2026 Cambridge IAL Chemistry (9701) Practice Paper with Answers

1. Write the balanced equation for the combustion of a hydrocarbon:
\[\text{C}_x\text{H}_y\text{(g)} + \left(x + \frac{y}{4}\right)\text{O}_2\text{(g)} \rightarrow x\text{CO}_2\text{(g)} + \frac{y}{2}\text{H}_2\text{O(l)}\]

2. The remaining \(50\text{ cm}^3\) of gas after cooling consists of unreacted \(\text{O}_2\) and product \(\text{CO}_2\) (water has condensed to a liquid with negligible volume).

3. Passing the mixture through \(\text{NaOH(aq)}\) absorbs all the \(\text{CO}_2\). The remaining \(10\text{ cm}^3\) is unreacted \(\text{O}_2\).

4. Volume of \(\text{CO}_2\) produced:
\[V(\text{CO}_2) = 50\text{ cm}^3 - 10\text{ cm}^3 = 40\text{ cm}^3\]
Since \(10\text{ cm}^3\) of \(\text{C}_x\text{H}_y\) produces \(40\text{ cm}^3\) of \(\text{CO}_2\):
\[x = \frac{40}{10} = 4\]

5. Volume of \(\text{O}_2\) reacted:
\[V(\text{O}_2\text{ used}) = 70\text{ cm}^3 - 10\text{ cm}^3 = 60\text{ cm}^3\]
\[10\left(x + \frac{y}{4}\right) = 60 \implies 4 + \frac{y}{4} = 6 \implies \frac{y}{4} = 2 \implies y = 8\]

Thus, the molecular formula is \(\text{C}_4\text{H}_8\).

[1 mark] for selecting option C (deducing x = 4 and y = 8 correctly from gas stoichiometry).

1. In \(\text{NH}_2^-\), the central nitrogen atom has 5 valence electrons + 1 extra electron from the charge = 6 electrons. It forms 2 single covalent bonds with H and has 2 lone pairs. Four electron pairs arrange tetrahedrally, giving a non-linear (bent) molecular geometry with a bond angle of approximately \(104.5^\circ\) (which is \(< 109.5^\circ\)).

2. In \(\text{H}_2\text{O}\), the central oxygen atom has 6 valence electrons, forming 2 single bonds with H and having 2 lone pairs. It also adopts a non-linear (bent) shape with a bond angle of approximately \(104.5^\circ\) (which is \(< 109.5^\circ\)).

3. Checking other options:
- \(\text{CO}_2\) is linear (\(180^\circ\)).
- \(\text{NO}_2^+\) is linear (\(180^\circ\)) and \(\text{NH}_4^+\) is tetrahedral (\(109.5^\circ\)).
- \(\text{BeCl}_2\) is linear (\(180^\circ\)) and \(\text{BF}_3\) is trigonal planar (\(120^\circ\)).

Therefore, A is the correct answer.

[1 mark] for identifying option A based on VSEPR theory (2 bond pairs, 2 lone pairs in both species).

Enthalpy change of reaction using bond enthalpies is given by:
\[\Delta H_r = \sum (\text{bond enthalpies of bonds broken}) - \sum (\text{bond enthalpies of bonds formed})\]

Bonds broken:
- \(1 \times (\text{C}=\text{C}) = 610\text{ kJ mol}^{-1}\)
- \(1 \times (\text{H}-\text{Cl}) = 431\text{ kJ mol}^{-1}\)
\[\text{Total energy absorbed to break bonds} = 610 + 431 = 1041\text{ kJ mol}^{-1}\]

Bonds formed:
- \(1 \times (\text{C}-\text{C}) = 350\text{ kJ mol}^{-1}\)
- \(1 \times (\text{C}-\text{H}) = 410\text{ kJ mol}^{-1}\)
- \(1 \times (\text{C}-\text{Cl}) = 340\text{ kJ mol}^{-1}\)
\[\text{Total energy released forming bonds} = 350 + 410 + 340 = 1100\text{ kJ mol}^{-1}\]

\[\Delta H_r = 1041 - 1100 = -59\text{ kJ mol}^{-1}\]

[1 mark] for correctly calculating bonds broken (1041) - bonds formed (1100) = -59 kJ mol-1.

1. When temperature increases from \(T_1\) to \(T_2\), the average kinetic energy of the molecules increases. The most probable energy increases, meaning the peak shifts to the right (to higher energy).

2. Because the total number of molecules (total area under the entire curve) remains constant, the curve flattens and the peak height decreases (becomes lower).

3. A greater proportion of molecules have energy greater than or equal to the activation energy \(E_{\text{a}}\), so the area under the curve to the right of \(E_{\text{a}}\) increases.

[1 mark] for selecting option A (peak shifts right and lower; area right of Ea increases).

1. In Experiment 1, \(\text{P}_4\text{O}_{10}\) is a solid acidic oxide that dissolves in and reacts with water to form phosphoric(V) acid, \(\text{H}_3\text{PO}_4\), giving a solution of approximately \(\text{pH } 2\):
\[\text{P}_4\text{O}_{10}\text{(s)} + 6\text{H}_2\text{O(l)} \rightarrow 4\text{H}_3\text{PO}_4\text{(aq)}\]
(\(\text{Al}_2\text{O}_3\) and \(\text{SiO}_2\) are insoluble; \(\text{MgO}\) is basic).

2. In Experiment 2, silicon tetrachloride, \(\text{SiCl}_4\), is a liquid chloride that hydrolyses vigorously in water to form a white precipitate of silicon dioxide (or hydrated silica) and steamy fumes of \(\text{HCl}\), leaving a strongly acidic solution (\(\text{pH } 1\)):
\[\text{SiCl}_4\text{(l)} + 2\text{H}_2\text{O(l)} \rightarrow \text{SiO}_2\text{(s)} + 4\text{HCl(aq)}\]

Therefore, \(\mathbf{X} = \text{P}_4\text{O}_{10}\) and \(\mathbf{Y} = \text{SiCl}_4\).

[1 mark] for identifying P4O10 as the acidic oxide and SiCl4 as the liquid chloride forming a precipitate.

1. The cream precipitate formed with \(\text{AgNO}_3\text{(aq)}\) is \(\text{AgBr}\), which identifies the halide ion in \(\mathbf{W}\) as bromide (\(\text{Br}^-\)). Thus, \(\mathbf{W}\) is \(\text{KBr}\).

2. Potassium bromide reacts with concentrated sulfuric acid in two stages:
- Acid-base reaction: \(\text{KBr} + \text{H}_2\text{SO}_4 \rightarrow \text{KHSO}_4 + \text{HBr}\) (steamy fumes of \(\text{HBr}\)).
- Redox reaction: \(2\text{HBr} + \text{H}_2\text{SO}_4 \rightarrow \text{Br}_2 + \text{SO}_2 + 2\text{H}_2\text{O}\).

3. The reddish-brown vapour is \(\text{Br}_2\) and the choking colourless gas is sulfur dioxide, \(\text{SO}_2\), which is a reducing agent that turns acidified potassium dichromate(VI) from orange (\(\text{Cr}_2\text{O}_7^{2-}\)) to green (\(\text{Cr}^{3+}\)).

Therefore, \(\mathbf{W} = \text{KBr}\) and the choking gas is \(\text{SO}_2\).

[1 mark] for selecting option A.

1. Alkene \(\mathbf{Q}\) reacts with hot concentrated acidified \(\text{KMnO}_4\) to produce only a single organic compound \(\mathbf{R}\), so \(\mathbf{Q}\) must be a symmetrical alkene.

2. Compound \(\mathbf{R}\) is reduced by \(\text{NaBH}_4\) to a secondary alcohol, meaning \(\mathbf{R}\) must be a ketone. This indicates that both carbons of the double bond in \(\mathbf{Q}\) are bonded to two carbon groups (i.e. of the type \(\text{R}_2\text{C}=\text{CR}_2\)).

3. Since \(\mathbf{Q}\) has 6 carbon atoms and cleaves symmetrically into two identical ketone molecules, \(\mathbf{R}\) must have 3 carbon atoms. Thus, \(\mathbf{R}\) is propanone, \(\text{CH}_3\text{COCH}_3\).

4. Propanone contains the \(\text{CH}_3\text{C=O}\) group, giving a positive tri-iodomethane test (yellow precipitate of \(\text{CHI}_3\)).

5. The alkene that oxidatively cleaves to give propanone only is 2,3-dimethylbut-2-ene, \(\text{(CH}_3)_2\text{C}=\text{C(CH}_3)_2\).

[1 mark] for deducing that 2,3-dimethylbut-2-ene cleaves to form propanone, which meets all criteria.

1. Formation of an orange precipitate with 2,4-dinitrophenylhydrazine indicates that \(\mathbf{T}\) is a carbonyl compound (an aldehyde or a ketone).

2. A negative Tollens' reagent test indicates that \(\mathbf{T}\) is not an aldehyde, so it must be a ketone.
- This eliminates \(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{CHO}\) (pentanal) and \(\text{(CH}_3)_3\text{CCHO}\) (2,2-dimethylpropanal).

3. A negative tri-iodomethane test with alkaline aqueous iodine indicates that \(\mathbf{T}\) does not contain a methyl carbonyl group (\(\text{CH}_3\text{C=O}\)).
- This eliminates pentan-2-one (\(\text{CH}_3\text{COCH}_2\text{CH}_2\text{CH}_3\)), which is a methyl ketone.

4. Pentan-3-one, \(\text{CH}_3\text{CH}_2\text{COCH}_2\text{CH}_3\), is a symmetrical ketone without a \(\text{CH}_3\text{C=O}\) group, so it gives a negative tri-iodomethane test.

Therefore, \(\mathbf{T}\) is pentan-3-one.

[1 mark] for selecting option C (pentan-3-one).

The general balanced equation for the complete combustion of a hydrocarbon is:
$$\text{C}_x\text{H}_y + \left(x + \frac{y}{4}\right)\text{O}_2 \rightarrow x\text{CO}_2 + \frac{y}{2}\text{H}_2\text{O}$$

Using Avogadro's law, reacting gas volumes are directly proportional to stoichiometric mole ratios at constant temperature and pressure:
1. Ratio of \(\text{CO}_2\) to hydrocarbon:
$$x = \frac{V(\text{CO}_2)}{V(\text{C}_x\text{H}_y)} = \frac{80\text{ cm}^3}{20\text{ cm}^3} = 4$$

2. Ratio of \(\text{O}_2\) to hydrocarbon:
$$x + \frac{y}{4} = \frac{V(\text{O}_2)}{V(\text{C}_x\text{H}_y)} = \frac{130\text{ cm}^3}{20\text{ cm}^3} = 6.5$$

Substituting \(x = 4\):
$$4 + \frac{y}{4} = 6.5 \implies \frac{y}{4} = 2.5 \implies y = 10$$

Therefore, the molecular formula is \(\text{C}_4\text{H}_{10}\).

1 mark for correct deduction of the molecular formula \(\text{C}_4\text{H}_{10}\) (option C).

To determine bond angles, consider electron pair repulsion theory (VSEPR):
- \(\text{BF}_3\): Boron has 3 bonding pairs and 0 lone pairs. Shape is trigonal planar with bond angle \(120^\circ\).
- \(\text{NH}_4^+\): Nitrogen has 4 bonding pairs and 0 lone pairs. Shape is tetrahedral with bond angle \(109.5^\circ\).
- \(\text{H}_3\text{O}^+\): Oxygen has 3 bonding pairs and 1 lone pair. Shape is trigonal pyramidal with bond angle \(\approx 107^\circ\).
- \(\text{NH}_2^-\): Nitrogen has 2 bonding pairs and 2 lone pairs. Shape is non-linear (bent) with bond angle \(\approx 104.5^\circ\).

Lone pair-lone pair repulsion > lone pair-bond pair repulsion > bond pair-bond pair repulsion. Each lone pair reduces bond angle by approximately \(2.5^\circ\) from \(109.5^\circ\). Thus, \(\text{NH}_2^-\) has the smallest bond angle.

1 mark for identifying \(\text{NH}_2^-\) as having the smallest bond angle (option C).

Using Hess's Law and standard enthalpy changes of formation:
$$\Delta H^\circ_\text{r} = \sum \Delta H^\circ_\text{f}(\text{products}) - \sum \Delta H^\circ_\text{f}(\text{reactants})$$

Note that \(\Delta H^\circ_\text{f}[\text{O}_2(\text{g})] = 0\text{ kJ mol}^{-1}\) by definition for an element in its standard state.

$$\sum \Delta H^\circ_\text{f}(\text{products}) = [1 \times (-824)] + [4 \times (-297)] = -824 - 1188 = -2012\text{ kJ mol}^{-1}$$
$$\sum \Delta H^\circ_\text{f}(\text{reactants}) = [2 \times (-178)] + \left[\frac{11}{2} \times 0\right] = -356\text{ kJ mol}^{-1}$$

$$\Delta H^\circ_\text{r} = -2012 - (-356) = -2012 + 356 = -1656\text{ kJ mol}^{-1}$$

1 mark for calculating \(-1656\text{ kJ mol}^{-1}\) correctly (option A).

Bromide ions (\(\text{Br}^-\)) are strong enough reducing agents to reduce concentrated \(\text{H}_2\text{SO}_4\) to sulfur dioxide (\(\text{SO}_2\)).
- The steamy fumes are \(\text{HBr}\) gas.
- The dark brown liquid/red-brown vapour is \(\text{Br}_2\).
- The \(\text{SO}_2\) gas acts as a reducing agent and reduces orange \(\text{Cr}_2\text{O}_7^{2-}\) to green \(\text{Cr}^{3+}\).

In contrast:
- \(\text{Cl}^-\) is not oxidized by concentrated \(\text{H}_2\text{SO}_4\) and only produces steamy fumes of \(\text{HCl}\).
- \(\text{I}^-\) reduces \(\text{H}_2\text{SO}_4\) further to \(\text{S}\) and \(\text{H}_2\text{S}\), forming a purple vapour / dark grey solid (\(\text{I}_2\)).

1 mark for identifying bromide \(\text{Br}^-\) (option C).

1. Halving the volume increases the total pressure.
2. According to Le Chatelier's principle, an increase in pressure causes the position of equilibrium to shift in the direction that produces fewer moles of gas (to the right, where \(2\text{ mol}\) of gas forms \(1\text{ mol}\) of gas).
3. Because the equilibrium shifts to the right, the total number of gas molecules in the syringe decreases. (Statement B is correct)
4. \(K_c\) is temperature-dependent and remains constant at constant temperature (eliminates A).
5. Although the equilibrium shifts right, the reduction in volume causes the partial pressure/concentration of \(\text{NO}_2\) at the new equilibrium to be higher than in the initial equilibrium (eliminates C).
6. Higher concentrations of both species mean both forward and reverse reaction rates are higher than before compression (eliminates D).

1 mark for identifying that the total number of molecules in the syringe decreases (option B).

- Adding a catalyst provides an alternative reaction pathway with a lower activation energy (\(E_\text{a}\)). It does not affect the energy distribution of the molecules.
- Increasing the temperature increases the average kinetic energy of the particles, which causes the peak of the Boltzmann distribution curve to shift to the right and flatten (lower maximum height).
- Therefore, statement A is completely correct.

1 mark for correctly matching the effects of a catalyst and temperature on activation energy and the Boltzmann distribution (option A).

Let us systematically inspect all structural isomers of \(\text{C}_4\text{H}_8\text{Cl}_2\):

Straight chain (butane skeleton):
1. 1,1-dichlorobutane: \(\text{CH}_3\text{CH}_2\text{CH}_2\text{CHCl}_2\) — no chiral centre.
2. 1,2-dichlorobutane: \(\text{CH}_3\text{CH}_2\text{C}^*\text{H(Cl)CH}_2\text{Cl}\) — C2 is bonded to \(-\text{H}\), \(-\text{Cl}\), \(-\text{CH}_2\text{Cl}\), and \(-\text{CH}_2\text{CH}_3\) (chiral).
3. 1,3-dichlorobutane: \(\text{CH}_3\text{C}^*\text{H(Cl)CH}_2\text{CH}_2\text{Cl}\) — C2 is bonded to \(-\text{H}\), \(-\text{Cl}\), \(-\text{CH}_3\), and \(-\text{CH}_2\text{CH}_2\text{Cl}\) (chiral).
4. 1,4-dichlorobutane: \(\text{ClCH}_2\text{CH}_2\text{CH}_2\text{CH}_2\text{Cl}\) — no chiral centre.
5. 2,2-dichlorobutane: \(\text{CH}_3\text{CCl}_2\text{CH}_2\text{CH}_3\) — no chiral centre.
6. 2,3-dichlorobutane: \(\text{CH}_3\text{C}^*\text{H(Cl)}\text{C}^*\text{H(Cl)}\text{CH}_3\) — C2 and C3 are chiral centres (chiral).

Branched chain (methylpropane skeleton):
7. 1,1-dichloro-2-methylpropane: \((\text{CH}_3)_2\text{CHCHCl}_2\) — no chiral centre.
8. 1,2-dichloro-2-methylpropane: \((\text{CH}_3)_2\text{C(Cl)CH}_2\text{Cl}\) — no chiral centre.
9. 1,3-dichloro-2-methylpropane: \(\text{CH}_3\text{CH}(\text{CH}_2\text{Cl})_2\) — no chiral centre.

Thus, exactly 3 structural isomers (1,2-dichlorobutane, 1,3-dichlorobutane, and 2,3-dichlorobutane) possess chiral carbon atoms.

1 mark for identifying that 3 structural isomers contain a chiral centre (option B).

Deducing the functional group and structure:
1. Positive test with 2,4-DNPH (orange precipitate) indicates a carbonyl group (aldehyde or ketone).
2. Negative test with Tollens' reagent (no silver mirror) confirms \(\text{Q}\) is a ketone, not an aldehyde.
3. Positive tri-iodomethane (iodoform) test with alkaline aqueous iodine (yellow precipitate of \(\text{CHI}_3\)) indicates the presence of a \(\text{CH}_3\text{C=O}\) group (methyl ketone).

Among the options:
- Butanal is an aldehyde (would react with Tollens').
- 2-Methylpropanal is an aldehyde (would react with Tollens').
- But-3-en-2-ol is an alcohol (would not form an orange precipitate with 2,4-DNPH).
- Butanone (\(\text{CH}_3\text{COCH}_2\text{CH}_3\)) is a methyl ketone which fits all three criteria.

1 mark for identifying butanone (option B).

Using the ideal gas equation:
\[pV = nRT = \frac{m}{M_r}RT\]
Rearranging for relative molecular mass:
\[M_r = \frac{mRT}{pV}\]
Convert quantities to SI units:
- \(m = 0.120\text{ g}\)
- \(R = 8.31\text{ J K}^{-1}\text{ mol}^{-1}\)
- \(T = 127 + 273 = 400\text{ K}\)
- \(p = 1.00 \times 10^5\text{ Pa}\)
- \(V = 47.4\text{ cm}^3 = 47.4 \times 10^{-6}\text{ m}^3\)

\[M_r = \frac{0.120 \times 8.31 \times 400}{1.00 \times 10^5 \times 47.4 \times 10^{-6}} = \frac{398.88}{4.74} = 84.15 \approx 84.2\]

B is correct (1 mark).
- A corresponds to using temperature in Celsius without converting to Kelvin (\(127\text{ }^\circ\text{C}\)).
- C and D correspond to errors in unit conversions for pressure or volume.

A molecule is non-polar if its individual bond dipoles cancel out completely due to symmetrical geometry.
- \(\text{CHCl}_3\) is an asymmetric tetrahedral molecule; individual \(\text{C}-\text{H}\) and \(\text{C}-\text{Cl}\) bond dipoles do not cancel, so it has an overall dipole moment.
- \(\text{NF}_3\) has a trigonal pyramidal shape with a lone pair on nitrogen, so it has a permanent dipole.
- \(\text{SO}_2\) has a bent (non-linear) shape with a lone pair on sulfur, so its bond dipoles do not cancel.
- \(\text{BCl}_3\) has a symmetrical trigonal planar shape (bond angles \(120^\circ\)); the three identical \(\text{B}-\text{Cl}\) bond dipoles cancel out completely, giving a net dipole moment of zero.

D is correct (1 mark).

Using Hess's Law and standard enthalpies of combustion:
\[\Delta H^\circ_{\text{r}} = \sum \Delta H^\circ_{\text{c}}(\text{reactants}) - \sum \Delta H^\circ_{\text{c}}(\text{products})\]
\[\Delta H^\circ_{\text{r}} = [(-2058) + (-286)] - [-2219]\]
\[\Delta H^\circ_{\text{r}} = -2344 - (-2219) = -125\text{ kJ mol}^{-1}\]

A is correct (1 mark).
- B gives the wrong sign (products minus reactants).
- C and D result from adding rather than subtracting the enthalpy of combustion of propane.

When temperature increases:
1. The average kinetic energy of the particles increases, shifting the Boltzmann distribution curve to the right with a lower peak.
2. A significantly larger fraction of the colliding particles possess energy equal to or greater than the activation energy (\(E \ge E_{\text{a}}\)), leading to a much higher frequency of successful collisions.
3. The activation energy \(E_{\text{a}}\) is a characteristic barrier of the reaction pathway and is not changed by temperature.

C is correct (1 mark).
- A is incorrect because activation energy is independent of temperature.
- B is incorrect because the peak shifts right and becomes lower.
- D is incorrect because the average kinetic energy increases.

- \(\text{AlCl}_3\) hydrolyses in water to produce \([\text{Al}(\text{H}_2\text{O})_6]^{3+}\) and \(\text{Cl}^-\); the hydrated cation loses protons to form an acidic solution (\(\text{pH} \approx 2-3\)).
- \(\text{Al}_2\text{O}_3\) is an amphoteric oxide: it is insoluble in water but reacts with both acids (such as \(\text{HCl}\)) and bases (such as \(\text{NaOH}\)) to form soluble salts.
- \(\text{MgO}\) is basic and does not react with \(\text{NaOH}\).
- \(\text{SiO}_2\) is acidic and does not react with \(\text{HCl}\).
- \(\text{P}_4\text{O}_{10}\) is acidic and dissolves in water to form \(\text{H}_3\text{PO}_4\).

B is correct (1 mark).

- When \(\text{NaBr}\) reacts with concentrated \(\text{H}_2\text{SO}_4\), \(\text{Br}^-\) reduces \(\text{H}_2\text{SO}_4\) to \(\text{SO}_2\) while being oxidised to orange/brown \(\text{Br}_2\) vapour:
\[2\text{HBr} + \text{H}_2\text{SO}_4 \rightarrow \text{Br}_2\text{(g)} + \text{SO}_2\text{(g)} + 2\text{H}_2\text{O}\]
- \(\text{NaCl}\) does not reduce concentrated \(\text{H}_2\text{SO}_4\); only steamy fumes of \(\text{HCl}\) are evolved (no \(\text{Cl}_2\)).
- \(\text{NaI}\) is a stronger reducing agent and reduces \(\text{H}_2\text{SO}_4\) to a mixture of products including \(\text{SO}_2\), yellow solid \(\text{S}\), and bad-egg smelling gas \(\text{H}_2\text{S}\) (not \(\text{H}_2\text{S}\) only).

B is correct (1 mark).

Oxidative cleavage of alkenes by hot concentrated acidified \(\text{KMnO}_4\):
- A \(=\text{C(R)}_1\text{R}_2\) group is oxidised to a ketone \(\text{R}_1\text{COR}_2\).
- A \(=\text{CHR}\) group is oxidised to a carboxylic acid \(\text{RCOOH}\).
- A \(=\text{CH}_2\) group is oxidised to \(\text{CO}_2\) and \(\text{H}_2\text{O}\).

For 2-methylpent-2-ene, \(\text{(CH}_3\text{)}_2\text{C=CHCH}_2\text{CH}_3\):
- \(\text{(CH}_3\text{)}_2\text{C=}\) gives propanone, \(\text{CH}_3\text{COCH}_3\).
- \(=\text{CHCH}_2\text{CH}_3\) gives propanoic acid, \(\text{CH}_3\text{CH}_2\text{COOH}\).
Thus \(Y\) is 2-methylpent-2-ene.

A is correct (1 mark).
- B (3-methylpent-2-ene) gives butanone and ethanoic acid.
- C (2,3-dimethylpent-2-ene) gives propanone and butanone.
- D (hex-2-ene) gives ethanoic acid and butanoic acid.

1. Orange precipitate with 2,4-DNPH confirms the presence of a carbonyl group (aldehyde or ketone). This rules out propan-2-ol.
2. No reaction with Tollens' reagent confirms that \(Z\) is a ketone rather than an aldehyde. This rules out propanal.
3. Yellow precipitate with alkaline aqueous iodine (tri-iodomethane test) confirms the presence of a \(\text{CH}_3\text{C=O}\) (methyl carbonyl) group.
- Butanone is \(\text{CH}_3\text{COCH}_2\text{CH}_3\), which contains a \(\text{CH}_3\text{C=O}\) group.
- Pentan-3-one is \(\text{CH}_3\text{CH}_2\text{COCH}_2\text{CH}_3\), which does not contain a \(\text{CH}_3\text{C=O}\) group.
Therefore, \(Z\) is butanone.

C is correct (1 mark).

Using the ideal gas equation, \(pV = nRT\):

Convert units: \(V = 122\text{ cm}^3 = 122 \times 10^{-6}\text{ m}^3\), \(p = 1.01 \times 10^5\text{ Pa}\), \(T = 298\text{ K}\).

Calculate the amount in moles, \(n\):
\(n = \frac{pV}{RT} = \frac{1.01 \times 10^5 \times 122 \times 10^{-6}}{8.31 \times 298} = \frac{12.322}{2476.38} \approx 4.976 \times 10^{-3}\text{ mol}\)

Calculate the relative molecular mass, \(M_r\):
\(M_r = \frac{\text{mass}}{n} = \frac{0.280}{4.976 \times 10^{-3}} \approx 56.3\text{ g mol}^{-1}\)

An alkene has the general formula \(\text{C}_n\text{H}_{2n}\) with empirical formula \(\text{CH}_2\) (empirical formula mass \(= 14.0\)).

\(n = \frac{56.3}{14.0} \approx 4\), so the molecular formula is \(\text{C}_4\text{H}_8\).

C is correct (1 mark).
- A gives \(M_r = 28.0\).
- B gives \(M_r = 42.0\).
- C gives \(M_r = 56.1\) which matches the calculated molar mass from \(pV = nRT\).
- D gives \(M_r = 70.1\).

According to VSEPR theory, lone pairs repel more strongly than bonding pairs:
- In \(\text{H}_2\text{O}\), oxygen has 2 bonding pairs and 2 lone pairs. The electron pairs adopt a tetrahedral arrangement, but due to lone pair-lone pair and lone pair-bonding pair repulsions, the bond angle is reduced to approximately \(104.5^\circ\).
- In \(\text{NH}_3\), nitrogen has 3 bonding pairs and 1 lone pair, giving a trigonal pyramidal shape with a bond angle of \(107^\circ\).
- In \(\text{CH}_4\), carbon has 4 bonding pairs and 0 lone pairs, giving a regular tetrahedral shape with a bond angle of \(109.5^\circ\).
- In \(\text{NH}_4^+\), nitrogen has 4 bonding pairs and 0 lone pairs, giving a regular tetrahedral shape with a bond angle of \(109.5^\circ\).

Therefore, \(\text{H}_2\text{O}\) has the smallest bond angle.

A is correct (1 mark).
- A has bond angle \(104.5^\circ\).
- B has bond angle \(107^\circ\).
- C has bond angle \(109.5^\circ\).
- D has bond angle \(109.5^\circ\).

The formation equation for propane is:
\(3\text{C}(\text{graphite}) + 4\text{H}_2(\text{g}) \rightarrow \text{C}_3\text{H}_8(\text{g})\)

Using enthalpy changes of combustion:
\(\Delta H^\circ_\text{f} = \sum \Delta H^\circ_\text{c}(\text{reactants}) - \sum \Delta H^\circ_\text{c}(\text{products})\)

\(\Delta H^\circ_\text{f} = 3\Delta H^\circ_\text{c}(\text{C}) + 4\Delta H^\circ_\text{c}(\text{H}_2) - \Delta H^\circ_\text{c}(\text{C}_3\text{H}_8)\)
\(\Delta H^\circ_\text{f} = 3(-394) + 4(-286) - (-2220)\)
\(\Delta H^\circ_\text{f} = -1182 - 1144 + 2220 = -2326 + 2220 = -106\text{ kJ mol}^{-1}\).

B is correct (1 mark).
- A is obtained if stoichiometric multipliers are omitted: \((-394 - 286) - (-2220)\).
- B is the correct application of Hess's Law with stoichiometric multipliers.
- C has incorrect overall sign.
- D reverses products and reactants.

- \(\text{P}_4\text{O}_{10}\) is a white solid that dissolves in water to form phosphoric(V) acid, \(\text{H}_3\text{PO}_4\), giving an acidic solution of pH \(\approx 1-2\).
- \(\text{SiCl}_4\) is a colourless liquid that hydrolyses vigorously with water to form white solid hydrated silicon dioxide, \(\text{SiO}_2 \cdot x\text{H}_2\text{O}\), and misty white fumes of \(\text{HCl}\).
- \(\text{Al}_2\text{O}_3\) is insoluble in water and does not form an acidic solution.
- \(\text{MgCl}_2\) dissolves to give a neutral or very faintly acidic solution (pH \(\approx 6.5\)) without misty fumes.

A is correct (1 mark).
- In A, \(\text{P}_4\text{O}_{10}\) is a solid acidic oxide (pH 2) and \(\text{SiCl}_4\) is a liquid chloride undergoing vigorous hydrolysis to a white precipitate (\(\text{SiO}_2\)) and \(\text{HCl}\) fumes.
- B contains \(\text{Al}_2\text{O}_3\) which is insoluble.
- C contains \(\text{SO}_2\) which is a gas, not a solid oxide.
- D contains \(\text{MgCl}_2\) which forms a colourless solution without vigorous hydrolysis.

- Decreasing the volume of the reaction container increases the total pressure and the concentration of gaseous particles. This increases the collision frequency between reactant molecules, increasing the rate of reaction.
- The forward reaction proceeds with a decrease in the number of moles of gas (from 3 moles of gaseous reactants to 2 moles of gaseous product). By Le Chatelier's principle, an increase in pressure shifts the equilibrium position to the right, increasing the equilibrium yield of \(\text{SO}_3(\text{g})\).
- Increasing the temperature increases the rate, but because the forward reaction is exothermic, it decreases the equilibrium yield.
- Adding a catalyst increases the rate of reaction but has no effect on the equilibrium yield.

B is correct (1 mark).
- A increases rate but decreases equilibrium yield (exothermic reaction).
- B increases pressure, which increases collision frequency (rate) and shifts equilibrium to the side with fewer gas moles (yield).
- C increases rate but leaves yield unchanged.
- D does not increase the rate of the forward reaction.

Oxidation of alkenes with hot concentrated acidified \(\text{KMnO}_4\) causes oxidative cleavage of the \(\text{C}=\text{C}\) double bond:
- A \(=\text{CH}_2\) group is oxidised to \(\text{CO}_2 + \text{H}_2\text{O}\).
- A \(=\text{CHR}\) group is oxidised to a carboxylic acid, \(\text{RCOOH}\).
- A \(=\text{CR}_1\text{R}_2\) group is oxidised to a ketone, \(\text{R}_1\text{COR}_2\).

For propanone (\((\text{CH}_3)_2\text{C}=\text{O}\)) to be the only organic product, both sides of the double bond must be \(=C(\text{CH}_3)_2\).
Thus, the alkene must be 2,3-dimethylbut-2-ene, \((\text{CH}_3)_2\text{C}=\text{C}(\text{CH}_3)_2\).

B is correct (1 mark).
- A gives propanoic acid as the sole organic product.
- B gives 2 molecules of propanone as the sole organic product.
- C gives butan-2-one and ethanoic acid.
- D gives propanone and propanoic acid.

1. Formation of an orange precipitate with 2,4-dinitrophenylhydrazine indicates that \(Z\) is a carbonyl compound (aldehyde or ketone).
2. No reaction with Tollens' reagent confirms that \(Z\) is not an aldehyde; it must be a ketone.
3. Formation of a yellow precipitate of tri-iodomethane (\(\text{CHI}_3\)) with alkaline aqueous iodine indicates the presence of a methyl carbonyl group, \(\text{CH}_3\text{C}=\text{O}\).

Among the four-carbon isomers:
- Butanal and 2-methylpropanal are aldehydes (would react with Tollens' reagent).
- But-3-en-2-ol is an alcohol, not a carbonyl.
- Butan-2-one, \(\text{CH}_3\text{COCH}_2\text{CH}_3\), is a ketone containing the \(\text{CH}_3\text{C}=\text{O}\) group, matching all observations.

B is correct (1 mark).
- A is an aldehyde (positive Tollens' test).
- B is a methyl ketone (positive 2,4-DNPH, negative Tollens', positive iodoform test).
- C is an alcohol (negative 2,4-DNPH test).
- D is an aldehyde (positive Tollens' test).

1. Determine the number of chlorine atoms:
The ratio of the \([\text{M}+2]^+\) peak height to the \([\text{M}]^+\) peak height is \(\frac{12.0}{36.0} = \frac{1}{3}\).
Since the relative abundance of \(^{37}\text{Cl}\) to \(^{35}\text{Cl}\) is approximately \(1 : 3\), this \(3 : 1\) ratio of \([\text{M}]^+\) to \([\text{M}+2]^+\) confirms the presence of 1 chlorine atom.

2. Determine the number of carbon atoms, \(n\):
\[n = \frac{100 \times \text{abundance of }[\text{M}+1]^+}{1.1 \times \text{abundance of }[\text{M}]^+} = \frac{100 \times 1.2}{1.1 \times 36.0} = \frac{120}{39.6} \approx 3.03 \approx 3\]

Checking molecular mass for \(\text{C}_3\text{H}_7\text{Cl}\): \((3 \times 12) + (7 \times 1) + 35 = 36 + 7 + 35 = 78\), which matches \(m/e = 78\).

Therefore, the compound contains 3 carbon atoms and 1 chlorine atom.

B is correct (1 mark).
- A gives incorrect carbon count calculation.
- B correctly deduces 3 carbon atoms using \(n = \frac{100 [\text{M}+1]^+}{1.1 [\text{M}]^+}\) and 1 chlorine atom from \([\text{M}]^+ : [\text{M}+2]^+ = 3 : 1\).
- C would require an \([\text{M}+2]^+\) peak height of approximately \(66\%\) of \([\text{M}]^+\) and a further \([\text{M}+4]^+\) peak.
- D gives incorrect carbon count.

Using the ideal gas equation, \(pV = nRT = \left(\frac{m}{M_{\text{r}}}\right)RT\):

1. Convert units to SI:
- \(p = 1.02 \times 10^{5}\text{ Pa} - \)V = 88.0\text{ cm}^{3} = 88.0 \times 10^{-6}\text{ m}^{3}
- \(T = 368\text{ K} - \)R = 8.31\text{ J K}^{-1}\text{ mol}^{-1}
- \(m = 0.285\text{ g} 2. Calculate the amount in moles, \)n\):
\[ n = \frac{pV}{RT} = \frac{1.02 \times 10^{5} \times 88.0 \times 10^{-6}}{8.31 \times 368} = \frac{8.976}{3058.08} \approx 2.935 \times 10^{-3}\text{ mol} \]

3. Calculate the relative molecular mass, \(M_{\text{r}}\):
\[ M_{\text{r}} = \frac{m}{n} = \frac{0.285}{2.935 \times 10^{-3}} \approx 97.1 \]

4. Compare with the relative molecular masses of the options:
- \(\text{CH}_{2}\text{Cl}_{2} = 12.0 + 2(1.0) + 2(35.5) = 85.0 - \)\text{CH}_{3}\text{CH}_{2}\text{CH}_{2}\text{Cl} = 3(12.0) + 7(1.0) + 35.5 = 78.5
- \(\text{CH}_{2}=\text{CCl}_{2} = 2(12.0) + 2(1.0) + 2(35.5) = 97.0 - \)\text{CH}_{2}\text{ClCH}_{2}\text{Cl} = 2(12.0) + 4(1.0) + 2(35.5) = 99.0

Therefore, \(\mathbf{Y}\) is 1,1-dichloroethene, \(\text{CH}_{2}=\text{CCl}_{2}\).

C is correct (1 mark).
Calculates \(M_{\text{r}} \approx 97\text{ g mol}^{-1}\) and matches to 1,1-dichloroethene (\(M_{\text{r}} = 97.0\)).

The bond angle is determined by the electron-pair repulsion theory (VSEPR):
- In \(\text{NH}_{2}^{-}\), nitrogen has 2 bonding pairs and 2 lone pairs (total 4 pairs, non-linear shape). Lone pair–lone pair and lone pair–bonding pair repulsions reduce the bond angle to approximately \(104.5^{\circ}\).
- In \(\text{NH}_{3}\), nitrogen has 3 bonding pairs and 1 lone pair (total 4 pairs, trigonal pyramidal). Repulsion from one lone pair reduces the bond angle to approximately \(107.0^{\circ}\).
- In \(\text{NH}_{4}^{+}\), nitrogen has 4 bonding pairs and 0 lone pairs (total 4 pairs, regular tetrahedral shape). The bond angle is \(109.5^{\circ}\).
- In \(\text{NO}_{2}^{+}\), nitrogen has 2 double bonds and 0 lone pairs (linear shape, \(180^{\circ}\)).

Thus, the increasing order of bond angle is:
\[ \text{NH}_{2}^{-} (104.5^{\circ}) < \text{NH}_{3} (107^{\circ}) < \text{NH}_{4}^{+} (109.5^{\circ}) < \text{NO}_{2}^{+} (180^{\circ}) \]

B is correct (1 mark).
Recognises that greater number of lone pairs reduces bond angles for tetrahedral-based geometries, and linear \(\text{NO}_{2}^{+}\) has the largest bond angle.

The equation for the standard enthalpy change of formation of ethanol is:
\[ 2\text{C(graphite)} + 3\text{H}_{2}\text{(g)} + \tfrac{1}{2}\text{O}_{2}\text{(g)} \rightarrow \text{CH}_{3}\text{CH}_{2}\text{OH(l)} \]

Using Hess's Law with standard enthalpy changes of combustion:
\[ \Delta H^{\ominus}_{\text{f}} = \sum \Delta H^{\ominus}_{\text{c}}(\text{reactants}) - \sum \Delta H^{\ominus}_{\text{c}}(\text{products}) \]

\[ \Delta H^{\ominus}_{\text{f}} = [2 \times \Delta H^{\ominus}_{\text{c}}(\text{C}) + 3 \times \Delta H^{\ominus}_{\text{c}}(\text{H}_{2})] - [\Delta H^{\ominus}_{\text{c}}(\text{CH}_{3}\text{CH}_{2}\text{OH})] \]

\[ \Delta H^{\ominus}_{\text{f}} = [2(-394) + 3(-286)] - [-1368] \]
\[ \Delta H^{\ominus}_{\text{f}} = [-788 - 858] + 1368 = -1646 + 1368 = -278\text{ kJ mol}^{-1} \]

A is correct (1 mark).
Correctly applies \(\Delta H = \sum \Delta H_{\text{c}}(\text{reactants}) - \sum \Delta H_{\text{c}}(\text{products}) = [2(-394) + 3(-286)] - (-1368) = -278\text{ kJ mol}^{-1}\).

1. The reaction with concentrated sulfuric acid produces purple vapour (\(\text{I}_{2}\)) and the smell of bad eggs (\(\text{H}_{2}\text{S}\)), which shows that \(\text{I}^{-}\) is present. \(\text{I}^{-}\) is a sufficiently powerful reducing agent to reduce \(\text{H}_{2}\text{SO}_{4}\) to \(\text{H}_{2}\text{S}\).
2. Addition of \(\text{AgNO}_{3}\text{(aq)}\) gives a yellow precipitate of silver iodide, \(\text{AgI}\).
3. Silver iodide (\(\text{AgI}\)) is insoluble in both dilute and concentrated aqueous ammonia.

D is correct (1 mark).
Identifies halide as iodide (\(\text{I}^{-}\)) and recalls that \(\text{AgI}\) is insoluble in both dilute and concentrated aqueous ammonia.

1. At a higher temperature \(T_{2} > T_{1}\), the Boltzmann distribution curve broadens and flattens: the peak moves to a higher energy (to the right) and has a lower height (lower fraction of molecules).
2. The addition of a catalyst provides an alternative reaction pathway with a lower activation energy, so the activation energy decreases from \(E_{\text{a}}\).

Therefore, the peak moves to the right and lower down, and the activation energy decreases.

B is correct (1 mark).
Identifies that higher temperature shifts the Boltzmann distribution peak down and to the right, and the catalyst lowers \(E_{\text{a}}\).

Alkenes with formula \(\text{C}_{5}\text{H}_{10}\) undergo oxidative cleavage of the \(\text{C}=\text{C}\) bond with hot concentrated acidified \(\text{KMnO}_{4}\):
- A \(=\text{CH}_{2}\) group gives \(\text{CO}_{2} + \text{H}_{2}\text{O}\).
- A \(=\text{CHR}\) group gives a carboxylic acid, \(\text{RCOOH}\).
- A \(=\text{CR}_{1}\text{R}_{2}\) group gives a ketone, \(\text{R}_{1}\text{COR}_{2}\).

To produce a carboxylic acid, the alkene must contain at least one \(=\text{CHR}\) group where \(\text{R}\) is an alkyl group.

Let's check all unbranched and branched acyclic isomers (structural isomers only):
1. Pent-1-ene, \(\text{CH}_{3}\text{CH}_{2}\text{CH}_{2}\text{CH}=\text{CH}_{2}\): gives butanoic acid and \(\text{CO}_{2}\). (Gives a carboxylic acid)
2. Pent-2-ene, \(\text{CH}_{3}\text{CH}_{2}\text{CH}=\text{CHCH}_{3}\): gives propanoic acid and ethanoic acid. (Gives a carboxylic acid)
3. 2-Methylbut-1-ene, \(\text{CH}_{3}\text{CH}_{2}\text{C(CH}_{3})=\text{CH}_{2}\): gives butanone and \(\text{CO}_{2}\). (No carboxylic acid)
4. 2-Methylbut-2-ene, \(\text{CH}_{3}\text{CH}=\text{C(CH}_{3})_{2}\): gives ethanoic acid and propanone. (Gives a carboxylic acid)
5. 3-Methylbut-1-ene, \((\text{CH}_{3})_{2}\text{CHCH}=\text{CH}_{2}\): gives 2-methylpropanoic acid and \(\text{CO}_{2}\). (Gives a carboxylic acid)

Thus, 4 structural isomers give a carboxylic acid: pent-1-ene, pent-2-ene, 2-methylbut-2-ene, and 3-methylbut-1-ene.

C is correct (1 mark).
Identifies the 4 structural isomers containing a \(=\text{CHR}\) unit: pent-1-ene, pent-2-ene, 2-methylbut-2-ene, and 3-methylbut-1-ene.

Let's analyse the observations:
1. Reacts with sodium to give \(\text{H}_{2}\text{(g)}\) \(\rightarrow\) contains an \(-\text{OH}\) group (alcohol).
2. Orange precipitate with 2,4-DNPH \(\rightarrow\) contains a carbonyl group (\(\text{C}=\text{O}\), aldehyde or ketone).
3. No reaction with Tollens' reagent \(\rightarrow\) it is a ketone, not an aldehyde.
4. Yellow precipitate of \(\text{CHI}_{3}\) with alkaline aqueous iodine \(\rightarrow\) contains a methyl ketone group, \(\text{CH}_{3}\text{C}=\text{O}\), or a \(\text{CH}_{3}\text{CH(OH)}-\) group.

Let's evaluate the options:
- \(\text{CH}_{3}\text{COCH(OH)CH}_{3}\) (3-hydroxybutan-2-one): has an \(-\text{OH}\) group, a ketone group (reacts with 2,4-DNPH, no Tollens' reaction), and a \(\text{CH}_{3}\text{CO}-\) group (positive tri-iodomethane test). Molecular formula is \(\text{C}_{4}\text{H}_{8}\text{O}_{2}\). This matches all observations.
- \(\text{CH}_{3}\text{CH}_{2}\text{CH(OH)CHO}\): is an aldehyde (would react with Tollens' reagent).
- \(\text{CH}_{3}\text{COOCH}_{2}\text{CH}_{3}\): is an ester (does not react with Na or 2,4-DNPH).
- \(\text{HOCH}_{2}\text{CH}_{2}\text{COCH}_{3}\): has \(-\text{OH}\), ketone, and \(\text{CH}_{3}\text{CO}-\), but its structural formula is 4-hydroxybutan-2-one, which also satisfies some tests, but option B provides 3-hydroxybutan-2-one which possesses both a methyl ketone and secondary alcohol group.

B is correct (1 mark).
Deduces that \(\mathbf{Q}\) must contain an \(-\text{OH}\) group, a \(\text{CH}_{3}\text{CO}-\) ketone group, and no aldehyde group.

1. Determine the number of carbon atoms, \(n\), using the formula:
\[ n = \frac{100 \times \text{abundance of } [\text{M}+1]^{+}}{1.1 \times \text{abundance of } \text{M}^{+}} = \frac{100 \times 1.07}{1.1 \times 32.4} = \frac{107}{35.64} \approx 3.00 \]
So \(\mathbf{G}\) contains 3 carbon atoms.

2. Infra-red spectrum:
- The very broad band at \(2500\text{--}3000\text{ cm}^{-1}\) is characteristic of the \(\text{O}-\text{H}\) stretch in a carboxylic acid.
- The strong sharp absorption at \(1715\text{ cm}^{-1}\) is characteristic of the \(\text{C}=\text{O}\) carbonyl stretch.

3. An aliphatic carboxylic acid with 3 carbon atoms is propanoic acid, \(\text{CH}_{3}\text{CH}_{2}\text{COOH}\).
- Molecular mass \(M_{\text{r}} = 3(12.0) + 6(1.0) + 2(16.0) = 36 + 6 + 32 = 74\).

Therefore, \(\mathbf{G}\) is propanoic acid.

A is correct (1 mark).
Calculates \(n = 3\) carbon atoms from the \([\text{M}+1]^{+}\) abundance and identifies the carboxylic acid functional group from IR data.

Relative atomic mass (\(A_{\text{r}}\)) is the weighted mean mass of an atom of an element relative to \(\frac{1}{12}\) of the mass of an atom of carbon-12 (which is defined as exactly 12.00 unified atomic mass units).

M1 (1.0 mark): Weighted average / mean mass of an atom of an element. [R: mass of a mole/element alone without 'atom'] M2 (0.5 mark): Compared to / relative to \(\frac{1}{12}\) of the mass of a carbon-12 atom (or relative to a carbon-12 atom taken as exactly 12).

The standard enthalpy change of formation is the heat energy change under constant pressure when exactly 1 mole of a compound is formed from its constituent elements, with all reactants and products in their standard states at 298 K and 101 kPa.

M1 (1.0 mark): Enthalpy / heat energy change when 1 mole of a compound is formed from its constituent elements. M2 (0.5 mark): (all reactants and products) in their standard states under standard conditions (or 298 K and 101 kPa / 1 atm).

First ionisation energy is defined as the energy required to remove one mole of electrons from one mole of isolated gaseous atoms of an element, producing one mole of gaseous unipositive (1+) ions.

M1 (1.0 mark): Energy required / absorbed to remove one electron from each atom in one mole of atoms. M2 (0.5 mark): In the gaseous state / from gaseous atoms to form one mole of gaseous 1+ / unipositive ions.

Activation energy is the minimum kinetic energy that reacting species / colliding particles must possess in order to initiate a reaction / overcome the energy barrier to form products.

M1 (1.0 mark): Minimum energy required / needed by colliding particles / reactants. M2 (0.5 mark): To start a reaction / for a collision to be successful (or to lead to a reaction).

Electronegativity is the relative attraction of a bonded atom for the shared electron pair within a covalent chemical bond.

M1 (1.0 mark): The ability / power / tendency of an atom to attract a bonding / shared pair of electrons. M2 (0.5 mark): In a covalent bond / within a molecule.

In a dynamic equilibrium: (i) the forward and reverse reactions proceed at identical rates, and (ii) the macroscopic properties, including concentrations of all reacting species, remain constant while both reactions continue in a closed system.

M1 (1.0 mark): Rate of the forward reaction equals the rate of the reverse / backward reaction. M2 (0.5 mark): Concentrations of reactants and products remain constant (or no change in macroscopic properties) in a closed system.

Structural isomers are molecules that possess an identical molecular formula (same number and type of atoms) but differ in the sequence and manner in which atoms are joined (different connectivity / structural formula).

M1 (1.0 mark): Compounds / molecules with the same molecular formula. M2 (0.5 mark): But different structural formulas (or different connectivity / arrangement of atoms in the molecule).

A disproportionation reaction is a specific redox process where atoms of the same element simultaneously undergo both an increase in oxidation number (oxidation) and a decrease in oxidation number (reduction). In elemental chlorine \(\text{Cl}_2\), the oxidation state is 0. In the chlorate(I) ion \(\text{ClO}^-\), oxygen is -2, so chlorine has an oxidation state of +1.

M1 (1.0 mark): A reaction in which the same element / species is simultaneously oxidised and reduced (or both increases and decreases its oxidation number). M2 (0.5 mark): \(\text{Cl}\) in \(\text{Cl}_2 = 0\) AND \(\text{Cl}\) in \(\text{ClO}^- = +1\) [Both required for the 0.5 mark].

Step 1: The \(\pi\) electrons of the \(\text{C=C}\) double bond in propene attack the partially positive hydrogen atom of polar \(\text{H}^{\delta+}\text{--Br}^{\delta-}\). A curly arrow originates from the double bond to the \(\text{H}\) atom, and a second curly arrow shows the cleavage of the \(\text{H--Br}\) bond towards the \(\text{Br}\) atom.

Step 2: This forms the more stable secondary carbocation intermediate, \(\text{CH}_3\overset{+}{\text{C}}\text{HCH}_3\), and a bromide ion, \(:\!\text{Br}^-\).

Step 3: A curly arrow from a lone pair of electrons on \(:\!\text{Br}^-\to\) the positively charged central carbon atom (\(\text{C}^+\)) gives the product, 2-bromopropane (\(\text{CH}_3\text{CH(Br)CH}_3\)).

M1: Correct dipole shown on \(\text{H--Br}\) (\(\text{H}^{\delta+}\text{--Br}^{\delta-}\)) AND curly arrow from \(\text{C=C}\) double bond to \(\text{H}^{\delta+}\).
M2: Curly arrow showing the breaking of the \(\text{H--Br}\) bond to \(\text{Br}\) AND correct structure of the secondary carbocation intermediate (\(\text{CH}_3\overset{+}{\text{C}}\text{HCH}_3\)).
M3: Curly arrow from a lone pair on \(:\!\text{Br}^-\) to the \(\text{C}^+\) carbon atom of the carbocation to form \(\text{CH}_3\text{CH(Br)CH}_3\).

Step 1: The hydroxide ion behaves as a nucleophile. A lone pair of electrons on the oxygen of \(\text{OH}^-\) attacks the electron-deficient carbon atom attached to bromine. The \(\text{C--Br}\) bond is polar, with \(\text{C}^{\delta+}\) and \(\text{Br}^{\delta-}\).

Step 2: A curly arrow starts from the lone pair on \(\text{OH}^-\) to the \(\text{C}^{\delta+}\) atom from the side opposite to the bromine atom. Simultaneously, a curly arrow shows the movement of the electron pair in the \(\text{C--Br}\) bond onto the \(\text{Br}\) atom.

Step 3: The reaction passes through a single transition state where carbon is weakly bonded to both \(\text{OH}\) and \(\text{Br}\) (represented with dashed partial bonds) with three coplanar \(\text{C--H}\) and \(\text{C--CH}_3\) bonds, carrying an overall negative charge (\([\text{HO}\cdots\text{CH}_2(\text{CH}_3)\cdots\text{Br}]^-\)), yielding ethanol and \(\text{Br}^-\).

M1: Correct dipole on \(\text{C--Br}\) (\(\text{C}^{\delta+}\text{--Br}^{\delta-}\)) AND curly arrow from a lone pair on \(\text{OH}^-\) to the \(\text{C}^{\delta+}\) atom.
M2: Curly arrow from the \(\text{C--Br}\) bond to the \(\text{Br}\) atom.
M3: Correct structure of the transition state showing partial bonds (dashed lines) to \(\text{OH}\) and \(\text{Br}\), and an overall negative charge (\(-\) or \([\dots]^-\)).

Step 1: The cyanide ion, \(:\!\text{C}\equiv\text{N}^-\), acts as a nucleophile. A curly arrow starts from the lone pair of electrons on the carbon atom of \(:\!\text{CN}^-\) and points to the carbonyl carbon atom (\(\text{C}^{\delta+}\)).

Step 2: A curly arrow goes from the \(\text{C=O}\) double bond to the electronegative oxygen atom (\(\text{O}^{\delta-}\)). This forms the intermediate alkoxide tetrahedral ion, \(\text{CH}_3\text{CH}_2\text{CH}(\text{O}^-)\text{CN}\).

Step 3: A curly arrow from the lone pair on the intermediate's negatively charged oxygen (\(\text{O}^-\)) attacks the hydrogen atom of \(\text{HCN}\) (or \(\text{H}^+\)), with another arrow from the \(\text{H--CN}\) bond onto carbon, forming \(\text{CH}_3\text{CH}_2\text{CH(OH)CN}\) and regenerating the \(\text{CN}^-\) catalyst.

M1: Correct dipoles on \(\text{C=O}\) (\(\text{C}^{\delta+}\text{=O}^{\delta-}\)) AND curly arrow from the lone pair on the \(\text{C}\) of \(:\!\text{CN}^-\) to the \(\text{C}^{\delta+}\) atom.
M2: Curly arrow from the \(\text{C=O}\) bond onto the \(\text{O}\) atom AND correct intermediate structure \(\text{CH}_3\text{CH}_2\text{CH}(\text{O}^-)\text{CN}\).
M3: Curly arrow from a lone pair on \(\text{O}^-\) to the \(\text{H}\) of \(\text{HCN}\) (or \(\text{H}^+\)) AND curly arrow from \(\text{H--CN}\) bond to \(\text{CN}\) (if using \(\text{HCN}\)).

(a) Step 1 is an electrophilic addition of hydrogen bromide across the double bond of ethene. Reagent: \(\text{HBr(g)}\) (or concentrated aqueous \(\text{HBr}\)) at room temperature.
(b) Step 2 is a nucleophilic substitution extending the carbon chain by one carbon atom. Reagents: \(\text{KCN}\) (or \(\text{NaCN}\)) dissolved in ethanol (ethanolic), heated under reflux.
(c) Step 3 is the acid hydrolysis of a nitrile to a carboxylic acid. Reagents: dilute aqueous strong acid, such as \(\text{HCl(aq)}\) or \(\text{H}_2\text{SO}_4\text{(aq)}\), heated under reflux.

M1: \(\text{HBr(g)}\) / hydrogen bromide at room temperature (allow concentrated \(\text{HBr(aq)}\)). [Reject \(\text{Br}_2\)]
M2: \(\text{KCN}\) OR \(\text{NaCN}\) in ethanol / ethanolic (allow heating / reflux). [Reject \(\text{HCN}\)]
M3: Dilute \(\text{HCl(aq)}\) OR dilute \(\text{H}_2\text{SO}_4\text{(aq)}\) AND heat under reflux.

Step 1: The polar \(\text{C--Cl}\) bond undergoes heterolytic fission. A curly arrow starts at the \(\text{C--Cl}\) bond and moves onto the electronegative chlorine atom (\(\text{Cl}^{\delta-}\)).

Step 2: This rate-determining step produces the stable tertiary carbocation intermediate, \((\text{CH}_3)_3\text{C}^+\), and a chloride ion, \(\text{Cl}^-\).

Step 3: In the second step, the hydroxide ion acts as a nucleophile. A curly arrow begins at a lone pair of electrons on \(\text{OH}^-\) and attacks the positively charged carbon atom of the carbocation to give 2-methylpropan-2-ol, \((\text{CH}_3)_3\text{COH}\).

M1: Correct dipole on \(\text{C--Cl}\) (\(\text{C}^{\delta+}\text{--Cl}^{\delta-}\)) AND curly arrow from \(\text{C--Cl}\) bond to \(\text{Cl}\).
M2: Correct structure of the tertiary carbocation intermediate \((\text{CH}_3)_3\text{C}^+\) with the positive charge located on the central carbon atom.
M3: Curly arrow from a lone pair of electrons on \(\text{OH}^-\) to the \(\text{C}^+\) atom of the carbocation.

(a) Dehydration of an alcohol to an alkene is achieved using concentrated sulfuric acid (\(\text{H}_2\text{SO}_4\)) or concentrated phosphoric acid (\(\text{H}_3\text{PO}_4\)) with heating (approx. \(170\ ^\circ\text{C}\)), or by passing alcohol vapour over heated aluminium oxide (\(\text{Al}_2\text{O}_3\)) at \(300\text{--}350\ ^\circ\text{C}\).
(b) Oxidation of a secondary alcohol to a ketone requires an oxidising agent: acidified potassium dichromate(VI) (\(\text{K}_2\text{Cr}_2\text{O}_7 / \text{H}_2\text{SO}_4\)) heated under reflux (or acidified potassium manganate(VII)).
(c) Butan-2-ol contains the \(\text{CH}_3\text{CH(OH)--}\) group, so it gives a positive result with alkaline aqueous iodine (\(\text{I}_2 / \text{NaOH}\)), forming a pale yellow precipitate of tri-iodomethane, \(\text{CHI}_3\).

M1: Concentrated \(\text{H}_2\text{SO}_4\) OR concentrated \(\text{H}_3\text{PO}_4\) AND heat (allow \(\text{Al}_2\text{O}_3\) and heat / \(300\text{--}350\ ^\circ\text{C}\)).
M2: Acidified potassium dichromate(VI) / \(\text{K}_2\text{Cr}_2\text{O}_7 + \text{H}^+\) (or \(\text{KMnO}_4 + \text{H}^+\)) AND heat / reflux / warm.
M3: Observation: Yellow precipitate / solid / crystals AND Formula: \(\text{CHI}_3\).

(a) (i) The balanced equation is:
\[\text{P}_4\text{O}_{10}\text{(s)} + 6\text{H}_2\text{O(l)} \rightarrow 4\text{H}_3\text{PO}_4\text{(aq)}\]

(ii) Phosphoric(V) acid is a strong/moderately strong acid, so the pH is approximately \(0\) to \(2\).

(b) Step 1: Calculate the molar mass of \(\text{P}_4\text{O}_{10}\):
\[M_{\text{r}}(\text{P}_4\text{O}_{10}) = (4 \times 31.0) + (10 \times 16.0) = 124.0 + 160.0 = 284.0\text{ g mol}^{-1}\]

Step 2: Calculate the amount of \(\text{P}_4\text{O}_{10}\) dissolved:
\[n(\text{P}_4\text{O}_{10}) = \frac{1.42}{284.0} = 0.00500\text{ mol}\]

Step 3: Calculate the amount of \(\text{H}_3\text{PO}_4\) produced:
From the stoichiometry, \(1\text{ mol }\text{P}_4\text{O}_{10} \rightarrow 4\text{ mol }\text{H}_3\text{PO}_4\):
\[n(\text{H}_3\text{PO}_4) = 4 \times 0.00500 = 0.0200\text{ mol}\]

Step 4: Calculate the concentration in \(250\text{ cm}^3\) (\(0.250\text{ dm}^3\)):
\[[\text{H}_3\text{PO}_4] = \frac{0.0200\text{ mol}}{0.250\text{ dm}^3} = 0.0800\text{ mol dm}^{-3}\]

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

(a)(ii) [1 mark]
• Any value or range in the range \(0\text{ to }2\) (inclusive)

(b) [3 marks]
• M1: Calculation of \(n(\text{P}_4\text{O}_{10}) = \frac{1.42}{284.0} = 0.00500\text{ mol}\)
• M2: Calculation of \(n(\text{H}_3\text{PO}_4) = 4 \times 0.00500 = 0.0200\text{ mol}\) (allow ecf from incorrect \(M_r\) or stoichiometry)
• M3: Final concentration \(= \frac{0.0200}{0.250} = 0.0800\text{ mol dm}^{-3}\) (or \(0.08\text{ mol dm}^{-3}\), 2 or 3 s.f.)

(a) The balanced equation is:
\[\text{SiCl}_4\text{(l)} + 2\text{H}_2\text{O(l)} \rightarrow \text{SiO}_2\text{(s)} + 4\text{HCl(aq)}\]
(Accept \(\text{SiCl}_4 + 4\text{H}_2\text{O} \rightarrow \text{Si(OH)}_4 + 4\text{HCl}\))

(b) Observations:
1. Steamy / white fumes or misty gas evolved (due to \(\text{HCl}\)).
2. White precipitate / white solid formed (due to \(\text{SiO}_2\) or hydrated silica).

(c) Step 1: Calculate the molar mass of \(\text{SiCl}_4\):
\[M_{\text{r}}(\text{SiCl}_4) = 28.1 + (4 \times 35.5) = 170.1\text{ g mol}^{-1}\]

Step 2: Calculate the amount of \(\text{SiCl}_4\):
\[n = \frac{0.850}{170.1} = 4.997 \times 10^{-3}\text{ mol}\]

Step 3: Use the ideal gas equation \(pV = nRT\):
\[p = 102\text{ kPa} = 102\,000\text{ Pa}\]
\[V = \frac{nRT}{p} = \frac{(4.997 \times 10^{-3})(8.31)(373)}{102\,000} = 1.5185 \times 10^{-4}\text{ m}^3\]

Step 4: Convert volume to \(\text{cm}^3\):
\[V = 1.5185 \times 10^{-4} \times 10^6 = 152\text{ cm}^3\text{ (to 3 s.f.)}\]

(a) [1 mark]
• \(\text{SiCl}_4 + 2\text{H}_2\text{O} \rightarrow \text{SiO}_2 + 4\text{HCl}\) OR \(\text{SiCl}_4 + 4\text{H}_2\text{O} \rightarrow \text{Si(OH)}_4 + 4\text{HCl}\)

(b) [2 marks]
• M1: Steamy / misty / white fumes (or effervescence / sizzling / heat released)
• M2: White precipitate / white solid formed

(c) [2 marks]
• M1: Correct calculation of \(n = \frac{0.850}{170.1} = 4.997 \times 10^{-3}\text{ mol}\) AND correct application of \(pV = nRT\) with appropriate unit conversions (e.g. \(p = 102\,000\text{ Pa}\))
• M2: Final answer \(152\text{ cm}^3\) (accept \(151.8\text{--}152\text{ cm}^3\))

(a)
- In phosphorus, each 3p orbital is singly occupied (\(3p_x^1 3p_y^1 3p_z^1\) / half-filled subshell).
- In sulfur, one 3p orbital contains a pair of electrons (\(3p^4\)).
- The repulsion between the two paired electrons in the same 3p orbital reduces the energy needed to remove one electron.

(b) (i) Since the oxide is \(\text{E}_2\text{O}\), \(\text{E}\) is in Group 1 (oxidation state +1). Thus, the formula of the chloride is \(\text{ECl}\).

(ii) Step 1: Calculate the molar mass of \(\text{AgCl}\):
\[M_{\text{r}}(\text{AgCl}) = 108.0 + 35.5 = 143.5\text{ g mol}^{-1}\]

Step 2: Calculate the amount of \(\text{AgCl}\) precipitated:
\[n(\text{AgCl}) = \frac{2.870}{143.5} = 0.0200\text{ mol}\]

Step 3: From the stoichiometry \(\text{ECl} + \text{Ag}^+ \rightarrow \text{AgCl} + \text{E}^+\):
\[n(\text{E}) = n(\text{ECl}) = n(\text{AgCl}) = 0.0200\text{ mol}\]

Step 4: Calculate the relative atomic mass of \(\text{E}\):
\[A_{\text{r}}(\text{E}) = \frac{\text{mass}}{n} = \frac{0.460}{0.0200} = 23.0\]

Element \(\text{E}\) is sodium (\(\text{Na}\)).

(a) [2 marks]
• M1: Sulfur has paired electrons in one of its 3p orbitals (or sulfur has \(3p^4\) while phosphorus has \(3p^3\))
• M2: (Inter-electron) repulsion between the pair of electrons (in the same orbital) makes the outer electron easier to remove (less energy required)

(b)(i) [1 mark]
• \(\text{ECl}\)

(b)(ii) [2 marks]
• M1: Calculation of \(n(\text{AgCl}) = \frac{2.870}{143.5} = 0.0200\text{ mol}\) AND equating to \(n(\text{E}) = 0.0200\text{ mol}\)
• M2: Calculation of \(A_{\text{r}} = \frac{0.460}{0.0200} = 23.0\) AND identifying \(\text{E}\) as sodium / \(\text{Na}\)

(a) Ionic bonding (electrostatic attraction between oppositely charged ions).

(b) (i) Step 1: Calculate total initial moles of \(\text{HCl}\):
\[n(\text{HCl})_{\text{initial}} = 1.000 \times \frac{100.0}{1000} = 0.1000\text{ mol}\]

Step 2: Calculate moles of unreacted \(\text{HCl}\) from titration:
\[n(\text{HCl})_{\text{unreacted}} = n(\text{NaOH}) = 0.500 \times \frac{50.00}{1000} = 0.0250\text{ mol}\]

Step 3: Calculate moles of \(\text{HCl}\) reacted:
\[n(\text{HCl})_{\text{reacted}} = 0.1000 - 0.0250 = 0.0750\text{ mol}\]

(ii) Let the mass of \(\text{Na}_2\text{O}\) be \(x\text{ g}\). Then the mass of \(\text{MgO}\) is \((2.000 - x)\text{ g}\).

Amount of \(\text{Na}_2\text{O} = \frac{x}{62.0}\text{ mol}\)
Amount of \(\text{MgO} = \frac{2.000 - x}{40.3}\text{ mol}\)

From the stoichiometry, each mole of oxide consumes \(2\text{ mol}\) of \(\text{HCl}\):
\[2\left(\frac{x}{62.0}\right) + 2\left(\frac{2.000 - x}{40.3}\right) = 0.0750\]
\[\frac{x}{31.0} + \frac{2.000 - x}{20.15} = 0.0750\]

Multiply through by \(31.0 \times 20.15 = 624.65\):
\[20.15x + 31.0(2.000 - x) = 0.0750 \times 624.65 = 46.84875\]
\[20.15x + 62.0 - 31.0x = 46.84875\]
\[-10.85x = 46.84875 - 62.0 = -15.15125\]
\[x = \frac{15.15125}{10.85} = 1.3964\text{ g}\]

Calculate the percentage by mass of \(\text{Na}_2\text{O}\):
\[\%\text{ by mass} = \frac{1.3964}{2.000} \times 100 = 69.82\% \approx 69.8\%\]

(a) [1 mark]
• Ionic (bonding)

(b)(i) [1 mark]
• \(n(\text{HCl})_{\text{reacted}} = 0.1000 - 0.0250 = 0.0750\text{ mol}\)

(b)(ii) [3 marks]
• M1: Setting up correct simultaneous or single algebraic equation relating mass and moles of \(\text{HCl}\), e.g. \(2\left(\frac{x}{62.0}\right) + 2\left(\frac{2.000 - x}{40.3}\right) = 0.0750\)
• M2: Correct rearrangement and solution for mass of \(\text{Na}_2\text{O} = 1.396\text{ g}\) (or mass of \(\text{MgO} = 0.604\text{ g}\)) (ecf from (b)(i))
• M3: Final percentage \(= 69.8\%\) (accept \(69.8\text{--}70.0\%\), 3 s.f.)

(a) The balanced ionic equation is:
\[\text{Al}_2\text{O}_3\text{(s)} + 2\text{OH}^-\text{(aq)} + 3\text{H}_2\text{O(l)} \rightarrow 2[\text{Al(OH)}_4]^-\text{(aq)}\]

(b) The equation is:
\[\text{SO}_3\text{(g/l)} + \text{H}_2\text{O(l)} \rightarrow \text{H}_2\text{SO}_4\text{(aq)}\]

(c) Step 1: Calculate the amount of \(\text{MnO}_4^-\) used:
\[n(\text{MnO}_4^-) = 0.0200 \times \frac{16.40}{1000} = 3.280 \times 10^{-4}\text{ mol}\]

Step 2: Calculate the amount of \(\text{SO}_3^{2-}\) (and hence \(\text{SO}_2\)):
From the stoichiometry, \(2\text{ mol }\text{MnO}_4^- \equiv 5\text{ mol }\text{SO}_3^{2-}\):
\[n(\text{SO}_2) = n(\text{SO}_3^{2-}) = \frac{5}{2} \times 3.280 \times 10^{-4} = 8.200 \times 10^{-4}\text{ mol}\]

Step 3: Calculate the volume occupied by \(8.200 \times 10^{-4}\text{ mol}\) of \(\text{SO}_2\) using \(pV = nRT\):
\[V = \frac{nRT}{p} = \frac{(8.200 \times 10^{-4})(8.31)(298)}{101\,000} = 2.010 \times 10^{-5}\text{ m}^3 = 0.02010\text{ dm}^3\]

Step 4: Calculate the percentage by volume:
\[\%\text{ by volume} = \frac{0.02010\text{ dm}^3}{0.200\text{ dm}^3} \times 100 = 10.05\% \approx 10.1\%\]

(a) [1 mark]
• \(\text{Al}_2\text{O}_3 + 2\text{OH}^- + 3\text{H}_2\text{O} \rightarrow 2[\text{Al(OH)}_4]^-\)

(b) [1 mark]
• \(\text{SO}_3 + \text{H}_2\text{O} \rightarrow \text{H}_2\text{SO}_4\)

(c) [3 marks]
• M1: Calculation of \(n(\text{MnO}_4^-) = 3.280 \times 10^{-4}\text{ mol}\) AND \(n(\text{SO}_2) = \frac{5}{2} \times 3.280 \times 10^{-4} = 8.200 \times 10^{-4}\text{ mol}\)
• M2: Calculation of volume of \(\text{SO}_2 = \frac{nRT}{p} = 0.02010\text{ dm}^3\) (or \(20.10\text{ cm}^3\))
• M3: Final percentage by volume \(= \frac{0.02010}{0.200} \times 100 = 10.1\%\) (accept \(10.05\text{--}10.1\%\), 3 s.f.)

(a) Dative covalent (or coordinate) bonding, where chlorine donates a lone pair into an empty 3p orbital of aluminium.

(b) The equation is:
\[\text{AlCl}_3\text{(s)} + 3\text{H}_2\text{O(l)} \rightarrow \text{Al(OH)}_3\text{(s)} + 3\text{HCl(aq)}\]
(Accept \(\text{AlCl}_3 + 6\text{H}_2\text{O} \rightarrow [\text{Al(H}_2\text{O)}_6]^{3+} + 3\text{Cl}^-\))

(c) Step 1: Calculate the amount of \(\text{AgCl}\) formed:
\[n(\text{AgCl}) = \frac{1.608}{143.5} = 0.011206\text{ mol}\]

Step 2: Calculate the amount of \(\text{Cl}^-\) ions and \(\text{AlCl}_3\):
\[n(\text{Cl}^-) = 0.011206\text{ mol}\]
Since each formula unit of \(\text{AlCl}_3\) contains 3 chloride ions:
\[n(\text{AlCl}_3) = \frac{0.011206}{3} = 3.7352 \times 10^{-3}\text{ mol}\]

Step 3: Calculate the mass of pure \(\text{AlCl}_3\):
\[\text{mass of pure }\text{AlCl}_3 = (3.7352 \times 10^{-3}\text{ mol}) \times 133.5\text{ g mol}^{-1} = 0.49865\text{ g}\]

Step 4: Calculate the percentage purity:
\[\%\text{ purity} = \frac{0.49865}{0.534} \times 100 = 93.38\% \approx 93.4\%\]

(a) [1 mark]
• Dative (covalent) / coordinate (bonding)

(b) [1 mark]
• \(\text{AlCl}_3 + 3\text{H}_2\text{O} \rightarrow \text{Al(OH)}_3 + 3\text{HCl}\) OR \(\text{AlCl}_3 + 6\text{H}_2\text{O} \rightarrow [\text{Al(H}_2\text{O)}_6]^{3+} + 3\text{Cl}^-\)

(c) [3 marks]
• M1: Calculation of \(n(\text{AgCl}) = \frac{1.608}{143.5} = 0.01121\text{ mol}\)
• M2: Division by 3 and multiplication by 133.5 to get mass of pure \(\text{AlCl}_3 = 0.49865\text{ g}\)
• M3: Final percentage purity \(= \frac{0.49865}{0.534} \times 100 = 93.4\%\) (accept \(93.3\text{--}93.5\%\), 3 s.f.)

(a)
(i) Typical experimental mass data:
- Mass of container $+$ $\text{FA 2} = 12.45\text{ g}$
- Mass of empty container $= 9.35\text{ g}$
- Mass of $\text{FA 2}$ added $= 12.45 - 9.35 = 3.10\text{ g}$

(ii) Table of results:
- Headings: $\text{time / min}$ and $\text{temperature / }^\circ\text{C}$.
- Readings taken at $0.0, 0.5, 1.0, 1.5, 2.0, 2.5\text{ min}$ (e.g. constant at $21.0\,^\circ\text{C}$).
- No reading at $3.0\text{ min}$.
- Readings taken at $3.5, 4.0, 4.5, 5.0, 5.5, 6.0, 6.5, 7.0, 7.5, 8.0, 8.5, 9.0\text{ min}$.
- All temperature readings recorded to nearest $0.5\,^\circ\text{C}$ or consistent $1$ d.p.

(b)
- Graph plotted with temperature ($y$-axis) against time ($x$-axis).
- Linear, sensible scales covering $\ge 50\%$ of grid in both directions.
- Points plotted accurately to within $\frac{1}{2}$ small square.
- Initial line drawn horizontally through points from $t = 0.0$ to $2.5\text{ min}$ and extrapolated to $t = 3.0\text{ min}$ ($T_1 = 21.0\,^\circ\text{C}$).
- Straight cooling line drawn through points after maximum (e.g. $t = 5.0$ to $9.0\text{ min}$) and extrapolated back to $t = 3.0\text{ min}$ ($T_2 = 46.0\,^\circ\text{C}$).
- $\Delta T = T_2 - T_1 = 46.0 - 21.0 = 25.0\,^\circ\text{C}$.

(c)
(i) Mass of solution $m = 25.0\text{ cm}^3 \times 1.00\text{ g cm}^{-3} = 25.0\text{ g}$
$$q = m \times c \times \Delta T = 25.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 25.0\text{ K} = 2612.5\text{ J} = 2.61\text{ kJ}$$

(ii) Amount of $\text{CuSO}_4$:
$$n(\text{CuSO}_4) = c \times V = 0.800\text{ mol dm}^{-3} \times \frac{25.0}{1000}\text{ dm}^3 = 0.0200\text{ mol}$$

(iii) Enthalpy change:
$$\Delta H = -\frac{q}{n} = -\frac{2.6125\text{ kJ}}{0.0200\text{ mol}} = -130.6\text{ kJ mol}^{-1}$$
(or $-219\text{ kJ mol}^{-1}$ depending on exact $\Delta T$; using $\Delta T = 25.0\,^\circ\text{C}$, $\Delta H = -131\text{ kJ mol}^{-1}$ to 3 s.f.)

(d)
(i) $\text{Percentage uncertainty} = \frac{2 \times 0.5}{\Delta T} \times 100 = \frac{1.0}{25.0} \times 100 = 4.00\%$

(ii) Copper is a good thermal conductor (low insulating property), leading to greater heat loss to the surroundings; this would cause the measured maximum temperature to be lower and make the calculated value of $\Delta H$ less exothermic / less accurate.

(a) [5 marks]
- M1: Clear table format containing unambiguous headings with standard units (`time / min` or `/ s` and `temperature / °C` or `/ K`).
- M2: Balance readings recorded consistently to at least 2 decimal places with correct subtraction showing mass of FA 2.
- M3: All temperature readings recorded to a consistent precision (either all to the nearest $0.5\,^\circ\text{C}$ or all to $1$ d.p.).
- M4: Complete collection of data: initial readings from $0$ to $2.5\text{ min}$ at $0.5\text{ min}$ intervals, no reading at $3.0\text{ min}$, and post-addition readings from $3.5$ to $9.0\text{ min}$.
- M5: Accuracy mark awarded if candidate's maximum $\Delta T$ is within $\pm 2.0\,^\circ\text{C}$ of the Supervisor's value.

(b) [4 marks]
- M1: Axes correctly labelled with units (`temperature / °C` on $y$-axis, `time / min` on $x$-axis) and suitable linear scales where points occupy more than $50\%$ of the grid in both dimensions.
- M2: All points plotted correctly within $\frac{1}{2}$ small square.
- M3: Drawing of two distinct best-fit lines: horizontal/initial trend line and cooling line through points after maximum, with both extrapolated to $t = 3.0\text{ min}$.
- M4: Correct reading and deduction of $\Delta T$ at $t = 3.0\text{ min}$ from the extrapolated lines.

(c) [5 marks]
- (i) [2 marks]
- M1: Correct formula and substitution: $q = 25.0 \times 4.18 \times \Delta T$.
- M2: Correct evaluation of $q$ in J or kJ.
- (ii) [1 mark]
- M1: $n(\text{CuSO}_4) = 0.800 \times 0.0250 = 0.0200\text{ mol}$.
- (iii) [2 marks]
- M1: Correct calculation of $\Delta H = -\frac{q}{n \times 1000}$ in $\text{kJ mol}^{-1}$ (allowing ecf from (c)(i) and (c)(ii)).
- M2: Correct negative sign ($-$) explicitly stated AND answer quoted to 3 significant figures.

(d) [2 marks]
- (i) [1 mark]
- M1: Percentage uncertainty $= \frac{2 \times 0.5}{\Delta T} \times 100 = \frac{1.0}{\Delta T} \times 100$ correctly evaluated.
- (ii) [1 mark]
- M1: Copper is a good conductor of heat / poor insulator, so more heat is lost to surroundings, making $\Delta T$ lower and $\Delta H$ less exothermic / less accurate.

(a)
(i) Mass of hydrated salt = \(22.39 - 18.45 = 3.94\text{ g}\)
(ii) Mass of anhydrous \(\text{NiSO}_4\) = \(20.77 - 18.45 = 2.32\text{ g}\)
(iii) Mass of \(\text{H}_2\text{O}\) lost = \(22.39 - 20.77 = 1.62\text{ g}\) (or \(3.94 - 2.32 = 1.62\text{ g}\))

(b)
(i) \(M_{\text{r}}(\text{NiSO}_4) = 58.7 + 32.1 + (4 \times 16.0) = 154.8\text{ g mol}^{-1}\)
\(n(\text{NiSO}_4) = \frac{2.32}{154.8} = 0.014987\text{ mol} = 0.0150\text{ mol}\) (to 3 s.f.)

(ii) \(M_{\text{r}}(\text{H}_2\text{O}) = (2 \times 1.0) + 16.0 = 18.0\text{ g mol}^{-1}\)
\(n(\text{H}_2\text{O}) = \frac{1.62}{18.0} = 0.0900\text{ mol}\) (to 3 s.f.)

(iii) \(x = \frac{n(\text{H}_2\text{O})}{n(\text{NiSO}_4)} = \frac{0.0900}{0.014987} = 6.005 \approx 6\)
Formula: \(\text{NiSO}_4\cdot 6\text{H}_2\text{O}\)

(c)
(i) Heating to constant mass ensures that all water of crystallisation has been completely evaporated/driven off, ensuring the decomposition to anhydrous salt is complete.

(ii) The mass of \(\text{H}_2\text{O}\) lost is obtained from the difference between two readings (mass before heating and mass after heating):
\(\text{Total absolute uncertainty} = 2 \times 0.01\text{ g} = 0.02\text{ g}\)
\(\text{Percentage uncertainty} = \frac{0.02}{1.62} \times 100 = 1.23\%\) (or \(1.2\%\))

(iii) If \(\text{NiSO}_4\) decomposes, \(\text{SO}_3\) gas escapes, causing an extra mass loss. This makes the calculated mass (and moles) of water lost larger than it actually was, and the mass of residue smaller, resulting in a higher calculated value of \(x\).

(iv) Place a loose-fitting lid on the crucible during heating to prevent loss of solid due to spitting / allowing gas to escape while preventing mechanical loss.

(a) [2 marks]
- M1: Correct calculation of mass of hydrated salt (\(3.94\text{ g}\)) AND mass of anhydrous residue (\(2.32\text{ g}\)).
- M2: Correct calculation of mass of water lost (\(1.62\text{ g}\)). Both M1 and M2 require units (\(\text{g}\)) and values to 2 decimal places.

(b) [3 marks]
- M1 (b)(i): Correct calculation of moles of \(\text{NiSO}_4\) = \(0.0150\text{ mol}\) (accept \(0.01499\) to \(0.0150\), minimum 3 s.f.).
- M2 (b)(ii): Correct calculation of moles of \(\text{H}_2\text{O}\) = \(0.0900\text{ mol}\) (minimum 3 s.f.).
- M3 (b)(iii): Correct ratio \(x = \frac{0.0900}{0.0150} = 6\) stated as an integer (allow ecf from (b)(i) and (b)(ii)).

(c) [6 marks]
- M1 (c)(i): States that heating to constant mass ensures all water of crystallisation has been completely removed / complete dehydration [1].
- M2 (c)(ii): Identifies maximum absolute error as \(2 \times 0.01 = 0.02\text{ g}\) [1].
- M3 (c)(ii): Calculates percentage uncertainty as \(\frac{0.02}{1.62} \times 100 = 1.23\%\) (accept \(1.2\%\) or \(1.235\%\)) [1].
- M4 (c)(iii): Predicts that \(x\) will be higher / larger / overestimated [1].
- M5 (c)(iii): Explains that mass of \(\text{SO}_3\text{ (g)}\) lost is incorrectly counted as water lost (so apparent \(n(\text{H}_2\text{O})\) increases and/or apparent mass of anhydrous salt decreases) [1].
- M6 (c)(iv): Suggests placing a lid on the crucible (to prevent spitting/loss of solid) OR allowing the crucible to cool in a desiccator with lid to prevent reabsorption of atmospheric moisture [1].

(a) Tests on FA 5:
- (a)(i) Reaction with \(\text{NaOH(aq)}\): Addition of \(\text{NaOH(aq)}\) dropwise produces a grey-green precipitate of chromium(III) hydroxide, \(\text{Cr(OH)}_3\). Upon adding excess \(\text{NaOH(aq)}\), the amphoteric precipitate dissolves to give a dark green solution of \([\text{Cr(OH)}_6]^{3-}\) (or \([\text{Cr(OH)}_4]^-\)).
- (a)(ii) Reaction with \(\text{NH}_3\text{(aq)}\): Addition of \(\text{NH}_3\text{(aq)}\) produces a grey-green precipitate of \(\text{Cr(OH)}_3\), which is insoluble in excess aqueous ammonia.
- (a)(iii) Anion test: Add dilute \(\text{HNO}_3\) (or dilute \(\text{HCl}\)) followed by \(\text{Ba(NO}_3)_2\text{(aq)}\) (or \(\text{BaCl}_2\text{(aq)}\)). A white precipitate of barium sulfate (\(\text{BaSO}_4\)) is formed, confirming the presence of sulfate ions (\(\text{SO}_4^{2-}\)).

(b) Tests on FA 6:
- (b)(i) Reaction with \(\text{NaOH(aq)}\): A green precipitate of iron(II) hydroxide, \(\text{Fe(OH)}_2\), forms that is insoluble in excess \(\text{NaOH}\). On standing, it oxidises in air at the surface to form a brown/red-brown precipitate of \(\text{Fe(OH)}_3\).
- (b)(ii) Halide test: Acidification with \(\text{HNO}_3\) and addition of \(\text{AgNO}_3\text{(aq)}\) gives a yellow precipitate of silver iodide (\(\text{AgI}\)), which remains insoluble upon addition of aqueous ammonia.
- (b)(iii) Redox test: Hydrogen peroxide oxidises \(\text{I}^-\) to \(\text{I}_2\), turning the solution brown. Addition of starch indicator produces a characteristic deep blue-black colour.

(c) Identification of ions:
- FA 5 cation: \(\text{Cr}^{3+}\) (chromium(III))
- FA 5 anion: \(\text{SO}_4^{2-}\) (sulfate)
- FA 6 cation: \(\text{Fe}^{2+}\) (iron(II))
- FA 6 anion: \(\text{I}^-\) (iodide)

(d) Ionic Equations:
- (d)(i) \(\text{Cr}^{3+}\text{(aq)} + 3\text{OH}^-\text{(aq)} \rightarrow \text{Cr(OH)}_3\text{(s)}\)
- (d)(ii) \(\text{Ag}^+\text{(aq)} + \text{I}^-\text{(aq)} \rightarrow \text{AgI(s)}\)

(a) [5 marks total]
- (a)(i) [2 marks]:
- M1: Grey-green precipitate / green precipitate on dropwise addition of \(\text{NaOH(aq)}\).
- M2: Precipitate dissolves / is soluble in excess to form a dark green / green solution.
- (a)(ii) [1 mark]:
- M3: Grey-green precipitate / green precipitate, insoluble in excess \(\text{NH}_3\text{(aq)}\).
- (a)(iii) [2 marks]:
- M4: Reagents specified: dilute \(\text{HNO}_3\) (or dilute \(\text{HCl}\)) AND \(\text{Ba(NO}_3)_2\text{(aq)}\) (or \(\text{BaCl}_2\text{(aq)}\)). [R: use of sulfuric acid]
- M5: Observation: White precipitate formed.

(b) [4 marks total]
- (b)(i) [1 mark]:
- M6: Green precipitate, insoluble in excess \(\text{NaOH}\) AND turns brown / red-brown on standing.
- (b)(ii) [2 marks]:
- M7: Yellow precipitate with \(\text{AgNO}_3\text{(aq)}\).
- M8: Precipitate is insoluble in aqueous ammonia / \(\text{NH}_3\text{(aq)}\).
- (b)(iii) [1 mark]:
- M9: Brown / dark brown solution formed with \(\text{H}_2\text{O}_2\) AND turns blue-black with starch.

(c) [2 marks total]
- M10: Cation in FA 5 is \(\text{Cr}^{3+}\) (or chromium(III)) AND Anion in FA 5 is \(\text{SO}_4^{2-}\) (or sulfate).
- M11: Cation in FA 6 is \(\text{Fe}^{2+}\) (or iron(II)) AND Anion in FA 6 is \(\text{I}^-\) (or iodide).
(Allow ecf from incorrect observations if deductions are completely consistent with candidate observations)

(d) [2 marks total]
- (d)(i) [1 mark]:
- M12: \(\text{Cr}^{3+}\text{(aq)} + 3\text{OH}^-\text{(aq)} \rightarrow \text{Cr(OH)}_3\text{(s)}\) with correct state symbols.
- (d)(ii) [1 mark]:
- M13: \(\text{Ag}^+\text{(aq)} + \text{I}^-\text{(aq)} \rightarrow \text{AgI(s)}\) with correct state symbols.

(a)(i)
A transition element is a d-block element that forms one or more stable ions with an incompletely filled (partially filled) d-subshell.

(a)(ii)
* $\text{Co}^{2+}: [\text{Ar}]\, 3\text{d}^7$
* $\text{Co}^{3+}: [\text{Ar}]\, 3\text{d}^6$

(b)(i)
* Formula of $\mathbf{A}$: $[\text{CoCl}_4]^{2-}$
* Colour change: from pink to blue
* Geometry of $\mathbf{A}$: tetrahedral

(b)(ii)
1. In the presence of ligands, the electric field of the ligand lone pairs causes the five degenerate d-orbitals of the central metal ion to split into two sets of different energy levels separated by an energy gap, $\Delta E$.
2. An electron absorbs a photon of visible light of specific frequency/wavelength corresponding to $\Delta E = hf$ ($= \frac{hc}{\lambda}$) and is promoted from a lower energy d-orbital to a higher energy d-orbital (d-d transition).
3. The remaining wavelengths of light are transmitted or reflected, producing the complementary colour seen.

(c)(i)
A bidentate ligand is a molecule or ion that donates two lone pairs of electrons to the central metal ion to form two coordinate (dative covalent) bonds.

(c)(ii)
$$K_{\text{stab}} = \frac{[\text{Co}(\text{en})_3^{2+}]}{[\text{Co}(\text{H}_2\text{O})_6^{2+}][\text{en}]^3}$$
Units: $\frac{\text{mol dm}^{-3}}{(\text{mol dm}^{-3}) \times (\text{mol dm}^{-3})^3} = \frac{1}{(\text{mol dm}^{-3})^3} = \text{mol}^{-3}\text{ dm}^9$

(c)(iii)
In the substitution reaction, 4 reacting species ($1 \times [\text{Co}(\text{H}_2\text{O})_6]^{2+} + 3 \times \text{en}$) form 7 product species ($1 \times [\text{Co}(\text{en})_3]^{2+} + 6 \times \text{H}_2\text{O}$). This large increase in the number of independent particles leads to a positive system entropy change ($\Delta S^\theta > 0$).
Because $\Delta G^\theta = \Delta H^\theta - T\Delta S^\theta$, a positive $\Delta S^\theta$ makes $\Delta G^\theta$ significantly more negative, driving the equilibrium far to the right.

(c)(iv)
In the cis isomer, the two $\text{Cl}^-$ ligands occupy adjacent coordination sites (at $90^\circ$ to each other). The two bidentate $\text{en}$ ligands span adjacent equatorial-axial or equatorial-equatorial positions.
Drawing the two optical isomers shows non-superimposable mirror-image octahedra with wedge-and-dash bonds and overall charge $+1$.

(d)
1. Calculate the initial concentrations:
$$\text{Volume} = 500\text{ cm}^3 = 0.500\text{ dm}^3$$
$$[\text{Ni}^{2+}]_{\text{initial}} = \frac{0.0200\text{ mol}}{0.500\text{ dm}^3} = 0.0400\text{ mol dm}^{-3}$$
$$[\text{en}]_{\text{initial}} = \frac{0.150\text{ mol}}{0.500\text{ dm}^3} = 0.300\text{ mol dm}^{-3}$$

2. At equilibrium, assuming complete conversion of $\text{Ni}^{2+}$ into $[\text{Ni}(\text{en})_3]^{2+}$:
$$[\text{Ni}(\text{en})_3^{2+}]_{\text{eq}} \approx 0.0400\text{ mol dm}^{-3}$$
$$\text{Amount of en consumed} = 3 \times 0.0200\text{ mol} = 0.0600\text{ mol}$$
$$\text{Amount of en remaining} = 0.150\text{ mol} - 0.0600\text{ mol} = 0.0900\text{ mol}$$
$$[\text{en}]_{\text{eq}} = \frac{0.0900\text{ mol}}{0.500\text{ dm}^3} = 0.180\text{ mol dm}^{-3}$$

3. Substitute into the stability constant expression:
$$K_{\text{stab}} = \frac{[\text{Ni}(\text{en})_3^{2+}]}{[\text{Ni}(\text{H}_2\text{O})_6^{2+}][\text{en}]^3}$$
$$4.00 \times 10^{18} = \frac{0.0400}{[\text{Ni}(\text{H}_2\text{O})_6^{2+}] \times (0.180)^3}$$
$$(0.180)^3 = 5.832 \times 10^{-3}$$
$$[\text{Ni}(\text{H}_2\text{O})_6^{2+}] = \frac{0.0400}{4.00 \times 10^{18} \times 5.832 \times 10^{-3}} = \frac{0.0400}{2.3328 \times 10^{16}} = 1.7147 \times 10^{-18}\text{ mol dm}^{-3}$$
$$[\text{Ni}(\text{H}_2\text{O})_6^{2+}] = 1.71 \times 10^{-18}\text{ mol dm}^{-3}$$

(e)(i)
Ammonia is a stronger field ligand / forms more stable complexes with $\text{Co}^{3+}$ than with $\text{Co}^{2+}$ ($K_{\text{stab}}$ for $[\text{Co}(\text{NH}_3)_6]^{3+}$ is much higher than for $[\text{Co}(\text{NH}_3)_6]^{2+}$). This stabilises the $+3$ oxidation state relative to $+2$, making the reduction of $\text{Co}^{3+}$ to $\text{Co}^{2+}$ less favourable (equilibrium shifts left), thus lowering $E^\theta$.

(e)(ii)
* For the ammine system: $E^\theta_{\text{cell}} = E^\theta(\text{O}_2/\text{H}_2\text{O}) - E^\theta([\text{Co}(\text{NH}_3)_6]^{3+}/[\text{Co}(\text{NH}_3)_6]^{2+}) = +1.23 - (+0.11) = +1.12\text{ V} > 0$, so the oxidation is thermodynamically feasible.
* For the aqua system: $E^\theta_{\text{cell}} = +1.23 - (+1.82) = -0.59\text{ V} < 0$, so oxidation of $[\text{Co}(\text{H}_2\text{O})_6]^{2+}$ by $\text{O}_2$ is not feasible.

(a)(i)
* M1: A d-block element that forms one or more stable ions with an incomplete / partially filled d-subshell. [1]
(Reject 'd-orbital' instead of 'd-subshell')

(a)(ii)
* M1: $\text{Co}^{2+}: [\text{Ar}]\, 3\text{d}^7$ AND $\text{Co}^{3+}: [\text{Ar}]\, 3\text{d}^6$ (both required). [1]

(b)(i)
* M1: $[\text{CoCl}_4]^{2-}$ (charge must be correct) [1]
* M2: (from) pink to blue [1]
* M3: tetrahedral [1]

(b)(ii)
* M1: Ligands cause the d-orbitals to split into two sets of different energy levels / non-degenerate levels. [1]
* M2: An electron absorbs a photon of visible light (corresponding to $\Delta E = hf$) and is promoted from a lower d-orbital to a higher d-orbital. [1]
* M3: The transmitted light / remaining wavelengths give the complementary colour observed. [1]

(c)(i)
* M1: A species / ligand that donates two lone pairs of electrons to form two coordinate / dative covalent bonds to a central metal atom / ion. [1]

(c)(ii)
* M1: Correct expression: $K_{\text{stab}} = \frac{[\text{Co}(\text{en})_3^{2+}]}{[\text{Co}(\text{H}_2\text{O})_6^{2+}][\text{en}]^3}$ ($[\text{H}_2\text{O}]$ must be omitted) [1]
* M2: Units: $\text{mol}^{-3}\text{ dm}^9$ [1]

(c)(iii)
* M1: More particles / moles on the right-hand side (7 species from 4 species) so entropy of system increases / $\Delta S > 0$. [1]
* M2: Reference to $\Delta G = \Delta H - T\Delta S$, so a positive $\Delta S$ makes $\Delta G$ more negative / more thermodynamically feasible. [1]

(c)(iv)
* M1: Both isomers drawn as 3D octahedral structures with two $\text{Cl}$ ligands positioned adjacent (at $90^\circ$) to each other. [1]
* M2: The two structures are non-superimposable mirror images of each other, showing correct connectivity / bidentate loops and charge $+1$. [1]

(d)
* M1: Deduces $[\text{Ni}(\text{en})_3^{2+}] = 0.0400\text{ mol dm}^{-3}$ and equilibrium $[\text{en}] = \frac{0.150 - 3(0.0200)}{0.500} = 0.180\text{ mol dm}^{-3}$. [1]
* M2: Correct substitution into rearranged expression: $[\text{Ni}(\text{H}_2\text{O})_6^{2+}] = \frac{0.0400}{(4.00 \times 10^{18}) \times (0.180)^3}$. [1]
* M3: Evaluates answer correctly to 2 or 3 sig figs: $1.71 \times 10^{-18}\text{ mol dm}^{-3}$ (or $1.7 \times 10^{-18}$). [1]
(Allow ecf from incorrect equilibrium concentration of en in M1)

(e)(i)
* M1: $\text{Co}^{3+}$ forms a more stable complex with $\text{NH}_3$ than $\text{Co}^{2+}$ does / $K_{\text{stab}}$ of $[\text{Co}(\text{NH}_3)_6]^{3+}$ is much larger than for $[\text{Co}(\text{NH}_3)_6]^{2+}$, which stabilises the $+3$ oxidation state (making reduction less favourable). [1]

(e)(ii)
* M1: For the ammine complex, $E^\theta_{\text{cell}} = +1.23 - (+0.11) = +1.12\text{ V} > 0$ (feasible), whereas for the aqua complex, $E^\theta_{\text{cell}} = +1.23 - (+1.82) = -0.59\text{ V} < 0$ (not feasible). [1]

(a) (i) \(K_{\text{sp}}\) is the product of the concentrations of each ion present in a saturated solution of a sparingly soluble salt at a stated temperature, each raised to the power of its stoichiometric coefficient.

(ii) \(\text{Pb(IO}_3)_2(\text{s}) \rightleftharpoons \text{Pb}^{2+}(\text{aq}) + 2\text{IO}_3^-(\text{aq})\)
\(K_{\text{sp}} = [\text{Pb}^{2+}][\text{IO}_3^-]^2\)
Units: \((\text{mol dm}^{-3}) \times (\text{mol dm}^{-3})^2 = \text{mol}^3\text{ dm}^{-9}\)

(b) (i) Let solubility be \(s\text{ mol dm}^{-3}\).
\([\text{Pb}^{2+}] = s\) and \([\text{IO}_3^-] = 2s\)
\(K_{\text{sp}} = s(2s)^2 = 4s^3\)
\(4s^3 = 3.20 \times 10^{-13}\)
\(s^3 = 8.00 \times 10^{-14}\)
\(s = \sqrt[3]{8.00 \times 10^{-14}} = 4.31 \times 10^{-5}\text{ mol dm}^{-3}\)

(ii) In \(0.0500\text{ mol dm}^{-3}\text{ NaIO}_3\), \([\text{IO}_3^-] = 0.0500\text{ mol dm}^{-3}\) (assuming solubility \(s \ll 0.0500\)).
\(K_{\text{sp}} = [\text{Pb}^{2+}][\text{IO}_3^-]^2\)
\(3.20 \times 10^{-13} = s \times (0.0500)^2 = s \times (2.50 \times 10^{-3})\)
\(s = \frac{3.20 \times 10^{-13}}{2.50 \times 10^{-3}} = 1.28 \times 10^{-10}\text{ mol dm}^{-3}\)

(c) Total volume = \(25.0 + 75.0 = 100.0\text{ cm}^3\)
\([\text{Pb}^{2+}] = 1.60 \times 10^{-4} \times \frac{25.0}{100.0} = 4.00 \times 10^{-5}\text{ mol dm}^{-3}\)
\([\text{IO}_3^-] = 4.00 \times 10^{-5} \times \frac{75.0}{100.0} = 3.00 \times 10^{-5}\text{ mol dm}^{-3}\)
\(\text{Ionic Product (IP)} = [\text{Pb}^{2+}][\text{IO}_3^-]^2 = (4.00 \times 10^{-5}) \times (3.00 \times 10^{-5})^2\)
\(\text{IP} = (4.00 \times 10^{-5}) \times (9.00 \times 10^{-10}) = 3.60 \times 10^{-14}\text{ mol}^3\text{ dm}^{-9}\)
Since \(\text{IP} (3.60 \times 10^{-14}) < K_{\text{sp}} (3.20 \times 10^{-13})\), no precipitate forms.

(a)(i) Product of the concentrations of ions in a saturated solution (of a sparingly soluble salt), with each concentration raised to its stoichiometric power / coefficient (at a given temperature) [1]

(a)(ii) M1: \(K_{\text{sp}} = [\text{Pb}^{2+}][\text{IO}_3^-]^2\) [1]
M2: \(\text{mol}^3\text{ dm}^{-9}\) [1]

(b)(i) M1: \(K_{\text{sp}} = 4s^3\) [1]
M2: \(s = 4.31 \times 10^{-5}\text{ (mol dm}^{-3})\) (must be 3 sig figs) [1]

(b)(ii) M1: Setting up \(3.20 \times 10^{-13} = s \times (0.0500)^2\) [1]
M2: \(s = 1.28 \times 10^{-10}\text{ (mol dm}^{-3})\) [1]

(c) M1: Calculation of diluted concentrations: \([\text{Pb}^{2+}] = 4.00 \times 10^{-5}\) and \([\text{IO}_3^-] = 3.00 \times 10^{-5}\) [1]
M2: Calculation of ionic product = \(3.60 \times 10^{-14}\text{ (mol}^3\text{ dm}^{-9})\) [1]
M3: State 'No' / 'No precipitate' AND state that \(\text{ionic product} < K_{\text{sp}}\) (comparison required, ecf on IP value) [1]

(a) (i) Comparing Exp 1 and Exp 2: \([\text{I}^-]\) is constant \((0.030\text{ mol dm}^{-3})\), while \([\text{S}_2\text{O}_8^{2-}]\) doubles from \(0.040\) to \(0.080\text{ mol dm}^{-3}\). The rate doubles from \(1.44 \times 10^{-5}\) to \(2.88 \times 10^{-5}\text{ mol dm}^{-3}\text{ s}^{-1}\). Therefore, the reaction is first order with respect to \([\text{S}_2\text{O}_8^{2-}]\).

Comparing Exp 2 and Exp 3: \([\text{S}_2\text{O}_8^{2-}]\) is constant \((0.080\text{ mol dm}^{-3})\), while \([\text{I}^-]\) triples from \(0.030\) to \(0.090\text{ mol dm}^{-3}\). The rate triples from \(2.88 \times 10^{-5}\) to \(8.64 \times 10^{-5}\text{ mol dm}^{-3}\text{ s}^{-1}\). Therefore, the reaction is first order with respect to \([\text{I}^-]\).

(ii) \(\text{rate} = k[\text{S}_2\text{O}_8^{2-}][\text{I}^-]\)

(iii) Using Exp 1: \(1.44 \times 10^{-5} = k \times (0.040) \times (0.030)\)
\(k = \frac{1.44 \times 10^{-5}}{0.00120} = 0.0120\text{ dm}^3\text{ mol}^{-1}\text{ s}^{-1}\) (or \(1.20 \times 10^{-2}\text{ dm}^3\text{ mol}^{-1}\text{ s}^{-1}\))
Units: \(\frac{\text{mol dm}^{-3}\text{ s}^{-1}}{(\text{mol dm}^{-3})^2} = \text{dm}^3\text{ mol}^{-1}\text{ s}^{-1}\)

(b) (i) Both reactants (\(\text{S}_2\text{O}_8^{2-}\) and \(\text{I}^-\)) are negatively charged (anions) and strongly repel each other electrostatically. This repulsion results in a high activation energy, meaning very few collisions have energy \(E \ge E_{\text{a}}\).

(ii) Equation 1: \(2\text{Fe}^{3+}(\text{aq}) + 2\text{I}^-(\text{aq}) \rightarrow 2\text{Fe}^{2+}(\text{aq}) + \text{I}_2(\text{aq})\)
Equation 2: \(2\text{Fe}^{2+}(\text{aq}) + \text{S}_2\text{O}_8^{2-}(\text{aq}) \rightarrow 2\text{Fe}^{3+}(\text{aq}) + 2\text{SO}_4^{2-}(\text{aq})\)
(Note: the steps can be written in either order).

(iii) Transition metals can exist in variable oxidation states / readily change oxidation state between \(+3\) and \(+2\).

(a)(i) M1: Explanation for \([\text{S}_2\text{O}_8^{2-}]\): Exp 1 to 2, \([\text{S}_2\text{O}_8^{2-}]\) doubles and rate doubles (while \([\text{I}^-]\) is constant) so order is 1 [1]
M2: Explanation for \([\text{I}^-]\): Exp 2 to 3, \([\text{I}^-]\) triples and rate triples (while \([\text{S}_2\text{O}_8^{2-}]\) is constant) so order is 1 [1]
M3: Stating order wrt \([\text{S}_2\text{O}_8^{2-}] = 1\) and order wrt \([\text{I}^-] = 1\) [1]

(a)(ii) \(\text{rate} = k[\text{S}_2\text{O}_8^{2-}][\text{I}^-]\) (allow ecf from (a)(i)) [1]

(a)(iii) M1: Value \(k = 0.0120\) or \(1.20 \times 10^{-2}\) (allow ecf from (a)(ii)) [1]
M2: Units: \(\text{dm}^3\text{ mol}^{-1}\text{ s}^{-1}\) [1]

(b)(i) M1: Repulsion between ions of like charges / two negative ions / two anions [1]
M2: Results in high activation energy / very few particles possess \(E \ge E_{\text{a}}\) [1]

(b)(ii) M1: \(2\text{Fe}^{3+} + 2\text{I}^- \rightarrow 2\text{Fe}^{2+} + \text{I}_2\) [1]
M2: \(2\text{Fe}^{2+} + \text{S}_2\text{O}_8^{2-} \rightarrow 2\text{Fe}^{3+} + 2\text{SO}_4^{2-}\) [1]
(Equations can be in either order; accept single electron stoichiometric equations)

(b)(iii) Variable oxidation states / ability to change oxidation state [1]

(a) Temperature = \(298\text{ K}\) (or \(25\text{ }^\circ\text{C}\)); Pressure = \(101\text{ kPa}\) (or \(1\text{ atm}\)); Concentration of \(\text{H}^+(\text{aq}) = 1.00\text{ mol dm}^{-3}\).

(b) (i) \(E^\ominus_{\text{cell}} = E^\ominus(\text{cathode}) - E^\ominus(\text{anode}) = +0.80 - (-0.14) = +0.94\text{ V}\)

(ii) \(2\text{Ag}^+(\text{aq}) + \text{Sn}(\text{s}) \rightarrow 2\text{Ag}(\text{s}) + \text{Sn}^{2+}(\text{aq})\)

(iii) \(\Delta G^\ominus = -n F E^\ominus_{\text{cell}}\)
Here, \(n = 2\) electrons transferred per mole of \(\text{Sn}\) oxidised.
\(\Delta G^\ominus = -2 \times (9.65 \times 10^4\text{ C mol}^{-1}) \times (+0.94\text{ V}) = -181420\text{ J mol}^{-1} = -181.42\text{ kJ mol}^{-1}\)
\(\Delta G^\ominus = -181\text{ kJ mol}^{-1}\) (3 sig figs)

(c) (i) For \(\text{Ag}^+/\text{Ag}\), \(z = 1\), \(E^\ominus = +0.80\text{ V}\).
\(E = +0.80 + \frac{0.059}{1} \log_{10}(2.50 \times 10^{-3})\)
\(\log_{10}(2.50 \times 10^{-3}) = -2.602\)
\(E = 0.80 + (0.059 \times -2.602) = 0.80 - 0.1535 = +0.6465\text{ V} \approx +0.647\text{ V}\)

(ii) \(\text{Cl}^-\) ions react with \(\text{Ag}^+\) to precipitate insoluble \(\text{AgCl}(\text{s})\).
This causes \([\text{Ag}^+]\) to decrease significantly. By Le Chatelier's principle, the equilibrium \(\text{Ag}^+(\text{aq}) + \text{e}^- \rightleftharpoons \text{Ag}(\text{s})\) shifts to the left, releasing electrons, making the electrode potential \(E\) less positive / more negative.

(a) M1: \(T = 298\text{ K}\) AND \(p(\text{H}_2) = 101\text{ kPa}\) / \(1\text{ atm}\) [1]
M2: \([\text{H}^+] = 1.00\text{ mol dm}^{-3}\) (or \(0.50\text{ mol dm}^{-3}\text{ H}_2\text{SO}_4\)) [1]

(b)(i) \(+0.94\text{ V}\) (sign required) [1]

(b)(ii) \(2\text{Ag}^+ + \text{Sn} \rightarrow 2\text{Ag} + \text{Sn}^{2+}\) [1]

(b)(iii) M1: Use of \(\Delta G^\ominus = -nFE^\ominus\) with \(n = 2\) and \(F = 9.65 \times 10^4\) [1]
M2: \(-181\text{ kJ mol}^{-1}\) (allow \(-181.4\text{ kJ mol}^{-1}\); negative sign required) [1]

(c)(i) M1: Correct substitution: \(E = 0.80 + \frac{0.059}{1} \log_{10}(2.50 \times 10^{-3})\) [1]
M2: \(E = +0.647\text{ V}\) (allow \(+0.65\text{ V}\) or \(+0.646\text{ V}\)) [1]

(c)(ii) M1: Stating that \(E\) decreases / becomes less positive / shifts more negative [1]
M2: \(\text{Ag}^+\) precipitated as \(\text{AgCl}\), reducing \([\text{Ag}^+]\), shifting equilibrium \(\text{Ag}^+ + \text{e}^- \rightleftharpoons \text{Ag}\) to the left [1]

(a) The ratio of the concentrations of a solute distributed between two immiscible solvent layers at equilibrium at a specified temperature.

(b) (i) Amount of \(\text{NaOH} = 0.250 \times \frac{18.40}{1000} = 4.60 \times 10^{-3}\text{ mol}\)
Amount of \(\text{CH}_3\text{COOH}\) in \(10.0\text{ cm}^3 = 4.60 \times 10^{-3}\text{ mol}\)
\([\text{CH}_3\text{COOH}(\text{aq})] = \frac{4.60 \times 10^{-3}}{10.0 / 1000} = 0.460\text{ mol dm}^{-3}\)

(ii) Initial moles of \(\text{CH}_3\text{COOH} = 0.800 \times \frac{50.0}{1000} = 0.0400\text{ mol}\)
Moles remaining in aqueous layer = \(0.460 \times \frac{50.0}{1000} = 0.0230\text{ mol}\)
Moles extracted into ethoxyethane = \(0.0400 - 0.0230 = 0.0170\text{ mol}\)
\([\text{CH}_3\text{COOH}(\text{ethoxyethane})] = \frac{0.0170}{50.0 / 1000} = 0.340\text{ mol dm}^{-3}\)

(iii) \(K_{\text{partition}} = \frac{0.340}{0.460} = 0.7391 \approx 0.739\)

(c) Extraction 1: \(V_{\text{aq}} = 50.0\text{ cm}^3\), \(V_{\text{org}} = 25.0\text{ cm}^3\).
Initial moles = \(0.0400\text{ mol}\).
Let \(x\) be moles of \(\text{CH}_3\text{COOH}\) remaining in aqueous layer.
Moles in organic layer = \(0.0400 - x\).
\(K_{\text{partition}} = \frac{(0.0400 - x) / 25.0}{x / 50.0} = \frac{2(0.0400 - x)}{x} = 0.7391\)
\(2(0.0400 - x) = 0.7391x \Rightarrow 0.0800 - 2x = 0.7391x \Rightarrow 2.7391x = 0.0800\)
\(x = \frac{0.0800}{2.7391} = 0.02921\text{ mol}\)

Extraction 2: Start with \(0.02921\text{ mol}\) in \(50.0\text{ cm}^3\) aqueous layer with fresh \(25.0\text{ cm}^3\) ethoxyethane.
Let \(y\) be moles remaining in aqueous layer.
\(\frac{2(0.02921 - y)}{y} = 0.7391 \Rightarrow 2.7391y = 2 \times 0.02921 = 0.05842\)
\(y = \frac{0.05842}{2.7391} = 0.02133\text{ mol}\) (or \(y = 0.0400 \times (\frac{2}{2.7391})^2 = 0.0400 \times 0.5332 = 0.02133\text{ mol}\))

Mass of \(\text{CH}_3\text{COOH}\) remaining = \(0.02133\text{ mol} \times 60.0\text{ g mol}^{-1} = 1.280\text{ g}\) (Wait, let's re-verify: \(0.02133 \times 60.0 = 1.28\text{ g}\); with 1 extraction of 50 cm3: moles remaining = 0.0230 mol, mass = 1.38 g. Here with 2 extractions: 1.28 g remaining, so more is extracted!).
Mass = \(1.28\text{ g}\) (or \(0.01568\text{ mol}\) extracted; mass remaining = \(1.28\text{ g}\)).

(d) Each successive extraction establishes a new equilibrium with fresh solvent containing zero initial solute, allowing a greater cumulative amount/percentage of solute to be extracted compared to a single batch equilibrium.

(a) Ratio of the concentrations of a solute in two immiscible solvents at equilibrium at a constant temperature [1]

(b)(i) M1: Moles of \(\text{NaOH} = 4.60 \times 10^{-3}\text{ mol}\) [1]
M2: \([\text{CH}_3\text{COOH}(\text{aq})] = 0.460\text{ (mol dm}^{-3})\) [1]

(b)(ii) M1: Moles in organic layer \(= 0.0400 - 0.0230 = 0.0170\text{ mol}\) [1]
M2: \([\text{CH}_3\text{COOH}(\text{ethoxyethane})] = 0.340\text{ (mol dm}^{-3})\) [1]

(b)(iii) \(K_{\text{partition}} = 0.739\) (allow ecf from (b)(i) and (b)(ii)) [1]

(c) M1: Correct expression for first extraction leading to moles remaining \(= 0.0292\text{ mol}\) [1]
M2: Correct expression for second extraction leading to moles remaining \(= 0.0213\text{ mol}\) [1]
M3: Calculation of mass: \(0.02133 \times 60.0 = 1.28\text{ g}\) (accept 1.28 to 1.30 g depending on intermediate rounding) [1]

(d) Fresh solvent (with no solute) establishes a new equilibrium, extracting an additional quantity of solute / higher total percentage extracted [1]

(a) (i) Reagents for nitration of methylbenzene: concentrated $\text{HNO}_3$ and concentrated $\text{H}_2\text{SO}_4$. Temperature: between $50\,^\circ\text{C}$ and $60\,^\circ\text{C}$ (e.g. $55\,^\circ\text{C}$).
(ii) The $-\text{CH}_3$ group is 2,4-directing and activating; the 4-substituted isomer is 4-nitromethylbenzene (or 1-methyl-4-nitrobenzene / 4-nitrotoluene).

(b) Mechanism:
- Electrophile generation: $\text{HNO}_3 + 2\text{H}_2\text{SO}_4 \rightarrow \text{NO}_2^+ + \text{H}_3\text{O}^+ + 2\text{HSO}_4^-$ (or $\text{HNO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{NO}_2^+ + \text{H}_2\text{O} + \text{HSO}_4^-$).
- Attack on the ring: A curly arrow from the $\pi$-electron ring of methylbenzene to the nitrogen atom of $^+\text{NO}_2$.
- Intermediate: A non-aromatic cyclohexadienyl cation with the positive charge delocalised over the 5 ring carbons (horseshoe spanning $C_2$ to $C_6$, open at $C_4$) with both $-\text{NO}_2$ and $-\text{H}$ bonded to $C_4$.
- Loss of proton: A curly arrow from the $\text{C}-\text{H}$ bond at $C_4$ into the ring system to restore aromaticity, and recovery of catalyst: $\text{H}^+ + \text{HSO}_4^- \rightarrow \text{H}_2\text{SO}_4$.

(c) (i) X is 4-nitrobenzoic acid, $4\text{-}\text{O}_2\text{N-C}_6\text{H}_4\text{-COOH}$.
(ii) Y is 4-aminobenzoic acid, $4\text{-}\text{H}_2\text{N-C}_6\text{H}_4\text{-COOH}$.
(iii) Z is the amide formed by acylation of the amine group: $4\text{-}(\text{CH}_3\text{CONH})\text{-C}_6\text{H}_4\text{-COOH}$.

(d) (i) Nitrous acid, $\text{HNO}_2$ (or $\text{NaNO}_2 + \text{HCl}(\text{aq})$) at a temperature between $0\,^\circ\text{C}$ and $10\,^\circ\text{C}$ (e.g. $< 10\,^\circ\text{C}$ or $5\,^\circ\text{C}$).
(ii) Azo coupling occurs at the para-position to the $-\text{OH}$ group on the phenoxide ring: $\text{HO-C}_6\text{H}_4\text{-N=N-C}_6\text{H}_4\text{-COOH}$ (4-[(4-hydroxyphenyl)diazenyl]benzoic acid).

(a)(i) [M1] conc. $\text{HNO}_3$ AND conc. $\text{H}_2\text{SO}_4$ AND temperature $50\text{--}60\,^\circ\text{C}$ (or reflux at $55\,^\circ\text{C}$).
(a)(ii) [M1] 4-nitromethylbenzene / 1-methyl-4-nitrobenzene / 4-nitrotoluene (accept structural formula $4\text{-}\text{O}_2\text{N-C}_6\text{H}_4\text{CH}_3$).

(b) [M1] Correct balanced equation generating $\text{NO}_2^+$ electrophile.
[M2] Curly arrow from delocalised ring to $\text{NO}_2^+$ and correct intermediate showing horseshoe open at position 4 containing $+$.
[M3] Curly arrow from $\text{C}-\text{H}$ bond back to ring to restore aromaticity, and regeneration of $\text{H}_2\text{SO}_4$.

(c)(i) [M1] Correct structure for X: $4\text{-}\text{O}_2\text{N-C}_6\text{H}_4\text{-COOH}$.
(c)(ii) [M1] Correct structure for Y: $4\text{-}\text{H}_2\text{N-C}_6\text{H}_4\text{-COOH}$.
(c)(iii) [M1] Correct structure for Z: $4\text{-}(\text{CH}_3\text{CONH})\text{-C}_6\text{H}_4\text{-COOH}$ (amide formed on $-\text{NH}_2$, $-\text{COOH}$ intact).

(d)(i) [M1] $\text{NaNO}_2 + \text{HCl}$ / $\text{HNO}_2$ AND $T < 10\,^\circ\text{C}$ / between $0\,^\circ\text{C}$ and $10\,^\circ\text{C}$.
(d)(ii) [M1] Correct azo structure showing $-\text{N}=\text{N}-$ bridge linked to position 4 of the phenol ring and position 4 of the benzoic acid ring: $\text{HO-C}_6\text{H}_4\text{-N=N-C}_6\text{H}_4\text{-COOH}$.

(a) (i) Reagents for nucleophilic addition to an aldehyde to form a hydroxynitrile: $\text{HCN}$ in the presence of $\text{NaCN}$ (or $\text{KCN}$) as catalyst, or $\text{NaCN} + \text{dilute } \text{H}_2\text{SO}_4$, at room temperature.
(ii) Type of reaction: addition (nucleophilic addition).
(iii) Hydrolysis of nitrile: dilute aqueous strong acid, e.g. $\text{HCl}(\text{aq})$ or $\text{H}_2\text{SO}_4(\text{aq})$, heated under reflux.

(b) (i) A chiral centre is an asymmetric carbon atom bonded to four different atoms or groups of atoms (here: $-\text{H}$, $-\text{OH}$, $-\text{CH}_2\text{CH}_3$, and $-\text{COOH}$).
(ii) Central carbon C2 drawn with 3D tetrahedral representation (two standard solid bonds in plane, one wedged bond pointing forward, one dashed bond pointing back) reflected across a vertical mirror plane.

(c) (i) Condensation of the $-\text{OH}$ and $-\text{COOH}$ groups produces a polyester. Repeat unit:
$-\text{O}-\text{CH}(\text{CH}_2\text{CH}_3)-\text{C}(=\text{O})-$
(ii) Two molecules undergo mutual intermolecular esterification between the alcohol of one molecule and the carboxylic acid of the other, forming a 6-membered ring: 3,6-diethyl-1,4-dioxane-2,5-dione (a cyclic dilactone).

(d) (i) $\text{SOCl}_2$ substitutes both the alcoholic $-\text{OH}$ and the carboxylic $-\text{OH}$ with chlorine atoms: $\text{CH}_3\text{CH}_2\text{CH(Cl)COCl}$ (2-chlorobutanoyl chloride).
(ii) Acyl chlorides react rapidly with primary amines at room temperature via addition-elimination to form secondary amides, whereas secondary halogenoalkanes require much higher temperatures/reflux. Thus, under mild room-temperature conditions, only the acyl chloride group undergoes substitution: $\text{CH}_3\text{CH}_2\text{CH(Cl)CONHCH}_2\text{CH}_3$ (2-chloro-N-ethylbutanamide).

(a)(i) [M1] $\text{HCN} + \text{NaCN}$ / $\text{KCN}$ (or $\text{NaCN} + \text{dilute } \text{H}_2\text{SO}_4$) at room temperature / cold.
(a)(ii) [M1] Nucleophilic addition.
(a)(iii) [M1] $\text{HCl}(\text{aq})$ / $\text{H}_2\text{SO}_4(\text{aq})$ AND heat under reflux.

(b)(i) [M1] A carbon atom bonded to four different atoms or groups of atoms.
(b)(ii) [M1] Clear 3D tetrahedral stereochemical drawings (wedges and dashes) for both enantiomers around C2.
[M2] Enantiomers correctly shown as non-superimposable mirror images.

(c)(i) [M1] Repeat unit: $-\text{O}-\text{CH}(\text{CH}_2\text{CH}_3)-\text{CO}-$ (with continuous open valence bonds at each end).
(c)(ii) [M1] Correct skeletal structure of 3,6-diethyl-1,4-dioxane-2,5-dione (6-membered ring with alternating ester units and two ethyl groups).

(d)(i) [M1] $\text{CH}_3\text{CH}_2\text{CH(Cl)COCl}$ (both $-\text{OH}$ converted to $-\text{Cl}$).
(d)(ii) [M1] $\text{CH}_3\text{CH}_2\text{CH(Cl)CONHCH}_2\text{CH}_3$ (selective amidation of acyl chloride group).

(a) (i) Order of increasing basicity: ethanamide < phenylamine < ethylamine.
(ii) In ethylamine, the alkyl (ethyl) group is electron-donating by positive inductive effect ($+\text{I}$), which increases the electron density on the nitrogen atom and makes the lone pair more readily available to accept a proton. In phenylamine, the lone pair of electrons on the nitrogen atom is delocalised into the $\pi$-electron system of the benzene ring, reducing electron density on the nitrogen and making it less available for protonation.
(iii) In ethanamide, the lone pair of electrons on nitrogen is delocalised over the adjacent carbonyl $\text{C}=\text{O}$ system due to the strong electronegativity of the oxygen atom, making the lone pair completely unavailable to accept a proton.

(b) (i) Compound J is 2,4,6-tribromophenylamine ($2,4,6\text{-}\text{Br}_3\text{C}_6\text{H}_2\text{NH}_2$).
(ii) Compound K is the secondary amide formed by reaction with ethanoyl chloride: $N$-(2,4,6-tribromophenyl)ethanamide ($2,4,6\text{-}\text{Br}_3\text{C}_6\text{H}_2\text{NHCOCH}_3$).
(iii) In phenylamine, the lone pair of electrons on the nitrogen atom is delocalised into the benzene ring $\pi$-system, which significantly increases the electron density of the aromatic ring. This activates the ring towards electrophiles and polarises incoming $\text{Br}_2$ molecules more effectively, so no Lewis acid halogen carrier catalyst (such as $\text{FeBr}_3$ or $\text{AlBr}_3$) is required.

(c) (i) Repeat unit of Kevlar: $-\text{NH}-\text{C}_6\text{H}_4-\text{NH}-\text{CO}-\text{C}_6\text{H}_4-\text{CO}-$
(ii) Water, $\text{H}_2\text{O}$.
(iii) The polymer chains are rigid and planar due to the aromatic rings, enabling tight and regular chain packing. This facilitates extensive, strong intermolecular hydrogen bonding between parallel chains, as well as $\pi\text{--}\pi$ aromatic stacking interactions, giving high tensile strength and thermal stability.

(a)(i) [M1] ethanamide < phenylamine < ethylamine (or named in this exact order).
(a)(ii) [M1] Ethylamine: ethyl group is electron-donating / positive inductive effect increases electron density on N lone pair.
[M2] Phenylamine: lone pair on N is delocalised into the benzene ring $\pi$-system, reducing electron density / availability to accept $\text{H}^+$.
(a)(iii) [M1] Lone pair on N is delocalised over the $\text{C}=\text{O}$ / carbonyl group.

(b)(i) [M1] Correct formula/structure for J: 2,4,6-tribromophenylamine ($2,4,6\text{-}\text{Br}_3\text{C}_6\text{H}_2\text{NH}_2$).
(b)(ii) [M1] Correct formula/structure for K: $2,4,6\text{-}\text{Br}_3\text{C}_6\text{H}_2\text{NHCOCH}_3$.
(b)(iii) [M1] Lone pair on N delocalises into the ring $\pi$-cloud AND increases electron density of the ring (activates ring towards electrophilic attack).

(c)(i) [M1] Correct displayed/structural formula for repeat unit: $-\text{NH}-\text{C}_6\text{H}_4-\text{NH}-\text{CO}-\text{C}_6\text{H}_4-\text{CO}-$ (with open bonds at both ends).
(c)(ii) [M1] Water / $\text{H}_2\text{O}$.
(c)(iii) [M1] Rigid / planar polymer chains allow close packing AND extensive / strong intermolecular hydrogen bonding between chains (or $\pi\text{--}\pi$ stacking).

(a)(i) Any two suitable reasons for TMS:
1. It produces a single sharp signal that is well-separated (absorbs at \(\delta = 0\text{ ppm}\)) from almost all organic proton and carbon signals.
2. It is chemically unreactive / inert (does not react with the sample).
3. It is volatile / has a low boiling point, so it is easily removed from the sample after analysis.

(b)(i) The six equivalent protons producing the signal at \(\delta = 1.18\text{ ppm}\) are adjacent to a carbon atom bearing one hydrogen atom (\(n = 1\)). According to the \(n+1\) rule, the peak is split into \(1 + 1 = 2\) peaks (a doublet).

(b)(ii)
- Doublet at \(\delta = 1.18\text{ ppm}\) (6H) and septet at \(\delta = 2.56\text{ ppm}\) (1H) indicate an isopropyl group attached to a carbonyl group: \(-\text{CH}(\text{CH}_3)_2\).
- Singlet at \(\delta = 3.67\text{ ppm}\) (3H) indicates a methoxy methyl group attached directly to an oxygen atom: \(-\text{OCH}_3\).
- Putting the fragments together gives methyl 2-methylpropanoate:
\(\text{CH}_3\text{CH}(\text{CH}_3)\text{COOCH}_3\) or \((\text{CH}_3)_2\text{CHCOOCH}_3\).

(c)(i) There are 4 distinct carbon environments:
1. The two equivalent methyl carbons of the isopropyl group: \(-(\mathbf{C}\text{H}_3)_2\)
2. The methine carbon: \(-\mathbf{C}\text{H}-\)
3. The carbonyl carbon: \(-\mathbf{C}=\text{O}\)
4. The ester methoxy carbon: \(-\text{O}-\mathbf{C}\text{H}_3\)
Thus, 4 peaks are observed.

(c)(ii) Chemical shift range for an ester carbonyl carbon is \(\delta = 160\text{--}185\text{ ppm}\).

(c)(iii) Acid hydrolysis of methyl 2-methylpropanoate yields 2-methylpropanoic acid, \((\text{CH}_3)_2\text{CHCOOH}\), and methanol, \(\text{CH}_3\text{OH}\).

(a)(i) [2 marks]
- Award [1] each for any two of the following:
- Unreactive / chemically inert [1]
- Volatile / low boiling point / easily removed [1]
- Gives a single, sharp resonance peak [1]
- Absorbs at higher field / at \(\delta = 0\text{ ppm}\) / outside the range of most organic protons/carbons [1]
- Non-toxic / soluble in organic solvents [1]

(b)(i) [1 mark]
- Neighbouring / adjacent carbon has 1 proton (and splitting is \(n+1 = 1+1 = 2\)) [1]

(b)(ii) [2 marks]
- Identifies the \((\text{CH}_3)_2\text{CH}-\) / isopropyl group and the \(-\text{OCH}_3\) / methoxy group [1]
- Fully correct structure of methyl 2-methylpropanoate, \((\text{CH}_3)_2\text{CHCOOCH}_3\) / \(\text{CH}_3\text{CH}(\text{CH}_3)\text{COOCH}_3\) (displayed, structural, or skeletal) [1]

(c)(i) [1 mark]
- 4 (peaks) [1]

(c)(ii) [1 mark]
- 160–185 (ppm) (accept any range completely within 155–190 ppm) [1]

(c)(iii) [1 mark]
- 2-methylpropanoic acid / \((\text{CH}_3)_2\text{CHCOOH}\) AND methanol / \(\text{CH}_3\text{OH}\) (both required for [1])

(a)(i)
1. Calculate the amount in moles of 4-ethanoyloxybenzoic acid required:
$$n = \frac{5.40\text{ g}}{180.0\text{ g mol}^{-1}} = 0.0300\text{ mol}$$
2. Account for the $75.0\%$ yield:
$$\text{Theoretical moles needed} = \frac{0.0300}{0.750} = 0.0400\text{ mol}$$
3. Calculate the required mass of 4-hydroxybenzoic acid:
$$\text{Mass} = 0.0400\text{ mol} \times 138.0\text{ g mol}^{-1} = 5.52\text{ g}$$

(a)(ii)
* Hazard: Corrosive / causes severe skin and eye burns OR reacts violently with moisture/water releasing toxic/irritant acidic fumes of hydrogen chloride ($\text{HCl}$).
* Precaution: Carry out the reaction/dispensing in a fume cupboard / fume hood OR ensure all glassware is completely dry / use a drying tube.

(a)(iii)
1. Dissolve the crude solid in the minimum volume of hot/boiling solvent.
2. Carry out hot filtration (using a preheated funnel/fluted filter paper) to remove insoluble impurities.
3. Allow the solution to cool slowly to room temperature / place in an ice bath to crystallise.
4. Separate crystals by filtration under reduced pressure (Büchner funnel), wash with a small volume of ice-cold solvent, and dry in a desiccator / oven below its melting point.

(b)(i)
* Concordant titres: Titrations 1, 2, and 3 (all within $\pm 0.10\text{ cm}^3$ of each other: $20.65\text{ cm}^3$, $20.55\text{ cm}^3$, $20.60\text{ cm}^3$).
* Mean titre:
$$\text{Mean titre} = \frac{20.65 + 20.55 + 20.60}{3} = 20.60\text{ cm}^3$$

(b)(ii)
1. Moles of $\text{NaOH}$ in mean titre:
$$n(\text{NaOH}) = 20.60 \times 10^{-3}\text{ dm}^3 \times 0.0380\text{ mol dm}^{-3} = 7.828 \times 10^{-4}\text{ mol}$$
2. Moles of 4-ethanoyloxybenzoic acid in $25.00\text{ cm}^3$ pipette sample $= 7.828 \times 10^{-4}\text{ mol}$
3. Moles in the total $250.0\text{ cm}^3$ volumetric flask:
$$n(\text{total}) = 7.828 \times 10^{-4} \times 10 = 7.828 \times 10^{-3}\text{ mol}$$
4. Mass of pure 4-ethanoyloxybenzoic acid:
$$m = 7.828 \times 10^{-3}\text{ mol} \times 180.0\text{ g mol}^{-1} = 1.4090\text{ g}$$
5. Percentage purity:
$$\%\text{ purity} = \frac{1.4090\text{ g}}{1.440\text{ g}} \times 100 = 97.85\% \approx 97.9\%$$

(c)(i)
* A titre requires two burette readings, each with an uncertainty of $\pm 0.05\text{ cm}^3$, giving a total uncertainty of $\pm 0.10\text{ cm}^3$.
$$\%\text{ uncertainty} = \frac{0.10}{20.60} \times 100 = 0.485\%\text{ (or } 0.49\%\text{)}$$

(c)(ii)
* Effect: The calculated percentage purity would be lower / underestimated.
* Explanation: The weighed solid contains residual water/solvent, meaning the mass of actual 4-ethanoyloxybenzoic acid present is less than $1.440\text{ g}$, which requires a smaller titre of $\text{NaOH}$ and hence yields a smaller calculated mass of pure product.

(a)(i) [2 marks]
* M1: Calculation of theoretical moles needed: $\frac{5.40}{180.0} \div 0.750 = 0.0400\text{ mol}$ [1]
* M2: Final mass: $0.0400 \times 138.0 = 5.52\text{ g}$ [1]

(a)(ii) [2 marks]
* M1: Hazard: Corrosive / causes burns / toxic/acidic fumes of $\text{HCl}$ / violent reaction with water [1]
* M2: Precaution: Use a fume cupboard / ensure apparatus is dry / use calcium chloride drying tube [1]
(Reject: wearing gloves/goggles/lab coat as given in question)

(a)(iii) [4 marks]
* M1: Dissolve the solid in the minimum volume of hot/boiling solvent [1]
* M2: Hot filtration to remove insoluble impurities [1]
* M3: Cool solution to allow recrystallisation [1]
* M4: Filter using Büchner funnel / under reduced pressure, wash with cold solvent AND dry [1]

(b)(i) [2 marks]
* M1: Selection of titrations 1, 2, and 3 as concordant (within $0.10\text{ cm}^3$) [1]
* M2: Correct calculation of mean titre: $20.60\text{ cm}^3$ [1]

(b)(ii) [3 marks]
* M1: Moles of $\text{NaOH}$ in mean titre $= 7.828 \times 10^{-4}\text{ mol}$ AND moles in $250.0\text{ cm}^3 = 7.828 \times 10^{-3}\text{ mol}$ [1]
* M2: Mass of pure acid $= 7.828 \times 10^{-3} \times 180.0 = 1.409\text{ g}$ [1]
* M3: Percentage purity $= 97.9\%$ (accept $97.8\%$ to $97.9\%$, must be 3 sig figs) [1]

(c)(i) [1 mark]
* M1: $\frac{2 \times 0.05}{20.60} \times 100 = 0.485\%$ or $0.49\%$ [1]

(c)(ii) [2 marks]
* M1: State: Calculated purity is lower / decreases [1]
* M2: Explain: Mass recorded contains water/solvent, so actual amount of acid present is lower, leading to smaller volume/moles of $\text{NaOH}$ reacting [1]

(a) (i) To avoid any loss of nitrogen gas upon mixing, the solution containing the benzenediazonium chloride and water/acid can be kept separate before sealing the system. This is done by placing one component in a small sample/ignition tube upright inside the conical flask containing the other component (or using a dropping funnel / side-arm divided flask). The bung connected to the gas syringe is tightly fitted, and the flask is then tilted or shaken to mix the reactants and simultaneously start the stopclock.

(ii) Benzenediazonium chloride and phenol are toxic and corrosive to skin; precautions include wearing chemical-resistant gloves, eye protection (goggles), and carrying out operations in a well-ventilated fume cupboard.

(iii) Temperature significantly affects reaction rate via the rate constant \(k\). It must be kept constant at \(35.0\,^\circ\text{C}\) using a thermostatically controlled water bath.

(b) (i)
For Experiment 3:
$$\log_{10}(0.120) = -0.9208... \approx -0.92$$
$$\log_{10}(0.302) = -0.5199... \approx -0.52$$

For Experiment 5:
$$\log_{10}(0.200) = -0.6989... \approx -0.70$$
$$\log_{10}(0.501) = -0.3001... \approx -0.30$$

(ii) Comparing \(\log_{10}(R_0) = n\log_{10}[\text{C}_6\text{H}_5\text{N}_2\text{Cl}] + \log_{10}(k)\) with \(y = mx + c\):
- Gradient \(m = n\) (the order of reaction with respect to benzenediazonium chloride).
- \(y\)-intercept \(c = \log_{10}(k)\).

(iii)
$$\text{Gradient} = \frac{\Delta y}{\Delta x} = \frac{-0.30 - (-1.00)}{-0.70 - (-1.40)} = \frac{+0.70}{+0.70} = 1.00$$
Therefore, the order of reaction \(n = 1\).

(iv)
Using \(R_0 = k[\text{C}_6\text{H}_5\text{N}_2\text{Cl}]^1\) and Experiment 1 data:
$$k = \frac{R_0}{[\text{C}_6\text{H}_5\text{N}_2\text{Cl}]} = \frac{0.100\,\text{cm}^3\,\text{s}^{-1}}{0.040\,\text{mol}\,\text{dm}^{-3}} = 2.50\,\text{cm}^3\,\text{s}^{-1}\,\text{mol}^{-1}\,\text{dm}^3$$
Alternatively, if expressing rate in \(\text{mol}\,\text{dm}^{-3}\,\text{s}^{-1}\), units simplify to \(\text{s}^{-1}\).

(c) (i) Plunger of the gas syringe sticking due to friction, or a slight gas leak around the delivery tube/bung.

(ii)
1. Use a digital gas pressure sensor connected to a data logger to continuously and automatically log gas production with high temporal resolution.
2. Carry out repeat trials at each concentration to identify anomalies and calculate a mean initial rate.

(a)(i) [2 marks]
- M1: Placing one reactant in an ignition/small test tube inside the conical flask OR using a dropping funnel / divided flask.
- M2: Inserting the stopper/bung firmly BEFORE mixing/swirling the contents.

(a)(ii) [1 mark]
- M1: Identifies toxicity/corrosiveness of phenol / diazonium compound AND states wearing gloves / using a fume cupboard / eye protection.

(a)(iii) [1 mark]
- M1: Temperature identified AND maintained with a thermostatically controlled water bath.

(b)(i) [2 marks]
- M1: Correct values calculated for Experiment 3 (-0.92 and -0.52).
- M2: Correct values calculated for Experiment 5 (-0.70 and -0.30) AND all values recorded consistently to 2 decimal places.

(b)(ii) [1 mark]
- M1: Gradient = order of reaction (or n) AND y-intercept = \(\log_{10}(k)\).

(b)(iii) [2 marks]
- M1: Correct working for gradient showing \(\frac{\Delta y}{\Delta x}\) giving \(1.00\) (accept range \(0.95 - 1.05\)).
- M2: Order stated as 1 (or first order) consistent with calculated gradient.

(b)(iv) [2 marks]
- M1: Correct numerical calculation of \(k = 2.50\) (allow ecf from (b)(iii)).
- M2: Correct units: \(\text{cm}^3\,\text{mol}^{-1}\,\text{dm}^3\,\text{s}^{-1}\) OR \(\text{s}^{-1}\).

(c)(i) [1 mark]
- M1: Syringe plunger sticking / friction in syringe barrel OR gas leak / loose stopper.

(c)(ii) [2 marks]
- M1: Use of a data logger / gas pressure sensor / video recording with timer to reduce human reaction time error.
- M2: Perform repeats at each concentration and calculate an average/mean rate to minimise random errors.