2023 IB DP Chemistry 模擬試題連答案詳解

1. The reducing agent is the species that undergoes oxidation (it causes another species to be reduced while itself being oxidized).
2. Aluminum (\(\text{Al}\)) starts with an oxidation state of 0 and goes to \(+3\) in \(\text{Al}_2\text{O}_3\). Since its oxidation state increases, \(\text{Al}\) is oxidized and is therefore the reducing agent.
3. Manganese (\(\text{Mn}\)) in \(\text{MnO}_2\) has an oxidation state of \(+4\) (since each oxygen has an oxidation state of \(-2\)). In elemental manganese (\(\text{Mn}\)), the oxidation state is 0. Thus, the oxidation state of Mn changes from \(+4\) to \(0\).

Award [1] for correct choice B.
No partial credit.

1. The standard reduction potential of the silver half-cell is more positive (\(+0.80\text{ V}\)) than that of the zinc half-cell (\(-0.76\text{ V}\)). Therefore, silver ions are reduced at the cathode: \(\text{Ag}^+(\text{aq}) + e^- \rightarrow \text{Ag}(\text{s})\). This causes solid silver to deposit on the electrode, increasing its mass.
2. Zinc is oxidized at the anode: \(\text{Zn}(\text{s}) \rightarrow \text{Zn}^{2+}(\text{aq}) + 2e^-\).
3. Electrons flow from the anode (Zn) to the cathode (Ag) via the external circuit.
4. Anions from the salt bridge migrate towards the anode to neutralize the accumulation of positive zinc ions.

Award [1] for correct choice C.
No partial credit.

An amine is classified based on the number of carbon atoms directly bonded to the nitrogen atom:
- A primary amine has one carbon bonded to nitrogen (e.g., \(\text{(CH}_3)_3\text{CNH}_2\) is primary because the nitrogen is attached to only one carbon, even though that carbon is tertiary).
- A secondary amine has two carbons bonded to nitrogen (e.g., \(\text{(CH}_3\text{CH}_2)_2\text{NH}\) and \(\text{CH}_3\text{CH}_2\text{NHCH}_3\)).
- A tertiary amine has three carbons bonded to nitrogen (e.g., \(\text{(CH}_3)_3\text{N}\)).

Award [1] for correct choice C.
No partial credit.

Ketones contain a carbonyl group (\(\text{C}=\text{O}\)) bonded to two alkyl groups.
- \(\text{CH}_3\text{CH}_2\text{CHO}\) contains a carbonyl group bonded to one alkyl group and a hydrogen atom, making it an aldehyde (propanal).
- \(\text{CH}_3\text{COCH}_3\) contains a carbonyl group bonded to two methyl groups, making it a ketone (propanone).
- \(\text{CH}_3\text{COOCH}_3\) is an ester (methyl ethanoate).
- \(\text{CH}_3\text{CH}_2\text{OCH}_3\) is an ether (methoxyethane).

Award [1] for correct choice B.
No partial credit.

- \(\text{SF}_4\) has 5 electron domains (4 bonding, 1 non-bonding) around the sulfur atom, resulting in a see-saw molecular geometry.
- \(\text{NH}_4^+\) has 4 electron domains (4 bonding, 0 non-bonding) around the nitrogen atom, resulting in a tetrahedral molecular geometry.
- \(\text{XeF}_4\) has 6 electron domains (4 bonding, 2 non-bonding) around the xenon atom, resulting in a square planar molecular geometry.
- \(\text{CO}_3^{2-}\) has 3 electron domains (3 bonding, 0 non-bonding) around the carbon atom, resulting in a trigonal planar molecular geometry.

Award [1] for correct choice B.
No partial credit.

The temperature change is given by \(\Delta T = \frac{q}{m c}\).
1. The heat released (\(q\)) is proportional to the number of moles reacting. Halving the volumes of both reactants halves the number of moles reacting (from 0.050 mol to 0.025 mol), so the heat released is halved (\(q_{\text{new}} = \frac{1}{2}q\)).
2. The total mass of the solution (\(m\)) is also halved (from 100 g to 50 g, assuming constant density and specific heat capacity), so \(m_{\text{new}} = \frac{1}{2}m\).
3. Therefore, \(\Delta T_{\text{new}} = \frac{q_{\text{new}}}{m_{\text{new}} c} = \frac{\frac{1}{2}q}{\frac{1}{2}m c} = \frac{q}{m c} = \Delta T\).

Award [1] for correct choice C.
No partial credit.

- Increasing the concentration of reactants increases the frequency of collisions, thereby increasing the rate of reaction. Thus, increasing the concentration of \(\text{HCl}\) (Option D) increases the reaction rate.
- Increasing the size of the solid pieces decreases the surface area available for reaction, which decreases the rate (Option A is incorrect).
- Decreasing the concentration of acid decreases the collision frequency, which decreases the rate (Option B is incorrect).
- Decreasing the temperature decreases the average kinetic energy of the particles and the frequency of successful collisions, which decreases the rate (Option C is incorrect).

Award [1] for correct choice D.
No partial credit.

- First ionization energy decreases down Group 17 because the outer electrons are in shells further from the nucleus, experience greater shielding, and are less strongly attracted to the nucleus.
- Boiling point increases down the group because the larger molecules have more electrons and stronger London dispersion forces.
- Density increases down the group.
- Atomic radius increases down the group as more electron shells are occupied.

Award [1] for correct choice A.
No partial credit.

1. Identify the species losing and gaining electrons: \(\text{Fe}^{2+}\) is oxidized to \(\text{Fe}^{3+}\) (loss of 1 electron) and acts as the reducing agent. \(\text{MnO}_4^-\) is reduced to \(\text{Mn}^{2+}\) and acts as the oxidizing agent. 2. Determine the oxidation state of manganese: In \(\text{MnO}_4^-\), \(x + 4(-2) = -1 \Rightarrow x = +7\). In \(\text{Mn}^{2+}\), the oxidation state is +2. Thus, the oxidation state of manganese decreases from +7 to +2, making option D correct.

[1] Award 1 mark for the correct answer (D).

1. Analyze the structure: The \(\text{-COOCH}_3\) group is an ester. The \(\text{-OH}\) group is attached directly to the benzene (aromatic) ring, which classifies it as a phenol group. 2. Distinguish between alcohol and phenol: Standard IUPAC classification separates an alcohol (where \(\text{-OH}\) is on an aliphatic carbon) from a phenol (where \(\text{-OH}\) is directly on a benzene ring). Thus, the correct pair of functional groups is ester and phenol.

[1] Award 1 mark for the correct answer (B).

1. Analyze the shapes: \(\text{H}_2\text{O}\) has 2 bonding pairs and 2 lone pairs, which gives a bent shape with a bond angle of \(104.5^\circ\). \(\text{CO}_3^{2-}\) has 3 charge centres and no lone pairs on the carbon, giving a trigonal planar geometry with a bond angle of exactly \(120^\circ\). \(\text{PCl}_3\) has 3 bonding pairs and 1 lone pair, giving a trigonal pyramidal shape with a bond angle of approximately \(107^\circ\). \(\text{NH}_4^+\) has 4 bonding pairs and no lone pairs, giving a tetrahedral shape with a bond angle of \(109.5^\circ\). Therefore, \(\text{CO}_3^{2-}\) is the correct choice.

[1] Award 1 mark for the correct answer (B).

1. Total mass of solution = \(50.0\text{ g} + 50.0\text{ g} = 100.0\text{ g}\). 2. Heat released in Joules = \(100.0 \times c \times \Delta T\). In kilojoules, this is \(\frac{100.0 \times c \times \Delta T}{1000}\). 3. Moles of reactant reacted = \(1.00\text{ mol dm}^{-3} \times 0.0500\text{ dm}^3 = 0.050\text{ mol}\). 4. Enthalpy of neutralization is exothermic, so we include a negative sign: \(\Delta H = -\frac{100 \times c \times \Delta T}{0.050 \times 1000}\text{ kJ mol}^{-1}\).

[1] Award 1 mark for the correct answer (A).

Average kinetic energy is directly proportional to absolute temperature. Therefore, only an increase in temperature will increase the average kinetic energy of the particles. An increase in temperature also increases the rate of reaction by increasing the frequency of collisions and the proportion of particles with energy greater than the activation energy. None of the other changes affect the average kinetic energy of the reactant particles.

[1] Award 1 mark for the correct answer (C).

A conjugate acid-base pair consists of two species that differ by exactly one proton (\(\text{H}^+\)). In this reaction, \(\text{H}_2\text{PO}_4^-\) acts as an acid and donates a proton to become \(\text{HPO}_4^{2-}\). Thus, \(\text{H}_2\text{PO}_4^-\) and \(\text{HPO}_4^{2-}\) represent a conjugate acid-base pair.

[1] Award 1 mark for the correct answer (A).

Across Period 3, the nuclear charge increases while inner-shell shielding remains relatively constant. This increases the attraction of the nucleus for the outer electrons, meaning: 1. Electronegativity (the ability to attract a bonding pair) increases. 2. First ionization energy (the energy required to remove an outer electron) generally increases. Thus, both properties generally increase.

[1] Award 1 mark for the correct answer (B).

1. Mass of anhydrous \(\text{MgSO}_4 = 2.41\text{ g}\). 2. Moles of \(\text{MgSO}_4 = \frac{2.41\text{ g}}{120.4\text{ g mol}^{-1}} \approx 0.0200\text{ mol}\). 3. Mass of water lost = \(4.93\text{ g} - 2.41\text{ g} = 2.52\text{ g}\). 4. Moles of \(\text{H}_2\text{O} = \frac{2.52\text{ g}}{18.0\text{ g mol}^{-1}} = 0.140\text{ mol}\). 5. Mole ratio of water to anhydrous salt = \(\frac{0.140\text{ mol}}{0.0200\text{ mol}} = 7\). Thus, \(x = 7\).

[1] Award 1 mark for the correct answer (C).

In the reaction, \(\text{H}_2\text{C}_2\text{O}_4\) acts as the reducing agent because it reduces manganese and itself gets oxidized. The oxidation state of carbon in oxalic acid (\(\text{H}_2\text{C}_2\text{O}_4\)) can be calculated as: \(2(+1) + 2(x) + 4(-2) = 0 \Rightarrow 2x = +6 \Rightarrow x = +3\). The oxidation state of carbon in carbon dioxide (\(\text{CO}_2\)) is +4. Therefore, carbon is oxidized from +3 to +4.

Award [1] for the correct choice B. Award [0] for incorrect answers.

Zinc has a more negative standard reduction potential (\(-0.76\text{ V}\)) than nickel (\(-0.26\text{ V}\)), which means zinc is more easily oxidized and acts as the anode. Nickel is reduced at the cathode: \(\text{Ni}^{2+}(aq) + 2\text{e}^- \rightarrow \text{Ni}(s)\). Since nickel ions are consumed in the reduction reaction, their concentration decreases over time. Electrons flow from anode (Zn) to cathode (Ni) through the external circuit. The cell potential is \(-0.26 - (-0.76) = +0.50\text{ V}\).

Award [1] for the correct choice C. Award [0] for incorrect answers.

The IUPAC name '3-hydroxybutanoic acid' indicates the presence of a carboxylic acid group (\(-\text{COOH}\)), which contains the carboxyl functional group, and a hydroxyl group (\(-\text{OH}\)) on the third carbon, which is an alcohol functional group. Therefore, the correct combination is carboxyl and alcohol.

Award [1] for the correct choice B. Award [0] for incorrect answers.

\(\text{SO}_2\) has a central sulfur atom with three electron domains (one single/coordinate bond, one double bond, and one lone pair), which gives it a trigonal planar electron domain geometry. The repulsion from the lone pair reduces the \(\text{O}-\text{S}-\text{O}\) bond angle slightly below \(120^\circ\) (to approximately \(119^\circ\)). In contrast, \(\text{H}_3\text{O}^+\) and \(\text{NH}_3\) are trigonal pyramidal with bond angles around \(107^\circ\), and \(\text{CO}_2\) is linear with a bond angle of \(180^\circ\).

Award [1] for the correct choice B. Award [0] for incorrect answers.

The total volume of the mixture is \(50.0 + 50.0 = 100.0\text{ cm}^3\). With a density of \(1.00\text{ g cm}^{-3}\), the mass \(m = 100.0\text{ g}\). The heat evolved is \(q = m \times c \times \Delta T = 100 \times c \times \Delta T\) Joules. The number of moles of acid/base reacting is \(n = 1.00\text{ mol dm}^{-3} \times 0.0500\text{ dm}^3 = 0.050\text{ mol}\). The enthalpy change of neutralization in \(\text{kJ mol}^{-1}\) is given by \(-\frac{q}{1000 \times n} = -\frac{100 \times c \times \Delta T}{1000 \times 0.050}\).

Award [1] for the correct choice A. Award [0] for incorrect answers.

When the temperature of a gas sample decreases, the average kinetic energy of the molecules decreases, shifting the peak of the distribution curve to the left (lower energy). Since the total number of particles (the area under the curve) remains constant, the peak must become higher to maintain the same area.

Award [1] for the correct choice B. Award [0] for incorrect answers.

An amphiprotic species can both donate and accept a proton. \(\text{H}_2\text{PO}_4^-\) can donate a proton to form \(\text{HPO}_4^{2-}\) (acting as an acid) or accept a proton to form \(\text{H}_3\text{PO}_4\) (acting as a base). The other species can only act as either an acid (\(\text{NH}_4^+\)) or a base (\(\text{CO}_3^{2-}\), \(\text{CH}_3\text{COO}^-\)).

Award [1] for the correct choice B. Award [0] for incorrect answers.

Across Period 3, the oxides of elements change from basic (\(\text{Na}_2\text{O}\), \(\text{MgO}\)) to amphoteric (\(\text{Al}_2\text{O}_3\)) to acidic (\(\text{SiO}_2\), \(\text{P}_4\text{O}_{10}\), \(\text{SO}_3\), \(\text{Cl}_2\text{O}_7\)). Option A is incorrect because atomic radius decreases across a period. Option C is incorrect because electronegativity increases across a period. Option D is incorrect because first ionization energy generally increases across a period.

Award [1] for the correct choice B. Award [0] for incorrect answers.

In the reactant \(\text{SO}_3^{2-}\), the oxidation state of sulfur (S) is +4 (since \(x + 3(-2) = -2\) gives \(x = +4\)). In the product \(\text{SO}_4^{2-}\), the oxidation state of sulfur is +6 (since \(x + 4(-2) = -2\) gives \(x = +6\)). An increase in oxidation state represents oxidation. Therefore, sulfur in \(\text{SO}_3^{2-}\) is oxidized, and its oxidation state changes from +4 to +6.

Award [1] for the correct answer (B). [0] for other options. [Note: Option A is incorrect because manganese is reduced, not oxidized. Option C is incorrect because sulfur is oxidized, not reduced. Option D is incorrect because oxygen maintains its oxidation state of -2 throughout the reaction.]

The compound \(\text{CH}_3\text{CH}_2\text{COOCH}_3\) contains the ester functional group \(-\text{COO}-\). The alkyl group attached directly to the oxygen atom is a methyl group (\(-\text{CH}_3\)), and the acyl chain contains three carbons (propanoyl group, derived from propanoic acid). Therefore, its IUPAC name is methyl propanoate and its functional group class is ester.

Award [1] for the correct option (A). [0] for any incorrect options. [Note: B is incorrect because propyl methanoate has the structure \(\text{HCOOCH}_2\text{CH}_2\text{CH}_3\). C is incorrect because the class is an ester, not an ether. D is incorrect because ethyl ethanoate has the formula \(\text{CH}_3\text{COOCH}_2\text{CH}_3\).]

Down Group 17, the first ionization energy decreases. This is because as you go down the group, additional electron shells are added, increasing the atomic radius and shielding effect. Consequently, the electrostatic attraction between the positively charged nucleus and the outermost valence electrons decreases, making it easier to remove an electron.

Award [1] for the correct option (B). [0] for any other option. [Note: Option A is incorrect because ionization energy decreases down the group. Option C is incorrect because ionization energy decreases, not increases. Option D is incorrect because electronegativity decreases down the group, not increases, and that is not the primary explanation for ionization energy trends.]

First, calculate the heat energy absorbed by the water: \(q = m \cdot c \cdot \Delta T\). Here, \(m = 50.0\text{ g}\), \(c = 4.18\text{ J g}^{-1}\_text{ K}^{-1}\), and \(\Delta T = 40.0 - 20.0 = 20.0\text{ K}\). So, \(q = 50.0 \times 4.18 \times 20.0 = 1000 \times 4.18 = 4180\text{ J} = 4.18\text{ kJ}\). Next, calculate the moles of ethanol burned: \(n = \frac{\text{mass}}{M_r} = \frac{0.46\text{ g}}{46.0\text{ g mol}^{-1}} = 0.010\text{ mol}\). Finally, the experimental enthalpy of combustion is: \(\Delta H_c = -\frac{q}{n} = -\frac{4.18\text{ kJ}}{0.010\text{ mol}} = -418\text{ kJ mol}^{-1}\). The negative sign indicates an exothermic combustion reaction.

Award [1] for the correct option (C). [0] for any other option. [Note: Option A, B, and D are incorrect due to misplacement of the decimal point or incorrect calculation of moles/heat energy conversion from J to kJ.]

Increasing the temperature increases the average kinetic energy of the reactant particles. This shifts the Maxwell-Boltzmann energy distribution curve to the right (higher energy values) and flattens it, so a larger fraction of particles possess kinetic energy equal to or greater than the activation energy (\(E_a\)).

Award [1] for the correct option (C). [0] for other options. [Note: Adding a catalyst (A) lowers the activation energy barrier but does not shift the distribution of molecular kinetic energies. Increasing reactant concentration (B) or decreasing particle size (D) increases collision frequency but does not change the kinetic energy distribution of the particles.]

A conjugate acid-base pair consists of two species that differ by exactly one hydrogen ion (\(\text{H}^+\)). In this reaction, \(\text{H}_2\text{PO}_4^-\)(aq) acts as a Brønsted-Lowry acid by donating a proton to form its conjugate base \(\text{HPO}_4^{2-}\)(aq). Thus, \(\text{H}_2\text{PO}_4^-\) and \(\text{HPO}_4^{2-}\) form a conjugate acid-base pair.

Award [1] for the correct option (C). [0] for other options. [Note: Options A and D represent reactant-reactant or reactant-product species that are not conjugate pairs because they do not differ by a single proton. Option B represents two species on the product side that do not form a conjugate pair with each other.]

\((a)\) For chlorine in \(\text{ClO}_3^-\): let \(x\) be the oxidation state of Cl. \(x + 3(-2) = -1 \implies x = +5\). For sulfur in \(\text{SO}_2\): let \(y\) be the oxidation state of S. \(y + 2(-2) = 0 \implies y = +4\). \((b)\) Reduction of \(\text{ClO}_3^-\): Balance oxygen atoms with water: \(\text{ClO}_3^- \rightarrow \text{Cl}^- + 3\text{H}_2\text{O}\). Balance hydrogen atoms with protons: \(\text{ClO}_3^- + 6\text{H}^+ \rightarrow \text{Cl}^- + 3\text{H}_2\text{O}\). Balance charges with electrons: \(\text{ClO}_3^- + 6\text{H}^+ + 6e^- \rightarrow \text{Cl}^- + 3\text{H}_2\text{O}\). \((c)\) Oxidation of \(\text{SO}_2\): Balance oxygen atoms with water: \(\text{SO}_2 + 2\text{H}_2\text{O} \rightarrow \text{SO}_4^{2-}\). Balance hydrogen atoms with protons: \(\text{SO}_2 + 2\text{H}_2\text{O} \rightarrow \text{SO}_4^{2-} + 4\text{H}^+\). Balance charges with electrons: \(\text{SO}_2 + 2\text{H}_2\text{O} \rightarrow \text{SO}_4^{2-} + 4\text{H}^+ + 2e^-\). \((d)\) Combine half-equations by multiplying the oxidation half-equation by 3 to balance electrons: \(3\text{SO}_2 + 6\text{H}_2\text{O} \rightarrow 3\text{SO}_4^{2-} + 12\text{H}^+ + 6e^-\). Adding the two half-equations and simplifying water and protons on both sides yields: \(\text{ClO}_3^-(\text{aq}) + 3\text{SO}_2(\text{g}) + 3\text{H}_2\text{O}(\text{l}) \rightarrow \text{Cl}^-(\text{aq}) + 3\text{SO}_4^{2-}(\text{aq}) + 6\text{H}^+(\text{aq})\). \((e)\) Sulfur dioxide, \(\text{SO}_2\), is the reducing agent because the oxidation state of sulfur increases from +4 to +6.

\((a)\) [2 marks] Award [1] for Cl = +5 (or V) and [1] for S = +4 (or IV). \((b)\) [2 marks] Award [2] for correct equation: \(\text{ClO}_3^- + 6\text{H}^+ + 6e^- \rightarrow \text{Cl}^- + 3\text{H}_2\text{O}\). Award [1] if formulas are correct but unbalanced. \((c)\) [2 marks] Award [2] for correct equation: \(\text{SO}_2 + 2\text{H}_2\text{O} \rightarrow \text{SO}_4^{2-} + 4\text{H}^+ + 2e^-\). Award [1] if formulas are correct but unbalanced. \((d)\) [2 marks] Award [2] for correct overall equation: \(\text{ClO}_3^- + 3\text{SO}_2 + 3\text{H}_2\text{O} \rightarrow \text{Cl}^- + 3\text{SO}_4^{2-} + 6\text{H}^+\). Award [1] if coefficients are partially correct. State symbols are not required. \((e)\) [0.33 marks] Award [0.33] for \(\text{SO}_2\) (or sulfur dioxide / sulfur(IV) oxide).

\((a)\) The full structural formula must show all atoms and all bonds individually (including O-H and C-H bonds): \(\text{H}-\text{C}(\text{H})_2-\text{C}(\text{H})(\text{OH})-\text{C}(\text{H})_2-\text{C}(=\text{O})-\text{O}-\text{H}\). \((b)\) It is a secondary alcohol because the carbon atom that is directly bonded to the hydroxyl group (\(-\text{OH}\)) is bonded to two other carbon atoms. \((c)\) The two functional groups are the carboxyl group (representing carboxylic acid) and the hydroxyl group (representing alcohol). \((d)\) 3-hydroxybutanoic acid contains highly polar \(\text{O}-\text{H}\) bonds, which allow its molecules to form strong intermolecular hydrogen bonds. Ethyl ethanoate is an ester and has no \(\text{O}-\text{H}\) bonds, meaning it cannot form hydrogen bonds with other ester molecules; it only experiences weaker dipole-dipole and London dispersion forces. Thus, more energy is required to break the forces in 3-hydroxybutanoic acid, resulting in a higher boiling point.

\((a)\) [2 marks] Award [2] for a fully drawn structure showing all bonds. Award [1] if O-H bonds are shown condensed as -OH or if a single atom is missing. \((b)\) [2 marks] Award [1] for 'secondary'. Award [1] for stating that the carbon atom carrying the hydroxyl group is bonded to two other carbon atoms. \((c)\) [2 marks] Award [1] for 'hydroxyl' (or alcohol) and [1] for 'carboxyl' (or carboxylic acid). Do not accept 'carbonyl'. \((d)\) [2.33 marks] Award [1] for stating that 3-hydroxybutanoic acid can form hydrogen bonds. Award [1] for stating that ethyl ethanoate cannot form hydrogen bonds / has only dipole-dipole and London dispersion forces. Award [0.33] for stating that hydrogen bonds are stronger than dipole-dipole forces and require more energy to break.

\((a)\) For \(\text{NF}_3\): Nitrogen has 5 valence electrons and each fluorine has 7, totaling 26 electrons. There are 3 single bonds to F, 1 lone pair on the N, and 3 lone pairs on each F. For \(\text{BF}_3\): Boron has 3 valence electrons and each fluorine has 7, totaling 24 electrons. There are 3 single bonds to F, 0 lone pairs on B (incomplete octet exception), and 3 lone pairs on each F. \((b)\) \(\text{NF}_3\) has 4 electron domains (3 bonding, 1 non-bonding) around N, so its electron domain geometry is tetrahedral, and its molecular geometry is trigonal pyramidal. \(\text{BF}_3\) has 3 electron domains (3 bonding) around B, so its electron domain geometry is trigonal planar, and its molecular geometry is trigonal planar. \((c)\) \(\text{BF}_3\) is non-polar because it is highly symmetrical, so the polar bond dipoles cancel out. \(\text{NF}_3\) is polar because it is asymmetrical due to the lone pair on nitrogen, which means the polar bond dipoles do not cancel out.

\((a)\) [3 marks] Award [1.5] for correct Lewis structure of \(\text{NF}_3\) (lone pair on N must be shown). Award [1.5] for correct Lewis structure of \(\text{BF}_3\) (no lone pairs on B). Fluorine lone pairs must be present on both structures. \((b)\) [4 marks] Award [1] for each correct geometry: \(\text{NF}_3\) electron domain = tetrahedral; \(\text{NF}_3\) molecular = trigonal pyramidal; \(\text{BF}_3\) electron domain = trigonal planar; \(\text{BF}_3\) molecular = trigonal planar. \((c)\) [1.33 marks] Award [0.5] for stating \(\text{BF}_3\) is non-polar and \(\text{NF}_3\) is polar. Award [0.83] for the reason: symmetry in \(\text{BF}_3\) allows dipoles to cancel, whereas asymmetry in \(\text{NF}_3\) prevents dipole cancellation.

\((a)\) Temperature rise, \(\Delta T = 35.6 - 20.2 = 15.4\text{ K}\). Mass of solution, \(m = 50.0\text{ cm}^3 \times 1.00\text{ g cm}^{-3} = 50.0\text{ g}\). Heat energy: \(q = m c \Delta T = 50.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 15.4\text{ K} = 3218.6\text{ J} \approx 3220\text{ J}\). \((b)\) Amount of \(\text{CuSO}_4\): \(n = C \times V = 0.400\text{ mol dm}^{-3} \times 0.0500\text{ dm}^3 = 0.0200\text{ mol}\). \((c)\) Enthalpy change, \(\Delta H = -\frac{q}{n \times 1000} = -\frac{3218.6}{20.0} = -160.93\text{ kJ mol}^{-1} \approx -161\text{ kJ mol}^{-1}\). The sign is negative because the temperature increased (exothermic). \((d)\) Major systematic error: Heat loss to the surroundings / polystyrene cup / thermometer. Since heat is lost, the measured \(\Delta T\) is smaller than theoretical, resulting in a calculated value of \(\Delta H\) that is less exothermic (less negative) than the actual literature value.

\((a)\) [2 marks] Award [1] for \(\Delta T = 15.4\text{ K}\). Award [1] for \(q = 3220\text{ J}\) (or \(3.22\text{ kJ}\); accept 3218.6 J). \((b)\) [2 marks] Award [2] for \(0.0200\text{ mol}\). Award [1] if volume was not converted to \(\text{dm}^3\). \((c)\) [3 marks] Award [1] for dividing \(q\) by \(n\). Award [1] for the value \(161\) (or \(160.9\)). Award [1] for negative sign. Accept error carried forward (ECF) from (a) and (b). \((d)\) [1.33 marks] Award [0.5] for identifying a valid error (e.g., heat loss to surroundings). Award [0.83] for describing the effect (makes \(\Delta T\) lower, so \(\Delta H\) is less exothermic / less negative).

\((a)\) and \((b)\) In the diagram, the y-axis represents the number/fraction of particles, and the x-axis represents kinetic energy. The curve for \(T_1\) starts at the origin, rises to a peak, and then tapers asymptotically towards the x-axis. \(E_a\) is marked as a vertical line on the right side of the curve. The curve for \(T_2\) has its peak lower and shifted to the right of the peak for \(T_1\), crossing the first curve only once and remaining above the \(T_1\) curve at high energy values. \((c)\) At temperature \(T_2\), the average kinetic energy of the molecules is higher. As seen from the diagram, a significantly larger fraction of molecules have energy equal to or greater than the activation energy (\(E \ge E_a\)). Therefore, a much higher proportion of collisions are successful (lead to reaction), which increases the reaction rate. (Also, particles move faster, which increases overall collision frequency, though this effect is minor compared to the increase in successful collisions).

\((a)\) [3 marks] Award [1] for correct axes labels (y: number/fraction of particles; x: kinetic energy/energy). Award [1] for a correct asymmetric distribution curve starting at origin. Award [1] for clearly marking \(E_a\). \((b)\) [2 marks] Award [1] for peak of \(T_2\) shifted to the right and lower than \(T_1\). Award [1] for the \(T_2\) curve being above \(T_1\) curve at high energy values. \((c)\) [3.33 marks] Award [1] for stating that temperature increases average kinetic energy. Award [1] for explaining that a larger fraction of particles have energy \(\ge E_a\) (larger shaded area under the curve to the right of \(E_a\)). Award [1] for stating that this leads to a higher frequency of successful collisions. Award [0.33] for mentioning that collision frequency also increases.

\((a)\) Going across Period 3 from sodium to chlorine, the number of protons in the nucleus increases, which increases the nuclear charge. Electrons are added to the same main energy level (third shell), so the shielding effect remains approximately constant. As a result, the effective nuclear charge increases, which pulls the outer electrons closer to the nucleus, decreasing the atomic radius. \((b)\) (i) \(\text{Na}_2\text{O}(\text{s}) + \text{H}_2\text{O}(\text{l}) \rightarrow 2\text{NaOH}(\text{aq})\) (ii) \(\text{P}_4\text{O}_{10}(\text{s}) + 6\text{H}_2\text{O}(\text{l}) \rightarrow 4\text{H}_3\text{PO}_4(\text{aq})\) (accept \(\text{P}_2\text{O}_5 + 3\text{H}_2\text{O} \rightarrow 2\text{H}_3\text{PO}_4\)). \((c)\) Sodium oxide is basic and phosphorus(V) oxide is acidic. The general trend for Period 3 oxides is a transition from basic (metallic oxides: \(\text{Na}_2\text{O}\), \(\text{MgO}\)) to amphoteric (metalloid boundary oxide: \(\text{Al}_2\text{O}_3\)) to acidic (non-metallic oxides: \(\text{SiO}_2\), \(\text{P}_4\text{O}_{10}\), \(\text{SO}_3\)).

\((a)\) [3 marks] Award [1] for stating nuclear charge increases / more protons. Award [1] for stating shielding remains constant / electrons are in the same shell. Award [1] for stating attraction between nucleus and outer shell electrons increases. \((b)\) [3 marks] (i) Award [1.5] for correctly balanced equation: \(\text{Na}_2\text{O} + \text{H}_2\text{O} \rightarrow 2\text{NaOH}\). (ii) Award [1.5] for correctly balanced equation: \(\text{P}_4\text{O}_{10} + 6\text{H}_2\text{O} \rightarrow 4\text{H}_3\text{PO}_4\) (or equivalent for \(\text{P}_2\text{O}_5\)). \((c)\) [2.33 marks] Award [1] for classifying \(\text{Na}_2\text{O}\) as basic and \(\text{P}_4\text{O}_{10}\) as acidic. Award [1] for describing the trend (basic to amphoteric to acidic). Award [0.33] for linking basicity to metals and acidity to non-metals.

(a) (i)
Percentage uncertainty in volume of water:
\(\frac{0.5}{50.0} \times 100 = 1.0\%\)

Percentage uncertainty in mass of ammonium chloride:
\(\frac{0.01}{5.35} \times 100 = 0.19\%\) (accept 0.2%)

(a) (ii)
The mass of ammonium chloride contributes the least because it has a significantly smaller percentage uncertainty (0.19%) compared to the volume of water (1.0%).

(b) (i)
\(q = m \cdot c \cdot \Delta T\)
\(q = 50.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 6.8\text{ K} = 1421.2\text{ J} = 1.42\text{ kJ}\)
(Award full marks if total mass of solution, \(55.35\text{ g}\), is used: \(q = 55.35\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 6.8\text{ K} = 1573.3\text{ J} = 1.57\text{ kJ}\))

(b) (ii)
Molar mass of \(\text{NH}_4\text{Cl} = 14.01 + 4(1.01) + 35.45 = 53.50\text{ g mol}^{-1}\)
\(n = \frac{5.35\text{ g}}{53.50\text{ g mol}^{-1}} = 0.100\text{ mol}\)

\(\Delta H_{\text{sol}} = +\frac{q}{n}\)
If using \(1.42\text{ kJ}\):
\(\Delta H_{\text{sol}} = +\frac{1.42\text{ kJ}}{0.100\text{ mol}} = +14.2\text{ kJ mol}^{-1}\)
If using \(1.57\text{ kJ}\):
\(\Delta H_{\text{sol}} = +\frac{1.57\text{ kJ}}{0.100\text{ mol}} = +15.7\text{ kJ mol}^{-1}\)
(The positive sign is required as the temperature decreased, indicating an endothermic reaction).

(c)
Any one of:
- No heat was exchanged with the surroundings (or polystyrene cup is a perfect insulator).
- The heat capacity of the polystyrene cup is negligible.
- The solution was well-mixed and of uniform temperature.

Part (a)(i):
[1] for volume percentage uncertainty = 1.0%
[1] for mass percentage uncertainty = 0.19% (or 0.2%)

Part (a)(ii):
[1] for stating "mass of ammonium chloride" AND giving a reason based on its smaller percentage uncertainty.

Part (b)(i):
[1] for correct substitution of values into \(q = mc\Delta T\).
[1] for correct value of \(1.42\text{ kJ}\) or \(1.57\text{ kJ}\) (with or without sign at this stage, but must have correct unit of \(\text{kJ}\) or \(\text{J}\)).

Part (b)(ii):
[1] for \(n(\text{NH}_4\text{Cl}) = 0.100\text{ mol}\).
[1] for \(\Delta H_{\text{sol}} = +14.2\text{ kJ mol}^{-1}\) (or \(+15.7\text{ kJ mol}^{-1}\)). Note: Positive sign (+) is required for the mark. Accept answers with 2 or 3 significant figures.

Part (c):
[1] for any valid assumption, e.g., "no heat loss to/gain from the surroundings" OR "heat capacity of the cup is negligible" OR "the solution was well-stirred/uniform temperature". Do not accept "density/specific heat capacity is equal to water" as it is already excluded by the question.

(a)
Any two of:
- Measuring the loss of mass of the flask/reaction mixture over time (as gas escapes).
- Monitoring the pH of the mixture over time (as \(\text{H}^+\) is consumed).
- Measuring the electrical conductivity of the solution over time (as ion concentrations change).

(b) (i)
As the reaction proceeds, the concentration of the acid (reactants) decreases, leading to less frequent collisions per unit time.

(b) (ii)
Draw a tangent to the curve at \(t = 60\text{ s}\) and calculate the gradient (slope) of this tangent.

(c) (i)
At higher temperature, particles have higher average kinetic energy, so a greater fraction of collisions have energy greater than or equal to the activation energy (\(E \ge E_a\)). Additionally, the frequency of collisions increases. This results in a higher frequency of successful collisions.

(c) (ii)
The final volume of gas remains the same because the limiting reactant (calcium carbonate) is in the same quantity, and the stoichiometry of the reaction remains unchanged.

Part (a):
[1] for each correct continuous monitoring method up to [2] max.
- Accept: "change in mass" / "mass loss of reaction vessel".
- Accept: "monitoring pH" / "concentration of hydrogen ions".
- Accept: "electrical conductivity".
- Reject: "measuring temperature".

Part (b)(i):
[1] for stating concentration of reactants/acid decreases OR less frequent collisions.

Part (b)(ii):
[1] for "draw a tangent at 60 s" AND "calculate its gradient/slope".

Part (c)(i):
[1] for stating that the fraction of particles with \(E \ge E_a\) increases (or average kinetic energy increases).
[1] for stating that the frequency of successful collisions increases.

Part (c)(ii):
[1] for stating "the final volume remains the same" because the amount of limiting reactant (calcium carbonate) is the same.

The therapeutic index (\(TI\)) for animal trials is calculated using the formula: \(TI = \frac{LD_{50}}{ED_{50}}\). Substituting the given values: \(TI = \frac{450}{15} = 30\). A high therapeutic index means there is a large difference between the effective therapeutic dose and the lethal dose. Consequently, the drug has a wide margin of safety, and accidental overdoses are much less likely to cause severe toxicity or death.

Award 1 mark for calculating the correct numerical value of \(TI = 30\). Award 1 mark for stating that a high \(TI\) indicates a large difference or wide margin between the effective and lethal doses. Award 0.5 marks for linking this wide margin to safety, such as a low risk of accidental toxic overdose.

The chemical name of aspirin is acetylsalicylic acid. Salicylic acid is highly acidic and causes severe irritation, pain, and bleeding in the stomach lining. Converting the phenolic hydroxyl group into an ester group to form acetylsalicylic acid significantly reduces this direct acidity and local irritation, making it much safer and more tolerable for patients while retaining its analgesic and anti-inflammatory properties.

Award 1 mark for identifying the chemical name as acetylsalicylic acid (accept 2-acetyloxybenzoic acid). Award 1 mark for explaining that salicylic acid causes severe irritation, ulceration, or bleeding in the stomach. Award 0.5 marks for stating that acetylsalicylic acid is less irritating/damaging to the stomach tissue.

The four-membered beta-lactam ring contains \(sp^3\) hybridized carbon and nitrogen atoms which naturally prefer bond angles of \(109.5^\circ\) (and \(sp^2\) carbonyl carbon preferring \(120^\circ\)). Because they are constrained to a small ring, the actual bond angles are forced to be approximately \(90^\circ\), causing severe ring strain. This strain makes the ring highly unstable and highly reactive. The ring easily opens to covalently bind to and irreversibly deactivate the transpeptidase enzyme, which is responsible for cross-linking peptidoglycan chains during bacterial cell wall synthesis. Without a stable cell wall, the bacteria burst and die.

Award 1 mark for explaining that the four-membered ring has bond angles of about \(90^\circ\) (instead of \(109.5^\circ\) or \(120^\circ\)), causing severe ring strain. Award 1 mark for stating that the ring opens and covalently binds to and deactivates the transpeptidase enzyme (or penicillin-binding proteins). Award 0.5 marks for stating that this deactivation prevents bacterial cell wall synthesis / cross-linking.

Morphine contains two polar hydroxyl (\(-OH\)) groups, whereas in diamorphine, these hydroxyl groups are converted into ester (acetyl, \(-OCOCH_3\)) groups. This structural modification makes diamorphine significantly less polar and much more lipid-soluble (lipophilic) than morphine. Because the blood-brain barrier consists of a lipid-rich membrane, the lipophilic diamorphine is able to dissolve in and cross this barrier much more quickly and in higher concentrations, leading to a much stronger and faster analgesic effect in the brain.

Award 1 mark for identifying that morphine has polar hydroxyl groups while diamorphine has less polar ester/acetyl groups. Award 1 mark for stating that diamorphine is more lipid-soluble / lipophilic / hydrophobic. Award 0.5 marks for explaining that high lipid solubility allows diamorphine to penetrate or cross the blood-brain barrier much faster/more easily.

First, determine the molar mass of magnesium hydroxide, \(Mg(OH)_2\): \(M = 24.31 + 2 \times (16.00 + 1.01) = 58.33\text{ g mol}^{-1}\). Next, calculate the moles of \(Mg(OH)_2\): \(n = \frac{0.35\text{ g}}{58.33\text{ g mol}^{-1}} = 0.005999\text{ mol}\). The neutralization reaction is: \(Mg(OH)_2(s) + 2HCl(aq) \rightarrow MgCl_2(aq) + 2H_2O(l)\). Therefore, \(0.005999\text{ mol}\) of \(Mg(OH)_2\) neutralizes \(2 \times 0.005999\text{ mol} = 0.011998\text{ mol}\) of \(HCl\). The volume of \(HCl\) neutralized is: \(V = \frac{n}{c} = \frac{0.011998\text{ mol}}{0.10\text{ mol dm}^{-3}} = 0.11998\text{ dm}^3 = 119.98\text{ cm}^3 \approx 120\text{ cm}^3\).

Award 1 mark for calculating the moles of \(Mg(OH)_2\) correctly as \(6.0 \times 10^{-3}\text{ mol}\) (or exact value \(0.005999\text{ mol}\)). Award 1 mark for multiplying by 2 to find moles of \(HCl = 0.012\text{ mol}\) based on the 1:2 stoichiometry. Award 0.5 marks for calculating the correct final volume of \(120\text{ cm}^3\) (accept range \(118\text{ to }121\text{ cm}^3\)).

Antacids like calcium carbonate are weak bases that act directly via a chemical neutralization reaction with the hydrochloric acid already present in the stomach lumen. In contrast, H2-receptor antagonists like ranitidine act biologically; they bind to histamine H2-receptors on the parietal cells in the gastric glands of the stomach wall, thereby blocking the histamine signal that triggers these cells to produce and secrete hydrochloric acid in the first place.

Award 1 mark for stating that antacids work by chemically neutralizing the hydrochloric acid already present in the stomach. Award 1 mark for stating that H2-receptor antagonists block histamine receptors (or H2 receptors) on parietal cells to prevent/reduce the production/secretion of acid. Award 0.5 marks for explicitly highlighting the contrast between direct chemical neutralization vs biological prevention of secretion.

Both oseltamivir and zanamivir are designed as neuraminidase inhibitors. Inside an infected host cell, new influenza virus particles are synthesized and bud off the cell membrane, but remain tethered to the cell by sialic acid receptors. The viral enzyme neuraminidase cleaves this sialic acid linkage, releasing the newly formed virions. By binding to and inhibiting the active site of neuraminidase, these antiviral drugs prevent this cleavage. As a result, the new virus particles remain trapped on the surface of the host cell, preventing them from spreading and infecting adjacent healthy cells.

Award 1 mark for identifying the target enzyme as neuraminidase. Award 1 mark for explaining that inhibiting this enzyme prevents the cleavage of sialic acid links, trapping newly formed viral particles on the host cell surface. Award 0.5 marks for stating that this prevents the newly formed virus from spreading or infecting other host cells.

Low-level waste (LLW) is characterized by low levels of radioactivity and radionuclides with relatively short half-lives (e.g., medical tools, syringes, protective clothing, and diagnostic radiopharmaceuticals). High-level waste (HLW) has high levels of radioactivity and contains radionuclides with long half-lives (such as spent nuclear fuel or target materials used in reactors to produce medical isotopes). LLW is safely disposed of by storing it in secure, shielded containers until its radioactivity has decayed to safe background levels. Once decayed, it can be disposed of alongside ordinary non-radioactive waste (e.g., in landfills or incinerators).

Award 1 mark for distinguishing between LLW and HLW based on both the level of activity AND the length of the half-lives. Award 1 mark for describing LLW disposal: stored in shielded containers until its activity decays to safe levels. Award 0.5 marks for stating that after decay, the waste can be safely disposed of as normal/non-radioactive waste.