2024 Cambridge IAL Chemistry (9701) 模擬試題連答案詳解

An increase in pH corresponds to a decrease in the concentration of \(\text{H}^+\) ions. According to Le Chatelier's principle, a decrease in the concentration of a reactant shifts the position of equilibrium to the left. This decreases the tendency of the forward reduction reaction to occur, resulting in a less positive (decreased) electrode potential, \(E\).

1 mark for the correct option C. No partial credit.

In an octahedral field, the five d-orbitals split into two sets of different energy levels. When visible light is incident on the complex, an electron from a lower-energy d-orbital is promoted to a vacant higher-energy d-orbital by absorbing a photon of a specific frequency. The transmitted light (the complementary colour) is observed. Options A, C, and D are scientifically incorrect details of this process.

1 mark for the correct option B. No partial credit.

Comparing Experiments 1 and 2: doubling \([\text{A}]\) while keeping \([\text{B}]\) constant increases the rate by a factor of 4, so the reaction is second order with respect to \(\text{A}\). Comparing Experiments 2 and 3: doubling \([\text{B}]\) while keeping \([\text{A}]\) constant doubles the rate, so the reaction is first order with respect to \(\text{B}\). The rate equation is: \text{Rate} = \(k[\text{A}]^2[\text{B}]\). The overall order is 3. The units for \(k\) are: \(k = \frac{\text{Rate}}{[\text{A}]^2[\text{B}]} = \frac{\text{mol dm}^{-3}\text{ s}^{-1}}{(\text{mol dm}^{-3})^3} = \text{mol}^{-2}\text{ dm}^6\text{ s}^{-1}\).

1 mark for the correct option B. No partial credit.

To synthesize 3-bromobenzoic acid, the bromine and carboxylic acid groups must be in a meta-relationship. A methyl group (-\(\text{CH}_3\)) is ortho/para-directing, whereas a carboxylic acid group (-\(\text{COOH}\)) is meta-directing. Therefore, we must convert benzene to methylbenzene first using a Friedel-Crafts alkylation, oxidize the methyl group to the meta-directing carboxylic acid group using alkaline potassium manganate(VII) followed by acid, and finally perform electrophilic bromination, which will be directed to the 3- (meta) position.

1 mark for the correct option A. No partial credit.

Down Group 2, the solubility of hydroxides increases (magnesium hydroxide is sparingly soluble, barium hydroxide is much more soluble), whereas the solubility of sulfates decreases (magnesium sulfate is soluble, barium sulfate is practically insoluble).

1 mark for the correct option B. No partial credit.

Using the relation: \(\Delta H = \sum\text{bond energies of reactants} - \sum\text{bond energies of products}\). Let \(x\) be the average bond energy of the \(\text{N}-\text{H}\) bond.
Reaction involves breaking 1 \(\text{N}\equiv\text{N}\) bond and 3 \(\text{H}-\text{H}\) bonds, and forming 6 \(\text{N}-\text{H}\) bonds.
\(-92 = [945 + 3(436)] - 6x\)
\(-92 = 2253 - 6x\)
\(6x = 2345\)
\(x = 390.8\text{ kJ mol}^{-1} \approx 391\text{ kJ mol}^{-1}\).

1 mark for the correct option B. No partial credit.

The Henderson-Hasselbalch equation is: \(\text{pH} = \text{p}K_a + \log\left(\frac{[\text{conjugate base}]}{[\text{acid}]}\right)\).
First, \(\text{p}K_a = -\log(1.35 \times 10^{-5}) = 4.87\).
Since both components are mixed in equal volumes, the ratio of concentrations is equal to the ratio of their initial molarities multiplied by their volumes (or simply the ratio of their moles):
\(\frac{[\text{propanoate}]}{[\text{propanoic acid}]} = \frac{0.050}{0.100} = 0.50\).
\(\text{pH} = 4.87 + \log(0.50) = 4.87 - 0.30 = 4.57\).

1 mark for the correct option A. No partial credit.

By Avogadro's Law, gas volume ratios correspond to molar ratios.
\(1\text{ mol of } \text{C}_x\text{H}_y + 5\text{ mol of } \text{O}_2 \rightarrow 3\text{ mol of } \text{CO}_2 + z\text{ mol of } \text{H}_2\text{O}\).
From carbon balance: \(x = 3\).
From oxygen balance: \(2 \times 5 = (2 \times 3) + z \implies 10 = 6 + z \implies z = 4\). Therefore, \(4\text{ H}_2\text{O}\) are produced.
From hydrogen balance: \(y = 2 \times 4 = 8\).
The formula is \(\text{C}_3\text{H}_8\).

1 mark for the correct option B. No partial credit.

Using the Nernst equation: \( E = E^\ominus + \frac{0.059}{z} \log_{10} [\text{Ag}^+] \) Here, \( z = 1 \) because the reduction of silver(I) involves the transfer of one electron: \( \text{Ag}^+ + e^- \rightarrow \text{Ag} \). Substituting the given values: \( E = 0.80 + \frac{0.059}{1} \log_{10}(2.5 \times 10^{-3}) \) Since \( \log_{10}(2.5 \times 10^{-3}) \approx -2.602 \): \( E = 0.80 + 0.059 \times (-2.602) \approx 0.80 - 0.153 = +0.647\text{ V} \). This rounds to \( +0.65\text{ V} \).

Award 1 mark for the correct calculation showing \( E \approx +0.65\text{ V} \). Incorrect values arising from misinterpreting \( z \) or sign errors in the logarithm do not get the mark.

First, determine the orders of reaction:
- From Experiment 1 to 2, \( [\text{A}] \) doubles while \( [\text{B}] \) remains constant, and the rate increases by \( 4\times \) (since \( 8.0 \times 10^{-4} / 2.0 \times 10^{-4} = 4 \)). Thus, the reaction is second order with respect to \( \text{A} \).
- From Experiment 2 to 3, \( [\text{B}] \) doubles while \( [\text{A}] \) remains constant, and the rate doubles (since \( 1.6 \times 10^{-3} / 8.0 \times 10^{-4} = 2 \)). Thus, the reaction is first order with respect to \( \text{B} \).

The rate equation is: \( \text{Rate} = k[\text{A}]^2[\text{B}] \).

Using Experiment 1 data to calculate \( k \):
\( 2.0 \times 10^{-4}\text{ mol dm}^{-3}\text{ s}^{-1} = k \times (0.10\text{ mol dm}^{-3})^2 \times (0.10\text{ mol dm}^{-3}) \)
\( 2.0 \times 10^{-4} = k \times (1.0 \times 10^{-3}) \)
\( k = 0.20 \).

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

Award 1 mark for identifying the correct rate equation, determining the numerical rate constant value of 0.20, and assigning the correct unit \( \text{dm}^6\text{ mol}^{-2}\text{ s}^{-1} \).

In a transition metal complex, the ligands coordinate to the metal ion, splitting the five d-orbitals into two non-degenerate energy levels of different energies. Electrons in lower energy d-orbitals absorb light in the visible region (red-orange) to get promoted to a higher d-orbital (d-d transition). The unabsorbed light (the complementary colour, which is blue) is transmitted, rendering the solution blue. Emission is not responsible for the color of transition metal complexes, and copper(II) does not have a fully occupied d-subshell.

Award 1 mark for identifying that the d-orbitals split into non-degenerate levels, and that the absorption of light during d-d transition leaves the complementary blue light to be transmitted.

Step 1: Conversion of a primary alcohol (propan-1-ol) into a halogenoalkane (1-chloropropane) requires a halogenating agent such as phosphorus pentachloride (\( \text{PCl}_5 \)).
Step 2: Conversion of a halogenoalkane into a primary amine (propylamine) is achieved by heating with excess ammonia (\( \text{NH}_3 \)) dissolved in ethanol under pressure (in a sealed tube) to prevent the escape of ammonia gas.

Award 1 mark for choosing the option that contains correct reagents and reaction conditions for both steps (nucleophilic substitution of alcohol, then nucleophilic substitution of halogenoalkane).

The \( \text{Mg}^{2+} \) ion is smaller than the \( \text{Ba}^{2+} \) ion. Because both ions carry a \( 2+ \) charge, the magnesium ion has a much higher charge density. This allows it to more strongly polarise (distort) the electron cloud of the adjacent nitrate anion (\( \text{NO}_3^- \)), weakening the covalent \( \text{N-O} \) bonds within the anion and causing it to decompose at a lower thermal threshold.

Award 1 mark for the correct explanation based on ionic radius, charge density, and the polarization of the nitrate anion.

Iodide (\( \text{I}^- \)) is a powerful reducing agent. It reduces concentrated sulfuric acid (\( \text{H}_2\text{SO}_4 \)) to sulfur dioxide (\( \text{SO}_2 \)), sulfur (\( \text{S} \)), and hydrogen sulfide (\( \text{H}_2\text{S} \)), while itself being oxidised to iodine (\( \text{I}_2 \)).
- The acidic gas turning blue litmus paper red is \( \text{HI} \) or \( \text{SO}_2 \).
- The gas turning acidified potassium dichromate(VI) green is \( \text{SO}_2 \) (reducing \( \text{Cr(VI)} \) to \( \text{Cr(III)} \)).
- The purple vapor is \( \text{I}_2 \).
Fluoride and chloride do not reduce concentrated sulfuric acid. Bromide reduces it but only produces a red-brown vapor (\( \text{Br}_2 \)) and no purple vapor.

Award 1 mark for recognizing that only iodide can reduce concentrated sulfuric acid to yield a purple vapor (\( \text{I}_2 \)) alongside \( \text{SO}_2 \).

First calculate the initial moles:
\( n(\text{CH}_3\text{COOH}) = 0.0500 \times 0.100 = 0.00500\text{ mol} \)
\( n(\text{NaOH}) = 0.0500 \times 0.0400 = 0.00200\text{ mol} \)

Upon mixing, NaOH reacts with ethanoic acid:
\( \text{CH}_3\text{COOH} + \text{NaOH} \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O} \)

Remaining moles after neutralization:
\( n(\text{CH}_3\text{COOH})_{\text{rem}} = 0.00500 - 0.00200 = 0.00300\text{ mol} \)
\( n(\text{CH}_3\text{COO}^-)_{\text{formed}} = 0.00200\text{ mol} \)

Using the buffer equation:
\( \text{pH} = \text{p}K_{\text{a}} + \log_{10} \left( \frac{[\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]} \right) \)
\( \text{p}K_{\text{a}} = -\log_{10}(1.75 \times 10^{-5}) = 4.757 \)
\( \text{pH} = 4.757 + \log_{10}\left(\frac{0.00200}{0.00300}\right) = 4.757 + (-0.176) = 4.581 \approx 4.58 \).

Award 1 mark for the correct buffer pH calculation resulting in \( 4.58 \).

By Hess's Law, the enthalpy of reaction can be calculated from the enthalpies of combustion of reactants and products:
\( \Delta H_{\text{r}}^\ominus = \sum \Delta H_{\text{c}}^\ominus(\text{reactants}) - \sum \Delta H_{\text{c}}^\ominus(\text{products}) \)

Reactants: \( 1\text{ mol of } \text{C}_3\text{H}_4 \) and \( 2\text{ mol of } \text{H}_2 \)
\( \sum \Delta H_{\text{c}}^\ominus(\text{reactants}) = -1938 + 2 \times (-286) = -2510\text{ kJ mol}^{-1} \)

Products: \( 1\text{ mol of } \text{C}_3\text{H}_8 \)
\( \sum \Delta H_{\text{c}}^\ominus(\text{products}) = -2220\text{ kJ mol}^{-1} \)

\( \Delta H_{\text{r}}^\ominus = -2510 - (-2220) = -290\text{ kJ mol}^{-1} \).

Award 1 mark for the correct application of Hess's Law with correct stoichiometry of \( \text{H}_2 \) yielding \( -290\text{ kJ mol}^{-1} \).

The cell potential for the reaction \(2\text{Fe}^{3+} + 2\text{I}^- \rightarrow 2\text{Fe}^{2+} + \text{I}_2\) is calculated using \(E^\theta_{\text{cell}} = E^\theta_{\text{reduction}} - E^\theta_{\text{oxidation}}\). Here, \(\text{Fe}^{3+}\) is reduced and \(\text{I}^-\rightleftharpoons \text{I}_2\) is oxidized. Thus, \(E^\theta_{\text{cell}} = +0.77\text{ V} - (+0.54\text{ V}) = +0.23\text{ V}\). Since \(E^\theta_{\text{cell}}\) is positive, the reaction is thermodynamically feasible under standard conditions.

Award 1 mark for the correct calculation of \(E^\theta_{\text{cell}}\) (+0.23 V) and the correct determination of its feasibility (feasible).

Let the original rate be \(R_1 = k[A][B]^2\). The new concentrations are \([A]' = 2[A]\) and \([B]' = 0.5[B]\). The new rate is \(R_2 = k[A]'[B]'^2 = k(2[A])(0.5[B])^2 = k(2[A])(0.25[B]^2) = 0.5k[A][B]^2 = 0.5 R_1\). Thus, the rate is halved (multiplied by 0.5).

Award 1 mark for correctly substituting the concentration factors into the rate equation to deduce that the rate is multiplied by 0.5.

When ligands approach a copper(II) ion, the degenerate d-orbitals split into two sets of non-degenerate orbitals. Electrons absorb energy in the red-orange region of the visible spectrum to be promoted from the lower energy set of d-orbitals to the higher energy set (d-d transition). The complementary blue color of the light that is not absorbed is seen.

Award 1 mark for identifying that color arises from the absorption of light during d-d promotion of electrons, resulting in the transition of complementary blue light.

Step 1 is diazotisation of phenylamine, which requires nitrous acid (generated in situ from \(\text{NaNO}_2\) and dilute \(\text{HCl}\)) at a low temperature (below \(10\ ^\circ\text{C}\)) to prevent decomposition of the benzenediazonium ion. Step 2 is a coupling reaction with phenol dissolved in an alkaline solution (aqueous \(\text{NaOH}\)) kept cold.

Award 1 mark for identifying correct reagents and cold temperature conditions for both the diazotisation and coupling steps.

As Group 2 is descended from magnesium to barium: 1) The solubility of sulfates decreases due to the lattice energy and hydration energy trends. 2) The thermal stability of carbonates increases because the metal cation radius increases, which decreases its charge density and its polarizing power on the carbonate ion's electron cloud.

Award 1 mark for identifying the correct trends: decreasing solubility of sulfates and increasing thermal stability of carbonates.

Iodide ions reduce concentrated sulfuric acid to several products. The purple vapour is iodine, \(\text{I}_2(g)\). The yellow solid is elemental sulfur, \(\text{S}(s)\). The gas that reduces acidified potassium dichromate(VI) paper from orange (dichromate(VI) ions) to green (chromium(III) ions) is sulfur dioxide, \(\text{SO}_2\).

Award 1 mark for correctly identifying sulfur as the yellow solid and sulfur dioxide as the reducing gas that turns dichromate paper green.

Since the forward reaction is exothermic, an increase in temperature shifts the position of equilibrium in the endothermic direction (to the left), which decreases the concentration of products relative to reactants, hence decreasing \(K_c\). Additionally, increasing the temperature increases the kinetic energy of all reacting particles, leading to more frequent and energetic collisions, so the rate of the reverse reaction increases.

Award 1 mark for correctly determining that \(K_c\) decreases and the rate of the reverse reaction increases.

The peak at \(1720\text{ cm}^{-1}\) corresponds to the carbonyl bond (\(\text{C}=\text{O}\)). The broad band at \(2500\text{--}3000\text{ cm}^{-1}\) is characteristic of the \(\text{O}-\text{H}\) group of a carboxylic acid. With the formula \(\text{C}_3\text{H}_6\text{O}_2\), this must be propanoic acid, \(\text{CH}_3\text{CH}_2\text{COOH}\).

Award 1 mark for correctly matching the infrared spectrum data (\(\text{C}=\text{O}\) and carboxylic acid \(\text{O}-\text{H}\)) with propanoic acid.

Use the Nernst equation: \(E = E^\ominus + \frac{0.059}{z} \log \frac{[\text{Fe}^{3+}]}{[\text{Fe}^{2+}]}\). Here, \(z = 1\) (since 1 electron is transferred), \([\text{Fe}^{3+}] = 0.10\text{ mol dm}^{-3}\), and \([\text{Fe}^{2+}] = 1.0 \times 10^{-3}\text{ mol dm}^{-3}\). This gives the concentration ratio as \(\frac{0.10}{0.0010} = 100\). Since \(\log_{10}(100) = 2\), the equation becomes: \(E = +0.77 + 0.059 \times 2 = +0.77 + 0.118 = +0.888\text{ V}\), which rounds to \(+0.89\text{ V}\).

1 mark for the correct calculation of \(E = +0.89\text{ V}\). Reject \(+0.65\text{ V}\) (which is obtained by incorrectly inverting the log ratio) or \(+0.77\text{ V}\) (standard conditions).

First, determine the orders of reaction:
1. Comparing Exp 1 and Exp 2: when \([A]\) doubles, the rate doubles. Hence, the order with respect to \(A\) is 1.
2. Comparing Exp 1 and Exp 3: when \([B]\) doubles, the rate quadruples. Hence, the order with respect to \(B\) is 2.
3. Comparing Exp 1 and Exp 4: when \([C]\) doubles, the rate remains unchanged. Hence, the order with respect to \(C\) is 0.

Therefore, the rate equation is: \(\text{Rate} = k[A][B]^2\).
Substitute values from Experiment 1 to find \(k\):
\(2.0 \times 10^{-3}\text{ mol dm}^{-3}\text{ s}^{-1} = k (0.10\text{ mol dm}^{-3}) (0.10\text{ mol dm}^{-3})^2 = k (1.0 \times 10^{-3}\text{ mol}^3\text{ dm}^{-9})\)
\(k = 2.0\).
Units of \(k\) are: \(\frac{\text{mol dm}^{-3}\text{ s}^{-1}}{(\text{mol dm}^{-3})(\text{mol dm}^{-3})^2} = \text{dm}^6\text{ mol}^{-2}\text{ s}^{-1}\).

1 mark for identifying the correct orders of reaction (1st for A, 2nd for B, 0th for C), calculating the rate constant value as 2.0, and deriving the correct units as \(\text{dm}^6\text{ mol}^{-2}\text{ s}^{-1}\).

The addition of concentrated hydrochloric acid results in ligand exchange where the octahedral aquo-complex \([\text{Cu}(\text{H}_2\text{O})_6]^{2+}\) is converted into the tetrahedral chlorido-complex \([\text{CuCl}_4]^{2-}\). Changing the coordination number and geometry from octahedral to tetrahedral alters the splitting pattern and decreases the energy gap (\(\Delta\)) of the d-orbitals. This alters the wavelengths of light absorbed and therefore changes the complementary color observed. The oxidation state of copper remains \(+2\) in both complexes, and chloride is a weaker field ligand than water, so B is the only correct and accurate explanation.

1 mark for identifying that the change in geometry from octahedral to tetrahedral alters the d-orbital splitting, changing the energy gap and the color of the complex.

The target molecule is 3-bromobenzoic acid. The methyl group (\(-\text{CH}_3\)) on methylbenzene is ortho/para-directing (2,4-directing). If we brominate first (Option A), we obtain 2-bromotoluene and 4-bromotoluene, which on oxidation yield 2-bromobenzoic acid and 4-bromobenzoic acid. However, if we oxidize the methyl group first using alkaline \(\text{KMnO}_4(aq)\) followed by acidification (Option B), we obtain benzoic acid. The carboxylic acid group (\(-\text{COOH}\)) is meta-directing (3-directing). Subsequent electrophilic bromination using \(\text{Br}_2\) and \(\text{AlCl}_3\) catalyst will yield 3-bromobenzoic acid as the major product.

1 mark for selecting the sequence that first oxidizes the 2,4-directing methyl group to a 3-directing carboxylic acid group before performing the electrophilic aromatic substitution step.

The solubility of Group 2 hydroxides increases down the group, meaning \(\text{Mg(OH)}_2\) is the least soluble hydroxide and forms a heavy precipitate most readily. Conversely, the solubility of Group 2 sulfates decreases down the group, meaning \(\text{MgSO}_4\) is highly soluble and does not precipitate out in significant amounts compared to the highly insoluble \(\text{BaSO}_4\). Thus, magnesium nitrate (\(\text{Mg(NO}_3)_2\)) fits both observations.

1 mark for identifying magnesium nitrate as having the least soluble hydroxide (most precipitate with NaOH) and the most soluble sulfate (least precipitate with sulfuric acid) among the Group 2 metals.

Potassium bromide reacts with concentrated \(\text{H}_2\text{SO}_4\) to form misty fumes of \(\text{HBr}\). Because bromide is a moderate reducing agent, it reduces some of the concentrated \(\text{H}_2\text{SO}_4\) to sulfur dioxide, \(\text{SO}_2\) (an acidic, reducing gas that turns acidified potassium dichromate(VI) paper green), and is itself oxidized to bromine, \(\text{Br}_2\) (brown vapor). Hence, B is correct. Chloride and fluoride ions are not oxidized by concentrated \(\text{H}_2\text{SO}_4\), while iodide ions are strongly oxidized to produce purple iodine vapor, yellow solid sulfur, and rotten-egg smelling \(\text{H}_2\text{S}\).

1 mark for identifying the correct products and observations corresponding to the reduction of concentrated sulfuric acid by hydrogen bromide.

Construct an ICE table to find equilibrium moles:
- Initial: \(n(\text{N}_2\text{O}_4) = 1.00\text{ mol}\), \(n(\text{NO}_2) = 0\text{ mol}\)
- Change: Since \(0.60\text{ mol}\) of \(\text{NO}_2\) is formed, \(\frac{0.60}{2} = 0.30\text{ mol}\) of \(\text{N}_2\text{O}_4\) reacted.
- Equilibrium: \(n(\text{N}_2\text{O}_4) = 1.00 - 0.30 = 0.70\text{ mol}\); \(n(\text{NO}_2) = 0.60\text{ mol}\).
- Total moles: \(n_{total} = 0.70 + 0.60 = 1.30\text{ mol}\).

Find partial pressures:
\(p(\text{N}_2\text{O}_4) = \frac{0.70}{1.30} \times 2.00\text{ atm} = 1.077\text{ atm}\)
\(p(\text{NO}_2) = \frac{0.60}{1.30} \times 2.00\text{ atm} = 0.923\text{ atm}\)

Calculate \(K_p\):
\(K_p = \frac{p(\text{NO}_2)^2}{p(\text{N}_2\text{O}_4)} = \frac{(0.923)^2}{1.077} = 0.791\text{ atm} \approx 0.79\text{ atm}\).

1 mark for calculating the equilibrium mole values (0.70 and 0.60), finding total moles (1.30), calculating partial pressures, and correctly evaluating \(K_p = 0.79\text{ atm}\).

1. Calculate the number of moles of \(\text{Na}_2\text{CO}_3\) in the original sample:
\(n = \frac{4.24\text{ g}}{106.0\text{ g mol}^{-1}} = 0.0400\text{ mol}\).

2. Find the moles of \(\text{Na}_2\text{CO}_3\) present in \(25.0\text{ cm}^3\) of solution:
\(n_{25.0} = 0.0400\text{ mol} \times \frac{25.0\text{ cm}^3}{250\text{ cm}^3} = 0.00400\text{ mol}\).

3. Sodium carbonate dissociates completely into 3 ions per formula unit: \(\text{Na}_2\text{CO}_3(s) \rightarrow 2\text{Na}^+(aq) + \text{CO}_3^{2-}(aq)\).
Total moles of ions in the \(25.0\text{ cm}^3\) sample = \(0.00400\text{ mol} \times 3 = 0.0120\text{ mol}\).

4. Calculate the total number of ions:
\(N = 0.0120\text{ mol} \times 6.02 \times 10^{23}\text{ mol}^{-1} = 7.22 \times 10^{21}\) ions.

1 mark for calculating the correct moles of sodium carbonate in the aliquot, multiplying by 3 to find total moles of ions, and multiplying by the Avogadro constant to obtain \(7.22 \times 10^{21}\).

1. Apply the Nernst equation to the silver half-cell:
\(E = E^\ominus + \frac{0.059}{z} \log [\text{Ag}^+]\)
Here, \(z = 1\) and \([\text{Ag}^+] = 0.010\text{ mol dm}^{-3}\).
\(E = +0.80 + 0.059 \log(0.010) = 0.80 + 0.059(-2) = +0.68\text{ V}\).

2. Determine the cell potential under these conditions:
The iron half-cell remains under standard conditions, so its potential is \(E(\text{Fe}^{3+}/\text{Fe}^{2+}) = +0.77\text{ V}\).
Since \(+0.77\text{ V} > +0.68\text{ V}\), the iron half-cell is the cathode (reduction occurs here) and the silver half-cell is the anode (oxidation occurs here).
\(E_{\text{cell}} = E_{\text{reduction}} - E_{\text{oxidation}} = E(\text{Fe}^{3+}/\text{Fe}^{2+}) - E(\text{Ag}^+/\text{Ag}) = +0.77\text{ V} - (+0.68\text{ V}) = +0.09\text{ V}\).

1 mark for the correct option A. Method: use Nernst equation to find the new silver half-cell potential, then subtract it from the iron half-cell potential to find the positive overall cell potential.

- Comparing Exp 1 and Exp 2: doubling \([\text{P}]\) while keeping others constant doubles the rate. Order with respect to P is 1.
- Comparing Exp 1 and Exp 3: doubling \([\text{Q}]\) while keeping others constant quadruples the rate (from \(4.0 \times 10^{-4}\) to \(1.6 \times 10^{-3}\)). Order with respect to Q is 2.
- Comparing Exp 3 and Exp 4: doubling \([\text{R}]\) while keeping others constant does not change the rate. Order with respect to R is 0.
- Therefore, the overall order of reaction is \(1 + 2 + 0 = 3\).
- The rate equation is: \(\text{Rate} = k[\text{P}][\text{Q}]^2\).
- Solving for the units of \(k\): \([k] = \frac{\text{mol dm}^{-3}\text{ s}^{-1}}{(\text{mol dm}^{-3})(\text{mol dm}^{-3})^2} = \text{mol}^{-2}\text{ dm}^6\text{ s}^{-1}\), which is written as \(\text{dm}^6 \text{mol}^{-2} \text{s}^{-1}\).

1 mark for correct option A. Method: find individual orders, sum them to get overall order, and derive units of k.

In the presence of ligands, the d-orbitals of a transition metal ion split into two groups of different energy levels. When white light is incident on the complex, electrons absorb a specific frequency/wavelength of light in the visible range to transition from a lower-energy d-orbital to a higher-energy d-orbital (d-d transition). The light that is not absorbed is transmitted or reflected, which is the complementary color seen by the observer.

1 mark for the correct explanation (Option A). Options B, C, and D represent incorrect mechanisms of absorption/emission.

- Step 1 is a free-radical side-chain substitution of methylbenzene to form (chloromethyl)benzene. This requires \(\text{Cl}_2\) gas and ultraviolet (UV) light. (Using \(\text{AlCl}_3\) as in Option B would cause electrophilic substitution on the benzene ring instead).
- Step 2 is a nucleophilic substitution of the chlorine atom by a cyano group, which requires heating with \(\text{KCN}\) dissolved in ethanol under reflux.
- Step 3 is the acid hydrolysis of the nitrile group to form a carboxylic acid, requiring heating under reflux with a dilute aqueous acid such as \(\text{HCl}(\text{aq})\).

1 mark for the correct reagent/condition set (Option A). Reject options with ring halogenation, incorrect nucleophiles, or incorrect hydrolysis reagents.

As the group is descended, the size (cationic radius) of the \(M^{2+}\) ion increases while the charge remains \(+2\). This results in a lower charge density of the cation down the group. The lower the charge density, the less the cation polarizes and distorts the electron cloud of the nitrate anion (\(\text{NO}_3^-\)). Consequently, the covalent bonds within the nitrate ion are weakened less, which increases the thermal stability of the compound down the group (it requires a higher temperature to decompose).

1 mark for the correct trend and ionic polarization explanation (Option A). Reject options that state polarization increases or attribute the trend incorrectly to lattice energy or ionization energy.

- The observation of a purple vapor indicates the formation of iodine, \(\text{I}_2(\text{g})\). This means the halide ion is iodide, \(\text{I}^-\).
- In the reaction, iodide ions reduce the sulfur in concentrated sulfuric acid from \(+6\) to \(+4\) in sulfur dioxide, \(\text{SO}_2\) (and further to \(0\) in \(\text{S}\) and \(-2\) in \(\text{H}_2\text{S}\)).
- The gas that specifically turns acidified potassium dichromate(VI) from orange to green is sulfur dioxide, \(\text{SO}_2\), because it is a reducing agent (reducing \(\text{Cr}_2\text{O}_7^{2-}\) to green \(\text{Cr}^{3+}\) ions).

1 mark for identifying both the iodide ion and sulfur dioxide as the correct gas (Option A).

- In reaction C, three moles of gaseous reactants (\(2\text{H}_2 + \text{O}_2\)) react to form two moles of liquid water (\(2\text{H}_2\text{O}\)). This represents a massive reduction in disorder because the gaseous state has far higher entropy than the liquid state. Thus, the standard entropy change, \(\Delta S^\ominus\), is highly negative.
- In reactions A and B, there is an increase in the number of moles of gas, which leads to a positive entropy change (\(\Delta S^\ominus > 0\)).
- In reaction D, the solid ionic lattice dissolves into mobile, aqueous hydrated ions, which increases disorder (\(\Delta S^\ominus > 0\)).

1 mark for the correct reaction showing a reduction in entropy (Option C).

1. Find the amount in moles of \(\text{Al}_2(\text{SO}_4)_3\) in solution:
\(n = \text{concentration} \times \text{volume}\)
\(n = 0.120\text{ mol dm}^{-3} \times \left(\frac{25.0}{1000}\right)\text{ dm}^3 = 3.00 \times 10^{-3}\text{ mol}\).

2. Determine the number of ions per formula unit of \(\text{Al}_2(\text{SO}_4)_3\):
\(\text{Al}_2(\text{SO}_4)_3(\text{aq}) \rightarrow 2\text{Al}^{3+}(\text{aq}) + 3\text{SO}_4^{2-}(\text{aq})\).
Each formula unit dissociates into \(2 + 3 = 5\) ions.

3. Calculate the total moles of ions in the solution:
\(n_{\text{ions}} = 5 \times 3.00 \times 10^{-3}\text{ mol} = 1.50 \times 10^{-2}\text{ mol}\).

4. Calculate the total number of ions:
\(N = n_{\text{ions}} \times L = 1.50 \times 10^{-2}\text{ mol} \times 6.02 \times 10^{23}\text{ mol}^{-1} = 9.03 \times 10^{21}\) ions.

1 mark for the correct calculation yielding Option A. Common mistakes: forgetting the dissociation factor 5 yields Option B; calculating only sulfate ions yields Option C; calculating only aluminium ions yields Option D.

(a) The salt bridge completes the electrical circuit and allows the migration of ions between the two half-cells to maintain charge neutrality. Potassium nitrate, \(\text{KNO}_3\), is a suitable electrolyte because neither the potassium ions nor the nitrate ions react with or precipitate the ions in either half-cell.

(b) The standard cell potential is given by:
\(E^\ominus_{\text{cell}} = E^\ominus_{\text{reduction}} - E^\ominus_{\text{oxidation}}\)
Since \(E^\ominus(\text{Ag}^+/\text{Ag}) = +0.80\text{ V}\) is more positive than \(E^\ominus(\text{Fe}^{3+}/\text{Fe}^{2+}) = +0.77\text{ V}\), the silver half-cell undergoes reduction and the iron half-cell undergoes oxidation under standard conditions.
\(E^\ominus_{\text{cell}} = +0.80 - (+0.77) = +0.03\text{ V}\)
Spontaneous equation:
\(\text{Fe}^{2+}(\text{aq}) + \text{Ag}^+(\text{aq}) \rightleftharpoons \text{Fe}^{3+}(\text{aq}) + \text{Ag}(\text{s})\)

(c)(i) For the \(\text{Ag}^+/\text{Ag}\) electrode, \(z = 1\) (1 electron transferred). The oxidised species is \(\text{Ag}^+(\text{aq})\) and the reduced species is \(\text{Ag}(\text{s})\) (which has activity = 1).
Using the Nernst equation:
\(E = E^\ominus + 0.059\log [\text{Ag}^+]\)
\(E = +0.80 + 0.059\log(0.050) = +0.80 + 0.059(-1.301) = 0.80 - 0.0768 = +0.723\text{ V}\)

(ii) The iron half-cell remains under standard conditions, so its potential is \(+0.77\text{ V}\). Comparing the two electrode potentials:
\(E(\text{Fe}^{3+}/\text{Fe}^{2+}) = +0.77\text{ V}\) and \(E(\text{Ag}^+/\text{Ag}) = +0.723\text{ V}\).
Since \(E(\text{Fe}^{3+}/\text{Fe}^{2+})\) is now more positive, the iron half-cell will undergo reduction and the silver half-cell will undergo oxidation.
The new cell potential for the spontaneous reaction is:
\(E_{\text{cell}} = 0.77 - 0.723 = +0.047\text{ V}\)
The spontaneous reaction is reversed compared to standard conditions, and is now:
\(\text{Fe}^{3+}(\text{aq}) + \text{Ag}(\text{s}) \rightleftharpoons \text{Fe}^{2+}(\text{aq}) + \text{Ag}^+(\text{aq})\)

(a) [1 mark] for stating that the salt bridge completes the circuit / maintains electrical neutrality.
[1 mark] for naming a suitable electrolyte (e.g., potassium nitrate / KNO3 or ammonium nitrate / NH4NO3) (Reject: sodium chloride / KCl due to precipitation of AgCl).

(b) [1 mark] for calculation of standard cell potential: +0.03 V.
[1 mark] for correct overall equation with correct state symbols: Fe2+(aq) + Ag+(aq) -> Fe3+(aq) + Ag(s).

(c)(i) [1 mark] for substitution of correct values into the Nernst equation (z = 1, [Ag+] = 0.050).
[1 mark] for calculating the log term correctly (-1.30).
[1 mark] for final value: +0.723 V (or +0.72 V).

(ii) [1 mark] for calculating the new cell potential: 0.77 - 0.723 = +0.047 V (or -0.047 V with appropriate explanation of sign).
[1 mark] for predicting that the spontaneous reaction is reversed.
[1 mark] for explaining that the Fe3+/Fe2+ half-cell has a higher potential than the Ag+/Ag half-cell under these conditions, so Fe3+ is reduced and Ag is oxidised.

(a)(i) Order with respect to propanone:
Compare Experiments 1 and 2: \([\text{I}_2]\) and \([\text{H}^+]\) are constant. \([\text{CH}_3\text{COCH}_3]\) is doubled (0.40 to 0.80), and the initial rate is doubled (\(1.20 \times 10^{-5}\) to \(2.40 \times 10^{-5}\)). Since the rate increases by a factor of \(2^1\), the order with respect to propanone is 1.
(ii) Order with respect to iodine:
Compare Experiments 1 and 3: \([\text{CH}_3\text{COCH}_3]\) and \([\text{H}^+]\) are constant. \([\text{I}_2]\) is doubled (0.020 to 0.040), and the initial rate remains constant (\(1.20 \times 10^{-5}\)). Since the rate is unaffected by changing the concentration, the order with respect to iodine is 0.
(iii) Order with respect to \(\text{H}^+\):
Compare Experiments 1 and 4: \([\text{CH}_3\text{COCH}_3]\) and \([\text{I}_2]\) are constant. \([\text{H}^+]\) is doubled (0.40 to 0.80), and the initial rate is doubled (\(1.20 \times 10^{-5}\) to \(2.40 \times 10^{-5}\)). Since the rate increases by a factor of \(2^1\), the order with respect to \(\text{H}^+\) is 1.

(b)(i) Rate = \(k[\text{CH}_3\text{COCH}_3][\text{H}^+]\)

(ii) Using data from Experiment 1:
\(1.20 \times 10^{-5} = k (0.40)(0.40)\)
\(1.20 \times 10^{-5} = 0.16 k\)
\(k = \frac{1.20 \times 10^{-5}}{0.16} = 7.50 \times 10^{-5}\)
Units: \(\text{rate} / ([\text{CH}_3\text{COCH}_3][\text{H}^+]) = (\text{mol dm}^{-3}\text{ s}^{-1}) / (\text{mol dm}^{-3} \times \text{mol dm}^{-3}) = \text{dm}^3\text{ mol}^{-1}\text{ s}^{-1}\)

(c) A consistent mechanism must have a rate-determining step involving only species that appear in the rate equation (propanone and \(\text{H}^+\)).
Step 1 (slow, rate-determining step): \(\text{CH}_3\text{COCH}_3 + \text{H}^+ \rightarrow [\text{CH}_3\text{C(OH)CH}_3]^+\) (protonation/enolisation)
Step 2 (fast): \([\text{CH}_3\text{C(OH)CH}_3]^+ + \text{I}_2 \rightarrow \text{CH}_3\text{COCH}_2\text{I} + \text{H}^+ + \text{I}^-\)

(a) [3 marks] for correctly identifying orders: 1st order for propanone, 0th order for iodine, and 1st order for H+.
[1 mark] for clear logical reasoning comparing the correct pairs of experiments (Expts 1 & 2 for propanone; Expts 1 & 3 for iodine; Expts 1 & 4 for H+).

(b)(i) [1 mark] for correct rate equation: Rate = k[CH3COCH3][H+] (must match deduced orders).
(ii) [1 mark] for correct substitution of values from Experiment 1.
[1 mark] for calculating the value of k = 7.50 x 10^-5 (accept 7.5 x 10^-5).
[1 mark] for correct units: dm3 mol^-1 s^-1.

(c) [1 mark] for stating that the slow (rate-determining) step must involve propanone and H+, but not iodine.
[1 mark] for showing a fast step in which the intermediate reacts with iodine.

(a) A ligand is a molecule or ion that has at least one lone pair of electrons which it donates to a central transition metal ion to form a coordinate (dative covalent) bond.

(b)(i) When excess concentrated ammonia is added, four water ligands are replaced by ammonia ligands. The formula of complex \(\mathbf{A}\) is \([\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}\). The geometry of this complex is distorted octahedral (where the four equatorial Cu-N bonds are shorter and stronger than the two axial Cu-O bonds).
(ii) This reaction is a ligand substitution (or ligand replacement) reaction.
Equation: \([\text{Cu}(\text{H}_2\text{O})_6]^{2+}(\text{aq}) + 4\text{NH}_3(\text{aq}) \rightleftharpoons [\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}(\text{aq}) + 4\text{H}_2\text{O}(\text{l})\)

(c) In a free transition metal gaseous ion, all five 3d orbitals are degenerate (have the same energy). However, in the presence of ligands, the d-orbitals split into two groups of different energy levels (e.g., three lower-energy and two higher-energy orbitals in an octahedral environment). An electron in a lower d-orbital can absorb a specific frequency of electromagnetic radiation in the visible spectrum and be promoted to a higher-energy d-orbital (a d-d transition). The energy change corresponds to \(\Delta E = h\nu\). The unabsorbed wavelengths of light are transmitted (not absorbed) and combine to give the complementary colour that we observe.

(a) [1 mark] for defining ligand as a species containing a lone pair of electrons.
[1 mark] for explaining that they bind via coordinate / dative covalent bonds.

(b)(i) [1 mark] for formula: [Cu(NH3)4(H2O)2]2+.
[1 mark] for geometry: distorted octahedral / octahedral.
(ii) [1 mark] for stating ligand substitution / ligand exchange.
[1 mark] for correct balanced equation.

(c) [1 mark] for stating that the 3d orbitals split into two energy levels (by ligands).
[1 mark] for explaining that an electron absorbs energy / light and is promoted from a lower to a higher d-orbital (d-d transition).
[1 mark] for relating the absorbed energy to a specific frequency/wavelength in the visible spectrum (ΔE = hν).
[1 mark] for stating that the remaining wavelengths are transmitted / seen as the complementary colour.

(a) A successful synthetic route to obtain 3-bromobenzoic acid is:

Step 1: Convert benzene into methylbenzene.
- Reagents & Conditions: Chloromethane (\(\text{CH}_3\text{Cl}\)) and anhydrous aluminium chloride catalyst (\(\text{AlCl}_3\)) at room temperature.
- Product: Methylbenzene (toluene), \(\text{C}_6\text{H}_5\text{CH}_3\).

Step 2: Oxidise methylbenzene to benzoic acid.
- Reagents & Conditions: Alkaline potassium manganate(VII) (\(\text{KMnO}_4\)), heated under reflux, followed by acidification with dilute hydrochloric acid or sulfuric acid.
- Product: Benzoic acid, \(\text{C}_6\text{H}_5\text{COOH}\).

Step 3: Brominate benzoic acid.
- Reagents & Conditions: Bromine (\(\text{Br}_2\)) and iron(III) bromide catalyst (\(\text{FeBr}_3\)) or iron filings.
- Product: 3-bromobenzoic acid.

(b) The order of steps is crucial because of the directing effects of the substituents on the benzene ring:
- The carboxylic acid group (\(-\text{COOH}\)) formed in Step 2 is an electron-withdrawing group and is meta-directing (or 3-directing). Thus, when bromination is performed on benzoic acid, the electrophilic bromine is directed to position 3, yielding the desired 3-bromobenzoic acid.
- Conversely, the methyl group (\(-\text{CH}_3\)) is electron-donating and is an ortho/para-director (or 2,4-director). If bromination were carried out immediately after Step 1 (on methylbenzene), the bromine would be directed to position 2 or 4, yielding 2-bromomethylbenzene and 4-bromomethylbenzene. Subsequent oxidation would yield 2-bromobenzoic acid or 4-bromobenzoic acid, not 3-bromobenzoic acid.

(c) The bromination step is an electrophilic substitution reaction.

(a) Step 1:
[1 mark] for reagents/conditions: CH3Cl and anhydrous AlCl3.
[1 mark] for product: methylbenzene.
Step 2:
[1 mark] for reagents/conditions: alkaline KMnO4, heat/reflux followed by acid (H+).
[1 mark] for product: benzoic acid.
Step 3:
[1 mark] for reagents/conditions: Br2 and FeBr3 (or AlBr3 or Fe).
[1 mark] for product: 3-bromobenzoic acid.

(b) [1 mark] for identifying -COOH as a meta-director (or 3-director).
[1 mark] for identifying -CH3 as an ortho/para-director (or 2,4-director).
[1 mark] for explaining that if bromination occurred first, 2-bromobenzoic acid or 4-bromobenzoic acid would be formed.

(c) [1 mark] for stating electrophilic substitution.

(a) Trend: Solubility of Group 2 sulfates decreases down the group (MgSO4 is soluble, BaSO4 is highly insoluble).

Explanation: As you go down the group, the ionic radius of the Group 2 cation increases (Mg2+ < Ca2+ < Sr2+ < Ba2+). Since the ionic charge remains +2, the charge density of the cation decreases.
Consequently, both the lattice energy of the sulfate compound and the hydration enthalpy of the cation decrease down the group.
However, because the sulfate ion (\(\text{SO}_4^{2-}\)) is very large, the increase in cation size has only a small percentage effect on the total distance between cation and anion, so the lattice energy decreases only slightly.
On the other hand, the hydration enthalpy of the cation decreases much more rapidly because the hydration of a cation depends strongly on its charge-to-size ratio.
Since the enthalpy of solution is given by: \(\Delta H_{\text{sol}} \approx \Delta H_{\text{hyd}} - \Delta H_{\text{latt}}\), and \(\Delta H_{\text{hyd}}\) becomes significantly less exothermic while \(\Delta H_{\text{latt}}\) decreases only slightly, the overall enthalpy of solution becomes more endothermic (or less exothermic) down the group. This decrease in thermodynamic favourability causes the solubility of the sulfates to decrease down the group.

(b)(i) Thermal decomposition of magnesium nitrate:
\(2\text{Mg}(\text{NO}_3)_2(\text{s}) \rightarrow 2\text{MgO}(\text{s}) + 4\text{NO}_2(\text{g}) + \text{O}_2(\text{g})\)

(ii) The barium ion (\(\text{Ba}^{2+}\)) is larger than the magnesium ion (\(\text{Mg}^{2+}\)), resulting in a much lower charge density. Therefore, \(\text{Ba}^{2+}\) has less polarizing power and is less able to polarize (distort) the electron cloud of the nitrate ion (\(\text{NO}_3^-\)). The distortion in magnesium nitrate weakens the covalent N-O bonds within the nitrate ion, making it easy to decompose upon heating. In barium nitrate, the nitrate ion is polarized much less, meaning the N-O bonds remain strong, and a much higher temperature (more thermal energy) is required to decompose the compound.

(a) [1 mark] for stating that the solubility of sulfates decreases down the group.
[1 mark] for stating that both lattice energy and hydration enthalpy of the cations decrease down the group.
[1 mark] for explaining that the hydration enthalpy decreases more rapidly than the lattice energy (due to large size of the sulfate ion).
[1 mark] for stating that the enthalpy of solution becomes more endothermic (or less exothermic).
[1 mark] for linking these enthalpy changes to the increasing size / decreasing charge density of the cation.

(b)(i) [1 mark] for correct balanced equation: 2Mg(NO3)2 -> 2MgO + 4NO2 + O2.
(ii) [1 mark] for comparing cation size / charge density: Ba2+ is larger and has lower charge density than Mg2+.
[1 mark] for stating that Ba2+ has less polarizing power / polarizes the nitrate ion less.
[1 mark] for explaining that less polarization means the N-O bonds within the nitrate ion are weakened less.
[1 mark] for concluding that more thermal energy / higher temperature is needed to break the bonds in barium nitrate.

(a) The dissociation equation is: \(\text{CH}_3\text{CH}_2\text{COOH}(\text{aq}) \rightleftharpoons \text{CH}_3\text{CH}_2\text{COO}^-(\text{aq}) + \text{H}^+(\text{aq})\)
\(K_a = \frac{[\text{CH}_3\text{CH}_2\text{COO}^-][\text{H}^+]}{[\text{CH}_3\text{CH}_2\text{COOH}]}\)

(b) For a weak acid, we make the approximations:
1. \([\text{H}^+] \approx [\text{CH}_3\text{CH}_2\text{COO}^-]\)
2. The equilibrium concentration of propanoic acid is approximately the same as its initial concentration (\(0.120\text{ mol dm}^{-3}\)).

\(K_a = \frac{[\text{H}^+]^2}{[\text{CH}_3\text{CH}_2\text{COOH}]}\)
\([\text{H}^+]^2 = K_a \times [\text{CH}_3\text{CH}_2\text{COOH}] = (1.35 \times 10^{-5}) \times 0.120 = 1.62 \times 10^{-6}\text{ mol}^2\text{ dm}^{-6}\)
\([\text{H}^+] = \sqrt{1.62 \times 10^{-6}} = 1.273 \times 10^{-3}\text{ mol dm}^{-3}\)
\(\text{pH} = -\log[\text{H}^+] = -\log(1.273 \times 10^{-3}) = 2.895 \approx 2.89\) (to 2 decimal places).

(c)(i) A buffer solution is a solution that resists or minimizes changes in pH when small amounts of acid (\(\text{H}^+\)) or alkali (\(\text{OH}^-\)) are added.

(ii) Total volume after mixing = \(50.0 + 50.0 = 100.0\text{ cm}^3\).
Because the volume has doubled, the concentration of both components is halved:
\([\text{CH}_3\text{CH}_2\text{COOH}] = 0.120 / 2 = 0.0600\text{ mol dm}^{-3}\)
\([\text{CH}_3\text{CH}_2\text{COO}^-] = 0.0800 / 2 = 0.0400\text{ mol dm}^{-3}\)

Alternatively, we can use the moles directly:
\(n(\text{acid}) = 0.0500 \times 0.120 = 0.00600\text{ mol}\)
\(n(\text{salt}) = 0.0500 \times 0.0800 = 0.00400\text{ mol}\)

Using the buffer equation:
\([\text{H}^+] = K_a \times \frac{[\text{acid}]}{[\text{salt}]} = 1.35 \times 10^{-5} \times \frac{0.0600}{0.0400} = 2.025 \times 10^{-5}\text{ mol dm}^{-3}\)
\(\text{pH} = -\log(2.025 \times 10^{-5}) = 4.693 \approx 4.69\)

(iii) Added \(\text{H}^+\) ions are consumed by reacting with the conjugate base (propanoate ions) in the buffer:
\(\text{CH}_3\text{CH}_2\text{COO}^-(\text{aq}) + \text{H}^+(\text{aq}) \rightarrow \text{CH}_3\text{CH}_2\text{COOH}(\text{aq})\)

(a) [1 mark] for correct Ka expression (charges and formulas must be correct).

(b) [1 mark] for showing [H+]^2 = Ka x [acid] or [H+] = sqrt(Ka x [acid]).
[1 mark] for calculating [H+] = 1.27 x 10^-3 mol dm^-3.
[1 mark] for pH = 2.89 (must be to 2 decimal places).

(c)(i) [1 mark] for defining buffer as a solution that resists/minimizes changes in pH when small amounts of acid/alkali are added.
(ii) [1 mark] for determining correct concentrations in the mixture (acid = 0.0600 and salt = 0.0400 mol dm^-3) or corresponding moles.
[1 mark] for correct substitution into the buffer formula.
[1 mark] for final pH value of 4.69.
(iii) [1 mark] for identifying the reacting species: CH3CH2COO- and H+.
[1 mark] for correct products and balanced equation: CH3CH2COO-(aq) + H+(aq) -> CH3CH2COOH(aq).

For a typical student run with a mean titer of 25.00 cm3: (a) Results recorded to 2 d.p. (b) Mean titer = 25.00 cm3. (c) Moles of thiosulfate = \(0.100 \times \frac{25.00}{1000} = 0.00250\) mol. (d) From the stoichiometry of the equations, \(2 \text{ moles of } \text{S}_2\text{O}_3^{2-} \equiv 1 \text{ mole of } \text{I}_2 \equiv 2 \text{ moles of } \text{Cu}^{2+}\). Therefore, moles of \(\text{Cu}^{2+}\) = moles of thiosulfate = 0.00250 mol. (e) Concentration of FA 1 = \(\frac{0.00250}{0.0250} = 0.100\) mol dm\(^{-3}\). (f) In 250 cm3 of FA 1, total moles = \(0.100 \times 0.250 = 0.0250\) mol. Formula mass of \(\text{CuSO}_4 \cdot x\text{H}_2\text{O} = \frac{6.24}{0.0250} = 249.6\) g mol\(^{-1}\). Formula mass of anhydrous \(\text{CuSO}_4\) = 159.6 g mol\(^{-1}\). Mass of water = \(249.6 - 159.6 = 90.0\) g mol\(^{-1}\). \(x = \frac{90.0}{18.0} = 5\). (g) At high iodine concentrations, starch forms an extremely stable, irreversible complex with iodine, which reduces the accuracy of the end-point.

1. Award 1 mark for titration table with appropriate headings and units. 2. Award 1 mark for recording all burette readings to 0.05 cm3. 3. Award 1 mark for obtaining concordant titers within 0.10 cm3 of each other. 4. Award 1 mark for correct calculation of mean titer. 5. Award 1.5 marks for calculating moles of thiosulfate = (mean titer/1000) * 0.100 (0.5 marks for formula, 1 mark for calculation with 3 significant figures). 6. Award 1.5 marks for showing moles of Cu2+ equal to moles of thiosulfate based on the 1:1 overall ratio. 7. Award 1.5 marks for calculating concentration of Cu2+ = moles of Cu2+ in 25.0 cm3 / 0.0250. 8. Award 3 marks for calculation of x: 1 mark for total moles in 250 cm3 (concentration * 0.250), 1 mark for calculated Mr = 6.24 / total moles, 1 mark for x = (Mr - 159.6)/18 = 5 (nearest integer). 9. Award 2 marks for stating that high concentration of iodine forms an irreversible insoluble complex with starch, which prevents a sharp end-point.

Typical student readings: Experiment 1: Initial Temp = 21.5 \(^{\circ}\)C, Max Temp = 47.0 \(^{\circ}\)C, \(dT = +25.5\) K. Experiment 2: Initial Temp = 21.0 \(^{\circ}\)C, Min Temp = 20.2 \(^{\circ}\)C, \(dT = -0.8\) K. (a) Readings recorded correctly to 0.5 \(^{\circ}\)C. (b) \(q_1 = m c dT = 50.0 \times 4.18 \times 25.5 = 5329.5\) J. (c) Moles of \(\text{MgCl}_2 = \frac{4.76}{95.3} = 0.0500\) mol. Enthalpy of solution \(\Delta H_1 = -\frac{5329.5}{1000 \times 0.0500} = -106.7\) kJ mol\(^{-1}\). (d) \(q_2 = 50.0 \times 4.18 \times (-0.8) = -167.2\) J. (e) Moles of \(\text{MgCl}_2 \cdot 6\text{H}_2\text{O} = \frac{10.16}{203.3} = 0.0500\) mol. Enthalpy of solution \(\Delta H_2 = \frac{167.2}{1000 \times 0.0500} = +3.35\) kJ mol\(^{-1}\). (f) By Hess's Law, \(\Delta H_{\text{hyd}} = \Delta H_1 - \Delta H_2 = -106.7 - (+3.35) = -110.05\) kJ mol\(^{-1}\). (g) Add a lid to the polystyrene cup to reduce heat transfer with the surroundings, or use a thermometer with smaller graduations (e.g., 0.1 \(^{\circ}\)C).

1. Award 1 mark for a table presenting mass of solid, initial temp, final temp, and temperature change with units. 2. Award 1 mark for recording all temperatures to the nearest 0.5 °C. 3. Award 1 mark for calculating correct temperature changes (Experiment 1 increases, Experiment 2 decreases). 4. Award 1.5 marks for calculating q1 correctly with 3 significant figures. 5. Award 2.5 marks for enthalpy change of solution 1: 1 mark for calculating moles of MgCl2 (mass / 95.3), 1 mark for calculating enthalpy change (q1 / moles), 1 mark for the correct negative sign and units. 6. Award 1.5 marks for calculating q2 correctly with 3 significant figures. 7. Award 2.5 marks for enthalpy change of solution 2: 1 mark for calculating moles of MgCl2.6H2O (mass / 203.3), 1 mark for calculating enthalpy change (q2 / moles), 1 mark for correct sign (positive) and units. 8. Award 1.5 marks for using Hess's law correctly to find hydration enthalpy: 0.5 marks for the relationship dH_hyd = dH1 - dH2, 1 mark for calculating the final value with correct sign and units. 9. Award 1 mark for suggesting a valid improvement: use of a lid/insulating material or a thermometer with 0.1 °C resolution.

Part (a):
- Test 1 (FA 1 + NaOH): Addition of NaOH dropwise forms a red-brown precipitate, which is insoluble in excess NaOH. This is characteristic of iron(III) ions, \( \text{Fe}^{3+} \).
- Test 2 (FA 1 + NH3): Dropwise addition of aqueous ammonia also yields a red-brown precipitate, insoluble in excess ammonia, confirming the presence of \( \text{Fe}^{3+} \).
- Test 3 (FA 1 + HNO3 + AgNO3): Addition of silver nitrate yields a white precipitate of silver chloride (\( \text{AgCl} \)), which completely dissolves upon addition of dilute ammonia to form a colourless solution. This confirms the presence of chloride ions, \( \text{Cl}^- \).
- Test 4 (FA 2 + HNO3 + AgNO3): Addition of silver nitrate yields a yellow precipitate of silver iodide (\( \text{AgI} \)), which is insoluble in aqueous ammonia. This confirms the presence of iodide ions, \( \text{I}^- \).

Part (b):
- Test 1: When yellow-brown \( \text{Fe}^{3+} \) is mixed with colourless \( \text{I}^- \), a redox reaction occurs, yielding a dark brown / red-brown solution (due to the formation of iodine, \( \text{I}_2 \) or triiodide, \( \text{I}_3^- \)).
- Test 2: Addition of starch indicator to the mixture yields a blue-black colour, confirming the presence of free iodine (\( \text{I}_2 \)).
- Test 3: Shaking with cyclohexane extracts the non-polar iodine into the upper organic layer, which turns purple / violet / pink.
- Test 4: Addition of excess NaOH decolourises the brown iodine solution (as iodine undergoes disproportionation in alkaline medium to form colourless \( \text{I}^- \) and \( \text{IO}_3^- \)) and precipitates a green / dirty-green precipitate of iron(II) hydroxide, \( \text{Fe(OH)}_2 \), which does not dissolve in excess NaOH. This confirms that \( \text{Fe}^{3+} \) has been reduced to \( \text{Fe}^{2+} \).

Part (c):
- (i) The reaction is a redox (reduction-oxidation) reaction.
- (ii) The balanced ionic equation with state symbols is: \( 2\text{Fe}^{3+}(\text{aq}) + 2\text{I}^-(\text{aq}) \rightarrow 2\text{Fe}^{2+}(\text{aq}) + \text{I}_2(\text{aq}) \) (or with \( \text{I}_2(\text{s}) \)).
- (iii) The green precipitate is \( \text{Fe(OH)}_2 \), formed by the reaction of the reduced iron(II) ions with hydroxide ions: \( \text{Fe}^{2+}(\text{aq}) + 2\text{OH}^-(\text{aq}) \rightarrow \text{Fe(OH)}_2(\text{s}) \). The fading/decolourisation of the brown iodine colour occurs because iodine reacts with sodium hydroxide (disproportionation) to form colourless iodide and iodate(I) or iodate(V) ions.

Part (a): [5 marks]
- 1 mark: Test 1 and Test 2 both record a red-brown precipitate, insoluble in excess.
- 1 mark: Test 3 records a white precipitate that dissolves in dilute ammonia.
- 1 mark: Test 4 records a yellow precipitate that is insoluble in ammonia.
- 1 mark: Identifies both cation (\( \text{Fe}^{3+} \)) and anion (\( \text{Cl}^- \)) in FA 1.
- 1 mark: Identifies anion in FA 2 as iodide (\( \text{I}^- \)).

Part (b): [4 marks]
- 1 mark: Test 1 records a brown / red-brown solution or dark precipitate.
- 1 mark: Test 2 records a blue-black / dark blue mixture.
- 1 mark: Test 3 records a purple / violet / pink organic (upper) layer.
- 1 mark: Test 4 records a green / dirty-green / grey-green precipitate (and/or decolourisation of the brown solution).

Part (c): [4 marks]
- 1 mark: Identifies the reaction type as redox / reduction-oxidation / electron transfer.
- 1 mark: Balanced equation: \( 2\text{Fe}^{3+} + 2\text{I}^- \rightarrow 2\text{Fe}^{2+} + \text{I}_2 \) (or half-integer equivalent).
- 1 mark: Correct state symbols on all species in the ionic equation.
- 1 mark: Explains that \( \text{Fe}^{3+} \) was reduced to \( \text{Fe}^{2+} \) which reacts with \( \text{OH}^- \) to form green \( \text{Fe(OH)}_2 \) precipitate, and/or iodine reacts with alkaline NaOH to form colourless products.

(a) The potential of a half-cell compared to a standard hydrogen electrode (S.H.E.) at \(298\text{ K}\), \(1\text{ atm}\) pressure, and with an ion concentration of \(1.00\text{ mol dm}^{-3}\).

(b)(i) \(Fe^{3+}/Fe^{2+}\) half-cell: Platinum electrode. \(Ag^+/Ag\) half-cell: Silver electrode.
(ii) \(E^{\theta}_{cell} = E^{\theta}_{cathode} - E^{\theta}_{anode} = 0.80 - 0.77 = +0.03\text{ V}\).
(iii) \(Fe^{2+}(aq) + Ag^+(aq) \rightarrow Fe^{3+}(aq) + Ag(s)\).

(c)(i) Using the Nernst equation for \(Ag^+(aq) + e^- \rightleftharpoons Ag(s)\), where \(z = 1\) and \([\text{oxidised}] = [Ag^+]\):
\(E = E^{\theta} + 0.059\log[Ag^+]\)
\(E = 0.80 + 0.059\log(0.025) = 0.80 + 0.059(-1.602) = 0.80 - 0.095 = +0.705\text{ V}\) (or \(+0.71\text{ V}\)).
(ii) Since the potential of the silver half-cell decreases to \(+0.705\text{ V}\), the \(Fe^{3+}/Fe^{2+}\) half-cell (\(+0.77\text{ V}\)) is now the cathode:
\(E_{cell} = 0.77 - 0.705 = +0.065\text{ V}\) (or \(+0.06\text{ V}\)).

Part (a): 2 marks
- 1 mark for 'potential relative to a standard hydrogen electrode'
- 1 mark for specifying standard conditions (298 K, 1 atm, 1.00 mol dm^-3)

Part (b): 4 marks
- 1 mark for Platinum electrode for Fe3+/Fe2+ and 1 mark for Silver electrode for Ag+/Ag
- 1 mark for E_cell = +0.03 V
- 1 mark for Fe2+(aq) + Ag+(aq) -> Fe3+(aq) + Ag(s) (allow equilibrium arrow, reject if state symbols missing or incorrect)

Part (c): 5 marks
- 1 mark for substituting values correctly into Nernst equation: E = 0.80 + 0.059 * log(0.025)
- 1 mark for log(0.025) = -1.60
- 1 mark for E = +0.71 V (or +0.705 V)
- 1 mark for recognizing that the Fe3+/Fe2+ half-cell is now the cathode (higher potential)
- 1 mark for E_cell = +0.06 V (or +0.065 V)

(a)(i) \([Cu(H_2O)_6]^{2+}(aq) + 4NH_3(aq) \rightleftharpoons [Cu(NH_3)_4(H_2O)_2]^{2+}(aq) + 4H_2O(l)\)
(ii) \(K_{stab} = \frac{[[Cu(NH_3)_4(H_2O)_2]^{2+}]}{[[Cu(H_2O)_6]^{2+}][NH_3]^4}\). Units: \(\text{dm}^{12} \text{mol}^{-4}\).

(b)(i) \(K_{stab} = 10^{13.1} = 1.26 \times 10^{13}\).
(ii) The very high value of \(K_{stab}\) indicates that the tetrammine complex is far more stable than the hexaaqua complex, and the equilibrium position lies heavily to the right.

(c)(i) Formula: \([CuCl_4]^{2-}\). Geometry: Tetrahedral.
(ii) The color changes from deep-blue to yellow (or green-yellow). The coordination number changes because the chloride ligand is much larger than the water or ammonia molecules, so steric hindrance prevents six chloride ligands from coordinating around the copper(II) center.
(iii) \([Cu(NH_3)_4(H_2O)_2]^{2+} + 4Cl^- + 4H^+ \rightarrow [CuCl_4]^{2-} + 4NH_4^+ + 2H_2O\)

Part (a): 3 marks
- 1 mark for the correct balanced equilibrium equation.
- 1 mark for the correct K_stab expression (water must be excluded).
- 1 mark for the correct units: dm^12 mol^-4.

Part (b): 2 marks
- 1 mark for calculating 1.26 x 10^13.
- 1 mark for stating that the tetrammine complex is much more stable / equilibrium lies far to the right.

Part (c): 6 marks
- 1 mark for formula [CuCl_4]^2-.
- 1 mark for tetrahedral geometry.
- 1 mark for color change: deep-blue to yellow/green.
- 1 mark for explaining steric hindrance / Cl- is too large.
- 1 mark for stating that only 4 chloride ligands can fit.
- 1 mark for the balanced equation with acid converting NH3 to NH4+.

(a) Comparing Experiments 1 and 2: \([I^-]\) is constant. When \([S_2O_8^{2-}]\) is doubled, the rate doubles. Therefore, the reaction is first-order with respect to \(S_2O_8^{2-}\).
Comparing Experiments 1 and 3: \([S_2O_8^{2-}]\) is constant. When \([I^-]\) is doubled, the rate doubles. Therefore, the reaction is first-order with respect to \(I^-\).

(b) \(\text{Rate} = k[S_2O_8^{2-}][I^-]\)

(c) Using Experiment 1:
\(1.2 \times 10^{-5} = k \times 0.010 \times 0.015\)
\(k = \frac{1.2 \times 10^{-5}}{1.5 \times 10^{-4}} = 0.080\)
Units: \(\text{dm}^3\text{ mol}^{-1}\text{ s}^{-1}\)

(d) Step 1 (Slow / Rate-determining step):
\(S_2O_8^{2-} + I^- \rightarrow S_2O_8I^{3-}\)
Step 2 (Fast):
\(S_2O_8I^{3-} + I^- \rightarrow 2SO_4^{2-} + I_2\)
(Any logical two-step sequence containing the species in the rate equation in the slow step is acceptable).

(e) Both reactant ions are negatively charged. There is a high electrostatic repulsion between them, resulting in a high activation energy. \(Fe^{2+}\) or \(Fe^{3+}\) ions can act as catalysts.

Part (a): 4 marks
- 1 mark for stating first-order wrt S2O8^2-
- 1 mark for explanation (concentration doubles, rate doubles in Exp 1 vs 2)
- 1 mark for stating first-order wrt I-
- 1 mark for explanation (concentration doubles, rate doubles in Exp 1 vs 3)

Part (b): 1 mark
- 1 mark for Rate = k[S2O8^2-][I-]

Part (c): 2 marks
- 1 mark for calculation of k = 0.080
- 1 mark for units: dm^3 mol^-1 s^-1

Part (d): 2 marks
- 1 mark for correct slow step involving S2O8^2- and I-
- 1 mark for correct fast step resulting in the correct overall equation

Part (e): 2 marks
- 1 mark for repulsion of like (negative) charges
- 1 mark for stating Fe2+ or Fe3+

(a) \(CH_3CH_2COOH + H_2O \rightleftharpoons CH_3CH_2COO^- + H_3O^+\) (or with \(H^+\))
\(K_a = \frac{[CH_3CH_2COO^-][H^+]}{[CH_3CH_2COOH]}\)

(b) \([H^+] = \sqrt{K_a \times [CH_3CH_2COOH]} = \sqrt{1.35 \times 10^{-5} \times 0.120} = \sqrt{1.62 \times 10^{-6}} = 1.27 \times 10^{-3}\text{ mol dm}^{-3}\)
\(\text{pH} = -\log_{10}(1.27 \times 10^{-3}) = 2.89\)
Assumptions: The dissociation of the weak acid is negligible relative to its initial concentration; autoionization of water is negligible.

(c)(i) Initial moles:
\(n(HA) = 0.0500 \times 0.120 = 6.00 \times 10^{-3}\text{ mol}\)
\(n(OH^-) = 0.0300 \times 0.100 = 3.00 \times 10^{-3}\text{ mol}\)
Reaction: \(HA + OH^- \rightarrow A^- + H_2O\)
Remaining \(n(HA) = 6.00 \times 10^{-3} - 3.00 \times 10^{-3} = 3.00 \times 10^{-3}\text{ mol}\)
Formed \(n(A^-) = 3.00 \times 10^{-3}\text{ mol}\)
Total volume = \(80.0\text{ cm}^3 = 0.0800\text{ dm}^3\)
\([HA] = \frac{3.00 \times 10^{-3}}{0.0800} = 0.0375\text{ mol dm}^{-3}\)
\([A^-] = \frac{3.00 \times 10^{-3}}{0.0800} = 0.0375\text{ mol dm}^{-3}\)

(ii) Since \([HA] = [A^-]\):
\(\text{pH} = \text{p}K_a + \log_{10}\left(\frac{[A^-]}{[HA]}\right) = -\log_{10}(1.35 \times 10^{-5}) + \log(1) = 4.87\).

(iii) When \(H^+\) is added, the large reservoir of conjugate base, propanoate ions (\(CH_3CH_2COO^-\)), reacts with the added \(H^+\) to form weak undissociated propanoic acid: \(CH_3CH_2COO^- + H^+ \rightarrow CH_3CH_2COOH\), keeping the \([H^+]\) concentration relatively constant.

Part (a): 2 marks
- 1 mark for correct equilibrium equation.
- 1 mark for correct Ka expression.

Part (b): 3 marks
- 1 mark for [H+] = 1.27 x 10^-3 mol dm^-3
- 1 mark for pH = 2.89
- 1 mark for stating that dissociation is negligible / [HA]_initial ≈ [HA]_equilibrium.

Part (c): 6 marks
- 1 mark for calculating initial moles of HA and NaOH.
- 1 mark for determining moles of remaining acid and conjugate base (both 3.00 x 10^-3 mol).
- 1 mark for concentrations: [HA] = [A^-] = 0.0375 mol dm^-3.
- 2 marks for pH of buffer = 4.87 (allow 1 mark for correct method with a minor math error).
- 1 mark for stating that added H+ reacts with CH3CH2COO^- to form CH3CH2COOH.

(a)(i) Reagents: Chloromethane (\(CH_3Cl\)). Conditions: Anhydrous aluminum chloride (\(AlCl_3\)) catalyst, room temperature.
(ii) Reagents: Alkaline potassium manganate(VII) (\(KMnO_4\)). Conditions: Heat under reflux, followed by acidification with dilute sulfuric or hydrochloric acid.
(iii) Reagents: Concentrated nitric acid (\(HNO_3\)) and concentrated sulfuric acid (\(H_2SO_4\)). Conditions: Heat above \(50^{\circ}\text{C}\).

(b) The carboxylic acid group (\(-COOH\)) is an electron-withdrawing group because of the electronegative carbonyl group. It withdraws electron density from the benzene ring, especially from the 2- and 4-positions, making the 3-position (meta-position) relatively more nucleophilic / directing electrophilic attack to the 3-position.

(c) Reagent: Ethanol (\(C_2H_5OH\)). Conditions: Heat under reflux with concentrated sulfuric acid catalyst.

(d) Reagents: Tin (\(Sn\)) and concentrated hydrochloric acid (\(HCl\)), followed by addition of aqueous sodium hydroxide (\(NaOH(aq)\)) to liberate the free amine. Reaction type: Reduction.

Part (a): 6 marks
- 1 mark for CH3Cl and 1 mark for anhydrous AlCl3 (or FeCl3) for Step 1.
- 1 mark for alkaline KMnO4 and 1 mark for acidification for Step 2 (accept acid/heat/reflux).
- 1 mark for conc. HNO3 + conc. H2SO4 and 1 mark for heating/reflux above 50°C for Step 3.

Part (b): 1 mark
- 1 mark for describing the -COOH group as electron-withdrawing / meta-directing.

Part (c): 2 marks
- 1 mark for ethanol.
- 1 mark for concentrated H2SO4 and heat/reflux.

Part (d): 2 marks
- 1 mark for Sn + conc. HCl followed by NaOH.
- 1 mark for identifying the reaction type as reduction.

(a) A gas (\(CO_2\)) is produced from a solid reactant (\(ZnCO_3\)). Gases have much higher disorder (entropy) than solids, so entropy increases.

(b) \(\Delta H^{\theta}_{r} = \sum \Delta H_f^{\theta}(\text{products}) - \sum \Delta H_f^{\theta}(\text{reactants})\)
\(\Delta H^{\theta}_{r} = [(-348) + (-394)] - [-812] = -742 + 812 = +70\text{ kJ mol}^{-1}\).

(c) \(\Delta S^{\theta}_{r} = \sum S^{\theta}(\text{products}) - \sum S^{\theta}(\text{reactants})\)
\(\Delta S^{\theta}_{r} = [44 + 214] - 82 = 258 - 82 = +176\text{ J K}^{-1}\text{ mol}^{-1}\).

(d) \(\Delta G^{\theta}_{r} = \Delta H^{\theta}_{r} - T\Delta S^{\theta}_{r}\)
Convert \(\Delta S\) to \(\text{kJ K}^{-1}\text{ mol}^{-1}\): \(+176 / 1000 = +0.176\text{ kJ K}^{-1}\text{ mol}^{-1}\).
\(\Delta G^{\theta}_{r} = +70 - (298 \times 0.176) = +70 - 52.45 = +17.55\text{ kJ mol}^{-1}\) (or \(+17.6\text{ kJ mol}^{-1}\)).
Since \(\Delta G^{\theta}_{r}\) is positive, the reaction is NOT feasible at \(298\text{ K}\).

(e) For the reaction to be feasible, \(\Delta G \le 0\).
\(\Delta H - T\Delta S \le 0 \implies T \ge \frac{\Delta H}{\Delta S}\)
\(T \ge \frac{70}{0.176} = 397.7\text{ K}\) (rounds to \(398\text{ K}\)).

Part (a): 1 mark
- 1 mark for: production of a gas increases system disorder.

Part (b): 2 marks
- 1 mark for correct mathematical setup.
- 1 mark for +70 kJ mol^-1 (must include '+' sign).

Part (c): 2 marks
- 1 mark for correct mathematical setup.
- 1 mark for +176 J K^-1 mol^-1 (must include '+' sign).

Part (d): 3 marks
- 1 mark for conversion of units of ΔS (division by 1000) or ΔH.
- 1 mark for calculation of ΔG = +17.6 kJ mol^-1 (or +17.55).
- 1 mark for stating reaction is not feasible because ΔG > 0.

Part (e): 3 marks
- 1 mark for setting ΔG = 0 / recognizing T = ΔH / ΔS.
- 1 mark for calculation: 70 / 0.176.
- 1 mark for 398 K (or 397.7 K) with correct units.

(a) In an octahedral complex, the six ligands approach the central metal ion along the axes (x, y, z). This causes greater electrostatic repulsion with the d-orbitals lying directly on the axes (\(d_{x^2-y^2}\) and \(d_{z^2}\)) than those lying between the axes (\(d_{xy}\), \(d_{yz}\), \(d_{xz}\)). Thus, the five degenerate d-orbitals split into two groups: three at a lower energy level and two at a higher energy level.

(b) Transition metal complexes are colored because d-orbitals are split in energy. When white light passes through a solution, d-electrons absorb a specific frequency of light corresponding to the energy gap \(\Delta E = h\nu\). This energy promotes an electron from a lower d-orbital to a higher d-orbital (d-to-d transition). The light that is not absorbed is transmitted, and the observer sees the complementary color of the absorbed light.

(c) Different ligands cause different extents of d-orbital splitting. \(CN^-\cap\) is a stronger field ligand than \(H_2O\), so it causes a larger energy gap (\(\Delta E\)) between the split d-orbitals. A larger \(\Delta E\) means light of higher frequency (shorter wavelength, e.g., blue-violet) is absorbed, resulting in a complementary pale-yellow color being transmitted instead of pale-green.

(d) The electronic configuration of \(Fe^{2+}\) is \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^6\). It has 4 unpaired electrons in its 3d subshell. Unpaired electrons have net magnetic spins that align with an external magnetic field, making the ion paramagnetic.

Part (a): 3 marks
- 1 mark for specifying ligand approach along axes causing unequal repulsion.
- 1 mark for stating splitting of 5 degenerate d-orbitals into two higher-energy and three lower-energy levels.
- 1 mark for naming the d-orbitals involved (dx^2-y^2/dz^2 higher, dxy/dyz/dxz lower).

Part (b): 4 marks
- 1 mark for absorbing a photon/frequency of white light.
- 1 mark for d-to-d transition / electron promotion.
- 1 mark for relationship ΔE = hν.
- 1 mark for seeing/transmitting the complementary color.

Part (c): 2 marks
- 1 mark for stating CN- causes a larger d-orbital splitting than H2O.
- 1 mark for explaining that a larger energy gap absorbs higher energy/frequency light, changing the transmitted (complementary) color.

Part (d): 2 marks
- 1 mark for correct electronic configuration: [Ar] 3d^6 or 1s^2 2s^2 2p^6 3s^2 3p^6 3d^6.
- 1 mark for stating it has unpaired electrons which align with external fields.

(a) Ethanamide < phenylamine < ethylamine.

(b)(i) In ethylamine, the ethyl group is electron-releasing (inductive effect), which increases the electron density on the nitrogen atom and makes the lone pair more available to coordinate with a proton (\(H^+\)). In phenylamine, the lone pair on the nitrogen atom is delocalised into the \(\pi\) system of the benzene ring, making it much less available to accept a proton.
(ii) In ethanamide, the lone pair on the nitrogen atom is strongly delocalised into the adjacent electronegative carbonyl group (\(C=O\)). This delocalisation is far more effective than into the benzene ring, making the nitrogen atom in ethanamide virtually non-basic (neutral).

(c)(i) Equation: \(CH_3COCl + CH_3CH_2NH_2 \rightarrow CH_3CONHCH_2CH_3 + HCl\).
Functional group: Secondary Amide (N-substituted amide).
(ii) Nucleophilic addition-elimination (or nucleophilic substitution).

(d)(i) Temperature range: \(0^{\circ}\text{C}\) to \(10^{\circ}\text{C}\) (or \(0\) to \(5^{\circ}\text{C}\)). If the temperature is below \(0^{\circ}\text{C}\), the reaction is too slow. If it is above \(10^{\circ}\text{C}\), the diazonium salt is unstable and decomposes to form phenol and nitrogen gas.
(ii) Product: 4-(phenyldiazeyl)phenol (an azo dye; accept drawn structure \(C_6H_5-N=N-C_6H_4-OH\)).

Part (a): 1 mark
- 1 mark for correct order: ethanamide < phenylamine < ethylamine.

Part (b): 4 marks
- 1 mark for ethyl group being electron-releasing / +I effect.
- 1 mark for nitrogen lone pair delocalisation into the benzene ring in phenylamine.
- 2 marks for ethanamide: explaining very strong delocalisation of the nitrogen lone pair into the electronegative carbonyl group (C=O) making it neutral.

Part (c): 3 marks
- 1 mark for correct balanced equation for amide formation.
- 1 mark for identifying 'amide' as the functional group.
- 1 mark for naming reaction as nucleophilic addition-elimination.

Part (d): 3 marks
- 1 mark for temperature: 0°C to 10°C.
- 1 mark for explaining instability/decomposition above 10°C (forming phenol/N2).
- 1 mark for correct structure or name of 4-(phenyldiazeyl)phenol.

**(a)**
- The standard electrode potential, \(E^\ominus\), is the electromotive force (emf) of a cell consisting of the electrode in question on the right-hand side and a standard hydrogen electrode (SHE) on the left-hand side.
- Standard conditions must be specified: temperature of \(298\text{ K}\) (\(25^\circ\text{C}\)), concentration of \(1.00\text{ mol dm}^{-3}\) for all aqueous ions, and pressure of \(1\text{ bar}\) (\(100\text{ kPa}\) or \(1\text{ atm}\)) for any gaseous species.

**(b)**
- The species with the more positive \(E^\ominus\) value is reduced: \(\text{Ag}^{+}(\text{aq}) + \text{e}^- \rightarrow \text{Ag}(\text{s})\).
- The species with the less positive (more negative) \(E^\ominus\) value is oxidized: \(\text{Co}(\text{s}) \rightarrow \text{Co}^{2+}(\text{aq}) + 2\text{e}^-\).
- Multiplying the silver half-equation by 2 and combining them gives:
$$\text{Co}(\text{s}) + 2\text{Ag}^{+}(\text{aq}) \rightarrow \text{Co}^{2+}(\text{aq}) + 2\text{Ag}(\text{s})$$
- The standard cell potential is:
$$E^\ominus_{\text{cell}} = E^\ominus_{\text{reduction}} - E^\ominus_{\text{oxidation}} = +0.80\text{ V} - (-0.28\text{ V}) = +1.08\text{ V}$$

**(c)**
- The overall cell potential \(E_{\text{cell}}\) **decreases**.
- Reason: According to Le Chatelier's principle, increasing the concentration of \(\text{Co}^{2+}(\text{aq})\) shifts the position of the \(\text{Co}^{2+}(\text{aq}) + 2\text{e}^- \rightleftharpoons \text{Co}(\text{s})\) equilibrium to the right (towards the reduction side). This makes the electrode potential of this half-cell, \(E(\text{Co}^{2+}/\text{Co})\), more positive (less negative).
- Since \(E_{\text{cell}} = E(\text{Ag}^{+}/\text{Ag}) - E(\text{Co}^{2+}/\text{Co})\), a more positive cobalt half-cell potential decreases the difference between the two half-cells, leading to a smaller overall cell potential.

**(d)**
- From the half-equation \(\text{Co}^{2+}(\text{aq}) + 2\text{e}^- \rightleftharpoons \text{Co}(\text{s})\), the number of electrons transferred is \(z = 2\).
- The oxidised species is \(\text{Co}^{2+}(\text{aq})\) with a concentration of \(0.015\text{ mol dm}^{-3}\).
- Substituting into the Nernst equation:
$$E = E^\ominus + \frac{0.059}{z}\log [\text{oxidised species}]$$
$$E = -0.28 + \frac{0.059}{2}\log(0.015)$$
$$E = -0.28 + 0.0295 \times (-1.8239)$$
$$E = -0.28 - 0.0538 = -0.3338\text{ V}$$
- Rounding to 3 decimal places (or 2 decimal places consistent with standard potential precision) gives \(-0.334\text{ V}\) (or \(-0.33\text{ V}\)).

**(e)**
- Ammonia is a stronger ligand than water, forming stronger coordinate bonds and resulting in a larger stability constant (\(K_{\text{stab}}\)) for the ammine complex compared to the aqua complex.
- This stabilizes both oxidation states, but the stabilization of \(\text{Co}^{3+}\) (higher charge density) is significantly greater than that of \(\text{Co}^{2+}\).
- This thermodynamic stabilization of the oxidized form relative to the reduced form shifts the redox equilibrium towards oxidation, making the species easier to oxidize and thus making the reduction potential much less positive.

**(a)** [2 marks total]
- M1: Potential difference / voltage measured between a standard half-cell and a standard hydrogen electrode (SHE). [1]
- M2: Under standard conditions of 298 K, 1 mol dm^-3 solutions, and 1 bar / 1 atm pressure. [1]

**(b)** [2 marks total]
- M1: Correct balanced ionic equation:
\(\text{Co}(\text{s}) + 2\text{Ag}^{+}(\text{aq}) \rightarrow \text{Co}^{2+}(\text{aq}) + 2\text{Ag}(\text{s})\) [1]
- M2: Correct cell potential calculation: \(+1.08\text{ V}\) (must include sign and unit). [1]

**(c)** [2 marks total]
- M1: Cell potential / \(E_{\text{cell}}\) decreases. [1]
- M2: Increasing \([\text{Co}^{2+}]\) shifts the cobalt equilibrium to the right, making its reduction potential more positive (or less negative), which reduces the potential difference between the two half-cells. [1]

**(d)** [3 marks total]
- M1: Identifies \(z = 2\) correctly. [1]
- M2: Correct substitution: \(E = -0.28 + \frac{0.059}{2}\log(0.015)\). [1]
- M3: Correct evaluation to \(-0.334\text{ V}\) or \(-0.33\text{ V}\) (must include sign and unit). [1]

**(e)** [2 marks total]
- M1: Standard ammine complex is more stable than standard aqua complex / ammonia is a stronger ligand than water. [1]
- M2: The ammine ligand stabilizes the higher oxidation state (\(\text{Co}^{3+}\)) much more than the lower state (\(\text{Co}^{2+}\)), shifting the equilibrium to favor oxidation (hence lowering \(E^\ominus\)). [1]

(a) \(\text{N}_2\) is a gas that is only very sparingly soluble in water, so it escapes the reaction mixture immediately. The volume of gas evolved is directly proportional to the amount of reactant decomposed.
Independent variable: Time (\(t\))
Dependent variable: Volume of nitrogen gas collected (\(V_t\))

(b) A \(100\text{ cm}^3\) conical flask (containing the reaction mixture) is placed inside a temperature-controlled water bath set at \(45^\circ\text{C}\). The flask is sealed with a rubber bung connected to a delivery tube, which is connected to a \(150\text{ cm}^3\) or \(250\text{ cm}^3\) gas syringe (or inverted measuring cylinder in a trough of water) to measure the gas volume over time.

(c) Moles of \(\text{C}_6\text{H}_5\text{N}_2\text{Cl} = 0.0500\text{ dm}^3 \times 0.100\text{ mol dm}^{-3} = 5.00 \times 10^{-3}\text{ mol}\).
Since 1 mole of \(\text{C}_6\text{H}_5\text{N}_2\text{Cl}\) produces 1 mole of \(\text{N}_2\):
Moles of \(\text{N}_2\) produced = \(5.00 \times 10^{-3}\text{ mol}\).
Volume of \(\text{N}_2\) gas = \(5.00 \times 10^{-3}\text{ mol} \times 24.0\text{ dm}^3\text{ mol}^{-1} = 0.120\text{ dm}^3 = 120\text{ cm}^3\).

(d) \(V_{\infty}\) is the total volume of gas produced at completion, which is proportional to the initial concentration of benzenediazonium chloride, \([\text{C}_6\text{H}_5\text{N}_2\text{Cl}]_0\).
\(V_t\) is the volume of gas produced at time \(t\), which is proportional to the concentration of reactant that has decomposed, \(x\).
Therefore, \((V_{\infty} - V_t)\) is proportional to \([\text{C}_6\text{H}_5\text{N}_2\text{Cl}]_0 - x\), which represents the concentration of benzenediazonium chloride remaining at time \(t\), \([\text{C}_6\text{H}_5\text{N}_2\text{Cl}]_t\).

(e) (i) First-order reaction.
(ii) For a first-order reaction: \(\ln[A]_t = -kt + \ln[A]_0\). Since \((V_{\infty} - V_t)\) is proportional to \([A]_t\), the plot of \(\ln(V_{\infty} - V_t)\) vs \(t\) yields a straight line where the gradient \(= -k\). Hence, \(k = -\text{gradient}\). The units of \(k\) are \(\text{s}^{-1}\) (or \(\text{min}^{-1}\) depending on the unit of time plotted).

(f) Hazard: Phenol produced is corrosive and toxic (or benzenediazonium chloride is unstable/explosive when dry).
Precaution: Wear safety goggles and protective gloves to avoid skin contact / keep benzenediazonium chloride in solution at all times.

Part (a) [3 marks]:
- [1] Explains that nitrogen is an insoluble gas that escapes the mixture, so its volume reflects the extent of reaction.
- [1] Identifies the independent variable as time.
- [1] Identifies the dependent variable as the volume of nitrogen gas collected.

Part (b) [3 marks]:
- [1] Describes a reaction vessel (e.g., conical flask) placed inside a water bath at \(45^\circ\text{C}\).
- [1] Shows/describes a delivery tube connecting the flask to a gas syringe or inverted measuring cylinder in a water trough.
- [1] Specifies reasonable capacities: e.g., \(100\text{ cm}^3\) flask and \(150\text{ cm}^3\) or \(250\text{ cm}^3\) gas syringe.

Part (c) [2 marks]:
- [1] Correctly calculates moles of reactant: \(5.00 \times 10^{-3}\text{ mol}\).
- [1] Correctly calculates gas volume: \(120\text{ cm}^3\) (must include units).

Part (d) [2 marks]:
- [1] Links \(V_{\infty}\) to initial concentration AND \(V_t\) to amount reacted.
- [1] Explains that the difference \((V_{\infty} - V_t)\) is directly proportional to the concentration of remaining reactant.

Part (e) [3 marks]:
- [1] (i) States the reaction is first-order.
- [1] (ii) Explains that \(k = -\text{gradient}\).
- [1] States correct units of rate constant: \(\text{s}^{-1}\) (or \(\text{min}^{-1}\)).

Part (f) [2 marks]:
- [1] Identifies a valid hazard (e.g., phenol is corrosive/toxic OR benzenediazonium salt is explosive when dry).
- [1] Specifies a matching safety precaution (e.g., wear nitrile gloves / do not evaporate the solution to dryness).

(a) Calculation:
\(n = c \times V = 0.200\text{ mol dm}^{-3} \times 0.2500\text{ dm}^3 = 0.0500\text{ mol}\)
\(m = 0.0500\text{ mol} \times 249.6\text{ g mol}^{-1} = 12.48\text{ g}\) (or \(12.5\text{ g}\))

Preparation Method:
1. Weigh exactly \(12.48\text{ g}\) of hydrated copper(II) sulfate solid in a weighing boat.
2. Transfer the solid to a beaker and dissolve it completely in about \(100\text{ cm}^3\) of distilled water.
3. Transfer the solution quantitatively (including the beaker washings) into a \(250.0\text{ cm}^3\) volumetric flask.
4. Add distilled water up to the graduation mark (using a teat pipette for the final drops), stopper, and invert the flask several times to ensure a homogeneous solution.

(b) Dilution Calculation:
Using \(C_1 V_1 = C_2 V_2\):
\(V_1 = \frac{C_2 V_2}{C_1} = \frac{0.020\text{ mol dm}^{-3} \times 100.0\text{ cm}^3}{0.200\text{ mol dm}^{-3}} = 10.0\text{ cm}^3\)

Method:
Use a burette to measure precisely \(10.0\text{ cm}^3\) of the stock \(0.200\text{ mol dm}^{-3}\) \(\text{CuSO}_4\) solution into a \(100.0\text{ cm}^3\) volumetric flask. Add distilled water to the flask until the bottom of the meniscus is aligned with the graduation mark, then stopper and invert to mix.

(c) The diagram must include:
- Two separate beakers: one containing a zinc electrode immersed in \(1.0\text{ mol dm}^{-3}\) \(\text{ZnSO}_4(\text{aq})\) and the other containing a copper electrode immersed in \(0.020\text{ mol dm}^{-3}\) \(\text{CuSO}_4(\text{aq})\).
- A high-resistance voltmeter connected in parallel/series between the zinc and copper electrodes.
- A salt bridge dipping into both solutions.
- Clearly marked polarity: Zinc is the negative terminal (anode) and Copper is the positive terminal (cathode).

(d) (i) The anomalous cell potential is at \(0.010\text{ mol dm}^{-3}\) (measured cell potential is \(1.01\text{ V}\), which is lower than expected based on the trend where it should be around \(1.04\text{ V}\)).
(ii) Possible experimental error: The \(0.010\text{ mol dm}^{-3}\) solution was over-diluted during preparation (e.g. water was added past the graduation mark), or the copper electrode was contaminated/not cleaned with emery paper between runs.

(e) (i) According to the equation: \(E_{\text{cell}} = E^\theta_{\text{cell}} + m \log [\text{Cu}^{2+}]\). When \(\log [\text{Cu}^{2+}] = 0\), \(E_{\text{cell}} = E^\theta_{\text{cell}}\). Therefore, the student should plot \(E_{\text{cell}}\) vs \(\log [\text{Cu}^{2+}]\), draw a line of best fit (excluding the anomalous point), and extrapolate the line to find the y-intercept (where \(\log [\text{Cu}^{2+}] = 0\)). The value of this y-intercept is equal to \(E^\theta_{\text{cell}}\).
(ii) Percentage error = \(\frac{|\text{Experimental Value} - \text{Theoretical Value}|}{\text{Theoretical Value}} \times 100\%\)
\(\text{Percentage error} = \frac{|0.0310 - 0.0296|}{0.0296} \times 100\% = \frac{0.0014}{0.0296} \times 100\% \approx 4.73\%\) (or \(4.7\%\)).

(f) Electrolyte: Aqueous potassium nitrate (\(\text{KNO}_3\)) or sodium nitrate (\(\text{NaNO}_3\)).
Purpose: To complete the circuit by allowing ions to flow, maintaining electrical neutrality in both half-cells without forming precipitates with \(\text{Zn}^{2+}\) or \(\text{Cu}^{2+}\) ions.

Part (a) [3 marks]:
- [1] Correctly calculates mass of \(\text{CuSO}_4\cdot 5\text{H}_2\text{O}\) as \(12.48\text{ g}\) (or \(12.5\text{ g}\)) showing working.
- [1] Describes dissolution of solid in a beaker with distilled water before transferring to the volumetric flask.
- [1] Specifies the quantitative transfer with washings and making up to the mark with distilled water followed by mixing.

Part (b) [2 marks]:
- [1] Calculates required volume of stock solution as \(10.0\text{ cm}^3\) showing working.
- [1] Describes using a burette to deliver this volume into a \(100.0\text{ cm}^3\) volumetric flask and filling to the mark with distilled water.

Part (c) [3 marks]:
- [1] Drawing shows two separate beakers with electrodes, voltmeter in circuit, and a salt bridge connecting solutions.
- [1] Labels solutions and electrodes correctly (including concentration: Zn in \(1.0\text{ mol dm}^{-3}\) \(\text{Zn}^{2+}\), Cu in \(0.020\text{ mol dm}^{-3}\) \(\text{Cu}^{2+}\)).
- [1] Labels Zn electrode as the negative terminal (and/or Cu as positive).

Part (d) [2 marks]:
- [1] Identifies \(0.010\text{ mol dm}^{-3}\) as the anomalous concentration.
- [1] Suggests a plausible reason (e.g. over-dilution of the solution, or electrode contamination).

Part (e) [3 marks]:
- [1] Explains that the y-intercept represents the standard cell potential (when \(\log[\text{Cu}^{2+}] = 0\)).
- [1] Explains how to find this by drawing a line of best fit (excluding the anomaly) and extrapolating it to \(\log[\text{Cu}^{2+}] = 0\).
- [1] Correctly calculates the percentage error as \(4.73\%\) (or \(4.7\%\)).

Part (f) [2 marks]:
- [1] Suggests potassium nitrate or sodium nitrate (accept ammonium nitrate) as the salt bridge electrolyte.
- [1] Explains that it allows ion migration to complete the circuit and maintain charge neutrality without reacting to form precipitates.