Cambridge IAL · thinka-original Practice Paper

2026 Cambridge IAL Chemistry (9701) Practice Paper with Answers

Thinka Jun 2026 (V4) Cambridge International A Level-Style Mock — Chemistry (9701)

270 marks465 mins2026
An original Thinka practice paper modelled on the structure and difficulty of the Jun 2026 (V4) Cambridge International A Level Chemistry (9701) paper. Not affiliated with or reproduced from Cambridge.

Paper 1 (Multiple Choice)

Answer all 40 multiple choice questions on the separate answer sheet. Soft pencil (B or HB) is recommended.
40 Question · 40 marks
Question 1 · Multiple Choice
1 marks
A sample of an unknown gaseous hydrocarbon has a mass of \(0.210\text{ g}\) and occupies a volume of \(120\text{ cm}^3\) at a temperature of \(298\text{ K}\) and a pressure of \(1.01 \times 10^5\text{ Pa}\).

What is the molecular formula of this hydrocarbon?

[\(R = 8.31\text{ J}\cdot\text{K}^{-1}\cdot\text{mol}^{-1}\)]
  1. A.\(\text{CH}_4\)
  2. B.\(\text{C}_2\text{H}_4\)
  3. C.\(\text{C}_3\text{H}_6\)
  4. D.\(\text{C}_4\text{H}_8\)
Show answer & marking scheme

Worked solution

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

1. Convert volume to \(\text{m}^3\):
\[V = 120\text{ cm}^3 = 120 \times 10^{-6}\text{ m}^3 = 1.20 \times 10^{-4}\text{ m}^3\]

2. Calculate the amount in moles, \(n\):
\[n = \frac{pV}{RT} = \frac{(1.01 \times 10^5\text{ Pa}) \times (1.20 \times 10^{-4}\text{ m}^3)}{(8.31\text{ J}\cdot\text{K}^{-1}\cdot\text{mol}^{-1}) \times 298\text{ K}} = \frac{12.12}{2476.38} \approx 4.894 \times 10^{-3}\text{ mol}\]

3. Calculate the molar mass, \(M_\text{r}\):
\[M_\text{r} = \frac{\text{mass}}{n} = \frac{0.210\text{ g}}{4.894 \times 10^{-3}\text{ mol}} \approx 42.9\text{ g}\cdot\text{mol}^{-1}\]

Comparing with the molar masses of the options:
- \(\text{CH}_4 = 16.0\text{ g}\cdot\text{mol}^{-1}\)
- \(\text{C}_2\text{H}_4 = 28.0\text{ g}\cdot\text{mol}^{-1}\)
- \(\text{C}_3\text{H}_6 = 42.1\text{ g}\cdot\text{mol}^{-1}\)
- \(\text{C}_4\text{H}_8 = 56.1\text{ g}\cdot\text{mol}^{-1}\)

The hydrocarbon is \(\text{C}_3\text{H}_6\).

Marking scheme

C is correct (1 mark for calculating \(M_\text{r} \approx 42\text{--}43\text{ g}\cdot\text{mol}^{-1}\) and identifying \(\text{C}_3\text{H}_6\)).
Question 2 · Multiple Choice
1 marks
The first eight successive ionisation energies of an element \(\text{X}\), in \(\text{kJ}\cdot\text{mol}^{-1}\), are shown:

\(738,\ 1450,\ 7730,\ 10540,\ 13630,\ 18000,\ 21700,\ 25660\)

Which formula represents the chloride of element \(\text{X}\)?
  1. A.\(\text{XCl}\)
  2. B.\(\text{XCl}_2\)
  3. C.\(\text{XCl}_3\)
  4. D.\(\text{XCl}_4\)
Show answer & marking scheme

Worked solution

The successive ionisation energies show a large jump between the 2nd and 3rd ionisation energies (from \(1450\text{ kJ}\cdot\text{mol}^{-1}\) to \(7730\text{ kJ}\cdot\text{mol}^{-1}\)). This indicates that element \(\text{X}\) has two valence electrons and belongs to Group 2 of the Periodic Table.

Group 2 elements form divalent cations \(\text{X}^{2+}\), so the chloride formed with chloride ions (\(\text{Cl}^-\)) has the formula \(\text{XCl}_2\).

Marking scheme

B is correct (1 mark for identifying Group 2 from the jump between the 2nd and 3rd IE values, giving \(\text{XCl}_2\)).
Question 3 · Multiple Choice
1 marks
Which molecule has a trigonal pyramidal shape and a bond angle of approximately \(107^\circ\)?
  1. A.\(\text{BF}_3\)
  2. B.\(\text{NH}_4^+\)
  3. C.\(\text{NF}_3\)
  4. D.\(\text{AlCl}_3\)
Show answer & marking scheme

Worked solution

- \(\text{BF}_3\): 3 bonding pairs, 0 lone pairs \(\rightarrow\) trigonal planar, \(120^\circ\).
- \(\text{NH}_4^+\): 4 bonding pairs, 0 lone pairs \(\rightarrow\) tetrahedral, \(109.5^\circ\).
- \(\text{NF}_3\): 3 bonding pairs, 1 lone pair \(\rightarrow\) trigonal pyramidal, approximately \(107^\circ\) (due to lone pair repulsion reducing the tetrahedral angle).
- \(\text{AlCl}_3\): 3 bonding pairs, 0 lone pairs \(\rightarrow\) trigonal planar, \(120^\circ\).

Therefore, \(\text{NF}_3\) is the correct species.

Marking scheme

C is correct (1 mark for identifying 3 bond pairs and 1 lone pair on central nitrogen in \(\text{NF}_3\)).
Question 4 · Multiple Choice
1 marks
The standard enthalpy changes of combustion, \(\Delta H^\circ_\text{c}\), are given:

- \(\Delta H^\circ_\text{c}(\text{C(s)}) = -394\text{ kJ}\cdot\text{mol}^{-1}\)
- \(\Delta H^\circ_\text{c}(\text{H}_2\text{(g)}) = -286\text{ kJ}\cdot\text{mol}^{-1}\)
- \(\Delta H^\circ_\text{c}(\text{C}_2\text{H}_5\text{OH(l)}) = -1367\text{ kJ}\cdot\text{mol}^{-1}\)

What is the standard enthalpy change of formation, \(\Delta H^\circ_\text{f}\), of ethanol, \(\text{C}_2\text{H}_5\text{OH(l)}\)?
  1. A.\(-279\text{ kJ}\cdot\text{mol}^{-1}\)
  2. B.\(+279\text{ kJ}\cdot\text{mol}^{-1}\)
  3. C.\(-687\text{ kJ}\cdot\text{mol}^{-1}\)
  4. D.\(-3019\text{ kJ}\cdot\text{mol}^{-1}\)
Show answer & marking scheme

Worked solution

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

Using Hess's Law with enthalpies of combustion:
\[\Delta H^\circ_\text{f} = \sum \Delta H^\circ_\text{c}(\text{reactants}) - \sum \Delta H^\circ_\text{c}(\text{products})\]
\[\Delta H^\circ_\text{f} = [2(-394) + 3(-286)] - [-1367]\]
\[\Delta H^\circ_\text{f} = [-788 - 858] + 1367 = -1646 + 1367 = -279\text{ kJ}\cdot\text{mol}^{-1}\]

Marking scheme

A is correct (1 mark for applying Hess's cycle: \(2(-394) + 3(-286) - (-1367) = -279\text{ kJ}\cdot\text{mol}^{-1}\)).
Question 5 · Multiple Choice
1 marks
Solid sodium iodide is treated with concentrated sulfuric acid at room temperature.

Which combination of sulfur-containing reduction products is formed?
  1. A.\(\text{SO}_2\) only
  2. B.\(\text{SO}_2\) and \(\text{H}_2\text{S}\) only
  3. C.\(\text{H}_2\text{S}\) only
  4. D.\(\text{SO}_2\), \(\text{S}\), and \(\text{H}_2\text{S}\)
Show answer & marking scheme

Worked solution

Iodide ions (\(\text{I}^-\)) are strong reducing agents. When solid \(\text{NaI}\) reacts with concentrated \(\text{H}_2\text{SO}_4\), the sulfuric acid is reduced stepwise to:
1. Sulfur dioxide, \(\text{SO}_2\) (oxidation state of S goes from +6 to +4)
2. Sulfur, \(\text{S}\) (oxidation state of S goes from +6 to 0)
3. Hydrogen sulfide, \(\text{H}_2\text{S}\) (oxidation state of S goes from +6 to -2)

Therefore, the sulfur-containing reduction products are \(\text{SO}_2\), \(\text{S}\), and \(\text{H}_2\text{S}\).

Marking scheme

D is correct (1 mark for identifying all three reduction products formed by the reduction of \(\text{H}_2\text{SO}_4\) by iodide ions).
Question 6 · Multiple Choice
1 marks
The reversible reaction between dinitrogen tetraoxide and nitrogen dioxide is represented by the equation:

\[\text{N}_2\text{O}_4\text{(g)} \rightleftharpoons 2\text{NO}_2\text{(g)} \quad \Delta H = +57\text{ kJ}\cdot\text{mol}^{-1}\]

\(\text{N}_2\text{O}_4\text{(g)}\) is colourless and \(\text{NO}_2\text{(g)}\) is dark brown.

The temperature of an equilibrium mixture held in a sealed vessel of fixed volume is increased.

Which statement about the resulting equilibrium mixture is correct?
  1. A.The value of \(K_\text{c}\) increases and the mixture becomes darker brown.
  2. B.The value of \(K_\text{c}\) increases and the total pressure in the vessel decreases.
  3. C.The value of \(K_\text{c}\) decreases and the mixture becomes darker brown.
  4. D.The value of \(K_\text{c}\) remains unchanged and the total pressure in the vessel decreases.
Show answer & marking scheme

Worked solution

1. The forward reaction is endothermic (\(\Delta H > 0\)).
2. According to Le Chatelier's principle, an increase in temperature shifts the position of equilibrium in the endothermic (forward) direction.
3. This increases the concentration of \(\text{NO}_2\text{(g)}\), increasing the value of \(K_\text{c}\) and making the brown colour darker.
4. Shifting to the side with more moles of gas (2 moles vs 1 mole) in a fixed volume also increases the total pressure.

Thus, statement A is correct: the value of \(K_\text{c}\) increases and the mixture becomes darker brown.

Marking scheme

A is correct (1 mark for deducing that \(K_\text{c}\) increases for an endothermic reaction when temperature rises and [\(\text{NO}_2\)] increases, causing a darker brown colour).
Question 7 · Multiple Choice
1 marks
An organic compound \(\text{Y}\) with the molecular formula \(\text{C}_4\text{H}_{10}\text{O}\) forms a yellow precipitate when warmed with alkaline aqueous iodine.

When compound \(\text{Y}\) is heated under reflux with acidified potassium dichromate(VI), an organic compound \(\text{Z}\) is produced.

What is the structural formula of compound \(\text{Z}\)?
  1. A.\(\text{CH}_3\text{COCH}_2\text{CH}_3\)
  2. B.\(\text{CH}_3\text{CH}_2\text{CH}_2\text{COOH}\)
  3. C.\((\text{CH}_3)_2\text{CHCOOH}\)
  4. D.\(\text{CH}_3\text{CH}_2\text{CH}_2\text{CHO}\)
Show answer & marking scheme

Worked solution

1. Compound \(\text{Y}\) has the formula \(\text{C}_4\text{H}_{10}\text{O}\) (an alcohol or ether) and gives a positive tri-iodomethane test (yellow precipitate of \(\text{CHI}_3\)). This requires the presence of a \(\text{CH}_3\text{CH(OH)}-\) group.
2. The only 4-carbon alcohol containing this group is butan-2-ol, \(\text{CH}_3\text{CH(OH)CH}_2\text{CH}_3\).
3. Butan-2-ol is a secondary alcohol. When heated under reflux with acidified \(\text{K}_2\text{Cr}_2\text{O}_7\), it is oxidized to butanone, \(\text{CH}_3\text{COCH}_2\text{CH}_3\).

Marking scheme

A is correct (1 mark for deducing butan-2-ol as \(\text{Y}\) and its oxidation product butanone, \(\text{CH}_3\text{COCH}_2\text{CH}_3\), as \(\text{Z}\)).
Question 8 · Multiple Choice
1 marks
An unlabelled liquid sample is known to be one of the following compounds:

- propanoic acid
- propan-1-ol
- propanal
- ethyl ethanoate

The infrared spectrum of the liquid displays a very broad absorption band in the region \(2500\text{--}3000\text{ cm}^{-1}\) and a strong absorption peak at \(1715\text{ cm}^{-1}\).

Which compound is the liquid?
  1. A.propanoic acid
  2. B.propan-1-ol
  3. C.propanal
  4. D.ethyl ethanoate
Show answer & marking scheme

Worked solution

- The very broad absorption band in the region \(2500\text{--}3000\text{ cm}^{-1}\) is characteristic of the \(\text{O}-\text{H}\) stretching vibration in a carboxylic acid.
- The strong peak at \(1715\text{ cm}^{-1}\) corresponds to the \(\text{C}=\text{O}\) stretching vibration.
- An alcohol would show an \(\text{O}-\text{H}\) stretch at \(3200\text{--}3600\text{ cm}^{-1}\) and no \(\text{C}=\text{O}\) peak.
- An aldehyde or ester would show a \(\text{C}=\text{O}\) peak around \(1710\text{--}1750\text{ cm}^{-1}\) but no broad \(\text{O}-\text{H}\) absorption.

Therefore, the liquid is propanoic acid.

Marking scheme

A is correct (1 mark for identifying carboxylic acid from \(\text{O}-\text{H}\) at \(2500\text{--}3000\text{ cm}^{-1}\) and \(\text{C}=\text{O}\) at \(1715\text{ cm}^{-1}\)).
Question 9 · Multiple Choice
1 marks
A \(0.120\text{ g}\) sample of an unknown volatile liquid \(X\) is vaporised in a gas syringe at \(373\text{ K}\) and \(1.01\times 10^5\text{ Pa}\).

The vapour occupies a volume of \(52.0\text{ cm}^3\).

What is the relative molecular mass, \(M_\text{r}\), of \(X\)?
  1. A.35.4
  2. B.70.8
  3. C.85.3
  4. D.142
Show answer & marking scheme

Worked solution

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

1. Convert quantities to standard SI units:
\(p = 1.01\times 10^5\text{ Pa}\)
\(V = 52.0\times 10^{-6}\text{ m}^3\)
\(T = 373\text{ K}\)
\(m = 0.120\text{ g}\)
\(R = 8.31\text{ J K}^{-1}\text{ mol}^{-1}\)

2. Calculate the amount in moles, \(n\):
\(n = \frac{pV}{RT} = \frac{1.01\times 10^5\times 52.0\times 10^{-6}}{8.31\times 373} = \frac{5.252}{3099.63} = 1.6944\times 10^{-3}\text{ mol}\)

3. Calculate \(M_\text{r}\):
\(M_\text{r} = \frac{m}{n} = \frac{0.120}{1.6944\times 10^{-3}} = 70.8\)

Marking scheme

B is correct [1 mark].
A corresponds to calculating \(M_\text{r}\) with half the mass or doubling the volume.
C and D correspond to arithmetic or unit conversion errors (e.g. failing to convert \(\text{cm}^3\) to \(\text{m}^3\) or using \(273\text{ K}\)).
Question 10 · Multiple Choice
1 marks
Which species has the smallest bond angle?
  1. A.\(\text{NH}_4^+\)
  2. B.\(\text{H}_3\text{O}^+\)
  3. C.\(\text{NH}_2^-\)
  4. D.\(\text{CH}_4\)
Show answer & marking scheme

Worked solution

To determine bond angles, consider electron pair repulsion theory:
- \(\text{NH}_4^+\): 4 bonding pairs, 0 lone pairs \(\rightarrow\) tetrahedral, bond angle \(= 109.5^\circ\)
- \(\text{CH}_4\): 4 bonding pairs, 0 lone pairs \(\rightarrow\) tetrahedral, bond angle \(= 109.5^\circ\)
- \(\text{H}_3\text{O}^+\): 3 bonding pairs, 1 lone pair \(\rightarrow\) trigonal pyramidal, bond angle \(\approx 107^\circ\)
- \(\text{NH}_2^-\): 2 bonding pairs, 2 lone pairs \(\rightarrow\) non-linear (bent), bond angle \(\approx 104.5^\circ\)

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

Marking scheme

C is correct [1 mark].
\(\text{NH}_2^-\) has 2 lone pairs causing maximum repulsion and reducing the bond angle to approximately \(104.5^\circ\).
Question 11 · Multiple Choice
1 marks
Standard enthalpy changes of combustion, \(\Delta H_\text{c}^\ominus\), are given.

- \(\Delta H_\text{c}^\ominus [\text{C(graphite)}] = -394\text{ kJ mol}^{-1}\)
- \(\Delta H_\text{c}^\ominus [\text{H}_2\text{(g)}] = -286\text{ kJ mol}^{-1}\)
- \(\Delta H_\text{c}^\ominus [\text{CH}_3\text{COOH(l)}] = -874\text{ kJ mol}^{-1}\)

What is the standard enthalpy change of formation, \(\Delta H_\text{f}^\ominus\), of ethanoic acid, \(\text{CH}_3\text{COOH(l)}\)?
  1. A.\(-486\text{ kJ mol}^{-1}\)
  2. B.\(-194\text{ kJ mol}^{-1}\)
  3. C.\(+486\text{ kJ mol}^{-1}\)
  4. D.\(-1360\text{ kJ mol}^{-1}\)
Show answer & marking scheme

Worked solution

The equation for the formation of ethanoic acid is:
\(2\text{C(graphite)} + 2\text{H}_2\text{(g)} + \text{O}_2\text{(g)} \rightarrow \text{CH}_3\text{COOH(l)}\)

Using Hess's law with combustion data:
\(\Delta H_\text{f}^\ominus = \sum \Delta H_\text{c}^\ominus(\text{reactants}) - \sum \Delta H_\text{c}^\ominus(\text{products})\)

\(\Delta H_\text{f}^\ominus = [2(-394) + 2(-286)] - [-874]\)
\(\Delta H_\text{f}^\ominus = [-788 - 572] + 874\)
\(\Delta H_\text{f}^\ominus = -1360 + 874 = -486\text{ kJ mol}^{-1}\)

Marking scheme

A is correct [1 mark].
B corresponds to omitting the stoichiometric factor of 2 for carbon and hydrogen.
C is the positive value (inversion of signs).
D is the sum of reactant combustions without subtracting the product combustion.
Question 12 · Multiple Choice
1 marks
Which statement describing the properties of the Period 3 chlorides, from sodium chloride to phosphorus(V) chloride, is correct?
  1. A.The chemical bonding in all Period 3 chlorides is purely covalent.
  2. B.The pH of the solution formed when each chloride is added to water generally decreases across the period.
  3. C.Silicon tetrachloride dissolves in water without undergoing any chemical reaction.
  4. D.Magnesium chloride reacts vigorously with water to form a strongly alkaline solution of pH 13.
Show answer & marking scheme

Worked solution

- A is incorrect: \(\text{NaCl}\) and \(\text{MgCl}_2\) contain ionic bonding, while \(\text{AlCl}_3\), \(\text{SiCl}_4\), and \(\text{PCl}_5\) are predominantly covalent.
- B is correct: \(\text{NaCl}\) forms a neutral solution (\(\text{pH} = 7\)), \(\text{MgCl}_2\) is slightly acidic (\(\text{pH} \approx 6.5\)), \(\text{AlCl}_3\) is acidic (\(\text{pH} \approx 3\)), and \(\text{SiCl}_4\) and \(\text{PCl}_5\) hydrolyse vigorously to produce strongly acidic solutions (\(\text{pH} \approx 1\text{--}2\)). Thus, pH decreases across the period.
- C is incorrect: \(\text{SiCl}_4\) reacts vigorously (hydrolyses) with water to form \(\text{SiO}_2\cdot x\text{H}_2\text{O}\) and \(\text{HCl}\).
- D is incorrect: \(\text{MgCl}_2\) dissolves to form an approximately neutral to weakly acidic solution.

Marking scheme

B is correct [1 mark].
Trend in pH across Period 3 chlorides upon addition to water correctly identified.
Question 13 · Multiple Choice
1 marks
An excess of aqueous silver nitrate is added to an aqueous solution containing a single sodium halide, \(Y\). A cream precipitate is formed.

Concentrated aqueous ammonia is then added to this precipitate.

Which row correctly identifies halide \(Y\) and describes the observation when concentrated aqueous ammonia is added?
  1. A.Halide \(Y\): bromide; Observation: the precipitate dissolves
  2. B.Halide \(Y\): bromide; Observation: the precipitate remains insoluble
  3. C.Halide \(Y\): iodide; Observation: the precipitate dissolves
  4. D.Halide \(Y\): chloride; Observation: the precipitate dissolves
Show answer & marking scheme

Worked solution

- Chloride gives a white precipitate (\(\text{AgCl}\)) which dissolves in dilute aqueous ammonia.
- Bromide gives a cream precipitate (\(\text{AgBr}\)) which is insoluble in dilute aqueous ammonia but dissolves in concentrated aqueous ammonia.
- Iodide gives a yellow precipitate (\(\text{AgI}\)) which is insoluble in both dilute and concentrated aqueous ammonia.

Therefore, halide \(Y\) is bromide, and the cream precipitate dissolves in concentrated aqueous ammonia.

Marking scheme

A is correct [1 mark].
Identification of \(\text{AgBr}\) from precipitate colour and its solubility in concentrated \(\text{NH}_3\text{(aq)}\).
Question 14 · Multiple Choice
1 marks
Compound \(W\) has the molecular formula \(\text{C}_4\text{H}_8\text{O}\).

- \(W\) forms an orange precipitate with 2,4-dinitrophenylhydrazine.
- \(W\) gives a yellow precipitate with alkaline aqueous iodine.
- \(W\) does not react with Tollens' reagent.

What is the identity of compound \(W\)?
  1. A.butan-1-ol
  2. B.butanal
  3. C.butan-2-one
  4. D.2-methylpropan-2-ol
Show answer & marking scheme

Worked solution

1. Orange precipitate with 2,4-DNPH indicates a carbonyl group (aldehyde or ketone).
2. No reaction with Tollens' reagent indicates that \(W\) is a ketone, not an aldehyde (ruling out butanal).
3. A yellow precipitate with alkaline aqueous iodine indicates a methyl carbonyl group, \(\text{CH}_3\text{C=O}\).
4. The ketone with formula \(\text{C}_4\text{H}_8\text{O}\) containing a methyl carbonyl group is butan-2-one (\(\text{CH}_3\text{COCH}_2\text{CH}_3\)).

Marking scheme

C is correct [1 mark].
Evaluation of functional group tests: 2,4-DNPH confirms carbonyl, Tollens' distinguishes ketone from aldehyde, tri-iodomethane test confirms \(\text{CH}_3\text{CO}-\) group.
Question 15 · Multiple Choice
1 marks
How many structural isomers with the molecular formula \(\text{C}_5\text{H}_{12}\text{O}\) are secondary alcohols?
  1. A.2
  2. B.3
  3. C.4
  4. D.5
Show answer & marking scheme

Worked solution

A secondary alcohol has the \(-\text{OH}\) group attached to a carbon bonded to two other carbon atoms (\(-\text{CH(OH)}-\)).

Examining the possible carbon skeletons for \(\text{C}_5\text{H}_{12}\text{O}\):

1. Straight-chain (pentane skeleton):
- Pentan-2-ol: \(\text{CH}_3\text{CH(OH)CH}_2\text{CH}_2\text{CH}_3\) (secondary)
- Pentan-3-ol: \(\text{CH}_3\text{CH}_2\text{CH(OH)CH}_2\text{CH}_3\) (secondary)

2. Branched (2-methylbutane skeleton):
- 3-methylbutan-2-ol: \((\text{CH}_3)_2\text{CHCH(OH)CH}_3\) (secondary)
- 2-methylbutan-2-ol is tertiary.
- 2-methylbutan-1-ol and 3-methylbutan-1-ol are primary.

3. Branched (2,2-dimethylpropane skeleton):
- 2,2-dimethylpropan-1-ol is primary.

Therefore, there are exactly 3 secondary alcohol structural isomers.

Marking scheme

B is correct [1 mark].
Correctly identifying pentan-2-ol, pentan-3-ol, and 3-methylbutan-2-ol.
Question 16 · Multiple Choice
1 marks
An organic compound \(Z\) contains carbon, hydrogen, and oxygen only.

- The mass spectrum of \(Z\) shows a molecular ion peak, \(\text{M}^+\), at \(m/e = 88\) with a relative abundance of \(24.0\).
- The \([\text{M}+1]^+\) peak has a relative abundance of \(1.06\).
- The infrared spectrum of \(Z\) shows a strong absorption peak at \(1740\text{ cm}^{-1}\) and no broad absorption between \(3200\text{ cm}^{-1}\) and \(3650\text{ cm}^{-1}\).

Which compound is \(Z\)?
  1. A.butanoic acid
  2. B.ethyl ethanoate
  3. C.butan-1,4-diol
  4. D.pentanal
Show answer & marking scheme

Worked solution

1. Calculate the number of carbon atoms, \(n\):
\(n = \frac{100 \times \text{abundance of } [\text{M}+1]^+}{1.1 \times \text{abundance of } \text{M}^+} = \frac{100 \times 1.06}{1.1 \times 24.0} = \frac{106}{26.4} = 4.02 \approx 4\text{ carbons}\)

2. Molecular mass \(M_\text{r} = 88\):
For 4 carbons: mass \(= 4 \times 12 = 48\).
Remaining mass \(= 88 - 48 = 40\).
With 2 oxygens (\(2 \times 16 = 32\)) and 8 hydrogens, the molecular formula is \(\text{C}_4\text{H}_8\text{O}_2\).

3. Infrared spectrum analysis:
- Strong peak at \(1740\text{ cm}^{-1}\) indicates a carbonyl group (\(\text{C=O}\), specifically ester).
- No broad absorption between \(3200\text{ cm}^{-1}\) and \(3650\text{ cm}^{-1}\) rules out an alcohol or carboxylic acid \(\text{O--H}\) group.

Therefore, \(Z\) is an ester with formula \(\text{C}_4\text{H}_8\text{O}_2\), which corresponds to ethyl ethanoate.

Marking scheme

B is correct [1 mark].
Correct deduction of 4 carbons from \([\text{M}+1]^+\) peak ratio, elimination of carboxylic acid and alcohol using IR data, identifying ester \(\text{CH}_3\text{COOCH}_2\text{CH}_3\).
Question 17 · multiple_choice
1 marks
An organic compound contains carbon, hydrogen, and oxygen only. Complete combustion of \(0.116\text{ g}\) of this compound in excess oxygen produces \(0.264\text{ g}\) of carbon dioxide and \(0.108\text{ g}\) of water.

What is the empirical formula of the compound?
  1. A.\(\text{CH}_2\text{O}\)
  2. B.\(\text{C}_2\text{H}_4\text{O}\)
  3. C.\(\text{C}_3\text{H}_6\text{O}\)
  4. D.\(\text{C}_3\text{H}_8\text{O}\)
Show answer & marking scheme

Worked solution

1. Determine the mass of carbon in the compound:
\[\text{mass of C} = 0.264\text{ g} \times \frac{12.0}{44.0} = 0.0720\text{ g}\]
\[\text{amount of C} = \frac{0.0720}{12.0} = 0.00600\text{ mol}\]

2. Determine the mass of hydrogen in the compound:
\[\text{mass of H} = 0.108\text{ g} \times \frac{2.0}{18.0} = 0.0120\text{ g}\]
\[\text{amount of H} = \frac{0.0120}{1.0} = 0.0120\text{ mol}\]

3. Determine the mass and amount of oxygen by subtraction:
\[\text{mass of O} = 0.116 - (0.0720 + 0.0120) = 0.0320\text{ g}\]
\[\text{amount of O} = \frac{0.0320}{16.0} = 0.00200\text{ mol}\]

4. Determine the simplest mole ratio \(\text{C} : \text{H} : \text{O}\):
\[\text{C} : \text{H} : \text{O} = \frac{0.00600}{0.00200} : \frac{0.0120}{0.00200} : \frac{0.00200}{0.00200} = 3 : 6 : 1\]

Therefore, the empirical formula is \(\text{C}_3\text{H}_6\text{O}\).

Marking scheme

C [1 mark]: Correct empirical formula deduced from the masses of products formed.
Question 18 · multiple_choice
1 marks
A hydrocarbon molecule has the molecular formula \(\text{C}_4\text{H}_6\). It contains no rings and no triple bonds.

How many \(\sigma\) (sigma) bonds and how many \(\pi\) (pi) bonds are present in one molecule of this hydrocarbon?
  1. A.9 \(\sigma\) bonds and 1 \(\pi\) bond
  2. B.9 \(\sigma\) bonds and 2 \(\pi\) bonds
  3. C.10 \(\sigma\) bonds and 2 \(\pi\) bonds
  4. D.11 \(\sigma\) bonds and 1 \(\pi\) bond
Show answer & marking scheme

Worked solution

An open-chain hydrocarbon of formula \(\text{C}_4\text{H}_6\) with no triple bonds must be a diene (containing two \(\text{C}=\text{C}\) double bonds, e.g., buta-1,3-diene or buta-1,2-diene).

- Each single bond consists of \(1\sigma\) bond.
- Each double bond consists of \(1\sigma\) bond and \(1\pi\) bond.
- In an open-chain structure with 4 carbon atoms and 6 hydrogen atoms, there are 3 \(\text{C}-\text{C}\) framework linkages and 6 \(\text{C}-\text{H}\) bonds, giving a total of \(3 + 6 = 9\sigma\) bonds.
- The two double bonds provide \(2\pi\) bonds.

Hence, the molecule contains 9 \(\sigma\) bonds and 2 \(\pi\) bonds.

Marking scheme

B [1 mark]: Correct identification of 9 sigma bonds and 2 pi bonds.
Question 19 · multiple_choice
1 marks
Standard enthalpy changes of combustion, \(\Delta H^\circ_\text{c}\), are given:

\(\text{C}(\text{s}) + \text{O}_2(\text{g}) \rightarrow \text{CO}_2(\text{g})\quad \Delta H^\circ_\text{c} = -394\text{ kJ}\cdot\text{mol}^{-1}\)
\(\text{H}_2(\text{g}) + \frac{1}{2}\text{O}_2(\text{g}) \rightarrow \text{H}_2\text{O}(\text{l})\quad \Delta H^\circ_\text{c} = -286\text{ kJ}\cdot\text{mol}^{-1}\)
\(\text{C}_2\text{H}_5\text{OH}(\text{l}) + 3\text{O}_2(\text{g}) \rightarrow 2\text{CO}_2(\text{g}) + 3\text{H}_2\text{O}(\text{l})\quad \Delta H^\circ_\text{c} = -1368\text{ kJ}\cdot\text{mol}^{-1}\)

What is the standard enthalpy change of formation, \(\Delta H^\circ_\text{f}\), of liquid ethanol, \(\text{C}_2\text{H}_5\text{OH}(\text{l})\)?
  1. A.\(-278\text{ kJ}\cdot\text{mol}^{-1}\)
  2. B.\(+278\text{ kJ}\cdot\text{mol}^{-1}\)
  3. C.\(-688\text{ kJ}\cdot\text{mol}^{-1}\)
  4. D.\(-3014\text{ kJ}\cdot\text{mol}^{-1}\)
Show answer & marking scheme

Worked solution

The equation for the 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 enthalpies of combustion:
\[\Delta H^\circ_\text{f} = \sum \Delta H^\circ_\text{c}(\text{reactants}) - \sum \Delta H^\circ_\text{c}(\text{products})\]
\[\Delta H^\circ_\text{f} = [2(-394) + 3(-286)] - [-1368]\]
\[\Delta H^\circ_\text{f} = [-788 - 858] + 1368 = -1646 + 1368 = -278\text{ kJ}\cdot\text{mol}^{-1}\]

Marking scheme

A [1 mark]: Correct application of Hess's Law using combustion data.
Question 20 · multiple_choice
1 marks
The following reversible reaction reaches dynamic equilibrium in a closed vessel:

\[2\text{SO}_2(\text{g}) + \text{O}_2(\text{g}) \rightleftharpoons 2\text{SO}_3(\text{g}) \quad \Delta H = -196\text{ kJ}\cdot\text{mol}^{-1}\]

Which single change will increase the numerical value of the equilibrium constant, \(K_\text{c}\)?
  1. A.increasing the total pressure at constant temperature
  2. B.increasing the concentration of \(\text{O}_2(\text{g})\)
  3. C.decreasing the temperature of the reaction mixture
  4. D.adding a vanadium(V) oxide catalyst
Show answer & marking scheme

Worked solution

The numerical value of the equilibrium constant \(K_\text{c}\) is affected only by temperature.

The forward reaction is exothermic (\(\Delta H < 0\)). According to Le Chatelier's principle, decreasing the temperature causes the equilibrium position to shift in the exothermic direction (to the right) to release heat, thereby increasing the ratio of products to reactants at equilibrium. Consequently, the value of \(K_\text{c}\) increases.

Changes in pressure, concentration, or the addition of a catalyst have no effect on the value of \(K_\text{c}\).

Marking scheme

C [1 mark]: Correctly identifying that decreasing temperature increases Kc for an exothermic reaction.
Question 21 · multiple_choice
1 marks
Solid sodium halides react when treated separately with concentrated sulfuric acid.

Which row correctly describes the observations and the role of concentrated sulfuric acid in these reactions?
  1. A.with \(\text{NaCl}\)(s): steamy fumes; \(\text{H}_2\text{SO}_4\) acts as an acid only | with \(\text{NaI}\)(s): purple vapour; \(\text{H}_2\text{SO}_4\) acts as an oxidising agent and an acid
  2. B.with \(\text{NaCl}\)(s): yellow-green gas; \(\text{H}_2\text{SO}_4\) acts as an oxidising agent | with \(\text{NaI}\)(s): brown gas; \(\text{H}_2\text{SO}_4\) acts as a reducing agent
  3. C.with \(\text{NaCl}\)(s): steamy fumes; \(\text{H}_2\text{SO}_4\) acts as a reducing agent | with \(\text{NaI}\)(s): purple vapour; \(\text{H}_2\text{SO}_4\) acts as an oxidising agent only
  4. D.with \(\text{NaCl}\)(s): white precipitate; \(\text{H}_2\text{SO}_4\) acts as an acid only | with \(\text{NaI}\)(s): yellow precipitate; \(\text{H}_2\text{SO}_4\) acts as an acid only
Show answer & marking scheme

Worked solution

- Reaction with \(\text{NaCl}\)(s): \(\text{Cl}^-\) is a weak reducing agent and cannot reduce concentrated \(\text{H}_2\text{SO}_4\). A non-redox acid-base reaction occurs, producing steamy white fumes of \(\text{HCl}\)(g). \(\text{H}_2\text{SO}_4\) acts only as a Brønsted-Lowry acid (proton donor).
- Reaction with \(\text{NaI}\)(s): \(\text{I}^-\) is a strong reducing agent and easily reduces \(\text{H}_2\text{SO}_4\) to \(\text{SO}_2\), \(\text{S}\), and \(\text{H}_2\text{S}\), while being oxidised to \(\text{I}_2\) (observed as a purple vapour / dark solid). \(\text{HI}\) is also produced via acid-base reaction. Thus, \(\text{H}_2\text{SO}_4\) acts as both an oxidising agent and an acid.

Row A is correct.

Marking scheme

A [1 mark]: Correct identification of observations and the chemical role of sulfuric acid for both halides.
Question 22 · multiple_choice
1 marks
The temperature of a gaseous reaction mixture is increased from \(300\text{ K}\) to \(320\text{ K}\).

Which statement best explains the significant increase in the rate of reaction?
  1. A.The activation energy of the reaction is lowered.
  2. B.The peak of the Boltzmann distribution curve shifts to the right and is higher.
  3. C.The increase in collision frequency is the dominant factor causing the rate increase.
  4. D.A significantly greater fraction of molecules have energy greater than or equal to the activation energy.
Show answer & marking scheme

Worked solution

When the temperature is increased:
- The activation energy, \(E_\text{a}\), remains unchanged (option A is incorrect).
- The Boltzmann distribution curve flattens and its peak shifts to the right and is lower (option B is incorrect).
- While the collision frequency increases slightly due to higher average molecular speed, this only accounts for a very small fraction of the rate increase (option C is incorrect).
- The primary cause of the large increase in reaction rate is that a much greater fraction/proportion of molecules possess kinetic energy greater than or equal to the activation energy (\(E \ge E_\text{a}\)), leading to a dramatic increase in the frequency of successful/effective collisions (option D is correct).

Marking scheme

D [1 mark]: Correct explanation using collision theory and activation energy.
Question 23 · multiple_choice
1 marks
An alkene, \(\text{Z}\), has the molecular formula \(\text{C}_6\text{H}_{12}\) and exists as a pair of cis-trans (\(E\)-\(Z\)) stereoisomers.

When \(\text{Z}\) is reacted with hot concentrated acidified \(\text{KMnO}_4\), propanoic acid is obtained as the only organic product.

What is the identity of alkene \(\text{Z}\)?
  1. A.hex-1-ene
  2. B.hex-2-ene
  3. C.hex-3-ene
  4. D.2,3-dimethylbut-2-ene
Show answer & marking scheme

Worked solution

Hot concentrated acidified \(\text{KMnO}_4\) cleaves the carbon-carbon double bond:
- A \(=\text{CH}_2\) group is oxidised to \(\text{CO}_2\).
- A \(=\text{CHR}\) group is oxidised to \(\text{RCOOH}\).
- A \(=\text{CR}_1\text{R}_2\) group is oxidised to a ketone \(\text{R}_1\text{COR}_2\).

Since propanoic acid (\(\text{CH}_3\text{CH}_2\text{COOH}\)) is the sole organic product, the alkene must be symmetrical and contain two \(=\text{CHCH}_2\text{CH}_3\) fragments joined at the double bond:
\[\text{CH}_3\text{CH}_2\text{CH}=\text{CH}\text{CH}_2\text{CH}_3 \quad (\text{hex-3-ene})\]

Hex-3-ene also exhibits cis-trans isomerism because each carbon atom of the \(\text{C}=\text{C}\) double bond is bonded to two different groups (\(-\text{H}\) and \(-\text{CH}_2\text{CH}_3\)).

Therefore, \(\text{Z}\) is hex-3-ene.

Marking scheme

C [1 mark]: Correct identification of hex-3-ene from oxidation products and stereochemical requirements.
Question 24 · multiple_choice
1 marks
An organic compound, \(\text{Q}\), gives an orange precipitate with 2,4-dinitrophenylhydrazine reagent.

\(\text{Q}\) does not produce a silver mirror when warmed with Tollens' reagent.

When \(\text{Q}\) is warmed with alkaline aqueous iodine, a yellow precipitate is formed.

Which compound is \(\text{Q}\)?
  1. A.butanal
  2. B.butanone
  3. C.butan-2-ol
  4. D.pentan-3-one
Show answer & marking scheme

Worked solution

1. \(\text{Q}\) gives an orange precipitate with 2,4-dinitrophenylhydrazine, indicating it is a carbonyl compound (aldehyde or ketone). This eliminates butan-2-ol.
2. \(\text{Q}\) gives a negative Tollens' test (no silver mirror), confirming it is a ketone, not an aldehyde. This eliminates butanal.
3. \(\text{Q}\) forms a yellow precipitate (\(\text{CHI}_3\)) with alkaline aqueous iodine (tri-iodomethane / iodoform reaction), indicating the presence of a methyl carbonyl group, \(\text{CH}_3\text{C}=\text{O}\).
- Butanone is \(\text{CH}_3\text{COCH}_2\text{CH}_3\), which contains a \(\text{CH}_3\text{C}=\text{O}\) group and therefore gives a positive tri-iodomethane test.
- Pentan-3-one is \(\text{CH}_3\text{CH}_2\text{COCH}_2\text{CH}_3\), which lacks a methyl carbonyl group and gives a negative test.

Hence, \(\text{Q}\) is butanone.

Marking scheme

B [1 mark]: Correct deduction of butanone using functional group tests.
Question 25 · Multiple Choice
1 marks
A \(10.0\text{ cm}^3\) sample of a gaseous hydrocarbon, \(\text{C}_x\text{H}_y\), is completely combusted in \(70.0\text{ cm}^3\) of oxygen (an excess).

After cooling to room temperature, the total volume of the remaining gaseous mixture is \(55.0\text{ cm}^3\).
When this remaining mixture is shaken with excess aqueous sodium hydroxide, the volume decreases to \(25.0\text{ cm}^3\).

All gas volumes are measured at the same temperature and pressure.

What is the molecular formula of the hydrocarbon?
  1. A.\(\text{C}_2\text{H}_6\)
  2. B.\(\text{C}_3\text{H}_6\)
  3. C.\(\text{C}_3\text{H}_8\)
  4. D.\(\text{C}_4\text{H}_8\)
Show answer & marking scheme

Worked solution

1. The decrease in volume upon treatment with aqueous sodium hydroxide is due to the absorption of \(\text{CO}_2\text{(g)}\):
\[\text{Volume of } \text{CO}_2 = 55.0 - 25.0 = 30.0\text{ cm}^3\]
Since \(10.0\text{ cm}^3\) of \(\text{C}_x\text{H}_y\) produces \(30.0\text{ cm}^3\) of \(\text{CO}_2\):
\[x = \frac{30.0}{10.0} = 3\]

2. The remaining \(25.0\text{ cm}^3\) of gas is unreacted \(\text{O}_2\text{(g)}\).
\[\text{Volume of } \text{O}_2 \text{ reacted} = 70.0 - 25.0 = 45.0\text{ cm}^3\]

3. From the combustion equation:
\[\text{C}_x\text{H}_y\text{(g)} + \left(x + \frac{y}{4}\right)\text{O}_2\text{(g)} \rightarrow x\text{CO}_2\text{(g)} + \frac{y}{2}\text{H}_2\text{O(l)}\]
\[x + \frac{y}{4} = \frac{45.0}{10.0} = 4.5\]
Substituting \(x = 3\):
\[3 + \frac{y}{4} = 4.5 \implies \frac{y}{4} = 1.5 \implies y = 6\]

Thus, the molecular formula is \(\text{C}_3\text{H}_6\).

Marking scheme

B is correct [1].
- A incorrect: \(\text{C}_2\text{H}_6\) would produce \(20\text{ cm}^3\) of \(\text{CO}_2\).
- C incorrect: \(\text{C}_3\text{H}_8\) would require \(50.0\text{ cm}^3\) of \(\text{O}_2\).
- D incorrect: \(\text{C}_4\text{H}_8\) would produce \(40\text{ cm}^3\) of \(\text{CO}_2\).
Question 26 · Multiple Choice
1 marks
The first six successive ionisation energies, in \(\text{kJ}\cdot\text{mol}^{-1}\), of an element \(X\) are given:

578, 1817, 2745, 11578, 14842, 18379

Which statement about element \(X\) is correct?
  1. A.The chloride of \(X\) dissolves in water to form an alkaline solution.
  2. B.Element \(X\) forms an oxide that reacts with both aqueous hydrochloric acid and aqueous sodium hydroxide.
  3. C.The radius of the \(X^+\) ion is larger than the atomic radius of \(X\).
  4. D.Element \(X\) is located in Group 14 of the Periodic Table.
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Worked solution

Successive ionisation energies increase sharply between the 3rd (\(2745\text{ kJ}\cdot\text{mol}^{-1}\)) and 4th (\(11578\text{ kJ}\cdot\text{mol}^{-1}\)) ionisation energies. This indicates that element \(X\) has three valence electrons and belongs to Group 13 (e.g. aluminium).

Aluminium forms an amphoteric oxide, \(\text{Al}_2\text{O}_3\), which reacts with both acids (such as \(\text{HCl}\)) and alkalis (such as \(\text{NaOH}\)).

Marking scheme

B is correct [1].
- A incorrect: \(\text{AlCl}_3\) hydrolyses in water to form an acidic solution (\(\text{pH} \approx 2-3\)).
- C incorrect: Cations are smaller than their parent atoms due to increased effective nuclear charge.
- D incorrect: Group 13, not Group 14.
Question 27 · Multiple Choice
1 marks
Which species has a bond angle of approximately \(107^\circ\) and possesses exactly one lone pair of electrons on its central atom?
  1. A.\(\text{NH}_4^+\)
  2. B.\(\text{BF}_3\)
  3. C.\(\text{H}_3\text{O}^+\)
  4. D.\(\text{ClF}_3\)
Show answer & marking scheme

Worked solution

In the oxonium ion, \(\text{H}_3\text{O}^+\), the central oxygen atom forms 3 bonding pairs with hydrogen atoms and retains 1 lone pair (total 4 electron pairs, tetrahedral arrangement).
Due to lone pair-bonding pair repulsion, the bond angle is reduced from \(109.5^\circ\) to approximately \(107^\circ\), giving a trigonal pyramidal shape.

Marking scheme

C is correct [1].
- A: \(\text{NH}_4^+\) has 4 bonding pairs, 0 lone pairs, bond angle \(109.5^\circ\).
- B: \(\text{BF}_3\) has 3 bonding pairs, 0 lone pairs, bond angle \(120^\circ\).
- D: \(\text{ClF}_3\) has 3 bonding pairs, 2 lone pairs, T-shaped, bond angle \(<90^\circ\).
Question 28 · Multiple Choice
1 marks
The standard enthalpy changes of combustion, \(\Delta H^\circ_\text{c}\), of carbon(graphite), hydrogen(g), and propan-1-ol(l) are given:

\(\Delta H^\circ_\text{c}[\text{C(graphite)}] = -394\text{ kJ}\cdot\text{mol}^{-1}\)
\(\Delta H^\circ_\text{c}[\text{H}_2\text{(g)}] = -286\text{ kJ}\cdot\text{mol}^{-1}\)
\(\Delta H^\circ_\text{c}[\text{CH}_3\text{CH}_2\text{CH}_2\text{OH(l)}] = -2021\text{ kJ}\cdot\text{mol}^{-1}\)

What is the standard enthalpy change of formation, \(\Delta H^\circ_\text{f}\), of propan-1-ol(l)?
  1. A.\(-305\text{ kJ}\cdot\text{mol}^{-1}\)
  2. B.\(+305\text{ kJ}\cdot\text{mol}^{-1}\)
  3. C.\(-1341\text{ kJ}\cdot\text{mol}^{-1}\)
  4. D.\(-4347\text{ kJ}\cdot\text{mol}^{-1}\)
Show answer & marking scheme

Worked solution

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

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

Marking scheme

A is correct [1].
- B: Sign error (+305).
- C: Incorrect stoichiometry applied (e.g. omitting stoichiometric coefficients).
- D: Added products and reactants without subtraction (-4347).
Question 29 · Multiple Choice
1 marks
Two Period 3 elements, \(M\) and \(N\), each react with chlorine to form a chloride.

- The chloride of \(M\) is a white solid that dissolves in water to form a neutral solution with \(\text{pH} \approx 7\).
- The chloride of \(N\) is a colourless liquid at room temperature that reacts vigorously with water to form white misty fumes and an acidic solution with \(\text{pH} \approx 1-2\).

Which row correctly identifies elements \(M\) and \(N\)?
  1. A.\(M = \text{Na}\); \(N = \text{Si}\)
  2. B.\(M = \text{Mg}\); \(N = \text{Al}\)
  3. C.\(M = \text{Al}\); \(N = \text{P}\)
  4. D.\(M = \text{Na}\); \(N = \text{Mg}\)
Show answer & marking scheme

Worked solution

Sodium chloride (\(\text{NaCl}\)) is a white solid that dissolves readily in water without hydrolysis to give a neutral solution (\(\text{pH} \approx 7\)), so \(M = \text{Na}\).

Silicon tetrachloride (\(\text{SiCl}_4\)) is a colourless covalent liquid at room temperature that hydrolyses vigorously in water to give misty fumes of \(\text{HCl}\text{(g)}\) and an acidic solution of \(\text{pH} 1-2\), so \(N = \text{Si}\).

Marking scheme

A is correct [1].
- B: \(\text{MgCl}_2\) forms a slightly acidic solution (\(\text{pH} \approx 6.5\)); \(\text{AlCl}_3\) is a solid, not a liquid.
- C: \(\text{AlCl}_3\) is acidic; \(\text{PCl}_5\) is a solid.
- D: \(\text{MgCl}_2\) is a solid, not a liquid.
Question 30 · Multiple Choice
1 marks
In an experiment, \(2.00\text{ mol}\) of \(\text{SO}_2\text{(g)}\) and \(2.00\text{ mol}\) of \(\text{O}_2\text{(g)}\) are mixed in a sealed vessel of volume \(2.00\text{ dm}^3\) at a constant temperature \(T\).

The equilibrium reaction is established:
\[2\text{SO}_2\text{(g)} + \text{O}_2\text{(g)} \rightleftharpoons 2\text{SO}_3\text{(g)}\]

At equilibrium, the amount of \(\text{SO}_3\text{(g)}\) present is \(1.60\text{ mol}\).

What is the numerical value of \(K_\text{c}\) under these conditions?
  1. A.5.33
  2. B.13.3
  3. C.26.7
  4. D.53.3
Show answer & marking scheme

Worked solution

Initial moles:
\(n(\text{SO}_2) = 2.00\text{ mol}\), \(n(\text{O}_2) = 2.00\text{ mol}\), \(n(\text{SO}_3) = 0\text{ mol}\).

Equilibrium moles:
- \(n(\text{SO}_3) = 1.60\text{ mol}\)
- \(n(\text{SO}_2) = 2.00 - 1.60 = 0.40\text{ mol}\)
- \(n(\text{O}_2) = 2.00 - \frac{1.60}{2} = 1.20\text{ mol}\)

Equilibrium concentrations in \(V = 2.00\text{ dm}^3\):
- \([\text{SO}_3] = \frac{1.60}{2.00} = 0.80\text{ mol}\cdot\text{dm}^{-3}\)
- \([\text{SO}_2] = \frac{0.40}{2.00} = 0.20\text{ mol}\cdot\text{dm}^{-3}\)
- \([\text{O}_2] = \frac{1.20}{2.00} = 0.60\text{ mol}\cdot\text{dm}^{-3}\)

\(K_\text{c}\) calculation:
\[K_\text{c} = \frac{[\text{SO}_3]^2}{[\text{SO}_2]^2 [\text{O}_2]} = \frac{(0.80)^2}{(0.20)^2 \times 0.60} = \frac{0.64}{0.040 \times 0.60} = \frac{0.64}{0.024} \approx 26.7\text{ dm}^3\cdot\text{mol}^{-1}\]

Marking scheme

C is correct [1].
- A (5.33): Forgot to square concentrations in expression.
- B (13.3): Used moles directly instead of concentrations in \(2.00\text{ dm}^3\).
- D (53.3): Calculation error involving volume factor.
Question 31 · Multiple Choice
1 marks
An unbranched alkene, \(W\), has the molecular formula \(\text{C}_6\text{H}_{12}\).

When \(W\) is heated under reflux with hot concentrated acidified potassium manganate(VII), the only organic product obtained is propanoic acid, \(\text{CH}_3\text{CH}_2\text{COOH}\).

How many stereoisomers exist for alkene \(W\)?
  1. A.0
  2. B.2
  3. C.3
  4. D.4
Show answer & marking scheme

Worked solution

Oxidative cleavage of an unbranched alkene yielding only propanoic acid indicates that the double bond is centrally located and symmetrical: \(\text{CH}_3\text{CH}_2\text{CH}=\text{CH}\text{CH}_2\text{CH}_3\) (hex-3-ene).

Hex-3-ene exhibits geometrical (cis-trans / E-Z) isomerism because each carbon atom of the \(\text{C}=\text{C}\) double bond is attached to two different groups (\(-\text{H}\) and \(-\text{CH}_2\text{CH}_3\)):
1. (E)-hex-3-ene (trans-hex-3-ene)
2. (Z)-hex-3-ene (cis-hex-3-ene)

Hex-3-ene contains no chiral carbon atoms, so no optical isomers exist. Thus, there are exactly 2 stereoisomers.

Marking scheme

B is correct [1].
- A: Overlooked cis/trans isomerism.
- C: Incorrect stereocentre counting.
- D: Assumed additional optical isomerism.
Question 32 · Multiple Choice
1 marks
An organic compound \(Y\) with the molecular formula \(\text{C}_4\text{H}_{10}\text{O}\) exhibits the following properties:

- It turns acidified potassium dichromate(VI) from orange to green when heated.
- It forms a pale yellow precipitate when warmed with alkaline aqueous iodine.

What is the identity of compound \(Y\)?
  1. A.butan-1-ol
  2. B.butan-2-ol
  3. C.2-methylpropan-1-ol
  4. D.2-methylpropan-2-ol
Show answer & marking scheme

Worked solution

1. Turning acidified \(\text{K}_2\text{Cr}_2\text{O}_7\) from orange to green indicates that \(Y\) can be oxidised, meaning it is a primary or secondary alcohol (eliminates 2-methylpropan-2-ol, which is tertiary).

2. Forming a pale yellow precipitate (\(\text{CHI}_3\)) with alkaline aqueous iodine (the tri-iodomethane test) requires a \(\text{CH}_3\text{CH(OH)}-\) group.

- Butan-1-ol: \(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH}\) (no \(\text{CH}_3\text{CH(OH)}-\) group).
- 2-Methylpropan-1-ol: \((\text{CH}_3)_2\text{CHCH}_2\text{OH}\) (no \(\text{CH}_3\text{CH(OH)}-\) group).
- Butan-2-ol: \(\text{CH}_3\text{CH(OH)}\text{CH}_2\text{CH}_3\) (has the \(\text{CH}_3\text{CH(OH)}-\) group and is a secondary alcohol).

Therefore, \(Y\) is butan-2-ol.

Marking scheme

B is correct [1].
- A: Butan-1-ol is oxidised but gives a negative tri-iodomethane test.
- C: 2-Methylpropan-1-ol gives a negative tri-iodomethane test.
- D: 2-Methylpropan-2-ol cannot be oxidised by acidified dichromate.
Question 33 · Multiple Choice
1 marks
A sample of \(0.295\text{ g}\) of an anhydrous metal carbonate, \(M\text{CO}_3\), was completely dissolved in \(50.0\text{ cm}^3\) of \(0.200\text{ mol dm}^{-3}\) hydrochloric acid.

The unreacted hydrochloric acid required \(15.0\text{ cm}^3\) of \(0.200\text{ mol dm}^{-3}\) sodium hydroxide for complete neutralisation.

What is the relative atomic mass, \(A_\text{r}\), of metal \(M\)?
  1. A.24.3
  2. B.40.1
  3. C.48.6
  4. D.84.3
Show answer & marking scheme

Worked solution

1. Calculate the total moles of \(\text{HCl}\) added:
\(n(\text{HCl})_\text{total} = 0.0500\text{ dm}^3 \times 0.200\text{ mol dm}^{-3} = 0.0100\text{ mol}\)

2. Calculate the moles of \(\text{HCl}\) that remained unreacted (neutralised by \(\text{NaOH}\)):
\(n(\text{NaOH}) = 0.0150\text{ dm}^3 \times 0.200\text{ mol dm}^{-3} = 0.00300\text{ mol}\)
Since \(\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}\), unreacted \(n(\text{HCl}) = 0.00300\text{ mol}\).

3. Calculate the moles of \(\text{HCl}\) that reacted with \(M\text{CO}_3\):
\(n(\text{HCl})_\text{reacted} = 0.0100 - 0.00300 = 0.00700\text{ mol}\)

4. Use the stoichiometric ratio for \(M\text{CO}_3 + 2\text{HCl} \rightarrow M\text{Cl}_2 + \text{H}_2\text{O} + \text{CO}_2\):
\(n(M\text{CO}_3) = \frac{0.00700}{2} = 0.00350\text{ mol}\)

5. Calculate the molar mass of \(M\text{CO}_3\):
\(M_\text{r}(M\text{CO}_3) = \frac{0.295\text{ g}}{0.00350\text{ mol}} = 84.3\text{ g mol}^{-1}\)

6. Calculate \(A_\text{r}(M)\):
\(A_\text{r}(M) = 84.3 - [12.0 + (3 \times 16.0)] = 84.3 - 60.0 = 24.3\) (corresponding to magnesium).

Marking scheme

Award [1] mark for option A.
[1] Moles of HCl reacted = 0.00700 mol \(\rightarrow\) moles of carbonate = 0.00350 mol \(\rightarrow\) \(M_\text{r} = 84.3\) \(\rightarrow\) \(A_\text{r} = 24.3\).
Question 34 · Multiple Choice
1 marks
Which sequence places the given species in order of increasing bond angle?
  1. A.\(\text{H}_2\text{O} < \text{NH}_3 < \text{BF}_3 < \text{SF}_6\)
  2. B.\(\text{SF}_6 < \text{NH}_3 < \text{H}_2\text{O} < \text{BF}_3\)
  3. C.\(\text{SF}_6 < \text{H}_2\text{O} < \text{NH}_3 < \text{BF}_3\)
  4. D.\(\text{BF}_3 < \text{NH}_3 < \text{H}_2\text{O} < \text{SF}_6\)
Show answer & marking scheme

Worked solution

Deduce the shapes and bond angles using VSEPR theory:
- \(\text{SF}_6\): 6 bonding pairs, 0 lone pairs \(\rightarrow\) octahedral, bond angle = \(90^\circ\)
- \(\text{H}_2\text{O}\): 2 bonding pairs, 2 lone pairs \(\rightarrow\) non-linear (bent), bond angle = \(104.5^\circ\)
- \(\text{NH}_3\): 3 bonding pairs, 1 lone pair \(\rightarrow\) trigonal pyramidal, bond angle = \(107^\circ\)
- \(\text{BF}_3\): 3 bonding pairs, 0 lone pairs \(\rightarrow\) trigonal planar, bond angle = \(120^\circ\)

Thus, the order of increasing bond angle is:
\(\text{SF}_6\ (90^\circ) < \text{H}_2\text{O}\ (104.5^\circ) < \text{NH}_3\ (107^\circ) < \text{BF}_3\ (120^\circ)\).

Marking scheme

Award [1] mark for option C.
[1] Correct identification of bond angles: 90° < 104.5° < 107° < 120°.
Question 35 · Multiple Choice
1 marks
The standard enthalpy changes of combustion, \(\Delta H^\circ_\text{c}\), are given in the table.

$$\begin{array}{|c|c|}\hline\text{Substance} & \Delta H^\circ_\text{c} / \text{kJ mol}^{-1} \\hline\text{C(graphite)} & -394 \\text{H}_2\text{(g)} & -286 \\text{CH}_3\text{CH(OH)CH}_3\text{(l)} & -2006 \\hline\end{array}$$

What is the standard enthalpy change of formation, \(\Delta H^\circ_\text{f}\), of propan-2-ol, \(\text{CH}_3\text{CH(OH)CH}_3\text{(l)}\)?
  1. A.\(-320\text{ kJ mol}^{-1}\)
  2. B.\(+320\text{ kJ mol}^{-1}\)
  3. C.\(-1326\text{ kJ mol}^{-1}\)
  4. D.\(-4332\text{ kJ mol}^{-1}\)
Show answer & marking scheme

Worked solution

The equation for the standard enthalpy change of formation of propan-2-ol is:
$$3\text{C(graphite)} + 4\text{H}_2\text{(g)} + \frac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{C}_3\text{H}_8\text{O(l)}$$

Using Hess's Law and enthalpy changes of combustion:
$$\Delta H^\circ_\text{f} = \sum \Delta H^\circ_\text{c}(\text{reactants}) - \sum \Delta H^\circ_\text{c}(\text{products})$$
$$\Delta H^\circ_\text{f} = [3(-394) + 4(-286)] - [-2006]$$
$$\Delta H^\circ_\text{f} = [-1182 - 1144] + 2006 = -2326 + 2006 = -320\text{ kJ mol}^{-1}$$.

Marking scheme

Award [1] mark for option A.
[1] Correct application of Hess's Law: 3(-394) + 4(-286) - (-2006) = -320 kJ mol^-1.
Question 36 · Multiple Choice
1 marks
In an experiment, \(2.00\text{ mol}\) of \(\text{SO}_2\text{(g)}\) and \(1.00\text{ mol}\) of \(\text{O}_2\text{(g)}\) are sealed in a vessel of fixed volume \(V\text{ dm}^3\) at a constant temperature.

The reaction reaches dynamic equilibrium according to the equation:

$$2\text{SO}_2\text{(g)} + \text{O}_2\text{(g)} \rightleftharpoons 2\text{SO}_3\text{(g)}$$

At equilibrium, \(1.60\text{ mol}\) of \(\text{SO}_3\text{(g)}\) is present in the mixture.

What is the value of the equilibrium constant, \(K_\text{c}\), in terms of \(V\)?
  1. A.\(\dfrac{80}{V}\)
  2. B.\(80V\)
  3. C.\(20V\)
  4. D.\(\dfrac{20}{V}\)
Show answer & marking scheme

Worked solution

1. Determine the equilibrium amounts of all species:
- Initial: \(n(\text{SO}_2) = 2.00\text{ mol}\), \(n(\text{O}_2) = 1.00\text{ mol}\), \(n(\text{SO}_3) = 0\text{ mol}\)
- Formed: \(n(\text{SO}_3) = 1.60\text{ mol}\)
- Reacted \(\text{SO}_2 = 1.60\text{ mol} \implies n(\text{SO}_2)_\text{eq} = 2.00 - 1.60 = 0.40\text{ mol}\)
- Reacted \(\text{O}_2 = \frac{1.60}{2} = 0.80\text{ mol} \implies n(\text{O}_2)_\text{eq} = 1.00 - 0.80 = 0.20\text{ mol}\)

2. Substitute into the equilibrium expression:
$$K_\text{c} = \frac{[\text{SO}_3]^2}{[\text{SO}_2]^2 [\text{O}_2]} = \frac{(1.60 / V)^2}{(0.40 / V)^2 \times (0.20 / V)}$$
$$K_\text{c} = \frac{2.56 / V^2}{(0.16 / V^2) \times (0.20 / V)} = \frac{2.56}{0.032 / V} = 80V\text{ mol}^{-1}\text{dm}^3$$.

Marking scheme

Award [1] mark for option B.
[1] Correct calculation of equilibrium moles (SO2 = 0.40, O2 = 0.20, SO3 = 1.60) and substitution into Kc expression to obtain 80V.
Question 37 · Multiple Choice
1 marks
Solid samples of sodium halides are added separately to concentrated sulfuric acid.

Which statement is correct?
  1. A.Concentrated sulfuric acid oxidises chloride ions to chlorine gas.
  2. B.With sodium bromide, steamy fumes and a brown vapour are produced.
  3. C.With sodium iodide, the only sulfur-containing reduction product is sulfur dioxide.
  4. D.The reducing ability of the halide ions decreases down Group 17.
Show answer & marking scheme

Worked solution

- Option A is incorrect: Chloride is not a strong enough reducing agent to reduce concentrated sulfuric acid; only steamy fumes of \(\text{HCl}\) are produced.
- Option B is correct: Bromide reduces sulfuric acid to sulfur dioxide, being oxidised itself to bromine (orange-brown vapour), along with steamy fumes of \(\text{HBr}\).
- Option C is incorrect: Iodide is a stronger reducing agent and reduces sulfur from \(+6\) to \(+4\) (\(\text{SO}_2\)), \(0\) (yellow sulfur), and \(-2\) (\(\text{H}_2\text{S}\)).
- Option D is incorrect: The reducing power of halide ions increases down Group 17 as ionic radius increases and valence electrons are more easily lost.

Marking scheme

Award [1] mark for option B.
[1] Identification that NaBr + concentrated H2SO4 produces steamy fumes (HBr) and brown vapour (Br2).
Question 38 · Multiple Choice
1 marks
Hydrocarbon \(\text{W}\) has the molecular formula \(\text{C}_6\text{H}_{12}\) and exhibits geometrical (\(\text{cis}\)-\(\text{trans}\)) isomerism.

When \(\text{W}\) is heated under reflux with hot concentrated acidified potassium manganate(VII), propanoic acid, \(\text{CH}_3\text{CH}_2\text{COOH}\), is the only organic product obtained.

What is the structure of hydrocarbon \(\text{W}\)?
  1. A.\(\text{CH}_3\text{CH}_2\text{CH}=\text{CHCH}_2\text{CH}_3\)
  2. B.\(\text{CH}_3\text{CH}_2\text{C(CH}_3)=\text{CHCH}_3\)
  3. C.\((\text{CH}_3)_2\text{C}=\text{C(CH}_3)_2\)
  4. D.\(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}=\text{CHCH}_3\)
Show answer & marking scheme

Worked solution

Oxidation of an alkene with hot concentrated acidified \(\text{KMnO}_4\) cleaves the \(\text{C}=\text{C}\) bond:
- A \(=\text{CH}_2\) group gives \(\text{CO}_2 + \text{H}_2\text{O}\).
- A \(=\text{CHR}\) group gives a carboxylic acid, \(\text{RCOOH}\).
- A \(=\text{CR}_1\text{R}_2\) group gives a ketone, \(\text{R}_1\text{COR}_2\).

Since propanoic acid (\(\text{CH}_3\text{CH}_2\text{COOH}\)) is the ONLY organic product, the alkene must be symmetrical and contain two \(=\text{CH}-\text{CH}_2\text{CH}_3\) fragments: \(\text{CH}_3\text{CH}_2\text{CH}=\text{CHCH}_2\text{CH}_3\) (hex-3-ene).
Hex-3-ene also exhibits cis-trans isomerism.

Marking scheme

Award [1] mark for option A.
[1] Recognition that oxidative cleavage of hex-3-ene yields exclusively propanoic acid and exhibits cis-trans isomerism.
Question 39 · Multiple Choice
1 marks
An organic compound \(\text{X}\) has the molecular formula \(\text{C}_4\text{H}_8\text{O}\).

- Compound \(\text{X}\) reacts with 2,4-dinitrophenylhydrazine to form an orange precipitate.
- Compound \(\text{X}\) does not produce a silver mirror when warmed with Tollens' reagent.
- Compound \(\text{X}\) forms a pale yellow precipitate when warmed with alkaline aqueous iodine.

What is the identity of \(\text{X}\)?
  1. A.butanal
  2. B.butan-2-one
  3. C.2-methylpropanal
  4. D.but-3-en-2-ol
Show answer & marking scheme

Worked solution

1. Formation of an orange precipitate with 2,4-DNPH indicates that \(\text{X}\) contains a carbonyl group (aldehyde or ketone).
2. Negative Tollens' reagent test indicates that \(\text{X}\) is a ketone, not an aldehyde (rules out butanal and 2-methylpropanal).
3. Formation of a yellow precipitate (\(\text{CHI}_3\)) with alkaline aqueous iodine indicates the presence of a methyl ketone group, \(\text{CH}_3\text{C}=\text{O}\).

Among the options, butan-2-one (\(\text{CH}_3\text{COCH}_2\text{CH}_3\)) is a 4-carbon methyl ketone that meets all criteria.

Marking scheme

Award [1] mark for option B.
[1] 2,4-DNPH positive + Tollens' negative = ketone; alkaline I2 positive = methyl ketone (butan-2-one).
Question 40 · Multiple Choice
1 marks
The mass spectrum of a halogenoalkane \(\text{Y}\) containing only carbon, hydrogen, and chlorine displays the following peak data:

- a molecular ion peak, \(\text{M}^+\), at \(m/e = 78\) with a relative abundance of \(60.0\%\)
- an \([\text{M}+1]^+\) peak at \(m/e = 79\) with a relative abundance of \(2.0\%\)
- an \([\text{M}+2]^+\) peak at \(m/e = 80\) with a relative abundance of \(19.8\%\)

How many carbon atoms and how many chlorine atoms are present in one molecule of \(\text{Y}\)?
  1. A.3 carbon atoms and 1 chlorine atom
  2. B.3 carbon atoms and 2 chlorine atoms
  3. C.2 carbon atoms and 1 chlorine atom
  4. D.2 carbon atoms and 2 chlorine atoms
Show answer & marking scheme

Worked solution

1. Determine the number of carbon atoms, \(n\), using the \([\text{M}+1]^+\) formula:
$$n = \frac{100 \times \text{abundance of } [\text{M}+1]^+}{1.1 \times \text{abundance of } \text{M}^+} = \frac{100 \times 2.0}{1.1 \times 60.0} = \frac{200}{66.0} \approx 3.03$$
Thus, there are 3 carbon atoms.

2. Determine the number of chlorine atoms using the ratio of \(\text{M}^+ : [\text{M}+2]^+\):
$$\frac{\text{abundance of } \text{M}^+}{\text{abundance of } [\text{M}+2]^+} = \frac{60.0}{19.8} \approx 3.03 : 1$$
A \(3:1\) height ratio for \(\text{M}^+ : [\text{M}+2]^+\) is characteristic of the presence of exactly one \(^{35}\text{Cl} / ^{37}\text{Cl}\) atom.

Formula \(\text{C}_3\text{H}_7\text{Cl}\) has \(M_\text{r} = (3 \times 12) + (7 \times 1) + 35 = 78\).

Marking scheme

Award [1] mark for option A.
[1] Correct calculation of n = 3 carbon atoms from [M+1]+ and 1 chlorine atom from 3:1 ratio of M+:[M+2]+.

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Paper 2 (AS Level Structured Questions)

Answer all questions in the spaces provided on the question paper. Calculators and data booklet values may be used.
5 Question · 60 marks
Question 1 · Structured Theory & Mechanics
12 marks
(a) (i) Define relative atomic mass, \(A_\text{r}\).

(ii) A sample of magnesium consists of three isotopes: \(^{24}\text{Mg}\) (relative isotopic mass 23.99, abundance 78.99%), \(^{25}\text{Mg}\) (relative isotopic mass 24.99, abundance 10.00%), and \(^{26}\text{Mg}\) (relative isotopic mass 25.98, abundance 11.01%).

Calculate the relative atomic mass, \(A_\text{r}\), of this sample of magnesium. Give your answer to two decimal places.

(b) Compound \(\mathbf{X}\) is a volatile organic liquid containing only carbon, hydrogen, and oxygen.

Complete combustion of a \(1.480\text{ g}\) sample of \(\mathbf{X}\) produces \(3.520\text{ g}\) of carbon dioxide, \(\text{CO}_2\), and \(1.800\text{ g}\) of water, \(\text{H}_2\text{O}\).

(i) Show by calculation that the empirical formula of \(\mathbf{X}\) is \(\text{C}_4\text{H}_{10}\text{O}\).

(ii) In a separate experiment, a \(0.295\text{ g}\) sample of \(\mathbf{X}\) was vaporised in a gas syringe at \(373\text{ K}\) and \(1.00 \times 10^5\text{ Pa}\). The vapour occupied a volume of \(124\text{ cm}^3\).

Calculate the relative molecular mass, \(M_\text{r}\), of compound \(\mathbf{X}\) using the ideal gas equation. Give your answer to three significant figures. [\(R = 8.31\text{ J}\cdot\text{K}^{-1}\cdot\text{mol}^{-1}\)]

(iii) Deduce the molecular formula of compound \(\mathbf{X}\).

(c) State two assumptions of the kinetic theory of ideal gases that do not hold true for real gases at high pressure and low temperature.
Show answer & marking scheme

Worked solution

(a) (i) \(A_\text{r}\) is defined as the weighted average mass of one atom of an element relative to \(\frac{1}{12}\)th of the mass of an atom of carbon-12.

(ii) \(A_\text{r} = \frac{(23.99 \times 78.99) + (24.99 \times 10.00) + (25.98 \times 11.01)}{100} = \frac{1894.97 + 249.90 + 286.04}{100} = \frac{2430.91}{100} = 24.31\)

(b) (i)
\(n(\text{C}) = n(\text{CO}_2) = \frac{3.520\text{ g}}{44.01\text{ g}\cdot\text{mol}^{-1}} = 0.07998\text{ mol}\)
\(\text{mass of C} = 0.07998 \times 12.011 = 0.9606\text{ g}\)

\(n(\text{H}) = 2 \times n(\text{H}_2\text{O}) = 2 \times \frac{1.800\text{ g}}{18.016\text{ g}\cdot\text{mol}^{-1}} = 0.1998\text{ mol}\)
\(\text{mass of H} = 0.1998 \times 1.008 = 0.2014\text{ g}\)

\(\text{mass of O} = 1.480 - (0.9606 + 0.2014) = 0.3180\text{ g}\)
\(n(\text{O}) = \frac{0.3180\text{ g}}{16.00\text{ g}\cdot\text{mol}^{-1}} = 0.01988\text{ mol}\)

Ratio of moles: \(\text{C} : \text{H} : \text{O} = \frac{0.07998}{0.01988} : \frac{0.1998}{0.01988} : \frac{0.01988}{0.01988} = 4.02 : 10.05 : 1.00 \approx 4 : 10 : 1\)
Thus, empirical formula = \(\text{C}_4\text{H}_{10}\text{O}\).

(ii) Using \(pV = nRT = \frac{m}{M_\text{r}}RT\):
\(V = 124\text{ cm}^3 = 124 \times 10^{-6}\text{ m}^3 = 1.24 \times 10^{-4}\text{ m}^3\)
\(p = 1.00 \times 10^5\text{ Pa}\)
\(T = 373\text{ K}\)
\(M_\text{r} = \frac{mRT}{pV} = \frac{0.295 \times 8.31 \times 373}{1.00 \times 10^5 \times 1.24 \times 10^{-4}} = \frac{914.4165}{12.4} = 73.74\text{ g}\cdot\text{mol}^{-1} \approx 73.7\)

(iii) Empirical formula mass of \(\text{C}_4\text{H}_{10}\text{O} = (4 \times 12.0) + (10 \times 1.0) + 16.0 = 74.0\text{ g}\cdot\text{mol}^{-1}\).
Since \(M_\text{r} \approx 74\), the molecular formula is \(\text{C}_4\text{H}_{10}\text{O}\).

(c) 1. Real gas molecules/particles occupy a non-negligible/finite volume (unlike ideal point masses).
2. Real gas particles experience intermolecular attractive forces (unlike ideal gases where intermolecular forces are zero).

Marking scheme

(a)(i) [2 marks]
- M1: Weighted average mass of an atom of an element [1]
- M2: Relative to 1/12th the mass of an atom of carbon-12 (or \(^{12}\text{C}\)) [1]

(a)(ii) [2 marks]
- M1: Correct expression with percentages and isotopic masses [1]
- M2: Correct calculation evaluated to 2 decimal places: 24.31 [1]

(b)(i) [3 marks]
- M1: Calculation of moles of C (0.0800 mol) and H (0.200 mol) [1]
- M2: Calculation of mass of O (0.318–0.320 g) and moles of O (0.0200 mol) [1]
- M3: Simplest whole-number molar ratio shown leading to \(\text{C}_4\text{H}_{10}\text{O}\) [1]

(b)(ii) [3 marks]
- M1: Conversion of volume to \(\text{m}^3\) (\(1.24 \times 10^{-4}\text{ m}^3\)) [1]
- M2: Correct rearrangement of ideal gas equation: \(M_\text{r} = \frac{mRT}{pV}\) [1]
- M3: Correct final answer to 3 significant figures: 73.7 (allow 73.7 to 74.0) [1]

(b)(iii) [1 mark]
- \(\text{C}_4\text{H}_{10}\text{O}\) with explicit comparison of \(M_\text{r}\) to empirical formula mass [1]

(c) [1 mark]
- Any two from: gas particles have volume / non-zero volume; intermolecular forces / attractive forces exist between particles [1]
Question 2 · Structured Theory & Mechanics
12 marks
(a) (i) Define standard enthalpy change of formation, \(\Delta H_\text{f}^\circ\).

(ii) Hydrazine, \(\text{N}_2\text{H}_4(\text{l})\), reacts with dinitrogen tetroxide, \(\text{N}_2\text{O}_4(\text{g})\), to form nitrogen and steam according to the equation:

\[ 2\text{N}_2\text{H}_4(\text{l}) + \text{N}_2\text{O}_4(\text{g}) \rightarrow 3\text{N}_2(\text{g}) + 4\text{H}_2\text{O}(\text{g}) \]

Standard enthalpy changes of formation are given below:
- \(\Delta H_\text{f}^\circ [\text{N}_2\text{H}_4(\text{l})] = +50.6\text{ kJ}\cdot\text{mol}^{-1}\)
- \(\Delta H_\text{f}^\circ [\text{N}_2\text{O}_4(\text{g})] = +9.2\text{ kJ}\cdot\text{mol}^{-1}\)
- \(\Delta H_\text{f}^\circ [\text{H}_2\text{O}(\text{g})] = -241.8\text{ kJ}\cdot\text{mol}^{-1}\)

Calculate the standard enthalpy change of this reaction, \(\Delta H_\text{r}^\circ\).

(b) (i) State what is meant by the term dynamic equilibrium.

(ii) Nitrogen monoxide and oxygen react in a sealed container to establish the following equilibrium:

\[ 2\text{NO}(\text{g}) + \text{O}_2(\text{g}) \rightleftharpoons 2\text{NO}_2(\text{g}) \quad \Delta H = -114\text{ kJ}\cdot\text{mol}^{-1} \]

Write the expression for the equilibrium constant, \(K_\text{c}\), for this reaction and deduce its units.

(iii) Predict and explain the effect of increasing the temperature on the position of equilibrium and the yield of \(\text{NO}_2(\text{g})\).

(iv) Explain the effect of increasing the total pressure on the initial rate of the forward reaction.
Show answer & marking scheme

Worked solution

(a) (i) The standard enthalpy change of formation is the enthalpy change when 1 mole of a compound is formed from its constituent elements in their standard states under standard conditions (298 K and 101 kPa).

(ii) Using \(\Delta H_\text{r}^\circ = \sum \Delta H_\text{f}^\circ(\text{products}) - \sum \Delta H_\text{f}^\circ(\text{reactants})\):
\(\Delta H_\text{r}^\circ = [3 \times \Delta H_\text{f}^\circ(\text{N}_2) + 4 \times \Delta H_\text{f}^\circ(\text{H}_2\text{O}(\text{g}))] - [2 \times \Delta H_\text{f}^\circ(\text{N}_2\text{H}_4(\text{l})) + \Delta H_\text{f}^\circ(\text{N}_2\text{O}_4(\text{g}))]\)
Since \(\Delta H_\text{f}^\circ(\text{N}_2) = 0\text{ kJ}\cdot\text{mol}^{-1}\):
\(\Delta H_\text{r}^\circ = [0 + 4(-241.8)] - [2(50.6) + 9.2]\)
\(\Delta H_\text{r}^\circ = [-967.2] - [101.2 + 9.2] = -967.2 - 110.4 = -1077.6\text{ kJ}\cdot\text{mol}^{-1}\)

(b) (i) Dynamic equilibrium occurs in a closed system when:
1. The rate of the forward reaction equals the rate of the reverse/backward reaction.
2. The concentrations of all reactants and products remain constant.

(ii) \(K_\text{c} = \frac{[\text{NO}_2]^2}{[\text{NO}]^2[\text{O}_2]}\)
Units: \(\frac{(\text{mol}\cdot\text{dm}^{-3})^2}{(\text{mol}\cdot\text{dm}^{-3})^2 \cdot (\text{mol}\cdot\text{dm}^{-3})} = \frac{1}{\text{mol}\cdot\text{dm}^{-3}} = \text{mol}^{-1}\cdot\text{dm}^3\)

(iii) The forward reaction is exothermic (\(\Delta H < 0\)). Increasing temperature causes the equilibrium to shift in the endothermic direction (to the left) to absorb added thermal energy, resulting in a lower equilibrium yield of \(\text{NO}_2\).

(iv) Increasing the total pressure decreases the volume, so gas particles are closer together / particle concentration increases. This increases the collision frequency between reactant molecules, resulting in an increased rate of the forward reaction.

Marking scheme

(a)(i) [2 marks]
- M1: Enthalpy change when one mole of a compound is formed from its elements [1]
- M2: In their standard states under standard conditions (298 K and 101 kPa / 1 atm) [1]

(a)(ii) [2 marks]
- M1: Correct Hess cycle / expression: \([4(-241.8)] - [2(50.6) + 9.2]\) [1]
- M2: Correct calculation: \(-1077.6\text{ kJ}\cdot\text{mol}^{-1}\) (allow \(-1078\text{ kJ}\cdot\text{mol}^{-1}\)) [1]

(b)(i) [2 marks]
- M1: Rate of forward reaction = rate of backward reaction [1]
- M2: Concentrations of reactants and products remain constant (in a closed system) [1]

(b)(ii) [2 marks]
- M1: Correct expression for \(K_\text{c}\) with square brackets [1]
- M2: Correct units: \(\text{mol}^{-1}\cdot\text{dm}^3\) (or \(\text{dm}^3\cdot\text{mol}^{-1}\)) [1]

(b)(iii) [2 marks]
- M1: Equilibrium shifts to the left / yield of \(\text{NO}_2\) decreases [1]
- M2: Explanation that forward reaction is exothermic, so equilibrium opposes increase in temperature by favoring endothermic backward reaction [1]

(b)(iv) [2 marks]
- M1: Particles are closer together / higher concentration of gas molecules [1]
- M2: Frequency of collisions (or collisions per unit time) increases, leading to higher rate [1]
Question 3 · Structured Theory & Mechanics
12 marks
(a) Across Period 3 of the Periodic Table, elements display regular trends in electronic configuration, bonding, and physical properties.

(i) Explain why the first ionisation energy of sulfur is lower than that of phosphorus.

(ii) Describe and explain the trend in melting point from sodium to aluminium.

(b) Phosphorus and sulfur react with chlorine to form covalent chlorides.

(i) Solid phosphorus(V) chloride, \(\text{PCl}_5\), exists as a covalent molecule in the gas phase.
Draw the 3D shape of a gaseous \(\text{PCl}_5\) molecule, state the bond angles present, and give the name of its molecular geometry.

(ii) Write a balanced chemical equation for the reaction of solid phosphorus(V) chloride with water, and state the approximate pH of the resulting solution.

(c) Chlorine is used extensively in water treatment.

(i) Define the term disproportionation.

(ii) Write an ionic equation for the reaction of chlorine gas with cold, dilute aqueous sodium hydroxide.
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Worked solution

(a) (i) Phosphorus has the valence configuration \(3\text{s}^2 3\text{p}_x^1 3\text{p}_y^1 3\text{p}_z^1\) (all three 3p orbitals are singly occupied). Sulfur has the valence configuration \(3\text{s}^2 3\text{p}_x^2 3\text{p}_y^1 3\text{p}_z^1\). The pairing of two electrons in one 3p orbital in sulfur creates inter-electron spin repulsion, which makes it require less energy to remove the outermost electron despite the higher nuclear charge.

(ii) Trend: Melting point increases from \(\text{Na} < \text{Mg} < \text{Al}\).
Explanation: Metallic bonding strength increases from Na to Al because:
1. The charge on the metal cation increases (\(\text{Na}^+\) to \(\text{Mg}^{2+}\) to \(\text{Al}^{3+}\)).
2. The ionic radius decreases from \(\text{Na}^+\) to \(\text{Al}^{3+}\), giving a higher charge density.
3. The number of delocalised electrons per atom contributing to the metallic lattice increases from 1 to 3, resulting in stronger electrostatic attraction between the metal cations and delocalised electrons.

(b) (i) Shape: Trigonal bipyramidal.
Drawing: Central P atom surrounded by 5 Cl atoms (3 equatorial in a planar triangle, 2 axial above and below the plane).
Bond angles: \(90^\circ\) (axial-equatorial), \(120^\circ\) (equatorial-equatorial), and \(180^\circ\) (axial-axial).

(ii) Equation: \(\text{PCl}_5(\text{s}) + 4\text{H}_2\text{O}(\text{l}) \rightarrow \text{H}_3\text{PO}_4(\text{aq}) + 5\text{HCl}(\text{aq})\)
(Allow \(\text{PCl}_5 + \text{H}_2\text{O} \rightarrow \text{POCl}_3 + 2\text{HCl}\) if limited water specified).
Approximate pH: \(0\) to \(2\) (strongly acidic).

(c) (i) Disproportionation is a redox reaction in which the same element is simultaneously oxidised and reduced.

(ii) \(\text{Cl}_2(\text{g}) + 2\text{OH}^-(\text{aq}) \rightarrow \text{Cl}^-(\text{aq}) + \text{ClO}^-(\text{aq}) + \text{H}_2\text{O}(\text{l})\)

Marking scheme

(a)(i) [2 marks]
- M1: Sulfur has a pair of electrons in a 3p orbital / \(3\text{p}^4\) vs \(3\text{p}^3\) in phosphorus [1]
- M2: Repulsion between the paired electrons reduces the energy required to remove an electron [1]

(a)(ii) [3 marks]
- M1: State that melting point increases from Na to Al [1]
- M2: Increase in number of delocalised electrons / increasing cation charge (\(\text{Na}^+\) to \(\text{Al}^{3+}\)) and decreasing ionic radius [1]
- M3: Stronger electrostatic attraction between metal cations and delocalised electron sea / stronger metallic bonding [1]

(b)(i) [3 marks]
- M1: Correct 3D diagram showing trigonal bipyramidal arrangement (3 equatorial bonds, 2 axial bonds) [1]
- M2: Bond angles: \(90^\circ\) and \(120^\circ\) (allow \(180^\circ\)) [1]
- M3: Name of shape: trigonal bipyramidal [1]

(b)(ii) [2 marks]
- M1: Correct balanced equation: \(\text{PCl}_5 + 4\text{H}_2\text{O} \rightarrow \text{H}_3\text{PO}_4 + 5\text{HCl}\) [1]
- M2: Approximate pH in range \(0\) to \(2\) [1]

(c)(i) [1 mark]
- Simultaneous oxidation and reduction of the same element [1]

(c)(ii) [1 mark]
- \(\text{Cl}_2 + 2\text{OH}^- \rightarrow \text{Cl}^- + \text{ClO}^- + \text{H}_2\text{O}\) [1]
Question 4 · Structured Theory & Mechanics
12 marks
(a) Halogenoalkanes are important synthetic intermediates undergoing substitution and elimination reactions.

(i) 2-bromo-2-methylpropane, \((\text{CH}_3)_3\text{CBr}\), reacts with aqueous sodium hydroxide to form 2-methylpropan-2-ol via an \(\text{S}_\text{N}1\) mechanism.

Outline the complete two-step mechanism for this reaction.
Include all relevant dipoles, curly arrows, charges, and lone pairs.

(ii) Explain why tertiary halogenoalkanes react predominantly via an \(\text{S}_\text{N}1\) mechanism rather than an \(\text{S}_\text{N}2\) mechanism.

(b) 1-bromobutane, \(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{Br}\), reacts with hot ethanolic potassium hydroxide to produce but-1-ene.

(i) State the role of the hydroxide ion, \(\text{OH}^-\), in this elimination reaction.

(ii) When but-1-ene is reacted with hydrogen bromide, \(\text{HBr}\), two isomeric bromoalkanes, \(\mathbf{A}\) and \(\mathbf{B}\), are formed.

Identify the major organic product and explain why it is formed in greater yield by referring to the relative stabilities of the carbocation intermediates.

(c) Describe a chemical test to distinguish between 1-chlorobutane and 1-iodobutane using aqueous silver nitrate dissolved in ethanol. State the observations for each compound and explain the difference in the rate of precipitate formation.
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Worked solution

(a) (i) Step 1: Heterolytic fission of the carbon-bromine bond:
- Partial charges: \(\text{C}^{\delta+} - \text{Br}^{\delta-}\).
- Curly arrow from the \(\text{C}-\text{Br}\) bond to the \(\text{Br}\) atom.
- Formation of the planar tertiary carbocation intermediate \((\text{CH}_3)_3\text{C}^+\) and bromide ion \(\text{Br}^-\).

Step 2: Nucleophilic attack on the carbocation:
- Curly arrow from a lone pair of electrons on \(:\text{OH}^-\) to the positively charged carbon atom \(\text{C}^+\).
- Formation of 2-methylpropan-2-ol, \((\text{CH}_3)_3\text{COH}\).

(ii) Tertiary halogenoalkanes react via \(\text{S}_\text{N}1\) because:
1. The tertiary carbocation intermediate is highly stable due to the positive inductive effect (electron-donating effect) of three methyl / alkyl groups.
2. Three bulky alkyl groups around the tertiary carbon atom create significant steric hindrance, preventing backside attack by the nucleophile (which would be necessary for an \(\text{S}_\text{N}2\) mechanism).

(b) (i) The hydroxide ion acts as a base (or proton / \(\text{H}^+\) acceptor).

(ii) Major product: 2-bromobutane, \(\text{CH}_3\text{CH}_2\text{CH}(\text{Br})\text{CH}_3\).
Explanation: Electrophilic addition of \(\text{H}^+\) to but-1-ene can produce either a primary carbocation (\(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2^+\)) or a secondary carbocation (\(\text{CH}_3\text{CH}_2\text{CH}^+\text{CH}_3\)). The secondary carbocation is more stable than the primary carbocation because it is stabilized by the positive inductive effect of two electron-releasing alkyl groups (compared to only one in the primary carbocation), lowering the activation energy for its formation.

(c) Test & Observations:
- Add ethanolic \(\text{AgNO}_3(\text{aq})\) and warm in a water bath.
- 1-iodobutane produces a yellow precipitate of \(\text{AgI}\) almost immediately / rapidly.
- 1-chlorobutane produces a white precipitate of \(\text{AgCl}\) very slowly / after prolonged heating.
Explanation: The \(\text{C}-\text{I}\) bond is longer and weaker (has a lower bond enthalpy: \(240\text{ kJ}\cdot\text{mol}^{-1}\)) than the \(\text{C}-\text{Cl}\) bond (\(340\text{ kJ}\cdot\text{mol}^{-1}\)), so the \(\text{C}-\text{I}\) bond breaks much faster to release \(\text{I}^-\).

Marking scheme

(a)(i) [4 marks]
- M1: Correct dipole \(\text{C}^{\delta+} - \text{Br}^{\delta-}\) and curly arrow from \(\text{C}-\text{Br}\) bond to \(\text{Br}\) [1]
- M2: Correct structure of tertiary carbocation \((\text{CH}_3)_3\text{C}^+\) and \(\text{Br}^-\) [1]
- M3: Curly arrow from lone pair on \(:\text{OH}^-\) to \(\text{C}^+\) of carbocation [1]
- M4: Correct structure of product \((\text{CH}_3)_3\text{COH}\) [1]

(a)(ii) [2 marks]
- M1: Tertiary carbocation is stable due to the positive inductive (+I) effect / electron release of three alkyl groups [1]
- M2: Steric hindrance from three alkyl groups prevents attack by nucleophile in \(\text{S}_\text{N}2\) [1]

(b)(i) [1 mark]
- Base / proton acceptor [1]

(b)(ii) [3 marks]
- M1: Major product identified as 2-bromobutane (name or correct structure) [1]
- M2: Formed via secondary carbocation intermediate (compared to primary) [1]
- M3: Secondary carbocation is more stable due to positive inductive effect of two alkyl groups (releasing electrons towards \(\text{C}^+\)) [1]

(c) [2 marks]
- M1: 1-iodobutane gives yellow ppt rapidly, 1-chlorobutane gives white ppt slowly [1]
- M2: \(\text{C}-\text{I}\) bond is weaker / has lower bond energy than \(\text{C}-\text{Cl}\) bond, so breaks more readily [1]
Question 5 · Structured Theory & Mechanics
12 marks
(a) Compound \(\mathbf{W}\) has the molecular formula \(\text{C}_4\text{H}_8\text{O}\).

- Compound \(\mathbf{W}\) reacts with 2,4-dinitrophenylhydrazine (2,4-DNPH) to give an orange precipitate.
- Compound \(\mathbf{W}\) shows no reaction when warmed with Tollens' reagent.

(i) Deduce the structural formula of \(\mathbf{W}\) and state its IUPAC systematic name.

(ii) State the characteristic absorption range in the infrared spectrum that confirms the presence of the carbonyl functional group in \(\mathbf{W}\).

(iii) In the mass spectrum of \(\mathbf{W}\), a major fragmentation peak appears at \(m/e = 43\). Give the formula of the fragment ion responsible for this peak.

(b) Compound \(\mathbf{W}\) can be synthesised by the oxidation of a four-carbon alcohol, \(\mathbf{V}\).

(i) Draw the displayed formula of alcohol \(\mathbf{V}\), and state the reagents and conditions required to convert \(\mathbf{V}\) into \(\mathbf{W}\).

(ii) State the type of reaction occurring in (b)(i).

(c) Propanal, \(\text{CH}_3\text{CH}_2\text{CHO}\), reacts with hydrogen cyanide, \(\text{HCN}\), in the presence of a catalytic amount of \(\text{KCN}\) to form a hydroxynitrile.

(i) Complete the mechanism for this reaction.
Include all relevant lone pairs, dipoles, charges, and curly arrows, as well as the structures of the intermediate and the organic product.

(ii) Explain the role of \(\text{KCN}\) in this mechanism.
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Worked solution

(a) (i) Reaction with 2,4-DNPH confirms a carbonyl group (aldehyde or ketone). No reaction with Tollens' reagent confirms \(\mathbf{W}\) is a ketone. With 4 carbons, \(\mathbf{W}\) must be butanone (or butan-2-one).
Structural formula: \(\text{CH}_3\text{COCH}_2\text{CH}_3\)
Systematic name: butanone (or butan-2-one).

(ii) The characteristic absorption for \(\text{C}=\text{O}\) in ketones is in the range \(1670 - 1740\text{ cm}^{-1}\).

(iii) The peak at \(m/e = 43\) corresponds to acylium ion \([\text{CH}_3\text{CO}]^+\) (mass \(15 + 28 = 43\)) or propyl carbocation \([\text{C}_3\text{H}_7]^+\) (mass \(36 + 7 = 43\)). Formula: \([\text{CH}_3\text{CO}]^+\) or \([\text{C}_3\text{H}_7]^+\).

(b) (i) Alcohol \(\mathbf{V}\) is butan-2-ol, \(\text{CH}_3\text{CH(OH)CH}_2\text{CH}_3\).
Reagents: Potassium dichromate(VI), \(\text{K}_2\text{Cr}_2\text{O}_7\), and dilute sulfuric acid, \(\text{H}_2\text{SO}_4\).
Conditions: Heat under reflux (or distillation).

(ii) Oxidation (or redox).

(c) (i) Mechanism for nucleophilic addition:
1. Nucleophilic attack: \(\text{CN}^-\) has a lone pair on the carbon atom \(:\text{C}\equiv\text{N}^-\). The carbonyl group has dipole \(\text{C}^{\delta+} = \text{O}^{\delta-}\). A curly arrow goes from the lone pair on \(:\text{C}\) to the carbonyl \(\text{C}^{\delta+}\). A second curly arrow goes from the \(\text{C}=\text{O}\) double bond to the \(\text{O}\) atom.
2. Intermediate formation: Tetrahedral alkoxide intermediate \(\text{CH}_3\text{CH}_2\text{CH}(\text{O}^-)\text{CN}\) is formed.
3. Protonation: A curly arrow goes from a lone pair on the negative oxygen \(:\text{O}^-\) to the \(\text{H}\) atom of \(\text{HCN}\) (or \(\text{H}^+\)), and the \(\text{H}-\text{CN}\) bond pair moves onto the \(\text{C}\) atom of \(\text{CN}\) to regenerate \(\text{CN}^-\).
4. Product: 2-hydroxybutanenitrile, \(\text{CH}_3\text{CH}_2\text{CH(OH)CN}\).

(ii) \(\text{KCN}\) provides the \(\text{CN}^-\) nucleophile (since \(\text{HCN}\) is a very weak acid that dissociates poorly in solution) and acts as a catalyst because \(\text{CN}^-\) is regenerated in the second step.

Marking scheme

(a)(i) [2 marks]
- M1: Correct structural formula: \(\text{CH}_3\text{COCH}_2\text{CH}_3\) [1]
- M2: IUPAC name: butanone / butan-2-one [1]

(a)(ii) [1 mark]
- \(1670 - 1740\text{ cm}^{-1}\) (or \(\text{C}=\text{O}\) absorption value within this range) [1]

(a)(iii) [1 mark]
- \([\text{CH}_3\text{CO}]^+\) or \([\text{C}_3\text{H}_7]^+\) / \([\text{CH}_3\text{CH}_2\text{CH}_2]^+\) (charge required) [1]

(b)(i) [2 marks]
- M1: Fully displayed formula of butan-2-ol [1]
- M2: Acidified potassium dichromate(VI) / \(\text{K}_2\text{Cr}_2\text{O}_7 + \text{H}_2\text{SO}_4\) and heat / reflux [1]

(b)(ii) [1 mark]
- Oxidation [1]

(c)(i) [4 marks]
- M1: Correct dipole \(\text{C}^{\delta+}=\text{O}^{\delta-}\) and curly arrow from lone pair on \(:\text{C}\) of \(:\text{CN}^-\) to carbonyl carbon [1]
- M2: Curly arrow from \(\text{C}=\text{O}\) double bond to oxygen atom [1]
- M3: Correct structure of intermediate \(\text{CH}_3\text{CH}_2\text{CH}(\text{O}^-)\text{CN}\) with negative charge on oxygen [1]
- M4: Curly arrow from lone pair on \(:\text{O}^-\) to \(\text{H}\) of \(\text{HCN}\) (or \(\text{H}^+\)) and curly arrow from \(\text{H}-\text{CN}\) bond to \(\text{CN}\), forming \(\text{CH}_3\text{CH}_2\text{CH(OH)CN}\) [1]

(c)(ii) [1 mark]
- Provides \(\text{CN}^-\) ions / increases concentration of nucleophile / acts as a catalyst (as \(\text{HCN}\) is a weak acid) [1]

Paper 3 (Advanced Practical Skills)

Carry out practical experiments in quantitative titration, gravimetric thermal analysis, and qualitative inorganic test-tube observations.
3 Question · 39.99 marks
Question 1 · Practical Laboratory Investigation
13.33 marks
In this experiment, you are to determine the percentage purity by mass of hydrated iron(II) sulfate crystals, \(\text{FeSO}_4 \cdot 7\text{H}_2\text{O}\) (molar mass \(277.9\text{ g}\cdot\text{mol}^{-1}\)), in an impure commercial sample labelled FA 1.

You are provided with:
* FA 1: An aqueous solution prepared by dissolving \(14.20\text{ g}\) of impure hydrated iron(II) sulfate crystals in dilute sulfuric acid and making up to \(500.0\text{ cm}^3\) in a volumetric flask.
* FA 2: \(0.0200\text{ mol}\cdot\text{dm}^{-3}\) aqueous potassium manganate(VII), \(\text{KMnO}_4\).

(a) A candidate pipettes \(25.0\text{ cm}^3\) of FA 1 into a conical flask and titrates it with FA 2 from a burette until the first permanent pale pink colour appears. The recorded titration data are shown below:

$$\begin{array}{|l|c|c|c|}
\hline
\text{Titration number} & 1 & 2 & 3 \\
\hline
\text{Final burette reading } / \text{ cm}^3 & 24.35 & 48.40 & 24.10 \\
\hline
\text{Initial burette reading } / \text{ cm}^3 & 0.00 & 24.35 & 0.05 \\
\hline
\text{Titre } / \text{ cm}^3 & 24.35 & 24.05 & 24.05 \\
\hline
\end{array}$$

From the table above, select suitable titres and calculate the mean titre of FA 2. Show which titres you used in your calculation.

(b) Calculate the amount, in moles, of \(\text{MnO}_4^-\text{(aq)}\) present in your mean titre.

(c) The ionic equation for the redox titration is:
$$\text{MnO}_4^-(\text{aq}) + 5\text{Fe}^{2+}(\text{aq}) + 8\text{H}^+(\text{aq}) \rightarrow \text{Mn}^{2+}(\text{aq}) + 5\text{Fe}^{3+}(\text{aq}) + 4\text{H}_2\text{O}(\text{l})$$

(i) Calculate the amount, in moles, of \(\text{Fe}^{2+}\) present in the \(25.0\text{ cm}^3\) sample of FA 1.
(ii) Calculate the total amount, in moles, of \(\text{Fe}^{2+}\) present in the \(500.0\text{ cm}^3\) volumetric flask.

(d) Using your answer to (c)(ii), calculate the mass of pure \(\text{FeSO}_4 \cdot 7\text{H}_2\text{O}\) in the original sample and deduce the percentage purity by mass of the commercial sample. Give your answer to 3 significant figures.

(e) Each individual reading of the burette has an uncertainty of \(\pm 0.05\text{ cm}^3\). Calculate the maximum percentage apparatus error in the titre volume obtained in Titration 2.
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Worked solution

(a) Mean Titre Calculation:
Concordant titres are within \(0.10\text{ cm}^3\). Titrations 2 and 3 are concordant (both \(24.05\text{ cm}^3\)). Titration 1 is a rough/non-concordant trial.
$$\text{Mean titre} = \frac{24.05 + 24.05}{2} = 24.05\text{ cm}^3$$

**(b) Moles of \(\text{MnO}_4^-\):**
$$n(\text{MnO}_4^-) = c \times V = 0.0200\text{ mol}\cdot\text{dm}^{-3} \times \frac{24.05}{1000}\text{ dm}^3 = 4.810 \times 10^{-4}\text{ mol}$$

**(c) Moles of \(\text{Fe}^{2+}\):**
(i) From stoichiometry, \(1\text{ mol } \text{MnO}_4^-\) reacts with \(5\text{ mol } \text{Fe}^{2+}\):
$$n(\text{Fe}^{2+})_{\text{pipette}} = 5 \times 4.810 \times 10^{-4} = 2.405 \times 10^{-3}\text{ mol}$$
(ii) In the total \(500.0\text{ cm}^3\) flask:
$$n(\text{Fe}^{2+})_{\text{total}} = 2.405 \times 10^{-3} \times \left(\frac{500.0}{25.0}\right) = 2.405 \times 10^{-3} \times 20 = 0.04810\text{ mol}$$

(d) Mass and Percentage Purity:
$$M_r(\text{FeSO}_4 \cdot 7\text{H}_2\text{O}) = 277.9\text{ g}\cdot\text{mol}^{-1}$$
$$\text{Mass of pure } \text{FeSO}_4 \cdot 7\text{H}_2\text{O} = 0.04810\text{ mol} \times 277.9\text{ g}\cdot\text{mol}^{-1} = 13.367\text{ g}$$
$$\text{Percentage purity} = \left(\frac{13.367\text{ g}}{14.20\text{ g}}\right) \times 100\% = 94.134\% \approx 94.1\%$$

(e) Percentage Uncertainty:
A titre requires two burette readings (initial and final), so total maximum absolute uncertainty is \(2 \times 0.05 = 0.10\text{ cm}^3\).
$$\text{Percentage uncertainty} = \left(\frac{0.10}{24.05}\right) \times 100\% = 0.4158\% \approx 0.42\%$$

Marking scheme

M1 (2 marks):
* Selects titres 2 and 3 and calculates mean titre as \(24.05\text{ cm}^3\) with correct units [1].
* Correctly specifies which titres were used and provides working [1].

M2 (2 marks):
* Correct calculation of \(n(\text{MnO}_4^-) = 4.81 \times 10^{-4}\text{ mol}\) [1].

M3 (3 marks):
* Multiplies \(n(\text{MnO}_4^-)\) by 5 to obtain \(2.405 \times 10^{-3}\text{ mol}\) for \(25.0\text{ cm}^3\) [1].
* Multiplies by 20 to obtain \(0.0481\text{ mol}\) in \(500.0\text{ cm}^3\) [2].

M4 (4 marks):
* Multiplies moles by \(277.9\) to find mass = \(13.37\text{ g}\) [2].
* Calculates \((13.37 / 14.20) \times 100 = 94.1\%\) (awarded only if given to 3 significant figures) [2].

M5 (2.33 marks):
* Accounts for two readings: \(\text{total uncertainty} = 0.10\text{ cm}^3\) [1].
* Correct percentage uncertainty calculation: \(0.42\%\) (or \(0.416\%\)) [1.33].
Question 2 · Practical Laboratory Investigation
13.33 marks
In this experiment, you are to determine the value of \(x\) in hydrated zinc sulfate, \(\text{ZnSO}_4 \cdot x\text{H}_2\text{O}\), by measuring the mass loss upon thermal decomposition.

$$\text{ZnSO}_4 \cdot x\text{H}_2\text{O}(\text{s}) \xrightarrow{\Delta} \text{ZnSO}_4(\text{s}) + x\text{H}_2\text{O}(\text{g})$$

(a) A sample of hydrated zinc sulfate was placed in a dry, pre-weighed porcelain crucible with a lid and heated using a Bunsen burner. The table of results obtained is shown below:

$$\begin{array}{|l|c|}
\hline
\text{Measurement} & \text{Mass } / \text{ g} \\
\hline
\text{Mass of crucible + lid} & 18.42 \\
\hline
\text{Mass of crucible + lid + } \text{ZnSO}_4 \cdot x\text{H}_2\text{O} & 22.73 \\
\hline
\text{Mass of crucible + lid + residue after 1st heating} & 20.91 \\
\hline
\text{Mass of crucible + lid + residue after 2nd heating} & 20.84 \\
\hline
\text{Mass of crucible + lid + residue after 3rd heating} & 20.84 \\
\hline
\end{array}$$

Using the data from the table:
(i) Calculate the initial mass of \(\text{ZnSO}_4 \cdot x\text{H}_2\text{O}\) heated.
(ii) Calculate the mass of anhydrous \(\text{ZnSO}_4\) residue formed.
(iii) Calculate the mass of water, \(\text{H}_2\text{O}\), driven off.

(b) Use your results from (a) to:
(i) Calculate the amount, in moles, of anhydrous \(\text{ZnSO}_4\) formed. [\(M_r(\text{ZnSO}_4) = 161.5\)]
(ii) Calculate the amount, in moles, of \(\text{H}_2\text{O}\) lost. [\(M_r(\text{H}_2\text{O}) = 18.0\)]
(iii) Determine the empirical formula and the integer value of \(x\) in \(\text{ZnSO}_4 \cdot x\text{H}_2\text{O}\).

(c) State why the crucible and residue were heated a second and third time.

(d) A student noticed that some solid 'spat' out of the crucible during vigorous initial heating before the lid was placed on.
State and explain the effect this loss of solid would have on the calculated value of \(x\).
Show answer & marking scheme

Worked solution

(a) Mass Determinations:
(i) Mass of hydrated salt = \(22.73 - 18.42 = 4.31\text{ g}\)
(ii) Mass of anhydrous \(\text{ZnSO}_4 = 20.84 - 18.42 = 2.42\text{ g}\)
(iii) Mass of \(\text{H}_2\text{O} = 22.73 - 20.84 = 1.89\text{ g}\) (or \(4.31 - 2.42 = 1.89\text{ g}\))

**(b) Mole Calculations and Value of \(x\):**
(i) Moles of \(\text{ZnSO}_4\):
$$n(\text{ZnSO}_4) = \frac{2.42\text{ g}}{161.5\text{ g}\cdot\text{mol}^{-1}} = 0.014985\text{ mol} \approx 0.0150\text{ mol}$$
(ii) Moles of \(\text{H}_2\text{O}\):
$$n(\text{H}_2\text{O}) = \frac{1.89\text{ g}}{18.0\text{ g}\cdot\text{mol}^{-1}} = 0.1050\text{ mol}$$
(iii) Ratio \(x\):
$$x = \frac{n(\text{H}_2\text{O})}{n(\text{ZnSO}_4)} = \frac{0.1050}{0.014985} = 7.007 \approx 7$$
Empirical formula: \(\text{ZnSO}_4 \cdot 7\text{H}_2\text{O}\), so \(x = 7\).

(c) Purpose of Reheating:
To heat to constant mass to ensure all water of crystallisation has been completely removed.

(d) Effect of Solid Spitting:
If solid spits out, the final mass of the crucible + residue will be smaller than it should be. Thus, the recorded mass loss (assumed to be solely water) will be artificially larger, leading to a calculated mole amount of water that is too high, which increases / overestimates the calculated value of \(x\).

Marking scheme

M1 (3 marks):
* (a)(i) \(4.31\text{ g}\) [1]
* (a)(ii) \(2.42\text{ g}\) (uses constant mass \(20.84\text{ g}\)) [1]
* (a)(iii) \(1.89\text{ g}\) [1]

M2 (4 marks):
* (b)(i) \(n(\text{ZnSO}_4) = 2.42 / 161.5 = 0.0150\text{ mol}\) (minimum 3 sig figs) [1.5]
* (b)(ii) \(n(\text{H}_2\text{O}) = 1.89 / 18.0 = 0.105\text{ mol}\) [1.5]
* (b)(iii) Computes ratio \(0.105 / 0.0150 = 7.0\) and deduces integer \(x = 7\) [1]

M3 (2.33 marks):
* (c) States 'to heat to constant mass' OR 'to ensure all water of crystallisation has been removed / reaction is complete' [2.33]

M4 (4 marks):
* (d) States value of \(x\) is higher / overestimated [2]
* Explains that lost solid contributes to a greater calculated mass loss / apparent water loss [2]
Question 3 · Practical Laboratory Investigation
13.33 marks
You are provided with two aqueous solutions, FA 3 and FA 4, each containing one cation and one anion from the Qualitative Analysis Notes.

(a) Carry out the tests described in the table below and deduce the observations. Write your observations in full, specifying the colour and state (precipitate/solution) of any products formed.

$$\begin{array}{|l|l|l|}
\hline
\textbf{Test} & \textbf{Observations for FA 3} & \textbf{Observations for FA 4} \\
\hline
\text{(i) Add aqueous sodium hydroxide} & & \\
\quad \text{dropwise, then in excess.} & & \\
\hline
\text{(ii) To the mixture from (i),} & & \\
\quad \text{warm gently with a piece of} & & \\
\quad \text{aluminium foil and test any} & & \\
\quad \text{gas evolved with damp red} & & \\
\quad \text{litmus paper.} & & \\
\hline
\text{(iii) Add aqueous ammonia} & & \\
\quad \text{dropwise, then in excess.} & & \\
\hline
\text{(iv) Add dilute nitric acid} & & \\
\quad \text{followed by aqueous barium nitrate.} & & \\
\hline
\text{(v) Add dilute nitric acid} & & \\
\quad \text{followed by aqueous silver nitrate,} & & \\
\quad \text{then add excess aqueous ammonia.} & & \\
\hline
\end{array}$$

Given observations:
* For FA 3:
- In (i): Green precipitate formed, insoluble in excess \(\text{NaOH}\).
- In (ii): No gas evolved; damp red litmus paper remains red.
- In (iii): Green precipitate formed, insoluble in excess \(\text{NH}_3\).
- In (iv): No precipitate formed.
- In (v): Cream/off-white precipitate formed, partially soluble/dissolves only in concentrated ammonia.
* For FA 4:
- In (i): No precipitate; on gentle warming without aluminium, no gas.
- In (ii): Effervescence; gas turns damp red litmus paper blue.
- In (iii): No precipitate.
- In (iv): Dense white precipitate formed, insoluble in excess dilute nitric acid.
- In (v): No precipitate.

(b) Identify the ions present in FA 3 and FA 4:
* Cation in FA 3:
* Anion in FA 3:
* Cation in FA 4:
* Anion in FA 4:

(c) Write an ionic equation, including state symbols, for the reaction between the cation in FA 3 and aqueous sodium hydroxide in test (a)(i).

(d) Write an ionic equation, including state symbols, for the precipitation reaction observed for FA 4 in test (a)(iv).
Show answer & marking scheme

Worked solution

(a) Analysis of Observations:
* FA 3:
- Green precipitate insoluble in excess \(\text{NaOH}\) and insoluble in excess \(\text{NH}_3\) identifies the cation as \(\text{Fe}^{2+}\).
- Cream precipitate with \(\text{AgNO}_3\text{(aq)}\) that is insoluble in dilute \(\text{NH}_3\) identifies the anion as \(\text{Br}^-\) (bromide).
* FA 4:
- No precipitate with \(\text{NaOH}\) or \(\text{NH}_3\); reduction with \(\text{Al}\) foil and warm \(\text{NaOH}\) produces \(\text{NH}_3\) gas (turns red litmus blue), identifying the anion as \(\text{NO}_3^-\) (nitrate).
- White precipitate with acidified \(\text{Ba(NO}_3)_2\) identifies \(\text{SO}_4^{2-}\) (sulfate). (Note: as an inorganic salt solution having one cation and nitrate/sulfate).
- Where FA 4 is an ammonium/metal salt:
- Cation: \(\text{NH}_4^+\) or \(\text{Na}^+\) / \(\text{K}^+\).
- Given test (iv) gives dense white precipitate: Anion = \(\text{SO}_4^{2-}\); Test (ii) gives ammonia with \(\text{Al}/\text{OH}^-\): Anion = \(\text{NO}_3^-\).
(Here, FA 4 contains \(\text{NH}_4^+\) and \(\text{SO}_4^{2-}\) or \(\text{Al}/\text{OH}^-\) reduction showing \(\text{NO}_3^-\)).

(b) Identified Ions:
* Cation in FA 3: \(\text{Fe}^{2+}\) (iron(II))
* Anion in FA 3: \(\text{Br}^-\) (bromide)
* Cation in FA 4: \(\text{NH}_4^+\) (or unprecipitated group 1 ion)
* Anion in FA 4: \(\text{SO}_4^{2-}\) (sulfate) [or \(\text{NO}_3^-\)]

(c) Ionic Equation for FA 3:
$$\text{Fe}^{2+}(\text{aq}) + 2\text{OH}^-(\text{aq}) \rightarrow \text{Fe(OH)}_2(\text{s})$$

(d) Ionic Equation for FA 4:
$$\text{Ba}^{2+}(\text{aq}) + \text{SO}_4^{2-}(\text{aq}) \rightarrow \text{BaSO}_4(\text{s})$$

Marking scheme

M1 (4 marks):
* Correctly links green precipitate insoluble in excess to \(\text{Fe}^{2+}\) in FA 3 [1].
* Correctly links cream precipitate with \(\text{AgNO}_3\) to \(\text{Br}^-\) in FA 3 [1].
* Correctly identifies \(\text{NH}_3\) gas evolution upon \(\text{Al}/\text{NaOH}\) warming to \(\text{NO}_3^-\) or \(\text{NH}_4^+\) in FA 4 [1].
* Correctly links white precipitate with \(\text{Ba}^{2+}\) to \(\text{SO}_4^{2-}\) in FA 4 [1].

M2 (4 marks):
* Identification of cation in FA 3: \(\text{Fe}^{2+}\) / iron(II) [1]
* Identification of anion in FA 3: \(\text{Br}^-\) / bromide [1]
* Identification of cation in FA 4: \(\text{NH}_4^+\) / \(\text{Na}^+\) [1]
* Identification of anion in FA 4: \(\text{SO}_4^{2-}\) / sulfate [1]

M3 (2.67 marks):
* \(\text{Fe}^{2+}(\text{aq}) + 2\text{OH}^-(\text{aq}) \rightarrow \text{Fe(OH)}_2(\text{s})\)
- Correct species and balancing [1.67]
- Correct state symbols `(aq)` and `(s)` [1.0]

M4 (2.66 marks):
* \(\text{Ba}^{2+}(\text{aq}) + \text{SO}_4^{2-}(\text{aq}) \rightarrow \text{BaSO}_4(\text{s})\)
- Correct species and balancing [1.66]
- Correct state symbols `(aq)` and `(s)` [1.0]

Paper 4 (A Level Structured Questions)

Answer all structured questions covering advanced physical, inorganic, organic, and analytical chemistry.
11 Question · 110 marks
Question 1 · Advanced Structured Theory & Calculation
10 marks
Calcium fluoride, \(\text{CaF}_2\), is a sparingly soluble ionic salt found naturally as the mineral fluorite.

(a) Define the term standard enthalpy change of atomisation.

(b) Relevant thermodynamic data for calcium and fluorine are given in the table below:

$$\begin{array}{|l|c|}
\hline
\text{Thermodynamic quantity} & \text{Value / } \text{kJ}\cdot\text{mol}^{-1} \\
\hline
\text{Standard enthalpy change of formation of } \text{CaF}_2(\text{s}) & -1220 \\
\text{Standard enthalpy change of atomisation of } \text{Ca}(\text{s}) & +178 \\
\text{First ionisation energy of calcium} & +590 \\
\text{Second ionisation energy of calcium} & +1145 \\
\text{Bond enthalpy of } \text{F}_2(\text{g}) & +158 \\
\text{First electron affinity of fluorine} & -328 \\
\hline
\end{array}$$

(i) Calculate the lattice energy, \(\Delta H_{\text{latt}}^\theta\), of calcium fluoride, \(\text{CaF}_2(\text{s})\).

(ii) The standard enthalpy changes of hydration of the gaseous ions are given below:
$$\Delta H_{\text{hyd}}^\theta(\text{Ca}^{2+}(\text{g})) = -1650 \text{ kJ}\cdot\text{mol}^{-1}$$
$$\Delta H_{\text{hyd}}^\theta(\text{F}^-(\text{g})) = -506 \text{ kJ}\cdot\text{mol}^{-1}$$

Calculate the standard enthalpy change of solution, \(\Delta H_{\text{sol}}^\theta\), of \(\text{CaF}_2(\text{s})\).

(c) The decomposition of solid calcium carbonate is an endothermic process:
$$\text{CaCO}_3(\text{s}) \rightarrow \text{CaO}(\text{s}) + \text{CO}_2(\text{g}) \quad \Delta H^\theta = +178 \text{ kJ}\cdot\text{mol}^{-1}$$

The standard entropies, \(S^\theta\), of the species involved are:
$$\begin{array}{|c|c|}
\hline
\text{Species} & S^\theta / \text{J}\cdot\text{K}^{-1}\cdot\text{mol}^{-1} \\
\hline
\text{CaCO}_3(\text{s}) & 92.9 \\
\text{CaO}(\text{s}) & 39.7 \\
\text{CO}_2(\text{g}) & 213.6 \\
\hline
\end{array}$$

(i) Calculate the standard entropy change, \(\Delta S^\theta\), for this reaction in \(\text{J}\cdot\text{K}^{-1}\cdot\text{mol}^{-1}\).

(ii) Calculate the minimum temperature, in \(\text{K}\), at which this decomposition becomes spontaneous under standard pressure. State any assumption made.
Show answer & marking scheme

Worked solution

(a) Standard enthalpy change of atomisation is the enthalpy change when 1 mole of gaseous atoms is formed from the element in its standard state under standard conditions (298 K, 101 kPa).

(b)(i) Using the Born-Haber cycle for \(\text{CaF}_2(\text{s})\):
$$\Delta H_{\text{f}}^\theta(\text{CaF}_2) = \Delta H_{\text{at}}^\theta(\text{Ca}) + \text{IE}_1(\text{Ca}) + \text{IE}_2(\text{Ca}) + \text{Bond Enthalpy}(\text{F}_2) + 2 \times \text{EA}_1(\text{F}) + \Delta H_{\text{latt}}^\theta(\text{CaF}_2)$$
$$-1220 = +178 + 590 + 1145 + 158 + 2(-328) + \Delta H_{\text{latt}}^\theta$$
$$-1220 = 2071 - 656 + \Delta H_{\text{latt}}^\theta = 1415 + \Delta H_{\text{latt}}^\theta$$
$$\Delta H_{\text{latt}}^\theta = -1220 - 1415 = -2635 \text{ kJ}\cdot\text{mol}^{-1}$$

(b)(ii) Enthalpy of solution:
$$\Delta H_{\text{sol}}^\theta = \Delta H_{\text{hyd}}^\theta(\text{Ca}^{2+}) + 2\Delta H_{\text{hyd}}^\theta(\text{F}^-) - \Delta H_{\text{latt}}^\theta(\text{CaF}_2)$$
$$\Delta H_{\text{sol}}^\theta = -1650 + 2(-506) - (-2635) = -1650 - 1012 + 2635 = -27 \text{ kJ}\cdot\text{mol}^{-1}$$

(c)(i) Standard entropy change:
$$\Delta S^\theta = \sum S^\theta(\text{products}) - \sum S^\theta(\text{reactants})$$
$$\Delta S^\theta = (39.7 + 213.6) - 92.9 = 253.3 - 92.9 = +160.4 \text{ J}\cdot\text{K}^{-1}\cdot\text{mol}^{-1}$$

(c)(ii) For spontaneity, \(\Delta G^\theta \le 0\). At equilibrium / boundary temperature:
$$\Delta H^\theta - T\Delta S^\theta = 0 \implies T = \frac{\Delta H^\theta}{\Delta S^\theta}$$
$$T = \frac{178 \times 1000}{160.4} = 1109.7 \text{ K} \approx 1110 \text{ K}$$
Assumption: \(\Delta H^\theta\) and \(\Delta S^\theta\) remain constant with changes in temperature.

Marking scheme

(a) M1: Enthalpy change when 1 mole of gaseous atoms is formed [1]
M2: from the element in its standard state (under standard conditions) [1]

(b)(i) M1: Correctly setting up Born-Haber expression: \(\Delta H_{\text{latt}}^\theta = \Delta H_f^\theta - [\Delta H_{\text{at}}^\theta + \text{IE}_1 + \text{IE}_2 + E(\text{F-F}) + 2\text{EA}_1]\) [1]
M2: Correct substitution with factor of 2 for electron affinity [1]
M3: Value = \(-2635 \text{ kJ}\cdot\text{mol}^{-1}\) (with negative sign) [1]

(b)(ii) M1: Correct expression \(\Delta H_{\text{sol}}^\theta = \sum \Delta H_{\text{hyd}}^\theta - \Delta H_{\text{latt}}^\theta\) [1]
M2: Value = \(-27 \text{ kJ}\cdot\text{mol}^{-1}\) [ecf from (b)(i)] [1]

(c)(i) M1: Correct calculation: \(39.7 + 213.6 - 92.9 = +160.4 \text{ J}\cdot\text{K}^{-1}\cdot\text{mol}^{-1}\) [1]

(c)(ii) M1: Converting \(\Delta H\) to J or \(\Delta S\) to kJ and using \(T = \frac{\Delta H}{\Delta S}\) [1]
M2: \(T = 1110 \text{ K}\) (allow 1109.7 K) [1]
M3: Assumption: \(\Delta H\) and \(\Delta S\) are independent of temperature [1]
Question 2 · Advanced Structured Theory & Calculation
10 marks
Propanoic acid, \(\text{CH}_3\text{CH}_2\text{COOH}\), is a weak monobasic carboxylic acid.

(a) (i) Write the expression for the acid dissociation constant, \(K_{\text{a}}\), of propanoic acid.

(ii) At \(298\text{ K}\), a \(0.150\text{ mol}\cdot\text{dm}^{-3}\) solution of propanoic acid has a \(\text{pH}\) of \(2.85\). Calculate the value of \(K_{\text{a}}\) for propanoic acid at this temperature. State the units of \(K_{\text{a}}\).

(b) A buffer solution was prepared by adding \(50.0\text{ cm}^3\) of \(0.200\text{ mol}\cdot\text{dm}^{-3}\text{ NaOH}(\text{aq})\) to \(100.0\text{ cm}^3\) of \(0.300\text{ mol}\cdot\text{dm}^{-3}\text{ CH}_3\text{CH}_2\text{COOH}(\text{aq})\).

Assume \(K_{\text{a}} = 1.34 \times 10^{-5}\text{ mol}\cdot\text{dm}^{-3}\) for propanoic acid.

Calculate the \(\text{pH}\) of this buffer solution at \(298\text{ K}\).

(c) Propanoic acid partitions between ethoxyethane and water. In an experiment, \(100.0\text{ cm}^3\) of an aqueous solution containing \(0.0450\text{ mol}\) of propanoic acid was shaken with \(40.0\text{ cm}^3\) of ethoxyethane until equilibrium was reached at \(298\text{ K}\).

The partition coefficient is defined as:
$$K_{\text{partition}} = \frac{[\text{CH}_3\text{CH}_2\text{COOH}(\text{ethoxyethane})]}{[\text{CH}_3\text{CH}_2\text{COOH}(\text{aq})]} = 3.50$$

Calculate the mass, in \(\text{g}\), of propanoic acid that remains extracted into the organic layer (ethoxyethane).
Show answer & marking scheme

Worked solution

(a)(i) \(K_{\text{a}} = \frac{[\text{CH}_3\text{CH}_2\text{COO}^-][\text{H}^+]}{[\text{CH}_3\text{CH}_2\text{COOH}]}\)

(a)(ii) \([\text{H}^+] = 10^{-\text{pH}} = 10^{-2.85} = 1.413 \times 10^{-3}\text{ mol}\cdot\text{dm}^{-3}\)
Assuming \([\text{H}^+] = [\text{CH}_3\text{CH}_2\text{COO}^-]\) and \([\text{CH}_3\text{CH}_2\text{COOH}]_{\text{eq}} \approx 0.150\text{ mol}\cdot\text{dm}^{-3}\):
$$K_{\text{a}} = \frac{(1.413 \times 10^{-3})^2}{0.150} = \frac{1.995 \times 10^{-6}}{0.150} = 1.33 \times 10^{-5}\text{ mol}\cdot\text{dm}^{-3}$$

(b) Initial moles of \(\text{CH}_3\text{CH}_2\text{COOH} = 0.100 \times 0.300 = 0.0300\text{ mol}\)
Moles of \(\text{NaOH} = 0.0500 \times 0.200 = 0.0100\text{ mol}\)
Reaction: \(\text{CH}_3\text{CH}_2\text{COOH} + \text{OH}^- \rightarrow \text{CH}_3\text{CH}_2\text{COO}^- + \text{H}_2\text{O}\)
Remaining \(\text{CH}_3\text{CH}_2\text{COOH} = 0.0300 - 0.0100 = 0.0200\text{ mol}\)
Moles of \(\text{CH}_3\text{CH}_2\text{COO}^-\) formed \(= 0.0100\text{ mol}\)
Total volume \(= 150.0\text{ cm}^3\).
$$[\text{H}^+] = K_{\text{a}} \times \frac{[\text{acid}]}{[\text{salt}]} = 1.34 \times 10^{-5} \times \frac{0.0200 / 0.150}{0.0100 / 0.150} = 1.34 \times 10^{-5} \times 2 = 2.68 \times 10^{-5}\text{ mol}\cdot\text{dm}^{-3}$$
$$\text{pH} = -\log_{10}(2.68 \times 10^{-5}) = 4.57$$

(c) Let \(x\) be the moles of propanoic acid extracted into ethoxyethane.
Then moles in aqueous layer \(= 0.0450 - x\).
$$[\text{CH}_3\text{CH}_2\text{COOH}(\text{org})] = \frac{x}{0.0400} = 25x$$
$$[\text{CH}_3\text{CH}_2\text{COOH}(\text{aq})] = \frac{0.0450 - x}{0.100} = 10(0.0450 - x)$$
$$K_{\text{partition}} = \frac{25x}{10(0.0450 - x)} = 3.50$$
$$25x = 35(0.0450 - x) = 1.575 - 35x$$
$$60x = 1.575 \implies x = 0.02625\text{ mol}$$
Relative molecular mass of propanoic acid (\(\text{C}_3\text{H}_6\text{O}_2\)):
$$M_{\text{r}} = 3(12.0) + 6(1.0) + 2(16.0) = 74.0$$
$$\text{Mass} = 0.02625 \times 74.0 = 1.9425 \approx 1.94 \text{ to } 1.95\text{ g}$$
(Using \(M_r=74.08\), mass \(= 1.945\text{ g} \approx 1.95\text{ g}\)).

Marking scheme

(a)(i) M1: \(K_a = \frac{[\text{CH}_3\text{CH}_2\text{COO}^-][\text{H}^+]}{[\text{CH}_3\text{CH}_2\text{COOH}]}\) [1]
(a)(ii) M1: \([\text{H}^+] = 10^{-2.85} = 1.413 \times 10^{-3}\text{ mol}\cdot\text{dm}^{-3}\) [1]
M2: \(K_a = 1.33 \times 10^{-5}\) and units \(\text{mol}\cdot\text{dm}^{-3}\) [1]

(b) M1: Calculate moles of acid \(= 0.0300\) and \(\text{OH}^- = 0.0100\) [1]
M2: Calculate equilibrium moles: acid \(= 0.0200\), conjugate base \(= 0.0100\) [1]
M3: Apply Henderson-Hasselbalch or \([\text{H}^+] = K_a \times \frac{0.0200}{0.0100} = 2.68 \times 10^{-5}\) [1]
M4: \(\text{pH} = 4.57\) [1]

(c) M1: Set up partition equation: \(\frac{x / 0.0400}{(0.0450 - x) / 0.100} = 3.50\) [1]
M2: Solve for moles in organic phase \(x = 0.02625\text{ mol}\) [1]
M3: Calculate mass \(= 0.02625 \times 74.0 = 1.94\) or \(1.95\text{ g}\) [1]
Question 3 · Advanced Structured Theory & Calculation
10 marks
Peroxodisulfate(\(\text{VI}\)) ions, \(\text{S}_2\text{O}_8^{2-}\), oxidise iodide ions to iodine in aqueous solution according to the equation:
$$\text{S}_2\text{O}_8^{2-}(\text{aq}) + 2\text{I}^-(\text{aq}) \rightarrow 2\text{SO}_4^{2-}(\text{aq}) + \text{I}_2(\text{aq})$$

(a) The rate of this reaction was investigated at \(298\text{ K}\) by measuring initial rates with different initial concentrations of the reactants. The results are shown:

$$\begin{array}{|c|c|c|c|}
\hline
\text{Experiment} & [\text{S}_2\text{O}_8^{2-}] / \text{mol}\cdot\text{dm}^{-3} & [\text{I}^-] / \text{mol}\cdot\text{dm}^{-3} & \text{Initial rate} / \text{mol}\cdot\text{dm}^{-3}\cdot\text{s}^{-1} \\
\hline
1 & 0.0400 & 0.0300 & 1.44 \times 10^{-4} \\
2 & 0.0800 & 0.0300 & 2.88 \times 10^{-4} \\
3 & 0.0800 & 0.0900 & 8.64 \times 10^{-4} \\
\hline
\end{array}$$

(i) Deduce the order of reaction with respect to \(\text{S}_2\text{O}_8^{2-}\) and with respect to \(\text{I}^-\). Explain your reasoning.

(ii) Write the rate equation for this reaction.

(iii) Calculate the rate constant, \(k\), for this reaction at \(298\text{ K}\). State its units.

(b) In a separate experiment carried out with a large excess of iodide ions, \([\text{I}^-]\) remains effectively constant at \(0.500\text{ mol}\cdot\text{dm}^{-3}\). The half-life, \(t_{1/2}\), for the decay of \([\text{S}_2\text{O}_8^{2-}]\) is observed to be constant.

Calculate the half-life, in seconds, under these pseudo-first-order conditions.

(c) The uncatalysed reaction between \(\text{S}_2\text{O}_8^{2-}\) and \(\text{I}^-\) is slow because it has a high activation energy. \(\text{Fe}^{2+}(\text{aq})\) acts as a homogeneous catalyst.

(i) State why two negatively charged ions have a high activation energy for reaction.

(ii) Write two ionic equations to show how \(\text{Fe}^{2+}(\text{aq})\) catalyses this reaction.
Show answer & marking scheme

Worked solution

(a)(i) Comparing Experiment 1 and 2: \([\text{I}^-]\) is constant; \([\text{S}_2\text{O}_8^{2-}]\) doubles (from \(0.0400\) to \(0.0800\)), rate doubles (from \(1.44 \times 10^{-4}\) to \(2.88 \times 10^{-4}\)). Thus, order with respect to \([\text{S}_2\text{O}_8^{2-}]\) is 1.
Comparing Experiment 2 and 3: \([\text{S}_2\text{O}_8^{2-}]\) is constant; \([\text{I}^-]\) triples (from \(0.0300\) to \(0.0900\)), rate triples (from \(2.88 \times 10^{-4}\) to \(8.64 \times 10^{-4}\)). Thus, order with respect to \([\text{I}^-]\) is 1.

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

(a)(iii) Using Experiment 1:
$$k = \frac{\text{Rate}}{[\text{S}_2\text{O}_8^{2-}][\text{I}^-]} = \frac{1.44 \times 10^{-4}}{(0.0400)(0.0300)} = \frac{1.44 \times 10^{-4}}{1.20 \times 10^{-3}} = 0.120$$
Units: \(\frac{\text{mol}\cdot\text{dm}^{-3}\cdot\text{s}^{-1}}{(\text{mol}\cdot\text{dm}^{-3})^2} = \text{dm}^3\cdot\text{mol}^{-1}\cdot\text{s}^{-1}\).

(b) Under pseudo-first-order conditions:
$$k' = k[\text{I}^-] = 0.120 \times 0.500 = 0.0600\text{ s}^{-1}$$
$$t_{1/2} = \frac{\ln 2}{k'} = \frac{0.693}{0.0600} = 11.55\text{ s} \approx 11.6\text{ s}$$

(c)(i) The reaction is between two negatively charged anions, which repel each other electrostatically, requiring significant energy to bring them close together.
(c)(ii) Step 1: \(2\text{Fe}^{2+}(\text{aq}) + \text{S}_2\text{O}_8^{2-}(\text{aq}) \rightarrow 2\text{Fe}^{3+}(\text{aq}) + 2\text{SO}_4^{2-}(\text{aq})\)
Step 2: \(2\text{Fe}^{3+}(\text{aq}) + 2\text{I}^-(\text{aq}) \rightarrow 2\text{Fe}^{2+}(\text{aq}) + \text{I}_2(\text{aq})\)

Marking scheme

(a)(i) M1: Order wrt \([\text{S}_2\text{O}_8^{2-}] = 1\) with clear justification from exp 1 and 2 [1]
M2: Order wrt \([\text{I}^-] = 1\) with clear justification from exp 2 and 3 [1]

(a)(ii) M1: Correct rate equation: \(\text{Rate} = k[\text{S}_2\text{O}_8^{2-}][\text{I}^-]\) [1]

(a)(iii) M1: Numerical value \(k = 0.120\) (allow 0.12) [1]
M2: Correct units: \(\text{dm}^3\cdot\text{mol}^{-1}\cdot\text{s}^{-1}\) [1]

(b) M1: Calculation of apparent first-order rate constant \(k' = 0.0600\text{ s}^{-1}\) [1]
M2: \(t_{1/2} = \frac{0.693}{0.0600} = 11.55\text{ s}\) or \(11.6\text{ s}\) [1]

(c)(i) M1: Electrostatic repulsion between negatively charged ions [1]

(c)(ii) M1: Equation 1: \(2\text{Fe}^{2+} + \text{S}_2\text{O}_8^{2-} \rightarrow 2\text{Fe}^{3+} + 2\text{SO}_4^{2-}\) [1]
M2: Equation 2: \(2\text{Fe}^{3+} + 2\text{I}^- \rightarrow 2\text{Fe}^{2+} + \text{I}_2\) [1]
Question 4 · Advanced Structured Theory & Calculation
10 marks
Cobalt is a transition element that forms complex ions in different oxidation states.

(a) State the electronic configuration of the cobalt(\(\text{II}\)) ion, \(\text{Co}^{2+}\).

(b) When concentrated hydrochloric acid is added to a pink aqueous solution containing \([\text{Co}(\text{H}_2\text{O})_6]^{2+}\), the solution turns blue as the tetrahedral complex \([\text{CoCl}_4]^{2-}\) is formed.

(i) Write an equation for this equilibrium reaction.

(ii) State the coordination number and geometry of cobalt in \([\text{Co}(\text{H}_2\text{O})_6]^{2+}\) and in \([\text{CoCl}_4]^{2-}\).

(iii) Explain why transition metal complexes are coloured, making reference to d-orbitals, ligands, and visible light absorption.

(c) The stability constant, \(K_{\text{stab}}\), for the complex ion formed between \(\text{Co}^{2+}\) and 1,2-diaminoethane (abbreviated as \(\text{en}\), \(\text{H}_2\text{NCH}_2\text{CH}_2\text{NH}_2\)) is \(1.0 \times 10^{14}\text{ dm}^9\cdot\text{mol}^{-3}\) at \(298\text{ K}\).

$$[\text{Co}(\text{H}_2\text{O})_6]^{2+} + 3\text{en} \rightleftharpoons [\text{Co}(\text{en})_3]^{2+} + 6\text{H}_2\text{O}$$

(i) Write the expression for \(K_{\text{stab}}\) for this equilibrium.

(ii) In terms of entropy change, explain why \([\text{Co}(\text{en})_3]^{2+}\) is significantly more stable than \([\text{Co}(\text{NH}_3)_6]^{2+}\) (the chelate effect).
Show answer & marking scheme

Worked solution

(a) Cobalt atomic number = 27 (\(1s^2 2s^2 2p^6 3s^2 3p^6 3d^7 4s^2\)). \(\text{Co}^{2+}\) loses the two 4s electrons: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^7\) or \([\text{Ar}]3d^7\).

(b)(i) \([\text{Co}(\text{H}_2\text{O})_6]^{2+} + 4\text{Cl}^- \rightleftharpoons [\text{CoCl}_4]^{2-} + 6\text{H}_2\text{O}\)

(b)(ii) \([\text{Co}(\text{H}_2\text{O})_6]^{2+}\): Coordination number = 6, shape = octahedral.
\([\text{CoCl}_4]^{2-}\): Coordination number = 4, shape = tetrahedral.

(b)(iii) When ligands coordinate to the transition metal ion, the electric field of their lone pairs splits the five degenerate 3d orbitals into two sets of non-degenerate energy levels separated by energy gap \(\Delta E\).
An electron absorbs a photon of visible light whose energy matches the gap (\(\Delta E = h\nu\)) and is promoted from a lower d-orbital to a higher d-orbital (d-d transition).
The remaining wavelengths/frequencies of light are transmitted/reflected, producing the observed complementary colour.

(c)(i) \(K_{\text{stab}} = \frac{[[\text{Co}(\text{en})_3]^{2+}]}{[[\text{Co}(\text{H}_2\text{O})_6]^{2+}][\text{en}]^3}\)

(c)(ii) In the substitution reaction, 4 particles (1 complex ion + 3 bidentate \(\text{en}\) molecules) produce 7 particles (1 complex ion + 6 uncoordinated water molecules). This increase in the number of independent particles leads to an increase in entropy (\(\Delta S^\theta > 0\)). Because \(\Delta G^\theta = \Delta H^\theta - T\Delta S^\theta\), a large positive \(T\Delta S^\theta\) term makes \(\Delta G^\theta\) much more negative, driving the equilibrium far to the right.

Marking scheme

(a) M1: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^7\) or \([\text{Ar}] 3d^7\) [1]

(b)(i) M1: Balanced equilibrium equation: \([\text{Co}(\text{H}_2\text{O})_6]^{2+} + 4\text{Cl}^- \rightleftharpoons [\text{CoCl}_4]^{2-} + 6\text{H}_2\text{O}\) [1]

(b)(ii) M1: Coordination numbers: 6 and 4 [1]
M2: Shapes: octahedral and tetrahedral [1]

(b)(iii) M1: Splitting of d-orbitals into two energy levels by ligands [1]
M2: Absorption of visible light / photons promotes an electron from lower to higher d-orbital (\(\Delta E = hf\)) [1]
M3: Transmitted / reflected light gives complementary colour [1]

(c)(i) M1: Correct expression for \(K_{\text{stab}}\) (water omitted from expression) [1]

(c)(ii) M1: Increase in number of particles / 4 particles become 7 particles [1]
M2: Entropy increases (\(\Delta S > 0\)) leading to more negative \(\Delta G\) / higher stability [1]
Question 5 · Advanced Structured Theory & Calculation
10 marks
Standard electrode potentials can be used to predict the feasibility of redox reactions.

(a) Standard reduction potentials at \(298\text{ K}\) are given below:
$$\begin{array}{rcll}
\text{Fe}^{3+}(\text{aq}) + \text{e}^- & \rightleftharpoons & \text{Fe}^{2+}(\text{aq}) & E^\theta = +0.77\text{ V} \\
\text{Ag}^+(\text{aq}) + \text{e}^- & \rightleftharpoons & \text{Ag}(\text{s}) & E^\theta = +0.80\text{ V} \\
\text{Cu}^{2+}(\text{aq}) + 2\text{e}^- & \rightleftharpoons & \text{Cu}(\text{s}) & E^\theta = +0.34\text{ V}
\end{array}$$

(i) Define standard electrode potential.

(ii) An electrochemical cell is constructed under standard conditions using the \(\text{Fe}^{3+}/\text{Fe}^{2+}\) half-cell and the \(\text{Ag}^+/\text{Ag}\) half-cell. Calculate the standard cell potential, \(E_{\text{cell}}^\theta\), and state which electrode is the positive terminal.

(iii) Write the overall ionic equation for the cell reaction that occurs spontaneously under standard conditions.

(b) The electrode potential of a non-standard half-cell can be calculated using the Nernst equation:
$$E = E^\theta + \frac{0.059}{z}\log_{10}\left(\frac{[\text{oxidised form}]}{[\text{reduced form}]}\right)$$

Calculate the electrode potential, \(E\), of a copper half-cell when \([\text{Cu}^{2+}] = 0.00200\text{ mol}\cdot\text{dm}^{-3}\) at \(298\text{ K}\).

(c) In an industrial electroplating experiment, a metal object is coated with copper by electrolysing aqueous copper(II) sulfate. A steady current of \(3.50\text{ A}\) was passed through the cell for \(45.0\text{ minutes}\).

Calculate the mass of copper, in \(\text{g}\), deposited on the electrode. \([F = 9.65 \times 10^4\text{ C}\cdot\text{mol}^{-1}; A_{\text{r}}(\text{Cu}) = 63.5]\)
Show answer & marking scheme

Worked solution

(a)(i) Standard electrode potential is the electromotive force (emf) of a half-cell compared with a standard hydrogen electrode (SHE) under standard conditions (\(298\text{ K}\), \(101\text{ kPa}\), and ion concentrations of \(1.00\text{ mol}\cdot\text{dm}^{-3}\)).

(a)(ii) \(E_{\text{cell}}^\theta = E^\theta(\text{cathode}) - E^\theta(\text{anode}) = +0.80 - (+0.77) = +0.03\text{ V}\).
The silver half-cell has the more positive potential (\(+0.80\text{ V}\)), so the silver electrode is the positive terminal (cathode).

(a)(iii) \(\text{Fe}^{2+}(\text{aq}) + \text{Ag}^+(\text{aq}) \rightarrow \text{Fe}^{3+}(\text{aq}) + \text{Ag}(\text{s})\)

(b) For \(\text{Cu}^{2+}(\text{aq}) + 2\text{e}^- \rightleftharpoons \text{Cu}(\text{s})\), \(z = 2\), \([\text{oxidised}] = [\text{Cu}^{2+}] = 0.00200\text{ mol}\cdot\text{dm}^{-3}\), and the concentration of the solid is taken as 1:
$$E = 0.34 + \frac{0.059}{2}\log_{10}(0.00200)$$
$$\log_{10}(0.00200) = -2.6990$$
$$E = 0.34 + 0.0295(-2.6990) = 0.34 - 0.0796 = +0.2604\text{ V} \approx +0.260\text{ V}$$

(c) Quantity of charge passed:
$$Q = I \times t = 3.50\text{ A} \times (45.0 \times 60\text{ s}) = 3.50 \times 2700 = 9450\text{ C}$$
Moles of electrons:
$$n(\text{e}^-) = \frac{Q}{F} = \frac{9450}{9.65 \times 10^4} = 0.09793\text{ mol}$$
Since \(\text{Cu}^{2+} + 2\text{e}^- \rightarrow \text{Cu}\), \(n(\text{Cu}) = \frac{0.09793}{2} = 0.04896\text{ mol}\).
Mass of copper deposited:
$$\text{mass} = 0.04896 \times 63.5 = 3.109\text{ g} \approx 3.11\text{ g}$$

Marking scheme

(a)(i) M1: Potential difference between half-cell and standard hydrogen electrode [1]
M2: Measured under standard conditions (298 K, 101 kPa, 1.00 mol dm^-3) [1]

(a)(ii) M1: \(E_{\text{cell}}^\theta = +0.03\text{ V}\) [1]
M2: Ag (or \(\text{Ag}^+/\text{Ag}\)) is the positive electrode / cathode [1]

(a)(iii) M1: Correct equation: \(\text{Fe}^{2+} + \text{Ag}^+ \rightarrow \text{Fe}^{3+} + \text{Ag}\) [1]

(b) M1: Correct use of \(z = 2\) and \(\log_{10}(0.00200) = -2.70\) in Nernst equation [1]
M2: \(E = +0.260\text{ V}\) (or +0.26 V) [1]

(c) M1: Calculation of \(Q = 3.50 \times 2700 = 9450\text{ C}\) [1]
M2: Moles of \(\text{Cu} = \frac{9450}{2 \times 9.65 \times 10^4} = 0.04896\text{ mol}\) [1]
M3: Mass \(= 0.04896 \times 63.5 = 3.11\text{ g}\) [1]
Question 6 · Advanced Structured Theory & Calculation
10 marks
Benzene and its derivatives undergo characteristic electrophilic substitution reactions.

(a) Benzene reacts with a mixture of concentrated nitric acid and concentrated sulfuric acid at \(50-55^\circ\text{C}\) to form nitrobenzene.

(i) Write an equation for the generation of the electrophile, \(\text{NO}_2^+\), from the two concentrated acids.

(ii) Describe the mechanism for the nitration of benzene. In your answer, include curly arrows to show the movement of electron pairs, the intermediate ion structure, and the regenerator of the catalyst.

(b) Nitrobenzene can be converted into phenylamine, \(\text{C}_6\text{H}_5\text{NH}_2\).

(i) State the reagents and conditions needed to reduce nitrobenzene to phenylamine in the laboratory.

(ii) Explain why phenylamine is a weaker base than ethylamine, \(\text{CH}_3\text{CH}_2\text{NH}_2\).

(c) Phenylamine is used to synthesise azo dyes. In the first step, phenylamine is treated with nitrous acid, \(\text{HNO}_2\), at \(5^\circ\text{C}\) to form a diazonium salt.

(i) State how nitrous acid is prepared in situ.

(ii) Draw the structural formula of the azo dye formed when benzenediazonium chloride reacts with alkaline phenol, \(\text{C}_6\text{H}_5\text{O}^-\).
Show answer & marking scheme

Worked solution

(a)(i) \(\text{HNO}_3 + 2\text{H}_2\text{SO}_4 \rightarrow \text{NO}_2^+ + 2\text{HSO}_4^- + \text{H}_3\text{O}^+\) (or \(\text{HNO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{NO}_2^+ + \text{HSO}_4^- + \text{H}_2\text{O}\)).

(a)(ii) Mechanism steps:
1. A curly arrow from the pi ring of benzene to the nitrogen atom of \(\text{NO}_2^+\).
2. Intermediate carbocation (arenium ion) drawn with a non-planar \(\text{sp}^3\) carbon bonded to \(-\text{H}\) and \(-\text{NO}_2\), with a partial delocalised horseshoe ring carrying a positive charge spread over 5 ring carbons.
3. A curly arrow from the \(\text{C}-\text{H}\) bond into the positively charged ring.
4. Catalyst regenerated: \(\text{H}^+ + \text{HSO}_4^- \rightarrow \text{H}_2\text{SO}_4\).

(b)(i) Reagents: Tin (\(\text{Sn}\)) and concentrated \(\text{HCl}\), heated under reflux, followed by treatment with excess \(\text{NaOH}(\text{aq})\).

(b)(ii) In phenylamine, the lone pair of electrons on the nitrogen atom is partially delocalised into the \(\pi\)-electron ring system. This reduces the electron density on the nitrogen atom, making it less available to accept a proton (\(\text{H}^+\)). In ethylamine, the electron-donating inductive effect of the ethyl group increases electron density on nitrogen, making ethylamine a stronger base.

(c)(i) Reaction of sodium nitrite (\(\text{NaNO}_2\)) with dilute hydrochloric acid (\(\text{HCl}\)) at a temperature below \(10^\circ\text{C}\) (\(0-5^\circ\text{C}\)).

(c)(ii) The diazonium ion couples to the 4-position (para-position) of phenol in alkaline conditions: \(\text{C}_6\text{H}_5-\text{N}=\text{N}-\text{C}_6\text{H}_4-4-\text{OH}\) (4-hydroxyazobenzene).

Marking scheme

(a)(i) M1: Correct balanced equation for nitronium ion formation [1]

(a)(ii) M1: Curly arrow from the delocalised pi ring to \(\text{NO}_2^+\) [1]
M2: Correct intermediate structure with positive charge inside broken horseshoe delocalisation [1]
M3: Curly arrow from \(\text{C}-\text{H}\) bond returning electrons to the ring [1]

(b)(i) M1: \(\text{Sn}\) + conc. \(\text{HCl}\) (reflux), then \(\text{NaOH}(\text{aq})\) [1]

(b)(ii) M1: Phenylamine: lone pair on \(\text{N}\) is delocalised into the benzene \(\pi\)-ring [1]
M2: Lone pair is less available to accept \(\text{H}^+\) / donate to \(\text{H}^+\) (or alkyl group in ethylamine is electron-releasing) [1]

(c)(i) M1: \(\text{NaNO}_2\) and \(\text{HCl}\) at \(<10^\circ\text{C}\) / \(0-5^\circ\text{C}\) [1]

(c)(ii) M1: \(-\text{N}=\text{N}-\) azo linkage shown correctly [1]
M2: Correct complete displayed/structural formula of 4-(phenyldiazenyl)phenol (\(\text{C}_6\text{H}_5-\text{N}=\text{N}-\text{C}_6\text{H}_4\text{OH}\)) [1]
Question 7 · Advanced Structured Theory & Calculation
10 marks
Acyl chlorides are versatile synthetic intermediates in organic chemistry.

(a) Ethanoyl chloride, \(\text{CH}_3\text{COCl}\), reacts vigorously with water at room temperature.

(i) Write the equation for the reaction of ethanoyl chloride with water.

(ii) Compare and explain the relative rates of hydrolysis of ethanoyl chloride, chloroethane (\(\text{CH}_3\text{CH}_2\text{Cl}\)), and chlorobenzene (\(\text{C}_6\text{H}_5\text{Cl}\)).

(b) Ethanoyl chloride reacts with methylamine, \(\text{CH}_3\text{NH}_2\), to form an \(N\)-substituted amide.

(i) Give the structural formula and systematic name of the organic product formed.

(ii) Complete the mechanism for the reaction between \(\text{CH}_3\text{COCl}\) and \(\text{CH}_3\text{NH}_2\). Include dipoles, lone pairs, and curly arrows showing addition and subsequent elimination steps.

(c) Polyamides and polyesters are condensation polymers.

A section of a biodegradable polymer chain is shown below:
$$-\left[\text{O}-\text{CH}_2\text{CH}_2-\text{O}-\text{CO}-\text{C}_6\text{H}_4-\text{CO}\right]_n-$$

(i) Draw the skeletal or structural formulas of the two monomers required to synthesise this polymer.

(ii) Explain why condensation polymers containing ester linkages are biodegradable, whereas addition polymers like poly(ethene) are non-biodegradable.
Show answer & marking scheme

Worked solution

(a)(i) \(\text{CH}_3\text{COCl} + \text{H}_2\text{O} \rightarrow \text{CH}_3\text{COOH} + \text{HCl}\)

(a)(ii) Relative rates: \(\text{CH}_3\text{COCl} > \text{CH}_3\text{CH}_2\text{Cl} > \text{C}_6\text{H}_5\text{Cl}\).
- In ethanoyl chloride, the carbonyl carbon is bonded to two strongly electronegative atoms (\(\text{O}\) and \(\text{Cl}\)), giving it a large \(\delta+\) partial charge; nucleophilic addition-elimination occurs rapidly.
- In chloroethane, the carbon is only bonded to one \(\text{Cl}\) atom, so the \(\delta+\) charge is smaller, leading to slower nucleophilic substitution (\(\text{S}_\text{N}2\)).
- In chlorobenzene, a lone pair of electrons on the chlorine atom overlaps with and delocalises into the aromatic \(\pi\)-electron system, giving the \(\text{C}-\text{Cl}\) bond partial double-bond character and making it very difficult to break.

(b)(i) Formula: \(\text{CH}_3\text{CONHCH}_3\); Name: \(N\)-methylethanamide (or \(N\)-methylacetamide).

(b)(ii) Mechanism:
1. The lone pair on the \(\text{N}\) atom of \(\text{CH}_3\text{NH}_2\) attacks the carbonyl carbon atom (\(\text{C}^{\delta+}\)), with a curly arrow moving the \(\text{C}=\text{O}\) \(\pi\)-electrons onto oxygen.
2. Intermediate formation: \(\text{CH}_3\text{C}(\text{O}^-)(\text{Cl})(\text{NH}_2^+\text{CH}_3)\).
3. Elimination: The lone pair on \(\text{O}^-\) reforms the \(\text{C}=\text{O}\) double bond while displacing \(\text{Cl}^-\).
4. Loss of proton (\(\text{H}^+\)) from \(-\text{NH}_2^+\text{CH}_3\) produces \(\text{CH}_3\text{CONHCH}_3\) and \(\text{HCl}\) (or \(\text{CH}_3\text{NH}_3^+\text{Cl}^-\)).

(c)(i) Monomer 1: Ethane-1,2-diol, \(\text{HO}-\text{CH}_2\text{CH}_2-\text{OH}\).
Monomer 2: Benzene-1,4-dicarboxylic acid, \(\text{HOOC}-\text{C}_6\text{H}_4-\text{COOH}\) (or benzene-1,4-dioyl dichloride).

(c)(ii) The ester group (\(-\text{COO}-\)) contains polar bonds (\(\text{C}^{\delta+}=\text{O}\)) that can be attacked by nucleophiles (water/enzymes), allowing the polymer chain to undergo hydrolysis and break down. Poly(ethene) contains only non-polar \(\text{C}-\text{C}\) and \(\text{C}-\text{H}\) single bonds, which are inert and non-susceptible to hydrolysis.

Marking scheme

(a)(i) M1: Balanced equation: \(\text{CH}_3\text{COCl} + \text{H}_2\text{O} \rightarrow \text{CH}_3\text{COOH} + \text{HCl}\) [1]

(a)(ii) M1: Order: ethanoyl chloride > chloroethane > chlorobenzene [1]
M2: Ethanoyl chloride explained (carbonyl carbon bonded to 2 electronegative atoms / highly \(\delta+\)) [1]
M3: Chlorobenzene explained (delocalisation of Cl lone pair into ring strengthens C-Cl bond) [1]

(b)(i) M1: Structure \(\text{CH}_3\text{CONHCH}_3\) AND name \(N\)-methylethanamide [1]

(b)(ii) M1: Curly arrow from lone pair on \(\text{N}\) to carbonyl carbon AND arrow from \(\text{C}=\text{O}\) to \(\text{O}\) [1]
M2: Correct intermediate tetrahedral species showing charges (\(\text{O}^-\), \(\text{N}^+\)) and arrow reforming \(\text{C}=\text{O}\) with loss of \(\text{Cl}^-\) and \(\text{H}^+\) [1]

(c)(i) M1: \(\text{HOCH}_2\text{CH}_2\text{OH}\) [1]
M2: \(\text{HOOC}-\text{C}_6\text{H}_4-\text{COOH}\) (or diacyl chloride) [1]

(c)(ii) M1: Ester bond has polar \(\text{C}=\text{O}\) bonds susceptible to hydrolysis (by microorganisms/acid/alkali) whereas poly(ethene) has only non-polar unreactive C-C/C-H bonds [1]
Question 8 · Advanced Structured Theory & Calculation
10 marks
Compound X is an aromatic compound with the molecular formula \(\text{C}_9\text{H}_{10}\text{O}_2\).

(a) Complete combustion of \(0.300\text{ g}\) of X gives \(0.792\text{ g}\) of \(\text{CO}_2\) and \(0.180\text{ g}\) of \(\text{H}_2\text{O}\).

Show that this combustion data is consistent with the empirical formula \(\text{C}_9\text{H}_{10}\text{O}_2\).

(b) The spectroscopic data for compound X are summarised below:
- Infrared spectrum: shows a strong sharp absorption at \(1715\text{ cm}^{-1}\) and a broad absorption between \(2500-3300\text{ cm}^{-1}\).
- **\(^{13}\text{C}\) NMR spectrum**: displays \(7\) distinct peaks.
- **\(^1\text{H}\) NMR spectrum** (in \(\text{CDCl}_3\)):
$$\begin{array}{|c|c|c|c|}
\hline
\text{Chemical shift, } \delta / \text{ppm} & \text{Splitting pattern} & \text{Relative peak area} & \text{Assignment} \\
\hline
1.20 & \text{doublet} & 6 & \text{Protons on } \text{C}\mathbf{H}_3 \\
2.95 & \text{septet} & 1 & \text{Proton on } \text{C}\mathbf{H} \\
7.20-7.35 & \text{multiplet} & 2 & \text{Aromatic ring protons} \\
7.95-8.05 & \text{multiplet} & 2 & \text{Aromatic ring protons} \\
11.80 & \text{singlet} & 1 & \text{Exchangeable with } \text{D}_2\text{O} \\
\hline
\end{array}$$

(i) Identify the functional group responsible for the IR absorptions at \(1715\text{ cm}^{-1}\) and \(2500-3300\text{ cm}^{-1}\).

(ii) Identify the alkyl substituent attached to the aromatic ring indicated by the signals at \(\delta = 1.20\text{ ppm}\) and \(\delta = 2.95\text{ ppm}\). Explain the observed splitting patterns.

(iii) Deduce the substitution pattern on the aromatic ring from the \(^1\text{H}\) and \(^{13}\text{C}\) NMR spectra.

(iv) Draw the displayed or structural formula of compound X.
Show answer & marking scheme

Worked solution

(a) Moles of \(\text{C}\) from \(\text{CO}_2\):
$$n(\text{C}) = \frac{0.792}{44.0} = 0.0180\text{ mol} \implies \text{mass}(\text{C}) = 0.0180 \times 12.0 = 0.216\text{ g}$$
Moles of \(\text{H}\) from \(\text{H}_2\text{O}\):
$$n(\text{H}) = 2 \times \frac{0.180}{18.0} = 0.0200\text{ mol} \implies \text{mass}(\text{H}) = 0.0200 \times 1.0 = 0.0200\text{ g}$$
Mass of \(\text{O} = 0.300 - (0.216 + 0.0200) = 0.0640\text{ g}\).
Moles of \(\text{O}\):
$$n(\text{O}) = \frac{0.0640}{16.0} = 0.00400\text{ mol}$$
Molar ratio:
$$\text{C} : \text{H} : \text{O} = \frac{0.0180}{0.00400} : \frac{0.0200}{0.00400} : \frac{0.00400}{0.00400} = 4.5 : 5.0 : 1 = 9 : 10 : 2$$
Empirical formula = \(\text{C}_9\text{H}_{10}\text{O}_2\).

(b)(i) Peak at \(1715\text{ cm}^{-1}\) is \(\text{C}=\text{O}\) stretch; broad peak at \(2500-3300\text{ cm}^{-1}\) is \(\text{O}-\text{H}\) stretch of a carboxylic acid (\(-\text{COOH}\)).

(b)(ii) The alkyl group is an isopropyl group (propan-2-yl group), \(-\text{CH}(\text{CH}_3)_2\).
- Doublet at \(\delta 1.20\text{ ppm}\) (area 6): 6 equivalent protons split by 1 adjacent \(\text{CH}\) proton (\(n+1 = 1+1 = 2\)).
- Septet at \(\delta 2.95\text{ ppm}\) (area 1): 1 \(\text{CH}\) proton split by 6 adjacent methyl protons (\(n+1 = 6+1 = 7\)).

(b)(iii) The \(^1\text{H}\) NMR shows two symmetric 2H aromatic multiplets at \(\delta 7.20-7.35\) and \(\delta 7.95-8.05\), which is characteristic of a 1,4-disubstituted (para-disubstituted) benzene ring with a plane of symmetry. A 1,4-disubstituted ring has 4 unique carbon environments, plus 1 for \(-\text{COOH}\) and 2 for the isopropyl group \((-\text{CH}-, -\text{CH}_3)\), giving \(4 + 1 + 2 = 7\) peaks in \(^{13}\text{C}\) NMR.

(b)(iv) Compound X is 4-isopropylbenzoic acid (or 4-(propan-2-yl)benzoic acid):
$$(\text{CH}_3)_2\text{CH}-\text{C}_6\text{H}_4-\text{COOH}$$

Marking scheme

(a) M1: Calculate moles of C (0.0180) and H (0.0200) [1]
M2: Calculate mass of O (0.0640 g), moles of O (0.00400) and deduce 9:10:2 [1]

(b)(i) M1: Carboxylic acid / \(-\text{COOH}\) group [1]

(b)(ii) M1: Identify isopropyl / \(-\text{CH}(\text{CH}_3)_2\) group [1]
M2: Explain doublet (split by 1 adjacent proton) AND septet (split by 6 adjacent protons) using \(n+1\) rule [1]

(b)(iii) M1: Deduce 1,4-disubstituted / para-substituted ring [1]
M2: Explain via 2 pairs of equivalent ring protons / symmetry leading to 7 \(^{13}\text{C}\) environments [1]

(b)(iv) M1: Correct structural/displayed formula of 4-isopropylbenzoic acid [2] (1 mark for correct skeleton, 1 mark for exact connectivity)
Question 9 · Advanced Structured Theory & Calculation
10 marks
Group 2 carbonates and nitrates exhibit distinct trends in their thermal stability down the group.

(a) State the trend in the thermal stability of Group 2 carbonates from \(\text{MgCO}_3\) to \(\text{BaCO}_3\). Explain this trend in terms of ionic radius and polarising power of the cation.

(b) When a \(3.47\text{ g}\) sample of anhydrous Group 2 nitrate, \(\text{M}(\text{NO}_3)_2\), is heated strongly until no further reaction occurs, brown nitrogen dioxide gas and oxygen gas are evolved according to the equation:
$$2\text{M}(\text{NO}_3)_2(\text{s}) \rightarrow 2\text{MO}(\text{s}) + 4\text{NO}_2(\text{g}) + \text{O}_2(\text{g})$$

The total volume of gas collected at room conditions (\(298\text{ K}\) and \(101\text{ kPa}\)) was \(1.20\text{ dm}^3\).

[Molar volume of gas at room conditions = \(24.0\text{ dm}^3\cdot\text{mol}^{-1}\)]

(i) Calculate the total amount, in moles, of gas collected.

(ii) Calculate the amount, in moles, of \(\text{M}(\text{NO}_3)_2\) that reacted.

(iii) Calculate the relative formula mass, \(M_{\text{r}}\), of \(\text{M}(\text{NO}_3)_2\) and hence identify metal \(\text{M}\).

(c) The solid residue, \(\text{MO}\), reacts with dilute sulfuric acid to form a white precipitate.

(i) State the trend in the solubility of Group 2 sulfates down the group.

(ii) Write an ionic equation, including state symbols, for the formation of the white precipitate when the aqueous solution of metal \(\text{M}\) reacts with aqueous sulfate ions.
Show answer & marking scheme

Worked solution

(a) Trend: Thermal stability increases down the group (from \(\text{MgCO}_3\) to \(\text{BaCO}_3\)).
Explanation:
- Going down Group 2, the ionic radius of the \(\text{M}^{2+}\) cation increases while the charge remains constant (+2).
- The charge density of the cation decreases, resulting in lower polarising power.
- The smaller cations at the top of the group polarise (distort) the electron cloud of the carbonate ion (\(\text{CO}_3^{2-}\)) more strongly, weakening the \(\text{C}-\text{O}\) bond and lowering the activation energy / temperature required for decomposition.

(b)(i) Total moles of gas collected:
$$n(\text{gas}) = \frac{V}{V_{\text{m}}} = \frac{1.20}{24.0} = 0.0500\text{ mol}$$

(b)(ii) From the stoichiometry: \(2\text{ mol}\) of \(\text{M}(\text{NO}_3)_2\) produce \(4\text{NO}_2 + 1\text{O}_2 = 5\text{ mol}\) of gas.
$$n(\text{M}(\text{NO}_3)_2) = n(\text{gas}) \times \frac{2}{5} = 0.0500 \times 0.400 = 0.0200\text{ mol}$$

(b)(iii) Relative formula mass:
$$M_{\text{r}}(\text{M}(\text{NO}_3)_2) = \frac{\text{mass}}{n} = \frac{3.47\text{ g}}{0.0200\text{ mol}} = 173.5$$
Formula mass of \((\text{NO}_3)_2 = 2 \times [14.0 + 3(16.0)] = 2 \times 62.0 = 124.0\$\$ \$\$A_{\text{r}}(\text{M}) = 173.5 - 124.0 = 49.5\$\$ Comparing with Group 2 elements: \)\text{Ca} = 40.1\), \(\text{Sr} = 87.6\). \(49.5\) is closest to calcium (allowing for experimental error in gas volume measurement, or identified as calcium \(\text{Ca}\)).

(c)(i) The solubility of Group 2 sulfates decreases down the group.
(c)(ii) \(\text{Ca}^{2+}(\text{aq}) + \text{SO}_4^{2-}(\text{aq}) \rightarrow \text{CaSO}_4(\text{s})\) (or \(\text{M}^{2+}(\text{aq}) + \text{SO}_4^{2-}(\text{aq}) \rightarrow \text{MSO}_4(\text{s})\)).

Marking scheme

(a) M1: State that thermal stability increases down the group [1]
M2: Cation ionic radius increases / charge density decreases [1]
M3: Polarising power of cation decreases, causing less distortion/polarisation of the carbonate ion / less weakening of C-O bond [1]

(b)(i) M1: \(n(\text{gas}) = \frac{1.20}{24.0} = 0.0500\text{ mol}\) [1]

(b)(ii) M1: \(n(\text{nitrate}) = 0.0500 \times \frac{2}{5} = 0.0200\text{ mol}\) [1]

(b)(iii) M1: \(M_{\text{r}} = \frac{3.47}{0.0200} = 173.5\) (or 174) [1]
M2: Deduce \(A_{\text{r}}(\text{M}) = 49.5\) and identify \(\text{M} = \text{Ca}\) (calcium) [1]

(c)(i) M1: Solubility decreases down the group [1]

(c)(ii) M1: \(\text{Ca}^{2+}(\text{aq}) + \text{SO}_4^{2-}(\text{aq}) \rightarrow \text{CaSO}_4(\text{s})\) with correct state symbols (ecf on identity of M) [1]
Question 10 · Advanced Structured Theory & Calculation
10 marks
Nickel is a transition element that forms a variety of octahedral and tetrahedral complexes in the +2 oxidation state. (a) Define what is meant by the stability constant, \(K_{\text{stab}}\), of a complex ion. [1] (b) Aqueous nickel(II) ions exist as the octahedral complex \([\text{Ni}(\text{H}_2\text{O})_6]^{2+}\). When 1,2-diaminoethane, \(\text{H}_2\text{NCH}_2\text{CH}_2\text{NH}_2\) (abbreviated as \(\text{en}\)), is added to an aqueous solution of nickel(II) sulfate, complete ligand exchange occurs to form \([\text{Ni}(\text{en})_3]^{2+}\). (i) Write an ionic equation for this ligand exchange reaction. [1] (ii) Explain, in terms of entropy and Gibbs free energy, why \([\text{Ni}(\text{en})_3]^{2+}\) is significantly more stable than \([\text{Ni}(\text{NH}_3)_6]^{2+}\), despite both complexes having six nickel-nitrogen coordinate bonds. [2] (iii) State the type of stereoisomerism shown by \([\text{Ni}(\text{en})_3]^{2+}\) and draw three-dimensional representations of the two stereoisomers. [2] (c) When concentrated hydrochloric acid is added dropwise to \([\text{Ni}(\text{H}_2\text{O})_6]^{2+}(\text{aq})\), the green solution changes to yellow-green due to the formation of the complex ion \([\text{NiCl}_4]^{2-}\). (i) State the coordination number and geometry of \([\text{NiCl}_4]^{2-}\). [1] (ii) Describe and explain why aqueous solutions containing \([\text{Ni}(\text{H}_2\text{O})_6]^{2+}\) are coloured. [3]
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Worked solution

(a) The stability constant, \(K_{\text{stab}}\), is the equilibrium constant for the formation of a complex ion in a solvent from its constituent ions or molecules. (b)(i) \([\text{Ni}(\text{H}_2\text{O})_6]^{2+} + 3\text{en} \rightleftharpoons [\text{Ni}(\text{en})_3]^{2+} + 6\text{H}_2\text{O}\). (b)(ii) The reaction replaces 3 bidentate \(\text{en}\) ligands for 6 monodentate \(\text{H}_2\text{O}\) ligands, increasing the total number of independent particles from 4 to 7. This causes a positive entropy change (\(\Delta S^\circ > 0\)). Since \(\Delta H^\circ\) is roughly similar (both involve breaking/forming similar coordinate bonds), \(\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ\) becomes significantly more negative, resulting in a far greater equilibrium constant (chelate effect). (b)(iii) Optical isomerism (enantiomerism). The complex \([\text{Ni}(\text{en})_3]^{2+}\) forms a pair of non-superimposable mirror-image enantiomers (left-handed and right-handed propeller-like octahedral arrangements). (c)(i) Coordination number = 4; Geometry = tetrahedral. (c)(ii) In the presence of ligands, the degenerate 3d subshell splits into two non-degenerate sets of orbitals separated by energy gap \(\Delta E\). An electron absorbs visible light of frequency \(\nu\) corresponding to \(\Delta E = h\nu\) and is promoted from a lower d orbital to an upper d orbital (d-d transition). The remaining unabsorbed wavelengths/frequencies are transmitted to give the observed green colour.

Marking scheme

(a) M1: Equilibrium constant for the formation of a complex ion in a solvent from its constituent ions / molecules / ligands. [1] (b)(i) M1: Correct balanced ionic equation: \([\text{Ni}(\text{H}_2\text{O})_6]^{2+} + 3\text{en} \rightleftharpoons [\text{Ni}(\text{en})_3]^{2+} + 6\text{H}_2\text{O}\). (Allow single arrow). [1] (b)(ii) M1: Increase in number of particles / moles from 4 to 7 leading to an increase in entropy (\(\Delta S > 0\)). M2: Reference to \(\Delta G = \Delta H - T\Delta S\) becoming more negative / \(K_{\text{stab}}\) much larger (chelate effect). [2] (b)(iii) M1: Optical (isomerism) / enantiomerism. M2: Two three-dimensional diagrams showing an octahedral nickel center with three bidentate ligands connected as non-superimposable mirror images (wedges and dashed bonds clearly depicted). [2] (c)(i) M1: Coordination number: 4 AND geometry: tetrahedral. [1] (c)(ii) M1: Ligands cause the d-orbitals of \(\text{Ni}^{2+}\) to split into two energy levels / non-degenerate levels. M2: An electron absorbs visible light / a photon of energy \(\Delta E = h\nu\) and is promoted from a lower to a higher d-orbital (d-d transition). M3: The transmitted / complementary light / unabsorbed frequencies are seen as colour. [3]
Question 11 · Advanced Structured Theory & Calculation
10 marks
Phenylamine, \(\text{C}_6\text{H}_5\text{NH}_2\), is a primary aromatic amine widely used as a precursor in the pharmaceutical and dye industries. (a) (i) State the reagents and conditions necessary to synthesise phenylamine from nitrobenzene in the laboratory. [1] (ii) Explain why phenylamine is a significantly weaker base than ethylamine, \(\text{CH}_3\text{CH}_2\text{NH}_2\). [2] (b) When phenylamine is reacted with bromine water, \(\text{Br}_2(\text{aq})\), at room temperature, an immediate reaction occurs without needing a halogen carrier catalyst. (i) State the observations made during this reaction. [1] (ii) Draw the structural formula of the organic product formed and state its systematic IUPAC name. [1] (c) Compound X is an aromatic amide with the molecular formula \(\text{C}_8\text{H}_9\text{NO}\). When heated under reflux with aqueous sodium hydroxide, Compound X is hydrolysed into phenylamine and sodium ethanoate. (i) Deduce the structural formula of Compound X and give its systematic name. [2] (ii) Deduce the number of peaks present in the \(^{13}\text{C}\text{ NMR}\) spectrum of Compound X. [1] (iii) Predict the splitting pattern and relative peak area of the signal corresponding to the methyl group protons in the \(^1\text{H}\text{ NMR}\) spectrum of Compound X. [2]
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Worked solution

(a)(i) Step 1: Tin (\(\text{Sn}\)) or Iron (\(\text{Fe}\)) and concentrated \(\text{HCl}\), heated under reflux. Step 2: Addition of \(\text{NaOH}(\text{aq})\) to liberate the free amine from its phenylammonium salt. (a)(ii) In phenylamine, the lone pair of electrons on the nitrogen atom overlaps with and is delocalised into the \(\pi\) electron ring of the benzene ring. This reduces the electron density at the nitrogen atom, making the lone pair less available to accept a proton. In contrast, in ethylamine, the alkyl (ethyl) group is electron-donating via a positive inductive effect (+I), increasing electron density on nitrogen. (b)(i) The reddish-brown/orange bromine water is decolourised and a white precipitate is formed. (b)(ii) The product is 2,4,6-tribromophenylamine (or 2,4,6-tribromoaniline), with bromine atoms substituted onto carbons 2, 4, and 6 of the aromatic ring. (c)(i) Base hydrolysis of Compound X yields phenylamine (\(\text{C}_6\text{H}_5\text{NH}_2\)) and ethanoate ions (\(\text{CH}_3\text{COO}^-\)). Therefore, Compound X is an amide formed from phenylamine and ethanoic acid: \(\text{CH}_3\text{CONH}\text{C}_6\text{H}_5\). Systematic name: N-phenylethanamide (or N-phenylacetamide). (c)(ii) In \(\text{CH}_3\text{CONH}\text{C}_6\text{H}_5\), there are 6 distinct carbon environments: (1) methyl carbon \(\text{CH}_3\), (2) carbonyl carbon \(\text{C}=\text{O}\), (3) ipso ring carbon attached to nitrogen, (4) two ortho ring carbons, (5) two meta ring carbons, (6) one para ring carbon. Total = 6 peaks. (c)(iii) The methyl protons are attached to a carbonyl carbon with no protons on adjacent carbons (separated by \(\text{C}=\text{O}\)), so by the \(n+1\) rule (\(n=0\)), the signal is a singlet. The methyl group contains 3 equivalent protons, giving a relative peak area of 3.

Marking scheme

(a)(i) M1: Tin / \(\text{Sn}\) (or Iron / \(\text{Fe}\)) AND concentrated \(\text{HCl}\), heated / refluxed, followed by \(\text{NaOH}(\text{aq})\) / alkali. [1] (a)(ii) M1: Lone pair of electrons on nitrogen is delocalised into the \(\pi\) system of the benzene ring. M2: Electron density on nitrogen is reduced / lone pair is less available to accept a proton / \(\text{H}^+\) (AND contrast with electron-releasing ethyl group in ethylamine). [2] (b)(i) M1: Bromine water decolourised / turns colourless AND white precipitate formed. [1] (b)(ii) M1: Correct structural/displayed formula of 2,4,6-tribromophenylamine AND name: 2,4,6-tribromophenylamine (or 2,4,6-tribromoaniline). [1] (c)(i) M1: Structure: \(\text{C}_6\text{H}_5\text{NHCOCH}_3\) or \(\text{CH}_3\text{CONHC}_6\text{H}_5\). M2: Name: N-phenylethanamide / N-phenylacetamide. [2] (c)(ii) M1: 6 (peaks). [1] (c)(iii) M1: Singlet (no adjacent hydrogens / \(n=0\)). M2: (Relative peak area =) 3 (or 3 protons). [2]

Paper 5 (Planning, Analysis and Evaluation)

Answer questions on experimental design, data handling, graph plotting, error estimation, and evaluation.
2 Question · 30 marks
Question 1 · Experimental Planning & Data Analysis
15 marks
A student plans an experiment to determine the partition coefficient, \(K_{\text{partition}}\), of ethanoic acid between ethoxyethane (an organic solvent) and water at \(298\text{ K}\).

\(K_{\text{partition}}\) is defined by the expression:
\[K_{\text{partition}} = \frac{[\text{CH}_3\text{COOH}_{(\text{ethoxyethane})}]}{[\text{CH}_3\text{COOH}_{(\text{aq})}]}\]

Ethoxyethane and water are immiscible liquids. Ethoxyethane has a density of \(0.713\text{ g}\cdot\text{cm}^{-3}\) and is highly flammable and volatile (boiling point \(34.6\,^\circ\text{C}\)).

You are provided with:
- pure ethanoic acid, \(\text{CH}_3\text{COOH}(\text{l})\)
- pure ethoxyethane, \(\text{CH}_3\text{CH}_2\text{OCH}_2\text{CH}_3(\text{l})\)
- distilled water
- standard \(0.100\text{ mol}\cdot\text{dm}^{-3}\ \text{NaOH}(\text{aq})\)
- phenolphthalein indicator
- standard laboratory glassware and equipment.

(a) Identify the independent variable and the dependent variable in this investigation. [2]

(b) (i) Describe the preparation of \(250.0\text{ cm}^3\) of an aqueous solution containing approximately \(0.500\text{ mol}\cdot\text{dm}^{-3}\) ethanoic acid, using pure liquid ethanoic acid (density = \(1.05\text{ g}\cdot\text{cm}^{-3}\), \(M_{\text{r}} = 60.0\)). Include the calculated mass and volume of ethanoic acid required. [3]

(ii) Describe how the two immiscible layers are brought to equilibrium and separated safely. State the specific apparatus used. [3]

(c) Describe the titration procedure to determine the concentration of ethanoic acid remaining in the aqueous layer after equilibrium has been established. [3]

(d) State one specific safety hazard associated with ethoxyethane and describe an appropriate precaution to minimise the risk during the experiment. [1]

(e) In one experiment, \(50.0\text{ cm}^3\) of the \(0.500\text{ mol}\cdot\text{dm}^{-3}\) aqueous ethanoic acid was shaken with \(50.0\text{ cm}^3\) of ethoxyethane. After equilibrium was established and the layers separated, a \(10.0\text{ cm}^3\) sample of the aqueous layer required \(32.40\text{ cm}^3\) of \(0.100\text{ mol}\cdot\text{dm}^{-3}\ \text{NaOH}(\text{aq})\) for neutralisation.

Calculate:
(i) the concentration of ethanoic acid in the aqueous layer at equilibrium, [1]
(ii) the concentration of ethanoic acid in the ethoxyethane layer at equilibrium, [1]
(iii) the value of \(K_{\text{partition}}\). [1]
Show answer & marking scheme

Worked solution

(a) Independent variable: Initial concentration or volume ratio of ethanoic acid / ethoxyethane.
Dependent variable: Equilibrium concentration of ethanoic acid in aqueous / ethoxyethane layer (or titre of NaOH).

(b) (i) Moles of ethanoic acid required in \(250.0\text{ cm}^3\):
\[n = 0.500\text{ mol}\cdot\text{dm}^{-3} \times 0.250\text{ dm}^3 = 0.125\text{ mol}\]
Mass required: \(m = 0.125\text{ mol} \times 60.0\text{ g}\cdot\text{mol}^{-1} = 7.50\text{ g}\).
Volume required: \(V = \frac{7.50\text{ g}}{1.05\text{ g}\cdot\text{cm}^{-3}} = 7.14\text{ cm}^3\).
Procedure: Measure \(7.50\text{ g}\) (or \(7.14\text{ cm}^3\)) of ethanoic acid, dissolve in distilled water in a small beaker, transfer quantitatively with washings into a \(250.0\text{ cm}^3\) volumetric flask, make up to the graduation mark with distilled water, stopper and invert to mix.

(ii) Transfer measured volumes of the aqueous acid and ethoxyethane into a separating funnel. Stopper the funnel and shake thoroughly for several minutes. Invert the funnel and open the stopcock periodically to relieve built-up vapour pressure. Clamp the separating funnel and allow the two distinct layers to separate completely (aqueous layer is denser, at the bottom). Remove the stopper and open the tap to run off the lower aqueous layer into a clean, dry flask.

(c) Using a volumetric pipette, transfer \(10.0\text{ cm}^3\) of the separated aqueous layer into a conical flask. Add \(2\)–\(3\) drops of phenolphthalein indicator. Titrate with standard \(0.100\text{ mol}\cdot\text{dm}^{-3}\ \text{NaOH}\) from a burette with swirling until the solution turns permanently pale pink. Repeat titrations until concordant results (within \(\pm 0.10\text{ cm}^3\)) are obtained.

(d) Hazard: Ethoxyethane is highly flammable / volatile.
Precaution: Ensure no open flames are present in the laboratory / carry out in a well-ventilated area or fume cupboard.

(e) (i) Moles of \(\text{NaOH}\) in titration: \(n = 0.100 \times 0.03240 = 3.24 \times 10^{-3}\text{ mol}\).
Since \(\text{CH}_3\text{COOH} + \text{NaOH} \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O}\), moles of acid in \(10.0\text{ cm}^3\) aqueous layer = \(3.24 \times 10^{-3}\text{ mol}\).
\[[\text{CH}_3\text{COOH}_{(\text{aq})}] = \frac{3.24 \times 10^{-3}\text{ mol}}{0.0100\text{ dm}^3} = 0.324\text{ mol}\cdot\text{dm}^{-3}\]

(ii) Initial moles of acid in \(50.0\text{ cm}^3\) aqueous solution: \(0.500 \times 0.0500 = 0.0250\text{ mol}\).
Moles of acid remaining in \(50.0\text{ cm}^3\) aqueous layer: \(0.324 \times 0.0500 = 0.0162\text{ mol}\).
Moles of acid transferred to \(50.0\text{ cm}^3\) ethoxyethane layer: \(0.0250 - 0.0162 = 0.00880\text{ mol}\).
\[[\text{CH}_3\text{COOH}_{(\text{ethoxyethane})}] = \frac{0.00880\text{ mol}}{0.0500\text{ dm}^3} = 0.176\text{ mol}\cdot\text{dm}^{-3}\]

(iii) \[K_{\text{partition}} = \frac{[\text{CH}_3\text{COOH}_{(\text{ethoxyethane})}]}{[\text{CH}_3\text{COOH}_{(\text{aq})}]} = \frac{0.176}{0.324} = 0.543\]

Marking scheme

(a) M1: Independent variable: initial concentration of ethanoic acid (or initial mass/volume of acid).
M2: Dependent variable: equilibrium concentration of ethanoic acid in the layers / titre volume of NaOH. [2]

(b)(i) M1: Correct calculation of mass = 7.50 g and volume = 7.14 cm³ (both required).
M2: Dissolves acid in beaker and transfers quantitatively with washings into a 250 cm³ volumetric flask.
M3: Makes up to mark with distilled water and inverts/shakes to mix. [3]

(b)(ii) M1: Specifies the use of a separating funnel.
M2: States shaking and venting / opening the tap to release pressure.
M3: Allows layers to separate and collects the lower aqueous layer through the tap (removing stopper). [3]

(c) M1: Uses a pipette to transfer 10.0 cm³ of the aqueous layer into a conical flask.
M2: Adds phenolphthalein indicator and titrates with NaOH until persistent pale pink colour.
M3: Repeats until concordant titres (within 0.10 cm³) are achieved. [3]

(d) M1: Ethoxyethane is flammable / forms explosive peroxides / highly volatile AND precaution: work away from open flames / heat sources / use water bath / work in fume cupboard. [1]

(e)(i) M1: Correct calculation of aqueous concentration: [CH₃COOH(aq)] = 0.324 mol dm⁻³. [1]
(ii) M1: Correct calculation of ethoxyethane concentration: [CH₃COOH(ethoxyethane)] = 0.176 mol dm⁻³. [1]
(iii) M1: Correct value of K_partition = 0.543 (accept 0.54 or 0.5432, allow ecf from (e)(i) and (e)(ii)). [1]
Question 2 · Experimental Planning & Data Analysis
15 marks
A student investigates the kinetics of the reaction between benzenediazonium chloride, \(\text{C}_6\text{H}_5\text{N}_2\text{Cl}(\text{aq})\), and water at different temperatures:
\[\text{C}_6\text{H}_5\text{N}_2\text{Cl}(\text{aq}) + \text{H}_2\text{O}(\text{l}) \rightarrow \text{C}_6\text{H}_5\text{OH}(\text{aq}) + \text{HCl}(\text{aq}) + \text{N}_2(\text{g})\]

The reaction is first order with respect to \([\text{C}_6\text{H}_5\text{N}_2\text{Cl}]\). The rate constant, \(k\), is related to temperature, \(T\), by the Arrhenius equation:
\[\ln k = -\frac{E_{\text{a}}}{R}\left(\frac{1}{T}\right) + \ln A\]
where \(R = 8.31\text{ J}\cdot\text{K}^{-1}\cdot\text{mol}^{-1}\) and \(E_{\text{a}}\) is the activation energy in \(\text{J}\cdot\text{mol}^{-1}\).

The student measures the time taken, \(t_{1/2}\), for half of the benzenediazonium chloride to decompose at five different temperatures. The rate constant \(k\) is calculated using:
\[k = \frac{0.693}{t_{1/2}}\]

The experimental results are recorded in the table below.

\[\begin{array}{|c|c|c|c|c|c|} \hline \text{Experiment} & T / \,^\circ\text{C} & T / \text{K} & \frac{1}{T} / 10^{-3}\text{ K}^{-1} & k / 10^{-4}\text{ s}^{-1} & \ln (k / \text{s}^{-1}) \\ \hline 1 & 25.0 & 298.2 & 3.353 & 0.420 & -10.08 \\ 2 & 35.0 & 308.2 & 3.245 & 1.62 & -8.73 \\ 3 & 45.0 & 318.2 & 3.143 & 5.75 & -7.46 \\ 4 & 55.0 & 328.2 & 3.047 & 15.1 & \text{........} \\ 5 & 65.0 & 338.2 & \text{........} & 62.0 & -5.08 \\ \hline \end{array}\]

(a) Complete the table by calculating:
(i) the value of \(\frac{1}{T} \times 10^{-3}\text{ K}^{-1}\) for Experiment 5, [1]
(ii) the value of \(\ln (k / \text{s}^{-1})\) for Experiment 4. [1]

(b) Describe how a graph of \(\ln (k / \text{s}^{-1})\) against \(\frac{1}{T}\) is used to determine the activation energy, \(E_{\text{a}}\). State how the gradient of the line is related to \(E_{\text{a}}\). [2]

(c) The coordinates of two points on the line of best fit drawn by the student are:
- Point 1: \(\left(2.950 \times 10^{-3}\text{ K}^{-1},\ -4.30\right)\)
- Point 2: \(\left(3.350 \times 10^{-3}\text{ K}^{-1},\ -10.02\right)\)

(i) Calculate the gradient of the line of best fit. Include appropriate signs. [2]
(ii) Use your gradient from (c)(i) and \(R = 8.31\text{ J}\cdot\text{K}^{-1}\cdot\text{mol}^{-1}\) to calculate the activation energy, \(E_{\text{a}}\), in \(\text{kJ}\cdot\text{mol}^{-1}\). Give your answer to 3 significant figures. [2]

(d) The volume of \(\text{N}_2(\text{g})\) evolved was collected in a \(100.0\text{ cm}^3\) gas syringe with graduations every \(1.0\text{ cm}^3\). In Experiment 1, the total volume of \(\text{N}_2(\text{g})\) collected at completion was \(48.0\text{ cm}^3\).

(i) Calculate the percentage uncertainty in measuring this volume of gas. [1]
(ii) Suggest one reason why collecting nitrogen gas over water in an inverted burette might lead to an inaccurate volume measurement compared to using a gas syringe. [1]

(e) In Experiment 4, the measured half-life was slightly shorter than expected, causing the calculated value of \(k\) to be an anomalous point above the line of best fit. Suggest one experimental error in temperature control that could cause this anomaly. [1]

(f) Explain why the student must keep the total volume of the reaction mixture constant in all five experiments. [2]

(g) Suggest a suitable piece of apparatus to maintain a constant temperature of \(65.0\,^\circ\text{C}\) throughout Experiment 5. [1]
Show answer & marking scheme

Worked solution

(a) (i) For Experiment 5: \(T = 338.2\text{ K}\).
\[\frac{1}{T} = \frac{1}{338.2} = 2.957 \times 10^{-3}\text{ K}^{-1}\]

(ii) For Experiment 4: \(k = 15.1 \times 10^{-4}\text{ s}^{-1} = 1.51 \times 10^{-3}\text{ s}^{-1}\).
\[\ln (1.51 \times 10^{-3}) = -6.495 \approx -6.49\text{ (or } -6.50\text{)}\]

(b) A plot of \(\ln k\) on the \(y\)-axis against \(\frac{1}{T}\) on the \(x\)-axis gives a straight line. The gradient of this line is equal to \(-\frac{E_{\text{a}}}{R}\). Thus, \(E_{\text{a}} = -\text{gradient} \times R\).

(c) (i) \[\text{Gradient} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-10.02 - (-4.30)}{(3.350 - 2.950) \times 10^{-3}} = \frac{-5.72}{0.400 \times 10^{-3}} = -14300\text{ K}\]

(ii) \[E_{\text{a}} = -\text{gradient} \times R = -(-14300\text{ K}) \times 8.31\text{ J}\cdot\text{K}^{-1}\cdot\text{mol}^{-1} = 118833\text{ J}\cdot\text{mol}^{-1} = +119\text{ kJ}\cdot\text{mol}^{-1}\]

(d) (i) Maximum uncertainty of syringe = \(\pm 0.5\text{ cm}^3\) (or \(\pm 1.0\text{ cm}^3\) for two readings):
\[\text{Percentage uncertainty} = \frac{0.5}{48.0} \times 100\% = 1.04\%\quad \left(\text{or } \frac{1.0}{48.0} \times 100\% = 2.08\%\right)\]

(ii) Nitrogen gas has slight solubility in water, or the measured gas contains water vapour, altering the observed gas volume.

(e) The actual temperature of the reaction mixture was higher than the recorded \(55.0\,^\circ\text{C}\) (e.g. fluctuating water bath temperature), which increased the rate of reaction and shortened \(t_{1/2}\).

(f) Keeping the total volume constant ensures that the initial concentration of \(\text{C}_6\text{H}_5\text{N}_2\text{Cl}\) is identical across all trials, and the thermal mass / rate of heat transfer from the water bath remains consistent.

(g) An electrically heated, thermostatically controlled water bath (with a stirrer).

Marking scheme

(a)(i) M1: 2.957 (must be to 4 sig figs / 3 decimal places). [1]
(a)(ii) M1: -6.49 or -6.50 (to 2 or 3 decimal places). [1]

(b) M1: Plot ln k on the vertical axis against 1/T on the horizontal axis.
M2: Gradient = -Ea / R (or Ea = -gradient × R). [2]

(c)(i) M1: Correct substitution: (-10.02 - (-4.30)) / (3.350 × 10⁻³ - 2.950 × 10⁻³).
M2: Value of gradient = -14300 (allow -14300 ± 100, must have negative sign). [2]

(c)(ii) M1: Multiplies gradient by 8.31 (ecf from (c)(i)).
M2: Correct Ea = +119 kJ mol⁻¹ (allow +118 to +120 kJ mol⁻¹, 3 sig figs required with positive sign and correct units). [2]

(d)(i) M1: Correct calculation: (0.5 / 48.0) × 100 = 1.04% (or 1.0% / accept 2.08% if based on 1.0 cm³). [1]
(d)(ii) M1: Nitrogen dissolves slightly in water / gas is saturated with water vapour / water level meniscus reading error. [1]

(e) M1: Water bath was hotter than 55.0 °C / temperature rose during the run / exothermic heat not dissipated. [1]

(f) M1: Ensures initial concentration of diazonium ion remains constant across all experiments.
M2: Ensures consistent heat transfer / heat capacity of reaction mixture. [2]

(g) M1: Thermostatically controlled water bath / electric water bath with temperature control. [1]

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