2023 Cambridge IAS-Level Chemistry (9701) Practice Paper with Answers

1. Calculate the mass of carbon in the compound from the mass of \(\text{CO}_2\):
\(n(\text{CO}_2) = \frac{0.300\text{ g}}{44.0\text{ g mol}^{-1}} = 0.006818\text{ mol}\)
\(\text{Mass of C} = 0.006818\text{ mol} \times 12.0\text{ g mol}^{-1} = 0.08182\text{ g}\)

2. Calculate the mass of hydrogen in the compound from the mass of \(\text{H}_2\text{O}\):
\(n(\text{H}_2\text{O}) = \frac{0.123\text{ g}}{18.0\text{ g mol}^{-1}} = 0.006833\text{ mol}\)
\(n(\text{H atoms}) = 2 \times 0.006833\text{ mol} = 0.01367\text{ mol}\)
\(\text{Mass of H} = 0.01367\text{ mol} \times 1.0\text{ g mol}^{-1} = 0.01367\text{ g}\)

3. Calculate the mass of oxygen by subtraction:
\(\text{Mass of O} = 0.150\text{ g} - 0.08182\text{ g} - 0.01367\text{ g} = 0.05451\text{ g}\)
\(n(\text{O atoms}) = \frac{0.05451\text{ g}}{16.0\text{ g mol}^{-1}} = 0.003407\text{ mol}\)

4. Determine the simplest molar ratio:
\(\text{C} = \frac{0.006818}{0.003407} \approx 2\)
\(\text{H} = \frac{0.01367}{0.003407} \approx 4\)
\(\text{O} = \frac{0.003407}{0.003407} = 1\)

The empirical formula is \(\text{C}_2\text{H}_4\text{O}\).

Award 1 mark for the correct choice B.
- Method: Correctly calculating the moles of carbon, hydrogen, and oxygen.
- Accuracy: Correctly identifying the ratio 2:4:1, leading to option B.

Option C is correct because the thermal stability of Group 2 carbonates increases down the group. This is due to the increasing ionic radius of the Group 2 cation, which reduces its charge density and its ability to polarise the carbonate anion, making thermal decomposition more difficult. Therefore, magnesium carbonate decomposes at a lower temperature than calcium carbonate.
- Option A is incorrect because sulfate solubility decreases down Group 2; hence, barium sulfate is much less soluble than magnesium sulfate.
- Option B is incorrect because hydroxide solubility increases down Group 2; hence, calcium hydroxide is less soluble than strontium hydroxide.
- Option D is incorrect because first ionisation energy decreases down Group 2 due to increased atomic shielding and distance from the nucleus.

Award 1 mark for the correct choice C.
- Method: Correctly identifying the periodic and group trends for Group 2 thermal stability, ionisation energy, and solubilities of sulfates/hydroxides.

Let us analyze the reaction of each alkene with \(\text{HBr}\):
- **but-2-ene** (Option A) is a symmetrical alkene. Reaction with \(\text{HBr}\) yields only one product (2-bromobutane) and does not produce a distinct major and minor product.
- **but-1-ene** (Option B) is unsymmetrical. According to Markovnikov's rule, the major product is formed via the more stable secondary carbocation, giving 2-bromobutane, \(\text{CH}_3\text{CH}_2\text{CH(Br)CH}_3\). This carbon-2 has four different groups attached (\(-\text{H}\), \(-\text{CH}_3\), \(-\text{CH}_2\text{CH}_3\), and \(-\text{Br}\)) and is therefore chiral. The minor product is 1-bromobutane, which is achiral. This fits the description.
- **2-methylbut-2-ene** (Option C) reacts to give 2-bromo-2-methylbutane as the major product, which is achiral since the central carbon has two identical methyl groups.
- **propene** (Option D) reacts to give 2-bromopropane as the major product, which is achiral.

Award 1 mark for the correct choice B.
- Method: Apply Markovnikov's rule to determine the major product and identify which product has a chiral center.

The reaction in a catalytic converter is: \(2\text{CO}(g) + 2\text{NO}(g) \rightarrow 2\text{CO}_2(g) + \text{N}_2(g)\).
- Option A is incorrect: although nitrogen monoxide acts as an oxidizing agent, the nitrogen in \(\text{NO}\) is reduced (oxidation state changes from +2 to 0).
- Option B is incorrect: carbon monoxide is oxidized, not reduced, to carbon dioxide.
- Option C is correct as it shows the balanced stoichiometric equation for the process.
- Option D is incorrect: nitrogen monoxide is formed in car engines by the direct reaction between atmospheric nitrogen and oxygen at high temperatures (\(\text{N}_2 + \text{O}_2 \rightarrow 2\text{NO}\)).

Award 1 mark for the correct choice C.
- Method: Recall the reaction occurring in a catalytic converter and analyze the redox properties and origins of the gases.

1. The molecular formula \(\text{C}_5\text{H}_{12}\text{O}\) belongs to a saturated aliphatic alcohol.
2. Heating under reflux with acidified \(\text{K}_2\text{Cr}_2\text{O}_7\) oxidises primary and secondary alcohols.
3. The product reacts with 2,4-dinitrophenylhydrazine (2,4-DNPH), which indicates it contains a carbonyl group (it is either an aldehyde or a ketone).
4. The product does not react with Fehling's solution, which shows that it is a ketone, not an aldehyde.
5. Since ketones are formed from the oxidation of secondary alcohols, \(Z\) must be a secondary alcohol.
6. Analyzing the options:
- 2-methylbutan-2-ol (A) is a tertiary alcohol (no oxidation).
- 3-methylbutan-2-ol (B) is a secondary alcohol (oxidised to 3-methylbutan-2-one, a ketone).
- 2,2-dimethylpropan-1-ol (C) and pentan-1-ol (D) are primary alcohols (oxidised to carboxylic acids, which do not react with 2,4-DNPH).

Award 1 mark for the correct choice B.
- Method: Deduce the functional group of the product (ketone) from its reactions and determine the class of the starting alcohol (secondary).

The successive ionisation energies show a very large jump between the 3rd and 4th ionisation energies (from 2745 to 11578 \(\text{kJ mol}^{-1}\)). This indicates that the fourth electron is being removed from an inner electronic shell, meaning element \(Q\) has three valence electrons. Since \(Q\) is in Period 3, it is aluminium (\(\text{Al}\)).
- Option A is incorrect because aluminium oxide is amphoteric, not purely basic.
- Option B is incorrect because aluminium does not react with cold water due to its protective oxide layer.
- Option C is incorrect because aluminium oxide is a giant ionic structure, not giant covalent (which describes \(\text{SiO}_2\)).
- Option D is correct because aluminium chloride (\(\text{Al}_2\text{Cl}_6\)) hydrolyses in water to form a highly acidic solution containing \([\text{Al}(\text{H}_2\text{O})_6]^{3+}\) which partially dissociates to release \(\text{H}^+\) ions, resulting in a pH of approximately 3.

Award 1 mark for the correct choice D.
- Method: Determine the group of element \(Q\) from the successive ionisation energy values, identify it as Aluminium, and evaluate its physical and chemical properties.

To calculate the enthalpy change of the reaction:
\(\Delta H = \sum(\text{bond energies of bonds broken}) - \sum(\text{bond energies of bonds formed})\)

Bonds broken:
- 1 \(\text{C}-\text{H}\) bond = \(413\text{ kJ mol}^{-1}\)
- 1 \(\text{Cl}-\text{Cl}\) bond = \(242\text{ kJ mol}^{-1}\)
Total energy absorbed = \(413 + 242 = 655\text{ kJ mol}^{-1}\)

Bonds formed:
- 1 \(\text{C}-\text{Cl}\) bond = \(346\text{ kJ mol}^{-1}\)
- 1 \(\text{H}-\text{Cl}\) bond = \(432\text{ kJ mol}^{-1}\)
Total energy released = \(346 + 432 = 778\text{ kJ mol}^{-1}\)

Enthalpy change:
\(\Delta H = 655 - 778 = -123\text{ kJ mol}^{-1}\)

Award 1 mark for the correct choice A.
- Method: Sum the bond energies of bonds broken and subtract the bond energies of bonds formed.

When concentrated sulfuric acid is added to solid potassium iodide (\(\text{KI}\)):
- Iodide ions (\(\text{I}^-\)) are very strong reducing agents and easily reduce concentrated \(\text{H}_2\text{SO}_4\).
- The iodide ions are oxidized to iodine, \(\text{I}_2\), which is seen as a purple vapor.
- The sulfuric acid is reduced to several products, including sulfur dioxide (\(\text{SO}_2\)).
- Sulfur dioxide is a reducing agent that reduces the orange dichromate(VI) ions (\(\text{Cr}_2\text{O}_7^{2-}\)) to green chromium(III) ions (\(\text{Cr}^{3+}\)), causing the test paper to turn from orange to green.
- Chloride and fluoride ions are not strong enough reducing agents to reduce sulfuric acid; they only undergo acid-base reactions to form misty white fumes of \(\text{HCl}\) and \(\text{HF}\).
- Bromide ions reduce sulfuric acid to form red-brown bromine vapor (\(\text{Br}_2\)), not a purple vapor.

Award 1 mark for the correct choice D.
- Method: Match the observations (purple vapor and reducing gas \(\text{SO}_2\)) to the reactions of Group 17 halide ions with concentrated sulfuric acid.

First, find the moles of carbon dioxide gas evolved: \(n(\text{CO}_2) = \frac{120}{24000} = 0.00500\text{ mol}\). The reaction equation is: \(\text{MCO}_3 + 2\text{HCl} \rightarrow \text{MCl}_2 + \text{H}_2\text{O} + \text{CO}_2\). The mole ratio of \(\text{MCO}_3\) to \(\text{CO}_2\) is 1:1, so \(n(\text{MCO}_3) = 0.00500\text{ mol}\). Next, calculate the molar mass of \(\text{MCO}_3\): \(M_r(\text{MCO}_3) = \frac{0.4215\text{ g}}{0.00500\text{ mol}} = 84.3\text{ g mol}^{-1}\). Subtract the formula mass of the carbonate ion, \(\text{CO}_3^{2-}\) (which is \(12.0 + 3 \times 16.0 = 60.0\text{ g mol}^{-1}\)), to find the relative atomic mass of metal M: \(A_r(\text{M}) = 84.3 - 60.0 = 24.3\). This corresponds to magnesium.

1 mark for the correct option A. Method: Calculate gas moles (0.00500 mol), use 1:1 ratio to find molar mass of carbonate (84.3 g/mol), and subtract 60.0 g/mol to get 24.3.

The thermal stability of Group 2 nitrates increases down the group because the cationic radius increases (from \(\text{Mg}^{2+}\) to \(\text{Ba}^{2+}\)), reducing the polarising power of the metal cation and making the nitrate ion more stable to heat. Therefore, magnesium nitrate decomposes at a lower temperature than barium nitrate. The decomposition of both nitrates follows the stoichiometry: \(2\text{M}(\text{NO}_3)_2 \rightarrow 2\text{MO} + 4\text{NO}_2 + \text{O}_2\). Since equal moles of both nitrates are completely decomposed, they will produce the exact same volume of gas.

1 mark for the correct option C. Barium nitrate requires a higher temperature for decomposition due to lower polarising power of the larger barium ion. Equal moles decompose to produce identical molar amounts of gas products.

Hot, concentrated, acidified \(\text{KMnO}_4\) cleaves the alkene double bond entirely. A double bond carbon bonded to two alkyl groups is oxidised to a ketone. Since the only organic product is propanone (which has three carbons), the starting alkene must be a symmetrical six-carbon alkene with two identical \((\text{CH}_3)_2\text{C}=\) units. Thus, the alkene is 2,3-dimethylbut-2-ene, \((\text{CH}_3)_2\text{C}=\text{C}(\text{CH}_3)_2\).

1 mark for the correct option B. Cleaving 2,3-dimethylbut-2-ene results in two identical propanone molecules. Other options yield aldehydes/carboxylic acids or different mixtures.

Catalytic converters facilitate the reaction between toxic carbon monoxide and nitrogen monoxide to produce harmless nitrogen and carbon dioxide gas: \(2\text{CO} + 2\text{NO} \rightarrow 2\text{CO}_2 + \text{N}_2\). Option A represents the formation of pollutant NO in the engine cylinder. Option C represents the oxidation of sulfur dioxide, which is not the primary mechanism of a catalytic converter, and option D is the reaction of carbon dioxide with water to form carbonic acid.

1 mark for the correct option B. Catalytic converters catalyze the redox reaction between carbon monoxide and nitrogen oxides to form nitrogen and carbon dioxide.

The equation for the formation of pentane is: \(5\text{C(s)} + 6\text{H}_2(\text{g}) \rightarrow \text{C}_5\text{H}_{12}(\text{l})\). According to Hess's Law, \(\Delta H^\theta_f = \sum \Delta H^\theta_c(\text{reactants}) - \sum \Delta H^\theta_c(\text{products})\). Therefore, \(\Delta H^\theta_f = [5 \times (-393.5) + 6 \times (-285.8)] - [-3509.0] = [-1967.5 - 1714.8] + 3509.0 = -3682.3 + 3509.0 = -173.3\text{ kJ mol}^{-1}\).

1 mark for the correct option A. Method: \(5 \times \Delta H^\theta_c(\text{C}) + 6 \times \Delta H^\theta_c(\text{H}_2) - \Delta H^\theta_c(\text{pentane})\). Correct numerical substitute yields -173.3.

Let us analyze the carbon atoms in \(\text{CH}_3\text{CH(OH)CH(CH}_3)\text{CH}=\text{CH}_2\). The carbon with the \(-\text{OH}\) group is bonded to four different groups: \(-\text{H}\), \(-\text{OH}\), \(-\text{CH}_3\), and \(-\text{CH(CH}_3)\text{CH}=\text{CH}_2\) (Chiral Center 1). The carbon with the methyl branch is bonded to four different groups: \(-\text{H}\), \(-\text{CH}_3\), \(-\text{CH}=\text{CH}_2\), and \(-\text{CH(OH)CH}_3\) (Chiral Center 2). There are no other chiral carbon atoms. The alkene group \(-\text{CH}=\text{CH}_2\) contains two hydrogen atoms on the terminal carbon, so it does not exhibit cis-trans (geometrical) isomerism. The total number of stereoisomers is therefore \(2^n = 2^2 = 4\).

1 mark for the correct option B. Identify exactly 2 chiral carbons and determine that only optical isomerism is possible with no cis-trans isomerism, giving 2^2 = 4 stereoisomers.

The change in colour of the acidified potassium dichromate(VI) from orange to green indicates that Z is oxidized, so it must be a primary or secondary alcohol (ruling out tertiary alcohols like 2-methylpropan-2-ol). Since the oxidation product does not react with Fehling's solution, the product must be a ketone (as aldehydes from primary alcohols would react with Fehling's solution). A ketone is produced by the oxidation of a secondary alcohol. Among the options, butan-2-ol is the only secondary alcohol. It oxidizes to butanone, which is a ketone and does not react with Fehling's solution.

1 mark for the correct option B. Identify Z as a secondary alcohol since it is oxidized to a ketone, which yields a negative Fehling's test. Butan-2-ol is the only secondary alcohol among the options.

The large jump in successive ionisation energies occurs between the 3rd and 4th ionisation energies (from 2745 to 11578 \(\text{kJ mol}^{-1}\)). This indicates that the 4th electron is removed from an inner core shell, meaning the element has 3 valence electrons and belongs to Group 13. In Period 3, the Group 13 element is aluminium (Al). The oxide of Al is \(\text{Al}_2\text{O}_3\), which corresponds to the formula \(\text{T}_2\text{O}_3\).

1 mark for the correct option C. Deduce that T is a Group 13 element due to the large jump between IE3 and IE4, then identify the stable oxide formula as T2O3.

1. Find the number of moles of carbon atoms produced:
\(n(\text{CO}_2) = \frac{1.76\text{ g}}{44.0\text{ g mol}^{-1}} = 0.040\text{ mol}\)
Since each \(\text{CO}_2\) molecule contains one carbon atom, there are \(0.040\text{ mol}\) of carbon atoms in \(0.010\text{ mol}\ of \)X\).
Therefore, the number of carbon atoms per molecule of \(X = \frac{0.040}{0.010} = 4\).

2. Find the number of moles of hydrogen atoms produced:
\(n(\text{H}_2\text{O}) = \frac{0.90\text{ g}}{18.0\text{ g mol}^{-1}} = 0.050\text{ mol}\)
Since each \(\text{H}_2\text{O}\) molecule contains two hydrogen atoms, there are \(2 \times 0.050 = 0.10\text{ mol}\) of hydrogen atoms in \(0.010\text{ mol}\) of \(X\).
Therefore, the number of hydrogen atoms per molecule of \(X = \frac{0.10}{0.010} = 10\).

3. The molecular formula of \(X\) is \(\text{C}_4\text{H}_{10}\).

Award 1 mark for selecting the correct option (C). No marks are awarded for incorrect options.

1. Thermal stability of Group 2 nitrates increases down the group because the ionic radius of the cation increases, which decreases its charge density and its polarizing power on the nitrate anion.
2. Since \(Y(\text{NO}_3)_2\) decomposes at a lower temperature than \(Z(\text{NO}_3)_2\), \(Y\) is positioned above \(Z\) in Group 2.
3. The solubilities of Group 2 sulfates decrease down the group. Since \(Y\) is higher in the group than \(Z\), the sulfate of \(Y\) is more soluble in water than the sulfate of \(Z\).
4. Therefore, statement B is correct.
- Statement A is incorrect because ionic radius increases down the group, so \(Y^{2+}\) has a smaller ionic radius than \(Z^{2+}\).
- Statement C is incorrect because the solubility of Group 2 hydroxides increases down the group, so the hydroxide of \(Y\) is less soluble than that of \(Z\).
- Statement D is incorrect because the basicity of oxides increases down the group, so the oxide of \(Y\) is less basic than that of \(Z\).

Award 1 mark for selecting the correct option (B). No marks are awarded for incorrect options.

1. Hot, concentrated, acidified \(\text{KMnO}_4\) cleaves the carbon-carbon double bond (\(\text{C=C}\)) of alkenes.
2. The reaction yields a single organic product, which means the alkene must be symmetrical.
3. The single product reacts with 2,4-dinitrophenylhydrazine to form an orange precipitate, indicating it is a carbonyl compound (aldehyde or ketone).
4. The product does not react with Tollens' reagent, indicating it is a ketone, not an aldehyde.
5. Let's analyze the oxidation products for each option:
- **2,3-dimethylbut-2-ene** (\(\text{(CH}_3)_2\text{C=C(CH}_3)_2\)) is symmetrical and cleaves to form two molecules of propanone (\(\text{CH}_3\text{COCH}_3\)). Propanone is a ketone, which reacts with 2,4-DNPH but not Tollens' reagent. This fits all observations.
- **hex-3-ene** (\(\text{CH}_3\text{CH}_2\text{CH=CHCH}_2\text{CH}_3\)) is symmetrical but cleaves to form propanal (an aldehyde), which would react with Tollens' reagent.
- **4-methylpent-2-ene** is unsymmetrical and cleaves to form two different products (ethanal and 2-methylpropanal).
- **2-methylpent-2-ene** is unsymmetrical and cleaves to form propanone and propanal.

Award 1 mark for selecting the correct option (A). No marks are awarded for incorrect options.

1. In this atmospheric reaction, both the reactants (\(\text{SO}_2\)) and the catalyst (\(\text{NO}_2\)) are in the gas phase. Therefore, nitrogen dioxide acts as a homogeneous catalyst, making option A incorrect.
2. The first step of the catalytic cycle is:
\(\text{SO}_2\text{(g)} + \text{NO}_2\text{(g)} \rightarrow \text{SO}_3\text{(g)} + \text{NO}\text{(g)}\)
In this reaction, \(\text{NO}_2\) is reduced to \(\text{NO}\) because the oxidation state of nitrogen decreases from +4 (in \(\text{NO}_2\)) to +2 (in \(\text{NO}\)). This makes option B correct and option C incorrect.
3. Sulfur dioxide is oxidized (not reduced) to sulfur trioxide because the oxidation state of sulfur increases from +4 to +6. This makes option D incorrect.

Award 1 mark for selecting the correct option (B). No marks are awarded for incorrect options.

Let's determine the oxidation product of each diol under reflux with excess acidified potassium dichromate(VI). Primary alcohol groups (\(\text{-CH}_2\text{OH}\)) are oxidized to carboxylic acids (\(\text{-COOH}\)), secondary alcohol groups (\(\text{-CH(OH)-}\)) are oxidized to ketones (\(\text{-CO-}\)), and tertiary alcohol groups are not oxidized.

- **A: butane-1,2-diol** (\(\text{HOCH}_2\text{-CH(OH)-CH}_2\text{CH}_3\)):
Contains one primary and one secondary alcohol group. Oxidation gives \(\text{HOOC-CO-CH}_2\text{CH}_3\) (\(\text{C}_4\text{H}_6\text{O}_3\)).

- **B: butane-1,3-diol** (\(\text{HOCH}_2\text{-CH}_2\text{-CH(OH)-CH}_3\)):
Contains one primary and one secondary alcohol group. Oxidation gives \(\text{HOOC-CH}_2\text{-CO-CH}_3\) (\(\text{C}_4\text{H}_6\text{O}_3\)).

- **C: butane-1,4-diol** (\(\text{HOCH}_2\text{-CH}_2\text{-CH}_2\text{-CH}_2\text{OH}\)):
Contains two primary alcohol groups. Both are oxidized to carboxylic acid groups, forming the dicarboxylic acid \(\text{HOOC-CH}_2\text{-CH}_2\text{-COOH}\) (butanedioic acid). The molecular formula of this product is \(\text{C}_4\text{H}_6\text{O}_4\). This matches the question.

- **D: butane-2,3-diol** (\(\text{CH}_3\text{-CH(OH)-CH(OH)-CH}_3\)):
Contains two secondary alcohol groups. Oxidation gives the diketone \(\text{CH}_3\text{-CO-CO-CH}_3\) (\(\text{C}_4\text{H}_6\text{O}_2\)).

Thus, compound \(V\) must be butane-1,4-diol.

Award 1 mark for selecting the correct option (C). No marks are awarded for incorrect options.

1. Write the equation for the formation of liquid ethanol:
\(2\text{C(s)} + 3\text{H}_2\text{(g)} + \frac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{CH}_3\text{CH}_2\text{OH(l)}\)

2. According to Hess's Law, when using enthalpy change of combustion data:
\(\Delta H_f^\ominus = \sum \Delta H_c^\ominus(\text{reactants}) - \sum \Delta H_c^\ominus(\text{products})\)

3. Substitute the values into the equation:
\(\Delta H_f^\ominus = [2 \times \Delta H_c^\ominus(\text{C(s)}) + 3 \times \Delta H_c^\ominus(\text{H}_2\text{(g)})] - [\Delta H_c^\ominus(\text{CH}_3\text{CH}_2\text{OH(l)})]\)
\(\Delta H_f^\ominus = [2(-393.5) + 3(-285.8)] - [-1367.3]\)
\(\Delta H_f^\ominus = [-787.0 - 857.4] + 1367.3\)
\(\Delta H_f^\ominus = -1644.4 + 1367.3 = -277.1\text{ kJ mol}^{-1}\).

Therefore, option A is correct.

Award 1 mark for selecting the correct option (A). No marks are awarded for incorrect options.

1. Analyze the successive ionisation energies to find the number of valence electrons:
- There is a relatively small increase between the 1st, 2nd, and 3rd ionisation energies.
- There is a very large jump between the 3rd and 4th ionisation energies (\(2745 \rightarrow 11578\text{ kJ mol}^{-1}\)). This indicates that the 4th electron is removed from a shell closer to the nucleus.
- Therefore, element \(D\) has 3 valence electrons and is in Group 13.

2. Since \(D\) is in Period 3 and Group 13, it is Aluminium (\(\text{Al}\)).
- Aluminium forms an oxide with the formula \(\text{Al}_2\text{O}_3\), which corresponds to \(D_2\text{O}_3\). Hence, B is correct.
- \(\text{Al}_2\text{O}_3\) has a giant ionic structure, not a giant molecular structure, so A is incorrect.
- \(D\) is in Group 13, not Group 14, so C is incorrect.
- Aluminium does not react rapidly with cold water because of its protective oxide layer, so D is incorrect.

Award 1 mark for selecting the correct option (B). No marks are awarded for incorrect options.

1. Let's analyze the reaction of solid sodium halides with concentrated sulfuric acid:
- **Fluoride and Chloride** are weak reducing agents. They undergo acid-base reactions to form \(\text{HF}\) and \(\text{HCl}\) respectively, but do not reduce sulfuric acid to \(\text{SO}_2\) or form halogens. Thus, no red/brown vapor or gas turning dichromate green is observed.
- **Bromide** is a stronger reducing agent. It is oxidized to bromine gas, \(\text{Br}_2\), which is a dark red-brown vapour. Concurrently, sulfuric acid is reduced to sulfur dioxide, \(\text{SO}_2\).
- **Sulfur dioxide** is a reducing agent that reduces the orange dichromate(VI) ion (\(\text{Cr}_2\text{O}_7^{2-}\)) to the green chromium(III) ion (\(\text{Cr}^{3+}\)), turning the paper green.
- **Iodide** is an even stronger reducing agent and is oxidized to purple iodine vapour (\(\text{I}_2\)), reducing sulfuric acid mainly to hydrogen sulfide (\(\text{H}_2\text{S}\), which has a rotten-egg smell) and yellow solid sulfur.

Therefore, the halide present is bromide (Option C).

Award 1 mark for selecting the correct option (C). No marks are awarded for incorrect options.

First, calculate the mass of carbon in the \(\text{CO}_2\) produced: \(\text{Mass of C} = 0.220\text{ g} \times \frac{12.0}{44.0} = 0.060\text{ g}\), which corresponds to \(\text{Moles of C} = \frac{0.060}{12.0} = 0.0050\text{ mol}\). Next, calculate the mass of hydrogen in the \(\text{H}_2\text{O}\) produced: \(\text{Mass of H} = 0.135\text{ g} \times \frac{2.0}{18.0} = 0.015\text{ g}\), which corresponds to \(\text{Moles of H} = \frac{0.015}{1.0} = 0.015\text{ mol}\). Now find the mass and moles of oxygen by subtraction: \(\text{Mass of O} = 0.115\text{ g} - (0.060\text{ g} + 0.015\text{ g}) = 0.040\text{ g}\), which corresponds to \(\text{Moles of O} = \frac{0.040}{16.0} = 0.0025\text{ mol}\). Determine the simplest molar ratio by dividing by the smallest number of moles (0.0025): \(\text{C} = 2\), \(\text{H} = 6\), \(\text{O} = 1\). Therefore, the empirical formula of \(Z\) is \(\text{C}_2\text{H}_6\text{O}\).

1 mark for the correct option B. Method: calculate moles of C, H, and O from the mass of combustion products, find the simplest ratio.

The solubility of Group 2 hydroxides increases down the group. Magnesium hydroxide is sparingly soluble and precipitates out as a white solid when sodium hydroxide is added to a solution of magnesium ions. Barium hydroxide is highly soluble, so no precipitate forms when dilute sodium hydroxide is added to a solution of barium ions. Thus, Test-tube 1 forms a white precipitate, while Test-tube 2 does not.

1 mark for the correct option A. Recall of hydroxide solubility trend down Group 2.

During electrophilic addition of \(\text{HBr}\) to an alkene, the electron-rich \(\pi\) bond of the double bond attacks the electrophilic (delta-positive) hydrogen atom of the \(\text{HBr}\) molecule. This forms a carbocation intermediate and a bromide ion. The major product, 2-bromobutane, is formed via the more stable secondary carbocation intermediate, while the minor product, 1-bromobutane, is formed via the less stable primary carbocation intermediate. The bromide ion acts as a nucleophile, not an electrophile. Thus, statement B is correct.

1 mark for identifying the correct mechanistic description in option B.

In the catalytic oxidation of sulfur dioxide in the atmosphere, nitrogen dioxide (\(\text{NO}_2\)) first reacts with sulfur dioxide to produce sulfur trioxide and nitrogen monoxide (\(\text{NO}\)): \(\text{SO}_2 + \text{NO}_2 \rightarrow \text{SO}_3 + \text{NO}\). The \(\text{NO}\) is then oxidized by atmospheric oxygen to regenerate the catalyst, \(\text{NO}_2\): \(2\text{NO} + \text{O}_2 \rightarrow 2\text{NO}_2\). Therefore, the regeneration step is represented by \(2\text{NO} + \text{O}_2 \rightarrow 2\text{NO}_2\).

1 mark for option B, identifying the regeneration of the \(\text{NO}_2\) catalyst.

The starting compound \(W\) has two oxygen atoms. The product has three oxygen atoms and four fewer hydrogen atoms. This indicates that one primary alcohol group has been oxidized to a carboxylic acid (\(\text{-CH}_2\text{OH}\) to \(\text{-COOH}\), gaining one O and losing two H) and one secondary alcohol group has been oxidized to a ketone (\(\text{-CH(OH)-}\) to \(\text{-CO-}\), losing two H). \(\text{HOCH}_2\text{CH}_2\text{CH}_2\text{CH(OH)CH}_3\) contains one primary alcohol and one secondary alcohol, making it the correct structure. \(\text{HOCH}_2\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_2\text{OH}\) (two primary alcohols) would oxidize to \(\text{C}_5\text{H}_8\text{O}_4\). \(\text{CH}_3\text{CH(OH)CH}_2\text{CH(OH)CH}_3\) (two secondary alcohols) would oxidize to \(\text{C}_5\text{H}_8\text{O}_2\). \(\text{HOCH}_2\text{CH}_2\text{C(OH)(CH}_3)_2\) (one primary, one tertiary alcohol) would oxidize to \(\text{C}_5\text{H}_{10}\text{O}_3\).

1 mark for option A, matching the structural changes in oxidation with the diol classification.

The total mass of the mixture is \(100.0\text{ g}\). The temperature change is \(28.2 - 21.5 = 6.7\text{ K}\). The heat energy released is \(q = m \times c \times \Delta T = 100.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 6.7\text{ K} = 2800.6\text{ J} = 2.80\text{ kJ}\). The number of moles of water formed is \(0.0500\text{ mol}\) (since \(50.0\text{ cm}^3\) of \(1.00\text{ mol dm}^{-3}\) \(\text{NaOH}\) and \(\text{HCl}\) react). The enthalpy change of neutralisation is \(-\frac{q}{n} = -\frac{2.80\text{ kJ}}{0.0500\text{ mol}} = -56.0\text{ kJ mol}^{-1}\).

1 mark for option B. Method: q = mc delta T, find moles of water, calculate enthalpy change.

The successive ionisation energies show a massive increase between the 5th and 6th ionisation energies (from 6274 to 21269 \(\text{kJ mol}^{-1}\)). This indicates that element \(X\) has five valence electrons, which means it belongs to Group 15. The Period 3 element in Group 15 is phosphorus. Phosphorus reacts with oxygen to form phosphorus(V) oxide, which has the empirical formula \(\text{P}_2\text{O}_5\) (represented here as \(X_2O_5\)).

1 mark for option C, identifying the group of the element and matching its oxide formula.

When sodium bromide (\(\text{NaBr}\)) reacts with concentrated sulfuric acid, hydrogen bromide (\(\text{HBr}\)) is initially produced. \(\text{HBr}\) is oxidized by the concentrated sulfuric acid to form bromine (orange-brown vapor) and sulfur dioxide (\(\text{SO}_2\)) gas. \(\text{SO}_2\) is a reducing agent that decolorizes acidified potassium manganate(VII) solution. No purple vapor (iodine) is produced. Sodium iodide (\(\text{NaI}\)) would produce purple iodine vapor. Sodium chloride (\(\text{NaCl}\)) and sodium fluoride (\(\text{NaF}\)) only undergo acid-base reactions to form \(\text{HCl}\) and \(\text{HF}\), which are not oxidized to produce \(\text{SO}_2\).

1 mark for option C, recognizing the reactions of halides with concentrated sulfuric acid.

First, let's find the volume of \(CO_2\) produced and excess \(O_2\) remaining. Let the volume of unreacted \(O_2\) be the gas remaining after passing through \(NaOH(aq)\), which is \(50\text{ cm}^3\). Therefore, the volume of \(CO_2\) produced is: \(110\text{ cm}^3 - 50\text{ cm}^3 = 60\text{ cm}^3\). The volume of \(O_2\) reacted is: \(150\text{ cm}^3 - 50\text{ cm}^3 = 100\text{ cm}^3\). Since \(20\text{ cm}^3\) of the hydrocarbon reacted to produce \(60\text{ cm}^3\) of \(CO_2\), we have \(x = 60 / 20 = 3\). The general equation for combustion of a hydrocarbon is \(C_xH_y + (x + y/4)O_2 \rightarrow xCO_2 + (y/2)H_2O\). The ratio of hydrocarbon reacted to oxygen reacted is \(20 : 100 = 1 : 5\). Therefore, \(x + y/4 = 5\). Since \(x = 3\), \(3 + y/4 = 5\), which gives \(y = 8\). Thus, the molecular formula of the hydrocarbon is \(C_3H_8\).

1 mark for the correct option C.

The thermal decomposition of a Group 2 nitrate is: \(2M(NO_3)_2(s) \rightarrow 2MO(s) + 4NO_2(g) + O_2(g)\). From the stoichiometry, \(2\text{ moles}\) of \(M(NO_3)_2\) produce \(5\text{ moles}\) of gaseous products. The total moles of gas collected at r.t.p. is \(n = 1.20\text{ dm}^3 / 24.0\text{ dm}^3\text{ mol}^{-1} = 0.0500\text{ mol}\). The moles of \(M(NO_3)_2\) decomposed is \(0.0500 \times (2/5) = 0.0200\text{ mol}\). The molar mass of the nitrate is \(2.97\text{ g} / 0.0200\text{ mol} = 148.5\text{ g mol}^{-1}\). The formula mass of the two nitrate groups is \(2 \times 62.0 = 124.0\text{ g mol}^{-1}\). Therefore, the relative atomic mass of \(M\) is \(148.5 - 124.0 = 24.5\), which corresponds closely to magnesium (\(A_r = 24.3\)).

1 mark for the correct option A.

Cleavage of the double bond with hot, concentrated \(KMnO_4\) produces propanone, \((CH_3)_2C=O\) (a 3-carbon ketone), and a branched-chain carboxylic acid \(Z\). Since the alkene has 7 carbons, \(Z\) must have \(7 - 3 = 4\) carbons. The only branched 4-carbon carboxylic acid is 2-methylpropanoic acid, \((CH_3)_2CHCOOH\). Reconstructing the alkene from propanone and 2-methylpropanoic acid gives \((CH_3)_2C=CH-CH(CH_3)_2\). The longest carbon chain containing the double bond has 5 carbons, and there are methyl substituents at positions 2 and 4. This is 2,4-dimethylpent-2-ene.

1 mark for the correct option B.

The reaction occurring is \(2NO(g) + 2CO(g) \rightarrow N_2(g) + 2CO_2(g)\). Carbon monoxide is oxidized to carbon dioxide (so A is incorrect). Nitrogen monoxide is reduced to nitrogen gas (so B is incorrect). In \(NO\), the oxidation state of N is +2. In \(N_2\), the oxidation state of N is 0. This is a reduction where the oxidation state changes from +2 to 0 (so C is correct). Catalysts increase the rate of reaction but do not affect the yield or equilibrium position (so D is incorrect).

1 mark for the correct option C.

The equation for the formation of propanoic acid is: \(3C(s) + 3H_2(g) + O_2(g) \rightarrow CH_3CH_2COOH(l)\). According to Hess's Law using combustion values: \(\Delta H_f^\theta = [3 \times \Delta H_c^\theta(C) + 3 \times \Delta H_c^\theta(H_2)] - \Delta H_c^\theta(CH_3CH_2COOH) = [3(-394) + 3(-286)] - (-1527) = [-1182 - 858] + 1527 = -2040 + 1527 = -513\text{ kJ mol}^{-1}\).

1 mark for the correct option A.

Let's analyze the dehydration products of the options. \(CH_3CH_2CH_2CH_2CH_2OH\) (pentan-1-ol) yields only pent-1-ene. \(CH_3CH_2CH(OH)CH_2CH_3\) (pentan-3-ol) is symmetrical and yields only pent-2-ene (which exists as cis and trans stereoisomers, so 2 isomeric alkenes). \(CH_3CH_2CH(OH)CH_3\) (pentan-2-ol) is asymmetrical and can form pent-1-ene, cis-pent-2-ene, and trans-pent-2-ene. This is a mixture of 3 isomeric alkenes. \((CH_3)_2CHCH(OH)CH_3\) (3-methylbutan-2-ol) can only form 3-methylbut-1-ene and 2-methylbut-2-ene, neither of which has stereoisomers, producing only 2 isomers. Therefore, \(W\) must be pentan-2-ol.

1 mark for the correct option C.

Concentrated sulfuric acid can act as an oxidizing agent. Fluoride (\(F^-\)) and chloride (\(Cl^-\)) ions are weak reducing agents and are not oxidized; they only undergo acid-base reactions to form \(HF\) and \(HCl\) gas respectively. Bromide (\(Br^-\)) is a stronger reducing agent and is oxidized to bromine gas (\(Br_2\)) while sulfuric acid is reduced to sulfur dioxide (\(SO_2\)). Thus, only \(NaBr\) has its halide ion oxidized.

1 mark for the correct option B.

The molecular formula \(C_4H_8\) indicates one degree of unsaturation (either one double bond or one ring). Let's list the structural isomers: Alkenes: (1) but-1-ene, (2) but-2-ene, (3) 2-methylpropene. Cycloalkanes: (4) cyclobutane, (5) methylcyclopropane. Although but-2-ene exhibits stereoisomerism (cis/trans), these are not structural isomers. Therefore, there are exactly 5 structural isomers.

1 mark for the correct option C.

(a)
\[A_r = \frac{(24 \times 79.0) + (25 \times 10.0) + (26 \times 11.0)}{100}\]
\[A_r = \frac{1896 + 250 + 286}{100} = 24.32\]
From the Periodic Table, the element with an \(A_r\) of 24.32 is Magnesium (Mg).

(b)(i)
The cream-colored precipitate with aqueous silver nitrate indicates that the halide ion \(X^-\) is bromide, \(\text{Br}^-\). The precipitate is silver bromide, \(\text{AgBr}\).
\[M_r(\text{AgBr}) = 107.9 + 79.9 = 187.8\text{ g mol}^{-1}\]
\[n(\text{AgBr}) = \frac{7.512}{187.8} = 0.0400\text{ mol}\]
Since there is 1 mole of \(\text{Br}^-\) in 1 mole of \(\text{AgBr}\):
\[n(\text{Br}^-) = 0.0400\text{ mol}\]

(b)(ii)
The formula of the salt is \(\text{MgBr}_2 \cdot n\text{H}_2\text{O}\). Each mole of hydrated salt contains 2 moles of bromide ions.
\[n(\text{MgBr}_2 \cdot n\text{H}_2\text{O}) = \frac{0.0400}{2} = 0.0200\text{ mol}\]
\[M_r(\text{MgBr}_2 \cdot n\text{H}_2\text{O}) = \frac{\text{mass}}{\text{moles}} = \frac{5.842\text{ g}}{0.0200\text{ mol}} = 292.1\text{ g mol}^{-1}\]
\[M_r(\text{MgBr}_2) = 24.3 + (2 \times 79.9) = 184.1\text{ g mol}^{-1}\]
\[M_r(n\text{H}_2\text{O}) = 292.1 - 184.1 = 108.0\text{ g mol}^{-1}\]
\[n = \frac{108.0}{18.0} = 6\]

(c)(i)
\[n(\text{Mg}) = \frac{0.1215}{24.3} = 0.00500\text{ mol}\]
\[n(\text{HCl}) = 25.0 \times 10^{-3} \times 1.00 = 0.0250\text{ mol}\]
From the chemical equation, 1 mole of \(\text{Mg}\) reacts with 2 moles of \(\text{HCl}\).
\[\text{Moles of HCl required} = 2 \times 0.00500 = 0.0100\text{ mol}\]
Since \(0.0250\text{ mol} > 0.0100\text{ mol}\), hydrochloric acid is in excess.

(c)(ii)
Magnesium is the limiting reactant, so the moles of hydrogen gas produced depends on the moles of magnesium.
\[n(\text{H}_2) = n(\text{Mg}) = 0.00500\text{ mol}\]
\[\text{Volume of H}_2 = 0.00500\text{ mol} \times 24000\text{ cm}^3\text{ mol}^{-1} = 120\text{ cm}^3\]

(a) [3 Marks]
- 1 Mark: Correct working shown for relative atomic mass calculation.
- 1 Mark: 24.32 (must be quoted to 2 decimal places).
- 1 Mark: Identifies the element as magnesium / Mg.

(b)(i) [3 Marks]
- 1 Mark: Identifies the halide as bromide / \(\text{Br}^-\) (based on cream-colored precipitate).
- 1 Mark: Correct molar mass calculation for \(\text{AgBr}\) (187.8) and division to find moles of precipitate (0.0400 mol).
- 1 Mark: Correctly deduces that moles of \(\text{Br}^-\) is 0.0400 mol.

(b)(ii) [3 Marks]
- 1 Mark: Deduces moles of hydrated salt = 0.0200 mol (moles of halide divided by 2).
- 1 Mark: Calculates molar mass of the hydrated salt = 292.1 \(\text{g mol}^{-1}\).
- 1 Mark: Calculates water mass of crystallization (108.0) and determines \(n = 6\).

(c)(i) [3 Marks]
- 1 Mark: Calculates moles of Mg = 0.00500 mol.
- 1 Mark: Calculates moles of HCl = 0.0250 mol.
- 1 Mark: Uses the 1:2 reacting ratio to show that 0.0100 mol of HCl is required, hence HCl is in excess.

(c)(ii) [3 Marks]
- 1 Mark: Deduces moles of \(\text{H}_2\) produced = 0.00500 mol.
- 1 Mark: Multiplies moles of \(\text{H}_2\) by 24000 (or 24.0 with subsequent unit conversion).
- 1 Mark: Final answer 120 \(\text{cm}^3\) (or 0.120 \(\text{dm}^3\)) with correct units.

(a)
Strontium has an atomic number of 38.
Full electron configuration: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^2 4p^6 5s^2\).
It is located in Group 2 and Period 5.

(b)
As you go down Group 2:
1. Atomic radius increases because there are more filled electron shells.
2. The outer electrons experience increased shielding from inner shells.
3. The electrostatic attraction between the positive nucleus and the outermost electrons becomes weaker, requiring less energy to remove the first electron, despite the increase in nuclear charge.

(c)(i)
Magnesium nitrate decomposes to form magnesium oxide, nitrogen dioxide gas, and oxygen gas:
\[2\text{Mg(NO}_3)_2(\text{s}) \rightarrow 2\text{MgO(s)} + 4\text{NO}_2(\text{g}) + \text{O}_2(\text{g})\]

(c)(ii)
- Trend: Thermal stability increases down the group.
- Explanation: Down the group, the ionic radius of the Group 2 cation (\(\text{M}^{2+}\)) increases while its charge remains the same (+2).
- This results in a decrease in charge density of the cation down the group.
- Consequently, the smaller magnesium cation has a higher polarizing power and distorts the electron cloud of the nitrate anion (specifically weakening the covalent \(\text{N}-\text{O}\) bonds) more than the larger barium cation does. Thus, magnesium nitrate decomposes at a lower temperature.

(d)(i)
- Trend: Solubility of Group 2 hydroxides increases down the group.
- Explanation: Both lattice energy and hydration energy decrease down the group as the size of the cation increases. However, the lattice energy of the hydroxide decrease more rapidly than the hydration energy of the cation. This causes the enthalpy change of solution, \(\Delta H_{\text{sol}}\), to become more exothermic (or less endothermic) down the group, increasing solubility.

(d)(ii)
- Test: Add a few drops of aqueous sodium sulfate, \(\text{Na}_2\text{SO}_4\text{(aq)}\), to separate test tubes containing saturated solutions of calcium hydroxide and barium hydroxide.
- Observation with barium hydroxide: A dense, thick white precipitate of barium sulfate, \(\text{BaSO}_4\), is formed immediately.
- Observation with calcium hydroxide: No precipitate (or only a very faint precipitate) is formed, because calcium sulfate is much more soluble than barium sulfate.

(a) [2 Marks]
- 1 Mark: Correct full electron configuration \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^2 4p^6 5s^2\) (accept \([\text{Kr}] 5s^2\)).
- 1 Mark: Group 2 and Period 5.

(b) [3 Marks]
- 1 Mark: Atomic radius increases / more shells down the group.
- 1 Mark: Shielding increases.
- 1 Mark: Weaker electrostatic attraction between the positive nucleus and outermost electrons (so less energy needed to remove the electron).

(c)(i) [2 Marks]
- 1 Mark: Correct products: \(\text{MgO}\), \(\text{NO}_2\), and \(\text{O}_2\).
- 1 Mark: Fully balanced equation with correct states or formulas.

(c)(ii) [4 Marks]
- 1 Mark: States that thermal stability increases down the group.
- 1 Mark: States that cationic radius increases (while charge remains +2).
- 1 Mark: Explains that charge density of the cation decreases.
- 1 Mark: Explains that there is less polarization / distortion of the nitrate anion / nitrate electron cloud.

(d)(i) [2 Marks]
- 1 Mark: Solubility of hydroxides increases down the group.
- 1 Mark: Explains that both lattice energy and hydration energy decrease, but lattice energy decreases more rapidly down the group (making dissolution more thermodynamically favorable).

(d)(ii) [2 Marks]
- 1 Mark: Identifies adding aqueous sodium sulfate / \(\text{Na}_2\text{SO}_4\text{(aq)}\) (or other soluble sulfate).
- 1 Mark: Saturated barium hydroxide forms a dense/thick white precipitate AND saturated calcium hydroxide forms no precipitate / faint precipitate.

(a)(i)
Alkene \(A\) is 3-methylpent-2-ene, \(\text{CH}_3\text{CH}=\text{C(CH}_3)\text{CH}_2\text{CH}_3\).
- In the (Z)-isomer, the higher priority groups on C2 (\(\text{CH}_3\)) and C3 (\(\text{CH}_2\text{CH}_3\)) are on the same side of the double bond.
- In the (E)-isomer, the higher priority groups on C2 (\(\text{CH}_3\)) and C3 (\(\text{CH}_2\text{CH}_3\)) are on opposite sides of the double bond.
Skeletal structures should show:
- \((E)\)-3-methylpent-2-ene: The main carbon chain forms a zig-zag trans-like pattern, with the methyl group at C3 trans to the hydrogen at C2.
- \((Z)\)-3-methylpent-2-ene: The main carbon chain has the ethyl group at C3 cis to the methyl group at C2.

(a)(ii)
Stereoisomerism is possible because:
1. There is restricted rotation around the \(\text{C}=\text{C}\) double bond due to the presence of the \(\pi\) bond.
2. Each carbon atom of the \(\text{C}=\text{C}\) double bond is attached to two different groups: C2 is attached to \(-\text{H}\) and \(-\text{CH}_3\); C3 is attached to \(-\text{CH}_3\) and \(-\text{CH}_2\text{CH}_3\).

(b)(i)
Reaction with cold, dilute, alkaline \(\text{KMnO}_4\) adds two hydroxyl groups across the double bond to form a diol.
Structure of \(B\): \(\text{CH}_3\text{CH(OH)C(OH)(CH}_3)\text{CH}_2\text{CH}_3\) (3-methylpentane-2,3-diol).
Functional group: Diol (or alcohol / hydroxyl).

(b)(ii)
The purple color of the manganate(VII) solution is decolourised, forming a brown precipitate of \(\text{MnO}_2\) (accept "purple to colorless" or "purple to brown precipitate").

(c)(i)
Cleavage of 3-methylpent-2-ene (\(\text{CH}_3\text{CH}=\text{C(CH}_3)\text{CH}_2\text{CH}_3\)) with hot, concentrated, acidified \(\text{KMnO}_4\) yields:
- From \(\text{CH}_3\text{CH}=\): ethanoic acid, \(\text{CH}_3\text{COOH}\) (carboxylic acid \(D\))
- From \(=\text{C(CH}_3)\text{CH}_2\text{CH}_3\): butanone, \(\text{CH}_3\text{COCH}_2\text{CH}_3\) (ketone \(C\))

(c)(ii)
- Reagent: Aqueous sodium carbonate (\(\text{Na}_2\text{CO}_3\)) or sodium hydrogencarbonate (\(\text{NaHCO}_3\)).
- Observation with \(D\) (ethanoic acid): Effervescence / bubbles of carbon dioxide gas evolved (which turn limewater cloudy).
- Observation with \(C\) (butanone): No observable change / no reaction.
[Alternative test: Add 2,4-dinitrophenylhydrazine (2,4-DNPH). Ketone \(C\) forms an orange/yellow precipitate; carboxylic acid \(D\) shows no reaction.]

(d)(i)
When \(\text{HBr}\) reacts with 3-methylpent-2-ene:
- The electrophile \(\text{H}^+\) adds to the double bond to form the more stable carbocation intermediate.
- C2 is a secondary carbon (bonded to 1 hydrogen) and C3 is a tertiary carbon (bonded to 0 hydrogens).
- Addition of \(\text{H}^+\) to C2 forms a stable tertiary carbocation on C3.
- Subsequent attack of \(\text{Br}^-\) on the tertiary carbocation at C3 yields 3-bromo-3-methylpentane.
Structure of the major product: \(\text{CH}_3\text{CH}_2\text{C(Br)(CH}_3)\text{CH}_2\text{CH}_3\).

(a)(i) [4 Marks]
- 1 Mark: Correct skeletal structure of (E)-3-methylpent-2-ene.
- 1 Mark: Correct skeletal structure of (Z)-3-methylpent-2-ene.
- 1 Mark: Correctly matches (E)- prefix to the E-structure.
- 1 Mark: Correctly matches (Z)- prefix to the Z-structure.

(a)(ii) [2 Marks]
- 1 Mark: States that there is restricted rotation around the C=C double bond.
- 1 Mark: States that each carbon atom of the C=C bond is bonded to two different atoms/groups.

(b)(i) [2 Marks]
- 1 Mark: Correct structural formula of 3-methylpentane-2,3-diol (\(\text{CH}_3\text{CH(OH)C(OH)(CH}_3)\text{CH}_2\text{CH}_3\)).
- 1 Mark: Identifies the functional group as diol / alcohol / hydroxyl.

(b)(ii) [1 Mark]
- 1 Mark: Purple to colorless OR purple to brown precipitate.

(c)(i) [2 Marks]
- 1 Mark: Correct structure of ketone C: butanone (\(\text{CH}_3\text{COCH}_2\text{CH}_3\)).
- 1 Mark: Correct structure of carboxylic acid D: ethanoic acid (\(\text{CH}_3\text{COOH}\)).

(c)(ii) [3 Marks]
- 1 Mark: Identifies a suitable reagent (e.g., \(\text{Na}_2\text{CO}_3\text{(aq)}\) / \(\text{NaHCO}_3\text{(aq)}\) OR 2,4-DNPH).
- 1 Mark: Correct observation for compound D (effervescence / bubbles for carbonate; no precipitate with 2,4-DNPH).
- 1 Mark: Correct observation for compound C (no effervescence for carbonate; orange/yellow precipitate with 2,4-DNPH).

(d)(i) [1 Mark]
- 1 Mark: Correct structural formula of 3-bromo-3-methylpentane (\(\text{CH}_3\text{CH}_2\text{C(Br)(CH}_3)\text{CH}_2\text{CH}_3\)).

(a)(i)
Inside a car engine, the temperature and pressure are extremely high. This provides the activation energy required for nitrogen and oxygen gases from the air to react together.
Equation:
\[\text{N}_2(\text{g}) + \text{O}_2(\text{g}) \rightarrow 2\text{NO}(\text{g})\]

(a)(ii)
Nitrogen monoxide acts as a homogeneous catalyst. It is first oxidised by atmospheric oxygen to form nitrogen dioxide:
\[\text{NO}(\text{g}) + \frac{1}{2}\text{O}_2(\text{g}) \rightarrow \text{NO}_2(\text{g})\]
Then, nitrogen dioxide oxidises sulfur dioxide to sulfur trioxide, while being reduced back to nitrogen monoxide:
\[\text{NO}_2(\text{g}) + \text{SO}_2(\text{g}) \rightarrow \text{SO}_3(\text{g}) + \text{NO}(\text{g})\]
Because \(\text{NO}\) is regenerated at the end, it acts as a catalyst.

(b)(i)
Using the ideal gas equation: \(pV = nRT\)
Convert volume to \(\text{m}^3\):
\[V = 1.50 \times 10^4\text{ dm}^3 = 15.0\text{ m}^3\]
Convert pressure: \(p = 1.01 \times 10^5\text{ Pa}\)
Temperature: \(T = 350\text{ K}\)
\[n(\text{SO}_2) = \frac{pV}{RT} = \frac{1.01 \times 10^5\text{ Pa} \times 15.0\text{ m}^3}{8.31\text{ J K}^{-1}\text{ mol}^{-1} \times 350\text{ K}}\]
\[n(\text{SO}_2) = \frac{1515000}{2908.5} = 520.89\text{ mol}\]

From the balanced equation, 1 mole of \(\text{CaCO}_3\) reacts with 1 mole of \(\text{SO}_2\).
\[n(\text{CaCO}_3) = 520.89\text{ mol}\]

Molar mass of \(\text{CaCO}_3\):
\[M_r = 40.1 + 12.0 + (3 \times 16.0) = 100.1\text{ g mol}^{-1}\]
\[\text{Mass of CaCO}_3 = 520.89\text{ mol} \times 100.1\text{ g mol}^{-1} = 52141\text{ g}\]
Convert to kg:
\[52141\text{ g} = 52.1\text{ kg} \text{ (to 3 sig. figs.)}\]

(b)(ii)
Equation:
\[2\text{CaSO}_3(\text{s}) + \text{O}_2(\text{g}) \rightarrow 2\text{CaSO}_4(\text{s})\]
Or:
\[\text{CaSO}_3(\text{s}) + \frac{1}{2}\text{O}_2(\text{g}) \rightarrow \text{CaSO}_4(\text{s})\]
- In \(\text{CaSO}_3\), the oxidation state of sulfur is +4.
- In \(\text{CaSO}_4\), the oxidation state of sulfur is +6.
- Change: Increases from +4 to +6 (increase of 2).

(c)(i)
- Test: Add aqueous sodium hydroxide, \(\text{NaOH(aq)}\), to the solid sample or solution, and warm the mixture gently.
- Observation: A pungent-smelling gas (ammonia) is evolved that turns damp red litmus paper blue.
- Ionic Equation:
\[\text{NH}_4^+(\text{aq}) + \text{OH}^-(\text{aq}) \rightarrow \text{NH}_3(\text{g}) + \text{H}_2\text{O}(\text{l})\]

(a)(i) [2 Marks]
- 1 Mark: Explains that nitrogen and oxygen from the air react together under high temperatures / pressures in the car engine.
- 1 Mark: Correct balanced equation: \(\text{N}_2(\text{g}) + \text{O}_2(\text{g}) \rightarrow 2\text{NO}(\text{g})\).

(a)(ii) [3 Marks]
- 1 Mark: Correct equation for oxidation of NO: \(\text{NO} + \frac{1}{2}\text{O}_2 \rightarrow \text{NO}_2\).
- 1 Mark: Correct equation for oxidation of \(\text{SO}_2\): \(\text{NO}_2 + \text{SO}_2 \rightarrow \text{SO}_3 + \text{NO}\).
- 1 Mark: Explains that NO is a catalyst because it is regenerated at the end of the second step / is not consumed overall.

(b)(i) [5 Marks]
- 1 Mark: Converts volume correctly to \(\text{m}^3\) (\(15.0\text{ m}^3\)).
- 1 Mark: Rearranges ideal gas equation and substitutes values correctly.
- 1 Mark: Calculates moles of \(\text{SO}_2\) = 520.89 mol (accept 521 mol).
- 1 Mark: Identifies 1:1 mole ratio and calculates molar mass of \(\text{CaCO}_3\) = 100.1 \(\text{g mol}^{-1}\).
- 1 Mark: Final answer 52.1 kg (or 52100 g) with correct unit and 3 significant figures.

(b)(ii) [2 Marks]
- 1 Mark: Correct balanced equation for the oxidation of calcium sulfite.
- 1 Mark: States that the oxidation state of sulfur increases from +4 to +6 (or change of +2).

(c)(i) [3 Marks]
- 1 Mark: States the reagent and conditions: warm with aqueous sodium hydroxide, \(\text{NaOH(aq)}\).
- 1 Mark: States correct observation: a gas is produced which turns damp red litmus paper blue.
- 1 Mark: Correct ionic equation: \(\text{NH}_4^+ + \text{OH}^- \rightarrow \text{NH}_3 + \text{H}_2\text{O}\).

Let us assume a representative student mean titre of \(24.00\text{ cm}^3\) of **FA 2**.

(a) Table of titration results should include initial and final burette readings, and the titre volume for each run, showing clear headers with units (/ \(\text{cm}^3\)).
(b) Mean titre = \(24.00\text{ cm}^3\) (calculated using concordant titres).
(c) Calculations:
(i) \(n(\text{MnO}_4^-) = 0.0200 \times \frac{24.00}{1000} = 4.80 \times 10^{-4}\text{ mol}\).
(ii) Complete ionic equation: \(\text{MnO}_4^-(\text{aq}) + 5\text{Fe}^{2+}(\text{aq}) + 8\text{H}^+(\text{aq}) \rightarrow \text{Mn}^{2+}(\text{aq}) + 5\text{Fe}^{3+}(\text{aq}) + 4\text{H}_2\text{O}(\text{l})\).
(iii) \(n(\text{Fe}^{2+}) \text{ in } 25.0\text{ cm}^3 = 5 \times 4.80 \times 10^{-4} = 2.40 \times 10^{-3}\text{ mol}\).
(iv) \(n(\text{Fe}^{2+}) \text{ in } 250\text{ cm}^3 = 10 \times 2.40 \times 10^{-3} = 2.40 \times 10^{-2}\text{ mol}\).
(v) \(M_r(\text{FeSO}_4 \cdot x\text{H}_2\text{O}) = \frac{6.67\text{ g}}{2.40 \times 10^{-2}\text{ mol}} = 277.9\text{ g mol}^{-1}\).
(vi) Molar mass of anhydrous \(\text{FeSO}_4 = 55.8 + 32.1 + 4(16.0) = 151.9\text{ g mol}^{-1}\).
\(M_r(\text{water of crystallisation}) = 277.9 - 151.9 = 126.0\text{ g mol}^{-1}\).
\(x = \frac{126.0}{18.0} = 7.00\). To the nearest integer, \(x = 7\).
(d) Evaluation:
(i) Manganate(VII) is a strong oxidising agent that oxidises chloride ions (\(\text{Cl}^-\)) in hydrochloric acid to chlorine gas (\(\text{Cl}_2\)). This side-reaction consumes extra **FA 2**, resulting in an erroneously large titre volume. Sulfuric acid contains sulfate ions (\(\text{SO}_4^{2-}\)), which cannot be oxidised further by manganate(VII).
(ii) \(\text{Percentage uncertainty} = \frac{0.06}{25.0} \times 100\% = 0.24\%\).

### Part (a) [8 marks]
- **Table format (1 mark)**: Table is well-organized with clear column/row headings and units: Initial burette reading / \(\text{cm}^3\), Final burette reading / \(\text{cm}^3\), Titre / \(\text{cm}^3\).
- **Precision (2 marks)**: All burette readings (including initial readings of 0.00) are recorded to the nearest 0.05 \(\text{cm}^3\).
- **Concordancy (1 mark)**: At least two titres are within 0.10 \(\text{cm}^3\).
- **Accuracy (4 marks)**: Average student titre is compared to the supervisor's expected value (24.00 \(\text{cm}^3\)):
- 4 marks: within 0.20 \(\text{cm}^3\)
- 3 marks: within 0.21 to 0.30 \(\text{cm}^3\)
- 2 marks: within 0.31 to 0.50 \(\text{cm}^3\)
- 1 mark: within 0.51 to 1.00 \(\text{cm}^3\)

### Part (b) [1 mark]
- Correct calculation of mean titre using only concordant titres. Must be expressed to 2 decimal places with correct units.

### Part (c) [8 marks]
- **(i) (1 mark)**: Correctly calculates \(n(\text{MnO}_4^-) = \text{mean titre} \times 10^{-3} \times 0.0200\) to 3 or 4 significant figures.
- **(ii) (1 mark)**: Correctly completes the products as \(\text{Mn}^{2+}(\text{aq}) + 5\text{Fe}^{3+}(\text{aq}) + 4\text{H}_2\text{O}(\text{l})\).
- **(iii) (1 mark)**: Multiplies moles of manganate(VII) by 5.
- **(iv) (1 mark)**: Multiplies moles of \(\text{Fe}^{2+}\) in 25.0 \(\text{cm}^3\) by 10.
- **(v) (2 marks)**: Divides 6.67 by the answer to (iv) to find \(M_r\). (Award 1 mark for correct working but minor arithmetic error).
- **(vi) (2 marks)**: Subtracts 151.9 from the calculated \(M_r\) and divides by 18.0 to get \(x\) as a whole number. (Award 1 mark for correct method leading to a non-integer value).

### Part (d) [3 marks]
- **(i) (1 mark)**: States that hydrochloric acid contains \(\text{Cl}^-\), which would be oxidised to \(\text{Cl}_2\) by \(\text{MnO}_4^-\).
- **(ii) (2 marks)**: Correctly calculates percentage uncertainty as 0.24%. (Award 1 mark for working: \(\frac{0.06}{25.0} \times 100\)).

(a) Observation table:
| Test | FB 1 | FB 2 | FB 3 |
|---|---|---|---|
| **Test 1 (NaOH)** | Pink/blue precipitate, insoluble in excess. | White precipitate, insoluble in excess. | No precipitate. Upon warming, a pungent gas is evolved which turns damp red litmus paper blue. |
| **Test 2 (NH3)** | Pink/blue precipitate, dissolves in excess to give a yellow-brown/brown solution. | White precipitate, insoluble in excess. | No precipitate/no reaction. |
| **Test 3 (AgNO3 + NH3)** | White precipitate, soluble in aqueous ammonia. | No precipitate/no reaction. | Yellow precipitate, insoluble in aqueous ammonia. |
| **Test 4 (BaCl2 + HCl)** | No precipitate/no reaction. | White precipitate, insoluble in dilute hydrochloric acid. | No precipitate/no reaction. |

(b) Identifications and Evidence:
(i) **FB 1**: Cation is cobalt(II), \(\text{Co}^{2+}\) (evidence: pink/blue precipitate with NaOH and \(\text{NH}_3\), soluble in excess \(\text{NH}_3\) to form a brown solution). Anion is chloride, \(\text{Cl}^-\) (evidence: white precipitate with \(\text{AgNO}_3\) that is soluble in \(\text{NH}_3\)).
(ii) **FB 2**: Cation is magnesium, \(\text{Mg}^{2+}\) (evidence: white precipitate with both NaOH and \(\text{NH}_3\), both insoluble in excess). Anion is sulfate, \(\text{SO}_4^{2-}\) (evidence: white precipitate with \(\text{BaCl}_2\) which is insoluble in dilute HCl).
(iii) **FB 3**: Cation is ammonium, \(\text{NH}_4^+\) (evidence: no precipitate with NaOH, but gas evolved on heating turns red litmus blue). Anion is iodide, \(\text{I}^-\) (evidence: yellow precipitate with \(\text{AgNO}_3\) which is insoluble in ammonia).

(c) Ionic Equation:
\(\text{Ag}^+(\text{aq}) + \text{I}^-(\text{aq}) \rightarrow \text{AgI}(\text{s})\)

### Part (a) [12 marks total]
- **FB 1 observations (4 marks)**:
- 1 mark: Pink/blue precipitate with NaOH, insoluble in excess.
- 1 mark: Pink/blue precipitate with \(\text{NH}_3\), dissolving in excess to a yellow-brown/brown solution.
- 1 mark: White precipitate with \(\text{AgNO}_3\).
- 1 mark: White precipitate dissolving in ammonia AND no reaction with \(\text{BaCl}_2\).
- **FB 2 observations (4 marks)**:
- 1 mark: White precipitate with NaOH, insoluble in excess.
- 1 mark: White precipitate with \(\text{NH}_3\), insoluble in excess.
- 1 mark: White precipitate with \(\text{BaCl}_2\).
- 1 mark: White precipitate insoluble in dilute HCl AND no reaction with \(\text{AgNO}_3\).
- **FB 3 observations (4 marks)**:
- 1 mark: No precipitate with NaOH AND gas evolved on warming turns damp red litmus blue.
- 1 mark: No reaction with \(\text{NH}_3\).
- 1 mark: Yellow precipitate with \(\text{AgNO}_3\).
- 1 mark: Yellow precipitate insoluble in ammonia AND no reaction with \(\text{BaCl}_2\).

### Part (b) [6 marks total]
- **(i) (2 marks)**: Correctly identifies \(\text{Co}^{2+}\) and \(\text{Cl}^-\). Cites specific, consistent observations as proof (e.g. pink ppt with NaOH dissolving in excess ammonia to brown solution, and white ppt with \(\text{AgNO}_3\) soluble in ammonia).
- **(ii) (2 marks)**: Correctly identifies \(\text{Mg}^{2+}\) and \(\text{SO}_4^{2-}\). Cites specific, consistent observations as proof (white ppt with both NaOH and ammonia, white ppt with \(\text{Ba}^{2+}\) insoluble in acid).
- **(iii) (2 marks)**: Correctly identifies \(\text{NH}_4^+\) and \(\text{I}^-\). Cites specific, consistent observations as proof (litmus paper turning blue on warming with NaOH, and yellow ppt with \(\text{Ag}^+\) insoluble in ammonia).

### Part (c) [2 marks total]
- **1 mark**: Correct chemical species and balancing: \(\text{Ag}^+ + \text{I}^- \rightarrow \text{AgI}\).
- **1 mark**: Correct state symbols: \(\text{Ag}^+(\text{aq}) + \text{I}^-(\text{aq}) \rightarrow \text{AgI}(\text{s})\).