Cambridge A Level Chemistry 9701 — 2022 Oct/Nov Paper 4 · Variant 2

*6410435209* CHEMISTRY 9701/42 Paper 4 A Level Structured Questions October/November 2022  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. IB22 11_9701_42/3RP © UCLES 2022 [Turn over This document has 24 pages. Cambridge International AS & A Level

2 9701/42/O/N/22 © UCLES 2022 1 Calcium chloride, CaCl 2, is an ionic solid. The values of some energy changes are shown in Table 1.1. Table 1.1 energy change value / kJ mol–1 lattice energy, , CaCl 2(s) –2237 standard enthalpy change of atomisation of calcium +193 first ionisation energy of calcium +590 second ionisation energy of calcium +1150 standard enthalpy change of atomisation of chlorine +121 first electron affinity of chlorine –364 (a) Define lattice energy. … … [1] (b) Use the data in Table 1.1 to calculate the standard enthalpy change of formation, , of calcium chloride. It may be helpful to draw an energy cycle. Show all your working.  (CaCl 2(s)) = … kJ mol–1 [2]

3 9701/42/O/N/22 © UCLES 2022 [Turn over (c) Three possible values for the first electron affinity of bromine are shown in Table 1.2. One of them is correct. Place a tick by the correct value. Explain your choice. Table 1.2 possible values place one tick (✓) in this column –342 kJ mol–1 –364 kJ mol–1 –386 kJ mol–1 explanation … … [1] (d) The enthalpy change of hydration of the chloride ion can be calculated using the lattice energy of calcium chloride and the data shown in Table 1.3. Table 1.3 energy change value / kJ mol–1 standard enthalpy change of solution of CaCl 2(s) –83 standard enthalpy change of hydration of Ca2+(g) –1650 (i) Define the following terms. enthalpy change of solution … … enthalpy change of hydration … …  [2] (ii) Calculate the standard enthalpy change of hydration of the chloride ion, Cl –(g). It may be helpful to draw an energy cycle. Show all your working.  (Cl –(g)) = … kJ mol–1 [2]

4 9701/42/O/N/22 © UCLES 2022 (e) Calcium fluoride, CaF2(s), can be synthesised directly from its elements. The value of (CaF2(s)) is –1214 kJ mol–1. (i) Predict the sign of the entropy change, ∆S o, for this synthesis. Explain your answer. The sign of the entropy change is … . explanation … …  [1] (ii) Use the value of (CaF2(s)) given in (e) and your answer to (e)(i) to predict how the feasibility for this synthesis will change with increasing temperature. … … … … [2]  [Total: 11]

5 9701/42/O/N/22 © UCLES 2022 [Turn over 2 (a) Nitrogen monoxide, NO, reacts with ozone, O3. NO(g) + O3(g) → NO2(g) + O2(g) This reaction is first order with respect to both NO and O3. At 298 K, the rate constant k = 11 500 mol–1 dm3 s–1. (i) Complete the rate equation for this reaction. rate = …  [1] (ii) A reaction is carried out in which the initial concentrations of NO and O3 are both 1.20 × 10–6 mol dm–3. Calculate the initial rate of the reaction. State its units.  rate of reaction = … units = … [2] (iii) The reaction described in (a)(ii) is monitored over a period of time. Predict whether or not the graph of [NO] against time, under these conditions, shows that the reaction has a constant half-life. Explain your answer. prediction … explanation … … [1] (b) Nitrous oxide, N2O, decomposes into its elements. 2N2O(g) → 2N2(g) + O2(g) At a high temperature, a small amount of platinum wire is added to a large amount of nitrous oxide. The reaction follows zero order kinetics. The platinum wire behaves as a catalyst. (i) Sketch a graph, on the axes below, of reaction rate against time for the catalysed decomposition of N2O under these conditions. reaction rate time  [1]

6 9701/42/O/N/22 © UCLES 2022 (ii) Sketch a graph, on the axes below, of [N2O] against time for this reaction. [N2O] time  [1] (iii) Platinum behaves as a heterogeneous catalyst in this reaction. Describe the mode of action of a heterogeneous catalyst. … … … … [3] (iv) Suggest a reason why this reaction has zero order kinetics when the amount of nitrous oxide is large and the amount of platinum is small. … … [1]  [Total: 10]

7 9701/42/O/N/22 © UCLES 2022 [Turn over Question 3 starts on the next page.

8 9701/42/O/N/22 © UCLES 2022 3 Data should be selected from Table 3.1 in order to answer some parts of this question. Table 3.1 electrode reaction E o / V Mg2+ + 2e– Mg –2.38 Mn2+ + 2e– Mn –1.18 Mn3+ + e– Mn2+ +1.49 MnO2 + 4H+ + 2e– Mn2+ + 2H2O +1.23 MnO4 – + e– MnO4 2– +0.56 MnO4 – + 4H+ + 3e– MnO2 + 2H2O +1.67 MnO4 – + 8H+ + 5e– Mn2+ + 4H2O +1.52 (a) An electrochemical cell can be constructed from a Mg2+ / Mg half-cell and a MnO4 – / Mn2+ half‑cell. The standard cell potential of this cell can be calculated using the standard electrode potentials of the two half-cells. (i) Define standard electrode potential. Include details of the standard conditions used. … … … … [2] (ii) Complete the diagram below to show an electrochemical cell constructed from a Mg2+ / Mg half‑cell and a MnO4 – / Mn2+ half-cell. Label your diagram.  [3] (iii) Use a positive (+) sign and a negative (–) sign to identify the polarity of each of the two electrodes in your diagram. Use an arrow and the symbol ‘e’ to show the direction of electron flow in the external circuit. [1]

9 9701/42/O/N/22 © UCLES 2022 [Turn over (iv) Calculate the standard cell potential, , of this cell.  = … V [1] (v) Construct an equation for the cell reaction. … [1] (vi) Predict how the cell reaction will change, if at all, when the solution in the Mg2+ / Mg half‑cell is diluted by the addition of a large volume of water. Explain your answer. … … … [1] (b) A molten magnesium salt is electrolysed for 15.0 minutes by a constant current. 4.75 × 1022 magnesium atoms are produced at the cathode. Calculate the value of the current used.  current = … A [2]  [Total: 11]

10 9701/42/O/N/22 © UCLES 2022 4 (a) The value of the solubility product, Ksp, of iron(III) hydroxide, Fe(OH)3, is given by the following expression. Ksp = [Fe3+][OH–]3 = 2.0 × 10–39 mol4 dm–12 (i) Calculate the solubility of Fe(OH)3 in water.  solubility = … mol dm–3 [1] (ii) Calculate the solubility of Fe(OH)3 in 0.010 mol dm–3 barium hydroxide, Ba(OH)2(aq).  solubility = … mol dm–3 [2] (iii) Fe(OH)3 is less soluble in Ba(OH)2(aq) than it is in pure water. Name this effect. … [1] (b) The numerical value of the Ka of HBrO is 2.00 × 10–9. X is a solution of HBrO which contains 4.00 × 10–3 mol of HBrO in 100 cm3 of solution. In this solution the following equilibrium is established in which there are two conjugate acid-base pairs. HBrO + H2O BrO– + H3O+ (i) Define conjugate acid-base pair. … … [1] (ii) Identify the two conjugate acid-base pairs shown in the equation above. pair one … … acid base pair two … … acid base  [1]

11 9701/42/O/N/22 © UCLES 2022 [Turn over (iii) Calculate the pH of solution X. Show all your working.  pH = … [2] (iv) A solution containing 2.00 × 10–3 mol of NaOH is added to solution X. A buffer solution is formed. Calculate the pH of this buffer solution.  pH = … [1]  [Total: 9]

12 9701/42/O/N/22 © UCLES 2022 5 Copper is a transition element. It forms a wide variety of compounds. (a) Define transition element. … … [1] (b) An aqueous solution of copper(II) sulfate, CuSO4, contains [Cu(H2O)6]2+ complex ions. If an excess of concentrated hydrochloric acid is added to this solution a ligand exchange reaction occurs and [CuCl 4]2– complex ions are formed. (i) Complete Table 5.1 to state the geometry, the coordination number of copper, and one bond angle in each of the two complex ions. Table 5.1 complex ion geometry coordination number bond angle [Cu(H2O)6]2+ [CuCl 4]2–  [3] (ii) In an isolated Cu2+ ion the d-orbitals are all degenerate. In a complex ion such as [Cu(H2O)6]2+ the d-orbitals are non-degenerate. Define degenerate and non-degenerate in this context. degenerate … non-degenerate …  [1] (iii) Explain why the solutions of the two complex ions in Table 5.1 are different colours. … … [1] (c) Cu2+ forms a complex ion containing water molecules and ethanedioate ions, C2O4 2–, as ligands. The formula of the complex is [Cu(C2O4)2(H2O)2]2–. The ethanedioate ion is a bidentate ligand. (i) Explain what is meant by bidentate. … … [1]

13 9701/42/O/N/22 © UCLES 2022 [Turn over (ii) There are three stereoisomers with the formula [Cu(C2O4)2(H2O)2]2–. Complete the three-dimensional diagrams to show these three stereoisomers. stereoisomer 1 Cu stereoisomer 2 Cu stereoisomer 3 Cu  [2] (iii) Use your answer to (c)(ii) to answer this question. Stereoisomers 1, 2 and 3 show two different types of isomerism. Name these two types of isomerism. For each type of isomerism identify the pair of stereoisomers that demonstrate this isomerism. type of isomerism pair of stereoisomers and and  [2]

14 9701/42/O/N/22 © UCLES 2022 (d) A solution contains 3.70 g of Na2[Cu(C2O4)2(H2O)2] dissolved in 100 cm3 of solution. A 25.0 cm3 sample of this solution is warmed and then oxidised by 0.0100 mol dm–3 acidified potassium manganate(VII). The equation for the redox reaction is shown. 5C2O4 2– + 2MnO4 – + 16H+ → 10CO2 + 2Mn2+ + 8H2O Calculate the minimum volume of 0.0100 mol dm–3 acidified potassium manganate(VII) needed to oxidise all of the ethanedioate ions, C2O4 2–, in the 25.0 cm3 sample. Show all your working. [Mr: Na2[Cu(C2O4)2(H2O)2], 321.5]  minimum volume = … cm3 [3] (e) Copper(II) nitrate, Cu(NO3)2, and barium nitrate, Ba(NO3)2, both decompose when heated. Copper(II) nitrate decomposes at a lower temperature than barium nitrate. Suggest a reason for this difference. Explain your answer. … … … … [2]  [Total: 16]

15 9701/42/O/N/22 © UCLES 2022 [Turn over 6 An aqueous solution of cobalt(II) chloride is a pink colour due to the presence of [Co(H2O)6]2+ complex ions. (a) (i) Complete Table 6.1 to describe what is observed when the named reagent is added to an aqueous solution of cobalt(II) chloride. Table 6.1 reagent colour of cobalt-containing product state of cobalt-containing product NaOH(aq) an excess of conc. HCl  [2] (ii) Write an equation for the reaction between [Co(H2O)6]2+ ions and NaOH(aq). … [1] (iii) Write an equation for the reaction between [Co(H2O)6]2+ ions and an excess of conc. HCl. … [1] (b) (i) Define stability constant. … … [1] (ii) Write an expression for the stability constant, Kstab, of the [Co(NH3)6]2+ complex ion. Kstab =  [1] (iii) Give the units of the stability constant, Kstab, of the [Co(NH3)6]2+ complex ion.  units = … [1] (iv) The numerical value of the stability constant, Kstab, of the [Co(NH3)6]2+ complex ion is 7.7 × 104. In an aqueous solution the concentration of the [Co(NH3)6]2+ complex ion is 0.0740 mol dm–3 and the concentration of NH3 is 0.480 mol dm–3 at equilibrium. Calculate the concentration of [Co(H2O)6]2+ in this solution.  concentration = … mol dm–3 [1]  [Total: 8]

16 9701/42/O/N/22 © UCLES 2022 7 (a) The structural and displayed formulae of three aromatic compounds, A, B and C, are shown in Fig. 7.1. C6H5CH2CH2COOH A HO O OH CH3C6H4OH B C6H5CH2CHCl CO2H C HO O Cl Fig. 7.1 Compare the relative acidities of A, B and C. … > … > … most acidic least acidic Explain your answer. … … … … [3] (b) Methylbenzene, C6H5CH3, can be made from benzene by an electrophilic substitution reaction. (i) Identify a compound that reacts with benzene to form methylbenzene. Identify the catalyst used. compound … catalyst …  [1] (ii) The first step in the reaction is the generation of the CH3 + electrophile. Write an equation for the reaction that generates this electrophile. … [1]

17 9701/42/O/N/22 © UCLES 2022 [Turn over (iii) Describe the mechanism for the reaction between benzene and the CH3 + electrophile. Include all relevant curly arrows and charges. CH3 +  [3] (iv) Identify a suitable reagent to oxidise methylbenzene to benzoic acid, C6H5COOH. Write an equation for this reaction using [O] to represent one atom of oxygen from the oxidising agent. reagent … equation …  [2] (v) Methylbenzene and benzoic acid are both nitrated with a mixture of concentrated nitric acid and sulfuric acid to give mononitrated products. The structural formulae of these products are CH3C6H4NO2 and HOOCC6H4NO2 respectively. Draw the structures of these two products.  [1]

18 9701/42/O/N/22 © UCLES 2022 (c) A reaction scheme is shown in Fig. 7.2. CH3C6H4NO2 CH3C6H4NH2 reaction 1 reaction 2 CH3C6H4OH CH3C6H4N2 +Cl – dye molecule Y reaction 3 reaction 4 Fig. 7.2 (i) Describe the reagents and conditions to produce CH3C6H4N2 +Cl – from CH3C6H4NH2 in reaction 2. reagents … conditions …  [1] (ii) Describe how CH3C6H4OH can be produced from CH3C6H4N2 +Cl – in reaction 3. … [1] (iii) Draw the structure of the dye molecule Y formed when CH3C6H4N2 +Cl – and CH3C6H4OH react together in reaction 4. Describe the conditions for this reaction. structure conditions …  [2]  [Total: 15]

19 9701/42/O/N/22 © UCLES 2022 [Turn over 8 When answering this question it should be assumed that together all the hydrogen atoms in a benzene ring result in a single unsplit peak at δ = 7.2 in a proton (1H) NMR spectrum. The structures of five isomeric ketones, P, Q, R, S and T are given. P C6H5COCH(CH3)2 S C6H5CH2CH2COCH3 Q C6H5COCH2CH2CH3 T C6H5CH(CH3)COCH3 R C6H5CH2COCH2CH3 (a) Identify all the chiral carbon atoms on the structures above. Label each chiral carbon atom with an asterisk (*). [1] (b) The proton (1H) NMR spectrum of one of the five isomers, P, Q, R, S or T, is shown in Fig. 8.1. 11 10 9 8 7 6 5 / ppm 4 3 2 1 0 Fig. 8.1 (i) Identify which of the compounds P, Q, R, S or T gives this spectrum. Draw the displayed formula of the compound you have identified. Identify the protons responsible for the peaks at δ = 3.7, δ = 2.5 and δ = 1.0 on the structure you have drawn.  [2] (ii) Name the splitting pattern of the peak at δ = 3.7. Explain why it has this splitting pattern. … … [1]

20 9701/42/O/N/22 © UCLES 2022 (c) Choose from the letters P, Q, R, S and T to identify: (i) the two compounds that each have a doublet peak in the proton (1H) NMR spectrum … [1] (ii) the compound with only three peaks in its proton (1H) NMR spectrum. … [1] (d) Suggest a suitable solvent that should be used for obtaining the spectrum shown in Fig. 8.1. … [1] (e) The proton (1H) NMR spectrum of compound T is compared in the presence of D2O and in the absence of D2O. Describe any difference between the two spectra. Explain your answer. … … [1] (f) Complete Table 8.1 below to give the number of peaks in the carbon-13 NMR spectrum of each compound. Table 8.1 compound number of peaks compound number of peaks  [2]  [Total: 10]

21 9701/42/O/N/22 © UCLES 2022 [Turn over 9 Tyrosine and lysine, shown in Fig. 9.1, are naturally occurring amino acids found in proteins. HO OH O NH2 tyrosine OH O NH2 H2N lysine Fig. 9.1 (a) The isoelectric point of lysine is 9.47. Draw the structure of lysine at the stated pH in the boxes below. lysine at pH 7.00 lysine at pH 9.47 lysine at pH 12.00  [2] (b) When ethanoic acid is treated with PCl 5 product D is formed. When D is added to tyrosine two different isomeric products, E and F, are formed. E has an ester linkage, F does not. Draw the structures of D, E and F in the boxes below. D E F  [3]

22 9701/42/O/N/22 © UCLES 2022 (c) Complete Table 9.1 by drawing the structure of the organic product formed when tyrosine reacts with each named reagent. Ignore the directing effect of the –CH2CHNH2COOH substituent. Table 9.1 reagent structure an excess of Br2(aq) an excess of NaOH(aq) an excess of HNO3(aq)  [3] (d) A mixture of tyrosine and lysine can be separated by thin-layer chromatography. Under certain conditions the Rf value of lysine is 0.14 and the Rf value of tyrosine is 0.45. (i) Explain what is meant by Rf value. … … [1] (ii) Suggest an explanation for the difference in Rf values. … … [1]  [Total: 10]

23 9701/42/O/N/22 © UCLES 2022 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/O/N/22 © UCLES 2022 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 11 printed pages. © UCLES 2022 [Turn over Cambridge International AS & A Level CHEMISTRY 9701/42 Paper 4 A Level Structured Questions October/November 2022 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 2022 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 2022 © UCLES 2022 Page 2 of 11 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 October/November 2022 © UCLES 2022 Page 3 of 11 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 2022 © UCLES 2022 Page 4 of 11 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 2022 © UCLES 2022 Page 5 of 11 Question Answer Marks 1(a) (energy change) when one mole of ionic solid is formed from gaseous ions 1 1(b) (–2237 + 193 + 590 + 1150 + (2  121) – (2  364)) [1] = –790 [1] 2 1(c) –342 and Br atom has larger radius 1 1(d)(i) energy change when one mole dissolves in water [1] energy change when one mole of gaseous ions dissolves in water [1] 2 1(d)(ii) (–2237 – 83 + 1650) / 2 [1] = –335 [1] 2 1(e)(i) negative and reduction in number of gas molecules 1 1(e)(ii) TS becomes more negative [1] less feasible AND G becomes positive [1] 2 Question Answer Marks 2(a)(i) rate = k[NO][O3] 1 2(a)(ii) 1.66  10–8 [1] mol dm–3 s–1 [1] 2 2(a)(iii) not constant AND overall second order / not overall first order 1 2(b)(i) graph is straight line clearly parallel to x-axis 1 2(b)(ii) graph is straight line with negative gradient 1

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2022 © UCLES 2022 Page 6 of 11 Question Answer Marks 2(b)(iii) adsorption of reactants onto catalyst surface [1] bonds in reactants weaken [1] reaction occurs followed by desorption of products [1] 3 2(b)(iv) all active sites on catalyst surface are occupied 1 Question Answer Marks 3(a)(i) the voltage produced by a half-cell compared with a standard hydrogen electrode [1] 1 mol dm–3, 298 K, 1 atm [1] 2 3(a)(ii) Mg wire and Pt wire [1] voltmeter, salt bridge, complete circuit [1] solutes Mg2+ and MnO4–, Mn2+, H+ [1] 3 3(a)(iii) Mg is minus, Pt is plus arrow points towards MnO4– / Mn2+ half-cell 1 3(a)(iv) 3.90 V 1 3(a)(v) 5Mg + 2MnO4– +16H+ → 5Mg2+ + 2Mn2+ + 8H2O 1 3(a)(vi) no change and dilution will make Mg2+ / Mg potential even more negative 1 3(b) either: 4.75  1022  2  1.60  10–19 = 15 200 C OR 2  96 500  (4.75  1022) / 6.02  1023 = 15 228 C [1] 15 200 / (15  60) = 16.9 A OR 15 228 / (15  60) = 16.9 A [1] 2

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2022 © UCLES 2022 Page 7 of 11 Question Answer Marks 4(a)(i) 27 x4 = 2  10–39 x = 9.28  10–11 mol dm–3 1 4(a)(ii) [Fe3+][0.02]3 = 2  10–39 [1] [Fe3+] = 2.5  10–34 [1] 2 4(a)(iii) common ion (effect) 1 4(b)(i) two species that differ by one H+ ion 1 4(b)(ii) HBrO BrO– H3O+ H2O 1 4(b)(iii) [acid] = 4  10–2 [H+]2 = 8.0  10–11 [1] pH = 5.05 [1] 2 4(b)(iv) 8.70 1 Question Answer Marks 5(a) a d-block element that forms one or more stable ions with incomplete d-orbitals 1 5(b)(i) geometry CN bond angle [Cu(H2O)6]2+ octahedral 6 90 or 180 [CuCl4]2– tetrahedral 4 109 to 110 3 5(b)(ii) degenerate – of the same energy non-degenerate – not of the same energy / different energy 1 5(b)(iii) different frequency / wavelength / photon absorbed 1

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2022 © UCLES 2022 Page 8 of 11 Question Answer Marks 5(c)(i) forms 2 dative / coordinate bonds 1 5(c)(ii) one isomer correct with correct 3D [1] three isomers different with correct 3D [1] 2 5(c)(iii) e.g. 1 and 2 optical [1] e.g. 1 or 2 and 3 cis-trans / geometric [1] 2 5(d) 0.0230 moles ethanedioate in whole 100 cm3 0.00575 moles ethanedioate in 25 cm3 sample [1] 0.00230 moles of MnO4– needed [1] 230 cm3 [1] 3 5(e) Cu2+ is smaller / has a smaller radius / than Ba2+ [1] Cu2+ polarises / distorts the anion more [1] 2 Question Answer Marks 6(a)(i) blue or pink s / solid blue aq / aqueous two cells correct [1] two more cells correct [1] 2 6(a)(ii) [Co(H2O)6]2+ + 2OH– → Co(OH)2 + 6H2O or [Co(H2O)6]2+ + 2OH– → Co(H2O)4(OH)2 + 2H2O 1 6(a)(iii) [Co(H2O)6]2+ + 4Cl– → [CoCl4]2– + 6H2O 1 6(b)(i) the equilibrium constant for the formation of a complex ion in a solvent OR the equilibrium constant for the formation of a complex ion from its constituent ions 1

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2022 © UCLES 2022 Page 9 of 11 Question Answer Marks 6(b)(ii) Kstab = [[Co(NH3)6]2+] / [[Co(H2O)6]2+][NH3]6 or [[Co(NH3)6]2+] / [Co2+][NH3]6 1 6(b)(iii) mol–6 dm18 1 6(b)(iv) 7.86  10–5 1 Question Answer Marks 7(a) C > A > B [1] chlorine and C=O are electronegative / withdraw charge and this causes greatest weakening of O–H bond or greatest stabilisation of the anion [1] 2nd oxygen / C=O is electronegative / withdraws charge and this weakens O–H bond or stabilises anion [1] 3 7(b)(i) chloromethane aluminium chloride 1 7(b)(ii) CH3Cl + AlCl3 → AlCl4– + CH3+ 1 7(b)(iii) curly arrow from within benzene to CH3+ [1] positively charged intermediate [1] curly arrow from C–H bond into ring, C6H5CH3, H+ [1] 3 7(b)(iv) hot alkaline KMnO4 [1] C6H5CH3 + 3[O] → C6H5CO2H + H2O OR C6H5CH3 + 3[O] + OH– → C6H5CO2– + 2H2O [1] 2 7(b)(v) 2 or 4-nitromethylbenzene and 3-nitrobenzoic acid 1 7(c)(i) HNO2, T between 0° and 10°C 1 7(c)(ii) warm / T>10°C and H2O 1

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2022 © UCLES 2022 Page 10 of 11 Question Answer Marks 7(c)(iii) CH3C6H4–N=N–C6H3(CH3)OH [1] T between 0° and 10°C and NaOH(aq) [1] 2 Question Answer Marks 8(a) C* marked on CH of T and nowhere else 1 8(b)(i) R – C6H5CH2COCH2CH3 [1] 3.7 is C6H5CH2COCH2CH3 2.5 is C6H5CH2COCH2CH3 1.0 is C6H5CH2COCH2CH3 [1] 2 8(b)(ii) singlet and no H on neighbouring C 1 8(c)(i) P and T 1 8(c)(ii) P 1 8(d) CDCl3 or CCl4 1 8(e) no difference and no protons that exchange with D 1 8(f) 6 5 [1] 4 3 [1] 2

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED October/November 2022 © UCLES 2022 Page 11 of 11 Question Answer Marks 9(a) pH 7 +H3N(CH2)4CHNH2COOH pH 9.47 +H3N(CH2)4CHNH2COO– [1] pH 12 H2N(CH2)4CHNH2COO– [1] 2 9(b) D ethanoyl chloride [1] E correct phenyl ester [1] F correct amide [1] 3 9(c) Br substitutes at both 2 and 6 positions [1] both phenol and COOH groups deprotonated [1] NO2 substitutes at one or both of 2 and 6 positions [1] 3 9(d)(i) distance travelled by amino acid divided by distance travelled by solvent [1] 1 9(d)(ii) tyr is more soluble in the solvent used or lys is more attracted to the stationary phase used 1