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To answer this question, let us analyze the factors influencing the first ionisation energies of the pairs:

* **Mg and Al**: The first ionisation energy of \( \text{Mg} \) (\( 1s^2 2s^2 2p^6 3s^2 \)) is higher than that of \( \text{Al} \) (\( 1s^2 2s^2 2p^6 3s^2 3p^1 \)) because the electron removed from \( \text{Al} \) is in a higher energy, more shielded \( 3p \) orbital than the \( 3s \) orbital of \( \text{Mg} \).
* **N and C**: The first ionisation energy of \( \text{N} \) is higher than that of \( \text{C} \) primarily due to the greater nuclear charge of nitrogen with a similar amount of shielding.
* **P and S**: \( \text{P} \) has the electronic configuration \( [\text{Ne}]\,3s^2 3p^3 \) (three unpaired electrons in \( 3p \) orbitals). \( \text{S} \) has the configuration \( [\text{Ne}]\,3s^2 3p^4 \). The fourth \( 3p \) electron in sulfur must pair up in one of the \( 3p \) orbitals. This pairing results in mutual spin-pair repulsion, making this electron easier to remove than one of the unpaired electrons in phosphorus. Therefore, phosphorus has a higher first ionisation energy than sulfur due to spin-pair repulsion in a \( p \)-orbital.
* **S and Cl**: The first ionisation energy of \( \text{S} \) is lower than that of \( \text{Cl} \) because chlorine has a higher nuclear charge.

Award 1 mark for the correct option C.
* Reject A: Magnesium and aluminium difference is due to subshell energy differences (s vs p).
* Reject B: Nitrogen and carbon difference is due to nuclear charge.
* Reject D: Sulfur has a lower, not higher, first ionisation energy than chlorine.

To find the standard enthalpy change of formation of propan-1-ol, we construct a Hess cycle based on the standard enthalpy changes of combustion.

The chemical equation for the formation of propan-1-ol is:
\( 3\text{C(s)} + 4\text{H}_2(\text{g}) + \frac{1}{2}\text{O}_2(\text{g}) \rightarrow \text{C}_3\text{H}_7\text{OH}(\text{l}) \)

According to Hess's law:
\( \Delta H_{\text{f}}^\theta [\text{C}_3\text{H}_7\text{OH}(\text{l})] = \sum \Delta H_{\text{c}}^\theta [\text{reactants}] - \sum \Delta H_{\text{c}}^\theta [\text{products}] \)
\( \Delta H_{\text{f}}^\theta [\text{C}_3\text{H}_7\text{OH}(\text{l})] = [3 \times \Delta H_{\text{c}}^\theta [\text{C(s)}] + 4 \times \Delta H_{\text{c}}^\theta [\text{H}_2(\text{g})]] - \Delta H_{\text{c}}^\theta [\text{C}_3\text{H}_7\text{OH}(\text{l})] \)

Substitute the given values:
\( \Delta H_{\text{f}}^\theta = [3 \times (-393.5) + 4 \times (-285.8)] - (-2021.0) \)
\( \Delta H_{\text{f}}^\theta = [-1180.5 - 1143.2] + 2021.0 \)
\( \Delta H_{\text{f}}^\theta = -2323.7 + 2021.0 = -302.7\text{ kJ mol}^{-1} \)

Award 1 mark for the correct option B.
* Incorrect option A (+302.7) arises from reversing the sign of the overall calculation.
* Incorrect option C (-2700.3) is obtained by adding the three given values directly without multiplying by stoichiometric coefficients.
* Incorrect option D (-4344.7) arises from adding the combustion of propan-1-ol rather than subtracting it.

We use the ideal gas equation: \( pV = nRT = \frac{m}{M_{\text{r}}}RT \)

First, convert all quantities to SI units:
* Temperature, \( T = 99.0 + 273.15 = 372.15\text{ K} \) (using \( 372\text{ K} \) is also acceptable and yields the same result)
* Pressure, \( p = 101\text{ kPa} = 101 \times 10^3\text{ Pa} \)
* Volume, \( V = 84.0\text{ cm}^3 = 84.0 \times 10^{-6}\text{ m}^3 \)
* Mass, \( m = 0.281\text{ g} \)

Rearranging the ideal gas equation to solve for \( M_{\text{r}} \):
\( M_{\text{r}} = \frac{mRT}{pV} \)
\( M_{\text{r}} = \frac{0.281 \times 8.31 \times 372}{(101 \times 10^3) \times (84.0 \times 10^{-6})} \)
\( M_{\text{r}} = \frac{868.31}{8.484} \approx 102.35 \approx 102 \)

Award 1 mark for the correct option D.
* Option A (27.2) is obtained by failing to convert the temperature from Celsius to Kelvin.
* Option C (84.0) is a distractor matching the volume numerical value.

The thermal stability of Group 2 carbonates increases down the group as the cationic radius increases. A larger cation has a lower charge density and a weaker polarizing effect on the carbonate ion's electron cloud, making the carbonate more thermally stable.

* Magnesium carbonate, \( \text{MgCO}_3 \), decomposes easily at the temperature of a Bunsen burner flame (approx. \( 500-600\text{ }^\circ\text{C} \)) to form magnesium oxide and carbon dioxide:
\( \text{MgCO}_3(\text{s}) \rightarrow \text{MgO}(\text{s}) + \text{CO}_2(\text{g}) \)
* Barium carbonate, \( \text{BaCO}_3 \), is extremely thermally stable due to the large size and low polarizing power of the \( \text{Ba}^{2+} \) ion. It does not decompose at Bunsen burner temperatures (it requires temperatures above \( 1200\text{ }^\circ\text{C} \)).

Therefore, only \( \text{MgCO}_3 \) decomposes, and the solid residue left in the crucible is a mixture of \( \text{MgO} \) and undecomposed \( \text{BaCO}_3 \).

Award 1 mark for the correct option B.
* Reject A: Magnesium carbonate decomposes under strong heating.
* Reject C: Barium carbonate does not decompose under ordinary laboratory Bunsen heating.
* Reject D: This describes the incorrect reverse stability trend.

First, we determine the number of structural isomers of aldehydes of formula \( \text{C}_5\text{H}_{10}\text{O} \). Since it is an aldehyde, it must contain a \( -\text{CHO} \) group. The remaining alkyl group must have 4 carbon atoms (\( -\text{C}_4\text{H}_9 \)).

There are 4 different butyl isomers, which give rise to 4 aldehyde structural isomers:
1. **Pentanal**: \( \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{CHO} \) (no chiral carbon)
2. **2-Methylbutanal**: \( \text{CH}_3\text{CH}_2\text{C}^*\text{H(CH}_3)\text{CHO} \). The C2 carbon is bonded to four different groups: \( -\text{H} \), \( -\text{CH}_3 \), \( -\text{CH}_2\text{CH}_3 \), and \( -\text{CHO} \). Hence, it contains a chiral carbon.
3. **3-Methylbutanal**: \( (\text{CH}_3)_2\text{CHCH}_2\text{CHO} \) (no chiral carbon)
4. **2,2-Dimethylpropanal**: \( (\text{CH}_3)_3\text{CCHO} \) (no chiral carbon)

Therefore, there are exactly 4 structural isomers of aldehydes, and only 1 of these (2-methylbutanal) contains a chiral carbon atom.

Award 1 mark for the correct option A.
* Reject options C and D as there are only 4 aldehyde structural isomers of this formula.
* Reject option B because only 1 of the 4 isomers has a chiral carbon.

Nucleophilic addition of HCN to a carbonyl compound forms a hydroxynitrile. Let us look at the products of each reaction:

* **propanone + HCN**: forms \( (\text{CH}_3)_2\text{C(OH)(CN)} \) (2-hydroxy-2-methylpropanenitrile). The central carbon is attached to two identical methyl groups, so it is not chiral.
* **pentan-3-one + HCN**: forms \( (\text{CH}_3\text{CH}_2)_2\text{C(OH)(CN)} \) (2-ethyl-2-hydroxybutanenitrile). The central carbon is attached to two identical ethyl groups, so it is not chiral.
* **methanal + HCN**: forms \( \text{CH}_2\text{(OH)(CN)} \) (hydroxyacetonitrile). The central carbon is attached to two identical hydrogen atoms, so it is not chiral.
* **butanone + HCN**: forms \( \text{CH}_3\text{-C}^*\text{(OH)(CN)-CH}_2\text{CH}_3 \) (2-hydroxy-2-methylbutanenitrile). The central carbon is chiral because it is attached to four distinct groups: a methyl group (\( -\text{CH}_3 \)), an ethyl group (\( -\text{CH}_2\text{CH}_3 \)), a hydroxyl group (\( -\text{OH} \)), and a nitrile group (\( -\text{CN} \)).

Award 1 mark for the correct option D.
* Reject A: The product has two identical methyl groups.
* Reject B: The product has two identical ethyl groups.
* Reject C: The product has two identical hydrogen atoms.

We analyze the behaviour of Period 3 chlorides with water:

* **Sodium chloride (\( \text{NaCl} \))**: dissolves in water to form a neutral solution of hydrated sodium and chloride ions (pH 7).
* **Magnesium chloride (\( \text{MgCl}_2 \))**: undergoes very slight hydrolysis to yield a weakly acidic solution (pH 6.5).
* **Aluminium chloride (\( \text{AlCl}_3 \))**: undergoes substantial hydrolysis to form an acidic solution (pH 3).
* **Silicon tetrachloride (\( \text{SiCl}_4 \))**: reacts vigorously with water (complete hydrolysis) to form silicon dioxide and dense white fumes of hydrogen chloride gas, resulting in a strongly acidic solution (pH 1.5–2.0).

Therefore, \( X \) is sodium and \( Y \) is silicon.

Award 1 mark for the correct option C.
* Reject A: Magnesium chloride does not react vigorously to release white fumes and has a pH around 6.5, not 2.
* Reject B: Magnesium chloride does not form a pH 7 solution.
* Reject D: Magnesium chloride does not form a pH 7 solution.

1. **Determine the number of carbon atoms (\( n \)) using the \( [M+1] \) peak ratio:**
Using the standard formula:
\( n = \frac{100}{1.1} \times \frac{\text{Abundance of } [M+1]^+}{\text{Abundance of } [M]^+} \)
Given that \( [M]^+ : [M+1]^+ = 100 : 4.4 \):
\( n = \frac{100}{1.1} \times \frac{4.4}{100} = 4 \)
There are 4 carbon atoms in a molecule of \( Z \).

2. **Determine the number of chlorine atoms using the \( [M+2] \) peak ratio:**
The \( [M]^+ : [M+2]^+ \) ratio of \( 3 : 1 \) is characteristic of a compound containing exactly one chlorine atom (due to the \( 3:1 \) natural abundance ratio of \( ^{35}\text{Cl} \) to \( ^{37}\text{Cl} \)).

3. **Find the molecular formula:**
Since the compound is a saturated monochloroalkane containing 4 carbon atoms, its general formula is \( \text{C}_n\text{H}_{2n+1}\text{Cl} \). For \( n = 4 \), the molecular formula is \( \text{C}_4\text{H}_9\text{Cl} \).

Award 1 mark for the correct option C.
* Reject A: This formula contains only 3 carbon atoms.
* Reject B: This represents an unsaturated compound (alkene or ring) with formula \( \text{C}_4\text{H}_7\text{Cl} \).
* Reject D: This formula contains 5 carbon atoms.

The successive ionisation energies show a very large jump between the 4th and 5th ionisation energies (from 4356 to 16091 \\text{kJ mol}^{-1}). This indicates that the outer shell contains 4 electrons, so element \\text{X} is in Group 14. In Period 3, Group 14 is silicon (\\text{Si}). Silicon tetrachloride, \\text{SiCl}_4, is a simple molecular covalent chloride that hydrolyses violently in water to form silicon dioxide and hydrogen chloride, producing a strongly acidic solution: \\text{SiCl}_4(\\text{l}) + 2\\text{H}_2\\text{O}(\\text{l}) \\rightarrow \\text{SiO}_2(\\text{s}) + 4\\text{HCl}(\\text{aq}). Therefore, statement C is correct. Option A is incorrect because silicon is a metalloid/semiconductor, not a highly conductive metal. Option B is incorrect because silicon dioxide (\\text{SiO}_2) has a giant covalent structure. Option D is incorrect because silicon has the outer electronic configuration 3\\text{s}^2 3\\text{p}^2.

1 mark for identifying the correct option C based on the massive jump between the 4th and 5th ionisation energies indicating Group 14 (Silicon).

Using the ideal gas equation, pV = nRT, where n = m / M_r. Rearranging this gives M_r = mRT / (pV). Convert the given values to SI units: mass m = 0.370 g; temperature T = 350 K; pressure p = 1.00 \\times 10^5 Pa; volume V = 145 cm^3 = 145 \\times 10^{-6} m^3. Substituting these values: M_r = (0.370 \\times 8.31 \\times 350) / (1.00 \\times 10^5 \\times 145 \\times 10^{-6}) = 1076.145 / 14.5 = 74.2. This is closest to 74.0, which corresponds to option C.

1 mark for the correct calculation of M_r using the ideal gas equation with appropriate SI unit conversions, choosing option C.

The chemical equation for the formation of propanone is: 3\\text{C(s)} + 3\\text{H}_2\\text{(g)} + 0.5\\text{O}_2\\text{(g)} \\rightarrow \\text{CH}_3\\text{COCH}_3\\text{(l)}. Using Hess's law and enthalpies of combustion: \\Delta H_{\\text{f}}^{\\theta} = 3 \\times \\Delta H_{\\text{c}}^{\\theta}[\\text{C(s)}] + 3 \\times \\Delta H_{\\text{c}}^{\\theta}[\\text{H}_2\\text{(g)}] - \\Delta H_{\\text{c}}^{\\theta}[\\text{CH}_3\\text{COCH}_3\\text{(l)}] = 3(-394) + 3(-286) - (-1786) = -1182 - 858 + 1786 = -2040 + 1786 = -254 kJ mol^{-1}. Therefore, option A is correct.

1 mark for correctly applying Hess's Law with correct stoichiometry of reactants and products, obtaining -254 kJ mol^{-1} (Option A).

To balance the redox equation: Chromium in Cr_2O_7^{2-} is in the +6 oxidation state and is reduced to Cr^{3+} (+3 state). Since there are 2 Cr atoms, the total decrease in oxidation state is 2 \\times 3 = 6. Sulfur in SO_2 is in the +4 oxidation state and is oxidised to SO_4^{2-} (+6 state), an increase of 2. To balance the electron transfer, 3 moles of SO_2 must react with 1 mole of Cr_2O_7^{2-}, so a = 3 and c = 3. The unbalanced equation becomes: Cr_2O_7^{2-} + 3SO_2 + bH^+ -> 2Cr^{3+} + 3SO_4^{2-} + dH_2O. Balancing charges: left side has charge (-2) + b(+); right side has charge 2(+3) + 3(-2) = 0. Therefore, -2 + b = 0, which gives b = 2. This is perfectly balanced with d = 1. Thus, option A is correct.

1 mark for correctly balancing the redox equation via oxidation states or half-equations to find the H^+ coefficient of 2 (Option A).

Down Group 2 from magnesium to barium: The solubility of sulfates decreases (magnesium sulfate is highly soluble, while barium sulfate is insoluble), making statement D correct. First ionisation energy decreases because of increased atomic radius and shielding, making A incorrect. The solubility of hydroxides increases down the group, making B incorrect. The thermal stability of carbonates increases down the group (due to decreasing polarising power of the larger cation), so they decompose less easily, making C incorrect.

1 mark for identifying the correct trend of decreasing solubility of Group 2 sulfates down the group (Option D).

The structural isomers with the molecular formula C_3H_6Cl_2 are: (1) 1,1-dichloropropane: Cl_2CH-CH_2-CH_3, (2) 1,2-dichloropropane: ClCH_2-CH(Cl)-CH_3, (3) 1,3-dichloropropane: ClCH_2-CH_2-CH_2Cl, (4) 2,2-dichloropropane: CH_3-CCl_2-CH_3. No other structural arrangements of the atoms are possible. Since the question asks specifically for structural isomers, we do not count optical stereoisomers. Therefore, there are exactly 4 structural isomers, which corresponds to option B.

1 mark for correctly identifying all 4 structural isomers without duplicates or stereoisomer counting, choosing option B.

Compound X reacts with 2,4-DNPH, showing it contains a carbonyl group (aldehyde or ketone). It does not react with Tollens' reagent, showing it is a ketone, not an aldehyde. Since X contains 4 carbon atoms, X must be butanone (CH_3COCH_2CH_3). When butanone is reduced with NaBH_4, the carbonyl group is converted into a secondary alcohol group, forming butan-2-ol (CH_3CH(OH)CH_2CH_3). Thus, compound Y is butan-2-ol, which corresponds to option B.

1 mark for identifying X as butanone and its reduction product Y as butan-2-ol (Option B).

When potassium chloride (KCl) reacts with concentrated sulfuric acid, only hydrogen chloride (HCl) gas is produced via an acid-base reaction: KCl(s) + H_2SO_4(l) -> KHSO_4(s) + HCl(g). HCl is an acidic gas that reacts with ammonia to form dense white fumes of NH_4Cl(s). Since chloride ions cannot be oxidized by concentrated sulfuric acid, no other gases (such as sulfur dioxide, which is a reducing agent and would decolorize potassium manganate(VII)) are produced. In contrast, bromide and iodide ions reduce sulfuric acid to produce SO_2 and other products, meaning multiple gases are evolved. Therefore, Y^- is Cl^-, which is option B.

1 mark for recognizing that chloride undergoes only an acid-base reaction with concentrated sulfuric acid to yield a single non-reducing acidic gas (Option B).

There is a very large increase between the sixth (\(8500\,\text{kJ}\,\text{mol}^{-1}\)) and seventh (\(27100\,\text{kJ}\,\text{mol}^{-1}\)) ionisation energies. This indicates that the seventh electron is removed from an inner quantum shell, meaning there are 6 electrons in the outer shell of \(X\). Therefore, \(X\) is sulfur (\(\text{S}\)).

- Option A is incorrect because the highest oxide of sulfur is sulfur trioxide (\(\text{SO}_3\)), which reacts with water to form sulfuric acid, a strongly acidic solution.
- Option B is correct because sulfur (\(1s^2 2s^2 2p^6 3s^2 3p^4\)) is immediately preceded by phosphorus (\(1s^2 2s^2 2p^6 3s^2 3p^3\)) in Period 3. Sulfur has a lower first ionisation energy than phosphorus because the electron removed from sulfur comes from a paired \(3p\) orbital, where spin-pair repulsion makes it easier to remove.
- Option C is incorrect because atomic radius decreases across Period 3, meaning sodium has a much larger atomic radius than sulfur.
- Option D is incorrect because sulfur chlorides (such as \(\text{SCl}_2\) or \(\text{S}_2\text{Cl}_2\)) are simple molecular substances.

1 mark for the correct option B.

We use the ideal gas equation: \(pV = nRT\).

Since \(n = \frac{m}{M}\), where \(m\) is mass and \(M\) is the molar mass:
\(pV = \frac{m}{M}RT \implies M = \frac{m}{V} \frac{RT}{p} = \rho \frac{RT}{p}\)

Convert density to SI units:
\(\rho = 1.80\,\text{g}\,\text{dm}^{-3} = 1.80\,\text{kg}\,\text{m}^{-3}\)

Now, substitute the values into the equation:
\(M = \frac{1.80 \times 8.31 \times 310}{1.10 \times 10^5} = 0.04216\,\text{kg}\,\text{mol}^{-1} = 42.2\,\text{g}\,\text{mol}^{-1}\)

Let's check the molar masses of the options:
- \(\text{C}_2\text{H}_6\): \(2 \times 12.0 + 6 \times 1.0 = 30.0\,\text{g}\,\text{mol}^{-1}\)
- \(\text{C}_3\text{H}_6\): \(3 \times 12.0 + 6 \times 1.0 = 42.0\,\text{g}\,\text{mol}^{-1}\)
- \(\text{C}_3\text{H}_8\): \(3 \times 12.0 + 8 \times 1.0 = 44.0\,\text{g}\,\text{mol}^{-1}\)
- \(\text{C}_4\text{H}_{10}\): \(4 \times 12.0 + 10 \times 1.0 = 58.0\,\text{g}\,\text{mol}^{-1}\)

The calculated molar mass corresponds to \(\text{C}_3\text{H}_6\).

1 mark for the correct option B.

The chemical equation for the standard enthalpy change of formation of dimethyl ether is:
\(2\text{C(graphite)} + 3\text{H}_2\text{(g)} + \frac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{CH}_3\text{OCH}_3\text{(g)}

Using Hess's Law with enthalpies of combustion:
\)\Delta H_{\text{f}}^{\ominus} = \sum \Delta H_{\text{c}}^{\ominus}(\text{reactants}) - \sum \Delta H_{\text{c}}^{\ominus}(\text{products})\)

\(\Delta H_{\text{f}}^{\ominus} = [2 \times \Delta H_{\text{c}}^{\ominus}(\text{C}) + 3 \times \Delta H_{\text{c}}^{\ominus}(\text{H}_2)] - \Delta H_{\text{c}}^{\ominus}(\text{CH}_3\text{OCH}_3)\)

Substitute the values:
\(\Delta H_{\text{f}}^{\ominus} = [2 \times (-393.5) + 3 \times (-285.8)] - (-1460.0)\)
\(\Delta H_{\text{f}}^{\ominus} = [-787.0 - 857.4] + 1460.0\)
\(\Delta H_{\text{f}}^{\ominus} = -1644.4 + 1460.0 = -184.4\,\text{kJ}\,\text{mol}^{-1}\)

1 mark for the correct option A.

Both magnesium compounds decompose upon strong heating to produce magnesium oxide, \(\text{MgO}\), as the solid residue:

\(\text{Mg(NO}_3)_2\text{(s)} \rightarrow \text{MgO(s)} + 2\text{NO}_2\text{(g)} + \frac{1}{2}\text{O}_2\text{(g)}\)
\(\text{MgCO}_3\text{(s)} \rightarrow \text{MgO(s)} + \text{CO}_2\text{(g)}\)

Calculate the molar masses of the substances:
- \(M_{\text{r}}(\text{MgO}) = 24.3 + 16.0 = 40.3\,\text{g}\,\text{mol}^{-1}\)
- \(M_{\text{r}}(\text{Mg(NO}_3)_2) = 24.3 + 2 \times (14.0 + 3 \times 16.0) = 148.3\,\text{g}\,\text{mol}^{-1}\)
- \(M_{\text{r}}(\text{MgCO}_3) = 24.3 + 12.0 + 3 \times 16.0 = 84.3\,\text{g}\,\text{mol}^{-1}\)

Let \(w\) be the mass of \(\text{Mg(NO}_3)_2\) in the mixture. Then the mass of \(\text{MgCO}_3\) is \((10.00 - w)\,\text{g}\).

The moles of \(\text{MgO}\) produced is:
\(n(\text{MgO}) = \frac{\text{mass of residue}}{M_{\text{r}}(\text{MgO})} = \frac{3.75}{40.3} = 0.09305\,\text{mol}\)

Using the stoichiometry of the reactions, 1 mole of each reactant produces 1 mole of \(\text{MgO}\):
\(n(\text{MgO}) = \frac{w}{148.3} + \frac{10.00 - w}{84.3} = 0.09305\)

Solve for \(w\):
\(0.006743 w + 0.11862 - 0.011862 w = 0.09305\)
\(-0.005119 w = -0.02557\)
\(w = 4.995\,\text{g} \approx 5.00\,\text{g}\)

1 mark for the correct option B.

The value of the equilibrium constant, \(K_{\text{c}}\), depends solely on the temperature. Therefore, changing concentrations, pressures, or adding a catalyst will not change \(K_{\text{c}}\).

For an exothermic reaction (\(\Delta H < 0\)), decreasing the temperature shifts the equilibrium in the forward (exothermic) direction to release heat. This increases the concentration of products and decreases the concentration of reactants at equilibrium, thus increasing the value of \(K_{\text{c}}\).

1 mark for the correct option A.

The presence of \(M\) and \(M+2\) peaks in a \(3:1\) ratio indicates that the compound contains one chlorine atom (due to the natural abundances of \(^{35}\text{Cl}\) and \(^{37}\text{Cl}\)).

The molecular ion peak is at \(m/z = 78\) (using the \(^{35}\text{Cl}\) isotope). Subtracting the mass of \(^{35}\text{Cl}\):
\(78 - 35 = 43\)

This fragment with \(m/z = 43\) corresponds to the alkyl group, \([\text{C}_3\text{H}_7]^+\) (\(3 \times 12.0 + 7 \times 1.0 = 43\)).

Thus, the compound is chloropropane (\(\text{C}_3\text{H}_7\text{Cl}\)), which has a molecular mass of \(78\,\text{g}\,\text{mol}^{-1}\) (with \(^{35}\text{Cl}\)).

1 mark for the correct option B.

Let's analyse the structures of each option:

- **4-chloropent-2-ene**: \(\text{CH}_3-\text{CH}=\text{CH}-\text{CH(Cl)}-\text{CH}_3\).
- The \(\text{C}=\text{C}\) double bond at C2 has different groups attached to each carbon (\(\text{H}\) and \(\text{CH}_3\) on C2; \(\text{H}\) and \(\text{CH(Cl)CH}_3\) on C3), so it exhibits **cis-trans isomerism**.
- Carbon-4 is a **chiral carbon** because it is bonded to four different groups (\(-\text{H}\), \(-\text{Cl}\), \(-\text{CH}_3\), and \(-\text{CH}=\text{CHCH}_3\)), so it exists as a **pair of enantiomers**.
Therefore, this compound exhibits both forms of isomerism.

- **3-chloropent-2-ene**: \(\text{CH}_3-\text{CH}=\text{C(Cl)}-\text{CH}_2-\text{CH}_3\). It shows cis-trans isomerism, but lacks a chiral carbon, so it does not have enantiomers.

- **1-chloropent-2-ene**: \(\text{ClCH}_2-\text{CH}=\text{CH}-\text{CH}_2-\text{CH}_3\). It shows cis-trans isomerism but has no chiral carbon.

- **3-chlorobut-1-ene**: \(\text{CH}_2=\text{CH}-\text{CH(Cl)}-\text{CH}_3\). It has a chiral carbon (C3), but C1 of the double bond has two identical hydrogen atoms, so it does not show cis-trans isomerism.

1 mark for the correct option A.

Let's analyze the properties:
- Compound \(Z\) with formula \(\text{C}_5\text{H}_{12}\text{O}\) is a saturated acyclic alcohol.
- Acidified potassium dichromate(VI) under reflux oxidizes primary alcohols to carboxylic acids, secondary alcohols to ketones, and does not oxidize tertiary alcohols.
- The product does not react with Tollens' reagent, indicating it is not an aldehyde.
- The product reacts with 2,4-DNPH to form an orange precipitate, indicating it contains a carbonyl group (aldehyde or ketone). Since it is not an aldehyde, it must be a ketone.
- A ketone is formed by the oxidation of a secondary alcohol. Therefore, \(Z\) must be a secondary alcohol.

Let's classify the options:
- **2-methylbutan-2-ol** is a tertiary alcohol (resists oxidation).
- **3-methylbutan-1-ol** is a primary alcohol (oxidizes to 3-methylbutanoic acid under reflux, which does not react with 2,4-DNPH).
- **3-methylbutan-2-ol** is a secondary alcohol (oxidizes to 3-methylbutan-2-one, which is a ketone).
- **2,2-dimethylpropan-1-ol** is a primary alcohol (oxidizes to 2,2-dimethylpropanoic acid).

1 mark for the correct option C.

First, analyze the successive ionisation energies of element \(Y\). The values are \(1012, 1903, 2912, 4957, 6274\) for the first five electrons, and then a massive jump occurs to \(21269\) for the sixth electron. This indicates that the first five electrons are in the outer shell, and the sixth electron is removed from an inner core shell closer to the nucleus. Therefore, element \(Y\) has five valence electrons and is in Group 15. In Period 3, the Group 15 element is phosphorus (\(P\)).

Now we evaluate each option:
- **A is incorrect**: Phosphorus chloride (\(\text{PCl}_3\)) is a simple covalent molecular liquid, not an ionic solid, and it does not conduct electricity when molten.
- **B is incorrect**: Phosphorus exists as a solid at room temperature (e.g., white phosphorus is \(\text{P}_4\)), not a diatomic gas.
- **C is correct**: Phosphorus(V) oxide (\(\text{P}_4\text{O}_{10}\)) reacts vigorously with water to form phosphoric(V) acid (\(\text{H}_3\text{PO}_4\)), which is an acidic solution.
- **D is incorrect**: The element immediately preceding phosphorus is silicon. Phosphorus has a higher first ionisation energy than silicon due to its stable, half-filled \(3\text{p}\) subshell (\(3\text{s}^2 3\text{p}^3\)) and a higher nuclear charge.

[1 mark] - Awarded for identifying the element as phosphorus from the jump between the 5th and 6th ionisation energies, and correctly selecting Option C based on its chemical properties.

We use the ideal gas equation: \(pV = nRT\) or \(pV = \frac{m}{M_r}RT\).
Rearranging to find the molar mass (\(M_r\)):
\(M_r = \frac{m R T}{p V}\)

Convert all units to SI values:
- Mass, \(m = 0.116\text{ g}\)
- Temperature, \(T = 127 + 273.15 = 400.15\text{ K} \approx 400\text{ K}\)
- Pressure, \(p = 1.00 \times 10^5\text{ Pa}\)
- Volume, \(V = 66.5\text{ cm}^3 = 66.5 \times 10^{-6}\text{ m}^3\)

Substitute values:
\(M_r = \frac{0.116 \times 8.31 \times 400}{1.00 \times 10^5 \times 66.5 \times 10^{-6}}\)
\(M_r = \frac{385.584}{6.65} \approx 58.0\text{ g mol}^{-1}\)

Now, calculate the molar mass of each option:
- **A**: \(\text{C}_2\text{H}_4\text{O}\) has \(M_r = 2(12.0) + 4(1.0) + 16.0 = 44.0\text{ g mol}^{-1}\)
- **B**: \(\text{C}_3\text{H}_6\text{O}\) has \(M_r = 3(12.0) + 6(1.0) + 16.0 = 58.0\text{ g mol}^{-1}\)
- **C**: \(\text{C}_4\text{H}_8\text{O}\) has \(M_r = 4(12.0) + 8(1.0) + 16.0 = 72.0\text{ g mol}^{-1}\)
- **D**: \(\text{C}_4\text{H}_{10}\text{O}\) has \(M_r = 4(12.0) + 10(1.0) + 16.0 = 74.0\text{ g mol}^{-1}\)

Only option B matches the calculated molar mass of \(58.0\text{ g mol}^{-1}\).

[1 mark] - Correct calculation of the molar mass of the vapor to be 58.0 and matching this value to the molecular formula of propanone / option B.

According to Hess's Law, the standard enthalpy change of a reaction can be calculated from standard enthalpy changes of combustion using:
\(\Delta H_{\text{reaction}}^\ominus = \sum \Delta H_c^\ominus(\text{reactants}) - \sum \Delta H_c^\ominus(\text{products})\)

For the formation of ethanol:
\(\Delta H_f^\ominus[\text{C}_2\text{H}_5\text{OH}(l)] = 2\Delta H_c^\ominus[\text{C}(s)] + 3\Delta H_c^\ominus[\text{H}_2(g)] - \Delta H_c^\ominus[\text{C}_2\text{H}_5\text{OH}(l)]\)

Substitute the given values:
\(\Delta H_f^\ominus = 2(-394) + 3(-286) - (-1367)\)
\(\Delta H_f^\ominus = -788 - 858 + 1367\)
\(\Delta H_f^\ominus = -1646 + 1367 = -279\text{ kJ mol}^{-1}\)

- Option A is correct (\(-279\text{ kJ mol}^{-1}\)).
- Option B is incorrect due to a sign error (\(+279\text{ kJ mol}^{-1}\)).
- Option C is incorrect as it represents only the sum of combustion of the reactants (\(-1646\text{ kJ mol}^{-1}\)).
- Option D is incorrect as it is calculated by adding all terms as negative values (\(-3013\text{ kJ mol}^{-1}\)).

[1 mark] - Correct application of Hess's law to relate enthalpies of combustion to enthalpy of formation, and correct numerical calculation.

We must find all the structural isomers of \(\text{C}_4\text{H}_8\text{Cl}_2\) and check each for the presence of a chiral carbon (a carbon atom bonded to four different groups).

Let's test the structures based on the carbon skeletons:

1. **Butane skeleton (straight chain):**
- **1,1-dichlorobutane**: \(\text{CHCl}_2-\text{CH}_2-\text{CH}_2-\text{CH}_3\) (No chiral carbon)
- **1,2-dichlorobutane**: \(\text{CH}_2\text{Cl}-\text{CH(Cl)}-\text{CH}_2-\text{CH}_3\) (Chiral carbon at C2, attached to: \(-\text{H}\), \(-\text{Cl}\), \(-\text{CH}_2\text{Cl}\), and \(-\text{CH}_2\text{CH}_3\)) \(\rightarrow\) **Structure 1**
- **1,3-dichlorobutane**: \(\text{CH}_2\text{Cl}-\text{CH}_2-\text{CH(Cl)}-\text{CH}_3\) (Chiral carbon at C3, attached to: \(-\text{H}\), \(-\text{Cl}\), \(-\text{CH}_2\text{CH}_2\text{Cl}\), and \(-\text{CH}_3\)) \(\rightarrow\) **Structure 2**
- **1,4-dichlorobutane**: \(\text{CH}_2\text{Cl}-\text{CH}_2-\text{CH}_2-\text{CH}_2\text{Cl}\) (No chiral carbon)
- **2,2-dichlorobutane**: \(\text{CH}_3-\text{CCl}_2-\text{CH}_2-\text{CH}_3\) (No chiral carbon)
- **2,3-dichlorobutane**: \(\text{CH}_3-\text{CH(Cl)}-\text{CH(Cl)}-\text{CH}_3\) (Chiral carbons at C2 and C3, both attached to: \(-\text{H}\), \(-\text{Cl}\), \(-\text{CH}_3\), and \(-\text{CH(Cl)CH}_3\)) \(\rightarrow\) **Structure 3**

2. **Methylpropane skeleton (branched chain):**
- **1,1-dichloro-2-methylpropane**: \(\text{CHCl}_2-\text{CH(CH}_3)_2\) (No chiral carbon)
- **1,2-dichloro-2-methylpropane**: \(\text{CH}_2\text{Cl}-\text{C(Cl)(CH}_3)_2\) (No chiral carbon)
- **1,3-dichloro-2-methylpropane**: \(\text{CH}_2\text{Cl}-\text{CH(CH}_2\text{Cl})-\text{CH}_3\) (No chiral carbon)

Thus, there are exactly 3 structural isomers containing at least one chiral carbon atom.

[1 mark] - Correctly drawing or listing all structural isomers of dichlorobutane, identifying which ones have chiral carbon atoms, and arriving at the correct count of 3.

Let's review the trends of Group 2 elements from magnesium (Mg) to barium (Ba):

- **A is incorrect**: The solubility of Group 2 sulfates *decreases* down the group (magnesium sulfate is highly soluble, barium sulfate is virtually insoluble).
- **B is incorrect**: The thermal stability of Group 2 nitrates *increases* down the group because the ionic radius of the cation increases down the group, reducing its charge density and its polarizing effect on the nitrate anion. Therefore, they decompose *less easily* (requiring higher temperatures) down the group.
- **C is correct**: The first ionisation energy *decreases* down the group. This is because atomic radius and electron shielding increase, which outweighs the increase in nuclear charge, making the outermost electrons less strongly attracted by the nucleus.
- **D is incorrect**: The reactivity of Group 2 elements with water *increases* down the group because the outer s-electrons are lost more easily.

[1 mark] - Awarded for identifying the correct trend (first ionisation energy decreases down Group 2) and dismissing incorrect trend statements.

In the atmosphere, nitrogen monoxide (\(\text{NO}\)) acts as a catalyst for the oxidation of sulfur dioxide (\(\text{SO}_2\)) to sulfur trioxide (\(\text{SO}_3\)). The catalytic cycle involves two key reactions:

1) Oxidation of nitrogen monoxide by oxygen to form nitrogen dioxide:
\(\text{NO}(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{NO}_2(g)\) (Here, \(\text{NO}\) is consumed)

2) Oxidation of sulfur dioxide by nitrogen dioxide, which regenerates the catalyst nitrogen monoxide:
\(\text{NO}_2(g) + \text{SO}_2(g) \rightarrow \text{NO}(g) + \text{SO}_3(g)\) (Here, \(\text{NO}\) is regenerated)

Therefore, the reaction where nitrogen monoxide is regenerated is represented by Option A.

[1 mark] - Correctly identifying the reaction in which nitrogen dioxide oxidizes sulfur dioxide to form sulfur trioxide and regenerate nitrogen monoxide.

We can determine the number of carbon atoms, \(n\), in a molecule using the relative abundances of the \([M]^+\) and \([M+1]^+\) peaks in the mass spectrum:

\(n = \frac{\text{relative abundance of } [M+1]^+}{\text{relative abundance of } [M]^+} \times \frac{100}{1.1}\)

Given:
- Relative abundance of \([M]^+ = 68.2\%\)
- Relative abundance of \([M+1]^+ = 3.0\%\)

Substitute these values into the formula:
\(n = \frac{3.0}{68.2} \times \frac{100}{1.1}\)
\(n = 0.04399 \times 90.91 \approx 4.00\)

Therefore, there are exactly 4 carbon atoms present in a molecule of \(W\).

[1 mark] - Correct application of the [M+1]/[M] peak ratio formula to calculate the number of carbon atoms, rounding to the nearest integer (4).

Lactic acid, \(\text{CH}_3\text{CH(OH)COOH}\), contains two distinct functional groups that can undergo acid-base or redox reactions: an alcohol (\(-\text{OH}\)) group and a carboxylic acid (\(-\text{COOH}\)) group.

1. **Reaction with sodium metal (Reagent 1):**
Sodium metal is a highly reactive metal that reacts with both alcohols and carboxylic acids to form salts (alkoxides and carboxylates, respectively) and release hydrogen gas.
Therefore, both the hydroxyl group and the carboxyl group react to yield: \(\text{CH}_3\text{CH(ONa)COONa}\).

2. **Reaction with sodium hydrogencarbonate (Reagent 2):**
Sodium hydrogencarbonate is a weak base. It is strong enough to react with carboxylic acids to form sodium carboxylate salts, water, and carbon dioxide, but it is too weak to react with alcohols.
Therefore, only the carboxylic acid group reacts, leaving the alcohol group unchanged: \(\text{CH}_3\text{CH(OH)COONa}\).

This makes Option A the correct choice.

[1 mark] - Correctly identifying that both functional groups in lactic acid react with sodium metal, while only the carboxylic acid group reacts with sodium hydrogencarbonate.

The successive ionisation energies show a very large increase between the third and fourth ionisation energies (from 2745 to 11578 \(\text{kJ}\,\text{mol}^{-1}\)). This indicates that the fourth electron is removed from an inner quantum shell, meaning the element has three valence electrons and belongs to Group 13. In Period 3, this element is aluminium (\(\text{Al}\)). Aluminium reacts with hot aqueous sodium hydroxide to form sodium tetrahydroxoaluminate and hydrogen gas according to the equation: \(2\text{Al(s)} + 2\text{NaOH(aq)} + 6\text{H}_2\text{O(l)} \rightarrow 2\text{NaAl(OH)}_4\text{(aq)} + 3\text{H}_2\text{(g)}\). Aluminium oxide is amphoteric, not purely basic. Aluminium is in the p-block, and its anhydrous chloride, \(\text{Al}_2\text{Cl}_6\), has a simple molecular structure.

Award 1 mark for identifying the correct option D. Reject other options: A (amphoteric oxide), B (p-block element), C (simple molecular structure).

Using the ideal gas equation: \(pV = nRT = \frac{m}{M_r}RT\), we can rearrange to find the relative molecular mass: \(M_r = \frac{mRT}{pV}\). Convert all units to SI: \(T = 97 + 273 = 370\text{ K}\), \(V = 108 \times 10^{-6}\text{ m}^3\), \(m = 0.205\text{ g}\), and \(p = 1.01 \times 10^5\text{ Pa}\). This gives: \(M_r = \frac{0.205 \times 8.31 \times 370}{1.01 \times 10^5 \times 108 \times 10^{-6}} = \frac{630.3}{10.908} \approx 57.8\text{ g}\,\text{mol}^{-1}\). The molecular masses of the options are: ethanol (46.0), propanone (58.0), pentane (72.0), and methanol (32.0). The calculated value of 57.8 is closest to the molecular mass of propanone.

Award 1 mark for the correct option B. Award 0 marks for any other option.

The formation of a purple vapor indicates the presence of iodine gas (\(\text{I}_2\)), which is formed when iodide ions (\(\text{I}^-\)) are oxidised by concentrated sulfuric acid. The yellow solid is sulfur (\(\text{S}\)) and the choking gas is sulfur dioxide (\(\text{SO}_2\)), which are products of the reduction of sulfuric acid. Therefore, the halide ion in \(Y\) is \(\text{I}^-\), and sulfuric acid acts as an oxidising agent by oxidising the iodide ions to iodine.

Award 1 mark for selecting option A. Accept only A. Other options are incorrect because bromide does not produce a purple vapor (it produces orange-brown bromine vapor) and sulfuric acid acts as an oxidising agent, not a reducing agent or simple acid here.

The formula \(\text{C}_4\text{H}_7\text{Cl}\) has one double bond equivalent. Since it reacts with bromine in the dark, it must be an alkene. There are 8 acyclic structural isomers of \(\text{C}_4\text{H}_7\text{Cl}\): (1) 1-chlorobut-1-ene, (2) 2-chlorobut-1-ene, (3) 3-chlorobut-1-ene, (4) 4-chlorobut-1-ene, (5) 1-chlorobut-2-ene, (6) 2-chlorobut-2-ene, (7) 1-chloro-2-methylpropene, and (8) 3-chloro-2-methylpropene. Of these structural isomers, the following exhibit stereoisomerism: 1-chlorobut-1-ene (cis-trans), 3-chlorobut-1-ene (optical isomerism due to a chiral carbon at C3), 1-chlorobut-2-ene (cis-trans), and 2-chlorobut-2-ene (cis-trans). Thus, exactly 4 structural isomers exhibit stereoisomerism.

Award 1 mark for the correct answer C. Correct identification of the 4 isomers that exhibit stereoisomerism.

Going down Group 2, the size of the metal cation increases, and its charge density decreases. The magnesium ion, being smaller than the barium ion, has a higher charge density and is better able to polarise the large carbonate anion. This polarising action weakens the carbon-oxygen bonds within the carbonate ion, making magnesium carbonate less thermally stable and causing it to decompose at a lower temperature. The solubility of sulfates decreases down the group, the solubility of hydroxides increases down the group, and the thermal stability of nitrates increases down the group.

Award 1 mark for option B. Accept only B. All other choices represent incorrect trends or faulty explanations.

The formation of propanoic acid is represented by: \(3\text{C(s)} + 3\text{H}_2\text{(g)} + \text{O}_2\text{(g)} \rightarrow \text{CH}_3\text{CH}_2\text{COOH(l)}\). Using Hess's Law with enthalpies of combustion: \(\Delta H^\theta_\text{f} = 3\Delta H^\theta_\text{c}\text{(C)} + 3\Delta H^\theta_\text{c}\text{(H}_2\text{)} - \Delta H^\theta_\text{c}\text{(propanoic acid)}\). Substituting the given values: \(\Delta H^\theta_\text{f} = [3(-394) + 3(-286)] - (-1527) = [-1182 - 858] + 1527 = -2040 + 1527 = -513\text{ kJ}\,\text{mol}^{-1}\).

Award 1 mark for A. Award 0 marks for incorrect calculations (such as D which has the incorrect sign, or C which uses incorrect stoichiometry).

Bromine exists naturally as two main isotopes, \(^{79}\text{Br}\) and \(^{81}\text{Br}\), in approximately a 1:1 ratio. Therefore, any molecular ion containing one bromine atom will exhibit two peaks, \(M\) and \(M+2\), of roughly equal heights, separated by 2 units of \(m/z\). Here, the peak at 122 is due to \([\text{C}_3\text{H}_7^{79}\text{Br}]^+\) and the peak at 124 is due to \([\text{C}_3\text{H}_7^{81}\text{Br}]^+\). The fragment at \(m/z = 43\) corresponds to the propyl carbocation \([\text{C}_3\text{H}_7]^+\), formed by the loss of a neutral bromine radical, not a \([\text{Br}]^+\) ion, and is formed by both 1-bromopropane and 2-bromopropane, so it cannot uniquely identify the isomer.

Award 1 mark for option B. Option A is incorrect because both isomers yield the same fragment; C is incorrect because a neutral radical is lost, not a positive halide ion; D is incorrect because the M+1 peak is much smaller.

In the presence of sunlight (UV radiation), nitrogen oxides undergo complex photochemical reactions with unburnt hydrocarbons from fuel emissions to produce photochemical smog, of which peroxyacetyl nitrate (PAN) is a highly toxic, irritating component. Nitrogen dioxide acts as a catalyst in the oxidation of atmospheric sulfur dioxide to sulfur trioxide, not carbon monoxide. Sulfur dioxide in car engines comes from the combustion of sulfur impurities in the fuel. Nitrogen monoxide is formed in engines by the direct reaction of atmospheric nitrogen and oxygen at very high temperatures.

Award 1 mark for option B. Accept only B.

(a)(i) Magnesium nitrate decomposes on heating to produce magnesium oxide, nitrogen dioxide, and oxygen: \(2\text{Mg(NO}_3)_2\text{(s)} \rightarrow 2\text{MgO(s)} + 4\text{NO}_2\text{(g)} + \text{O}_2\text{(g)}\) (or the 1-mole equivalent). (a)(ii) Down Group 2, the cationic radius increases while the charge remains \(2+\). This leads to a decrease in charge density of the cation. Consequently, the larger cation has less polarizing effect on the nitrate anion, making the \(\text{N}-\text{O}\) bonds in the nitrate ion less weakened and therefore more stable to heat. (b)(i) The solubility of Group 2 hydroxides increases down the group. Both lattice energy and hydration energy decrease down the group, but lattice energy decreases more rapidly, making the enthalpy of solution more exothermic. (b)(ii) Adding aqueous sodium sulfate to magnesium chloride gives no visible reaction (magnesium sulfate is highly soluble). Barium chloride forms a thick white precipitate of barium sulfate. Ionic equation: \(\text{Ba}^{2+}\text{(aq)} + \text{SO}_4^{2-}\text{(aq)} \rightarrow \text{BaSO}_4\text{(s)}\). (c)(i) The misty gas is hydrogen chloride, \(\text{HCl}\). The reaction is an acid-base (or precipitation/non-redox) reaction. (c)(ii) Purple vapor is iodine, \(\text{I}_2\) (or \(\text{I}_2\text{(g)}\)); Yellow solid is sulfur, \(\text{S}\); Smell of bad eggs is hydrogen sulfide, \(\text{H}_2\text{S}\). (c)(iii) Iodide ions (\(\text{I}^-\)) have a larger ionic radius than chloride ions (\(\text{Chl}^-\)), so the valence electrons are further from the nucleus and more shielded, making them easier to lose. Thus, iodide is a much stronger reducing agent than chloride. Iodide can reduce sulfur in \(\text{H}_2\text{SO}_4\) from \(+6\) to \(0\) and \(-2\), whereas chloride is not a strong enough reducing agent to reduce sulfuric acid.

(a)(i) M1: Balanced equation with correct formulas of products: MgO, \(\text{NO}_2\), and \(\text{O}_2\). [1] M2: State symbols correct: \(\text{Mg(NO}_3)_2\text{(s)}\), \(\text{MgO(s)}\), \(\text{NO}_2\text{(g)}\), \(\text{O}_2\text{(g)}\). [1] (a)(ii) M1: Cation size increases down the group, so charge density decreases. [1] M2: Lower charge density means less polarization/distortion of the nitrate anion. [1] (b)(i) M1: Solubility increases down the group. [1] M2: Explanation in terms of lattice energy decreasing more rapidly than hydration energy. [1] (b)(ii) M1: Observation: no change for Mg, white precipitate for Ba. [1] M2: Ionic equation with state symbols: \(\text{Ba}^{2+}\text{(aq)} + \text{SO}_4^{2-}\text{(aq)} \rightarrow \text{BaSO}_4\text{(s)}\). [1] (c)(i) M1: Hydrogen chloride / \(\text{HCl}\). [1] M2: Acid-base / acid-carbonate / non-redox. [1] (c)(ii) M1: \(\text{I}_2\) / iodine. [1] M2: \(\text{S}\) / sulfur. [1] M3: \(\text{H}_2\text{S}\) / hydrogen sulfide. [1] (c)(iii) M1: Iodide is a stronger reducing agent than chloride. [1] M2: Reference to the ease of electron loss due to larger ionic radius / more shielding in iodide. [1]

(a)(i) Standard enthalpy change of formation is the enthalpy change when 1 mole of a compound is formed from its constituent elements in their standard states under standard conditions (100 kPa and 298 K). (a)(ii) Balanced formation equation: \(3\text{C(graphite)} + 4\text{H}_2\text{(g)} + \frac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{OH(l)}\). Hess's Law calculation: Using combustion data: \(\Delta H_f^\ominus = \sum \Delta H_c^\ominus(\text{reactants}) - \sum \Delta H_c^\ominus(\text{products})\). \(\Delta H_f^\ominus = [3 \times \Delta H_c^\ominus(\text{C}) + 4 \times \Delta H_c^\ominus(\text{H}_2)] - \Delta H_c^\ominus(\text{propan-1-ol})\). \(\Delta H_f^\ominus = [3 \times (-394) + 4 \times (-286)] - (-2021) = [-1182 - 1144] + 2021 = -2326 + 2021 = -305\text{ kJ mol}^{-1}\). (b)(i) \(K_c = \frac{[\text{CH}_3\text{OH}]}{[\text{CO}][\text{H}_2]^2}\). Units: \(\frac{\text{mol dm}^{-3}}{(\text{mol dm}^{-3})(\text{mol dm}^{-3})^2} = \text{mol}^{-2}\text{ dm}^6\) or \(\text{dm}^6\text{ mol}^{-2}\). (b)(ii) Calculate concentrations at equilibrium (Volume = 2.00 dm\(^3\)): \([\text{CO}] = 0.400 / 2.00 = 0.200\text{ mol dm}^{-3}\), \([\text{H}_2] = 0.300 / 2.00 = 0.150\text{ mol dm}^{-3}\), \([\text{CH}_3\text{OH}] = 0.150 / 2.00 = 0.075\text{ mol dm}^{-3}\). Substitute into \(K_c\) expression: \(K_c = \frac{0.075}{0.200 \times (0.150)^2} = \frac{0.075}{0.200 \times 0.0225} = \frac{0.075}{0.0045} = 16.67\text{ dm}^6\text{ mol}^{-2}\). Round to 3 sig figs: \(16.7\text{ dm}^6\text{ mol}^{-2}\). (b)(iii) If temperature is increased, the equilibrium constant \(K_c\) decreases. This is because the forward reaction is exothermic (\(\Delta H < 0\)). According to Le Chatelier's principle, an increase in temperature shifts the equilibrium position to the left (absorbing heat), which decreases product concentration and increases reactant concentrations, lowering \(K_c\).

(a)(i) M1: Enthalpy change when 1 mole of substance is formed from its elements. [1] M2: Elements in standard states under standard conditions. [1] (a)(ii) M1: Correct balanced formation equation with state symbols (C as solid/graphite, H2 as gas, O2 as gas, propanol as liquid). [1] M2: Correct application of Hess's Law cycle (sum of combustion reactants - product). [1] M3: Correct calculation showing \(-2326\). [1] M4: Final answer \(-305\text{ kJ mol}^{-1}\) (must have sign and units). [1] (b)(i) M1: Correct expression for \(K_c\). [1] M2: Correct units \(\text{dm}^6\text{ mol}^{-2}\) (or \(\text{mol}^{-2}\text{ dm}^6\)). [1] (b)(ii) M1: Divide all moles by 2.00 to find equilibrium concentrations. [1] M2: Correct substitution into the expression. [1] M3: Final calculated answer of 16.7 (or 16.67) to 3 significant figures. [1] (b)(iii) M1: \(K_c\) decreases. [1] M2: Because the forward reaction is exothermic, equilibrium shifts left to favor the endothermic direction. [1]

(a)(i) First ionisation energy is the energy required to remove one electron from each atom in one mole of gaseous atoms to form one mole of gaseous 1+ ions. (a)(ii) Element X is Aluminium (\(\text{Al}\)). The successive values show a relatively gradual increase from 1st to 3rd ionisation energy, followed by a very large increase (a jump of almost 9000 \(\text{kJ mol}^{-1}\)) to the 4th ionisation energy. This indicates that the 4th electron is being removed from a lower principal energy level (an inner shell), which is closer to the nucleus and experiences less shielding. Therefore, X has 3 valence electrons and is in Group 13. In Period 3, this element is aluminium. The electronic configuration of Al is \(1s^2 2s^2 2p^6 3s^2 3p^1\). (a)(iii) The \(p_y\) orbital should be sketched as a dumbbell shape centered on the origin, lying symmetrically along the vertical y-axis (with x and z axes clearly labeled and perpendicular). (b)(i) In \(\text{BF}_3\), boron has 3 bonding pairs and 0 lone pairs in its valence shell. To minimize repulsion, these pairs position themselves as far apart as possible, giving a trigonal planar shape with a bond angle of \(120^\t\). In \(\text{NF}_3\), nitrogen has 3 bonding pairs and 1 lone pair in its valence shell. Due to the greater repulsion of the lone pair compared to the bonding pairs, the bond angles are squeezed to \(107^\t\) (accept \(102^\t - 108^\t\)), giving a trigonal pyramidal shape. (b)(ii) \(\text{BF}_3\) is non-polar because it is highly symmetrical (trigonal planar shape). The individual polar \(\text{B}-\text{F}\) bonds are equal in magnitude and point in opposite directions, canceling out the overall molecular dipole. \(\text{NF}_3\) is polar because its trigonal pyramidal geometry is asymmetrical. The dipole moments of the \(\text{N}-\text{F}\) bonds do not cancel, resulting in a net molecular dipole. (b)(iii) Fluorine is highly electronegative, creating a strongly polar \(\text{H}-\text{F}\) bond. The small, highly electronegative F atom allows for strong intermolecular hydrogen bonding between \(\text{HF}\) molecules. Chlorine is less electronegative than fluorine and has a larger atomic size, so \(\text{HCl}\) molecules cannot form hydrogen bonds and only experience weaker permanent dipole-dipole forces, requiring less energy to overcome.

(a)(i) M1: Energy required to remove one electron from each gaseous atom in one mole. [1] M2: Equation shown or mention of gaseous state of atoms and 1+ ions. [1] (a)(ii) M1: Identity: Aluminium (Al). [1] M2: Large jump between 3rd and 4th ionisation energies indicates 3 outer electrons. [1] M3: Outer electrons are in the third shell; inner shell is much harder to ionise. [1] M4: Configuration: \(1s^2 2s^2 2p^6 3s^2 3p^1\). [1] (a)(iii) M1: Correct dumbbell shape aligned along the y-axis. [1] (b)(i) M1: \(\text{BF}_3\): trigonal planar, \(120^\circ\). [1] M2: \(\text{NF}_3\): trigonal pyramidal, \(107^\t\) (accept range \(102^\t - 108^\t\)). [1] M3: VSEPR theory explanation: Electron pairs repel to be as far apart as possible. [1] M4: Lone pair-bond pair repulsion is greater than bond pair-bond pair repulsion. [1] (b)(ii) M1: \(\text{BF}_3\) is symmetrical, so bond dipoles cancel out. [1] M2: \(\text{NF}_3\) is asymmetrical (due to lone pair), so dipoles do not cancel. [1] (b)(iii) M1: \(\text{HF}\) forms intermolecular hydrogen bonds. [1] M2: \(\text{HCl}\) only forms weaker permanent dipole-dipole intermolecular forces. [1]

(a)(i) The molecular formula \(\text{C}_4\text{H}_{10}\text{O}\) represents a saturated monohydric alcohol. There are four alcohol isomers: butan-1-ol, butan-2-ol, 2-methylpropan-1-ol, and 2-methylpropan-2-ol. The two primary alcohols (where the \(-\text{OH}\) group is attached to a carbon bonded to only one other carbon) are: 1. Butan-1-ol (a straight chain of 4 carbons with the \(-\text{OH}\) on carbon-1): \(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH}\). 2. 2-Methylpropan-1-ol (a branched chain of 3 carbons with a methyl group on carbon-2 and the \(-\text{OH}\) on carbon-1): \((\text{CH}_3)_2\text{CHCH}_2\text{OH}\). (a)(ii) Primary alcohols oxidize fully under reflux to form carboxylic acids. Therefore, compound Z contains the carboxylic acid (or carboxyl) functional group. Acidified potassium dichromate(VI) is reduced from orange dichromate(VI) ions (\(\text{Cr}_2\text{O}_7^{2-}\)) to green chromium(III) ions (\(\text{Cr}^{3+}\)), giving a color change of orange to green. (a)(iii) Warm the compounds separately with acidified potassium dichromate(VI). The primary alcohol will undergo oxidation, changing the color from orange to green. The tertiary alcohol (2-methylpropan-2-ol) is resistant to oxidation because it lacks a hydrogen atom on the carbon bearing the \(-\text{OH}\) group, so the mixture will remain orange. (b)(i) To oxidize the secondary alcohol butan-2-ol to butanone, use acidified potassium dichromate(VI) (\(\text{K}_2\text{Cr}_2\text{O}_7 / \text{H}_2\text{SO}_4\)) and heat under reflux (or distillation, since ketones cannot be further oxidized easily). (b)(ii) Reaction with 2,4-DNPH gives a bright yellow, orange, or red crystalline precipitate, which confirms the presence of a carbonyl group (aldehyde or ketone). To identify the specific ketone (butanone), filter the precipitate, recrystallize it from a suitable solvent, dry it, and measure its melting point. This melting point is then compared to a table of known melting points of 2,4-DNPH derivatives. (b)(iii) The product butanone contains a carbonyl group. The strong, sharp peak in its IR spectrum is due to the \(\text{C}=\text{O}\) bond, which absorbs in the range of 1670–1740 \(\text{cm}^{-1}\). The reactant butan-2-ol was an alcohol, which has a broad, strong \(\text{O}-\text{H}\) absorption peak in the range of 3200–3600 \(\text{cm}^{-1}\); this peak is absent in the spectrum of butanone.

(a)(i) M1: Correct structure of butan-1-ol. [1] M2: Correct IUPAC name: butan-1-ol. [1] M3: Correct structure of 2-methylpropan-1-ol. [1] M4: Correct IUPAC name: 2-methylpropan-1-ol. [1] (a)(ii) M1: Carboxylic acid / carboxyl group. [1] M2: Orange to green. [1] (a)(iii) M1: Reagent: acidified potassium dichromate(VI) and heat. [1] M2: Observation: primary turns green AND tertiary stays orange. [1] (b)(i) M1: Reagent: acidified potassium dichromate(VI) (or \(\text{K}_2\text{Cr}_2\text{O}_7 / \text{H}^+\)). [1] M2: Condition: Heat / reflux (or distill). [1] (b)(ii) M1: Formation of orange/yellow/red precipitate with 2,4-DNPH. [1] M2: Measure melting point of the crystalline derivative and compare with literature values. [1] (b)(iii) M1: Carbonyl group / \(\text{C}=\text{O}\) bond. [1] M2: Range: 1670–1740 \(\text{cm}^{-1}\). [1] M3: Alcohol \(\text{O}-\text{H}\) stretch / peak in the range 3200–3600 \(\text{cm}^{-1}\) is missing. [1]