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Chemistry 9701/44 October/November 2025: question paper and mark scheme

Cambridge AS & A Level Chemistry (9701), Paper 4 A Level Structured Questions, October/November 2025. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.

9701_w25_qp_449701_w25_ms_449701/44/O/N/25

Paper
Paper 4 A Level Structured Questions
Total marks
100
Time
2 hours
Questions
8
Code on the paper
9701/44/O/N/25
9701_w25_ms_44.pdf

9701/44 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 119 marks

Question

(a) Define a transition element. …… [1] (b) The 3d orbitals in an isolated gaseous Cu2+ ion are degenerate. (i) Define the term degenerate. …… [1] (ii) Complete the electronic configuration of Cu2+. 1s2 …… [1] (c) (i) State the colours of the aqueous solutions for the two copper(II) complex ions shown. • [Cu(NH3)4(H2O)2]2+(aq) …… • [CuCl 4]2–(aq) …… [1] (ii) Explain why aqueous complex ions of transition elements are usually coloured. …… [3] (d) (i) When an excess of NH3(aq) is added to a solution of [CuCl 4]2–(aq), [Cu(NH3)4(H2O)2]2+(aq) is formed. State the type of reaction. Complete the equation for this reaction. State symbols are not required. type of reaction …… equation [CuCl 4]2– + …… [Cu(NH3)4(H2O)2]2+ + …… [2] * 0000800000003 * DFD 3 (ii) The [Cu(NH3)4(H2O)2]2+ complex ion shows stereoisomerism. Complete the three‑dimensional diagrams in Fig. 1.1 to show the two different stereoisomers of [Cu(NH3)4(H2O)2]2+. isomer 1 isomer 2 Cu Cu Fig. 1.1 [2] (iii) Deduce which stereoisomer in (d)(ii) is polar. Explain your answer. polar isomer …… explanation …… [1] (e) The dianion P can act as a tridentate ligand. P – – O O N O H O (i) Suggest how P can form three dative covalent bonds. …… [1] (ii) 2 moles of dianion P, C4H5NO42–, react with 1 mole of aqueous cobalt(III) ions, [Co(H2O)6]3+ to form 1 mole of complex ion Q. Deduce the formula and charge of Q. …… [1] * 0000800000004 * DFD 4 (f) Table 1.1 shows values for the stability constants, Kstab, of some silver(I) complexes. Table 1.1 complex value of Kstab [Ag(CN)2]–(aq) 1.1 × 1018 [Ag(NH3)2]+(aq) 1.2 × 107 [Ag(S2O3)2]3–(aq) 2.9 × 1013 (i) Define the stability constant of a complex. …… [1] (ii) Use the information in Table 1.1 to identify the most stable silver(I) complex. Explain your answer. most stable …… explanation …… [1] * 0000800000005 * DFD 5 (g) Sodium sulfite, Na2SO3, is used as a food preservative. A 3.75 g sample of impure Na2SO3 is dissolved in distilled water and made up to 250 cm3 in a volumetric flask. 10.0 cm3 of this solution requires 18.70 cm3 of acidified 0.0150 mol dm–3 MnO4–(aq) to reach the end‑point. The equation for the reaction is shown. 2MnO4– + 5SO32– + 6H+ 2Mn2+ + 5SO42– + 3H2O Calculate the percentage by mass of Na2SO3 in the sample. percentage by mass of Na2SO3 = …… [3] [Total: 19] * 0000800000006 * DFD 6

Mark scheme

1(a) forms (one or more) stable ions / compounds / oxidation states 1 with incomplete / partially filled (3)d orbital(s) / d shell / d sub-shell 1(b)(i) (orbitals) are at the same energy 1 1(b)(ii) (1s2) 2s2 2p6 3s2 3p6 3d9 1 1(c)(i) [Cu(NH3)4(H2O)2]2+(aq) deep / dark / royal blue 1 AND [CuCl4]2–(aq) yellow 1(c)(ii) M1: d orbital(s) of different energy / d-d splitting occurs / d sub-shell splits 3 M2: electron(s) promoted / excited M3: light / wavelength / frequency / photon absorbed AND complementary colour seen 1(d)(i) M1: ligand exchange / replacement / substitution / displacement 2 M2: [CuCl4]2– + 4NH3 + 2H2O → [Cu(NH3)4(H2O)2]2+ + 4Cl– 1(d)(ii) M1: one correct 3D structure of [Cu(H2O)2(NH3)4]2+ 2 M2: correct 3D stereoisomer structure of isomer 1 1(d)(iii) cis isomer AND dipoles do not cancel 1 1(e)(i) (two) oxygen and (one) nitrogen (can donate) three lone pair(s) of electrons (to the metal ion) 1 1(e)(ii) [Co(C4H5NO4)2]– 1 1(f)(i) equilibrium constant for the formation of the complex (ion) 1 in a solvent / water / solution OR from its constituent ions or molecules 1(f)(ii) [Ag(CN)2]–(aq) AND largest (value of) Kstab 1 1(g) M1 M2: any two [1], all four [2] 3 • moles MnO4– = 0.015 ×18.70/1000 = 2.805 × 10–4 • moles SO32– = 2.805 × 10–4 × 5/2 = 7.0125 × 10–4 (in 10 cm3) • moles SO32– = 1.753 × 10–2 (in 250 cm3) • mass Na2SO3 = 1.753 × 10–2 × 126.1 = 2.21 g M3: % purity = 100 × 2.21 / 3.75 = 58.9 – 59.0 % min 2sf

Question 218 marks

Question

(a) The Group 2 sulfates and the Group 2 chromates show similar trends in solubility. Suggest the trend in the solubility of the Group 2 chromates down the group. Explain your answer. …… [4] (b) Silver(I) chromate, Ag2CrO4, is sparingly soluble in water. (i) Write an ionic equation to show the equilibrium between solid Ag2CrO4 and its aqueous solution. Include state symbols. …… [1] (ii) The value of the solubility product, Ksp, of Ag2CrO4 is 1.12 × 10–12 at 298 K. Calculate the equilibrium concentration of Ag+, in mol dm–3, in a saturated solution of Ag2CrO4 at 298 K. equilibrium concentration of Ag+ = …… mol dm–3 [3] * 0000800000007 * DFD 7 (c) The hydrogenchromate ion, HCrO4–, is a weak acid. The pKa of HCrO4– is 6.49. (i) Calculate the pH of a 0.0250 mol dm–3 HCrO4– solution. pH = …… [2] (ii) HCrO4– can show amphoteric behaviour. State the formula of: • the conjugate acid of HCrO4– …… • the conjugate base of HCrO4–. …… [1] * 0000800000008 * DFD 8 (d) Table 2.1 shows some energy changes. Table 2.1 energy change value / kJ mol–1 first ionisation energy of silver +731 second ionisation energy of silver +2074 first ionisation energy of sulfur +1000 second ionisation energy of sulfur +2251 first electron affinity of sulfur –200 second electron affinity of sulfur +532 enthalpy change of atomisation of sulfur +279 enthalpy change of formation of silver(I) sulfide, Ag2S(s) –33 lattice energy of silver(I) sulfide, Ag2S(s) –2677 (i) Define the term first electron affinity. …… [1] (ii) Explain why the value for the second electron affinity of sulfur is positive. …… [1] (iii) Construct an equation for the lattice energy of Ag2S. Include state symbols. …… [1] * 0000800000009 * DFD 9 (iv) Calculate the enthalpy change of atomisation, ΔHat, in kJ mol–1, of silver using relevant data from Table 2.1. It may be helpful to draw a labelled Born–Haber cycle. Show your working. ΔHat of silver = …… kJ mol–1 [3] (e) Suggest how the magnitude for the lattice energy of Ag2S(s) differs from the lattice energy of Cu2S(s). Explain your answer. …… [1] [Total: 18] * 0000800000010 * DFD 10

Mark scheme

2(a) M1: (solubility) decreases down the group 4 M2: ΔHlatt and ΔHhyd decrease / become less exothermic / less negative M3: ΔHhyd decreases / changes more / dominant factor / becomes less exothermic by a larger extent OR ΔHlatt decreases / changes less / changes slower M4: ΔHsol becomes less exothermic / less negative OR ΔHsol becomes (more) endothermic / (more) positive 2(b)(i) Ag2CrO4(s) ⇌ 2Ag+(aq) + CrO42–(aq) 1 2(b)(ii) M1: Ksp = [Ag+]2 [CrO42-] 3 M2: √(1.12× 10−12 ÷ 4) = 6.54 × 10–5 3 M3: [Ag+] = 2 × 6.54 × 10–5 = 1.31 × 10–4 (mol dm–3) min 2sf 2(c)(i) Ka = 10–6.49 = 3.236 × 10–7 2 [H+]2 = 8.08984 × 10–9 M1: [H+] = 8.99 × 10–5 M2: pH = –log(8.99 × 10–5) = 4.05 min 2sf 2(c)(ii) conjugate acid H2CrO4 1 AND conjugate base CrO42– 2(d)(i) (the energy change / released when) one mole of gaseous atoms become 1– ions / gain one mole of electrons 1 2(d)(ii) due to repulsion between negative ion and incoming / gained electron 1 2(d)(iii) 2Ag+(g) S2–(g) → Ag2S(s) 1 2(d)(iv) M1: selection of correct six numbers only: –33, 731, 279, –200, 532, –2677 3 M2: use of × 2 as only multiplier with Ag M3: correct signs and evaluation of data –33 = 2ΔHat + (2 × 731) + 279 + (–200) + 532 + (–2677) 2ΔHat = 571 (kJ mol –1) ΔHat = +285.5 (kJ mol –1) 2(e) Ag2S smaller / less negative / less exothermic lattice neergy AND larger ionic radius (of cation) 1

Question 39 marks

Question

(a) Define the term entropy. …… [1] (b) (i) Place one tick (3) in each row of Table 3.1 to show the sign of the entropy change, ΔS, for each process. Table 3.1 process ΔS is negative ΔS is positive steam condensing into water solid KCl dissolving in water [1] (ii) Chlorine trifluoride, Cl F3, decomposes on heating into its elements, as shown. reaction 1 2Cl F3(g) Cl 2(g) + 3F2(g) Standard entropies are shown in Table 3.2. Table 3.2 substance Cl F3(g) Cl 2(g) F2(g) S⦵ / J K–1 mol–1 +281.6 +223.1 +203.0 Calculate the standard entropy change, ΔS⦵, in J K–1 mol–1, for reaction 1. ΔS⦵ for reaction 1 = …… J K–1 mol–1 [2] (c) Group 2 carbonates decompose on heating. The decomposition for one of the Group 2 carbonates, MCO3, is shown in reaction 2. reaction 2 MCO3(s) MO(s) + CO2(g) (i) Predict the sign of the entropy change, ΔS, for reaction 2. Explain your answer. …… [1] * 0000800000011 * DFD 11 (ii) The Gibbs equation is shown. ΔG⦵ = ΔH ⦵ – TΔS⦵ Fig. 3.1 shows values of the Gibbs free energy change, ΔG⦵, in kJ mol–1, at different temperatures, T, in K, for reaction 2. Assume ΔH ⦵ and ΔS⦵ values for this reaction remain constant over this temperature range. 200 150 ⦵ ΔG / kJ mol–1 100 50 0 0 200 400 600 800 1000 1200 1400 1600 –50 –100 T/K Fig. 3.1 Use the gradient and intercept on the y‑axis in Fig. 3.1 and the Gibbs equation to determine: • ΔS⦵, in J K–1 mol–1, for reaction 2 • the minimum temperature, T, in K, at which the reaction is feasible • ΔH ⦵, in kJ mol–1, for reaction 2. ΔS⦵ for reaction 2 = …… J K–1 mol–1 minimum temperature, T = …… K ΔH ⦵ for reaction 2 = …… kJ mol–1 [4] [Total: 9] * 0000800000012 * DFD 12

Mark scheme

3(a) number of possible arrangements of particles and energy in a system 1 3(b)(i) 1 process ΔS is negative ΔS is positive steam condensing into water ✓ solid KCl dissolving into water ✓ 3(b)(ii) M1: three numbers and correct multipliers used 2 M2: correct signs and evaluation ΔSo = (3 × 203.0) + (223.1) – (2 × 281.6) = (+)268.9 (J K–1 mol–1) 3(c)(i) positive 1 AND gas produced / on the right OR less / no gas on left 3(c)(ii) M1 and M2: calculation of gradient from the graph for ΔS 4 gradient = (–)140 / 900 = (–)0.1556 ΔSo = (–)0.1556 × 1000 = (+)155.6 ± 5 (J K–1 mol–1) M3: T = 1120 ± 5 (K) M4: ΔHo = (+)172 ± 10 (kJ mol–1)

Question 412 marks

Question

(a) Nitrogen monoxide, NO, reacts with hydrogen, as shown in reaction 3. reaction 3 2NO + 2H2 N2 + 2H2O (i) The rate equation for reaction 3 is shown. rate = k[H2][NO]2 Complete Table 4.1. Table 4.1 the order of reaction with respect to [H2] the order of reaction with respect to [NO] the overall order of the reaction [1] (ii) Predict how the initial rate for reaction 3 changes when the concentration of NO is halved. …… [1] (iii) Predict how the initial rate for reaction 3 changes when the concentrations of NO and H2 are both increased three times. …… [1] (iv) Suggest why reaction 3 is unlikely to proceed by a mechanism involving only a single step. …… [1] (v) Suggest equations for the three steps of the reaction mechanism for reaction 3. Each step involves a reaction between two molecules. step 1 …… …… step 2 …… + …… N2O + …… step 3 N2O + …… …… + …… [2] (vi) Suggest the role of N2O in this mechanism. Explain your reasoning. …… [1] * 0000800000013 * DFD 13 (b) Iodine, I2, reacts with thiosulfate ions, S2O32–, as shown in reaction 4. reaction 4 2S2O32– + I2 S4O62– + 2I– Reaction 4 is carried out in the presence of a large excess of I2. Under these conditions, the reaction is first order with respect to [S2O32–] and zero order with respect to [I2]. The half‑life, t1 , for reaction 4 is 720 s under certain conditions. 2 Calculate the value of the rate constant, k, for reaction 4. Include the units of k. k = …… units …… [1] (c) The reaction between iodide ions, I–(aq), and peroxydisulfate ions, S2O82–(aq), is catalysed by Co3+(aq). The mechanism is similar to the mechanism of this reaction when Fe3+(aq) is used as the catalyst. (i) State the type of catalysis that occurs in this reaction. Explain your reasoning. …… [1] (ii) Write two equations to show how Co3+(aq) catalyses this reaction. equation 1 …… equation 2 …… [2] (iii) Suggest why this reaction is slow in the absence of Co3+(aq). …… [1] [Total: 12] * 0000800000014 * DFD 14

Mark scheme

4(a)(i) 1 the order of reaction with respect to [H 2] 1 the order of reaction with respect to [NO] 2 overall order of the reaction 3 4(a)(ii) 1 1 rate × 4 4(a)(iii) rate × 27 1 4(a)(iv) a four-particle collision is unlikely 1 4(a)(v) step 1 2NO → N2O2 2 step 2 N2O2 + H2 → N2O + H2O step 3 N2O + H2 → N2 + H2O 4(a)(vi) intermediate formed (step 2) then used up later (step 3) 1 4(b) k = 0.693 / 720 = 9.625 × 10–4 min 2sf 1 AND units: s–1 4(c)(i) homogeneous 1 AND Co3+ / catalyst / it is in the same phase / same state as the reactants 4(c)(ii) M1: (equation 1) 2Co3+ + 2I– → 2Co2+ + I2 2 M2: (equation 2) 2Co2+ + S2O82– → 2Co3+ + 2SO42– 4(c)(iii) repulsion of two negative / same charge ions slows the reaction / raises EA 1

Question 513 marks

Question

(a) Describe and explain the shape of benzene. In your answer, include: • the shape and bond angle in the ring • the hybridisation of the carbon atoms • how orbital overlap forms σ and π bonds between the carbon atoms in the ring. …… [4] (b) Fig. 5.1 shows two reactions of benzoic acid. COOH COOH COOH reaction 5 reaction 6 O Fig. 5.1 (i) Suggest reagents and conditions for reaction 5 and for reaction 6 in Fig. 5.1. reaction 5 …… reaction 6 …… [2] (ii) State the type of reaction for reaction 5 in Fig. 5.1. …… [1] * 0000800000015 * DFD 15 (c) In the electrophilic substitution of arenes, different substituents can direct to different ring positions. (i) Describe the directing effect of the –CH2CH3 group. Explain your answer. …… [1] (ii) The alkylation of arenes uses a mixture of CH3CH2Br and FeBr3 to generate the CH3CH2+ electrophile. Write an equation for the formation of the CH3CH2+ electrophile. …… [1] (iii) Complete the mechanism in Fig. 5.2. Include all relevant curly arrows and charges. Draw the structure of the organic intermediate. COOH organic intermediate COOH + …… +CH CH 2 3 Fig. 5.2 [3] (iv) Write an equation to show how FeBr3 is regenerated after the reaction in Fig. 5.2. …… [1] [Total: 13] * 0000800000016 * DFD 16

Mark scheme

5(a) M1 and M2: any two [1], all three [2] 4 • (hexagonal ring) planar / (trigonal) planar • 120o • sp2 hybridised M3: p orbitals overlap sideways / laterally (with each other above and below the ring) forming π bonds M4: σ bonds form when sp 2 / hybridised orbitals overlap end-on-end / head on 5(b)(i) M1: (reaction 5) Pt / Ni + H2 (+ heat) 2 M2: (reaction 6) (CH3)2CHCOCl + Al Cl 3 (+ heat) 5(b)(ii) reduction / hydrogenation 1 5(c)(i) (CH2CH3 group) directs to 2-, 4- (and 6-) positions AND due to being an electron donating group 1 5(c)(ii) CH3CH2Br + FeBr3 → CH3CH2+ + FeBr4– 1 5(c)(iii) 3 organic intermediate M1: curly arrow from inside / touching the hexagon to (CH2) group / C+ charge M2: structure of the intermediate M3: curly arrow from C-H bond into / onto the ring AND formation / loss of H+ 5(c)(iv) H+ + FeBr4– → HBr + FeBr3 1

Question 69 marks

Question

(a) Compound Z is used in organic synthesis. Z O NH2 CN H 2N Complete Table 6.1 to show the number of sp, sp2 and sp3 hybridised carbon atoms present in one molecule of Z. Table 6.1 type of hybridisation sp sp2 sp3 number of carbon atoms [1] (b) Z can undergo different reactions, as shown in Fig. 6.1. excess HCl (aq) heat reaction 7 Z O NH2 CH3COCl CN reaction 8 H 2N excess LiAl H4 reaction 9 Fig. 6.1 (i) Name the two types of reaction occurring in reaction 7 in Fig. 6.1. …… and …… [1] (ii) Draw the structures of the organic products of reactions 7, 8 and 9 in Fig. 6.1. [4] * 0000800000017 * DFD 17 (c) Compound Z is dissolved in D2O and analysed by carbon‑13 NMR and proton (1H) NMR spectroscopy. (i) Predict the number of peaks in the carbon‑13 NMR spectrum of Z. …… [1] (ii) The proton (1H) NMR spectrum of Z in D2O gives three peaks for the proton environments, labelled a, b and c, as shown on Fig. 6.2. Z O NH2 C CH CN H 2N CH2 CH2 a b c Fig. 6.2 Complete Table 6.2 for the proton (1H) NMR spectrum of Z in D2O. Table 6.2 proton environment a b c name of splitting pattern chemical shift range, δ / ppm [2] Table 6.3 chemical shift environment of proton example range, δ / ppm alkane –CH3, –CH2–, >CH– 0.9–1.7 CH3–C=O, –CH2–C=O, alkyl next to C=O 2.2–3.0 >CH–C=O alkyl next to nitrile –CH2–CN 2.0–3.0 alkyl next to electronegative atom CH3–O, –CH2–O, –CH2–N 3.2– 4.0 attached to alkene =CHR 4.5–6.0 alkyl amine R–NH– 1.0–5.0 amide RCONHR 5.0–12.0 [Total: 9] * 0000800000018 * DFD 18

Mark scheme

6(a) sp = 1 AND sp2 = 1 AND sp3 = 3 1 6(b)(i) acid-base / neutralisation AND hydrolysis 1 6(b)(ii) * * 4 – – * reaction 7 * – * – – * – reaction 8 reaction 9 * * – – any two * [1], any three * [2], any five * [3], all six * [4] 6(c)(i) five / 5 1 6(c)(ii) 2 proton environment a b c name of splitting pattern doublet multiplet doublet chemical shift range, 2.2–3.0 3.2– 4.0 2.0–3.0 δ / ppm any three [1], all six [2]

Question 79 marks

Question

Phenylmethanol and 4‑methylphenol are isomers. phenylmethanol 4-methylphenol CH2OH OH (a) Complete Table 7.1 to show the relative acidities of benzoic acid (C6H5COOH), phenylmethanol, 4‑methylphenol and water. Explain your answer. Table 7.1 name of compound most acidic least acidic explanation …… [4] (b) 4‑methylphenol reacts readily with sodium. Complete the equation for this reaction. OH + …… [1] * 0000800000019 * DFD 19 (c) Under certain conditions, ethane‑1,2‑diol, HOCH2CH2OH, reacts with propane‑1,3‑dioic acid, HOOCCH2COOH, to form different organic products, as shown in Fig. 7.1. X, C5H6O4 reaction 10 O O HO + OH HO OH reaction 11 polymer Y Fig. 7.1 (i) X does not react with Na metal. Draw the structure of the organic product X, C5H6O4, shown in Fig. 7.1. [1] (ii) Reactions 10 and 11 in Fig. 7.1 are different types of reaction. Name the type of reaction for reaction 10 and for reaction 11. reaction 10 …… reaction 11 …… [1] (iii) Draw a section of polymer Y showing only one repeat unit. The new functional group formed should be displayed. [2] [Total: 9] * 0000800000020 * DFD 20

Mark scheme

7(a) M1: (most acidic) benzoic acid 4 4-methylphenol water (least acidic) phenylmethanol M2: correct link of acidity to weakens O—H / H+ more easily lost / anion stabilised / conjugate base stabilised M3 and M4: any two [2] • for benzoic acid: negative inductive effect of C=O / electronegative C=O / two electronegative O / –ve charge delocalised (across two O) / resonance in –COO– • for 4-methylphenol: as lone pair on oxygen delocalised into the ring / π system / delocalised system OR p orbital on oxygen overlaps with ring / π system / delocalised system • for phenylmethanol: positive inductive effect of alkyl group / R / CH2 7(b) 1 7(c)(i) 1 7(c)(ii) (reaction 10) addition–elimination / condensation / esterification 1 AND (reaction 11) addition–elimination / condensation (polymerisation) / esterification 7(c)(iii) 2 M1: ester linkage shown M2: rest of the structure correct with continuation bonds

Question 811 marks

Question

(a) Describe the difference in reactivity between ethanoyl chloride and chlorobenzene with water. Explain your answer. …… [2] (b) The structure of compound V is shown. V CN O O N O H O (i) Name all the functional groups in V. …… [2] (ii) Deduce the number of possible optical isomers for V. …… [1] (iii) Suggest one reason, other than better biological activity and lower dosage required, why it is beneficial to synthesise a single optical isomer of V for use as a drug. …… [1] * 0000800000021 * DFD 21 (c) A sample of V is hydrolysed with an excess of hot aqueous alkali. The products are isolated from the reaction mixture at pH 12. Draw the structures of the two organic products of the complete alkaline hydrolysis of V in Fig. 8.1. Fig. 8.1 [3] * 0000800000022 * DFD 22 (d) A polypeptide formed from four amino acids, A, B, C and D, is completely hydrolysed and then analysed by gas–liquid chromatography. The chromatogram produced is shown in Fig. 8.2. 58 B 42 intensity 28 D A 13 C time / minutes Fig. 8.2 The number above each peak represents the area under the peak. The area under each peak is proportional to the mass of the respective amino acid in the mixture. (i) Calculate the percentage by mass of amino acid A in the original mixture. percentage by mass of amino acid A = …… [1] (ii) The retention time for amino acid D is the longest. Explain why D has a longer retention time than the other amino acids A, B and C. …… [1] [Total: 11] * 0000800000023 * DFD 23 Important values, constants and standards molar gas constant R = 8.31 J K–1 mol–1 Faraday constant F = 9.65 × 104 C mol–1 Avogadro constant L = 6.02 × 1023 mol–1 electronic charge e = –1.60 × 10–19 C molar volume of gas Vm = 22.4 dm3 mol–1 at s.t.p. (101 kPa and 273 K) Vm = 24.0 dm3 mol–1 at room conditions ionic product of water Kw = 1.00 × 10–14 mol2 dm–6 (at 298 K (25 °C)) specific heat capacity of water c = 4.18 kJ kg–1 K–1 (4.18 J g–1 K–1) The Periodic Table of Elements Group 1 2 13 14 15 16 17 18 1 2 H He hydrogen helium Key 1.0 4.0 3 4 atomic number 5 6 7 8 9 10 Li Be atomic symbol B C N O F Ne lithium beryllium name boron carbon nitrogen oxygen fluorine neon 6.9 9.0 relative atomic mass 10.8 12.0 14.0 16.0 19.0 20.2 11 12 13 14 15 16 17 18 Na Mg Al Si P S Cl Ar sodium magnesium aluminium silicon phosphorus sulfur chlorine argon DFD 23.0 24.3 3 4 5 6 7 8 9 10 11 12 27.0 28.1 31.0 32.1 35.5 39.9 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 at www.cambridgeinternational.org after the live examination series. K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr potassium calcium scandium titanium vanadium chromium manganese iron cobalt nickel copper zinc gallium germanium arsenic selenium bromine krypton 39.1 40.1 45.0 47.9 50.9 52.0 54.9 55.8 58.9 58.7 63.5 65.4 69.7 72.6 74.9 79.0 79.9 83.8 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 24 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe rubidium strontium yttrium zirconium niobium molybdenum technetium ruthenium rhodium palladium silver cadmium indium tin antimony tellurium iodine xenon 85.5 87.6 88.9 91.2 92.9 95.9 – 101.1 102.9 106.4 107.9 112.4 114.8 118.7 121.8 127.6 126.9 131.3 55 56 57–71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 lanthanoids Cs Ba Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn caesium barium hafnium tantalum tungsten rhenium osmium iridium platinum gold mercury thallium lead bismuth polonium astatine radon 132.9 137.3 178.5 180.9 183.8 186.2 190.2 192.2 195.1 197.0 200.6 204.4 207.2 209.0 – – – 87 88 89–103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 actinoids Fr Ra Rf Db Sg Bh Hs Mt Ds Rg Cn Nh Fl Mc Lv Ts Og francium radium rutherfordium dubnium seaborgium bohrium hassium meitnerium darmstadtium roentgenium copernicium nihonium flerovium moscovium livermorium tennessine oganesson – – – – – – – – – – – – – – – – – 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 lanthanoids La Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu lanthanum cerium praseodymium neodymium promethium samarium europium gadolinium terbium dysprosium holmium erbium thulium ytterbium lutetium 138.9 140.1 140.9 144.2 – 150.4 152.0 157.3 158.9 162.5 164.9 167.3 168.9 173.1 175.0 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 actinoids Ac Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr actinium thorium protactinium uranium neptunium plutonium americium curium berkelium californium einsteinium fermium mendelevium nobelium lawrencium – 232.0 231.0 238.0 – – – – – – – – – – – Assessment International Education Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download To avoid the issue of disclosure of answer‑related information to candidates, all copyright acknowledgements are reproduced online in the Cambridge

Mark scheme

8(a) • chlorobenzene is less reactive than ethanoyl chloride 2 • p orbital / lone pair on Cl / –Cl will overlap / delocalise into the ring OR C of COCl has most electron deficient OR has an electronegative oxygen atom / two electronegative atoms / electron withdrawing C=O group • due to partial double C–Cl bond OR C–Cl bond strengthened (more) / stronger bond between Cl and benzene / ring OR C–Cl weakened in ROCl any two [1], all three [2] 8(b)(i) • nitrile 2 • amide • ketone / carbonyl • ester any two [1], all four [2] 8(b)(ii) 8 / eight 1 8(b)(iii) no need to separate optical isomers 1 8(c) 3 M1: amide bond hydrolysed and COO – and NH2 formed M2: nitrile bond hydrolysed and COO – formed M3: ester bond hydrolysed and COO – and OH formed 8(d)(i) percentage = 100 × 28 / (28 + 58 + 13 + 42) = 19.9 (%) min 2sf 1 8(d)(ii) D is attracted more strongly to the stationary phase / forms stronger intermolecular forces with the stationary phase 1

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