)(ii) M1 M2 any two bullets [1] or all four [2]  moles MnO4– = 0.025  22.40 / 1000 = 5.6  10–4  moles SO32– = 5.6  10–4  5 / 2 = 1.4  10–3 (in 25 cm3) ecf from (c)(i) and bullet 1  moles SO32– = 1.4  10–2 (in 250 cm3) ecf bullet 2  mass K2SO3 = 1.4  10–2  158.3 = 2.2162 g ecf bullet 3 OR moles K2SO3 (if 100% pure) = 3.40 ÷ 158.3 = 0.02148 M3 % purity = 100  2.2162 / 3.40 = 65.2 / 65.3 % ecf min 2sf OR % purity = 100  0.014 / 0.02148 = 65.2 / 65.3 % 3 1(d) S() tetrahedral 1 Question Answer Marks 2(a) Any two from:  they have variable / multiple oxidation states OWTTE  they behave as catalysts  they form complex ions / complexes  they form coloured compounds/ions 1 2(b)(i) (is a molecule or ion formed by a central) metal atom / metal ion bonded / surrounded by (one or more) ligands 1

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2023 © UCLES 2023 Page 8 of 19 Question Answer Marks 2(b)(ii) reagent added formula of species formed colour / state of species formed type of reaction an excess of NH3(aq) [Co(NH3)6]2+ brown solution ALLOW yellow- brown solution ligand exchange an excess of concentrated HCl [CoCl4]2– blue solution ligand exchange NaOH(aq) Co(OH)2 OR Co(OH)2(H2O)4 blue ppt. ALLOW pink ppt precipitation formula and colour / state: any two [1], any four [2], all six [3] type of reaction: all three correct [1] 4 2(c)(i) M1 (a species) that donates / uses two lone pairs M2 to form a two dative / coordinate bond to a metal atom / metal ion 2

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2023 © UCLES 2023 Page 9 of 19 Question Answer Marks 2(c)(ii) Any four from: Any two [1], any three [2], all four [3] 3 2(c)(iii) oxidation state of cobalt (+)3 AND type of stereoisomerism cis-trans OR geometric(al) OR optical 1

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2023 © UCLES 2023 Page 10 of 19 Question Answer Marks 3(a) energy change always positive always negative can be either negative or positive bond energy  enthalpy change of atomisation  enthalpy change of formation  1 3(b) M1 (enthalpy change) when one mole of gaseous atoms is produced IGNORE energy released M2 from its element(s) in its standard state / standard conditions / 298 K AND 1 atm 2 3(c) M1 use of correct six numbers only –31 / 285 / 731 / –141 / 798 / 496 M2 2  used correctly with Ag (2  285 (570) and 2  731 (1462)) AND 0.5 with O=O (496 (248)) M3 correct signs and evaluation –31 = (2  285) + (2  731) + (–141) + (798) + x + (0.5  496) x = –2968 kJ mol–1 3 3(d)  Ag2Se Ag2S Ag2O OWTTE least exothermic most exothermic  charge density of anion decreases down the group ORA / radius/size of anion increases down the group / Se2– largest radius / O2– smallest radius / O has smallest ionic radius  less attraction between the ions / ionic bond gets weaker (with Ag2Se) ORA Any two [1], all three [2] 2 3(e)(i) (Ksp = ) [Ag+]2[SO32–] [1] 1

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2023 © UCLES 2023 Page 11 of 19 Question Answer Marks 3(e)(ii) x = 3√1.5  10–14 / 4 = 1.55  10–5 (mol dm-3) [Ag+] = 1.55  10–5  2 = 3.11  10–5 (mol dm-3) min 2sf ecf (e)(i) 1 3(f)  feasibility / it increases as temperature increases ORA  S is positive / S is >0 / entropy change is positive (and H is positive) OR –TS becomes more negative / TS becomes more positive  as G becomes / is negative / G < 0 Any two [1], all three [2] 2 Question Answer Marks 4(a)(i) the sum/total of the power / exponent to which a concentration of a reactant is raised in the rate equation / law 1 4(a)(ii) M1 (expt 1 and 2) when [Fe3+]  3, rate  3 so first order w.r.t. Fe3+ M2 (expt 1 and 3) when [Fe3+]  2, [I–]  2, rate  8 so second order w.r.t. I– OR (expt 2 and 3) when [Fe3+] ÷ 1.5, [I–]  2, rate  2.7 so second order w.r.t. I– 2 4(a)(iii) (rate =) k[Fe3+][I–]2 [1] ecf (a)(ii) 1 4(a)(iv) k = rate / [Fe3+][I–]2 = (2.64  10–4) / (0.04  0.022) k = 16.5 min 2sf ecf units = mol-2 dm6 s–1 ecf 2 4(a)(v) (k and rate of reaction) both increase 1

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2023 © UCLES 2023 Page 12 of 19 Question Answer Marks 4(b)(i) step 1 as this has one mole of H2O2 and one mole of I– OR step 1 as correct stoichiometry / coefficients of 1 H2O2 and 1 I– OR step 1 as number of moles of each reactant are consistent with rate equation / their orders 1 4(b)(ii) step 1 I oxid no –1  +1 AND O oxid no –1  –2 OR step 3 I oxid no –1  0 AND I oxid no +1  0 1 4(b)(iii) intermediate AND formed (in step 2) and used up (in step 3) ALLOW oxidising agent (in step 3) AND oxidises I– (to I2) 1 Question Answer Marks 5(a)(i) M1 M2 M3 BOTH 3 5(a)(ii) M1 reaction 1: hydrogenation / reduction M2 mechanism 2: (free) radical substitution 2 5(b) the substitution product is stabilised by delocalization of -electrons / by -electrons in the ring OR the addition product is not stabilised by delocalisation (of -)electrons ALLOW addition product will remove -electron delocalised system 1

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2023 © UCLES 2023 Page 13 of 19 Question Answer Marks 5(c)(i) M1 curly arrow from inside the hexagon to S atom M2 structure of the intermediate M3 curly arrow from C-H bond into the ring AND formation/loss of H+ 3 5(c)(ii) 1 5(d)(i) CH3COOH + SOBr2  CH3COBr + HBr + SO2 1

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2023 © UCLES 2023 Page 14 of 19 Question Answer Marks 5(d)(ii) M1 acyl bromides > acyl chlorides > alkyl chlorides easiest hardest M2 M3 Any two from: Acyl bromides easiest  in acyl bromides C–Br is much weaker ORA AND due to less orbital overlap / Br having a larger atomic radii Acyl halides easier than alkyl chlorides  carbon in C–Cl / C–Br / C–X bond is more +/electron deficient OR C–Cl / C–Br bond is weaker (than C-Cl in alkyl chlorides) AND attached to an oxygen atom / two electronegative atoms / electron withdrawing C=O group ORA alkyl chlorides hardest  in alkyl chlorides C–Cl bond strengthened AND by positive inductive effect / electron donating effect of alkyl / R group 3 Question Answer Marks 6(a)(i) 5 / five 1 6(a)(ii) M1 benefit: higher biological efficiency / activity (of the drug) OR less side effects OR smaller dose required (as drug more potent) M2 disadvantage: the need to separate (a racemic mixture into a single stereoisomer) OR lower yield (of biologically active molecule/product) OR need a chiral catalyst / enzyme (in the synthesis so expensive) 2 6(b)(i) carboxylic acid/carboxyl, ester, amide, amine Any two [1], all four [2] 2

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2023 © UCLES 2023 Page 15 of 19 Question Answer Marks 6(b)(ii) 3  [1] for each structure 3 Question Answer Marks 7(a) M1 p-orbital / lone pair from O atom is overlaps / delocalised into the ring M2 greater  electron density around the ring OR makes the ring more electron rich / positions 2,4,6 more electron rich M3 polarises electrophiles more easily 3 7(b) iodobenzene AND as Br is more electronegative (than I / I+/I+ in the electrophile) 1 7(c)(i) C6H6O / C6H5OH + Na  C6H5O–Na+ + ½H2 1 7(c)(ii) 1

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2023 © UCLES 2023 Page 16 of 19 Question Answer Marks 7(c)(iii) M1 reaction 1: aqueous HNO3 / dilute HNO3 M2 reaction 2: alkaline/ NaOH AND C6H5N2+ (Cl–) OR alkaline / NaOH AND benzene / phenyl diazonium ion / salt 2 Question Answer Marks 8(a)  chlorobenzene is less reactive than chloroethane OWTTE  p-orbital / lone pair on Cl will overlap / delocalise into the ring  due to partial double C-Cl bond OWTTE OR C-Cl bond strengthened (more) Any two [1], all three [2] 2 8(b)(i) ethyl 3-chloropropanoate 1

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2023 © UCLES 2023 Page 17 of 19 Question Answer Marks 8(b)(ii) HO O ONa W O OH Y HO O O O H2N Z X OH 4  [1] for each structure 4 8(b)(iii) M1 step 1 NaCN / KCN in ethanol (heat) M2 step 2 H2 + Ni / H2 + Pt ALLOW LiAlH4 2 8(c)(i) (because solvent / D / CDCl3) doesn’t give a signal / peak / absorption ORA 1

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2023 © UCLES 2023 Page 18 of 19 Question Answer Marks 8(c)(ii) chemical shift δ / ppm environment of proton splitting pattern number of 1H atoms responsible for the peak 1.2 alkyl / alkane / (R-)CH(3) triplet 3 2.8 alkyl next to C=O / CH(2)C=O triplet 2 3.7 alkyl next to electronegative atom / CH(2)-Cl triplet 2 3.9 alkyl next to electronegative atom / CH(2)-O quartet / quadruplet 2 Any three [1], any six [2], any nine [3], all twelve [4] 4 8(c)(iii) ( = 3.9) three H on neighbouring / adjacent C / it’s next to a CH3 1

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2023 © UCLES 2023 Page 19 of 19 Question Answer Marks 9(a) potential difference / voltage between the two half-cells / two electrodes (in a cell) under standard conditions 1 9(b) salt bridge voltmeter complete circuit [wires/salt bridge in/touching both solutions] Cl2 Cl – good delivery system (no arrow required) Pt Pt Fe2+ and Fe3+ Any three [1], any six [2], all nine [3] 3 9(c) M1 Go ) = – nFEcello OR –2  96500  0.59  M2 Go = –2  96500  0.59 = –113870 J mol-1 Go = –114 kJ mol-1 min 3sf ecf 2 9(d)(i) E = 0.77 + (0.059 / 1) log (1/0.15) use of z = 1 E = 0.82 V 1 9(d)(ii) Ecell = 0.59 + 0.77 – (answer to (d)(i)) = 0.54 V ecf 1