2025 IB DP Chemistry 模擬試題連答案詳解

An oxidizing agent undergoes reduction. The stronger the oxidizing agent, the more positive its standard reduction potential. Comparing the reduction potentials, \( \text{Ag}^+(aq) \) has the highest standard reduction potential (+0.80 V) and is therefore the strongest oxidizing agent among the options. \( \text{Ag}(s) \) is a product of reduction and acts as a reducing agent, not an oxidizing agent.

Award 1 mark for the correct option (B).

The rate-determining step is the slow step (Step 2), so the rate law is rate = \( k_2[\text{N}_2\text{O}_2][\text{H}_2] \). Since \( \text{N}_2\text{O}_2 \) is an intermediate, its concentration can be expressed using the fast equilibrium in Step 1: \( K_c = \frac{[\text{N}_2\text{O}_2]}{[\text{NO}]^2} \), which gives \( [\text{N}_2\text{O}_2] = K_c[\text{NO}]^2 \). Substituting this back into the rate law yields \( \text{rate} = k_2 K_c [\text{NO}]^2 [\text{H}_2] = k'[\text{NO}]^2[\text{H}_2] \). This is consistent with the experimentally determined rate law, so statement B is correct.

Award 1 mark for the correct option (B).

For a weak monoprotic acid, the acid dissociation constant \( K_a \) is \( 10^{-\text{p}K_a} = 1.0 \times 10^{-5} \). Using the weak acid approximation, \( K_a \approx \frac{[\text{H}^+]^2}{[\text{HA}]} \). Therefore, \( 1.0 \times 10^{-5} = \frac{[\text{H}^+]^2}{0.10} \), which gives \( [\text{H}^+]^2 = 1.0 \times 10^{-6} \). Solving for \( [\text{H}^+] \) gives \( 1.0 \times 10^{-3}\text{ mol dm}^{-3} \). The pH is \( -\log_{10}[\text{H}^+] = -\log_{10}(1.0 \times 10^{-3}) = 3.00 \).

Award 1 mark for the correct option (B).

A carbonyl group is a \( \text{C}=\text{O} \) double bond, and a hydroxyl group is an \( -\text{OH} \) group. 3-hydroxybutanal is an aldehyde containing a carbonyl group at carbon-1 and a hydroxyl group at carbon-3. Methyl ethanoate is an ester (no hydroxyl group), propanone is a ketone (no hydroxyl group), and ethoxyethane is an ether (no carbonyl or hydroxyl groups).

Award 1 mark for the correct option (B).

To find the empirical formula, divide each percentage by its relative atomic mass to determine the molar ratio: Moles of \( \text{C} = 40.0 / 12.01 = 3.33 \); Moles of \( \text{H} = 6.7 / 1.01 = 6.63 \); Moles of \( \text{O} = 53.3 / 16.00 = 3.33 \). Dividing each value by the smallest value (3.33) yields a ratio of \( \text{C} : \text{H} : \text{O} = 1 : 2 : 1 \). Therefore, the empirical formula is \( \text{CH}_2\text{O} \).

Award 1 mark for the correct option (B).

Using the combined gas law, \( \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \). Convert temperatures to Kelvin: \( T_1 = 27 + 273 = 300\text{ K} \) and \( T_2 = 327 + 273 = 600\text{ K} \). Rearranging for \( V_2 \) gives \( V_2 = V_1 \times \frac{P_1}{P_2} \times \frac{T_2}{T_1} = 2.0\text{ dm}^3 \times \frac{100\text{ kPa}}{200\text{ kPa}} \times \frac{600\text{ K}}{300\text{ K}} = 2.0 \times 0.5 \times 2 = 2.0\text{ dm}^3 \). Therefore, the volume remains unchanged.

Award 1 mark for the correct option (B).

In \( \text{NH}_4^+ \), the central nitrogen atom has 4 bonding electron pairs and 0 lone pairs, giving a tetrahedral molecular geometry. \( \text{SF}_4 \) has 4 bonding pairs and 1 lone pair (see-saw molecular geometry). \( \text{XeF}_4 \) has 4 bonding pairs and 2 lone pairs (square planar molecular geometry). \( \text{CO}_3^{2-} \) has 3 electron domains around the central carbon atom (trigonal planar molecular geometry).

Award 1 mark for the correct option (C).

A reaction is spontaneous when the change in Gibbs free energy is negative (\( \Delta G^\theta < 0 \)). Using \( \Delta G^\theta = \Delta H^\theta - T\Delta S^\theta \), we set \( \Delta H^\theta - T\Delta S^\theta < 0 \). Convert \( \Delta H^\theta \) to \( \text{J mol}^{-1} \): \( -92000 - T(-198) < 0 \), which gives \( -92000 + 198T < 0 \), or \( 198T < 92000 \). Solving for temperature gives \( T < 464.6\text{ K} \). Therefore, the reaction is spontaneous only at temperatures below approximately 465 K.

Award 1 mark for the correct option (C).

Cobalt has an atomic number of 27, and its ground-state electron configuration is \( [\text{Ar}] 4\text{s}^2 3\text{d}^7 \). When forming the \( \text{Co}^{2+} \) ion, it loses the two \( 4\text{s} \) electrons, leaving the configuration as \( [\text{Ar}] 3\text{d}^7 \). According to Hund's rule, the seven electrons in the five \( 3\text{d} \) orbitals are distributed such that two orbitals are doubly occupied and three orbitals are singly occupied. Therefore, there are 3 unpaired electrons.

Award 1 mark for the correct option B. Award 0 marks for incorrect options.

Sulfur has 6 valence electrons and each of the 4 fluorine atoms contributes 1 electron for sharing, resulting in 4 bonding pairs and 1 lone pair on the sulfur atom (total of 5 electron domains). According to VSEPR theory, a 5-domain system with 1 lone pair adopts a see-saw molecular geometry.

Award 1 mark for the correct option C. Award 0 marks for incorrect options.

The half-reaction with the more positive reduction potential runs as reduction (cathode): \( \text{Fe}^{3+}(\text{aq}) + \text{e}^- \rightarrow \text{Fe}^{2+}(\text{aq}) \) with \( E^\ominus = +0.77\text{ V} \). The other half-reaction is reversed and runs as oxidation (anode): \( 2\text{I}^-(\text{aq}) \rightarrow \text{I}_2(\text{s}) + 2\text{e}^- \) with \( E^\ominus = +0.54\text{ V} \). The overall equation is \( 2\text{Fe}^{3+}(\text{aq}) + 2\text{I}^-(\text{aq}) \rightarrow 2\text{Fe}^{2+}(\text{aq}) + \text{I}_2(\text{s}) \). The standard cell potential is calculated as: \( E^\ominus_{\text{cell}} = E^\ominus_{\text{cathode}} - E^\ominus_{\text{anode}} = +0.77\text{ V} - 0.54\text{ V} = +0.23\text{ V} \).

Award 1 mark for the correct option A. Award 0 marks for incorrect options.

The overall order of the reaction is \( 1 + 2 = 3 \). The rate is in units of \( \text{mol dm}^{-3} \text{s}^{-1} \). Rearranging the rate expression for \( k \) gives: \( k = \frac{\text{Rate}}{[X][Y]^2} \). Substituting the units: \( k = \frac{\text{mol dm}^{-3} \text{s}^{-1}}{(\text{mol dm}^{-3})^3} = \text{mol}^{-2} \text{dm}^6 \text{s}^{-1} \).

Award 1 mark for the correct option B. Award 0 marks for incorrect options.

An acidic buffer solution can be prepared by reacting a weak acid with a limiting amount of a strong base. Here, \( 0.10\text{ mol dm}^{-3}\text{ CH}_3\text{COOH(aq)} \) reacts with \( 0.05\text{ mol dm}^{-3}\text{ NaOH(aq)} \) in equal volumes. Because \( \text{NaOH} \) is limiting, it reacts completely to form \( 0.05\text{ mol dm}^{-3} \) of \( \text{CH}_3\text{COO}^- \), leaving \( 0.05\text{ mol dm}^{-3} \) of unreacted \( \text{CH}_3\text{COOH} \). This partial neutralization results in a mixture containing both the weak acid and its conjugate base, which constitutes a buffer solution.

Award 1 mark for the correct option B. Award 0 marks for incorrect options.

The benzene ring is categorized as a phenyl/arene group. The hydroxyl group (\( -\text{OH} \)) directly attached to the benzene ring forms a phenol functional group. The group \( -\text{COOCH}_3 \) is an ester. Hence, the functional groups present are phenol, ester, and phenyl.

Award 1 mark for the correct option A. Award 0 marks for incorrect options.

From the ideal gas equation \( PV = nRT \), where \( n = \frac{m}{M} \), we can rewrite the equation as \( PV = \frac{m}{M}RT \). Since density \( \rho = \frac{m}{V} \), the equation becomes \( P = \frac{\rho RT}{M} \). Solving for molar mass: \( M = \frac{\rho RT}{P} \). Converting the density to SI units: \( 1.62\text{ g dm}^{-3} = 1.62\text{ kg m}^{-3} \). Thus, \( M = \frac{1.62 \times 8.31 \times 300}{1.00 \times 10^5} = 0.0403866\text{ kg mol}^{-1} \). Converting this back to grams per mole gives \( 40.4\text{ g mol}^{-1} \).

Award 1 mark for the correct option C. Award 0 marks for incorrect options.

The spontaneity of a reaction is determined by the Gibbs free energy change equation: \( \Delta G^\ominus = \Delta H^\ominus - T\Delta S^\ominus \). For the reaction to be spontaneous only at high temperatures, \( \Delta G^\ominus \) must be positive at low temperatures and negative at high temperatures. This is only possible if both \( \Delta H^\ominus \) and \( \Delta S^\ominus \) are positive. At low temperatures, the positive enthalpy term dominates, making \( \Delta G^\ominus > 0 \). At high temperatures, the entropy term \( -T\Delta S^\ominus \) dominates and overrides the positive \( \Delta H^\ominus \) value, making \( \Delta G^\ominus < 0 \).

Award 1 mark for the correct option B. Award 0 marks for incorrect options.

The overall cell reaction is \(\text{Ni}(\text{s}) + 2\text{Ag}^{+}(\text{aq}) \rightleftharpoons \text{Ni}^{2+}(\text{aq}) + 2\text{Ag}(\text{s})\). According to Le Chatelier's principle and the Nernst equation, decreasing the concentration of the products (\(\text{Ni}^{2+}\)) shifts the equilibrium position to the right, which increases the electromotive force (cell potential). Changing the size of solid electrodes does not affect their activity or the cell potential.

[1 mark] for selecting B. Award 0 marks for any other option.

Methylamine is a weak base, so its conjugate acid, \(\text{CH}_3\text{NH}_3^+\), is a weak acid. Option A is incorrect because a weak base only partially dissociates, meaning \([\text{OH}^-] < 0.10\text{ mol dm}^{-3}\). Option C is incorrect because \(\text{p}K_{\text{b}} = -\log_{10}(4.4 \times 10^{-4}) \approx 3.36\), which makes the \(\text{p}K_{\text{a}} = 14.00 - 3.36 = 10.64\). Option D is incorrect because although dilution increases the degree of ionization, the concentration of hydroxide ions decreases, lowering the pH.

[1 mark] for selecting B. Award 0 marks for any other option.

\(\text{CO}_2\) contains two double bonds, each consisting of one \(\sigma\) and one \(\pi\) bond. Because \(\text{CO}_2\) is linear, the individual bond dipoles cancel each other out, resulting in a net dipole moment of zero. \(\text{HCN}\) and \(\text{SO}_2\) also have both types of bonds but are polar molecules (non-zero net dipole). \(\text{CH}_2\text{Cl}_2\) is polar and contains only \(\sigma\) bonds.

[1 mark] for selecting B. Award 0 marks for any other option.

Step 1 is the slow step, so it is the rate-determining step. The rate law is determined by the reactants of this step: \(\text{rate} = k[\text{NO}_2]^2\). Since carbon monoxide (\(\text{CO}\)) is not involved in the rate-determining step, its concentration has no effect on the initial rate of reaction (the order with respect to \(\text{CO}\) is zero).

[1 mark] for selecting C. Award 0 marks for any other option.

The structural formula contains a hydroxyl group (\(-\text{OH}\)) on the second carbon atom, and an ester group (\(-\text{COO}-\)) connecting the fourth carbon to a methyl group. Therefore, the functional groups present are hydroxyl and ester.

[1 mark] for selecting A. Award 0 marks for any other option.

Assume \(100\text{ g}\) of the compound. Mass of \(\text{Fe} = 70.0\text{ g}\) and mass of \(\text{O} = 30.0\text{ g}\). Moles of \(\text{Fe} = 70.0 / 55.85 \approx 1.253\text{ mol}\). Moles of \(\text{O} = 30.0 / 16.00 = 1.875\text{ mol}\). Dividing both by the smaller value (1.253) gives a ratio of 1 to 1.5, which multiplies to 2 to 3. Hence, the empirical formula is \(\text{Fe}_2\text{O}_3\).

[1 mark] for selecting B. Award 0 marks for any other option.

The neutral cobalt atom has the ground-state electron configuration \([\text{Ar}] 3\text{d}^7 4\text{s}^2\). When forming a \(2+\) ion, transition metals lose their outer \(4s\) electrons first. Thus, the configuration of \(\text{Co}^{2+}\) is \([\text{Ar}] 3\text{d}^7 4\text{s}^0\).

[1 mark] for selecting A. Award 0 marks for any other option.

Spontaneity requires \(\Delta G^\theta = \Delta H^\theta - T\Delta S^\theta < 0\). Setting \(\Delta G^\theta = 0\) gives \(T = \Delta H^\theta / \Delta S^\theta = -125,000\text{ J mol}^{-1} / -400\text{ J K}^{-1}\text{ mol}^{-1} = 312.5\text{ K}\). Since both \(\Delta H^\theta\) and \(\Delta S^\theta\) are negative, the reaction is spontaneous at low temperatures (below \(312.5\text{ K}\)) where enthalpy dominates, and becomes non-spontaneous at high temperatures (above \(312.5\text{ K}\)) where the entropy term dominates.

[1 mark] for selecting B. Award 0 marks for any other option.

To determine the bond angle, we must look at the Lewis structures and the number of electron domains around the central nitrogen atom:

- \(\text{NO}_2^+\): The central nitrogen has 2 bonding domains and 0 lone pairs. It is linear with a bond angle of \(180^\circ\).
- \(\text{NO}_2\): The central nitrogen has 2 bonding domains and 1 single unpaired electron. It is bent with a bond angle of approximately \(134^\circ\).
- \(\text{NO}_2^-\): The central nitrogen has 2 bonding domains and 1 lone pair. The lone pair-bonding pair repulsion is stronger than the single electron-bonding pair repulsion, resulting in a bent shape with a bond angle of approximately \(115^\circ\).
- \(\text{NO}_3^-\): The central nitrogen has 3 bonding domains and 0 lone pairs. It is trigonal planar with a bond angle of \(120^\circ\).

Comparing these, \(\text{NO}_2^-\) has the smallest bond angle.

Award [1] for correct option C.
Award [0] for other options.

1. Compare standard electrode potentials: \(E^\theta(\text{Fe}^{2+}/\text{Fe}) = -0.45\text{ V}\) is more negative than \(E^\theta(\text{Ag}^+/\text{Ag}) = +0.80\text{ V}\).
2. The half-cell with the more negative potential undergoes oxidation: \(\text{Fe}(\text{s}) \to \text{Fe}^{2+}(\text{aq}) + 2\text{e}^-\). Oxidation always occurs at the anode, so \(\text{Fe}(\text{s})\) is the anode.
3. The half-cell with the more positive potential undergoes reduction: \(\text{Ag}^+(\text{aq}) + \text{e}^- \to \text{Ag}(\text{s})\). Reduction always occurs at the cathode, so \(\text{Ag}(\text{s})\) is the cathode.
4. Electrons flow through the external circuit from the anode (oxidation site) to the cathode (reduction site), which is from \(\text{Fe}(\text{s})\) to \(\text{Ag}(\text{s})\).

Award [1] for correct option B.
Award [0] for other options.

The rate-determining step is the slowest step in the reaction mechanism, which is Step 1:
\(\text{NO}_2(\text{g}) + \text{NO}_2(\text{g}) \to \text{NO}_3(\text{g}) + \text{NO}(\text{g})\).

The rate equation is derived directly from the reactants of this rate-determining step. Since two molecules of \(\text{NO}_2\) react in this step, the rate equation is:
\(\text{Rate} = k [\text{NO}_2]^2\).

Award [1] for correct option B.
Award [0] for other options.

Let's analyze the structural components of \(N\)-(4-hydroxyphenyl)acetamide (paracetamol):
1. The carbonyl group attached to nitrogen (\(-\text{NH}-\text{CO}-\text{CH}_3\)) is an **amide** group.
2. The benzene ring is an **arene** group.
3. The hydroxyl group (\(-\text{OH}\)) attached directly to the benzene ring is a **phenol** group.

Therefore, the functional groups present are amide, phenol, and arene.

Award [1] for correct option C.
Award [0] for other options.

A conjugate acid-base pair consists of two species that differ by exactly one proton (\(\text{H}^+\)).

- \(\text{H}_3\text{O}^+\) and \(\text{OH}^-\): differ by two protons (incorrect).
- \(\text{H}_2\text{PO}_4^-\) and \(\text{PO}_4^{3-}\): differ by two protons (incorrect).
- \(\text{NH}_4^+\) and \(\text{NH}_2^-\): differ by two protons (incorrect).
- \(\text{HSO}_4^-\) and \(\text{SO}_4^{2-}\): differ by exactly one proton (\(\text{HSO}_4^- \rightleftharpoons \text{H}^+ + \text{SO}_4^{2-}\)) (correct).

Award [1] for correct option D.
Award [0] for other options.

Let's determine the electron configuration of each transition metal ion:
- Manganese (\(\text{Mn}\), \(Z = 25\)): Neutral atom configuration is \([\text{Ar}] 4\text{s}^2 3\text{d}^5\). When forming the \(\text{Mn}^{2+}\) ion, the two electrons are removed from the outer \(4\text{s}\) orbital first, resulting in \([\text{Ar}] 3\text{d}^5\).
- Iron (\(\text{Fe}\), \(Z = 26\)): Neutral is \([\text{Ar}] 4\text{s}^2 3\text{d}^6\). \(\text{Fe}^{2+}\) is \([\text{Ar}] 3\text{d}^6\).
- Chromium (\(\text{Cr}\), \(Z = 24\)): Neutral is \([\text{Ar}] 4\text{s}^1 3\text{d}^5\). \(\text{Cr}^{3+}\) is \([\text{Ar}] 3\text{d}^3\).
- Cobalt (\(\text{Co}\), \(Z = 27\)): Neutral is \([\text{Ar}] 4\text{s}^2 3\text{d}^7\). \(\text{Co}^{3+}\) is \([\text{Ar}] 3\text{d}^6\).

Award [1] for correct option B.
Award [0] for other options.

According to the ideal gas equation: \(PV = nRT\), which can be rearranged to \(V = \frac{nRT}{P}\).

Let the initial state be \(V_1 = \frac{nRT_1}{P_1}\).
For the final state:
- Temperature \(T_2 = 2T_1\)
- Pressure \(P_2 = 0.5P_1\)

Substituting these values into the expression for the final volume \(V_2\):
\(V_2 = \frac{nR(2T_1)}{0.5P_1} = 4 \left( \frac{nRT_1}{P_1} \right) = 4V_1\).

Therefore, the new volume is \(4V\).

Award [1] for correct option D.
Award [0] for other options.

Lattice enthalpy depends on the charges of the ions and their ionic radii:
1. **Ionic Charges**: The electrostatic force of attraction is directly proportional to the product of the ionic charges (\(q_1 q_2\)).
- For \(\text{NaCl}\) and \(\text{LiF}\), the charges are \(+1\) and \(-1\) (product = 1).
- For \(\text{MgO}\) and \(\text{CaO}\), the charges are \(+2\) and \(-2\) (product = 4).
Therefore, \(\text{MgO}\) and \(\text{CaO}\) have much higher lattice enthalpies than \(\text{NaCl}\) and \(\text{LiF}\).
2. **Ionic Radii**: The electrostatic force of attraction is inversely proportional to the sum of the ionic radii. Smaller ions can get closer together, resulting in a stronger electrostatic attraction.
- \(\text{Mg}^{2+}\) is smaller than \(\text{Ca}^{2+}\) because it has fewer electron shells.
- Therefore, the attractions in \(\text{MgO}\) are stronger than in \(\text{CaO}\).

Thus, \(\text{MgO}\) has the highest lattice enthalpy.

Award [1] for correct option B.
Award [0] for other options.

The half-reactions are: 1) Reduction: \(\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{e}^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}\) and 2) Oxidation: \(\text{C}_2\text{H}_5\text{OH} + \text{H}_2\text{O} \rightarrow \text{CH}_3\text{COOH} + 4\text{H}^+ + 4\text{e}^-\). To equalize the electron transfer (common multiple of 12), we multiply the reduction half-reaction by 2 and the oxidation half-reaction by 3: \(2\text{Cr}_2\text{O}_7^{2-} + 28\text{H}^+ + 12\text{e}^- \rightarrow 4\text{Cr}^{3+} + 14\text{H}_2\text{O}\) and \(3\text{C}_2\text{H}_5\text{OH} + 3\text{H}_2\text{O} \rightarrow 3\text{CH}_3\text{COOH} + 12\text{H}^+ + 12\text{e}^-\). Combining these and simplifying the \(\text{H}^+\) and \(\text{H}_2\text{O}\) terms yields: \(2\text{Cr}_2\text{O}_7^{2-} + 3\text{C}_2\text{H}_5\text{OH} + 16\text{H}^+ \rightarrow 4\text{Cr}^{3+} + 3\text{CH}_3\text{COOH} + 11\text{H}_2\text{O}\). Thus, the coefficient of \(\text{H}^+\) is 16.

Award [1] for the correct answer C. Deduct 0 marks for any other response.

In \(\text{ClF}_3\), chlorine has 7 valence electrons, three of which form single covalent bonds with fluorine atoms, leaving two non-bonding lone pairs. The electron-domain geometry is trigonal bipyramidal (5 domains), and the molecular geometry with two lone pairs in equatorial positions is T-shaped.

Award [1] for the correct answer A.

The reaction is third-order overall. The units for the rate constant \(k\) are calculated from: \(k = \frac{\text{Rate}}{[\text{A}]^2[\text{B}]} = \frac{\text{mol}\,\text{dm}^{-3}\,\text{s}^{-1}}{(\text{mol}\,\text{dm}^{-3})^3} = \text{dm}^6\,\text{mol}^{-2}\,\text{s}^{-1}\).

Award [1] for the correct answer C.

For pure water, \([\text{H}^+] = [\text{OH}^-]\), which means the water is neutral by definition. \([\text{H}^+] = \sqrt{K_{\text{w}}} = \sqrt{5.48 \times 10^{-14}} = 2.34 \times 10^{-7}\,\text{mol}\,\text{dm}^{-3}\). Thus, \(\text{pH} = -\log_{10}(2.34 \times 10^{-7}) = 6.63\). Even though the pH is less than 7, the water is neutral because the concentrations of hydronium and hydroxide ions remain equal.

Award [1] for the correct answer C.

Paramagnetism requires the presence of one or more unpaired electrons. \(\text{Zn}^{2+}\) has configuration \([\text{Ar}]3\text{d}^{10}\) (all paired), \(\text{Sc}^{3+}\) has configuration \([\text{Ar}]\) (all paired), \(\text{Cu}^+\) has configuration \([\text{Ar}]3\text{d}^{10}\) (all paired), and \(\text{Fe}^{3+}\) has configuration \([\text{Ar}]3\text{d}^5\) with five unpaired d-electrons, making it paramagnetic.

Award [1] for the correct answer D.

An amine's classification (primary, secondary, or tertiary) is based on the number of alkyl or aryl groups directly attached to the nitrogen atom. In \(\text{(CH}_3\text{)}_2\text{NCH}_2\text{CH}_3\), the nitrogen atom is bonded to three carbon atoms (two methyl groups and one ethyl group), classifying it as a tertiary amine. Option A is a primary amine because the nitrogen is attached to only one carbon atom.

Award [1] for the correct answer B.

In \(100\,\text{g}\) of the compound: Mass of \(\text{N} = 30.4\,\text{g}\), Mass of \(\text{O} = 100 - 30.4 = 69.6\,\text{g}\). Moles of \(\text{N} = \frac{30.4}{14.01} = 2.17\,\text{mol}\), Moles of \(\text{O} = \frac{69.6}{16.00} = 4.35\,\text{mol}\). Dividing by the smallest value gives the molar ratio: \(\text{N}: \frac{2.17}{2.17} = 1\) and \(\text{O}: \frac{4.35}{2.17} = 2.00\). Thus, the empirical formula is \(\text{NO}_2\).

Award [1] for the correct answer B.

Real gases behave most ideally under conditions where intermolecular attractive forces are minimized and the volume of the gas molecules is negligible compared to the total volume of the container. Low pressure ensures that the gas particles are widely spaced, minimizing the molecular volume contribution. High temperature ensures that the gas particles have high kinetic energy, which overcomes any intermolecular attractive forces during collisions.

Award [1] for the correct answer D.

(a) Sulfur dioxide (\(\text{SO}_2\)) is toxic/irritating to the respiratory system. Precaution: Carry out the reaction in a fume cupboard / ensure good ventilation.
(b) Let \(k_1 \propto 1/50.0 = 0.0200\text{ s}^{-1}\) and \(k_2 \propto 1/12.5 = 0.0800\text{ s}^{-1}\).
\(\frac{k_2}{k_1} = \frac{0.0800}{0.0200} = 4.00\)
\(\ln(4.00) = 1.386\)
Using the equation:
\(1.386 = \frac{E_a}{8.31} \times \left(\frac{1}{298} - \frac{1}{318}\right)\)
\(\frac{1}{298} - \frac{1}{318} = 3.356 \times 10^{-3} - 3.145 \times 10^{-3} = 2.11 \times 10^{-4}\text{ K}^{-1}\)
\(1.386 = \frac{E_a}{8.31} \times 2.11 \times 10^{-4}\)
\(E_a = \frac{1.386 \times 8.31}{2.11 \times 10^{-4}} = 54598\text{ J mol}^{-1} = 54.6\text{ kJ mol}^{-1}\).
(c) Percentage uncertainty = \(\frac{0.5}{25.0} \times 100\% = 2.0\%\).
(d) Microscale reduces the volume of reagents used, thereby minimizing waste and reducing the amount of toxic sulfur dioxide gas released, which aligns with the Green Chemistry principle of waste prevention.
(e) There will be no change in the rate of reaction because the order with respect to \(\text{HCl}\) is zero (meaning its concentration does not affect the rate).
(f) As temperature increases, the average kinetic energy of the particles increases, shifting the Maxwell-Boltzmann distribution curve to the right. Consequently, a significantly larger fraction of molecules have energy greater than or equal to the activation energy (\(E \ge E_a\)).

(a) [1 mark] Award 1 mark for identifying \(\text{SO}_2\) as toxic/harmful to the lungs/irritant AND specifying a fume hood/fume cupboard or well-ventilated area.
(b) [4 marks]
- Award 1 mark for correctly determining the ratio \(\frac{k_2}{k_1} = 4.0\) or \(\ln\left(\frac{k_2}{k_1}\right) = 1.39\).
- Award 1 mark for calculating \(\left(\frac{1}{298} - \frac{1}{318}\right) = 2.11 \times 10^{-4}\text{ K}^{-1}\).
- Award 1 mark for correct algebraic transposition to find \(E_a\) in Joules (\(54598\text{ J mol}^{-1}\)).
- Award 1 mark for converting to \(\text{kJ mol}^{-1}\) with correct significant figures (\(54.6\text{ kJ mol}^{-1}\)). Accept range \(54.0 - 55.0\).
(c) [1 mark] Award 1 mark for \(2.0\%\).
(d) [1 mark] Award 1 mark for mentioning minimization of waste/toxic gas production or aligning with Green Chemistry waste prevention.
(e) [0.75 marks] Award 0.75 marks for stating no change in rate because it is zero order.
(f) [1 mark] Award 1 mark for stating that a larger fraction of particles have energy \(\ge E_a\).

(a) Extrapolation corrects for the heat lost to the surroundings during the time the reaction is occurring, which prevents underestimating the maximum temperature rise.
(b) Mass of solution = \(50.0\text{ cm}^3 \times 1.00\text{ g cm}^{-3} = 50.0\text{ g}\).
\(q = m \cdot c \cdot \Delta T = 50.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 10.5\text{ K} = 2194.5\text{ J} \approx 2190\text{ J}\) (or \(2.19\text{ kJ}\)).
(c) First, determine the limiting reactant:
Moles of \(\text{Cu}^{2+} = C \times V = 0.200\text{ mol dm}^{-3} \times 0.0500\text{ dm}^3 = 0.0100\text{ mol}\).
Moles of \(\text{Mg} = \frac{0.500\text{ g}}{24.31\text{ g mol}^{-1}} = 0.0206\text{ mol}\).
Since the stoichiometry is 1:1, \(\text{Cu}^{2+}\) is the limiting reactant.
\(\Delta H = -\frac{q}{n(\text{limiting})} = -\frac{2194.5\text{ J}}{0.0100\text{ mol}} = -219450\text{ J mol}^{-1} = -219\text{ kJ mol}^{-1}\).
(d) Source 1: Heat loss to surroundings. Improvement: Use a lid on the polystyrene cup or place it in a beaker with cotton wool insulation.
Source 2: Heat capacity of the polystyrene cup/thermometer is ignored. Improvement: Determine the heat capacity of the calorimeter first and include it in the calculation.
(e) Use moderate concentrations of chemicals (such as \(0.200\text{ mol dm}^{-3}\)) rather than highly concentrated solutions, or ensure proper management and recycling of the copper metal produced rather than washing it down the sink.

(a) [1 mark] Award 1 mark for explaining that extrapolation compensates/corrects for heat lost to the surroundings during the reaction.
(b) [1.75 marks]
- Award 1 mark for correct calculation of mass (\(50.0\text{ g}\)) and formula \(q = mc\Delta T\).
- Award 0.75 marks for correct calculation of \(2194.5\text{ J}\) (or \(2190\text{ J}\) or \(2.19\text{ kJ}\)).
(c) [3 marks]
- Award 1 mark for calculating moles of \(\text{Cu}^{2+} = 0.0100\text{ mol}\) and showing it is the limiting reactant.
- Award 1 mark for dividing heat by moles of limiting reactant (\(\frac{2.19}{0.0100}\)).
- Award 1 mark for final value with correct negative sign and units (\(-219\text{ kJ mol}^{-1}\)).
(d) [2 marks] Award 1 mark for each valid systematic error coupled with a suitable improvement (up to 2 marks). Accept: Heat loss -> insulated lid/double cups; Heat absorbed by cup -> calibrate calorimeter.
(e) [1 mark] Award 1 mark for mentioning using dilute/safe concentrations of salt solutions to minimize hazard, or recycling the copper produced.

(a) Volumetric pipette (or bulb pipette).
(b) Reaction: \(\text{CH}_3\text{COOH} + \text{NaOH} \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O}\).
Moles of \(\text{NaOH} = 0.01850\text{ dm}^3 \times 0.100\text{ mol dm}^{-3} = 1.85 \times 10^{-3}\text{ mol}\).
Since stoichiometry is 1:1, moles of \(\text{CH}_3\text{COOH}\) in \(25.00\text{ cm}^3\) = \(1.85 \times 10^{-3}\text{ mol}\).
Concentration of diluted \(\text{CH}_3\text{COOH} = \frac{1.85 \times 10^{-3}\text{ mol}}{0.02500\text{ dm}^3} = 0.0740\text{ mol dm}^{-3}\).
Dilution factor = \(\frac{100.0\text{ cm}^3}{10.00\text{ cm}^3} = 10\).
Concentration of original vinegar = \(0.0740\text{ mol dm}^{-3} \times 10 = 0.740\text{ mol dm}^{-3}\).
(c) At the half-equivalence point, \([\text{CH}_3\text{COOH}] = [\text{CH}_3\text{COO}^-]\), so \(\text{pH} = \text{p}K_a\).
\(\text{p}K_a = 4.76\).
\(K_a = 10^{-\text{pH}} = 10^{-4.76} = 1.74 \times 10^{-5}\text{ mol dm}^{-3}\).
(d) At the equivalence point, the salt sodium ethanoate is formed. The ethanoate ion (\(\text{CH}_3\text{COO}^-\)) hydrolyzes in water to form hydroxide ions:
\(\text{CH}_3\text{COO}^-(\text{aq}) + \text{H}_2\text{O}(\text{l}) \rightleftharpoons \text{CH}_3\text{COOH}(\text{aq}) + \text{OH}^-(\text{aq})\).
The increase in \([\text{OH}^-]\) causes the pH to be greater than 7.
(e) A pH probe avoids the use of synthetic organic indicator dyes (such as phenolphthalein), which are often dissolved in toxic organic solvents (like ethanol) and can be hazardous to synthesize and dispose of.

(a) [1 mark] Award 1 mark for volumetric pipette (accept 'pipette', reject 'dropping pipette' or 'measuring cylinder').
(b) [3 marks]
- Award 1 mark for calculating moles of \(\text{NaOH} = 1.85 \times 10^{-3}\text{ mol}\).
- Award 1 mark for finding diluted concentration = \(0.0740\text{ mol dm}^{-3}\).
- Award 1 mark for correct undiluted concentration = \(0.740\text{ mol dm}^{-3}\) (consequent on dilution factor of 10).
(c) [1.75 marks]
- Award 0.75 marks for recognizing that \(\text{pH} = \text{p}K_a\) at half-equivalence.
- Award 1 mark for calculating \(K_a = 1.74 \times 10^{-5}\text{ mol dm}^{-3}\) (accept \(1.7 \times 10^{-5}\) to \(1.8 \times 10^{-5}\)).
(d) [2 marks]
- Award 1 mark for explaining that \(\text{CH}_3\text{COO}^-\) reacts with water to release \(\text{OH}^-\) ions / basic hydrolysis.
- Award 1 mark for a correct ionic equation: \(\text{CH}_3\text{COO}^- + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{COOH} + \text{OH}^-\) (must show equilibrium sign).
(e) [1 mark] Award 1 mark for explaining that pH probes eliminate the need for chemical indicator solutions, thus reducing the use of hazardous chemicals/solvents.

(a) Colorless to permanent pale pink. The reactant \(\text{MnO}_4^-\) (permanganate ion) acts as its own indicator.
(b) \(n(\text{MnO}_4^-) = C \times V = 0.0200\text{ mol dm}^{-3} \times 0.01520\text{ dm}^3 = 3.04 \times 10^{-4}\text{ mol}\).
(c) From the balanced equation, \(5\text{ moles of Fe}^{2+}\) react with \(1\text{ mole of MnO}_4^-\).
\(n(\text{Fe}^{2+}) = 5 \times n(\text{MnO}_4^-) = 5 \times 3.04 \times 10^{-4}\text{ mol} = 1.52 \times 10^{-3}\text{ mol}\).
Mass of \(\text{Fe} = 1.52 \times 10^{-3}\text{ mol} \times 55.85\text{ g mol}^{-1} = 0.08489\text{ g}\).
Percentage of iron by mass = \(\frac{0.08489\text{ g}}{0.560\text{ g}} \times 100\% = 15.2\%\) (to 3 significant figures).
(d) A titre volume is calculated as the difference between two readings (the final reading and the initial reading). Since each reading has an absolute uncertainty of \(\pm 0.05\text{ cm}^3\), the total absolute uncertainty is the sum of both: \(\pm 0.10\text{ cm}^3\).
Percentage uncertainty = \(\frac{0.10\text{ cm}^3}{15.20\text{ cm}^3} \times 100\% \approx 0.66\%\).
(e) Potassium dichromate(VI) contains chromium in the +6 oxidation state, which is a known carcinogen, mutagen, and highly toxic to aquatic life. Manganese(II) compounds produced by \(\text{KMnO}_4\) reduction are significantly less hazardous to health and the environment, aligning with the Green Chemistry principle of 'Designing safer chemicals' and reducing toxicity.

(a) [1 mark] Award 1 mark for stating colorless to pale/light pink AND that \(\text{MnO}_4^-\) (or potassium manganate(VII)) acts as the self-indicator.
(b) [1.75 marks]
- Award 1 mark for showing correct formula and conversion of volume to \(\text{dm}^3\) (\(0.01520\text{ dm}^3\)).
- Award 0.75 marks for correct moles = \(3.04 \times 10^{-4}\text{ mol}\).
(c) [3 marks]
- Award 1 mark for calculating moles of \(\text{Fe}^{2+} = 1.52 \times 10^{-3}\text{ mol}\) (applying the 1:5 ratio).
- Award 1 mark for calculating the mass of \(\text{Fe} = 0.0849\text{ g}\).
- Award 1 mark for the final percentage = \(15.2\%\) (accept range \(15.1\% - 15.3\%\)).
(d) [2 marks]
- Award 1 mark for explaining that two readings (initial and final) are taken, so their uncertainties are added (\(0.05 + 0.05 = 0.10\text{ cm}^3\)).
- Award 1 mark for calculating percentage uncertainty = \(0.66\%\) (accept \(0.7\%\)).
(e) [1 mark] Award 1 mark for stating that chromium(VI) is a carcinogen/highly toxic, whereas manganese species are less toxic/less hazardous.

Part (a)(i)
- Anode (Oxidation): \( \text{Mg(s)} \rightarrow \text{Mg}^{2+}(\text{aq}) + 2\text{e}^- \)
- Cathode (Reduction): \( \text{Fe}^{2+}(\text{aq}) + 2\text{e}^- \rightarrow \text{Fe(s)} \)
- Overall standard cell potential: \( E^\theta_{\text{cell}} = E^\theta_{\text{cathode}} - E^\theta_{\text{anode}} = -0.44\text{ V} - (-2.37\text{ V}) = +1.93\text{ V} \).
- State symbols must be correct for all species.

Part (a)(ii)
- The salt bridge completes the electrical circuit and maintains electrical neutrality in both half-cells by allowing ion flow.
- Electrons flow through the external circuit from the magnesium electrode (anode) to the iron electrode (cathode).
- The magnesium electrode is the negative electrode (anode) and the iron electrode is the positive electrode (cathode).
- Potassium ions (\( \text{K}^+ \)) migrate from the salt bridge into the iron half-cell (cathode) to replace the positive charge lost as \( \text{Fe}^{2+} \) ions are reduced.
- Nitrate ions (\( \text{NO}_3^- \)) migrate into the magnesium half-cell (anode) to balance the positive charge generated as \( \text{Mg}^{2+} \) ions are produced.

Part (b)(i)
- Anode product: Chlorine gas, \( \text{Cl}_2(\text{g}) \)
- Cathode product: Sodium liquid/solid, \( \text{Na(l/s)} \)
- Anode half-equation: \( 2\text{Cl}^-(\text{l}) \rightarrow \text{Cl}_2(\text{g}) + 2\text{e}^- \)
- Cathode half-equation: \( \text{Na}^+(\text{l}) + \text{e}^- \rightarrow \text{Na(l)} \)

Part (b)(ii)
- Anode product: Oxygen gas, \( \text{O}_2(\text{g}) \)
- Cathode product: Hydrogen gas, \( \text{H}_2(\text{g}) \)
- Explanation at Cathode: Both \( \text{H}_2\text{O} \) (or \( \text{H}^+ \)) and \( \text{Na}^+ \) compete for reduction. Water is reduced preferentially because its reduction potential is much less negative (more positive) than that of \( \text{Na}^+ \), meaning water is more easily reduced.
- Explanation at Anode: Both \( \text{H}_2\text{O} \) (or \( \text{OH}^- \)) and \( \text{Cl}^- \) compete for oxidation. Water is oxidized preferentially because in a dilute solution, the concentration of \( \text{Cl}^- \) is very low, making the oxidation of water more thermodynamically and kinetically favorable.

Part (c)
- The standard cell potential, \( E^\theta_{\text{cell}} \), is the maximum potential difference between two standard half-cells measured under standard conditions of 298 K, 1 mol dm\( ^{-3} \) ion concentrations, and 100 kPa pressure.
- It relates to spontaneity via \( \Delta G^\theta = -nFE^\theta_{\text{cell}} \). A spontaneous reaction has a positive standard cell potential (\( E^\theta_{\text{cell}} > 0 \)), which corresponds to a negative standard Gibbs free energy change (\( \Delta G^\theta < 0 \)).

Part (a)(i) [Total: 4 marks]
- [1] Correct anode half-equation including state symbols.
- [1] Correct cathode half-equation including state symbols.
- [1] Correct calculation of standard cell potential: \( +1.93\text{ V} \).
- [1] Correct state symbols throughout both half-equations.

Part (a)(ii) [Total: 5 marks]
- [1] Role of salt bridge: completes circuit AND maintains charge balance/neutrality.
- [1] Electron flow: from Mg to Fe through the external wire.
- [1] Polarities: Mg is negative, Fe is positive.
- [1] \( \text{K}^+ \) ions move into the iron (cathode) half-cell.
- [1] \( \text{NO}_3^- \) ions move into the magnesium (anode) half-cell.

Part (b)(i) [Total: 3 marks]
- [1] Correctly identifies both products: \( \text{Cl}_2 \) at anode, \( \text{Na} \) at cathode.
- [1] Correct anode equation: \( 2\text{Cl}^- \rightarrow \text{Cl}_2 + 2\text{e}^- \).
- [1] Correct cathode equation: \( \text{Na}^+ + \text{e}^- \rightarrow \text{Na} \).

Part (b)(ii) [Total: 4 marks]
- [1] Correctly identifies both products: \( \text{O}_2 \) at anode, \( \text{H}_2 \) at cathode.
- [1] Explains cathode selection: \( \text{H}_2\text{O}/\text{H}^+ \) is more easily reduced than \( \text{Na}^+ \) (or has a more positive reduction potential).
- [1] Explains anode selection: in dilute solution, water is oxidized preferentially over \( \text{Cl}^- \) because \( \text{Cl}^- \) concentration is too low.
- [1] Provides at least one supporting half-equation, e.g., \( 2\text{H}_2\text{O} + 2\text{e}^- \rightarrow \text{H}_2 + 2\text{OH}^- \) OR \( 2\text{H}_2\text{O} \rightarrow \text{O}_2 + 4\text{H}^+ + 4\text{e}^- \).

Part (c) [Total: 2 marks]
- [1] Definition: Potential difference under standard conditions.
- [1] Spontaneity link: \( \Delta G^\theta = -nFE^\theta_{\text{cell}} \), so spontaneous reactions have positive \( E^\theta \) and negative \( \Delta G^\theta \).

Part (a)(i)
- Determine order with respect to \( [\text{Cl}_2] \): Compare Run 1 and Run 2. \( [\text{NO}] \) is held constant, \( [\text{Cl}_2] \) doubles, and the initial rate doubles (from \( 1.20 \times 10^{-3} \) to \( 2.40 \times 10^{-3} \)). Therefore, the order of reaction with respect to \( [\text{Cl}_2] \) is 1.
- Determine order with respect to \( [\text{NO}] \): Compare Run 1 and Run 3. \( [\text{Cl}_2] \) is held constant, \( [\text{NO}] \) doubles, and the initial rate quadruples (from \( 1.20 \times 10^{-3} \) to \( 4.80 \times 10^{-3} \)). Therefore, the order of reaction with respect to \( [\text{NO}] \) is 2.
- Overall rate expression: \( \text{Rate} = k[\text{NO}]^2[\text{Cl}_2] \).

Part (a)(ii)
- Substitute values from Run 1 into the rate expression: \( 1.20 \times 10^{-3} \text{ mol dm}^{-3}\text{ s}^{-1} = k (0.10\text{ mol dm}^{-3})^2 (0.10\text{ mol dm}^{-3}) \).
- \( 1.20 \times 10^{-3} = k (1.00 \times 10^{-3}) \).
- \( k = 1.20 \).
- Units: \( k = \frac{\text{Rate}}{[\text{NO}]^2[\text{Cl}_2]} = \frac{\text{mol dm}^{-3}\text{ s}^{-1}}{(\text{mol dm}^{-3})^3} = \text{dm}^6\text{ mol}^{-2}\text{ s}^{-1} \).

Part (b)(i)
- The slow step (Step 2) is the rate-determining step (RDS), so the rate equation for this step is: \( \text{Rate} = k_2 [\text{NOCl}_2][\text{NO}] \).
- Since \( \text{NOCl}_2 \) is a reaction intermediate, its concentration must be expressed in terms of reactants. From the fast pre-equilibrium Step 1:
\( K_{\text{eq}} = \frac{[\text{NOCl}_2]}{[\text{NO}][\text{Cl}_2]} \Rightarrow [\text{NOCl}_2] = K_{\text{eq}}[\text{NO}][\text{Cl}_2] \).
- Substitute this expression into the rate equation for the slow step:
\( \text{Rate} = k_2 (K_{\text{eq}}[\text{NO}][\text{Cl}_2])[\text{NO}] = (k_2 K_{\text{eq}}) [\text{NO}]^2 [\text{Cl}_2] \).
- Letting \( k = k_2 K_{\text{eq}} \), the rate law becomes \( \text{Rate} = k[\text{NO}]^2[\text{Cl}_2] \), which matches the experimental rate expression.

Part (b)(ii)
- A sketch of potential energy vs. reaction coordinate showing:
- Two distinct activation peaks with a trough (local energy minimum) in between representing the intermediate \( \text{NOCl}_2 \).
- The second peak is higher than the first peak because Step 2 is the rate-determining step (higher activation energy, \( E_{a,2} \)).
- The energy of the products is lower than the reactants (exothermic reaction).
- Labels: Reactants (\( 2\text{NO} + \text{Cl}_2 \)), Products (\( 2\text{NOCl} \)), Intermediate (\( \text{NOCl}_2 + \text{NO} \)), Transition State 1 and 2, Overall \( \Delta H \) (negative), and \( E_{a} \) for the rate-determining step.

Part (c)
- A \( 10\text{ K} \) temperature increase causes the Maxwell-Boltzmann distribution curve to shift to the right and flatten.
- The peak of the curve shifts to a higher energy level, and more importantly, a much larger fraction of molecules possess kinetic energies greater than or equal to the activation energy (\( E \ge E_a \)).
- This results in a higher frequency of successful collisions, significantly increasing the rate constant, \( k \), and hence the reaction rate.

Part (a)(i) [Total: 4 marks]
- [1] Deduces order wrt \( \text{Cl}_2 \) is 1 with reasoning.
- [1] Deduces order wrt \( \text{NO} \) is 2 with reasoning.
- [1] Writes correct rate expression: \( \text{Rate} = k[\text{NO}]^2[\text{Cl}_2] \).
- [1] Logical path shown clearly.

Part (a)(ii) [Total: 3 marks]
- [1] Correct substitution of values into rate equation.
- [1] Value of \( k = 1.20 \) (accept 1.2).
- [1] Correct units: \( \text{dm}^6\text{ mol}^{-2}\text{ s}^{-1} \).

Part (b)(i) [Total: 4 marks]
- [1] Identifies slow step as the rate-determining step: \( \text{Rate} = k_2 [\text{NOCl}_2][\text{NO}] \).
- [1] Writes equilibrium expression for Step 1: \( K_{\text{eq}} = \frac{[\text{NOCl}_2]}{[\text{NO}][\text{Cl}_2]} \).
- [1] Rearranges to express intermediate concentration: \( [\text{NOCl}_2] = K_{\text{eq}}[\text{NO}][\text{Cl}_2] \).
- [1] Substitutes into RDS to show matching overall rate expression.

Part (b)(ii) [Total: 4 marks]
- [1] Correct axes: Y-axis as Potential Energy / Energy and X-axis as Reaction Coordinate / Progress of Reaction.
- [1] Double-peak profile with intermediate well, showing overall products at lower energy than reactants.
- [1] Shows second peak higher than first peak (consistent with step 2 being slower).
- [1] Clearly labels reactants, products, intermediate, \( \Delta H \), and \( E_a \) for the second step.

Part (c) [Total: 3 marks]
- [1] Maxwell-Boltzmann curve shifts to the right / has lower peak and wider distribution.
- [1] Identifies that a larger fraction / area under the curve is past the \( E_a \) boundary.
- [1] Mentions increased frequency/rate of successful/effective collisions (both words are required).

Part (a)(i)
- \( \text{XeF}_4 \): Xe has 8 valence electrons. 4 are shared in bonding domains, leaving 4 non-bonding electrons (2 lone pairs). Thus: 4 bonding domains, 2 lone pairs.
- \( \text{SF}_4 \): S has 6 valence electrons. 4 are shared in bonding domains, leaving 2 non-bonding electrons (1 lone pair). Thus: 4 bonding domains, 1 lone pair.
- \( \text{ClF}_3 \): Cl has 7 valence electrons. 3 are shared in bonding domains, leaving 4 non-bonding electrons (2 lone pairs). Thus: 3 bonding domains, 2 lone pairs.

Part (a)(ii)
- \( \text{XeF}_4 \): octahedral electron domain geometry; molecular geometry is square planar (the two lone pairs position themselves opposite to each other to minimize repulsion).
- \( \text{SF}_4 \): trigonal bipyramidal electron domain geometry; molecular geometry is seesaw / sawhorse (the lone pair occupies an equatorial position to minimize repulsion).
- \( \text{ClF}_3 \): trigonal bipyramidal electron domain geometry; molecular geometry is T-shaped (the two lone pairs occupy equatorial positions).

Part (a)(iii)
- \( \text{XeF}_4 \): Bond angles are exactly \( 90^\circ \) (and \( 180^\circ \)) because the two lone pairs are symmetrically placed opposite each other, canceling out any axial distortions.
- \( \text{SF}_4 \): Equatorial bond angles are compressed to \( < 120^\circ \) (actual \( \approx 102^\circ \)) and axial bond angles to \( < 90^\circ \) (actual \( \approx 173^\circ \) or \( < 180^\circ \)) due to lone pair-bonding pair repulsion.
- \( \text{ClF}_3 \): Bond angles are \( < 90^\circ \) (actual \( \approx 87.5^\circ \)) due to the strong repulsion from the two equatorial lone pairs.

Part (b)(i)
- A \( \sigma \) (sigma) bond is formed by the end-on (head-on) overlap of orbitals (either hybrid or atomic) along the internuclear axis. The electron density is concentrated directly between the nuclei of the bonded atoms.
- A \( \pi \) (pi) bond is formed by the sideways (parallel) overlap of unhybridized p-orbitals above and below the plane of the internuclear axis. Pi bonds restrict rotation around the bond.

Part (b)(ii)
- The carbon atoms in ethene are \( sp^2 \) hybridized.
- This leads to a trigonal planar geometry around each carbon atom with ideal bond angles of \( 120^\circ \).

Part (c)
- Carbon dioxide (\( \text{CO}_2 \)) has a linear molecular geometry. The two polar C=O bond dipoles point in equal and opposite directions, meaning they cancel each other out, resulting in a net molecular dipole of zero (non-polar).
- Sulfur dioxide (\( \text{SO}_2 \)) has a bent/V-shaped molecular geometry because of the lone pair on the sulfur atom. The polar S=O bond dipoles do not point in opposite directions and do not cancel out, resulting in a net molecular dipole (polar).

Part (a)(i) [Total: 3 marks]
- [1] \( \text{XeF}_4 \): 4 bonding domains and 2 lone pairs.
- [1] \( \text{SF}_4 \): 4 bonding domains and 1 lone pair.
- [1] \( \text{ClF}_3 \): 3 bonding domains and 2 lone pairs.

Part (a)(ii) [Total: 3 marks]
- [1] \( \text{XeF}_4 \): Square planar.
- [1] \( \text{SF}_4 \): Seesaw / sawhorse.
- [1] \( \text{ClF}_3 \): T-shaped.

Part (a)(iii) [Total: 3 marks]
- [1] \( \text{XeF}_4 \): \( 90^\circ \) (and/or \( 180^\circ \)).
- [1] \( \text{SF}_4 \): \( < 120^\circ \) and \( < 90^\circ \) (accept values in range \( 100\text{--}118^\circ \) and \( 85\text{--}89^\circ \)).
- [1] \( \text{ClF}_3 \): \( < 90^\circ \) (accept values in range \( 85\text{--}89^\circ \)).

Part (b)(i) [Total: 4 marks]
- [1] \( \sigma \) bond: formed by end-on / head-on overlap of orbitals.
- [1] \( \sigma \) bond: electron density along internuclear axis.
- [1] \( \pi \) bond: formed by sideways / parallel overlap of unhybridized p-orbitals.
- [1] \( \pi \) bond: electron density above and below the internuclear axis.

Part (b)(ii) [Total: 2 marks]
- [1] Hybridization: \( sp^2 \).
- [1] Geometry: Trigonal planar shape with bond angles of \( 120^\circ \).

Part (c) [Total: 3 marks]
- [1] State that \( \text{CO}_2 \) is linear and \( \text{SO}_2 \) is bent.
- [1] In \( \text{CO}_2 \), the bond dipoles are symmetrical and cancel out.
- [1] In \( \text{SO}_2 \), the bend shape is asymmetrical (due to lone pair repulsion) so dipoles do not cancel.

Part (a)(i)
- \( K_a = \frac{[\text{CH}_3\text{COO}^-][\text{H}_3\text{O}^+]}{[\text{CH}_3\text{COOH}]} \).
- Let \( x = [\text{H}_3\text{O}^+] = [\text{CH}_3\text{COO}^-] \). Then \( 1.80 \times 10^{-5} = \frac{x^2}{0.150 - x} \).
- Assuming \( x \ll 0.150 \), we approximate \( 1.80 \times 10^{-5} \approx \frac{x^2}{0.150} \).
- \( x^2 = 2.70 \times 10^{-6} \Rightarrow x = 1.64 \times 10^{-3} \text{ mol dm}^{-3} \).
- \( \text{pH} = -\log_{10}(1.64 \times 10^{-3}) = 2.78 \).
- Assumptions: 1) The ionization of the weak acid is negligible relative to its initial concentration. 2) The autoionization of water contributes a negligible amount of \( \text{H}_3\text{O}^+ \).

Part (a)(ii)
- \( n(\text{CH}_3\text{COOH}) = 0.0250\text{ dm}^3 \times 0.150\text{ mol dm}^{-3} = 3.75 \times 10^{-3}\text{ mol} \).
- \( n(\text{NaOH}) = 0.0125\text{ dm}^3 \times 0.150\text{ mol dm}^{-3} = 1.875 \times 10^{-3}\text{ mol} \).
- Reaction: \( \text{CH}_3\text{COOH} + \text{NaOH} \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O} \).
- Since \( n(\text{NaOH}) \) is half of \( n(\text{CH}_3\text{COOH}) \), exactly half of the ethanoic acid is neutralized.
- Remaining \( n(\text{CH}_3\text{COOH}) = 3.75 \times 10^{-3} - 1.875 \times 10^{-3} = 1.875 \times 10^{-3}\text{ mol} \).
- Formed \( n(\text{CH}_3\text{COO}^-) = 1.875 \times 10^{-3}\text{ mol} \).
- Because \( [\text{CH}_3\text{COOH}] = [\text{CH}_3\text{COO}^-] \), this is the half-equivalence point.
- Using Henderson-Hasselbalch: \( \text{pH} = \text{p}K_a + \log_{10}\left(\frac{[\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]}\right) = \text{p}K_a \).
- \( \text{pH} = -\log_{10}(1.80 \times 10^{-5}) = 4.74 \).

Part (b)
- Sketch showing standard weak acid-strong base titration curve profile:
- Starts at pH \( \approx 2.8 \).
- Rises slightly, then levels off to form a flat buffer region centered around pH \( 4.74 \) at \( 12.5 \text{ cm}^3 \).
- Steep vertical inflection occurs around \( 25.0 \text{ cm}^3 \).
- The equivalence point is at \( 25.0 \text{ cm}^3 \) and has a \( \text{pH} > 7 \) (typically around \( 8.7 \)).
- Explanation for pH > 7: The salt formed is sodium ethanoate. The conjugate base, \( \text{CH}_3\text{COO}^- \), hydrolyzes in water to produce hydroxide ions: \( \text{CH}_3\text{COO}^-(\text{aq}) + \text{H}_2\text{O(l)} \rightleftharpoons \text{CH}_3\text{COOH(aq)} + \text{OH}^-(\text{aq}) \).

Part (c)(i)
- Sulfurous acid formation: \( \text{SO}_2(\text{g}) + \text{H}_2\text{O(l)} \rightarrow \text{H}_2\text{SO}_3(\text{aq}) \)
- Nitric acid formation: \( 4\text{NO}_2(\text{g}) + \text{O}_2(\text{g}) + 2\text{H}_2\text{O(l)} \rightarrow 4\text{HNO}_3(\text{aq}) \) (or \( 2\text{NO}_2(\text{g}) + \text{H}_2\text{O(l)} \rightarrow \text{HNO}_2(\text{aq}) + \text{HNO}_3(\text{aq}) \)).

Part (c)(ii)
- Soil impact: Acid deposition leaches essential plant nutrients (such as \( \text{Ca}^{2+} \) and \( \text{Mg}^{2+} \)) and releases toxic \( \text{Al}^{3+} \) ions from minerals, which damages plant roots.
- Aquatic impact: Increased acidity and dissolved aluminium concentrations are highly toxic to fish, clogging their gills and disrupting reproductive cycles.
- Reduction: Switch to renewable energy resources / implement catalytic converters in vehicles / install industrial flue-gas desulfurization (scrubbers).

Part (a)(i) [Total: 4 marks]
- [1] Correct expression for \( K_a \).
- [1] Calculates \( [\text{H}_3\text{O}^+] = 1.64 \times 10^{-3} \text{ mol dm}^{-3} \) and \( \text{pH} = 2.78 \).
- [1] States assumption that ethanoic acid dissociation is negligible (or \( [\text{CH}_3\text{COOH}]_{\text{equilibrium}} \approx [\text{CH}_3\text{COOH}]_{\text{initial}} \)).
- [1] States assumption that water autoionization contribution is negligible.

Part (a)(ii) [Total: 4 marks]
- [1] Calculates moles of ethanoic acid and sodium hydroxide correctly.
- [1] Identifies that exactly half of the acid is neutralized to form a buffer system.
- [1] Identifies that at the half-equivalence point, \( [\text{acid}] = [\text{conjugate base}] \).
- [1] Correctly calculates \( \text{pH} = \text{p}K_a = 4.74 \).

Part (b) [Total: 4 marks]
- [1] Correctly shaped weak-acid/strong-base titration curve (starts at \( \approx 2.8 \), flat buffer region, steep rise, ends high at \( \approx 12.5 \)).
- [1] Labels half-equivalence point at \( 12.5 \text{ cm}^3 \) and pH \( 4.74 \).
- [1] Labels equivalence point at \( 25.0 \text{ cm}^3 \) with pH \( > 7 \).
- [1] Explains basic pH at equivalence point using salt hydrolysis reaction: \( \text{CH}_3\text{COO}^- + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{COOH} + \text{OH}^- \).

Part (c)(i) [Total: 3 marks]
- [1] Correct balanced equation for sulfurous acid formation.
- [1] Correct balanced equation for nitric acid formation.
- [1] All formulas and balancing coefficients are correct.

Part (c)(ii) [Total: 3 marks]
- [1] Soil impact: nutrient leaching or release of toxic aluminum ions.
- [1] Aquatic impact: damage to fish gills or inhibition of aquatic reproduction.
- [1] Reduction: scrubbers in power stations, catalytic converters, or alternate energy.

Part (a)(i)
- Primary alcohol: butan-1-ol (a straight chain of four carbons with the -OH group on the first carbon). Structural formula: \( \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH} \).
- Secondary alcohol: butan-2-ol (a straight chain of four carbons with the -OH group on the second carbon). Structural formula: \( \text{CH}_3\text{CH}_2\text{CH(OH)CH}_3 \).
- Tertiary alcohol: 2-methylpropan-2-ol (three carbons in a row, with a methyl group and an -OH group on the second carbon). Structural formula: \( (\text{CH}_3)_3\text{COH} \).
- Ether: ethoxyethane (or diethyl ether) (an oxygen atom connected to two ethyl groups). Structural formula: \( \text{CH}_3\text{CH}_2\text{OCH}_2\text{CH}_3 \).

Part (a)(ii)
- Primary alcohol name: butan-1-ol
- Secondary alcohol name: butan-2-ol
- Tertiary alcohol name: 2-methylpropan-2-ol
- Ether name: ethoxyethane (accept diethyl ether)

Part (b)(i)
- Butan-1-ol has a much higher boiling point than ethoxyethane.
- This is because butan-1-ol contains a polar hydroxyl (\( -\text{OH} \)) group, which allows it to form strong intermolecular hydrogen bonds between its molecules.
- Ethoxyethane is a polar molecule, but it lacks an oxygen-hydrogen (O-H) bond, so it cannot form hydrogen bonds with other ethoxyethane molecules.
- Instead, ethoxyethane is held together by weaker dipole-dipole attractions and London dispersion forces, which require less thermal energy to overcome.

Part (b)(ii)
- Both molecules contain the polar, hydrophilic hydroxyl (\( -\text{OH} \)) group, which can form hydrogen bonds with water molecules.
- However, butan-1-ol has a relatively small non-polar, hydrophobic alkyl tail (\( -\text{C}_4\text{H}_9 \)), which does not disrupt the hydrogen-bonding network of water significantly.
- Hexan-1-ol has a much larger hydrophobic alkyl tail (\( -\text{C}_6\text{H}_{13} \)). The non-polar chain disruptions outweigh the positive interactions of the single polar \( -\text{OH} \) group, reducing its solubility in water significantly.

Part (c)(i)
- Reagent: Acidified potassium dichromate(VI) solution (\( \text{H}^+/\text{Cr}_2\text{O}_7^{2-} \)) (acidified potassium manganate(VII) is also acceptable).
- Conditions: Reflux (this ensures the intermediate aldehyde is returned to the flask and fully oxidized to the carboxylic acid).
- Color change: The solution changes from orange (due to \( \text{Cr}_2\text{O}_7^{2-} \)) to green (due to \( \text{Cr}^{3+} \)).

Part (a)(i) [Total: 4 marks]
- [1] Correct structural formula for butan-1-ol.
- [1] Correct structural formula for butan-2-ol.
- [1] Correct structural formula for 2-methylpropan-2-ol.
- [1] Correct structural formula for ethoxyethane.
(Accept full structural formulas displaying all bonds, or condensed structural formulas if unambiguous).

Part (a)(ii) [Total: 4 marks]
- [1] IUPAC Name: butan-1-ol.
- [1] IUPAC Name: butan-2-ol.
- [1] IUPAC Name: 2-methylpropan-2-ol.
- [1] IUPAC Name: ethoxyethane (accept diethyl ether).

Part (b)(i) [Total: 4 marks]
- [1] Identifies that butan-1-ol has a higher boiling point than ethoxyethane.
- [1] Explains that butan-1-ol forms hydrogen bonds.
- [1] Explains that ethoxyethane forms dipole-dipole/London dispersion forces but cannot hydrogen-bond with itself.
- [1] Links intermolecular force strength to the amount of energy required to separate molecules.

Part (b)(ii) [Total: 3 marks]
- [1] Identifies the hydrophilic polar \( -\text{OH} \) group forming hydrogen bonds with water.
- [1] Identifies the non-polar, hydrophobic nature of the alkyl chains.
- [1] Explains that the longer alkyl chain in hexan-1-ol has a dominant hydrophobic effect over the polar group, reducing solubility.

Part (c)(i) [Total: 3 marks]
- [1] Reagent: acidified potassium dichromate(VI) / \( \text{H}^+ \) and \( \text{Cr}_2\text{O}_7^{2-} \).
- [1] Condition: reflux.
- [1] Color change: orange to green.