Cambridge A Level Chemistry 9701 — 2020 May/June Paper 4 · Variant 2

*3731431728* CHEMISTRY 9701/42 Paper 4 A Level Structured Questions May/June 2020  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 [ ]. IB20 06_9701_42/3RP © UCLES 2020 [Turn over This document has 24 pages. Blank pages are indicated. Cambridge International AS & A Level

2 9701/42/M/J/20 © UCLES 2020 Answer all the questions in the spaces provided. 1 EDTA4–, is a polydentate ligand. (a) (i) Explain what is meant by the term polydentate ligand. … … … [2] (ii) When a solution containing EDTA4– is added to a solution containing [Cd(H2O)6]2+ a new complex is formed, [CdEDTA]2–. equilibrium 1 [Cd(H2O)6]2+ + EDTA4– [CdEDTA]2– + 6H2O Circle, on the structure of EDTA4–, the six atoms that form bonds with the metal ion. NCH2CH2N EDTA4– CH2C CH2C CCH2 CCH2 O– O O O– –O –O O O  [1] (iii) Write an expression for the stability constant, Kstab1, for equilibrium 1, and state its units. Kstab1 =  units = …  [2]

3 9701/42/M/J/20 © UCLES 2020 [Turn over (b) Cadmium ions form complexes with methylamine, CH3NH2, and with 1,2‑diaminoethane, H2NCH2CH2NH2, as shown in equilibriums 2 and 3. 1,2‑diaminoethane is shown as en. equilibrium 2 [Cd(H2O)6]2+ + 4CH3NH2 [Cd(CH3NH2)4(H2O)2]2+ + 4H2O Kstab2 = 3.60 × 106 equilibrium 3 [Cd(H2O)6]2+ + 2en [Cd(en)2(H2O)2]2+ + 4H2O Kstab3 = 4.20 × 1010 An equilibrium is set up between these two complexes as shown in equilibrium 4. equilibrium 4 [Cd(CH3NH2)4(H2O)2]2+ + 2en [Cd(en)2(H2O)2]2+ + 4CH3NH2 ∆H o = +0.840 kJ mol–1 ∆S o = +80.9 J K–1 mol–1 (i) Keq4 is the equilibrium constant for equilibrium 4. Write an expression for Keq4 in terms of Kstab2 and Kstab3. Keq4 =  [1] (ii) Calculate the value of the standard Gibbs free energy change, ∆G o, for equilibrium 4 at 298 K.  ∆G o = … kJ mol–1 [2] (iii) State how the value of ∆G o changes as the temperature increases. Explain your answer. … … … [1]  [Total: 9]

4 9701/42/M/J/20 © UCLES 2020 2 (a) Describe and explain how the solubility of the Group 2 sulfates varies down the group. … … … … … … [4] (b) The trend in the decomposition temperatures of Group 2 peroxides, MO2, is similar to that of Group 2 carbonates. Suggest which of barium peroxide, BaO2, and calcium peroxide, CaO2, will decompose at the lower temperature. Explain your answer. … … … [2] (c) Magnesium iodate(V), Mg(IO3)2, decomposes when heated to form magnesium oxide, oxygen and iodine. Construct an equation for this reaction. … [1] (d) Calcium iodate(V), Ca(IO3)2, is sparingly soluble in water. The concentration of its saturated solution is 5.6 × 10–3 mol dm–3 at 298 K. (i) Write an expression for the solubility product, Ksp, of Ca(IO3)2, and state its units. Ksp =  units = … [2] (ii) Calculate the numerical value for Ksp Ca(IO3)2 at 298 K.  Ksp = … [1]

5 9701/42/M/J/20 © UCLES 2020 [Turn over (iii) When a few cm3 of concentrated Ca(NO3)2(aq) is added to a saturated solution of Ca(IO3)2 a white precipitate forms. Identify the white precipitate and give an explanation for this observation. … … … [2] (e) Iodised salt is sodium chloride mixed with a small amount of sodium iodate(V), NaIO3. ● ● 50.00 g of iodised salt is dissolved in distilled water and the solution made up to 250 cm3 in a volumetric flask with distilled water. ● ● 50.0 cm3 of this solution is pipetted into an excess of aqueous acidified potassium iodide. IO3 – + 5I– + 6H+ → 3I2 + 3H2O ● ● The iodine produced requires 12.40 cm3 of 0.00200 mol dm–3 aqueous sodium thiosulfate solution for complete reaction. I2 + 2S2O3 2– → 2I– + S4O6 2– Calculate the mass of sodium iodate(V) present in 50.00 g of iodised salt.  mass of NaIO3 = … g [3] (f) The half-equation for the reduction of iodate(V) ions is shown. IO3 – + 6H+ + 5e– → 1 2I2 + 3H2O E o = +1.19 V Use data from the Data Booklet to predict whether a reaction is feasible when aqueous solutions of acidified iodate(V) ions and bromide ions are mixed. Explain your answer. … … [1]

6 9701/42/M/J/20 © UCLES 2020 (g) Iodate(V) ions react with sulfite ions in acidic solution at pH 5.00 as shown. IO3 – + 3SO3 2– → I– + 3SO4 2– The initial rate of reaction was found to be first order with respect to IO3 –, first order with respect to SO3 2– and first order with respect to H+. (i) Write the rate equation for this reaction, stating the units of the rate constant, k. rate = … mol dm–3 s–1 units of k = …  [2] (ii) The rate of reaction depends on the pH of the solution. Assume all other concentrations remain the same. Use the expression x = rate at pH 5.00 rate at pH 4.00 to calculate the value of x.  x = … [1]  [Total: 19]

7 9701/42/M/J/20 © UCLES 2020 [Turn over 3 (a) Complete the electronic configuration of an isolated gaseous nickel(II) ion, Ni2+. 3d [Ar] 4s  [1] (b) Explain the origin of colour in transition element complexes. … … … … … … [4] (c) Hexaaquanickel(II) ions are green. They form a green precipitate with hydroxide ions, OH–, in equilibrium 1 and a blue complex with ammonia, NH3, in equilibrium 2. equilibrium 1 [Ni(H2O)6]2+ + 2OH– Ni(OH)2 + 6H2O green ppt. equilibrium 2 [Ni(H2O)6]2+ + 6NH3 [Ni(NH3)6]2+ + 6H2O blue Use Le Chatelier’s principle to suggest explanations for the following observations. (i) Explain why when aqueous NH3 is added dropwise to [Ni(H2O)6]2+ a green precipitate is formed. … … … [1] (ii) Explain why when a large excess of aqueous NH3 is added to [Ni(H2O)6]2+, the green precipitate dissolves and a blue solution is formed. … … … [1]

8 9701/42/M/J/20 © UCLES 2020 (d) The complex ion [NiBr2(CN)2]2– shows stereoisomerism. Draw diagrams to show the two isomers of [NiBr2(CN)2]2–. Name the type of stereoisomerism. isomer 1 isomer 2 type of stereoisomerism …  [2]  [Total: 9]

9 9701/42/M/J/20 © UCLES 2020 [Turn over 4 (a) (i) When benzene undergoes nitration a nitro group substitutes at a carbon atom. State the shape (geometry) around the substituted carbon atom ● ● in the benzene molecule, … ● ● in the intermediate complex, … ● ● in the nitrobenzene product. …  [2] (ii) Naphthalene, C10H8, is an arene hydrocarbon. naphthalene When naphthalene undergoes nitration, a mixture of two organic compounds is formed. Each compound contains one nitro group. Suggest the structures of these compounds. structure 1 structure 2  [1] (b) Naphthalene can be oxidised under certain conditions to phthalic anhydride, C8H4O3, carbon dioxide and water. Construct an equation for this reaction. Use [O] to represent an atom of oxygen from the oxidising agent. … [1]

10 9701/42/M/J/20 © UCLES 2020 (c) The indicator, phenolphthalein, can be synthesised from phthalic anhydride and phenol under certain conditions. O phthalic anhydride phenolphthalein phenol O O O + H2O 2 + O OH HO OH Deduce the type of reaction shown by this equation. … [1]

11 9701/42/M/J/20 © UCLES 2020 [Turn over (d) (i) Name the functional groups, in addition to the benzene ring, present in a phenolphthalein molecule. … [1] (ii) Phenolphthalein reacts separately with the two reagents shown in the table. Complete the table by: ● ● drawing the structures of the organic products formed (part of the structure has been given for you) ● ● stating the types of reaction. reagent organic product structure type of reaction an excess of hot NaOH(aq) an excess of Br2(aq)  [4]

12 9701/42/M/J/20 © UCLES 2020 (e) Phenolphthalein is an indicator and is represented by the formula HIn. Phenolphthalein, HIn, is a weak acid. HIn H+ + In– The Ka value for phenolphthalein is 5.0 × 10–10 mol dm–3 at 298 K. This indicator changes colour at a pH of approximately 8.8. Calculate the ratio [In–] [HIn] at pH 8.8.  ratio [In–] [HIn] = … [2] (f) Methyl orange is another acid-base indicator. Its structure in aqueous solution at pH 4.4 is shown. N N SO3 –Na+ N methyl orange H3C H3C (i) On the structure of methyl orange, circle the bond or bonds that make this compound a dye. [1] The colour of this indicator changes between pH 3.2 and pH 4.4. (ii) Suggest the structure of methyl orange at pH 3.0. Assume the –SO3 –Na+ group is unreactive.  [1]

13 9701/42/M/J/20 © UCLES 2020 [Turn over (g) Methyl orange can be synthesised as shown. N N SO3 –Na+ SO3 –Na+ N methyl orange H3C H3C N O2N H3C H3C step 2 S step 1 R (i) Deduce the identities of compounds R and S and draw their structures in the boxes. [2] (ii) Suggest reagents and conditions for step 1 and step 2. step 1 … step 2 …  [3]  [Total: 19]

14 9701/42/M/J/20 © UCLES 2020 5 (a) Define the term partition coefficient, Kpc. … … … [2] (b) Kpc of benzoic acid between octan‑1‑ol and water is 79.4. (i) A solution of 0.400 g of benzoic acid in 25.0 cm3 octan‑1‑ol is shaken with 125 cm3 of water. Calculate the mass of benzoic acid extracted into the water layer.  mass of benzoic acid extracted = … g [2] (ii) Kpc of benzophenone, C6H5COC6H5, between octan‑1‑ol and water is different from the value of Kpc of benzoic acid given in (b)(i). Explain why. … … … [1]

15 9701/42/M/J/20 © UCLES 2020 [Turn over (c) Benzophenone can be synthesised from benzoic acid in two steps as shown. In step 1 compound J, a reactive reaction intermediate, is formed. Compound J then reacts with an organic compound, K, to form benzophenone. K J Al Cl 3 CO2H step 1 benzoic acid O benzophenone (i) Deduce the identities of organic compounds J and K and draw their structures in the boxes. [2] (ii) Suggest reagents and conditions for step 1. … [1]

16 9701/42/M/J/20 © UCLES 2020 (d) Benzophenone can also be synthesised in two steps from bromomethylbenzene. L, C13H12 step 3 bromomethylbenzene step 4 KMnO4, heat O benzophenone Br (i) Deduce the identity of compound L and draw its structure in the box. [1] (ii) Name the mechanism of step 3 and suggest reagents and conditions for step 3. mechanism of step 3 … reagents and conditions …  [2] (iii) Deduce the type of reaction in step 4. … [1] (e) (i) Deduce the number of peaks that would be present in the carbon‑13 NMR spectrum of benzophenone. number of peaks … [1] (ii) Identify two different environments of carbon atom that would result in different chemical shift ranges in this carbon‑13 NMR spectrum of benzophenone. environment of carbon atom chemical shift range (δ)  [2]  [Total: 15]

17 9701/42/M/J/20 © UCLES 2020 [Turn over 6 The class of polymers called polycarbonates are made by the reaction of carbonyl dichloride, COCl 2, with a diol. n HOROH + n COCl 2 + 2nHCl O a diol R O a polycarbonate C O n (a) (i) Deduce the type of polymerisation shown here. … [1] Nalgene® is a polycarbonate formed from the diol P and COCl 2. C CH3 H3C P HO OH (ii) Draw one repeat unit of Nalgene®.  [1] (iii) Nalgene® is a strong and tough polymer. Identify two types of intermolecular force that are responsible for these properties of Nalgene®. 1 … 2 …  [1] (b) Proteins are polymers of amino acids. Complete the table to show how the secondary and tertiary structures of proteins are stabilised. one intermolecular force responsible groups involved secondary structure tertiary structure  [2]

18 9701/42/M/J/20 © UCLES 2020 (c) Explain the significance of hydrogen bonding in DNA in relation to the accurate replication of genetic information. … … … [2] (d) Many polymers are degradable. State two different processes by which some polymers can be degraded. … … [1] (e) The cyclic peptide B is shown. O O O O O B O O N H H N HN HN HN NH NH HO Cyclic peptide B is broken into its monomers by heating under reflux with dilute hydrochloric acid. The amino acid threonine, Thr, and two other organic products are formed. OH threonine CO2H H2N

19 9701/42/M/J/20 © UCLES 2020 [Turn over (i) Draw the structures of the two other organic products formed.  [2] (ii) Using the 3-letter abbreviations for the amino acids as given in the Data Booklet, complete the sequence for the cyclic peptide, B. … … … … … … Thr  [1] (iii) Name two analytical techniques that could be used to separate these amino acids. … and … [1]  [Total: 12]

20 9701/42/M/J/20 © UCLES 2020 7 (a) (i) Define the term electron affinity. … … … [2] (ii) Define the term lattice energy. … … … [2] (b) Use the following data and relevant data from the Data Booklet to calculate a value for the enthalpy change of formation of zinc bromide, ZnBr2(s). You might find it helpful to construct an energy cycle. electron affinity of Br(g) = –325 kJ mol–1 enthalpy change of atomisation of Zn(s) = +131 kJ mol–1 enthalpy change of vaporisation of Br2(l) = +31 kJ mol–1 lattice energy of ZnBr2(s) = –2678 kJ mol–1  enthalpy change of formation of ZnBr2(s) = … kJ mol–1 [4]

21 9701/42/M/J/20 © UCLES 2020 [Turn over (c) The lattice energies of ZnBr2, ZnCl 2 and ZnO are shown. compound lattice energy / kJ mol–1 ZnBr2 –2678 ZnCl 2 –2734 ZnO –3971 (i) Explain why there is a difference between the lattice energies of ZnBr2 and ZnCl 2. … … [1] (ii) Explain why there is a difference between the lattice energies of ZnCl 2 and ZnO. … … [1]  [Total: 10]

22 9701/42/M/J/20 © UCLES 2020 8 (a) (i) Define the term standard cell potential. … … … [2] An electrochemical cell is set up to measure the standard electrode potential of a cell, , made of a Co3+ / Co2+ half-cell and a Cl 2 / Cl – half-cell. (ii) Complete the table with the substance used to make the electrode in each of these half‑cells. half-cell electrode Co3+ / Co2+ Cl 2 / Cl –  [1] (iii) Use data from the Data Booklet to calculate the .  = … V [1] (iv) Write the equation for the overall cell reaction. … [1] (b) A fuel cell is an electrochemical cell that can be used to generate electrical energy. A methanol-oxygen fuel cell can be used as an alternative to a hydrogen-oxygen fuel cell. When the cell operates, the carbon atoms in the methanol molecules are converted into carbon dioxide. CH3OH + H2O → CO2 + 6H+ + 6e– Calculate the volume of CO2, in cm3, formed when a current of 2.5 A is delivered by the cell for 30 minutes. Assume the cell is operated at room conditions.  volume of CO2 = … cm3 [2]  [Total: 7]

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24 9701/42/M/J/20 © UCLES 2020 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 14 printed pages. © UCLES 2020 [Turn over Cambridge International AS & A Level CHEMISTRY 9701/42 Paper 4 A Level Structured Questions May/June 2020 MARK SCHEME Maximum Mark: 100 Published Students did not sit exam papers in the June 2020 series due to the Covid-19 global pandemic. This mark scheme is published to support teachers and students and should be read together with the question paper. It shows the requirements of the exam. The answer column of the mark scheme shows the proposed basis on which Examiners would award marks for this exam. Where appropriate, this column also provides the most likely acceptable alternative responses expected from students. Examiners usually review the mark scheme after they have seen student responses and update the mark scheme if appropriate. In the June series, Examiners were unable to consider the acceptability of alternative responses, as there were no student responses to consider. Mark schemes should usually be read together with the Principal Examiner Report for Teachers. However, because students did not sit exam papers, there is no Principal Examiner Report for Teachers for the June 2020 series. Cambridge International will not enter into discussions about these mark schemes. Cambridge International is publishing the mark schemes for the June 2020 series for most Cambridge IGCSE™ and Cambridge International A & AS Level components, and some Cambridge O Level components.

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2020 © UCLES 2020 Page 2 of 14 Generic Marking Principles These general marking principles must be applied by all examiners when marking candidate answers. They should be applied alongside the specific content of the mark scheme or generic level descriptors for a question. Each question paper and mark scheme will also comply with these marking principles. GENERIC MARKING PRINCIPLE 1: Marks must be awarded in line with: • the specific content of the mark scheme or the generic level descriptors for the question • the specific skills defined in the mark scheme or in the generic level descriptors for the question • the standard of response required by a candidate as exemplified by the standardisation scripts. GENERIC MARKING PRINCIPLE 2: Marks awarded are always whole marks (not half marks, or other fractions). GENERIC MARKING PRINCIPLE 3: Marks must be awarded positively: • marks are awarded for correct/valid answers, as defined in the mark scheme. However, credit is given for valid answers which go beyond the scope of the syllabus and mark scheme, referring to your Team Leader as appropriate • marks are awarded when candidates clearly demonstrate what they know and can do • marks are not deducted for errors • marks are not deducted for omissions • answers should only be judged on the quality of spelling, punctuation and grammar when these features are specifically assessed by the question as indicated by the mark scheme. The meaning, however, should be unambiguous. GENERIC MARKING PRINCIPLE 4: Rules must be applied consistently e.g. in situations where candidates have not followed instructions or in the application of generic level descriptors.

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2020 © UCLES 2020 Page 3 of 14 GENERIC MARKING PRINCIPLE 5: Marks should be awarded using the full range of marks defined in the mark scheme for the question (however; the use of the full mark range may be limited according to the quality of the candidate responses seen). GENERIC MARKING PRINCIPLE 6: Marks awarded are based solely on the requirements as defined in the mark scheme. Marks should not be awarded with grade thresholds or grade descriptors in mind.

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2020 © UCLES 2020 Page 4 of 14 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 (see examples below) For questions that require n responses (e.g. State two reasons …): • The response should be read as continuous prose, even when numbered answer spaces are provided • Any response marked ignore in the mark scheme should not count towards n • Incorrect responses should not be awarded credit but will still count towards n • Read the entire response to check for any responses that contradict those that would otherwise be credited. Credit should not be awarded for any responses that are contradicted within the rest of the response. Where two responses contradict one another, this should be treated as a single incorrect response • Non-contradictory responses after the first n responses may be ignored even if they include incorrect science.

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2020 © UCLES 2020 Page 5 of 14 6 Calculation specific guidance Correct answers to calculations should be given full credit even if there is no working or incorrect working, unless the question states ‘show your working’. For questions in which the number of significant figures required is not stated, credit should be awarded for correct answers when rounded by the examiner to the number of significant figures given in the mark scheme. This may not apply to measured values. For answers given in standard form, (e.g. a × 10n) in which the convention of restricting the value of the coefficient (a) to a value between 1 and 10 is not followed, credit may still be awarded if the answer can be converted to the answer given in the mark scheme. Unless a separate mark is given for a unit, a missing or incorrect unit will normally mean that the final calculation mark is not awarded. Exceptions to this general principle will be noted in the mark scheme. 7 Guidance for chemical equations Multiples / fractions of coefficients used in chemical equations are acceptable unless stated otherwise in the mark scheme. State symbols given in an equation should be ignored unless asked for in the question or stated otherwise in the mark scheme.

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2020 © UCLES 2020 Page 6 of 14 Question Answer Marks 1(a)(i) M1 (a species) that donates/uses a many lone pairs/more than one lone pair M2 to form a dative/coordinate to a metal atom/metal ion/TM/TE/metal OR M1 (a species) that donates/uses lone pairs to form many/more than one M2 dative/coordinate bond to a metal atom/metal ion/TM/TE/metal 2 1(a)(ii) structure of EDTA any six atoms circled of 2 N & 4 O 1 1(a)(iii) M1 Kstab1 = [[CdEDTA]2-] [[Cd(H2O)6]2+][EDTA4-] M2 units = mol-1 dm3 2 1(b)(i) Keq4 = Kstab3/Kstab2 1 1(b)(ii) M1 ΔGo = ΔHo -TΔSo ΔGo = 0.84 – (298*0.0809) M2 ΔGo = -23.3 (kJ mol-1) 3sf min 2 1(b)(iii) more negative as TΔS increases OR more negative as ΔS is positive 1 Question Answer Marks 2(a) M1 solubility decreases down the group M2 ΔHlatt and ΔHhyd both become less exothermic / more endothermic M3 ΔHlatt changes less (than ΔHhyd as SO42- being larger than M2+) M4 ΔHsol becomes less exothermic / less negative 4

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2020 © UCLES 2020 Page 7 of 14 Question Answer Marks 2(b) M1 CaO2 and Ca2+ has a smaller ionic radii/ Ca2+ has a higher charge density M2 anion/O22- becomes more polarised /distorted 2 2(c) Mg(IO3)2  MgO + 2.5O2 + I2 1 2(d)(i) M1 Ksp = [Ca2+][IO3-]2 M2 units = mol3 dm-9 2 2(d)(ii) Ksp = 4 x (5.6 × 10–3)3 Ksp = 7.03 x 10–7 2sf min 1 2(d)(iii) M1 Ca(IO3)2 AND as solubility of Ca(IO3)2 decreases M2 due to common ion effect 2 2(e) M1 moles S2O32- = 0.002 x 12.4/1000 = 2.48 x 10-5 moles of I2 = 1.24 x 10-5 M2 moles of IO3− = 4.13 x 10-6 in 50 cm3 moles of IO3− = 2.07 x 10-5 in 250 cm3 mass of NaIO3 = 2.07 x 10-5 x 197.9 M3 mass of NaIO3 = 0.0041 3 2(f) It is feasible as the Ecell will be positive/+0.12 V 1 2(g)(i) M1 Rate = k[IO3−][SO32-][H+] M2 units = mol-2dm6s-1 2 2(g)(ii) 0.10 1

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2020 © UCLES 2020 Page 8 of 14 Question Answer Marks 3(a) 1 3(b) M1 d orbitals split into two levels/ lower and upper orbitals M2 electron(s) promoted / excited to a higher d–orbital M3 frequency of light absorbed M4 observed colour is complement of light absorbed 4 3(c)(i) (addition of NH3) increases [OH-] (due to ionisation of NH3 in water) and shifts equilibrium 1 to the right (forming Ni(OH)2) 1 3(c)(ii) (a large excess of NH3) shifts eqm 2 to the right (forming [Ni(NH3)6]2+) AND the [Ni2+]/[[Ni(H2O)6]2+] decreases and eqm 1 shifts to the left (causing the ppt to dissolve) 1 3(d) M1 two correct structures cis & trans for [NiBr2(CN)2]2- M2 type of stereoisomerism: cis-trans/geometric 2 Question Answer Marks 4(a)(i) • trigonal planar • tetrahedral. • trigonal planar Award one mark for two correct statements, award two marks for three correct statements 2

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2020 © UCLES 2020 Page 9 of 14 Question Answer Marks 4(a)(ii) Both structures required 1 4(b) C10H8 + 9[O] → C8H4O3 + 2CO2 + 2H2O 1 4(c) condensation/ addition-elimination 1 4(d)(i) phenol AND ester 1 4(d)(ii) M1 correct hydrolysis product of ester M2 (di)phenoxide salt M3 bromination of both phenol rings at position 2 or/and 6 M4 hydrolysis AND (electrophilic) substitution 4

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2020 © UCLES 2020 Page 10 of 14 Question Answer Marks 4(e) M1 [H+] = 10-8.8 = 1.585 x10-9 M2 [In-]/[HIn] = 5.0 x 10-10/1.585 x10-9 = 0.315 2 4(f)(i) bond circled between the two Ns, or N=N or –N=N– 1 4(f)(ii) or 1 4(g)(i) R H2N SO3 -Na+ S N SO3 -Na+ N + Award one mark for each correct structure 2 4(g)(ii) M1 step 1 Sn and HCl conc. and heat M2 step 2 NaNO2 and HCl and 0-10 oC 3 Question Answer Marks 5(a) M1 ratio of the concentration of a solute in the two immiscible solvents /liquids M2 at equilibrium 2

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2020 © UCLES 2020 Page 11 of 14 Question Answer Marks 5(b)(i) M1 79.4 = (0.4-x/25)/(x/125) M2 x = 0.0237 g [2] min 2sf 2 5(b)(ii) (higher as) benzophenone is more non-polar/more soluble in octan-1-ol ora 1 5(c)(i) Award one mark for each correct structure 2 5(c)(ii) step 1 PCl5 OR SOCl2 OR PCl3 + heat 1 5(d)(i) 1 5(d)(ii) M1 step 3 electrophilic substitution M2 step 3 benzene and AlCl3 (and heat) 2 5(d)(iii) step 4 oxidation 1 5(e)(i) 5 peaks 1 5(e)(ii) environment of carbon atom chemical shift range (δ) carbonyl / RCOR 190–220 arene / benzene 110–160 Award one mark for each correct for each row 2

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2020 © UCLES 2020 Page 12 of 14 Question Answer Marks 6(a)(i) condensation 1 6(a)(ii) 1 6(a)(iii) id-id forces/London forces AND permanent dipole-dipole forces 1 6(b) M1 (secondary structure by) hydrogen bonding between CO and NH groups M2 (tertiary structure by) interactions between R groups and one example of a named intermolecular force 2 6(c) M1 (hydrogen bonding between) base pairs M2 A with T and C with G 2 6(d) hydrolysis and by action of light/UV 1 6(e)(i) Award one mark for each correct structure 2

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2020 © UCLES 2020 Page 13 of 14 Question Answer Marks 6(e)(ii) correct labelling as shown 1 6(e)(iii) electrophoresis and thin-layer / paper chromatography [1] 1 Question Answer Marks 7(a)(i) • energy change • when one electron is added • to each atom /ion in one mole of • gaseous atoms /ions Award one mark for two correct statements. Award two marks for four correct statements 2 7(a)(ii) M1 energy change when 1 mole of an ionic compound is formed M2 from gas phase ions/ gaseous ions 2 Gly Phe Phe .Gly (Thr) Phe Phe

9701/42 Cambridge International AS & A Level – Mark Scheme PUBLISHED May/June 2020 © UCLES 2020 Page 14 of 14 Question Answer Marks 7(b) M1 use of data (with no multipliers) 31, 131, -2678 M2 extraction of data 908, 1730, 193 M3 use of (2 x-325) M4 evaluation of their expression correctly, as shown ΔHf(ZnBr2) = 131 + (908 +1730) + 193 + 31 + (2 x-325) + (-2678) = -335 kJ mol-1 [4] 4 7(c)(i) Br- is a largest ion/larger ion than Cl- so attraction between Br- and Zn2+ is smaller 1 7(c)(ii) O2- is a smallest ion/smaller ion than Cl- AND O2- has the highest charge/ higher charge than Cl- (so attraction between O2- and Zn2+ is larger) 1 Question Answer Marks 8(a)(i) M1 potential difference between two half-cells/two electrodes in a cell M2 under conditions of 1 atm., 298 K, (all) solutions being 1 mol dm-3 2 8(a)(ii) both platinum 1 8(a)(iii) Eocell = 1.82 – 1.36 = (+)0.46 V 1 8(a)(iv) 2Co3+ + 2Cl-  Cl2 + 2Co2+ 1 8(b) M1 Q= 2.5 x 30 x 60 C = 4500 C AND 96500 OR 579000 seen moles of CO2= 4500/579000 = 7.8 x10-3 or 7.77 x 10-3 M2 volume of CO2 = 7.77 x 10-3 x 24000 = 187 cm3 2