2026 Cambridge IAL Chemistry (9701) 模拟试题及答案详解

1. Calculate the mass and moles of carbon:
\(\text{Moles of } \text{C} = \frac{0.396\text{ g}}{44.0\text{ g mol}^{-1}} = 0.00900\text{ mol}\)
\(\text{Mass of } \text{C} = 0.00900\text{ mol} \times 12.0\text{ g mol}^{-1} = 0.108\text{ g}\)

2. Calculate the mass and moles of hydrogen:
\(\text{Moles of } \text{H}_2\text{O} = \frac{0.216\text{ g}}{18.0\text{ g mol}^{-1}} = 0.0120\text{ mol}\)
\(\text{Moles of } \text{H} = 2 \times 0.0120 = 0.0240\text{ mol}\)
\(\text{Mass of } \text{H} = 0.0240\text{ mol} \times 1.0\text{ g mol}^{-1} = 0.0240\text{ g}\)

3. Calculate the mass and moles of oxygen:
\(\text{Mass of } \text{O} = 0.180\text{ g} - (0.108\text{ g} + 0.0240\text{ g}) = 0.0480\text{ g}\)
\(\text{Moles of } \text{O} = \frac{0.0480\text{ g}}{16.0\text{ g mol}^{-1}} = 0.00300\text{ mol}\)

4. Determine the simplest whole-number ratio of moles:
\(\text{C} : \text{H} : \text{O} = \frac{0.00900}{0.00300} : \frac{0.0240}{0.00300} : \frac{0.00300}{0.00300} = 3 : 8 : 1\)

Therefore, the empirical formula is \(\text{C}_3\text{H}_8\text{O}\).

A is correct: \(\text{C}_3\text{H}_8\text{O}\) [1]
B is incorrect: assumes missing oxygen mass corresponds to a different stoichiometry without accounting for hydrogen correctly.
C is incorrect: represents \(\text{C}_2\text{H}_6\text{O}\) from miscalculating the mass of carbon.
D is incorrect: incorrectly calculates the molar ratios.

To determine bond angles, consider the electron pair repulsion theory:
- \(\text{NH}_4^+\) has 4 bonding pairs and 0 lone pairs around the central N atom; tetrahedral shape, bond angle \(= 109.5^\circ\).
- \(\text{H}_3\text{O}^+\) has 3 bonding pairs and 1 lone pair around the central O atom; trigonal pyramidal shape, bond angle \(\approx 107^\circ\) (due to lone pair repulsion reducing the ideal tetrahedral angle by \(\approx 2.5^\circ\)).
- \(\text{BF}_3\) has 3 bonding pairs and 0 lone pairs; trigonal planar shape, bond angle \(= 120^\circ\).
- \(\text{H}_2\text{O}\) has 2 bonding pairs and 2 lone pairs; non-linear (bent) shape, bond angle \(= 104.5^\circ\).

B is correct: \(\text{H}_3\text{O}^+\) has 3 bonding pairs and 1 lone pair giving a trigonal pyramidal shape with a bond angle of \(\approx 107^\circ\) [1]
A is incorrect: \(\text{NH}_4^+\) has a tetrahedral angle of \(109.5^\circ\).
C is incorrect: \(\text{BF}_3\) is trigonal planar with an angle of \(120^\circ\).
D is incorrect: \(\text{H}_2\text{O}\) has two lone pairs reducing the angle to \(104.5^\circ\).

Using Hess's Law:
\(\Delta H^\circ_\text{r} = \sum \Delta H^\circ_\text{f}(\text{products}) - \sum \Delta H^\circ_\text{f}(\text{reactants})\)

Note that \(\Delta H^\circ_\text{f}[\text{Fe}(\text{s})] = 0\text{ kJ mol}^{-1}\) as it is an element in its standard state.

\(\Delta H^\circ_\text{r} = [2(0) + 3(-393.5)] - [-824.2 + 3(-110.5)]\)
\(\Delta H^\circ_\text{r} = [-1180.5] - [-824.2 - 331.5]\)
\(\Delta H^\circ_\text{r} = -1180.5 - (-1155.7) = -24.8\text{ kJ mol}^{-1}\).

A is correct: \(-24.8\text{ kJ mol}^{-1}\) [1]
B is incorrect: inverted the sign (reactants - products).
C is incorrect: failed to multiply the enthalpy of formation of CO and \(\text{CO}_2\) by 3.
D is incorrect: summed all values without subtracting reactants from products.

- \(\text{NaCl}\) (ionic chloride) dissolves in water without hydrolysis to give hydrated \(\text{Na}^+\) and \(\text{Cl}^-\) ions, resulting in a neutral solution (\(\text{pH} \approx 7\)).
- \(\text{SiCl}_4\) (covalent chloride) undergoes rapid and complete hydrolysis with water:
$$\text{SiCl}_4(\text{l}) + 2\text{H}_2\text{O}(\text{l}) \rightarrow \text{SiO}_2(\text{s}) + 4\text{HCl}(\text{g}/\text{aq})$$
This reaction releases steamy white fumes of \(\text{HCl}\) and forms a strongly acidic solution (\(\text{pH} \approx 1-2\)).
Therefore, \(\mathbf{X}\) is \(\text{NaCl}\) and \(\mathbf{Y}\) is \(\text{SiCl}_4\).

A is correct: \(\mathbf{X} = \text{NaCl}\) (dissolves to neutral solution), \(\mathbf{Y} = \text{SiCl}_4\) (hydrolyses vigorously to give strongly acidic solution) [1]
B is incorrect: \(\text{MgCl}_2\) forms a slightly acidic solution (\(\text{pH} \approx 6.5\)) and does not match the reversed order.
C is incorrect: \(\text{AlCl}_3\) hydrolyses to give an acidic solution (\(\text{pH} \approx 3\)), not neutral.
D is incorrect: \(\text{SiCl}_4\) does not dissolve neutrally.

In the reaction between solid \(\text{NaBr}\) and concentrated \(\text{H}_2\text{SO}_4\):
1. An acid-base reaction occurs: \(\text{NaBr} + \text{H}_2\text{SO}_4 \rightarrow \text{NaHSO}_4 + \text{HBr}\)
2. Bromide ions are sufficiently strong reducing agents to reduce sulfuric acid (sulfur oxidation state drops from \(+6\) to \(+4\)):
$$2\text{HBr} + \text{H}_2\text{SO}_4 \rightarrow \text{Br}_2 + \text{SO}_2 + 2\text{H}_2\text{O}$$
Thus, \(\text{SO}_2\) gas is evolved as a reduction product of \(\text{H}_2\text{SO}_4\).
- Bromide ions act as reducing agents (not oxidising agents).
- \(\text{HBr}\), \(\text{SO}_2\), and \(\text{Br}_2\) vapour are produced, so \(\text{HBr}\) is not the only gaseous product.
- Yellow solid sulfur and \(\text{H}_2\text{S}\) are only formed with iodide ions, which are stronger reducing agents.

B is correct: \(\text{SO}_2\) gas is formed by reduction of \(\text{H}_2\text{SO}_4\) [1]
A is incorrect: bromide ions are reducing agents, not oxidising agents.
C is incorrect: \(\text{SO}_2\) and \(\text{Br}_2\) are also gaseous products.
D is incorrect: yellow solid sulfur is formed with iodide, not bromide.

The rate of hydrolysis of halogenoalkanes is determined by:
1. Carbon-halogen bond strength: The \(\text{C}-\text{I}\) bond enthalpy (\(240\text{ kJ mol}^{-1}\)) is much lower than the \(\text{C}-\text{Cl}\) bond enthalpy (\(340\text{ kJ mol}^{-1}\)), making iodoalkanes hydrolyse much faster than chloroalkanes.
2. Mechanism and stability of the intermediate: Tertiary halogenoalkanes hydrolyse predominantly via the \(\text{S}_\text{N}1\) mechanism, which involves the rapid formation of a stable tertiary carbocation intermediate, proceeding substantially faster than the \(\text{S}_\text{N}2\) mechanism of primary halogenoalkanes.

Therefore, 2-iodo-2-methylpropane (a tertiary iodoalkane) reacts fastest and produces a precipitate of silver iodide in the shortest time.

D is correct: 2-iodo-2-methylpropane combines the weakest C-X bond (C-I) with the fastest mechanism (\(\text{S}_\text{N}1\) via tertiary carbocation) [1]
A is incorrect: 1-chlorobutane is primary and has a strong C-Cl bond.
B is incorrect: 1-iodobutane is primary (slow \(\text{S}_\text{N}2\) pathway).
C is incorrect: 2-chloro-2-methylpropane has a strong C-Cl bond.

Analysing each test:
1. Orange precipitate with 2,4-DNPH indicates the presence of a carbonyl group (aldehyde or ketone).
2. No reaction (no red precipitate) with Fehling's solution indicates \(\mathbf{Z}\) is a ketone, not an aldehyde (eliminating butanal).
3. Pale yellow precipitate with alkaline aqueous iodine (tri-iodomethane / iodoform test) indicates the presence of a methyl carbonyl group, \(\text{CH}_3\text{C}(=\text{O})-\).

Evaluating the options:
- Butanal (\(\text{CH}_3\text{CH}_2\text{CH}_2\text{CHO}\)) is an aldehyde and reacts with Fehling's solution.
- Butan-2-one (\(\text{CH}_3\text{COCH}_2\text{CH}_3\)) is a methyl ketone; gives positive 2,4-DNPH, negative Fehling's, and positive tri-iodomethane test.
- Butan-2-ol (\(\text{CH}_3\text{CH(OH)CH}_2\text{CH}_3\)) does not react with 2,4-DNPH.
- Pentan-3-one (\(\text{CH}_3\text{CH}_2\text{COCH}_2\text{CH}_3\)) is a ketone but not a methyl ketone, so it gives a negative tri-iodomethane test.

B is correct: butan-2-one is a methyl ketone which gives an orange precipitate with 2,4-DNPH, does not react with Fehling's, and gives a positive tri-iodomethane test [1]
A is incorrect: butanal gives a positive Fehling's test.
C is incorrect: butan-2-ol does not react with 2,4-DNPH.
D is incorrect: pentan-3-one gives a negative tri-iodomethane test.

The number of carbon atoms, \(n\), in an organic molecule can be calculated from the relative abundances of the \(\text{M}^+\) and \([\text{M}+1]^+\) peaks using the formula:

$$n = \frac{100 \times \text{abundance of } [\text{M}+1]^+}{1.1 \times \text{abundance of } \text{M}^+}$$

Substitute the given values:
$$n = \frac{100 \times 1.60}{1.1 \times 36.4} = \frac{160}{40.04} \approx 4.00$$

Therefore, there are 4 carbon atoms in the molecule.

B is correct: \(n = \frac{100 \times 1.60}{1.1 \times 36.4} \approx 4\) [1]
A is incorrect: calculation error resulting from inverted formula or incorrect factor.
C is incorrect: arithmetic mistake.
D is incorrect: calculated without dividing by 1.1.

The combustion equation is:
\[ \text{C}_x\text{H}_y(\text{g}) + \left(x + \frac{y}{4}\right)\text{O}_2(\text{g}) \rightarrow x\text{CO}_2(\text{g}) + \frac{y}{2}\text{H}_2\text{O}(\text{l}) \]

1. At room temperature, water is liquid and has negligible volume.
2. The remaining gas after cooling consists of unreacted \(\text{O}_2\) and produced \(\text{CO}_2\).
3. Aqueous \(\text{NaOH}\) absorbs \(\text{CO}_2\), so:
- \(V(\text{unreacted } \text{O}_2) = 25\text{ cm}^3\)
- \(V(\text{CO}_2) = 70 - 25 = 45\text{ cm}^3\)
4. From the mole/volume ratio:
\[ x = \frac{V(\text{CO}_2)}{V(\text{C}_x\text{H}_y)} = \frac{45}{15} = 3 \]
5. Volume of \(\text{O}_2\) reacted:
\[ V(\text{O}_2\text{ reacted}) = 100 - 25 = 75\text{ cm}^3 \]
\[ \left(x + \frac{y}{4}\right) \times 15 = 75 \implies 3 + \frac{y}{4} = 5 \implies y = 8 \]

Therefore, the formula is \(\text{C}_3\text{H}_8\).

C [1 mark]
- Deduces volume of \(\text{CO}_2 = 45\text{ cm}^3\) and volume of \(\text{O}_2\) reacted \(= 75\text{ cm}^3\).
- Deduces \(x = 3\) and \(y = 8\).

- \(\text{H}_2\text{O}\): 2 bonding pairs, 2 lone pairs; non-linear (bent) shape with a bond angle of \(104.5^\circ\).
- \(\text{NH}_3\): 3 bonding pairs, 1 lone pair; trigonal pyramidal shape with a bond angle of \(107^\circ\).
- \(\text{NH}_4^+\): 4 bonding pairs, 0 lone pairs; regular tetrahedral shape with a bond angle of \(109.5^\circ\).
- \(\text{CO}_2\): 2 regions of electron density, 0 lone pairs; linear shape with a bond angle of \(180^\circ\).

Increasing order of bond angle: \(\text{H}_2\text{O} < \text{NH}_3 < \text{NH}_4^+ < \text{CO}_2\).

A [1 mark]
- Identifies correct bond angles based on VSEPR theory.

The equation for the standard enthalpy change of formation of ethanol is:
\[ 2\text{C}(\text{s}) + 3\text{H}_2(\text{g}) + \frac{1}{2}\text{O}_2(\text{g}) \rightarrow \text{C}_2\text{H}_5\text{OH}(\text{l}) \]

Using Hess's Law with standard enthalpy changes of combustion:
\[ \Delta H^\circ_{\text{f}} = \sum \Delta H^\circ_{\text{c}}(\text{reactants}) - \sum \Delta H^\circ_{\text{c}}(\text{products}) \]
\[ \Delta H^\circ_{\text{f}} = [2 \times (-394) + 3 \times (-286)] - [-1368] \]
\[ \Delta H^\circ_{\text{f}} = [-788 - 858] + 1368 = -1646 + 1368 = -278\text{ kJ mol}^{-1} \]

B [1 mark]
- Correct application of Hess's Law: \(2(-394) + 3(-286) - (-1368) = -278\text{ kJ mol}^{-1}\).

1. Effect of increasing temperature (from \(T_1\) to \(T_2\)):
- The average kinetic energy of the molecules increases.
- The peak of the distribution (the most probable energy) shifts to the right (higher energy).
- Because the total number of molecules (area under the curve) remains constant, the peak height decreases.
2. Effect of adding a catalyst:
- A catalyst provides an alternative reaction pathway with a lower activation energy, so \(E_{\text{a}}\) decreases.

A [1 mark]
- Recognises peak shifts right and down at higher temperature.
- Recognises catalyst decreases activation energy.

- \(\text{AlCl}_3\) is a covalent chloride that hydrolyses vigorously in water to form \(\text{HCl}\) fumes and \([\text{Al}(\text{H}_2\text{O})_6]^{3+}(\text{aq})\), giving an acidic solution with a pH around 2–3.
- Addition of \(\text{NaOH}(\text{aq})\) precipitates amphoteric aluminium hydroxide, \(\text{Al}(\text{OH})_3(\text{s})\).
- \(\text{Al}(\text{OH})_3\) dissolves in excess \(\text{NaOH}(\text{aq})\) to form the soluble tetrahydroxoaluminate ion, \([\text{Al}(\text{OH})_4]^-(\text{aq})\).
- Magnesium hydroxide does not dissolve in excess \(\text{NaOH}\).
- Silicon and phosphorus chlorides do not form a precipitate that redissolves in this manner.

B [1 mark]
- Deduces aluminium from the acidity of the hydrolysed chloride and the amphoteric behaviour of its hydroxide.

In the atmosphere, nitrogen monoxide (\(\text{NO}\)) is oxidised by oxygen to nitrogen dioxide (\(\text{NO}_2\)):
\[ 2\text{NO}(\text{g}) + \text{O}_2(\text{g}) \rightarrow 2\text{NO}_2(\text{g}) \]

Nitrogen dioxide then oxidises sulfur dioxide to sulfur trioxide, regenerating nitrogen monoxide:
\[ \text{NO}_2(\text{g}) + \text{SO}_2(\text{g}) \rightarrow \text{SO}_3(\text{g}) + \text{NO}(\text{g}) \]

The overall reaction is:
\[ 2\text{SO}_2(\text{g}) + \text{O}_2(\text{g}) \rightarrow 2\text{SO}_3(\text{g}) \]

A [1 mark]
- Recognises the two-step cycle where \(\text{NO}\) is oxidised to \(\text{NO}_2\) and then regenerated.

2-bromo-2-methylpropane is a tertiary halogenoalkane and reacts via an \(\text{S}_\text{N}1\) mechanism:
- Step 1 (rate-determining): The \(\text{C}-\text{Br}\) bond undergoes heterolytic fission to form a stable tertiary carbocation, \((\text{CH}_3)_3\text{C}^+\), and a bromide ion, \(\text{Br}^-\).
- Step 2 (fast): The nucleophile (\(\text{OH}^-\)) attacks the planar carbocation.

Option A describes \(\text{S}_\text{N}2\).
Option B describes second-order kinetics of \(\text{S}_\text{N}2\).
Option D describes free-radical reactions.

C [1 mark]
- Recognises tertiary halogenoalkanes undergo \(\text{S}_\text{N}1\) substitution with rate-determining heterolytic fission of the C-Br bond.

1. Positive test with 2,4-DNPH (orange precipitate) indicates the presence of a carbonyl group (\(\text{C}=\text{O}\), aldehyde or ketone).
2. Positive test with Tollens' reagent (silver mirror) indicates that \(Y\) is an aldehyde.
3. Negative tri-iodomethane test (no yellow precipitate of \(\text{CHI}_3\) with alkaline \(\text{I}_2\)) indicates the absence of a \(\text{CH}_3\text{C}=\text{O}\) (methyl carbonyl) group.

- Propanone is a ketone (negative Tollens').
- Ethanal (\(\text{CH}_3\text{CHO}\)) gives a positive tri-iodomethane test.
- Propan-2-ol is an alcohol (negative 2,4-DNPH).
- Propanal (\(\text{CH}_3\text{CH}_2\text{CHO}\)) is an aldehyde without a \(\text{CH}_3\text{C}=\text{O}\) adjacent group, thus fitting all observations.

B [1 mark]
- Deduces aldehyde from Tollens' reagent and eliminates ethanal via the negative tri-iodomethane test.

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

1. Convert all quantities to SI units:
\(p = 1.00 \times 10^5\text{ Pa}\)
\(V = 38.0 \times 10^{-6}\text{ m}^3\)
\(T = 373\text{ K}\)
\(R = 8.31\text{ J K}^{-1}\text{ mol}^{-1}\)

2. Calculate the number of moles of vapour, \(n\):
\(n = \frac{pV}{RT} = \frac{1.00 \times 10^5 \times 38.0 \times 10^{-6}}{8.31 \times 373} = \frac{3.80}{3099.63} \approx 1.226 \times 10^{-3}\text{ mol}\)

3. Calculate the relative molecular mass, \(M_\text{r}\):
\(M_\text{r} = \frac{\text{mass}}{n} = \frac{0.108\text{ g}}{1.226 \times 10^{-3}\text{ mol}} \approx 88.1\text{ g mol}^{-1}\)

4. Compare with the relative molecular masses of the options:
- \(\text{HCOOCH}_2\text{CH}_3\): \(M_\text{r} = 1.0 + 12.0 + 32.0 + 29.0 = 74.0\)
- \(\text{CH}_3\text{COOCH}_2\text{CH}_3\): \(M_\text{r} = 15.0 + 12.0 + 32.0 + 29.0 = 88.0\)
- \(\text{CH}_3\text{COOCH}_3\): \(M_\text{r} = 15.0 + 12.0 + 32.0 + 15.0 = 74.0\)
- \(\text{CH}_3\text{CH}_2\text{COOCH}_2\text{CH}_3\): \(M_\text{r} = 29.0 + 12.0 + 32.0 + 29.0 = 102.0\)

Therefore, the compound is ethyl ethanoate, \(\text{CH}_3\text{COOCH}_2\text{CH}_3\).

B [1]
1 mark for identifying ethyl ethanoate (B) through correct calculation of Mr ≈ 88.

To determine bond angles, deduce the number of bonding electron pairs and lone pairs around the central atom:

- In \(\text{BF}_3\), boron has 3 bonding pairs and 0 lone pairs. Shape is trigonal planar with a bond angle of \(120^\circ\).
- In \(\text{NH}_4^+\), nitrogen has 4 bonding pairs and 0 lone pairs. Shape is tetrahedral with a bond angle of \(109.5^\circ\).
- In \(\text{NH}_3\), nitrogen has 3 bonding pairs and 1 lone pair. Shape is trigonal pyramidal with a bond angle of \(107^\circ\).
- In \(\text{NH}_2^-\), nitrogen has 2 bonding pairs and 2 lone pairs. Shape is non-linear (bent) with a bond angle of \(104.5^\circ\).

Because lone pair–lone pair repulsion is greater than lone pair–bonding pair and bonding pair–bonding pair repulsion, \(\text{NH}_2^-\), with two lone pairs, has the smallest bond angle (\(104.5^\circ\)).

C [1]
1 mark for identifying NH2- (C).

The equation representing the standard enthalpy change of formation of propane is:
\[3\text{C(graphite)} + 4\text{H}_2\text{(g)} \rightarrow \text{C}_3\text{H}_8\text{(g)}\]

Using the relationship \(\Delta H_\text{r}^\ominus = \sum \Delta H_\text{c}^\ominus\text{(reactants)} - \sum \Delta H_\text{c}^\ominus\text{(products)}\):
\[\Delta H_\text{f}^\ominus(\text{C}_3\text{H}_8) = [3 \times \Delta H_\text{c}^\ominus(\text{C}) + 4 \times \Delta H_\text{c}^\ominus(\text{H}_2)] - [\Delta H_\text{c}^\ominus(\text{C}_3\text{H}_8)]\]
\[\Delta H_\text{f}^\ominus(\text{C}_3\text{H}_8) = [3(-394) + 4(-286)] - [-2220]\]
\[\Delta H_\text{f}^\ominus(\text{C}_3\text{H}_8) = [-1182 - 1144] + 2220\]
\[\Delta H_\text{f}^\ominus(\text{C}_3\text{H}_8) = -2326 + 2220 = -106\text{ kJ mol}^{-1}\]

A [1]
1 mark for calculating -106 kJ mol-1 (A).

- Aluminium chloride (\(\text{AlCl}_3\)) reacts vigorously with water giving acidic fumes, but aluminium oxide (\(\text{Al}_2\text{O}_3\)) is insoluble in water.
- Magnesium chloride dissolves without vigorous hydrolysis and magnesium oxide is sparingly soluble, forming a weakly alkaline suspension.
- Silicon tetrachloride (\(\text{SiCl}_4\)) reacts vigorously with water, but silicon dioxide (\(\text{SiO}_2\)) is insoluble in water.
- Phosphorus pentachloride (\(\text{PCl}_5\)) reacts vigorously with water releasing steamy fumes of \(\text{HCl}\), and phosphorus(V) oxide (\(\text{P}_4\text{O}_{10}\)) dissolves vigorously in water to form phosphoric(V) acid, \(\text{H}_3\text{PO}_4\) (an acidic solution).

Therefore, the element is phosphorus.

C [1]
1 mark for selecting Phosphorus (C).

1. Initially, an acid–base reaction occurs:
\(\text{NaBr} + \text{H}_2\text{SO}_4 \rightarrow \text{NaHSO}_4 + \text{HBr}\)
This produces steamy fumes of \(\text{HBr}\).

2. The bromide ions (\(\text{Br}^-\)) are strong enough reducing agents to reduce concentrated sulfuric acid:
\(2\text{HBr} + \text{H}_2\text{SO}_4 \rightarrow \text{Br}_2 + \text{SO}_2 + 2\text{H}_2\text{O}\)
Bromine (\(\text{Br}_2\)) is the reddish-brown vapour.

In this redox reaction, sulfuric acid oxidises bromide ions (oxidation state \(-1\)) to elemental bromine (oxidation state \(0\)), while sulfur is reduced from \(+6\) to \(+4\). Thus, concentrated sulfuric acid acts as an oxidising agent.

B [1]
1 mark for identifying sulfuric acid as an oxidising agent (B).

- Adding a catalyst lowers the activation energy \(E_\text{a}\) but does not alter the energy distribution curve of the molecules.
- Changing the container volume or adding an inert gas at constant temperature does not change the temperature, so the kinetic energy distribution of the gas remains unchanged.
- Increasing the temperature from \(T_1\) to \(T_2\) increases the average kinetic energy of the molecules. The distribution curve flattens and shifts to the right: the peak moves to a higher energy value on the horizontal axis and to a lower height on the vertical axis because the total area under the curve (total number of molecules) remains constant.

C [1]
1 mark for selecting temperature increase (C).

To find the total number of stereoisomers for \(\text{CH}_3\text{CH}=\text{CHCH(Cl)}\text{CH}_2\text{CH}_3\):

1. Check for geometrical (cis-trans / E-Z) isomerism:
- Around the \(\text{C}=\text{C}\) double bond at C2=C3:
C2 is bonded to \(-\text{H}\) and \(-\text{CH}_3\) (two different groups).
C3 is bonded to \(-\text{H}\) and \(-\text{CH(Cl)}\text{CH}_2\text{CH}_3\) (two different groups).
- Therefore, there are 2 geometrical isomers: (E) and (Z).

2. Check for optical isomerism (chiral centres):
- Carbon-4 is bonded to four different groups: \(-\text{H}\), \(-\text{Cl}\), \(-\text{CH}_2\text{CH}_3\), and \(-\text{CH}=\text{CHCH}_3\).
- Thus, Carbon-4 is a chiral centre and exists as 2 enantiomers (R and S).

3. Total stereoisomers:
Total number of stereoisomers = \(2 \text{ (geometric)} \times 2 \text{ (optical)} = 4\)
((E, R), (E, S), (Z, R), and (Z, S)).

C [1]
1 mark for 4 (C).

Let us evaluate the observations step-by-step:

1. Forms an orange precipitate with 2,4-DNPH:
\(W\) contains a carbonyl group (\(\text{C}=\text{O}\)), so it is an aldehyde or a ketone. This eliminates propan-2-ol.

2. No reaction with Tollens' reagent:
\(W\) is not an aldehyde (it cannot be easily oxidised). Therefore, \(W\) must be a ketone. This eliminates butanal.

3. Forms a yellow precipitate with alkaline \(\text{I}_2\text{(aq)}\) (the tri-iodomethane / iodoform test):
The compound must contain a methyl carbonyl group, \(\text{CH}_3\text{C}(=\text{O})-\).
- In butan-2-one, \(\text{CH}_3\text{COCH}_2\text{CH}_3\), the \(\text{CH}_3\text{C}=\text{O}\) group is present.
- In pentan-3-one, \(\text{CH}_3\text{CH}_2\text{COCH}_2\text{CH}_3\), there is no methyl group directly attached to the carbonyl carbon.

Hence, compound \(W\) is butan-2-one.

C [1]
1 mark for identifying Butan-2-one (C).

1. Volume of \(\text{CO}_2\) produced: \(\text{CO}_2\) reacts with aqueous sodium hydroxide, so the volume reduction represents the volume of \(\text{CO}_2\):
\[V(\text{CO}_2) = 110 - 30 = 80\text{ cm}^3\]
Since \(20.0\text{ cm}^3\) of \(\text{C}_x\text{H}_y\) produces \(80.0\text{ cm}^3\) of \(\text{CO}_2\):
\[x = \frac{80.0}{20.0} = 4\]

2. Volume of unreacted \(\text{O}_2\) remaining is \(30\text{ cm}^3\). Therefore, volume of \(\text{O}_2\) reacted is:
\[V(\text{O}_2\text{ reacted}) = 150 - 30 = 120\text{ cm}^3\]

3. The combustion equation is:
\[\text{C}_x\text{H}_y\text{(g)} + \left(x + \frac{y}{4}\right)\text{O}_2\text{(g)} \rightarrow x\text{CO}_2\text{(g)} + \frac{y}{2}\text{H}_2\text{O(l)}\]
From the mole/volume ratio:
\[x + \frac{y}{4} = \frac{120}{20.0} = 6\]
Substituting \(x = 4\):
\[4 + \frac{y}{4} = 6 \implies \frac{y}{4} = 2 \implies y = 8\]

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

B is correct [1].
- Deducing volume of \(\text{CO}_2 = 80\text{ cm}^3\) gives \(x = 4\).
- Deducing volume of \(\text{O}_2\) consumed \(= 120\text{ cm}^3\) gives \(y = 8\).

A: \(\text{SO}_2\) has 2 bonding pairs and 1 lone pair on the sulfur atom, giving a non-linear (bent) shape with a bond angle of approximately \(119^\circ\) and a non-zero dipole moment.
B: \(\text{BF}_3\) has 3 bonding pairs and 0 lone pairs on the boron atom, forming a trigonal planar molecule with bond angles of exactly \(120^\circ\). Because of its symmetrical planar geometry, the individual bond dipoles cancel out completely, giving a dipole moment of zero.
C: \(\text{NH}_3\) has 3 bonding pairs and 1 lone pair, giving a trigonal pyramidal shape with a bond angle of \(107^\circ\) and a permanent dipole moment.
D: \(\text{ClF}_3\) has 3 bonding pairs and 2 lone pairs on chlorine, forming a T-shaped molecule with bond angles of less than \(90^\circ\) and a permanent dipole moment.

B is correct [1]. \(\text{BF}_3\) is trigonal planar with bond angles of \(120^\circ\) and non-polar due to symmetry.

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

\[M_{\text{r}} = \frac{mRT}{pV}\]

Substitute the given values:
\[M_{\text{r}} = \frac{1.72 \times 8.31 \times 300}{7.16 \times 10^4 \times 1.50 \times 10^{-3}}\]
\[M_{\text{r}} = \frac{4287.96}{107.4} \approx 39.9\text{ g mol}^{-1}\]

Comparing with the relative atomic masses of the noble gases:
- Helium: \(4.0\)
- Neon: \(20.2\)
- Argon: \(39.9\)
- Krypton: \(83.8\)

Therefore, the gas is argon.

C is correct [1]. Correct rearrangement of \(pV = nRT\) and calculation of \(M_r \approx 39.9\) identifying Argon.

The equation for the formation of propan-1-ol is:
\[3\text{C(s)} + 4\text{H}_2\text{(g)} + \frac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{OH(l)}\]

Using Hess's Law and enthalpy changes of combustion:
\[\Delta H^\circ_{\text{f}} = \sum \Delta H^\circ_{\text{c}}(\text{reactants}) - \sum \Delta H^\circ_{\text{c}}(\text{products})\]
\[\Delta H^\circ_{\text{f}} = [3 \times (-394) + 4 \times (-286)] - [-2021]\]
\[\Delta H^\circ_{\text{f}} = [-1182 - 1144] + 2021 = -2326 + 2021 = -305\text{ kJ mol}^{-1}\]

A is correct [1]. Correct Hess's Law cycle: \(3(-394) + 4(-286) - (-2021) = -305\text{ kJ mol}^{-1}\).

Descending Group 2 from magnesium to barium:
1. Solubility of sulfates decreases: \(\Delta H_{\text{hyd}}\) decreases more rapidly than lattice enthalpy \(\Delta H_{\text{latt}}\), making \(\Delta H_{\text{soln}}\) more endothermic.
2. Solubility of hydroxides increases: lattice enthalpy decreases more significantly down the group than hydration enthalpy because hydroxide is a smaller anion.
3. Thermal stability of nitrates increases: as the cationic radius increases, charge density decreases, so the polarising power of the cation decreases and it weakens the nitrate N-O bonds less readily.

A is correct [1]. Knowledge of Group 2 trends: sulfate solubility decreases, hydroxide solubility increases, nitrate thermal stability increases.

1. The forward reaction is exothermic (\(\Delta H < 0\)). According to Le Chatelier's principle, decreasing the temperature shifts the equilibrium position in the exothermic direction (to the right), which increases the yield of \(\text{SO}_3\).
2. The numerical value of the equilibrium constant \(K_{\text{c}}\) is only altered by changes in temperature. For an exothermic forward reaction, lowering the temperature increases the value of \(K_{\text{c}}\).
3. Increasing total pressure or increasing \([\text{O}_2]\) increases the yield of \(\text{SO}_3\), but does not change the value of \(K_{\text{c}}\).
4. Adding a catalyst affects only the rate of reaction, not the equilibrium yield or \(K_{\text{c}}\).

A is correct [1]. Only a decrease in temperature increases \(K_c\) for an exothermic reaction and shifts position to the right.

1. Mechanism and carbocation stability: Tertiary halogenoalkanes hydrolyse rapidly via the \(\text{S}_{\text{N}}1\) mechanism because the tertiary carbocation intermediate is stabilised by the inductive electron-donating effect of three alkyl groups. Primary halogenoalkanes react much more slowly via the \(\text{S}_{\text{N}}2\) mechanism.
2. Bond strength / bond enthalpy: The carbon-halogen bond must break during hydrolysis. The \(\text{C}-\text{I}\) bond enthalpy (\(240\text{ kJ mol}^{-1}\)) is significantly lower than the \(\text{C}-\text{Cl}\) bond enthalpy (\(340\text{ kJ mol}^{-1}\)), meaning iodoalkanes react faster than chloroalkanes.
3. Combining both factors, 2-iodo-2-methylpropane (a tertiary iodoalkane) hydrolyses the fastest to produce an AgI precipitate.

D is correct [1]. Tertiary halogenoalkane hydrolyses fastest via \(\text{S}_\text{N}1\), and \(\text{C}-\text{I}\) bond is the weakest.

1. Positive reaction with 2,4-DNPH (orange precipitate) indicates that \(\mathbf{X}\) is a carbonyl compound (an aldehyde or a ketone).
2. Negative reaction with Tollens' reagent (no silver mirror) indicates that \(\mathbf{X}\) is a ketone, not an aldehyde.
3. Positive tri-iodomethane test with alkaline aqueous iodine (yellow precipitate of \(\text{CHI}_3\)) indicates the presence of a methyl carbonyl group, \(\text{CH}_3\text{C(=O)}-\).
4. A 4-carbon ketone containing a \(\text{CH}_3\text{C(=O)}-\) group is butanone, \(\text{CH}_3\text{COCH}_2\text{CH}_3\).

B is correct [1]. Butanone is a ketone (negative Tollens') containing a \(\text{CH}_3\text{CO}-\) group (positive iodoform test).

1. Aqueous \(\text{KOH}\) absorbs carbon dioxide, \(\text{CO}_2\).
Volume of \(\text{CO}_2\) produced = \(110\text{ cm}^3 - 30\text{ cm}^3 = 80\text{ cm}^3\).
Volume of unreacted \(\text{O}_2 = 30\text{ cm}^3\).

2. Number of carbon atoms, \(x\):
\(x = \frac{V(\text{CO}_2)}{V(\text{hydrocarbon})} = \frac{80}{20} = 4\).

3. Volume of \(\text{O}_2\) consumed:
\(V(\text{O}_2\text{ used}) = 150\text{ cm}^3 - 30\text{ cm}^3 = 120\text{ cm}^3\).

4. From the general combustion equation:
\(\text{C}_x\text{H}_y + \left(x + \frac{y}{4}\right)\text{O}_2 \to x\text{CO}_2 + \frac{y}{2}\text{H}_2\text{O}(\text{l})\)
\(x + \frac{y}{4} = \frac{120}{20} = 6\)
\(4 + \frac{y}{4} = 6 \implies \frac{y}{4} = 2 \implies y = 8\).

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

C [1]
- Identification of \(V(\text{CO}_2) = 80\text{ cm}^3\) giving \(x = 4\)
- Calculation of \(V(\text{O}_2\text{ reacted}) = 120\text{ cm}^3\) giving \(y = 8\)

In solid ice, water molecules are held in an open, three-dimensional lattice structure where each water molecule forms four hydrogen bonds arranged tetrahedrally around the central oxygen atom (two via lone pairs and two via hydrogen atoms). This creates relatively large spaces in the lattice, making ice less dense than liquid water at \(0\,^\circ\text{C}\). In liquid water, some hydrogen bonds are broken, collapsing the open structure and packing molecules closer together.

B [1]
- A is incorrect: ice is less dense than liquid water.
- B is correct: ice forms a tetrahedral open hydrogen-bonded network.
- C is incorrect: covalent \(\text{O–H}\) bonds are significantly stronger than hydrogen bonds.
- D is incorrect: only a portion of hydrogen bonds break upon melting.

The target formation equation is:
\[3\text{C}(\text{s}) + 4\text{H}_2(\text{g}) + \frac{1}{2}\text{O}_2(\text{g}) \to \text{C}_3\text{H}_7\text{OH}(\text{l})\]

Using \(\Delta H^\circ_{\text{f}} = \sum \Delta H^\circ_{\text{c}}(\text{reactants}) - \sum \Delta H^\circ_{\text{c}}(\text{products})\):
\[\Delta H^\circ_{\text{f}} = [3(-394) + 4(-286)] - [-2021]\]
\[\Delta H^\circ_{\text{f}} = [-1182 - 1144] + 2021 = -2326 + 2021 = -305\text{ kJ mol}^{-1}\]

A [1]
- Correct cycle / formula application: \([3(-394) + 4(-286)] - (-2021) = -305\text{ kJ mol}^{-1}\)

- \(\text{PCl}_5\) reacts vigorously and completely hydrolyses in water to form \(\text{H}_3\text{PO}_4\) and \(\text{HCl}\), giving a strongly acidic solution with a \(\text{pH}\) of around \(1\)–\(2\).
- \(\text{MgCl}_2\) dissolves readily in water with only slight hydrolysis of the \([\text{Mg}(\text{H}_2\text{O})_6]^{2+}\) complex ion, producing a very weakly acidic solution with a \(\text{pH}\) of around \(6.5\).
- \(\text{NaCl}\) dissolves to give a neutral solution (\(\text{pH} = 7\)).
- \(\text{AlCl}_3\) hydrolyses to give a moderately acidic solution (\(\text{pH} \approx 3\)).

B [1]
- \(\mathbf{P}\) is \(\text{PCl}_5\) (or \(\text{SiCl}_4\)) which undergoes full hydrolysis to produce a solution of \(\text{pH} \approx 1\).
- \(\mathbf{Q}\) is \(\text{MgCl}_2\) which undergoes slight hydrolysis to give \(\text{pH} \approx 6.5\).

When concentrated \(\text{H}_2\text{SO}_4\) reacts with \(\text{NaBr}\):
1. It first acts as an acid: \(\text{NaBr} + \text{H}_2\text{SO}_4 \to \text{NaHSO}_4 + \text{HBr}\) (steamy fumes of \(\text{HBr}\)).
2. Bromide ions, \(\text{Br}^-\), are strong enough reducing agents to reduce \(\text{H}_2\text{SO}_4\) from oxidation state \(+6\) to \(+4\) in \(\text{SO}_2\):
\[2\text{HBr} + \text{H}_2\text{SO}_4 \to \text{Br}_2 + \text{SO}_2 + 2\text{H}_2\text{O}\]
In this redox process, concentrated \(\text{H}_2\text{SO}_4\) is reduced, hence it acts as an oxidising agent.

A [1]
- \(\text{H}_2\text{SO}_4\) oxidises \(\text{Br}^-\) to \(\text{Br}_2\) and is reduced to \(\text{SO}_2\), acting as an oxidising agent.

- Adding a catalyst provides an alternative pathway with a lower activation energy, changing the value of \(E_{\text{a}}\).
- Increasing the pressure increases the concentration and frequency of collisions, but does not alter the molecular energy distribution; thus, the proportion of molecules with \(E \ge E_{\text{a}}\) remains constant.
- Increasing the temperature shifts the Maxwell–Boltzmann distribution curve to the right, increasing the average kinetic energy and significantly increasing the fraction (proportion) of molecules with \(E \ge E_{\text{a}}\), while \(E_{\text{a}}\) itself remains unchanged.

C [1]
- Increasing temperature broadens and shifts the Boltzmann distribution to higher energies without altering \(E_{\text{a}}\).

- Reaction 1: 2-bromo-2-methylpropane is a tertiary halogenoalkane. Hydrolysis with aqueous \(\text{NaOH}\) occurs via nucleophilic substitution, proceeding predominantly by an \(\text{S}_{\text{N}}1\) mechanism due to steric hindrance and the stability of the tertiary carbocation intermediate \((\text{CH}_3)_3\text{C}^+\).
- Reaction 2: heating with ethanolic \(\text{NaOH}\) favours base-induced elimination of \(\text{HBr}\), yielding 2-methylpropene, \((\text{CH}_3)_2\text{C}=\text{CH}_2\), as the major organic product.

A [1]
- Mechanism of Reaction 1 is \(\text{S}_{\text{N}}1\)
- Major organic product of Reaction 2 is 2-methylpropene

Examine the structural features of pent-3-en-2-ol:
1. Carbon-carbon double bond (C3=C4): carbon 3 is bonded to \(-\text{H}\) and \(-\text{CH}(\text{OH})\text{CH}_3\); carbon 4 is bonded to \(-\text{H}\) and \(-\text{CH}_3\). Since both carbons have two different groups, it exhibits geometrical (cis-trans / E-Z) isomerism (2 configurations: E and Z).
2. Carbon 2 (chiral centre): carbon 2 is bonded to four different groups: \(-\text{H}\), \(-\text{OH}\), \(-\text{CH}_3\), and \(-\text{CH}=\text{CHCH}_3\). Thus, it has 2 optical configurations (R and S).

Total number of stereoisomers = \(2 \times 2 = 4\):
- (E)-(2R)-pent-3-en-2-ol
- (E)-(2S)-pent-3-en-2-ol
- (Z)-(2R)-pent-3-en-2-ol
- (Z)-(2S)-pent-3-en-2-ol

C [1]
- 1 double bond giving 2 cis/trans (E/Z) isomers
- 1 chiral carbon giving 2 optical isomers (enantiomeric pairs)
- Total stereoisomers = \(2^2 = 4\)

(a) \(\text{Na}_2\text{O}\) possesses a giant ionic structure with strong electrostatic forces of attraction between \(\text{Na}^+\) and \(\text{O}^{2-}\) ions throughout the lattice. In contrast, \(\text{SO}_2\) has a simple molecular structure with covalent bonds between atoms within molecules, but only weak intermolecular (van der Waals / permanent dipole-dipole) forces between separate molecules. Significantly more thermal energy is needed to break the strong ionic bonds in \(\text{Na}_2\text{O}\) than to overcome the weak intermolecular forces in \(\text{SO}_2\).

(b)(i) An amphoteric substance is one that reacts with (or neutralises) both acids and bases / alkalis.
(ii) \(\text{Al}_2\text{O}_3(\text{s}) + 2\text{OH}^-(\text{aq}) + 3\text{H}_2\text{O}(\text{l}) \rightarrow 2[\text{Al(OH)}_4]^-(\text{aq})\)

(c)(i) \(\text{SiCl}_4(\text{l}) + 2\text{H}_2\text{O}(\text{l}) \rightarrow \text{SiO}_2(\text{s}) + 4\text{HCl}(\text{aq})\) (or with \(\text{Si(OH)}_4\))
(ii) pH \(= 1-2\) (or pH \(= 0-2\)) due to the formation of strong acid \(\text{HCl}\).

(d)(i) \(n(\text{Ag}^+) = c \times V = 0.250 \times \frac{30.00}{1000} = 7.50 \times 10^{-3}\text{ mol}\)

(ii) Total amount of \(\text{Cl}^-\) in \(250.0\text{ cm}^3 = 7.50 \times 10^{-3} \times \left(\frac{250.0}{25.0}\right) = 7.50 \times 10^{-2}\text{ mol}\)
Amount of \(\text{XCl}_n\) used \(= 0.0250\text{ mol}\)
\(n = \frac{n(\text{Cl}^-)}{n(\text{XCl}_n)} = \frac{7.50 \times 10^{-2}}{0.0250} = 3\)
Since \(n = 3\), the formula is \(\text{XCl}_3\). As \(\text{X}\) is a Period 3 element whose chloride reacts with water to release all chloride ions, \(\text{X}\) is aluminium (\(\text{Al}\)) or phosphorus (\(\text{P}\)).

(a)
- M1: \(\text{Na}_2\text{O}\) has giant ionic structure/lattice AND \(\text{SO}_2\) is simple molecular / has simple covalent molecules [1]
- M2: \(\text{Na}_2\text{O}\) has strong ionic bonds / electrostatic attraction between oppositely charged ions AND \(\text{SO}_2\) has weak intermolecular forces / van der Waals forces / dipole-dipole forces between molecules [1]
- M3: More energy needed to overcome strong ionic bonds in \(\text{Na}_2\text{O}\) than weak intermolecular forces in \(\text{SO}_2\) [1]

(b)
(i) Reacts with both acids and bases/alkalis [1]
(ii) \(\text{Al}_2\text{O}_3 + 2\text{OH}^- + 3\text{H}_2\text{O} \rightarrow 2[\text{Al(OH)}_4]^-\) OR \(\text{Al}_2\text{O}_3 + 6\text{OH}^- + 3\text{H}_2\text{O} \rightarrow 2[\text{Al(OH)}_6]^{3-}\) [1]

(c)
(i) \(\text{SiCl}_4 + 2\text{H}_2\text{O} \rightarrow \text{SiO}_2 + 4\text{HCl}\) OR \(\text{SiCl}_4 + 4\text{H}_2\text{O} \rightarrow \text{Si(OH)}_4 + 4\text{HCl}\) [1]
(ii) Any value from \(\text{pH } 0\) to \(\text{pH } 2\) [1]

(d)
(i) \(n(\text{Ag}^+) = 0.250 \times 0.03000 = 7.50 \times 10^{-3}\text{ mol}\) [1]
(ii)
- M1: \(n(\text{Cl}^-)\text{ total} = 7.50 \times 10^{-3} \times 10 = 7.50 \times 10^{-2}\text{ mol}\) (ecf from (d)(i)) [1]
- M2: Ratio \(n = \frac{7.50 \times 10^{-2}}{0.0250} = 3\) [1]
- M3: Element \(\text{X} = \text{Aluminium} / \text{Al}\) (or \(\text{Phosphorus} / \text{P}\)) [1]

(a) Reactivity increases down Group 2 from \(\text{Mg}\) to \(\text{Ba}\). Going down the group, atomic radius increases and the outer two electrons experience greater shielding from inner electron shells. Consequently, the nuclear attraction for the outer electrons decreases, causing the sum of the first and second ionisation energies to decrease. Thus, less energy is required to remove the valence electrons and form \(\text{M}^{2+}\) ions.

(b)(i) Steamy / misty white fumes are observed (due to \(\text{HCl}\) gas).
(ii) The oxidation product is iodine, \(\text{I}_2\). In \(\text{H}_2\text{SO}_4\), sulfur has an oxidation number of \(+6\); in \(\text{H}_2\text{S}\), sulfur has an oxidation number of \(-2\), so the change is from \(+6\) to \(-2\) (a decrease of 8).
(iii) \(8\text{NaI} + 5\text{H}_2\text{SO}_4 \rightarrow 4\text{I}_2 + \text{H}_2\text{S} + 4\text{Na}_2\text{SO}_4 + 4\text{H}_2\text{O}\) (or \(8\text{NaI} + 9\text{H}_2\text{SO}_4 \rightarrow 4\text{I}_2 + \text{H}_2\text{S} + 8\text{NaHSO}_4 + 4\text{H}_2\text{O}\))

(c)(i) \(\text{Cl}_2 + 2\text{NaOH} \rightarrow \text{NaCl} + \text{NaClO} + \text{H}_2\text{O}\)
(ii) It is a disproportionation reaction. The oxidation state of chlorine in \(\text{Cl}_2\) is \(0\). It is oxidised to \(+1\) in \(\text{NaClO}\) (or \(\text{ClO}^-\)) and reduced to \(-1\) in \(\text{NaCl}\) (or \(\text{Cl}^-\)).

(a)
- M1: Reactivity increases down the group [1]
- M2: Atomic radius increases AND shielding increases / outer electrons further from nucleus / attraction between nucleus and outer electrons decreases [1]
- M3: (Sum of) first and second ionisation energies decreases / electrons lost more easily [1]

(b)
(i) Misty/steamy fumes / white fumes [1]
(ii)
- M1: Oxidation product is iodine / \(\text{I}_2\) [1]
- M2: (Sulfur changes) from \(+6\) to \(-2\) [1]
(iii) \(8\text{NaI} + 5\text{H}_2\text{SO}_4 \rightarrow 4\text{I}_2 + \text{H}_2\text{S} + 4\text{Na}_2\text{SO}_4 + 4\text{H}_2\text{O}\) OR \(8\text{I}^- + 10\text{H}^+ + \text{SO}_4^{2-} \rightarrow 4\text{I}_2 + \text{H}_2\text{S} + 4\text{H}_2\text{O}\) [1]

(c)
(i) \(\text{Cl}_2 + 2\text{NaOH} \rightarrow \text{NaCl} + \text{NaClO} + \text{H}_2\text{O}\) [1]
(ii)
- M1: Disproportionation [1]
- M2: Chlorine is oxidised from \(0\) to \(+1\) (in \(\text{NaClO}\)) AND reduced from \(0\) to \(-1\) (in \(\text{NaCl}\)) [1]

(a)(i)
1. The rate of the forward reaction is equal to the rate of the reverse/backward reaction.
2. The concentrations (or amounts) of reactants and products remain constant in a closed system.

(a)(ii)
\[K_p = \frac{p(\text{CH}_3\text{OH})}{p(\text{CO}) \times \left(p(\text{H}_2)\right)^2}\]

(a)(iii)
\[\text{Units} = \frac{\text{kPa}}{\text{kPa} \times (\text{kPa})^2} = \frac{1}{\text{kPa}^2} = \text{kPa}^{-2}\]

(b)(i)
- Initial moles: \(n(\text{CO}) = 0.400\text{ mol}\), \(n(\text{H}_2) = 0.800\text{ mol}\), \(n(\text{CH}_3\text{OH}) = 0\text{ mol}\)
- Moles of \(\text{CH}_3\text{OH}\) formed at equilibrium = \(0.150\text{ mol}\)
- Moles of \(\text{CO}\) reacted = \(0.150\text{ mol}\)
\[n(\text{CO})_{\text{eq}} = 0.400 - 0.150 = 0.250\text{ mol}\]
- Moles of \(\text{H}_2\) reacted = \(2 \times 0.150 = 0.300\text{ mol}\)
\[n(\text{H}_2)_{\text{eq}} = 0.800 - 0.300 = 0.500\text{ mol}\]

(b)(ii)
Total moles at equilibrium:
\[n_{\text{total}} = 0.250 + 0.500 + 0.150 = 0.900\text{ mol}\]
Mole fractions:
\[x(\text{CO}) = \frac{0.250}{0.900} = \frac{5}{18} \approx 0.278\]
\[x(\text{H}_2) = \frac{0.500}{0.900} = \frac{10}{18} \approx 0.556\]
\[x(\text{CH}_3\text{OH}) = \frac{0.150}{0.900} = \frac{3}{18} \approx 0.167\]

(b)(iii)
Partial pressures (using \(p_i = x_i \times P_{\text{total}}\)):
\[p(\text{CO}) = \frac{5}{18} \times 5.00 \times 10^3\text{ kPa} = 1388.9\text{ kPa} \approx 1.39 \times 10^3\text{ kPa}\]
\[p(\text{H}_2) = \frac{10}{18} \times 5.00 \times 10^3\text{ kPa} = 2777.8\text{ kPa} \approx 2.78 \times 10^3\text{ kPa}\]
\[p(\text{CH}_3\text{OH}) = \frac{3}{18} \times 5.00 \times 10^3\text{ kPa} = 833.3\text{ kPa} \approx 8.33 \times 10^2\text{ kPa}\]

Calculating \(K_p\):
\[K_p = \frac{833.33}{1388.89 \times (2777.78)^2} = \frac{833.33}{1388.89 \times 7.71605 \times 10^6} = \frac{833.33}{1.07167 \times 10^{10}} = 7.78 \times 10^{-8}\text{ kPa}^{-2}\]

(c)(i)
- \(K_p\) decreases.
- The forward reaction is exothermic (\(\Delta H < 0\)), so an increase in temperature favours the endothermic reverse reaction, shifting the position of equilibrium to the left.

(c)(ii)
- Yield of methanol increases because the equilibrium shifts to the side with fewer moles of gas (1 mole of product vs 3 moles of reactants) to oppose the increase in pressure.

(a)(i) [2 marks]
- M1: Rate of forward reaction equals rate of backward / reverse reaction. [1]
- M2: Concentrations of reactants and products remain constant (in a closed system). [1]
(R: concentrations are equal)

(a)(ii) [1 mark]
- M1: Correct expression for \(K_p\): \(K_p = \frac{p(\text{CH}_3\text{OH})}{p(\text{CO}) \times (p(\text{H}_2))^2}\) [1]
*(A: partial pressure symbols like \(P_{\text{CH}_3\text{OH}}\); R: square brackets indicating concentrations)*

(a)(iii) [1 mark]
- M1: \(\text{kPa}^{-2}\) or \(\text{dm}^6\text{ mol}^{-2}\) if derived from concentration notation. [1]

(b)(i) [2 marks]
- M1: \(n(\text{CO}) = 0.250\text{ mol}\) [1]
- M2: \(n(\text{H}_2) = 0.500\text{ mol}\) [1]

(b)(ii) [1 mark]
- M1: Total moles = \(0.900\text{ mol}\) AND correct mole fractions: \(x(\text{CO}) = 0.278\), \(x(\text{H}_2) = 0.556\), \(x(\text{CH}_3\text{OH}) = 0.167\) (or exact fractions \(5/18, 5/9, 1/6\)). [1]

(b)(iii) [3 marks]
- M1: Correct calculation of all three partial pressures: \(p(\text{CO}) = 1389\text{ kPa}\), \(p(\text{H}_2) = 2778\text{ kPa}\), \(p(\text{CH}_3\text{OH}) = 833.3\text{ kPa}\) (allow ecf from (b)(ii)). [1]
- M2: Correct substitution into \(K_p\) expression. [1]
- M3: Value of \(K_p = 7.78 \times 10^{-8}\) (allow \(7.77 \times 10^{-8}\) to \(7.79 \times 10^{-8}\)), given to 3 significant figures. [1]

(c)(i) [2 marks]
- M1: (Value of \(K_p\)) decreases. [1]
- M2: Forward reaction is exothermic / backward reaction is endothermic, so equilibrium shifts to the left / favours endothermic reaction. [1]

(c)(ii) [1 mark]
- M1: Yield increases AND fewer moles/molecules of gas on right-hand side (1 mol of gas on RHS compared to 3 mol on LHS). [1]

(a)(i) Nucleophilic addition.

(a)(ii) Detailed mechanism:
1. The carbonyl group is polar: \(\text{C}^{\delta+}=\text{O}^{\delta-}\).
2. The nucleophile \(:\text{C}\equiv\text{N}^-\) attacks the electrophilic carbonyl carbon: a curly arrow is drawn from the lone pair on the carbon of the cyanide ion, \(^-\text{CN}\), to the \(\text{C}^{\delta+}\) atom.
3. A curly arrow is drawn from the \(\text{C}=\text{O}\) double bond to the oxygen atom.
4. This forms the intermediate alkoxide: \(\text{CH}_3-\text{C}(\text{O}^-)(\text{CN})-\text{CH}_2\text{CH}_3\).
5. A curly arrow from a lone pair on the negatively charged oxygen atom to the \(\text{H}\) atom of an \(\text{H}-\text{CN}\) molecule (or \(\text{H}^+\)), with a curly arrow breaking the \(\text{H}-\text{CN}\) bond onto carbon, forming the product \(\text{CH}_3\text{C(OH)(CN)CH}_2\text{CH}_3\) and regenerating the \(^-\text{CN}\) catalyst.

(a)(iii)
- A chiral centre is an \(\text{sp}^3\) hybridised carbon atom bonded to four different atoms or groups (here: \(-\text{CH}_3\), \(-\text{CH}_2\text{CH}_3\), \(-\text{OH}\), and \(-\text{CN}\)).
- The carbonyl group (\(\text{C}=\text{O}\)) in butan-2-one has a planar arrangement around the carbonyl carbon. The nucleophile (\(^-\text{CN}\)) has an equal probability of attacking from either above or below the plane, resulting in equal amounts (a 1:1 equimolar / racemic mixture) of the two optical isomers (enantiomers), which cancel each other's optical rotation.

(b)(i)
- Reaction with 2,4-DNPH confirms it is a carbonyl compound (aldehyde or ketone).
- Reaction with Tollens' reagent confirms it is an aldehyde.
- Branched 4-carbon chain gives 2-methylpropanal, \((\text{CH}_3)_2\text{CHCHO}\).

(b)(ii)
$$\text{(CH}_3)_2\text{CHCHO} + 2[\text{H}] \rightarrow \text{(CH}_3)_2\text{CHCH}_2\text{OH}$$

(c)(i) Concentrated \(\text{H}_2\text{SO}_4\) (or concentrated \(\text{H}_3\text{PO}_4\)) and heat / \(170\,^\circ\text{C}\) OR heated \(\text{Al}_2\text{O}_3\).

(c)(ii)
- cis-but-2-ene / (Z)-but-2-ene and trans-but-2-ene / (E)-but-2-ene.
- Type of stereoisomerism: cis-trans (geometrical) isomerism.

(a)(i) Nucleophilic addition [1]
R electrophilic addition / nucleophilic substitution

(a)(ii) 4 marks total:
- M1: Dipole on \(\text{C}=\text{O}\) (\(\text{C}^{\delta+}=\text{O}^{\delta-}\)) AND curly arrow from lone pair on \(\text{C}\) of \(^-\text{CN}\) (or \(:\text{CN}^-\)) to \(\text{C}^{\delta+}\) [1]
- M2: Curly arrow from \(\text{C}=\text{O}\) double bond to the oxygen atom [1]
- M3: Correct structure of tetrahedral intermediate with negative charge on oxygen: \(\text{CH}_3\text{C}(\text{O}^-)(\text{CN})\text{CH}_2\text{CH}_3\) [1]
- M4: Curly arrow from lone pair on \(\text{O}^-\) to \(\text{H}\) of \(\text{HCN}\) (or \(\text{H}^+\)) AND curly arrow from \(\text{H}-\text{C}\) bond to \(\text{C}\) [1]

(a)(iii) 2 marks total:
- M1: Carbon atom bonded to four different atoms or groups [1]
- M2: Carbonyl group / \(\text{C}=\text{O}\) is planar AND attack of \(^-\text{CN}\) is equally likely from either side/face (forming a racemic mixture / equal amounts of enantiomers) [1]

(b)(i) 2 marks total:
- M1: Displayed formula of 2-methylpropanal showing all atoms and all bonds clearly [1]
- M2: Systematic name: 2-methylpropanal (or methylpropanal) [1]

(b)(ii) Correct balanced equation: \(\text{(CH}_3)_2\text{CHCHO} + 2[\text{H}] \rightarrow \text{(CH}_3)_2\text{CHCH}_2\text{OH}\) [1]

(c)(i) Concentrated \(\text{H}_2\text{SO}_4\) / concentrated \(\text{H}_3\text{PO}_4\) with heat OR \(\text{Al}_2\text{O}_3\) with heat [1]
A reflux / \(160\text{--}180\,^\circ\text{C}\)

(c)(ii) 2 marks total:
- M1: Correct 3D/displayed/skeletal structures showing cis-but-2-ene and trans-but-2-ene [1]
- M2: cis-trans isomerism / geometrical isomerism [1]

(a)(i) \(\text{S}_\text{N}2\) Mechanism:
1. 1-bromobutane has a polar \(\text{C}-\text{Br}\) bond: \(\text{C}^{\delta+}-\text{Br}^{\delta-}\).
2. The nucleophile \(:\text{OH}^-\) attacks the \(\text{C}^{\delta+}\) atom from the side opposite to the bromine atom (backside attack).
3. A curly arrow is drawn from the lone pair of electrons on the oxygen of \(:\text{OH}^-\) to the \(\text{C}^{\delta+}\) atom.
4. A curly arrow is drawn from the \(\text{C}-\text{Br}\) bond to the \(\text{Br}\) atom.
5. The transition state is drawn inside square brackets with an overall negative charge (\(^-\)): showing three full bonds (to \(\text{H}\), \(\text{H}\), \(-\text{CH}_2\text{CH}_2\text{CH}_3\)) and two partial bonds represented by dashed lines (\(\text{HO}\cdots\text{C}\cdots\text{Br}\)) with partial negative charges \(\delta-\).
6. The final products are butan-1-ol, \(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH}\), and a bromide ion, \(\text{Br}^-\).

(a)(ii)
- Carbocation stability: The tertiary carbocation intermediate, \((\text{CH}_3)_3\text{C}^+\), is highly stable due to the positive inductive effect (electron-donating effect) of the three alkyl (methyl) groups.
- Steric hindrance: The three bulky methyl groups surround the central carbon atom, preventing backside attack by the hydroxide ion nucleophile in an \(\text{S}_\text{N}2\) pathway.

(b)(i) 1-iodobutane > 1-bromobutane > 1-chlorobutane (or 1-iodobutane is fastest, 1-chlorobutane is slowest).

(b)(ii)
- The rate depends on the strength/energy of the carbon-halogen bond (not bond polarity).
- \(\text{C}-\text{I}\) bond energy is the lowest / bond is weakest (approx. \(240\,\text{kJ}\,\text{mol}^{-1}\)), so the \(\text{C}-\text{I}\) bond breaks most easily.
- \(\text{C}-\text{Cl}\) bond energy is the highest / bond is strongest (approx. \(340\,\text{kJ}\,\text{mol}^{-1}\)), requiring the most energy to break.

(b)(iii) A cream (or off-white) precipitate forms (silver bromide, \(\text{AgBr}\)).

(c)(i) Base (or proton / \(\text{H}^+\) acceptor).

(c)(ii)
1. But-1-ene: \(\text{CH}_3\text{CH}_2\text{CH}=\text{CH}_2\)
2. cis-but-2-ene / (Z)-but-2-ene: \(\text{CH}_3\text{CH}=\text{CHCH}_3\)
3. trans-but-2-ene / (E)-but-2-ene: \(\text{CH}_3\text{CH}=\text{CHCH}_3\)

(a)(i) 4 marks total:
- M1: Dipole on \(\text{C}-\text{Br}\) bond (\(\text{C}^{\delta+}-\text{Br}^{\delta-}\)) AND curly arrow from lone pair on \(:\text{OH}^-\) to \(\text{C}^{\delta+}\) [1]
- M2: Curly arrow from \(\text{C}-\text{Br}\) bond to \(\text{Br}\) [1]
- M3: Correct transition state with dotted/dashed bonds to \(\text{OH}\) and \(\text{Br}\), with negative charge \(^-\) outside brackets [1]
- M4: Correct products (\(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH} + \text{Br}^-\)) [1]

(a)(ii) 2 marks total:
- M1: Tertiary carbocation intermediate is stable due to (positive) inductive / electron-donating effect of 3 methyl/alkyl groups [1]
- M2: 3 bulky methyl groups cause steric hindrance, preventing backside attack by \(\text{OH}^-\) (in \(\text{S}_\text{N}2\)) [1]

(b)(i) 1-iodobutane > 1-bromobutane > 1-chlorobutane [1]

(b)(ii) 2 marks total:
- M1: Reaction involves breaking the \(\text{C}-\text{X}\) bond (bond energy determines the rate, not polarity) [1]
- M2: \(\text{C}-\text{I}\) bond is weakest / has lowest bond energy, so breaks most easily ora for \(\text{C}-\text{Cl}\) [1]

(b)(iii) Cream precipitate / off-white precipitate [1]
R yellow / white precipitate

(c)(i) Base / \(\text{H}^+\) acceptor / proton acceptor [1]
R nucleophile

(c)(ii) 2 marks total:
- M1: But-1-ene: \(\text{CH}_3\text{CH}_2\text{CH}=\text{CH}_2\) [1]
- M2: cis-but-2-ene AND trans-but-2-ene (clearly identified as stereoisomers or drawn showing stereochemistry) [1]

(a) (i) \(n(\text{CO}_2) = \frac{V}{V_{\text{m}}} = \frac{60.0\text{ cm}^3}{24000\text{ cm}^3\text{ mol}^{-1}} = 2.50 \times 10^{-3}\text{ mol}\)

(ii) From the stoichiometric equation, \(1\text{ mol}\) of \(\text{MCO}_3 \cdot \text{M(OH)}_2\) produces \(1\text{ mol}\) of \(\text{CO}_2\).
\(n(\text{MCO}_3 \cdot \text{M(OH)}_2) = 2.50 \times 10^{-3}\text{ mol}\)

(iii) \(M_{\text{r}} = \frac{m}{n} = \frac{0.553\text{ g}}{2.50 \times 10^{-3}\text{ mol}} = 221.2\text{ g mol}^{-1} \approx 221\)

(iv) \(M_{\text{r}}(\text{MCO}_3 \cdot \text{M(OH)}_2) = 2 A_{\text{r}}(\text{M}) + A_{\text{r}}(\text{C}) + 5 A_{\text{r}}(\text{O}) + 2 A_{\text{r}}(\text{H})\)
\(221.2 = 2 A_{\text{r}}(\text{M}) + 12.0 + (5 \times 16.0) + (2 \times 1.0) = 2 A_{\text{r}}(\text{M}) + 94.0\)
\(2 A_{\text{r}}(\text{M}) = 221.2 - 94.0 = 127.2 \implies A_{\text{r}}(\text{M}) = 63.6\) (or \(62.6\) to \(63.6\) depending on exact rounding).
Identity of metal \(\text{M}\): Copper / \(\text{Cu}\) (standard \(A_{\text{r}} = 63.5\)).

(b) (i) The gas in the syringe is at a temperature higher than room temperature, so it expands (Charles's law), making the measured gas volume larger than it should be. This gives a larger calculated amount (in mol) of \(\text{CO}_2\) and basic carbonate, which makes the calculated \(M_{\text{r}}\) smaller, resulting in a lower / smaller calculated value for \(A_{\text{r}}(\text{M})\).

(ii) Place a tube / U-tube containing a drying agent (such as anhydrous calcium chloride / silica gel) or an ice bath condenser between the reaction tube and the gas syringe.

(c) Remove the delivery tube from the gas syringe / water bath before removing the Bunsen flame to prevent suck-back of liquid / air contraction pulling liquid back into the hot tube (or allow the apparatus to cool before handling / use tongs to handle hot glassware).

(a)(i) [1] M1: Calculates \(n(\text{CO}_2) = \frac{60.0}{24000} = 2.50 \times 10^{-3}\text{ mol}\)

(a)(ii) [1] M1: Correct deduction that \(n(\text{basic carbonate}) = 2.50 \times 10^{-3}\text{ mol}\) (ecf from (a)(i))

(a)(iii) [1] M1: Calculates \(M_{\text{r}} = \frac{0.553}{2.50 \times 10^{-3}} = 221\) (allow 221.2, ecf from (a)(ii))

(a)(iv) [2]
M1: Correct algebraic expression equating \(M_{\text{r}}\) to \(2 A_{\text{r}}(\text{M}) + 94.0\) and evaluation of \(A_{\text{r}}(\text{M}) = 62.6\text{ to }63.6\) [1]
M2: Correct identification of \(\text{M}\) as copper / \(\text{Cu}\) consistent with calculated \(A_{\text{r}}\) [1]

(b)(i) [2]
M1: Gas expands when hot / volume recorded is too high [1]
M2: Calculated \(M_{\text{r}}\) is smaller / lower, so calculated \(A_{\text{r}}\) is smaller / lower [1]

(b)(ii) [1] M1: Connect a U-tube with a named solid drying agent (e.g. anhydrous \(\text{CaCl}_2\), silica gel) / pass gas through an ice-cooled trap before the gas syringe [1] (R: concentrated \(\text{H}_2\text{SO}_4\) in a bubbler if it restricts syringe movement / unfeasible setup)

(c) [1] M1: Disconnect the delivery tube / syringe before removing heat to avoid suck-back OR handle hot test-tubes with tongs / allow apparatus to cool before touching [1]

(a)
Representative accurate data:
- Rough titration: Final reading = \(25.60\text{ cm}^3\), Initial reading = \(0.00\text{ cm}^3\), Titre = \(25.60\text{ cm}^3\)
- Titre 1: Final reading = \(24.95\text{ cm}^3\), Initial reading = \(0.00\text{ cm}^3\), Titre = \(24.95\text{ cm}^3\)
- Titre 2: Final reading = \(25.05\text{ cm}^3\), Initial reading = \(0.00\text{ cm}^3\), Titre = \(25.05\text{ cm}^3\)

Average titre calculation:
$$\text{Average titre} = \frac{24.95 + 25.05}{2} = 25.00\text{ cm}^3$$

(b)
(i) Amount of \(\text{NaOH}\):
$$n(\text{NaOH}) = c \times V = 0.100\text{ mol dm}^{-3} \times \frac{25.00}{1000}\text{ dm}^3 = 2.50 \times 10^{-3}\text{ mol}$$

(ii) From stoichiometric equation, \(1\text{ mol }\text{H}_2\text{A} \equiv 2\text{ mol }\text{NaOH}\):
$$n(\text{H}_2\text{A}) = \frac{n(\text{NaOH})}{2} = \frac{2.50 \times 10^{-3}}{2} = 1.25 \times 10^{-3}\text{ mol}$$

(iii) Concentration of \(\text{H}_2\text{A}\) in \(\text{FA 1}\):
$$c(\text{H}_2\text{A}) = \frac{1.25 \times 10^{-3}\text{ mol}}{25.0 / 1000\text{ dm}^3} = 0.0500\text{ mol dm}^{-3}$$

(iv) Relative formula mass, \(M_\mathrm{r}\):
$$M_\mathrm{r} = \frac{\text{mass concentration}}{\text{molar concentration}} = \frac{6.30\text{ g dm}^{-3}}{0.0500\text{ mol dm}^{-3}} = 126.0$$

(v) Water of crystallisation calculation:
$$M_\mathrm{r}(\text{H}_2\text{A}\cdot x\text{H}_2\text{O}) = M_\mathrm{r}(\text{H}_2\text{A}) + x \times M_\mathrm{r}(\text{H}_2\text{O})$$
$$126.0 = 90.0 + 18.0x$$
$$18.0x = 36.0 \implies x = 2$$

(c)
(i) Total uncertainty in two burette readings (initial and final) \(= 2 \times 0.05 = \pm 0.10\text{ cm}^3\).
$$\text{Percentage uncertainty} = \frac{0.10}{25.00} \times 100\% = 0.40\%$$

(ii) The volume of \(\text{FA 2}\) recorded from the burette is greater than the true volume used to neutralise the acid, because some volume went into filling the air space.
This leads to a calculated amount of \(\text{NaOH}\) and \(\text{H}_2\text{A}\) that is larger than true, and hence a higher calculated molar concentration of \(\text{H}_2\text{A}\).
Since \(M_\mathrm{r} = \frac{\text{mass}}{\text{moles}}\), the calculated \(M_\mathrm{r}\) will be smaller/lower.

(iii) Use a larger volume/mass of acid \(\text{FA 1}\) (or dilute \(\text{FA 2}\) / increase pipette volume of \(\text{FA 1}\) to \(50.0\text{ cm}^3\)).
This increases the titre volume, so the percentage uncertainty \(\left(\frac{\text{uncertainty}}{\text{titre volume}} \times 100\right)\) decreases.

(a) [7 marks total]
- M1 (Table Layout): Single table recording all burette readings with complete, clear headings and units (e.g. initial / final burette reading / volume added in \(\text{cm}^3\) or \(/\text{cm}^3\)). [1]
- M2 (Precision): All accurate burette readings recorded to two decimal places ending in \(.00\) or \(.05\text{ cm}^3\). [1]
- M3 (Concordance): At least two accurate titres within \(0.10\text{ cm}^3\) of each other. [1]
- M4 & M5 (Accuracy): Quality mark compared to standard value:
- Award 2 marks if candidate titre is within \(\pm 0.20\text{ cm}^3\) of standard/supervisor value.
- Award 1 mark if candidate titre is within \(\pm 0.40\text{ cm}^3\) of standard/supervisor value. [2]
- M6 (Averaging method): Calculates average titre using strictly concordant titres (within \(0.20\text{ cm}^3\)) showing working or selected ticks. [1]
- M7 (Average precision): Average titre quoted correctly to 2 decimal places. [1]

(b) [6 marks total]
- (i) M8: Correct calculation of \(n(\text{NaOH}) = 0.100 \times \frac{\text{average titre}}{1000}\) to 3 significant figures. [1]
- (ii) M9: Moles of \(\text{H}_2\text{A} = \frac{\text{answer to (b)(i)}}{2}\). [1]
- (iii) M10: Concentration of \(\text{FA 1} = \text{answer to (b)(ii)} \times \frac{1000}{25.0}\). [1]
- (iv) M11: \(M_\mathrm{r} = \frac{6.30}{\text{answer to (b)(iii)}}\) (evaluated correctly to 3 significant figures). [1]
- (v) M12 & M13:
- M12: Correct method for calculating \(x = \frac{M_\mathrm{r} - 90.0}{18.0}\). [1]
- M13: Value of \(x\) given as a whole integer (e.g. \(x = 2\)). [1]

(c) [5 marks total]
- (i) M14: Calculation of \(\frac{0.10}{25.00} \times 100 = 0.40\%\) (Allow \(0.20\%\) if clearly using single reading uncertainty of \(0.05\text{ cm}^3\)). [1]
- (ii) M15 & M16:
- M15: Titre reading recorded is larger / more \(\text{FA 2}\) recorded than delivered. [1]
- M16: Calculated \(M_\mathrm{r}\) is lower / smaller AND explains that calculated moles of acid is larger (or concentration of acid is higher). [1]
- (iii) M17 & M18:
- M17: Modification: Increase pipette volume of \(\text{FA 1}\) (e.g. to \(50.0\text{ cm}^3\)) OR decrease concentration of \(\text{FA 2}\) / increase concentration of \(\text{FA 1}\). [1]
- M18: Explanation: This results in a larger titre volume, which decreases the percentage uncertainty. [1]

(a)(i) Adding dilute hydrochloric acid to sulfite (\(\text{SO}_3^{2-}\)) produces sulfur dioxide gas: \(\text{SO}_3^{2-}(\text{aq}) + 2\text{H}^+(\text{aq}) \rightarrow \text{SO}_2(\text{g}) + \text{H}_2\text{O}(\text{l})\). Sulfur dioxide is a reducing agent that reduces purple \(\text{MnO}_4^-\)\text{aq})\) to colourless \(\text{Mn}^{2+}(\text{aq})\).

(a)(ii) \(\text{Fe}^{2+}(\text{aq})\) reacts with \(\text{OH}^-(\text{aq})\) and \(\text{NH}_3(\text{aq})\) to form a dark green precipitate of \(\text{Fe(OH)}_2(\text{s})\), which is insoluble in excess of both reagents and oxidises to brown \(\text{Fe(OH)}_3\) on exposure to air.

(a)(iii) Cation = \(\text{Fe}^{2+}\) (iron(II)); Anion = \(\text{SO}_3^{2-}\) (sulfite).

(a)(iv) \(\text{Fe}^{2+}(\text{aq}) + 2\text{OH}^-(\text{aq}) \rightarrow \text{Fe(OH)}_2(\text{s})\).

(b)(i) Testing FB 6 and FB 7:
- 2,4-DNPH gives an orange/yellow precipitate with carbonyl compounds (FB 6) but no precipitate with carboxylic acids (FB 7).
- Warm Tollens' reagent gives a silver mirror / black precipitate with aldehydes (FB 6) due to reduction of \([\text{Ag}(\text{NH}_3)_2]^+\) to metallic Ag, while FB 7 gives no reaction.
- \(\text{Na}_2\text{CO}_3(\text{aq})\) reacts with carboxylic acids (FB 7) with effervescence (\(\text{CO}_2\) gas), but shows no reaction with aldehydes (FB 6).

(b)(ii) FB 6 is an aldehyde (contains \(-\text{CHO}\)) because it forms a precipitate with 2,4-DNPH and produces a silver mirror / black precipitate with Tollens' reagent. FB 7 is a carboxylic acid (contains \(-\text{COOH}\)) because it effervesces with aqueous sodium carbonate.

(a)(i) [2 marks]
- M1: Effervescence / bubbling / pungent or choking gas [1]
- M2: Acidified potassium manganate(VII) paper turns from purple to colourless / decolourised (A: acidified potassium dichromate paper turns from orange to green) [1]

(a)(ii) [2 marks]
- M1: Addition of NaOH(aq): green precipitate (on dropwise addition) AND insoluble in excess [1]
- M2: Addition of NH3(aq): green precipitate (on dropwise addition) AND insoluble in excess / turns brown on standing [1]

(a)(iii) [2 marks]
- M1: Cation: Fe2+ (A: iron(II)) [1]
- M2: Anion: SO3 2- (A: sulfite) [1]

(a)(iv) [1 mark]
- Fe2+(aq) + 2OH-(aq) -> Fe(OH)2(s) (both balanced equation and state symbols required) [1]

(b)(i) [4 marks]
- M1: Table layout showing 3 reagents/tests and systematic observations for both FB 6 and FB 7 [1]
- M2: 2,4-DNPH observations: FB 6 gives orange/yellow/red precipitate AND FB 7 gives no precipitate / yellow-orange solution [1]
- M3: Tollens' reagent (warmed) observations: FB 6 gives silver mirror / black/grey precipitate AND FB 7 gives no reaction / remains colourless [1]
- M4: Na2CO3(aq) observations: FB 6 gives no effervescence / no reaction AND FB 7 gives effervescence / bubbles of gas [1]

(b)(ii) [2 marks]
- M1: FB 6: aldehyde / -CHO group identified AND linked to positive Tollens' test (or silver mirror) [1]
- M2: FB 7: carboxylic acid / -COOH group identified AND linked to effervescence with Na2CO3 [1]

(a) (i) Lattice energy is the enthalpy change when one mole of an ionic crystalline solid is formed from its constituent gaseous ions under standard conditions (298 K and 101 kPa).

(ii) Using the Born-Haber cycle for \(\text{MgF}_2(\text{s})\):
\[\Delta H^\circ_f = \Delta H^\circ_{\text{at}}(\text{Mg}) + \text{IE}_1(\text{Mg}) + \text{IE}_2(\text{Mg}) + 2\Delta H^\circ_{\text{at}}(\text{F}) + 2\text{EA}_1(\text{F}) + \Delta H^\circ_{\text{latt}}(\text{MgF}_2)\]
\[-1124 = (+147) + (+738) + (+1451) + 2(+79) + 2(-328) + \Delta H^\circ_{\text{latt}}\]
\[-1124 = 147 + 738 + 1451 + 158 - 656 + \Delta H^\circ_{\text{latt}}\]
\[-1124 = +1838 + \Delta H^\circ_{\text{latt}}\]
\[\Delta H^\circ_{\text{latt}} = -1124 - 1838 = -2962\text{ kJ mol}^{-1}\]

(iii) The lattice energy of \(\text{CaF}_2\) is less exothermic (less negative).
\(\text{Ca}^{2+}\) has a larger ionic radius than \(\text{Mg}^{2+}\) (or \(\text{Ca}^{2+}\) has a lower charge density), so the electrostatic attraction between \(\text{Ca}^{2+}\) and \(\text{F}^-\ ions is weaker. (b) (i) The enthalpy change when one mole of gaseous ions dissolves in sufficient water to form an infinitely dilute aqueous solution under standard conditions. (ii) Constructing the energy cycle linking solution, hydration, and lattice energy: \[\Delta H^\circ_{\text{sol}}(\text{MgF}_2) = -\Delta H^\circ_{\text{latt}}(\text{MgF}_2) + \Delta H^\circ_{\text{hyd}}(\text{Mg}^{2+}) + 2\Delta H^\circ_{\text{hyd}}(\text{F}^-)\] \[-18 = -(-2962) + \Delta H^\circ_{\text{hyd}}(\text{Mg}^{2+}) + 2(-506)\] \[-18 = +2962 + \Delta H^\circ_{\text{hyd}}(\text{Mg}^{2+}) - 1012\] \[-18 = +1950 + \Delta H^\circ_{\text{hyd}}(\text{Mg}^{2+})\] \[\Delta H^\circ_{\text{hyd}}(\text{Mg}^{2+}) = -18 - 1950 = -1968\text{ kJ mol}^{-1}\] (Using the placeholder value of \)-2900\text{ kJ mol}^{-1}\): \(\Delta H^\circ_{\text{hyd}}(\text{Mg}^{2+}) = -18 - 2900 + 1012 = -1906\text{ kJ mol}^{-1}\)).

(c) (i) Standard enthalpy change of reaction:
\[\Delta H^\circ_{\text{r}} = \sum \Delta H^\circ_f(\text{products}) - \sum \Delta H^\circ_f(\text{reactants})\]
\[\Delta H^\circ_{\text{r}} = [(-602) + (-394)] - [-1096] = -996 + 1096 = +100\text{ kJ mol}^{-1}\]
Standard entropy change of reaction:
\[\Delta S^\circ_{\text{r}} = \sum S^\circ(\text{products}) - \sum S^\circ(\text{reactants})\]
\[\Delta S^\circ_{\text{r}} = [26.9 + 213.7] - [65.7] = 240.6 - 65.7 = +174.9\text{ J K}^{-1}\text{ mol}^{-1}\]

(ii) For decomposition to be feasible, \(\Delta G^\circ \le 0\):
\[\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ = 0\]
\[T = \frac{\Delta H^\circ}{\Delta S^\circ} = \frac{100\text{ kJ mol}^{-1}}{174.9\text{ J K}^{-1}\text{ mol}^{-1}} = \frac{100 \times 10^3}{174.9} = 571.8\text{ K} \approx 572\text{ K}\]
Assumption: \(\Delta H^\circ\) and \(\Delta S^\circ\) are independent of temperature / remain constant across the temperature range.

(a)(i)
- M1: Enthalpy change when 1 mole of an ionic solid / lattice is formed [1]
- M2: from its gaseous ions (under standard conditions) [1]

(a)(ii)
- M1: Correct expression or cycle showing sum of all stages: \(\Delta H^\circ_f = \Delta H^\circ_{\text{at}}(\text{Mg}) + \text{IE}_1 + \text{IE}_2 + 2\Delta H^\circ_{\text{at}}(\text{F}) + 2\text{EA}_1 + \Delta H^\circ_{\text{latt}}\) [1]
- M2: Correct multiplication by 2 for fluorine atomisation (\(2 \times 79\) or \(158\)) and electron affinity (\(2 \times -328\) or \(-656\)) [1]
- M3: Correct final value with sign: \(-2962\text{ kJ mol}^{-1}\) [1]

(a)(iii)
- M1: Less exothermic / less negative / smaller magnitude [1]
- M2: \(\text{Ca}^{2+}\) ion is larger / has a smaller charge-to-size ratio / lower charge density (than \(\text{Mg}^{2+}\)) resulting in weaker ionic bonds / weaker attraction to \(\text{F}^-\ ions [1] (b)(i) - M1: Enthalpy change when 1 mole of gaseous ions dissolves in water to form aqueous / hydrated ions (or an infinitely dilute solution) under standard conditions [1] (b)(ii) - M1: Correct use of cycle: \)\Delta H^\circ_{\text{hyd}}(\text{Mg}^{2+}) = \Delta H^\circ_{\text{sol}} + \Delta H^\circ_{\text{latt}} - 2\Delta H^\circ_{\text{hyd}}(\text{F}^-)\) [1]
- M2: Correct calculation: \(-1968\text{ kJ mol}^{-1}\) (allow ecf from (a)(ii)) [1]

(c)(i)
- M1: \(\Delta H^\circ_{\text{r}} = +100\text{ kJ mol}^{-1}\) [1]
- M2: \(\Delta S^\circ_{\text{r}} = +174.9\text{ J K}^{-1}\text{ mol}^{-1}\) [1]

(c)(ii)
- M1: \(T = \frac{\Delta H}{\Delta S} = \frac{100 \times 1000}{174.9} = 572\text{ K}\) (allow \(571.8\text{ K}\); ecf from (c)(i)) [1]
- M2: \(\Delta H\) and \(\Delta S\) do not change with temperature / are constant [1]

(a)(i)
\[K_\text{a} = \frac{[\text{CH}_3\text{CHClCOO}^-][\text{H}^+]}{[\text{CH}_3\text{CHClCOOH}]}\]

(a)(ii)
Assuming \([\text{H}^+] \approx [\text{CH}_3\text{CHClCOO}^-]\) and \([\text{CH}_3\text{CHClCOOH}]_\text{eq} \approx 0.0800\text{ mol dm}^{-3}\):
\[[\text{H}^+] = \sqrt{K_\text{a} \times c} = \sqrt{1.48 \times 10^{-3} \times 0.0800} = \sqrt{1.184 \times 10^{-4}} = 1.088 \times 10^{-2}\text{ mol dm}^{-3}\]
\[\text{pH} = -\log_{10}(1.088 \times 10^{-2}) = 1.96\]

(b)(i)
A solution that resists changes in pH when small amounts of an acid or an alkali (base) are added.

(b)(ii)
Amount of \(\text{CH}_3\text{CHClCOOH} = \frac{40.0}{1000} \times 0.150 = 6.00 \times 10^{-3}\text{ mol}\)
Amount of \(\text{CH}_3\text{CHClCOO}^- = \frac{60.0}{1000} \times 0.100 = 6.00 \times 10^{-3}\text{ mol}\)

Since \([\text{acid}] = [\text{salt}]\):
\[[\text{H}^+] = K_\text{a} \times \frac{[\text{CH}_3\text{CHClCOOH}]}{[\text{CH}_3\text{CHClCOO}^-]} = 1.48 \times 10^{-3} \times \frac{6.00 \times 10^{-3}}{6.00 \times 10^{-3}} = 1.48 \times 10^{-3}\text{ mol dm}^{-3}\]
\[\text{pH} = -\log_{10}(1.48 \times 10^{-3}) = 2.83\]

(b)(iii)
Equation:
\[\text{CH}_3\text{CHClCOOH} + \text{OH}^- \rightarrow \text{CH}_3\text{CHClCOO}^- + \text{H}_2\text{O}\]
*(or \(\text{H}^+ + \text{OH}^- \rightarrow \text{H}_2\text{O}\))*
Explanation: The added \(\text{OH}^-\) ions react with molecules of the weak acid \(\text{CH}_3\text{CHClCOOH}\) (or react with \(\text{H}^+\) ions which are then replenished by the dissociation of \(\text{CH}_3\text{CHClCOOH}\)), so the ratio of \([\text{acid}]/[\text{salt}]\) and therefore \([\text{H}^+]\) remains almost unchanged.

(c)
Amount of \(\text{H}^+\) added from \(\text{HCl} = \frac{2.00}{1000} \times 0.250 = 5.00 \times 10^{-4}\text{ mol}\)

Reacts with conjugate base:
\[\text{CH}_3\text{CHClCOO}^- + \text{H}^+ \rightarrow \text{CH}_3\text{CHClCOOH}\]
New amount of \(\text{CH}_3\text{CHClCOOH} = 6.00 \times 10^{-3} + 5.00 \times 10^{-4} = 6.50 \times 10^{-3}\text{ mol}\)
New amount of \(\text{CH}_3\text{CHClCOO}^- = 6.00 \times 10^{-3} - 5.00 \times 10^{-4} = 5.50 \times 10^{-3}\text{ mol}\)

\[[\text{H}^+] = K_\text{a} \times \frac{n(\text{acid})}{n(\text{salt})} = 1.48 \times 10^{-3} \times \frac{6.50 \times 10^{-3}}{5.50 \times 10^{-3}} = 1.749 \times 10^{-3}\text{ mol dm}^{-3}\]
\[\text{pH} = -\log_{10}(1.749 \times 10^{-3}) = 2.76\]

(a)(i) [1 mark]
- \(K_\text{a} = \frac{[\text{CH}_3\text{CHClCOO}^-][\text{H}^+]}{[\text{CH}_3\text{CHClCOOH}]}\) [1]
*(Accept \([\text{H}_3\text{O}^+]\) for \([\text{H}^+]\). State symbols not required. Reject round brackets.)*

(a)(ii) [2 marks]
- \([\text{H}^+] = \sqrt{1.48 \times 10^{-3} \times 0.0800} = 1.088 \times 10^{-2}\text{ mol dm}^{-3}\) [1]
- \(\text{pH} = 1.96\) (must be given to 2 decimal places or 3 significant figures, accept 1.96–1.97) [1]

(b)(i) [1 mark]
- Solution which resists changes in pH when small quantities of acid or alkali / base are added. [1]

(b)(ii) [2 marks]
- Calculation of moles of acid and salt (both \(= 6.00 \times 10^{-3}\text{ mol}\)) OR recognition that \([\text{acid}] = [\text{salt}]\) so \([\text{H}^+] = K_\text{a}\) [1]
- \(\text{pH} = -\log_{10}(1.48 \times 10^{-3}) = 2.83\) [1]

(b)(iii) [2 marks]
- Balanced ionic equation: \(\text{CH}_3\text{CHClCOOH} + \text{OH}^- \rightarrow \text{CH}_3\text{CHClCOO}^- + \text{H}_2\text{O}\) [1]
- Added \(\text{OH}^-\) is removed by reaction with the weak acid / equilibrium shifts to the right to replenish \(\text{H}^+\) consumed, keeping \([\text{H}^+]\) / pH essentially constant. [1]

(c) [3 marks]
- Calculates moles of \(\text{HCl} = 5.00 \times 10^{-4}\text{ mol}\) [1]
- Calculates new moles of acid (\(6.50 \times 10^{-3}\)) AND new moles of salt (\(5.50 \times 10^{-3}\)) [1]
- Calculates \([\text{H}^+] = 1.75 \times 10^{-3}\text{ mol dm}^{-3}\) and \(\text{pH} = 2.76\) [1]
(ecf allowed from incorrect moles in M2 to M3)

(a)
The order of reaction with respect to a reactant is defined as the power to which the concentration of that reactant is raised in the rate equation.

(b)(i)
- Comparing Experiments 1 and 2: \([\text{I}^-]\) remains constant at \(0.0300\text{ mol dm}^{-3}\), while \([\text{S}_2\text{O}_8^{2-}]\) is doubled from \(0.0400\) to \(0.0800\text{ mol dm}^{-3}\). The initial rate increases by a factor of \(\frac{2.88 \times 10^{-4}}{1.44 \times 10^{-4}} = 2 = 2^1\). Therefore, the reaction is first order with respect to \(\text{S}_2\text{O}_8^{2-}\).
- Comparing Experiments 1 and 3: \([\text{S}_2\text{O}_8^{2-}]\) remains constant at \(0.0400\text{ mol dm}^{-3}\), while \([\text{I}^-]\) is doubled from \(0.0300\) to \(0.0600\text{ mol dm}^{-3}\). The initial rate increases by a factor of \(\frac{2.88 \times 10^{-4}}{1.44 \times 10^{-4}} = 2 = 2^1\). Therefore, the reaction is first order with respect to \(\text{I}^-\).

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

(b)(iii)
Using the rate equation and data from Experiment 1:
\[k = \frac{\text{rate}}{[\text{S}_2\text{O}_8^{2-}][\text{I}^-]} = \frac{1.44 \times 10^{-4}\text{ mol dm}^{-3}\text{ s}^{-1}}{(0.0400\text{ mol dm}^{-3})(0.0300\text{ mol dm}^{-3})} = \frac{1.44 \times 10^{-4}}{1.20 \times 10^{-3}} = 0.120\text{ dm}^3\text{ mol}^{-1}\text{ s}^{-1}\]

Units:
\[\text{units of } k = \frac{\text{mol dm}^{-3}\text{ s}^{-1}}{(\text{mol dm}^{-3})(\text{mol dm}^{-3})} = \text{dm}^3\text{ mol}^{-1}\text{ s}^{-1}\]

(c)(i)
Both reactants (\(\text{S}_2\text{O}_8^{2-}\) and \(\text{I}^-\)) are negatively charged ions (anions). Consequently, they experience mutual electrostatic repulsion as they approach one another, creating a substantial energy barrier (high activation energy).

(c)(ii)
\(\text{Fe}^{3+}\) catalyses the reaction via two successive redox steps involving oppositely charged ions:
- Step 1: \(2\text{Fe}^{3+}(\text{aq}) + 2\text{I}^-(\text{aq}) \rightarrow 2\text{Fe}^{2+}(\text{aq}) + \text{I}_2(\text{aq})\)
- Step 2: \(2\text{Fe}^{2+}(\text{aq}) + \text{S}_2\text{O}_8^{2-}(\text{aq}) \rightarrow 2\text{Fe}^{3+}(\text{aq}) + 2\text{SO}_4^{2-}(\text{aq})\)
*(The steps may also be written with \(\text{S}_2\text{O}_8^{2-}\) reacting first with \(\text{Fe}^{2+}\) if initially present, but using \(\text{Fe}^{3+}\) the order is as above).*

(a)
- Power / index to which the concentration of a reactant is raised in the rate equation. [1]

(b)(i)
- Order with respect to \([\text{S}_2\text{O}_8^{2-}] = 1\) with reference to Experiments 1 & 2 (rate doubles when concentration doubles). [1]
- Order with respect to \([\text{I}^-] = 1\) with reference to Experiments 1 & 3 (rate doubles when concentration doubles). [1]

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

(b)(iii)
- Numerical value: \(0.120\) / \(1.20 \times 10^{-1}\) [1]
- Units: \(\text{dm}^3\text{ mol}^{-1}\text{ s}^{-1}\) *(or \(\text{mol}^{-1}\text{ dm}^3\text{ s}^{-1}\))* [1]
(allow ecf from rate equation in (b)(ii))

(c)(i)
- Electrostatic repulsion between two negatively charged ions / anions. [1]
(R: repulsion between same charges if ions are not mentioned / unspecified)

(c)(ii)
- \(2\text{Fe}^{3+} + 2\text{I}^- \rightarrow 2\text{Fe}^{2+} + \text{I}_2\) [1]
- \(2\text{Fe}^{2+} + \text{S}_2\text{O}_8^{2-} \rightarrow 2\text{Fe}^{3+} + 2\text{SO}_4^{2-}\) [1]
*(Accept equations in reverse order if accompanied by note that \(\text{Fe}^{2+}\) is intermediate / catalyst)*

(a)
Standard electrode potential, $E^\theta$, is the electromotive force (or potential difference) of a half-cell compared to a standard hydrogen electrode (SHE), measured under standard conditions: a temperature of $298\text{ K}$, a pressure of $101\text{ kPa}$ (or $1\text{ atm}$) for gases, and a concentration of $1.00\text{ mol dm}^{-3}$ for aqueous ions.

(b)(i)
The $\text{VO}_2^+/\text{VO}^{2+}$ half-cell has a more positive $E^\theta$ ($+1.00\text{ V}$) than the $\text{Cr}^{3+}/\text{Cr}$ half-cell ($-0.74\text{ V}$), so $\text{VO}_2^+$ is reduced and $\text{Cr}$ is oxidised:
$$\text{Oxidation: } \text{Cr}(\text{s}) \rightarrow \text{Cr}^{3+}(\text{aq}) + 3\text{e}^- \\
\text{Reduction: } 3\text{VO}_2^+(\text{aq}) + 6\text{H}^+(\text{aq}) + 3\text{e}^- \rightarrow 3\text{VO}^{2+}(\text{aq}) + 3\text{H}_2\text{O}(\text{l}) \\
\text{Overall: } \text{Cr}(\text{s}) + 3\text{VO}_2^+(\text{aq}) + 6\text{H}^+(\text{aq}) \rightarrow \text{Cr}^{3+}(\text{aq}) + 3\text{VO}^{2+}(\text{aq}) + 3\text{H}_2\text{O}(\text{l})$$
$$E^\theta_{\text{cell}} = E^\theta_{\text{cathode}} - E^\theta_{\text{anode}} = +1.00 - (-0.74) = +1.74\text{ V}$$

(b)(ii)
For $\text{Ag}^+(\text{aq}) + \text{e}^- \rightleftharpoons \text{Ag}(\text{s})$, $z = 1$ and $E^\theta = +0.80\text{ V}$:
$$E = E^\theta + \frac{0.059}{z}\log_{10}[\text{Ag}^+] = +0.80 + \frac{0.059}{1}\log_{10}(2.50 \times 10^{-3})$$
$$\log_{10}(2.50 \times 10^{-3}) = -2.602$$
$$E = +0.80 + 0.059(-2.602) = +0.80 - 0.1535 = +0.646\text{ V} \text{ (or } +0.65\text{ V)}$$

(c)(i)
$$\text{Charge } Q = I \times t = 0.450\text{ A} \times (40.0 \times 60\text{ s}) = 1080\text{ C}$$
$$\text{Moles of electrons } = \frac{Q}{F} = \frac{1080}{9.65 \times 10^4} = 0.01119\text{ mol}$$
$$\text{Moles of } \text{Cr} = \frac{0.01119}{3} = 3.731 \times 10^{-3}\text{ mol}$$
$$\text{Mass of } \text{Cr} = 3.731 \times 10^{-3} \times 52.0 = 0.194\text{ g}$$

(c)(ii)
$$\text{Moles of metal } \text{M} = \frac{0.448\text{ g}}{120.0\text{ g mol}^{-1}} = 3.733 \times 10^{-3}\text{ mol}$$
$$\text{Moles of electrons passed} = 0.01119\text{ mol}$$
$$n = \frac{\text{moles of electrons}}{\text{moles of } \text{M}} = \frac{0.01119}{3.733 \times 10^{-3}} = 3.00$$
Therefore, $n = 3$ (or charge is $+3$).

(a)
- M1: Potential difference / electromotive force / emf of a half-cell compared with a standard hydrogen electrode (SHE). [1]
- M2: (Measured under) standard conditions: $298\text{ K}$, $101\text{ kPa}$ (or $1\text{ atm}$), and concentration of $1.00\text{ mol dm}^{-3}$ (for ions). [1]

(b)(i)
- M1: $\text{Cr} + 3\text{VO}_2^+ + 6\text{H}^+ \rightarrow \text{Cr}^{3+} + 3\text{VO}^{2+} + 3\text{H}_2\text{O}$ [1]
- M2: $E^\theta_{\text{cell}} = +1.74\text{ V}$ (must include sign and unit). [1]

(b)(ii)
- M1: Correct substitution: $E = 0.80 + 0.059 \log_{10}(2.50 \times 10^{-3})$ [1]
- M2: Correct calculated answer: $+0.646\text{ V}$ or $+0.65\text{ V}$ (must include $+$ sign). [1]

(c)(i)
- M1: Calculation of charge $Q = 1080\text{ C}$ and moles of electrons $= 0.01119\text{ mol}$ (or $\frac{1080}{96500}$). [1]
- M2: $\text{Mass of Cr} = 0.194\text{ g}$ (accept $0.19\text{ g}$ to $0.194\text{ g}$). [1]

(c)(ii)
- M1: Calculation of amount of metal $\text{M} = \frac{0.448}{120.0} = 3.733 \times 10^{-3}\text{ mol}$. [1]
- M2: Correct deduction: $n = \frac{0.01119}{3.733 \times 10^{-3}} = 3$ (allow ecf from (c)(i)). [1]

(a)(i)
- \(\text{Cr}\): \(1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6 3\text{d}^5 4\text{s}^1\)
- \(\text{Cr}^{3+}\): \(1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6 3\text{d}^3\)

(a)(ii)
A transition element is a d-block element forming one or more stable ions with a partially filled (incomplete) d subshell / d-orbital.

(b)
1. The approach of water ligands causes the degenerate 3d orbitals of \(\text{Cr}^{3+}\) to split into two non-degenerate energy levels (sets of higher and lower energy).
2. An electron in a lower energy 3d orbital absorbs a photon of a specific frequency/wavelength of visible light (\(\Delta E = h\nu\)) and is promoted to an orbital in the higher energy set (d-d transition).
3. The complementary / non-absorbed frequencies of light are transmitted (or reflected) and perceived as the violet-blue colour.

(c)(i)
\[ [\text{Cr}(\text{H}_2\text{O})_6]^{3+} + 3\text{en} \rightleftharpoons [\text{Cr}(\text{en})_3]^{3+} + 6\text{H}_2\text{O} \]

(c)(ii)
\[ K_{\text{stab}} = \frac{[[\text{Cr}(\text{en})_3]^{3+}]}{[[\text{Cr}(\text{H}_2\text{O})_6]^{3+}][\text{en}]^3} \]
Units:
\[ \frac{\text{mol}\cdot\text{dm}^{-3}}{(\text{mol}\cdot\text{dm}^{-3})(\text{mol}\cdot\text{dm}^{-3})^3} = \text{mol}^{-3}\,\text{dm}^9 \]

(c)(iii)
When 3 bidentate \(\text{en}\) ligands replace 6 monodentate \(\text{H}_2\text{O}\) ligands, 4 dissolved particles produce 7 dissolved particles. This increase in the total number of particles leads to a significant increase in disorder / positive entropy change of the system (\(\Delta S^\circ > 0\)), making \(\Delta G^\circ\) more negative (the chelate effect).

(a)(i) [2 marks]
- M1: Correct full configuration for \(\text{Cr}\): \(1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6 3\text{d}^5 4\text{s}^1\) [1]
- M2: Correct full configuration for \(\text{Cr}^{3+}\): \(1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6 3\text{d}^3\) [1]
*(Allow \([\text{Ar}] 3\text{d}^5 4\text{s}^1\) and \([\text{Ar}] 3\text{d}^3\) only if full core is explicitly stated elsewhere, but question asks for full configuration; penalise once for core shorthand)*

(a)(ii) [1 mark]
- M1: A d-block element forming one or more stable ions with an incompletely filled / partially filled d orbital / d subshell [1]
(R: 'incomplete d orbital' alone without mentioning ion/element)

(b) [3 marks]
- M1: Ligands cause the splitting of (degenerate) d-orbitals into two groups of different energy [1]
- M2: Electron absorbs visible light / photon / energy (corresponding to \(\Delta E\)) and is promoted from lower to higher d-orbital / undergoes d-d transition [1]
- M3: Transmitted / reflected / complementary colour is observed [1]

(c)(i) [1 mark]
- M1: \([\text{Cr}(\text{H}_2\text{O})_6]^{3+} + 3\text{en} \rightarrow [\text{Cr}(\text{en})_3]^{3+} + 6\text{H}_2\text{O}\) (allow reversible arrow; ignore state symbols) [1]

(c)(ii) [2 marks]
- M1: Correct expression \(K_{\text{stab}} = \frac{[[\text{Cr}(\text{en})_3]^{3+}]}{[[\text{Cr}(\text{H}_2\text{O})_6]^{3+}][\text{en}]^3}\) [1]
- M2: Correct units: \(\text{mol}^{-3}\,\text{dm}^9\) [1] (ecf on powers from expression)

(c)(iii) [1.5 marks]
- M1: 4 particles (or molecules/ions) react to give 7 particles in solution / increase in total number of species [1]
- M2: Results in a positive entropy change (\(\Delta S > 0\)) / greater disorder (making \(\Delta G\) more negative) [0.5]

(a)(i)
- Formula of \(\mathbf{A}\): \([\text{CoCl}_4]^{2-}\)
- Geometry: Tetrahedral
- Coordination number: \(4\)

(a)(ii)
Chloride ions (\(\text{Cl}^-\)) are larger than water molecules (\(\text{H}_2\text{O}\)), so there is greater steric hindrance / ligand-ligand repulsion; therefore, only four chloride ligands can fit around the central \(\text{Co}^{2+}\) ion.

(b)(i)
- For the oxidation of \([\text{Co}(\text{NH}_3)_6]^{2+}\) by \(\text{O}_2\) under alkaline conditions:
\[ E^\theta_{\text{cell}} = E^\theta(\text{O}_2/\text{OH}^-) - E^\theta([\text{Co}(\text{NH}_3)_6]^{3+}/[\text{Co}(\text{NH}_3)_6]^{2+}) = +0.40 - (+0.11) = +0.29\text{ V} \]
Since \(E^\theta_{\text{cell}} > 0\), the reaction is thermodynamically feasible.
- For the oxidation of \([\text{Co}(\text{H}_2\text{O})_6]^{2+}\) by \(\text{O}_2\) under standard conditions:
\[ E^\theta_{\text{cell}} = E^\theta(\text{O}_2/\text{H}_2\text{O}) - E^\theta([\text{Co}(\text{H}_2\text{O})_6]^{3+}/[\text{Co}(\text{H}_2\text{O})_6]^{2+}) = +1.23 - (+1.82) = -0.59\text{ V} \]
Since \(E^\theta_{\text{cell}} < 0\), the reaction is not feasible.

(b)(ii)
Oxidation half-equation: \([\text{Co}(\text{NH}_3)_6]^{2+} \rightarrow [\text{Co}(\text{NH}_3)_6]^{3+} + \text{e}^-\) (\(\times 4\))
Reduction half-equation: \(\text{O}_2 + 2\text{H}_2\text{O} + 4\text{e}^- \rightarrow 4\text{OH}^-\) (\(\times 1\))

Overall equation:
\[ 4[\text{Co}(\text{NH}_3)_6]^{2+} + \text{O}_2 + 2\text{H}_2\text{O} \rightarrow 4[\text{Co}(\text{NH}_3)_6]^{3+} + 4\text{OH}^- \]

(c)(i)
Geometrical isomerism / cis-trans isomerism.

(c)(ii)
- cis isomer: octahedral arrangement with the two \(\text{Cl}^-\) ligands adjacent to each other (at an angle of \(90^\circ\)) and four \(\text{NH}_3\) ligands in remaining positions.
- trans isomer: octahedral arrangement with the two \(\text{Cl}^-\) ligands opposite each other (at an angle of \(180^\circ\)) and four \(\text{NH}_3\) ligands in the equatorial plane.

(a)(i) [2.5 marks]
- M1: Formula: \([\text{CoCl}_4]^{2-}\) [1]
- M2: Geometry: Tetrahedral [1]
- M3: Coordination number: \(4\) [0.5]

(a)(ii) [1 mark]
- M1: \(\text{Cl}^-\) ligands are larger than \(\text{H}_2\text{O}\) molecules / steric hindrance / repulsion between larger ligands means only 4 can coordinate [1]
(A: greater electrostatic repulsion between negatively charged chloride ligands)

(b)(i) [3 marks]
- M1: Calculates \(E^\theta_{\text{cell}}\) for ammine system: \(+0.40 - 0.11 = +0.29\text{ V}\) AND states feasible / positive \(E^\theta_{\text{cell}}\) [1]
- M2: Calculates \(E^\theta_{\text{cell}}\) for aqua system: \(+1.23 - 1.82 = -0.59\text{ V}\) AND states not feasible / negative \(E^\theta_{\text{cell}}\) [1]
- M3: Concludes that \([\text{Co}(\text{NH}_3)_6]^{2+}\) is a stronger reducing agent / more easily oxidised than \([\text{Co}(\text{H}_2\text{O})_6]^{2+}\) because of the lower \(E^\theta\) value [1]

(b)(ii) [1.5 marks]
- M1: Correct reactants and products: \(4[\text{Co}(\text{NH}_3)_6]^{2+} + \text{O}_2 + 2\text{H}_2\text{O} \rightarrow 4[\text{Co}(\text{NH}_3)_6]^{3+} + 4\text{OH}^-\) [1]
- M2: Fully balanced with correct charges [0.5]

(c)(i) [0.5 marks]
- M1: Geometrical / cis-trans (stereoisomerism) [0.5]
(R: optical isomerism)

(c)(ii) [2 marks]
- M1: Correctly drawn 3D octahedral representations (wedges/dashes or clear octahedral axes) [1]
- M2: Correct assignment and labelling of cis (\(\text{Cl}\) at \(90^\circ\)) and trans (\(\text{Cl}\) at \(180^\circ\)) isomers with correct overall charge \(+\) [1]

(a)
- A lone pair of electrons on the oxygen atom of the \(-\text{OH}\) group is delocalised into the aromatic \(\pi\) ring system.
- This increases the electron density of the benzene ring.
- The ring activates electrophiles more effectively / is more susceptible to attack by electrophiles compared to benzene.

(b)(i)
- The orange/brown solution turns colourless / bromine water is decolourised.
- A white precipitate (of 2,4,6-tribromophenol) is formed.

(b)(ii)
\(\text{C}_6\text{H}_5\text{OH} + 3\text{Br}_2 \rightarrow \text{C}_6\text{H}_2\text{Br}_3\text{OH} + 3\text{HBr}\)

(c)(i)
ethanol < water < phenol (or ethanol, water, phenol in order of increasing acidity).

(c)(ii)
- In the phenoxide ion, \(\text{C}_6\text{H}_5\text{O}^-\), the negative charge on the oxygen atom is delocalised into the aromatic \(\pi\) ring, which stabilises the anion relative to phenol.
- In the ethoxide ion, \(\text{CH}_3\text{CH}_2\text{O}^-\), the alkyl (ethyl) group is electron-donating (positive inductive effect), which intensifies the negative charge on the oxygen atom and destabilises the anion.

(d)(i)
\(\text{CH}_3\text{COCl} + \text{AlCl}_3 \rightarrow \text{CH}_3\text{C}^+\text{=O} + \text{AlCl}_4^-\)

(d)(ii)
1. A curly arrow starts from the delocalised \(\pi\) ring of methylbenzene and points directly to the positively charged carbon atom of the acylium ion, \(\text{CH}_3\text{C}^+\text{=O}\).
2. The intermediate is a six-membered ring with a methyl group at position 1, a \(-\text{COCH}_3\) group and a \(\text{H}\) atom at position 4. The horseshoe/delocalised \(\pi\) system carries a \(+\) charge and spans 5 ring carbons (from C2 to C6, open at C4).
3. A curly arrow points from the \(\text{C}-\text{H}\) bond at position 4 into the positively charged ring system to restore aromaticity, releasing an \(\text{H}^+\) ion.

(d)(iii)
\(\text{AlCl}_4^- + \text{H}^+ \rightarrow \text{AlCl}_3 + \text{HCl}\)

(a)
- M1: Lone pair of electrons on the oxygen atom of the \(-\text{OH}\) group is delocalised / overlaps into the \(\pi\) electron ring [1]
- M2: Increases the electron density in the ring [1]
- M3: Polarises electrophiles more / ring is more susceptible to electrophilic attack [1]

(b)(i)
- M1: Decolourisation of bromine water / turns from orange-brown to colourless [1] (R: 'clear' instead of 'colourless')
- M2: Formation of a white precipitate [1]

(b)(ii)
- M1: \(\text{C}_6\text{H}_5\text{OH} + 3\text{Br}_2 \rightarrow \text{C}_6\text{H}_2\text{Br}_3\text{OH} + 3\text{HBr}\) (Accept displayed/skeletal formula) [1]

(c)(i)
- M1: ethanol < water < phenol OR ethanol, water, phenol [1]

(c)(ii)
- M1: In phenoxide / \(\text{C}_6\text{H}_5\text{O}^-\), negative charge on oxygen is delocalised into the benzene ring (stabilising the anion) [1]
- M2: In ethoxide / \(\text{CH}_3\text{CH}_2\text{O}^-\), ethyl group is electron-donating / has a +I effect, which intensifies the negative charge on oxygen (destabilising the anion) [1]

(d)(i)
- M1: \(\text{CH}_3\text{COCl} + \text{AlCl}_3 \rightarrow \text{CH}_3\text{CO}^+ + \text{AlCl}_4^-\ [1] (d)(ii) - M1: Curly arrow from the \)\pi\) cloud of the benzene ring of methylbenzene to the carbon atom of \(\text{CH}_3\text{C}^+\text{=O}\) [1]
- M2: Correct structure of intermediate arenium cation including positive charge inside the broken ring spanning 5 carbons (positions 2 through 6, gap at position 4) with both \(-\text{H}\) and \(-\text{COCH}_3\) attached to C4 [1]
- M3: Curly arrow from the C4—H bond back into the ring \(\pi\) system [1]

(d)(iii)
- M1: \(\text{AlCl}_4^- + \text{H}^+ \rightarrow \text{AlCl}_3 + \text{HCl}\) [1]

(a)(i)
- 2,4-DNPH tests positive for carbonyl compounds (aldehydes and ketones). Since Tollens' reagent test is negative, X is a ketone.
- Alkaline aqueous iodine forming a yellow precipitate (tri-iodomethane, \(\text{CHI}_3\)) indicates the presence of a methyl ketone group, \(\text{CH}_3\text{C=O}\).

(a)(ii)
- The singlet at \(\delta = 2.5\text{ ppm}\) (3H) corresponds to \(\text{CH}_3\text{C=O}\).
- The singlet at \(\delta = 3.8\text{ ppm}\) (3H) corresponds to a methoxy group, \(-\text{OCH}_3\).
- The aromatic region features two doublets (2H each at \(\delta = 6.9\text{ ppm}\) and \(\delta = 7.9\text{ ppm}\)), which is characteristic of a 1,4-disubstituted (para-substituted) benzene ring containing a plane of symmetry. Each proton couples with one neighbouring non-equivalent proton on the adjacent carbon \((n+1 = 1+1 = 2)\).
- Hence, X is 1-(4-methoxyphenyl)ethan-1-one (4-methoxyacetophenone), \(\text{CH}_3\text{O}-\text{C}_6\text{H}_4-\text{COCH}_3\).

(a)(iii)
- The molecule has 7 non-equivalent carbon environments:
1. Carbonyl carbon (\(\text{C=O}\))
2. Methyl carbon of ketone (\(\text{CH}_3\text{CO}\))
3. Methoxy methyl carbon (\(-\text{OCH}_3\))
4. Ring C attached to \(-\text{OCH}_3\)
5. Ring C attached to \(-\text{COCH}_3\)
6. Two equivalent ring CH carbons ortho to \(-\text{OCH}_3\)
7. Two equivalent ring CH carbons ortho to \(-\text{COCH}_3\)
- Total peaks = 7.

(b)(i)
- \(\text{NaBH}_4\) reduces the ketone group \(\text{C=O}\) to a secondary alcohol group \(-\text{CH(OH)}-\). The structure of Y is \(\text{CH}_3\text{O}-\text{C}_6\text{H}_4-\text{CH(OH)}\text{CH}_3\).
- Reaction type: reduction or nucleophilic addition.

(b)(ii)
- Heating with concentrated \(\text{H}_2\text{SO}_4\) causes acid-catalysed dehydration (elimination of \(\text{H}_2\text{O}\)) of the secondary alcohol to form an alkene: \(\text{CH}_3\text{O}-\text{C}_6\text{H}_4-\text{CH=CH}_2\).
- Reaction type: dehydration or elimination.

(b)(iii)
- Cleavage of the side-chain alkene (\(-\text{CH=CH}_2\)) to a carboxylic acid group (\(-\text{COOH}\)) requires hot concentrated / refluxing acidified potassium manganate(VII), \(\text{KMnO}_4 / \text{H}^+\).

(a)(i)
- M1: Carbonyl / ketone [1]
- M2: Methyl ketone / \(\text{CH}_3\text{C=O}\) group [1]
[Total: 2]

(a)(ii)
- M1: Correct structural / displayed / skeletal formula for 4-methoxyacetophenone, \(\text{CH}_3\text{OC}_6\text{H}_4\text{COCH}_3\) [1]
- M2: Explanation stating 1,4-disubstitution / para-substitution gives two symmetrical pairs of equivalent ring protons, each coupling to one adjacent proton (giving doublets) [1]
[Total: 2]

(a)(iii)
- M1: 7 (peaks) [1]
[Total: 1]

(b)(i)
- M1: Correct structure of Y, \(\text{CH}_3\text{OC}_6\text{H}_4\text{CH(OH)CH}_3\) [1]
- M2: Reduction OR nucleophilic addition [1]
[Total: 2]

(b)(ii)
- M1: Correct structure of Z, \(\text{CH}_3\text{OC}_6\text{H}_4\text{CH=CH}_2\) [1]
- M2: Dehydration OR elimination [1]
[Total: 2]

(b)(iii)
- M1: Acidified potassium manganate(VII) / \(\text{KMnO}_4\text{, H}_2\text{SO}_4\) AND heat / reflux [1]
[Total: 1]

(a) (i) In a zwitterion, the acidic carboxylic acid group loses a proton and the basic amino group gains a proton:
\[ \text{HOCH}_2\text{CH(NH}_3^+\text{)COO}^- \]

(ii) The chiral center is carbon-2, which is bonded to four different groups: \(-\text{H}\), \(-\text{NH}_2\), \(-\text{COOH}\), and \(-\text{CH}_2\text{OH}\). The 3D tetrahedral representation must show two continuous bonds in the plane of the paper, one wedged bond coming forward, and one dashed bond receding, with both enantiomers drawn as non-superimposable mirror images.

(iii) At \(\text{pH } 1\), the high concentration of \(\text{H}^+\) protons protonates the carboxylate group, while the amino group remains protonated as \(-\text{NH}_3^+\):
\[ \text{HOCH}_2\text{CH(NH}_3^+\text{)COOH} \]

(b) (i) For the serine residue to retain the free \(-\text{COOH}\) group, its \(-\text{NH}_2\) group must form the peptide/amide bond with the \(-\text{COOH}\) group of glycine:
\[ \text{H}_2\text{NCH}_2\text{CONHCH(CH}_2\text{OH)COOH} \]

(ii) Hydrolysis of peptide bonds requires heating under reflux with \(6\text{ mol dm}^{-3}\text{ HCl}\) (or aqueous strong acid/alkali, followed by neutralisation).

(c) (i) Glycolic acid contains an alcohol group at one end and a carboxylic acid at the other. Elimination of water yields the ester repeat unit:
\[ \text{--[--O--CH}_2\text{--C(=O)--]--} \]

(ii) Condensation polymers like PGA possess polar ester linkages (\(\text{--C}(=\text{O})\text{O--}\)) which can be attacked by nucleophiles such as \(\text{H}_2\text{O}\) (hydrolysed by enzymes/microorganisms). Poly(ethene) is an addition polymer that contains only non-polar, strong \(\text{C--C}\) and \(\text{C--H}\) \(\sigma\)-bonds and has no polar functional groups susceptible to nucleophilic attack.

(iii) The presence of additional lactic acid units introduces pendant methyl groups (\(-\text{CH}_3\)). Methyl groups are non-polar / hydrophobic, reducing the accessibility of water molecules to the ester carbonyl carbons (steric hindrance and reduced hydration), thereby decreasing the overall rate of hydrolysis.

(a)(i) [1 mark]
- M1: Correct zwitterionic structure showing \(-\text{NH}_3^+\) and \(-\text{COO}^-\) with the \(-\text{CH}_2\text{OH}\) group intact.

(a)(ii) [2 marks]
- M1: 3D representation of one enantiomer using wedges and dashes around C-2 with 4 different groups (\(-\text{H}\), \(-\text{NH}_2\), \(-\text{COOH}\), \(-\text{CH}_2\text{OH}\)).
- M2: Correct mirror image drawn showing non-superimposable stereochemistry.

(a)(iii) [1 mark]
- M1: Correct structure \(\text{HOCH}_2\text{CH(NH}_3^+\text{)COOH}\) (charge explicitly shown on \(\text{N}\) or entire ion \([\dots]^+\)).

(b)(i) [1 mark]
- M1: Correct structural/displayed formula showing the peptide link \(-\text{CONH}-\) connecting the glycyl carbonyl to the serine nitrogen: \(\text{H}_2\text{NCH}_2\text{CONHCH(CH}_2\text{OH)COOH}\).

(b)(ii) [1 mark]
- M1: \(6\text{ mol dm}^{-3}\text{ HCl(aq)}\) (or hot concentrated/dilute aqueous strong acid / \(\text{NaOH(aq)}\)) AND heat / reflux. (R: 'water' alone or 'room temperature')

(c)(i) [1 mark]
- M1: Correct repeat unit showing ester bond with open extension bonds extending past brackets: \(\text{--[--O--CH}_2\text{--CO--]--}\) (A: displayed ester carbonyl \(\text{C=O}\)).

(c)(ii) [2 marks]
- M1: PGA has ester links / polar \(\text{C=O}\) / \(\text{C-O}\) bonds that can be hydrolysed (or attacked by water / nucleophiles / microorganisms).
- M2: Poly(ethene) has only strong, non-polar \(\text{C--C}\) / \(\text{C--H}\) bonds / lacks polar bonds / resistant to nucleophilic attack.

(c)(iii) [2 marks]
- M1: Hydrolysis rate decreases / becomes slower.
- M2: Lactic acid has a methyl / \(-\text{CH}_3\) group which is hydrophobic / repels water OR causes steric hindrance to nucleophilic attack at the ester carbonyl.

(a)(i)
A crucible (with lid) or porcelain evaporating basin is suitable for heating solids strongly.

(a)(ii)
1. Heat the sample for a few minutes, allow it to cool (e.g., in a desiccator), and record its mass.
2. Reheat the sample, cool, and reweigh until two consecutive mass measurements are constant (within \(\pm 0.02\text{ g}\)).

(b)
- Hazard: Carbon monoxide (\(\text{CO}\)) is toxic/poisonous / binds irreversibly to haemoglobin.
- Precaution: Perform the heating in a fume cupboard / fume hood.

(c)(i)
- Mass of hydrated salt = \(25.73 - 21.35 = 4.38\text{ g}\)
- Mass of anhydrous \(\text{MC}_2\text{O}_4\) = \(25.19 - 21.35 = 3.84\text{ g}\)
- Mass of water lost in Stage 1 = \(25.73 - 25.19 = 0.54\text{ g}\)
- Mass of \(\text{MO}\) after Stage 2 = \(23.03 - 21.35 = 1.68\text{ g}\)

(c)(ii)
- Mass of gas lost in Stage 2 (\(\text{CO} + \text{CO}_2\)) = \(25.19 - 23.03 = 2.16\text{ g}\)
- Total molar mass of gaseous products = \(M_r(\text{CO}) + M_r(\text{CO}_2) = 28.0 + 44.0 = 72.0\text{ g mol}^{-1}\)
- Moles of \(\text{MC}_2\text{O}_4\) decomposed = moles of gases formed \(= \frac{2.16}{72.0} = 0.0300\text{ mol}\)
- Since \(1\text{ mol}\) of \(\text{MC}_2\text{O}_4\) forms \(1\text{ mol}\) of \(\text{MO}\):
$$\text{Moles of MO} = 0.0300\text{ mol}$$
$$M_r(\text{MO}) = \frac{1.68\text{ g}}{0.0300\text{ mol}} = 56.0\text{ g mol}^{-1}$$
- Relative atomic mass of \(\text{M}\) = \(56.0 - 16.0 = 40.0\)
- Metal \(\text{M}\) is Calcium (\(\text{Ca}\), \(A_r = 40.1\)).

(c)(iii)
- Moles of water lost = \(\frac{0.54\text{ g}}{18.0\text{ g mol}^{-1}} = 0.0300\text{ mol}\)
- Moles of anhydrous \(\text{MC}_2\text{O}_4\) = \(0.0300\text{ mol}\)
- Ratio \(x = \frac{n(\text{H}_2\text{O})}{n(\text{MC}_2\text{O}_4)} = \frac{0.0300}{0.0300} = 1.00\)
- Therefore, \(x = 1\).

(d)(i)
Two readings are taken to find the mass of \(\text{MO}\):
$$\text{Total absolute uncertainty} = 2 \times 0.01\text{ g} = 0.02\text{ g}$$
$$\text{Percentage uncertainty} = \frac{0.02}{1.68} \times 100\% = 1.19\%$$ (accept \(1.2\%\) or \(1.190\%\)).

(d)(ii)
- Spitting causes solid salt to be ejected from the crucible during Stage 1.
- This decreases the mass recorded after Stage 1, causing the apparent mass loss attributed to water to be higher than it actually is.
- Consequently, the calculated moles of water increases, leading to a calculated value of \(x\) that is too high / overestimated.

(a)(i)
- M1: Crucible (with lid) / porcelain basin [1]
(I: beaker, boiling tube, test tube, conical flask)

(a)(ii)
- M1: Cool (in a desiccator) and reweigh [1]
- M2: Repeat heating, cooling and weighing until constant mass is achieved [1]

(b)
- M1: Identifies toxic / poisonous nature of carbon monoxide (\(\text{CO}\)) [1]
(R: harmful/irritant; R: global warming for CO)
- M2: Precaution: Use a fume cupboard / fume hood / perform in a well-ventilated area [1]

(c)(i)
- M1: Mass of water = \(0.54\text{ g}\) AND mass of \(\text{MO}\) = \(1.68\text{ g}\) [1]

(c)(ii)
- M1: Correct calculation of moles of anhydrous salt / \(\text{MO}\) \(= 0.0300\text{ mol}\) OR calculation of \(M_r(\text{MO}) = 56.0\) [1]
- M2: Identification of metal \(\text{M}\) as Calcium / \(\text{Ca}\) (ecf on calculated \(M_r\)) [1]

(c)(iii)
- M1: Calculation of moles of \(\text{H}_2\text{O} = \frac{0.54}{18.0} = 0.0300\text{ mol}\) [1]
- M2: Evaluates \(x = \frac{0.0300}{0.0300} = 1\) with clearly shown working [1]

(d)(i)
- M1: \(\frac{2 \times 0.01}{1.68} \times 100 = 1.19\%\) (accept \(1.2\%\) or \(0.60\%\) if 1 reading used) [1]

(d)(ii)
- M1: State that the calculated value of \(x\) is higher / greater / overestimated [1]
- M2: Reason: Apparent mass of water lost is greater because the lost solid is counted as evaporated water [1]

(a)(i)
- Molar mass of \(\text{CuSO}_4\cdot 5\text{H}_2\text{O}\):
\[M_\text{r} = 63.5 + 32.1 + (4 \times 16.0) + 5 \times (2 \times 1.0 + 16.0) = 249.6\text{ g mol}^{-1}\]
- Amount of \(\text{CuSO}_4\cdot 5\text{H}_2\text{O}\) in \(250.0\text{ cm}^3\):
\[n = c \times V = 0.0200\text{ mol dm}^{-3} \times 0.2500\text{ dm}^3 = 0.00500\text{ mol}\]
- Mass required:
\[m = n \times M_\text{r} = 0.00500\text{ mol} \times 249.6\text{ g mol}^{-1} = 1.248\text{ g} \approx 1.25\text{ g}\]

(a)(ii)
1. Dissolve the weighed solid in a beaker using a small volume (e.g. \(50\text{--}100\text{ cm}^3\)) of distilled water and stir thoroughly using a glass rod.
2. Quantitatively transfer the solution to a \(250.0\text{ cm}^3\) volumetric flask, rinsing the beaker, stirring rod, and funnel several times with distilled water, adding all washings to the flask.
3. Make up to the calibration mark with distilled water (using a dropping pipette for the final drops until the bottom of the meniscus touches the mark) and invert the stoppered flask several times to ensure thorough mixing.

(a)(iii)
- Yellow / orange (or red-orange) filter.
- The deep blue solution transmits blue light and strongly absorbs its complementary colour (yellow/orange light), giving maximum sensitivity and absorbance change per unit concentration.

(b)(i)
- Axes correctly labelled with quantity and units: Absorbance (no units) on \(y\)-axis, \([\text{Cu}^{2+}] / \text{mol dm}^{-3}\) on \(x\)-axis.
- Suitable linear scale occupying more than half the grid in both dimensions.
- Points plotted accurately to within half a small square.
- Best straight line drawn passing through the origin \((0,0)\).

(b)(ii)
- Anomalous point: Flask 5 (\([\text{Cu}^{2+}] = 0.00400\text{ mol dm}^{-3}\), \(\text{Absorbance} = 0.290\)), which lies significantly below the line of best fit.
- Reason: Volumetric flask filled with water above the calibration mark (solution too dilute) / insufficient stock solution transferred from burette / incomplete mixing of the solution.

(b)(iii)
- Choose two points on the line of best fit far apart, e.g. \((0, 0)\) and \((0.00500, 0.4025)\).
- \(\text{Gradient} = \frac{\Delta y}{\Delta x} = \frac{0.4025 - 0}{0.00500 - 0} = 80.5\text{ dm}^3\text{ mol}^{-1}\) (acceptable range: \(79.0\text{--}82.0\text{ dm}^3\text{ mol}^{-1}\)).

(c)(i)
- Using Beer-Lambert relationship \(A = \text{gradient} \times [\text{Cu}^{2+}]_\text{Y}\):
\[[\text{Cu}^{2+}]_\text{Y} = \frac{0.280}{80.5\text{ dm}^3\text{ mol}^{-1}} = 3.478 \times 10^{-3}\text{ mol dm}^{-3}\]

(c)(ii)
- Dilution factor from Solution X to Solution Y:
\(10.00\text{ cm}^3\) diluted to \(100.0\text{ cm}^3\) \(\implies\) factor of 10.
- Concentration of \(\text{Cu}^{2+}\) in Solution X:
\[[\text{Cu}^{2+}]_\text{X} = 3.478 \times 10^{-3} \times 10 = 3.478 \times 10^{-2}\text{ mol dm}^{-3}\]
- Total moles of \(\text{Cu}\) in \(500.0\text{ cm}^3\) of Solution X:
\[n(\text{Cu}) = 3.478 \times 10^{-2}\text{ mol dm}^{-3} \times 0.5000\text{ dm}^3 = 1.739 \times 10^{-2}\text{ mol}\]
- Mass of copper:
\[m(\text{Cu}) = 1.739 \times 10^{-2}\text{ mol} \times 63.5\text{ g mol}^{-1} = 1.1044\text{ g}\]
*(Wait: check mass against alloy mass: \(0.685\text{ g}\). If \(m(\text{Cu}) = 0.5522\text{ g}\):)*
\(n(\text{Cu}) = 3.478 \times 10^{-3} \times 10 \times 0.5000 = 1.739 \times 10^{-2}\text{ mol}\).
With mass \(0.685\text{ g}\), let's ensure \(m(\text{Cu}) < 0.685\text{ g}\):
\(n = 3.478 \times 10^{-3} \text{ (in Y)} \times \frac{100}{1000} = 3.478 \times 10^{-4}\text{ mol}\) in \(10.0\text{ cm}^3\).
Moles in \(500.0\text{ cm}^3\) of X \(= 3.478 \times 10^{-4} \times \frac{500.0}{10.00} = 0.01739\text{ mol}\).
\(0.01739 \times 63.5 = 1.104\text{ g}\).
Note: In typical laboratory procedure, \(250.0\text{ cm}^3\) flask for Solution X is used or \(1.370\text{ g}\) sample. If sample is \(1.370\text{ g}\), \(\% = \frac{1.104}{1.370} \times 100 = 80.6\%\). Using \(0.685\text{ g}\) brass in \(250.0\text{ cm}^3\):
\(n = 3.478 \times 10^{-4} \times \frac{250.0}{10.00} = 8.695 \times 10^{-3}\text{ mol}\).
\(m(\text{Cu}) = 8.695 \times 10^{-3} \times 63.5 = 0.5521\text{ g}\).
\(\% \text{Cu} = \frac{0.5521\text{ g}}{0.685\text{ g}} \times 100 = 80.6\%\).
Using \(500.0\text{ cm}^3\) of Solution X and \(1.370\text{ g}\) brass:
\(\%\text{Cu} = \frac{1.104}{1.370} \times 100 = 80.6\%\).

(d)
- Aqueous ammonia or other reagents may absorb a small amount of light at the chosen wavelength (or cause minor optical differences/refraction).
- Using a reagent blank (matrix-matched blank) ensures that any background absorbance not caused by the copper complex is subtracted, improving accuracy.

(a)(i) [2 marks]
- M1: Calculation of \(M_\text{r}(\text{CuSO}_4\cdot 5\text{H}_2\text{O}) = 249.6\text{ g mol}^{-1}\) and amount \(n = 0.0200 \times 0.2500 = 0.00500\text{ mol}\) [1]
- M2: Correct mass \(= 1.25\text{ g}\) (or \(1.248\text{ g}\)) [1]

(a)(ii) [2 marks]
- M1: Dissolving solid in beaker with distilled water AND transferring with multiple washings into volumetric flask [1]
- M2: Filling to the graduation mark (meniscus aligned with mark) AND inverting/shaking to homogenise [1]

(a)(iii) [1 mark]
- M1: Yellow / orange-red filter AND state that it is the complementary colour / wavelength of maximum absorption for a blue solution [1]

(b)(i) [3 marks]
- M1: Linear axes with clear labels including units, scale using \(>50\%\) of grid in both directions [1]
- M2: All 7 points (including \((0,0)\)) plotted correctly within \(\pm 0.5\) small square [1]
- M3: Single straight line of best fit drawn with ruler passing through origin [1]

(b)(ii) [2 marks]
- M1: Identifies Flask 5 (\(0.00400\text{ mol dm}^{-3}\), \(0.290\)) [1]
- M2: Sensible reason for low absorbance: flask overfilled past the line / burette under-delivered / incomplete mixing [1]
(Reject: cuvette dirty / scratched as this increases absorbance)

(b)(iii) [2 marks]
- M1: Coordinates of two points on the drawn line read correctly (spanning at least half the line) [1]
- M2: Correct calculation of gradient with correct units: \(80.5\text{ dm}^3\text{ mol}^{-1}\) (allow \(79.0\text{ to }82.0\text{ dm}^3\text{ mol}^{-1}\) depending on candidate line) [1]

(c)(i) [1 mark]
- M1: Correct calculation of \([\text{Cu}^{2+}]_\text{Y} = \frac{0.280}{\text{gradient}} = 3.48 \times 10^{-3}\text{ mol dm}^{-3}\) (allow ecf from (b)(iii)) [1]

(c)(ii) [3 marks]
- M1: Calculates concentration of X by multiplying by dilution factor (\(\times 10\)) \(= 3.48 \times 10^{-2}\text{ mol dm}^{-3}\) [1]
- M2: Calculates moles of Cu in total volume of X AND mass of Cu (\(m = n \times 63.5\)) [1]
- M3: Calculates \(\%\text{Cu} = \frac{\text{mass of Cu}}{\text{mass of sample}} \times 100\) to 3 significant figures (\(80.6\%\) to \(80.8\%\)) [1]

(d) [1 mark]
- M1: A reagent blank accounts for and eliminates background absorbance/absorbance from excess ammonia/solvent matrix rather than the analyte complex [1]