Cambridge A Level Chemistry 9701 — 2021 Oct/Nov Paper 4 · Variant 2

1 Radium is a Group 2 element. The predicted lattice energy, , of radium sulfide, RaS, is –2612 kJ mol–1. (a) Define . .................................................................................................................................................... .............................................................................................................................................. [2] Some data relating to radium and sulfur are listed. Select relevant data from this list for use in your answers to parts (b) to (e). process value / kJ mol–1 enthalpy change for Ra(s) → Ra2+(g) + 2e– +1619 first ionisation energy of sulfur +1000 second ionisation energy of sulfur +2260 first electron affinity of sulfur –200 second electron affinity of sulfur +532 enthalpy change for S8(s) + 2e– → S2–(g) +555 lattice energy of RaS(s) –2612 (b) Write an equation for the process corresponding to the second electron affinity of sulfur. Include state symbols. .............................................................................................................................................. [1] (c) Sulfur exists as S8 molecules in the solid state. Use the data in this question to calculate the enthalpy change for the reaction S8(s) → 8S(g). enthalpy change = .............................. kJ mol–1 [3] (d) Calculate the standard enthalpy change of formation, , of radium sulfide. standard enthalpy change, = .............................. kJ mol–1 [2] (e) (i) State the two major factors that affect the numerical magnitude of a lattice energy. ............................................................................................................................................. ....................................................................................................................................... [2] (ii) For each factor you have identified in (e)(i), state whether it tends to make the lattice energy of radium sulfide more or less exothermic than that of sodium chloride. Explain your answer. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ....................................................................................................................................... [2] (iii) The lattice energies of sodium chloride, NaCl, and radium sulfide, RaS, are –771 kJ mol–1 and –2612 kJ mol–1, respectively. Identify the dominant factor in determining the relative numerical magnitudes of the lattice energies of radium sulfide and sodium chloride. Explain your answer. ............................................................................................................................................. ....................................................................................................................................... [1] [Total: 13]

Mark scheme: 1(a) • enthalpy / energy change / given out / evolved / released • one mole is formed / made [1] • of compound / solid / lattice / crystal • (from) gaseous ions [1] 2 1(b)b S–(g) + e– → S2–(g) [1] 1 1(c)c (555 + 200 – 532 = 223, 223 × 8 = 1784) M1 selecting correct data 555, 200, 532 only, (ignore signs and multipliers) [1] M2 evaluation to give +223 [1] M3 multiplying M2 by 8 and evaluation ans (+) 1784 [1] 3 1(d) (1619 + 555 – 2612 = –438) M1 selecting correct data 1619 555 2612 only, (ignore signs and multipliers) [1] M2 evaluation to give –438 [1] 2 1(e)(i) ionic radius / size / sum of ionic radii [1] ionic charge / product of ionic charges [1] 2 1(e)(ii) M1 (size tends to make ΔHolatt of radium sulfide) less exothermic since the ions are larger [1] M2 (charge tends to make ΔHolatt of radium sulfide) less exothermic since the ions are more highly charged [1] 2 1(e)(iii) (ionic) charge (since) AND ΔHolatt of radium sulfide is more exothermic [1] 1

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2 Ethoxyethane, C2H5OC2H5, can dissolve both in water and in octan-1-ol. The expression and numerical value for the partition coefficient of ethoxyethane between water and octan-1-ol are given. Water and octan-1-ol are immiscible. concentration of C2H5OC2H5 in octan-1-ol Kpc = = 6.760 at 20 °C concentration of C2H5OC2H5 in water (a) In an experiment, octan-1-ol at 20 °C is added to a solution of ethoxyethane in water at 20 °C. The mixture is analysed immediately and a value of Kpc is calculated. The calculation is performed correctly; the value calculated is 5.625. Explain why the value calculated is less than 6.760. .................................................................................................................................................... .............................................................................................................................................. [2] (b) A second experiment is performed and the value of Kpc is found to be 6.760. The concentration of ethoxyethane in the octan-1-ol layer is 7.62 g dm–3. (i) Calculate the concentration, in g dm–3, of ethoxyethane in the aqueous layer. .............................. g dm–3 [1] (ii) 100 cm3 of the octan-1-ol layer is taken and shaken with 100 cm3 of water. Calculate the maximum amount, in mol, of ethoxyethane that can be extracted into the water. .............................. mol [3] (c) An aqueous solution of lead(II) nitrate is mixed with an aqueous solution of sodium iodide. A yellow precipitate of lead(II) iodide is formed and is filtered out, leaving solution X. The concentration of Pb2+ in solution X is 5.68 × 10–3 mol dm–3. The concentration of I– in solution X is 4.20 × 10–4 mol dm–3. (i) Use these data to calculate a value for the solubility product, Ksp, of lead(II) iodide. State the units of Ksp. Ksp = .............................. units = .............................. [2] (ii) Potassium iodide is very soluble in water. Describe and explain what is seen if a few drops of saturated potassium iodide solution are added to a portion of solution X. ............................................................................................................................................. ............................................................................................................................................. ....................................................................................................................................... [2] [Total: 10]

Mark scheme: 2(a) M1 reference to equilibrium not being established / not at equilibrium [1] M2 concentration in water too high or conc in octan-1-ol too low owtte [1] 2 2(b)(i) 1.1 / 1.13 / 1.127 / 1.1272 [1] 1 2(b)(ii) M1 6.760 = (0.762 – x) / x [1] M2 7.76x = 0.762 x = 0.0982 g (correct value scores 2) [1] M3 0.0982 / 74 = 0.0013 / 0.00133 mol [1] 3 2(c)(i) 1.0x10–9 [1] mol3 dm–9 [1] 2 2(c)(ii) yellow precipitate / yellow solid / yellow crystals [1] common ion effect / Ksp has been exceeded [1] 2

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3 Pure water is a very poor conductor of electricity. However, when hydrogen chloride gas is dissolved in water, ions are formed and a current flows during electrolysis. HCl (g) + aq → H+(aq) + Cl –(aq) The overall change after electrolysis is that hydrogen chloride gas is converted into hydrogen and chlorine. 2HCl (g) → H2(g) + Cl 2(g) When a current of 3.10 A is passed through the solution for Y minutes, 351 cm3 of chlorine are produced at the anode, measured under room conditions. (a) Calculate the number of chlorine molecules produced during the electrolysis. number of chlorine molecules = .............................. [2] (b) Calculate the total number of electrons transferred to produce this number of chlorine molecules. total number of electrons = .............................. [1] (c) Calculate the quantity of charge, in coulombs, of the total number of electrons calculated in (b). quantity of charge = .............................. C [1] (d) Calculate the time, Y, in minutes, for which the current flows. Y = .............................. minutes [1] (e) The standard entropies, S o, of three species are given in the table. species S o / J K–1 mol–1 HCl (g) +187 H2(g) +131 Cl 2(g) +223 (i) Calculate ∆S o for the reaction 2HCl (g) → H2(g) + Cl 2(g). ∆S o = .............................. J K–1 mol–1 [1] (ii) ∆H o for the reaction 2HCl (g) → H2(g) + Cl 2(g) is +185 kJ mol–1. Calculate ∆G o for this reaction at 298 K. ∆G o = .............................. kJ mol–1 [2] (iii) Predict the effect of increasing temperature on the spontaneity of this reaction. Explain your answer. ............................................................................................................................................. ....................................................................................................................................... [1] [Total: 9]

Mark scheme: 3(a) 0.351 / 24 = 0.015 (mol) [1] 0.015 × 6.02 × 1023 = 9.0 × 1021 / 8.8 × 1021 [1] 3(b) 1.76 × 1022 / 1.8 × 1022 [1] 1 3(c) 2817 / 2816 / 2820 / 2800 C [1] 1 3(d) 15 / 15.1 / 15.05 / 15.15 minutes [1] 1 3(e)(i) –20 [1] 1 Question Answer Marks 3(e)(ii) states / uses correct Gibbs equation [1] answer = 190 / 191 / 190.0 [1] 2 3(e)(iii) Becomes less feasible / less spontaneous / AND because ΔS is negative / TΔS becomes more negative / –TΔS becomes more positive [1] 1

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4 Separate samples of 0.01 mol of magnesium nitrate and 0.01 mol of strontium nitrate are heated until completely decomposed to the metal oxide, nitrogen dioxide and oxygen. (a) State which of these two Group 2 nitrates requires the higher temperature before it begins to decompose. Explain your answer. .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .............................................................................................................................................. [2] (b) After decomposition is complete the 0.01 mol sample of magnesium oxide is taken and increasing amounts of water are added to it, with stirring, until no solid remains. This procedure is repeated with the 0.01 mol sample of strontium oxide. Identify the sample to which most water must be added to cause all the solid to dissolve. Explain your answer by reference to the solubilities of the products formed when water is added to the oxides. You should refer to relevant energy terms in your answer. .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .............................................................................................................................................. [3] (c) The nitrogen dioxide given off by the decomposition of 0.0100 mol of strontium nitrate is dissolved in water. The oxidising agent H2O2(aq) is then added to give 150.0 cm3 of a solution in which nitric acid, HNO3, is the only nitrogen-containing product. (i) Calculate the concentration, in mol dm–3, of HNO3 in the 150.0 cm3 of solution. concentration = .............................. mol dm–3 [1] (ii) The HNO3 present in 25.0 cm3 of this solution is neutralised using 0.125 mol dm–3 NaOH(aq). Calculate the minimum volume, in cm3, of NaOH(aq) needed. Give your answer to three significant figures. volume = .............................. cm3 [1] [Total: 7]

Mark scheme: 4(a) strontium nitrate AND because of larger cationic radius [1] NO3– / nitrate ion / anion is less distorted / polarised OR N–O or N=O bond less weakened / distorted [1] 4(b) M1 magnesium oxide requires more water to dissolve AND because Sr(OH)2 is more soluble [1] M2 lattice and hydration energies less for Sr(OH)2 [1] M3 lattice energy is dominant factor / change in lattice energy is greater [1] 3 4(c)(i) 0.13 [1] 1 4(c)(ii) 26.7 cm3 must be 3SF [1] 1

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5 Transition elements form complexes. (a) Molybdenum, Mo, forms an octahedral complex consisting of one Mo atom surrounded by carbon monoxide, CO, molecules. CO is a monodentate ligand. Iron forms an octahedral complex consisting of one Fe3+ and a number of cyanide, CN–, ions. CN– is a monodentate ligand. (i) Define the term monodentate ligand. ............................................................................................................................................. ....................................................................................................................................... [1] (ii) Complete the table by stating the formulae and charges of the complexes described. formula charge molybdenum complex iron(III) complex [2] (iii) Draw a three-dimensional diagram to show the shape of this iron(III) complex. Label one 180° bond angle on your diagram. Fe [1] (b) An excess of aqueous ammonia is added to dilute copper(II) sulfate solution. A dark blue complex, [Cu(NH3)4(H2O)2]2+, is formed. (i) Write an ionic equation for this reaction. ....................................................................................................................................... [1] (ii) Explain the origin of colour in copper(II) complexes. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ....................................................................................................................................... [4] (c) An excess of concentrated hydrochloric acid is added to the dark blue solution of [Cu(NH3)4(H2O)2]2+. A new complex, Z, is formed. The colour of the solution changes. (i) Write an equation for the formation of Z from the solution of [Cu(NH3)4(H2O)2]2+. Include the formula and charge of Z. ....................................................................................................................................... [2] (ii) Name the type of reaction when Z forms from [Cu(NH3)4(H2O)2]2+. ....................................................................................................................................... [1] (iii) State the geometry of Z. ....................................................................................................................................... [1] (iv) State the colour of a solution of Z. ....................................................................................................................................... [1] (v) Explain why the colour of a solution of [Cu(NH3)4(H2O)2]2+ is different from the colour of a solution of Z. You should refer to the energies of the orbitals involved in your answer. ............................................................................................................................................. ....................................................................................................................................... [1] [Total: 15]

Mark scheme: 5(a)(i) donates one lone pair / forms one dative bond AND to metal atom / to metal ion [1] 1 5(a)(ii) Mo(CO)6 0 / no charge / neutral / zero Fe(CN)6 3– both formulae [1] both charges [1] 2 5(a)(iii) either of these renditions, with a CN on each bond: AND 180o bond angle between two opposite bonds [1] 1 5(b)(i) [Cu(H2O)6]2+ + 4NH3 → [Cu(NH3)4(H2O)2]2+ + 4H2O OR Cu2+ + 2H2O + 4NH3 → [Cu(NH3)4(H2O)2]2+ [1] 1 5(b)(ii) M1 d-orbitals split (could be in diagram) [1] M2 electrons transition to higher orbital / electrons promoted or excited [1] M3 wavelength / frequency / light / photon / hf / hn absorbed [1] M4 colour seen / reflected / transmitted is complement of colour absorbed [1] 4 Question Answer Marks 5(c)(i) [Cu(NH3)4(H2O)2]2+ + 4HCl → CuCl 42– + 2H2O + 4NH4+ OR [Cu(NH3)4(H2O)2]2+ + 4Cl – (+ 4H+) → CuCl 42– + 2H2O + 4NH3 (+ 4H+) M1 formula and charge of CuCl 42– [1] M2 rest of equation [1] 2 5(c)(ii) ligand exchange / substitution / replacement / displacement [1] 1 5(c)(iii) tetrahedral [1] 1 5(c)(iv) yellow [1] 1 5(c)(v) d-orbital splitting changes / ΔE changes [1] 1

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6 An excess of aqueous ammonia is added to a solution containing the complex ion [Co(H2O)6]2+. [Co(H2O)6]2+ + 6NH3 [Co(NH3)6]2+ + 6H2O (a) Complete the sentence to describe the colour change that will be seen during this reaction. The colour changes from ................................................ to ���������������������������������������������� . [1] (b) Write an expression for the stability constant, Kstab, of [Co(NH3)6]2+. Kstab = [1] (c) The numerical value of Kstab of [Co(NH3)6]2+ is 7.7 × 104. What deduction about the properties of [Co(NH3)6]2+ and [Co(H2O)6]2+ can be made from this Kstab value? .............................................................................................................................................. [1] (d) Oxygen can oxidise [Co(NH3)6]2+ to [Co(NH3)6]3+ under standard conditions in alkaline solutions. [Co(NH3)6]3+ + e– [Co(NH3)6]2+ E o = +0.10 V (i) Use this information and the Data Booklet to calculate the value for this oxidation of [Co(NH3)6]2+. ............................................................................................................................................. ............................................................................................................................................. = .............................. V [1] (ii) Write an ionic equation for this oxidation of [Co(NH3)6]2+. ....................................................................................................................................... [1] (iii) Predict, by selecting suitable data from the Data Booklet, whether oxygen can oxidise Co2+(aq) in acidic solution, in the absence of ammonia. Explain your answer. ............................................................................................................................................. ....................................................................................................................................... [2] [Total: 7]

Mark scheme: 6(a) red or pink to yellow / straw / brown / orange [1] 1 6(b) [[Co(NH3)6]2+] / [[Co(H2O)6]2+][NH3]6 [1] 1 6(c) Co(NH3)62+ is more stable (than Co(H2O)62+) [1] 1 6(d)(i) Eocell = + 0.30 [1] 1 6(d)(ii) 4[Co(NH3)6]2+ + O2 + 2H2O ⇌ 4[Co(NH3)6]3+ + 4OH– [1] 1 6(d)(iii) M1 No, because 1.82 V and 1.23 V [1] M2 Eocell = –0.59 / 1.23 lower than 1.82 / 1.82 greater than 1.2 [1] 2

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7 The structure of benzene-1,3-dicarboxylic acid is shown. benzene-1,3-dicarboxylic acid O HO OH O (a) State the empirical formula of benzene-1,3-dicarboxylic acid. .............................................................................................................................................. [1] (b) Benzene-1,3-dicarboxylic acid is an isomer of benzene-1,4-dicarboxylic acid. These two isomers can be distinguished by carbon-13 (13C) NMR spectroscopy. State the number of peaks in the carbon-13 (13C) NMR spectrum of each compound. benzene-1,3-dicarboxylic acid benzene-1,4-dicarboxylic acid [2] (c) Benzene-1,3-dicarboxylic acid can be made by the two-step synthesis shown below. benzene-1,3-dicarboxylic acid compound P compound Q O O step 1 step 2 C9H8O3 HO HO OH O (i) Name compound P. ....................................................................................................................................... [1] (ii) Explain why the major product of this two-step synthesis is benzene-1,3-dicarboxylic acid and not benzene-1,4-dicarboxylic acid. ............................................................................................................................................. ....................................................................................................................................... [1] (iii) The reagents used for step 1 are CH3COCl and Al Cl 3. These reagents give rise to + CH3C=O ions which react with compound P. Name the mechanism of this reaction. ....................................................................................................................................... [1] + (iv) Draw the mechanism of the reaction of CH3C=O ions with compound P. Include all relevant curly arrows and charges, the structure of the intermediate and the structure of compound Q. intermediate compound Q + CH3C=O O HO [3] (v) Suggest a reagent and conditions to convert compound Q into benzene‑1,3‑dicarboxylic acid, in step 2. ....................................................................................................................................... [1] [Total: 10]

Mark scheme: 7(a) C4H3O2 [1] 1 7(b) benzene-1,3-dicarboxylic acid 5 [1] benzene-1,4-dicarboxylic acid 3 [1] 2 7(c)(i) benzoic acid [1] 1 7(c)(ii) COOH directs 3 position [1] 1 7(c)(iii) electrophilic substitution [1] 1 7(c)(iv) M1 curly arrow from within hexagon towards CH3C+=O [1] M2 correct intermediate [1] M3 curly arrow from C–H bond into hexagon and correct product Q [1] 3 7(c)(v) MnO4– / KMnO4 / manganateVII / permanganate aq / H+ / acidified / OH– then acid / alkaline then acid heat / boil / reflux / T>50o OR alkaline iodine followed by acidification [1] 1

8 Alanine, H2NCH(CH3)CO2H, and glutamic acid, H2NCH(CH2CH2CO2H)CO2H, are two naturally occurring amino acids. (a) H2NCH(CH3)CO2H exists as two optical isomers. Draw three-dimensional structures of these two optical isomers. [2] (b) The proton (1H) NMR spectrum of either alanine in D2O or glutamic acid in D2O is shown. 9 8 7 6 5 4 3 2 1 0 / ppm State whether this is the spectrum of alanine in D2O or the spectrum of glutamic acid in D2O. Explain your answer by reference to the number of peaks and splitting patterns. .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .............................................................................................................................................. [3] (c) The mass spectrum of glutamic acid, H2NCH(CH2CH2CO2H)CO2H, is obtained. (i) State the m/e value of the molecular ion peak in this spectrum. ....................................................................................................................................... [1] (ii) The spectrum has peaks with m/e values of 88 and 131. Draw the structures of the ions responsible for these peaks. m/e structure of ion 88 131 [2] (d) At pH 11 alanine exists as H2NCH(CH3)CO2– ions and glutamic acid exists as H2NCH(CH2CH2CO2–)CO2– ions. A mixture of alanine and glutamic acid at pH 11 is subjected to electrophoresis. (i) State how the mixture can be maintained at pH 11 during electrophoresis. ....................................................................................................................................... [1] (ii) Draw a fully labelled diagram for the apparatus that would be used to carry out this electrophoresis. Your diagram should include the position of the mixture of alanine and glutamic acid at the start of the electrophoresis experiment. [2] (iii) Identify the electrode that each amino acid travels towards during electrophoresis at pH 11. alanine ................................................................................................................................. glutamic acid ....................................................................................................................... [1] (iv) In a particular electrophoresis experiment at pH 11, the glutamic acid travels 3.4 cm. Alanine travels a shorter distance. Explain the factors that account for the difference in the distances travelled. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ....................................................................................................................................... [2] [Total: 14]

Mark scheme: 8(a) M1 one diagram correct [1] M2 both diagrams correct 3D and different [1] 2 Question Answer Marks 8(b) alanine because glutamic acid would have more than two / three peaks / absorbns / proton environments [1] reason why alanine has a doublet given as one neighbouring proton [1] glutamic acid would have (two) triplet(s) OR a multiplet [1] reason why alanine has a quartet / quadruplet given as three neighbouring protons [1] 3 8(c)(i) 147 [1] 1 8(c)(ii) 88 = [H2NCH(CH2)CO2H]+ [1] 131 = [CH(CH2CH2CO2H)CO2H]+ [1] 2 8(d)(i) use of buffer [1] 1 8(d)(ii) • correct circuit including DC power supply • paper or gel labelled [1] • sample towards the middle of the paper / gel OR on cathode side [1] 2 8(d)(iii) anode / positive / + AND anode / positive / + [1] 1 8(d)(iv) M1 ala is –1 and glu is –2 [1] M2 ala is lighter / has lower Mr [1] 2

9 Butylamine, CH3CH2CH2CH2NH2, can be synthesised from different organic compounds by using suitable reagents. Each reaction involves one step. (a) Complete the table to describe three different syntheses. ● One of the three syntheses should involve a nucleophilic substitution reaction. ● The starting organic compound for each synthesis should contain a different functional group. ● A different reagent should be used for each synthesis. starting organic compound reagent and conditions [6] (b) Compare and explain the relative basicities of ammonia, butylamine and phenylamine. .............................................. > ............................................. > ............................................. most basic least basic .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .............................................................................................................................................. [4] [Total: 10]

Mark scheme: 9(a) organic starting material reagent and conditions 1-butyl halide, e.g. CH3CH2CH2CH2Br NH3 under pressure or heated in sealed tube [1] + [1] butanenitrile CH3CH2CH2CN H2 and Ni or Pt / LiAlH4 / Na + ethanol [1] + [1] butanamide CH3CH2CH2CONH2 LiAlH4 / Na + ethanol [1] + [1] 6 9(b) M1 butylamine > ammonia > phenylamine [1] M2 basicity related to ability of lp to accept proton / H+ [1] M3 butylamine is stronger because of positive inductive effect of alkyl group / C4H9 [1] M4 phenylamine is weaker because lp on N is delocalised into ring [1] 4 C6H5O– / C6H5O–Na+ / C6H5ONa [1] 4

10 (a) Complete the table to show the structure of the organic product from each reaction of phenol, C6H5OH. reaction reaction mixture structure of organic product 1 phenol + NaOH(aq) 2 phenol + Na(s) + N N 3 phenol + (aq) + NaOH, at 4 °C an excess of phenol + O O 4 C C Cl Cl [4] (b) Identify two reactions from the table in which ethanol would behave in a similar way to phenol. .............................................................................................................................................. [1] [Total: 5]

Mark scheme: 10(a) C6H5O– / C6H5O–Na+ / C6H5ONa [1] C6H5N2C6H4O– or C6H5N2C6H4OH or [1] C6H5OCOC6H4CO2C6H5 [1] 10(b) 2 and 4 [1] 1

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