Cambridge A Level Chemistry 9701 — 2016 May/June Paper 4 · Variant 2

READ THESE INSTRUCTIONS FIRST Write your Centre number, candidate number and name on all the work you hand in. Write in dark blue or black pen. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. DO NOT WRITE IN ANY BARCODES. Answer all questions. Electronic calculators may be used. You may lose marks if you do not show your working or if you do not use appropriate units. A Data Booklet is provided. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. CHEMISTRY 9701/42 Paper 4 A Level Structured Questions May/June 2016 2 hours Candidates answer on the Question Paper. Additional Materials: Data Booklet This document consists of 20 printed pages. [Turn over IB16 06_9701_42/5RP © UCLES 2016 *1230666070* Cambridge International Examinations Cambridge International Advanced Subsidiary and Advanced Level

2 9701/42/M/J/16 © UCLES 2016 Answer all questions in the spaces provided. 1 (a) Magnesium nitrate, Mg(NO3)2, is very soluble in water. When a hot saturated solution of magnesium nitrate is cooled, crystals of the hydrate, Mg(NO3)2.6H2O, are formed. In the crystals, six water molecules bond to each Mg2+ ion, and some of these water molecules are also bonded to the nitrate ions. (i) Suggest the type of bonding that occurs between H2O and Mg2+, … H2O and NO3 –. … [2] (ii) Describe the arrangement of the water molecules around the Mg2+ ion. … [1] (iii) Describe in detail what you would observe when crystals of Mg(NO3)2.6H2O are heated in a boiling tube, gently at first and then more strongly. Write equations for any reactions that occur. … … … … … … … … [4] (iv) Calculate the percentage loss in mass when Mg(NO3)2.6H2O is heated strongly to constant mass. percentage loss = … % [2]

3 9701/42/M/J/16 © UCLES 2016 [Turn over (b) Explain why the Group 2 nitrates become more stable to heat down the group. … … … … … [2] (c) Magnesium nitrate and silver nitrate, AgNO3, decompose on heating to produce the same gases. Silver nitrate also produces silver metal during decomposition. Write an equation for the decomposition of AgNO3. … [1] [Total: 12]

4 9701/42/M/J/16 © UCLES 2016 2 Ethanoic acid is a weak acid. (a) Explain what is meant by the term weak acid. … … [1] (b) The pKa values of four acids are listed below. acid structural formula pKa 1 CH3CO2H 4.8 2 CH3CH2CO2H 4.9 3 CH3CHCl CO2H 2.8 4 CH2Cl CH2CO2H 4.0 (i) State the mathematical relationship between pKa and the acid dissociation constant Ka. … [1] (ii) With reference to acidity, explain the difference in pKa values between • acid 1 and acid 2, … … • acid 2 and acid 3, … … • acid 3 and acid 4. … … [3]

5 9701/42/M/J/16 © UCLES 2016 [Turn over (c) (i) Draw a fully labelled diagram of the equipment needed to measure the voltage of an electrochemical cell consisting of the standard hydrogen electrode and the standard Cu / Cu2+ electrode. [4] (ii) For the cell drawn in (i), calculate the and state which electrode is positive. = … identity of the positive electrode … [1] (d) A monobasic acid, D, has Ka = 1.23 × 10–5 mol dm–3. (i) Calculate the pH of a 0.100 mol dm–3 solution of D. pH = … [2] (ii) An electrochemical cell similar to the one you have drawn in (c)(i) was set up using a 0.100 mol dm–3 solution of D in the hydrogen electrode instead of the standard solution. Use the data and the Nernst equation, E = E o + 0.059 log [H+(aq)], to calculate the new E cell in this experiment. E cell = … V [2] [Total: 14]

6 9701/42/M/J/16 © UCLES 2016 3 (a) 2-bromopropane can be used to synthesise methylethylamine and 2-methylpropylamine. Br 2-bromopropane 2-methylpropylamine NH2 methylethylamine reaction 1 reaction 2 reaction 3 X NH2 (i) Draw the structure of the intermediate X in the box above. [1] (ii) Suggest reagents and conditions for • reaction 1, … • reaction 2, … • reaction 3. … [3] (b) (i) Write an equation showing why aqueous solutions of ethylamine are alkaline. … [1] (ii) Compare the basicities of ethylamine and ammonia. Explain your answer. … … … … [2]

7 9701/42/M/J/16 © UCLES 2016 [Turn over (c) Solutions containing mixtures of amines and their salts are buffer solutions. (i) Explain what is meant by the term buffer solution. … … [1] (ii) Write two equations to show how a solution containing a mixture of CH3NH2 and CH3NH3Cl acts as a buffer. … … [2] [Total: 10]

8 9701/42/M/J/16 © UCLES 2016 4 (a) There are two isomeric complexes with the formula Pt(NH3)2Cl 2, one of which is an anti-cancer drug. (i) Draw diagrams to show the three-dimensional structures of the two isomers. isomer 1 isomer 2 [2] (ii) Comment on the polarity of the two isomers of Pt(NH3)2Cl 2. Explain your answer. … … [1] Oxaloplatin is another successful anti-cancer drug in which the stereochemistry around the platinum atom is the same as that in Pt(NH3)2Cl 2. Pt NH2 NH2 O O O O C C oxaloplatin (iii) Explain why there are no isomers of oxaloplatin. … … [1]

9 9701/42/M/J/16 © UCLES 2016 [Turn over (b) Only one structure of the complex [Ni(R3P)2Cl 2] is known. (R = CH3, R3P is a monodentate ligand) (i) What does this indicate about the stereochemistry around the nickel atom? … [1] (ii) Draw a three-dimensional diagram showing the structure of this complex. [1] [Total: 6]

10 9701/42/M/J/16 © UCLES 2016 5 Cadmium ions form complexes with primary amines and with 1,2-diaminoethane. Cd2+(aq) + 4CH3NH2(aq) [Cd(CH3NH2)4]2+(aq) Kstab = 3.6 × 106 equilibrium I Cd2+(aq) + 2H2NCH2CH2NH2(aq) [Cd(H2NCH2CH2NH2)2]2+(aq) Kstab = 4.2 × 1010 equilibrium II (a) (i) Write an expression for the stability constant, Kstab, for equilibrium I, and state its units. Kstab = units … [2] Cadmium ions are poisonous and need to be removed from some water supplies. This is often done by adding a complexing agent. (ii) In a sample of ground water the concentration of Cd2+(aq) is 1.00 × 10–4 mol dm–3. Calculate the concentration of CH3NH2(aq) needed to reduce the concentration of Cd2+(aq) in this dilute solution by a factor of one thousand. concentration of CH3NH2(aq) = … mol dm–3 [2]

11 9701/42/M/J/16 © UCLES 2016 [Turn over (b) Values for ∆H o and ∆G o for equilibria I and II, and the value of ∆S o for equilibrium I, are given in the table below. All values are at a temperature of 298 K. equilibrium ∆H o / kJ mol–1 ∆G o / kJ mol–1 ∆S o / J K–1 mol–1 I –57.3 –37.4 –66.8 II –56.5 –60.7 to be calculated (i) Suggest a reason why the ∆H o values for the two equilibria are very similar. … … [1] (ii) Calculate ∆S o for equilibrium II. ∆S o = … J K–1 mol–1 [1] (iii) Suggest a reason for the difference between the ∆S o you have calculated for equilibrium II and that for equilibrium I given in the table. … … … [1] (iv) Which of the two complexes is the more stable? Give a reason for your answer. … … [1] [Total: 8]

12 9701/42/M/J/16 © UCLES 2016 6 Esterases are enzymes that hydrolyse esters. R C + H2O + HO–R' O O R' R C O O H Enzymes can be quite specific in the structures of the substrates they act upon. For example, an esterase isolated from the mould Aspergillus niger will hydrolyse phenyl ethanoate, CH3CO2C6H5, but not its isomer methyl benzoate, C6H5CO2CH3. (a) Outline how enzymes catalyse reactions, and explain their specificity. Use diagrams in your answer where appropriate. … … … [3]

13 9701/42/M/J/16 © UCLES 2016 [Turn over (b) Sample bottles of each of the isomers phenyl ethanoate and methyl benzoate have lost their labels and so have been named isomer A and isomer B. (i) The carbon-13 NMR spectra of isomers A and B contain the following peaks. isomer A isomer B δ 52 δ 26 δ 128 δ 122 δ 129 δ 126 δ 130 δ 129 δ 133 δ 151 δ 167 δ 169 The identity of the compound responsible for each spectrum can be deduced by studying the chemical shifts (δ) of the peaks in the spectra. Use the Data Booklet to assign the correct peaks to the labelled carbon atoms in the structures of the isomers below. Write each value next to the relevant carbon atom and hence deduce the identity of each isomer. C CH3 O O methyl benzoate is isomer … O CH3 C O phenyl ethanoate is isomer … … … … … [2] (ii) These two isomers are difficult to distinguish chemically. Describe a method of converting them to suitable products in step 1 which can then be tested in step 2. You should state the reagents and conditions for each step, and any observations you would make. step 1 … … … step 2 … … … [3] [Total: 8]

14 9701/42/M/J/16 © UCLES 2016 7 (a) Amino acids can be separated by electrophoresis. (i) Draw a labelled diagram of the apparatus used to separate a mixture by electrophoresis. [3] (ii) Explain the principles of the separation of amino acids by electrophoresis. … … … [2] (b) Electrophoresis is usually carried out in a buffer solution. Given three buffers, with pH values of 2.0, 7.0 and 12.0, suggest, with a reason, which buffer would be the most suitable for the separation of the following amino acid mixtures. Your reasons should refer to the structure of each molecule. (The structures of these amino acids are given in the Data Booklet.) (i) Asp and Val buffer pH … reason … … (ii) Lys and Ser buffer pH … reason … … (iii) Tyr and Phe buffer pH … reason … … [3]

15 9701/42/M/J/16 © UCLES 2016 [Turn over (c) (i) Draw the structure of the dipeptide Gly-Ser, showing the peptide bond in full. [2] The infra-red spectrum of Gly-Ser is shown below. 100 50 0 4000 3000 2000 1500 1000 500 transmittance wavenumber / cm–1 E F G (ii) Use the Data Booklet to identify the bond in the molecule of Gly-Ser that is responsible for each of the peaks indicated on the above infra-red spectrum. E … F … G … [2] [Total: 12]

16 9701/42/M/J/16 © UCLES 2016 8 (a) Describe and explain the trend in the solubility of the hydroxides down Group 2. … … … … [3] (b) Calcium reacts vigorously with HCl (aq) producing H2(g). Ca(s) + 2HCl (aq) → CaCl 2(aq) + H2(g) (i) How would you expect the enthalpy change for this reaction to compare with the enthalpy change for the reaction where HNO3(aq) is used in place of HCl but all other conditions are the same? Explain your answer. … … [1]

17 9701/42/M/J/16 © UCLES 2016 [Turn over (ii) The ionic equation for this reaction is shown. Ca(s) + 2H+(aq) → Ca2+(aq) + H2(g) ∆H o = x kJ mol–1 Construct a fully labelled Hess’ Law cycle to connect each side of this equation to the relevant gas phase ions. Use your cycle, the following data, and data from the Data Booklet, to calculate a value for x. standard enthalpy of atomisation of Ca(s), (Ca) +178 kJ mol–1 standard enthalpy of hydration of Ca2+(g), (Ca2+) –1576 kJ mol–1 standard enthalpy of hydration of H+(g), (H+) –1090 kJ mol–1 x = … kJ mol–1 [4] (c) The standard enthalpy change for the reaction between Ca(s) and CH3CO2H(aq) is less negative than x by 2 kJ mol–1. Suggest an explanation for this. … … [2] [Total: 10]

18 9701/42/M/J/16 © UCLES 2016 9 The anti-inflammatory drug ibuprofen can be synthesised from benzene via the following six steps. step 1 step 5 step 2 step 3 H J step 4 OH CN CO2H step 6 CO2H ibuprofen (a) Draw circles around any chiral carbon atoms in the above five formulae. [1] (b) Suggest the structures of compounds H and J and draw them in the boxes above. [2] (c) Suggest reagents and conditions for steps 1-6. step 1 … step 2 … step 3 … step 4 … step 5 … step 6 … [6] (d) Name the mechanism of step 1 and state the type of reaction for step 6. step 1 … step 6 … [2] [Total: 11]

19 9701/42/M/J/16 © UCLES 2016 [Turn over 10 (a) (i) Complete the electronic configuration of the iron atom. Fe 1s22s22p6… [1] (ii) In some of its complexes, the Fe3+ ion has only one unpaired electron in its d orbitals. Using the symbols ↑ and ↓ to represent electrons of opposite spins, complete the following diagram to show the d orbital electronic configuration of this Fe3+ ion. 3d … … … … … energy [1] (b) A solution containing a mixture of Sn2+(aq) and Sn4+(aq) is added to a solution containing a mixture of Fe2+(aq) and Fe3+(aq). Use E o data from the Data Booklet to predict the reaction that might take place when the two solutions are mixed, and write an equation for the reaction. … … … [2]

20 9701/42/M/J/16 © UCLES 2016 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 International Examinations Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download at www.cie.org.uk after the live examination series. Cambridge International Examinations is part of the Cambridge Assessment Group. Cambridge Assessment is the brand name of University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge. (c) Hexaaquairon(III) ions are pale violet. They form a colourless complex with fluoride ions, F–, equilibrium 1, and a deep-red complex with thiocyanate ions, SCN–, equilibrium 2. [Fe(H2O)6]3+ + F– [Fe(H2O)5F]2+ + H2O equilibrium 1 Kstab = 2.0 × 105 mol–1 dm3 violet colourless [Fe(H2O)6]3+ + SCN– [Fe(H2O)5SCN]2+ + H2O equilibrium 2 Kstab = 1.0 × 103 mol–1 dm3 violet deep-red (i) Predict and explain the sequence of colour changes you would observe in each of the following experiments. • A few drops of KSCN(aq) are added to 5 cm3 of Fe3+(aq), followed by a few drops of KF(aq). … … … • A few drops of KF(aq) are added to 5 cm3 of Fe3+(aq), followed by a few drops of KSCN(aq). … … … [4] (ii) What type of reaction is occurring during the experiments in (i)? … [1] [Total: 9]

® IGCSE is the registered trademark of Cambridge International Examinations. This document consists of 12 printed pages. © UCLES 2016 [Turn over Cambridge International Examinations Cambridge International Advanced Subsidiary and Advanced Level CHEMISTRY 9701/42 Paper 4 A Level Structured Questions May/June 2016 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 will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes for the May/June 2016 series for most Cambridge IGCSE®, Cambridge International A and AS Level components and some Cambridge O Level components.

Page 2 Mark Scheme Syllabus Paper Cambridge International AS/A Level – May/June 2016 9701 42 © Cambridge International Examinations 2016 Question Answer Marks 1 (a) (i) dative (covalent) or coordinate Hydrogen / H (boding) 2 (ii) octahedral 1 (iii) Mg(NO3)2.6H2O → Mg(NO3)2 + 6H2O Mg(NO3)2 → MgO + 2NO2 + 1 2 O2 any three of (solid) dissolves / turns to liquid condensation on tube white solid (forms / remains) brown fumes (evolved) gas formed that relights a glowing splint 4 (iv) Mr values: Mg(NO3)2.6H2O = 256.3 MgO = 40.3 or (loss in molar mass = 256.3 – 40.3 =) 216 percentage loss = 100 × 216 / 256.3 = 84.3 / 84.4% 2 (b) (cat)-ionic radius / ion size increases (down the group) less polarisation / distortion of nitrate ion / NO3 – 2 (c) 2AgNO3 → 2Ag + 2NO2 + O2 1 [Total: 12] 2 (a) (i) (an acid that is) partially / incompletely ionised / dissociated 1 (b) (i) pKa = –logKa or Ka = 10–pKa 1

Page 3 Mark Scheme Syllabus Paper Cambridge International AS/A Level – May/June 2016 9701 42 © Cambridge International Examinations 2016 Question Answer Marks (ii) ethanoic acid (1) is more acidic than propanoic acid (2) due to smaller electron-donating (R / alkyl) group / less electron-donating (R / alkyl) group(s) 2-chloropropanoic acid (3) is more acidic than propanoic acid (2) due to electron-withdrawing / electronegative (Cl / chlorine) atom 2-chloropropanoic acid (3) is more acidic than 3¬-chloropropanoic acid (4) since the Cl / chlorine / electronegative atom is closer to the CO2 – / acid 3 (c) (i) V H2(g) H+(aq) Cu2+(aq) Cu Pt salt bridge M1: voltmeter / V and salt bridge labelled M2: Cu and Cu2+ / CuSO4 (any soluble Cu(II) salt) M3: H2 (arrow in) and H+ / HCl / H2SO4 / any mineral acid M4 Pt and one solution at 1 M / 1 mol dm–3 OR H2 at 1 atm 4 (ii) Eo cell = 0.34 (V) and (Cu2+) / Cu is the positive electrode 1 d (i) Ka = 1.23 × 10–5 [H+] = √(Ka.c) = √(1.23 × 10–5 × 0.1) = 1.11 × 10–3 mol dm–3 pH = 3.0 (2.96) ecf from [H+] 2

Page 4 Mark Scheme Syllabus Paper Cambridge International AS/A Level – May/June 2016 9701 42 © Cambridge International Examinations 2016 Question Answer Marks (ii) E = 0.0 + 0.059log(1.11 × 10–3) OR = –0.17(4)V so new Ecell = 0.34 + 0.17 = 0.51V ecf from (d)(i) 2 [Total: 14] 3 (a) (i) (CH3)2CHCN 1 (ii) reaction 1: NH3 (in ethanol) under pressure (+ heat) or heat NH3 in a sealed tube reaction 2: KCN / NaCN and heat / reflux (in ethanol) reaction 3: H2 + Ni or LiAl H4 3 (b) (i) CH3CH2NH2 + H2O → CH3CH2NH3 + (+) OH– 1 (ii) ethylamine is more basic than ammonia… because of electron-donating (alkyl / ethyl / R) group (in ethylamine) which makes the lone pair (on N) more available for donation or the lone pair (on N) more available for a proton / H+ 2 (c) (i) A solution which resists / minimises / roughly maintains changes in pH when (small amounts of) H+ or OH– are added 1 (ii) CH3NH2 + H+ → CH3NH3 + CH3NH3Cl + OH → CH3NH2 + H2O + Cl 2 [Total: 10]

Page 5 Mark Scheme Syllabus Paper Cambridge International AS/A Level – May/June 2016 9701 42 © Cambridge International Examinations 2016 Question Answer Marks 4 (a) (i) NH3 Pt Cl Cl NH3 NH3 Pt Cl NH3 Cl (cis) (trans) 2 (ii) cis is (more) polar due to both Cl(δ–) on same side or cis is (more) polar as dipoles do not cancel / unsymmetrical or trans is non-polar as it is bond dipoles cancel 1 (iii) (This can only be cis) its mirror image is the same / superimposable or the distance between two coordinating nitrogens / oxygens is too small to bond trans or difficult for the NH2 and O to change places (since 5-memebered rings can only bridge adjacent positions) 1 (b) (i) It’s not square planar or it’s tetrahedral 1 (ii) must be 3D structure (i.e. tetrahedral-like) R3P PR3 Ni Cl Cl R3P Cl Ni Cl PR3 etc or 1 [Total: 6]

Page 6 Mark Scheme Syllabus Paper Cambridge International AS/A Level – May/June 2016 9701 42 © Cambridge International Examinations 2016 Question Answer Marks 5 (a) (i) Kstab = [Cd(CH3NH2)4 2+] [Cd2+] [CH3NH2]4 units: mol–4 dm12 2 (ii) Cd2+ + 4CH3NH2 ⇌ [Cd(CH3NH2)4]2+ at start: 1 × 1–4 0 at eqm: 1× 10–7 y 1 × 10–4 – 1 × 10–7 or 9.99 × 10–5 or 1.0 × 10–4 9.99 × 10–5 / (y4 × 10–7)= 3.6 × 106 and y = 4√(9.99 × 10–5) / (1 × 10–7 × 3.6 × 106) = 0.129 / 0.13 2 (b) (i) (each complex is formed by) making (4 ×)N-Cd bonds and breaking (6 ×) O-Cd bonds or same types of / similar bonds forming / breaking or same number of bonds forming / breaking 1 (ii) ∆S = (∆H – ∆G) / T = (60.7 – 56.5) × 1000 / 298 = (+)14 / (+)14.1 1 (iii) fewer moles (of solutes) are forming (one mole of) the complex (so less loss of disorder) or one en displaces two H2O whereas one CH3NH2 only displaces one H2O 1 (iv) The [Cd(H2NCH2CH2NH2)2]2+ / equilibrium 2 complex (is more stable) because: either Kstab is greater or ∆Go is more negative. 1 [Total: 8]

Page 7 Mark Scheme Syllabus Paper Cambridge International AS/A Level – May/June 2016 9701 42 © Cambridge International Examinations 2016 Question Answer Marks 6 (a) essential mark M1 the reactants / substrate has a shape complementary / specific to active site – can be awarded from a labelled diagram as below or diagrams showing this specificity clearly any two of M2: reactants / substrate binds to / fits into the active site of the enzyme M3: (Interaction with site) causes a specific bond to be weakened, (which breaks) or lowers activation energy M4: forms an E-S complex M5: products released from enzyme / active site (products) 3 (b) (i) δ 26 is CH3-CO δ 52 is CH3-O δ 169 is CH3CO δ 167 is phenyl-CO Phenyl ethanoate is B methyl benzoate is A M1 = any two correct δ linked to phenylethanoate / methyl benzoate M2 = the rest correct 2 (ii) heat with H3O+ (to hydrolyse the ester) then add Br2(aq) / bromine water decolourises/gives white ppt. (with phenol from B) 3 [Total: 8]

Page 8 Mark Scheme Syllabus Paper Cambridge International AS/A Level – May/June 2016 9701 42 © Cambridge International Examinations 2016 Question Answer Marks 7 (a) (i) labelled with M1: DC power supply + and – / battery / cell / + and – sign (on cell / electrodes) with a complete circuit M2: buffer solution / electrolyte labelled M3: (amino acid) mixture / x on (filter) paper / gel / agarose 3 (ii) direction of movement related to charge (of amino acids) distance travelled depends on charge / Mr (of amino acids) 2 (b) (i) Asp + Val: pH 12 because Asp will be –CH2COO– (R-group) moves further (to positive electrode than Val) or pH 12 Asp more negative so moves further (to positive electrode) or pH 12 because Asp has a charge of 2– but Val has a charge of 1– or best at pH 7 because Asp will be negatively charged (anionic) but Val neutral 1 (ii) Lys + Ser: pH 2 because Lys will be (CH2)4NH3 + (R-group) moves further (to negative electrode than Ser) or pH 2 Lys more positive so moves further (to negative electrode) or pH 2 because Lys has a charge of 2+ and Ser has a charge of 1+ or pH 7 because Lys is positively charged (cationic) but Ser neutral / zwitterionic 1

Page 9 Mark Scheme Syllabus Paper Cambridge International AS/A Level – May/June 2016 9701 42 © Cambridge International Examinations 2016 Question Answer Marks (iii) Tyr + Phe: pH 12 because Tyr will be C6H5CH2O– (R-group) moves further / more / faster (to positive electrode than Phe) or pH12 because Tyr has a charge of 2– but Phe has a charge of 1– 1 (c) (i) H2N CH2 C N H CH CO2H O CH2OH M1: for –CONH– as shown above M2: for rest of molecule and correct connectivity of the bonds 2 (ii) from the IR spectrum • E is O-H or N-H (allow NH2) • F is C=O • G is C-O 2 [Total: 12] 8 (a) M1: solubility increases (down the group) M2: because lattice energy decreases faster than does ∆Hhyd M3:∆Hsol / enthalpy of solution becomes more exothermic / less endothermic 3 (b) (i) Should be the same / similar (enthalpy change), as (both acids) are fully ionised / strong acids 1

Page 10 Mark Scheme Syllabus Paper Cambridge International AS/A Level – May/June 2016 9701 42 © Cambridge International Examinations 2016 Question Answer Marks (ii) x = ∆Hat(Ca) + IE(1) + IE(2) – 2∆Hhyd(H+) + ∆Hhyd(Ca2+) – 2IE(H) – E(H-H) x = 178 + 590 + 1150 + 2(1090) – 1576 – 2(1310) – 436 x = –534 kJ mol–1 4 (c) CH3CO2H is incompletely ionised / weak acid / weaker acid enthalpy change of ionisation (of CH3COOH) is +2 kJ mol–1 or energy needed to ionise / dissociate (CH3COOH) 2 [Total: 10] 9 (a) CO2H CN OH 1

Page 11 Mark Scheme Syllabus Paper Cambridge International AS/A Level – May/June 2016 9701 42 © Cambridge International Examinations 2016 Question Answer Marks (b) J1 J2 2 (c) step 1: (CH3)2CHCH2Cl + Al Cl 3 (+ heat) step 2: CH3COCl + Al Cl 3 (+ heat) step 3: HCN + NaCN or HCN + base or HCN + CN– (steps 4 and 5 could be reversed on J) If J1 step 4 then step 5 J2 step 5 then step 4 step 4: H3O+ + heat / aqueous HCl + heat step 5: conc H2SO4 + heat / conc H3PO4 + heat or Al 2O3 + heat step 6: H2 + Ni (+ heat) 6 (d) step 1: electrophilic substitution or alkylation step 6: reduction / hydrogenation / addition 2 [Total: 11]

Page 12 Mark Scheme Syllabus Paper Cambridge International AS/A Level – May/June 2016 9701 42 © Cambridge International Examinations 2016 Question Answer Marks 10 (a) (i) Fe is …3s23p63d64s2 1 (ii) 1 (b) Eo values: Sn4+ / Sn2+ = +0.15(V); Fe3+ / Fe2+ = +0.77(V) or Eo cell = +0.62 (V) (Sn2+ will reduce Fe3+) Sn2+ + 2Fe3+ → 2Fe2+ 2 (c) (i) essential mark Kstab / stability: [Fe(H2O)5F]2+ > [Fe(H2O)5SCN]+ (> [Fe(H2O)6]2+) observations (violet) → deep-red (deep-red) → colourless (violet) → colourless which stays colourless / does not change 4 (ii) ligand displacement / exchange / substitution 1 [Total: 9]