2026 Edexcel IAS-Level Chemistry (XCH11) 模擬試題連答案詳解

The successive ionization energies show a massive increase between the 5th and the 6th ionization energies (from \(6274\text{ kJ mol}^{-1}\) to \(21269\text{ kJ mol}^{-1}\)). This indicates that the 6th electron is being removed from an inner quantum shell, meaning there are 5 electrons in the outer shell of element \(X\). Thus, \(X\) is in Group 15 (Group 5). A Group 15 element in Period 3 (Phosphorus) forms a hydride with the formula \(XH_3\).

Award 1 mark for the correct option (B).

Let the probability of selecting a \(^{79}\text{Br}\) atom be \(p\) and a \(^{81}\text{Br}\) atom be \(q = 1-p\).
The peak at \(m/z = 158\) corresponds to \(^{79}\text{Br}\)-\(^{79}\text{Br}\) with probability \(p^2\).
The peak at \(m/z = 160\) corresponds to \(^{79}\text{Br}\)-\(^{81}\text{Br}\) and \(^{81}\text{Br}\)-\(^{79}\text{Br}\) with probability \(2pq\).
The peak at \(m/z = 162\) corresponds to \(^{81}\text{Br}\)-\(^{81}\text{Br}\) with probability \(q^2\).
Given the relative abundances of \(1 : 2 : 1\), we have:
\(p^2 : 2pq : q^2 = 1 : 2 : 1\)
This matches the expansion of \((p + q)^2 = p^2 + 2pq + q^2\) where \(p = q = 0.5\).
Thus, the isotopic ratio \(^{79}\text{Br} : ^{81}\text{Br}\) is \(1 : 1\).

Award 1 mark for the correct option (A).

In \(\text{H}_3\text{O}^+\), the central oxygen atom has 5 valence electrons (6 minus 1 for the positive charge) plus 3 electrons from the hydrogen atoms, giving 8 electrons (4 pairs). These consist of 3 bonding pairs and 1 lone pair, resulting in a trigonal pyramidal shape. The repulsion from the lone pair reduces the tetrahedral bond angle from \(109.5^\circ\) to approximately \(107^\circ\).

Award 1 mark for the correct option (C).

According to Fajans' rules, covalent character in an ionic compound is maximized when the cation is small and highly charged (high polarizing power) and the anion is large (high polarizability).
- Comparing cations: \(\text{Mg}^{2+}\) is smaller than \(\text{Ba}^{2+}\), so \(\text{Mg}^{2+}\) has greater polarizing power.
- Comparing anions: \(\text{I}^{-}\) is larger than \(\text{Cl}^{-}\), so \(\text{I}^{-}\) is more polarizable.
Therefore, \(\text{MgI}_2\) has the greatest degree of covalent character.

Award 1 mark for the correct option (A).

1. Calculate moles of carbon:
\(n(\text{CO}_2) = \frac{17.6}{44.0} = 0.40\text{ mol}\)
Since each mole of \(\text{CO}_2\) contains 1 mole of \(\text{C}\) atoms, \(n(\text{C}) = 0.40\text{ mol}\).

2. Calculate moles of hydrogen:
\(n(\text{H}_2\text{O}) = \frac{7.2}{18.0} = 0.40\text{ mol}\)
Since each mole of \(\text{H}_2\text{O}\) contains 2 moles of \(\text{H}\) atoms, \(n(\text{H}) = 0.80\text{ mol}\).

3. Determine \(x\) and \(y\) in the molecular formula \(\text{C}_x\text{H}_y\) (for \(0.10\text{ mol}\) of hydrocarbon):
\(x = \frac{0.40}{0.10} = 4\)
\(y = \frac{0.80}{0.10} = 8\)

Thus, the molecular formula is \(\text{C}_4\text{H}_8\).

Award 1 mark for the correct option (C).

The balanced equation for the neutralization is:
\(2\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}\)

1. Calculate moles of \(\text{H}_2\text{SO}_4\):
\(n(\text{H}_2\text{SO}_4) = 0.0500\text{ mol dm}^{-3} \times \frac{20.0}{1000}\text{ dm}^3 = 1.00 \times 10^{-3}\text{ mol}\)

2. Calculate moles of \(\text{NaOH}\):
\(n(\text{NaOH}) = 2 \times 1.00 \times 10^{-3} = 2.00 \times 10^{-3}\text{ mol}\)

3. Calculate concentration of \(\text{NaOH}\) in \(\text{mol dm}^{-3}\):
\([\text{NaOH}] = \frac{2.00 \times 10^{-3}\text{ mol}}{0.0250\text{ dm}^3} = 0.0800\text{ mol dm}^{-3}\)

4. Convert concentration to \(\text{g dm}^{-3}\):
\(\text{Concentration} = 0.0800\text{ mol dm}^{-3} \times 40.0\text{ g mol}^{-1} = 3.20\text{ g dm}^{-3}\)

Award 1 mark for the correct option (C).

To name the alkane:
1. Find the longest continuous carbon chain. The longest chain contains 5 carbon atoms (pentane).
2. Identify the substituents. The substituents are a methyl group and an ethyl group.
3. Number the carbon chain from the end that gives the lowest locants. Numbering from left-to-right gives substituents at positions 2 and 3 (2-methyl and 3-ethyl). Numbering from right-to-left gives substituents at positions 3 and 4 (3-ethyl and 4-methyl). Therefore, we number from left-to-right (locants 2,3 instead of 3,4).
4. List substituents alphabetically: 'ethyl' comes before 'methyl'. This gives 3-ethyl-2-methylpentane.

Note: '3-isopropylpentane' is incorrect because choosing the chain with the maximum number of substituents (2 substituents instead of 1) is preferred under IUPAC rules.

Award 1 mark for the correct option (A).

In the free radical substitution mechanism:
- Option A represents an initiation step (homolytic fission of the Br-Br bond by UV light).
- Option B represents a termination step (two free radicals combining to form a stable molecule).
- Option C is incorrect as the reaction of a halogen radical with an alkane produces an alkyl radical and hydrogen halide (e.g., \(\text{CH}_3\text{CH}_3 + \text{Br}^\bullet \rightarrow \text{CH}_3\text{CH}_2^\bullet + \text{HBr}\)), not hydrogen radicals.
- Option D represents a correct propagation step, where an ethyl radical reacts with a bromine molecule to form bromoethane and regenerates the bromine radical.

Award 1 mark for the correct option (D).

There is a very large jump between the third and fourth ionization energies (from 2745 to 11578 kJ mol^{-1}). This indicates that the fourth electron is being removed from an inner shell which experiences much less shielding and stronger electrostatic attraction to the nucleus. Therefore, the element has three valence electrons and belongs to Group 3 (Group 13) of the periodic table. The Period 3 element in Group 13 is aluminium.

[1] C - Aluminium. 1 mark for identifying the correct element based on the position of the largest jump in successive ionization energies.

Let the hydrocarbon be C_xH_y. The volume decrease of 40 cm^{3} when passed through NaOH is due to the absorption of CO_2. Since 10 cm^{3} of hydrocarbon produces 40 cm^{3} of CO_2, x = 4. The initial volume of reactants (excluding excess oxygen) is 10 + 10(x + y/4) cm^{3}, and the final volume of products (excluding excess oxygen) is 10x cm^{3} of CO_2 (since H_2O is a liquid at RTP). The volume decrease is: Delta V = (10 + 10x + 2.5y) - 10x = 10 + 2.5y. Given Delta V = 30 cm^{3}: 10 + 2.5y = 30, which gives y = 8. Thus, the molecular formula is C_4H_8.

[1] D - C_4H_8. 1 mark for calculating the values of x and y using volume ratios and the contraction formula.

The hydronium ion, H_3O^{+}, has three bonding pairs of electrons and one lone pair of electrons around the central oxygen atom (giving a total of 4 electron pairs, which adopt a tetrahedral arrangement). Due to the greater repulsion of the lone pair compared to the bonding pairs, the H-O-H bond angle is reduced from the tetrahedral angle of 109.5 degrees to approximately 107 degrees, resulting in a trigonal pyramidal shape.

[1] A - H_3O^{+}. 1 mark for identifying that the presence of 3 bonding pairs and 1 lone pair results in a bond angle of 107 degrees.

During the propagation stage, a radical reacts with a non-radical molecule to produce a new radical and a new molecule. The first propagation step involves a chlorine radical abstracting a hydrogen atom from ethane to form an ethyl radical and hydrogen chloride: C_2H_6 + Cl^{\bullet} \rightarrow C_2H_5^{\bullet} + HCl. Option A is initiation, Option B is incorrect as H^{\bullet} is not formed, and Option D is a termination step.

[1] C. 1 mark for identifying the correct radical-molecule reaction that propagates the chain reaction.

For an alkene to show E-Z stereoisomerism, each carbon in the double bond must be attached to two different groups. In 3-methylhex-3-ene, Carbon-3 is bonded to a methyl group (-CH_3) and an ethyl group (-CH_2CH_3), while Carbon-4 is bonded to a hydrogen atom (-H) and an ethyl group (-CH_2CH_3). Since both carbons are bonded to two different groups, E-Z stereoisomerism is possible. In the other options, at least one of the double-bonded carbons is attached to two identical groups.

[1] C - 3-methylhex-3-ene. 1 mark for identifying the alkene with two different groups on both double-bonded carbons.

The equation for the formation of propane is: 3C(s) + 4H_2(g) \rightarrow C_3H_8(g). Using Hess's law: \Delta_f H^{\ominus} = 3 \times \Delta_c H^{\ominus}[C] + 4 \times \Delta_c H^{\ominus}[H_2] - \Delta_c H^{\ominus}[C_3H_8]. Calculation: \Delta_f H^{\ominus} = 3(-393.5) + 4(-285.8) - (-2219.2) = -1180.5 - 1143.2 + 2219.2 = -104.5 kJ mol^{-1}.

[1] A - -104.5 kJ mol^{-1}. 1 mark for the correct application of Hess's law to calculate the formation enthalpy.

Down Group 2, the ionic radius of the M^{2+} cation increases, which decreases its charge density and polarizing power. Consequently, the larger cation causes less distortion (polarization) of the electron cloud of the nitrate anion, making the nitrate ion more stable and requiring more heat energy to decompose.

[1] B. 1 mark for the correct explanation linking increased ionic radius to decreased polarizing power and increased thermal stability.

Tertiary halogenoalkanes react predominantly via an S_N1 mechanism because they form highly stable tertiary carbocations due to the inductive electron-donating effect of the three alkyl groups, and the steric hindrance prevents direct back-side nucleophilic attack (S_N2). 2-chloro-2-methylpropane is a tertiary halogenoalkane, whereas 1-chlorobutane and chloromethane are primary, and 2-chlorobutane is secondary.

[1] C - 2-chloro-2-methylpropane. 1 mark for identifying the tertiary halogenoalkane as the one reacting via S_N1.

There is a very large increase in ionization energy between the 5th and the 6th ionization energies (from 6274 to 21269 \(\text{kJ mol}^{-1}\)). This indicates that the 6th electron is removed from an inner quantum shell, meaning there are 5 electrons in the outermost shell. Therefore, the element belongs to Group 15 (Group 5). Since it is in Period 3, the element is phosphorus.

[1] C - Phosphorus. Award 1 mark for the correct option. Reject other options because they belong to other groups (Mg in Group 2, Si in Group 14, S in Group 16) which would show jumps after 2, 4, and 6 electrons respectively.

First, calculate the amount in moles of \(\text{CO}_2\) gas: \(n(\text{CO}_2) = \frac{1.80\text{ dm}^3}{24.0\text{ dm}^3\text{ mol}^{-1}} = 0.0750\text{ mol}\). Second, determine the moles of \(\text{NaHCO}_3\) required using the stoichiometric ratio (2:1): \(n(\text{NaHCO}_3) = 2 \times 0.0750\text{ mol} = 0.150\text{ mol}\). Third, calculate the mass of \(\text{NaHCO}_3\): \(m = 0.150\text{ mol} \times 84.0\text{ g mol}^{-1} = 12.6\text{ g}\).

[1] C - 12.6 g. Correct application of gas volume to moles conversion and stoichiometry (2:1 ratio). Incorrect options: A represents a 1:2 ratio; B represents a 1:1 ratio; D represents doubling the final correct mass.

Let's analyze the bond angles of each molecule using Electron Pair Repulsion Theory: \(\text{BeCl}_2\) has 2 bonding pairs and 0 lone pairs, so it is linear with a bond angle of \(180^\circ\). \(\text{BCl}_3\) has 3 bonding pairs and 0 lone pairs, so it is trigonal planar with a bond angle of \(120^\circ\). \(\text{CCl}_4\) has 4 bonding pairs and 0 lone pairs, so it is tetrahedral with a bond angle of \(109.5^\circ\). \(\text{PCl}_3\) has 3 bonding pairs and 1 lone pair, so the shape is trigonal pyramidal. Because lone pairs repel more than bonding pairs, the bond angle is reduced below the tetrahedral angle of \(109.5^\circ\) to approximately \(100^\circ\) to \(107^\circ\). Therefore, \(\text{PCl}_3\) has the smallest bond angle.

[1] D - PCl3. Award 1 mark for the correct option. Distractors represent linear (180), trigonal planar (120), and tetrahedral (109.5) shapes respectively.

In free radical substitution: Option A is an initiation step where homolytic fission occurs under UV light. Option B is incorrect as a free hydrogen radical (\(\text{H}\cdot\)) is not formed in propagation. Option C is a correct propagation step, where a chlorine radical abstracts a hydrogen atom from butane to form a butyl radical and hydrogen chloride. Option D is a termination step where two radicals combine to form a stable molecule.

[1] C - correct propagation equation. A is initiation, B is chemically incorrect, D is termination.

(a) The relative atomic mass is defined as the weighted average mass of an atom of an element relative to one-twelfth of the mass of a carbon-12 atom.

(b) Relative atomic mass:
\(A_r = \frac{(24 \times 78.99) + (25 \times 10.00) + (26 \times 11.01)}{100} = \frac{1895.76 + 250.00 + 286.26}{100} = \frac{2432.02}{100} = 24.3202 \approx 24.32\)
Comparing with the Periodic Table, the element is Magnesium (\(\text{Mg}\)).

(c) Phosphorus has the outer electron configuration \(3s^2 3p^3\) with three singly occupied 3p orbitals. Sulfur has the configuration \(3s^2 3p^4\) with one doubly occupied 3p orbital. The repulsion between the two electrons in the same 3p orbital (spin-pair repulsion) makes it easier to remove the outer electron from sulfur than from the half-filled p-subshell of phosphorus.

(d) The third ionization energy represents the removal of the third electron from gaseous \(\text{Mg}^{2+}\) ions:
\(\text{Mg}^{2+}(\text{g}) \rightarrow \text{Mg}^{3+}(\text{g}) + \text{e}^-\)

(e) The third electron in magnesium is removed from the \(2\text{p}\) subshell (an inner shell), whereas the first two electrons were removed from the outer \(3\text{s}\) subshell. The \(2\text{p}\) electrons are much closer to the nucleus, experience significantly less shielding, and are therefore held much more strongly by electrostatic attraction.

(a)
- MP1: Weighted average mass of an atom of an element (1)
- MP2: Relative to 1/12th of the mass of an atom of carbon-12 (1)

(b)
- MP1: Correct expression for calculation: \(\frac{(24 \times 78.99) + (25 \times 10.00) + (26 \times 11.01)}{100}\) (1)
- MP2: Value calculated as 24.32 (must be 2 decimal places) (1)
- MP3: Identity is Magnesium / Mg (1)

(c)
- MP1: Phosphorus has 3p subshell with singly occupied orbitals AND sulfur has a paired electron in a 3p orbital (1)
- MP2: Reference to spin-pair repulsion in sulfur (1)
- MP3: Repulsion makes it easier to remove the electron in sulfur / requires less energy (1)

(d)
- MP1: Correct species: \(\text{Mg}^{2+}\) as reactant and \(\text{Mg}^{3+} + \text{e}^-\)\ (or similar) as products (1)
- MP2: Correct state symbols: \((\text{g})\) on both \(\text{Mg}^{2+}\) and \(\text{Mg}^{3+}\) (1)

(e)
- MP1: Electron is removed from an inner shell / \(2\text{p}\) subshell / shell closer to the nucleus (1)
- MP2: Experienced less shielding / stronger electrostatic attraction from the nucleus (1)

(a)(i) \(\text{BF}_3\) has 3 bonding pairs of electrons and 0 lone pairs around the central boron atom. To minimize repulsion, these electron pairs move as far apart as possible, resulting in a trigonal planar shape with a bond angle of \(120^\circ\).
(ii) \(\text{NF}_3\) has 3 bonding pairs and 1 lone pair of electrons around the central nitrogen atom. This gives a tetrahedral arrangement of electron pairs, but because lone pairs repel more strongly than bonding pairs, the bond angle is compressed to \(107^\circ\) (accept \(106^\circ - 108^\circ\)) and the shape is trigonal pyramidal.

(b) Both molecules contain highly polar bonds because fluorine is much more electronegative than boron and nitrogen. However, \(\text{BF}_3\) is highly symmetrical (trigonal planar), which causes the individual bond dipoles to cancel out, resulting in a net dipole moment of zero. In contrast, \(\text{NF}_3\) is unsymmetrical (trigonal pyramidal due to the lone pair on the nitrogen), so the bond dipoles do not cancel out, leading to a permanent net molecular dipole.

(c) In graphite, each carbon atom is covalently bonded to three others in planar hexagonal sheets, leaving one delocalized electron per carbon atom. These delocalized electrons are free to move between the layers and conduct electricity. In diamond, each carbon atom is covalently bonded to four other carbon atoms in a giant tetrahedral network. All valence electrons are fixed/localized in localized covalent bonds, meaning there are no free-moving charged particles to conduct electricity.

(a)(i)
- MP1: Trigonal planar shape and \(120^\circ\) bond angle (1)
- MP2: Boron has 3 bonding pairs and 0 lone pairs (1)
- MP3: Electron pairs repel equally to minimize repulsion / maximize separation (1)

(a)(ii)
- MP1: Trigonal pyramidal shape and \(107^\circ\) (allow \(106^\circ - 108^\circ\)) bond angle (1)
- MP2: Nitrogen has 3 bonding pairs and 1 lone pair (1)
- MP3: Lone pair-bonding pair repulsion is greater than bonding pair-bonding pair repulsion (1)

(b)
- MP1: State that both molecules have polar bonds because of the difference in electronegativity between F and B/N (1)
- MP2: \(\text{BF}_3\) is symmetrical, so bond dipoles cancel (1)
- MP3: \(\text{NF}_3\) is unsymmetrical, so bond dipoles do not cancel / net permanent dipole exists (1)

(c)
- MP1: Graphite conducts electricity but diamond does not (1)
- MP2: In graphite, each carbon has 3 bonds and 1 delocalized electron free to move along layers (1)
- MP3: In diamond, each carbon has 4 bonds, all electrons are localized in covalent bonds / no delocalized electrons (1)

(a)(i) Divide the percentages by the respective relative atomic masses:
- Carbon: \(\frac{85.7}{12.0} = 7.142\text{ mol}\)
- Hydrogen: \(\frac{14.3}{1.0} = 14.300\text{ mol}\)

Divide by the smallest value (7.142):
- Ratio of \(\text{C} : \text{H} = 1 : 2\)
- Empirical formula is \(\text{CH}_2\).

(a)(ii) Use the ideal gas equation: \(PV = nRT\)
Rearrange to solve for \(n\):
\(n = \frac{PV}{RT} = \frac{101 \times 10^3\text{ Pa} \times 8.98 \times 10^{-4}\text{ m}^3}{8.31\text{ J mol}^{-1}\text{ K}^{-1} \times 298\text{ K}} = \frac{90.698}{2476.38} = 0.036626\text{ mol}\)

Now find the molar mass (\(M\)):
\(M = \frac{\text{mass}}{n} = \frac{1.55\text{ g}}{0.036626\text{ mol}} = 42.32\text{ g mol}^{-1}\)

The empirical formula mass of \(\text{CH}_2 = 12.0 + 2(1.0) = 14.0\text{ g mol}^{-1}\).
Determine the molecular formula multiple:
\(\frac{42.32}{14.0} \approx 3\)
Therefore, the molecular formula is \(\text{C}_3\text{H}_6\).

(b)(i) Atom economy is calculated as:
\(\text{\% Atom Economy} = \frac{\text{Molar mass of desired product}}{\text{Total molar mass of all reactants}} \times 100\)
- Molar mass of reactants: \(2 \times \text{NH}_3 + \text{NaClO} = (2 \times 17.0) + 74.5 = 108.5\text{ g mol}^{-1}\)
- Molar mass of desired product (\(\text{N}_2\text{H}_4\)) = \(32.0\text{ g mol}^{-1}\)
- \(\text{\% Atom Economy} = \frac{32.0}{108.5} \times 100 = 29.49\% \approx 29.5\%\)

(b)(ii) Calculate initial moles of \(\text{NH}_3\):
\(n(\text{NH}_3) = \text{concentration} \times \text{volume} = 2.50\text{ mol dm}^{-3} \times 0.100\text{ dm}^3 = 0.250\text{ mol}\)

From the equation, 2 moles of \(\text{NH}_3\) produce 1 mole of \(\text{N}_2\text{H}_4\):
\(n(\text{N}_2\text{H}_4)_{\text{theoretical}} = \frac{0.250}{2} = 0.125\text{ mol}\)

Theoretical mass of hydrazine:
\(m_{\text{theoretical}} = 0.125\text{ mol} \times 32.0\text{ g mol}^{-1} = 4.00\text{ g}\)

Percentage yield:
\(\text{\% Yield} = \frac{\text{Actual Mass}}{\text{Theoretical Mass}} \times 100 = \frac{3.25\text{ g}}{4.00\text{ g}} \times 100 = 81.25\% \approx 81.3\%\)

(a)(i)
- MP1: Calculates moles of C and H correctly (7.14 and 14.3) (1)
- MP2: Deduces empirical formula \(\text{CH}_2\) (1)

(a)(ii)
- MP1: Correctly substitutes values into ideal gas equation: \(n = \frac{101000 \times 8.98 \times 10^{-4}}{8.31 \times 298}\) (1)
- MP2: Obtains moles \(n = 0.0366\text{ mol}\) (1)
- MP3: Obtains molar mass of approximately \(42.3\text{ g mol}^{-1}\) (allow \(42.0 - 43.0\)) (1)
- MP4: Determines molecular formula \(\text{C}_3\text{H}_6\) based on empirical formula mass (1)

(b)(i)
- MP1: Calculates total mass of reactants = \(2(17.0) + 74.5 = 108.5\) (1)
- MP2: Expresses atom economy fraction: \(\frac{32.0}{108.5}\) (1)
- MP3: Evaluates to \(29.5\%\) (allow \(29.49\%\)) (1)

(b)(ii)
- MP1: Calculates moles of ammonia as \(0.250\text{ mol}\) and theoretical moles of hydrazine as \(0.125\text{ mol}\) (1)
- MP2: Calculates theoretical mass of hydrazine as \(4.00\text{ g}\) (1)
- MP3: Calculates percentage yield as \(81.3\%\) (allow \(81.25\%\)) (1)

(a)(i) Heat energy calculation:
\(q = m c \Delta T\)
\(m = 150\text{ g}\)
\(c = 4.18\text{ J g}^{-1}\ ^\circ\text{C}^{-1}\)
\(\Delta T = 58.5 - 21.5 = 37.0\ ^\circ\text{C}\)

\(q = 150 \times 4.18 \times 37.0 = 23200\text{ J} = 23.2\text{ kJ}\) (or \(23.199\text{ kJ}\))

(a)(ii) Moles of ethanol burned:
\(n = \frac{\text{mass}}{M_r} = \frac{0.920\text{ g}}{46.0\text{ g mol}^{-1}} = 0.0200\text{ mol}\)

Enthalpy of combustion:
\(\Delta H = -\frac{q}{n} = -\frac{23.199\text{ kJ}}{0.0200\text{ mol}} = -1160\text{ kJ mol}^{-1}\) (to 3 significant figures)

(a)(iii) Common reasons for the discrepancies in simple colorimetry experiments include:
- Heat loss to the air / surrounding copper calorimeter.
- Incomplete combustion of ethanol (forming carbon monoxide or carbon / soot).
- Evaporation of ethanol from the wick while burning.
- Non-standard conditions.

(b) Target equation for formation of pentane:
\(5\text{C(s)} + 6\text{H}_2(\text{g}) \rightarrow \text{C}_5\text{H}_{12}(\text{l})\)

Using a Hess's Law cycle based on combustion data:
\(\Delta H_f^\theta = \sum \Delta H_c^\theta(\text{reactants}) - \sum \Delta H_c^\theta(\text{products})\)
\(\Delta H_f^\theta = [5 \times \Delta H_c^\theta(\text{C(s)}) + 6 \times \Delta H_c^\theta(\text{H}_2(\text{g}))] - [\Delta H_c^\theta(\text{C}_5\text{H}_{12}(\text{l}))]\)

Substitute the values:
\(\Delta H_f^\theta = [5 \times (-393.5) + 6 \times (-285.8)] - [-3509.0]\)
\(\Delta H_f^\theta = [-1967.5 - 1714.8] + 3509.0\)
\(\Delta H_f^\theta = -3682.3 + 3509.0 = -173.3\text{ kJ mol}^{-1}\)

(a)(i)
- MP1: Correctly calculates temperature difference \(\Delta T = 37.0\ ^\circ\text{C}\) (1)
- MP2: Calculates heat transfer correctly: \(q = 23.2\text{ kJ}\) (or \(23200\text{ J}\)) (1)

(a)(ii)
- MP1: Calculates moles of ethanol = \(0.0200\text{ mol}\) (1)
- MP2: Divides energy by moles (e.g., \(\frac{23.2}{0.0200}\)) (1)
- MP3: Evaluation to \(-1160\text{ kJ mol}^{-1}\) (must have negative sign and be 3 s.f.) (1)

(a)(iii)
- MP1: Heat lost to surroundings / beaker (1)
- MP2: Incomplete combustion of fuel (1)
(Accept: evaporation of alcohol from wick)

(b)
- MP1: Writes correct target equation for enthalpy of formation: \(5\text{C(s)} + 6\text{H}_2(\text{g}) \rightarrow \text{C}_5\text{H}_{12}(\text{l})\) (1)
- MP2: Constructs cycle or writes algebraic expression: \(\Delta H_f = 5\Delta H_c(\text{C}) + 6\Delta H_c(\text{H}_2) - \Delta H_c(\text{pentane})\) (1)
- MP3: Correctly multiplies values: \(5 \times (-393.5) = -1967.5\) AND \(6 \times (-285.8) = -1714.8\) (1)
- MP4: Sets up full subtraction: \(-3682.3 - (-3509.0)\) (1)
- MP5: Final answer of \(-173.3\text{ kJ mol}^{-1}\) (including minus sign and correct units) (1)

(a) The thermal stability of Group 2 nitrates increases down the group. As you go down the group, the ionic radius of the Group 2 cation (\(M^{2+}\)) increases while its charge remains the same (\(+2\)). This results in a decrease in charge density of the cation down the group. Consequently, the larger cation has a weaker polarizing effect on the nitrate anion, distorting its electron cloud and the \(\text{N}-\text{O}\) bonds to a lesser extent. Since the nitrate ion is less polarized, more heat energy is required to decompose the nitrate down the group.

(b) Thermal decomposition of calcium nitrate produces calcium oxide, nitrogen dioxide, and oxygen gas:
\(2\text{Ca(NO}_3)_2(\text{s}) \rightarrow 2\text{CaO(s)} + 4\text{NO}_2(\text{g}) + \text{O}_2(\text{g})\)

(c)(i) The solution changes from colorless to brown (due to the formation of aqueous iodine, \(\text{I}_2(\text{aq})\)).
(ii) Ionic equation:
\(\text{Cl}_2(\text{aq}) + 2\text{I}^-(\text{aq}) \rightarrow 2\text{Cl}^-(\text{aq}) + \text{I}_2(\text{aq})\)
(iii) Chlorine is a smaller atom than iodine, with less electron shielding. This means chlorine has a greater attraction for electrons and is a stronger oxidizing agent than iodine. Therefore, chlorine can displace/oxidize iodide ions into iodine.
(iv) Reaction of chlorine with cold, dilute NaOH:
\(\text{Cl}_2(\text{aq}) + 2\text{NaOH}(\text{aq}) \rightarrow \text{NaCl}(\text{aq}) + \text{NaClO}(\text{aq}) + \text{H}_2\text{O}(\text{l})\)

The oxidation state of chlorine changes from \(0\) in \(\text{Cl}_2\) to \(-1\) in \(\text{NaCl}\) (reduction) and \(+1\) in \(\text{NaClO}\) (oxidation).

(a)
- MP1: Trend: Thermal stability increases down the group (1)
- MP2: Ionic radius of \(M^{2+}\) cation increases / charge density decreases (1)
- MP3: Polarizing power of the cation decreases / polarization of nitrate anion decreases (1)
- MP4: Less weakening of the \(\text{N}-\text{O}\) bond, so more energy is required to break it (1)

(b)
- MP1: Correct formulae for products (\(\text{CaO}\), \(\text{NO}_2\), \(\text{O}_2\)) and reactants (1)
- MP2: Balanced equation and correct state symbols: \(2\text{Ca(NO}_3)_2(\text{s}) \rightarrow 2\text{CaO(s)} + 4\text{NO}_2(\text{g}) + \text{O}_2(\text{g})\) (1)

(c)(i)
- MP1: Colorless to brown / orange / yellow (1)

(c)(ii)
- MP1: \(\text{Cl}_2 + 2\text{I}^- \rightarrow 2\text{Cl}^- + \text{I}_2\) (ignore state symbols) (1)

(c)(iii)
- MP1: Chlorine is a stronger oxidizing agent than iodine (or chlorine is more easily reduced) (1)
- MP2: Because chlorine has a smaller atomic radius / less shielding, so there is a stronger attraction for incoming electrons (1)

(c)(iv)
- MP1: Correct balanced chemical equation: \(\text{Cl}_2 + 2\text{NaOH} \rightarrow \text{NaCl} + \text{NaClO} + \text{H}_2\text{O}\) (1)
- MP2: State that the oxidation state of chlorine changes from 0 to -1 and +1 (1)