2025 IB DP Chemistry 模擬試題連答案詳解

Carbon monoxide, \(\text{CO}\), contains a triple carbon-oxygen bond (bond order = 3), which is the shortest. Carbon dioxide, \(\text{CO}_2\), has double bonds (bond order = 2). The carbonate ion, \(\text{CO}_3^{2-}\), has resonance structures yielding three equivalent carbon-oxygen bonds with a bond order of approximately 1.33, making them the longest of the three. Therefore, the order of increasing length is \(\text{CO} < \text{CO}_2 < \text{CO}_3^{2-}\).

Award [1] for the correct choice: A.

A conjugate acid-base pair consists of two species that differ by only a single hydrogen ion, \(\text{H}^+\). Here, \(\text{H}_2\text{PO}_4^{-}\) acts as an acid (donating a proton) to form its conjugate base, \(\text{HPO}_4^{2-}\).

Award [1] for the correct choice: A.

The largest jump in ionization energy occurs between the 3rd and 4th ionization energies (from 2745 to 11577 \(\text{kJ}\,\text{mol}^{-1}\)). This indicates that the first three electrons are relatively easy to remove because they are in the outer shell, whereas the fourth electron is removed from a complete inner shell. Therefore, the element has 3 valence electrons and belongs to Group 13.

Award [1] for the correct choice: C.

In \(\text{SO}_4^{2-}\), the oxidation state of sulfur is +6. In \(\text{S}_2\text{O}_3^{2-}\), the average oxidation state of sulfur is +2. In \(\text{SO}_3^{2-}\), the oxidation state of sulfur is +4. In \(\text{H}_2\text{S}\), the oxidation state of sulfur is -2. Thus, sulfur has the highest oxidation state in \(\text{SO}_4^{2-}\).

Award [1] for the correct choice: A.

Increasing the temperature increases the average kinetic energy of the particles, which causes the peak of the Maxwell-Boltzmann distribution curve to shift to the right (higher energy). Because the total area under the curve represents the total number of particles (which remains constant), the curve must flatten, meaning the peak height decreases (becomes lower).

Award [1] for the correct choice: C.

First, calculate the heat energy absorbed by the water: \(q = m c \Delta T = 100.0\text{ g} \times 4.18\text{ J}\,\text{g}^{-1}\,\text{K}^{-1} \times 24.0\text{ K} = 10032\text{ J} = 10.032\text{ kJ}\). Since the reaction is exothermic (combustion), \(\Delta H = -\frac{q}{n} = -\frac{10.032\text{ kJ}}{0.0200\text{ mol}} = -501.6\text{ kJ}\,\text{mol}^{-1} \approx -502\text{ kJ}\,\text{mol}^{-1}\).

Award [1] for the correct choice: A.

Only a change in temperature can change the value of the equilibrium constant \(K_c\). For an exothermic reaction (\(\Delta H < 0\)), lowering the temperature shifts the equilibrium in the forward direction, which increases the concentration of products and decreases the concentration of reactants, thereby increasing the value of \(K_c\).

Award [1] for the correct choice: B.

The molecular formula \(\text{C}_5\text{H}_{12}\) represents pentane. Its three structural isomers are pentane (straight chain), 2-methylbutane (branched chain), and 2,2-dimethylpropane (highly branched chain).

Award [1] for the correct choice: B.

1. Find the electron configuration of \(\text{Cu}^{2+}\):
- Ground state \(\text{Cu}\) is \([\text{Ar}] 3d^{10} 4s^1\).
- When forming \(\text{Cu}^{2+}\), electrons are removed first from the \(4s\) orbital and then the \(3d\) orbital, giving \([\text{Ar}] 3d^9\).
- According to Hund's rule, a \(3d^9\) configuration has 4 filled \(3d\) orbitals and 1 half-filled orbital, meaning it has exactly 1 unpaired electron.

2. Determine the number of unpaired electrons in each option:
- \(\text{Ni}^{2+}\): \([\text{Ar}] 3d^8\) has 2 unpaired electrons.
- \(\text{Fe}^{3+}\): \([\text{Ar}] 3d^5\) has 5 unpaired electrons.
- \(\text{V}^{4+}\): \([\text{Ar}] 3d^1\) has 1 unpaired electron.
- \(\text{Co}^{2+}\): \([\text{Ar}] 3d^7\) has 3 unpaired electrons.

Therefore, \(\text{V}^{4+}\) also has exactly 1 unpaired electron, matching \(\text{Cu}^{2+}\).

[1 mark] awarded for identifying that both \(\text{Cu}^{2+}\) and \(\text{V}^{4+}\) have exactly one unpaired electron in their ground state (Option C).

1. Determine the number of electron domains and lone pairs on the central atom of each molecule:
- \(\text{SF}_4\): The central sulfur atom has 6 valence electrons, and each of the 4 fluorine atoms shares 1 electron, making a total of 10 valence shell electrons (5 electron pairs). This corresponds to 4 bonding pairs and 1 lone pair. The molecular geometry is seesaw.
- \(\text{XeF}_4\): The central xenon atom has 8 valence electrons + 4 from fluorine = 12 electrons (6 electron pairs). This corresponds to 4 bonding pairs and 2 lone pairs. The molecular geometry is square planar.
- \(\text{SiF}_4\): The central silicon atom has 4 valence electrons + 4 from fluorine = 8 electrons (4 electron pairs). This corresponds to 4 bonding pairs and 0 lone pairs. The molecular geometry is tetrahedral.
- \(\text{ClF}_4^-\): The central chlorine atom has 7 valence electrons + 1 (charge) + 4 from fluorine = 12 electrons (6 electron pairs). This corresponds to 4 bonding pairs and 2 lone pairs. The molecular geometry is square planar.

[1 mark] awarded for selecting \(\text{SF}_4\) as having seesaw molecular geometry based on VSEPR theory (Option A).

According to the Brønsted–Lowry theory, a conjugate acid-base pair consists of two species that differ by exactly one proton (\(\text{H}^+\)).
- \(\text{H}_2\text{PO}_4^-\rightleftharpoons \text{HPO}_4^{2-} + \text{H}^+\), so \(\text{H}_2\text{PO}_4^-\)/\(\text{HPO}_4^{2-}\) is a conjugate acid-base pair.
- \(\text{H}_2\text{CO}_3 \rightleftharpoons \text{HCO}_3^- + \text{H}^+\), so \(\text{H}_2\text{CO}_3\)/\(\text{HCO}_3^-\) is a conjugate acid-base pair.

Comparing with the options, \(\text{H}_2\text{PO}_4^-\)/\(\text{HPO}_4^{2-}\) is the correct pair.

[1 mark] awarded for selecting the pair that differs by exactly one proton (Option C).

1. For a spontaneous electrochemical cell, the overall standard cell potential \(E^\theta_{\text{cell}}\) must be positive.
2. The half-reaction with the more positive reduction potential will undergo reduction (cathode):
\(\text{Ce}^{4+}(aq) + \text{e}^- \rightarrow \text{Ce}^{3+}(aq) \quad E^\theta_{\text{reduction}} = +1.61\text{ V}\)
3. The half-reaction with the less positive reduction potential will be reversed and undergo oxidation (anode):
\(\text{Fe}^{2+}(aq) \rightarrow \text{Fe}^{3+}(aq) + \text{e}^- \quad E^\theta_{\text{oxidation}} = +0.77\text{ V}\)
4. Calculate the standard cell potential:
\(E^\theta_{\text{cell}} = E^\theta_{\text{reduction}} - E^\theta_{\text{oxidation}} = +1.61\text{ V} - 0.77\text{ V} = +0.84\text{ V}\).

[1 mark] awarded for correctly identifying cathode and anode and calculating standard cell potential to be +0.84 V (Option B).

1. Determine the order with respect to \(\text{A}\):
Compare Experiments 1 and 2 where \([\text{B}]\) is kept constant at \(0.10\text{ mol dm}^{-3}\). As \([\text{A}]\) doubles from \(0.10\) to \(0.20\text{ mol dm}^{-3}\), the initial rate doubles from \(2.0 \times 10^{-4}\) to \(4.0 \times 10^{-4}\text{ mol dm}^{-3}\text{ s}^{-1}\). Since \(2^x = 2\), the order with respect to \(\text{A}\) is 1 (first-order).

2. Determine the order with respect to \(\text{B}\):
Compare Experiments 2 and 3 where \([\text{A}]\) is kept constant at \(0.20\text{ mol dm}^{-3}\). As \([\text{B}]\) doubles from \(0.10\) to \(0.20\text{ mol dm}^{-3}\), the initial rate quadruples from \(4.0 \times 10^{-4}\) to \(1.6 \times 10^{-3}\text{ mol dm}^{-3}\text{ s}^{-1}\). Since \(2^y = 4\), the order with respect to \(\text{B}\) is 2 (second-order).

3. Write the overall rate expression:
\(\text{Rate} = k[\text{A}][\text{B}]^2\).

[1 mark] awarded for determining the orders of reaction for each reactant correctly to yield the rate law (Option C).

1. Construct an ICE table with initial concentration of \([\text{HF}] = 1.0\text{ mol dm}^{-3}\):
- Initial: \([\text{HF}] = 1.0\), \([\text{H}_2] = 0\), \([\text{F}_2] = 0\)
- Change: \([\text{HF}] = -2x\), \([\text{H}_2] = +x\), \([\text{F}_2] = +x\)
- Equilibrium: \([\text{HF}] = 1.0 - 2x\), \([\text{H}_2] = x\), \([\text{F}_2] = x\)

2. Set up the equilibrium expression:
\(K_c = \frac{[\text{H}_2][\text{F}_2]}{[\text{HF}]^2} = \frac{x^2}{(1.0 - 2x)^2} = 4.0\)

3. Take the square root of both sides:
\(\frac{x}{1.0 - 2x} = 2.0\)

4. Solve for \(x\):
\(x = 2.0(1.0 - 2x) = 2.0 - 4.0x\)
\(5.0x = 2.0\)
\(x = 0.40\text{ mol dm}^{-3}\).

Since \([\text{H}_2]_{\text{eq}} = x\), the equilibrium concentration is \(0.40\text{ mol dm}^{-3}\).

[1 mark] awarded for the correct set up and calculation of the equilibrium concentration of \(\text{H}_2\) (Option B).

1. Calculate the heat gained by the water:
\(q_{\text{water}} = m_{\text{water}} \cdot c_{\text{water}} \cdot \Delta T_{\text{water}}\}
\)q_{\text{water}} = 100.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times (21.0^\circ\text{C} - 20.0^\circ\text{C}) = 418\text{ J}\)

2. Assuming no heat loss to the surroundings, heat lost by the metal equals heat gained by the water:
\(q_{\text{metal}} = q_{\text{water}} = 418\text{ J}\)

3. Calculate the specific heat capacity of the metal (\(c_{\text{metal}}\)):
\(q_{\text{metal}} = m_{\text{metal}} \cdot c_{\text{metal}} \cdot (T_{\text{initial}} - T_{\text{final}})\)
\(418\text{ J} = 5.00\text{ g} \times c_{\text{metal}} \times (100.0^\circ\text{C} - 21.0^\circ\text{C})\)
\(418 = 5.00 \times c_{\text{metal}} \times 79.0\)
\(418 = 395.0 \times c_{\text{metal}}\)
\(c_{\text{metal}} = \frac{418}{395.0} \approx 1.06\text{ J g}^{-1}\text{ K}^{-1}\).

[1 mark] awarded for the correct thermodynamic calculation using heat exchange relations (Option B).

1. Analyze the molecular structure:
- The benzene ring \(\text{C}_6\text{H}_4\) has a hydroxyl group (\(-\text{OH}\)) attached directly to it. A hydroxyl group directly attached to an aromatic ring forms a **phenol** functional group.
- The ester group (\(-\text{COOCH}_3\)) attached to the benzene ring is an **ester** functional group.

2. Comparing to the choices, Phenol and Ester correctly describe these functional groups.

[1 mark] awarded for correctly classifying the aromatic hydroxyl group as a phenol and the carboxyl derivative as an ester (Option B).

Acrylonitrile has the structural formula \(\text{H}_2\text{C}=\text{CH}-\text{C}\equiv\text{N}\). Let's count the bonds: 1. Carbon-hydrogen (\(\text{C}-\text{H}\)) bonds: There are three single covalent bonds, each representing 1 \(\sigma\) bond (3 \(\sigma\) bonds in total). 2. Carbon-carbon (\(\text{C}=\text{C}\)) double bond: Consists of 1 \(\sigma\) bond and 1 \(\pi\) bond. 3. Carbon-carbon (\(\text{C}-\text{C}\)) single bond: Consists of 1 \(\sigma\) bond. 4. Carbon-nitrogen (\(\text{C}\equiv\text{N}\)) triple bond: Consists of 1 \(\sigma\) bond and 2 \(\pi\) bonds. Summing these up: Total \(\sigma\) bonds = \(3 + 1 + 1 + 1 = 6\); Total \(\pi\) bonds = \(1 + 2 = 3\). Therefore, the molecule has 6 \(\sigma\) and 3 \(\pi\) bonds.

Award 1 mark for the correct choice (A). [1] Correctly identifies that every single bond is 1 \(\sigma\) bond, a double bond has 1 \(\sigma\) and 1 \(\pi\) bond, and a triple bond has 1 \(\sigma\) and 2 \(\pi\) bonds, leading to the total of 6 \(\sigma\) and 3 \(\pi\) bonds.

1. For a strong acid, dissociation is complete (100%). When it is diluted 10-fold, the concentration of hydrogen ions, \([\text{H}^+]\), decreases by exactly a factor of 10. Therefore, the new pH will be: \(\text{pH}_{\text{new}} = \text{pH}_{\text{old}} - \log_{10}(0.1) = 3.0 + 1.0 = 4.0\). 2. For a weak acid, it exists in equilibrium: \(\text{HA(aq)} \rightleftharpoons \text{H}^+(\text{aq}) + \text{A}^-(\text{aq})\). When diluted 10-fold, the equilibrium shifts to the right (according to Le Chatelier's principle) to oppose the decrease in concentration, resulting in increased dissociation of \(\text{HA}\). This extra dissociation produces more \(\text{H}^+\) ions than would be expected from simple dilution alone. Consequently, \([\text{H}^+]\) decreases by less than a factor of 10, meaning the pH increases by less than 1.0 unit. Thus, the pH of the weak acid will be between 3.0 and 4.0.

Award 1 mark for the correct choice (B). [1] Correctly identifies that dilution of a strong acid by a factor of 10 increases pH by 1, and dilution of a weak acid shifts the equilibrium to favor dissociation, meaning pH increases by less than 1 unit.

1. The atomic number of cobalt (\(\text{Co}\)) is 27. Its ground-state electron configuration is: \(\text{Co}: [\text{Ar}] 3\text{d}^7 4\text{s}^2\). 2. When forming a transition metal cation, electrons are lost from the outermost \(4\text{s}\) subshell first. Thus, for \(\text{Co}^{2+}\), two electrons are removed from the \(4\text{s}\) orbital: \(\text{Co}^{2+}: [\text{Ar}] 3\text{d}^7\). 3. According to Hund's rule, we distribute the 7 electrons in the five degenerate \(3\text{d}\) orbitals to maximize spin before pairing them: First 5 electrons go into separate orbitals with parallel spins: \(\uparrow\) \(\uparrow\) \(\uparrow\) \(\uparrow\) \(\uparrow\). The remaining 2 electrons pair up: \(\uparrow\downarrow\) \(\uparrow\downarrow\) \(\uparrow\) \(\uparrow\) \(\uparrow\). This leaves exactly 3 orbitals with single, unpaired electrons.

Award 1 mark for the correct choice (C). [1] Correctly writes the electron configuration of \(\text{Co}^{2+}\) as \([\text{Ar}] 3\text{d}^7\) (noting that \(4\text{s}\) electrons are lost first) and applies Hund's rule to find 3 unpaired electrons.

Let \(A_r(\text{M})\) be the relative atomic mass of the metal \(\text{M}\). The empirical formula of the oxide is \(\text{MO}_2\). The molar mass of \(\text{MO}_2\) can be represented as: \(M_r(\text{MO}_2) = A_r(\text{M}) + 2 \times A_r(\text{O}) = A_r(\text{M}) + 2 \times 16.0 = A_r(\text{M}) + 32.0\). We are given that the mass percentage of the metal \(\text{M}\) is 60.0%. Therefore: \(\frac{A_r(\text{M})}{A_r(\text{M}) + 32.0} \times 100 = 60.0\). Solving for \(A_r(\text{M})\): \(\frac{A_r(\text{M})}{A_r(\text{M}) + 32.0} = 0.60 \implies A_r(\text{M}) = 0.60 \times (A_r(\text{M}) + 32.0) \implies A_r(\text{M}) = 0.60 A_r(\text{M}) + 19.2 \implies 0.40 A_r(\text{M}) = 19.2 \implies A_r(\text{M}) = 48.0\).

Award 1 mark for the correct choice (B). [1] Correctly sets up the equation for the mass percentage of \(\text{M}\) using the empirical formula \(\text{MO}_2\) and solves for \(A_r(\text{M}) = 48.0\).

1. Enthalpy change of neutralization is an intensive property and is defined per mole of water formed. 2. In the first experiment: Volume of reaction mixture = \(100.0 \text{ cm}^3\), corresponding to a mass \(m_1 \approx 100.0 \text{ g}\). Amount of reactants reacted = \(0.0500 \text{ mol}\). Heat released is \(q_1 = m_1 \cdot c \cdot \Delta T_1 = 100.0 \times c \times 6.0\). 3. In the second experiment: Volume of reaction mixture = \(50.0 \text{ cm}^3\), corresponding to a mass \(m_2 \approx 50.0 \text{ g}\) (half the initial mass). Amount of reactants reacted = \(0.0250 \text{ mol}\) (half the initial amount). Since the amount of reactant is halved, the heat released (\(q_2\)) is also halved: \(q_2 = 0.5 \cdot q_1\). 4. Using the formula \(q = m \cdot c \cdot \Delta T\): \(\Delta T_2 = \frac{q_2}{m_2 \cdot c} = \frac{0.5 \cdot q_1}{0.5 \cdot m_1 \cdot c} = \Delta T_1 = 6.0 \text{ }^\circ\text{C}\). The temperature change remains the same at \(6.0 \text{ }^\circ\text{C}\).

Award 1 mark for the correct choice (B). [1] Deduces that both the heat energy released and the total volume/mass of the mixture are halved, so the temperature change remains identical.

1. Identify the species being oxidized: The oxidation state of iron increases from +2 in \(\text{Fe}^{2+}\) to +3 in \(\text{Fe}^{3+}\). Since oxidation is an increase in oxidation state, \(\text{Fe}^{2+}\) is oxidized, making it the reducing agent. 2. Determine the change in oxidation state of manganese: In the permanganate ion (\(\text{MnO}_4^-\)), let \(x\) be the oxidation state of Mn: \(x + 4(-2) = -1 \implies x = +7\). In the manganese(II) ion (\(\text{Mn}^{2+}\)), the oxidation state of Mn is +2. Therefore, the oxidation state of Mn changes from +7 to +2. This matches option A.

Award 1 mark for the correct choice (A). [1] Correctly identifies \(\text{Fe}^{2+}\) as the reducing agent (since it undergoes oxidation) and calculates the oxidation state of Mn in \(\text{MnO}_4^-\) as +7 and in \(\text{Mn}^{2+}\) as +2.

Let the initial concentrations be \([\text{A}]_0\) and \([\text{B}]_0\). The initial rate is: \(\text{Rate}_1 = k[\text{A}]_0^2[\text{B}]_0\). When \([\text{A}]\) is doubled and \([\text{B}]\) is halved, the new concentrations are \([\text{A}]_{\text{new}} = 2[\text{A}]_0\) and \([\text{B}]_{\text{new}} = 0.5[\text{B}]_0\). Substituting these into the rate expression gives: \(\text{Rate}_2 = k(2[\text{A}]_0)^2(0.5[\text{B}]_0) = k(4[\text{A}]_0^2)(0.5[\text{B}]_0) = 2 \cdot \left(k[\text{A}]_0^2[\text{B}]_0\right) = 2 \cdot \text{Rate}_1\). Therefore, the rate of reaction increases by a factor of 2.

Award 1 mark for the correct choice (B). [1] Correctly evaluates the mathematical effect on the rate when \([\text{A}]\) is squared and \([\text{B}]\) is raised to the power of 1, yielding a net factor of \(2^2 \times 0.5 = 2\).

Let's analyze the structure of ethyl 2-hydroxypropanoate: \(\text{CH}_3-\text{CH}(\text{OH})-\text{C}(=\text{O})-\text{O}-\text{CH}_2-\text{CH}_3\). 1. The molecule contains an \(-\text{OH}\) group bonded to a saturated carbon atom. This represents a hydroxyl group (characteristic of an alcohol). 2. The molecule also contains a \(-\text{C}(=\text{O})-\text{O}-\text{}\) group linking carbon chains. This represents an ester group. Therefore, the two functional groups present in this molecule are a hydroxyl group and an ester group.

Award 1 mark for the correct choice (B). [1] Correctly identifies the hydroxyl group (\(-\text{OH}\)) and the ester group (\(-\text{COO}-\)) within the structural formula of ethyl 2-hydroxypropanoate.

Cobalt has an atomic number of 27, so a neutral cobalt atom has the ground-state electron configuration of \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^7\) or \([\text{Ar}] 4s^2 3d^7\). When transition metals form cations, they lose electrons from the outermost \(4s\) subshell first before losing any from the \(3d\) subshell. Therefore, removing two electrons to form \(\text{Co}^{2+}\) results in the configuration \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^7\).

Award [1] for the correct choice A. Award [0] for any other choice.

First, calculate the heat energy change of the water: \(q = m c \Delta T\), where \(m = 100.0\text{ g}\), \(c = 4.2\text{ J g}^{-1}\text{ K}^{-1}\), and \(\Delta T = 16.0 - 20.0 = -4.0\text{ K}\). This gives \(q = 100.0 \times 4.2 \times (-4.0) = -1680\text{ J}\) or \(-1.68\text{ kJ}\). Since the temperature of the surroundings decreased, the reaction is endothermic, so \(\Delta H > 0\) and \(q_{\text{reaction}} = +1.68\text{ kJ}\). Next, calculate the moles of ammonium nitrate: \(n = \frac{5.0\text{ g}}{80.0\text{ g mol}^{-1}} = 0.0625\text{ mol}\). Finally, calculate the molar enthalpy change of solution: \(\Delta H_{\text{sol}} = \frac{+1.68\text{ kJ}}{0.0625\text{ mol}} = +26.88\text{ kJ mol}^{-1}\), which rounds to \(+27\text{ kJ mol}^{-1}\).

Award [1] for the correct choice A. Award [0] for any other choice.

The central sulfur atom (group 16) has 6 valence electrons. In \(\text{SF}_4\), it forms 4 single covalent bonds with 4 fluorine atoms, using 4 of its valence electrons. This leaves \(6 - 4 = 2\) non-bonding valence electrons, which constitute 1 non-bonding (lone) pair. Therefore, there are 4 bonding pairs and 1 lone pair around the central sulfur atom.

Award [1] for the correct choice B. Award [0] for any other choice.

Using the ICE (Initial, Change, Equilibrium) method: For \(\text{PCl}_5\), initial moles = \(0.20\), change = \(-0.10\), and equilibrium moles = \(0.10\). For both \(\text{PCl}_3\) and \(\text{Cl}_2\), initial moles = \(0\), change = \(+0.10\), and equilibrium moles = \(0.10\). The volume is \(2.0\text{ dm}^3\). The equilibrium concentrations are: \([\text{PCl}_5] = \frac{0.10}{2.0} = 0.050\text{ mol dm}^{-3}\), \([\text{PCl}_3] = \frac{0.10}{2.0} = 0.050\text{ mol dm}^{-3}\), and \([\text{Cl}_2] = \frac{0.10}{2.0} = 0.050\text{ mol dm}^{-3}\). The equilibrium constant expression is \(K_c = \frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]} = \frac{(0.050)(0.050)}{0.050} = 0.050\text{ mol dm}^{-3}\).

Award [1] for the correct choice B. Award [0] for any other choice.

A conjugate acid-base pair consists of two species that differ by exactly one proton (\(\text{H}^+\)). \(\text{H}_2\text{PO}_4^-\text{(aq)}\) is the acid and \(\text{HPO}_4^{2-}\text{(aq)}\) is its conjugate base, because removing one \(\text{H}^+\) from \(\text{H}_2\text{PO}_4^-\implies \text{HPO}_4^{2-}\). Thus, they form a conjugate acid-base pair.

Award [1] for the correct choice B. Award [0] for any other choice.

The zinc half-cell has a more negative standard reduction potential, indicating that zinc is a stronger reducing agent and will undergo oxidation at the anode (\(\text{Zn(s)} \to \text{Zn}^{2+}\text{(aq)} + 2e^-\)). Silver ions have a more positive standard reduction potential and will undergo reduction at the cathode (\(\text{Ag}^+\text{(aq)} + e^- \to \text{Ag(s)}\)). The standard cell potential is calculated as \(E^\theta_{\text{cell}} = E^\theta_{\text{cathode}} - E^\theta_{\text{anode}} = +0.80\text{ V} - (-0.76\text{ V}) = +1.56\text{ V}\). Thus, the zinc electrode is the anode and \(E^\theta_{\text{cell}} = +1.56\text{ V}\).

Award [1] for the correct choice B. Award [0] for any other choice.

Compare Experiments 1 and 2: \([\text{B}]\) is kept constant while \([\text{A}]\) doubles. The rate doubles from \(2.0 \times 10^{-4}\) to \(4.0 \times 10^{-4}\text{ mol dm}^{-3}\text{ s}^{-1}\). This means the reaction is first-order with respect to \(\text{A}\). Compare Experiments 2 and 3: \([\text{A}]\) is kept constant while \([\text{B}]\) doubles. The rate increases by a factor of 4 from \(4.0 \times 10^{-4}\) to \(1.6 \times 10^{-3}\text{ mol dm}^{-3}\text{ s}^{-1}\). This means the reaction is second-order with respect to \(\text{B}\). Therefore, the overall rate expression is \(\text{Rate} = k[\text{A}][\text{B}]^2\).

Award [1] for the correct choice C. Award [0] for any other choice.

A reaction is spontaneous when \(\Delta G^\theta < 0\). Using the relation \(\Delta G^\theta = \Delta H^\theta - T\Delta S^\theta\), we can find the transition temperature where \(\Delta G^\theta = 0\): \(-80.0 \times 10^3\text{ J mol}^{-1} - T(-200\text{ J K}^{-1}\text{ mol}^{-1}) = 0\), which simplifies to \(-80000 + 200T = 0\), so \(T = \frac{80000}{200} = 400\text{ K}\). Since \(\Delta H^\theta\) is negative (favourable) and \(\Delta S^\theta\) is negative (unfavourable), the reaction is spontaneous only at lower temperatures where the \(T\Delta S^\theta\) term does not outweigh the \(\Delta H^\theta\) term. Therefore, the reaction is spontaneous only at temperatures below \(400\text{ K}\).

Award [1] for the correct choice C. Award [0] for any other choice.

To determine the hybridization of the central atom, we count the number of electron domains (bonding domains + lone pairs) around it:
- In the carbonate ion, \( \text{CO}_3^{2-} \), the central carbon atom has 3 bonding domains (resonance structures distribute the bonds, giving three equivalent domains) and 0 lone pairs, giving 3 electron domains. This corresponds to \( \text{sp}^2 \) hybridization.
- In sulfur dioxide, \( \text{SO}_2 \), the central sulfur atom has 2 bonding domains and 1 lone pair, giving 3 electron domains. This corresponds to \( \text{sp}^2 \) hybridization.
- In ozone, \( \text{O}_3 \), the central oxygen atom has 2 bonding domains and 1 lone pair, giving 3 electron domains. This corresponds to \( \text{sp}^2 \) hybridization.

Since all three species have a central atom with 3 electron domains, they are all \( \text{sp}^2 \) hybridized.

Award [1] for the correct answer (D).
- Award [0] for any other response.

For a weak monoprotic acid, we use the approximation:
\( K_a \approx \frac{[\text{H}^+]^2}{[\text{HA}]_i} \)

Substituting the given values:
\( 1.0 \times 10^{-6} = \frac{[\text{H}^+]^2}{0.010} \)
\( [\text{H}^+]^2 = 1.0 \times 10^{-6} \times 1.0 \times 10^{-2} = 1.0 \times 10^{-8} \text{ mol}^2 \text{ dm}^{-6} \)
\( [\text{H}^+] = \sqrt{1.0 \times 10^{-8}} = 1.0 \times 10^{-4} \text{ mol dm}^{-3} \)

Now, calculate pH:
\( \text{pH} = -\log_{10}[\text{H}^+] = -\log_{10}(1.0 \times 10^{-4}) = 4.0 \)

Award [1] for the correct answer (B).
- Award [0] for any other response.

To find the empirical formula, determine the molar ratio of the elements:
- Moles of C: \( \frac{40.0 \text{ g}}{12.0 \text{ g mol}^{-1}} = 3.33 \text{ mol} \)
- Moles of H: \( \frac{6.7 \text{ g}}{1.0 \text{ g mol}^{-1}} = 6.70 \text{ mol} \)
- Moles of O: \( \frac{53.3 \text{ g}}{16.0 \text{ g mol}^{-1}} = 3.33 \text{ mol} \)

Divide each value by the smallest number of moles (3.33):
- C: \( \frac{3.33}{3.33} = 1 \)
- H: \( \frac{6.70}{3.33} \approx 2 \)
- O: \( \frac{3.33}{3.33} = 1 \)

The empirical formula is \( \text{CH}_2\text{O} \).

Award [1] for the correct answer (B).
- Award [0] for any other response.

1. Find the heat energy released (\( q \)):
Total volume of reaction mixture = \( 50.0 + 50.0 = 100.0 \text{ cm}^3 \)
Mass (\( m \)) = \( 100.0 \text{ g} \) (since density is \( 1.00 \text{ g cm}^{-3} \))
Temperature change (\( \Delta T \)) = \( 26.8 - 20.0 = 6.8 \text{ K} \)
\( q = m c \Delta T = 100.0 \text{ g} \times 4.18 \text{ J g}^{-1} \text{ K}^{-1} \times 6.8 \text{ K} = 2842.4 \text{ J} = 2.84 \text{ kJ} \)

2. Determine the moles of water formed:
\( n(\text{HCl}) = 1.0 \text{ mol dm}^{-3} \times 0.0500 \text{ dm}^3 = 0.050 \text{ mol} \)
\( n(\text{NaOH}) = 1.0 \text{ mol dm}^{-3} \times 0.0500 \text{ dm}^3 = 0.050 \text{ mol} \)
Since they react in a 1:1 ratio, \( 0.050 \text{ mol} \) of \( \text{H}_2\text{O} \) is produced.

3. Calculate the molar enthalpy change:
\( \Delta H_n = -\frac{q}{n} = -\frac{2.8424 \text{ kJ}}{0.050 \text{ mol}} = -56.8 \text{ kJ mol}^{-1} \)

Award [1] for the correct answer (C).
- Award [0] for any other response.

The neutral iron atom (Fe, atomic number 26) has the ground-state electron configuration:
\( 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^6 \)

When transition metals form cations, they lose electrons from the outermost shell (the \( 4s \) subshell) first before losing electrons from the \( 3d \) subshell.
Therefore, to form the \( \text{Fe}^{2+} \) ion, two electrons are removed from the \( 4s \) subshell, resulting in the configuration:
\( 1s^2 2s^2 2p^6 3s^2 3p^6 3d^6 \)

Award [1] for the correct answer (B).
- Award [0] for any other response.

Let the original rate be \( R_1 = k[\text{NO}]^2[\text{O}_2] \).

If the concentration of \( \text{NO} \) is doubled, it becomes \( 2[\text{NO}] \).
If the concentration of \( \text{O}_2 \) is halved, it becomes \( 0.5[\text{O}_2] \).

The new rate, \( R_2 \), is given by:
\( R_2 = k(2[\text{NO}])^2(0.5[\text{O}_2]) \)
\( R_2 = k \times 4[\text{NO}]^2 \times 0.5[\text{O}_2] \)
\( R_2 = 2 \left( k[\text{NO}]^2[\text{O}_2] \right) = 2R_1 \)

Thus, the rate increases by a factor of 2.

Award [1] for the correct answer (B).
- Award [0] for any other response.

1. Analyze half-cells: The standard reduction potential of the silver half-cell is more positive (\( +0.80 \text{ V} \)) than that of the iron half-cell (\( -0.44 \text{ V} \)). Thus, \( \text{Ag}^+ \) ions undergo reduction at the cathode, and solid \( \text{Fe} \) undergoes oxidation at the anode.
2. Cell Potential: \( E^\theta_{\text{cell}} = E^\theta_{\text{reduction}} - E^\theta_{\text{oxidation}} = +0.80 - (-0.44) = +1.24 \text{ V} \).
3. Electron flow: Electrons flow from the anode (iron electrode, site of oxidation) to the cathode (silver electrode, site of reduction).
4. Mass change: At the silver cathode, \( \text{Ag}^+\text{(aq)} + \text{e}^- \rightarrow \text{Ag(s)} \) occurs, depositing solid silver onto the electrode and increasing its mass.

Award [1] for the correct answer (D).
- Award [0] for any other response.

1. Identify the principal functional group: The compound contains an aldehyde group (\( -\text{CHO} \)) and a hydroxyl group (\( -\text{OH} \)). The aldehyde has higher priority, so the compound is classified as an aldehyde, and the suffix is "-anal".
2. Determine the longest carbon chain containing the principal group: There are 5 carbons, so the parent name is pentanal.
3. Number the chain starting from the aldehyde carbon as C1:
- C1: \( \text{-CHO} \)
- C2: \( \text{-CH}_2- \)
- C3: \( \text{-CH}_2- \)
- C4: \( \text{-CH(OH)}- \)
- C5: \( \text{-CH}_3 \)
4. Identify and locate substituents: There is a hydroxyl group on C4. Under IUPAC rules, when \( -\text{OH} \) is a substituent, it is named "hydroxy".
Therefore, the systematic name is 4-hydroxypentanal.

Award [1] for the correct answer (B).
- Award [0] for any other response.

(i) The rate of reaction decreases over time as the reactants are consumed. This is represented on a volume-time graph by a curve whose gradient (slope) decreases over time, eventually becoming horizontal (zero gradient) when the reaction is complete.
(ii) Moles of \(\text{HCl} = 0.0500\text{ dm}^3 \times 1.00\text{ mol dm}^{-3} = 0.0500\text{ mol}\). From the balanced equation, \(2\text{ mol}\) of \(\text{HCl}\) produces \(1\text{ mol}\) of \(\text{CO}_2\). Therefore, moles of \(\text{CO}_2 = 0.0500 / 2 = 0.0250\text{ mol}\). Volume of \(\text{CO}_2\) at STP \(= 0.0250\text{ mol} \times 22.7\text{ dm}^3\text{ mol}^{-1} = 0.5675\text{ dm}^3 = 568\text{ cm}^3\).
(iii) Pipette percentage uncertainty: \((0.06 / 50.0) \times 100 = 0.12\%\). Syringe percentage uncertainty: \((0.5 / 60.0) \times 100 = 0.83\%\).
(iv) Improvements include: 1. Use a water bath to maintain a constant temperature throughout the experiment. 2. Ensure all ground-glass joints are lubricated with vaseline to prevent any gas leakage. 3. Saturate the solution/water (if collecting over water) with carbon dioxide beforehand so no gas dissolves during the experiment (or use a gas syringe directly, ensuring the plunger moves smoothly with zero friction).
(v) If the reaction is first-order with respect to \(\text{HCl}\), doubling the concentration of \(\text{HCl}\) will double the initial rate of reaction.

(i) Rate decreases as concentration of reactants decreases [1 mark]. Curve gradient decreases / slope flattens [0.66 marks].
(ii) Moles of HCl calculated correctly: \(0.0500\text{ mol}\) [1 mark]. Moles of CO2 calculated correctly: \(0.0250\text{ mol}\) [1 mark]. Volume of CO2 calculated as \(568\text{ cm}^3\) (or \(0.568\text{ dm}^3\)) [1 mark].
(iii) Pipette percentage uncertainty = \(0.12\%\) [1 mark]. Syringe percentage uncertainty = \(0.83\%\) [1 mark].
(iv) Any three valid improvements, e.g.: thermostat bath [1 mark]; check/seal leaks [1 mark]; smooth plunger movement / avoid friction [1 mark].
(v) State that the initial rate of reaction doubles [1 mark]. Explain that rate is directly proportional to concentration for a first-order reaction [1 mark].

(i) \(\Delta T = 27.8^\circ\text{C} - 21.4^\circ\text{C} = 6.4^\circ\text{C}\) (or \(6.4\text{ K}\)). Mass of solution \(m = 50.0\text{ g} + 50.0\text{ g} = 100.0\text{ g}\). \(q = m c \Delta T = 100.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 6.4\text{ K} = 2675.2\text{ J} = 2.68\text{ kJ}\). The thermometer uncertainty is \(\pm 0.1^\circ\text{C}\) per reading, so for a temperature difference, absolute uncertainty \(= 0.1 + 0.1 = \pm 0.2^\circ\text{C}\). Relative uncertainty in \(\Delta T = 0.2 / 6.4 = 0.03125\) (or \(3.13\%\)). Absolute uncertainty in \(q = 2.6752\text{ kJ} \times 0.03125 = \pm 0.08\text{ kJ}\). Thus, \(q = 2.68 \pm 0.08\text{ kJ}\).
(ii) Moles of \(\text{H}_2\text{O}\) formed \(= n(\text{HCl}) = 0.0500\text{ dm}^3 \times 1.00\text{ mol dm}^{-3} = 0.0500\text{ mol}\). \(\Delta H_{\text{n}} = -q / n = -2.6752\text{ kJ} / 0.0500\text{ mol} = -53.5\text{ kJ mol}^{-1}\).
(iii) Percentage error \(= (| -53.5 - (-57.1) | / 57.1) \times 100 = (3.6 / 57.1) \times 100 = 6.30\%\). Physical reasons: 1. Heat loss to the surroundings. 2. Heat absorbed by the polystyrene cup and thermometer themselves.
(iv) Ethanoic acid is a weak acid that is only partially dissociated in solution. Some heat energy released during neutralization is absorbed to complete the dissociation of ethanoic acid. Therefore, the temperature rise will be smaller, and the calculated \(\Delta H_{\text{n}}\) will be less exothermic (less negative) than for \(\text{HCl}\).

(i) Calculation of \(q = 2.68\text{ kJ}\) [1 mark]. Absolute uncertainty of temperature change identified as \(\pm 0.2^\circ\text{C}\) [0.66 marks]. Calculation of absolute uncertainty in \(q\) as \(\pm 0.08\text{ kJ}\) [1 mark].
(ii) Calculation of moles of acid/base/water as \(0.0500\text{ mol}\) [1 mark]. Correct formula used for \(\Delta H_{\text{n}}\) [1 mark]. Final answer of \(-53.5\text{ kJ mol}^{-1}\) (including negative sign) [1 mark].
(iii) Percentage error calculated as \(6.30\%\) (accept \(6.31\%\)) [1 mark]. Suggesting heat loss to surroundings [1 mark] and heat capacity of cup/thermometer neglected [1 mark].
(iv) State that temperature change is lower [1 mark]. State that \(\Delta H_{\text{n}}\) is less exothermic / less negative [1 mark]. Explain that energy is required to dissociate the weak ethanoic acid [1 mark].

(i) Potassium manganate(VII) is a deep purple solution, whereas the reduced manganese(II) ions, \(\text{Mn}^{2+}\), are practically colorless. As long as \(\text{Fe}^{2+}\) is present, the purple colour of the added titrant is immediately discharged. At the end-point, the first tiny excess of \(\text{MnO}_4^-\text{(aq)}\) remains unreacted, causing the mixture to turn permanently pale pink.
(ii) Moles of \(\text{MnO}_4^-\text{(aq)} = 0.0150\text{ mol dm}^{-3} \times 0.01120\text{ dm}^3 = 1.68 \times 10^{-4}\text{ mol}\). According to the stoichiometric ratio, \(1\text{ mol}\) of \(\text{MnO}_4^-\text{(aq)}\) reacts with \(5\text{ mol}\) of \(\text{Fe}^{2+}\text{(aq)}\). Moles of \(\text{Fe}^{2+}\text{(aq)}\) in the aliquot \(= 5 \times 1.68 \times 10^{-4}\text{ mol} = 8.40 \times 10^{-4}\text{ mol}\).
(iii) The total volume of the dissolved tablet was \(100.0\text{ cm}^3\). Moles of \(\text{Fe}^{2+}\) in the tablet \(= 8.40 \times 10^{-4}\text{ mol} \times (100.0 / 10.00) = 8.40 \times 10^{-3}\text{ mol}\). Mass of \(\text{Fe}^{2+}\) in the tablet \(= 8.40 \times 10^{-3}\text{ mol} \times 55.85\text{ g mol}^{-1} = 0.469\text{ g}\) (or \(469\text{ mg}\)).
(iv) Oxidizing agent: \(\text{MnO}_4^-\text{(aq)}\) (manganate(VII) ion). Reducing agent: \(\text{Fe}^{2+}\text{(aq)}\) (iron(II) ion). Oxidation half-equation: \(\text{Fe}^{2+}(\text{aq}) \rightarrow \text{Fe}^{3+}(\text{aq}) + \text{e}^-\). Reduction half-equation: \(\text{MnO}_4^-(\text{aq}) + 8\text{H}^+(\text{aq}) + 5\text{e}^- \rightarrow \text{Mn}^{2+}(\text{aq}) + 4\text{H}_2\text{O}(\text{l})\).

(i) Identify that \(\text{MnO}_4^-\text{(aq)}\) is purple and \(\text{Mn}^{2+}\text{(aq)}\) is colorless/pale pink [1 mark]. State that the end-point is reached when one drop of \(\text{MnO}_4^-\text{(aq)}\) imparts a permanent pale pink color [0.66 marks].
(ii) Moles of \(\text{MnO}_4^-\text{(aq)}\) calculated correctly as \(1.68 \times 10^{-4}\text{ mol}\) [1 mark]. Correct mole ratio (5:1) identified [1 mark]. Moles of \(\text{Fe}^{2+}\text{(aq)}\) in aliquot calculated as \(8.40 \times 10^{-4}\text{ mol}\) [1 mark].
(iii) Scaling factor of 10 applied correctly to get \(8.40 \times 10^{-3}\text{ mol}\) of total \(\text{Fe}^{2+}\) [1 mark]. Using molar mass of \(55.85\text{ g mol}^{-1}\) [1 mark]. Final mass calculated as \(0.469\text{ g}\) (or \(469\text{ mg}\)) [1 mark].
(iv) Correct identification of oxidizing and reducing agents [1 mark]. Oxidation half-equation with state symbols: \(\text{Fe}^{2+}(\text{aq}) \rightarrow \text{Fe}^{3+}(\text{aq}) + \text{e}^-\) [1 mark]. Reduction half-equation with state symbols: \(\text{MnO}_4^-(\text{aq}) + 8\text{H}^+(\text{aq}) + 5\text{e}^- \rightarrow \text{Mn}^{2+}(\text{aq}) + 4\text{H}_2\text{O}(\text{l})\) [2 marks].

(a) A Br%f8nsted-Lowry acid is a proton (\( \text{H}^+ \)) donor (e.g., \( \text{HCl} \)). A Lewis acid is an electron-pair acceptor (e.g., \( \text{BF}_3 \)). (b) \( K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} \). Since \( [\text{H}^+] = [\text{A}^-] \) and assuming \( [\text{HA}]_{\text{eq}} \approx [\text{HA}]_{\text{initial}} \), we have: \( 1.35 \times 10^{-5} = \frac{[\text{H}^+]^2}{0.150} \). Therefore, \( [\text{H}^+] = \sqrt{1.35 \times 10^{-5} \times 0.150} = 1.42 \times 10^{-3} \text{ mol dm}^{-3} \). \( \text{pH} = -\log_{10}(1.42 \times 10^{-3}) = 2.85 \). Assumptions: (1) Water autoionization is negligible; (2) Dissociation of propanoic acid is negligible compared to its initial concentration. (c)(i) \( \text{C}_2\text{H}_5\text{COOH}(aq) + \text{NaOH}(aq) \rightarrow \text{C}_2\text{H}_5\text{COONa}(aq) + \text{H}_2\text{O}(l) \). (ii) Moles of acid = \( 0.0500 \times 0.150 = 0.00750 \text{ mol} \). Moles of NaOH = \( 0.0500 \times 0.0800 = 0.00400 \text{ mol} \). NaOH is the limiting reactant, so moles of propanoate ions formed = \( 0.00400 \text{ mol} \). Remaining moles of propanoic acid = \( 0.00750 - 0.00400 = 0.00350 \text{ mol} \). Total volume = \( 100.0 \text{ cm}^3 = 0.100 \text{ dm}^3 \). Final concentrations: \( [\text{C}_2\text{H}_5\text{COOH}] = 0.00350 / 0.100 = 0.0350 \text{ mol dm}^{-3} \); \( [\text{C}_2\text{H}_5\text{COO}^-] = 0.00400 / 0.100 = 0.0400 \text{ mol dm}^{-3} \). (iii) Using Henderson-Hasselbalch equation: \( \text{pH} = \text{p}K_a + \log_{10}\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) \). \( \text{p}K_a = -\log_{10}(1.35 \times 10^{-5}) = 4.87 \). \( \text{pH} = 4.87 + \log_{10}\left(\frac{0.0400}{0.0350}\right) = 4.87 + 0.06 = 4.93 \).

(a) 1 mark for Br%f8nsted-Lowry definition, 1 mark for Lewis definition, 1 mark for Br%f8nsted-Lowry example, 1 mark for Lewis example. (b) 1 mark for setting up equilibrium equation, 1 mark for calculating \( [\text{H}^+] \), 1 mark for correct pH value (2.85), 2 marks for stating both assumptions (1 mark each). (c)(i) 1 mark for balanced equation. (ii) 1 mark for correct initial moles, 1 mark for remaining moles of acid, 1 mark for moles of salt formed, 1 mark for final concentrations. (iii) 1 mark for calculating \( \text{p}K_a \), 1 mark for correct substitution of ratio, 2 marks for final pH calculation (4.93).

(a) Across Period 3, nuclear charge increases due to more protons, while shielding remains relatively constant as electrons are added to the same energy level. Consequently, the effective nuclear charge increases, attracting the outer electrons more strongly and making them harder to remove. (b)(i) \( 1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^1 \). (ii) \( 1s^2 2s^2 2p^6 3s^2 3p^6 3d^9 \). (c) The fully filled 3d subshell (\( 3d^{10} \)) and half-filled 4s subshell (\( 4s^1 \)) offer extra stability by minimizing electron-electron repulsion, making this lower in energy than the expected \( 3d^9 4s^2 \) configuration. (d) Phosphorus has a half-filled 3p subshell (\( 3p^3 \)) with singly occupied orbitals. Sulfur has a \( 3p^4 \) configuration, meaning two electrons must pair up in one 3p orbital. The repulsion between these paired electrons makes it easier to remove one of them, resulting in a lower first ionization energy. (e)(i) \( f = \frac{E}{h} = \frac{3.03 \times 10^{-19}}{6.63 \times 10^{-34}} = 4.57 \times 10^{14} \text{ Hz} \). (ii) \( \lambda = \frac{c}{f} = \frac{3.00 \times 10^8}{4.57 \times 10^{14}} = 6.56 \times 10^{-7} \text{ m} \). (f) A continuous spectrum shows all wavelengths without gaps. A line spectrum consists of discrete, sharp lines at specific wavelengths. Line spectra provide evidence of discrete levels because electrons only emit light of specific, quantized energies when transitioning between these fixed energy levels.

(a) 1 mark for increasing nuclear charge, 1 mark for constant shielding, 1 mark for stronger attraction of valence electrons. (b)(i) 1 mark for correct Cu configuration. (ii) 1 mark for correct Cu2+ configuration. (c) 1 mark for noting the stability of the filled d-subshell, 1 mark for noting the symmetric charge distribution. (d) 1 mark for mentioning the orbital diagrams of P and S, 1 mark for noting the spin-pairing repulsion in S, 1 mark for concluding that less energy is needed to remove the paired electron. (e)(i) 1 mark for formula, 1 mark for value with units. (ii) 1 mark for formula, 1 mark for wavelength value. (f) 1 mark for continuous spectrum description, 1 mark for line spectrum description, 2 marks for linking transitions between discrete levels to discrete emission wavelengths.

(a)(i) SF4 has 34 valence electrons. S is central with 4 single bonds to F and 1 lone pair. Geometry: see-saw. Bond angles: <90 degrees and <120 degrees. (ii) CO3^2- has 24 valence electrons. C has one double bond to O and two single bonds to O (with formal negative charges). Geometry: trigonal planar. Bond angle: 120 degrees. (b) The carbonate ion is represented by three equivalent resonance structures with a double bond and two single bonds rotating among the oxygen atoms. Experimentally, all three C-O bonds are identical in length, lying between a single and a double bond (bond order of 1.33), showing delocalization of pi electrons. (c) Two resonance structures of SCN^-: Structure A: [S=C=N]^- with formal charges S = 0, C = 0, N = -1. Structure B: [S-C#N]^- with formal charges S = 0, C = 0, N = -1 (or [S#C-N]^- with S = +1, C = 0, N = -2). Comparing Structure A [S=C=N]^- and Structure B [S-C#N]^-: both have formal charges localized on N, but Structure B [S-C#N]^- has a single bond on the more electronegative N or S? Nitrogen is more electronegative than sulfur (N = 3.0, S = 2.5). Thus, the negative charge is preferred on N. Structure B [S-C#N]^- has S = 0, C = 0, N = -1, which is highly stable. Structure A also has S = 0, C = 0, N = -1. Both are main contributors, but Structure B is the major contributor because the triple bond between C and N is very stable, or accept discussion of electronegativity where negative formal charge resides on N. (d) BF3 is trigonal planar; its dipoles are symmetric and cancel out, giving a net dipole of zero. NF3 is trigonal pyramidal with a lone pair on N; the dipoles do not cancel, resulting in a net dipole moment.

(a)(i) 1 mark for correct Lewis structure (showing 10 electrons around S), 1 mark for geometry (see-saw), 2 marks for correct angles (allow <90 and <120 or similar). (ii) 1 mark for correct Lewis structure (including brackets and 2- charge), 1 mark for geometry (trigonal planar), 2 marks for angle (120 degrees). (b) 1 mark for stating that pi electrons are delocalized, 1 mark for drawing/describing the resonance hybrid, 1 mark for explaining that all bonds are equal, 1 mark for stating the bond order is 1.33 / bond length is intermediate between C-O and C=O. (c) 1 mark for drawing Structure A [S=C=N]-, 1 mark for drawing Structure B [S-C#N]-, 1 mark for calculating correct formal charges for both structures, 1 mark for identifying the most stable structure (Structure B or A with sound reasoning based on electronegativity / formal charge minimization). (d) 1 mark for BF3 geometry symmetry / dipole cancellation, 1 mark for NF3 asymmetry / lone pair preventing dipole cancellation.

(a) Oxidation half-reaction: \( \text{H}_2\text{C}_2\text{O}_4 \rightarrow 2\text{CO}_2 + 2\text{H}^+ + 2\text{e}^- \). Reduction half-reaction: \( \text{MnO}_4^- + 8\text{H}^+ + 5\text{e}^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} \). Multiply oxidation by 5 and reduction by 2 to balance electrons: \( 5\text{H}_2\text{C}_2\text{O}_4 \rightarrow 10\text{CO}_2 + 10\text{H}^+ + 10\text{e}^- \) and \( 2\text{MnO}_4^- + 16\text{H}^+ + 10\text{e}^- \rightarrow 2\text{Mn}^{2+} + 8\text{H}_2\text{O} \). Summing and simplifying: \( 2\text{MnO}_4^-(aq) + 5\text{H}_2\text{C}_2\text{O}_4(aq) + 6\text{H}^+(aq) \rightarrow 2\text{Mn}^{2+}(aq) + 10\text{CO}_2(g) + 8\text{H}_2\text{O}(l) \). (b)(i) Cell reaction: \( \text{Zn}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu}(s) \). \( E^\ominus_{\text{cell}} = E^\ominus_{\text{cathode}} - E^\ominus_{\text{anode}} = +0.34 - (-0.76) = +1.10 \text{ V} \). (ii) Electrons flow from the zinc electrode (anode) to the copper electrode (cathode) in the external circuit. The copper electrode is the positive electrode (cathode). (iii) The salt bridge completes the circuit and maintains electrical neutrality by allowing ions to flow. A suitable salt is potassium nitrate (\( \text{KNO}_3 \)) or sodium chloride (\( \text{NaCl} \)). (iv) \( \Delta G^\ominus = -nFE^\ominus \). Here, \( n = 2 \). \( \Delta G^\ominus = -2 \times 96485 \text{ C mol}^{-1} \times 1.10 \text{ V} = -212267 \text{ J mol}^{-1} = -212 \text{ kJ mol}^{-1} \). (c) Voltaic cell converts chemical energy into electrical energy; its anode is negative. Electrolytic cell converts electrical energy into chemical energy; its anode is positive.

(a) 1 mark for correct oxidation half-reaction, 1 mark for correct reduction half-reaction, 1 mark for multiplying to balance electrons, 1 mark for combining half-reactions, 1 mark for correct final balanced equation. (b)(i) 2 marks for correct cell reaction (including states or not), 1 mark for correct potential (+1.10 V). (ii) 1 mark for direction of electron flow (Zn to Cu), 1 mark for identifying cathode as Cu (positive electrode). (iii) 1 mark for completing circuit/maintaining neutrality, 1 mark for identifying a non-reactive salt (like KNO3 or KCl). (iv) 1 mark for formula \( \Delta G^\ominus = -nFE^\ominus \), 1 mark for identifying n=2, 1 mark for correct calculation in Joules, 1 mark for final answer in kJ mol^-1 (-212 kJ mol^-1). (c) 1 mark for energy conversion difference, 1 mark for sign of anode difference.

(a)(i) Family 1: Carboxylic acids. Isomer: \( \text{CH}_3\text{CH}_2\text{CH}_2\text{COOH} \) (butanoic acid). Family 2: Esters. Isomer: \( \text{CH}_3\text{COOCH}_2\text{CH}_3 \) (ethyl ethanoate). Family 3: Hydroxy-aldehydes or hydroxy-ketones. Isomer: \( \text{CH}_3\text{COCH}_2\text{CH}_2\text{OH} \) (4-hydroxybutan-2-one). (ii) Add sodium hydrogencarbonate solution (\( \text{NaHCO}_3 \)) to both isomers. The carboxylic acid (butanoic acid) will produce bubbles of carbon dioxide gas (effervescence), while the ester (ethyl ethanoate) will show no reaction. (b)(i) The structural formula is ethyl ethanoate, \( \text{CH}_3\text{COOCH}_2\text{CH}_3 \). Reason: 1. The singlet at 2.0 ppm (3H) corresponds to a methyl group next to a carbonyl group (\( \text{CH}_3\text{-CO-} \)). 2. The triplet at 1.3 ppm (3H) corresponds to a methyl group next to a methylene group (\( \text{-CH}_2\text{-CH}_3 \)). 3. The quartet at 4.1 ppm (2H) corresponds to a methylene group next to a methyl group (\( \text{-CH}_2\text{-CH}_3 \)) shifted downfield because it is bonded to oxygen (\( \text{-O-CH}_2\text{-} \)). This shows an ethyl group adjacent to an oxygen atom. (ii) IUPAC name: Ethyl ethanoate. In mass spectrometry, fragmentation of ethyl ethanoate can yield a stable acylium ion, \( [\text{CH}_3\text{CO}]^+ \), at \( m/z = 43 \), or an ethoxy ion, \( [\text{OCH}_2\text{CH}_3]^- \) / \( [\text{C}_2\text{H}_5\text{O}]^+ \), at \( m/z = 45 \). Most common fragment is \( [\text{CH}_3\text{CO}]^+ \) at \( m/z = 43 \).

(a)(i) 2 marks for each family (1 mark for structural formula, 1 mark for correct IUPAC name), up to a maximum of 6 marks. (ii) 1 mark for identifying suitable reagent (e.g., NaHCO3 or Na2CO3 or reactive metal like Na), 1 mark for observing bubbles/effervescence with carboxylic acid, 1 mark for observing no visible reaction/odor with ester. (ii) 1 mark for correct structural formula of ethyl ethanoate. 3 marks for splitting/integration explanation: 1 mark for linking singlet (3H) to CH3-CO-, 1 mark for linking triplet (3H) to -CH3 next to -CH2-, 1 mark for linking quartet (2H) to -CH2- next to -CH3 bonded to O. 1 mark for explaining overall consistency. (b) 1 mark for IUPAC name (ethyl ethanoate), 2 marks for fragment (1 mark for m/z = 43, 1 mark for formula [CH3CO]+ with positive charge).