Cambridge A Level Chemistry 9701 — 2021 Feb/March Paper 4 · Variant 2

*6430441026* CHEMISTRY 9701/42 Paper 4 A Level Structured Questions February/March 2021  2 hours You must answer on the question paper. You will need: Data booklet 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, use appropriate units and use an appropriate number of significant figures. INFORMATION ● The total mark for this paper is 100. ● The number of marks for each question or part question is shown in brackets [ ]. IB21 03_9701_42/2RP © UCLES 2021 [Turn over This document has 28 pages. Any blank pages are indicated. Cambridge International AS & A Level

2 9701/42/F/M/21 © UCLES 2021 Answer all the questions in the spaces provided. 1 (a) The most common oxidation states of cobalt are +2 and +3. Complete the electronic configurations of the following free ions. ● Co2+ [Ar] ���������������������������������������������������������������������������������������������������������������������������� ● Co3+ [Ar] ����������������������������������������������������������������������������������������������������������������������������  [1] (b) Co2+ and Co3+ both form complexes with edta4–. half-equation E o / V Co3+ + e– Co2+ +1.82 O2 + 4H+ + 4e– 2H2O +1.23 [Co(edta)]– + e– [Co(edta)]2– +0.38 Co2+ + 2e– Co –0.28 Use the data in the table to predict what happens, if anything, when separate aqueous solutions of Co3+ and [Co(edta)]– are left to stand in the air. aqueous solution of Co3+ … … aqueous solution of [Co(edta)]– … …  [3]

3 9701/42/F/M/21 © UCLES 2021 [Turn over (c) Hydrated cobalt(II) nitrate, Co(NO3)2•6H2O, is a red solid that behaves like hydrated magnesium nitrate, Mg(NO3)2•6H2O, when heated. Describe in detail what you would expect to observe when crystals of Co(NO3)2•6H2O are heated in a boiling tube, gently at first and then more strongly. … … … … [2] (d) Explain why the thermal stability of the Group 2 nitrates increases down the group. … … … … [2]  [Total: 8]

4 9701/42/F/M/21 © UCLES 2021 2 (a) Iron(II) compounds are generally only stable in neutral, non-oxidising conditions. It is difficult to determine the lattice energy of FeO experimentally. (i) Use data from the Data Booklet and this Born–Haber cycle to calculate the lattice energy, ∆H latt, of FeO(s) in kJ mol–1. Fe2+(g) + O(g) + 2e– Fe2+(g) + O–(g) + e– Fe+(g) + O(g) + e– Fe(g) + O(g) FeO(s) Fe2+(g) + O2–(g) Fe(g) + 2O2(g) 1 Fe(s) + 2O2(g) 1 –141 kJ mol–1 –272 kJ mol–1 +798 kJ mol–1 +416 kJ mol–1 H latt  ∆H lattFeO(s) = … kJ mol–1 [2]

5 9701/42/F/M/21 © UCLES 2021 [Turn over (ii) Most naturally occurring samples of iron(II) oxide are found as the mineral wüstite. Wüstite has formula Fe20Ox. It contains both Fe2+ and Fe3+ ions. 90% of the iron is present as Fe2+ and 10% is present as Fe3+. Deduce the value of x.  x = … [1] (iii) State and explain how the lattice energy of FeO(s) compares to the lattice energy of CaO(s). … … … … … [2]

6 9701/42/F/M/21 © UCLES 2021 (b) Heating of FeO results in the formation of Fe3O4, as shown. reaction 1 4FeO → Fe + Fe3O4 Each formula unit of Fe3O4 contains one Fe2+ and two Fe3+ ions. (i) Show how reaction 1 can be described as a disproportionation reaction. … … … [1] Fe3O4(l) can be electrolysed using inert electrodes to form Fe. (ii) Write the half-equation for the reaction that occurs at the anode during the electrolysis of Fe3O4(l). … [1] (iii) Calculate the maximum mass of iron metal formed when Fe3O4(l) is electrolysed for six hours using a current of 50 A. Assume the one Fe2+ and two Fe3+ ions are discharged at the same rate.  mass of iron = … g [3]

7 9701/42/F/M/21 © UCLES 2021 [Turn over (c) LiFePO4 can be used in lithium-ion rechargeable batteries. When the cell is charging, lithium reacts with a graphite electrode to form LiC6. When the cell is discharging, the half-equations for the two processes that occur are as follows. anode half-equation LiC6 → 6C + Li+ + e– cathode half-equation Li+ + FePO4 + e– → LiFePO4 (i) State one possible advantage of developing cells such as lithium-ion rechargeable batteries. … [1] (ii) Use the cathode half-equation to determine the change, if any, in oxidation states of lithium and iron at the cathode during discharging. metal change in oxidation state during discharging from to lithium iron  [1] (iii) Write the equation for the overall reaction that occurs when this cell is discharging. … [1]  [Total: 13]

8 9701/42/F/M/21 © UCLES 2021 3 Iodates are compounds that contain the IO3 – anion. (a) The IO3 – anion is shown. O O O– I Explain, with reference to the qualitative model of electron-pair repulsion, why the IO3 – anion has a pyramidal shape. … … … [1] (b) The reaction of iodine and hot aqueous sodium hydroxide is similar to that of chlorine and hot aqueous sodium hydroxide. Sodium iodate, NaIO3, is formed as one of the products. Suggest an equation for the reaction of iodine and hot aqueous sodium hydroxide. … [1]

9 9701/42/F/M/21 © UCLES 2021 [Turn over (c) The decomposition of hydrogen peroxide, H2O2, is catalysed by acidified IO3 –. H2O2 reduces acidified IO3 – as shown. 5H2O2 + 2H+ + 2IO3 – → I2 + 5O2 + 6H2O This reaction is followed by the oxidation of I2 by H2O2. half-equation E o / V H2O2 + 2H+ + 2e– 2H2O +1.77 IO3 – + 6H+ + 5e– 1 2I2 + 3H2O +1.19 O2 + 2H+ + 2e– H2O2 +0.68 (i) Use the data to show that the separate reactions of H2O2 with IO3 – and with I2 are both feasible under standard conditions. In your answer, give the equation for the reaction of H2O2 with I2. … … … … … … [3] (ii) Write the overall equation for the decomposition of H2O2 catalysed by acidified IO3 –. … [1]

10 9701/42/F/M/21 © UCLES 2021 (d) A student collects some data for the reaction of H2O2 with acidified IO3 –, as shown in the table. experiment [H2O2] / mol dm–3 [IO3 –] / mol dm–3 [H+] / mol dm–3 initial rate of reaction / mol dm–3 s–1 1 0.0500 0.0700 0.025 1.47 × 10–5 2 0.100 0.0700 0.050 2.94 × 10–5 3 0.100 0.140 0.025 5.88 × 10–5 4 0.150 0.140 0.025 8.82 × 10–5 (i) Use the data to determine the order of reaction with respect to [H2O2], [IO3 –] and [H+]. Show your reasoning. order with respect to [H2O2] = … … … … order with respect to [IO3 –] = … … … … order with respect to [H+] = … … … …  [3] (ii) Use your answer to (d)(i) to write the rate equation for this reaction. rate = ���������������������������������������������������������������������������������������������������������������������������� [1]

11 9701/42/F/M/21 © UCLES 2021 [Turn over (iii) Calculate the value of the rate constant, k, using data from experiment 4 and your answer to (d)(ii). Give the units of k.  k = …  units = …  [2] (e) Pb(IO3)2 is only sparingly soluble in water at 25 °C. The solubility product, Ksp, of Pb(IO3)2 is 3.69 × 10–13 mol3 dm–9 at 25 °C. (i) Write an expression for the solubility product of Pb(IO3)2. Ksp =  [1] (ii) Calculate the solubility, in mol dm–3, of Pb(IO3)2 at 25 °C.  solubility = … mol dm–3 [2]

12 9701/42/F/M/21 © UCLES 2021 (f) NH4IO3 is an unstable compound that readily decomposes when warmed. The decomposition reaction is shown. NH4IO3(s) → 1 2N2(g) + 1 2O2(g) + 1 2I2(g) + 2H2O(l) ∆H = –154.6 kJ mol–1 (i) Use the data in the table to calculate the entropy change of reaction, ∆S, of the decomposition of NH4IO3(s). compound S / J K–1 mol–1 NH4IO3(s) 42 N2(g) 192 O2(g) 205 I2(g) 261 H2O(l) 70  ∆S = … J K–1 mol–1 [2] (ii) This reaction is feasible at all temperatures. Explain why, using the data in (f) and your answer to (f)(i). … … … [1]  [Total: 18]

13 9701/42/F/M/21 © UCLES 2021 [Turn over 4 The transition elements are able to form stable complexes with a wide range of molecules and ions. (a) State the meaning of transition element. … … … [1] (b) The d orbitals in an isolated transition metal ion are degenerate. In complexes, the d orbitals occupy two energy levels. (i) Complete the diagram to show the arrangement of d orbital energy levels in octahedral and in tetrahedral complexes. energy octahedral complex isolated transition metal ion degenerate d orbitals tetrahedral complex  [1] (ii) Sketch the shape of two d orbitals: ● one d orbital from the lower energy level in an octahedral complex ● one d orbital from the higher energy level in an octahedral complex. Use the axes below. z lower energy level x y z higher energy level x y  [2]

14 9701/42/F/M/21 © UCLES 2021 (c) Edds4– and edta4– are polydentate ligands that form octahedral complexes with Fe3+(aq). edds4– H N N H –O2C CO2 – CO2 – CO2 – edta4– N N CO2 – CO2 – CO2 – CO2 – The formulae of the complexes are [Fe(edds)]– and [Fe(edta)]– respectively. (i) On the diagram of edds4–, circle each atom that forms a bond to the Fe3+ ion in [Fe(edds)]–.  [1] (ii) [Fe(edds)]– is red and [Fe(edta)]– is yellow. Explain why the two complexes have different colours. … … … … … [2] (iii) When edds4–(aq) is added to Fe3+(aq), the following reaction occurs. [Fe(H2O)6]3+(aq) + edds4–(aq) [Fe(edds)]–(aq) + 6H2O(l) State the type of reaction that occurs. … [1]

15 9701/42/F/M/21 © UCLES 2021 [Turn over (iv) Write an expression for the stability constant, Kstab, of [Fe(edds)]–(aq). Kstab =  [1] (v) The table shows the values for the stability constants, Kstab, of both complexes. complex Kstab / mol–1 dm3 [Fe(edds)]– 3.98 × 1020 [Fe(edta)]– 1.26 × 1025 Predict which of the [Fe(edds)]– and [Fe(edta)]– complexes is more stable. Explain your answer with reference to the Kstab value for each complex. … … … [1] (vi) When an excess of edta4–(aq) is added to [Fe(edds)]–(aq), the following equilibrium is established. [Fe(edds)]–(aq) + edta4–(aq) [Fe(edta)]–(aq) + edds4–(aq) Calculate the equilibrium constant, Kc, for this equilibrium, using the Kstab values given in the table in (c)(v).  Kc = … [1]  [Total: 11]

16 9701/42/F/M/21 © UCLES 2021 5 (a) Carboplatin and satraplatin are used as anticancer drugs instead of cisplatin. carboplatin satraplatin Pt NH3 N H2 O O O O Cl Cl O O O NH3 NH3 O Pt (i) Describe the action of cisplatin as an anticancer drug. … … … [2] (ii) Suggest the geometry of the platinum centre in the carboplatin complex. … [1] (iii) Suggest why carboplatin does not show cis-trans isomerism. … … … [1] (iv) Satraplatin is a neutral complex, containing the ligands CH3CO2 –, C6H11NH2, Cl – and NH3. Deduce the oxidation state of platinum in satraplatin. … [1]

17 9701/42/F/M/21 © UCLES 2021 [Turn over (b) Compound M is made from 1,3-dimethylbenzene in a two-step synthesis. 1,3-dimethylbenzene C8H6O4 Cl Cl O M L O step 1 step 2 (i) Draw the structure of L.  [1] (ii) Suggest reactants and conditions for each step of this synthesis. step 1 … step 2 …  [2] (iii) Write an equation for step 2. … [1] (iv) A student investigates a possible synthesis of M directly from benzene using COCl 2 in the presence of an Al Cl 3 catalyst. Benzene initially reacts with COCl 2 as shown. reaction 1 COCl 2 + Al Cl 3 → Al Cl 4 – + Cl C O reaction 2 C6H6 + Cl C O → C6H5COCl + H+ Reaction 2 is the electrophilic substitution of Cl C O for H+ in benzene. Suggest a mechanism for reaction 2.  [3]  [Total: 12] + + +

18 9701/42/F/M/21 © UCLES 2021 6 Fumaric acid is a naturally occurring dicarboxylic acid. HO2C CO2H fumaric acid (a) Identify the products of the reaction between fumaric acid and an excess of hot, concentrated, acidified manganate(VII). … [1] (b) Fumaric acid can form addition and condensation polymers. (i) Draw the repeat unit of the addition polymer poly(fumaric acid).  [1] (ii) Draw the repeat unit of the polyester formed when fumaric acid reacts with ethane‑1,2‑diol, (CH2OH)2. The ester bond should be shown fully displayed.  [2] (iii) Explain why polyesters normally biodegrade more readily than polyalkenes. … … … [1]

19 9701/42/F/M/21 © UCLES 2021 [Turn over (c) Fumaric acid reacts with cold, dilute, acidified manganate(VII) to form compound P. P CO2H OH OH HO2C Only three stereoisomers of P exist. One of the stereoisomers is shown. CO2H OH OH H H HO2C Complete the three-dimensional diagrams in the boxes to show the other two stereoisomers of P. HO2C HO2C  [2]

20 9701/42/F/M/21 © UCLES 2021 (d) The enzyme fumarase catalyses the reaction of fumarate ions, C4H2O4 2–, with water to form malate ions, C4H4O5 2–. C4H2O4 2– + H2O C4H4O5 2– Describe, with the aid of a suitably labelled diagram, how an enzyme such as fumarase can catalyse a reaction. … … … …  [3]  [Total: 10]

21 9701/42/F/M/21 © UCLES 2021 [Turn over 7 Proline (Pro) is a naturally occurring amino acid. proline O OH N H (a) Proline is often found bonded to glycine (Gly) in a protein. (i) Draw the dipeptide Pro-Gly. The peptide bond must be shown fully displayed.  [2] (ii) Name the type of reaction that forms a dipeptide from two amino acids. … [1] (iii) Proline is able to form a poly(proline) peptide chain. A section of a poly(proline) chain is shown. N O N O O N Suggest why the secondary structure of poly(proline) cannot be stabilised by hydrogen bonding. … … [1]

22 9701/42/F/M/21 © UCLES 2021 (b) The reaction scheme shows several reactions of proline. proline O OH N H prolinol OH N H reaction 1 reaction 3 reaction 2 CH3COCl R C7H11NO3 Q NaOH(aq) (i) Write an equation for the reaction of proline with NaOH(aq) in reaction 1. C4H7NHCO2H + ����������������������������������������������������������������������������������������������������������� [1] (ii) Proline has a secondary amine functional group. Secondary amines react with acyl chlorides. For example, dimethylamine reacts with RCOCl according to the following equation. NH + RCOCl H3C H3C NCOR + HCl H3C H3C dimethylamine Suggest the skeletal structure of R, C7H11NO3, the product of reaction 2.  [1] (iii) Suggest the reagent required for reaction 3. … [1]

23 9701/42/F/M/21 © UCLES 2021 [Turn over (c) Proline was first synthesised in the laboratory using a multi-stage synthetic route. In stage 1, CH2(CO2C2H5)2 and CH2=CHCN react to form a single product U. stage 1 CH2(CO2C2H5)2 + CH2 H C U CHCN CO2C2H5 CO2C2H5 CH2CH2CN (i) Name all the functional groups present in the reactants of stage 1. CH2(CO2C2H5)2 … CH2=CHCN …  [2] (ii) Suggest the type of reaction that occurs in stage 1. … [1] In stage 2, U reacts with reagent V to form W. stage 2 H C U CO2C2H5 CO2C2H5 CH2CH2CN H C W CO2C2H5 CO2C2H5 CH2CH2CH2NH2 reagent V (iii) Suggest a suitable reagent V. … [1] Stage 3 takes place in the presence of an acid catalyst. X and Y are the only products of the reaction. stage 3 H C O X + Y W CO2C2H5 CO2C2H5 CO2C2H5 CH2CH2CH2NH2 HN (iv) Suggest the type of reaction that occurs in stage 3. … [1] (v) Deduce the identity of Y. … [1]

24 9701/42/F/M/21 © UCLES 2021 After several further stages, Z is produced. CO2H NH2 Cl Z In the final stage of the synthesis, Z reacts via a nucleophilic substitution mechanism to form proline. (vi) Complete the diagram to describe the reaction mechanism of the final stage. Draw curly arrows, ions and charges, partial charges and lone pairs of electrons, as appropriate. Draw the structure of any organic intermediate ion. CO2H proline N H CO2H NH2 Cl Z  [3] (vii) Identify with an asterisk (*) the chiral centre in proline. CO2H N H  [1]

25 9701/42/F/M/21 © UCLES 2021 [Turn over (d) Part of the structure of gelatin is shown. N C C O O C O C H H N H N H N H C H N H C C O N C O H N N H C O H C N H C C O O H C H H CH3 CH2 CH2 CH2 H C H2C H2C CO2 – NH C NH2 NH2 + C O Identify the number of amino acid units in the structure shown. … [1] (e) (i) At pH 6.5, proline exists in aqueous solution as a zwitterion. Draw the structure of the zwitterion of proline. Explain how the zwitterion of proline forms. … … … [2]

26 9701/42/F/M/21 © UCLES 2021 (ii) The isoelectric point of an amino acid is the pH at which it exists as a zwitterion. Three of the amino acids in gelatin are proline, alanine and glutamic acid. Their isoelectric points are shown. amino acid isoelectric point proline 6.5 alanine 6.0 H2N CO2H glutamic acid 3.1 H2N CO2H CO2H CO2H N H A mixture of these amino acids was analysed by electrophoresis using a buffer solution at pH 4.0. Draw and label three spots on the diagram of the electropherogram to indicate the likely position of each of these three species after electrophoresis. Explain your answer. + – mixture applied here … … … … … … … … [4]

27 9701/42/F/M/21 © UCLES 2021 [Turn over (f) The weak acid ACES is a compound that can be used to make a buffer solution for electrophoresis experiments. OH H N O ACES H2N S O O The anion of the sodium salt of ACES, C4H9N2O4SNa, is a strong base. A buffer solution is prepared by the following steps. ● 3.50 g of C4H9N2O4SNa is dissolved in 100 cm3 of distilled water. ● 50.0 cm3 of 0.200 mol dm–3 dilute hydrochloric acid is added to the solution. ● The resulting mixture is transferred to a 250.0 cm3 volumetric flask, and the solution made up to the mark. C4H9N2O4SNa reacts with HCl with a 1 : 1 stoichiometry. The pKa of ACES is 6.88 at 298 K. Calculate the pH of the buffer solution formed at 298 K. [Mr: C4H9N2O4SNa, 204.1]  pH = … [4]  [Total: 28]

28 9701/42/F/M/21 © UCLES 2021 BLANK PAGE Permission to reproduce items where third-party owned material protected by copyright is included has been sought and cleared where possible. Every reasonable effort has been made by the publisher (UCLES) to trace copyright holders, but if any items requiring clearance have unwittingly been included, the publisher will be pleased to make amends at the earliest possible opportunity. 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. Cambridge Assessment International Education is part of the Cambridge Assessment Group. Cambridge Assessment is the brand name of the University of Cambridge Local Examinations Syndicate (UCLES), which itself is a department of the University of Cambridge.

This document consists of 13 printed pages. © UCLES 2021 [Turn over Cambridge International AS & A Level CHEMISTRY 9701/42 Paper 4 A Level Structured Questions March 2021 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 March 2021 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 March 2021 © UCLES 2021 Page 2 of 13 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 March 2021 © UCLES 2021 Page 3 of 13 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 March 2021 © UCLES 2021 Page 4 of 13 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 March 2021 © UCLES 2021 Page 5 of 13 Question Answer Marks 1(a) Co2+ = [Ar] 3d7 (4s0) Co3+ = [Ar] 3d6 (4s0) 1 1(b) M1/2: Any two of: • Co3+ is reduced Co2+ • oxygen gas/O2 is evolved • E of Co3+ greater than E of O2 M3: no change (to [Co(edta)]–) / not feasible OWTTE 3 1(c) Any two of VISUAL observations: • condensation on tube / steam evolved • brown fumes / brown gas evolved • O2 formed that relights a glowing splint • (solid) dissolves / turns to liquid 2 1(d) M1: cationic radius / ion size increases (down the group) M2: less polarisation / distortion of nitrate ion / anion / NO3– 2 Question Answer Marks 2(a)(i) M1 the only number extracted: 762, 1560, 496 M2 correct multiplier, other four numbers used and calculation to the answer –272 = +416 + ½(496) + 762 + 1560 –141 + 798 + ΔHlattice ∴ ΔHlattice = –3915 (kJ mol–1) ecf 2 2(a)(ii) 20 × [0.9(+2) + 0.1(+3)] –2x = 0 ∴ x = 21 1 2(a)(iii) • FeO more exothermic/more negative • Fe2+ has smaller radius/higher charge density (also same charge) • greater attraction/ stronger ionic bonds (between Fe2+ and O2–) All three for two marks 2

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED March 2021 © UCLES 2021 Page 6 of 13 Question Answer Marks 2(b)(i) • Fe2+ reduced to Fe OR oxid no. Fe +2 → 0 • Fe2+ oxidised to Fe3+ (in Fe3O4) OR oxid no. Fe +2 → +3 BOTH bullets required 1 2(b)(ii) 2O2– → O2 + 4e– 1 2(b)(iii) M1: coulombs and correct use of ÷ 96500 M2: correct use of 3 and 8 M3: correct use of 55.8 and answer M1: Q = It = 50 × 6 × 602 OR 1.08 × 106 C AND no. of faraday = 1.08 × 106 ÷ 96500 OR 11.2 / 11.19 mol e– M2: Fe2+ + 2Fe3+ + 8e– → 3Fe ∴ moles of Fe = 3 / 8 × M1 = 4.20 mol Fe ecf M3: mass of Fe = 55.8 × M2 = 234.2 g ecf 3sf min 3 2(c)(i) Any one of: small size / compact, low mass, high voltage OWTTE 1 2(c)(ii) Li from +1 to +1 Fe from +3 to +2 1 2(c)(iii) LiC6 + FePO4 → LiFePO4 + 6C 1 Question Answer Marks 3(a) 3 bonding-pair centres and one lone pair (on iodine) 1 3(b) 3I2 + 6NaOH → NaIO3 + 5NaI + 3H2O 1

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED March 2021 © UCLES 2021 Page 7 of 13 Question Answer Marks 3(c)(i) M1: E⦵cell for IO3– / H2O2 = –0.68 + 1.19 = +0.51 (∴ feasible) M2: E⦵cell for H2O2 / I2 = +1.77 – 1.19 = +0.58 (∴ feasible) M3: 5H2O2 + I2 → 4H2O + 2IO3– + 2H+ 3 3(c)(ii) 2H2O2 → 2H2O + O2 1 3(d)(i) M1: first order w.r.t. H2O2 AND change in conc. × 1.5 gives increase rate × 1.5 (expts 3 / 4) M2: first order w.r.t. IO3– AND change in conc. × 2 gives increase rate × 2 (as reaction first order w.r.t. H2O2) (expts 1 / 3) M3: zeroth order w.r.t. H+ AND change in conc. has no effect on rate (expts 1 / 3 / 4 and 2) 3 3(d)(ii) rate = k[H2O2][IO3–] ecf 1 3(d)(iii) M1: k = 8.82 × 10–5 ÷ (0.150 × 0.140) = 4.20 × 10–3 min 2sf ecf M2: mol–1 dm3 s–1 ecf 2 3(e)(i) Ksp = [Pb2+][IO3–]2 1 3(e)(ii) M1: 3.69 × 10–13 = x(2x)2 OR x = ∛(3.69 × 10–13 ÷ 4) M2: = 4.5(2) × 10–5 (mol dm–3) min 2sf ecf 2 3(f)(i) M1: ΔS = ½(192) + ½(205) + ½(261) + 2(70) – 42 M2: (+)427 (J K–1 mol–1) ecf 2 3(f)(ii) ΔG (always) negative because • ΔH < 0 / negative OR exothermic AND • ΔS > 0 / positive OR –TΔS < 0 for all T 1 Question Answer Marks 4(a) (element that forms one or more stable) ions with incomplete/ partially filled 3d-orbitals/d-subshell 1 4(b)(i) Oh AND Td 1

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED March 2021 © UCLES 2021 Page 8 of 13 Question Answer Marks 4(b)(ii) M1: lower energy level (in between axes) OR OR M2: higher energy level (on the axes) OR 2 4(c)(i) Circles round both N atoms and all four O– 1 4(c)(ii) M1: (d–d) energy gap / ΔE is different M2: different frequency / wavelength (of light) absorbed 2 4(c)(iii) ligand exchange / substitution / displacement / replacement 1 4(c)(iv) ( ) ( ) – stab 3+ 4– 2 6 Fe edds K = Fe H O edds            1 4(c)(v) [Fe(edta)]– is more stable as it has the higher Kstab 1 4(c)(vi) 25 4 stab c 20 stab K (edta) 1.26×10 K = = 3.17 × 10 K (edds) 3.98×10 = (31658) min 2sf 1

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED March 2021 © UCLES 2021 Page 9 of 13 Question Answer Marks 5(a)(i) M1: (cisplatin) can bond / bind with DNA / (nitrogenous) base A, T, C, G etc. M2: which prevents replication (of the DNA / strand) OR prevents cell division / prevents mitosis OR prevents transcription (and formation of mRNA) 2 5(a)(ii) square planar 1 5(a)(iii) the distance between two coordinating oxygens is too small to bond trans OR atoms in a bidentate ligand can only bond 90° not 180° 1 5(a)(iv) +4 1 5(b)(i) L 1 5(b)(ii) M1: heat / reflux with acidified / alkaline KMnO4 (then acidify) M2: PCl5 OR SOCl2 / (heat with) PCl3 2 5(b)(iii) C8H6O4 + 2PCl5 → C8H4O2Cl2 + 2POCl3 + 2HCl OR C8H6O4 + 2SOCl2 → C8H4O2Cl2 + 2SO2 + 2HCl OR 3C8H6O4 + 2PCl3 → 3 C8H4O2Cl2 + 2H3PO3 1 5(b)(iv) M1: curly arrow from inside hexagon to C of electrophile M2: correct intermediate M3: curly arrow from C—H bond AND formation/loss of H+ 3

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED March 2021 © UCLES 2021 Page 10 of 13 Question Answer Marks 6(a) CO2 and H2O / in words 1 6(b)(i) if more than one unit drawn ALLOW one repeat unit identified 1 6(b)(ii) M1: presence of an ester group from the diol and COOH OR presence of an ester group from the fumaric acid and OH M2: rest of repeat unit including ‘dangling’ bonds 2 6(b)(iii) C—C bonds are non-polar / polyalkenes cannot be hydrolysed OR polyesters / they can be broken down by hydrolysis 1 6(c) 2

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED March 2021 © UCLES 2021 Page 11 of 13 Question Answer Marks 6(d) M1: (can be in words or diagram) substrate shape is complementary to active site M2: (can be in words or diagram) the substrate bind / bonds / fits (into the active site) M3: (can be in words or diagram) products are released 3 Question Answer Marks 7(a)(i) OR M1: peptide link shown M2: rest of Pro–Gly correct 2 7(a)(ii) condensation ALLOW substitution / addition–elimination 1 7(a)(iii) there is no H attached to the N 1 7(b)(i) (C4H7NHCO2H +) NaOH → C4H7NHCO2Na + H2O 1 7(b)(ii) skeletal only 1

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED March 2021 © UCLES 2021 Page 12 of 13 Question Answer Marks 7(b)(iii) LiAlH4 1 7(c)(i) CH2(CO2C2H5)2: ● (di)ester CH2=CHCN: ● alkene ● nitrile/cyanide All three correct for two marks 2 7(c)(ii) addition 1 7(c)(iii) H2/ Ni OR H2/ Pt OR H2/ Pd 1 7(c)(iv) condensation / (nucleophilic) substitution / elimination 1 7(c)(v) ethanol / C2H5OH / CH3CH2OH 1 7(c)(vi) M1/2: All four correct: • lone pair on NH2 • curly arrow from N: to C of C—Cl • correct dipole on C—Cl • curly arrow from C—Cl to Cl M3: intermediate = OR 3 7(c)(vii) Asterisk on *CHCO2H 1

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED March 2021 © UCLES 2021 Page 13 of 13 Question Answer Marks 7(d) 9 1 7(e)(i) M1: M2: proton / H+ transferred from carboxylic acid to amine 2 7(e)(ii) M1: glutamic acid towards + end (from the diagram) M2: proline and alanine towards – end (from the diagram) M3: Glu moves towards positive (pole) as negatively charged / contains a COO– OR Pro/Ala move towards negative (pole) as positively charged / contains a NH2+ / contains a NH3+ M4: Ala moves farther than Pro because of lower Mr/ size (with positive charge) ORA 4 7(f) M1: initial amount of C4H9N2O4SNa = 3.50 / 204.1 OR 0.0171(48) mol AND amount of HCl added = 0.200 × 50.0 / 1000 OR 0.0100 mol M2: equilibrium amount of C4H9N2O4SNa = 0.0171(48) – 0.0100 OR 0.0071(48) mol AND equilibrium amount of ACES = 0.0100 mol ecf M3: Ka = 10–6.88 = 1.32 ×10–7 (mol dm–3) [H+] = (1.32 × 10–7)0.01 / 0.0071(48) = 1.86 × 10–7 OR 1.8465 × 10–7 ecf M4: pH = –log(1.86 × 10–7) = 6.73 3sf min ecf 4