2023 AQA IAL Chemistry (9620) Practice Paper with Answers

[Part a] Standard definition: The enthalpy change when 1 mole of an ionic crystal lattice is formed from its constituent gaseous ions under standard conditions. [Part b] Applying Hess's Law: \(\Delta H_{f}^{\ominus} = \Delta H_{at}^{\ominus}(\text{Ca}) + \text{IE}_1(\text{Ca}) + \text{IE}_2(\text{Ca}) + 2 \times \Delta H_{at}^{\ominus}(\text{Cl}) + 2 \times \text{EA}_1(\text{Cl}) + \Delta H_{L}^{\ominus}\). Substituting the values: \(-796 = 178 + 590 + 1145 + 2(121) + 2(-348) + \Delta H_{L}^{\ominus}\). \(-796 = 178 + 590 + 1145 + 242 - 696 + \Delta H_{L}^{\ominus}\). \(-796 = 1459 + \Delta H_{L}^{\ominus}\). \Delta H_{L}^{\ominus} = -796 - 1459 = -2255\text{ kJ mol}^{-1}. [Part c] Calcium chloride is almost purely ionic because the calcium ion has relatively low polarizing power and the chloride ion has low polarizability, resulting in experimental and theoretical values that align well. In zinc iodide, the zinc ion (\(\text{Zn}^{2+}\)) has high polarizing power due to high charge density, and the iodide ion (\(\text{I}^{-}\)) is highly polarizable due to its large size. This leads to electron density being drawn between the nuclei, giving the bonding significant covalent character and causing a large difference between experimental and theoretical lattice enthalpies.

[Part a] 2 marks: 1 mark for '1 mole of solid ionic lattice is formed from gaseous ions', 1 mark for 'under standard conditions'. [Part b] 6 marks: 1 mark for correct expression/Born-Haber cycle setup; 1 mark for multiplying atomisation of chlorine by 2 (242); 1 mark for multiplying electron affinity of chlorine by 2 (-696); 1 mark for summing the endothermic steps correctly (1459); 1 mark for correct arithmetic process; 1 mark for correct final answer with units (-2255 kJ mol^-1). [Part c] 3.67 marks: 1 mark for stating calcium chloride is almost purely ionic; 1.67 marks for explaining zinc iodide has covalent character due to polarization of the iodide ion by the Zn2+ ion; 1 mark for stating polarization is caused by high charge density of Zn2+ or large size of I-.

[Part a] \(\Delta H^{\ominus} = \sum \Delta H_{f}^{\ominus}(\text{products}) - \sum \Delta H_{f}^{\ominus}(\text{reactants}) = [0 + (-111)] - [-348 + 0] = -111 + 348 = +237\text{ kJ mol}^{-1}\). [Part b] \(\Delta S^{\ominus} = \sum S^{\ominus}(\text{products}) - \sum S^{\ominus}(\text{reactants}) = [42 + 198] - [44 + 6] = 240 - 50 = +190\text{ J K}^{-1}\text{ mol}^{-1}\). [Part c] Reaction is feasible when \(\Delta G^{\ominus} \le 0\). At the temperature boundary: \(\Delta G^{\ominus} = \Delta H^{\ominus} - T\Delta S^{\ominus} = 0\). \(T = \frac{\Delta H^{\ominus}}{\Delta S^{\ominus}} = \frac{237 \times 10^3\text{ J mol}^{-1}}{190\text{ J K}^{-1}\text{ mol}^{-1}} = 1247.37\text{ K}\). Rounding to three significant figures gives 1250 K (or 1247 K). [Part d] Referencing \(\Delta G = \Delta H - T\Delta S\), since both \(\Delta H\) and \(\Delta S\) are positive, the \(-T\Delta S\) term is negative. At low temperatures, \(|T\Delta S| < |\Delta H|\), so \(\Delta G\) is positive and the reaction is not feasible. At high temperatures, the \(-T\Delta S\) term dominates, making \(\Delta G\) negative and the reaction feasible.

[Part a] 2 marks: 1 mark for correct method of calculation (products - reactants); 1 mark for correct answer with sign (+237 kJ mol^-1). [Part b] 3 marks: 1 mark for correct expression of S (products - reactants); 1 mark for correct values substituted (240 - 50); 1 mark for correct final answer with units (+190 J K^-1 mol^-1). [Part c] 4 marks: 1 mark for stating condition of feasibility (DG <= 0); 1 mark for rearranging equation to T = DH/DS; 1 mark for converting DH to J (237000) or DS to kJ (0.190); 1 mark for correct temperature value (1247 K to 1250 K). [Part d] 2.67 marks: 1 mark for stating that both DH and DS are positive; 1.67 marks for explaining that at low T, DH dominates so DG is positive, but at high T, -T*DS dominates making DG negative.

[Part a] Concentrated sulfuric acid acts as an acid catalyst (or proton donor). [Part b] \(\text{CH}_3\text{CH(OH)CH}_2\text{CH}_3 \to \text{CH}_3\text{CH=CHCH}_3 + \text{H}_2\text{O}\) (molecular formula \(\text{C}_4\text{H}_9\text{OH} \to \text{C}_4\text{H}_8 + \text{H}_2\text{O}\) is also acceptable). [Part c] The mechanism proceeds as follows: 1) Lone pair of electrons on the oxygen of the OH group attacks the \(\text{H}^+\) ion. 2) The \(\text{C}-\text{O}\) bond breaks as a water molecule is lost, forming a secondary carbocation: \(\text{CH}_3\text{CH}^+\text{CH}_2\text{CH}_3\). 3) A \(\text{C}-\text{H}\) bond on carbon-3 (or carbon-1) breaks, releasing a proton (\(\text{H}^+\)) and donating the pair of electrons to the carbon-carbon bond to form the double bond. [Part d] The three isomers are: 1) But-1-ene, 2) E-but-2-ene (trans-but-2-ene), 3) Z-but-2-ene (cis-but-2-ene). [Part e] Stereoisomerism (specifically E-Z isomerism) arises in but-2-ene because of the restricted rotation about the carbon-carbon double bond, and the fact that both carbons of the double bond are bonded to two different groups (a methyl group and a hydrogen atom). In contrast, but-1-ene has one of its double-bonded carbons bonded to two identical hydrogen atoms, preventing E-Z isomerism.

[Part a] 1 mark: Catalyst / proton donor. [Part b] 1 mark: Balanced equation with correct formulas. [Part c] 4 marks: 1 mark for curly arrow from lone pair on O to H+; 1 mark for curly arrow from C-O bond to O+; 1 mark for correct structure of secondary carbocation; 1 mark for curly arrow from C-H bond to adjacent C-C bond to reform H+. [Part d] 3.67 marks: 1.67 marks for identifying/naming the three isomers (but-1-ene, E-but-2-ene, Z-but-2-ene); 2 marks for drawing clear structural/displayed formulas of all three. [Part e] 2 marks: 1 mark for mentioning restricted rotation of the double bond; 1 mark for stating that both C atoms in the double bond of but-2-ene are bonded to two different groups, whereas but-1-ene has one carbon bonded to two identical H atoms.

[Part a](i) Fermentation of glucose requires yeast (enzymes), anaerobic conditions (no oxygen), and a temperature of \(30-40^{\circ}\text{C}\). The equation is: \(\text{C}_6\text{H}_{12}\text{O}_6 \to 2\text{C}_2\text{H}_5\text{OH} + 2\text{CO}_2\). (ii) Industrial hydration of ethene requires a concentrated phosphoric acid (\(\text{H}_3\text{PO}_4\)) catalyst absorbed on a solid support, a temperature of \(300^{\circ}\text{C}\), and a pressure of \(60-70\text{ atm}\). (iii) Advantage of fermentation: Uses renewable resources (crops containing glucose) and has low energy inputs. Disadvantage: It is a batch process (slow and labor-intensive) and produces a dilute/impure solution that requires fractional distillation to concentrate. [Part b](i) To fully oxidize propan-1-ol to propanoic acid, use acidified potassium dichromate(VI) (\(\text{K}_2\text{Cr}_2\text{O}_7\) in dilute \(\text{H}_2\text{SO}_4\)), heat under reflux, and use an excess of the oxidizing agent. (ii) A suitable chemical test is adding sodium hydrogencarbonate (\(\text{NaHCO}_3\)) solution. Propanoic acid reacts to produce carbon dioxide gas, causing effervescence/bubbling. Propanal does not react (no change). (Alternatively, Tollens' reagent can be used: propanal forms a silver mirror; propanoic acid shows no reaction).

[Part a](i) 3 marks: 2 marks for any two conditions (yeast / anaerobic / temperature 30-40 °C); 1 mark for correct equation. (ii) 2 marks: 1 mark for phosphoric acid catalyst; 1 mark for high temperature (250-350 °C) / high pressure (50-100 atm). (iii) 2 marks: 1 mark for a valid advantage (e.g. renewable, low energy); 1 mark for a valid disadvantage (e.g. batch process, slow, dilute/impure). [Part b](i) 2.67 marks: 1.67 marks for acidified potassium dichromate(VI) (or K2Cr2O7 and H2SO4); 1 mark for reflux and/or excess oxidizing agent. (ii) 2 marks: 1 mark for a suitable reagent (e.g. NaHCO3 / Na2CO3 or Tollens' reagent); 1 mark for correct observations for both compounds.

[Part a](i) Silicon dioxide (\(\text{SiO}_2\)) has a macromolecular (giant covalent) structure with strong covalent bonds throughout the giant lattice. To melt it, many strong covalent bonds must be broken, requiring a large amount of energy. Sulfur trioxide (\(\text{SO}_3\)) has a simple molecular structure with weak intermolecular van der Waals forces (and dipole-dipole interactions) between molecules. Melting sulfur trioxide only requires overcoming these weak intermolecular forces, which requires very little energy. (ii) Sodium oxide reacts with water to form sodium hydroxide: \(\text{Na}_2\text{O}(\text{s}) + \text{H}_2\text{O}(\text{l}) \to 2\text{NaOH}(\text{aq})\). The solution is strongly alkaline, with a pH of 13-14. (iii) Phosphorus(V) oxide reacts with water to form phosphoric(V) acid: \(\text{P}_4\text{O}_{10}(\text{s}) + 6\text{H}_2\text{O}(\text{l}) \to 4\text{H}_3\text{PO}_4(\text{aq})\). The solution is strongly acidic, with a pH of 1-2. [Part b](i) When silicon tetrachloride is added to water, it undergoes rapid hydrolysis. Observations: Vigorous reaction, release of misty white fumes of hydrogen chloride gas (\(\text{HCl}\)), and the formation of a white solid precipitate (silicon dioxide/hydrated silica, \(\text{SiO}_2\)). The equation is: \(\text{SiCl}_4(\text{l}) + 2\text{H}_2\text{O}(\text{l}) \to \text{SiO}_2(\text{s}) + 4\text{HCl}(\text{g})\) (or \(\text{SiCl}_4 + 4\text{H}_2\text{O} \to \text{Si(OH)}_4 + 4\text{HCl}\)).

[Part a](i) 4 marks: 1 mark for describing silicon dioxide as macromolecular/giant covalent with strong covalent bonds; 1 mark for explaining high energy is needed to break these bonds; 1 mark for describing sulfur trioxide as simple molecular with weak intermolecular forces; 1 mark for explaining low energy is needed to overcome these forces. (ii) 2 marks: 1 mark for correct balanced equation; 1 mark for pH 13 or 14. (iii) 2 marks: 1 mark for correct balanced equation; 1 mark for pH 1 or 2. [Part b](i) 3.67 marks: 1.67 marks for observations (1 mark for white fumes, 0.67 marks for white solid); 2 marks for correct balanced equation.

[Part a](i) Convert percentages to moles: Moles of C = 40.0 / 12.0 = 3.33 mol. Moles of H = 6.7 / 1.0 = 6.7 mol. Moles of O = 53.3 / 16.0 = 3.33 mol. Divide by the smallest value (3.33): C : H : O = 1 : 2 : 1. Empirical formula is \(\text{CH}_2\text{O}\). (ii) Apply ideal gas equation: \(pV = nRT \Rightarrow n = \frac{pV}{RT}\). Convert variables to SI units: \(p = 101 \times 10^3\text{ Pa}\), \(V = 71.8 \times 10^{-6}\text{ m}^3\), \(T = 373\text{ K}\). \(n = \frac{101 \times 10^3 \times 71.8 \times 10^{-6}}{8.31 \times 373} = \frac{7.2518}{3099.63} = 2.3396 \times 10^{-3}\text{ mol}\). Calculate \(M_{\text{r}}\): \(M_{\text{r}} = \frac{\text{mass}}{n} = \frac{0.210}{2.3396 \times 10^{-3}} = 89.76 \approx 90.0\text{ g mol}^{-1}\). Empirical mass of \(\text{CH}_2\text{O} = 12.0 + 2(1.0) + 16.0 = 30.0\text{ g mol}^{-1}\). Ratio \(\frac{M_{\text{r}}}{\text{empirical mass}} = \frac{90.0}{30.0} = 3\). Thus, the molecular formula is \(\text{C}_3\text{H}_6\text{O}_3\). [Part b] 1) Calculate moles of \(\text{H}_2\text{SO}_4\): \(n(\text{H}_2\text{SO}_4) = c \times V = 0.0500 \times \frac{18.40}{1000} = 9.20 \times 10^{-4}\text{ mol}\). 2) Determine moles of \(\text{NaOH}\) reacting (using 1:2 stoichiometry): \(n(\text{NaOH}) = 2 \times 9.20 \times 10^{-4} = 1.84 \times 10^{-3}\text{ mol}\). 3) Calculate concentration of \(\text{NaOH}\): \(c(\text{NaOH}) = \frac{n}{V} = \frac{1.84 \times 10^{-3}}{0.0250} = 0.0736\text{ mol dm}^{-3}\).

[Part a](i) 2 marks: 1 mark for working out molar ratio (3.33 : 6.7 : 3.33); 1 mark for correct empirical formula CH2O. (ii) 4.67 marks: 1 mark for converting pressure and volume to SI units (Pa and m^3); 1 mark for calculating correct moles of gas (2.34 x 10^-3); 1 mark for calculating Mr = 90.0; 1.67 marks for deducing the molecular formula C3H6O3. [Part b] 5 marks: 1 mark for converting titre to dm^3; 1 mark for calculating moles of H2SO4 (9.20 x 10^-4); 1 mark for identifying the 1:2 reacting ratio; 1 mark for calculating moles of NaOH (1.84 x 10^-3); 1 mark for final concentration of NaOH (0.0736 mol dm^-3) to 3 sig figs.

(a) The first electron affinity is the enthalpy change when 1 mole of gaseous 1- ions is formed from 1 mole of gaseous atoms by the addition of 1 mole of electrons.

(b) Using a Born-Haber cycle:
\(\Delta H_f\text{(CaO)} = \Delta H_{at}\text{(Ca)} + IE_1\text{(Ca)} + IE_2\text{(Ca)} + \Delta H_{at}\text{(O)} + EA_1\text{(O)} + EA_2\text{(O)} + \Delta H_{L,form}\text{(CaO)}\)

Substituting the values:
\(-635 = 178 + 590 + 1145 + 248 - 141 + 798 + \Delta H_{L,form}\)
\(-635 = 2818 + \Delta H_{L,form}\)
\(\Delta H_{L,form} = -635 - 2818 = -3453\\ \text{kJ\\ mol}^{-1}\)

(c) The second electron affinity involves adding a negatively charged electron to a negatively charged oxide ion (\(\text{O}^-\)). Energy is required to overcome the electrostatic repulsion between the negative ion and the incoming electron, making the process endothermic.

(a)
M1: Enthalpy change when 1 mole of gaseous 1- ions is formed [1 mark]
M2: from 1 mole of gaseous atoms by gaining 1 mole of electrons [1 mark]

(b)
M1: Clear expression of Born-Haber cycle (algebraic or diagrammatic) [1 mark]
M2: Correct substitution of values into the cycle expression [2 marks] (deduct 1 mark for each transcription or sign error)
M3: Correct arithmetic simplification: \(-635 = 2818 + \Delta H_{L,form}\) [1 mark]
M4: Final calculated value of \(-3453\) [1 mark]
M5: Correct negative sign and unit \(\text{kJ\\ mol}^{-1}\) [1 mark]

(c)
M1: Oxide ion (\(\text{O}^-\) intermediate) is negatively charged [1 mark]
M2: Energy is required to overcome repulsion between the negative ion and the incoming negative electron [1 mark]

(a) \(\Delta S^\theta = \sum S^\theta\text{(products)} - \sum S^\theta\text{(reactants)}\)
\(\Delta S^\theta = [(2 \times 27.3) + (3 \times 213.6)] - [87.4 + (3 \times 197.6)]\)
\(\Delta S^\theta = [54.6 + 640.8] - [87.4 + 592.8]\)
\(\Delta S^\theta = 695.4 - 680.2 = +15.2\\ \text{J\\ K}^{-1}\text{mol}^{-1}\)

(b) \(\Delta G^\theta = \Delta H^\theta - T\Delta S^\theta\)
Converting \(\Delta S^\theta\) to \(\text{kJ\\ K}^{-1}\text{mol}^{-1}\): \(15.2 / 1000 = 0.0152\\ \text{kJ\\ K}^{-1}\text{mol}^{-1}\)
\(\Delta G^\theta = -24.8 - (298 \times 0.0152)\)
\(\Delta G^\theta = -24.8 - 4.53 = -29.33\\ \text{kJ\\ mol}^{-1}\) (accept \(-29.3\\ \text{kJ\\ mol}^{-1}\))
Since \(\Delta G^\theta < 0\), the reaction is feasible at \(298\\ \text{K}\).

(c) Because \(\Delta H\) is negative (exothermic) and \(\Delta S\) is positive (entropy increases), the term \(-T\Delta S\) is negative at all temperatures. Since both \(\Delta H\) and \(-T\Delta S\) are negative, \(\Delta G\) will be negative at all temperatures. Therefore, the reaction is feasible at all temperatures and becomes increasingly feasible (more negative \(\Delta G\)) as temperature increases.

(a)
M1: For showing a correct sum of product entropies minus reactant entropies [1 mark]
M2: For standard arithmetic evaluation of both parts (695.4 and 680.2) [1 mark]
M3: Correct value of \(+15.2\\ \text{J\\ K}^{-1}\text{mol}^{-1}\) (sign must be present) [1 mark]

(b)
M1: Convert entropy change to \(\text{kJ\\ K}^{-1}\text{mol}^{-1}\) by dividing by 1000 [1 mark]
M2: Correct substitution into the free energy equation resulting in \(-29.3\\ \text{kJ\\ mol}^{-1}\) (accept range \(-29.3\) to \(-29.4\)) [1 mark]
M3: Clear statement that the reaction is feasible because \(\Delta G < 0\) [1 mark]

(c)
M1: Mentions \(\Delta H\) is negative / exothermic [1 mark]
M2: Mentions \(\Delta S\) is positive / disorder increases [1 mark]
M3: Explains that \(-T\Delta S\) remains negative at all temperatures [1 mark]
M4: Concludes that \(\Delta G\) is always negative, so the reaction remains feasible at all temperatures (or becomes more feasible as temperature increases) [1 mark]

(a) Observation: The solution remains orange / no color change.
Explanation: Tertiary alcohols do not have a hydrogen atom attached to the carbon atom that is bonded to the hydroxyl (\(\text{-OH}\)) group, meaning they cannot be oxidized without breaking carbon-carbon bonds.

(b) Since A is a primary alcohol (butan-1-ol), excess oxidizing agent under reflux converts it completely to a carboxylic acid.
IUPAC Name: Butanoic acid.
Structure: skeletal showing 4 carbons and a carboxylic acid group: \(\text{CH}_3\text{CH}_2\text{CH}_2\text{COOH}\).

(c) (i) \(\text{C}_4\text{H}_{10}\text{O} \rightarrow \text{C}_4\text{H}_8 + \text{H}_2\text{O}\)

(ii) Dehydration of butan-2-ol can eliminate a hydrogen atom from either carbon-1 or carbon-3, yielding three isomers:
1. But-1-ene: \(\text{CH}_2\text{=CH-CH}_2\text{-CH}_3\)
2. (E)-but-2-ene: with methyl groups trans/opposite across the double bond.
3. (Z)-but-2-ene: with methyl groups cis/together on the same side of the double bond.

(iii) Type of stereoisomerism: E-Z isomerism (or stereoisomerism/geometric isomerism).
Explanation: It arises because there is restricted rotation around the carbon-carbon double bond (\(\text{C=C}\)) and there are two different groups attached to each of the carbon atoms in the double bond (a hydrogen and a methyl group).

(a)
M1: Observation: remains orange / no color change [1 mark]
M2: Reason: no hydrogen atom on the carbon bonded to the -OH group [1 mark]

(b)
M1: IUPAC Name: Butanoic acid [1 mark]
M2: Skeletal structure of butanoic acid drawn correctly [1 mark]

(c)(i)
M1: Correct molecular equation: \(\text{C}_4\text{H}_{10}\text{O} \rightarrow \text{C}_4\text{H}_8 + \text{H}_2\text{O}\) [1 mark]

(c)(ii)
M1: Correct structure of but-1-ene [1 mark]
M2: Correct structure of (E)-but-2-ene [1 mark]
M3: Correct structure of (Z)-but-2-ene [1 mark]

(c)(iii)
M1: Identifies E-Z isomerism / geometric isomerism [1 mark]
M2: Explains restriction of rotation around double bond AND two different groups on each carbon of the double bond [1 mark]

(a) (i) Industrial conditions:
* Temperature: \(300^\circ\text{C}\) (accept range \(280 - 320^\circ\text{C}\))
* Pressure: \(60 - 70\\ \text{atm}\)
* Catalyst: Concentrated phosphoric acid (\(\text{H}_3\text{PO}_4\))

(ii) Mechanism for hydration of ethene:
1. Step 1: Protonation of ethene. Curly arrow from the \(\text{C=C}\) double bond to the \(\text{H}^+\) ion from the catalyst. This forms a carbocation intermediate (\(\text{CH}_3\text{CH}_2^+\)).
2. Step 2: Nucleophilic attack of water. Curly arrow from the lone pair on the oxygen of a water molecule (\(\text{H}_2\text{O}\)) to the positively charged carbon of the carbocation. This forms the protonated ethanol intermediate (\(\text{CH}_3\text{CH}_2\text{OH}_2^+\)).
3. Step 3: Deprotonation. Curly arrow from the \(\text{O-H}\) bond in \(\text{CH}_3\text{CH}_2\text{OH}_2^+\)\\ to the oxygen atom, regenerating the \(\text{H}^+\) catalyst and producing ethanol.

(b) (i) Equation:
\(\text{C}_6\text{H}_{12}\text{O}_6 \rightarrow 2\text{C}_2\text{H}_5\text{OH} + 2\text{CO}_2\)

(ii) Advantages:
* Fermentation uses renewable resources (sugars from crops like sugarcane or sugar beet) whereas ethene is obtained from non-renewable crude oil.
* Fermentation uses much lower temperatures and pressures (around \(35^\circ\text{C}\) and atmospheric pressure), which requires far less energy input and cheaper equipment.

(a)(i)
M1: Temperature: \(300^\circ\text{C}\) (accept \(280 - 320^\circ\text{C}\)) [1 mark]
M2: Pressure: \(60 - 70\\ \text{atm}\) (or \(6\text{-}7\\ \text{MPa}\)) [1 mark]
M3: Catalyst: Concentrated phosphoric acid / \(\text{H}_3\text{PO}_4\) [1 mark]

(a)(ii)
M1: Curly arrow from \(\text{C=C}\) to \(\text{H}^+\) [1 mark]
M2: Correct carbocation intermediate structure \(\text{CH}_3\text{CH}_2^+\) [1 mark]
M3: Curly arrow from oxygen lone pair on \(\text{H}_2\text{O}\) to the carbocation carbon [1 mark]
M4: Correct protonated intermediate and curly arrow from the \(\text{O-H}\) bond back to oxygen [1 mark]

(b)(i)
M1: Balanced equation: \(\text{C}_6\text{H}_{12}\text{O}_6 \rightarrow 2\text{C}_2\text{H}_5\text{OH} + 2\text{CO}_2\) [1 mark]

(b)(ii)
M1: Reference to renewable raw materials / sustainable feedstock [1 mark]
M2: Reference to lower energy consumption / operating cost / mild conditions [1 mark]

(a) (i) \(\text{P}_4\text{O}_{10} + 6\text{H}_2\text{O} \rightarrow 4\text{H}_3\text{PO}_4\)

(ii) pH of phosphoric acid solution: 0 to 2 (accept any value in this range).

(iii) \(\text{SO}_3 + \text{H}_2\text{O} \rightarrow \text{H}_2\text{SO}_4\)

(b) (i) Observations: Steamy/white fumes (of hydrogen chloride gas) and a white solid/precipitate (of silicon dioxide or silicic acid).
Equation: \(\text{SiCl}_4 + 2\text{H}_2\text{O} \rightarrow \text{SiO}_2 + 4\text{HCl}\) (or \(\text{SiCl}_4 + 4\text{H}_2\text{O} \rightarrow \text{Si(OH)}_4 + 4\text{HCl}\))

(ii) NaCl is a giant ionic lattice. In water, it simply dissolves; the polar water molecules hydrate the sodium and chloride ions, breaking the ionic lattice but without any chemical reaction (hydrolysis). The solution is neutral (pH = 7).
SiCl4 is a simple molecular structure with covalent bonds. Silicon has empty 3d orbitals which can accept lone pairs of electrons from water molecules, initiating a hydrolysis reaction. This reaction breaks covalent bonds to form HCl, resulting in a strongly acidic solution (pH = 0 to 2).

(a)(i)
M1: Correctly balanced equation: \(\text{P}_4\text{O}_{10} + 6\text{H}_2\text{O} \rightarrow 4\text{H}_3\text{PO}_4\) [1 mark]

(a)(ii)
M1: pH value of 0, 1 or 2 [1 mark]

(a)(iii)
M1: Equation: \(\text{SO}_3 + \text{H}_2\text{O} \rightarrow \text{H}_2\text{SO}_4\) [1 mark]

(b)(i)
M1: White fumes / steamy fumes [1 mark]
M2: White solid / white precipitate [1 mark]
M3: Equation: \(\text{SiCl}_4 + 2\text{H}_2\text{O} \rightarrow \text{SiO}_2 + 4\text{HCl}\) (or balanced with \(\text{Si(OH)}_4\)) [1 mark]

(b)(ii)
M1: NaCl has ionic bonding/giant ionic structure, dissolves without reacting, pH = 7 [2 marks] (1 mark for bonding, 1 mark for dissolving/pH)
M2: SiCl4 is simple molecular covalent, reacts/hydrolyses, because silicon has vacant d-orbitals [1 mark]
M3: Forms strongly acidic solution / pH 0-2 due to HCl production [1 mark]

(a) (i) Functional group isomerism (as they have different functional groups, aldehyde vs ketone).

(ii) Test: Add Tollens' reagent (or Fehling's solution) and warm.
* Observation with propanal: Silver mirror (or red precipitate for Fehling's).
* Observation with propanone: No visible change (remains colorless/blue).

(b) (i) IUPAC Name: 2-hydroxybutanenitrile.

(ii) Mechanism:
1. Step 1: The nucleophile \(\text{CN}^-\) attacks the carbonyl carbon. A curly arrow is drawn from the lone pair on the carbon of \(\text{CN}^-\) to the carbonyl carbon (showing \(\delta+\) on carbon, \(\delta-\) on oxygen). A second curly arrow goes from the double bond of \(\text{C=O}\) to the oxygen atom.
2. Step 2: The intermediate \(\text{CH}_3\text{CH}_2\text{CH(O}^-\)\\text{CN}\) is formed. A curly arrow is drawn from the lone pair on the negative oxygen atom to a proton (\(\text{H}^+\) or \(\text{HCN}\)), forming the alcohol group.

(iii) The carbonyl carbon of propanal has a planar geometry. The nucleophile (\(\text{CN}^-\) ion) can attack the planar carbonyl group with equal probability from either side (above or below the plane). This results in an equimolar (50:50) mixture of the two optical isomers (enantiomers), known as a racemic mixture. The optical rotations of the enantiomers cancel each other out, making the mixture optically inactive.

(a)(i)
M1: Functional group isomerism [1 mark]

(a)(ii)
M1: Tollens' reagent / Fehling's solution [1 mark]
M2: Silver mirror with propanal / Red precipitate with propanal [1 mark]
M3: No change/reaction with propanone [1 mark]

(b)(i)
M1: 2-hydroxybutanenitrile (correct spelling) [1 mark]

(b)(ii)
M1: Arrow from lone pair of \(\text{CN}^-\) to carbonyl carbon [1 mark]
M2: Arrow from \(\text{C=O}\) bond to oxygen, showing correct partial charges [1 mark]
M3: Correct intermediate structure and final arrow from \(\text{O}^-\) to \(\text{H}^+\) to form product [1 mark]

(b)(iii)
M1: The carbonyl group / C=O is planar [1 mark]
M2: Attack can occur from either side with equal probability, forming a racemic mixture [1 mark]

(a)
* Ligand: An atom, ion, or molecule that can donate a lone pair of electrons to a central metal ion to form a coordinate (dative covalent) bond.
* Coordination number: The total number of coordinate bonds formed between the central metal ion and the ligands.

(b) (i)
* Shape: Octahedral
* Bond angle(s): \(90^\circ\) (and \(180^\circ\))

(ii) Equation:
\([\text{Cu(H}_2\text{O)}_6]^{2+} + 4\text{Cl}^- \rightarrow [\text{CuCl}_4]^{2-} + 6\text{H}_2\text{O}\)

(iii)
* Color: Yellow-green / green / yellow.
* Shape: Tetrahedral.
* Explanation: Chloride ligands (\(\text{Cl}^-\)) are larger than water molecules (\(\text{H}_2\text{O}\)). Consequently, fewer chloride ligands can fit around the copper(II) ion due to steric hindrance and repulsion between the negative ligands, reducing the coordination number from 6 to 4.

(c) Haemoglobin is a biologically important complex. The transition metal ion present in haemoglobin is iron (\(\text{Fe}^{2+}\)).

(a)
M1: Ligand: donor of lone pair to form coordinate bond [1 mark]
M2: Coordination number: number of coordinate bonds to metal ion [1 mark]

(b)(i)
M1: Octahedral shape [1 mark]
M2: \(90^\circ\) (or \(90^\circ\) and \(180^\circ\)) bond angle [1 mark]

(b)(ii)
M1: Correct formulae for reactants and products [1 mark]
M2: Correct balancing and overall charges (e.g. \(4\text{Cl}^-\) and \([\text{CuCl}_4]^{2-}\)) [1 mark]

(b)(iii)
M1: Color: Yellow-green / green / yellow [1 mark]
M2: Shape: Tetrahedral [1 mark]
M3: Reason: Chloride ligands are larger than water ligands / steric hindrance [1 mark]

(c)
M1: Haemoglobin (accept myoglobin) and Iron (\(\text{Fe}\)) [1 mark]

(a) The enthalpy change when 1 mole of solid ionic compound (calcium sulfide) is formed from its constituent gaseous ions under standard conditions. Equation: \(Ca^{2+}(g) + S^{2-}(g) \rightarrow CaS(s)\).

(b)(i) \(S(s) \rightarrow S(g)\)
(ii) \(S^-(g) + e^- \rightarrow S^{2-}(g)\)

(c) Using the Born-Haber cycle calculation:
\(\Delta H_f^\ominus(CaS) = \Delta H_{at}^\ominus(Ca) + IE_1(Ca) + IE_2(Ca) + \Delta H_{at}^\ominus(S) + EA_1(S) + EA_2(S) + \Delta H_{L,form}^\ominus(CaS)\)
\(-482 = 178 + 590 + 1145 + 279 - 200 + 640 + \Delta H_{L,form}^\ominus(CaS)\)
\(-482 = 2432 + \Delta H_{L,form}^\ominus(CaS)\)
\(\Delta H_{L,form}^\ominus(CaS) = -482 - 2432 = -2914\text{ kJ mol}^{-1}\).

(d) The experimental value is more exothermic (larger negative value) than the theoretical value. This is because calcium sulfide has a degree of covalent character (due to polarization of the sulfide ion by the calcium ion), which makes the bonding stronger than purely ionic bonding.

(a) Award 1 mark for: enthalpy change when 1 mole of solid ionic compound is formed from its constituent gaseous ions. Award 1 mark for: reference to standard conditions / standard states.

(b)(i) \(S(s) \rightarrow S(g)\) - 1 mark. (Reject if no state symbols or incorrect state symbols).
(b)(ii) \(S^-(g) + e^- \rightarrow S^{2-}(g)\) - 1 mark.

(c) Award marks as follows:
- 1 mark for correct algebraic expression or cycle setup.
- 1 mark for correct summation of positive terms: \(178 + 590 + 1145 + 279 + 640 = 2832\) (or total sum including \(EA_1\) as \(2432\)).
- 1 mark for rearranging to find \(\Delta H_{L,form}\): \(-482 - 2432\).
- 1 mark for correct final value: \(-2914\text{ kJ mol}^{-1}\) (must include sign and unit).

(d) Award 1 mark for stating the experimental value is more exothermic / larger negative. Award 1 mark for explaining that the compound has covalent character / polarisation of the sulfide ion occurs.

(a) \(\Delta S^\ominus = \sum S^\ominus(\text{products}) - \sum S^\ominus(\text{reactants})\)
\(\Delta S^\ominus = [S^\ominus(CO) + 3 \times S^\ominus(H_2)] - [S^\ominus(CH_4) + S^\ominus(H_2O)]\)
\(\Delta S^\ominus = [198 + 3(131)] - [186 + 189]\)
\(\Delta S^\ominus = 591 - 375 = +216\text{ J K}^{-1}\text{ mol}^{-1}\).

(b) \(\Delta G^\ominus = \Delta H^\ominus - T\Delta S^\ominus\)
Convert \(\Delta S^\ominus\) to \(\text{kJ K}^{-1}\text{ mol}^{-1}\): \(216 / 1000 = 0.216\text{ kJ K}^{-1}\text{ mol}^{-1}\).
\(\Delta G^\ominus = 206 - (298 \times 0.216)\)
\(\Delta G^\ominus = 206 - 64.368 = +141.632\text{ kJ mol}^{-1}\) (rounds to \(+142\text{ kJ mol}^{-1}\)).
Since \(\Delta G^\ominus\) is positive, the reaction is not feasible at \(298\text{ K}\).

(c) For the reaction to be feasible, \(\Delta G^\ominus \le 0\).
Therefore, \(T \ge \frac{\Delta H^\ominus}{\Delta S^\ominus}\).
\(T = \frac{206 \times 1000}{216} = 953.7\text{ K}\) (rounds to \(954\text{ K}\)).

(d) The reaction has a very high activation energy, which prevents it from occurring at a noticeable rate under these conditions.

(a) Award 1 mark for correct formula or correct substitution of values.
Award 1 mark for \(591\) and \(375\).
Award 1 mark for correct final answer: \(+216\text{ J K}^{-1}\text{ mol}^{-1}\) (must include sign and units).

(b) Award 1 mark for converting \(\Delta S^\ominus\) to kJ or \(\Delta H^\ominus\) to J.
Award 1 mark for calculation of \(+142\text{ kJ mol}^{-1}\) (or \(+141.6\text{ kJ mol}^{-1}\)).
Award 1 mark for concluding that the reaction is NOT feasible because \(\Delta G\) is positive.

(c) Award 1 mark for setting \(\Delta G^\ominus = 0\) or stating \(T = \frac{\Delta H}{\Delta S}\).
Award 1 mark for correct substitution: \(206 \times 1000 / 216\).
Award 1 mark for correct value: \(954\text{ K}\) (accept range \(953.7 - 954\text{ K}\)).

(d) Award 1 mark for reference to high activation energy.

(a) Reagent: Concentrated sulfuric acid (\(H_2SO_4\)) or concentrated phosphoric acid (\(H_3PO_4\)). Conditions: Heat (or temperature in range \(150 - 180^\circ\text{C}\)).

(b) The three isomers are:
1. But-1-ene: \(CH_2=CHCH_2CH_3\)
2. (E)-but-2-ene (trans-but-2-ene)
3. (Z)-but-2-ene (cis-but-2-ene)
Explanation: Dehydration involves removing a hydrogen atom from a carbon adjacent to the C-OH carbon. The hydrogen can be removed from C1 to form but-1-ene, or from C3 to form but-2-ene. But-2-ene exists as stereoisomers (E/Z isomers) due to restricted rotation about the C=C double bond and the fact that each carbon in the double bond is bonded to two different groups (a methyl group and a hydrogen atom).

(c) Mechanism:
- Step 1: The oxygen lone pair of butan-2-ol is protonated by an \(H^+\) ion from the acid catalyst to form a protonated alcohol.
- Step 2: The C-O bond breaks, releasing water (\(H_2O\)) and forming a secondary carbocation (\(CH_3CH^+CH_2CH_3\)).
- Step 3: A water molecule or hydrogen sulfate ion acts as a base and removes a proton (\(H^+\)) from C3 (or C1), with the electrons from the C-H bond moving to form the C=C double bond, yielding but-2-ene.

(a) Award 1 mark for: concentrated sulfuric acid / concentrated phosphoric acid (reject dilute acid). Award 1 mark for: heat / hot / temp of \(170^\circ\text{C}\).

(b) Award 2 marks for drawing/naming all three isomers correctly (but-1-ene, E-but-2-ene, Z-but-2-ene). (Deduct 1 mark if only two are drawn).
Award 1 mark for explaining that hydrogen can be removed from C1 or C3.
Award 1 mark for explaining that but-2-ene exhibits E/Z (or stereoisomerism) due to restricted rotation around C=C.

(c) Award marks for the mechanism as follows:
- 1 mark for a curly arrow from lone pair on the O of the OH group to the \(H^+\) ion.
- 1 mark for a curly arrow from the C-O bond to the positive O atom in the protonated intermediate to form the carbocation.
- 1 mark for drawing the structure of the carbocation intermediate: \(CH_3CH^+CH_2CH_3\).
- 1 mark for a curly arrow from the C-H bond on C3 to the adjacent C-C single bond, forming the double bond.

(a) Equation: \(CH_2=CH_2(g) + H_2O(g) \rightleftharpoons CH_3CH_2OH(g)\)
Catalyst: Concentrated phosphoric acid (\(H_3PO_4\)) supported on silica.
Temperature: \(300^\circ\text{C}\) (accept \(250 - 350^\circ\text{C}\)).
Pressure: \(60 - 70\text{ atm}\) (accept \(6 - 7\text{ MPa}\)).

(b) Equation: \(C_6H_{12}O_6(aq) \rightarrow 2CH_3CH_2OH(aq) + 2CO_2(g)\)
Conditions: Yeast catalyst, temperature around \(35 - 40^\circ\text{C}\) (not exceeding \(45^\circ\text{C}\) to prevent enzyme denaturation), and anaerobic conditions (absence of oxygen) to prevent oxidation of ethanol to ethanoic acid.

(c)(i) Rate: Hydration is rapid / fast; fermentation is slow.
(ii) Purity: Hydration yields high-purity ethanol; fermentation yields low-purity aqueous solution (requires distillation to concentrate).
(iii) Sustainability: Hydration uses non-renewable crude oil fractions (ethene); fermentation uses renewable agricultural resources (glucose / crops).

(a) Award marks as follows:
- 1 mark for correct equation with state symbols: \(CH_2=CH_2(g) + H_2O(g) \rightleftharpoons CH_3CH_2OH(g)\).
- 1 mark for concentrated phosphoric acid (\(H_3PO_4\)) catalyst.
- 1 mark for temperature: \(300^\circ\text{C}\) (accept \(250 - 350^\circ\text{C}\)).
- 1 mark for pressure: \(60 - 70\text{ atm}\).

(b) Award marks as follows:
- 1 mark for correct equation: \(C_6H_{12}O_6 \rightarrow 2C_2H_5OH + 2CO_2\).
- 1 mark for temperature of \(30 - 42^\circ\text{C}\).
- 1 mark for anaerobic / absence of oxygen.

(c) Award marks as follows:
- (i) 1 mark for stating hydration is fast AND fermentation is slow.
- (ii) 1 mark for stating hydration yields pure product AND fermentation yields impure product.
- (iii) 1 mark for stating hydration uses non-renewable resources AND fermentation uses renewable resources.

(a)(i) Equation: \(Na_2O(s) + H_2O(l) \rightarrow 2NaOH(aq)\) (or \(2Na^+(aq) + 2OH^-(aq)\)). pH: 13 or 14.
(ii) Equation: \(P_4O_{10}(s) + 6H_2O(l) \rightarrow 4H_3PO_4(aq)\). pH: 0, 1, or 2.

(b)(i) \(SO_2 + 2NaOH \rightarrow Na_2SO_3 + H_2O\) (or \(SO_2 + NaOH \rightarrow NaHSO_3\)).
(ii) \(SO_3 + H_2O \rightarrow H_2SO_4\).

(c)(i) The giant covalent structure has very strong covalent bonds in a 3D macromolecular network, which require a massive amount of energy to break, and water molecules cannot solvate/hydrate the atoms.
(ii) \(SiO_2 + 2NaOH \rightarrow Na_2SiO_3 + H_2O\).

(d)(i) \(Al_2O_3 + 6HCl \rightarrow 2AlCl_3 + 3H_2O\) (or ionic: \(Al_2O_3 + 6H^+ \rightarrow 2Al^{3+} + 3H_2O\)).
(ii) \(Al_2O_3 + 2NaOH + 3H_2O \rightarrow 2Na[Al(OH)_4]\) (or \(Al_2O_3 + 2NaOH \rightarrow 2NaAlO_2 + H_2O\)).

(a)(i) Award 1 mark for correct equation. Award 1 mark for pH 13 or 14.
(a)(ii) Award 1 mark for correct equation. Award 1 mark for pH 0, 1, or 2.

(b)(i) Award 1 mark for correct equation: \(SO_2 + 2NaOH \rightarrow Na_2SO_3 + H_2O\) (allow ionic equivalents).
(b)(ii) Award 1 mark for: \(SO_3 + H_2O \rightarrow H_2SO_4\).

(c)(i) Award 1 mark for reference to strong covalent bonds in macromolecular / giant covalent structure which are difficult to break.
(c)(ii) Award 1 mark for: \(SiO_2 + 2NaOH \rightarrow Na_2SiO_3 + H_2O\).

(d)(i) Award 1 mark for: \(Al_2O_3 + 6HCl \rightarrow 2AlCl_3 + 3H_2O\).
(d)(ii) Award 1 mark for: \(Al_2O_3 + 2NaOH + 3H_2O \rightarrow 2Na[Al(OH)_4]\) (accept \(2NaAlO_2 + H_2O\)).

(a) Bonding, structure, and pH:
- \(NaCl\): Ionic bonding, giant ionic lattice. Dissolves to form a solution with pH approximately 7.
- \(AlCl_3\): Ionic with covalent character (or covalent dimer in gas phase), giant ionic / layer lattice. Hydrolyses to form an acidic solution with pH approximately 3.
- \(SiCl_4\): Covalent bonding, simple molecular structure. Hydrolyses vigorously to form a strongly acidic solution with pH approximately 1-2.
- \(PCl_5\): Covalent bonding, simple molecular structure (or ionic in solid state \([PCl_4]^+[PCl_6]^-\)). Hydrolyses vigorously to form a strongly acidic solution with pH approximately 1-2.

(b)(i) \(SiCl_4(l) + 2H_2O(l) \rightarrow SiO_2(s) + 4HCl(aq)\) (accept \(SiCl_4(l) + 4H_2O(l) \rightarrow H_4SiO_4(aq) + 4HCl(aq)\)).
(ii) \(PCl_5(s) + 4H_2O(l) \rightarrow H_3PO_4(aq) + 5HCl(aq)\).

(c) Sodium chloride is a simple ionic lattice containing stable \(Na^+\) and \(Cl^-\). These ions are simply hydrated by water molecules because \(Na^+\) has a relatively low charge density and does not polarise water sufficiently to cause hydrolysis. Silicon tetrachloride has a central silicon atom with vacant 3d-orbitals that can accept a lone pair of electrons from water molecules, initiating the nucleophilic attack that leads to rapid hydrolysis.

(a) Award marks as follows:
- 1 mark for \(NaCl\): ionic + pH 7.
- 1 mark for \(AlCl_3\): covalent/ionic with covalent character + pH 3.
- 1 mark for \(SiCl_4\): covalent/simple molecular + pH 1 or 2.
- 1 mark for \(PCl_5\): covalent/simple molecular + pH 1 or 2.

(b)(i) Award 1 mark for correct formulas. Award 1 mark for correct state symbols: \(SiCl_4(l) + 2H_2O(l) \rightarrow SiO_2(s) + 4HCl(aq)\).
(b)(ii) Award 1 mark for correct formulas. Award 1 mark for correct state symbols: \(PCl_5(s) + 4H_2O(l) \rightarrow H_3PO_4(aq) + 5HCl(aq)\).

(c) Award 1 mark for: \(Na^+\) has low charge density and only hydration occurs / no vacant d-orbitals in period 2 (like Na level is filled or Na ion has no reaction). Award 1 mark for: Silicon has vacant 3d orbitals which can accept lone pairs of electrons from water molecules to allow coordinate bonding/hydrolysis.

(a)(i) A ligand is an atom, ion or molecule that donates a lone pair of electrons to a central metal ion to form a coordinate (dative covalent) bond.
(ii) Co-ordination number is the total number of coordinate bonds formed between the central metal ion and the ligands.

(b)(i) \(90^\circ\) (accept \(180^\circ\)).
(ii) \([Cu(H_2O)_6]^{2+}(aq) + 4Cl^-(aq) \rightleftharpoons [CuCl_4]^{2-}(aq) + 6H_2O(l)\).
(iii) Shape: Tetrahedral.
Explanation: Chloride ligands (\(Cl^-\)) are larger than water molecules (\(H_2O\)). Only four chloride ligands can fit around the central copper(II) ion due to steric hindrance (ligand-ligand repulsion).

(c) Equation: \([Cr(H_2O)_6]^{3+} + 6NH_3 \rightarrow [Cr(NH_3)_6]^{3+} + 6H_2O\).
Appearance change: Violet/green solution turns into a green precipitate first (of \(Cr(OH)_3\)), which then redissolves in excess ammonia to form a purple solution (containing \([Cr(NH_3)_6]^{3+}\)).

(a)(i) Award 1 mark for: species donating a lone pair of electrons to a metal ion.
(a)(ii) Award 1 mark for: total number of coordinate bonds formed with the central metal ion.

(b)(i) Award 1 mark for: \(90^\circ\) (accept \(90^\circ\) and \(180^\circ\)).
(b)(ii) Award 2 marks: 1 mark for correct reactants and products, 1 mark for balanced equation.
(b)(iii) Award 1 mark for: tetrahedral shape.
Award 2 marks for explanation: 1 mark for identifying chloride ions as larger than water; 1 mark for steric hindrance / ligand-ligand repulsion prevents 6 ligands fitting.

(c) Award 1 mark for equation: \([Cr(H_2O)_6]^{3+} + 6NH_3 \rightarrow [Cr(NH_3)_6]^{3+} + 6H_2O\).
Award 1 mark for color change: violet/green solution to purple solution.

(a) Reagent: Tollens' reagent (or Fehling's solution / Benedict's solution).
Observation with propanal: Silver mirror formed on the tube wall (or red precipitate with Fehling's).
Observation with propanone: No visible change (remains colorless/blue).

(b)(i) Product: Propan-1-ol.
Reaction type: Nucleophilic addition (accept reduction).
(ii) Mechanism:
1. Curly arrow from the lone pair of the hydride ion (\(H^-\)) to the carbonyl carbon (\(C=O\)).
2. Curly arrow from the C=O double bond to the oxygen atom.
3. Correct intermediate drawn: \(CH_3CH_2CH(O^-)H\) showing a negative charge on the oxygen.
4. Curly arrow from the lone pair on the negative oxygen to the hydrogen of water/proton source, forming \(CH_3CH_2CH_2OH\).

(c)(i) \(HCN\) is a weak acid and dissociates poorly, resulting in a very low concentration of the \(CN^-\)\ nucleophile. \(KCN\) is ionic and dissociates completely to provide a high concentration of the \(CN^-\)\ nucleophile, increasing the rate of reaction.

(a) Award 1 mark for correct reagent: Tollens' reagent (or Fehling's / Benedict's).
Award 1 mark for propanal observation: silver mirror / black precipitate (or red precipitate with Fehling's).
Award 1 mark for propanone observation: no change / remains colorless (or remains blue).

(b)(i) Award 1 mark for: propan-1-ol (reject propanol).
Award 1 mark for: nucleophilic addition / reduction.
(b)(ii) Award marks for mechanism:
- 1 mark for curly arrow from lone pair on \(H^-\)\ to the carbonyl carbon.
- 1 mark for curly arrow from double bond of C=O to oxygen.
- 1 mark for correct structure of intermediate (including charge on oxygen).
- 1 mark for curly arrow from intermediate oxygen lone pair to \(H^+\) / \(H_2O\).

(c)(i) Award 1 mark for: \(KCN\) increases the concentration of \(CN^-\)\ nucleophile / \(HCN\) alone is too weak of an acid.

(a) The enthalpy change when 1 mole of an ionic crystal lattice is separated into its constituent gaseous ions under standard conditions.
(b) \(\Delta H^\ominus_f(\text{CaO}) = \Delta H^\ominus_{\text{at}}(\text{Ca}) + 1\text{st IE}(\text{Ca}) + 2\text{nd IE}(\text{Ca}) + \Delta H^\ominus_{\text{at}}(\text{O}) + 1\text{st EA}(\text{O}) + 2\text{nd EA}(\text{O}) - \Delta H^\ominus_{L,\text{diss}}\)
(c) Substituting the values:
\(-635 = 178 + 590 + 1145 + 249 - 141 + 798 - \Delta H^\ominus_{L,\text{diss}}\)
\(-635 = 2819 - \Delta H^\ominus_{L,\text{diss}}\)
\(\Delta H^\ominus_{L,\text{diss}} = 2819 + 635 = +3454\ \text{kJ mol}^{-1}\)
(d) Calcium oxide is almost purely ionic and fits the ionic model well, so the theoretical and experimental values are very similar. Zinc oxide has a significant degree of covalent character due to the polarization of the oxide ion by the small, highly polarizing \(\text{Zn}^{2+}\) ion, which makes the actual bonding stronger than predicted by the purely ionic model.

(a) M1: Enthalpy change when 1 mole of solid ionic lattice is converted to gaseous ions. (1)
M2: Under standard conditions. (1)
(b) M1: Expression showing correct summation of terms. (1)
M2: Correct signs for lattice dissociation enthalpy subtraction or correct equation rearranged for lattice enthalpy. (1)
(c) M1: Correct setup of equation with values: \(-635 = 178 + 590 + 1145 + 249 - 141 + 798 - \Delta H^\ominus_{L,\text{diss}}\). (1)
M2: Sum of terms = +2819. (1)
M3: Final answer: \(+3454\ \text{kJ mol}^{-1}\) (with correct sign and units). (1) (Allow 2 marks for -3454 or missing sign, but full 3 marks only for positive value with correct unit).
(d) M1: CaO has almost pure ionic bonding / matches ionic model. (1)
M2: ZnO has covalent character / polarization of the anion. (1)
M3: Due to the high charge density / polarizing power of \(\text{Zn}^{2+}\) compared to \(\text{Ca}^{2+}\). (1)

(a) \(\Delta G = \Delta H - T\Delta S\). For a reaction to be feasible, \(\Delta G \le 0\).
(b) \(\Delta S^\ominus = \sum S^\ominus(\text{products}) - \sum S^\ominus(\text{reactants})\)
\(\Delta S^\ominus = [2 \times 27.3 + 3 \times 213.6] - [87.4 + 3 \times 197.6]\)
\(\Delta S^\ominus = [54.6 + 640.8] - [87.4 + 592.8] = 695.4 - 680.2 = +15.2\ \text{J K}^{-1}\text{ mol}^{-1}\)
(c) \(\Delta G^\ominus = \Delta H^\ominus - T\Delta S^\ominus\)
Convert \(\Delta S^\ominus\) to \(\text{kJ K}^{-1}\text{ mol}^{-1}\): \(15.2 / 1000 = 0.0152\ \text{kJ K}^{-1}\text{ mol}^{-1}\)
\(\Delta G^\ominus = -24.8 - (298 \times 0.0152) = -24.8 - 4.53 = -29.33\ \text{kJ mol}^{-1}\)
Since \(\Delta G^\ominus\) is negative (-29.33 \(\text{kJ mol}^{-1}\)), the reaction is feasible.
(d) Since \(\Delta H^\ominus\) is negative (exothermic) and \(\Delta S^\ominus\) is positive (increase in disorder), the term \(-T\Delta S^\ominus\) is always negative at all temperatures. Therefore, \(\Delta G^\ominus\) is negative at all temperatures.

(a) M1: \(\Delta G = \Delta H - T\Delta S\). (1)
M2: Feasible when \(\Delta G \le 0\) (or negative). (1)
(b) M1: Correct calculation of product entropy: \(2 \times 27.3 + 3 \times 213.6 = 695.4\). (1)
M2: Correct calculation of reactant entropy: \(87.4 + 3 \times 197.6 = 680.2\). (1)
M3: Final difference: \(+15.2\ \text{J K}^{-1}\text{ mol}^{-1}\). (1)
(c) M1: Divides \(\Delta S^\ominus\) by 1000 to convert to \(\text{kJ K}^{-1}\text{ mol}^{-1}\). (1)
M2: Calculates \(\Delta G^\ominus = -29.3\ \text{kJ mol}^{-1}\) (allow -29.3 to -29.4). (1)
M3: States reaction is feasible because \(\Delta G^\ominus < 0\). (1)
(d) M1: Explains that \(\Delta H\) is negative and \(\Delta S\) is positive. (1)
M2: Concludes that \(\Delta G\) will always remain negative regardless of the value of T. (1)

(a) Catalyst: Concentrated sulfuric acid (\(\text{H}_2\text{SO}_4\)) or concentrated phosphoric acid (\(\text{H}_3\text{PO}_4\)).
Equation: \(\text{CH}_3\text{CH(OH)CH}_2\text{CH}_2\text{CH}_3 \rightarrow \text{CH}_3\text{CH}=\text{CHCH}_2\text{CH}_3 + \text{H}_2\text{O}\).
(b) The three isomers are: pent-1-ene, cis-pent-2-ene ((Z)-pent-2-ene), and trans-pent-2-ene ((E)-pent-2-ene).
(c) Mechanism steps:
1. Protonation of the -OH group by \(\text{H}^+\) to form a protonated alcohol intermediate.
2. Loss of a water molecule to form a secondary carbocation intermediate.
3. Elimination of a proton from adjacent carbon (C3) by a base (such as \(\text{HSO}_4^-\)) to form the double bond and regenerate the acid catalyst.
(d) Pent-2-ene has a carbon-carbon double bond (\(\text{C}=\text{C}\)) which restricts rotation. Additionally, each carbon of the double bond is bonded to two different groups: one carbon is bonded to -H and -\(\text{CH}_3\), and the other carbon is bonded to -H and -\(\text{CH}_2\text{CH}_3\).

(a) M1: Concentrated sulfuric acid / \(\text{H}_2\text{SO}_4\) OR concentrated phosphoric acid / \(\text{H}_3\text{PO}_4\). (1)
M2: Correct equation with formula of pentan-2-ol and any pentene + water. (1)
(b) M1: Skeletal structure of pent-1-ene. (1)
M2: Skeletal structure of (E)-pent-2-ene. (1)
M3: Skeletal structure of (Z)-pent-2-ene. (1)
(c) M1: Curly arrow from lone pair on O of -OH to \(\text{H}^+\) (or to H of \(\text{H}_2\text{SO}_4\) with arrow breaking O-H bond). (1)
M2: Correct structure of protonated alcohol and curly arrow from C-O bond to positive O. (1)
M3: Correct secondary carbocation structure (\(\text{CH}_3\text{CH}^+\text{CH}_2\text{CH}_2\text{CH}_3\)). (1)
M4: Curly arrow from C-H bond on C3 (or adjacent carbon) to C-C bond, forming the double bond and releasing \(\text{H}^+\). (1)
(d) M1: Restricted rotation about \(\text{C}=\text{C}\) double bond AND each carbon atom of the double bond is attached to two different groups. (1)

(a) Reagent: Acidified potassium dichromate(VI) (\(\text{K}_2\text{Cr}_2\text{O}_7 / \text{H}_2\text{SO}_4\)).
Conditions: Heat (warm).
Observations: Butan-2-ol turns the solution from orange to green. 2-methylpropan-2-ol shows no color change (remains orange).
(b) Since Y reacts with sodium carbonate to produce gas, Y must be a carboxylic acid. Thus, X must be a primary alcohol, which is butan-1-ol. Y is butanoic acid.
Equation: \(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH} + 2[\text{O}] \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{COOH} + \text{H}_2\text{O}\).
(c) The IR spectrum of X contains a broad O-H (alcohol) absorption peak at \(3230-3550\ \text{cm}^{-1}\). The IR spectrum of Y has a very broad O-H (acid) absorption at \(2500-3000\ \text{cm}^{-1}\) and a strong C=O absorption at \(1680-1750\ \text{cm}^{-1}\). Complete conversion is confirmed when the alcohol O-H absorption peak at \(3230-3550\ \text{cm}^{-1}\) completely disappears and only the carboxylic acid peaks remain.

(a) M1: Acidified potassium dichromate(VI) / \(\text{K}_2\text{Cr}_2\text{O}_7 + \text{H}_2\text{SO}_4\) and heat. (1)
M2: Observation for butan-2-ol: turns from orange to green. (1)
M3: Observation for 2-methylpropan-2-ol: remains orange / no change. (1)
(b) M1: X is butan-1-ol; Y is butanoic acid. (Both correct for 2 marks, 1 correct for 1 mark) (2)
M2: Balanced equation: \(\text{C}_4\text{H}_9\text{OH} + 2[\text{O}] \rightarrow \text{C}_3\text{H}_7\text{COOH} + \text{H}_2\text{O}\). (2)
(c) M1: Mention loss of alcohol O-H peak at \(3230-3550\ \text{cm}^{-1}\). (1)
M2: Mention appearance of C=O peak at \(1680-1750\ \text{cm}^{-1}\) (or carboxylic acid O-H at \(2500-3000\ \text{cm}^{-1}\)). (1)
M3: Conversion is complete when the peak at \(3230-3550\ \text{cm}^{-1}\) is entirely absent. (1)

(a) Silicon dioxide has a giant covalent (macromolecular) structure. Melting it requires breaking many strong covalent bonds between Si and O atoms, which requires a large amount of energy. Phosphorus(V) oxide has a simple molecular structure with weak van der Waals forces between the \(\text{P}_4\text{O}_{10}\) molecules, which require very little energy to overcome.
(b) Equation: \(\text{Na}_2\text{O}(s) + \text{H}_2\text{O}(l) \rightarrow 2\text{NaOH}(aq)\)
pH: 13-14 (accept 14).
(c) Equation: \(\text{P}_4\text{O}_{10} + 12\text{NaOH} \rightarrow 4\text{Na}_3\text{PO}_4 + 6\text{H}_2\text{O}\)
(d) Equation: \(\text{SO}_3(g) + \text{H}_2\text{O}(l) \rightarrow \text{H}_2\text{SO}_4(aq)\)
Sulfur trioxide reacts with water to form sulfuric acid (\(\text{H}_2\text{SO}_4\)), which is a strong acid and fully dissociates in water to release a high concentration of hydrogen/hydronium ions (\(\text{H}^+\)/\(\text{H}_3\text{O}^+\)).

(a) M1: \(\text{SiO}_2\) is giant covalent / macromolecular and \(\text{P}_4\text{O}_{10}\) is simple molecular. (1)
M2: Melting \(\text{SiO}_2\) requires breaking strong covalent bonds. (1)
M3: Melting \(\text{P}_4\text{O}_{10}\) only requires overcoming weak van der Waals forces (intermolecular forces) between molecules. (1)
(b) M1: \(\text{Na}_2\text{O} + \text{H}_2\text{O} \rightarrow 2\text{NaOH}\). (1)
M2: pH = 13 or 14. (1)
(c) M1: Correct reactants and products (\(\text{P}_4\text{O}_{10} + \text{NaOH} \rightarrow \text{Na}_3\text{PO}_4 + \text{H}_2\text{O}\)). (1)
M2: Fully balanced equation: \(\text{P}_4\text{O}_{10} + 12\text{NaOH} \rightarrow 4\text{Na}_3\text{PO}_4 + 6\text{H}_2\text{O}\). (1)
(d) M1: \(\text{SO}_3 + \text{H}_2\text{O} \rightarrow \text{H}_2\text{SO}_4\). (1)
M2: Sulfuric acid is a strong acid that fully dissociates. (1)
M3: High concentration of \(\text{H}^+\) / \(\text{H}_3\text{O}^+\) ions in solution. (1)

(a) Observations: Violent reaction, release of white fumes/mist (\(\text{HCl}\)), and formation of a white precipitate/solid (\(\text{SiO}_2\)).
Equation: \(\text{SiCl}_4(l) + 2\text{H}_2\text{O}(l) \rightarrow \text{SiO}_2(s) + 4\text{HCl}(g)\).
(b) Dimer structure: Two \(\text{AlCl}_3\) units linked by coordinate covalent bonds from a chlorine atom of each unit to the aluminum atom of the other. Arrows point from bridging Cl to Al.
In the solid state, aluminum chloride has localized covalent bonding with no free ions to carry charge. When molten, it dissociates into mobile ions (\(\text{Al}^{3+}\) and \(\text{Cl}^-\)), allowing it to conduct electricity.
(c) \(\text{NaCl}\) contains ionic bonding and simply dissolves in water to form a neutral solution (pH = 7) as the ions are hydrated. \(\text{PCl}_5\) is a covalent chloride that reacts vigorously with water (hydrolyzes) to form phosphoric acid and hydrochloric acid, which release high concentrations of \(\text{H}^+\) ions, resulting in a strongly acidic solution (pH = 1-2).

(a) M1: White fumes/mist and white solid/precipitate. (1)
M2: Correct equation: \(\text{SiCl}_4 + 2\text{H}_2\text{O} \rightarrow \text{SiO}_2 + 4\text{HCl}\). (1)
M3: Correct state symbols: \(\text{SiCl}_4(l)\), \(\text{H}_2\text{O}(l)\), \(\text{SiO}_2(s)\), \(\text{HCl}(g)\). (1)
(b) M1: Correct structural formula of \(\text{Al}_2\text{Cl}_6\) showing bridging chlorines. (1)
M2: Coordinate bonds correctly indicated by arrows pointing from bridging chlorine atoms to aluminum atoms. (1)
M3: Solid state has localized covalent bonds / no free ions to carry charge. (1)
M4: Molten aluminum chloride contains mobile ions that can move and carry charge. (1)
(c) M1: \(\text{NaCl}\) dissolves to form a neutral solution (pH 7). (1)
M2: \(\text{PCl}_5\) hydrolyzes to form a strongly acidic solution (pH 1-2). (1)
M3: Difference is due to \(\text{NaCl}\) being ionic (simple hydration of ions) while \(\text{PCl}_5\) is covalent (allowing attack of water on P atom causing hydrolysis). (1)

(a) Mechanism: Nucleophilic addition.
Steps:
1. Nucleophilic attack by cyanide ion (\(:\text{CN}^-\)) on the carbonyl carbon (\(\text{C}^{\delta+}\)). Curly arrow from the lone pair on the carbon of \(:\text{CN}^-\)
to the carbonyl carbon. Curly arrow from the \(\text{C}=\text{O}\) double bond to the oxygen atom.
2. Formation of an intermediate with a negative charge on oxygen (\(\text{CH}_3\text{CH}_2\text{CH}(\text{O}^-)\text{CN}\)).
3. Protonation of the intermediate: Curly arrow from the lone pair on the negative oxygen atom to a hydrogen ion (\(\text{H}^+\)) to form the OH group.
(b) The carbonyl carbon in propanal has a planar shape around the \(\text{C}=\text{O}\) group. The nucleophile (\(:\text{CN}^-\)) is equally likely to attack the planar carbonyl carbon from either above or below. This produces an equimolar mixture (racemic mixture / racemate) of both enantiomers, which rotate plane-polarized light by equal amounts in opposite directions, canceling out the overall optical activity.
(c) HCN is a very weak acid and dissociates poorly, providing a very low concentration of the nucleophile, \(\text{CN}^-\). KCN is ionic and fully dissociates, providing a much higher concentration of \(\text{CN}^-\), which increases the rate of reaction. Furthermore, HCN is a highly toxic, volatile gas, whereas KCN can be handled safely in solution.

(a) M1: Nucleophilic addition. (1)
M2: Curly arrow from lone pair on carbon of \(:\text{CN}^-\)- to the carbonyl carbon. (1)
M3: Curly arrow from \(\text{C}=\text{O}\) double bond to O. (1)
M4: Correct intermediate structure with negative charge on O. (1)
M5: Curly arrow from lone pair on \(\text{O}^-\)- to \(\text{H}^+\). (1)
(b) M1: Carbonyl carbon is planar. (1)
M2: Attack by \(\text{CN}^-\) is equally likely from above or below. (1)
M3: Leads to an equimolar mixture of two enantiomers (racemate) which cancel each other's optical rotation. (1)
(c) M1: HCN is a weak acid / dissociates poorly (or KCN provides a much higher concentration of the nucleophile \(\text{CN}^-\)). (1)
M2: Safety reason: HCN is a toxic gas, whereas KCN is safer / easier to control in solution with acid. (1)

(a) The uncatalyzed reaction involves the collision of two negatively charged ions. These ions repel each other electrostatically, leading to a high activation energy and a very slow rate of reaction.
Equations for catalyzed reaction:
1) \(2\text{Fe}^{2+}(aq) + \text{S}_2\text{O}_8^{2-}(aq) \rightarrow 2\text{Fe}^{3+}(aq) + 2\text{SO}_4^{2-}(aq)\)
2) \(2\text{Fe}^{3+}(aq) + 2\text{I}^-(aq) \rightarrow 2\text{Fe}^{2+}(aq) + \text{I}_2(aq)\)
(b) Equation: \([\text{Co}(\text{H}_2\text{O})_6]^{2+} + 4\text{Cl}^- \rightarrow [\text{CoCl}_4]^{2-} + 6\text{H}_2\text{O}\)
Color change: Pink to blue.
Shape of product: Tetrahedral.
Reason for change in coordination number: The chloride ligands are much larger than the water ligands. Therefore, fewer chloride ligands (only 4 instead of 6) can fit around the central cobalt(II) ion due to steric hindrance / repulsion between the larger ligands.

(a) M1: Both reactant ions are negatively charged so they repel each other. (1)
M2: This results in a high activation energy. (1)
M3: Equation 1: \(2\text{Fe}^{2+} + \text{S}_2\text{O}_8^{2-} \rightarrow 2\text{Fe}^{3+} + 2\text{SO}_4^{2-}\). (1)
M4: Equation 2: \(2\text{Fe}^{3+} + 2\text{I}^- \rightarrow 2\text{Fe}^{2+} + \text{I}_2\). (1)
(b) M1: Correct reactant and product formulas (\([\text{Co}(\text{H}_2\text{O})_6]^{2+}\) and \([\text{CoCl}_4]^{2-}\)). (1)
M2: Correctly balanced equation. (1)
M3: Color change: Pink to blue. (1)
M4: Shape: Tetrahedral. (1)
M5: Chloride ions are larger than water molecules (or have a negative charge, causing more mutual repulsion). (1)
M6: Only four chloride ions can fit around the cobalt ion. (1)

(a) \(2\text{Cu}^{2+}(aq) + 4\text{I}^-(aq) \rightarrow 2\text{CuI}(s) + \text{I}_2(aq)\)

(b) \(2\text{S}_2\text{O}_3^{2-}(aq) + \text{I}_2(aq) \rightarrow \text{S}_4\text{O}_6^{2-}(aq) + 2\text{I}^-(aq)\)

(c) Titrate the mixture until it becomes pale yellow/straw-colored. Then add a few drops of starch indicator, which turns the mixture blue-black. Continue titration dropwise until the blue-black color completely disappears, leaving a white/off-white precipitate (or cloudy creamy suspension).

(d)
1. Moles of \(\text{S}_2\text{O}_3^{2-}\) used: \(n = 0.02180\text{ dm}^3 \times 0.100\text{ mol dm}^{-3} = 2.180 \times 10^{-3}\text{ mol}\).
2. From the stoichiometry of the equations, \(2\text{Cu}^{2+} \equiv \text{I}_2 \equiv 2\text{S}_2\text{O}_3^{2-}\), so there is a 1:1 molar ratio between \(\text{Cu}^{2+}\) and \(\text{S}_2\text{O}_3^{2-}\).
Therefore, moles of \(\text{Cu}^{2+}\) in the \(25.0\text{ cm}^3\) aliquot = \(2.180 \times 10^{-3}\text{ mol}\).
3. Moles of \(\text{Cu}^{2+}\) in the original \(250.0\text{ cm}^3\) solution: \(n(\text{total}) = 2.180 \times 10^{-3} \times 10 = 0.02180\text{ mol}\).
4. Mass of copper in the alloy: \(m = 0.02180\text{ mol} \times 63.5\text{ g mol}^{-1} = 1.3843\text{ g}\).
5. Percentage by mass of copper: \(\text{Percentage} = \frac{1.3843\text{ g}}{2.80\text{ g}} \times 100\% = 49.44\% \approx 49.4\%\).

(a) M1: Correct equation: \(2\text{Cu}^{2+}(aq) + 4\text{I}^-(aq) \rightarrow 2\text{CuI}(s) + \text{I}_2(aq)\) (state symbols not required). [1 mark]

(b) M1: Correct equation: \(2\text{S}_2\text{O}_3^{2-}(aq) + \text{I}_2(aq) \rightarrow \text{S}_4\text{O}_6^{2-}(aq) + 2\text{I}^-(aq)\) (state symbols not required). [1 mark]

(c)
M1: Titrate to a pale yellow/straw color before adding starch. [1 mark]
M2: Addition of starch turns the solution blue-black. [1 mark]
M3: Endpoint is when the blue-black color completely disappears, leaving a white/off-white precipitate (accept 'turns colorless' or 'creamy precipitate remains'). [1 mark]

(d)
M1: Moles of \(\text{S}_2\text{O}_3^{2-} = 2.180 \times 10^{-3}\text{ mol}\). [1 mark]
M2: State or use 1:1 ratio between \(\text{Cu}^{2+}\) and \(\text{S}_2\text{O}_3^{2-}\) to find \(n(\text{Cu}^{2+}) = 2.180 \times 10^{-3}\text{ mol}\) in 25.0 cm3. [1 mark]
M3: Scaling up to 250 cm3: \(n(\text{total Cu}^{2+}) = 0.02180\text{ mol}\). [1 mark]
M4: Mass of copper = \(0.02180 \times 63.5 = 1.384\text{ g}\) (allow consequential error from incorrect moles). [1 mark]
M5: Percentage by mass = 49.4% (must be to 3 significant figures; accept 49.5% if 63.55 is used for Ar of Cu). [1 mark]

(a) Sodium thiosulfate reacts immediately with any iodine produced: \(\text{I}_2 + 2\text{S}_2\text{O}_3^{2-} \rightarrow 2\text{I}^- + \text{S}_4\text{O}_6^{2-}\). This prevents iodine from reacting with starch. Once all the sodium thiosulfate is consumed, the iodine accumulates and reacts with starch, turning the solution blue-black. The time taken for this color change is inversely proportional to the initial rate of reaction.

(b) The amount of thiosulfate is very small compared to the reactants. Therefore, only a very small, negligible fraction of the reactants is consumed before the blue-black color appears, meaning their concentrations remain virtually constant during this period.

(c)
- Comparing Exp 1 and Exp 2: \([\text{H}_2\text{O}_2]\) is doubled, while \([\text{I}^-]\) remains constant. The time taken halves from 44.0 s to 22.0 s. Since \(\text{Rate} \propto \frac{1}{t}\), the rate doubles. Therefore, the order with respect to \(\text{H}_2\text{O}_2\) is 1.
- Comparing Exp 1 and Exp 3: \([\text{I}^-]\) is doubled, while \([\text{H}_2\text{O}_2]\) remains constant. The time taken halves from 44.0 s to 22.0 s. Thus, the rate doubles. Therefore, the order with respect to \(\text{I}^-\) is 1.

(d)
Rate equation: \(\text{Rate} = k[\text{H}_2\text{O}_2][\text{I}^-]\)
Using Experiment 1:
\(\text{Rate} = \frac{1}{44.0} = 0.0227\text{ s}^{-1}\)
\(0.0227 = k \times 0.040 \times 0.050\)
\(k = \frac{0.0227}{0.0020} = 11.4\text{ dm}^3\text{ mol}^{-1}\text{ s}^{-1}\) (or \(11.36\)).

(a)
M1: Sodium thiosulfate reacts with iodine as soon as it is formed (converting it back to iodide). [1 mark]
M2: When all thiosulfate is used up, free iodine reacts with starch to turn the solution blue-black, allowing the time taken for a fixed amount of reaction to be measured. [1 mark]

(b)
M1: The amount of thiosulfate added is very small. [1 mark]
M2: Hence, only a tiny/insignificant fraction of the reactants is used up by the time the color change occurs. [1 mark]

(c)
M1: In Exp 1 and 2, when \([\text{H}_2\text{O}_2]\) doubles (with \([\text{I}^-]\) constant), the time halves, so rate doubles. [1 mark]
M2: Therefore, order with respect to \(\text{H}_2\text{O}_2\) is 1. [1 mark]
M3: In Exp 1 and 3, when \([\text{I}^-]\) doubles (with \([\text{H}_2\text{O}_2]\) constant), the time halves, so rate doubles. [1 mark]
M4: Therefore, order with respect to \(\text{I}^-\) is 1. [1 mark]

(d)
M1: \(\text{Rate} = k[\text{H}_2\text{O}_2][\text{I}^-]\) (consequential on orders from part c). [1 mark]
M2: \(k = 11.4\) (accept range \(11.3 - 11.4\)) and units \(\text{dm}^3\text{ mol}^{-1}\text{ s}^{-1}\). [1 mark]

(a) The diagram must show:
1. A round-bottomed or pear-shaped flask containing the reaction mixture, heated with a heat source (e.g., electric heating mantle or water bath/sand bath).
2. A still head containing a thermometer (with bulb level with the side-arm).
3. A condenser with water flowing in at the bottom and out at the top, connected to a receiver adapter leading to a collection vessel (e.g., conical flask or measuring cylinder).
*(Note: The system must not be sealed/closed, i.e., there must be an opening to the air at the receiver).*

(b) Purification steps:
1. Transfer the crude distillate to a separating funnel.
2. Add sodium hydrogencarbonate solution to neutralize any phosphoric acid. Shake, releasing pressure regularly by opening the tap.
3. Allow the mixture to settle into two layers. Discard the lower aqueous layer, retaining the upper organic layer.
4. Wash the organic layer with water (or saturated NaCl solution) in the separating funnel, and separate again, discarding the aqueous layer.
5. Transfer the cyclohexene layer to a clean conical flask and add an anhydrous drying agent (such as anhydrous calcium chloride, \(\text{CaCl}_2\), or anhydrous magnesium sulfate, \(\text{MgSO}_4\)). Shake and leave until the liquid becomes clear.
6. Decant or filter the dry liquid into a clean flask and perform a final redistillation, collecting the fraction that boils close to the boiling point of cyclohexene (e.g., \(81-85^\circ\text{C}\)).

(c) Add a few drops of bromine water. The orange/brown solution will decolorize (turn colorless).

(a)
M1: Flask with heat source and thermometer correctly positioned (bulb level with side arm). [1 mark]
M2: Downward condenser with correct water flow direction (in at bottom, out at top). [1 mark]
M3: Completed collection system that is open to the air (not sealed). [1 mark]

(b)
M1: Use a separating funnel to separate the layers. [1 mark]
M2: Add sodium hydrogencarbonate (or sodium carbonate) solution to neutralize acidic impurities (and release pressure/gas). [1 mark]
M3: Discard the aqueous layer (retaining the organic/upper layer). [1 mark]
M4: Add an anhydrous drying agent (e.g., anhydrous \(\text{CaCl}_2\), \(\text{MgSO}_4\), or \(\text{Na}_2\text{SO}_4\)) to dry the organic layer until clear. [1 mark]
M5: Decant/filter and redistill, collecting the fraction boiling between \(81-85^\circ\text{C}\). [1 mark]

(c)
M1: Reagent: Bromine water (or bromine in organic solvent, or alkaline \(\text{KMnO}_4\)). [1 mark]
M2: Observation: Turns from orange/brown to colorless (or purple to colorless/brown ppt for permanganate). [1 mark]

For a reaction to be feasible, the Gibbs free energy change must be negative: \(\Delta G^\ominus = \Delta H^\ominus - T\Delta S^\ominus < 0\). Rearranging this inequality gives \(T > \frac{\Delta H^\ominus}{\Delta S^\ominus}\). Converting the enthalpy change to joules: \(T > \frac{135000}{285} \approx 473.68\text{ K}\). Therefore, the reaction becomes feasible above \(474\text{ K}\).

Award 1 mark for the correct calculation showing that feasibility is achieved above 474 K (Choice A).

The octahedral complex ion \(\text{[Co(H_2NCH_2CH_2NH_2)_3]^{3+}}\) contains three bidentate ligands and has no plane of symmetry, which allows it to exist as two non-superimposable mirror images (enantiomers). The trans-isomers in options B and C possess a plane of symmetry and are achiral. Option D is a square planar complex and does not exhibit optical isomerism.

Award 1 mark for identifying that only the tris-ethylenediamine complex is chiral (Choice A).

The reaction of propanal with HCN/KCN is a nucleophilic addition reaction. The first step is the nucleophilic attack of the cyanide ion, \(\text{CN}^-\), on the electron-deficient carbonyl carbon. The product, 2-hydroxybutanenitrile, is formed as a racemic mixture because the planar carbonyl group allows equal probability of attack from above or below the plane, meaning it is optically inactive.

Award 1 mark for identifying the correct mechanical description of the nucleophilic attack (Choice A).

Sulfur trioxide (\(\text{SO}_3\)) reacts vigorously with water to form sulfuric acid (\(\text{H}_2\text{SO}_4\)), a strong diprotic acid with a typical pH around 0-1. Phosphorus(V) oxide (\(\text{P}_4\text{O}_{10}\)) reacts to form phosphoric acid (\(\text{H}_3\text{PO}_4\)), a weaker triprotic acid with a pH of around 1-2. Silicon dioxide (\(\text{SiO}_2\)) is giant covalent and insoluble in water, so the resulting pH is neutral (7). The correct order of increasing pH is \(\text{SO}_3 < \text{P}_4\text{O}_{10} < \text{SiO}_2\).

Award 1 mark for ordering the three oxides correctly from lowest to highest pH (Choice A).

First calculate the moles of CO2 using \(pV = nRT\). Convert pressure: \(100\text{ kPa} = 100 \times 10^3\text{ Pa}\); convert volume: \(348\text{ cm}^3 = 348 \times 10^{-6}\text{ m}^3\). \(n = \frac{pV}{RT} = \frac{100 \times 10^3 \times 348 \times 10^{-6}}{8.31 \times 298} \approx 0.01405\text{ mol}\). Because the molar ratio of CaCO3 to CO2 is 1:1, moles of CaCO3 decomposed is also \(0.01405\text{ mol}\). Finally, \(\text{mass} = n \times M_{\text{r}} = 0.01405 \times 100.1 \approx 1.41\text{ g}\).

Award 1 mark for the correct calculation of mass using the ideal gas equation (Choice A).

Iron(III) chloride dissolves in water to form the hexaaqua complex ion \(\text{[Fe(H_2O)_6]^{3+}}\). Due to the high charge-to-size ratio (charge density) of the iron(III) ion, the oxygen-hydrogen bonds in the coordinated water molecules are strongly polarized. This allows a water molecule from the solvent to act as a base and accept a proton, forming hydronium ions (\(\text{H_3O^+}\)) according to: \(\text{[Fe(H_2O)_6]^{3+}(aq)} + \text{H_2O(l)} \rightleftharpoons \text{[Fe(H_2O)_5(OH)]^{2+}(aq)} + \text{H_3O^+(aq)}\).

Award 1 mark for choosing the correct equilibrium equation representing the hydrolysis of the hexaaquairon(III) ion (Choice A).

Using a Born-Haber cycle: \(\Delta H_{\text{f}}^\ominus = \Delta H_{\text{at}}^\ominus(\text{Mg}) + 1\text{st IE}(\text{Mg}) + 2\text{nd IE}(\text{Mg}) + \Delta H_{\text{diss}}^\ominus(\text{Cl}_2) + 2 \times \text{EA}(\text{Cl}) - \Delta H_{\text{L,diss}}^\ominus\). Substituting the data: \(-642 = 148 + 738 + 1451 + 242 + 2(-349) - \Delta H_{\text{L,diss}}^\ominus\). This simplifies to: \(-642 = 2579 - 698 - \Delta H_{\text{L,diss}}^\ominus \implies -642 = 1881 - \Delta H_{\text{L,diss}}^\ominus\). Rearranging gives: \(\Delta H_{\text{L,diss}}^\ominus = 1881 + 642 = +2523\text{ kJ mol}^{-1}\).

Award 1 mark for correct calculation of the lattice dissociation enthalpy of MgCl2 (Choice A).

Pentan-3-one has the symmetrical structure \(\text{CH}_3\text{CH}_2\text{COCH}_2\text{CH}_3\). Upon reaction with \(\text{HCN}\), it forms 2-ethyl-2-hydroxybutanenitrile: \(\text{CH}_3\text{CH}_2\text{C(OH)(CN)CH}_2\text{CH}_3\). The central carbon is bonded to a hydroxyl group, a nitrile group, and two identical ethyl groups. Since it does not contain a carbon atom with four different groups attached, it lacks a chiral centre and is optically inactive.

Award 1 mark for identifying the symmetrical ketone that yields an achiral product (Choice C).

1. Calculate the standard entropy change, \(\Delta S^\ominus\):
\(\Delta S^\ominus = \sum S^\ominus(\text{products}) - \sum S^\ominus(\text{reactants})\)
\(\Delta S^\ominus = 2 \times 240 - 304 = 480 - 304 = +176\text{ J K}^{-1}\text{ mol}^{-1}\)

2. For a reaction to be feasible, \(\Delta G^\ominus < 0\).
Since \(\Delta G^\ominus = \Delta H^\ominus - T\Delta S^\ominus\), we have:
\(\Delta H^\ominus - T\Delta S^\ominus < 0 \implies T > \frac{\Delta H^\ominus}{\Delta S^\ominus}\)

3. Substitute the values (remembering to convert \(\Delta H^\ominus\) to \(\text{J mol}^{-1}\)):
\(T > \frac{58.0 \times 10^3}{176} \approx 330\text{ K}\).

[1 mark] Correctly identifies 330 K as the minimum temperature.

The entropy change of a ligand substitution reaction is primarily determined by the change in the total number of species in solution.
- In reaction a: 7 species react to form 7 species (\(\Delta n = 0\)).
- In reaction b: 2 species react to form 2 species (\(\Delta n = 0\)).
- In reaction c: 5 species react to form 7 species (\(\Delta n = +2\)).
- In reaction d: 4 species react to form 7 species (\(\Delta n = +3\)), since \(\text{H}_2\text{NCH}_2\text{CH}_2\text{NH}_2\) (1,2-diaminoethane) is a bidentate ligand. This significant increase in the number of particles in solution leads to the most positive entropy change (the chelate effect).

[1 mark] Correctly identifies reaction d.

- \(\text{SiO}_2\) has a giant covalent macromolecular structure and is completely insoluble in water (thus option a is incorrect).
- Phosphorus(V) oxide (\(\text{P}_4\text{O}_{10}\)) reacts vigorously and exothermically with water to form phosphoric(V) acid (\(\text{H}_3\text{PO}_4\)), which is a strong acid and gives a solution with pH 1-2 (thus option b is correct).
- Aluminium oxide is completely insoluble in water (thus option c is incorrect).
- Sulfur dioxide (\(\text{SO}_2\)) reacts with water to form sulfurous acid (sulfuric(IV) acid, \(\text{H}_2\text{SO}_3\)), which is a weak acid (thus option d is incorrect).

[1 mark] Correctly identifies option B.

- If \(X\) is butanone, reduction yields butan-2-ol, which dehydrates to form a mixture of but-1-ene and but-2-ene (which shows E/Z isomerism).
- If \(X\) is butanal, reduction yields butan-1-ol, which dehydrates to form a mixture of but-1-ene and but-2-ene.
- If \(X\) is 2-methylpropanal, reduction yields 2-methylpropan-1-ol, \(\text{(CH}_3)_2\text{CH-CH}_2\text{OH}\). Dehydration of this primary alcohol removes \(\text{-OH}\) from C1 and \(\text{-H}\) from C2, giving 2-methylpropene, \(\text{(CH}_3)_2\text{C=CH}_2\). This is the only possible alkene that can form from this alcohol, and since one double-bonded carbon has two identical hydrogen atoms, it does not display E/Z isomerism.
- Pentan-3-one has the wrong molecular formula (\(\text{C}_5\text{H}_{10}\text{O}\)).
Therefore, the correct compound is 2-methylpropanal.

[1 mark] Correctly identifies 2-methylpropanal.

1. Moles of \(\text{NaOH}\) in titration:
\(n(\text{NaOH}) = C \times V = 0.100\text{ mol dm}^{-3} \times 0.0200\text{ dm}^3 = 0.00200\text{ mol}\)

2. Since the carboxylic acid \(W\) is monoprotic, the moles of \(W\) in the titrated \(25.0\text{ cm}^3\) portion is also \(0.00200\text{ mol}\).

3. Total moles of \(W\) in \(250\text{ cm}^3\):
\(n_{\text{total}}(W) = 0.00200 \times \frac{250}{25.0} = 0.0200\text{ mol}\)

4. Since oxidation of a primary alcohol to a carboxylic acid is in a \(1:1\) molar ratio, the moles of alcohol \(V\) originally present is \(0.0200\text{ mol}\).

5. Calculate the molar mass of alcohol \(V\):
\(M_r = \frac{\text{mass}}{\text{moles}} = \frac{0.92\text{ g}}{0.0200\text{ mol}} = 46.0\text{ g mol}^{-1}\)

6. Ethanol (\(\text{CH}_3\text{CH}_2\text{OH}\)) has a molecular formula of \(\text{C}_2\text{H}_6\text{O}\), with \(M_r = 2 \times 12.0 + 6 \times 1.0 + 16.0 = 46.0\text{ g mol}^{-1}\), matching alcohol \(V\).

[1 mark] Correctly identifies Ethanol.

The thermodynamic feasibility is determined by calculating the standard cell potential \(E^\ominus_{\text{cell}} = E^\ominus_{\text{reduction}} - E^\ominus_{\text{oxidation}}\). If \(E^\ominus_{\text{cell}} > 0\), the reaction is feasible.
- For a: \(E^\ominus_{\text{cell}} = E^\ominus(\text{Ag}^+/\text{Ag}) - E^\ominus(\text{Fe}^{2+}/\text{Fe}) = +0.80 - (-0.44) = +1.24\text{ V}\) (feasible).
- For b: \(E^\ominus_{\text{cell}} = E^\ominus(\text{Fe}^{3+}/\text{Fe}^{2+}) - E^\ominus(\text{Fe}^{2+}/\text{Fe}) = +0.77 - (-0.44) = +1.21\text{ V}\) (feasible).
- For c: \(E^\ominus_{\text{cell}} = E^\ominus(\text{Fe}^{3+}/\text{Fe}^{2+}) - E^\ominus(\text{I}_2/\text{I}^-) = +0.77 - 0.54 = +0.23\text{ V}\) (feasible).
- For d: \(E^\ominus_{\text{cell}} = E^\ominus(\text{Fe}^{3+}/\text{Fe}^{2+}) - E^\ominus(\text{Ag}^+/\text{Ag}) = +0.77 - 0.80 = -0.03\text{ V}\) (not feasible since \(E^\ominus_{\text{cell}} < 0\)).

[1 mark] Correctly identifies process d.

The rate of nucleophilic substitution in halogenoalkanes is determined by the carbon-halogen bond strength (bond enthalpy). The weaker the C-X bond, the lower the activation energy, and the faster the reaction.
- The bond enthalpies decrease in the order: \(\text{C-F} > \text{C-Cl} > \text{C-Br} > \text{C-I}\).
- Therefore, the C-I bond is the easiest to break, making 1-iodobutane the most reactive. Although the C-I bond is the least polar, the bond enthalpy factor dominates over the bond polarity factor in determining substitution rates.

[1 mark] Correctly identifies option C.

1. Moles of \(\text{MnO}_4^-\):
\(n(\text{MnO}_4^-) = C \times V = 0.0200\text{ mol dm}^{-3} \times 0.0250\text{ dm}^3 = 5.00 \times 10^{-4}\text{ mol}\)

2. From the stoichiometry of the reaction, \(2\text{ moles of } \text{MnO}_4^-\text{ react with } 5\text{ moles of } \text{H}_2\text{O}_2\):
\(n(\text{H}_2\text{O}_2) = \frac{5}{2} \times n(\text{MnO}_4^-) = 2.5 \times 5.00 \times 10^{-4}\text{ mol} = 1.25 \times 10^{-3}\text{ mol}\)

3. Volume of \(0.0500\text{ mol dm}^{-3}\) \(\text{H}_2\text{O}_2\) needed:
\(V = \frac{n}{C} = \frac{1.25 \times 10^{-3}\text{ mol}}{0.0500\text{ mol dm}^{-3}} = 0.0250\text{ dm}^3 = 25.0\text{ cm}^3\).

[1 mark] Correctly calculates and identifies 25.0 cm3.

According to the Born-Haber cycle:
\(\Delta H^\theta_f(\text{CaCl}_2) = \Delta H^\theta_{\text{at}}(\text{Ca}) + 1\text{st IE}(\text{Ca}) + 2\text{nd IE}(\text{Ca}) + \Delta H^\theta_{\text{diss}}(\text{Cl}_2) + 2 \times \text{EA}(\text{Cl}) - \Delta H^\theta_L(\text{diss})\)

Rearranging to find the lattice enthalpy of dissociation (\(\Delta H^\theta_L(\text{diss})\)):
\(\Delta H^\theta_L(\text{diss}) = -\Delta H^\theta_f(\text{CaCl}_2) + \Delta H^\theta_{\text{at}}(\text{Ca}) + 1\text{st IE}(\text{Ca}) + 2\text{nd IE}(\text{Ca}) + \Delta H^\theta_{\text{diss}}(\text{Cl}_2) + 2 \times \text{EA}(\text{Cl})\)

Substitute the values:
\(\Delta H^\theta_L(\text{diss}) = -(-796) + 178 + 590 + 1145 + 242 + 2(-349)\)
\(\Delta H^\theta_L(\text{diss}) = 796 + 178 + 590 + 1145 + 242 - 698 = +2253\text{ kJ mol}^{-1}\)

1 mark for correct calculation and selection of option A.
0 marks for any other response.

Dehydration of pentan-2-ol, \(\text{CH}_3\text{CH(OH)}\text{CH}_2\text{CH}_2\text{CH}_3\), occurs via elimination of a water molecule. Elimination can take place from either adjacent carbon:
1. Removing hydrogen from carbon-1 gives pent-1-ene (1 structural isomer).
2. Removing hydrogen from carbon-3 gives pent-2-ene. Pent-2-ene displays geometric (E/Z) isomerism, and thus exists as both (E)-pent-2-ene and (Z)-pent-2-ene (2 stereoisomers).

Thus, a total of 3 isomeric alkenes are formed.

1 mark for correct choice of 3 isomers (Option B).
0 marks for any incorrect response.

Aluminium oxide (\(\text{Al}_2\text{O}_3\)) is amphoteric, meaning it reacts with both acids (such as HCl) and bases (such as NaOH) to form salt and water.

- Option A is incorrect because sulfur dioxide dissolves in water to form sulfurous acid, which is an acidic solution (pH around 1-2).
- Option C is incorrect because silicon dioxide is a macromolecular structure and is completely insoluble in water.
- Option D is incorrect because sodium oxide has a giant ionic lattice structure, not molecular covalent.

1 mark for identifying that aluminium oxide is amphoteric (Option B).
0 marks for incorrect options.

Propanal (\(\text{CH}_3\text{CH}_2\text{CHO}\)) reacts with potassium cyanide followed by acid to form 2-hydroxybutanenitrile (\(\text{CH}_3\text{CH}_2\text{CH(OH)CN}\)) by nucleophilic addition.

The carbonyl carbon of propanal is planar, allowing the cyanide ion to attack with equal probability from either side of the plane. This results in the formation of equal amounts of both enantiomers, producing a racemic mixture which is optically inactive.

1 mark for selecting Option B.
0 marks for incorrect options.

Both reactant ions (\(\text{S}_2\text{O}_8^{2-}\) and \(\text{I}^-\)) are negatively charged. Like charges repel each other, which creates a high activation energy barrier for the uncatalysed reaction, resulting in a very slow rate of reaction.

1 mark for correct explanation (Option A).
0 marks for any other choice.

First, calculate the moles of \(\text{CO}_2\):
\(n = \frac{\text{mass}}{M_r} = \frac{5.50}{44.0} = 0.125\text{ mol}\)

Convert pressure from kPa to Pa:
\(p = 120 \times 10^3\text{ Pa} = 120,000\text{ Pa}\)

Use the ideal gas equation, \(pV = nRT\), to calculate volume \(V\) in \(\text{m}^3\):
\(V = \frac{nRT}{p} = \frac{0.125 \times 8.31 \times 350}{120,000} = 0.0030297\text{ m}^3\)

Convert \(\text{m}^3\) to \(\text{dm}^3\) (multiply by 1000):
\(V = 0.0030297 \times 1000 = 3.03\text{ dm}^3\)

1 mark for the correct volume calculation (Option A).
0 marks for any other response.

Calculate \(\Delta S^\theta\) for the reaction:
\(\Delta S^\theta = \sum S^\theta(\text{products}) - \sum S^\theta(\text{reactants})\)
\(\Delta S^\theta = (39.7 + 213.6) - 92.9 = 160.4\text{ J K}^{-1}\text{mol}^{-1}\)

Convert \(\Delta H^\theta\) to \(\text{J mol}^{-1}\):
\(\Delta H^\theta = 178 \times 10^3\text{ J mol}^{-1} = 178,000\text{ J mol}^{-1}\)

For a reaction to be feasible, \(\Delta G^\theta \le 0\). The transition occurs where \(\Delta G^\theta = 0\):
\(T = \frac{\Delta H^\theta}{\Delta S^\theta} = \frac{178,000}{160.4} = 1109.7\text{ K} \approx 1110\text{ K}\)

1 mark for correct entropy change calculation and temperature calculation (Option A).
0 marks for any incorrect answer.

The rate of hydrolysis of halogenoalkanes is determined by the bond enthalpy of the carbon-halogen bond. The C-I bond is the longest and weakest of all carbon-halogen bonds listed, meaning it requires the least energy to break. Therefore, 1-iodobutane reacts the fastest.

1 mark for identifying 1-iodobutane (Option C) as the fastest reacting halogenoalkane.
0 marks for other choices.

In this reaction, the larger chloride ligand replaces the smaller water ligands around the \(Co^{2+}\) central metal ion. Due to steric hindrance, only four chloride ligands can fit around the cobalt ion, resulting in a change in coordination number from 6 to 4, and a change in shape from octahedral to tetrahedral. The equation for the reaction is: \([Co(H_2O)_6]^{2+}(aq) + 4Cl^-(aq) \rightleftharpoons [CoCl_4]^{2-}(aq) + 6H_2O(l)\). The starting hexaaquacobalt(II) complex is pink, and the newly formed tetrachorocobaltate(II) complex, \([CoCl_4]^{2-}\), is blue. Thus, the correct option is A.

1 mark: Correctly identifies formula as \([CoCl_4]^{2-}\), coordination number as 4, and color as blue.

Metal ions with a \(3+\) charge, such as \(Al^{3+}\) and \(Fe^{3+}\), have a high charge-to-size ratio (high charge density). This makes them strongly polarizing. They polarize the O-H bonds in the coordinated water molecules of the hexaaqua ion, releasing protons and making the solution distinctly acidic: \([Al(H_2O)_6]^{3+}(aq) + H_2O(l) \rightleftharpoons [Al(H_2O)_5(OH)]^{2+}(aq) + H_3O^+(aq)\). When carbonate ions (\(CO_3^{2-}\)) are added, they act as a base, accepting these protons. This shifts the acid-base equilibria further, resulting in the precipitation of the insoluble aluminium hydroxide and the liberation of carbon dioxide gas: \(2[Al(H_2O)_6]^{3+}(aq) + 3CO_3^{2-}(aq) \rightarrow 2[Al(H_2O)_3(OH)_3](s) + 3CO_2(g) + 3H_2O(l)\). In contrast, \(2+\) metal ions (like \(Cu^{2+}\), \(Fe^{2+}\), and \(Co^{2+}\)) have a lower charge density, are less acidic, and undergo a simple precipitation reaction to form the metal carbonate (e.g., \(CuCO_3\)), with no gas evolved. Therefore, aqueous aluminium sulfate (containing \(Al^{3+}\)) is the correct choice.

1 mark: Selects the correct option (C) and understands that \(3+\) ions undergo hydrolysis reactions with carbonate to form a hydroxide precipitate and \(CO_2\) gas.

In the preparation of the nitrating mixture, concentrated sulfuric acid acts as a stronger acid than concentrated nitric acid, protonating the nitric acid. The protonated nitric acid then loses a molecule of water to yield the nitronium ion (\(NO_2^+\)), which is the active electrophile. The overall equation for this protonation and subsequent dehydration is: \(\text{HNO}_3 + 2\text{H}_2\text{SO}_4 \rightarrow \text{NO}_2^+ + 2\text{HSO}_4^- + \text{H}_3\text{O}^+\). Thus, option B is the correct representation.

1 mark: Correctly identifies the balanced equation showing the role of \(H_2SO_4\) as an acid and the generation of \(NO_2^+\) and \(H_3O^+\).

The strength of a base depends on the availability of the lone pair of electrons on the nitrogen atom to accept a proton (\(H^+\)). In diethylamine (a secondary amine), there are two ethyl groups which release electron density toward the nitrogen atom via the positive inductive effect. This increases the electron density on the nitrogen atom, making its lone pair more available than in ethylamine. In ethylamine (a primary amine), there is only one ethyl group releasing electron density via the inductive effect. Ammonia has no alkyl groups, so there is no inductive effect to increase the electron density on the nitrogen atom. In phenylamine (an aromatic amine), the lone pair of electrons on the nitrogen atom is delocalized into the \(\pi\)-system of the benzene ring, making it significantly less available to accept a proton. Therefore, the order of basic strength is: diethylamine > ethylamine > ammonia > phenylamine. Diethylamine is the strongest base, making D the correct answer.

1 mark: Correctly identifies diethylamine as the strongest base due to the positive inductive effect of two alkyl groups.

The repeating unit is a polyamide, as indicated by the amide linkage (\(-\text{CO}-\text{NH}-\)). This is formed from a diamine and a dicarboxylic acid. By breaking the amide bonds in the repeating unit: 1. The diamine part is \(-\text{NH}-\text{CH}_2-\text{CH}_2-\text{NH}-\). Adding hydrogen atoms to the nitrogen atoms gives the monomer: \(\text{H}_2\text{N}-\text{CH}_2-\text{CH}_2-\text{NH}_2\) (ethane-1,2-diamine). 2. The dicarboxylic acid part is \(-\text{CO}-\text{CH}_2-\text{CH}_2-\text{CO}-\). Adding hydroxyl (\(-\text{OH}\)) groups to the carbonyl carbons gives the monomer: \(\text{HOOC}-\text{CH}_2-\text{CH}_2-\text{COOH}\) (butanedioic acid, containing four carbon atoms in total). Therefore, the monomers are ethane-1,2-diamine and butanedioic acid. This corresponds to option B.

1 mark: Correctly deduces the structure of the diamine and dicarboxylic acid monomers from the polyamide repeating unit.

To find the correct isomer, we count the number of unique carbon environments for each option: 1. Ethyl ethanoate (\(\text{CH}_3\text{COOCH}_2\text{CH}_3\)) has 4 different carbon environments (4 peaks). 2. Methyl propanoate (\(\text{CH}_3\text{CH}_2\text{COOCH}_3\)) has 4 different carbon environments (4 peaks). 3. Propyl methanoate (\(\text{HCOOCH}_2\text{CH}_2\text{CH}_3\)) has 4 different carbon environments (4 peaks). 4. Isopropyl methanoate (\(\text{HCOOCH}(\text{CH}_3)_2\)) has 3 different carbon environments: the methanoate carbon, the CH carbon, and the two equivalent methyl carbons attached to the CH group (3 peaks). Therefore, isopropyl methanoate is the correct option (D).

1 mark: Correctly identifies isopropyl methanoate as having exactly three carbon environments due to the symmetry of the two methyl groups.