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(a) A Brønsted–Lowry acid is a proton (\(\text{H}^+\)) donor. The dissociation of a weak acid HA is: \(\text{HA(aq)} \rightleftharpoons \text{H}^+\text{(aq)} + \text{A}^-\text{(aq)}\). The acid dissociation constant is: \(K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}\). (b) For a weak acid, we assume \([\text{H}^+] \approx [\text{A}^-]\) and that dissociation is negligible, so \([\text{HA}] \approx 0.120\text{ mol dm}^{-3}\). \(K_a = \frac{[\text{H}^+]^2}{[\text{HA}]} \Rightarrow [\text{H}^+] = \sqrt{K_a \times [\text{HA}]} = \sqrt{1.35 \times 10^{-5} \times 0.120} = 1.273 \times 10^{-3}\text{ mol dm}^{-3}\). \(\text{pH} = -\log_{10}(1.273 \times 10^{-3}) = 2.895 \approx 2.89\). (c)(i) Initial moles of \(\text{CH}_3\text{CH}_2\text{COOH} = 0.0500\text{ dm}^3 \times 0.120\text{ mol dm}^{-3} = 6.00 \times 10^{-3}\text{ mol}\). Moles of \(\text{NaOH}\) added = \(0.0250\text{ dm}^3 \times 0.100\text{ mol dm}^{-3} = 2.50 \times 10^{-3}\text{ mol}\). \(\text{CH}_3\text{CH}_2\text{COOH}\) reacts with \(\text{NaOH}\) in a 1:1 ratio: Moles of \(\text{CH}_3\text{CH}_2\text{COOH}\) remaining = \(6.00 \times 10^{-3} - 2.50 \times 10^{-3} = 3.50 \times 10^{-3}\text{ mol}\). Moles of \(\text{CH}_3\text{CH}_2\text{COO}^-\text{Na}^+\) formed = \(2.50 \times 10^{-3}\text{ mol}\). Total volume of mixture = \(50.0 + 25.0 = 75.0\text{ cm}^3 = 0.0750\text{ dm}^3\). \([\text{CH}_3\text{CH}_2\text{COOH}] = \frac{3.50 \times 10^{-3}\text{ mol}}{0.0750\text{ dm}^3} = 0.0467\text{ mol dm}^{-3}\). \([\text{CH}_3\text{CH}_2\text{COO}^-] = \frac{2.50 \times 10^{-3}\text{ mol}}{0.0750\text{ dm}^3} = 0.0333\text{ mol dm}^{-3}\). (ii) \(\text{pH} = \text{p}K_a + \log_{10}\left(\frac{[\text{CH}_3\text{CH}_2\text{COO}^-]}{[\text{CH}_3\text{CH}_2\text{COOH}]}\right)\). \(\text{p}K_a = -\log_{10}(1.35 \times 10^{-5}) = 4.870\). \(\text{pH} = 4.870 + \log_{10}\left(\frac{0.0333}{0.0467}\right) = 4.870 - 0.147 = 4.723 \approx 4.72\).

(a) [2 marks total]: 1 mark for defining Brønsted–Lowry acid as a proton donor; 1 mark for correct expression of \(K_a\). (b) [3 marks total]: 1 mark for using the weak acid approximation equation; 1 mark for calculating \([\text{H}^+] = 1.27 \times 10^{-3}\text{ mol dm}^{-3}\); 1 mark for calculating pH = 2.89 (accept 2.9). (c)(i) [4.5 marks total]: 1 mark for calculating initial moles of acid and NaOH; 1 mark for calculating remaining moles of acid; 1 mark for calculating moles of propanoate ions; 1.5 marks for dividing by total volume (0.075 dm3) to find both concentrations correctly. (c)(ii) [3 marks total]: 1 mark for calculating \(\text{p}K_a = 4.87\); 1 mark for setting up the buffer equation with correct concentrations; 1 mark for final pH value of 4.72 (accept 4.7).

(a)(i) Comparing Expt 1 and Expt 2: \([\text{I}_2]\) and \([\text{H}^+]\) are kept constant. When \([\text{CH}_3\text{COCH}_3]\) is doubled (from 0.40 to 0.80), the rate doubles (from \(1.20 \times 10^{-5}\) to \(2.40 \times 10^{-5}\)). Thus, the order with respect to propanone is 1. Comparing Expt 1 and Expt 3: \([\text{CH}_3\text{COCH}_3]\) and \([\text{H}^+]\) are kept constant. When \([\text{I}_2]\) is doubled (from 0.005 to 0.010), the rate remains unchanged (\(1.20 \times 10^{-5}\)). Thus, the order with respect to iodine is 0. Comparing Expt 1 and Expt 4: \([\text{CH}_3\text{COCH}_3]\) and \([\text{I}_2]\) are kept constant. When \([\text{H}^+]\) is doubled (from 0.05 to 0.10), the rate doubles (from \(1.20 \times 10^{-5}\) to \(2.40 \times 10^{-5}\)). Thus, the order with respect to \([\text{H}^+]\) is 1. (ii) The rate equation is: \(\text{Rate} = k[\text{CH}_3\text{COCH}_3][\text{H}^+]\). (iii) Using Expt 1 data: \(1.20 \times 10^{-5} = k (0.40)(0.05) \Rightarrow k = \frac{1.20 \times 10^{-5}}{0.020} = 6.00 \times 10^{-4}\). Units of \(k\): \(\text{mol dm}^{-3}\text{ s}^{-1} / (\text{mol dm}^{-3} \times \text{mol dm}^{-3}) = \text{mol}^{-1}\text{ dm}^3\text{ s}^{-1}\). (b) A plausible mechanism involves: Step 1 (slow, rate-determining step): \(\text{CH}_3\text{COCH}_3 + \text{H}^+ \rightarrow \text{CH}_3\text{C(OH)}^+ \text{CH}_3\) (or rearrangement to enol intermediate). Step 2 (fast step): \(\text{enol} + \text{I}_2 \rightarrow \text{CH}_3\text{COCH}_2\text{I} + \text{H}^+ + \text{I}^-\).

(a)(i) [4.5 marks total]: 1.5 marks for deducing first-order for propanone with full explanation; 1.5 marks for deducing zero-order for iodine with full explanation; 1.5 marks for deducing first-order for H+ with full explanation. (a)(ii) [1 mark]: Correct rate equation corresponding to deduced orders. (a)(iii) [3 marks total]: 2 marks for correct calculation of \(k = 6.00 \times 10^{-4}\); 1 mark for correct units (mol-1 dm3 s-1). (b) [4 marks total]: 2 marks for Step 1 (must include reactants in rate equation, labelled as 'slow' or 'rate-determining step'); 2 marks for Step 2 (must show correct stoichiometry of final products, and be labelled 'fast').

(a) Lattice energy is the enthalpy change when one mole of an ionic crystalline solid is formed from its constituent gaseous ions under standard conditions. First electron affinity is the enthalpy change when one mole of gaseous atoms gains one mole of electrons to form one mole of singly charged gaseous negative ions. (b) By Hess's Law: \(\Delta H_{\text{f}}^{\ominus}[\text{CaS(s)}] = \Delta H_{\text{at}}^{\ominus}[\text{Ca(s)}] + \text{IE}_1[\text{Ca(g)}] + \text{IE}_2[\text{Ca(g)}] + \Delta H_{\text{at}}^{\ominus}[\text{S(s)}] + \text{EA}_1[\text{S(g)}] + \text{EA}_2[\text{S(g)}] + \Delta H_{\text{latt}}^{\ominus}[\text{CaS(s)}]\). Substituting values: \(-482 = +178 + 590 + 1145 + 279 + (-200) + 640 + \Delta H_{\text{latt}}^{\ominus}\). \(-482 = 2632 + \Delta H_{\text{latt}}^{\ominus}\). \(\Delta H_{\text{latt}}^{\ominus} = -482 - 2632 = -3114\text{ kJ mol}^{-1}\). (c) The first electron affinity is exothermic because the incoming electron is attracted to the positively charged nucleus of the neutral sulfur atom, releasing energy. The second electron affinity is endothermic because an electron is being added to a negatively charged \(\text{S}^-\text{(g)}\) ion. Energy must be supplied to overcome the strong electrostatic repulsion between the negative ion and the incoming negative electron.

(a) [3 marks total]: 1.5 marks for lattice energy definition (must specify 1 mole of solid formed from gaseous ions under standard conditions); 1.5 marks for first electron affinity definition (must specify 1 mole of gaseous atoms gaining 1 mole of electrons). (b) [5.5 marks total]: 2 marks for setting up a correct Born-Haber cycle equation; 2.5 marks for correct substitution and calculations; 1 mark for correct sign and units (-3114 kJ mol-1). (c) [4 marks total]: 2 marks for explaining first electron affinity (attraction between nucleus and incoming electron); 2 marks for explaining second electron affinity (repulsion between negative ion and incoming electron, requiring energy input).

(a) A ligand is an atom, ion, or molecule that can donate a lone pair of electrons to a central metal atom or ion. A coordinate bond is formed when the ligand donates its non-bonding lone pair of electrons into an empty d- or hybrid orbital of the transition metal ion. (b) The electronic configuration of a copper atom is \([\text{Ar}] 4\text{s}^1 3\text{d}^{10}\). To form a \(\text{Cu}^{2+}\) ion, it loses the \(4\text{s}\) electron first and then one \(3\text{d}\) electron, yielding: \(1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6 3\text{d}^9\) (or \([\text{Ar}] 3\text{d}^9\)). (c)(i) \([\text{Cu}(\text{H}_2\text{O})_6]^{2+}\text{(aq)} + 4\text{Cl}^-\text{(aq)} \rightleftharpoons [\text{CuCl}_4]^{2-}\text{(aq)} + 6\text{H}_2\text{O(l)}\). (ii) The coordination number changes from 6 to 4. The geometry changes from octahedral to tetrahedral. (iii) Concentrated HCl provides a very high concentration of \(\text{Cl}^-\text{(aq)}\) ions. This shifts the position of the equilibrium to the right-hand side (Le Chatelier's principle), even though water molecules are typically stronger ligands than chloride ions (higher \(K_{\text{stab}}\)). (iv) A 3D diagram representing a tetrahedral geometry around the central \(\text{Cu}\) atom, with four bonded \(\text{Cl}\) ligands showing wedges and dashes to illustrate depth, and a total charge of \(2-\).

(a) [2 marks total]: 1 mark for ligand definition (must mention lone pair donor); 1 mark for coordinate bond explanation (overlap of lone pair orbital with empty metal orbital). (b) [1.5 marks]: Correct electronic configuration of Cu2+. (c)(i) [2 marks]: Balanced equation showing reversible arrow and correct charges. (c)(ii) [2 marks]: 1 mark for stating coordination number change 6 to 4; 1 mark for stating geometry change octahedral to tetrahedral. (c)(iii) [3 marks]: 1.5 marks for mentioning high concentration of Cl- shifts equilibrium right; 1.5 marks for linking to the relative stability of complexes (Kstab). (c)(iv) [2 marks]: 3D tetrahedral drawing with wedge, dash, solid lines and correct outer charge.

(a) A bidentate ligand is a species that donates two lone pairs of electrons from different atoms within the same molecule to form two coordinate bonds with a single transition metal ion. A common neutral example is 1,2-diaminoethane, \(\text{H}_2\text{N-CH}_2\text{-CH}_2\text{-NH}_2\). (b)(i) Since there are two bidentate ligands (each donating 2 lone pairs) and two monodentate chloride ligands, the coordination number is \(2 \times 2 + 2 = 6\). (ii) It exhibits geometrical (cis-trans) isomerism because the two chloride ligands can be situated adjacent to each other at a \(90^\circ\) angle (cis) or opposite to each other at a \(180^\circ\) angle (trans). It also exhibits optical isomerism because the cis-isomer is asymmetric (non-superimposable on its mirror image). (iii) The trans-isomer has chloride ligands opposite (top and bottom), and the two \(\text{en}\) ligands wrapped around the equatorial plane. The cis-isomer has chloride ligands next to each other (e.g., top and adjacent equatorial positions), and its mirror image cannot be superimposed. (iv) The cis-isomer is optically active. Optical activity can be demonstrated by passing plane-polarised light through a solution of the isolated enantiomer; the isomer will rotate the plane of polarisation.

(a) [2 marks total]: 1 mark for bidentate ligand definition (donates 2 lone pairs to form 2 coordinate bonds); 1 mark for neutral example (1,2-diaminoethane). (b)(i) [1 mark]: Coordination number = 6. (b)(ii) [3 marks total]: 1.5 marks for explaining geometrical (cis-trans) isomerism; 1.5 marks for explaining optical isomerism. (b)(iii) [4 marks total]: 2 marks for clear 3D drawing of trans-isomer; 2 marks for cis-isomer with 3D representation. (b)(iv) [2.5 marks total]: 1.5 marks for identifying the cis-isomer as optically active; 1 mark for explaining rotation of plane-polarised light.

(a) In benzene, each carbon atom is \(\text{sp}^2\) hybridised and forms three \(\sigma\) bonds (two with adjacent carbon atoms and one with a hydrogen atom). This results in a planar hexagonal ring with bond angles of \(120^\circ\). Each carbon atom has one unhybridised p-orbital containing a single electron perpendicular to the plane of the ring. These six p-orbitals overlap sideways both above and below the plane to form a continuous ring of delocalised \(\pi\) electrons. (b)(i) The generation of the electrophile: \(\text{HNO}_3 + 2\text{H}_2\text{SO}_4 \rightarrow \text{NO}_2^+ + \text{H}_3\text{O}^+ + 2\text{HSO}_4^-\) (or \(\text{HNO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{NO}_2^+ + \text{HSO}_4^- + \text{H}_2\text{O}\)). (ii) Electrophilic substitution mechanism: A curly arrow goes from the delocalised \(\pi\) ring of benzene to the \(\text{NO}_2^+\) ion. A carbocation intermediate is formed containing a positive charge inside a partial ring of delocalisation (broken ring over 5 carbons, open towards the sp3 carbon). A curly arrow goes from the C-H bond of the intermediate back to the ring to reform the delocalised system, releasing \(\text{H}^+\). (c)(i) The methyl group is an electron-donating group and is 2,4-directing. Therefore, the products are 2-nitromethylbenzene (2-nitrotoluene) and 4-nitromethylbenzene (4-nitrotoluene). (ii) Alkaline \(\text{KMnO}_4\) followed by acid oxidises the alkyl side chain completely to form benzoic acid, \(\text{C}_6\text{H}_5\text{COOH}\).

(a) [4.5 marks total]: 1.5 marks for describing sp2 hybridisation, planar hexagonal shape and 120-degree angles; 1.5 marks for describing C-C and C-H sigma bonds; 1.5 marks for describing the continuous delocalised pi system formed by sideways overlap of p-orbitals. (b)(i) [2 marks]: Balanced equation showing generation of NO2+. (b)(ii) [4 marks total]: 1 mark for correct curly arrow from benzene ring to NO2+; 1.5 marks for correct structure of intermediate (horseshoe showing + charge); 1.5 marks for curly arrow from C-H bond back into the ring. (c)(i) [1 mark]: Correct structures of 2-nitromethylbenzene and 4-nitromethylbenzene. (c)(ii) [1 mark]: Correct structure or formula of benzoic acid.

(a)(i) Step 1 is the nitration of methylbenzene. Reagents: concentrated nitric acid (\(\text{HNO}_3\)) and concentrated sulfuric acid (\(\text{H}_2\text{SO}_4\)). Conditions: temperature held around \(50-60\ ^∘\text{C}\). (ii) Step 2 is the oxidation of the methyl side chain to a carboxylic acid group. Reagents: alkaline potassium manganate(VII) (\(\text{KMnO}_4\)), followed by acidification with dilute hydrochloric acid. Conditions: heat under reflux. (iii) Step 3 is the reduction of the nitro group to an amine group. Reagents: tin (\(\text{Sn}\)) and concentrated hydrochloric acid, followed by addition of aqueous sodium hydroxide. Conditions: heat under reflux. (b) The intermediate for Step 1 is 4-nitromethylbenzene, which contains a benzene ring with a \(-\text{CH}_3\) group at carbon 1 and a \(-\text{NO}_2\) group at carbon 4. The intermediate for Step 2 is 4-nitrobenzoic acid, containing a \(-\text{COOH}\) group at carbon 1 and a \(-\text{NO}_2\) group at carbon 4. (c) The alternative pathway would fail because the potassium manganate(VII) used to oxidise the methyl group is a very strong oxidising agent. It would also easily oxidise the highly reactive primary amine group (\(-\text{NH}_2\)) of 4-methylphenylamine, leading to unwanted side-products or decomposition. (d) This is an esterification reaction: \(\text{H}_2\text{N-C}_6\text{H}_4\text{-COOH} + \text{CH}_3\text{CH}_2\text{OH} \rightleftharpoons \text{H}_2\text{N-C}_6\text{H}_4\text{-COOCH}_2\text{CH}_3 + \text{H}_2\text{O}\).

(a)(i) [2 marks]: 1 mark for concentrated HNO3 + concentrated H2SO4; 1 mark for temperature of 50-60 °C. (a)(ii) [2 marks]: 1 mark for alkaline KMnO4 and heat under reflux; 1 mark for acidification step. (a)(iii) [2 marks]: 1 mark for Sn and concentrated HCl with reflux; 1 mark for subsequent addition of NaOH. (b) [2 marks]: 1 mark for drawing/naming 4-nitromethylbenzene; 1 mark for drawing/naming 4-nitrobenzoic acid. (c) [2.5 marks]: Explanation that amine groups are easily oxidised by KMnO4, hence side-reactions/destruction of the amine group would occur. (d) [2 marks]: Correct balanced equation with structures of reactants and products.

(a) Order of increasing acidity: ethanol < phenol < benzoic acid. Explanation: Ethoxide ion (conjugate base of ethanol) is destabilised because the ethyl group is electron-donating, concentrating negative charge on the oxygen atom, making it more basic. Phenoxide ion (conjugate base of phenol) is stabilised because the lone pair of electrons on the oxygen atom overlaps with the \(\pi\) system of the benzene ring, delocalising the negative charge and making it less likely to protonate. Benzoate ion (conjugate base of benzoic acid) is highly stabilised because the negative charge is delocalised over two highly electronegative oxygen atoms in the carboxylate group, making it the most stable conjugate base and thus benzoic acid the strongest acid of the three. (b)(i) Phenol reacts rapidly with bromine water at room temperature to form a white precipitate of 2,4,6-tribromophenol. Equation: \(\text{C}_6\text{H}_5\text{OH} + 3\text{Br}_2 \rightarrow \text{C}_6\text{H}_2\text{Br}_3\text{OH} + 3\text{HBr}\). (ii) Phenol reacts with dilute nitric acid at room temperature to form a mixture of 2-nitrophenol and 4-nitrophenol. Equation: \(\text{C}_6\text{H}_5\text{OH} + \text{HNO}_3 \rightarrow \text{C}_6\text{H}_4(\text{NO}_2)\text{OH} + \text{H}_2\text{O}\). (c) Test: Add neutral iron(III) chloride solution, \(\text{FeCl}_3\text{(aq)}\). Observation with phenol: A purple solution/complex forms. Observation with cyclohexanol: No color change (remains yellow-orange).

(a) [5 marks total]: 1 mark for correct order of acidity; 1 mark for explaining ethoxide destabilisation (electron-donating alkyl group); 1.5 marks for explaining phenoxide stabilisation (delocalisation of lone pair into benzene ring); 1.5 marks for explaining benzoate stabilisation (delocalisation over two electronegative oxygen atoms). (b)(i) [3.5 marks total]: 2 marks for correct balanced equation; 1.5 marks for describing observations (decolourisation of bromine and formation of a white precipitate). (b)(ii) [2 marks]: Correct equation showing nitration at position 2 or 4 under mild conditions. (c) [2 marks total]: 1 mark for choosing a suitable test reagent (e.g., FeCl3 or Bromine water); 1 mark for correct observations for both compounds.