2023 HKDSE Chemistry Answers & Marking Scheme

\( \text{BF}_3 \), \( \text{C}_2\text{H}_4 \), and \( \text{CO}_3^{2-} \) are all planar species. \( \text{H}_3\text{O}^+ \) has 3 bonding pairs and 1 lone pair on the oxygen atom, giving it a trigonal pyramidal shape which is non-planar.

Award 1 mark for the correct option. No marks for incorrect options or multiple answers.

(1) is incorrect because pH depends on both acid strength and concentration. A dilute strong acid can have the same pH as a concentrated weak acid. (2) is correct because \([\text{H}^+] = 10^{-\text{pH}}\). For X, \([\text{H}^+] = 10^{-2} \text{ mol dm}^{-3}\); for Y, \([\text{H}^+] = 10^{-4} \text{ mol dm}^{-3}\). Ratio = 100. (3) is incorrect because a concentrated weak acid of pH 4.0 might have a higher total concentration of ionizable hydrogen ions than a dilute strong acid of pH 2.0, thus requiring more base for complete neutralization.

Award 1 mark for the correct option. No marks for incorrect options or multiple answers.

The oxidation numbers of Cl are: in \( \text{NaClO} \), +1; in \( \text{ClO}_2 \), +4; in \( \text{KClO}_3 \), +5; in \( \text{Cl}_2\text{O}_7 \), +7. Therefore, \( \text{Cl}_2\text{O}_7 \) has the highest oxidation number of +7.

Award 1 mark for the correct option. No marks for incorrect options or multiple answers.

Find the longest carbon chain, which has 5 carbons (pentane). Numbering from either end gives substituents at positions 2 and 4. According to IUPAC rules, when locants are identical from both ends, alphabetical priority decides the lower locant. Since 'chloro' comes before 'methyl' alphabetically, the chloro group is assigned position 2 and the methyl group is assigned position 4. Thus, the name is 2-chloro-4-methylpentane.

Award 1 mark for the correct option. No marks for incorrect options or multiple answers.

Metal X is the least reactive as it does not react with dilute HCl (below hydrogen in the reactivity series). Metal Y can displace X, so Y is more reactive than X (Y > X). The oxide of Z cannot be reduced by heating with carbon (meaning Z is highly reactive, above Al), whereas the oxide of Y can be reduced by carbon (meaning Y is less reactive, below Al). Thus, Z is more reactive than Y (Z > Y). The ascending order of reactivity is X < Y < Z.

Award 1 mark for the correct option. No marks for incorrect options or multiple answers.

The target reaction is the reverse of the given reaction, multiplied by \( \frac{1}{2} \). Therefore, the new equilibrium constant \( K_c' = \left( \frac{1}{K_c} \right)^{1/2} = \frac{1}{\sqrt{4.0 \times 10^2}} = \frac{1}{20} = 0.05 = 5.0 \times 10^{-2} \text{ mol}^{1/2}\text{dm}^{-3/2} \).

Award 1 mark for the correct option. No marks for incorrect options or multiple answers.

(1) is incorrect because only temperature can increase the average kinetic energy of reactant molecules. (2) is correct because a catalyst decreases the activation energy of both the forward and backward reactions by the same amount, leaving the overall enthalpy change (\( \Delta H \)) unaltered. (3) is incorrect because a catalyst accelerates both forward and backward reactions equally; it helps reach equilibrium faster but does not affect the position of equilibrium or change the equilibrium yield.

Award 1 mark for the correct option. No marks for incorrect options or multiple answers.

The standard enthalpy change of formation of \( \text{CO(g)} \) refers to the reaction: \( \text{C(graphite)} + \frac{1}{2}\text{O}_2(\text{g}) \rightarrow \text{CO}(\text{g}) \). According to Hess's Law, this is obtained by subtracting the second reaction from the first reaction: \( \Delta H_f = -393.5 \text{ kJ mol}^{-1} - (-283.0 \text{ kJ mol}^{-1}) = -110.5 \text{ kJ mol}^{-1} \).

Award 1 mark for the correct option. No marks for incorrect options or multiple answers.

A is incorrect because atomic radius decreases across a period due to increased effective nuclear charge. B is incorrect because electronegativity increases across the period from Li to F. C is correct because the first ionization energy generally increases due to increasing nuclear charge and decreasing atomic radius. D is incorrect because Neon (Ne) is an inert gas and does not form stable compounds like oxides under normal conditions.

Award 1 mark for the correct option. No marks for incorrect options or multiple answers.

The structural isomers of butyl bromide (\( \text{C}_4\text{H}_9\text{Br} \)) can be derived from two carbon skeletons: butane and methylpropane. From butane, we have 1-bromobutane and 2-bromobutane. From methylpropane, we have 1-bromo-2-methylpropane and 2-bromo-2-methylpropane. Thus, there are 4 structural isomers in total.

Award 1 mark for the correct option. No marks for incorrect options or multiple answers.

(1) \(\text{H}_3\text{O}^+\) has 3 bonding pairs and 1 lone pair on the central oxygen atom, giving it a trigonal pyramidal shape. (2) \(\text{BF}_3\) has 3 bonding pairs and 0 lone pairs on the central boron atom, giving it a trigonal planar shape. (3) \(\text{NH}_3\) has 3 bonding pairs and 1 lone pair on the central nitrogen atom, giving it a trigonal pyramidal shape. Therefore, both (1) and (3) are trigonal pyramidal.

Award 1 mark for the correct option C. Incorrect options are based on failing to recognize that \(\text{H}_3\text{O}^+\) is trigonal pyramidal due to the lone pair on the oxygen atom, or confusing \(\text{BF}_3\) (trigonal planar) with trigonal pyramidal.

Metal \(Y\) is highly reactive as it reacts with cold water (e.g., sodium or calcium). Metal \(X\) is moderately reactive because it can be extracted by carbon reduction (e.g., zinc or iron). Metal \(Z\) is unreactive with dilute acid but reacts with heating in oxygen, showing it is less reactive than hydrogen but still oxidizable (e.g., copper). Thus, the descending order of reactivity is \(Y > X > Z\).

Award 1 mark for Option A. Incorrect options reverse or scramble the positions of the metals based on their chemical properties.

The combustion of propane is represented by: \(\text{C}_3\text{H}_8(g) + 5\text{O}_2(g) \rightarrow 3\text{CO}_2(g) + 4\text{H}_2\text{O}(l)\). By definition, \(\Delta H_c^\theta = 3 \Delta H_f^\theta[\text{CO}_2(g)] + 4 \Delta H_f^\theta[\text{H}_2\text{O}(l)] - \Delta H_f^\theta[\text{C}_3\text{H}_8(g)]\). Substituting the values: \(-2220.0 = 3(-393.5) + 4(-285.8) - \Delta H_f^\theta[\text{C}_3\text{H}_8(g)]\) which gives \(-2220.0 = -1180.5 - 1143.2 - \Delta H_f^\theta[\text{C}_3\text{H}_8(g)]\) and thus \(\Delta H_f^\theta[\text{C}_3\text{H}_8(g)] = -103.7\text{ kJ mol}^{-1}\).

Award 1 mark for Option B. Option C has the incorrect sign. Options A and D represent common errors where stoichiometric coefficients are misapplied or products and reactants are confused.

Disproportionation is a redox reaction in which the same element is simultaneously oxidized and reduced. In B, nitrogen in \(\text{NO}_2\) has an oxidation state of \(+4\). In the products, nitrogen in \(\text{HNO}_3\) has an oxidation state of \(+5\) (oxidized) and nitrogen in \(\text{NO}\) has an oxidation state of \(+2\) (reduced). Hence, nitrogen undergoes disproportionation.

Award 1 mark for Option B. In option A, nitrogen undergoes comproportionation (from \(-3\) and \(+5\) to \(+1\)). In C, nitrogen is only oxidized. In D, nitrogen is only reduced.

The equilibrium constant \(K_c\) is affected ONLY by temperature. Since the forward reaction is endothermic (\(\Delta H > 0\)), increasing the temperature will shift the equilibrium position to the right, thereby increasing the concentration of products and decreasing the concentration of reactants. This results in an increased value of \(K_c\). Changes in pressure, concentration, or addition of a catalyst do not alter the value of \(K_c\).

Award 1 mark for Option C. Options A, B, and D do not change the equilibrium constant \(K_c\) because it is a temperature-dependent constant.

Compound \(X\) has the molecular formula \(\text{C}_4\text{H}_8\text{O}_2\) and reacts with \(\text{NaHCO}_3\) to release \(\text{CO}_2\) gas (which turns limewater milky). This indicates that \(X\) contains a carboxylic acid functional group (\(-\text{COOH}\)). Therefore, \(X\) must be butanoic acid. Ethyl ethanoate, methyl propanoate, and ethyl methanoate are structural isomers of butanoic acid but they are esters, which do not show acidic properties to react with hydrogencarbonates.

Award 1 mark for Option B. Options A, C, and D are esters with the same molecular formula but do not react with sodium hydrogencarbonate.

The chemical equation for complete neutralization is: \(\text{H}_2\text{A}(aq) + 2\text{NaOH}(aq) \rightarrow \text{Na}_2\text{A}(aq) + 2\text{H}_2\text{O}(l)\). The number of moles of \(\text{H}_2\text{A} = 0.0250\text{ dm}^3 \times 0.100\text{ mol dm}^{-3} = 0.00250\text{ mol}\). According to the stoichiometric ratio, the number of moles of \(\text{NaOH}\) required \(= 2 \times 0.00250\text{ mol} = 0.00500\text{ mol}\). Thus, the concentration of the \(\text{NaOH}\) solution \(= \frac{0.00500\text{ mol}}{0.0200\text{ dm}^3} = 0.250\text{ mol dm}^{-3}\).

Award 1 mark for Option C. Option B is a common distractor if the 1:2 mole ratio of acid to alkali is incorrectly taken as 1:1. Option A is obtained if the ratio is reversed.

\(\text{HNO}_3\) and \(\text{HCl}\) are strong monoprotic acids. At \(0.1\text{ mol dm}^{-3}\), they fully dissociate to give \([\text{H}^+] = 0.1\text{ mol dm}^{-3}\) (pH = 1). \(\text{CH}_3\text{COOH}\) is a weak acid and only partially dissociates, giving \([\text{H}^+] < 0.1\text{ mol dm}^{-3}\) (pH > 1). \(\text{H}_2\text{SO}_4\) is a strong diprotic acid. Its first dissociation is complete, giving \(0.1\text{ mol dm}^{-3}\) of \(\text{H}^+\), and its second dissociation further contributes to the concentration of \(\text{H}^+\), making \([\text{H}^+] > 0.1\text{ mol dm}^{-3}\). Therefore, \(0.1\text{ mol dm}^{-3}\ \text{H}_2\text{SO}_4(aq)\) has the highest concentration of hydrogen ions and the lowest pH value.

Award 1 mark for Option B. Strong diprotic acids release more hydrogen ions than monoprotic acids of the same molarity, thus giving a lower pH.

(1) Transition metals typically have giant metallic structures with strong metallic bonding because they have more delocalized electrons, resulting in high melting points and densities. (2) They have partially filled d-orbitals which allow d-d electronic transitions that absorb specific wavelengths of visible light, leading to colored ions and compounds. (3) They have d-electrons in their inner shell that can participate in bonding, enabling them to exhibit variable oxidation states. All three statements are correct.

Award 1 mark for Option D. All listed characteristics are fundamental properties of transition metals in the DSE syllabus.

Enantiomerism requires the presence of a chiral carbon atom (a carbon bonded to four different atoms or groups of atoms). In 2-chlorobutane, the second carbon (C-2) is bonded to: (1) a hydrogen atom (\(-\text{H}\)), (2) a chlorine atom (\(-\text{Cl}\)), (3) a methyl group (\(-\text{CH}_3\)), and (4) an ethyl group (\(-\text{CH}_2\text{CH}_3\)). Since all four attached groups are different, C-2 is chiral, making 2-chlorobutane optically active. The other compounds listed do not possess a chiral carbon.

Award 1 mark for Option A. Other options are achiral because they contain carbon atoms bonded to at least two identical groups.

(1) \( \text{SF}_4 \) has a see-saw molecular shape (derived from a trigonal bipyramidal electron-pair geometry with one lone pair), which is asymmetrical. Thus, the polar S-F bonds do not cancel out, resulting in a net dipole moment.
(2) \( \text{BF}_3 \) has a symmetrical trigonal planar molecular shape. The polar B-F bonds cancel each other out completely, resulting in a net dipole moment of zero.
(3) \( \text{trans-1,2-dichloroethene} \) has a highly symmetrical planar shape. The dipoles of the two polar C-Cl bonds point in opposite directions and cancel out, as do the C-H bonds, resulting in a net dipole moment of zero.

Award 1 mark for the correct answer D. No partial marks.

- From (1), X is more reactive than Y (X > Y).
- From (2), X oxide cannot be reduced by carbon, but W oxide can, which indicates that X is more reactive than W (X > W).
- From (3), both react with dilute HCl, but W reacts more vigorously than Y, which indicates W is more reactive than Y (W > Y).
- From (4), Z is very unreactive since its oxide can be decomposed by heat alone (Z is lowest in reactivity).
Combining these results, the descending order of reactivity is: X > W > Y > Z.

Award 1 mark for the correct answer A. No partial marks.

When chlorine gas reacts with hot concentrated sodium hydroxide solution, the chemical equation is:
\( 3\text{Cl}_2(\text{g}) + 6\text{NaOH}(\text{aq}) \rightarrow 5\text{NaCl}(\text{aq}) + \text{NaClO}_3(\text{aq}) + 3\text{H}_2\text{O}(\text{l}) \)
Chlorine in \( \text{NaCl} \) has an oxidation number of -1.
Chlorine in \( \text{NaClO}_3 \) has an oxidation number of +5.
Therefore, the oxidation numbers of chlorine in the products are -1 and +5.

Award 1 mark for the correct answer C. No partial marks.

(1) Correct. Both are monoprotic acids and contain \( 0.010\text{ mol} \) of ionisable hydrogen atoms. Thus, both require the same number of moles (and hence the same volume) of NaOH for complete neutralisation.
(2) Incorrect. \( \text{HCl} \) is a strong acid which completely ionises, while \( \text{CH}_3\text{COOH} \) is a weak acid which only partially ionises. Thus, \( \text{HCl(aq)} \) has a much higher concentration of ions and higher electrical conductivity.
(3) Correct. Since zinc is in excess, the total yield of hydrogen gas depends only on the total number of moles of ionisable hydrogen atoms in the acids, which is identical for both solutions.

Award 1 mark for the correct answer B. No partial marks.

1. Initial moles of HCl = \( 0.0500\text{ dm}^3 \times 0.500\text{ mol dm}^{-3} = 0.0250\text{ mol} \)
2. Moles of NaOH used = \( 0.0285\text{ dm}^3 \times 0.100\text{ mol dm}^{-3} = 0.00285\text{ mol} \)
3. Excess moles of HCl = \( 0.00285\text{ mol} \)
4. Moles of HCl reacted with \( \text{CaCO}_3 \) = \( 0.0250 - 0.00285 = 0.02215\text{ mol} \)
5. Since \( \text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{CO}_2 + \text{H}_2\text{O} \),
moles of \( \text{CaCO}_3 \) = \( \frac{0.02215}{2} = 0.011075\text{ mol} \)
6. Mass of \( \text{CaCO}_3 \) = \( 0.011075\text{ mol} \times 100.1\text{ g mol}^{-1} = 1.1086\text{ g} \)
7. Percentage by mass = \( \frac{1.1086\text{ g}}{1.25\text{ g}} \times 100\% = 88.69\% \approx 88.7\% \)

Award 1 mark for the correct answer C. No partial marks.

- Step 1: Addition of hydrogen bromide (e.g., \( \text{HBr(g)} \) or concentrated \( \text{HBr(aq)} \)) to propene yields 2-bromopropane as the major product.
- Step 2: Nucleophilic substitution using aqueous sodium hydroxide (\( \text{NaOH(aq)} \)) under heat converts 2-bromopropane to propan-2-ol.
- Step 3: Oxidation of the secondary alcohol propan-2-ol to the ketone propanone requires acidified potassium dichromate solution (\( \text{K}_2\text{Cr}_2\text{O}_7\text{/H}^+(\text{aq}) \)) under heat.

Award 1 mark for the correct answer B. No partial marks.

The molecular formula \( \text{C}_4\text{H}_8 \) represents compounds with one double bond or one ring.
Acyclic alkenes (3 structural isomers):
1. But-1-ene: \( \text{CH}_2=\text{CH-CH}_2\text{-CH}_3 \)
2. But-2-ene: \( \text{CH}_3\text{-CH}=\text{CH-CH}_3 \) (Note: cis- and trans-but-2-ene are stereoisomers, but they are the same structural isomer)
3. 2-methylpropene: \( \text{CH}_2=\text{C(CH}_3)_2 \)

Cyclic alkanes (2 structural isomers):
4. Cyclobutane
5. Methylcyclopropane

Total = 3 + 2 = 5 structural isomers.

Award 1 mark for the correct answer C. No partial marks.

The formation reaction of ethyne is:
\( 2\text{C(graphite)} + \text{H}_2(\text{g}) \rightarrow \text{C}_2\text{H}_2(\text{g}) \)

According to Hess's Law, the enthalpy change of formation is:
\( \Delta H_f^\theta = 2 \times \Delta H_c^\theta[\text{C(graphite)}] + \Delta H_c^\theta[\text{H}_2(\text{g})] - \Delta H_c^\theta[\text{C}_2\text{H}_2(\text{g})] \)
\( \Delta H_f^\theta = 2(-393.5) + (-285.8) - (-1299.6) \)
\( \Delta H_f^\theta = -787.0 - 285.8 + 1299.6 = +226.8\text{ kJ mol}^{-1} \)

Award 1 mark for the correct answer B. No partial marks.

Let's set up the equilibrium mole table:
Reaction: \( \text{PCl}_5(\text{g}) \rightleftharpoons \text{PCl}_3(\text{g}) + \text{Cl}_2(\text{g}) \)
Initial moles: \( \text{PCl}_5 = 1.0 \), \( \text{PCl}_3 = 0 \), \( \text{Cl}_2 = 0 \)
Change in moles: \( \text{PCl}_5 = -0.60 \), \( \text{PCl}_3 = +0.60 \), \( \text{Cl}_2 = +0.60 \)
Equilibrium moles: \( \text{PCl}_5 = 0.40 \), \( \text{PCl}_3 = 0.60 \), \( \text{Cl}_2 = 0.60 \)

At equilibrium, the concentration of each gas (volume \( V = 2.0\text{ dm}^3 \)):
\( [\text{PCl}_5] = \frac{0.40\text{ mol}}{2.0\text{ dm}^3} = 0.20\text{ mol dm}^{-3} \)
\( [\text{PCl}_3] = \frac{0.60\text{ mol}}{2.0\text{ dm}^3} = 0.30\text{ mol dm}^{-3} \)
\( [\text{Cl}_2] = \frac{0.60\text{ mol}}{2.0\text{ dm}^3} = 0.30\text{ mol dm}^{-3} \)

\( K_c = \frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]} = \frac{0.30 \times 0.30}{0.20} = 0.45\text{ mol dm}^{-3} \)

Award 1 mark for the correct answer A. No partial marks.

(1) Suitable. Hydroxide ions (\( \text{OH}^- \)), which have high ionic mobility, are replaced by ethanoate ions (\( \text{CH}_3\text{COO}^- \)), which have much lower ionic mobility. Thus, there is a measurable decrease in the electrical conductivity of the mixture over time.
(2) Suitable. As \( \text{OH}^- \) ions are consumed in the reaction, the pH of the mixture decreases significantly, which can be monitored using a pH meter.
(3) Not suitable. The reaction occurs entirely in the liquid phase and no gas is produced or consumed. The change in volume of the liquid mixture is negligible and cannot be used to monitor the rate.

Award 1 mark for the correct answer C. No partial marks.

I. In \(\text{Cr}_2\text{O}_7^{2-}\), the oxidation state of \(\text{Cr}\) is +6. In \(\text{Cr}^{3+}\), it is +3. This is a reduction (decrease in oxidation number).
II. In \(\text{H}_2\text{O}_2\), the oxidation state of \(\text{O}\) is -1. In \(\text{O}_2\), it is 0. This is an oxidation (increase in oxidation number).
III. In both \(\text{NO}_2\) and \(\text{N}_2\text{O}_4\), the oxidation state of \(\text{N}\) is +4. There is no change in oxidation state.

Award 1 mark for the correct option (A). No partial marks.

In \(\text{H}_3\text{O}^+\), the central oxygen atom has 5 valence electrons (6 - 1 from positive charge). It forms three single covalent bonds with hydrogen atoms and has one lone pair of electrons. According to VSEPR theory, a species with 3 bond pairs and 1 lone pair has a trigonal pyramidal shape. \(\text{BF}_3\) and \(\text{CO}_3^{2-}\) are trigonal planar, while \(\text{NH}_4^+\) is tetrahedral.

Award 1 mark for selecting B. No partial marks.

The value of the equilibrium constant \(K_c\) depends solely on the temperature. Since the forward reaction is exothermic (\(\Delta H < 0\)), lowering the temperature of the container shifts the equilibrium position to the right, which increases the concentrations of products and decreases those of reactants. Consequently, the value of \(K_c\) increases. Changes in pressure, volume, concentration, or addition of a catalyst do not alter the value of \(K_c\).

Award 1 mark for selecting B. No partial marks.

Since \(X\) undergoes alkaline hydrolysis (reaction with NaOH) to form sodium ethanoate and ethanol, \(X\) must be an ester formed from ethanoic acid (giving sodium ethanoate) and ethanol (giving ethanol). Thus, \(X\) is ethyl ethanoate, with the chemical formula \(\text{CH}_3\text{COOCH}_2\text{CH}_3\).

Award 1 mark for the correct answer B. No partial marks.

The equation for the combustion of ethanoic acid is:
\(\text{CH}_3\text{COOH(l)} + 2\text{O}_2\text{(g)} \rightarrow 2\text{CO}_2\text{(g)} + 2\text{H}_2\text{O(l)}\)
Using Hess's law, we construct this targeted equation from the given steps:
- Reverse Eq (1): \(\text{CH}_3\text{COOH(l)} \rightarrow 2\text{C(s)} + 2\text{H}_2\text{(g)} + \text{O}_2\text{(g)}\) (\(-\Delta H_1\))
- Multiply Eq (2) by 2: \(2\text{C(s)} + 2\text{O}_2\text{(g)} \rightarrow 2\text{CO}_2\text{(g)}\) (\(2\Delta H_2\))
- Multiply Eq (3) by 2: \(2\text{H}_2\text{(g)} + \text{O}_2\text{(g)} \rightarrow 2\text{H}_2\text{O(l)}\) (\(2\Delta H_3\))
Adding these equations together gives the overall combustion equation. Thus, the total enthalpy change is \(2\Delta H_2 + 2\Delta H_3 - \t\Delta H_1\).

Award 1 mark for option B. No partial marks.

The neutralisation equation is:
\(\text{M}_2\text{CO}_3(\text{aq}) + 2\text{HCl}(\text{aq}) \rightarrow 2\text{MCl}(\text{aq}) + \text{CO}_2(\text{g}) + \text{H}_2\text{O}(\text{l})\)
Number of moles of \(\text{HCl}\) used in the titration = \(0.100 \times 0.0200 = 0.00200\text{ mol}\).
Since 1 mole of \(\text{M}_2\text{CO}_3\) reacts with 2 moles of \(\text{HCl}\):
Number of moles of \(\text{M}_2\text{CO}_3\) in \(25.0\text{ cm}^3\) = \(0.00200 / 2 = 0.00100\text{ mol}\).
Number of moles of \(\text{M}_2\text{CO}_3\) in the total \(250.0\text{ cm}^3\) solution = \(0.00100 \times (250.0 / 25.0) = 0.0100\text{ mol}\).
Molar mass of \(\text{M}_2\text{CO}_3 = 1.38\text{ g} / 0.0100\text{ mol} = 138.0\text{ g mol}^{-1}\).
Let \(x\) be the relative atomic mass of \(\text{M}\):
\(2x + 12.0 + 3(16.0) = 138.0\)
\(2x + 60.0 = 138.0\)
\(2x = 78.0 \Rightarrow x = 39.0\).
Therefore, \(\text{M}\) is Potassium (\(\text{K}\), relative atomic mass \(39.1\)).

Award 1 mark for selecting C. No partial marks.

(a) Drawing shows central carbon atom double-bonded to oxygen and single-bonded to two chlorine atoms in a flat, trigonal arrangement with wedge/dash or flat lines. The shape name is trigonal planar. (b) Oxygen and chlorine are more electronegative than carbon, resulting in polar C=O and C-Cl bonds. Because the trigonal planar shape is asymmetrical, the dipoles do not cancel each other out, resulting in a net dipole moment.

(a) 1 mark for correct 3D representation showing trigonal planar geometry with a double bond to O and two single bonds to Cl. 1 mark for stating 'trigonal planar'. (b) 1 mark for explaining that O and Cl are more electronegative than C, leading to polar bonds. 1 mark for explaining that the asymmetrical shape prevents the bond dipoles from canceling, making it polar.

(a) The blue solution gradually fades (or becomes colorless), and a reddish-brown solid is deposited on the surface of the magnesium. (b) Ionic equation: \(\text{Zn}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu}(s)\). It is a redox reaction because zinc atoms lose electrons (oxidation) to form zinc ions, while copper(II) ions gain electrons (reduction) to form copper atoms.

(a) 1 mark for 'blue solution fades/colorless'. 1 mark for 'red-brown solid'. (b) 1 mark for correct ionic equation with state symbols. 1 mark for explaining that Zn loses electrons (oxidation) and Cu2+ gains electrons (reduction).

(a) Phenolphthalein is suitable because it is a weak acid-strong base titration. The color changes from colorless to pale pink. (b) Moles of NaOH = \(0.120 \times 0.01850 = 0.00222\text{ mol}\). Since ethanoic acid and NaOH react in a 1:1 ratio, moles of CH3COOH in 25.00 mL diluted solution = 0.00222 mol. Concentration of diluted acid = \(0.00222 / 0.02500 = 0.0888\text{ mol dm}^{-3}\). Since the dilution factor is 10 (from 25.00 mL to 250.0 mL), the original concentration = \(0.0888 \times 10 = 0.888\text{ mol dm}^{-3}\).

(a) 1 mark for phenolphthalein. 1 mark for 'colorless to pale/light pink' (reject deep pink/purple). (b) 1 mark for calculating diluted concentration (0.0888 mol dm-3) or showing dilution calculation. 1 mark for correct final answer of 0.888 mol dm-3 with unit.

(a) The two carboxylic acids are butanoic acid, \(\text{CH}_3\text{CH}_2\text{CH}_2\text{COOH}\), and 2-methylpropanoic acid, \((\text{CH}_3)_2\text{CHCOOH}\). (b) The ester must be a methyl ester to yield methanol. With 4 carbons in total, the ester is methyl propanoate, \(\text{CH}_3\text{CH}_2\text{COOCH}_3\).

(a) 1 mark for drawing both structures correctly. 1 mark for both systematic names ('butanoic acid' and '2-methylpropanoic acid'). (b) 1 mark for drawing the structure of methyl propanoate correctly. 1 mark for 'methyl propanoate'.

(a) The solution turns from orange to yellow. Adding NaOH introduces hydroxide ions, which neutralize the hydrogen ions to form water, decreasing [H+]. By Le Chatelier's principle, the equilibrium shifts to the left to produce more hydrogen ions, leading to an increase in yellow chromate ions. (b) The equilibrium position shifts to the right (product side). Adding HCl increases [H+]. By Le Chatelier's principle, the system shifts to the right to consume the added H+ ions.

(a) 1 mark for stating color change 'orange to yellow'. 1 mark for explaining that OH- reacts with H+, decreasing [H+], shifting equilibrium to the left. (b) 1 mark for stating 'shifts to the right'. 1 mark for explaining that adding H+ shifts equilibrium to the right to consume H+.

(a) Equation: \(3\text{C}(graphite) + 4\text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{C}_3\text{H}_7\text{OH}(l)\). (b) Using Hess's law: \(\Delta H_f^\ominus[\text{C}_3\text{H}_7\text{OH}(l)] = 3 \times \Delta H_c^\ominus[\text{C}] + 4 \times \Delta H_c^\ominus[\text{H}_2] - \Delta H_c^\ominus[\text{C}_3\text{H}_7\text{OH}]\). Calculation: \(\Delta H_f^\ominus = 3(-393.5) + 4(-285.8) - (-2021.0) = -1180.5 - 1143.2 + 2021.0 = -302.7\text{ kJ mol}^{-1}\).

(a) 1 mark for correct balanced chemical equation with state symbols. (b) 1 mark for showing correct cycle setup or algebraic formula. 1 mark for calculating correct intermediate numbers. 1 mark for correct final answer of -302.7 kJ mol-1 with sign and units.

(a) Chlorine gas is formed at the anode. Half-equation: \(2\text{Cl}^-(aq) \rightarrow \text{Cl}_2(g) + 2e^-\). (b) At the cathode, hydrogen ions (from the self-ionization of water) are preferentially discharged to form hydrogen gas: \(2\text{H}^+(aq) + 2e^- \rightarrow \text{H}_2(g)\). As hydrogen ions are continuously consumed, the concentration of hydroxide ions remaining in the solution increases, making the solution around the cathode alkaline.

(a) 1 mark for chlorine gas (or Cl2). 1 mark for correct half-equation. (b) 1 mark for stating that hydrogen ions are preferentially discharged to form hydrogen gas. 1 mark for explaining that the loss of H+ leaves an excess of OH- ions, making the solution alkaline.

(a) The graph should show two curves starting from the origin (0,0) and leveling off at the exact same final volume of carbon dioxide. The curve for 'powder' must have a steeper initial gradient than 'large chips' and reach the plateau earlier. (b) Powdered calcium carbonate has a much larger total surface area exposed to the acid. This increases the contact area and results in a higher frequency of collisions between reactant particles (calcium carbonate and H+ ions), which leads to a higher frequency of effective collisions per unit time, resulting in a faster initial reaction rate.

(a) 1 mark for both curves starting at the origin and leveling off at the same final volume. 1 mark for the 'powder' curve being steeper and leveling off earlier than the 'large chips' curve. (b) 1 mark for stating that powder has a larger surface area. 1 mark for explaining that this increases the frequency of collisions / effective collisions between reactant particles.

(a) Silicon dioxide has a giant covalent network structure. Each silicon atom is strongly bonded covalently to four oxygen atoms, and each oxygen to two silicon atoms. To melt silicon dioxide, a vast amount of energy is required to break these strong covalent bonds throughout the giant network. (b) Carbon dioxide has a simple molecular structure. Individual CO2 molecules are held together by weak intermolecular forces (Van der Waals' forces), while the carbon and oxygen atoms within each molecule are held by strong covalent bonds. Only a very small amount of energy is needed to overcome these weak intermolecular forces, causing it to sublime at low temperature.

(a) 1 mark for identifying 'giant covalent network structure'. 1 mark for stating that a large amount of energy is required to break strong covalent bonds. (b) 1 mark for identifying 'simple molecular structure'. 1 mark for stating that only a small amount of energy is needed to overcome weak intermolecular forces / Van der Waals' forces.

(a) Monomer structure shows a C=C double bond with one hydrogen atom on one carbon replaced by a phenyl ring (benzene ring), \(\text{CH}_2=\text{CH}(\text{C}_6\text{H}_5)\). (b) Repeating unit structure shows a single C-C bond with open bonds extending from both carbons, and the phenyl ring attached to one carbon, \(-[\text{CH}_2-\text{CH}(\text{C}_6\text{H}_5)]-\). (c) Environmental problem: Polystyrene is non-biodegradable, persisting in landfills for hundreds of years and causing severe terrestrial/marine plastic pollution. Green solution: Promote catalytic pyrolysis to depolymerize polystyrene back into monomers for recycling, or replace polystyrene packaging with biodegradable materials (such as starch-based foam).

(a) 1 mark for correct structure of phenylethene showing C=C bond and phenyl ring. (b) 1 mark for correct structure of the repeating unit with open bonds and single bonds. (c) 1 mark for a valid environmental problem (e.g., non-biodegradability, landfill accumulation, toxic gases on combustion). 1 mark for a corresponding green solution (e.g., recycling/pyrolysis, replacing with biodegradable plastics/starch-based packaging).

(a) \(\text{CaCO}_3\text{(s)} + 2\text{HCl(aq)} \rightarrow \text{CaCl}_2\text{(aq)} + \text{CO}_2\text{(g)} + \text{H}_2\text{O(l)}\)
(b) From colorless to pale pink.
(c) Boiling removes dissolved carbon dioxide / carbonic acid from the mixture. If not removed, the dissolved carbon dioxide would react with \(\text{NaOH}\) during the titration, causing an overestimation of the remaining acid and thus underestimating the percentage of \(\text{CaCO}_3\).
(d)
Moles of \(\text{NaOH}\) used = \(0.125\text{ mol dm}^{-3} \times 0.02240\text{ dm}^3 = 0.00280\text{ mol}\)
Moles of excess \(\text{HCl}\) in \(25.0\text{ cm}^3\) aliquot = \(0.00280\text{ mol}\)
Moles of excess \(\text{HCl}\) in \(250.0\text{ cm}^3\) volumetric flask = \(0.00280 \times 10 = 0.0280\text{ mol}\)
Moles of initial \(\text{HCl}\) added = \(1.00\text{ mol dm}^{-3} \times 0.0500\text{ dm}^3 = 0.0500\text{ mol}\)
Moles of \(\text{HCl}\) reacted with \(\text{CaCO}_3\) = \(0.0500 - 0.0280 = 0.0220\text{ mol}\)
Moles of \(\text{CaCO}_3\) in eggshell = \(0.0220 / 2 = 0.0110\text{ mol}\)
Mass of \(\text{CaCO}_3\) = \(0.0110\text{ mol} \times 100.1\text{ g mol}^{-1} = 1.101\text{ g}\)
Percentage by mass of \(\text{CaCO}_3\) = \((1.101\text{ g} / 1.50\text{ g}) \times 100\% = 73.4\%\)
(e) Source of error: The eggshell may contain other acid-reactive impurities (such as \(\text{MgCO}_3\)), which would lead to a higher calculated percentage by mass of \(\text{CaCO}_3\); OR incomplete drying of the eggshell sample, which would lead to a lower calculated percentage by mass.

(a) Correct balanced equation: \(\text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{CO}_2 + \text{H}_2\text{O}\) (1)
(b) Correct color change: colorless to pale pink / pink (1)
(c) State that boiling removes dissolved carbon dioxide (1); explain that carbon dioxide reacts with sodium hydroxide to cause titration error (1)
(d) Correct calculation of moles of \(\text{NaOH}\) / excess \(\text{HCl}\) in aliquot (1); correct calculation of excess \(\text{HCl}\) in volumetric flask (1); correct calculation of reacted \(\text{HCl}\) (1); correct stoichiometry ratio (1:2) and moles of \(\text{CaCO}_3\) (1); correct mass and final percentage with 3 significant figures (1)
(e) Identify a plausible source of error (e.g., acid-reacting impurities or incomplete drying) (1); correctly state its effect on the final percentage (1)

(a)
1. Since A is an ester with molecular formula \(\text{C}_5\text{H}_{10}\text{O}_2\), its acid hydrolysis produces carboxylic acid B and alcohol C.
2. Oxidation of alcohol C yields ketone D, which implies that C must be a secondary alcohol.
3. The simplest secondary alcohol is propan-2-ol (3 carbons), so C is \(\text{CH}_3\text{CH(OH)CH}_3\) and D is propanone (\(\text{CH}_3\text{COCH}_3\)).
4. Since A has 5 carbons and C has 3 carbons, B must have 2 carbons. Thus, B is ethanoic acid (\(\text{CH}_3\text{COOH}\)).
5. Ester A, formed from B and C, is isopropyl ethanoate (or 1-methylethyl ethanoate, \(\text{CH}_3\text{COOCH(CH}_3)_2\)).

(b) Add sodium hydrogencarbonate solution to separate samples of B and D. B (ethanoic acid) will react to produce effervescence (colorless gas that turns limewater cloudy), whereas D (propanone) will show no observable change.

(c)
(i) Esterification (or condensation).
(ii) Concentrated sulfuric acid acts as a catalyst (to increase the reaction rate) and a dehydrating agent (to absorb water and shift the equilibrium position to the right to increase ester yield).
(iii) The reactants (propan-2-ol) and products (isopropyl ethanoate) are highly volatile and flammable. Direct heating with a Bunsen burner poses a severe fire hazard, whereas a water bath provides safe, even heating.

(a) Deduce C is a secondary alcohol from the fact that its oxidation yields a ketone (1); Deduce that C has 3 carbons and B has 2 carbons (1); Correct structural formula/name for B (ethanoic acid) (1); Correct structural formula/name for C (propan-2-ol) (1); Correct structural formula/name for A (isopropyl ethanoate) and D (propanone) (1)
(b) State a suitable reagent (e.g. \(\text{NaHCO}_3\text{(aq)}\) or \(\text{Na}_2\text{CO}_3\text{(aq)}\) or Mg ribbon) (1); Describe correct observations for B and D (1)
(c) (i) Esterification / condensation (1)
(ii) Catalyst (1); Dehydrating agent (1)
(iii) To prevent the ignition of volatile and flammable reactants/products by open flames (1)

(a) \(K_c = \frac{[\text{SO}_3]^2}{[\text{SO}_2]^2 [\text{O}_2]}\)
Unit: \(\text{dm}^3\text{ mol}^{-1}\) (or \(\text{mol}^{-1}\text{ dm}^3\))

(b)
Let us calculate the equilibrium moles using an ICE table:
Reaction: \(2\text{SO}_2\text{(g)} + \text{O}_2\text{(g)} \rightleftharpoons 2\text{SO}_3\text{(g)}\)
Initial moles: \(\text{SO}_2 = 0.60\), \(\text{O}_2 = 0.40\), \(\text{SO}_3 = 0\)
Change in moles of \(\text{O}_2\) = \(0.16 - 0.40 = -0.24\text{ mol}\)
Change in moles of \(\text{SO}_2\) = \(2 \times (-0.24) = -0.48\text{ mol}\)
Change in moles of \(\text{SO}_3\) = \(2 \times (+0.24) = +0.48\text{ mol}\)

Equilibrium moles:
\(\text{SO}_2 = 0.60 - 0.48 = 0.12\text{ mol}\)
\(\text{O}_2 = 0.16\text{ mol}\)
\(\text{SO}_3 = 0.48\text{ mol}\)

Since the volume of the vessel is \(2.0\text{ dm}^3\), the equilibrium concentrations are:
\([\text{SO}_2] = 0.12\text{ mol} / 2.0\text{ dm}^3 = 0.060\text{ mol dm}^{-3}\)
\([\text{O}_2] = 0.16\text{ mol} / 2.0\text{ dm}^3 = 0.080\text{ mol dm}^{-3}\)
\([\text{SO}_3] = 0.48\text{ mol} / 2.0\text{ dm}^3 = 0.24\text{ mol dm}^{-3}\)

Substitute concentrations into the \(K_c\) expression:
\(K_c = \frac{(0.24)^2}{(0.060)^2 \times (0.080)} = \frac{0.0576}{0.0036 \times 0.080} = \frac{0.0576}{0.000288} = 200\)

(c)
(i) Position of equilibrium: Shifts to the right (product side).
Explanation: Decreasing volume increases the pressure of the system. According to Le Chatelier's Principle, the system shifts to decrease the pressure by moving towards the side with fewer number of gaseous moles (from 3 moles on the left to 2 moles on the right).
Effect on \(K_c\): No change, because temperature is kept constant.

(ii) Position of equilibrium: Shifts to the left (reactant side).
Explanation: The forward reaction is exothermic (\(\Delta H < 0\)). According to Le Chatelier's Principle, increasing temperature shifts the equilibrium in the endothermic direction (reverse reaction) to absorb heat.
Effect on \(K_c\): Decreases.

(a) Correct \(K_c\) expression (1); Correct unit (1)
(b) Correct calculation of change in moles of \(\text{O}_2\) and equilibrium moles of \(\text{SO}_2\) and \(\text{SO}_3\) (1); Correct calculation of all three equilibrium concentrations (1); Correct substitution into \(K_c\) expression (1); Correct answer of 200 (1)
(c) (i) Correctly state equilibrium shifts right and explain in terms of pressure/gas moles (1); Correctly state \(K_c\) is unchanged (1); Explain that \(K_c\) is temperature-dependent only (1)
(ii) Correctly state equilibrium shifts left because forward reaction is exothermic (1); Correctly state \(K_c\) decreases (1)

(a) Anode reaction (oxidation of methanol):
\(\text{CH}_3\text{OH(aq)} + \text{H}_2\text{O(l)} \rightarrow \text{CO}_2\text{(g)} + 6\text{H}^+\text{(aq)} + 6\text{e}^-\)
(b) Cathode reaction (reduction of oxygen):
\(\text{O}_2\text{(g)} + 4\text{H}^+\text{(aq)} + 4\text{e}^- \rightarrow 2\text{H}_2\text{O(l)}\)
(c) Overall cell reaction:
\(2\text{CH}_3\text{OH(aq)} + 3\text{O}_2\text{(g)} \rightarrow 2\text{CO}_2\text{(g)} + 4\text{H}_2\text{O(l)}\)
(d) Electrons flow from the anode (negative electrode) to the cathode (positive electrode) through the external circuit. At the anode, methanol undergoes oxidation and releases electrons; at the cathode, oxygen undergoes reduction and gains electrons.
(e)
Advantage: Methanol is a liquid at room temperature and pressure, which is much easier and safer to store, handle, and transport than gaseous hydrogen.
Disadvantage: DMFCs produce carbon dioxide (a greenhouse gas) during operation, whereas hydrogen fuel cells produce only water; OR DMFCs have lower efficiency due to fuel crossover through the proton exchange membrane.
(f)
Total charge transferred, \(Q = I \times t = 0.50\text{ A} \times (3.0 \times 3600\text{ s}) = 5400\text{ C}\)
Moles of electrons, \(n(\text{e}^-) = Q / F = 5400 / 96500 = 0.05596\text{ mol}\)
According to the anode half-equation, \(1\text{ mol}\) of \(\text{CH}_3\text{OH}\) releases \(6\text{ mol}\) of \(\text{e}^-\).
Moles of methanol required, \(n(\text{CH}_3\text{OH}) = 0.05596 / 6 = 0.009326\text{ mol}\)
Minimum mass of methanol = \(0.009326\text{ mol} \times 32.0\text{ g mol}^{-1} = 0.298\text{ g}\) (or \(0.30\text{ g}\))

(a) Correct species (1); Correctly balanced in acidic medium (1)
(b) Correct species (1); Correctly balanced in acidic medium (1)
(c) Correctly balanced overall reaction (1)
(d) Flow: anode to cathode in the external circuit (1); Reason: Oxidation releases electrons at the anode, reduction consumes electrons at the cathode (1)
(e) Any one valid advantage (e.g. liquid state, easier storage/transport) (1); Any one valid disadvantage (e.g. CO2 greenhouse gas, lower efficiency, membrane crossover) (1)
(f) Correct calculation of charge Q and moles of electrons (1); Correct calculation of mass of methanol using 1:6 ratio (1)

(a) (i) Comparing Exp 1 and Exp 2: [B] is constant. When [A] is doubled, the initial rate increases by a factor of 4. Since \(2^2 = 4\), the order with respect to A is 2. Comparing Exp 2 and Exp 3: [A] is constant. When [B] is doubled, the initial rate increases by a factor of 2. Since \(2^1 = 2\), the order with respect to B is 1. Therefore, Rate = \(k[A]^2[B]\).
(ii) Using Exp 1: \(1.2 \times 10^{-3} = k (0.10)^2 (0.10) \Rightarrow k = 1.2 \text{ dm}^6\text{ mol}^{-2}\text{ s}^{-1}\).
(iii) Using Arrhenius equation: \(\ln(k_2/k_1) = \frac{E_a}{R} (1/T_1 - 1/T_2)\) \(\Rightarrow \ln(11.8/1.2) = \frac{E_a}{8.31} (1/298 - 1/318)\) \(\Rightarrow 2.2858 = \frac{E_a}{8.31} (2.11 \times 10^{-4}) \Rightarrow E_a = 90022 \text{ J mol}^{-1} \approx 90.0 \text{ kJ mol}^{-1}\).

(b) (i) Molar mass of DMC = \(90.0 \text{ g mol}^{-1}\). Route 1: Total mass of reactants = \(2 \times 32.0 + 99.0 = 163.0 \text{ g}\). AE = \((90.0 / 163.0) \times 100\% = 55.2\%\). Route 2: Total mass of reactants = \(2 \times 32.0 + 44.0 = 108.0 \text{ g}\). AE = \((90.0 / 108.0) \times 100\% = 83.3\%\).
(ii) Route 2 uses non-toxic, abundant \(\text{CO}_2\) instead of highly toxic phosgene (\(\text{COCl}_2\)). It produces harmless water (\(\text{H}_2\text{O}\)) as by-product instead of toxic and corrosive hydrogen chloride (\(\text{HCl}\)). It also helps reduce greenhouse gases by utilizing carbon dioxide.
(iii) (1) Heterogeneous catalyst. It can be easily separated from liquid products/reactants by filtration. (2) Continually remove water from the reaction mixture, or increase the reaction pressure/temperature to shift equilibrium forward.

(a)(i) Deduce order of reaction with respect to A is 2 (1) and with respect to B is 1 (1). Deduce rate equation: Rate = \(k[A]^2[B]\) (1).
(a)(ii) Calculate \(k = 1.2\) (1). Correct unit: \(\text{dm}^6\text{ mol}^{-2}\text{ s}^{-1}\) (1).
(a)(iii) Correct Arrhenius equation substitution (1). Correct calculation of ratio and temperature terms (1). Correct calculation of \(E_a\) value (90.0) (1). Correct unit \(\text{kJ mol}^{-1}\) (1).
(b)(i) Correct AE calculation for Route 1: 55.2% (2) (1 mark for total reactant mass 163.0). Correct AE calculation for Route 2: 83.3% (2) (1 mark for total reactant mass 108.0).
(b)(ii) Give 1.5 marks for each of the TWO safety/environmental advantages (Max 3 marks). Must compare Route 1 vs Route 2.
(b)(iii)(1) State 'heterogeneous catalyst' (1). State ease of separation by filtration over distillation (1).
(b)(iii)(2) Suggest one valid action (e.g. removing water, increasing pressure, or increasing temperature) (1) with explanation using Le Chatelier's principle (1).

(a) (i) \(\text{MnO}_4^-(aq) + 5\text{Fe}^{2+}(aq) + 8\text{H}^+(aq) \rightarrow \text{Mn}^{2+}(aq) + 5\text{Fe}^{3+}(aq) + 4\text{H}_2\text{O}(l)\).
(ii) From colorless to pale pink.
(iii) Moles of \(\text{MnO}_4^-\text{ used} = 0.0200 \times (22.50/1000) = 4.50 \times 10^{-4} \text{ mol}\).
From the equation, moles of \(\text{Fe}^{2+}\) in 25.00 cm\(^3\) = \(5 \times 4.50 \times 10^{-4} = 2.25 \times 10^{-3} \text{ mol}\).
Moles of \(\text{Fe}^{2+}\) in 250.0 cm\(^3\) = \(2.25 \times 10^{-3} \times (250.0 / 25.00) = 2.25 \times 10^{-2} \text{ mol}\).
Mass of Fe = \(2.25 \times 10^{-2} \times 55.8 = 1.256 \text{ g}\).
Percentage by mass of iron = \((1.256 / 5.00) \times 100\% = 25.1\%\).
(iv) Hydrochloric acid contains chloride (\(\text{Cl}^-\)) ions, which can be oxidized by \(\text{MnO}_4^-\). This would consume extra \(\text{MnO}_4^-\), leading to an overestimation of the iron content.

(b) (i) Carboxylic acid group (\(-\text{COOH}\)) is absent.
(ii) The strong absorption at 1735 cm\(^{-1}\) indicates a carbonyl group (\(\text{C}=\text{O}\)). Lacking 3230-3670 cm\(^{-1}\) indicates the absence of an \(-\text{O}-\text{H}\) group. Therefore, the functional group is an ester group (\(-\text{COO}-\)).
(iii) Two possible structures: Ethyl ethanoate (\(\text{CH}_3\text{COOCH}_2\text{CH}_3\)) and Methyl propanoate (\(\text{CH}_3\text{CH}_2\text{COOCH}_3\)).
Species for \(m/z = 43\): \([\text{CH}_3\text{CO}]^+\).
Species for \(m/z = 59\): \([\text{CH}_3\text{COO}]^+\) or \([\text{COOCH}_3]^+\).
(iv) Both show 3 signals with integration ratio 3:2:3 and splitting patterns: a singlet (3H), a quartet (2H), and a triplet (3H).
For ethyl ethanoate: the singlet (\(\text{CH}_3\text{CO}-\)) is at \(\delta \approx 2.0 \text{ ppm}\), and the quartet (\(-\text{OCH}_2-\)) is highly deshielded at \(\delta \approx 4.1 \text{ ppm}\).
For methyl propanoate: the singlet (\(-\text{OCH}_3\)) is highly deshielded at \(\delta \approx 3.7 \text{ ppm}\), and the quartet (\(-\text{CH}_2\text{CO}-\)) is at \(\delta \approx 2.3 \text{ ppm}\). This chemical shift difference clearly distinguishes them.

(a)(i) Complete balance of elements (1) and charge balance (1). State symbols not strictly required but must be correct if written.
(a)(ii) State 'colorless to pale pink' / 'colorless to pale purple' (1). Reject: purple, dark pink.
(a)(iii) Correct calculation of moles of \(\text{MnO}_4^-\) (1). Correct mole ratio calculation to find moles of Fe in 250 cm\(^3\) (1). Calculate mass of Fe (1.256 g) (1). Correct percentage by mass 25.1% (1).
(a)(iv) State that \(\text{Cl}^-\)\n ions are oxidized to \(\text{Cl}_2\) (1). Explain that this consumes extra titrant, leading to overestimation of iron content (1).
(b)(i) Deduce carboxylic acid (\(-\text{COOH}\)) is absent (1).
(b)(ii) Correct linkage of 1735 cm\(^{-1}\) to \(\text{C}=\text{O}\) (1) and lack of 3230-3670 cm\(^{-1}\) to lack of \(\text{O}-\text{H}\) (1). Deduce functional group is ester (1).
(b)(iii) Propose two correct ester structures (1, 1). Identify both fragment ions (with positive charges!) (1, 1).
(b)(iv) State that both show 3 signals with integration 3:2:3 (singlet, quartet, triplet) (1). Explain the key difference: for ethyl ethanoate, the quartet is at \(\delta \approx 4.1 \text{ ppm}\) and the singlet is at \(\delta \approx 2.0 \text{ ppm}\) (1). For methyl propanoate, the singlet is at \(\delta \approx 3.7 \text{ ppm}\) and the quartet is at \(\delta \approx 2.3 \text{ ppm}\) (1). (1 mark for chemical shift/deshielding reasoning based on electronegative oxygen atom).