Cambridge A Level Chemistry 9701 — 2017 Oct/Nov Paper 4 · Variant 1

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 9701/41 Paper 4 A Level Structured Questions October/November 2017  2 hours Candidates answer on the Question Paper. Additional Materials: Data Booklet Cambridge International Examinations Cambridge International Advanced Subsidiary and Advanced Level This document consists of 17 printed pages and 3 blank pages. [Turn over IB17 11_9701_41/FP © UCLES 2017 *5794148263*

2 9701/41/O/N/17 © UCLES 2017 Answer all the questions in the spaces provided. 1 The compound nitrosyl bromide, NOBr, can be formed by the reaction shown. 2NO + Br2 2NOBr (a) Using oxidation numbers, explain why this reaction is a redox reaction. … … … [2] (b) Nitrosyl bromide contains a trivalent nitrogen atom. Draw the ‘dot-and-cross’ diagram for NOBr. Show outer electrons only. [2] (c) The rate of the reaction was measured at various concentrations of the two reactants, NO and Br2, and the following results were obtained. experiment [NO] / mol dm–3 [Br2] / mol dm–3 initial rate / mol dm–3 s–1 1 0.03 0.02 3.4 × 10–3 2 0.03 0.04 6.8 × 10–3 3 0.09 0.04 6.1 × 10–2 4 0.12 0.06 to be calculated The general form of the rate equation for this reaction is as follows. rate = k [NO]a[Br2]b (i) What is meant by the term order of reaction with respect to a particular reagent? … … [1]

3 9701/41/O/N/17 © UCLES 2017 [Turn over (ii) Use the data in the table to deduce the values of a and b in the rate equation. Show your reasoning. … … … … [2] (iii) Use the data in the table to calculate the initial rate for experiment 4. initial rate = … mol dm–3 s–1 [1] (iv) Use the results of experiment 1 to calculate the rate constant, k, for this reaction. Include the units of k. rate constant, k = … units … [2] (v) By considering the rate equation, explain why the rate decreases with decreasing temperature. … … [1] (d) The reaction between X and Y was studied. 2X + Y Z The following sequence of steps is a proposed mechanism for the reaction. step 1 2X V step 2 V + Y Z The general form of the rate equation for this reaction is as follows. rate = k [X]m[Y]n Step 1 is the slower step in the mechanism. Deduce the values of m and n in the rate equation. m = … n = … [1] [Total: 12]

4 9701/41/O/N/17 © UCLES 2017 2 (a) The table lists values of solubility products, Ksp, of some Group 2 carbonates. solubility product in water at 298 K, Ksp / mol2 dm–6 MgCO3 1.0 × 10–5 CaCO3 5.0 × 10–9 SrCO3 1.1 × 10–10 Use the data in the table to describe the trend in the solubility of the Group 2 carbonates down the group. … … [1] (b) (i) Write an equation to show the equilibrium for the solubility product for MgCO3. Include state symbols. … [1] (ii) With reference to your equation in (i), suggest what is observed when a few cm3 of concentrated Na2CO3(aq) are added to a saturated solution of MgCO3. Explain your answer. … … … [2] (c) Use the data in the table to calculate the solubility of MgCO3 in water at 298 K, in g dm–3. solubility of MgCO3 = … g dm–3 [2]

5 9701/41/O/N/17 © UCLES 2017 [Turn over (d) (i) Magnesium nitrate decomposes at a lower temperature than barium nitrate. Explain why. … … … [2] (ii) A sample of barium nitrate was heated strongly until no further change occurred. A white solid was formed. Write an equation for the action of heat on barium nitrate. … [1] (iii) When water was added to the white solid produced in (d)(ii), an alkaline solution was produced. Adding sulfuric acid to this solution produced a white precipitate. Write equations to explain these observations. … … [2] [Total: 11]

6 9701/41/O/N/17 © UCLES 2017 3 (a) Define the term standard cell potential. … … … [2] (b) (i) Draw a fully labelled diagram of the experimental set‑up you could use to measure the standard electrode potential of the Pb2+(aq) / Pb(s) electrode. Include the necessary chemicals. [4] (ii) The E o for a Pb2+(aq) / Pb(s) electrode is – 0.13 V. Suggest how the E for this electrode would differ from its E o value if the concentration of Pb2+(aq) ions is reduced. Indicate this by placing a tick () in the appropriate box in the table. more negative no change less negative Explain your answer. … … … [2]

7 9701/41/O/N/17 © UCLES 2017 [Turn over (c) Car batteries are made up of rechargeable lead-acid cells. Each cell consists of a negative electrode made of Pb metal and a positive electrode made of PbO2. The electrolyte is H2SO4(aq). When a lead-acid cell is in use, Pb2+ ions are precipitated out as PbSO4(s) at the negative electrode. Pb(s) + SO4 2–(aq) PbSO4(s) + 2e– (i) Calculate the mass of Pb that is converted to PbSO4 when a current of 0.40 A is delivered by the cell for 80 minutes. mass of Pb = … g [2] (ii) Complete the half-equation for the reaction taking place at the positive electrode. PbO2(s) + SO4 2–(aq) + … + … PbSO4(s) + … [1] (d) The diagrams show how the voltage across two different cells changes with time when each cell is used to provide an electric current. voltage / V time / hours lead-acid cell voltage / V time / hours H2 / O2 fuel cell Suggest a reason why ● the voltage of the lead-acid cell changes after several hours, … … ● the voltage of the fuel cell remains constant. … … [2] [Total: 13]

8 9701/41/O/N/17 © UCLES 2017 4 (a) Describe and explain how the density and melting point of cobalt compare to those of calcium. density of cobalt … explanation … … melting point of cobalt … explanation … … [3] (b) Transition metals can form complexes. What is meant by the term transition metal complex? … … [1] (c) (i) Cobalt can form the compounds [Co(NH3)5Br]SO4 and [Co(NH3)5SO4]Br. These two compounds are structural isomers. Define the term structural isomer. … … [1] (ii) Draw a three-dimensional diagram to show the structure of the ion [Co(NH3)5Br]2+. Name its shape. [Co(NH3)5Br]2+ shape … [1] (iii) State the type of bonding between the cobalt ion and NH3 groups in the [Co(NH3)5Br]2+ ion. … [1]

9 9701/41/O/N/17 © UCLES 2017 [Turn over (iv) State the oxidation number of cobalt in ● [Co(NH3)5Br]2+ oxidation number of Co = … ● [Co(NH3)5SO4]+ oxidation number of Co = … [1] (d) Solutions of the compounds [Co(NH3)5Br]SO4 and [Co(NH3)5SO4]Br can be distinguished from each other by simple chemical tests. Assume that any species bonded to the cobalt ion does not react in these tests. Complete the table with two different tests that could be used to positively identify each compound. Give the expected observation with each compound. test observation with [Co(NH3)5Br]SO4(aq) observation with [Co(NH3)5SO4]Br(aq) [2] (e) The two compounds [Co(NH3)5Br]SO4 and [Co(NH3)5SO4]Br are different colours. Explain why the colours of the two compounds are different. … … … [2] (f) Some transition metals and their compounds act as catalysts. The catalysis can be classified as heterogeneous or homogeneous. Complete the table by placing one tick () in each row to indicate the type of catalysis in each reaction. heterogeneous homogeneous Fe in the Haber process Fe2+ in the I– / S2O8 2– reaction NO2 in the oxidation of SO2 V2O5 in the Contact process [2] [Total: 14]

10 9701/41/O/N/17 © UCLES 2017 5 Compound P contains several functional groups. C C CN CN H Cl P (a) Name the functional groups present in P. … … [2] (b) Compound P can be polymerised. Draw a section of the polymer of P showing two repeat units. Name the type of polymerisation. type of polymerisation … [2]

11 9701/41/O/N/17 © UCLES 2017 [Turn over (c) Complete the following table to show the structures of the products formed and the type of organic reaction when P reacts with the four reagents. reagent structure(s) of product(s) type of organic reaction excess Br2(aq) excess hot, concentrated, acidified MnO4 –(aq) excess hot HCl (aq) excess H2 / Pt catalyst [8] [Total: 12]

12 9701/41/O/N/17 © UCLES 2017 6 (a) 4-nitromethylbenzene can be prepared via an electrophilic substitution reaction as shown. CH3 conc. HNO3 conc. H2SO4 methylbenzene CH3 NO2 H intermediate T CH3 NO2 4-nitromethylbenzene (i) This reaction also forms an isomer of 4-nitromethylbenzene as a by-product. Draw the structure of this by-product. [1] (ii) Write an equation for the reaction between HNO3 and H2SO4 that forms the electrophile for this reaction. … [1] (iii) Describe how the structure and bonding of the six-membered ring in intermediate T differs from that in methylbenzene. … … … … [3]

13 9701/41/O/N/17 © UCLES 2017 [Turn over (b) Benzocaine is used as a local anaesthetic. It can be synthesised from 4-nitromethylbenzene by the route shown. CH3 NO2 step 1 CH3 NH2 CH3 NHCOCH3 step 2 CO2CH2CH3 NH2 step 5 CO2H NH2 benzocaine W CO2H NHCOCH3 step 4 step 3 4-nitromethylbenzene (i) Give the systematic name of compound W. … [1] (ii) Suggest the reagents and conditions for steps 1– 5. step 1 … step 2 … step 3 … step 4 … step 5 … [6] (c) Suggest how the basicity of benzocaine would compare to that of ethylamine. Explain your answer. … … … [2]

14 9701/41/O/N/17 © UCLES 2017 (d) A sample of benzocaine, shown below, was analysed by proton NMR and carbon-13 NMR spectroscopy. (i) Predict the number of peaks that would be seen in the carbon-13 NMR spectrum. … [1] (ii) Benzocaine was dissolved in CDCl 3 and the proton NMR spectrum of this solution was recorded. 8 7 6 5 4 3 2 1 0 δ / ppm CO2CH2CH3 NH2 benzocaine Suggest why CDCl 3 and not CHCl 3 is used as the solvent when obtaining a proton NMR spectrum. … … [1] (iii) Use the Data Booklet and the spectrum in (d)(ii) to complete the table for the proton NMR spectrum of benzocaine. The actual chemical shifts, δ, for the four absorptions have been added. δ / ppm group responsible for the peak number of 1H atoms responsible for the peak splitting pattern 1.2 3.5 5.5 7.1–7.4 multiplet [4] (iv) Explain the splitting pattern for the absorption at δ1.2 ppm. … … [1]

15 9701/41/O/N/17 © UCLES 2017 [Turn over (v) The proton NMR spectrum of benzocaine dissolved in D2O was recorded. Suggest how this spectrum would differ from the spectrum in (d)(ii). Explain your answer. … … [1] (e) Benzocaine can also be used to synthesise the dyestuff S by the following route. step 1 step 2 NaOH(aq), R S OH phenol CO2CH2CH3 NH2 benzocaine (i) Suggest the reagents used for step 1. … [1] (ii) Suggest structures for compounds R and S and draw them in the boxes. [2] [Total: 25]

16 9701/41/O/N/17 © UCLES 2017 7 (a) Complete the following electronic structures. ● the iron atom, Fe 1s22s22p6 … ● the iron(III) ion, Fe3+ 1s22s22p6 … [1] (b) Solutions of iron(III) salts are acidic due to the equilibrium shown. [Fe(H2O)6]3+(aq) [Fe(H2O)5(OH)]2+(aq) + H+(aq) Ka = 8.9 × 10–4 mol dm–3 Calculate the pH of a 0.25 mol dm–3 FeCl 3 solution. pH = … [2] (c) The table shows numerical values of the stability constants for the following equilibrium where M can be one of the metal ions listed and L one of the ligands which replaces one H2O molecule. [M(H2O)6]n+(aq) + L–(aq) [M(H2O)5L](n–1)+(aq) + H2O(l) metal ion, M ligand, L stability constant, Kstab Fe3+ F– 1.0 × 106 Fe3+ Cl – 2.5 × 101 Fe3+ SCN– 9.0 × 102 Hg2+ Cl – 5.0 × 106 (i) What is meant by the term stability constant, Kstab? … … [1] (ii) Use the data in the table to predict the formula of the complex formed in the greatest amount when ● a solution containing equal concentrations of both F– and SCN– ions is added to Fe3+(aq), … ● a solution containing equal concentrations of both Fe3+ and Hg2+ ions is added to Cl –(aq). … [1]

17 9701/41/O/N/17 © UCLES 2017 [Turn over Ethanedioate ions, –O2CCO2 –, are bidentate ligands. The abbreviation ed 2– can be used to represent ethanedioate ions. (d) The complex [Fe(ed )2Cl 2]3– can be formed according to the equation shown. [Fe(H2O)4Cl 2]+(aq) + 2ed 2–(aq) [Fe(ed )2Cl 2]3–(aq) + 4H2O(l) Write the expression for the equilibrium constant, Kstab, and state its units. Kstab = units … [2] (e) [Fe(ed )2Cl 2]3– shows geometrical and optical isomerism. (i) Complete the three-dimensional diagrams to show the three stereoisomers of [Fe(ed )2Cl 2]3–. You may use –O O– to represent ed 2–. Fe isomer A Fe isomer B Fe isomer C [3] (ii) Give the letters of two isomers of [Fe(ed )2Cl 2]3– which are geometrical isomers of each other. … [1] (iii) Give the letters of the two isomers of [Fe(ed )2Cl 2]3– which show optical isomerism. … [1] (iv) Give the letter of the isomer which has no dipole moment. … [1] [Total: 13]

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® IGCSE is a registered trademark. This document consists of 13 printed pages. © UCLES 2017 [Turn over Cambridge Assessment International Education Cambridge International Advanced Subsidiary and Advanced Level CHEMISTRY 9701/41 Paper 4 A Level Structured Questions October/November 2017 MARK SCHEME Maximum Mark: 100 Published This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners’ meeting before marking began, which would have considered the acceptability of alternative answers. Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for Teachers. Cambridge International will not enter into discussions about these mark schemes. Cambridge International is publishing the mark schemes for the October/November 2017 series for most Cambridge IGCSE®, Cambridge International A and AS Level components and some Cambridge O Level components.

9701/41 Cambridge International AS/A Level – Mark Scheme PUBLISHED October/November 2017 © UCLES 2017 Page 2 of 13 Question Answer Marks 1(a) N +2 to +3 (and oxidised) 1 Br2 / Br 0 to –1 (and reduced) 1 1(b) 3 bonding pairs around N (in a structure involving NOBr) 1 rest of molecule correct 1 1(c)(i) the power to which a concentration of a reactant is raised in the rate equation 1 1(c)(ii) using expt. 2 and 3 a = 2 or [NO] 2nd order and conc × 3 rate × 9 or 6.1 × 10–2 / 6.8 × 10–3 = (0.09 / 0.03)a 1 using expt. 1 and 2 b = 1 or [Br2] 1st order and conc × 2 rate × 2 or 6.8 × 10–3 / 3.4 × 10–3 = (0.04 / 0.02)b 1 (c)(iii) initial rate = 0.16(32) 1 1(c)(iv) (0.0034 = k(0.03)2(0.02)) k = 188.9 1 mol–2 dm6 s–1 1 1(c)(v) k decreases (as rate decreases) 1

9701/41 Cambridge International AS/A Level – Mark Scheme PUBLISHED October/November 2017 © UCLES 2017 Page 3 of 13 Question Answer Marks 1(d) m = 2 and n = 0 1 Question Answer Marks 2(a) it / solubility decreases down the group and Ksp decreases 1 2(b)(i) MgCO3(s) ⇌ Mg2+(aq) + CO3 2–(aq) 1 2(b)(ii) (white) solid appears / precipitation (of MgCO3) 1 as [CO3 2–] increases shifting equilibrium to the LHS (precipitating out MgCO3) 1 2(c) solubility = √1.0 × 10-5 = 3.16 × 10–3 mol dm–3 1 solubility= 3.2 × 10–3 × 84.3 = 0.27 g dm–3 1 2(d)(i) Mg2+ ion is smaller than Ba2+ ion or ionic radii increase down group ora 1 (Mg2+) distorts / polarises / the anion / nitrate group / nitrate ion / NO3 (1)– / NO3 ion more easily (than Ba2+) ora 1 2(d)(ii) Ba(NO3)2 → BaO + 2NO2 + ½O2 1 2(d)(iii) BaO + H2O → Ba(OH)2 1 Ba(OH)2 + H2SO4 → BaSO4 + 2H2O 1

9701/41 Cambridge International AS/A Level – Mark Scheme PUBLISHED October/November 2017 © UCLES 2017 Page 4 of 13 Question Answer Marks 3(a) the potential difference between two half-cells / two electrodes (in a cell) 1 under standard conditions of 1 atm., 298 K, (all) solutions being 1 mol dm–3 1 3(b)(i) 8 marking points, any 2 points for each mark H2 / hydrogen correct delivery system for H2 Pb2+ (aq) Pb electrode Pt electrode H+(aq) solution salt bridge voltmeter / V labelled 4 3(b)(ii) more negative 1 shifts Pb2+ (+ 2e–) ⇋ Pb equilibrium / reaction to the left 1

9701/41 Cambridge International AS/A Level – Mark Scheme PUBLISHED October/November 2017 © UCLES 2017 Page 5 of 13 Question Answer Marks 3(c)(i) Q = 0.4 × 80 × 60 = 1920 C and use of 96500 / 193000 Moles of Pb = 1920 / 193000 = 9.95 × 10–3 Mass of Pb = 207.2 × 9.95 × 10–3 = 2.1 g OR Q = 0.4 × 80 × 60 = 1920 C and use of 1.6 × 10–19 / 1.2 × 1022 atoms Pb = 6 × 1021; moles of Pb =6 × 1021 / 6 × 1023 = 0.01 Mass of Pb = 207.2 × 0.01 = 2.1 g 2 3(c)(ii) PbO2(s) + SO4 2–(aq) + 4H+ + 2e– → PbSO4(s) + 2H2O 1 3(d) reagents / PbO2 / H2SO4 and used up / concentration decreases 1 as fuel / hydrogen is being continuously supplied / fuel has not run out 1 Question Answer Marks 4(a) density is higher and melting point is higher 1 (density) due to Ar being larger and smaller atomic radii or (Co) atoms / ions heavier and smaller 1 (melting point) due to stronger attraction to cations as more delocalised electrons 1 4(b) (a molecule or ion) formed by a central metal atom / ion surrounded by (one or more) ligands 1 4(c)(i) same number and type of atoms and different structural formula 1

9701/41 Cambridge International AS/A Level – Mark Scheme PUBLISHED October/November 2017 © UCLES 2017 Page 6 of 13 Question Answer Marks 4(c)(ii) octahedral AND 3D structure of [Co(NH3)5Br]2+ e.g. 1 4(c)(iii) co-ordinate / dative covalent 1 4(c)(iv) +3 for both 1 4(d) (HNO3) Ag+ / AgNO3 cream(–yellow) ppt. (of AgBr) and no reaction / white ppt. for other isomer 1 Ba(OH)2 / Ba2+(aq) / BaCl 2 / Ba(NO3)2 white ppt. (of BaSO4) and no reaction for other isomer 1 4(e) (d-d) energy gap / ∆E is different 1 absorb different wavelength / frequency (of light) 1 4(f) heterogeneous homogeneous Fe in the Haber process 9 Fe2+ in the I– / S2O8 2– reaction 9 NO2 in the oxidation of SO2 9 V2O5 in the Contact process 9 2

9701/41 Cambridge International AS/A Level – Mark Scheme PUBLISHED October/November 2017 © UCLES 2017 Page 7 of 13 Question Answer Marks 5(a) nitrile; alkene; chloro; benzene / arene 2 5(b) 1 addition (polymerisation) 1

9701/41 Cambridge International AS/A Level – Mark Scheme PUBLISHED October/November 2017 © UCLES 2017 Page 8 of 13 Question Answer Marks 5(c) reagent structure of product type of organic reaction excess Br2(aq) [1] (electrophilic) addition excess hot, conc. MnO4 –(aq) [1] + [1] oxidation excess hot, aqueous HCl [1] hydrolysis excess H2 / Pt catalyst both CH2NH2 formed [1] both arene and alkene reduced [1] reduction / hydrogenation structures [6] 2 correct for 1 mark total [2] 8

9701/41 Cambridge International AS/A Level – Mark Scheme PUBLISHED October/November 2017 © UCLES 2017 Page 9 of 13 Question Answer Marks 6(a)(i) 1 6(a)(ii) HNO3 + 2H2SO4 → H3O+ + NO2 + + 2HSO4 – 1 6(a)(iii) any three from: Point 1: bonds / electrons are partially delocalised in T or delocalised / π system / π bonding extends over only five carbons Point 2: four π-electrons in the (delocalised system of T) or methylbenzene has (two) more π-electrons / (two) more delocalised electrons Point 3: contains a carbon that is sp3 hybridised in T or (all the) carbons are sp2 hybridised in methylbenzene Point 4: one carbon has a bond angle of 109.5° / tetrahedral (in T) or (C-C) bond strengths / lengths are not all the same or not all the bond angles are 120° (in T) 3 6(b)(i) 4-aminobenzoic acid 1 6(b)(ii) step 1 Sn + HCl [1] concentrated / reflux / heat [1] step 2 CH3COCl [1] step 3 KMnO4 / manganate(VII) / MnO4 – (acidified / alkaline) and heat [1] step 4 aqueous HCl and heat [1] step 5 ethanol, H2SO4, concentrated / reflux / heat [1] 6

9701/41 Cambridge International AS/A Level – Mark Scheme PUBLISHED October/November 2017 © UCLES 2017 Page 10 of 13 Question Answer Marks 6(c) (benzocaine) is less (basic than ethylamine) AND lone pair (on N) is less available to accept a proton / H+ since (lone pair on N) is delocalised over the ring or phenyl ring is electron withdrawing group OR ethylamine is more basic (than benzocaine) AND lone pair (on N) is more available to accept a proton / H+ since ethyl / alkyl group is electron-donating group 2 6(d)(i) 7 peaks 1 6(d)(ii) CDCl3 will produce no signal in the spectrum or CHCl3 would produce a signal / would be detected 1 6(d)(iii) δ / ppm group responsible for the peak number of H atoms responsible for the peak splitting pattern 1.2 CH(3) 3 triplet 3.5 CH(2)O 2 quartet 5.5 NH2 2 singlet (broad) 7.1–7.4 H attached to aromatic / benzene ring 4 multiplet 4 6(d)(iv) neighbouring / adjacent carbon atom has two protons / H (attached to it) or there is an adjacent CH2(O) group 1 6(d)(v) peak at 5.5 / NH2 peak will disappear and NH2 / protons exchange / swap with deuterium 1

9701/41 Cambridge International AS/A Level – Mark Scheme PUBLISHED October/November 2017 © UCLES 2017 Page 11 of 13 Question Answer Marks 6(e)(i) NaNO2 + HCl or HNO2 1 6(e)(ii) structure of diazonium salt R 1 structure of azo dye S 1

9701/41 Cambridge International AS/A Level – Mark Scheme PUBLISHED October/November 2017 © UCLES 2017 Page 12 of 13 Question Answer Marks 7(a) Fe atom= (1s22s22p6)3s23p63d64s2 Fe3+ ion= (1s22s22p6)3s23p63d5 1 7(b) ([H+]2 = 8.9 × 10–4 × 0.25 or 2.225 × 10–4) [H+] = 0.0149 1 pH = –log(0.0149) = 1.83 1 7(c)(i) (Kstab is) the equilibrium constant for the formation of a complex (ion) (in a solvent from its constituent ions / molecules) 1 7(c)(ii) [Fe(H2O)5F]2+ and [Hg(H2O)5Cl]+ 1 7(d) Kstab = ] − 3 2 2 + 2 2 4 2 [Fe(ed) C [Fe(H O) C ][ed] l l 1 mol–2 dm6 1 7(e)(i) 3

9701/41 Cambridge International AS/A Level – Mark Scheme PUBLISHED October/November 2017 © UCLES 2017 Page 13 of 13 Question Answer Marks 7(e)(ii) any cis isomer and the trans isomer identified 1 7(e)(iii) both correct cis isomers identified 1 7(e)(iv) trans isomer identified 1