2023 Cambridge IAL Chemistry (9701) Practice Paper with Answers

Ammonia is a stronger ligand than water and lies higher in the spectrochemical series. Therefore, substituting water molecules with ammonia ligands increases the d-orbital splitting energy, \(\Delta E\).

Since the energy difference is related to the wavelength of absorbed light by the equation:

\(\Delta E = \frac{hc}{\lambda}\)

an increase in \(\Delta E\) results in a decrease in the wavelength of light absorbed, \(\lambda\). The solution appears deep blue because it now absorbs higher-energy yellow-orange light (shorter wavelength) rather than the lower-energy red-orange light (longer wavelength) absorbed by the hexaaqua complex.

Award 1 mark for identifying that:
- Ammonia is a stronger ligand causing a larger splitting energy (\(\Delta E\)).
- An increase in \(\Delta E\) corresponds to a decrease in the wavelength of light absorbed (\(\lambda\)).

In the coupling reaction, the benzenediazonium ion, \(\text{C}_6\text{H}_5\text{N}_2^+\), acts as a weak electrophile. Phenol is a relatively weak nucleophile, so it is dissolved in aqueous sodium hydroxide to deprotonate the hydroxyl group, forming the phenoxide ion, \(\text{C}_6\text{H}_5\text{O}^-\).

The phenoxide ion is a much stronger nucleophile than phenol because the negatively charged oxygen increases electron density in the benzene ring via resonance, facilitating rapid electrophilic substitution by the weak diazonium electrophile.

Award 1 mark for identifying both:
- The electrophile as \(\text{C}_6\text{H}_5\text{N}_2^+\).
- The purpose of aqueous \(\text{NaOH}\) as deprotonating phenol to the more reactive phenoxide ion.

Step 1: Methylbenzene is reacted with chlorine gas in the presence of UV light to undergo free-radical substitution on the side chain, producing chloromethylbenzene, \(\text{C}_6\text{H}_5\text{CH}_2\text{Cl}\).
Step 2: Chloromethylbenzene is heated under reflux with potassium cyanide (\(\text{KCN}\)) in ethanol. This undergoes nucleophilic substitution to yield phenylethanenitrile, \(\text{C}_6\text{H}_5\text{CH}_2\text{CN}\).
Step 3: Phenylethanenitrile is heated under reflux with dilute sulfuric acid, \(\text{H}_2\text{SO}_4(\text{aq})\), which hydrolyzes the nitrile group to form the carboxylic acid, \(\text{C}_6\text{H}_5\text{CH}_2\text{COOH}\).

Other options are incorrect:
- Option A uses a Lewis acid catalyst in Step 1, which leads to ring substitution (chlorination of the benzene ring).
- Option C uses \(\text{HCN}/\text{NaCN}\) in Step 2, which is for nucleophilic addition to carbonyls, not substitution.
- Option D oxidizes the methyl group to benzoic acid in Step 1, meaning the extra carbon atom cannot be easily added in subsequent steps to form phenylethanoic acid.

Award 1 mark for selecting the correct three-step sequence involving side-chain chlorination, nucleophilic substitution with cyanide, and acid hydrolysis of the nitrile.

For a reaction to be feasible, the standard Gibbs free energy change must be negative:

\(\Delta G^\ominus = \Delta H^\ominus - T\Delta S^\ominus < 0\)

Substitute the given values (converting enthalpy to \(\text{J mol}^{-1}\)):

\(178\,000\text{ J mol}^{-1} - T(161\text{ J K}^{-1}\text{ mol}^{-1}) < 0\)

\(178\,000 < 161\,T\)

\(T > \frac{178\,000}{161} \approx 1105.6\text{ K}\)

Therefore, the reaction is only feasible at temperatures above \(1106\text{ K}\).

Award 1 mark for converting units of enthalpy to J, applying the Gibbs equation, and concluding that the reaction is feasible above the calculated temperature of 1106 K.

1. Comparing Exp 1 and Exp 2: \([\text{A}]\) doubles, while \([\text{B}]\) and \([\text{C}]\) are constant. The rate doubles (\(1.2 \times 10^{-3}\) to \(2.4 \times 10^{-3}\)). Therefore, the reaction is first-order with respect to \([\text{A}]\).
2. Comparing Exp 1 and Exp 3: \([\text{B}]\) doubles, while \([\text{A}]\) and \([\text{C}]\) are constant. The rate quadruples (\(1.2 \times 10^{-3}\) to \(4.8 \times 10^{-3}\)). Therefore, the reaction is second-order with respect to \([\text{B}]\).
3. Comparing Exp 1 and Exp 4: \([\text{C}]\) doubles, while \([\text{A}]\) and \([\text{B}]\) are constant. The rate does not change. Therefore, the reaction is zero-order with respect to \([\text{C}]\).

Thus, the rate equation is:

\(\text{Rate} = k[\text{A}][\text{B}]^2\)

To find the units of \(k\):

\(k = \frac{\text{Rate}}{[\text{A}][\text{B}]^2}\)

Units = \(\frac{\text{mol dm}^{-3}\text{ s}^{-1}}{(\text{mol dm}^{-3})(\text{mol dm}^{-3})^2} = \text{dm}^6\text{ mol}^{-2}\text{ s}^{-1}\).

Award 1 mark for correctly determining the orders of reaction for A, B, and C, formulating the correct rate equation, and deriving the unit of \(k\).

The stability constant of the colorless hexafluoroferrate(III) complex, \([\text{FeF}_6]^{3-}\), is many orders of magnitude larger (\(1.0 \times 10^{15}\)) than that of the blood-red thiocyanatopentaaquairon(III) complex, \([\text{Fe}(\text{SCN})(\text{H}_2\text{O})_5]^{2+}\) (\(1.0 \times 10^3\)).

When fluoride ions are added in excess, they successfully and almost completely displace both the water and thiocyanate ligands from the iron(III) coordination sphere. This shifts the ligand exchange equilibrium to produce the highly thermodynamically stable \([\text{FeF}_6]^{3-}\) complex, resulting in a decrease in the concentration of the red complex, causing the blood-red color to fade.

Award 1 mark for explaining that the fluoride ligand successfully displaces the thiocyanate ligand due to the much larger stability constant of the fluoride complex.

At a highly alkaline pH of 12:
1. Both acidic carboxylic acid groups (\(-\text{COOH}\)) are fully deprotonated to form negative carboxylate groups (\(-\text{COO}^-\)).
2. The basic amino group remains in its neutral unprotonated form (\(-\text{NH}_2\)), as there is a very low concentration of hydrogen ions available to protonate it.

Therefore, the predominant species has a net charge of \(-2\) and has the structure: \(^-\text{OOC}-\text{CH}_2-\text{CH}(\text{NH}_2)-\text{COO}^-\).

Award 1 mark for identifying that both carboxylic acid groups are deprotonated and the amine group remains neutral at high pH.

According to Hess's Law, the enthalpy of formation can be equated to the sum of the enthalpy changes in the cycle:

\(\Delta H_{\text{f}}^\ominus = \Delta H_{\text{at}}^\ominus(\text{Mg}) + \text{IE}_1(\text{Mg}) + \text{IE}_2(\text{Mg}) + 2\Delta H_{\text{at}}^\ominus(\text{Cl}) + 2\text{EA}_1(\text{Cl}) + \Delta H_{\text{latt}}^\ominus\)

Substitute the given values into the equation:

\(-641 = +148 + 738 + 1451 + 2(+121) + 2(-349) + \Delta H_{\text{latt}}^\ominus\)

\(-641 = 148 + 738 + 1451 + 242 - 698 + \Delta H_{\text{latt}}^\ominus\)

\(-641 = 1881 + \Delta H_{\text{latt}}^\ominus\)

\(\Delta H_{\text{latt}}^\ominus = -641 - 1881 = -2522\text{ kJ mol}^{-1}\).

Award 1 mark for the correct calculation of the lattice energy, accounting for stoichiometric factors of 2 for both chlorine atomisation and chlorine electron affinity.

The purple color of the \([Ti(H_2O)_6]^{3+}\) complex is due to the absorption of light in the green-yellow region of the visible spectrum. The energy of this absorbed light matches the energy gap between the split d-orbitals, promoting a d-electron from a lower energy d-orbital to a higher energy d-orbital (d-d transition). The remaining non-absorbed light (the complementary color, which is purple) is transmitted or reflected, which is what we see.

1 mark for the correct option (A). Reject options suggesting emission of light for the absorption spectrum or incorrect electron transition directions.

To synthesize the orange azo dye 4-hydroxyphenylazobenzene, benzenediazonium chloride is coupled with phenol. This reaction must be carried out in alkaline conditions (such as in aqueous sodium hydroxide) at a low temperature (around \(5^\circ\text{C}\)) to prevent the thermal decomposition of the unstable diazonium ion into phenol and nitrogen gas.

1 mark for identifying the correct coupling partner (phenol) and the necessary alkaline, low-temperature conditions (A).

Step 1: Propan-1-ol is oxidized to propanal using acidified potassium dichromate(VI) with immediate distillation to prevent further oxidation. Step 2: Propanal reacts with hydrogen cyanide (using sodium cyanide as a catalyst/buffer to provide CN- nucleophiles) to undergo nucleophilic addition, forming 2-hydroxybutanenitrile. Step 3: Acid hydrolysis of the nitrile group using dilute hydrochloric acid under reflux yields 2-hydroxybutanoic acid.

1 mark for the correct synthetic route involving oxidation, nucleophilic addition, and acid hydrolysis (A).

Using Hess's Law for a Born-Haber cycle:
\(\Delta H^\theta_f = \Delta H^\theta_\text{at}(\text{Mg}) + \text{IE}_1(\text{Mg}) + \text{IE}_2(\text{Mg}) + E(\text{F-F}) + 2 \times \text{EA}(\text{F}) + \Delta H^\theta_\text{latt}\)

Note that forming 2 moles of fluoride ions from fluorine gas requires 1 mole of F-F bond enthalpies (since \(\text{F}_2 \rightarrow 2\text{F}\)) and 2 moles of electron affinities of F.

Substituting the values:
\(-1124 = 148 + 738 + 1451 + 158 + 2(-328) + \Delta H^\theta_\text{latt}\)
\(-1124 = 2495 - 656 + \Delta H^\theta_\text{latt}\)
\(-1124 = 1839 + \Delta H^\theta_\text{latt}\)
\(\Delta H^\theta_\text{latt} = -1124 - 1839 = -2963 \text{ kJ mol}^{-1}\)

1 mark for the correct calculation of lattice energy with the correct sign (-2963 kJ/mol) (A).

Comparing Exp 1 and Exp 2: [Q] is constant, [P] doubles, and the rate increases by a factor of 4 (\(1.2 \times 10^{-3} / 3.0 \times 10^{-4} = 4\)). Thus, the reaction is second order with respect to P.

Comparing Exp 1 and Exp 3: [P] is constant, [Q] doubles, and the rate increases by a factor of 2 (\(6.0 \times 10^{-4} / 3.0 \times 10^{-4} = 2\)). Thus, the reaction is first order with respect to Q.

The rate equation is: \(\text{Rate} = k[\text{P}]^2[\text{Q}]^1\).

Using the data from Exp 1 to calculate \(k\):
\(3.0 \times 10^{-4} \text{ mol dm}^{-3} \text{ s}^{-1} = k (0.10 \text{ mol dm}^{-3})^2 (0.10 \text{ mol dm}^{-3})\)
\(3.0 \times 10^{-4} = k (1.0 \times 10^{-3})\)
\(k = 0.30\).

The units of \(k\) are:
\(\text{units of } k = \frac{\text{mol dm}^{-3} \text{ s}^{-1}}{(\text{mol dm}^{-3})^3} = \text{mol}^{-2} \text{ dm}^6 \text{ s}^{-1}\).

1 mark for identifying the correct orders of reaction, calculating the numerical value of 0.30, and deriving the correct units (A).

The stability constant, \(K_\text{stab}\), is an equilibrium constant for the formation of a complex ion from its constituent ions or molecules in solvent. In aqueous solutions, the concentration of water is extremely high and remains essentially constant, so it is omitted from the equilibrium expression. Therefore, the expression is: \(K_\text{stab} = \frac{[[\text{CuCl}_4]^{2-}]}{[[\text{Cu(H}_2\text{O)}_6]^{2+}][\text{Cl}^-]^4}\).

1 mark for the correct stability constant expression excluding water (A).

At a highly acidic pH of 1, the high concentration of hydronium ions (\(\text{H}^+\)) shifts protonation equilibria to the left. The carboxylate group is protonated to form carboxylic acid (\(\text{-COOH}\)) and the amine group is protonated to form an ammonium ion (\(\text{-NH}_3^+\)). This results in a net positive charge on the molecule: \(\text{CH}_3\text{CH(NH}_3^+)\text{COOH}\).

1 mark for identifying the fully protonated cationic form of alanine at low pH (A).

1. Benzene undergoes electrophilic substitution with conc. \(\text{HNO}_3\) and conc. \(\text{H}_2\text{SO}_4\) at \(55^\circ\text{C}\) to form nitrobenzene (Compound X).
2. Nitrobenzene is reduced by Sn and concentrated \(\text{HCl}\) (followed by addition of aq. \(\text{NaOH}\)) to form phenylamine (Compound Y).
3. Phenylamine reacts with ethanoyl chloride (\(\text{CH}_3\text{COCl}\)) via nucleophilic addition-elimination to form N-phenylethanamide (Compound Z).

1 mark for correctly identifying phenylamine as Compound Y and N-phenylethanamide as Compound Z (A).

The complex \([\text{Co}(\text{en})_2\text{Cl}_2]^+\) can exist as both cis and trans isomers:
1. In the **trans-isomer**, the two chloride ligands are opposite to each other (at an angle of \(180^\circ\)). This molecule has a plane of symmetry, making it achiral and therefore optically inactive (ruling out options A, C, and D).
2. In the **cis-isomer**, the two chloride ligands are adjacent to each other (at an angle of \(90^\circ\)). This arrangement lacks a plane of symmetry (is asymmetric), making it chiral. Thus, the cis-isomer exists as a pair of non-superimposable mirror images (enantiomers) which are optically active (confirming option B).

Award 1 mark for identifying that the cis-isomer lacks a plane of symmetry and therefore exists as a pair of optically active enantiomers (option B).

1. **Determine the order with respect to A:** Comparing Exp 1 and Exp 2, \([\text{B}]\) is kept constant at \(0.10 \text{ mol dm}^{-3}\), while \([\text{A}]\) is doubled from \(0.10\) to \(0.20 \text{ mol dm}^{-3}\). The rate increases from \(1.2 \times 10^{-3}\) to \(4.8 \times 10^{-3} \text{ mol dm}^{-3} \text{ s}^{-1}\) (a four-fold increase: \(2^2 = 4\)). Therefore, the reaction is second order with respect to \(\text{A}\).
2. **Determine the order with respect to B:** Comparing Exp 1 and Exp 3, \([\text{A}]\) is kept constant at \(0.10 \text{ mol dm}^{-3}\), while \([\text{B}]\) is tripled from \(0.10\) to \(0.30 \text{ mol dm}^{-3}\). The rate remains unchanged at \(1.2 \times 10^{-3} \text{ mol dm}^{-3} \text{ s}^{-1}\). Therefore, the reaction is zero order with respect to \(\text{B}\).
3. **Rate Equation:** \(\text{Rate} = k[\text{A}]^2\)
4. **Determine the units of the rate constant, \(k\):**
\[k = \frac{\text{Rate}}{[\text{A}]^2} = \frac{\text{mol dm}^{-3} \text{ s}^{-1}}{(\text{mol dm}^{-3})^2} = \text{dm}^3 \text{ mol}^{-1} \text{ s}^{-1}\]

This perfectly matches option B.

Award 1 mark for deducing the correct rate equation (second order in A, zero order in B) and deriving the corresponding units for a second-order rate constant as \(\text{dm}^3 \text{ mol}^{-1} \text{ s}^{-1}\) (option B).

To synthesize an azo dye from phenylamine:
* **Step 1:** Diazotisation. Phenylamine is reacted with nitrous acid (\(\text{HNO}_2\)). Because nitrous acid is unstable, it must be generated *in situ* using sodium nitrite (\(\text{NaNO}_2\)) and an excess of dilute hydrochloric acid (\(\text{HCl}\)). The temperature must be kept below \(10^\circ\text{C}\) (typically \(5^\circ\text{C}\)) to prevent the benzenediazonium chloride from decomposing into phenol and nitrogen gas. Thus, option A is correct.
* Option B describes nitration conditions for benzene, which would not produce the diazonium ion.
* Option C uses alkaline conditions and too high a temperature, which does not form the diazonium salt.
* Option D does not provide the nitrous acid required for diazotisation.

Award 1 mark for selecting the correct reagents (\(\text{NaNO}_2\text{(aq)}\) and \(\text{HCl(aq)}\)) and low temperature conditions (\(5^\circ\text{C}\)) required for diazotisation (option A).

Using Hess's Law and a Born-Haber cycle for \(\text{CaCl}_2(\text{s})\):
\[\Delta H_{\text{f}}^{\ominus}[\text{CaCl}_2(\text{s})] = \Delta H_{\text{at}}^{\ominus}(\text{Ca}) + \text{IE}_1(\text{Ca}) + \text{IE}_2(\text{Ca}) + 2 \times \Delta H_{\text{at}}^{\ominus}(\text{Cl}) + 2 \times \text{EA}_1(\text{Cl}) + \Delta H_{\text{latt}}^{\ominus}[\text{CaCl}_2(\text{s})]\]

Substitute the given values into the equation:
\[-796 = +178 + 590 + 1145 + 2(+121) + 2(-349) + \Delta H_{\text{latt}}^{\ominus}\]
\[-796 = 178 + 590 + 1145 + 242 - 698 + \Delta H_{\text{latt}}^{\ominus}\]
\[-796 = 1457 + \Delta H_{\text{latt}}^{\ominus}\]
\[\Delta H_{\text{latt}}^{\ominus} = -796 - 1457 = -2253 \text{ kJ mol}^{-1}\]

Thus, the lattice energy of \(\text{CaCl}_2(\text{s})\) is \(-2253 \text{ kJ mol}^{-1}\) (Option A).

Award 1 mark for the correct calculation, ensuring the stoichiometric coefficients of 2 for both the atomisation and electron affinity of chlorine are correctly applied (option A).

To convert propan-1-ol (3 carbons) to butanoic acid (4 carbons), the carbon chain must be extended by one carbon atom. This is achieved using a nitrile intermediate:
* **Step 1:** Convert the alcohol to a halogenoalkane. Heating propan-1-ol with \(\text{HBr(aq)}\) forms 1-bromopropane:
\(\text{CH}_3\text{CH}_2\text{CH}_2\text{OH} + \text{HBr} \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{Br} + \text{H}_2\text{O}\).
* **Step 2:** Nucleophilic substitution to extend the carbon chain. Heating 1-bromopropane with \(\text{KCN}\) in ethanol under reflux yields butanenitrile:
\(\text{CH}_3\text{CH}_2\text{CH}_2\text{Br} + \text{CN}^- \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{CN} + \text{Br}^-\).
* **Step 3:** Acid hydrolysis of the nitrile. Heating butanenitrile with dilute hydrochloric acid under reflux yields butanoic acid:
\(\text{CH}_3\text{CH}_2\text{CH}_2\text{CN} + \text{HCl} + 2\text{H}_2\text{O} \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{COOH} + \text{NH}_4\text{Cl}\).

This matches option B exactly.

Award 1 mark for choosing the synthetic route that correctly converts propan-1-ol to 1-bromopropane, introduces a nitrile group to extend the chain, and hydrolyses it to butanoic acid (option B).

The stability constant, \(K_{\text{stab}}\), is an equilibrium constant for the formation of a complex ion in solution.
* The general expression is the ratio of the concentration of the products to the concentration of the reactants, each raised to the power of their stoichiometric coefficients.
* Because water (\(\text{H}_2\text{O}\)) is the solvent, its concentration remains extremely high and virtually constant during the reaction. Therefore, \([\text{H}_2\text{O}]\) is omitted from the \(K_{\text{stab}}\) expression.
* This yields:
\[K_{\text{stab}} = \frac{[[\text{CoCl}_4]^{2-}]}{[[\text{Co}(\text{H}_2\text{O})_6]^{2+}][\text{Cl}^-]^4}\]

This corresponds to option B.

Award 1 mark for selecting the correct \(K_{\text{stab}}\) expression where the concentration of water is omitted and reactants/products are in their correct stoichiometric powers (option B).

At different pH values, amino acids exist in different protonation states:
* At very low pH (highly acidic, e.g., pH 1), both the amino group and the carboxyl group are protonated: \(\text{CH}_3\text{CH}(\text{NH}_3^+)\text{COOH}\).
* At intermediate neutral pH (near the isoelectric point), the molecule exists as a zwitterion: \(\text{CH}_3\text{CH}(\text{NH}_3^+)\text{COO}^-\).
* At high pH (highly alkaline, e.g., pH 12), the excess hydroxide ions deprotonate both groups, leaving the carboxyl group as \(\text{COO}^-\) and the amino group as \(\text{NH}_2\). The resulting species is anionic: \(\text{CH}_3\text{CH}(\text{NH}_2)\text{COO}^-\).

Therefore, option C is correct.

Award 1 mark for identifying the correct anionic species formed by deprotonation of alanine in a strongly alkaline environment (option C).

Let's analyse the properties of compound \(\text{W}\):
1. It has the formula \(\text{C}_4\text{H}_9\text{NO}\).
2. It does not react with dilute acid/alkali at room temperature, indicating it does not contain highly reactive acidic (carboxylic acid) or basic (amine) functional groups. This rules out options A and C.
3. Upon refluxing with aqueous \(\text{NaOH}\) (alkaline hydrolysis), it undergoes decomposition to release an alkaline gas (which turns red litmus blue) and leaves a sodium carboxylate salt in solution. This behavior is characteristic of primary amides, which undergo hydrolysis to yield ammonia gas and a carboxylate salt:
\[\text{CH}_3\text{CH}_2\text{CH}_2\text{CONH}_2 + \text{NaOH} \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{COONa} + \text{NH}_3(\text{g})\]
The ammonia gas (\(\text{NH}_3\)) turns damp red litmus paper blue.
4. Butanamide (option B) has the molecular formula \(\text{C}_4\text{H}_9\text{NO}\) (4 carbons, 9 hydrogens, 1 nitrogen, 1 oxygen) and is a primary amide, which fits all the criteria.

Hence, option B is correct.

Award 1 mark for identifying that the properties described match those of an amide (butanamide) which undergoes alkaline hydrolysis to release ammonia gas (option B).

The octahedral complex ion [Co(en)2Cl2]+ exhibits stereoisomerism. Firstly, it displays cis-trans isomerism: 1. In the trans isomer, the two chloro ligands are directly opposite each other (bond angle 180 degrees). This isomer has a plane of symmetry and is therefore achiral (there is only 1 trans isomer). 2. In the cis isomer, the two chloro ligands are adjacent to each other (bond angle 90 degrees). The cis isomer lacks a plane of symmetry and is therefore chiral, existing as a pair of non-superimposable mirror images (enantiomers). This gives 2 cis isomers. Summing these, we have 1 trans isomer + 2 cis enantiomers = 3 stereoisomers in total.

Award 1 mark for identifying that there is 1 trans isomer and a pair of cis optical isomers (enantiomers), giving a total of 3 stereoisomers (Option B). Reject option A (ignores optical isomerism of cis) and option C (assumes trans is also chiral).

Base strength depends on the availability of the lone pair of electrons on the nitrogen atom to accept a proton. 1. Phenylamine is the weakest base because the nitrogen lone pair is partially delocalised into the benzene ring's pi-system, making it less available for donation. 2. Ammonia is stronger than phenylamine as it lacks this delocalising aromatic system. 3. Ethylamine is the strongest base of the three because the electron-releasing ethyl group (positive inductive effect) increases electron density on the nitrogen atom, making its lone pair more available to accept a proton. Thus, the order is: phenylamine < ammonia < ethylamine.

Award 1 mark for selecting the correct order of increasing basicity: phenylamine < ammonia < ethylamine (Option A).

Concentration is inversely proportional to volume (c = n / V). When the volume of the reaction vessel is halved, the concentration of all reactants is doubled. The new rate can be calculated as follows: rate_new = k (2[A]) (2[B])^2 = k (2[A]) (4[B]^2) = 8 k[A][B]^2. Therefore, the rate increases by a factor of 8.

Award 1 mark for the correct calculation showing that doubling concentrations for a first-order and second-order reactant increases the overall rate by a factor of 2^1 * 2^2 = 8 (Option C).

A reaction ceases to be feasible when the Gibbs free energy change, G, becomes positive. At the boundary of feasibility, G = H - T * S = 0, which means T = H / S. Converting H to Joules: H = -120000 J mol-1. Substituting the values: T = -120000 / -150 = 800 K. Since S is negative, the -T * S term is positive, making the reaction non-feasible (G > 0) at temperatures above 800 K.

Award 1 mark for converting units of H to Joules and correctly calculating the temperature boundary as 800 K (Option C). Reject Option A (no unit conversion, 0.8 K) and Option D.

Step 1 is a Friedel-Crafts alkylation: reacting benzene with chloromethane in the presence of an anhydrous aluminium chloride catalyst. Step 2 requires the oxidation of the methyl side-chain of methylbenzene to form benzoic acid. This requires a strong oxidizing agent, alkaline potassium manganate(VII), heated under reflux, followed by acidification with a dilute acid (acidified potassium dichromate is too weak to oxidize alkyl side chains). Step 3 is an esterification reaction: heating benzoic acid with ethanol in the presence of a concentrated sulfuric acid catalyst.

Award 1 mark for correctly identifying the correct reagents for the alkylation, alkyl side-chain oxidation, and esterification steps (Option B). Reject Option A because acidified dichromate cannot oxidize methylbenzene to benzoic acid.

The stability constant, Kstab, is the equilibrium constant for the formation of a complex ion from its component species in solution. Since water is the solvent and its concentration is extremely large and virtually constant, it is omitted from the Kstab expression. Therefore, Kstab = [[Fe(SCN)(H2O)5]2+] / ([[Fe(H2O)6]3+] * [SCN-]). The unit is determined by the concentration terms: (mol dm-3) / ((mol dm-3) * (mol dm-3)) = dm3 mol-1.

Award 1 mark for the correct Kstab expression (omitting water) and the correct calculated units of dm3 mol-1 (Option A).

N-ethylethanamide, CH3CONHCH2CH3, is a secondary amide. Hydrolysis under acidic conditions cleaves the amide C-N bond: 1. The acyl portion forms ethanoic acid, CH3COOH, which remains un-ionized in an acidic environment. 2. The amine portion forms ethylamine, CH3CH2NH2, which is basic and reacts with the excess hydrochloric acid to form the protonated ethylammonium cation, CH3CH2NH3+. Therefore, the organic products present in the mixture are CH3COOH and CH3CH2NH3+.

Award 1 mark for identifying that under acidic conditions the carboxylic acid is un-ionized (CH3COOH) and the amine is protonated (CH3CH2NH3+) (Option C). Reject Option A (amine cannot exist as a neutral molecule in acid) and Options B and D (carboxylate ion is only formed in alkaline hydrolysis).

Entropy is a measure of the disorder of a system. 1. Process A (H2O(g) -> H2O(l)) involves gas turning to liquid, representing a decrease in entropy (negative S). 2. Process B (CO2(s) -> CO2(g)) is sublimation, which turns a highly ordered solid directly into a highly disordered gas, creating a massive increase in entropy (highly positive S). 3. Process C (2NO2(g) -> N2O4(g)) reduces the number of gas moles, representing a decrease in entropy (negative S). 4. Process D (NaCl(s) -> NaCl(aq)) increases disorder but is far less disordered than producing a gas phase because ions remain constrained in solution. Thus, Process B has the most positive S.

Award 1 mark for identifying that the phase change from solid to gas has the largest positive entropy change (Option B). Reject option D (dissolving a solid has a lower positive entropy change than sublimation).

From Experiment 1 and 2, doubling [A] doubles the rate, so the reaction is first order with respect to A. From Experiment 1 and 3, doubling [B] quadruples the rate, so the reaction is second order with respect to B. The rate equation is Rate = k[A][B]^2. Rearranging gives k = Rate / ([A][B]^2). Substituting values from Experiment 1: k = 1.2 \times 10^{-4} / (0.10 \times 0.10^2) = 0.12. The unit of k is (mol dm^{-3} s^{-1}) / (mol dm^{-3} \times (mol dm^{-3})^2) = dm^6 mol^{-2} s^{-1}.

Award 1 mark for the correct calculation of the rate constant value (0.12) and determination of the correct units (dm^6 mol^-2 s^-1).

The desired reaction is obtained by subtracting Reaction 1 from Reaction 2. Therefore, the equilibrium constant for the new reaction is K_new = K_2 / K_1 = (1.0 \times 10^{16}) / (8.9 \times 10^2) = 1.12 \times 10^{13}. The units of K_new are (dm^{18} mol^{-6}) / (dm^3 mol^{-1}) = dm^{15} mol^{-5}.

Award 1 mark for dividing K2 by K1 to obtain 1.1 x 10^13 and correctly determining the units as dm^15 mol^-5.

In Step 1 (diazotisation), phenylamine is reacted with nitrous acid (generated in situ from sodium nitrate(III) and dilute hydrochloric acid) at temperatures below \(10\text{ }^\circ\text{C}\) (typically \(5\text{ }^\circ\text{C}\)) to prevent decomposition of the diazonium salt to phenol. In Step 2 (coupling), the benzenediazonium ion acts as an electrophile and reacts with 2-naphthol dissolved in an alkaline solution (which deprotonates 2-naphthol to form the highly reactive naphthoxide ion).

Award 1 mark for selecting the option with diazotisation at 5 degrees Celsius and coupling under alkaline conditions.

At a highly acidic pH of 1.0, both basic amino groups (one on the side-chain and one on the alpha-carbon) are fully protonated to form \(-NH_3^{+}\) groups. The acidic carboxyl group is in its protonated neutral form, \(-COOH\). Therefore, the overall species has a net charge of +2, as represented by \(^{+}H_3N-(CH_2)_4-CH(NH_3^{+})-COOH\).

Award 1 mark for correctly identifying the fully protonated form of lysine at pH 1.0.

Step 1 requires hot alkaline \(KMnO_4\) followed by acidification to oxidize the methyl group of methylbenzene to a carboxylic acid group (acidified \(K_2Cr_2O_7\) is not strong enough). Step 3 requires selective reduction of the nitro group without reducing the carboxylic acid group. Tin (Sn) and concentrated \(HCl\) selectively reduce the \(-NO_2\) group to \(-NH_3^{+}\), and addition of aqueous \(NaOH\) liberates the free amine \(-NH_2\) group. \(LiAlH_4\) is too strong and would reduce the \(-COOH\) group to a primary alcohol.

Award 1 mark for choosing the combination of alkaline KMnO4 for side-chain oxidation and Sn + HCl for selective nitro reduction.

Compound \(X\) (\(C_4H_9Br\)) is a halogenoalkane with 4 carbons. Reaction with \(KCN\) introduces an extra carbon atom, forming nitrile \(Y\) (\(C_5H_9N\)) with 5 carbons. Hydrolysis of \(Y\) with dilute sulfuric acid yields the 5-carbon carboxylic acid \(Z\) (pentanoic acid or its branched isomer 2-methylbutanoic acid). Reacting \(Z\) with \(PCl_5\) yields acyl chloride \(W\) (pentanoyl chloride or 2-methylbutanoyl chloride). Reacting \(W\) with ethylamine (\(C_2H_5NH_2\)) produces the amide, which, if unbranched, is \(N\)-ethylpentanamide.

Award 1 mark for tracing the carbon-chain length changes correctly (from 4 carbons in X to 5 in Y, Z, and W) and identifying the amide as N-ethylpentanamide.

For a reaction to be thermodynamically feasible, \(\Delta G^\ominus \le 0\). Since \(\Delta G^\ominus = \Delta H^\ominus - T\Delta S^\ominus\), the transition temperature \(T\) is found when \(\Delta G^\ominus = 0\), so \(T = \Delta H^\ominus / \Delta S^\ominus\). Converting \(\Delta H^\ominus\) to J: \(178 \times 10^3\text{ J mol}^{-1}\). Thus, \(T = 178000 / 160 = 1112.5\text{ K}\), which rounds to \(1113\text{ K}\).

Award 1 mark for converting kJ to J and correctly using the Gibbs free energy formula to calculate 1113 K.

The transition metal ion \(X^{3+}\) has the electronic configuration \([Ar] 3d^3\), meaning it has 18 (from Ar) + 3 = 21 electrons. The neutral transition metal atom must have 3 more electrons, so 24 electrons. The element with atomic number 24 is Chromium (Cr). Its ground-state configuration is \([Ar] 3d^5 4s^1\). When forming the \(Cr^{3+}\) ion, it loses the single \(4s\) electron and two \(3d\) electrons, leaving the configuration as \([Ar] 3d^3\).

Award 1 mark for correctly identifying Chromium as the transition metal that forms a 3+ ion with a 3d^3 configuration.

(a) Iron has atomic number 26. Fe atom: \( 1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6 3\text{d}^6 4\text{s}^2 \). The \( \text{Fe}^{3+} \) ion loses three electrons (two from 4s and one from 3d): \( 1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6 3\text{d}^5 \).

(b) Shape: octahedral; Coordination number: 6; Bond angle: \( 90^\circ \) (accept \( 180^\circ \) as well).

(c) (i) \( [\text{Fe}(\text{OH}_2)_6]^{3+} + \text{SCN}^- \rightarrow [\text{Fe}(\text{OH}_2)_5(\text{SCN})]^{2+} + \text{H}_2\text{O} \)
(ii) Ligand exchange (or ligand substitution).

(d) 1. In the presence of ligands, the d-orbitals split into two sets of non-degenerate energy levels with an energy gap, \( \Delta E \).
2. Electrons absorb energy from the visible spectrum and are promoted from the lower d-orbitals to the higher d-orbitals (d-d transition).
3. The colour transmitted/observed is the complementary colour of the wavelength absorbed (calculated via \( \Delta E = hf \)).
4. Different ligands produce different crystal field splitting energies (\( \Delta E \)), resulting in different wavelengths of light being absorbed, hence a change in colour.

(e) The stability constant, \( K_{\text{stab}} \), is the equilibrium constant for the formation of a complex ion from its component ions or molecules in a solvent. Since fluoride ions displace thiocyanate ions to form \( [\text{FeF}_6]^{3-} \), the equilibrium lies far to the right for the fluoride complex, showing that \( [\text{FeF}_6]^{3-} \) has a larger \( K_{\text{stab}} \) than \( [\text{Fe}(\text{OH}_2)_5(\text{SCN})]^{2+} \).

Part (a):
- 1 Mark: \( 1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6 3\text{d}^5 \) (or \( [\text{Ar}] 3\text{d}^5 \))

Part (b):
- 1 Mark: Octahedral shape
- 1 Mark: Coordination number = 6
- 1 Mark: Bond angle = \( 90^\circ \) (allow \( 180^\circ \))

Part (c):
- 1 Mark: Correct equation with balanced charges and species: \( [\text{Fe}(\text{OH}_2)_6]^{3+} + \text{SCN}^- \rightarrow [\text{Fe}(\text{OH}_2)_5(\text{SCN})]^{2+} + \text{H}_2\text{O} \)
- 1 Mark: Ligand exchange / ligand substitution

Part (d):
- 1 Mark: Ligands cause d-orbitals to split into two non-degenerate energy levels / groups of different energies.
- 1 Mark: An electron absorbs visible light / a photon and is promoted to a higher energy d-orbital / undergoes d-d transition.
- 1 Mark: The colour observed is the complementary colour of the absorbed wavelength.
- 1 Mark: Different ligands create a different energy gap (\( \Delta E \)), leading to absorption of different wavelengths / frequencies.

Part (e):
- 1 Mark: Correct definition of stability constant as the equilibrium constant for the formation of a complex ion from its constituent ions/ligands.
- 1 Mark: State that \( [\text{FeF}_6]^{3-} \) has the higher stability constant because the addition of fluoride successfully displaces thiocyanate ligands (equilibrium shifts to the fluoride-containing complex).

(a) Concentrated sulfuric acid acts as a catalyst and a proton donor (Brønsted-Lowry acid). Equation: \( \text{HNO}_3 + 2\text{H}_2\text{SO}_4 \rightarrow \text{NO}_2^+ + \text{H}_3\text{O}^+ + 2\text{HSO}_4^- \) (or equivalent balanced equation showing protonation of nitric acid followed by dehydration to yield the nitronium ion).

(b) (i) \( \text{C}_6\text{H}_5\text{NO}_2 + 6[\text{H}] \rightarrow \text{C}_6\text{H}_5\text{NH}_2 + 2\text{H}_2\text{O} \)
(ii) Sodium hydroxide (\( \text{NaOH}(\text{aq}) \)) is added to liberate the free amine from its protonated salt (\( \text{C}_6\text{H}_5\text{NH}_3^+\text{Cl}^- \)).

(c) (i) At temperatures above \( 10^\circ\text{C} \), benzenediazonium chloride is unstable and decomposes rapidly to form phenol and nitrogen gas (\( \text{N}_2 \)).
(ii) Sodium nitrite (\( \text{NaNO}_2 \)) and dilute hydrochloric acid (\( \text{HCl} \)).

(d) (i) The structure must show the azo group, \( -\text{N}=\text{N}- \), connecting the benzene ring to position 1 of the naphthalen-2-ol molecule: \( \text{C}_6\text{H}_5-\text{N}=\text{N}-\text{C}_{10}\text{H}_6\text{OH} \).
(ii) Electrophilic substitution.

(e) 1. Phenylamine is less basic than ethylamine.
2. In phenylamine, the lone pair of electrons on the nitrogen atom is delocalised into the benzene \( \pi \)-electron ring system, reducing its availability to accept a proton (\( \text{H}^+ \)).
3. In ethylamine, the alkyl (ethyl) group is electron-releasing via the positive inductive effect (\( +I \)), which increases the electron density on the nitrogen atom and makes the lone pair more available to accept a proton.

Part (a):
- 1 Mark: Role of sulfuric acid: catalyst / proton donor / acid.
- 1 Mark: Correct equation: \( \text{HNO}_3 + 2\text{H}_2\text{SO}_4 \rightarrow \text{NO}_2^+ + \text{H}_3\text{O}^+ + 2\text{HSO}_4^- \) (allow \( \text{HNO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{NO}_2^+ + \text{H}_2\text{O} + \text{HSO}_4^- \))

Part (b):
- 1 Mark: Correct balanced equation with \( 6[\text{H}] \) and \( 2\text{H}_2\text{O} \): \( \text{C}_6\text{H}_5\text{NO}_2 + 6[\text{H}] \rightarrow \text{C}_6\text{H}_5\text{NH}_2 + 2\text{H}_2\text{O} \)
- 1 Mark: Aqueous sodium hydroxide / NaOH(aq) (or any strong base)

Part (c):
- 1 Mark: Prevent decomposition of the diazonium salt (to phenol and nitrogen)
- 1 Mark: Sodium nitrite (\( \text{NaNO}_2 \)) AND hydrochloric acid (\( \text{HCl} \)) (or any strong mineral acid)

Part (d):
- 2 Marks: Correctly drawn skeletal/structural formula of the azo dye (1 mark for \( -\text{N}=\text{N}- \) linkage, 1 mark for correct connection to naphthalene-2-ol at C1 adjacent to the -OH group).
- 1 Mark: Electrophilic substitution.

Part (e):
- 1 Mark: Phenylamine is a weaker base than ethylamine (or vice versa).
- 1 Mark: Explanation for phenylamine: Nitrogen lone pair delocalises into the benzene ring, making it less available to accept a proton.
- 1 Mark: Explanation for ethylamine: Ethyl group has a positive inductive effect / is electron-donating, making the nitrogen lone pair more available to coordinate with a proton.

(a) Reagents: Acidified potassium dichromate(VI) (\( \text{K}_2\text{Cr}_2\text{O}_7 / \text{H}^+ \)).
Conditions: Distillation immediately (heating under distillation setup so the more volatile ethanal distils off before further oxidation to ethanoic acid).

(b) (i) Mechanism:
1. Dipole shown on ethanal carbon-oxygen bond: \( \text{C}^{\delta+}=\text{O}^{\delta-} \).
2. Curly arrow from the lone pair on the carbon of the cyanide ion (\( :\text{CN}^- \)) to the carbonyl carbon (\( \text{C}^{\delta+} \)).
3. Curly arrow from the \( \text{C}=\text{O} \) double bond to the oxygen atom.
4. Intermediate drawn: \( \text{CH}_3\text{CH}(\text{CN})\text{O}^- \) with a negative charge on the oxygen atom and a lone pair on the oxygen.
5. Curly arrow from the negative oxygen atom's lone pair to the \( \text{H} \) of an \( \text{HCN} \) molecule (or to \( \text{H}^+ \)), showing regeneration of the cyanide catalyst.
(ii) NaCN provides a high concentration of the nucleophile, \( \text{CN}^- \), and acts as a catalyst.

(c) (i) Misty/steamy white fumes (of \( \text{HCl} \) gas).
(ii) \( \text{CH}_3\text{CH}(\text{Cl})\text{CN} + 2\text{NH}_3 \rightarrow \text{CH}_3\text{CH}(\text{NH}_2)\text{CN} + \text{NH}_4\text{Cl} \)
(iii) Dilute hydrochloric acid (or dilute sulfuric acid) and heat under reflux.

(d) 1. 2-aminopropanoic acid contains a chiral carbon (asymmetric carbon atom) bonded to four different groups: \( -\text{H} \), \( -\text{CH}_3 \), \( -\text{NH}_2 \), and \( -\text{COOH} \).
2. Enantiomers can be distinguished using a polarimeter: each optical isomer rotates the plane of plane-polarised monochromatic light by equal angles in opposite directions (one clockwise, the other anticlockwise).

Part (a):
- 1 Mark: Acidified potassium dichromate(VI) / \( \text{K}_2\text{Cr}_2\text{O}_7 \) and dilute \( \text{H}_2\text{SO}_4 \).
- 1 Mark: Heat and distill off immediately.

Part (b):
- 1 Mark: Correct dipole on carbonyl group: \( \text{C}^{\delta+}=\text{O}^{\delta-} \).
- 1 Mark: Curly arrow from carbon of \( :\text{CN}^- \) to the carbonyl carbon AND curly arrow from \( \text{C}=\text{O} \) pi bond to oxygen.
- 1 Mark: Correct intermediate structure \( \text{CH}_3\text{CH}(\text{CN})\text{O}^- \) showing negative charge on oxygen.
- 1 Mark: Curly arrow from \( \text{O}^- \) lone pair to the proton of \( \text{HCN} \) / \( \text{H}^+ \).
- 1 Mark: Catalyst / source of nucleophilic cyanide ions (\( \text{CN}^- \)).

Part (c):
- 1 Mark: Misty/steamy white fumes.
- 1 Mark: Balanced equation: \( \text{CH}_3\text{CH}(\text{Cl})\text{CN} + 2\text{NH}_3 \rightarrow \text{CH}_3\text{CH}(\text{NH}_2)\text{CN} + \text{NH}_4\text{Cl} \).
- 1 Mark: Dilute acid (e.g., \( \text{HCl}(\text{aq}) \)) and heating under reflux.

Part (d):
- 1 Mark: Mentions chiral center / asymmetric carbon atom bonded to 4 different groups (\( -\text{H} \), \( -\text{CH}_3 \), \( -\text{NH}_2 \), \( -\text{COOH} \)).
- 1 Mark: Distinguish by passing plane-polarised light: they rotate the plane in opposite directions.

(a) Standard lattice energy of formation is the enthalpy change when 1 mole of an ionic crystalline solid is formed from its constituent gaseous ions under standard conditions (298 K, 1 bar).

(b) Using Hess's law to write the Born-Haber energy balance:
\( \Delta H_{\text{f}}^\ominus[\text{CaCl}_2(\text{s})] = \Delta H_{\text{at}}^\ominus[\text{Ca}(\text{s})] + I_1[\text{Ca}] + I_2[\text{Ca}] + 2 \times \Delta H_{\text{at}}^\ominus[\text{Cl}_2(\text{g})] + 2 \times EA_1[\text{Cl}] + \Delta H_{\text{lattice}}^\ominus[\text{CaCl}_2(\text{s})] \)

Substitute the values:
\( -796 = 178 + 590 + 1145 + 2(121) + 2(-349) + \Delta H_{\text{lattice}}^\ominus \)
\( -796 = 178 + 590 + 1145 + 242 - 698 + \Delta H_{\text{lattice}}^\ominus \)
\( -796 = 1457 + \Delta H_{\text{lattice}}^\ominus \)
\( \Delta H_{\text{lattice}}^\ominus = -796 - 1457 = -2253\ \text{kJ\ mol}^{-1} \).

(c) 1. The \( \text{Ca}^{2+} \) ion has a higher charge than the \( \text{Na}^+ \) ion (+2 vs +1).
2. The \( \text{Ca}^{2+} \) ion has a higher charge density (due to smaller/comparable radius and higher charge), which results in stronger electrostatic attractions between the oppositely charged ions in the lattice.

(d) (i) \( \Delta S_{\text{sys}}^\ominus \) is positive (increase in entropy) because the highly ordered crystalline solid lattice breaks down into highly disordered, mobile hydrated ions in solution.
(ii) Relationship:
\( \Delta H_{\text{sol}}^\ominus = -\Delta H_{\text{lattice}}^\ominus + \Delta H_{\text{hyd}}^\ominus[\text{Ca}^{2+}] + 2\Delta H_{\text{hyd}}^\ominus[\text{Cl}^-] \)
(Note: if lattice energy is defined as formation from gaseous ions: \( \Delta H_{\text{sol}} = -\Delta H_{\text{lattice}} + \sum \Delta H_{\text{hyd}} \))
Substitute values:
\( -81.3 = -(-2253) + \Delta H_{\text{hyd}}^\ominus[\text{Ca}^{2+}] + 2(-378) \)
\( -81.3 = 2253 + \Delta H_{\text{hyd}}^\ominus[\text{Ca}^{2+}] - 756 \)
\( -81.3 = 1497 + \Delta H_{\text{hyd}}^\ominus[\text{Ca}^{2+}] \)
\( \Delta H_{\text{hyd}}^\ominus[\text{Ca}^{2+}] = -81.3 - 1497 = -1578.3\ \text{kJ\ mol}^{-1} \).

Part (a):
- 1 Mark: Enthalpy change when 1 mole of an ionic solid is formed.
- 1 Mark: From its constituent gaseous ions under standard conditions.

Part (b):
- 1 Mark: Correct cycle setup showing multiplication of chlorine's atomisation energy and electron affinity by 2.
- 1 Mark: Correct algebraic rearrangement showing: \( \Delta H_{\text{lattice}} = -796 - (178 + 590 + 1145 + 242 - 698) \).
- 1 Mark: Correct value with sign and unit: \( -2253\ \text{kJ\ mol}^{-1} \).

Part (c):
- 1 Mark: Calcium ion has a higher charge / double the charge of the sodium ion (\( \text{Ca}^{2+} \) vs \( \text{Na}^+ \)).
- 1 Mark: Stronger electrostatic forces of attraction between cations and anions in \( \text{CaCl}_2 \) due to higher charge density of \( \text{Ca}^{2+} \).

Part (d):
- 1 Mark: \( \Delta S_{\text{sys}}^\ominus \) is positive / greater than zero.
- 1 Mark: Explanation: Solid lattice breaks down to mobile ions in solution, increasing disorder.
- 1 Mark: Correct algebraic setup: \( \Delta H_{\text{sol}} = -\Delta H_{\text{lattice}} + \Delta H_{\text{hyd}}(\text{Ca}^{2+}) + 2\Delta H_{\text{hyd}}(\text{Cl}^-) \).
- 1 Mark: Correct substitution using candidate's lattice energy.
- 1 Mark: Final value: \( -1578.3\ \text{kJ\ mol}^{-1} \) (or consistent with part (b) value).

(a) (i)
- Comparing Experiments 1 and 2: \( [\text{H}_2] \) is held constant, and \( [\text{NO}] \) is doubled (from 0.10 to 0.20). The rate increases by a factor of 4 (\( \frac{4.80 \times 10^{-3}}{1.20 \times 10^{-3}} = 4 \)). Since \( 2^2 = 4 \), the reaction is second order with respect to \( [\text{NO}] \).
- Comparing Experiments 1 and 3: \( [\text{NO}] \) is held constant, and \( [\text{H}_2] \) is doubled (from 0.10 to 0.20). The rate doubles (\( \frac{2.40 \times 10^{-3}}{1.20 \times 10^{-3}} = 2 \)). Since \( 2^1 = 2 \), the reaction is first order with respect to \( [\text{H}_2] \).
(ii) Rate equation: \( \text{Rate} = k[\text{NO}]^2[\text{H}_2] \)
(iii) Calculate \( k \) using Experiment 1:
\( 1.20 \times 10^{-3} = k(0.10)^2(0.10) \)
\( 1.20 \times 10^{-3} = k(0.001) \)
\( k = 1.20 \)
Units of \( k \): \( \frac{\text{mol\ dm}^{-3}\ \text{s}^{-1}}{(\text{mol\ dm}^{-3})^3} = \text{mol}^{-2}\ \text{dm}^6\ \text{s}^{-1} \).

(b) Proposed mechanism:
Step 1 (slow): \( 2\text{NO} + \text{H}_2 \rightarrow \text{N}_2\text{O} + \text{H}_2\text{O} \) (Rate-determining step, involving two molecules of NO and one of H2).
Step 2 (fast): \( \text{N}_2\text{O} + \text{H}_2 \rightarrow \text{N}_2 + \text{H}_2\text{O} \).
Overall equation: \( 2\text{NO} + 2\text{H}_2 \rightarrow \text{N}_2 + 2\text{H}_2\text{O} \).
(Alternatively, accept Step 1: \( 2\text{NO} \rightleftharpoons \text{N}_2\text{O}_2 \) (fast) followed by Step 2: \( \text{N}_2\text{O}_2 + \text{H}_2 \rightarrow \text{N}_2\text{O} + \text{H}_2\text{O} \) (slow), and Step 3: \( \text{N}_2\text{O} + \text{H}_2 \rightarrow \text{N}_2 + \text{H}_2\text{O} \) (fast)).

(c) As temperature increases, the average kinetic energy of the particles increases, so they move faster. Consequently, there is a much larger fraction of colliding molecules with kinetic energy greater than or equal to the activation energy (\( E_a \)). This leads to a significantly higher frequency of successful collisions.

Part (a):
- 1 Mark: Correct reasoning for order with respect to \( [\text{NO}] \).
- 1 Mark: State that reaction is 2nd order with respect to \( [\text{NO}] \).
- 1 Mark: Correct reasoning for order with respect to \( [\text{H}_2] \).
- 1 Mark: State that reaction is 1st order with respect to \( [\text{H}_2] \).
- 1 Mark: Correct rate equation: \( \text{Rate} = k[\text{NO}]^2[\text{H}_2] \) (must be consistent with deduced orders).
- 1 Mark: Correct value: \( k = 1.20 \).
- 1 Mark: Correct units: \( \text{mol}^{-2}\ \text{dm}^6\ \text{s}^{-1} \).

Part (b):
- 1 Mark: Two step equations that sum to the overall reaction stoichiometry.
- 1 Mark: Step 1 identified as the slow / rate-determining step.
- 1 Mark: The slow step has reactants matching the orders in the rate equation (i.e. \( 2\text{NO} + \text{H}_2 \)).

Part (c):
- 1 Mark: Increased temperature leads to a much larger fraction of molecules having energy \( \ge \text{activation energy} \).
- 1 Mark: This results in a higher frequency of successful/effective collisions.

**(a) Titration Results Table:**
An exemplary table must show:
- Correctly labelled columns with appropriate units, e.g., 'Final burette reading / \(\text{cm}^3\)', 'Initial burette reading / \(\text{cm}^3\)', 'Volume of FA 2 added / \(\text{cm}^3\)'.
- All burette readings recorded to \(0.05\text{ cm}^3\).
- At least two concordant titres within \(0.10\text{ cm}^3\) of each other.

Assuming the concordant titres obtained are \(12.50\text{ cm}^3\) and \(12.50\text{ cm}^3\):
Average titre volume of FA 2 = \(12.50\text{ cm}^3\).

**(c) Calculations:**
(i) \(\text{Moles of } \text{KMnO}_4 = \text{concentration} \times \text{volume} = 0.0200 \times \frac{12.50}{1000} = 2.50 \times 10^{-4}\text{ mol}\).
(ii) From the equation, \(1\text{ mol}\) of \(\text{MnO}_4^-\) reacts with \(5\text{ mol}\) of \(\text{Fe}^{2+}\).
\(\text{Moles of } \text{Fe}^{2+} = 5 \times 2.50 \times 10^{-4}\text{ mol} = 1.25 \times 10^{-3}\text{ mol}\).
(iii) \(\text{Concentration of } \text{Fe}^{2+} \text{ in FA 1} = \frac{1.25 \times 10^{-3} \text{ mol}}{25.0/1000 \text{ dm}^3} = 0.0500\text{ mol dm}^{-3}\).
(iv) Since \(1\text{ mol}\) of \(\text{(NH}_4\text{)}_2\text{Fe(SO}_4\text{)}_2 \cdot x\text{H}_2\text{O}\) yields \(1\text{ mol}\) of \(\text{Fe}^{2+}\), the concentration of the hydrated salt is \(0.0500\text{ mol dm}^{-3}\).
\(\text{Molar mass } (M_r) = \frac{\text{Mass in } 1\text{ dm}^3}{\text{Concentration}} = \frac{19.60\text{ g dm}^{-3}}{0.0500\text{ mol dm}^{-3}} = 392.0\text{ g mol}^{-1}\).
(v) The formula mass of anhydrous \(\text{(NH}_4\text{)}_2\text{Fe(SO}_4\text{)}_2\) is:
\(M_r = 2 \times [14.0 + (4 \times 1.0)] + 55.8 + 2 \times [32.1 + (4 \times 16.0)] = 36.0 + 55.8 + 192.2 = 284.0\).
Mass of water of crystallisation = \(392.0 - 284.0 = 108.0\text{ g mol}^{-1}\).
\(18.0x = 108.0 \implies x = \frac{108.0}{18.0} = 6\).

**(d) Explanation:**
Hydrochloric acid contains \(\text{Cl}^-\) ions. \(\text{MnO}_4^-\) is a very strong oxidising agent that oxidises \(\text{Cl}^-\) to \(\text{Cl}_2\) gas. This secondary redox reaction consumes extra \(\text{MnO}_4^-\) ions, leading to an artificially high titre volume and a calculated value of \(x\) that is lower than the actual value.

Total Marks: 13.33 marks.
- **[1 mark]** Table of results is clearly constructed with appropriate headings and units. All burette readings (including rough) are recorded to \(0.05\text{ cm}^3\).
- **[2 marks]** Accuracy: Concordant titres are within \(0.10\text{ cm}^3\) of the reference value (typically \(12.50\text{ cm}^3\)). Give 2 marks if within \(\pm 0.10\text{ cm}^3\); 1 mark if within \(\pm 0.20\text{ cm}^3\).
- **[1 mark]** Calculation of average volume of FA 2: calculated correctly to 2 decimal places using concordant values only.
- **[1 mark]** Moles of \(\text{KMnO}_4\) (c)(i): correctly calculated as \(\text{volume} \times 0.0200 / 1000\).
- **[1 mark]** Moles of \(\text{Fe}^{2+}\) (c)(ii): answer to (i) multiplied by 5, showing clear working.
- **[1 mark]** Concentration of \(\text{Fe}^{2+}\) (c)(iii): answer to (ii) divided by \(0.0250\), expressed to 3 significant figures.
- **[2 marks]** Relative formula mass (c)(iv): mass concentration (\(19.60\)) divided by concentration of \(\text{Fe}^{2+}\) from (iii). Deduct 1 mark if calculation has a minor arithmetic error.
- **[2 marks]** Value of \(x\) (c)(v): correctly calculates anhydrous formula mass (\(284.0\)), subtracts this from \(M_r\) and divides by \(18.0\) to get a whole number.
- **[1.33 marks]** Answer to (d): identifies that hydrochloric acid contains \(\text{Cl}^-\) which is oxidised by \(\text{MnO}_4^-\) (0.67 marks), resulting in more manganate(VII) being used than is required for \(\text{Fe}^{2+}\) alone (0.66 marks).

**(b) Results Table:**
An exemplary table contains:
- Masses of weighing bottles before and after addition, mass of FA 3 and FA 4 added.
- Initial and final temperatures of the solution for both reactions.
- All temperatures recorded to \(0.5^\circ\text{C}\).
- All masses recorded to \(0.01\text{ g}\) (or \(0.001\text{ g}\) depending on the balance used).

**Sample Data:**
Reaction 1 (Anhydrous \(\text{Na}_2\text{CO}_3\)):
- Mass of FA 3 = \(3.18\text{ g}\) (which is \(3.18 / 106.0 = 0.0300\text{ mol}\)).
- Initial temperature = \(21.0^\circ\text{C}\)
- Maximum temperature = \(25.0^\circ\text{C}\)
- \(\Delta T = +4.0^\circ\text{C}\).

Reaction 2 (Hydrated \(\text{Na}_2\text{CO}_3 \cdot 10\text{H}_2\text{O}\)):
- Mass of FA 4 = \(8.58\text{ g}\) (which is \(8.58 / 286.0 = 0.0300\text{ mol}\)).
- Initial temperature = \(21.0^\circ\text{C}\)
- Minimum temperature = \(17.0^\circ\text{C}\)
- \(\Delta T = -4.0^\circ\text{C}\).

**(c) Calculations:**
(i) \(q_1 = m \times c \times \Delta T = 50.0 \times 4.18 \times 4.0 = 836\text{ J}\) (or \(0.836\text{ kJ}\)).
(ii) \(\text{Moles of } \text{Na}_2\text{CO}_3 = \frac{3.18}{106.0} = 0.0300\text{ mol}\).
Since the reaction is exothermic:
\(\Delta H_1 = -\frac{0.836\text{ kJ}}{0.0300\text{ mol}} = -27.9\text{ kJ mol}^{-1}\).
(iii) \(q_2 = m \times c \times \Delta T = 50.0 \times 4.18 \times (-4.0) = -836\text{ J}\) (or \(-0.836\text{ kJ}\)).
(iv) \(\text{Moles of } \text{Na}_2\text{CO}_3 \cdot 10\text{H}_2\text{O} = \frac{8.58}{286.0} = 0.0300\text{ mol}\).
Since the reaction is endothermic:
\(\Delta H_2 = +\frac{0.836\text{ kJ}}{0.0300\text{ mol}} = +27.9\text{ kJ mol}^{-1}\).
(v) By constructing the Hess's Law cycle:
\(\text{Na}_2\text{CO}_3(s) + 10\text{H}_2\text{O}(l) \xrightarrow{\Delta H_{hyd}} \text{Na}_2\text{CO}_3 \cdot 10\text{H}_2\text{O}(s)\)
Both react with hydrochloric acid to yield the same final products.
Thus:
\(\Delta H_{hyd} + \Delta H_2 = \Delta H_1 \implies \Delta H_{hyd} = \Delta H_1 - \Delta H_2\)
\(\Delta H_{hyd} = -27.9 - (+27.9) = -55.8\text{ kJ mol}^{-1}\).

Total Marks: 13.33 marks.
- **[1 mark]** Table structure is logical, containing all required mass and temperature measurements with correct column headers and units.
- **[2 marks]** Balance readings are recorded consistently to 2 or 3 decimal places (1 mark), and all thermometer readings are recorded to the nearest \(0.5^\circ\text{C}\) (1 mark).
- **[2 marks]** Accuracy of temperature changes: compares candidates' values with supervisor values. 2 marks if temperature change matches reference within \(\pm 0.5^\circ\text{C}\); 1 mark if within \(\pm 1.0^\circ\text{C}\).
- **[1 mark]** (c)(i): Correct calculation of \(q_1 = m c \Delta T_1\) using \(m = 50.0\text{ g}\) and \(c = 4.18\).
- **[1 mark]** (c)(ii): Correct calculation of \(\text{moles of } \text{Na}_2\text{CO}_3\), and calculation of \(\Delta H_1 = -q_1 / \text{moles}\), including the negative sign.
- **[1 mark]** (c)(iii): Correct calculation of \(q_2 = m c \Delta T_2\) using \(m = 50.0\text{ g}\) and \(c = 4.18\).
- **[1 mark]** (c)(iv): Correct calculation of \(\text{moles of } \text{Na}_2\text{CO}_3 \cdot 10\text{H}_2\text{O}\), and calculation of \(\Delta H_2 = -q_2 / \text{moles}\), showing a positive value.
- **[2 marks]** (c)(v): Correct construction of Hess's Law cycle or expression showing \(\Delta H_{hyd} = \Delta H_1 - \Delta H_2\) (1 mark) and final value calculated with correct sign and unit of \(\text{kJ mol}^{-1}\) (1 mark).
- **[2.33 marks]** Evaluation: Suggest two ways to reduce experimental heat loss in this procedure (e.g., using a lid, using double nesting cups / copper calorimeter, or using a vacuum flask).

**Expected Observations:**

| Test | Observation with FA 6 (\(\text{CrCl}_3\)) | Observation with FA 7 (\(\text{CoSO}_4\)) | Observation with FA 8 (\(\text{NH}_4\text{I}\)) |
| :--- | :--- | :--- | :--- |
| **(i)** \(\text{NaOH}\) | Green precipitate, soluble in excess to form a dark green solution. | Blue/grey precipitate, insoluble in excess. | No precipitate. On heating, pungent gas evolved. |
| **(ii)** \(\text{NH}_3\) | Grey-green precipitate, insoluble in excess (or slightly soluble to violet/pink). | Blue precipitate, soluble in excess to give a yellow-brown/brown solution. | No precipitate. |
| **(iii)** \(\text{Ba}^{2+}\) | No reaction / no precipitate. | White precipitate, insoluble in dilute nitric acid. | No reaction / no precipitate. |
| **(iv)** \(\text{AgNO}_3\) | White precipitate, soluble in dilute ammonia. | No reaction / no precipitate. | Yellow precipitate, insoluble in dilute and concentrated ammonia. |
| **(v)** \(\text{Al} + \text{NaOH}\) | No gas evolved that turns litmus blue. | No gas evolved that turns litmus blue. | Pungent gas evolved (ammonia) that turns damp red litmus paper blue. |

**(a) Identification:**
- FA 6 contains \(\text{Cr}^{3+}\) (chromium(III)) and \(\text{Cl}^-\) (chloride).
- FA 7 contains \(\text{Co}^{2+}\) (cobalt(II)) and \(\text{SO}_4^{2-}\) (sulfate).
- FA 8 contains \(\text{NH}_4^+\) (ammonium) and \(\text{I}^-\) (iodide).

**(b) Ionic Equations:**
(i) Reaction of \(\text{Cr}^{3+}\) with excess \(\text{OH}^-\):
\(\text{Cr}^{3+}(aq) + 6\text{OH}^-(aq) \rightarrow [\text{Cr}(\text{OH})_6]^{3-}(aq)\) (or \(\text{Cr}(\text{OH})_3(s) + 3\text{OH}^-(aq) \rightarrow [\text{Cr}(\text{OH})_6]^{3-}(aq)\)).
(ii) Reaction of \(\text{I}^-\) with \(\text{Ag}^+\):
\(\text{Ag}^+(aq) + \text{I}^-(aq) \rightarrow \text{AgI}(s)\).

Total Marks: 13.33 marks.
- **[4 marks]** Recording observations in tests (i) and (ii):
- FA 6: green ppt, soluble in excess NaOH to green solution; grey-green ppt insoluble in excess NH3 (1 mark).
- FA 7: blue/grey ppt, insoluble in excess NaOH; blue ppt, soluble in excess NH3 to yellow-brown solution (2 marks).
- FA 8: no ppt with either NaOH or NH3 (1 mark).
- **[3 marks]** Recording observations in tests (iii), (iv), and (v):
- Test (iii): FA 7 gives a white precipitate insoluble in acid; others show no reaction (1 mark).
- Test (iv): FA 6 gives a white ppt soluble in NH3; FA 8 gives a yellow ppt insoluble in NH3 (1 mark).
- Test (v): FA 8 yields a gas that turns damp red litmus paper blue (1 mark).
- **[3 marks]** Correct identification of all 3 cations and 3 anions (0.5 marks per ion identified with consistent evidence).
- Cations: \(\text{Cr}^{3+}\), \(\text{Co}^{2+}\), \(\text{NH}_4^+\).
- Anions: \(\text{Cl}^-\), \(\text{SO}_4^{2-}\), \(\text{I}^-\).
- **[3.33 marks]** Equations in (b):
- (i) Correctly balanced ionic equation for the formation of the soluble complex (1.67 marks):
e.g., \(\text{Cr}^{3+}(aq) + 6\text{OH}^-(aq) \rightarrow [\text{Cr}(\text{OH})_6]^{3-}(aq)\) or \(\text{Cr}^{3+}(aq) + 4\text{OH}^-(aq) \rightarrow [\text{Cr}(\text{OH})_4]^-(aq)\).
- (ii) Correctly balanced ionic equation for precipitation of silver iodide (1.66 marks) with state symbols:
\(\text{Ag}^+(aq) + \text{I}^-(aq) \rightarrow \text{AgI}(s)\).

(a) Lattice energy is the enthalpy change when 1 mole of an ionic crystal/solid is formed from its constituent gaseous ions under standard conditions.

(b) Magnesium ions (\(\text{Mg}^{2+}\)) have a higher charge (+2 vs +1) and a smaller ionic radius than sodium ions (\(\text{Na}^+\)). This means the charge density of \(\text{Mg}^{2+}\) is significantly higher, resulting in a much stronger electrostatic attraction to the chloride ions in the lattice, making the lattice energy of \(\text{MgCl}_2\) more exothermic.

(c) The enthalpy change of hydration is the enthalpy change when 1 mole of gaseous ions is completely dissolved in water to form infinitely dilute hydrated ions under standard conditions.

(d) Using Hess's law, the dissolution process can be represented by:
\(\Delta H_{\text{sol}}^{\ominus} = -\Delta H_{\text{latt}}^{\ominus} + \Delta H_{\text{hyd}}^{\ominus}(\text{Mg}^{2+}) + 2 \times \Delta H_{\text{hyd}}^{\ominus}(\text{Cl}^-)\)
\(\Delta H_{\text{sol}}^{\ominus} = -(-2526) + (-1920) + 2 \times (-364)\)
\(\Delta H_{\text{sol}}^{\ominus} = 2526 - 1920 - 728 = -122\text{ kJ mol}^{-1}\)

(e) The hydration enthalpy of \(\text{Ca}^{2+}\) is less exothermic (less negative) than that of \(\text{Mg}^{2+}\). This is because the ionic radius of \(\text{Ca}^{2+}\) is larger than that of \(\text{Mg}^{2+}\), resulting in a lower charge density and thus weaker electrostatic attraction to the lone pairs on water molecules.

(a) 1 mark for '1 mole of ionic solid formed from gaseous ions'; 1 mark for 'under standard conditions'.

(b) 1 mark for stating \(\text{Mg}^{2+}\) has a higher charge / smaller ionic radius / higher charge density than \(\text{Na}^+\); 1 mark for linking this to stronger electrostatic attractions.

(c) 1 mark for '1 mole of gaseous ions dissolved in water'; 1 mark for 'to form an infinitely dilute solution under standard conditions'.

(d) 1 mark for correct algebraic formula; 1 mark for multiplying the chloride hydration enthalpy by 2; 1 mark for correct substitution; 1 mark for correct value (-122); 0.5 marks for correct units (\(\text{kJ mol}^{-1}\)).

(e) 1 mark for predicting it is less exothermic; 1 mark for explanation involving the larger ionic radius of calcium ions.

(a) (i) Comparing Exp 1 and Exp 2: \([\text{I}^-]\) is constant. Doubling \([\text{S}_2\text{O}_8^{2-}]\) from \(0.0100\) to \(0.0200\text{ mol dm}^{-3}\) doubles the rate from \(1.25 \times 10^{-5}\) to \(2.50 \times 10^{-5}\text{ mol dm}^{-3}\text{ s}^{-1}\). Thus, the order is 1.
(ii) Comparing Exp 1 and Exp 3: \([\text{S}_2\text{O}_8^{2-}]\) is constant. Doubling \([\text{I}^-]\) from \(0.0200\) to \(0.0400\text{ mol dm}^{-3}\) doubles the rate from \(1.25 \times 10^{-5}\) to \(2.50 \times 10^{-5}\text{ mol dm}^{-3}\text{ s}^{-1}\). Thus, the order is 1.

(b) The rate equation is: \(\text{Rate} = k[\text{S}_2\text{O}_8^{2-}][\text{I}^-]\)
Using data from Exp 1:
\(1.25 \times 10^{-5} = k \times 0.0100 \times 0.0200\)
\(k = \frac{1.25 \times 10^{-5}}{2.00 \times 10^{-4}} = 0.0625\text{ (or } 6.25 \times 10^{-2}\text{)}\)
Units: \(\text{mol}^{-1}\text{ dm}^3\text{ s}^{-1}\) (or \(\text{dm}^3\text{ mol}^{-1}\text{ s}^{-1}\)).

(c) (i) Both reacting species, \(\text{S}_2\text{O}_8^{2-}\) and \(\text{I}^-\), are negatively charged. Therefore, there is strong electrostatic repulsion between them, creating a high potential energy barrier (high activation energy) to bring them together.
(ii) \(\text{Fe}^{2+}\) reduces peroxodisulfate and is oxidized to \(\text{Fe}^{3+}\):
\(\text{S}_2\text{O}_8^{2-}(\text{aq}) + 2\text{Fe}^{2+}(\text{aq}) \to 2\text{SO}_4^{2-}(\text{aq}) + 2\text{Fe}^{3+}(\text{aq})\)
Then, \(\text{Fe}^{3+}\) oxidizes iodide to iodine, regenerating the catalyst:
\(2\text{Fe}^{3+}(\text{aq}) + 2\text{I}^-(\text{aq}) \to 2\text{Fe}^{2+}(\text{aq}) + \text{I}_2(\text{aq})\)

(a) (i) 1 mark for stating that doubling concentration doubles rate when the other reactant is constant; 0.5 marks for concluding first order.
(ii) 1 mark for stating that doubling concentration doubles rate when the other reactant is constant; 0.5 marks for concluding first order.

(b) 1 mark for correct rate equation; 1.5 marks for correct calculation of \(k = 0.0625\); 1 mark for correct units (\(\text{dm}^3\text{ mol}^{-1}\text{ s}^{-1}\)).

(c) (i) 1 mark for identifying both ions are negatively charged; 1 mark for stating they repel each other.
(ii) 2 marks for the first balanced equation; 2 marks for the second balanced equation.

(a) (i) Since only \(1.00\text{ mol}\) of \(\text{AgCl}\) is precipitated per mole of \(X\), only one chloride ion is outside the coordination sphere. Therefore, the formula of the complex compound is \([\text{Co}(\text{NH}_3)_4\text{Cl}_2]\text{Cl}\), and the complex ion is \([\text{Co}(\text{NH}_3)_4\text{Cl}_2]^+\).
- Coordination number is 6 (four ammonia and two chloride ligands directly bonded to cobalt).
- Oxidation state of Co is +3 (since the overall charge of the complex ion is +1, and the two chlorido ligands each carry a -1 charge: \(x + 2(-1) = +1 \implies x = +3\)).
(ii) The other two chlorine atoms are covalently coordinate-bonded as ligands directly to the central cobalt ion inside the coordination sphere, so they do not dissociate into free ions in solution.

(b) The isomerism is cis-trans (geometric) isomerism.
- In the cis isomer, the two chloride ligands are located adjacent to each other (90° apart).
- In the trans isomer, the two chloride ligands are located opposite to each other (180° apart).
Both should be drawn using 3D representations (solid wedge and dashed bond notation) with octahedral geometry.

(c) (i) The colour changes from pink to blue.
(ii) Equation: \([\text{Co}(\text{H}_2\text{O})_6]^{2+}(\text{aq}) + 4\text{Cl}^-(\text{aq}) \rightleftharpoons [\text{CoCl}_4]^{2-}(\text{aq}) + 6\text{H}_2\text{O}(\text{l})\)
The shape of the \([\text{CoCl}_4]^{2-}\) complex is tetrahedral.

(a) (i) 1 mark for correct complex ion formula \([\text{Co}(\text{NH}_3)_4\text{Cl}_2]^+\); 1 mark for coordination number = 6; 1.5 marks for oxidation state +3.
(ii) 1 mark for explaining that these chlorines are acting as ligands inside the coordination sphere and cannot dissociate.

(b) 1 mark for identifying cis-trans (or geometric) isomerism; 1.5 marks for correct 3D drawing of the cis-isomer; 1.5 marks for correct 3D drawing of the trans-isomer.

(c) (i) 1 mark for 'pink to blue'.
(ii) 2 marks for the correct balanced equation; 1 mark for identifying the tetrahedral shape of \([\text{CoCl}_4]^{2-}\).

(a) The stability constant expression is:
\(K_{\text{stab}} = \frac{[[\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}]}{[[\text{Cu}(\text{H}_2\text{O})_6]^{2+}][\text{NH}_3]^4}\)
(Water is omitted from the expression as its concentration is extremely large and remains virtually constant).

(b) 1,2-diaminoethane is a bidentate ligand, whereas ammonia is a monodentate ligand.
During the coordination with 1,2-diaminoethane, 3 reactant species (1 complex ion + 2 ligands) form 5 product species (1 complex ion + 4 water molecules). This leads to an increase in the number of free particles in solution, resulting in a large positive entropy change (\(\Delta S^{\ominus} > 0\)). For the ammonia reaction, 5 reactant particles produce 5 product particles (no net increase in particles). Since \(\Delta G^{\ominus} = \Delta H^{\ominus} - T\Delta S^{\ominus}\), a more positive \(\Delta S^{\ominus}\) results in a more negative \(\Delta G^{\ominus}\) and hence a much larger \(K_{\text{stab}}\). This is known as the chelate effect.

(c) When dropwise aqueous ammonia is added, a pale blue precipitate of copper(II) hydroxide, \(\text{Cu}(\text{OH})_2\), forms. On adding excess ammonia, this precipitate dissolves to form a deep blue (or dark blue) solution containing \([\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}\).

(d) The ionic equation is:
\(2\text{Cu}^{2+}(\text{aq}) + 4\text{I}^-(\text{aq}) \to 2\text{CuI}(\text{s}) + \text{I}_2(\text{aq})\)
- White precipitate: copper(I) iodide, \(\text{CuI}\)
- Brown solution: iodine, \(\text{I}_2\) (or triiodide, \(\text{I}_3^-\))

(a) 2 marks for the correct expression (deduct 1 mark if water is included in the expression).

(b) 1 mark for noting that 1,2-diaminoethane is bidentate and ammonia is monodentate; 1 mark for stating that the reaction increases the number of particles in solution (3 to 5); 1 mark for explaining that this causes an increase in entropy (\(\Delta S^{\ominus} > 0\)); 0.5 marks for linking to a more negative \(\Delta G^{\ominus}\) / greater stability.

(c) 1 mark for pale blue precipitate; 1 mark for identifying it as \(\text{Cu}(\text{OH})_2\); 1 mark for precipitate dissolving; 1 mark for deep blue solution.

(d) 2 marks for the correct balanced ionic equation; 1 mark for identifying \(\text{CuI}\) as the white precipitate and \(\text{I}_2\) as the brown solution.

(a) Basicity depends on the availability of the lone pair of electrons on the nitrogen atom to accept a proton.
- In phenylamine, the lone pair of electrons on the nitrogen atom overlaps with the pi-delocalized electron system of the benzene ring. This delocalization decreases the electron density on the nitrogen atom, making the lone pair less available to coordinate with a proton.
- In ethylamine, the ethyl group is electron-releasing (positive inductive effect), which increases the electron density on the nitrogen atom, making its lone pair more available to accept a proton.

(b) Step 1: Benzene to Nitrobenzene (intermediate)
- Reagents: Concentrated \(\text{HNO}_3\) and concentrated \(\text{H}_2\text{SO}_4\)
- Conditions: Maintain temperature at \(50-55^{\circ}\text{C}\)
Step 2: Nitrobenzene to Phenylamine
- Reagents: Tin (\(\text{Sn}\)) and concentrated \(\text{HCl}\), followed by addition of aqueous sodium hydroxide (\(\text{NaOH}\))
- Conditions: Heat under reflux

(c) (i) Reagents: Sodium nitrite (\(\text{NaNO}_2\)) and dilute hydrochloric acid (\(\text{HCl}\)) (which react in situ to produce nitrous acid, \(\text{HNO}_2\)).
- Temperature: \(0-10^{\circ}\text{C}\) (or below \(10^{\circ}\text{C}\)).
(ii) The structure of the product (4-hydroxyphenylazobenzene) is:
\(\text{C}_6\text{H}_5-\text{N}=\text{N}-\text{C}_6\text{H}_4\text{OH}\) with the hydroxyl group in the para-position (position 4) relative to the azo linkage.

(a) 1 mark for basicity depending on lone pair availability; 1 mark for stating that phenylamine's nitrogen lone pair is delocalized into the benzene pi-system; 1 mark for stating this makes it less available; 1 mark for explaining that ethyl group increases electron density on nitrogen via positive inductive effect.

(b) 1 mark for identifying nitrobenzene as the intermediate; 1 mark for Step 1 reagents (conc. \(\text{HNO}_3\) + conc. \(\text{H}_2\text{SO}_4\)); 0.5 marks for Step 1 temperature (\(50-55^{\circ}\text{C}\)); 1 mark for Step 2 reagents (\(\text{Sn}\) + conc. \(\text{HCl}\) then \(\text{NaOH}\)); 1 mark for Step 2 conditions (heat/reflux).

(c) (i) 1 mark for reagents (\(\text{NaNO}_2\) and \(\text{HCl}\)); 1 mark for temperature (\(0-10^{\circ}\text{C}\)).
(ii) 1 mark for showing correct azo linkage (\(-\text{N}=\text{N}-\)) connecting two benzene rings; 1 mark for placing the hydroxyl (\(-\text{OH}\)) group in the para-position on the second ring.

(a) (i) Highly acidic: \(\text{CH}_3\text{CH}(\text{NH}_3^+)\text{COOH}\)
(ii) Highly alkaline: \(\text{CH}_3\text{CH}(\text{NH}_2)\text{COO}^-\)
(iii) Zwitterion: \(\text{CH}_3\text{CH}(\text{NH}_3^+)\text{COO}^-

(b) (i) Serine has its chiral carbon at C2. The four groups attached are: \)-\text{H}\), \(-\text{NH}_2\), \(-\text{COOH}\), and \(-\text{CH}_2\text{OH}\). The 3D drawings must show a central carbon with two bonds in the page plane, one wedged bond pointing forward, and one dashed bond pointing backward, with the second drawing being a non-superimposable mirror image of the first.
(ii) A pure enantiomer will rotate the plane of plane-polarized light by a specific angle in either a clockwise or anticlockwise direction.

(c) (i) The two dipeptides are Ala-Ser and Ser-Ala:
- Ala-Ser: \(\text{H}_2\text{N-CH(CH}_3\text{)-CO-NH-CH(CH}_2\text{OH)-COOH}\)
- Ser-Ala: \(\text{H}_2\text{N-CH(CH}_2\text{OH)-CO-NH-CH(CH}_3\text{)-COOH}\)
(Structures must be shown skeletally as requested).
(ii) Condensation (polymerisation/reaction).
(iii) Heat under reflux with moderately concentrated hydrochloric acid (e.g. \(6\text{ mol dm}^{-3}\text{ HCl}\)).

(a) (i) 1 mark for protonating the amine group only (\(-\text{NH}_3^+\)).
(ii) 1 mark for deprotonating the carboxylic acid group only (\(-\text{COO}^-\)).
(iii) 1 mark for showing both charges (\(-\text{NH}_3^+\)) and (\(-\text{COO}^-\)) correctly.

(b) (i) 1 mark for correctly identifying the chiral carbon; 2 marks for two clearly drawn mirror-image 3D tetrahedral structures (using wedge/dash notation).
(ii) 1 mark for 'rotates plane-polarized light'; 0.5 marks for specifying 'clockwise or anticlockwise'.

(c) (i) 1.5 marks for the correct skeletal structure of Ala-Ser; 1.5 marks for the correct skeletal structure of Ser-Ala.
(ii) 1 mark for 'condensation'.
(iii) 1 mark for stating 'heat/reflux with aqueous hydrochloric acid'.

(a) (i) - Intermediate 1 (from Step 1): Propanal, \(\text{CH}_3\text{CH}_2\text{CHO}\)
- Intermediate 2 (from Step 2): 2-hydroxybutanenitrile, \(\text{CH}_3\text{CH}_2\text{CH(OH)CN}\)

(ii) - Step 1 (oxidation of propan-1-ol to propanal): Reagents: Acidified potassium dichromate(VI), \(\text{K}_2\text{Cr}_2\text{O}_7 / \text{H}^+\). Conditions: Heat and distil immediately.
- Step 2 (nucleophilic addition to form hydroxynitrile): Reagents: Hydrogen cyanide, \(\text{HCN}\), with a small amount of \(\text{NaCN}\) or \(\text{KCN}\) (or alkaline buffer). Conditions: Room temperature / cold.
- Step 3 (acid hydrolysis of nitrile to carboxylic acid): Reagents: Dilute hydrochloric acid (\(\text{HCl}\)) or dilute sulfuric acid (\(\text{H}_2\text{SO}_4\)). Conditions: Heat under reflux.

(b) (i) The repeat unit is formed via esterification between the \(-\text{OH}\) and \(-\text{COOH}\) groups of adjacent monomer molecules:
\(-\text{O}-\text{CH}(\text{CH}_2\text{CH}_3)-\text{CO}-\)
(ii) Polyester.
(iii) The polymer contains ester linkages which can be easily broken down by hydrolysis, catalysed by moisture, dilute acids/bases, or bacterial enzymes in the environment.

(a) (i) 1 mark for propanal; 1 mark for 2-hydroxybutanenitrile.
(ii) Step 1: 1 mark for acidified potassium dichromate(VI); 1 mark for heat and distil.
Step 2: 1 mark for \(\text{HCN}\) + \(\text{NaCN}\) (or \(\text{KCN}\)); 1 mark for room temperature / cold.
Step 3: 1 mark for dilute acid (e.g. \(\text{HCl}\)); 1 mark for heat under reflux.

(b) (i) 2 marks for the correct repeat unit with open ester bonds at both ends.
(ii) 1 mark for 'polyester'.
(iii) 1 mark for identifying the presence of hydrolysable ester linkages; 0.5 marks for noting they can be broken down by water/acid/enzymes.

(a) (i) Vigorous/rapid reaction, misty fumes of hydrogen chloride gas, and the test tube becomes hot (exothermic reaction).
(ii) In benzoyl chloride, the carbonyl carbon is bonded directly to the benzene ring.
- The lone pair of electrons on the chlorine atom can overlap with the pi-delocalized electron system of the aromatic ring, strengthening the \(\text{C}-\text{Cl}\) bond.
- Additionally, the carbonyl carbon is less susceptible to nucleophilic attack because the benzene ring is bulky (steric hindrance) and electron-donating by resonance, reducing the partial positive charge (\(\delta+\)) on the carbonyl carbon.
- In ethanoyl chloride, the methyl group is weakly electron-donating, and the highly electronegative chlorine and oxygen atoms draw electron density away from the carbonyl carbon, making it highly electrophilic and easily attacked by nucleophiles like water.

(b) (i) \(\text{C}_6\text{H}_5\text{COCl} + \text{CH}_3\text{CH}_2\text{NH}_2 \to \text{C}_6\text{H}_5\text{CONHCH}_2\text{CH}_3 + \text{HCl}\)
(Or: \(\text{C}_6\text{H}_5\text{COCl} + 2\text{CH}_3\text{CH}_2\text{NH}_2 \to \text{C}_6\text{H}_5\text{CONHCH}_2\text{CH}_3 + \text{CH}_3\text{CH}_2\text{NH}_3^+\text{Cl}^-\))
(ii) N-ethylbenzamide

(c) (i) Phosphorus pentachloride, \(\text{PCl}_5\) (or phosphorus trichloride, \(\text{PCl}_3\), or thionyl chloride, \(\text{SOCl}_2\)).
(ii) The product is N-phenylbenzamide. Its skeletal structure consists of two benzene rings connected by an amide linkage: \(\text{C}_6\text{H}_5-\text{CONH}-\text{C}_6\text{H}_5\).

(a) (i) 1 mark for misty fumes; 1 mark for mixture getting hot / vigorous reaction.
(ii) 1 mark for stating that the C-Cl bond is stronger in benzoyl chloride; 1 mark for mentioning conjugation/overlap of chlorine lone pair with the aromatic pi-system; 1 mark for steric hindrance / reduced \(\delta+\)
charge of the carbonyl carbon; 1 mark for explaining that ethanoyl chloride has a highly electrophilic carbonyl carbon susceptible to nucleophilic attack.

(b) (i) 2 marks for a fully balanced chemical equation (accept either equation with \(\text{HCl}\) or salt formation).
(ii) 1.5 marks for 'N-ethylbenzamide'.

(c) (i) 1 mark for \(\text{PCl}_5\), \(\text{PCl}_3\), or \(\text{SOCl}_2\).
(ii) 2 marks for the correct skeletal structure of N-phenylbenzamide (1 mark if drawn non-skeletally but chemically correct).

1. **Variables**:
- Independent variable: the concentration / volume of iodide ions, \(\text{I}^-(\text{aq})\).
- Dependent variable: the time taken, \(t\), for the blue-black color to appear (which is inversely proportional to the initial rate of reaction).

2. **Reagent Table (Total Volume = \(50.0\text{ cm}^3\))**:
To vary \(\text{I}^-\), we change the volume of \(\text{KI}\)(aq). To keep the concentrations of \(\text{S}_2\text{O}_8^{2-}\) and \(\text{S}_2\text{O}_3^{2-}\) constant, we must keep their volumes constant across all three experiments, and use distilled water to make the total volume up to \(50.0\text{ cm}^3\).

Let's use a fixed volume of \(10.0\text{ cm}^3\) of \(\text{K}_2\text{S}_2\text{O}_8\), \(5.0\text{ cm}^3\) of \(\text{Na}_2\text{S}_2\text{O}_3\), and \(1.0\text{ cm}^3\) of starch indicator. This leaves \(34.0\text{ cm}^3\) to be split between \(\text{KI}\) and distilled water:

- **Experiment 1**:
- Volume of \(0.200\text{ mol dm}^{-3}\) \(\text{KI}\) = \(20.0\text{ cm}^3\)
- Volume of \(0.100\text{ mol dm}^{-3}\) \(\text{K}_2\text{S}_2\text{O}_8\) = \(10.0\text{ cm}^3\)
- Volume of \(0.00500\text{ mol dm}^{-3}\) \(\text{Na}_2\text{S}_2\text{O}_3\) = \(5.0\text{ cm}^3\)
- Volume of starch indicator = \(1.0\text{ cm}^3\)
- Volume of distilled water = \(14.0\text{ cm}^3\)

- **Experiment 2**:
- Volume of \(0.200\text{ mol dm}^{-3}\) \(\text{KI}\) = \(15.0\text{ cm}^3\)
- Volume of \(0.100\text{ mol dm}^{-3}\) \(\text{K}_2\text{S}_2\text{O}_8\) = \(10.0\text{ cm}^3\)
- Volume of \(0.00500\text{ mol dm}^{-3}\) \(\text{Na}_2\text{S}_2\text{O}_3\) = \(5.0\text{ cm}^3\)
- Volume of starch indicator = \(1.0\text{ cm}^3\)
- Volume of distilled water = \(19.0\text{ cm}^3\)

- **Experiment 3**:
- Volume of \(0.200\text{ mol dm}^{-3}\) \(\text{KI}\) = \(10.0\text{ cm}^3\)
- Volume of \(0.100\text{ mol dm}^{-3}\) \(\text{K}_2\text{S}_2\text{O}_8\) = \(10.0\text{ cm}^3\)
- Volume of \(0.00500\text{ mol dm}^{-3}\) \(\text{Na}_2\text{S}_2\text{O}_3\) = \(5.0\text{ cm}^3\)
- Volume of starch indicator = \(1.0\text{ cm}^3\)
- Volume of distilled water = \(24.0\text{ cm}^3\)

*(Any other consistent set of volumes is acceptable as long as the volume of KI is systematically varied, the volumes of peroxodisulfate, thiosulfate, and starch are kept constant, and the total volume of all components equals exactly 50.0 cm3)*.

3. **Chemical Role of Thiosulfate and Starch**:
- **Sodium thiosulfate (\(\text{Na}_2\text{S}_2\text{O}_3\))** reacts instantaneously with any iodine (\(\text{I}_2\)) generated in the main reaction, reducing it back to iodide (\(\text{I}^-\)): \(\text{I}_2 + 2\text{S}_2\text{O}_3^{2-} \rightarrow 2\text{I}^- + \text{S}_4\text{O}_6^{2-}\). This prevents iodine from reacting with starch until all thiosulfate has been consumed.
- **Starch indicator** serves to detect the exact moment when the thiosulfate is completely exhausted. At this point, the free iodine remaining reacts with the starch to form a highly visible, dark blue-black complex.

4. **Experimental Procedure and Rate Calculation**:
- Measure the volumes of \(\text{KI}\), \(\text{Na}_2\text{S}_2\text{O}_3\), starch indicator, and distilled water using a burette/measuring cylinder and add them to a beaker.
- In a separate container, measure the required volume of \(\text{K}_2\text{S}_2\text{O}_8\).
- Add the peroxodisulfate solution to the beaker containing the other reagents, and start the stopwatch immediately upon mixing. Stir the mixture continuously.
- Stop the stopwatch at the exact instant the solution turns blue-black. Record the time, \(t\).
- Because the same concentration of thiosulfate is used in each run, the same amount of iodine is produced before the color change occurs. Therefore, the initial rate of reaction is directly proportional to \(1/t\).

5. **Safety Hazard and Precaution**:
- Hazard: Potassium peroxodisulfate is a strong oxidizing agent / skin sensitizer / irritant.
- Precaution: Wear protective gloves and safety goggles when handling the solid or solutions, and wash spills immediately with water.

Total 10 marks:
- 1. Variables (2 marks):
- 1 mark for correctly identifying the independent variable as the concentration / volume of iodide ions.
- 1 mark for identifying the dependent variable as the time taken for the color change / rate of reaction.

- 2. Volume Table (3 marks):
- 1 mark for choosing a range of KI volumes (at least three values, e.g., 20.0, 15.0, 10.0 cm3).
- 1 mark for keeping the volumes of peroxodisulfate, thiosulfate, and starch indicator constant across all three trials.
- 1 mark for calculating the volume of distilled water correctly such that the total volume of each reaction mixture is exactly 50.0 cm3.

- 3. Role of Reagents (2 marks):
- 1 mark for explaining that thiosulfate reacts with the iodine as it is produced, delaying the starch-iodine color reaction until all thiosulfate is consumed.
- 1 mark for explaining that starch indicates the endpoint by turning blue-black when excess free iodine is present.

- 4. Procedure & Rate Representation (2 marks):
- 1 mark for a practical mixing procedure, stating that the stopwatch is started at the moment of mixing and stopped when the blue-black color appears.
- 1 mark for stating that initial rate is represented by or proportional to 1/t.

- 5. Hazard & Precaution (1 mark):
- 1 mark for identifying potassium peroxodisulfate as an oxidizing agent / irritant and recommending safety goggles or protective gloves.

1. **Mass Calculation**:
- Number of moles of \(\text{CaCl}_2\) required:
\(n = c \times V = 0.800\text{ mol dm}^{-3} \times \frac{50.0}{1000}\text{ dm}^3 = 0.0400\text{ mol}\)
- Mass of \(\text{CaCl}_2\) required:
\(m = n \times M_r = 0.0400\text{ mol} \times 111.0\text{ g mol}^{-1} = 4.44\text{ g}\).

2. **Detailed Step-by-Step Procedure**:
- Accurately weigh a dry weighing bottle on the balance, add approximately \(4.44\text{ g}\) of anhydrous \(\text{CaCl}_2\), and record the exact mass of the bottle + solid.
- Using a pipette (or a volumetric measuring cylinder), transfer exactly \(50.0\text{ cm}^3\) of distilled water into the expanded polystyrene cup.
- Place the thermometer through the hole in the lid of the polystyrene cup and place the cup inside a beaker for stability.
- Record the temperature of the water every minute from \(t = 0\) min to \(t = 3\) min.
- At exactly \(t = 4\) min, add the anhydrous \(\text{CaCl}_2\) powder from the weighing bottle into the cup. Place the lid on immediately and stir the mixture continuously using the thermometer or magnetic stirrer. Do *not* record a temperature at \(t = 4\) min.
- Reweigh the empty weighing bottle to determine the exact mass of solid transferred by difference.
- Record the temperature of the solution every minute from \(t = 5\) min to \(t = 10\) min.
- Plot a graph of Temperature (y-axis) against Time (x-axis).
- Draw a line of best fit through the pre-addition points (0-3 mins) and extrapolate it forward to the 4th minute. Draw a line of best fit through the cooling points (5-10 mins) and extrapolate it back to the 4th minute. The vertical distance between these two lines at \(t = 4\) min gives the corrected maximum temperature change, \(\Delta T\).

3. **Processing Results and Assumptions**:
- The heat energy released, \(q\), in Joules is given by: \(q = m_{\text{solution}} \times c \times \Delta T\)
- Assuming density of the solution is \(1.00\text{ g cm}^{-3}\), the mass of the solution is \(50.0\text{ g}\).
- Assuming the specific heat capacity, \(c\), of the solution is equal to that of pure water, \(4.18\text{ J g}^{-1} \text{K}^{-1}\):
\(q = 50.0 \times 4.18 \times \Delta T\text{ (J)}\)
- Convert energy to kilojoules: \(q_{\text{kJ}} = \frac{q}{1000}\)
- The molar enthalpy of solution, \(\Delta H_{\text{sol}}\), is calculated by dividing the heat released by the number of moles of \(\text{CaCl}_2\) dissolved (\(n = 0.0400\text{ mol}\)):
\(\Delta H_{\text{sol}} = -\frac{q_{\text{kJ}}}{n}\text{ kJ mol}^{-1}\) (the negative sign indicates an exothermic reaction).
- **Two major assumptions**:
1. The density of the calcium chloride solution is \(1.00\text{ g cm}^{-3}\) (same as water).
2. The specific heat capacity of the calcium chloride solution is \(4.18\text{ J g}^{-1} \text{K}^{-1}\) (same as water).

4. **Systematic Error and Improvement**:
- **Systematic Error**: Heat loss from the polystyrene cup to the cooler surroundings during the dissolving process, which lowers the maximum recorded temperature.
- **Improvement**: Use a double-walled polystyrene cup (nesting one cup inside another to create an extra insulating air layer), or use a vacuum-insulated flask, or surround the cup with cotton wool insulation inside the beaker.

Total 10 marks:
- 1. Calculation (2 marks):
- 1 mark for calculating the moles of calcium chloride needed = 0.0400 mol.
- 1 mark for calculating the mass = 4.44 g (must show units).

- 2. Procedure (4 marks):
- 1 mark for specifying the measurement of water volume using a precise instrument (pipette/measuring cylinder) and weighing the solid by difference.
- 1 mark for describing the timing sequence: recording initial temperature of water at regular intervals (e.g., every minute up to 3 mins), adding solid at 4 mins (no reading), and recording temperature from 5 mins onwards.
- 1 mark for stirring the mixture after adding the solid to ensure complete dissolution.
- 1 mark for describing the extrapolation method on a temperature-time graph to find the corrected temperature change (\(\Delta T\)) at the time of mixing (4th minute).

- 3. Calculation & Assumptions (3 marks):
- 1 mark for the correct formula for heat energy: \(q = m c \Delta T\) and using mass of solution = 50.0 g.
- 1 mark for the correct formula to find molar enthalpy: \(\Delta H_{\text{sol}} = -q / (1000 \times n)\) with negative sign included.
- 1 mark for stating both key assumptions (density is 1.00 g/cm3 AND specific heat capacity is 4.18 J/g K, or negligible heat capacity of the container).

- 4. Error & Improvement (1 mark):
- 1 mark for identifying heat loss as a systematic error and suggesting a valid method of additional insulation (e.g., nesting cups, lid, or vacuum flask).

1. **Temperature Control in Diazotisation**:
- **Why 0–5 °C**: Benzenediazonium chloride is thermally unstable; above \(5^\circ\text{C}\), it rapidly decomposes to phenol and nitrogen gas (\(\text{C}_6\text{H}_5\text{N}_2^+ + \text{H}_2\text{O} \rightarrow \text{C}_6\text{H}_5\text{OH} + \text{N}_2 + \text{H}^+\)). Below \(0^\circ\text{C}\), the rate of reaction is too slow, and the aqueous reaction mixture may freeze.
- **Maintenance & Monitoring**: Perform the reaction in a beaker immersed in an ice-water bath. Monitor the temperature throughout by keeping a thermometer directly inside the reaction mixture, adding ice to the bath as necessary to keep the temperature strictly between \(0^\circ\text{C}\) and \(5^\circ\text{C}\).

2. **Preparation & Role of NaOH in Coupling**:
- **Preparation**: Dissolve the 2-naphthol in an aqueous solution of sodium hydroxide.
- **Role of NaOH**: 2-naphthol is virtually insoluble in water, but reacting it with NaOH deprotonates the hydroxyl group to form the highly soluble sodium 2-naphtholate salt (or naphthoxide ion, \(\text{C}_{10}\text{H}_7\text{O}^-\)). The negatively charged oxygen increases the electron density on the aromatic ring, making it a much stronger nucleophile (activated ring) which is capable of undergoing electrophilic substitution (coupling) with the weak electrophile benzenediazonium ion.

3. **Isolation of Crude Dye**:
- Once the reaction is complete, the crude benzenazo-2-naphthol precipitates out of solution as an orange-red solid.
- Cool the mixture thoroughly in ice to ensure complete precipitation.
- Isolate the solid by filtration under reduced pressure (vacuum filtration) using a Buchner funnel and flask.
- Wash the collected solid residue with a small volume of cold distilled water to remove soluble salts (like NaCl) and unreacted starting materials.

4. **Recrystallisation Process**:
- Dissolve the crude solid in the **minimum volume of hot solvent** (typically ethanol or an ethanol-water mixture) to create a saturated solution.
- Perform a **hot filtration** through a pre-heated funnel to remove any insoluble impurities.
- Allow the hot filtrate to cool slowly to room temperature, and then place it in an ice bath. This slow cooling promotes the formation of large, pure crystals of the dye while keeping soluble impurities dissolved in the cold solvent.
- Filter the pure crystals under reduced pressure using a Buchner funnel.
- Wash the crystals with a small amount of **cold solvent**.
- Dry the crystals in a desiccator or a warm oven.

5. **Purity & Identity Verification**:
- Measure the melting point of the dried crystals using a melting point apparatus.
- A **sharp** melting point (occurring over a narrow range of 1-2 °C) indicates a pure sample. Comparing this experimental melting point with the literature value (Sudan I = 131–133 °C) confirms its identity.

Total 10 marks:
- 1. Temperature Control (2 marks):
- 1 mark for stating that above 5 °C the benzenediazonium salt decomposes / forms phenol, and below 0 °C the reaction is too slow / mixture freezes.
- 1 mark for specifying the use of an ice-water bath and monitoring with a thermometer placed directly in the reaction mixture.

- 2. Role of Sodium Hydroxide (2 marks):
- 1 mark for stating that NaOH reacts with 2-naphthol to form the soluble sodium 2-naphtholate salt / naphtholate ion.
- 1 mark for explaining that the phenoxide/naphtholate ion is a stronger nucleophile / has an activated ring required for coupling with the weak electrophile.

- 3. Isolation (2 marks):
- 1 mark for specifying filtration under reduced pressure / suction filtration (using a Buchner funnel/flask).
- 1 mark for washing the residue with cold distilled water to remove soluble impurities.

- 4. Recrystallisation (3 marks):
- 1 mark for dissolving the crude solid in the minimum volume of hot solvent.
- 1 mark for cooling the filtrate slowly to room temperature and then in ice to recrystallise.
- 1 mark for filtering the pure crystals under suction, washing with a small amount of cold solvent, and drying.

- 5. Purity and Identity (1 mark):
- 1 mark for suggesting a melting point determination: a sharp melting point that matches the literature value confirms purity and identity.