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Chemistry 9701/42 May/June 2019: question paper and mark scheme

Cambridge AS & A Level Chemistry (9701), Paper 4 A Level Structured Questions, May/June 2019. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.

9701_s19_qp_429701_s19_ms_429701/42/M/J/19

Paper
Paper 4 A Level Structured Questions
Total marks
100
Questions
8
Code on the paper
9701/42/M/J/19
9701_s19_ms_42.pdf

9701/42 mark scheme, question by question

Each question as it's printed, and the mark scheme's points for it, as text. Diagrams, tables and graphs are in the PDFs above.

Question 117 marks

Question

(a) (i) Complete the electronic configuration of the copper(II) ion. 1s22s22p6 . …… [1] (ii) State the colour of the solutions containing the following ions. ● [Cu(H2O)6]2+(aq) …… ● [CuCl 4]2–(aq) …… [1] (iii) Octahedral complexes of Cu2+ with different ligands can have different colours. Explain why. . …… . …… . …… . …… . …… [2] (b) Copper(I) and silver(I) salts are colourless. Suggest why. . …… . …… . …… . …… [2] (c) Consider the following two equilibria and associated data values at 298 K. AgBr(s) Ag+(aq) + Br –(aq) equilibrium 1 Ksp = 5.0 × 10–13 mol2 dm–6 Ag+(aq) + 2NH3(aq) [Ag(NH3)2]+(aq) equilibrium 2 Kstab = 1.7 × 107 mol–2 dm6 The equilibrium constant for equilibrium 1 is the solubility product, Ksp, of AgBr(s). The equilibrium constant for equilibrium 2 is the stability constant, Kstab, for the formation of [Ag(NH3)2]+(aq). (i) Calculate the solubility of AgBr at 298 K in mol dm–3. solubility of AgBr = …… mol dm–3 [1] (ii) Use Le Chatelier’s principle as applied to equilibria 1 and 2 to suggest why AgBr(s) dissolves in concentrated NH3(aq). . …… . …… . …… . …… [2] (iii) Use equilibria 1 and 2 to construct an equation for the reaction of AgBr(s) with concentrated NH3(aq). This is equilibrium 3. . …… equilibrium 3 [1] Write an expression for the equilibrium constant of equilibrium 3, Keq3, in terms of Ksp for (iv) equilibrium 1 and Kstab for equilibrium 2. Keq3 = [1] (d) Define the term standard electrode potential, E o. . …… . …… . …… [1] (e) (i) Complete and label the diagram to show how the standard electrode potential, E o, of Ag+(aq) / Ag(s) could be measured under standard conditions. [4] (ii) Use the Data Booklet to label the diagram in (e)(i) to show ●● which is the positive electrode, ●● the direction of electron flow in the external circuit when a current flows. [1] [Total: 17]

Mark scheme

1(a)(i) 3s23p63d9 [1] 1 1(a)(ii) [Cu(H2O)6]2+ (pale) blue 1 CuCl42- yellow both [1] 1(a)(iii) M1 energy gap / ∆E is different (for the ligands) [1] 2 M2 different frequency / wavelength of light absorbed / transmitted / reflected [1] 1(b) M1 (Cu+/Ag+) d-shell is full / complete OR d-orbitals are full [1] 2 M2 no electrons can be promoted [1] 1(c)(i) solubility = √5.0 × 10–13 = 7.1 × 10–7 (mol dm–3) [1] min 2sf 1 1(c)(ii) M1 (in conc. NH3) [NH3] increases and equilibrium 2 shifts to the right [1] 2 M2 [Ag+] decreases and equilibrium 1 shifts to the right [1] 1(c)(iii) AgBr + 2NH3 ⇌ [Ag(NH3)2]+ + Br– [1] 1 1(c)(iv) Keq3 = Ksp × Kstab [1] ALLOW Keq3= [Ag(NH3)2+][Br–]/[NH3]2 1 1(d) The potential difference when a half-cell is connected to a (standard) hydrogen electrode under standard conditions [1] 1 OR the potential difference / voltage / EMF between a hydrogen electrode and another half-cell under standard conditions [1] 1(e)(i) salt bridge y voltmeter / V y 4 Ag y Ag+ (or soluble silver salt) y Pt y H2 (and delivery correct) + H+ (or named strong acid) y 1 atm. (pressure) y 1 mol dm–3 (and 298 K) y mark as y 9 y 9 y 9 y 9 [4] 1(e)(ii) Ag electrode labelled and arrow (in the external circuit moving towards this electrode) [1] 1

Question 29 marks

Question

(a) Group 2 carbonates decompose when heated. Write an equation for the decomposition of the carbonate ion. . …… [1] (b) Describe and explain how the thermal stability of the Group 2 carbonates changes with increasing atomic number. . …… . …… . …… . …… . …… [3] (c) Lead(II) carbonate, PbCO3, and zinc carbonate, ZnCO3, decompose on heating in a similar way to calcium carbonate, CaCO3. State relevant data from the Data Booklet and use it to predict the order of thermal stability of these three carbonates. data . …… . …… (most stable) …… > …… > …… (least stable) [2] (d) Dolomite contains the double carbonate of calcium and magnesium, CaMg(CO3)2, and some impurities. When 0.642 g of a sample of dolomite reacts with an excess of hydrochloric acid, 125.0 cm3 of CO2 is formed under room conditions. Calculate the percentage of CaMg(CO3)2 in the sample of dolomite. Show all your working. Assume that none of the impurities react with HCl. % of CaMg(CO3)2 in dolomite = …… % [3] [Total: 9]

Mark scheme

2(a) CO32– → O2– + CO2 [1] 1 2(b) Increases (with increasing atomic number / implied) [1] 3 cationic radius / ion size increases (down the group) [1] less polarisation / distortion of anion/CO32– [1] 2(c) (Pb2+) 0.120 nm; (Ca2+) 0.099 nm; (Zn2+) 0.075 nm [1] 2 (most stable) PbCO3 > CaCO3 > ZnCO3 (least stable) [1] ECF from atomic radii 2(d) amount of CO2 = 125 / 24000 = 5.21 × 10–3 mol [1] 3 CaMg(CO3)2 :CO2 1:2 amount of carbonate = 2.60(4) × 10–3 mol [1] ECF mass of carbonate = 184(.4) × 2.60(4) × 10–3 = 0.480 g % of CaMg(CO3)2 = 100 × 0.480 / 0.642 = 74.8 % [1] ECF

Question 313 marks

Question

(a) Sketch the shape of a d orbital. [1] (b) (i) Explain what is meant by the term transition element. . …… . …… [1] Transition elements can form complex ions which contain ligands. (ii) Name the type of bonding that occurs between a ligand and a transition element. . …… [1] (c) Give the formulae of two oxides of iron. State the oxidation number of iron in each compound. . …… . …… [1] (d) CO and CN– are monodentate ligands. Complete the table for the following two complexes. co-ordination formula of charge of metal ion ligand number complex ion complex ion Ni2+ CO 4 Fe3+ CN– 3- [2] (e) Transition element complexes can exhibit stereoisomerism. [Cu(H2O)4(NH3)2]2+ and Pt(NH3)2Cl 2 show the same type of isomerism. (i) Name this type of isomerism. . …… [1] (ii) Complete the three-dimensional diagrams of the two isomers for [Cu(H2O)4(NH3)2]2+ and Pt(NH3)2Cl 2. Cu Cu Pt Pt [2] (f) Copper can form complexes with the ligands ammonia and en, H2NCH2CH2NH2, as shown. [Cu(H2O)6]2+(aq) + en(aq) [Cu(H2O)4(en)]2+(aq) + 2H2O(l) Kstab = 3.98 × 1010 equilibrium 4 [Cu(H2O)6]2+(aq) + 2NH3(aq) [Cu(H2O)4(NH3)2]2+(aq) + 2H2O(l) Kstab = 5.01 × 107 equilibrium 5 (i) Write an expression for the stability constant, Kstab, for equilibrium 5. State its units. Kstab = units = …… [2] (ii) The standard entropy change, ΔS o, for equilibrium 4 is +23 J K–1 mol–1 and for equilibrium 5 is –8.4 J K–1 mol–1. Suggest an explanation for this difference by reference to both equilibria. . …… . …… . …… [1] (iii) Of the three copper complexes in equilibria 4 and 5, state the formula of the copper complex that is the most stable and explain your choice. copper complex …… explanation …… . …… [1] [Total: 13] Question 4 starts on the next page.

Mark scheme

3(a) 1 any diagram [1] 3(b)(i) (elements) forming one or more (stable) ions with incomplete / partially filled d orbital(s) / sub-shell [1] 1 3(b)(ii) dative covalent / coordinate [1] 1 3(c) FeO and +2 1 Fe2O3 and +3 all [1] ALLOW Fe3O4 and +3 and +2 3(d) 2 co-ordination formula of complex charge of complex metal ion ligand number ion ion Ni2+ CO 4 Ni(CO)4 2+ Fe3+ CN- 6 Fe(CN)6 3- mark as y 9 y 9 [2] 3(e)(i) cis-trans isomerism [1] ALLOW geometric(al) 1 3(e)(ii) 2 one correct pair [1] two correct pairs [2] 3(f)(i) [Cu(H2O)4 (NH3 )22+ ] 2 Kstab = [1] units = dm6 mol–2 [1] ecf from M1 [Cu(H2O)6 2+ ][NH3 ]2 3(f)(ii) equilibrium 4 has a (net) increase in moles of product / 2 moles goes to 3 moles whereas equilibrium 5 has same number of 1 moles of reactants and products / 3 moles vs 3 moles [1] 3(f)(iii) [Cu(H2O)4(en)]2+ and (equilibrium) constant / Kstab is the largest / highest [1] 1 ALLOW [Cu(H2O)4(en)]2+ and constant / Kstab of eqm 4 is greater / higher

Question 411 marks

Question

The initial rate of reaction for propanone and iodine in acid solution is measured in a series of experiments at a constant temperature. H+ catalyst CH3COCH3 + I2 CH3COCH2I + HI The rate equation was determined experimentally to be as shown. rate = k [CH3COCH3][H+] (a) State the order of reaction with respect to ● CH3COCH3 . …… ● I2 …… ● H+ . …… and state the overall order of this reaction. …… [2] (b) The rate of this reaction is 5.40 × 10–3 mol dm–3 s–1 when ● the concentration of CH3COCH3 is 1.50 × 10–2 mol dm–3 ● the concentration of I2 is 1.25 × 10–2 mol dm–3 ● the concentration of H+ is 7.75 × 10–1 mol dm–3. (i) Calculate the rate constant, k, for this reaction. State the units of k. k = …… units = …… [2] (ii) Complete the table by placing one tick () in each row to describe the effect of decreasing the temperature on the rate constant and on the rate of reaction. decreases no change increases rate constant rate of reaction [1] (c) From the results, a graph is produced which shows how the concentration of I2 changes during the reaction. [I2] 0 0 time Describe how this graph could be used to determine the initial rate of the reaction. . …… . …… . …… [2] (d) On the axes below, sketch a graph to show how the initial rate changes with different initial concentrations of CH3COCH3 in this reaction. rate 0 0 [CH3COCH3] [1] (e) The rate of a reaction between metal ions was studied. The following three-step mechanism has been suggested for this reaction. Step 1 is the rate-determining step. step 1 Ce4+ + Mn2+ Ce3+ + Mn3+ step 2 Ce4+ + Mn3+ Ce3+ + Mn4+ step 3 Mn4+ + Tl + Tl 3+ + Mn2+ (i) Explain the meaning of the term rate-determining step. . …… . …… [1] (ii) Use this mechanism to ● determine the overall equation for this reaction . …… ● suggest the role of Mn2+ ions in this mechanism. Explain your answer. . …… . …… . …… [2] [Total: 11]

Mark scheme

4(a) CH3COCH3 = 1 2 I2 = 0 H+ = 1 overall order = 2 M1 3 orders [1] M2 overall order based on their M1 [1] 4(b)(i) k = 5.40 × 10–3 / (1.50 × 10–2 × 7.75 × 10–1) 2 k = 0.46(452) [1] dm3 mol–1 s–1 [1] 2sf min 4(b)(ii) 1 decreases no change increases rate constant 9 rate of reaction 9 both [1] 4(c) draw a tangent at time, t=0 [1] 2 measure the gradient / slope of the tangent [1] 4(d) straight line graph starting at 0,0 and showing rate α [CH3COCH3] [1] 1 4(e)(i) slowest step / reaction (in the mechanism) [1] 1 4(e)(ii) 2Ce4+ + Tl+ → Tl3+ + 2Ce3+ [1] 2 catalyst and (used in step 1 and) regenerated / reformed in step 3 / end of the reaction [1]

Question 59 marks

Question

(a) Complete the table by placing one tick () in each row to indicate the sign of each type of energy change under standard conditions. always always either negative energy change positive negative or positive lattice energy enthalpy change of neutralisation [1] (b) Define, in words, the term enthalpy change of solution. . …… . …… [1] (c) The following enthalpy changes are given. enthalpy change value / kJ mol–1 standard enthalpy change of formation, , for K3PO4(s) –2035 standard enthalpy change, ∆H o, for P(s) + 2O2(g) + 3e– PO43–(aq) –1284 standard enthalpy change, ∆H o, for K(s) K+(aq) + e– –251 Determine the standard enthalpy change of solution of potassium phosphate, K3PO4(s). It may be helpful to draw a labelled energy cycle. = …… kJ mol–1 [3] (d) Some lattice energy values are shown in the table. lattice energy compound value / kJ mol–1 CaBr2(s) –2176 KBr(s) –679 Suggest an explanation for why CaBr2 is more exothermic than KBr. . …… . …… . …… [2] (e) For a particular gas phase reaction the variation in standard Gibbs free energy change, ΔG o, with temperature is shown. Assume standard enthalpy change, ΔH o, and standard entropy change, ΔS o, remain constant with temperature. ∆G o / kJ mol–1 0 350 550 T/K (i) Write the equation that relates ΔG o to ΔH o and ΔS o. . …… [1] (ii) Use this equation to explain why ΔG o becomes less positive as temperature increases in this reaction. . …… . …… [1] [Total: 9] Question 6 starts on the next page.

Mark scheme

5(a) 1 always always either negative energy change positive negative or positive lattice energy 9 enthalpy of neutralisation 9 both [1] 5(b) (energy change) when 1 mole of solute is dissolved in an infinite amount of water to form a dilute solution 1 5(c) calculation of ∆Hosol with –251, –1284 and –2035 only and two correct signs [1] 3 calculation of ∆Hosol with –251, –1284 and –2035 only and correct signs OR calculation of ∆Hosol with (–251 × 3), –1284 and –2035 only and two correct signs [2] ∆Hosol = (3 × –251) + (–1284) – (–2035) = –2 (kJ mol–1) [3] 5(d) Ca2+ have a higher charge / greater charge density [1] ora 2 stronger electrostatic forces between Br– and Ca2+ [1] 5(e)(i) ∆Go = ∆Ho – T∆So [1] 1 5(e)(ii) T∆S is more positive 1 OR –T∆S becomes more negative [1]

Question 616 marks

Question

(a) By reference to the formation of σ and π bonds, describe and explain the shape of a benzene molecule, C6H6. . …… . …… . …… . …… . …… . …… . …… . …… [3] (b) 2,3-dimethylphenylamine can be prepared from 1,2-dimethylbenzene in two steps as shown. 1,2-dimethylbenzene M 2,3-dimethylphenylamine CH3 CH3 CH3 CH3 CH3 CH3 HNO3 / H2SO4 step 1 step 2 NO2 NH2 Step 1 is catalysed by H2SO4. (i) Write an equation to show how H2SO4 generates the electrophile during step 1. . …… [1] (ii) Draw the mechanism of the reaction between this electrophile and 1,2-dimethylbenzene to form M. Include all relevant curly arrows and charges. intermediate M CH3 CH3 CH3 CH3 NO2 [3] (iii) Write an equation to show how the H2SO4 catalyst is reformed. . …… [1] (iv) For step 2, suggest the reagents and conditions and name the type of reaction. ● reagents and conditions …… ● type of reaction …… [2] (c) The drug mefenamic acid can be made using 2,3-dimethylphenylamine in an excess of 2‑chlorobenzoic acid. 2,3-dimethylphenylamine 2-chlorobenzoic acid mefenamic acid CH3 CO2H CH3 CH3 Cl CH3 + + HCl NH2 N H CO2H (i) Deduce the molecular formula of mefenamic acid. . …… [1] (ii) Name the functional groups, apart from the benzene ring, in mefenamic acid. . …… [1] (iii) Calculate the maximum mass of mefenamic acid that could be formed from 5.00 g of 2,3‑dimethylphenylamine in this reaction. Give your answer to three significant figures. mass of mefenamic acid = …… g [2] (d) The position of substitution in the electrophilic substitution of arenes can be explained based on the stability of the intermediate cations formed in the first step. The example given involves the bromination of methylbenzene. CH3 CH3 CH3 Br + H 2-position 3-position + Br H methylbenzene 4-position CH3 + H Br Use this information and your knowledge about the stability of cations to suggest why the CH3 group directs incoming electrophiles to the 2- and 4-positions in preference to the 3-position. . …… . …… . …… . …… [2] [Total: 16]

Mark scheme

6(a) any three points from: 3 × [1] 3 • bond angle = 120° and shape is (hexagonal ring) planar / (trigonal) planar • carbons are sp2 hybridised • contains delocalised electrons in the π bonds / system • sp2 orbitals between C-H / C-C overlap to form σ bonds • a p orbital from each carbon atom overlap sideways with each other above and below the ring forming π bonds ALLOW labelled diagrams for bullets 1–5 6(b)(i) HNO3 + H2SO4 → HSO4- + H2O + NO2+ 1 or HNO3 + 2H2SO4 → 2HSO4- + H3O+ + NO2+ [1] 6(b)(ii) 3 first curly arrow to N of NO2+ [1] correct intermediate [1] 2nd curly arrow and H+ formed / lost [1] 6(b)(iii) HSO4- + H+ → H2SO4 [1] 1 6(b)(iv) Sn + conc. HCl (+ heat) [1] 2 reduction [1] IGNORE redox 6(c)(i) C15H15NO2 [1] 1 6(c)(ii) amine and carboxylic acid both [1] 1 6(c)(iii) amount of 2,3-dimethylphenylamine = 5.00 / 121 = 0.0413 mol [1] 2 amount of mefenamic acid = 0.0413 mol mass of mefenamic acid = 0.0413 × 241 = 9.96 / 9.95 g 3sf required [1] ECF 6(d) 3o carbocations are more stable than 2o carbocations [1] 2 due to the methyl group acting as an electron donating group (leading to an increase in electron density on the cation stabilising it) [1]

Question 715 marks

Question

(a) Amino acids can be separated by thin-layer chromatography. A mixture of amino acids is analysed using this technique. The chromatogram obtained is shown, drawn to scale. The table shows some Rf values for different amino acids in the solvent used. solvent front 5 4 amino acid Rf value A alanine 0.40 3 glutamic acid 0.29 2 B leucine 0.71 1 valine 0.61 baseline cm 0 (i) Use the chromatogram and the Rf values to deduce the amino acid responsible for spot A and spot B. amino acid responsible for spot A …… amino acid responsible for spot B …… [1] (ii) A second chromatogram of the same mixture is taken using a more polar solvent. Predict the effect on the Rf values of the amino acids. Explain your reasoning. . …… . …… . …… . …… [1] (b) Glycine, H2NCH2CO2H, is the simplest amino acid. (i) Complete the equations to show the acid-base properties of glycine. H2NCH2CO2H(aq) + HCl (aq) …… H2NCH2CO2H(aq) + NaOH(aq) . …… [2] (ii) In aqueous solution, amino acids exist as zwitterions. Draw the zwitterionic structure of glycine. Explain how the zwitterion for glycine is formed. . …… . …… . …… [2] (c) Apart from glycine, all naturally occurring amino acids have a chiral centre and exhibit stereoisomerism. Draw the two stereoisomers of alanine, CH3CH(NH2)CO2H. C C [1] (d) The amino acid alanine can be synthesised from 2‑chloropropanoic acid, CH3CHCl CO2H. (i) State the reagents and conditions and name the mechanism for this reaction. reagents and conditions …… name of mechanism …… [2] (ii) State and explain the relative acidities of trichloroethanoic acid, chloroethanoic acid and ethanoic acid. . …… . …… . …… . …… . …… [3] (e) Serine, HOCH2CH(NH2)CO2H, can react with alanine, CH3CH(NH2)CO2H, to form three different structural isomers, each with the molecular formula C6H12N2O4. Draw the structures of these three structural isomers. isomer 1 (C6H12N2O4) isomer 2 (C6H12N2O4) isomer 3 (C6H12N2O4) [3] [Total: 15]

Mark scheme

7(a)(i) A= leucine 1 B= glutamic acid both [1] 7(a)(ii) greater and more soluble in the solvent / mobile phase 1 OR greater and form more H-bonds with the solvent [1] 7(b)(i) H2NCH2CO2H + HCl → Cl–H3N+CH2CO2H [1] 2 H2NCH2CO2H + NaOH → H2NCH2CO2–Na+ + H2O [1] 7(b)(ii) H3N+CH2CO2– [1] 2 Proton is transferred from the CO2H group to the NH2 group [1] 7(c) CO2H CO2H 1 C C H H H2N NH2 CH3 H3C two non-superimposable mirror images for alanine drawn [1] 7(d)(i) NH3 (in ethanol) heat in a sealed tube [1] 2 nucleophilic substitution [1] 7(d)(ii) acidity of Cl3CCO2H > ClCH2CO2H > CH3CO2H [1] 3 any two of: Cl is electronegative / electron withdrawing group AND Cl3CCO2H has more / 3 Cl groups [1] weakens O-H bond so more likely to ionise / dissociate OR negative charge on anion is more stabilised OR charge / electron density on COO– decreases so anion is (more) stabilised [1] CH3 is electron donating so O-H bond is stronger so less likely to ionise in CH3CO2H OR CH3CO2H has no -I group so O-H bond is stronger and less likely to ionise [1] 7(e) O 3 H2N CO2H O NH2 HO H N CO2H H N CO2H H2N H2N O O OH One mark for each structure. [1] [1] [1]

Question 810 marks

Question

Compound R is shown. R CH3 HO Cl CH3 (a) State the systematic name of compound R. . …… [1] (b) (i) R is dissolved in CDCl 3 and analysed using carbon‑13 and proton NMR spectroscopy. ● Predict the number of peaks that are seen in the carbon-13 NMR spectrum of R. . …… ● Predict the number of peaks that are seen in the proton NMR spectrum of R. . …… [2] (ii) A separate sample of R is dissolved in D2O. The proton NMR spectrum of this solution shows one less peak than is obtained in CDCl 3. Explain why. . …… . …… [1] (c) Compound R reacts separately with the four reagents shown in the table. Complete the table by ● drawing the structures of the organic products formed, ● stating the type of reaction. reagent organic product structure type of reaction Na CH3COCl Br2(aq) + N N Cl – [6] [Total: 10]

Mark scheme

8(a) 4-chloro-3,5-dimethylphenol OR 3,5-dimethyl-4-chlorophenol [1] 1 ALLOW 2,6-dimethyl-4-hydroxychlorobenzene and 2-chloro-5-hydroxy-1,3-dimethylbenzene 8(b)(i) carbon-13 NMR = 5 peaks [1] 2 proton NMR = 3 peaks [1] 8(b)(ii) OH proton had disappeared due to proton exchange with D / D2O [1] 1 ALLOW OH + D2O → OD + HOD 8(c) reagent organic product structure type of reaction 6 redox Na Condensation CH3COCl (electrophilic) substitution Br2(aq) (electrophilic) substitution each structure [1] × 4 type of reaction y 9 y 9 [2]

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