Cambridge A Level Chemistry 9701 — 2023 Feb/March Paper 4 · Variant 2

This document has 28 pages. Any blank pages are indicated. [Turn over Cambridge International AS & A Level * 7 3 4 9 9 8 8 3 1 2 * DC (CE/SG) 311729/4 © UCLES 2023 CHEMISTRY 9701/42 Paper 4 A Level Structured Questions February/March 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/F/M/23 © UCLES 2023 1 (a) The most common zinc mineral contains zinc(II) sulfide, ZnS. (i) Complete the electrons in boxes diagram in Fig. 1.1 to show the electronic configuration of a zinc(II) ion. [Ar] 3d 4s 4p Fig. 1.1 [1] (ii) Complete Fig. 1.2 to show the Born–Haber diagram for the ionic solid ZnS. Include state symbols of relevant species. Zn2+(g) + S(g) + 2e– ZnS(s) Zn2+(g) + S–(g) + e– Zn2+(g) + S2–(g) IE2 IE1 ΔHf ΔHlatt ΔHat(S) EA2 EA1 ΔHat(Zn) Fig. 1.2 [3]

3 9701/42/F/M/23 © UCLES 2023 [Turn over (iii) Describe the trend in the first electron affinity of the Group 16 elements S to Te. Explain your answer. … … … … [2] (iv) Explain why the lattice energy, ∆Hlatt, of ZnO is more exothermic than that of ZnS. … … … … [2] (b) Zinc metal can be obtained in a two-step process as shown. step 1 2ZnS(s) + 3O2(g) 2ZnO(s) + 2SO2(g) step 2 ZnO(s) + C(s) Zn(l) + CO(g) The reactions are carried out at 800 °C. (i) Predict the sign of the entropy change, ∆S o , of the reaction in step 1. Explain your answer. … … [1] (ii) Use the data in Table 1.1 to calculate ∆S o of the reaction shown in step 2. Table 1.1 chemical ZnO(s) C(s) Zn(l) CO(g) S o / J K–1 mol–1 43.7 5.7 50.8 197.7 ∆S o = … J K–1 mol–1 [1]

4 9701/42/F/M/23 © UCLES 2023 (iii) An equation for the direct reduction of ZnS by carbon is shown. 2ZnS(s) + C(s) 2Zn(l) + CS2(g) ∆H o = +733 kJ mol–1 ∆S o = +218 J K–1 mol–1 This reaction is not feasible at 800 °C. Calculate ∆G o for this reaction at 800 °C. ∆G o = … kJ mol–1 [2] (c) Zn(NO3)2 undergoes thermal decomposition when heated. The reaction is similar to the thermal decomposition of Group 2 nitrates. (i) Construct an equation for the thermal decomposition of Zn(NO3)2. … [1] (ii) The radii of some Group 2 cations and Zn2+ are shown in Table 1.2. Table 1.2 cation Mg2+ Ca2+ Sr2+ Ba2+ Zn2+ radius / pm 65 99 113 135 74 State and explain the trend in thermal stability of the Group 2 nitrates down the group. … … … [2] (iii) Use Table 1.2 to suggest which Group 2 nitrates are less thermally stable than zinc nitrate. … [1] [Total: 16]

5 9701/42/F/M/23 © UCLES 2023 [Turn over 2 Hypophosphorous acid is an inorganic acid. The conjugate base of hypophosphorous acid is H2PO2 –. (a) Give the formula of hypophosphorous acid. … [1] (b) H2PO2 – is a strong reducing agent. It can be used to reduce metal cations without the need for electrolysis. equation 1 HPO3 2– + 2H2O + 2e– H2PO2 – + 3OH– E o = –1.57 V (i) In an experiment, an alkaline HPO3 2–/H2PO2 – half-cell is constructed with [H2PO2 –] = 0.050 mol dm–3. All other ions are at their standard concentration. Predict how the value of E of this half-cell differs from its E o value. Explain your answer. … … … … [2] (ii) The Cr3+/Cr half-cell has a standard electrode potential of –0.74 V. An electrochemical cell consists of an alkaline HPO3 2–/H2PO2 – half-cell and a Cr3+/Cr half-cell. Calculate the standard cell potential, Ecell. Ecell = … V [1] o o

6 9701/42/F/M/23 © UCLES 2023 (iii) Complete the diagram in Fig. 2.1 to show how the standard electrode potential of the Cr3+/Cr half-cell can be measured relative to that of the standard hydrogen electrode. Identify the chemicals, conditions and relevant pieces of apparatus. Fig. 2.1 [3] (iv) Label Fig. 2.1 to show: • which is the positive electrode • the direction of electron flow in the external circuit. [1] (v) H2PO2 – reduces Ni2+ to Ni in alkaline conditions. Use equation 1 to construct the ionic equation for this reaction. equation 1 HPO3 2– + 2H2O + 2e– H2PO2 – + 3OH– … [1]

7 9701/42/F/M/23 © UCLES 2023 [Turn over (c) H2PO2 –(aq) reacts with OH–(aq). H2PO2 –(aq) + OH–(aq) HPO3 2–(g) + H2(g) Table 2.1 shows the results of a series of experiments used to investigate the rate of this reaction. Table 2.1 experiment [H2PO2 –(aq)] / mol dm–3 [OH–(aq)] / mol dm–3 volume of H2 produced in 60 s / cm3 1 0.40 2.00 6.4 2 0.80 2.00 12.8 3 1.20 1.00 4.8 (i) The volume of H2 was measured under room conditions. Use the molar volume of gas, Vm, and the data from experiment 1 to calculate the rate of reaction in mol dm–3 s–1. rate of reaction = … mol dm–3 s–1 [1] (ii) The rate equation was found to be: rate = k [H2PO2 –(aq)] [OH–(aq)]2 Show that the data in Table 2.1 is consistent with the rate equation. … … … … … [2] (iii) State the units of the rate constant, k, for the reaction. … [1]

8 9701/42/F/M/23 © UCLES 2023 (iv) The experiment is repeated using a large excess of OH–(aq). Under these conditions, the rate equation is: rate = k1 [H2PO2 –(aq)] k1 = 8.25 × 10–5 s–1 Calculate the value of the half-life, t1 2, of the reaction. t1 2 = … s [1] (v) Describe how an increase in temperature affects the value of the rate constant, k1. … … [1] (d) A student suggests that the reaction between H2PO2 –(aq) and OH–(aq) might happen more quickly in the presence of a heterogeneous catalyst. Describe the mode of action of a heterogeneous catalyst. … … … … [2] [Total: 17]

9 9701/42/F/M/23 © UCLES 2023 [Turn over 3 Vanadium is a transition element in Period 4 of the Periodic Table. (a) Define transition element. … … [1] (b) Vanadium shows typical chemical properties of transition elements, including variable oxidation states. (i) State two other typical chemical properties of transition elements. 1 … 2 … [1] (ii) Explain why transition elements have variable oxidation states. … … [1]

10 9701/42/F/M/23 © UCLES 2023 (c) VO2 + can be reduced to V2+ by C2O4 2– in acidic conditions. equation 2 2VO2 + + 3C2O4 2– + 8H+ 2V2+ + 6CO2 + 4H2O (i) In a titration, 25.00 cm3 of 0.0300 mol dm–3 VO2 +(aq) is added to 10 cm3 of dilute sulfuric acid. A solution of 0.0400 mol dm–3 C2O4 2–(aq) is then added from a burette until the end-point is reached. The titration is repeated and concordant results obtained, as shown in Table 3.1. Table 3.1 1 2 volume of C2O4 2–(aq) added / cm3 28.15 28.10 Show that these results are consistent with the stoichiometry of equation 2. [2] (ii) An excess of C2O4 2– reacts with VO2 + to form a mixture of two octahedral complex ions. The complex ions are stereoisomers of each other. Each complex ion contains a V2+ cation and three C2O4 2– ions. Complete the diagram to show the three-dimensional structure of one of the complex ions. Include the charge of the complex ion. Use O O to represent a C2O4 2– ion. V [2]

11 9701/42/F/M/23 © UCLES 2023 [Turn over (d) V2+(aq) can be oxidised by H2O2(aq). Table 3.2 gives some relevant data. Table 3.2 half-equation E o / V 1 H2O2(aq) + 2H+(aq) + 2e– 2H2O(l) +1.77 2 VO2 +(aq) + 2H+(aq) + e– VO2+(aq) + H2O(l) +1.00 3 VO2+(aq) + 2H+(aq) + e– V3+(aq) + H2O(l) +0.34 4 V3+(aq) + e– V2+(aq) –0.26 (i) Identify the vanadium species that forms when an excess of H2O2(aq) reacts with V2+(aq) under standard conditions. Explain your answer with reference to the data in Table 3.2. … … … [1] (ii) Concentrated acidified H2O2 can react with V2+ to form red VO2 3+ ions. VO2 3+ contains vanadium combined with the peroxide anion, O2 2–. Deduce the oxidation state of vanadium in VO2 3+. … [1] [Total: 9]

12 9701/42/F/M/23 © UCLES 2023 4 Ethylamine and phenylamine are primary amines. NH2 phenylamine ethylamine CH3CH2NH2 Fig. 4.1 These two compounds are synthesised by different methods. (a) Several methods can be used to form ethylamine. (i) Ethylamine forms when ethanamide, CH3CONH2, is reduced by LiAl H4. Write an equation for this reaction. Use [H] to represent one atom of hydrogen from the reducing agent. … [1] (ii) Ethylamine is a product of the reaction of bromoethane with ammonia. Name the mechanism of this reaction and state the conditions used. mechanism … conditions … [2] (iii) The reaction in (a)(ii) also forms secondary and tertiary amines. Suggest the identity of a secondary or tertiary amine formed by the reaction in (a)(ii). … [1]

13 9701/42/F/M/23 © UCLES 2023 [Turn over (b) Ethylamine is a weak base. State the relative basicities of ammonia, ethylamine and phenylamine. Explain your answer. … < … < … least basic most basic … … … … … … … [4] (c) Pure phenylamine, C6H5NH2, can be prepared from benzene in two steps. Draw the structure of the intermediate compound. Suggest reagents and conditions for each step. … … … … … [3]

14 9701/42/F/M/23 © UCLES 2023 (d) Fig. 4.2 shows some reactions of phenylamine. NH2 N2Cl OH N N phenylamine reaction 1 reaction 2 excess Br2(aq) room temperature reaction 4 W reaction 3 X Y benzenediazonium chloride Fig. 4.2

15 9701/42/F/M/23 © UCLES 2023 [Turn over (i) Draw the structure of W, the organic product of reaction 1. [1] (ii) State the reagents used in reaction 2. … [1] Benzenediazonium chloride, C6H5N2Cl, and X react together in reaction 4 to form Y, an azo compound. (iii) Name X, the organic product of reaction 3. … [1] (iv) State the necessary conditions for reaction 4 to occur. … [1] (v) Suggest a use for Y. … [1]

16 9701/42/F/M/23 © UCLES 2023 (e) Methylamine, CH3NH2, is another primary amine. CH3NH2 can act as a monodentate ligand. (i) Define monodentate ligand. … … … [2] (ii) Cu2+(aq) reacts with CH3NH2 to form [Cu(CH3NH2)2(H2O)4]2+. Draw three-dimensional diagrams to show the two geometrical isomers of [Cu(CH3NH2)2(H2O)4]2+. Cu Cu [2] (iii) State the coordination number of copper in [Cu(CH3NH2)2(H2O)4]2+. … [1] (f) Cd2+(aq) ions form tetrahedral complexes with CH3NH2, OH– and Cl – ions, as shown in equilibria 1, 2 and 3. equilibrium 1 Cd2+(aq) + 4CH3NH2(aq) [Cd(CH3NH2)4]2+(aq) Kstab = 3.3 × 106 equilibrium 2 Cd2+(aq) + 4OH–(aq) Cd(OH)4 2–(aq) Kstab = 5.0 × 108 equilibrium 3 Cd2+(aq) + 4Cl –(aq) CdCl 4 2–(aq) Kstab = 6.3 × 102 (i) Give the units of Kstab for equilibrium 1. … [1] (ii) Write an expression for Kstab for equilibrium 3. Kstab = [1]

17 9701/42/F/M/23 © UCLES 2023 [Turn over (iii) A solution of Cl –(aq) is added to Cd2+(aq) and allowed to reach equilibrium. The equilibrium concentrations are given. [Cd2+(aq)] = 0.043 mol dm–3 [Cl –(aq)] = 0.072 mol dm–3 Use your expression in (f)(ii) to calculate the concentration of CdCl 4 2–(aq) in the equilibrium mixture. [CdCl 4 2–(aq)] = … mol dm–3 [1] (iv) When CH3NH2(aq) is added to Cd2+(aq), a mixture of [Cd(CH3NH2)4]2+(aq) and [Cd(OH)4]2–(aq) forms. Suggest how the [Cd(OH)4]2–(aq) is formed. … … [1] (v) Cd2+(aq) exists as a complex ion, [Cd(H2O)6]2+(aq). Identify the most stable and the least stable of the complexes in Table 4.1 by placing one tick (3) in each column. Explain your answer. Table 4.1 complex most stable least stable [Cd(H2O)6]2+(aq) [Cd(OH)4]2–(aq) [Cd(CH3NH2)4]2+(aq) [CdCl4]2–(aq) explanation … … … [2] [Total: 27]

18 9701/42/F/M/23 © UCLES 2023 5 Tulobuterol is used in some medicines. Cl OH H N tulobuterol Fig. 5.1 (a) Tulobuterol contains a benzene ring in its structure. Describe and explain the shape of benzene. In your answer, include: • the bond angle between carbon atoms • the hybridisation of the carbon atoms • how orbital overlap forms v and r bonds between the carbon atoms. … … … … [3]

19 9701/42/F/M/23 © UCLES 2023 [Turn over (b) In a synthesis of tulobuterol, the first step involves the formation of chlorobenzene. Benzene reacts with Cl2 in the presence of an Al Cl3 catalyst. Cl step 1 Cl2 and AlCl3 Fig. 5.2 (i) Write an equation to show how Cl2 reacts with Al Cl3 to generate an electrophile. … [1] (ii) Complete the mechanism in Fig. 5.3 for the reaction of benzene with the electrophile generated in (b)(i). Include all relevant curly arrows and charges. Draw the structure of the intermediate. Cl intermediate Fig. 5.3 [3]

20 9701/42/F/M/23 © UCLES 2023 (c) The second step of the synthesis involves the reaction of chlorobenzene with Cl COCH2Cl, also in the presence of an Al Cl 3 catalyst, forming compound Q. Cl Cl Cl Cl O O Cl Q step 2 and AlCl3 Fig. 5.4 (i) Name the mechanism of the reaction in step 2. … [1] (ii) Draw the structure of an isomer of Q that forms as an organic by-product of the reaction in step 2. [1] (iii) The reactants used in step 2 contain acyl chloride, alkyl chloride and aryl chloride functional groups. State and explain the relative ease of hydrolysis of acyl chlorides, alkyl chlorides and aryl chlorides. … , … , … easiest to hydrolyse hardest to hydrolyse … … … … … [3]

21 9701/42/F/M/23 © UCLES 2023 [Turn over (d) Tulobuterol is produced from Q as shown in Fig. 5.5. Cl O Cl Q Z step 3 step 4 Cl OH H N tulobuterol Fig. 5.5 Suggest reagents and conditions for steps 3 and 4. Draw the structure of Z in the box. step 3 … step 4 … Z [3] (e) The synthesis produces two enantiomers of tulobuterol. (i) Define enantiomers. … … … [1] (ii) Suggest one disadvantage of producing two enantiomers in this synthesis. … … [1] (iii) Suggest a method of adapting the synthesis to produce a single enantiomer. … [1]

22 9701/42/F/M/23 © UCLES 2023 Cl OH H N tulobuterol (f) (i) Predict the number of peaks that would be seen in the carbon-13 NMR spectrum of tulobuterol. … [1] (ii) The proton (1H) NMR spectrum of tulobuterol dissolved in D2O shows peaks in four different types of proton environment. The peak for the —CH2N— environment is a doublet in the chemical shift range d = 2.0–3.0 ppm. Give details for each of the other three peaks in the proton NMR spectrum of tulobuterol, to include: • chemical shift • environment of the proton • splitting pattern • number of 1H atoms responsible. Table 5.1 gives information about typical chemical shift values. … … … … … … … [3]

23 9701/42/F/M/23 © UCLES 2023 [Turn over Table 5.1 environment of proton example chemical shift range d / 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 [Total: 22]

24 9701/42/F/M/23 © UCLES 2023 6 A student uses thin-layer chromatography (TLC) to analyse a mixture containing different metal cations. The student repeats the experiment using different solvents. Fig. 6.1 shows the chromatogram obtained by the student using water as a solvent. solvent front 5 4 3 2 1 0 cm baseline M Fig. 6.1 (a) (i) Suggest a compound that could be used as the stationary phase in this experiment. … [1] (ii) Table 6.1 shows the Rf values for different metal cations when separated by TLC using water as a solvent. Table 6.1 cation Rf value (water) Cd2+(aq) 0.40 Co2+(aq) 0.77 Cu2+(aq) 0.32 Fe3+(aq) 0.12 Hg2+(aq) 0.23 Ni2+(aq) 0.75 Suggest the identity of the cation that causes the spot at M in Fig. 6.1. Explain your answer. … … [1]

25 9701/42/F/M/23 © UCLES 2023 [Turn over (b) The student repeats the experiment using butan-1-ol as a solvent. The metal cations do not travel as far up the TLC plate in this experiment. Suggest why the metal cations do not move as far up the TLC plate with butan-1-ol as a solvent. … … [1] (c) The student sprays the TLC plate in Fig. 6.1 with KSCN(aq). The colour of some of the spots changes, as some of the metal cations undergo a ligand exchange reaction. Identify the ligands involved in the ligand exchange reaction. … exchanges with … [1]

26 9701/42/F/M/23 © UCLES 2023 (d) In a third experiment, the pH of the mixture of metal ions is kept constant using a buffer solution. The student prepares the buffer solution by mixing 20.0 cm3 of 0.150 mol dm–3 KOH(aq) and 50.0 cm3 of 0.100 mol dm–3 C8H5O4K(aq). C8H5O4K is a weak carboxylic acid that has pKa = 5.40. OH O–K+ O O C8H5O4K Fig. 6.2 (i) Complete the equation for the reaction of C8H5O4K(aq) with KOH(aq). C8H5O4K + … [1] (ii) Calculate the pH of the buffer solution. Show all your working. pH = … [4] [Total: 9]

27 9701/42/F/M/23 © UCLES 2023 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)

28 9701/42/F/M/23 © UCLES 2023 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 – 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.

This document consists of 17 printed pages. © UCLES 2023 [Turn over Cambridge International AS & A Level CHEMISTRY 9701/42 Paper 4 A Level Structured Questions February/March 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 February/March 2023 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 February/March 2023 © UCLES 2023 Page 2 of 17 Generic Marking Principles These general marking principles must be applied by all examiners when marking candidate answers. They should be applied alongside the specific content of the mark scheme or generic level 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 February/March 2023 © UCLES 2023 Page 3 of 17 GENERIC MARKING PRINCIPLE 5: Marks should be awarded using the full range of marks defined in the mark scheme for the question (however; the use of the full mark range may be limited according to the quality of the candidate responses seen). GENERIC MARKING PRINCIPLE 6: Marks awarded are based solely on the requirements as defined in the mark scheme. Marks should not be awarded with grade thresholds or grade descriptors in mind. Science-Specific Marking Principles 1 Examiners should consider the context and scientific use of any keywords when awarding marks. Although keywords may be present, marks should not be awarded if the keywords are used incorrectly. 2 The examiner should not choose between contradictory statements given in the same question part, and credit should not be awarded for any correct statement that is contradicted within the same question part. Wrong science that is irrelevant to the question should be ignored. 3 Although spellings do not have to be correct, spellings of syllabus terms must allow for clear and unambiguous separation from other syllabus terms with which they may be confused (e.g. ethane / ethene, glucagon / glycogen, refraction / reflection). 4 The error carried forward (ecf) principle should be applied, where appropriate. If an incorrect answer is subsequently used in a scientifically correct way, the candidate should be awarded these subsequent marking points. Further guidance will be included in the mark scheme where necessary and any exceptions to this general principle will be noted. 5 ‘List rule’ guidance For questions that require n responses (e.g. State two reasons …): • The response should be read as continuous prose, even when numbered answer spaces are provided. • Any response marked ignore in the mark scheme should not count towards n. • Incorrect responses should not be awarded credit but will still count towards n. • Read the entire response to check for any responses that contradict those that would otherwise be credited. Credit should not be awarded for any responses that are contradicted within the rest of the response. Where two responses contradict one another, this should be treated as a single incorrect response. • Non-contradictory responses after the first n responses may be ignored even if they include incorrect science.

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED February/March 2023 © UCLES 2023 Page 4 of 17 6 Calculation specific guidance Correct answers to calculations should be given full credit even if there is no working or incorrect working, unless the question states ‘show your working’. For questions in which the number of significant figures required is not stated, credit should be awarded for correct answers when rounded by the examiner to the number of significant figures given in the mark scheme. This may not apply to measured values. For answers given in standard form (e.g. a  10n) in which the convention of restricting the value of the coefficient (a) to a value between 1 and 10 is not followed, credit may still be awarded if the answer can be converted to the answer given in the mark scheme. Unless a separate mark is given for a unit, a missing or incorrect unit will normally mean that the final calculation mark is not awarded. Exceptions to this general principle will be noted in the mark scheme. 7 Guidance for chemical equations Multiples / fractions of coefficients used in chemical equations are acceptable unless stated otherwise in the mark scheme. State symbols given in an equation should be ignored unless asked for in the question or stated otherwise in the mark scheme.

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED February/March 2023 © UCLES 2023 Page 5 of 17 Question Answer Marks 1(a)(i) [Ar] ↿⇂ ↿⇂ ↿⇂ ↿⇂ ↿⇂ 1 1(a)(ii) one ● for each of the eight species (including state symbol) on the correct line AND one ● e– any three [1] any six [2] all nine [3] 3 1(a)(iii) • EA becomes less negative/ less exothermic (down group / S to Te) • atomic radii increases OR outer shell gets farther from nucleus OR electron added at higher energy level OR more shielding (of outer shells) • less nuclear attraction OR less attraction for incoming/added electron any two [1] all three [2] 2 1(a)(iv) M1: O2– (has same charge but) smaller (radius than S2–) ORA OR oxygen has a smaller ion (than S2–) 1 M2: stronger ionic bond OR greater attraction between Zn2+ and O2– ORA 1

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED February/March 2023 © UCLES 2023 Page 6 of 17 Question Answer Marks 1(b)(i) S negative AND more moles / molecules of gaseous reactants ORA OR S negative AND moles / molecules of gas are reduced (in the reaction) 1 1(b)(ii) S = 50.8 + 197.7 – 43.7 – 5.7 = (+)199.1 (J K–1 mol–1) 1 1(b)(iii) G = H – TS ALLOW G = H – TS 1 = +733 – (800 + 273)  0.218 = (+)499.086 (kJ mol–1) min 3sf 1 1(c)(i) Zn(NO3)2 → ZnO + 2NO2 + ½O2 OR 2Zn(NO3)2 → 2ZnO + 4NO2 + O2 1 1(c)(ii) increases (in thermal stability down the group) AND (cat)ion(ic) radius / ion size increases (down the group) 1 less polarisation / less distortion of anion/ of nitrate ion/NO3– / NO32– OR less weakening of N—O bond 1 1(c)(iii) Mg(NO3)2 only ALLOW Mg2+ / magnesium 1 Question Answer Marks 2(a) H3PO2 1 2(b)(i) electrode potential E would become more positive / less negative (than E⦵) 1 lower [H2PO2–] AND shifts equilibrium to the right-hand side 1 2(b)(ii) +1.57 – 0.74 = (+)0.83 (V) 1

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED February/March 2023 © UCLES 2023 Page 7 of 17 Question Answer Marks 2(b)(iii) • Pt(s) / C / graphite • Cr(s) • H+(aq) / any strong acid • Cr3+(aq) • H2(g) • voltmeter / V • salt bridge labelled • conditions of 1 atm AND 1 mol dm–3 / 1 M / 1 mol / dm3 • other liquid level (salt bridge) and wire to electrode any three [1] any six [2] all nine [3] 3 2(b)(iv) Pt electrode positive AND flow of electrons anticlockwise (to the SHE) 1 2(b)(v) H2PO2– + 3OH– + Ni2+ → HPO32– + 2H2O + Ni 1 2(c)(i) (6.4 / 24000) ÷ 60 = 4.44  10–6 (mol dm-3 s–1) min 2sf 1 2(c)(ii) [H2PO2–] doubles / 2 from experiments 1 to 2 ORA volume of H2 produced doubles / 2 (∴ first order wrt [H2PO2–]) 1 [H2PO2–]  3 and [OH–]  ½ from experiments 1 to 3 ORA volume of H2 produced falls to ¾ original (if first order wrt [H2PO2–] then must be second order wrt [OH–]) ALLOW input data into rate equation and show k is the same k =2.8  10–6 / k = 6.7  10–2 (1 / 15) / k = 4 for all experiments [2] 1

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED February/March 2023 © UCLES 2023 Page 8 of 17 Question Answer Marks 2(c)(iii) mol–2 dm6 s–1 1 2(c)(iv) t½ = 0.693 / 8.25  10–5 = 8400 (s) OR t½ = In2 / 8.25  10–5 = 8401.8 (s) 1 2(c)(v) (k1) increases (with temperature) 1 2(d) • reactants adsorb (to surface of catalyst) • bonds (in reactant) weaken • (reaction occurs and the) products are desorbed OR reaction occurs and substances are desorbed any two [1] all three [2] 2 Question Answer Marks 3(a) (a d-block element forms one or more) stable ions with incomplete filled d-subshell 1 3(b)(i) • they behave as catalysts • they form complex ions / complexes • they form coloured compounds / salts / ions any two 1 3(b)(ii) the d and s sub-shells/orbitals are close/similar in energy 1 3(c)(i) 0.02500  0.0300 OR 7.50  10–4 mol VO2+ OR ½(28.15 + 28.10)/1000  0.0400 OR 1.13  10–3 mol C2O42– 1 Use of their values to show ratio of VO2+ : C2O42– = 1:1.5 ALLOW any viable approach 1

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED February/March 2023 © UCLES 2023 Page 9 of 17 Question Answer Marks 3(c)(ii) 3D structure 1 charge 1 3(d)(i) VO2+ AND E⦵ of H2O2 is largest / most positive value OR VO2+ AND E⦵ is less positive than H2O2 1 3(d)(ii) +5 1 Question Answer Marks 4(a)(i) CH3CONH2 + 4[H] → CH3CH2NH2 + H2O 1 4(a)(ii) nucleophilic substitution 1 (ammonia with) ethanol AND heat under pressure OR ethanol AND heat in a sealed tube 1 4(a)(iii) (CH3CH2)2NH OR (CH3CH2)3N 1

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED February/March 2023 © UCLES 2023 Page 10 of 17 Question Answer Marks 4(b) (least) phenylamine < ammonia < ethylamine (most) 1 explanation • (order of basicity) ability of base AND to accept a proton OR donate its lone pair (to a proton) phenylamine • lone pair / p-orbital from N delocalised / overlaps with (-)ring / benzene ethylamine • alkyl / ethyl group is electron donating group / +I group • increases electron density on N (ethylamine) ORA any two [1] any three [2] all four [3] 3 4(c) 1 concentrated HNO3 and H2SO4 (and 25–60 °C) 1 (reduction with) Sn and concentrated HCl (heat) 1 4(d)(i) 1 4(d)(ii) nitrous acid / HNO2 OR NaNO2 AND dilute HCl 1

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED February/March 2023 © UCLES 2023 Page 11 of 17 Question Answer Marks 4(d)(iii) phenol 1 4(d)(iv) NaOH / alkali 1 4(d)(v) dyestuffs / dyes 1 4(e)(i) species that uses one lone pair of electrons 1 that forms a single dative covalent bond to a central metal atom / ion 1 4(e)(ii) correct 3D cis isomer 1 correct 3D trans isomer 1 4(e)(iii) 6 1 4(f)(i) units = mol–4 dm12 1 4(f)(ii) (Kstab =) 2 4 4 2 CdC Cd C − + −           l l 1 4(f)(iii) [CdCl42–] = Kstab  0.043  0.0724 = 7.28 / 7.3  10–4 (mol dm–3) min 2sf 1

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED February/March 2023 © UCLES 2023 Page 12 of 17 Question Answer Marks 4(f)(iv) CH3NH2 (is basic so) reacts with water to produce OH– that reacts with Cd2+ OR CH3NH2 acts as a base and accepts a proton from Cd[H2O)6]2+ / water OR [Cd(H2O)6]2+ + 4CH3NH2 → [Cd(OH)4(H2O)2]2– + 4CH3NH3+ OWTTE 1 4(f)(v) Cd(OH)42– = most stable (2nd box) AND [Cd(H2O)6]2+ = least stable (first box) 1 Cd(OH)42– has highest Kstab (and all Kstab values given > 1) 1 Question Answer Marks 5(a) any three points from: • bond angle = 120° AND shape is hexagonal ring planar / trigonal planar • (carbons are) sp2 hybridised • contains delocalised electrons in the  bonds / system • (sp2 orbitals) overlap end-on-end/ head-on to form  bonds • a p orbital (from each carbon atom) overlaps sideways (with each other above and below the ring) forming  bonds 3 5(b)(i) Cl2 + AlCl3 → Cl+ + AlCl4– 1

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED February/March 2023 © UCLES 2023 Page 13 of 17 Question Answer Marks 5(b)(ii) + H+ M1 curly arrow from inside hexagon of aromatic ring to Cl + CON going to a lone pair on Cl M2 intermediate M3 curly arrow from C—H bond to inside the ring AND H+ 3 5(c)(i) electrophilic substitution 1

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED February/March 2023 © UCLES 2023 Page 14 of 17 Question Answer Marks 5(c)(ii) OR 4-substituted acyl / alkyl derivative ALLOW 3-substituted acyl / alkyl derivative 1

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED February/March 2023 © UCLES 2023 Page 15 of 17 Question Answer Marks 5(c)(iii) (most) acyl chloride > alkyl chloride > aryl chloride (least) 1 any two from: • acyl chlorides: carbon (in C—Cl) is more electron deficient AND it is also attached to an oxygen atom / two electronegative atoms OR C—Cl bond is weakest / weakened in acyl chlorides AND it is also attached to an oxygen atom / two electronegative atoms • aryl chlorides (no hydrolysis) C—Cl bond is part of delocalised system / partially double bond character so C—Cl bond is stronger OR lone pair on Cl delocalises with  ring so C—Cl bond is stronger • alkyl chlorides carbon atom has a smaller + AND due to the carbon being only attached to one electronegative atom OR C—Cl bond strengthened AND by electron donating effect / positive inductive effect of alkyl / R group 2 5(d) (CH3)3CNH2 (in ethanol) [substitution] } in either order 1 LiAlH4 OR NaBH4 [reduction ] } in either order 1 [substitution first] OR [reduction first] 1 5(e)(i) rotate the plane of polarised light in the opposite direction OR stereoisomers / molecules that are non-superimposable mirror images 1 5(e)(ii) need to separate the optical isomers to form the pure active isomer OR reduced / different biological activity of ‘other’ enantiomer OR lower yield of biologically active molecule / desired molecule 1 5(e)(iii) chiral catalyst OR use of an enzyme 1

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED February/March 2023 © UCLES 2023 Page 16 of 17 Question Answer Marks 5(f)(i) ten / 10 1 5(f)(ii) ( = 0.9–1.7) • 9H • singlet • –CH3 / alkane ( = 3.2–4.0) / ( = 2.3–3.0) • 1H • triplet • –CHO / alkyl next to electronegative atom OR Ar-CH / alkyl next to aromatic ring ( = 6.0–9.0) • 4H • multiplet • H–Ar / attached to aromatic ring any three [1] any six [2] all nine [3] 3 Question Answer Marks 6(a)(i) SiO2 OR Al2O3 OR silica OR alumina OWTTE 1 6(a)(ii) Cd2+ AND Rf of M (= 2/5) = 0.40 / same Rf as in the Table 6.1 1 6(b) metal cations are less soluble in butan-1-ol (than in water) OR metal cations weaker ion-dipole forces with butan-1-ol 1 6(c) H2O AND SCN– 1 6(d)(i) C8H5O4K + KOH → C8H4O4K2 + H2O OR C8H5O4K + OH– → C8H4O4K– + H2O 1

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED February/March 2023 © UCLES 2023 Page 17 of 17 Question Answer Marks 6(d)(ii) M1 initially: moles of KOH = 0.150  20.0 ÷ 1000 = 3.0  10–3 AND moles of acid HA = 0.100  50.0÷1000 = 5.0  10–3 1 M2 at equilibrium: moles of salt KA = 3.0  10–3 ecf AND moles of acid HA = 5.0  10–3 – moles of NaOH = 2.0  10–3 ecf 1 M3 Ka = 10–5.40 = 3.98  10–6 3.98  10–6 = [H+] [3.0  10–3] ÷ [2.0  10–3] [H+] = 2.65  10–6 ecf 1 M4 pH = –log(2.65  10–6) = 5.58 / 5.6 ecf on a calculated and identified [H+] ALLOW alternative approach using Henderson-Hasselbalch equation for M3 and M4 1