2023 Edexcel IAL Chemistry (YCH11) Practice Paper with Answers

1. Calculate the number of moles of aluminium sulfate:
\(\text{moles of Al}_2(\text{SO}_4)_3 = 25.0 \times 10^{-3}\text{ dm}^3 \times 0.040\text{ mol dm}^{-3} = 0.0010\text{ mol}\)

2. Determine the number of moles of ions per mole of formula unit:
Each formula unit of \(\text{Al}_2(\text{SO}_4)_3\) dissociates into \(2\text{Al}^{3+}\) and \(3\text{SO}_4^{2-}\) ions, which is a total of \(5\) ions.

3. Calculate the total moles of ions:
\(\text{moles of ions} = 0.0010\text{ mol} \times 5 = 0.0050\text{ mol}\)

4. Calculate the total number of ions:
\(\text{number of ions} = 0.0050\text{ mol} \times 6.02 \times 10^{23}\text{ mol}^{-1} = 3.01 \times 10^{21}\)

• D is the correct answer (1 mark)
• A represents the number of ions if there were only 1 ion per formula unit (0 marks)
• B represents the number of aluminium ions only (0 marks)
• C represents the number of sulfate ions only (0 marks)

1. Look for the largest ratio jump between successive ionisation energies.
- Ratio 2nd/1st = \(1817 / 578 \approx 3.1\)
- Ratio 3rd/2nd = \(2745 / 1817 \approx 1.5\)
- Ratio 4th/3rd = \(11578 / 2745 \approx 4.2\)
- Ratio 5th/4th = \(14831 / 11578 \approx 1.3\)

The largest jump occurs between the third and fourth ionisation energies. This indicates that the fourth electron is being removed from an inner quantum shell. Therefore, \(X\) has three valence electrons in its outer shell and forms a \(X^{3+}\) ion.

2. The oxide ion is \(\text{O}^{2-}\). To form a neutral compound, two \(X^{3+}\) ions combine with three \(\text{O}^{2-}\) ions, giving the empirical formula \(X_2\text{O}_3\).

• C is the correct answer (1 mark)
• A is incorrect as it represents a Group 1 element (0 marks)
• B is incorrect as it represents a Group 2 element (0 marks)
• D is incorrect as it represents a Group 4 element (0 marks)

The shape of a molecule is determined by the electron pairs surrounding the central atom:
- \(\text{NH}_3\): Nitrogen has 5 valence electrons, forms 3 bonding pairs and has 1 lone pair. This gives a trigonal pyramidal shape.
- \(\text{BF}_3\): Boron has 3 valence electrons, forms 3 bonding pairs and has 0 lone pairs. This gives a trigonal planar shape.
- \(\text{ClF}_3\): Chlorine has 7 valence electrons, forms 3 bonding pairs and has 2 lone pairs. This gives a T-shaped geometry.
- \(\text{PF}_3\): Phosphorus has 5 valence electrons, forms 3 bonding pairs and has 1 lone pair. This gives a trigonal pyramidal shape.

• B is the correct answer (1 mark)
• A, C and D are incorrect as they describe molecules with other geometries due to the presence of lone pairs (0 marks)

- Propagation steps always involve a reaction between a stable molecule and a radical to generate a new stable molecule and a new radical.
- Reaction A (\(\text{Cl}_2 \rightarrow 2\text{Cl}^\bullet\)) is an initiation step because it produces radicals from a neutral molecule.
- Reactions B (\(\text{CH}_3^\bullet + \text{Cl}^\bullet \rightarrow \text{CH}_3\text{Cl}\)) and D (\(\text{CH}_3^\bullet + \text{CH}_3^\bullet \rightarrow \text{C}_2\text{H}_6\)) are termination steps because they combine two radicals to form a stable product.
- Reaction C (\(\text{CH}_4 + \text{Cl}^\bullet \rightarrow \text{CH}_3^\bullet + \text{HCl}\)) is a propagation step because a chlorine radical reacts with methane to produce a methyl radical and hydrogen chloride.

• C is the correct answer (1 mark)
• A is incorrect as it is an initiation step (0 marks)
• B and D are incorrect as they are termination steps (0 marks)

For an alkene to exhibit geometric (\(E\)/\(Z\)) isomerism, each carbon atom of the double bond must be attached to two different groups.
- In \(\text{2-methylbut-2-ene}\), \((\text{CH}_3)_2\text{C}=\text{CHCH}_3\), one of the carbon atoms in the double bond is bonded to two identical methyl groups, so it cannot show isomerism.
- In \(\text{but-1-ene}\), \(\text{CH}_2=\text{CHCH}_2\text{CH}_3\), the terminal carbon is bonded to two identical hydrogen atoms.
- In \(\text{1,2-dichloroethene}\), \(\text{CHCl}=\text{CHCl}\), each carbon of the double bond is bonded to a hydrogen atom and a chlorine atom (two different groups), allowing both cis/trans and E/Z stereoisomers.
- In \(\text{2-methylpropene}\), \((\text{CH}_3)_2\text{C}=\text{CH}_2\), both double-bonded carbons are bonded to identical groups.

• C is the correct answer (1 mark)
• A, B and D are incorrect because they have at least one double-bonded carbon attached to two identical groups (0 marks)

1. Calculate the moles of ethanol reactant:
\(\text{moles of ethanol} = \frac{4.60\text{ g}}{46.0\text{ g mol}^{-1}} = 0.100\text{ mol}\)

2. The oxidation equation shows a 1:1 stoichiometry between ethanol and ethanoic acid, so the theoretical yield of ethanoic acid is \(0.100\text{ mol}\).

3. Calculate the theoretical mass of ethanoic acid:
\(\text{theoretical mass} = 0.100\text{ mol} \times 60.0\text{ g mol}^{-1} = 6.00\text{ g}\)

4. Calculate the percentage yield:
\(\text{percentage yield} = \frac{\text{actual mass}}{\text{theoretical mass}} \times 100\% = \frac{3.30\text{ g}}{6.00\text{ g}} \times 100\% = 55.0\%\)

• A is the correct answer (1 mark)
• B is the percentage if one mistakenly uses actual mass / ethanol reactant mass (0 marks)
• C and D are incorrect calculations (0 marks)

1. The electronic configuration of a neutral iron atom (\(\text{Fe}\), \(Z = 26\)) is \([\text{Ar}] 3\text{d}^6 4\text{s}^2\).
2. When transition metals form cations, they lose electrons from the \(4\text{s}\) subshell first, then the \(3\text{d}\) subshell.
3. For \(\text{Fe}^{3+}\), three electrons are removed: two from \(4\text{s}\) and one from \(3\text{d}\), resulting in \([\text{Ar}] 3\text{d}^5\).
- \(\text{Fe}^{2+}\) is \([\text{Ar}] 3\text{d}^6\).
- \(\text{Cr}^{3+}\) (from \(\text{Cr} = [\text{Ar}] 3\text{d}^5 4\text{s}^1\)) is \([\text{Ar}] 3\text{d}^3\).
- \(\text{Co}^{3+}\) (from \(\text{Co} = [\text{Ar}] 3\text{d}^7 4\text{s}^2\)) is \([\text{Ar}] 3\text{d}^6\).

• B is the correct answer (1 mark)
• A, C and D are incorrect configurations of the given ions (0 marks)

The degree of covalent character in an ionic compound is governed by Fajans' rules:
1. Polarization of the anion increases with increasing charge density of the cation (smaller ionic radius, higher charge).
2. Polarization increases with increasing size of the anion, as a larger anion has outer electrons less tightly held by its nucleus (more polarizable).

Comparing the ions:
- Cations: \(\text{Be}^{2+}\) has a smaller ionic radius and higher charge than \(\text{Li}^+\) and \(\text{Mg}^{2+}\), giving it the highest charge density and greatest polarizing power.
- Anions: \(\text{I}^-\) has a much larger ionic radius than \(\text{Cl}^-\), making it more easily polarized.

Therefore, beryllium iodide, \(\text{BeI}_2\), experiences the greatest polarization of the anion, leading to the greatest degree of covalent character.

• B is the correct answer (1 mark)
• A is incorrect because the lithium ion has less polarizing power than beryllium (0 marks)
• C is incorrect because the chloride ion is less polarizable than the iodide ion (0 marks)
• D is incorrect because the magnesium ion is larger and less polarizing than the beryllium ion (0 marks)

To find the number of moles (\(n\)), use the ideal gas equation: \(PV = nRT\). First, convert all values to SI units: Pressure, \(P = 100\text{ kPa} = 1.00 \times 10^5\text{ Pa}\); Volume, \(V = 250\text{ cm}^3 = 2.50 \times 10^{-4}\text{ m}^3\); Temperature, \(T = 27 + 273 = 300\text{ K}\). Rearranging the equation for \(n\): \(n = \frac{PV}{RT} = \frac{1.00 \times 10^5 \times 2.50 \times 10^{-4}}{8.31 \times 300} = \frac{25.0}{2493} \approx 1.00 \times 10^{-2}\text{ mol}\).

Correct option is A (1 mark). Option B is obtained by failing to convert temperature to Kelvin. Option C is obtained by failing to convert volume to \(\text{m}^3\) correctly. Option D is obtained by failing to convert both volume and pressure correctly.

An electronic configuration of \(1s^2 2s^2 2p^6 3s^2 3p^6\) represents a species with a total of 18 electrons. \(\text{S}^-\)\ has 17 electrons. \(\text{K}^+\)\ has 18 electrons, which gives it this noble gas configuration. \(\text{Ar}^+\)\ has 17 electrons. \(\text{Sc}^{2+}\)\ has 19 electrons with configuration \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^1\).

Correct option is B (1 mark). Other options represent species with different numbers of electrons.

Bond polarity is determined by the difference in electronegativity between the two bonded atoms. Electronegativity increases across Period 2 from carbon to fluorine (\(\text{C} < \text{N} < \text{O} < \text{F}\)). Therefore, the electronegativity difference is greatest for the \(\text{C}-\text{F}\) bond, making it the most polar.

Correct option is A (1 mark). Other options have smaller electronegativity differences as the elements are closer to fluorine in the periodic table.

The relative atomic mass (\(A_r\)) is the weighted average of the isotopic masses: \(A_r = \frac{(20 \times 90.48) + (21 \times 0.27) + (22 \times 9.25)}{100} = \frac{1809.60 + 5.67 + 203.50}{100} = 20.1877\). Rounding to two decimal places gives 20.19.

Correct option is B (1 mark). Option A is incorrect rounding; Option D is a simple average of the mass numbers.

A propagation step must involve a free radical reacting with a stable molecule to produce a new free radical and a new stable molecule. \(\text{CH}_4 + \text{Cl}^\bullet \rightarrow \text{CH}_3^\bullet + \text{HCl}\) fits this definition. Option A is initiation, while options C and D are termination steps where two radicals combine to form a stable molecule.

Correct option is B (1 mark). A is initiation; C and D are termination.

The reaction of propene with \(\text{HBr}\) proceeds via electrophilic addition. The addition of a proton (\(\text{H}^+\)) can form either a primary carbocation or a secondary carbocation. The secondary carbocation is more stable because the two electron-releasing alkyl groups reduce the charge density on the positive carbon atom (inductive effect). Thus, the reaction proceeds predominantly through the secondary carbocation to form 2-bromopropane as the major product.

Correct option is A (1 mark). B is incorrect as secondary is more stable than primary; C is incorrect terminology and mechanism; D is incorrect because propene is unsymmetrical.

Atom economy is calculated using the formula: Atom Economy = (Mass of desired product / Total mass of all reactants) x 100%. The desired product is ethanol. From the equation, 1 mol of glucose (180.0 g) produces 2 mol of ethanol (2 x 46.0 g = 92.0 g). Atom Economy = (92.0 / 180.0) x 100% = 51.11%.

Correct option is C (1 mark). Option A is obtained if the stoichiometry of 2 for ethanol is omitted. Option B is the atom economy for CO2. Option D is obtained by omitting the stoichiometry of CO2 in the products.

Generally, first ionization energy increases across Period 3. However, there is a dip from magnesium (Mg) to aluminium (Al). The outer electron of Al is in the 3p subshell, which is higher in energy and better shielded than the 3s subshell of Mg, making it easier to remove.

Correct option is B (1 mark). All other pairs show an increase in first ionization energy.

First, write the balanced equation for the reaction: \(\text{MgCO}_3(\text{s}) + 2\text{HCl}(\text{aq}) \rightarrow \text{MgCl}_2(\text{aq}) + \text{CO}_2(\text{g}) + \text{H}_2\text{O}(\text{l})\). Moles of \(\text{HCl} = 0.0250 \times 0.400 = 0.0100\text{ mol}\). Since \(\text{MgCO}_3\) is in excess, \(\text{HCl}\) is the limiting reactant. From the stoichiometry of the equation, 2 moles of \(\text{HCl}\) produce 1 mole of \(\text{CO}_2\). Therefore, moles of \(\text{CO}_2\) produced = \(0.0100 / 2 = 0.00500\text{ mol}\). Volume of \(\text{CO}_2\) produced = \(0.00500\text{ mol} \times 24000\text{ cm}^3\text{ mol}^{-1} = 120\text{ cm}^3\).

Award 1 mark for the correct option B. [AO1, AO2]

There is a massive increase in energy between the 3rd and 4th ionisation energies (from 2745 to 11578 \(\text{kJ mol}^{-1}\)). This indicates that the fourth electron is removed from an inner shell which is closer to the nucleus and experiences much stronger electrostatic attraction. Therefore, element \(\text{Y}\) has three electrons in its outer shell and is in Group 3.

Award 1 mark for the correct option C. [AO1, AO2]

Option A is the initiation step, which involves the homolytic fission of chlorine. Option B is a propagation step where a chlorine radical reacts with an ethane molecule to produce an ethyl radical and hydrogen chloride gas. Options C and D are termination steps where two free radicals combine to form a stable molecule.

Award 1 mark for the correct option B. [AO1, AO2]

For an alkene to show stereoisomerism, both of the carbon atoms in the double bond must be attached to two different groups. In pent-2-ene, carbon-2 is attached to \(-\text{H}\) and \(-\text{CH}_3\), and carbon-3 is attached to \(-\text{H}\) and \(-\text{CH}_2\text{CH}_3\). Thus, it has stereoisomers (E/Z or cis/trans). In all other options, at least one of the double-bonded carbon atoms has two identical groups (H atoms in propene and but-1-ene, and methyl groups in 2-methylbut-2-ene).

Award 1 mark for the correct option D. [AO1, AO2]

Part (a)
(i) \(\text{M}_2\text{CO}_3(\text{aq}) + 2\text{HCl}(\text{aq}) \rightarrow 2\text{MCl}(\text{aq}) + \text{CO}_2(\text{g}) + \text{H}_2\text{O}(\text{l})\)
(ii) \(\text{Moles of HCl} = 0.03040\text{ dm}^3 \times 0.100\text{ mol dm}^{-3} = 3.04 \times 10^{-3}\text{ mol}\)
(iii) Moles of \(\text{M}_2\text{CO}_3\) in \(25.0\text{ cm}^3 = 3.04 \times 10^{-3} / 2 = 1.52 \times 10^{-3}\text{ mol}\).
Moles of \(\text{M}_2\text{CO}_3\) in \(250\text{ cm}^3 = 1.52 \times 10^{-3} \times 10 = 0.0152\text{ mol}\).
(iv) Molar mass of \(\text{M}_2\text{CO}_3 = \text{mass} / \text{moles} = 2.10\text{ g} / 0.0152\text{ mol} = 138.16\text{ g mol}^{-1}\).
\(2 \times A_r(\text{M}) + 12.0 + 48.0 = 138.2 \implies 2 \times A_r(\text{M}) = 78.2 \implies A_r(\text{M}) = 39.1\text{ g mol}^{-1}\).
From the Periodic Table, the group 1 metal is Potassium (\(\text{K}\)).

Part (b)
(i) Mass of water lost = \(3.484\text{ g} - 2.764\text{ g} = 0.720\text{ g}\).
Moles of water = \(0.720\text{ g} / 18.0\text{ g mol}^{-1} = 0.0400\text{ mol}\).
Moles of anhydrous carbonate \(\text{K}_2\text{CO}_3 = 2.764\text{ g} / 138.2\text{ g mol}^{-1} = 0.0200\text{ mol}\).
Ratio of \(\text{H}_2\text{O} : \text{K}_2\text{CO}_3 = 0.0400 / 0.0200 = 2\). Thus, \(x = 2\).
(ii) Heating to constant mass ensures that all the water of crystallization has been completely driven off and the reaction is finished. This is confirmed by heating, cooling, weighing, and repeating until two consecutive weights are identical.

Part (a)
(i) [2 marks]: 1 mark for balanced equation with correct formulas, 1 mark for all correct state symbols.
(ii) [1 mark]: \(3.04 \times 10^{-3}\text{ mol}\).
(iii) [2 marks]: 1 mark for moles in \(25.0\text{ cm}^3 = 1.52 \times 10^{-3}\text{ mol}\), 1 mark for scaling up by 10 to get \(0.0152\text{ mol}\) in \(250\text{ cm}^3\).
(iv) [4 marks]: 1 mark for dividing 2.10 by their calculated moles, 1 mark for molar mass of \(138.2\text{ g mol}^{-1}\), 1 mark for calculating \(A_r = 39.1\), 1 mark for identifying Potassium (\(\text{K}\)).

Part (b)
(i) [4 marks]: 1 mark for mass of water = \(0.720\text{ g}\), 1 mark for moles of water = \(0.0400\text{ mol}\), 1 mark for moles of anhydrous salt = \(0.0200\text{ mol}\), 1 mark for the ratio and deducing \(x = 2\).
(ii) [2 marks]: 1 mark for ensuring all water of crystallization has been removed, 1 mark for explaining that it involves repeating the heating/weighing process until successive masses remain the same.

Part (a)
(i) Relative isotopic mass is the mass of an individual atom of an isotope relative to \(1/12\)th of the mass of a carbon-12 atom.
(ii) \(\text{Relative atomic mass} = \frac{(24 \times 78.99) + (25 \times 10.00) + (26 \times 11.01)}{100} = \frac{1895.76 + 250.00 + 286.26}{100} = 24.32\).

Part (b)
(i) \(\text{Mg}^{2+}(\text{g}) \rightarrow \text{Mg}^{3+}(\text{g}) + \text{e}^{-}\)
(ii) Successive ionization energies show a general increase because as each electron is removed, the remaining electrons experience a stronger pull from the nucleus. There is a small jump from the first to the second ionization energy, and then a very large jump from the second to the third ionization energy. This large jump indicates that the third electron is being removed from an inner quantum shell (closer to the nucleus, with significantly less shielding), providing evidence that there are 2 valence electrons and the configuration is 2, 8, 2.

Part (c)
- Electronic configuration of \(\text{Mg}^{2+}\): \(1\text{s}^2 2\text{s}^2 2\text{p}^6\).
- Isoelectronic noble gas: Neon (\(\text{Ne}\)).

Part (d)
Phosphorus has the outer configuration \(3\text{s}^2 3\text{p}^3\) with three singly-occupied orbitals. Sulfur has the configuration \(3\text{s}^2 3\text{p}^4\) with one paired set of electrons in a 3p orbital. The repulsion between these paired electrons (spin-pair repulsion) in sulfur makes it easier to remove its outer electron than the single electron in phosphorus, overriding the increase in nuclear charge.

Part (a)
(i) [2 marks]: 1 mark for mass of an atom of an isotope, 1 mark for relative to \(1/12\)th of the mass of carbon-12.
(ii) [2 marks]: 1 mark for correct mathematical setup, 1 mark for final calculated value of 24.32 (to 2 decimal places).

Part (b)
(i) [2 marks]: 1 mark for correct species and balancing (\(\text{Mg}^{2+} \rightarrow \text{Mg}^{3+} + \text{e}^{-}\)), 1 mark for correct gas state symbols \((\text{g})\) on both magnesium species.
(ii) [5 marks]: 1 mark for stating that ionization energy increases generally as more electrons are removed, 1 mark for identifying the large jump between the 2nd and 3rd ionization energies, 1 mark for explaining that the third electron is removed from a shell closer to the nucleus, 1 mark for mentioning the effect of decreased shielding, 1 mark for concluding this confirms 2 electrons in the outermost shell.

Part (c)
[2 marks]: 1 mark for \(1\text{s}^2 2\text{s}^2 2\text{p}^6\), 1 mark for identifying Neon.

Part (d)
[2 marks]: 1 mark for noting that sulfur has paired electrons in a 3p orbital, 1 mark for explaining that the repulsion between these paired electrons (spin-pair repulsion) makes the electron easier to remove.

Part (a)
(i) In graphite, each carbon atom is covalently bonded to three others, leaving one delocalized electron per carbon atom. These delocalized electrons are free to move throughout the layers and conduct electricity. In diamond, each carbon is bonded to four others in a giant tetrahedral structure. All electrons are localized in covalent bonds, so diamond cannot conduct electricity.
(ii) Similarity: Both have carbon atoms covalently bonded to three other carbons in hexagonal rings. Difference: Graphite is made of multiple stacked layers held together by weak intermolecular forces, while graphene is a single, one-atom-thick layer.

Part (b)
(i) The central carbon forms a double bond with oxygen (four shared electrons: two from C, two from O) and two single bonds with each chlorine (each chlorine shares two electrons: one from C, one from Cl). Carbon has zero lone pairs, oxygen has two lone pairs, and each chlorine has three lone pairs.
(ii) Shape: Trigonal planar. Bond angle: approximately \(120^\circ\) (accept values in the range \(111^\circ\) to \(120^\circ\)). Explanation: The central carbon atom has three regions of electron density (one double bond, two single bonds) and no lone pairs. To minimize repulsion, these three regions push as far apart as possible, leading to a trigonal planar arrangement.

Part (c)
Chlorine is more electronegative than carbon, creating polar bonds with dipoles pointing towards chlorine. \(\text{CCl}_4\) is symmetrical (tetrahedral), so the individual bond dipoles point in opposite directions and cancel out, giving no net molecular dipole. \(\text{CH}_2\text{Cl}_2\) is asymmetrical, meaning the individual bond dipoles do not cancel, resulting in a net molecular dipole.

Part (a)
(i) [3 marks]: 1 mark for stating graphite has delocalized electrons that are free to move, 1 mark for diamond having 4 covalent bonds per carbon, 1 mark for explaining diamond has no free/delocalized electrons.
(ii) [2 marks]: 1 mark for similarity (hexagonal rings / three bonds per carbon), 1 mark for difference (graphene is a single layer, graphite has multiple layers).

Part (b)
(i) [2 marks]: 1 mark for correct double bond to O and single bonds to Cl, 1 mark for showing all correct non-bonding pairs on O (4 electrons) and Cl (6 electrons on each).
(ii) [4 marks]: 1 mark for Trigonal planar, 1 mark for bond angle of \(120^\circ\) (or range \(111^\circ\) to \(120^\circ\)), 1 mark for stating carbon has 3 regions of electron density and no lone pairs, 1 mark for explaining that regions of electron density repel to achieve minimum repulsion.

Part (c)
[4 marks]: 1 mark for noting chlorine is more electronegative than carbon, 1 mark for stating \(\text{CCl}_4\) is symmetrical, 1 mark for stating that the dipoles in \(\text{CCl}_4\) cancel, 1 mark for explaining that \(\text{CH}_2\text{Cl}_2\) is asymmetrical so its dipoles do not cancel.

Part (a)
(i) (E)-but-2-ene is drawn as a linear zig-zag skeletal structure with methyl groups opposite. (Z)-but-2-ene is drawn as a U-shaped skeletal structure with methyl groups on the same side of the double bond. This stereoisomerism occurs because of the restricted rotation about the \(\text{C}=\text{C}\) double bond (due to the \(\pi\) bond) and because each carbon atom of the double bond is attached to two different groups (a methyl group and a hydrogen atom).
(ii) In the first step, a curly arrow goes from the double bond of but-2-ene to the \(\text{H}\) of \(\text{H}-\text{Br}\). The \(\text{H}-\text{Br}\) bond has a dipole with \(\delta+\) on \(\text{H}\) and \(\delta-\) on \(\text{Br}\). Another curly arrow goes from the \(\text{H}-\text{Br}\) bond to the \(\text{Br}\) atom. This forms a secondary carbocation intermediate (\(\text{CH}_3\text{CH}^+\text{CH}_2\text{CH}_3\)) and a bromide ion (\(\text{Br}^-\)). In the second step, a curly arrow goes from a lone pair on \(\text{Br}^-\) to the carbocation carbon to form the product 2-bromobutane.

Part (b)
(i) UV radiation provides the energy required for the homolytic fission of the chlorine-chlorine bond to form chlorine free radicals.
(ii)
- Initiation: \(\text{Cl}_2 \rightarrow 2\text{Cl}^\bullet\)
- Propagation 1: \(\text{CH}_4 + \text{Cl}^\bullet \rightarrow {}^\bullet\text{CH}_3 + \text{HCl}\)
- Propagation 2: \({}^\bullet\text{CH}_3 + \text{Cl}_2 \rightarrow \text{CH}_3\text{Cl} + \text{Cl}^\bullet\)
- Termination: \(2\text{Cl}^\bullet \rightarrow \text{Cl}_2\) or \({}^\bullet\text{CH}_3 + \text{Cl}^\bullet \rightarrow \text{CH}_3\text{Cl}\) or \(2{}^\bullet\text{CH}_3 \rightarrow \text{C}_2\text{H}_6\).
(iii) Ethane is formed during a termination step where two methyl radicals (\({}^\bullet\text{CH}_3\)) collide and combine: \(2{}^\bullet\text{CH}_3 \rightarrow \text{C}_2\text{H}_6\).

Part (a)
(i) [4 marks]: 1 mark for correct skeletal drawing of (E)-but-2-ene, 1 mark for correct skeletal drawing of (Z)-but-2-ene, 1 mark for explaining restricted rotation about the carbon-carbon double bond, 1 mark for stating that both carbons of the double bond have two different groups attached.
(ii) [4 marks]: 1 mark for arrow from double bond to \(\text{H}\) and arrow from \(\text{H}-\text{Br}\) bond to \(\text{Br}\), 1 mark for showing \(\delta+\) on \(\text{H}\) and \(\delta-\) on \(\text{Br}\), 1 mark for drawing the carbocation intermediate and \(\text{Br}^-\), 1 mark for arrow from lone pair of \(\text{Br}^-\) to carbon containing the positive charge.

Part (b)
(i) [1 mark]: Homolytic fission of the \(\text{Cl}-\text{Cl}\) bond / to create chlorine free radicals.
(ii) [4 marks]: 1 mark for correct initiation step, 1 mark for first propagation step, 1 mark for second propagation step, 1 mark for any correct termination step. (Radical dots are required for marks).
(iii) [2 marks]: 1 mark for explaining that methyl radicals combine, 1 mark for the equation: \(2{}^\bullet\text{CH}_3 \rightarrow \text{C}_2\text{H}_6\).

HF has hydrogen bonding between its molecules, which is the strongest type of intermolecular force, giving it the highest boiling temperature. For HCl, HBr, and HI, the dominant intermolecular forces are London forces, which increase in strength as the number of electrons in the halogen atom increases from Cl to I. Therefore, the boiling temperature increases from HCl to HI, with HF having the highest overall boiling temperature. The correct order is \( \text{HCl} < \text{HBr} < \text{HI} < \text{HF} \).

1 mark: Correctly identifies the order and the underlying chemical principles (hydrogen bonding in HF and increasing London forces from HCl to HI).

Going down Group 2, ionic radius increases while ionic charge remains 2+. This means the charge density of the cation decreases from \( \text{Mg}^{2+} \) to \( \text{Ca}^{2+} \). Consequently, \( \text{Ca}^{2+} \) has a lower polarising power and polarises the electron cloud of the nitrate ion less strongly, causing less weakening of the covalent N–O bonds within the nitrate group. Thus, calcium nitrate is more thermally stable and requires a higher temperature to decompose.

1 mark: Correctly identifies the relation between lower charge density, weaker polarization of the anion, and greater thermal stability.

The equation for the standard enthalpy of formation of ethanol is: \( 2\text{C}(s) + 3\text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{C}_2\text{H}_5\text{OH}(l) \). Using a Hess cycle with enthalpies of combustion: \( \Delta_f H^\theta = \sum \Delta_c H^\theta (\text{reactants}) - \sum \Delta_c H^\theta (\text{products}) \). Thus, \( \Delta_f H^\theta = [2 \times (-393.5) + 3 \times (-285.8)] - [-1367.3] = [-787.0 - 857.4] + 1367.3 = -1644.4 + 1367.3 = -277.1\text{ kJ mol}^{-1} \).

1 mark: Correct calculation of the enthalpy of formation with appropriate stoichiometry and sign.

The rate of hydrolysis of halogenoalkanes depends on the strength of the carbon-halogen bond. The C–I bond is the longest and has the lowest bond enthalpy compared to C–Br and C–Cl, meaning it is broken most easily. Therefore, 1-iodobutane reacts the fastest. The silver halide precipitate formed with iodide ions is silver iodide, \( \text{AgI} \), which is a yellow solid.

1 mark: Correctly identifies the fastest reacting halogenoalkane and the correct color of its silver halide precipitate.

A catalyst provides an alternative reaction pathway with a lower activation energy, which decreases the required activation energy (\(E_a\) decreases). However, since the temperature of the system is held constant, the distribution of kinetic energies of the gas molecules remains unchanged, meaning the Maxwell-Boltzmann distribution curve is unaffected.

1 mark: Correctly identifies that activation energy decreases and the Maxwell-Boltzmann distribution curve is unchanged.

Concentrated sulfuric acid acts as an oxidising agent. Iodide ions are strong reducing agents and reduce sulfuric acid (where sulfur is in the +6 state) to a mixture of sulfur dioxide, \( \text{SO}_2 \) (+4 state), sulfur, \( \text{S} \) (0 state), and hydrogen sulfide, \( \text{H}_2\text{S} \) (-2 state). Iodine, \( \text{I}_2 \), is an oxidation product of iodide, whereas \( \text{KHSO}_4 \) and \( \text{HI} \) are products of the non-redox acid-base reaction.

1 mark: Correctly identifies the reduction products of sulfuric acid.

The sharp IR peak at \( 1715\text{ cm}^{-1} \) indicates a carbonyl group (\( \text{C}=\text{O} \)). The absence of a broad peak in the range \( 2500 - 3300\text{ cm}^{-1} \) shows that no carboxylic acid is present. Heating a secondary alcohol (propan-2-ol) under reflux with acidified potassium dichromate(VI) oxidises it to a ketone (propanone), which matches these spectral properties. A primary alcohol (propan-1-ol) would oxidise under reflux to a carboxylic acid (propanoic acid), which would show a broad O-H absorption. Tertiary alcohols (methylpropan-2-ol) are resistant to oxidation.

1 mark: Correctly matches the spectral data of the oxidized product to the secondary alcohol starting material.

Percentage Atom Economy is given by \( \frac{\text{Molar mass of desired product}}{\text{Total molar mass of reactants}} \times 100\% \). Desired product is bromoethane, \( M = 108.9\text{ g mol}^{-1} \). Total mass of reactants = \( 46.0 + 102.9 + 98.1 = 247.0\text{ g mol}^{-1} \). Atom economy = \( \frac{108.9}{247.0} \times 100\% = 44.09\% \approx 44.1\% \).

1 mark: Correctly calculates percentage atom economy by mass using reactant stoichiometry.

Hydrogen fluoride (HF) has the highest boiling temperature due to strong intermolecular hydrogen bonding, which requires significant energy to overcome. For the remaining hydrogen halides (HCl, HBr, and HI), the dominant intermolecular forces are London forces. As the halogen atom increases in size down the group, the number of electrons increases, leading to stronger instantaneous dipole-induced dipole (London) forces. Thus, the boiling temperatures increase from HCl to HBr to HI.

1 mark for correct answer A. Reject all other options.

As you descend Group 2, the ionic radius of the \( \text{M}^{2+} \) cation increases while its charge remains constant. This leads to a lower charge density and weaker polarising power. Consequently, the larger cation causes less distortion (polarisation) of the nitrate anion's electron cloud, which makes the covalent bonds within the nitrate group less weakened and therefore more thermally stable. This results in an increase in decomposition temperature down the group.

1 mark for correct answer B. Reject all other options.

The rate of hydrolysis of halogenoalkanes is determined by the strength of the carbon-halogen bond. Bond enthalpy decreases down Group 7 because the halogen atom gets larger, leading to a longer and weaker carbon-halogen bond. Since the C-Cl bond has the highest bond enthalpy, it is the hardest to break, making 1-chlorobutane the slowest to hydrolyse. Conversely, the C-I bond has the lowest bond enthalpy and is easiest to break, so 1-iodobutane hydrolyses fastest.

1 mark for correct answer B. Reject all other options.

The equation for the standard enthalpy of formation of propene is: \( 3\text{C(s)} + 3\text{H}_2\text{(g)} \rightarrow \text{C}_3\text{H}_6\text{(g)} \). Using standard enthalpies of combustion: \( \Delta_f H^\theta = \sum \Delta_c H^\theta(\text{reactants}) - \sum \Delta_c H^\theta(\text{products}) \). Thus, \( \Delta_f H^\theta = [3 \times \Delta_c H^\theta(\text{C(s)}) + 3 \times \Delta_c H^\theta(\text{H}_2\text{(g)})] - \Delta_c H^\theta(\text{C}_3\text{H}_6\text{(g)}) = [3(-394) + 3(-286)] - (-2058) = [-1182 - 858] + 2058 = -2040 + 2058 = +18\text{ kJ mol}^{-1} \).

1 mark for correct answer B. Reject all other options.

When temperature increases, the average kinetic energy of the gas molecules increases. This shifts the overall energy distribution to higher values, meaning the peak of the curve moves to the right. Since the total number of molecules (and thus the total area under the curve) remains constant, the peak must become lower to accommodate the broader distribution of higher-energy molecules.

1 mark for correct answer C. Reject all other options.

The reaction of chlorine with cold, dilute aqueous sodium hydroxide is a disproportionation reaction: \( \text{Cl}_2 + 2\text{NaOH} \rightarrow \text{NaCl} + \text{NaClO} + \text{H}_2\text{O} \). The chlorine-containing products are sodium chloride (\( \text{NaCl} \)), where the oxidation state of chlorine is \( -1 \), and sodium chlorate(I) (\( \text{NaClO} \)), where the oxidation state of chlorine is \( +1 \).

1 mark for correct answer A. Reject all other options.

The molecular formula \( \text{C}_3\text{H}_6\text{O} \) corresponds to a degree of unsaturation (one double bond or ring). The strong peak at \( 1715\text{ cm}^{-1} \) indicates a carbonyl group (\( \text{C=O} \)). The absence of a broad band in the range \( 3200\text{--}3600\text{ cm}^{-1} \) shows that there is no hydroxyl group (\( \text{O-H} \)). Therefore, the compound is a ketone, propanone (\( \text{CH}_3\text{COCH}_3 \)). Propan-1-ol has the formula \( \text{C}_3\text{H}_8\text{O} \) and contains an \( \text{O-H} \) group. Propanoic acid and methyl ethanoate both have two oxygen atoms (\( \text{C}_3\text{H}_6\text{O}_2 \)).

1 mark for correct answer C. Reject all other options.

According to Le Chatelier's Principle, increasing the temperature of a system in equilibrium shifts the position of equilibrium in the endothermic direction to absorb the added thermal energy. Since the forward reaction is exothermic, the backward reaction is endothermic. Thus, equilibrium shifts to the left, decreasing the yield of \( \text{SO}_3 \). Since \( K_c = \frac{[\text{SO}_3]^2}{[\text{SO}_2]^2 [\text{O}_2]} \), a shift to the left decreases the concentration of products and increases the concentration of reactants, thereby decreasing the value of \( K_c \).

1 mark for correct answer A. Reject all other options.

1-bromobutane has the highest boiling temperature because of the relative strengths of its intermolecular forces. When comparing chlorine and bromine, bromine has a larger number of electrons, which leads to stronger London forces in 1-bromobutane than in 1-chlorobutane. Furthermore, when comparing straight-chain isomers to branched isomers, the straight-chain structure of 1-bromobutane allows for a larger surface area of contact between molecules, resulting in stronger London forces compared to the branched isomer 2-bromo-2-methylpropane.

1 mark: Correct option chosen (C).

Down Group 2, the thermal stability of the carbonates increases. This is because the ionic radius of the Group 2 cation increases down the group while carrying the same 2+ charge. As the cationic radius increases, its charge density and polarising power decrease. Consequently, the larger cation causes less distortion to the carbonate anion (specifically weakening the C-O bonds to a lesser extent), which makes the carbonate more thermally stable.

1 mark: Correct option chosen (A).

When the temperature of a gas mixture increases, the Maxwell-Boltzmann distribution curve flattens and shifts to the right. Because the peak of the curve shifts to the right, the most probable molecular energy (\(E_{mp}\)) increases. Additionally, because the entire distribution is shifted toward higher energies, a larger fraction of molecules have an energy greater than or equal to the activation energy (\(E_a\)), represented by an increased area under the curve to the right of \(E_a\).

1 mark: Correct option chosen (A).

The equation for the formation of propane is: \(3\text{C(graphite)} + 4\text{H}_2\text{(g)} \rightarrow \text{C}_3\text{H}_8\text{(g)}\). Using Hess's Law with standard enthalpies of combustion: \(\Delta_f H^\ominus = \sum \Delta_c H^\ominus(\text{reactants}) - \sum \Delta_c H^\ominus(\text{products})\). Therefore, \(\Delta_f H^\ominus = [3 \times (-394) + 4 \times (-286)] - [-2220] = [-1182 - 1144] + 2220 = -2326 + 2220 = -106 \text{ kJ mol}^{-1}\).

1 mark: Correct option chosen (A).

(a) Reagent: Aqueous sodium hydroxide, \(\text{NaOH}(aq)\) (or potassium hydroxide, \(\text{KOH}(aq)\)). Conditions: Heat under reflux.

(b) The mechanism is nucleophilic substitution (specifically \(\text{S}_\text{N}2\)):
- Draw 1-bromobutane showing the dipole on the carbon-bromine bond: \(\text{C}^{\delta+}-\text{Br}^{\delta-}\).
- Show a curly arrow starting from a lone pair on the oxygen of the hydroxide ion, \(\text{OH}^-\), pointing to the electron-deficient carbon atom (\(\text{C}^{\delta+}\)).
- Show a curly arrow starting from the middle of the C-Br bond pointing to the bromine atom.
- The products are butan-1-ol and a bromide ion, \(\text{Br}^-\).

(c) The C-I bond is weaker than the C-Br bond because iodine is a larger atom, resulting in a longer and weaker covalent bond (lower bond enthalpy). Therefore, the C-I bond breaks more easily / requires less energy to break during the hydrolysis reaction.

(d)(i) High-resolution mass spectrometry can measure the \(m/z\) ratio to several decimal places (e.g. 4 d.p.), which allows the exact molecular formula to be determined by distinguishing between compounds with the same integer mass but different atomic compositions.

(d)(ii) X is a tertiary alcohol with four carbons, so its structure must be 2-methylpropan-2-ol. Skeletal formula: a central carbon atom bonded to three methyl groups (represented as three lines forming a Y-shape) and one hydroxyl group (-\(\text{OH}\)). Name: 2-methylpropan-2-ol.

(d)(iii) Reagents: Potassium dichromate(VI) and dilute sulfuric acid (\(\text{K}_2\text{Cr}_2\text{O}_7\) / \(\text{H}_2\text{SO}_4\)). Observation: Solution remains orange / no color change (since tertiary alcohols cannot be oxidized).

(a) [2 marks]
- Aqueous sodium hydroxide / aqueous potassium hydroxide [1]
- Heat under reflux [1]
(b) [4 marks]
- Correct dipoles shown as \(\text{C}^{\delta+}-\text{Br}^{\delta-}\) on 1-bromobutane [1]
- Curly arrow from a lone pair on the oxygen of \(\text{OH}^-\)[1]
- Curly arrow from the C-Br bond to the Br atom [1]
- Correct products: butan-1-ol and \(\text{Br}^-\)[1]
(c) [2 marks]
- C-I bond is weaker / has a lower bond enthalpy than the C-Br bond [1]
- C-I bond breaks more easily / requires less energy to break [1]
(d)(i) [1 mark]
- Measures precise/accurate mass (to 4 decimal places) to determine the exact molecular formula [1]
(d)(ii) [2 marks]
- Correct skeletal formula of 2-methylpropan-2-ol [1]
- Correct IUPAC name: 2-methylpropan-2-ol [1]
(d)(iii) [2 marks]
- Reagents: (acidified) potassium dichromate(VI) / \(\text{K}_2\text{Cr}_2\text{O}_7\) and sulfuric acid / \(\text{H}_2\text{SO}_4\) [1]
- Observation: remains orange / no change [1]

(a) Trend: Thermal stability increases down the group.
Explanation: Down the group, the cationic radius (size of the Group 2 cation, \(\text{M}^{2+}\)) increases while the charge remains \(2+\). This causes the charge density of the cation to decrease. Consequently, the larger cation has less polarizing power and causes less distortion of the electron cloud of the carbonate ion (\(\text{CO}_3^{2-}\)), making the C-O bond more difficult to break.

(b)(i) Strong direct heating is required to decompose calcium carbonate, making it impossible to measure the heat energy transferred directly to a calorimeter/water.

(b)(ii) The Hess's Law cycle is:
\(\text{CaCO}_3(s) \xrightarrow{\Delta H_{\text{r}}} \text{CaO}(s) + \text{CO}_2(g)\)
Both reactant and product mixtures react with acid (\(2\text{HCl}(aq)\)) to form the same final system: \(\text{CaCl}_2(aq) + \text{H}_2\text{O}(l) + \text{CO}_2(g)\).
- Downward arrow from \(\text{CaCO}_3(s)\) to the final system represents \(\Delta H_1\).
- Downward arrow from \(\text{CaO}(s) + \text{CO}_2(g)\) to the final system represents \(\Delta H_2\).
Expression: \(\Delta H_{\text{r}} = \Delta H_1 - \Delta H_2\).

(b)(iii) Calculation:
- Step 1: Calculate heat energy transferred (\(q\)).
Using solution mass \(m = 50.0\text{ g}\):
\(q = m c \Delta T = 50.0 \times 4.18 \times 12.5 = 2612.5\text{ J} = 2.6125\text{ kJ}\)
(If using total mass \(m = 50.0 + 1.50 = 51.5\text{ g}\):
\(q = 51.5 \times 4.18 \times 12.5 = 2690.875\text{ J} = 2.690875\text{ kJ}\))
- Step 2: Calculate moles of \(\text{CaO}\).
\(n(\text{CaO}) = \frac{1.50\text{ g}}{56.1\text{ g mol}^{-1}} = 0.026738\text{ mol}\)
- Step 3: Determine molar enthalpy change (\(\Delta H_2\)).
Since the temperature rose, the reaction is exothermic, so \(\Delta H_2\) must be negative.
Using \(m = 50.0\text{ g}\):
\(\Delta H_2 = -\frac{2.6125}{0.026738} = -97.707\text{ kJ mol}^{-1} \approx -97.7\text{ kJ mol}^{-1}\)
Using \(m = 51.5\text{ g}\):
\(\Delta H_2 = -\frac{2.690875}{0.026738} = -100.64\text{ kJ mol}^{-1} \approx -101\text{ kJ mol}^{-1}\)

(a) [4 marks]
- Thermal stability increases down the group [1]
- Cation / ionic radius increases [1]
- Charge density of cation decreases [1]
- Less polarization / distortion of the carbonate ion (electron cloud) [1]
(b)(i) [1 mark]
- Strong heating is required (heat is constantly supplied) / cannot measure the temperature change of the solid directly [1]
(b)(ii) [3 marks]
- Correct cycle showing three states in a triangular relationship with correct arrow directions [2]
- \(\Delta H_{\text{r}} = \Delta H_1 - \Delta H_2\) [1]
(b)(iii) [5 marks]
- Calculation of heat transferred: \(q = 2612.5\text{ J}\) or \(2.6125\text{ kJ}\) (or \(2690.9\text{ J}\) / \(2.6909\text{ kJ}\)) [1]
- Moles of \(\text{CaO} = 0.02674\text{ mol}\) [1]
- Division of heat energy by moles [1]
- Negative sign indicating an exothermic reaction [1]
- Final value to 3 s.f.: \(-97.7\text{ kJ mol}^{-1}\) (or \(-101\text{ kJ mol}^{-1}\)) [1]

(a)(i) London forces / instantaneous dipole-induced dipole forces only.

(a)(ii) Propan-1-ol has hydrogen bonding (as well as London forces and permanent dipole-dipole forces) because of its polar -OH group. Butane has only London forces between its non-polar molecules. Hydrogen bonds are significantly stronger than London forces and require much more thermal energy to overcome.

(a)(iii) Ethane-1,2-diol has two hydroxyl (-OH) groups per molecule, whereas propan-1-ol has only one. This allows ethane-1,2-diol to form more hydrogen bonds per molecule, creating a more extensive network of hydrogen bonds that requires significantly more energy to break.

(b)(i) Propan-1-ol can form hydrogen bonds with water molecules, allowing it to mix fully.
Diagram requirements:
- A propan-1-ol molecule, \(\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}\), and a water molecule, \(\text{H}_2\text{O}\), are drawn showing the O-H bonds.
- Dipoles are shown on the O-H bonds of both molecules (\(\text{O}^{\delta-}\) and \(\text{H}^{\delta+}\)).
- Two lone pairs are shown on the oxygen atoms of both molecules.
- A dashed line (or dotted line) represents the hydrogen bond, drawn directly from a lone pair on the oxygen of one molecule (e.g., water) to the hydrogen atom (\(\text{H}^{\delta+}\)) of the O-H group on the other molecule (e.g., propan-1-ol).
- The angle of the \(\text{O}-\text{H}\cdots\text{O}\) system is linear (approx. \(180^{\circ}\)).

(b)(ii) As the carbon chain length increases, the size of the non-polar, hydrophobic hydrocarbon chain increases. This large non-polar region cannot form hydrogen bonds with water; it disrupts the existing hydrogen bonds between water molecules without contributing equivalent attractive forces, making dissolution energetically unfavorable.

(a)(i) [1 mark]
- London forces / instantaneous dipole-induced dipole forces [1]
(a)(ii) [3 marks]
- Propan-1-ol has hydrogen bonds [1]
- Butane only has London forces [1]
- Hydrogen bonds are stronger than London forces and require more energy to break [1]
(a)(iii) [2 marks]
- Ethane-1,2-diol has two -OH groups whereas propan-1-ol has one [1]
- More hydrogen bonds can be formed per molecule / more extensive hydrogen-bonding network [1]
(b)(i) [5 marks]
- Explanation: Propan-1-ol forms hydrogen bonds with water molecules [1]
- Diagram shows correct structures of propan-1-ol and water with correct dipoles (\(\text{O}^{\delta-}\) and \(\text{H}^{\delta+}\)) [1]
- Diagram shows lone pairs on both oxygen atoms [1]
- Diagram shows a dashed/dotted line representing a hydrogen bond from an oxygen lone pair to the hydrogen of the other molecule [1]
- Diagram shows a linear \(\text{O}-\text{H}\cdots\text{O}\) alignment (angle of \(180^{\circ}\) around the hydrogen bond) [1]
(b)(ii) [2 marks]
- The non-polar / hydrophobic hydrocarbon chain increases in length/size [1]
- The non-polar region disrupts hydrogen bonding between water molecules (and cannot form hydrogen bonds itself) [1]

### Worked Solution

**Part (a)**
* **Initiation:** \(\text{CF}_2\text{Cl}_2 \xrightarrow{uv} \bullet\text{CF}_2\text{Cl} + \text{Cl}\bullet\)
* **Propagation 1:** \(\text{Cl}\bullet + \text{O}_3 \rightarrow \text{ClO}\bullet + \text{O}_2\)
* **Propagation 2:** \(\text{ClO}\bullet + \text{O} \rightarrow \text{Cl}\bullet + \text{O}_2\) (or \(\text{ClO}\bullet + \text{O}_3 \rightarrow \text{Cl}\bullet + 2\text{O}_2\))
* **Explanation:** HFCs contain only C, H, and F (no chlorine). The \(C-F\) bond is extremely strong and cannot be homolytically cleaved by stratospheric UV light, so no radicals capable of catalytic ozone destruction are released.

**Part (b) (i)**
* Greenhouse gases contain polar bonds (such as \(C-F\) or \(C-H\)).
* The absorption of infrared radiation causes these bonds to undergo stretching or bending vibrations, which change the molecular dipole moment.
* The absorbed energy is subsequently re-emitted in all directions, including back toward the Earth's surface, trapping heat within the atmosphere.

**Part (b) (ii)**
* HFO-1234yf contains a highly reactive \(C=C\) double bond (alkene functional group), whereas HFC-134a has only \(C-C\) single bonds (saturated alkane derivative).
* The high electron density of the \(C=C\) double bond makes HFO-1234yf susceptible to rapid attack by naturally occurring hydroxyl (\(\bullet\text{OH}\)) radicals in the troposphere.

**Part (c) (i)**
* **Bonds Broken:**
\(1 \times (C-Cl) = 346 \text{ kJ mol}^{-1}\)
\(1 \times (H-F) = 562 \text{ kJ mol}^{-1}\)
Total energy input = \(346 + 562 = 908 \text{ kJ mol}^{-1}\)

* **Bonds Formed:**
\(1 \times (C-F) = 467 \text{ kJ mol}^{-1}\)
\(1 \times (H-Cl) = 431 \text{ kJ mol}^{-1}\)
Total energy released = \(467 + 431 = 898 \text{ kJ mol}^{-1}\)

* **Enthalpy Change Calculation:**
\(\Delta H^\theta = \sum(\text{bonds broken}) - \sum(\text{bonds formed})\)
\(\Delta H^\theta = 908 - 898 = +10 \text{ kJ mol}^{-1}\)

**Part (c) (ii)**
* Increasing the temperature shifts the equilibrium position to the right (the forward direction).
* Because the forward reaction is endothermic (\(\Delta H^\theta > 0\)), and according to Le Chatelier's principle, the system shifts to absorb the added heat.

**Part (d) (i)**
* **Step 1:** Warm/heat the sample with aqueous sodium hydroxide (\(\text{NaOH(aq)}\)) or water in ethanol to hydrolyse the halogenoalkane and release chloride ions.
* **Step 2:** Acidify the mixture with dilute nitric acid (\(\text{HNO}_3\)) and add aqueous silver nitrate (\(\text{AgNO}_3\)).
* **Observation:** A white precipitate (of silver chloride, \(\text{AgCl}\)) forms, which is soluble in dilute aqueous ammonia.

**Part (d) (ii)**
* The \(C-F\) bond is much stronger than the \(C-Cl\) bond, as indicated by its much higher bond enthalpy (\(467 \text{ kJ mol}^{-1}\) vs \(346 \text{ kJ mol}^{-1}\)).
* Consequently, the activation energy required to break the \(C-F\) bond during nucleophilic attack is significantly higher, rendering it kinetically inert under typical hydrolysis conditions.

**Part (e)**
* The fragment containing chlorine would be \([\text{CH}_2^{35}\text{Cl}]^+\) with \(m/z = 49\) (or \([\text{CF}_3\text{CH}_2^{35}\text{Cl}]^+\) at \(m/z = 118\) or \(^{35}\text{Cl}^+\) at \(m/z = 35\)).
* It is absent in the mass spectrum of \(\text{CF}_3\text{CH}_2\text{F}\) because that compound has no chlorine atoms.

**(a)** [Total: 4 marks]
* **M1:** Correct initiation equation with UV radiation (\(uv\) or \(h\nu\)) above/on the arrow: \(\text{CF}_2\text{Cl}_2 \xrightarrow{uv} \bullet\text{CF}_2\text{Cl} + \text{Cl}\bullet\). (1 mark)
* **M2:** Correct propagation steps: \(\text{Cl}\bullet + \text{O}_3 \rightarrow \text{ClO}\bullet + \text{O}_2\) AND \(\text{ClO}\bullet + \text{O} \rightarrow \text{Cl}\bullet + \text{O}_2\). (1 mark)
* **M3:** States that HFCs contain no chlorine atoms / only contain carbon, hydrogen, and fluorine. (1 mark)
* **M4:** States that the \(C-F\) bond is too strong to be broken by UV light in the stratosphere, preventing the formation of halogen radicals. (1 mark)

**(b) (i)** [Total: 3 marks]
* **M1:** Polar bonds undergo stretching/bending vibrations when absorbing IR. (1 mark)
* **M2:** This vibration results in a change in the dipole moment. (1 mark)
* **M3:** The absorbed energy is re-emitted/re-radiated back to the Earth's surface (trapping heat). (1 mark)

**(b) (ii)** [Total: 2 marks]
* **M1:** HFO-1234yf contains a reactive carbon-carbon double bond (\(C=C\)) / is an alkene (whereas HFC-134a has only single bonds). (1 mark)
* **M2:** The double bond is highly susceptible to attack by atmospheric hydroxyl (\(\bullet\text{OH}\)) radicals, leading to fast breakdown. (1 mark)

**(c) (i)** [Total: 3 marks]
* **M1:** Correctly identifies bonds broken (\(C-Cl\) and \(H-F\)) and formed (\(C-F\) and \(H-Cl\)), showing correct substitutions: bonds broken = \(346 + 562 = 908\); bonds formed = \(467 + 431 = 898\). (1 mark)
* **M2:** Correct calculation structure: \(\Delta H^\theta = 908 - 898\). (1 mark)
* **M3:** Final value of \(+10 \text{ kJ mol}^{-1}\) (must include positive sign and unit). (1 mark)

**(c) (ii)** [Total: 2 marks]
* **M1:** Equilibrium shifts to the right / forward direction. (1 mark)
* **M2:** Because the forward reaction is endothermic (or \(\Delta H > 0\)), so the system opposes the temperature increase by absorbing heat. (1 mark)

**(d) (i)** [Total: 3 marks]
* **M1:** Reagents and conditions for hydrolysis: Heat/warm with aqueous sodium hydroxide (or water/ethanol mixture). (1 mark)
* **M2:** Acidification and silver nitrate: Acidify with dilute nitric acid (\(\text{HNO}_3\)) and add aqueous silver nitrate (\(\text{AgNO}_3\)). Do NOT accept HCl. (1 mark)
* **M3:** Correct observation: White precipitate forms (which dissolves in dilute ammonia). (1 mark)

**(d) (ii)** [Total: 2 marks]
* **M1:** Identifies that the \(C-F\) bond is much stronger than the \(C-Cl\) bond / quotes the bond enthalpy values (467 vs 346 \(\text{kJ mol}^{-1}\)). (1 mark)
* **M2:** Explains that the higher bond enthalpy results in a much larger activation energy for the hydrolysis pathway. (1 mark)

**(e)** [Total: 2 marks]
* **M1:** Correct formula of a chlorine-containing fragment ion, including the positive charge (e.g., \([\text{CH}_2^{35}\text{Cl}]^+\), \([\text{CF}_3\text{CH}_2^{35}\text{Cl}]^+\), or \(^{35}\text{Cl}^+\)). (1 mark)
* **M2:** Correct corresponding \(m/z\) value (49 for \([\text{CH}_2^{35}\text{Cl}]^+\), 118 for \([\text{CF}_3\text{CH}_2^{35}\text{Cl}]^+\), or 35 for \(^{35}\text{Cl}^+\)). (1 mark)