Cambridge A Level Chemistry 9701 — 2022 Feb/March Paper 2 · Variant 2

1 Fig. 1.1 shows how first ionisation energies vary across Period 2. A ionisation energy Li Be B C N O F Ne element Fig. 1.1 (a) Construct an equation to represent the first ionisation energy of oxygen. Include state symbols. .............................................................................................................................................. [1] (b) (i) State and explain the general trend in first ionisation energies across Period 2. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ....................................................................................................................................... [3] (ii) Explain why ionisation energy A in Fig. 1.1 does not follow the general trend in first ionisation energies across Period 2. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ....................................................................................................................................... [2] (c) Element E is in Period 3 of the Periodic Table. The first eight ionisation energy values of E are shown in Table 1.1. Table 1.1 ionisation 1st 2nd 3rd 4th 5th 6th 7th 8th ionisation energy / kJ mol–1 577 1820 2740 11 600 14 800 18 400 23 400 27 500 Deduce the full electronic configuration of E. Explain your answer. full electronic configuration of E = .............................................................................................. explanation ................................................................................................................................. .................................................................................................................................................... .................................................................................................................................................... [3] [Total: 9]

Mark scheme: 1(a) O(g) → O+(g) + e– 1 1(b)(i) increase across period AND increased nuclear attraction for (valence / outer) electrons [1] increase in (positive) nuclear charge / number of protons (in the nucleus) [1] similar shielding (of outer electrons) [1] 3 1(b)(ii) spin-pair repulsion (of electrons) in (2)p orbital [1] outweighs increased nuclear charge [1] 2 1(c) 1s2 2s2 2p6 3s2 3p1 [1] greatest jump between 3rd and 4th ionisations [1] indicates three electrons in outer shell [1] 3

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2 Some oxides of elements in Period 3 are shown. Na2O Al 2O3 P4O6 P4O10 SO2 SO3 (a) Na reacts with O2 to form Na2O. Na is the reducing agent in this reaction. (i) Define reducing agent. ............................................................................................................................................. ....................................................................................................................................... [1] (ii) Write an equation for the reaction of Na2O with water. ....................................................................................................................................... [1] (b) Al 2O3 is an amphoteric oxide found in bauxite. (i) State what is meant by amphoteric. ............................................................................................................................................. ....................................................................................................................................... [1] (ii) Al 2O3 is purified from bauxite in several steps. The first step involves heating Al 2O3 with an excess of NaOH(aq). A colourless solution forms. Write an equation for this reaction. ....................................................................................................................................... [1] (iii) Al 2O3 is used as a catalyst in the dehydration of alcohols. State the effect of using Al 2O3 as a catalyst in the dehydration of alcohols. Use the Boltzmann distribution in Fig. 2.1 to help explain your answer. number of molecules energy Fig. 2.1 ............................................................................................................................................. ............................................................................................................................................. ....................................................................................................................................... [3] (c) P4O6 is a white solid that has a melting point of 24 °C. Solid P4O6 reacts with water to form H3PO3. (i) Deduce the type of structure and bonding shown by P4O6. Explain your answer. ............................................................................................................................................. ............................................................................................................................................. ....................................................................................................................................... [2] (ii) Determine the oxidation number of P in H3PO3. ....................................................................................................................................... [1] (iii) When P4O6(s) is heated with oxygen it forms P4O10(s). P4O6(s) + 2O2(g) → P4O10(s) ∆Hr = –1372 kJ mol–1 The enthalpy change of formation, ∆Hf, of P4O10(s) is –3012 kJ mol–1. Calculate the enthalpy change of formation, ∆Hf, of P4O6(s). ∆Hf of P4O6(s) = .............................. kJ mol–1 [1] (iv) Write an equation for the reaction of P4O10 with water. ....................................................................................................................................... [1] (d) SO2 and SO3 are found in the atmosphere. The oxidation of SO2 to SO3 in the atmosphere is catalysed by NO2. The first step of the catalytic oxidation is shown in equation 1. equation 1 SO2(g) + NO2(g) SO3(g) + NO(g) (i) Construct an equation to show how NO2 is regenerated in the catalytic oxidation of SO2. ....................................................................................................................................... [1] (ii) NO2 can also react with unburned hydrocarbons to form photochemical smog. State the product of this reaction that contributes to photochemical smog. ....................................................................................................................................... [1] (iii) Fig. 2.2 shows how the temperature of the atmosphere varies with height from the ground. 110 100 90 80 70 60 height / km 50 40 30 20 10 0 –120 –100 –80 –60 –40 –20 0 20 temperature / C Fig. 2.2 The equilibrium reaction in equation 1 has ∆Hr = –168 kJ mol–1. Suggest how the position of this equilibrium differs at a height of 20 km compared with a height of 50 km from the ground. Explain your answer. ............................................................................................................................................. ............................................................................................................................................. ....................................................................................................................................... [2] [Total: 16]

Mark scheme: 2(a)(i) species that donates electrons 1 2(a)(ii) Na2O + H2O → 2NaOH 1 2(b)(i) reacts with both acids and bases / shows both acidic and basic behaviour 1 2(b)(ii) Al2O3 + 2NaOH + 3H2O → 2NaAl(OH)4 1 2(b)(iii) two lines shown on diagram, e.g. EA and EA,cat [1] greater proportion of molecules with E ⩾ EA [1] frequency of effective collisions increases [1] 3 Question Answer Marks 2(c)(i) (structure =) simple/molecular, because it has a low melting/boiling point [1] (bonding =) covalent, because it is hydrolysed [1] 2 2(c)(ii) (+)3 / III 1 2(c)(iii) –1640 (kJ mol–1) 1 2(c)(iv) P4O10 + 6H2O → 4H3PO4 1 2(d)(i) NO + ½O2 → NO2 1 2(d)(ii) peroxyac(et)ylnitrate / PAN 1 2(d)(iii) position of equilibrium moves / farther to right (at 20 km) [1] (forward) reaction is exothermic AND temperature colder at 20 km (cf. 50 km) [1] 2

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3 The hydrogen halides HCl, HBr and HI are all colourless gases at room temperature. (a) The hydrogen halides can be formed by reacting the halogens with hydrogen. Describe and explain the relative reactivity of the halogens down the group when they react with hydrogen to form HCl, HBr and HI. .................................................................................................................................................... .................................................................................................................................................... .............................................................................................................................................. [2] (b) HCl is a product of several different reactions. Some of these are shown in Fig. 3.1. SiCl 4 NaCl reaction 1 reaction 2 H2O concentrated H2SO4 HCl Fig. 3.1 (i) Write an equation for reaction 1. ....................................................................................................................................... [1] (ii) In reaction 2, NaCl reacts with concentrated H2SO4 to form HCl and NaHSO4. When NaBr reacts with concentrated H2SO4, the products include Br2 and SO2. Identify the type(s) of reaction that occur in each case by completing Table 3.1. Explain the difference in these reactions. Table 3.1 reactants type(s) of reaction NaCl and concentrated H2SO4 NaBr and concentrated H2SO4 explanation .......................................................................................................................... ............................................................................................................................................. [3] (c) When heated with a Bunsen burner, HCl does not decompose, whereas HI forms H2 and I2. Explain the difference in the effect of heating on HCl and HI. .................................................................................................................................................... .............................................................................................................................................. [1] (d) The hydrogen halides dissolve in water to form strong Brønsted–Lowry acids. The concentration of a strong acid can be determined by titration. (i) State what is meant by strong Brønsted–Lowry acid. ............................................................................................................................................. ............................................................................................................................................. ....................................................................................................................................... [2] (ii) On Fig. 3.2, sketch the pH titration curves produced when: ● 0.1 mol dm–3 NaOH(aq) is added to 25 cm3 of 0.1 mol dm–3 HBr(aq), to excess ● 0.1 mol dm–3 NH3(aq) is added to 25 cm3 of 0.1 mol dm–3 HBr(aq), to excess. reaction of NaOH(aq) and HBr(aq) reaction of NH3(aq) and HBr(aq) 14 14 12 12 10 10 8 8 pH pH 6 6 4 4 2 2 0 0 0 5 10 15 20 25 30 35 40 45 0 5 10 15 20 25 30 35 40 45 volume of NaOH / cm3 volume of NH3 / cm3 Fig. 3.2 [3] (e) HBr reacts with propene to form two bromoalkanes, CH3CH2CH2Br and (CH3)2CHBr. (i) Complete the diagram to show the mechanism of the reaction of HBr and propene to form the major organic product. Include charges, dipoles, lone pairs of electrons and curly arrows, as appropriate. Draw the structures of the intermediate and the major organic product. H H H C C H C H H H Br [4] (ii) Explain why the two bromoalkanes are not produced in equal amounts by this reaction. ............................................................................................................................................. ............................................................................................................................................. ....................................................................................................................................... [2] (iii) The reaction of CH3CH2CH2Br and NaOH is different depending on whether water or ethanol is used as a solvent. Complete Table 3.2 to identify the organic and inorganic products of the reaction of CH3CH2CH2Br and NaOH in each solvent. Table 3.2 solvent organic product(s) inorganic product(s) water ethanol [2] [Total: 20]

Mark scheme: 3(a) M1: reaction less vigorous (down the group) M2: Any two of the following for one mark: • electronegativity decreases • less attractive to e– addition • weaker oxidising agent • greater nuclear charge outweighing increased shielding (ENC argument) 2 3(b)(i) SiCl4 + 2H2O → SiO2 + 4HCl 1 3(b)(ii) M1: All three correct for two marks: row 1 • acid–base row 2 • acid–base • redox M2: explanation H2SO4 is strong enough to oxidise / is an oxidising agent with NaBr / HBr / bromide 3 Question Answer Marks 3(c) H—Cl bond is stronger than H—I / BDE decreases down the group 1 3(d)(i) proton / H+ donor [1] fully dissociates (in aqueous solution / water / solvent) [1] 2 3(d)(ii) M1: correct sigmoid shape with vertical section at 25 cm3 for both M2: both curves show initial pH ⩽ 2 M3: (with NaOH) heading to pH ⩾ 12 (with NH3) heading to pH 8–12 3 3(e)(i) M1: curly arrow from C=C to H of HBr M2: correct dipole (δ+H—Brδ–) AND curly arrow from H—Br to Br M3: curly arrow from lone pair on :Br– to carbocation M4: correct intermediate AND product (2-bromopropane) 4 Question Answer Marks 3(e)(ii) 2° carbocation more stable (than 1°) (or reverse argument) [1] greater positive inductive effect of two methyl groups (cf. one ethyl) (or reverse argument) [1] 2 3(e)(iii) M1: (water solvent) CH3CH2CH2OH AND NaBr / sodium bromide / propan-1-ol M2: (ethanol solvent) CH3CHCH2 AND H2O / water / propene AND NaBr / sodium bromide 2

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4 Compounds J and K are found in plant oils. J K O OH H COOH C C CHO C C H3C CH3 H CH3 Fig. 4.1 (a) (i) Complete Table 4.1 to state what you would observe when J reacts with the reagents listed. Table 4.1 reagent observation with J 2,4-dinitrophenylhydrazine (2,4-DNPH) Tollens’ reagent sodium metal [3] (ii) J has two optical isomers. Draw the three-dimensional structures of the two optical isomers of J. [2] (b) K is used to make the addition polymer Perspex®. A synthesis of Perspex® is shown in Fig. 4.2. K M H COOH H COOCH3 reactant L C C C C Perspex® reaction 1 reaction 2 H CH3 H CH3 Fig. 4.2 (i) Identify L. State the conditions required for reaction 1. L = ....................................................................................................................................... conditions = ......................................................................................................................... [2] (ii) Draw one repeat unit of the addition polymer Perspex®. [2] (iii) Use information from Table 4.2 to suggest how the infrared spectra of M and Perspex® would differ. Explain your answer. ............................................................................................................................................. ....................................................................................................................................... [1] Table 4.2 bond functional group containing the bond characteristic infrared absorption range (in wavenumbers) / cm–1 C–O hydroxy, ester 1040–1300 C=C aromatic compound, alkene 1500–1680 C=O amide 1640–1690 carbonyl, carboxyl 1670–1740 ester 1710–1750 C≡N nitrile 2200–2250 C–H alkane 2850–3100 N–H amine, amide 3300–3500 O–H carboxyl 2500–3000 hydroxy 3200–3650 (iv) K can be made from propanone in the three-step synthesis shown in Fig. 4.3. propanone K O OH OH H COOH step 1 step 2 step 3 C H3C C C N H3C C COOH C C H3C CH3 H CH3 CH3 CH3 Fig. 4.3 Complete Table 4.3 to identify the reagent(s) used and the type of reaction in each step. Table 4.3 step reagent(s) type of reaction 1 2 3 Al 2O3 [5] [Total: 15]

Mark scheme: 4(a)(i) red / orange / yellow precipitate / ppt / solid [1] silver mirror / silver / grey solid / precipitate / ppt [1] effervescence / bubbling / fizzing [1] 3 4(a)(ii) 2 4(b)(i) L = CH3OH / methanol [1] conditions = acid(ic) / H+ / H2SO4 AND (heat under) reflux [1] 2 Question Answer Marks 4(b)(ii) carbon backbone with ‘dangling’ bonds [1] rest of structure correct [1] 2 4(b)(iii) Perspex® would not have absorption 1500–1680 cm–1 AND Perspex® does not have C=C 1 4(b)(iv) step 1 KCN / HCN OR NaCN / H2SO4 [1] addition [1] step 2 H+ / H2SO4(aq) [1] hydrolysis / substitution [1] step 3 elimination / dehydration [1] 5

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