2023 CCEA A-Level Chemistry 1110 模拟试题及答案详解

Lattice enthalpy of formation is defined as the enthalpy change when one mole of a solid ionic lattice is formed from its constituent ions in the gaseous state, under standard conditions.

1 mark: B. No credit for A, C or D, which describe different standard enthalpy definitions.

\( \Delta G = \Delta H - T\Delta S \). If \( \Delta H \) is negative and \( \Delta S \) is positive, both terms make \( \Delta G \) more negative regardless of the value of T, so the reaction is feasible at every temperature.

1 mark: B.

Doubling [NO] quadruples the rate, so the order with respect to NO is 2. Doubling \( [O_2] \) doubles the rate, so the order with respect to \( O_2 \) is 1. Overall order \( = 2 + 1 = 3 \).

1 mark: C.

\( K_c \) is products over reactants, each raised to the power of its stoichiometric coefficient: \( K_c = \dfrac{[NO_2]^2}{[N_2O_4]} \).

1 mark: B.

\( [H^+] = \sqrt{K_a c} = \sqrt{1.8 \times 10^{-5} \times 0.100} = \sqrt{1.8 \times 10^{-6}} = 1.34 \times 10^{-3} \text{ mol dm}^{-3} \). \( pH = -\log(1.34\times10^{-3}) = 2.87 \).

1 mark: A.

Restricted rotation about a C=C double bond, combined with each carbon of the double bond carrying two different groups, gives rise to E/Z (geometric) isomerism.

1 mark: B.

Fehling's solution contains \( Cu^{2+} \) ions in alkaline solution. Aldehydes reduce \( Cu^{2+} \) to a brick-red precipitate of \( Cu_2O \); ketones cannot be oxidised in this way and give no reaction.

1 mark: B.

Acyl chlorides are highly reactive towards nucleophiles because Cl is a good leaving group and strongly withdraws electron density from the carbonyl carbon. They hydrolyse rapidly, even violently, in cold water, unlike amides, esters or nitriles.

1 mark: C.

\( AlBr_3 \) polarises the \( Br-Br \) bond and accepts a bromide ion: \( Br_2 + AlBr_3 \rightarrow Br^+ + AlBr_4^- \). The electron-deficient \( Br^+ \) is the electrophile attacked by the benzene ring.

1 mark: B.

The nitro group withdraws electron density from the ring by both induction and resonance, deactivating the ring towards further electrophilic substitution and directing incoming electrophiles to the meta position, where the destabilising effect on the intermediate cation is smallest.

1 mark: C.

Lattice enthalpy of formation is the enthalpy change when one mole of a solid ionic compound is formed from its constituent ions in the gaseous state, under standard conditions. Bringing together oppositely charged gaseous ions to form a solid lattice releases energy because of the strong electrostatic (ionic) attraction between them, so the process is always exothermic.

1 mark: correct definition (1 mol solid formed from gaseous ions); 1 mark: reference to standard conditions/gaseous ions correctly stated; 1 mark: exothermic explained by electrostatic attraction between oppositely charged ions releasing energy. [3]

MgO has a more exothermic lattice enthalpy of formation than NaCl. \( Mg^{2+} \) and \( O^{2-} \) carry greater ionic charge than \( Na^+ \) and \( Cl^- \), and the ions are smaller, so the electrostatic attraction between the ions is much stronger, releasing more energy.

1 mark: more exothermic stated; 1 mark: valid reason given (greater ionic charge and/or smaller ionic radii in MgO increasing electrostatic attraction). [2]

One mole of gas is produced from none, so \( \Delta S \) is positive (increase in disorder). Since \( \Delta G = \Delta H - T\Delta S \), at low temperatures the \( T\Delta S \) term is small, so \( \Delta G \) remains positive (not feasible). As T increases, \( T\Delta S \) increases and eventually exceeds \( \Delta H \), making \( \Delta G \) negative, so the reaction becomes feasible above a threshold temperature.

1 mark: \( \Delta S \) is positive, correctly justified by gas formation; 1 mark: \( \Delta G = \Delta H - T\Delta S \) used/quoted correctly; 1 mark: correct explanation that \( T\Delta S \) must exceed \( \Delta H \) for \( \Delta G \) to become negative. [3]

\( \Delta S \) is negative. Three moles of gas react to form two moles of gas, so the number of particles (and disorder) of the system decreases.

1 mark: negative stated; 1 mark: correct justification (moles of gas decrease from 3 to 2). [2]

New rate \( \propto (\tfrac{1}{2}[P]) \times (3[Q])^2 = \tfrac{1}{2} \times 9 \times [P][Q]^2 = 4.5 \times [P][Q]^2 \). The rate therefore increases by a factor of 4.5.

1 mark: effect of halving [P] correctly identified (×0.5); 1 mark: effect of tripling [Q] correctly identified (×9, since squared); 1 mark: combined factor correctly calculated as ×4.5. [3]

Zero order with respect to X means that changing the concentration of X has no effect on the rate of reaction. For an overall zero order reaction, rate = k, so k has the same units as rate: \( \text{mol dm}^{-3}\text{s}^{-1} \).

1 mark: correct statement that rate is independent of [X]; 1 mark: correct units \( \text{mol dm}^{-3}\text{s}^{-1} \). [2]

There are 2 moles of gas on the reactant side (\( H_2 + I_2 \)) and 2 moles of gas on the product side (2HI). Increasing pressure favours the side with fewer gas moles, but since both sides have equal moles of gas, an increase in pressure affects the forward and reverse reactions equally, so the position of equilibrium does not shift.

1 mark: correct identification that both sides have 2 mol of gas; 1 mark: statement that increased pressure favours the side with fewer gas moles (Le Chatelier); 1 mark: correct conclusion that no shift occurs because the moles of gas are equal. [3]

A buffer solution resists changes in pH when small amounts of acid or base are added. The mixture contains a large reservoir of \( CH_3COO^- \) ions (from the salt) and undissociated \( CH_3COOH \). On addition of acid, most of the added \( H^+ \) ions are removed by reaction with the ethanoate ions: \( CH_3COO^- + H^+ \rightarrow CH_3COOH \), so the pH changes only slightly.

1 mark: correct definition of buffer solution (resists pH change on small addition of acid/base); 1 mark: recognition that the mixture contains a large reservoir of \( CH_3COO^- \) and \( CH_3COOH \); 1 mark: correct equation/explanation showing \( CH_3COO^- \) removing added \( H^+ \). [3]

\( K_w = [H^+][OH^-] \). The dissociation of water, \( H_2O \rightleftharpoons H^+ + OH^- \), is endothermic, so increasing temperature shifts the equilibrium to the right, increasing \( K_w \) and hence increasing \( [H^+] \). This lowers the pH. However, \( [H^+] \) still equals \( [OH^-] \) at every temperature, so the water remains neutral even though its pH is no longer 7.

1 mark: correct expression \( K_w=[H^+][OH^-] \); 1 mark: correct reasoning that dissociation of water is endothermic so \( K_w \) increases with T; 1 mark: correct conclusion that \( [H^+]=[OH^-] \) still holds so water remains neutral despite lower pH. [3]

An ideal primary standard should be of very high purity, have a known and stable chemical formula, not be hygroscopic or efflorescent, not react with or absorb atmospheric gases (e.g. \( CO_2 \), water vapour), and have a reasonably high molar mass to minimise weighing errors.

1 mark each for any two valid properties (e.g. high purity; stable formula/not hygroscopic; does not react with air; high Mr for accurate weighing). [2]

Optical isomers are non-superimposable mirror images of each other (enantiomers) that rotate the plane of plane-polarised light in opposite directions. A carbon atom is chiral when it is bonded to four different atoms or groups, since this gives rise to two possible non-superimposable arrangements.

1 mark: correct definition (non-superimposable mirror images); 1 mark: reference to rotation of plane-polarised light in opposite directions; 1 mark: chiral carbon correctly defined as bonded to four different groups. [3]

Warm each compound gently with Tollens' reagent (ammoniacal silver nitrate). Propanal, an aldehyde, is oxidised and reduces \( Ag^+ \) to metallic silver, giving a silver mirror on the inside of the test tube. Propanone, a ketone, cannot be oxidised in this way and gives no reaction/no silver mirror.

1 mark: correct reagent named (Tollens'/ammoniacal silver nitrate, or Fehling's); 1 mark: correct positive observation with propanal (silver mirror, or brick-red precipitate if Fehling's used); 1 mark: correct negative observation with propanone (no reaction). [3]

The mechanism is nucleophilic addition. The attacking species is the cyanide ion, \( CN^- \), which acts as the nucleophile, attacking the electrophilic carbonyl carbon of propanone.

1 mark: nucleophilic addition stated; 1 mark: \( CN^- \) correctly named as the nucleophile. [2]

Hydrogen gas is initially evolved faster with hydrochloric acid than with ethanoic acid of the same concentration. HCl is a strong acid and dissociates completely, giving a higher concentration of \( H^+ \) ions in solution. Ethanoic acid is a weak acid and only partially dissociates, giving a much lower \( [H^+] \), so the rate of reaction with magnesium (which depends on \( [H^+] \)) is slower.

1 mark: correct observation that HCl reacts faster initially; 1 mark: correct reasoning that HCl is fully dissociated / ethanoic acid is only partially dissociated; 1 mark: correct link between lower \( [H^+] \) and lower rate for ethanoic acid. [3]

\( 2CH_3CH_2COOH + Na_2CO_3 \rightarrow 2CH_3CH_2COONa + H_2O + CO_2 \). Effervescence (bubbles) of a colourless gas is observed as \( CO_2 \) is released.

1 mark: correctly balanced equation; 1 mark: correct observation (effervescence/bubbles of colourless gas). [2]

Ethanoic acid is treated with sulfur dichloride oxide (thionyl chloride, \( SOCl_2 \)), or alternatively \( PCl_5 \), at room temperature. The reaction is a substitution because the \( -OH \) group of the carboxylic acid is directly replaced by a \( -Cl \) atom, with the carbon skeleton unchanged.

1 mark: correct reagent (\( SOCl_2 \), or \( PCl_5 \)/\( PCl_3 \)); 1 mark: room temperature/mild conditions stated; 1 mark: correct explanation that \( -OH \) is replaced by \( -Cl \), i.e. substitution. [3]

Chlorine is more electronegative than oxygen and forms a weaker, more polarisable bond to the carbonyl carbon, making it a much better leaving group than an alkoxy (\( -OR \)) group. Chlorine also withdraws electron density from the carbonyl carbon more strongly by induction than \( -OR \), and unlike the oxygen lone pair in an ester, the chlorine lone pair donates less effectively into the carbonyl system, leaving the carbonyl carbon more electrophilic and so more readily attacked by nucleophiles.

1 mark: Cl is a better leaving group than \( OR^- \); 1 mark: acyl chloride carbonyl carbon is more electrophilic/electron-deficient; 1 mark: correct reasoning linking this to weaker electron donation from Cl lone pair into the carbonyl compared with O in an ester. [3]

Ethanoyl chloride reacts with excess ethylamine to give N-ethylethanamide, \( CH_3CONHCH_2CH_3 \), and a second molecule of ethylamine reacts with the HCl by-product to form ethylammonium chloride, \( CH_3CH_2NH_3^+Cl^- \).

1 mark: N-ethylethanamide correctly identified; 1 mark: ethylammonium chloride (or ethylamine hydrochloride) correctly identified as the second product. [2]

The six p-orbitals on the ring carbons overlap to form a delocalised \( \pi \) system spread over the whole ring, giving benzene extra thermodynamic stability (delocalisation/resonance energy) compared with a hypothetical structure with three localised double bonds. An addition reaction would break this delocalised system and convert some ring carbons to \( sp^3 \), destroying the stability, so substitution (which retains the intact delocalised ring in the product) is energetically favoured instead.

1 mark: reference to the delocalised \( \pi \) system over the ring; 1 mark: reference to the extra stability (delocalisation energy) this provides; 1 mark: correct explanation that addition would destroy this delocalisation/stability, whereas substitution preserves the ring. [3]

The methyl group is electron-donating by induction (positive inductive effect), pushing electron density into the ring, particularly at the ortho and para positions. This increases the electron density of the delocalised \( \pi \) system, making the ring more attractive to electrophiles, so methylbenzene reacts faster than benzene under the same conditions.

1 mark: \( CH_3 \) is electron-donating by induction; 1 mark: correct statement that ring electron density increases; 1 mark: correct link to increased attraction for electrophiles/faster reaction. [3]

Benzene is heated under reflux at about 50°C with a mixture of concentrated nitric acid and concentrated sulfuric acid. Concentrated sulfuric acid acts as a catalyst: it protonates nitric acid, which then loses water to generate the electrophile, the nitronium ion \( NO_2^+ \): \( HNO_3 + 2H_2SO_4 \rightarrow NO_2^+ + H_3O^+ + 2HSO_4^- \).

1 mark: correct reagents (conc. \( HNO_3 \) and conc. \( H_2SO_4 \)); 1 mark: correct temperature (approximately 50 °C); 1 mark: correct role of \( H_2SO_4 \) as catalyst generating the \( NO_2^+ \) electrophile. [3]

The oxygen lone pair on the \( -OH \) group of phenol is delocalised into the aromatic ring, increasing the electron density of the ring (especially at the ortho and para positions). This activates the ring strongly towards electrophilic substitution, allowing reaction with the weak electrophile \( Br_2 \) directly, without a catalyst.

1 mark: reference to the oxygen lone pair being donated/delocalised into the ring; 1 mark: correct conclusion that this increases ring electron density and activates it towards electrophilic substitution. [2]

By Hess's law (Born–Haber cycle): \( \Delta H_f^{\ominus} = \Delta H_{at}(Ca) + IE_1 + IE_2 + 2\Delta H_{at}(Cl) + 2EA(Cl) + LE \). \( -795 = 178 + 590 + 1145 + 2(121) + 2(-349) + LE \). Sum of known terms \( = 178+590+1145+242-698 = 1457 \). So \( LE = -795 - 1457 = -2252 \text{ kJ mol}^{-1} \).

1 mark: correct Hess's law cycle/equation set up with all six terms; 1 mark: atomisation and ionisation energy terms (for Ca and 2×Cl) summed correctly; 1 mark: electron affinity term for 2×Cl correctly doubled and included; 1 mark: correct final answer \( -2252 \text{ kJ mol}^{-1} \) (accept \( \pm 5 \)). [4]

Feasibility requires \( \Delta G \le 0 \), i.e. \( \Delta H \le T\Delta S \), so the minimum temperature is when \( \Delta H = T\Delta S \): \( T = \dfrac{\Delta H}{\Delta S} = \dfrac{126000 \text{ J mol}^{-1}}{218 \text{ J K}^{-1}\text{mol}^{-1}} = 578 \text{ K} \) (3 s.f.).

1 mark: condition for feasibility stated (\( \Delta G \le 0 \) or \( \Delta H = T\Delta S \)); 1 mark: \( \Delta H \) correctly converted to J mol⁻¹ (126000 J mol⁻¹); 1 mark: correct division method \( \Delta H / \Delta S \); 1 mark: correct final answer 578 K. [4]

Comparing Exp 1 and 2: [A] doubles, [B] constant, rate doubles \( \Rightarrow \) order 1 in A. Comparing Exp 2 and 3: [B] doubles, [A] constant, rate increases ×4 \( \Rightarrow \) order 2 in B. So rate \( = k[A][B]^2 \). Using Exp 1: \( 2.0\times10^{-3} = k(0.10)(0.10)^2 = k \times 1.0\times10^{-3} \), so \( k = 2.0 \text{ mol}^{-2}\text{dm}^6\text{s}^{-1} \).

1 mark: order 1 with respect to A correctly deduced with reasoning; 1 mark: order 2 with respect to B correctly deduced with reasoning; 1 mark: correct substitution to find k using any experiment; 1 mark: correct value and units, \( k = 2.0 \text{ mol}^{-2}\text{dm}^6\text{s}^{-1} \). [4]

\( pH = pK_a + \log\left(\dfrac{[\text{salt}]}{[\text{acid}]}\right) \). \( pK_a = -\log(1.8\times10^{-5}) = 4.745 \). \( \log\left(\dfrac{0.15}{0.20}\right) = \log(0.75) = -0.125 \). \( pH = 4.745 - 0.125 = 4.62 \) (2 d.p.).

1 mark: Henderson–Hasselbalch expression correctly quoted/used; 1 mark: \( pK_a \) correctly calculated (4.745, or equivalent unrounded); 1 mark: correct ratio \( \log(0.15/0.20) \) evaluated; 1 mark: correct final answer pH = 4.62. [4]

NaOH is a strong base, fully dissociated, so \( [OH^-] = 0.0500 \text{ mol dm}^{-3} \). \( pOH = -\log(0.0500) = 1.30 \). \( pH = 14.00 - 1.30 = 12.70 \).

1 mark: \( [OH^-] = 0.0500 \text{ mol dm}^{-3} \) correctly identified (full dissociation); 1 mark: pOH correctly calculated as 1.30; 1 mark: correct final answer pH = 12.70 using \( pH + pOH = 14 \). [3]

\( PCl_5 \) reacted \( = 0.400 - 0.150 = 0.250 \text{ mol} \), so at equilibrium \( n(PCl_3) = n(Cl_2) = 0.250 \text{ mol} \). Concentrations: \( [PCl_5] = 0.150/2.00 = 0.0750 \), \( [PCl_3] = [Cl_2] = 0.250/2.00 = 0.125 \text{ mol dm}^{-3} \). \( K_c = \dfrac{[PCl_3][Cl_2]}{[PCl_5]} = \dfrac{0.125\times0.125}{0.0750} = 0.208... \) Recalculating: \( 0.125^2 = 0.015625 \), \( /0.0750 = 0.208 \text{ mol dm}^{-3} \).

1 mark: correct equilibrium moles of \( PCl_3 \) and \( Cl_2 \) found (0.250 mol each); 1 mark: all concentrations correctly calculated by dividing by 2.00 dm³; 1 mark: correct \( K_c \) expression substituted and evaluated with correct units mol dm⁻³ (accept 0.208 mol dm⁻³, allow follow-through from candidate's concentrations). [3]

\( \Delta S^{\ominus} = \Sigma S^{\ominus}(\text{products}) - \Sigma S^{\ominus}(\text{reactants}) = [2\times193] - [192 + 3\times131] = 386 - (192+393) = 386 - 585 = -199 \text{ J K}^{-1}\text{mol}^{-1} \) (accept \( -198 \) to \( -199 \) depending on rounding).

1 mark: correct products total (\( 2\times193=386 \)); 1 mark: correct reactants total (\( 192+3\times131=585 \)); 1 mark: correct final answer, approximately \( -199 \text{ J K}^{-1}\text{mol}^{-1} \), with negative sign. [3]

\( \Delta H_f^{\ominus} = \Delta H_{at}(Na) + IE_1(Na) + \Delta H_{at}(Br) + EA(Br) + LE \). \( -361 = 107 + 496 + 112 + EA(Br) + (-742) \). Sum of known terms \( = 107+496+112-742 = -27 \). So \( -361 = -27 + EA(Br) \), giving \( EA(Br) = -361-(-27) = -334 \text{ kJ mol}^{-1} \).

1 mark: correct Born–Haber cycle equation set up with all five known terms plus EA as the unknown; 1 mark: known terms correctly summed (\( -27 \text{ kJ mol}^{-1} \)); 1 mark: correct final answer \( EA(Br) = -334 \text{ kJ mol}^{-1} \). [3]

A fixed mass of solid \( MnO_2 \) catalyst is added to a known volume and concentration of \( H_2O_2(aq) \) in a flask connected to a gas syringe, and a stopclock is started immediately. The volume of oxygen gas collected is recorded at regular time intervals (e.g. every 15 or 30 seconds) until the reaction is complete, and a graph of volume of \( O_2 \) against time is plotted. The initial rate is found from the gradient of the tangent to the curve at t = 0. The whole experiment is then repeated using several different starting concentrations of \( H_2O_2 \), keeping the mass of catalyst, temperature and total volume constant. The initial rate is plotted against initial \( [H_2O_2] \): a straight line through the origin shows the reaction is first order with respect to \( H_2O_2 \), while a curved graph indicates a different order, which can be confirmed by plotting rate against \( [H_2O_2]^2 \) or by comparing initial rates directly when concentration is doubled.

Band A (5–6 marks): a full, coherent, logically sequenced method including apparatus (gas syringe), procedure for measuring volume of O₂ vs time, method for finding initial rate from a tangent, repetition at different [H₂O₂] with other variables controlled, and a correct method for determining the order from the resulting data (e.g. rate vs concentration graph); fluent use of specialist terms, essentially free from error in spelling, punctuation and grammar. Band B (3–4 marks): most of the above points present but method less complete or less clearly sequenced; reasonable use of specialist terms. Band C (1–2 marks): only basic/fragmented description of the method, e.g. mentions gas collection but no clear method for determining order; limited use of specialist terms. Indicative content: gas syringe/collection of O₂; timed volume readings; tangent to find initial rate; repeat at varying [H₂O₂]; control of other variables; use of initial rate vs concentration graph to deduce order. [6]

A known volume of ethanoic acid of unknown concentration is measured into a beaker using a pipette, and a pH meter (or pH probe connected to a data logger) is calibrated and placed in the acid. Sodium hydroxide solution of known concentration is added from a burette in small, measured volumes (e.g. 1–2 cm³ at a time, reducing to 0.1–0.2 cm³ increments near the expected end point), with the solution stirred continuously, and the pH is recorded after each addition. A graph of pH against volume of NaOH added is plotted; the volume at the centre of the steep vertical portion of the curve is the equivalence volume, which can be read more precisely than by indicator colour change alone. Because a weak acid is titrated against a strong base, the pH at the equivalence point is greater than 7 and the steep part of the curve lies entirely within the alkaline range, so phenolphthalein (colour change range approximately pH 8.2–10.0) is a suitable indicator, whereas methyl orange (range approximately pH 3.1–4.4) would change colour too early and give an inaccurate end point. Once the equivalence volume is known, moles of NaOH used are calculated from its concentration, and since the acid reacts with the base in a 1:1 mole ratio, the concentration of the ethanoic acid can be calculated.

Band A (5–6 marks): full, logically ordered method including pipetting the acid, use of a pH meter, gradual/controlled addition of NaOH from a burette with stirring, plotting pH against volume, correct identification of the equivalence point from the steep part of the graph, correct identification of phenolphthalein with a valid reason linked to the pH range of the equivalence point, and a correct outline of the calculation of acid concentration from the equivalence volume; fluent use of specialist terms, essentially free from error. Band B (3–4 marks): most key steps present (pH meter, titration procedure, graph, indicator choice) but with some detail missing or reasoning less developed. Band C (1–2 marks): basic outline only, e.g. mentions titration and pH meter but no clear identification of the correct indicator with reasoning, or no clear method for finding the concentration. Indicative content: pipette known volume of acid; pH meter/probe; controlled addition of NaOH with stirring; pH vs volume graph; equivalence point from steep section; phenolphthalein justified by alkaline equivalence point; 1:1 mole ratio used to calculate acid concentration. [6]

Formation of the electrophile: \( HNO_3 + 2H_2SO_4 \rightarrow NO_2^+ + H_3O^+ + 2HSO_4^- \) (or the two-step equivalent \( HNO_3 + H_2SO_4 \rightarrow H_2NO_3^+ + HSO_4^- \), then \( H_2NO_3^+ \rightarrow NO_2^+ + H_2O \)). Mechanism: a pair of electrons from the delocalised ring attacks the electrophilic nitrogen of \( NO_2^+ \), forming a new \( C-N \) bond and breaking the ring's delocalisation over that carbon, giving a positively charged, non-aromatic arenium (Wheland) intermediate in which the positive charge is delocalised over the remaining five carbons; a curly arrow then shows a \( C-H \) bond on the substituted carbon breaking heterolytically, with the electron pair returning to re-form the aromatic ring and releasing \( H^+ \). Regeneration of catalyst: \( H^+ + HSO_4^- \rightarrow H_2SO_4 \), confirming that \( H_2SO_4 \) is a true catalyst (net consumption of only \( HNO_3 \), which acts as the source of the nitro group, water being the only other product formed overall).

1 mark: correct equation for generation of \( NO_2^+ \) from \( HNO_3 \) and \( H_2SO_4 \); 1 mark: curly arrow from the ring \( \pi \) system to the electrophile \( NO_2^+ \), forming the new \( C-N \) bond; 1 mark: correct structure/description of the arenium intermediate with the positive charge delocalised over the ring (aromaticity broken); 1 mark: curly arrow showing loss of \( H^+ \) and re-formation of the aromatic ring; 1 mark: correct equation showing regeneration of \( H_2SO_4 \) from \( H^+ \) and \( HSO_4^- \). [5]

Step 1: oxidise propan-1-ol to propanoic acid using excess acidified potassium dichromate(VI) (\( K_2Cr_2O_7/H_2SO_4 \)) under reflux; this is an oxidation. Step 2: convert propanoic acid to propanoyl chloride using thionyl chloride, \( SOCl_2 \), at room temperature (alternatively \( PCl_5 \)); this is a substitution reaction. Step 3: react propanoyl chloride with excess propylamine (\( CH_3CH_2CH_2NH_2 \)) at room temperature; this is a nucleophilic addition–elimination (condensation) reaction producing N-propylpropanamide, \( CH_3CH_2CONHCH_2CH_2CH_3 \), and propylammonium chloride as a by-product from the second equivalent of amine neutralising the HCl formed.

1 mark: Step 1 reagent/conditions correct (\( K_2Cr_2O_7/H_2SO_4 \), reflux, excess oxidant) with reaction type (oxidation); 1 mark: propanoic acid correctly identified as the Step 1 product; 1 mark: Step 2 reagent correct (\( SOCl_2 \) or \( PCl_5 \)) with reaction type (substitution), forming propanoyl chloride; 1 mark: Step 3 reagent correct (excess propylamine, room temperature); 1 mark: correct final product N-propylpropanamide named/structure given. [5]

The carbonyl carbon of butanone carries a partial positive charge, \( C^{\delta+}=O^{\delta-} \), because oxygen is more electronegative than carbon. The cyanide ion, \( CN^- \) (generated in small amount from \( HCN \), often with a trace of base/KCN catalyst), acts as a nucleophile and attacks the electrophilic carbonyl carbon; a curly arrow is drawn from the lone pair on carbon of \( CN^- \) to the carbonyl carbon, and a second curly arrow shows the \( C=O \) \( \pi \) bond breaking, with both electrons moving onto the oxygen. This gives a tetrahedral alkoxide intermediate bearing a negative charge on oxygen and a \( -CN \) group and two alkyl groups on the former carbonyl carbon. A third curly arrow shows the lone pair on the negatively charged oxygen attacking a hydrogen atom of \( HCN \), breaking the \( H-CN \) bond heterolytically (electrons to the departing \( CN^- \)) to protonate the alkoxide, giving the neutral hydroxynitrile product, 2-hydroxy-2-methylbutanenitrile, and regenerating \( CN^- \) for the cycle to continue.

1 mark: correct dipole shown on the carbonyl group (\( C^{\delta+}=O^{\delta-} \)); 1 mark: curly arrow from \( CN^- \) lone pair to the carbonyl carbon with a second arrow showing the \( C=O \) \( \pi \) electrons moving to oxygen; 1 mark: correct tetrahedral alkoxide intermediate structure with negative charge on oxygen; 1 mark: correct final protonation step (by HCN or \( H_2O/H^+ \)) giving the neutral hydroxynitrile product. [4]

Loss of \( \cdot CH_3 \) (mass 15) from the molecular ion (46) gives a fragment of mass \( 46 - 15 = 31 \), corresponding to \( CH_2=OH^+ \).

1 mark: C.

Ethanoic acid has two chemically distinct types of proton: the three equivalent \( CH_3 \) protons and the single \( OH \) proton.

1 mark: B.

Moles NaOH \( = 0.100 \times 0.0250 = 2.50\times10^{-3} \text{ mol} = \) moles HCl (1:1 ratio). Concentration \( = \dfrac{2.50\times10^{-3}}{0.0200} = 0.125 \text{ mol dm}^{-3} \).

1 mark: A.

Distance travelled by X \( = R_f \times \) distance travelled by solvent \( = 0.65 \times 8.0 = 5.2 \text{ cm} \).

1 mark: A.

Copper is \( [Ar]3d^{10}4s^1 \). Forming \( Cu^{2+} \) removes the single 4s electron and one 3d electron, giving \( [Ar]3d^9 \).

1 mark: A.

Excess \( Cl^- \) ligands, being larger than water, substitute to form the tetrahedral complex \( [CuCl_4]^{2-} \), which is yellow-green in concentrated solution.

1 mark: A.

\( E_{cell} = E_{cathode} - E_{anode} = 0.34 - (-0.76) = +1.10 \text{ V} \).

1 mark: A.

The ethyl group is electron-donating by induction, increasing electron density on the nitrogen lone pair compared with ammonia, so ethylamine is a stronger base. In phenylamine the lone pair is delocalised into the ring, reducing its availability, and amides are not basic because the lone pair is delocalised onto the carbonyl oxygen.

1 mark: C.

At the isoelectric point the amino group is protonated and the carboxylic acid group is deprotonated, giving a species with equal positive and negative charge and zero overall net charge: a zwitterion.

1 mark: C.

Each amide linkage forms with the elimination of a small molecule, water, from two different monomers. This loss of a small molecule on bond formation defines condensation polymerisation.

1 mark: B.

The peak at m/z = 45 corresponds to the \( COOH^+ \) fragment (mass of \( C+2O+H = 12+32+1 = 45 \)), formed by homolytic/heterolytic cleavage of the bond between the carboxyl carbon and the adjacent alkyl chain, giving the stable acylium-type fragment retaining the charge on the \( -COOH \) group.

1 mark: fragment correctly identified as \( COOH^+ \) (or equivalent, e.g. \( HOCO^+ \)); 1 mark: correct explanation that it arises from cleavage of the bond next to the carbonyl/carboxyl group. [2]

Methyl ethanoate has two distinct proton environments: the \( CH_3 \) attached to the carbonyl carbon, and the \( OCH_3 \) attached to the ester oxygen. This gives two peaks in the spectrum, in a 1:1 (3H:3H) integration ratio.

1 mark: two peaks correctly identified; 1 mark: correct environments identified (CO-\(CH_3\) and O-\(CH_3\)); 1 mark: correct integration ratio 1:1 (3H:3H). [3]

Spin–spin coupling occurs because the magnetic field experienced by a proton is slightly affected by the spin orientations of non-equivalent protons on adjacent carbon atoms, causing the NMR signal to split into a group of closely spaced peaks (a multiplet) rather than a single line. The \( CH_2 \) protons in \( CH_3CH_2Cl \) are adjacent to the 3 equivalent \( CH_3 \) protons, so by the n+1 rule the \( CH_2 \) signal is split into \( 3+1 = 4 \) lines, i.e. a quartet.

1 mark: correct explanation of spin–spin coupling (splitting due to non-equivalent protons on adjacent carbons); 1 mark: correct application of n+1 rule using n = 3 (adjacent \( CH_3 \) protons); 1 mark: correct multiplicity stated, quartet. [3]

TMS provides a reference point of zero on the chemical shift scale (\( \delta = 0 \)), against which the chemical shifts of all other protons in the sample are measured. It is suitable because all twelve of its protons are chemically equivalent, giving a single sharp peak, and it is chemically inert (does not react with the sample), non-toxic and volatile, so it is easily removed from the sample afterwards.

1 mark: correct statement of purpose (reference point, \( \delta=0 \)); 1 mark: any two valid reasons for suitability (e.g. single peak from 12 equivalent protons; chemically inert; volatile/low boiling point, easily removed). [2]

An accurately weighed mass of solid anhydrous sodium carbonate is dissolved in a small volume of distilled water in a beaker, stirring until fully dissolved. The resulting solution is transferred quantitatively (with thorough rinsing of the beaker and stirring rod) into a 250 cm³ volumetric flask via a funnel. Distilled water is then added, with swirling, up to just below the graduation mark, and finally a dropping pipette is used to add water drop by drop until the bottom of the meniscus sits exactly on the graduation line; the stopper is inserted and the flask inverted repeatedly to ensure thorough mixing.

1 mark: solid dissolved in a small volume of water and transferred quantitatively (with rinsings) to the volumetric flask; 1 mark: correct use of a 250 cm³ volumetric flask, made up to the graduation mark; 1 mark: correct detail on accurate meniscus reading (dropwise addition near the mark) and thorough mixing by inversion. [3]

A burette allows a variable, precisely controlled and measurable volume of titrant to be added dropwise, including very small increments near the end point, and its readings can be taken accurately from the graduated scale, whereas a measuring cylinder is far less precise and cannot be used to add liquid in small, controlled increments. A standard 50 cm³ burette is graduated in 0.10 cm³ divisions, so readings can be estimated to the nearest 0.05 cm³.

1 mark: correct reason relating to controlled/variable dropwise addition; 1 mark: correct reason relating to greater precision/accuracy of reading than a measuring cylinder; 1 mark: correct reading precision stated (to the nearest 0.05 cm³, from 0.10 cm³ graduations). [3]

At the end point the indicator changes from yellow to orange (or red). Sodium carbonate is a salt of a weak acid and a strong acid overall (via \( CO_2/H_2CO_3 \)), so the equivalence point of this titration lies on the acidic side of pH 7, within the colour-change range of methyl orange (approximately pH 3.1–4.4). Phenolphthalein changes colour only in the range approximately pH 8.2–10.0, which is well above the equivalence point, so it would give a false (early) end point if used here.

1 mark: correct colour change (yellow to orange/red); 1 mark: correct statement that the equivalence point of this titration is acidic (below pH 7); 1 mark: correct explanation that methyl orange's range matches the acidic equivalence point, unlike phenolphthalein's alkaline range. [3]

Two sources of random error are: (i) judging exactly when the indicator has permanently changed colour, which can be minimised by titrating slowly near the end point and comparing against a white background/a previously titrated 'blank' colour; (ii) parallax error when reading the burette scale, which can be minimised by reading the meniscus at eye level. In general, random errors are also reduced by repeating the titration until concordant results (within 0.10 cm³) are obtained and calculating a mean titre.

1 mark each for any two valid random errors with a correct method of minimising their effect (e.g. judging end-point colour — repeat until concordant; parallax error in burette reading — read at eye level). [2]

Covering the tank keeps the atmosphere inside saturated with solvent vapour. This prevents the solvent evaporating from the plate/paper as it rises, which would otherwise cause an uneven, distorted solvent front and unreliable, non-reproducible \( R_f \) values.

1 mark: correct reference to keeping the atmosphere saturated with solvent vapour; 1 mark: correct consequence explained (prevents evaporation/uneven solvent front, giving reliable/reproducible \( R_f \) values). [2]

In an octahedral complex, ligands cause the five degenerate 3d orbitals to split into two sets of different energy. Colour arises when an electron absorbs a photon of visible light and is promoted between these split d orbitals (a d–d transition), with the complementary colour transmitted/reflected being observed. \( Sc^{3+} \) has the electron configuration \( [Ar]3d^0 \) — an empty 3d subshell — so no d–d transition is possible and the ion is colourless. \( Fe^{3+} \), \( [Ar]3d^5 \), has a partially filled 3d subshell, so d–d transitions can occur, absorbing certain wavelengths of visible light and giving a coloured solution.

1 mark: correct reference to splitting of the d orbitals by ligands and colour arising from d–d electron transitions; 1 mark: correct explanation that \( Sc^{3+} \) has an empty (\( 3d^0 \)) subshell so no transition is possible; 1 mark: correct explanation that \( Fe^{3+} \) has a partially filled 3d subshell (\( 3d^5 \)) allowing d–d transitions. [3]

Transition metals such as manganese can readily exist in more than one stable oxidation state. This allows a transition metal ion or its compound to be temporarily oxidised or reduced during a reaction, forming an intermediate species with reactants, and then be regenerated once the reaction is complete. This provides an alternative reaction pathway with a lower activation energy than the uncatalysed reaction, increasing the rate of reaction without the catalyst itself being permanently consumed.

1 mark: correct reference to transition metals having variable/multiple oxidation states; 1 mark: correct explanation that the metal forms an intermediate by being oxidised/reduced then regenerated; 1 mark: correct link to an alternative pathway of lower activation energy, increasing rate. [3]

On dropwise addition, a pale blue precipitate of copper(II) hydroxide forms: \( Cu^{2+}(aq) + 2OH^-(aq) \rightarrow Cu(OH)_2(s) \) (the \( OH^- \) coming from aqueous ammonia acting as a weak base). On addition of excess concentrated ammonia, the pale blue precipitate dissolves to give a deep/royal blue solution, as ammonia ligands substitute for water/hydroxide to form the soluble complex ion: \( Cu(OH)_2(s) + 4NH_3(aq) \rightarrow [Cu(NH_3)_4(H_2O)_2]^{2+}(aq) + 2OH^-(aq) \).

1 mark: correct observation of pale blue precipitate forming with dropwise ammonia; 1 mark: correct observation that the precipitate dissolves in excess ammonia to give a deep blue solution; 1 mark: correct ionic equation for either stage (precipitate formation or dissolution to the ammine complex). [3]

In \( Cr_2O_7^{2-} \): letting the oxidation number of Cr be x, \( 2x + 7(-2) = -2 \Rightarrow 2x = 12 \Rightarrow x = +6 \). In \( CrO_4^{2-} \): \( x + 4(-2) = -2 \Rightarrow x = +6 \). Chromium is in the +6 oxidation state in both ions.

1 mark: +6 for \( Cr_2O_7^{2-} \); 1 mark: +6 for \( CrO_4^{2-} \). [2]

The standard electrode potential of a half-cell is the EMF (voltage) measured when that half-cell is connected, via a salt bridge, to a standard hydrogen electrode, under standard conditions (298 K, 100 kPa gas pressure, 1 mol dm⁻³ ion concentrations), with the standard hydrogen electrode itself assigned a potential of exactly 0 V. The standard hydrogen electrode consists of hydrogen gas at 100 kPa bubbled over a platinum electrode (coated in platinum black) immersed in a 1 mol dm⁻³ solution of \( H^+ \) ions.

1 mark: correct definition referring to EMF measured against a standard hydrogen electrode; 1 mark: correct reference to standard conditions (298 K, 100 kPa, 1 mol dm⁻³); 1 mark: correct description of the standard hydrogen electrode (Pt/platinised platinum, \( H_2 \) gas at 100 kPa, 1 mol dm⁻³ \( H^+ \), assigned 0 V). [3]

For \( Cl_2 \) oxidising \( Fe^{2+} \): \( E_{cell} = E^{\ominus}(Cl_2/Cl^-) - E^{\ominus}(Fe^{3+}/Fe^{2+}) = 1.36 - 0.77 = +0.59 \text{ V} \), which is positive, so the reaction is feasible. For \( I_2 \) oxidising \( Fe^{2+} \): \( E_{cell} = 0.54 - 0.77 = -0.23 \text{ V} \), which is negative, so this reaction is not feasible. \( Cl_2 \) has a more positive (higher) standard electrode potential than \( Fe^{3+}/Fe^{2+} \) and so is a stronger oxidising agent than \( Fe^{3+} \), able to oxidise \( Fe^{2+} \); \( I_2 \) has a lower/less positive potential than \( Fe^{3+}/Fe^{2+} \) and so is a weaker oxidising agent than \( Fe^{3+} \) and cannot oxidise \( Fe^{2+} \).

1 mark: correct \( E_{cell} \) calculated for the \( Cl_2 \) reaction (+0.59 V), positive, feasible; 1 mark: correct \( E_{cell} \) calculated for the \( I_2 \) reaction (−0.23 V), negative, not feasible; 1 mark: correct overall explanation linking sign of \( E_{cell} \) to relative oxidising strength of \( Cl_2 \)/\( I_2 \) compared with \( Fe^{3+} \). [3]

Standard conditions are a temperature of 298 K (25 °C), a gas pressure of 100 kPa for any gaseous species involved, and solute (ion) concentrations of 1 mol dm⁻³.

1 mark: temperature 298 K and gas pressure 100 kPa both stated; 1 mark: ion concentration of 1 mol dm⁻³ stated. [2]

The ethyl group in ethylamine is electron-donating by induction, pushing electron density onto the nitrogen atom and making its lone pair more available to accept a proton, so ethylamine is a stronger base than ammonia. In phenylamine, the nitrogen lone pair is partly delocalised into the aromatic ring (overlapping with the ring's \( \pi \) system), which reduces the electron density available on nitrogen to accept a proton, making phenylamine a weaker base than ammonia.

1 mark: correct explanation for ethylamine (electron-donating alkyl group increases electron density on N); 1 mark: correct explanation for phenylamine (lone pair delocalised into the ring, reducing availability); 1 mark: both correctly linked to basicity relative to ammonia (stronger/weaker respectively). [3]

Amino acids are amphoteric because they contain both a basic amino group (\( -NH_2 \)) and an acidic carboxylic acid group (\( -COOH \)) within the same molecule, so they can react as either an acid or a base. Acting as a base with HCl: \( H_2NCH_2COOH + HCl \rightarrow {}^+H_3NCH_2COOH \ Cl^- \), where the amino group is protonated.

1 mark: correct identification of both the \( -NH_2 \) and \( -COOH \) groups as the reason for amphoteric behaviour; 1 mark: correct equation/product showing protonation of the amino group; 1 mark: correct overall balanced equation including \( Cl^- \) counter-ion. [3]

A zwitterion is a species that carries both a positive and a negative charge simultaneously but has zero overall net charge. In glycine, the amino group is protonated to \( -NH_3^+ \) and the carboxylic acid group is deprotonated to \( -COO^- \), giving the zwitterion \( {}^+H_3NCH_2COO^- \).

1 mark: correct general definition (species with both + and − charge, zero net charge); 1 mark: correct structure/formula of the glycine zwitterion, \( {}^+H_3NCH_2COO^- \). [2]

Poly(alkene) plastics (e.g. poly(ethene), poly(propene)) are formed by addition polymerisation and have strong, unreactive C–C backbones, so they are not biodegradable and persist in landfill for a very long time, taking up space and causing environmental pollution. This can be addressed by sorting and recycling the plastic (melting and remoulding it into new products), or by incineration to recover energy (though this may require care to avoid releasing toxic combustion products).

1 mark: correct valid problem stated (e.g. non-biodegradable, persists in landfill); 1 mark: correct valid method of dealing with it (e.g. recycling, or incineration for energy recovery). [2]

Moles \( S_2O_3^{2-} = 0.100 \times 0.02140 = 2.140\times10^{-3} \text{ mol} \). Moles \( I_2 = \tfrac{1}{2}(2.140\times10^{-3}) = 1.070\times10^{-3} \text{ mol} \). Moles \( Cu^{2+} = 2\times(1.070\times10^{-3}) = 2.140\times10^{-3} \text{ mol} \) (in 25.0 cm³). Concentration \( = \dfrac{2.140\times10^{-3}}{0.0250} = 0.0856 \text{ mol dm}^{-3} \).

1 mark: moles of \( S_2O_3^{2-} \) correctly calculated; 1 mark: correct 2:1 ratio used to find moles \( I_2 \); 1 mark: correct 2:1 ratio used to find moles \( Cu^{2+} \) from \( I_2 \); 1 mark: correct final concentration 0.0856 mol dm⁻³. [4]

Moles \( MnO_4^- = 0.0200 \times 0.01865 = 3.73\times10^{-4} \text{ mol} \). Moles \( Fe^{2+} = 5 \times (3.73\times10^{-4}) = 1.865\times10^{-3} \text{ mol} \) (in 25.0 cm³). Concentration \( = \dfrac{1.865\times10^{-3}}{0.0250} = 0.0746 \text{ mol dm}^{-3} \).

1 mark: moles of \( MnO_4^- \) correctly calculated; 1 mark: correct 1:5 ratio identified from the equation; 1 mark: moles \( Fe^{2+} \) correctly calculated; 1 mark: correct final concentration 0.0746 mol dm⁻³. [4]

\( Cr_2O_7^{2-} \) is reduced (higher \( E^{\ominus} \)) while \( Fe^{2+} \) is oxidised (lower \( E^{\ominus} \)). \( E_{cell} = E^{\ominus}(\text{reduction}) - E^{\ominus}(\text{oxidation}) = 1.33 - 0.77 = +0.56 \text{ V} \). Since \( E_{cell} \) is positive, the reaction is feasible under standard conditions.

1 mark: correct identification of \( Cr_2O_7^{2-}/Cr^{3+} \) as the reduction (higher potential) half-reaction; 1 mark: correct subtraction method \( E_{cell}=E_{red}-E_{ox} \); 1 mark: correct value \( E_{cell}=+0.56 \text{ V} \); 1 mark: correct conclusion that the reaction is feasible (positive EMF). [4]

\( E_{cell} = E^{\ominus}(Ag^+/Ag) - E^{\ominus}(Pb^{2+}/Pb) = 0.80 - (-0.13) = +0.93 \text{ V} \). The half-cell with the more positive potential (\( Ag^+/Ag \)) undergoes reduction, and the half-cell with the more negative potential (\( Pb^{2+}/Pb \)) undergoes oxidation. Balancing electrons (multiply the Ag half-equation by 2): overall equation \( Pb(s) + 2Ag^+(aq) \rightarrow Pb^{2+}(aq) + 2Ag(s) \).

1 mark: correct EMF calculation method; 1 mark: correct value \( E_{cell}=+0.93 \text{ V} \); 1 mark: correct identification of Pb as oxidised (anode) and \( Ag^+ \) as reduced (cathode); 1 mark: correctly balanced overall equation with electrons balanced (2Ag⁺). [4]

With \( M_r = 74 \) and formula \( C_3H_6O_2 \), and two singlets each integrating for 3H (total 6H, matching the formula), Y must contain two isolated \( CH_3 \) groups with no adjacent protons on either side (since both signals are singlets, showing no spin–spin coupling). The peak at \( \delta \) 3.65 is typical of a \( CH_3 \) group attached directly to an ester oxygen (\( -OCH_3 \)), and the peak at \( \delta \) 2.05 is typical of a \( CH_3 \) group attached to a carbonyl carbon (\( CH_3CO- \)). This is consistent with methyl ethanoate, \( CH_3COOCH_3 \), in which the two \( CH_3 \) groups are separated by the ester linkage and cannot couple with each other.

1 mark: correct deduction that both signals are singlets, meaning no adjacent (coupling) protons on either methyl group; 1 mark: correct assignment of \( \delta \) 3.65 to \( OCH_3 \); 1 mark: correct assignment of \( \delta \) 2.05 to \( CH_3CO \)-; 1 mark: correct final structure, methyl ethanoate \( CH_3COOCH_3 \), consistent with \( M_r=74 \) and \( C_3H_6O_2 \). [4]

Moles EDTA \( = 0.0100 \times 0.01580 = 1.580\times10^{-4} \text{ mol} \). By the 1:1 ratio, moles \( (Ca^{2+}+Mg^{2+}) = 1.580\times10^{-4} \text{ mol} \) in 25.0 cm³. Concentration \( = \dfrac{1.580\times10^{-4}}{0.0250} = 6.32\times10^{-3} \text{ mol dm}^{-3} \). As \( CaCO_3 \) equivalent: \( 6.32\times10^{-3} \times 100 = 0.632 \text{ g dm}^{-3} = 632 \text{ mg dm}^{-3} \).

1 mark: moles of EDTA correctly calculated; 1 mark: correct 1:1 ratio used to find total moles of \( Ca^{2+}+Mg^{2+} \); 1 mark: correct concentration \( 6.32\times10^{-3} \text{ mol dm}^{-3} \); 1 mark: correct conversion to \( CaCO_3 \) equivalent, 632 mg dm⁻³. [4]

An accurately weighed mass of the impure iron(II) sulfate crystals is dissolved in dilute sulfuric acid (to prevent hydrolysis/oxidation of \( Fe^{2+} \)) and made up to exactly 250 cm³ in a volumetric flask, using a funnel and rinsings, then made up to the mark and mixed thoroughly. A 25.0 cm³ aliquot of this solution is pipetted into a conical flask and further acidified with dilute sulfuric acid. This is titrated against standardised potassium manganate(VII) solution added from a burette, swirling constantly; \( KMnO_4 \) is self-indicating, so the end point is the first permanent faint pink colour that persists for at least 15–30 seconds. The titration is repeated until at least two concordant titres (within 0.10 cm³) are obtained, and a mean titre calculated. Using the mean titre and the known concentration of \( KMnO_4 \), the moles of \( MnO_4^- \) used are calculated, then converted to moles of \( Fe^{2+} \) using the 1:5 mole ratio (\( MnO_4^- + 5Fe^{2+} + 8H^+ \rightarrow Mn^{2+} + 5Fe^{3+} + 4H_2O \)). This is scaled up by a factor of 10 (250/25) to find total moles of \( Fe^{2+} \), which is converted to a mass of \( FeSO_4 \) (or \( FeSO_4\cdot7H_2O \)) using its molar mass, and the percentage purity is found by dividing this mass by the original mass weighed out and multiplying by 100.

Band A (5–6 marks): a full, coherent, correctly sequenced method covering: accurate weighing and dissolving in acid; quantitative transfer to a 250 cm³ volumetric flask made up to the mark; pipetting a 25.0 cm³ aliquot; acidified titration against standard \( KMnO_4 \) to a self-indicating pink end point; repeating to concordance and taking a mean titre; and a clear, correct outline of the calculation (moles \( MnO_4^- \) → moles \( Fe^{2+} \) via 1:5 ratio → scale to 250 cm³ → mass → % purity); fluent use of specialist terms, essentially free from error. Band B (3–4 marks): most steps present but some detail (e.g. scaling factor, concordance, self-indicating end point) missing or method less clearly sequenced. Band C (1–2 marks): only a fragmented or partial method given, e.g. mentions titration against \( KMnO_4 \) but no clear volumetric-flask/dilution step or no coherent calculation method. Indicative content: dissolve weighed sample in dilute \( H_2SO_4 \); 250 cm³ volumetric flask; 25.0 cm³ aliquots; titrate vs standard \( KMnO_4 \); self-indicating pink end point; concordant titres/mean; 1:5 mole ratio; scale by 10; convert to mass and % purity. [6]

In a free transition metal ion, the five 3d orbitals are degenerate (of equal energy). When ligands such as water molecules bond to the central metal ion in an octahedral complex, their approach along the axes causes electrostatic repulsion that raises the energy of the d orbitals unequally, splitting them into two sets of different energy (a lower set of three and a higher set of two, separated by an energy gap \( \Delta E \)). If the metal ion has a partially filled d subshell, an electron in a lower-energy d orbital can absorb a photon of visible light with energy exactly equal to \( \Delta E \) and be promoted to a higher-energy d orbital — this is called a d–d transition. Because specific wavelengths (colours) of visible light are absorbed, the light transmitted or reflected, and therefore observed by the eye, is the complementary colour to that absorbed, so the complex appears coloured. Ammonia is a stronger-field ligand than water (higher in the spectrochemical series), so replacing water with ammonia increases the size of the splitting energy gap \( \Delta E \). A larger \( \Delta E \) means a higher-energy (shorter wavelength) photon must be absorbed for the d–d transition to occur, which shifts the wavelength of light absorbed, and therefore changes the complementary colour observed — for example, the pale blue \( [Cu(H_2O)_6]^{2+} \) becomes the much darker/more intense blue \( [Cu(NH_3)_4(H_2O)_2]^{2+} \) on ligand exchange.

Band A (5–6 marks): a complete, logically developed explanation covering degeneracy of free-ion d orbitals; splitting into two energy levels by the ligand field; the requirement for a partially filled d subshell; absorption of a specific wavelength of visible light promoting an electron (d–d transition); observed colour being complementary to the light absorbed; and a correct, well-reasoned explanation of how a stronger-field ligand (ammonia) increases \( \Delta E \), shifting the absorbed wavelength and hence changing the observed colour; fluent and accurate use of specialist terms. Band B (3–4 marks): most of the above ideas present (splitting, d–d transition, complementary colour) but the effect of changing the ligand is explained less fully or with minor inaccuracy. Band C (1–2 marks): basic/fragmented answer, e.g. states transition metals are coloured due to d-orbitals without clear reference to splitting or the electronic transition, and/or no meaningful attempt to explain the ligand-change effect. Indicative content: degenerate d orbitals in free ion; ligand field splits d orbitals into two sets; partially filled d subshell needed; d–d electron transition absorbs visible light of energy \( \Delta E \); observed colour is complementary to absorbed colour; \( NH_3 \) is a stronger-field ligand than \( H_2O \); stronger field increases \( \Delta E \); shifts wavelength absorbed and hence the colour seen. [6]

The signal at \( \delta \) 2.60, a singlet that exchanges with \( D_2O \), is characteristic of an \( -OH \) proton (broad/singlet because it does not couple with neighbouring protons, and exchangeable because \( OH \) protons swap rapidly with deuterium in \( D_2O \)). The signal at \( \delta \) 1.20 (6H, doublet) indicates two equivalent \( CH_3 \) groups, each coupled to a single adjacent proton (n+1 rule, n=1 gives a doublet). The signal at \( \delta \) 4.00 (1H, septet) indicates one proton adjacent to six equivalent protons (n+1 rule, n=6 gives a septet) and, at this relatively downfield shift, is attached to the carbon bearing the \( OH \) group. This pattern (two equivalent \( CH_3 \) doublets coupled to one \( CH \) septet, plus an exchangeable \( OH \)) is consistent with propan-2-ol, \( (CH_3)_2CHOH \).

1 mark: \( \delta \) 2.60 singlet, exchangeable with \( D_2O \), correctly assigned to \( -OH \); 1 mark: \( \delta \) 1.20 (6H, doublet) correctly assigned to two equivalent \( CH_3 \) groups coupled to one adjacent proton; 1 mark: \( \delta \) 4.00 (1H, septet) correctly assigned to the \( CH \) proton coupled to six equivalent protons; 1 mark: correct overall structure, propan-2-ol \( (CH_3)_2CHOH \), consistent with \( C_3H_8O \). [4]

Step 1: bromoethane is heated under reflux with potassium cyanide dissolved in ethanol, \( CH_3CH_2Br + CN^- \rightarrow CH_3CH_2CN + Br^- \), a nucleophilic substitution reaction, giving propanenitrile (note this extends the carbon chain by one carbon, since the nucleophile \( CN^- \) supplies a new C atom). Step 2: propanenitrile is reduced using lithium aluminium hydride, \( LiAlH_4 \), in dry ether (or alternatively using hydrogen gas with a nickel catalyst), giving propan-1-amine, \( CH_3CH_2CH_2NH_2 \). This nitrile route is preferred over direct reaction of the haloalkane with excess ammonia, since direct reaction of a haloalkane with ammonia tends to give a mixture of primary, secondary, tertiary amines and quaternary ammonium salt, whereas reduction of a nitrile gives only the single primary amine product.

1 mark: Step 1 reagent/conditions correct (KCN, ethanolic solution, reflux) with product propanenitrile; 1 mark: correct reasoning that this extends the carbon chain by one carbon (via nucleophilic substitution); 1 mark: Step 2 reagent/conditions correct (\( LiAlH_4 \) in dry ether, or \( H_2 \)/Ni catalyst); 1 mark: correct final product, propan-1-amine, with a valid reason why this route avoids a mixture of amines. [4]

The solution changes colour from pink to blue. This occurs because the smaller water ligands (coordination number 6, octahedral geometry) are substituted by the larger chloride ions, which can only fit four around the cobalt ion, giving a tetrahedral complex, \( [CoCl_4]^{2-} \) (coordination number 4). This change in coordination number and geometry alters the way the d orbitals are split by the ligand field (a smaller splitting energy in the tetrahedral complex), so a different wavelength of visible light is absorbed, changing the observed colour from pink to blue.

1 mark: correct colour change stated (pink to blue); 1 mark: correct reference to the change in coordination number/geometry (6, octahedral, to 4, tetrahedral); 1 mark: correct link to a change in d-orbital splitting energy causing the colour change. [3]

Each \( -OH \) group of the diol condenses with a \( -COOH \) group of the diacid, losing a molecule of water and forming an ester (\( -CO-O- \)) linkage at every join; this repeats to build up a long-chain polyester (poly(ethylene terephthalate), PET). The repeat unit can be represented as \( -[-O-CH_2CH_2-O-CO-C_6H_4-CO-]- \), where \( C_6H_4 \) represents the 1,4-disubstituted benzene ring from the diacid, with a molecule of water lost at each of the two ester linkages formed per repeat unit.

1 mark: correct repeat unit structure/description showing the diol and diacid units joined; 1 mark: ester linkage (\( -CO-O- \)) correctly identified at each join; 1 mark: correct reference to loss of water at each linkage (condensation polymerisation). [3]

(a) Titre 1 \( = 23.45-0.05=23.40 \text{ cm}^3 \); Titre 2 \( = 23.40-0.10=23.30 \text{ cm}^3 \); Titre 3 \( = 23.50-0.15=23.35 \text{ cm}^3 \). (b) The rough titre (24.60) is not concordant and is discarded. Titres 1, 2 and 3 (23.40, 23.30, 23.35) all lie within 0.10 cm³ of one another and are concordant. Mean \( = \dfrac{23.40+23.30+23.35}{3} = 23.35 \text{ cm}^3 \). (c) Moles NaOH \( = 0.150 \times 0.0250 = 3.75\times10^{-3} \text{ mol} \). Since \( NaOH + HCl \rightarrow NaCl + H_2O \) is 1:1, moles HCl \( = 3.75\times10^{-3} \text{ mol} \), in 23.35 cm³. Concentration \( = \dfrac{3.75\times10^{-3}}{0.02335} = 0.161 \text{ mol dm}^{-3} \) (3 s.f.).

(a) [3]: 1 mark each for Titre 1 = 23.40, Titre 2 = 23.30, Titre 3 = 23.35 cm³. (b) [3]: 1 mark for correctly excluding the rough titre; 1 mark for correctly identifying titres 1–3 as concordant (within 0.10 cm³); 1 mark for correct mean titre 23.35 cm³ (allow follow-through from candidate's own titre values). (c) [4]: 1 mark for moles NaOH correctly calculated (3.75×10⁻³ mol); 1 mark for correct 1:1 mole ratio applied to find moles HCl; 1 mark for correct method dividing by mean titre volume in dm³; 1 mark for correct final answer 0.161 mol dm⁻³ (accept follow-through from candidate's mean titre). [10]

(a) With NaOH added dropwise, an orange-brown gelatinous precipitate of iron(III) hydroxide forms immediately; on addition of excess NaOH, the precipitate does not dissolve/redissolve, remaining orange-brown and insoluble. (b) With aqueous ammonia added dropwise, the same orange-brown gelatinous precipitate of iron(III) hydroxide forms; in excess ammonia, the precipitate again does not dissolve (unlike, for example, copper(II) hydroxide, which forms a soluble ammine complex). (c) \( Fe^{3+}(aq) + 3OH^-(aq) \rightarrow Fe(OH)_3(s) \). (d) On adding potassium thiocyanate solution, an intense blood-red colouration forms; this confirms the presence of \( Fe^{3+} \) ions, which form the complex \( [Fe(SCN)(H_2O)_5]^{2+} \). (e) The \( Fe^{3+} \) ion, \( [Ar]3d^5 \), has a partially filled 3d subshell, so electrons can be promoted between the split d orbitals by absorbing visible light (a d–d transition), giving a coloured solution. The \( Zn^{2+} \) ion, \( [Ar]3d^{10} \), has a completely full 3d subshell, so no d–d transition is possible (there are no vacant d orbitals of the same subshell for an electron to be promoted into), and \( Zn^{2+} \) solutions are colourless. (f) The cation present in X is \( Fe^{3+} \).

(a) [3]: 1 mark orange-brown precipitate formed with dropwise NaOH; 1 mark precipitate described as gelatinous; 1 mark precipitate insoluble/unchanged in excess NaOH. (b) [3]: 1 mark orange-brown precipitate formed with dropwise ammonia; 1 mark correct comparison/consistency with (a); 1 mark precipitate insoluble in excess ammonia. (c) [2]: 1 mark correct species (\( Fe^{3+} \), \( OH^- \), \( Fe(OH)_3 \)); 1 mark correctly balanced ionic equation with state symbols. (d) [2]: 1 mark blood-red colouration observed; 1 mark correctly confirms \( Fe^{3+} \) present. (e) [3]: 1 mark reference to partially filled 3d subshell in \( Fe^{3+} \) allowing d–d transitions; 1 mark reference to full 3d¹⁰ subshell in \( Zn^{2+} \) preventing transitions; 1 mark correct overall link to absorption of visible light causing colour. (f) [2]: 1 mark \( Fe^{3+} \) stated; 1 mark consistent with all observations given. [15]

First, add a small amount of solid sodium carbonate (or aqueous sodium hydrogencarbonate) to a sample of each liquid: propanoic acid produces effervescence (bubbles of colourless gas, \( CO_2 \)), while propanal and propan-1-ol show no reaction, since neither is acidic enough to react. This identifies the propanoic acid. To distinguish the remaining two, warm samples of each with Tollens' reagent (or Fehling's solution): propanal, an aldehyde, is oxidised and gives a silver mirror with Tollens' reagent (or a brick-red precipitate with Fehling's), while propan-1-ol, an alcohol with no carbonyl group, gives no reaction with either reagent under these mild conditions.

1 mark: correct first test (sodium carbonate/hydrogencarbonate) applied to all three samples; 1 mark: correct positive observation for propanoic acid (effervescence/bubbles of \( CO_2 \)); 1 mark: correct second test named (Tollens' reagent or Fehling's solution, warmed); 1 mark: correct positive observation for propanal (silver mirror, or brick-red precipitate); 1 mark: correct negative observation for propan-1-ol (no reaction) confirming its identity by elimination. [5]

An accurately weighed mass of anhydrous sodium carbonate (weighed by difference on a balance reading to at least 2 decimal places) is dissolved in a small volume of distilled water and transferred quantitatively, with rinsings, into a 250 cm³ volumetric flask; distilled water is added up to the graduation mark (using a dropping pipette for the final additions), and the flask is stoppered and inverted repeatedly to mix, giving a standard solution of accurately known concentration. A 25.0 cm³ aliquot of this standard sodium carbonate solution is transferred, using a pipette and pipette filler, into a conical flask, and a few drops of methyl orange indicator are added. The sodium hydroxide solution is placed in a burette (rinsed first with a little of the NaOH solution) and titrated into the conical flask, swirling constantly, until the indicator just changes from yellow to orange, marking the end point. The titration is repeated until at least two concordant titres (within 0.10 cm³) are obtained, and the mean of the concordant titres is used, together with the known moles of sodium carbonate and the equation \( Na_2CO_3 + 2NaOH \rightarrow ... \) (more precisely \( Na_2CO_3 + 2HCl \), so here the reverse standardisation actually treats \( Na_2CO_3 \) as reacting via its basic carbonate protonation) to calculate the concentration of the sodium hydroxide solution.

1 mark: accurate weighing of \( Na_2CO_3 \) and dissolving in water; 1 mark: quantitative transfer (with rinsings) to a 250 cm³ volumetric flask made up to the mark; 1 mark: correct use of a pipette to measure a 25.0 cm³ aliquot into a conical flask; 1 mark: correct indicator named (methyl orange) with correct end-point colour change; 1 mark: sodium hydroxide correctly placed in a rinsed burette and titrated with swirling; 1 mark: repetition to concordance (within 0.10 cm³) and use of a mean titre in the calculation. [6]

Concentrated aqueous ammonia is added, with stirring, to a concentrated solution of copper(II) sulfate in a beaker, in a fume cupboard, until well in excess; the initial pale blue precipitate of \( Cu(OH)_2 \) formed with the first few drops redissolves to give a deep/royal blue solution of the complex \( [Cu(NH_3)_4(H_2O)_2]^{2+} \). A small volume of ethanol is then added carefully to this solution, which reduces the solubility of the complex salt and encourages crystals to form; the mixture is left to stand (e.g. in an ice bath, or covered, to slow evaporation and allow good crystals to grow) until crystallisation is complete. The crystals are then separated from the remaining solution by filtration under reduced pressure using a Büchner funnel and flask, and are washed with a small volume of ice-cold ethanol (which will not dissolve much of the product) to remove surface impurities without dissolving the crystals. Finally, the crystals are dried between sheets of filter paper or left in a desiccator at room temperature; they are not dried in a hot oven, since heating could decompose the complex or drive off the ammonia/water of crystallisation.

1 mark: excess concentrated ammonia added to \( CuSO_4(aq) \) with correct observation (deep blue solution forms); 1 mark: reference to carrying this out in a fume cupboard (ammonia is hazardous); 1 mark: ethanol added to reduce solubility and aid crystallisation; 1 mark: correct filtration method (Büchner funnel, reduced pressure); 1 mark: correct washing step (small volume of cold ethanol/solvent); 1 mark: correct drying method (desiccator or between filter paper, not a hot oven, to avoid decomposition). [6]

Moles \( S_2O_3^{2-} = 0.0500 \times 0.01985 = 9.925\times10^{-4} \text{ mol} \). Moles \( I_2 = \tfrac{1}{2}(9.925\times10^{-4}) = 4.9625\times10^{-4} \text{ mol} \), and by the 1:1 ratio in the first equation, moles \( ClO^- = 4.9625\times10^{-4} \text{ mol} \) (in the 25.0 cm³ portion). Scaling up to the full 250 cm³ diluted solution (×10): moles \( ClO^- = 4.9625\times10^{-3} \text{ mol} \), which was present in the original 10.0 cm³ of bleach. Concentration \( = \dfrac{4.9625\times10^{-3}}{0.0100} = 0.496 \text{ mol dm}^{-3} \) (3 s.f.).

1 mark: moles \( S_2O_3^{2-} \) and hence moles \( I_2 \) (using 2:1 ratio) correctly calculated; 1 mark: correct 1:1 ratio used to find moles \( ClO^- \) in the 25.0 cm³ portion; 1 mark: correct scaling by ×10 to the full 250 cm³ diluted solution; 1 mark: correct final concentration in the original bleach, 0.496 mol dm⁻³. [4]

Initial moles HCl \( = 0.500 \times 0.0500 = 0.0250 \text{ mol} \). Moles NaOH used (= moles unreacted HCl) \( = 0.400 \times 0.02260 = 9.04\times10^{-3} \text{ mol} \). Moles HCl that reacted with \( CaCO_3 \): \( 0.0250 - 9.04\times10^{-3} = 0.01596 \text{ mol} \). From \( CaCO_3 + 2HCl \rightarrow CaCl_2 + H_2O + CO_2 \), moles \( CaCO_3 = \dfrac{0.01596}{2} = 7.98\times10^{-3} \text{ mol} \). Mass \( CaCO_3 = 7.98\times10^{-3} \times 100 = 0.798 \text{ g} \). Percentage purity \( = \dfrac{0.798}{1.00}\times100 = 79.8\% \).

1 mark: initial moles HCl correctly calculated; 1 mark: moles unreacted HCl correctly calculated from moles NaOH; 1 mark: moles HCl reacted with \( CaCO_3 \) found and correct 2:1 ratio applied to find moles \( CaCO_3 \); 1 mark: correct final answer, 79.8% (allow follow-through). [4]

\( E_{cell} = E^{\ominus}(MnO_4^-/Mn^{2+}) - E^{\ominus}(Cl_2/Cl^-) = 1.51 - 1.36 = +0.15 \text{ V} \). Since this is positive, the reaction (\( MnO_4^- \) oxidising \( Cl^- \) to \( Cl_2 \)) is thermodynamically feasible. This means that if concentrated hydrochloric acid were used to acidify a potassium manganate(VII) titration, some of the \( MnO_4^- \) would be consumed oxidising the \( Cl^- \) ions (from the acid itself) to chlorine gas, rather than only reacting with the intended analyte. This side reaction would give an inaccurately high titre for \( KMnO_4 \), so dilute sulfuric acid (whose \( SO_4^{2-} \)/\( HSO_4^- \) ions are not readily oxidised by \( MnO_4^- \)) is used instead.

1 mark: correct EMF calculation method and value, +0.15 V; 1 mark: correct conclusion that the reaction is feasible (positive EMF); 1 mark: correct explanation that \( Cl^- \) from HCl would be oxidised by \( MnO_4^- \), consuming extra \( KMnO_4 \); 1 mark: correct link to an inaccurately high/incorrect titre, and dilute sulfuric acid given as the appropriate acid to use instead. [4]

By convention, the half-cell with the more negative electrode potential (here \( Ni^{2+}/Ni \), the anode/oxidation) is written on the left, and the half-cell with the more positive potential (here \( Ag^+/Ag \), the cathode/reduction) is written on the right, with a double vertical line representing the salt bridge: \( Ni(s)\,|\,Ni^{2+}(aq)\,||\,Ag^+(aq)\,|\,Ag(s) \). \( E_{cell} = E^{\ominus}(Ag^+/Ag) - E^{\ominus}(Ni^{2+}/Ni) = 0.80 - (-0.25) = +1.05 \text{ V} \).

1 mark: correct cell diagram with \( Ni \) half-cell on the left and \( Ag \) half-cell on the right; 1 mark: correct use of single and double vertical lines (phase boundary and salt bridge); 1 mark: correct EMF calculation method; 1 mark: correct value, +1.05 V. [4]

The single \( CHCl_2 \) proton is adjacent to the two equivalent protons of the \( CH_2Cl \) group, so by the n+1 rule (n=2) its signal is split into a triplet. The two equivalent \( CH_2Cl \) protons are adjacent to the single \( CHCl_2 \) proton, so by the n+1 rule (n=1) their signal is split into a doublet.

1 mark: correct multiplicity for \( CHCl_2 \) (triplet); 1 mark: correct multiplicity for \( CH_2Cl \) (doublet); 1 mark: correct reasoning/application of the n+1 rule shown for both. [3]

The absorption at 1735 cm⁻¹ is characteristic of a \( C=O \) stretch in an ester (typical range approximately 1735–1750 cm⁻¹). The absence of any broad absorption above 3000 cm⁻¹ rules out an \( O-H \) (carboxylic acid or alcohol) or \( N-H \) group. Two singlets in a 1:1 (3H:3H) ratio, with no coupling between them, indicate two isolated methyl groups with no protons on adjacent carbons. This is consistent with methyl ethanoate, \( CH_3COOCH_3 \), where the carbonyl group at 1735 cm⁻¹ and the two isolated \( CH_3 \) environments (\( CH_3CO- \) and \( -OCH_3 \)) match all the data.

1 mark: 1735 cm⁻¹ correctly assigned to an ester \( C=O \) group; 1 mark: correct reasoning that the absence of absorption above 3000 cm⁻¹ rules out \( O-H \)/\( N-H \); 1 mark: correct final structure (methyl ethanoate) consistent with two isolated \( CH_3 \) singlets in a 1:1 ratio. [3]

The halogen present is chlorine. Chlorine occurs naturally as two isotopes, \( ^{35}Cl \) and \( ^{37}Cl \), in a natural abundance ratio of approximately 3:1. Molecules containing \( ^{35}Cl \) give the lower mass molecular ion peak (m/z = 78) and those containing the heavier \( ^{37}Cl \) isotope give a peak two mass units higher (m/z = 80), with peak heights in the same 3:1 ratio as the natural isotopic abundance.

1 mark: halogen correctly identified as chlorine; 1 mark: correct reference to the two isotopes \( ^{35}Cl \) and \( ^{37}Cl \); 1 mark: correct explanation that the 3:1 peak ratio reflects the natural isotopic abundance ratio. [3]

Since \( E^{\ominus}(Zn^{2+}/Zn) = -0.76 \text{ V} \) is more negative than the standard hydrogen electrode (0 V by definition), zinc is written on the left (as the anode/oxidation half-cell) and the hydrogen electrode on the right (as the cathode/reduction half-cell): \( Zn(s)\,|\,Zn^{2+}(aq)\,||\,H^+(aq)\,|\,H_2(g)\,|\,Pt(s) \). By convention, the more negative electrode (zinc) is the negative terminal of the cell, and electrons flow from it, through the external circuit, to the more positive electrode (the hydrogen/platinum electrode).

1 mark: correct cell diagram with Zn half-cell on the left and hydrogen electrode (including Pt) on the right, correct use of single/double lines; 1 mark: zinc correctly identified as the negative electrode (more negative \( E^{\ominus} \)); 1 mark: correct statement that electrons flow externally from Zn to the hydrogen electrode. [3]

The salt bridge completes the electrical circuit between the two half-cells by allowing ions to flow between them, which maintains overall electrical neutrality in each half-cell as electrons flow through the external wire (ions migrate to balance the charge building up as the reactions proceed), without allowing the two electrode solutions to mix directly and react with each other. The ions used (e.g. \( K^+ \) and \( NO_3^- \)) should be chemically inert, i.e. unreactive with either electrode solution and not easily oxidised or reduced themselves, and should have high/similar ionic mobility so that they can migrate freely and quickly to maintain the charge balance.

1 mark: correct explanation that the salt bridge completes the circuit by allowing ion flow while keeping the two solutions from mixing directly; 1 mark: reference to maintaining electrical neutrality/charge balance in each half-cell; 1 mark: two valid properties given (e.g. chemically inert/unreactive with electrode solutions; high ionic mobility). [3]

In GLC, a gaseous mobile phase (carrier gas) carries the vaporised sample components through a column coated with, or packed with a solid support coated in, a stationary liquid phase. Components separate because they distribute (partition) differently between the mobile gas phase and the stationary liquid phase, depending on their relative volatility and their affinity/solubility for the stationary liquid: a component that spends more time dissolved in the stationary phase moves through the column more slowly and has a longer retention time, while a component that remains mostly in the gas phase moves through faster. Besides the identity of the stationary phase, the retention time of a given component is also affected by the temperature of the column (higher temperature increases volatility, generally decreasing retention time) or by the flow rate of the carrier gas.

1 mark: correct reference to partition of components between the mobile gas phase and stationary liquid phase; 1 mark: correct explanation that retention time depends on relative affinity/solubility for the stationary phase (or relative volatility); 1 mark: one valid additional factor correctly stated (e.g. column temperature, or carrier gas flow rate). [3]