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

1. Calculate the total molecular mass of the reactants:
Total mass of reactants = \(4 \times M_r(\text{NH}_3) + 5 \times M_r(\text{O}_2) = 4 \times (14.0 + 3 \times 1.0) + 5 \times (2 \times 16.0) = 4 \times 17.0 + 5 \times 32.0 = 68.0 + 160.0 = 228.0\text{ g mol}^{-1}\).

2. Calculate the mass of the desired product (nitrogen monoxide):
Mass of desired product = \(4 \times M_r(\text{NO}) = 4 \times (14.0 + 16.0) = 4 \times 30.0 = 120.0\text{ g mol}^{-1}\).

3. Calculate the percentage atom economy:
Percentage atom economy = \(\frac{120.0}{228.0} \times 100\% = 52.63\% \approx 52.6\%\).

Award 1.05 marks for selecting correct answer A.
Method marks:
- Correct calculation of reactant mass (228.0) and product mass (120.0).
- Application of the atom economy formula to get 52.6%.

There is a very large increase (jump) between the 5th and 6th ionization energies (from 6274 to 21269 \(\text{kJ mol}^{-1}\)). This indicates that the 6th electron is being removed from an inner quantum shell, and element \(X\) has 5 outer-shell (valence) electrons. In Period 3, the element with 5 valence electrons is phosphorus (Group 5 / 15). The hydride of phosphorus is phosphorus trihydride, which has the formula \(\text{XH}_3\) (\(\text{PH}_3\)).

Award 1.05 marks for selecting correct answer B.
Method marks:
- Identification of the jump between the 5th and 6th ionization energy, indicating 5 valence electrons.
- Deduced hydride formula based on valence group.

Boron trifluoride, \(\text{BF}_3\), has three bonding pairs of electrons and no lone pairs on the central boron atom. It has a trigonal planar shape with bond angles of exactly \(120^\circ\). Because of the symmetrical arrangement of the polar B-F bonds, the dipole moments cancel out completely, making the molecule non-polar.

Award 1.05 marks for selecting correct answer A.
Reject other options because:
- \(\text{PF}_3\) is trigonal pyramidal with bond angles less than \(109.5^\circ\) and is polar.
- \(\text{SF}_6\) is octahedral with bond angles of \(90^\circ\).
- \(\text{SO}_2\) is bent with a bond angle less than \(120^\circ\) and is polar.

A propagation step in free radical substitution must involve a radical reactant reacting with a neutral molecule to produce a new radical and a new neutral molecule. The correct step is \(\text{Cl}^\bullet + \text{CH}_3\text{CH}_2\text{CH}_3 \rightarrow \text{HCl} + \text{CH}_3\text{CH}_2\text{CH}_2^\bullet\), which produces a propyl radical and hydrogen chloride.

Award 1.05 marks for selecting correct answer C.
Reject other options because:
- A represents initiation.
- B is chemically incorrect as free hydrogen radicals (\(\text{H}^\bullet\)) are not produced in this mechanism.
- D represents termination.

The addition of \(\text{H}^+\) to the double bond of 2-methylbut-2-ene, \(\text{(CH}_3)_2\text{C=CHCH}_3\), can occur at carbon-3 to form the more stable tertiary carbocation, \(\text{(CH}_3)_2\text{C}^+\text{CH}_2\text{CH}_3\). This is more stable than the secondary carbocation, \(\text{(CH}_3)_2\text{CHCH}^+\text{CH}_3\), which would be formed if the \(\text{H}^+\) added to carbon-2. The tertiary carbocation intermediate leads to the major product, 2-bromo-2-methylbutane.

Award 1.05 marks for selecting correct answer A.
Method marks:
- Identification of tertiary carbocation as the more stable intermediate leading to the major product.

1. Moles of \(\text{H}_2\text{SO}_4 = \frac{18.40}{1000} \times 0.0500 = 9.20 \times 10^{-4}\text{ mol}\).
2. Moles of \(\text{NaOH}\) required = \(2 \times 9.20 \times 10^{-4} = 1.84 \times 10^{-3}\text{ mol}\).
3. Concentration of \(\text{NaOH} = \frac{1.84 \times 10^{-3}}{0.0250} = 0.0736\text{ mol dm}^{-3}\).

Award 1.05 marks for selecting correct answer B.
Method marks:
- Calculating moles of acid correctly.
- Using the correct 1:2 stoichiometry to find moles of sodium hydroxide.
- Dividing moles by volume to get the correct concentration.

The second ionization energy of magnesium is defined as the energy required to remove one mole of electrons from one mole of gaseous singly charged magnesium ions to form one mole of gaseous doubly charged magnesium ions. This is represented by: \(\text{Mg}^+(\text{g}) \rightarrow \text{Mg}^{2+}(\text{g}) + \text{e}^-\).

Award 1.05 marks for selecting correct answer B.
Reject other options because:
- A represents the sum of the first and second ionization energies.
- C features solid-state species.
- D is a combination of sublimation and double ionization.

Covalent character in an ionic compound is maximized when there is high polarization of the anion by the cation (Fajans' Rules). This occurs when:
1. The cation has a high positive charge and small ionic radius (high polarizing power).
2. The anion has a large ionic radius (high polarizability).

Comparing the cations: \(\text{Mg}^{2+}\) has a higher charge and smaller radius than \(\text{Na}^+\), hence higher polarizing power.
Comparing the anions: \(\text{I}^-\) is larger than \(\text{Cl}^-\), hence it is more polarizable.
Therefore, \(\text{MgI}_2\) has the greatest degree of covalent character.

Award 1.05 marks for selecting correct answer D.
Method marks:
- Identification of the cation with high polarizing power (\(\text{Mg}^{2+}\)).
- Identification of the anion with high polarizability (\(\text{I}^-\)).

The successive ionization energies show a massive increase between the 3rd and 4th ionization energy (from 2745 to 11578 \( \text{kJ mol}^{-1} \)). This indicates that the fourth electron is being removed from an inner quantum shell, closer to the nucleus and less shielded. Therefore, the element has three valence electrons and is in Group 13 (aluminum, in Period 3). The element forms a \( 3+ \) ion (\( X^{3+} \)), meaning it reacts with chlorine to form the chloride \( X\text{Cl}_3 \).

1 mark for identifying the correct option (C).

1. Calculate the amount of \( \text{CO}_2 \) produced: \( \text{Moles of CO}_2 = \frac{600\text{ cm}^3}{24000\text{ cm}^3\text{ mol}^{-1}} = 0.025\text{ mol} \). 2. Determine the molar mass of \( M\text{CO}_3 \): Since the mole ratio of \( M\text{CO}_3 : \text{CO}_2 \) is \( 1 : 1 \), \( \text{moles of } M\text{CO}_3 = 0.025\text{ mol} \). \( \text{Molar mass of } M\text{CO}_3 = \frac{2.10\text{ g}}{0.025\text{ mol}} = 84.0\text{ g mol}^{-1} \). 3. Calculate the relative atomic mass of \( M \): \( A_{\text{r}}(M) = 84.0 - A_{\text{r}}(\text{C}) - 3 \times A_{\text{r}}(\text{O}) = 84.0 - 12.0 - 48.0 = 24.0\text{ g mol}^{-1} \). This atomic mass corresponds to Magnesium (\( \text{Mg} \)).

1 mark for identifying the correct option (B).

The amide ion, \( \text{NH}_2^- \), has a central nitrogen atom with 2 bonding pairs and 2 lone pairs (total valence electrons = 5 from N + 2 from H + 1 from negative charge = 8 electrons = 4 pairs). This results in a bent geometry based on a tetrahedral electron-pair geometry. Due to strong lone pair-lone pair repulsion, the bond angle is compressed from the tetrahedral \( 109.5^\circ \) to approximately \( 104.5^\circ \), similar to water (\( \text{H}_2\text{O} \)).

1 mark for selecting the correct option (C).

1. Identify the longest continuous carbon chain: The chain contains 6 carbon atoms, so the parent name is hexane. 2. Number the carbon chain from the end that gives the substituents the lowest possible numbers (locants): Numbering from left to right gives methyl at C2 and ethyl at C4 (locants 2,4). Numbering from right to left gives ethyl at C3 and methyl at C5 (locants 3,5). Since 2,4 is lower than 3,5, we use numbering from left to right. 3. Assemble the name alphabetically: 'ethyl' comes before 'methyl'. Therefore, the IUPAC name is 4-ethyl-2-methylhexane.

1 mark for selecting the correct option (B).

For a molecule to exhibit geometric isomerism, both carbon atoms of the double bond must be attached to two different groups. In 3-methylhex-3-ene (\( \text{CH}_3\text{CH}_2\text{C}(\text{CH}_3)=\text{CHCH}_2\text{CH}_3 \)), C3 is attached to \( -\text{CH}_3 \) and \( -\text{CH}_2\text{CH}_3 \) (different), and C4 is attached to \( -\text{H} \) and \( -\text{CH}_2\text{CH}_3 \) (different). Therefore, it exhibits \( E \)/\( Z \) isomerism.

1 mark for selecting the correct option (C).

A sulfide ion, \( \text{S}^{2-} \), has 18 electrons (configuration \( 1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6 \)). A chloride ion, \( \text{Cl}^- \), also has 18 electrons and shares the exact same electronic configuration.

1 mark for identifying the correct option (A).

Atom economy is calculated as: \( \text{Atom Economy} = \frac{\text{Mass of desired product}}{\text{Total mass of all reactants}} \times 100\% \). Here, the desired product is ethanol: \( 2 \times 46.0 = 92.0\text{ g mol}^{-1} \). The total mass of the reactant glucose is \( 180.0\text{ g mol}^{-1} \). Therefore: \( \text{Atom Economy} = \frac{92.0}{180.0} \times 100\% = 51.1\% \).

1 mark for selecting the correct option (C).

A propagation step must involve a radical reactant reacting with a neutral molecule to produce another radical and a neutral molecule. The reaction \( \text{CH}_3^\bullet + \text{Cl}_2 \rightarrow \text{CH}_3\text{Cl} + \text{Cl}^\bullet \) perfectly matches this description and is one of the two propagation steps in this mechanism.

1 mark for selecting the correct option (C).

There is a very large increase between the third and fourth ionization energies (from \( 2745 \) to \( 11578 \text{ kJ mol}^{-1} \)). This indicates that the fourth electron is being removed from an inner quantum shell, which is closer to the nucleus and experiences less shielding. Thus, element \( \text{X} \) has three valence electrons in its outer shell and forms stable \( \text{X}^{3+} \) ions. The formula of its oxide is therefore \( \text{X}_2\text{O}_3 \).

1 mark for identifying the large increase between the 3rd and 4th ionization energies, concluding that the element forms \( \text{X}^{3+} \) ions, and selecting option B.

Calculate the mass of water lost upon heating:
\( \text{Mass of water lost} = 5.72 \text{ g} - 2.12 \text{ g} = 3.60 \text{ g} \)

Calculate the number of moles of anhydrous \( \text{Na}_2\text{CO}_3 \) and water:
\( n(\text{Na}_2\text{CO}_3) = \frac{2.12 \text{ g}}{106.0 \text{ g mol}^{-1}} = 0.020 \text{ mol} \)
\( n(\text{H}_2\text{O}) = \frac{3.60 \text{ g}}{18.0 \text{ g mol}^{-1}} = 0.20 \text{ mol} \)

Find the simplest whole-number ratio of water to anhydrous salt:
\( x = \frac{0.20 \text{ mol}}{0.020 \text{ mol}} = 10 \)

1 mark for calculating the moles of anhydrous salt and water, establishing the correct molar ratio, and selecting option D.

For an alkene to exhibit stereoisomerism, both carbon atoms involved in the double bond must be attached to two different groups.

- In 2-methylbut-2-ene, \( (\text{CH}_3)_2\text{C=CHCH}_3 \), one carbon of the double bond is bonded to two identical methyl groups.
- In 1,1-dichloroprop-1-ene, \( \text{Cl}_2\text{C=CHCH}_3 \), one carbon of the double bond is bonded to two identical chlorine atoms.
- In 3-methylpent-2-ene, \( \text{CH}_3\text{CH=C(CH}_3)\text{CH}_2\text{CH}_3 \), carbon-2 is bonded to \( -\text{H} \) and \( -\text{CH}_3 \) (different), while carbon-3 is bonded to \( -\text{CH}_3 \) and \( -\text{CH}_2\text{CH}_3 \) (different). Thus, it can exhibit \( E \)-\( Z \) isomerism.
- In 2-methylpent-1-ene, \( \text{CH}_2\text{=C(CH}_3)\text{CH}_2\text{CH}_2\text{CH}_3 \), the terminal carbon is bonded to two identical hydrogen atoms.

1 mark for identifying that both double-bonded carbons must have two different attached groups and correctly identifying 3-methylpent-2-ene.

(a) To transfer the dissolved solid, the student should pour the solution from the beaker through a clean funnel into the volumetric flask. The beaker, glass rod, and funnel must be rinsed multiple times with distilled water, and all the washings transferred into the flask. Distilled water is added until the level is just below the graduation mark, then added dropwise using a pipette until the bottom of the meniscus is exactly on the mark. Finally, the flask is stoppered and inverted several times to ensure it is thoroughly mixed and homogeneous.

(b) Moles of \(\text{HCl}\) = \(0.02500\text{ dm}^3 \times 0.100\text{ mol dm}^{-3} = 2.50 \times 10^{-3}\text{ mol}\).

(c) From the balanced equation, \(1\text{ mol}\) of \(\text{Na}_2\text{CO}_3\) reacts with \(2\text{ mol}\) of \(\text{HCl}\).
Moles of \(\text{Na}_2\text{CO}_3\) in \(25.0\text{ cm}^3\) sample = \(\frac{2.50 \times 10^{-3}}{2} = 1.25 \times 10^{-3}\text{ mol}\).
Moles of \(\text{Na}_2\text{CO}_3\) in \(250.0\text{ cm}^3\) solution = \(1.25 \times 10^{-3} \times 10 = 1.25 \times 10^{-2}\text{ mol}\).

(d) Molar mass of \(\text{Na}_2\text{CO}_3 \cdot n\text{H}_2\text{O}\) = \(\frac{3.58\text{ g}}{1.25 \times 10^{-2}\text{ mol}} = 286.4\text{ g mol}^{-1}\).
Molar mass of anhydrous \(\text{Na}_2\text{CO}_3 = (2 \times 23.0) + 12.0 + (3 \times 16.0) = 106.0\text{ g mol}^{-1}\).
Mass of \(n\text{H}_2\text{O}\) per mole = \(286.4 - 106.0 = 180.4\text{ g}\).
Since the molar mass of \(\text{H}_2\text{O}\) is \(18.0\text{ g mol}^{-1}\):
\(n = \frac{180.4}{18.0} = 10.02 \approx 10\).

(e) Methyl orange transitions in the acidic range (pH 3.1–4.4) which coincides with the equivalence point of the titration of a weak base (sodium carbonate) with a strong acid (hydrochloric acid). Phenolphthalein transitions in the basic range (pH 8.3–10.0) and would change colour too early before the equivalence point is reached.

Maximum 12 marks:
(a) 5 marks total:
- Transfer solution using a funnel [1]
- Rinse beaker, rod, and funnel with distilled water and add washings to the flask [1]
- Add distilled water dropwise near the mark [1]
- Bottom of meniscus aligned with the graduation line [1]
- Invert the stoppered flask several times to mix [1]

(b) 1 mark:
- \(2.50 \times 10^{-3}\text{ mol}\) (or \(0.0025\text{ mol}\)) [1]

(c) 2 marks total:
- Moles of carbonate in \(25.0\text{ cm}^3 = 1.25 \times 10^{-3}\text{ mol}\) [1]
- Moles of carbonate in \(250.0\text{ cm}^3 = 1.25 \times 10^{-2}\text{ mol}\) (allow consequential error from sample moles \(\times 10\)) [1]

(d) 3 marks total:
- Molar mass calculation: \(\frac{3.58}{\text{moles}} = 286.4\text{ g mol}^{-1}\) [1]
- Calculation of mass of water: \(286.4 - 106.0 = 180.4\text{ g}\) [1]
- Correct integer value \(n = 10\) [1]

(e) 1 mark:
- Methyl orange matches the pH at equivalence point of weak base/strong acid titration / phenolphthalein would change colour too early [1]

(a) Gaseous atoms of X are ionized by firing high-energy electrons at them from an electron gun. These high-speed electrons knock out an outer electron from each atom of X, creating positive ions (typically \(\text{X}^+\)). These ions are then accelerated by an electric field (or potential difference) towards a grid, ensuring all ions are accelerated to have the same kinetic energy.

(b) \(A_r = \frac{(24 \times 78.99) + (25 \times 10.00) + (26 \times 11.01)}{100} = \frac{1895.76 + 250.00 + 286.26}{100} = \frac{2432.02}{100} = 24.32\).
Comparing this relative atomic mass with the periodic table, element X is Magnesium (\(\text{Mg}\)).

(c) The second ionization energy of magnesium involves removing an electron from a gaseous \(1+\) ion:
\(\text{Mg}^{+}(\text{g}) \rightarrow \text{Mg}^{2+}(\text{g}) + \text{e}^{-}\)

(d) Magnesium has a higher first ionization energy than sodium because magnesium has a greater nuclear charge (12 protons vs 11 protons) and the outer electron is in the same outer shell (similar shielding), leading to a stronger attraction between the nucleus and the outer electron. Magnesium has a lower first ionization energy than neon because its outer electron is in a higher energy level (\(3s\) vs \(2p\)), which is further from the nucleus and experiences more shielding from inner electron shells, making it easier to remove.

Maximum 12 marks:
(a) 4 marks total:
- Electron gun / firing high-energy electrons [1]
- Knocks out an electron to form \(1+\) positive ions [1]
- Accelerated by an electric field / towards a negative plate [1]
- Ions receive the same kinetic energy [1]

(b) 3 marks total:
- Correct setup: \(\frac{(24 \times 78.99) + (25 \times 10.00) + (26 \times 11.01)}{100}\) [1]
- Relative atomic mass = 24.32 (must be 2 d.p.) [1]
- Identification: Magnesium / \(\text{Mg}\) [1]

(c) 2 marks total:
- Correct species: \(\text{Mg}^+(\text{g}) \rightarrow \text{Mg}^{2+}(\text{g}) + \text{e}^-\)\ (or similar layout, e.g., minus electron on left) [1]
- Correct state symbols: \((\text{g})\) on both \(\text{Mg}^+\) and \(\text{Mg}^{2+}\) [1]

(d) 3 marks total:
- Magnesium has more protons/greater nuclear charge than sodium, but outer electrons are in the same shell / experience same shielding [1]
- Stronger electrostatic attraction between nucleus and outer electron in Mg than Na [1]
- Magnesium's outer electron is in a higher energy shell/further from nucleus/more shielded than neon's outer electron [1]

(a)(i) The dot-and-cross diagram should show a central Silicon (Si) atom sharing 4 pairs of electrons with 4 Chlorine (Cl) atoms (one pair for each bond). Silicon will have 8 electrons in its outer shell (4 dots and 4 crosses). Each chlorine atom will have 8 electrons in its outer shell, comprising 1 shared pair and 3 lone pairs (6 non-bonding electrons).

(a)(ii) Shape: Tetrahedral
Bond angle: \(109.5^\circ\)
Explanation: The central silicon atom has four bonding pairs of electrons and zero lone pairs in its outer shell. According to electron pair repulsion theory, these four bonding pairs repel each other equally and move as far apart as possible to minimize repulsion, resulting in a tetrahedral geometry.

(b) Shape of \(\text{PCl}_4^{+}\): Tetrahedral (since phosphorus has 5 valence electrons, losing one to form the \(1+\) ion, leaving 4 valence electrons which form 4 bonding pairs with chlorine and 0 lone pairs).

(c)(i) Magnesium oxide consists of \(\text{Mg}^{2+}\) and \(\text{O}^{2-}\) ions, which have higher charges than the \(\text{Na}^{+}\) and \(\text{Cl}^{-}\) ions in sodium chloride. Additionally, \(\text{Mg}^{2+}\) and \(\text{O}^{2-}\) have smaller ionic radii than \(\text{Na}^{+}\) and \(\text{Cl}^{-}\). The electrostatic forces of attraction between the doubly charged, smaller ions in \(\text{MgO}\) are much stronger than those between the singly charged, larger ions in \(\text{NaCl}\), requiring far more energy to break the giant ionic lattice.

(c)(ii) Water is a polar molecule with a partial negative charge (\(\delta-\)) on the oxygen atom and a partial positive charge (\(\delta+\)) on the hydrogen atoms. When \(\text{NaCl}\) dissolves, the \(\delta-\) oxygen atoms of water molecules are electrostatically attracted to the positive sodium ions (\(\text{Na}^{+}\)), while the \(\delta+\) hydrogen atoms are attracted to the negative chloride ions (\(\text{Cl}^{-}\)). These ion-dipole forces release enough energy (hydration enthalpy) to overcome the ionic bonds within the giant lattice, allowing the ions to become hydrated and free to move in solution.

Maximum 12 marks:
(a)(i) 2 marks total:
- Correct 4 shared pairs of electrons between Si and Cl [1]
- Correct 6 non-bonding outer shell electrons shown on each of the 4 Cl atoms [1]

(a)(ii) 3 marks total:
- Shape: Tetrahedral AND Bond angle: \(109.5^\circ\) (both required) [1]
- Mentions 4 bonding pairs and 0 lone pairs around the silicon atom [1]
- States that electron pairs repel each other to get as far apart as possible to minimize repulsion [1]

(b) 1 mark:
- Tetrahedral [1]

(c)(i) 3 marks total:
- \(\text{Mg}^{2+}\) and \(\text{O}^{2-}\) have higher charges than \(\text{Na}^{+}\) and \(\text{Cl}^{-}\) [1]
- \(\text{Mg}^{2+}\) and \(\text{O}^{2-}\) are smaller than \(\text{Na}^{+}\) and \(\text{Cl}^{-}\) / have higher charge density [1]
- Stronger electrostatic attraction in magnesium oxide requires more energy to overcome [1]

(c)(ii) 3 marks total:
- Water is polar with \(\delta+\text{ H}\) and \(\delta-\text{ O}\) [1]
- Oxygen (\(\delta-\)) is attracted to \(\text{Na}^{+}\) and hydrogen (\(\delta+\)) is attracted to \(\text{Cl}^{-}\) [1]
- Form ion-dipole forces which release energy to break the ionic lattice [1]

(a)(i) The three structural isomers of \(\text{C}_5\text{H}_{12}\) are:
1. Pentane (a continuous chain of 5 carbon atoms in a zigzag skeletal representation).
2. 2-Methylbutane (a continuous chain of 4 carbon atoms with a single methyl branch on the second carbon).
3. 2,2-Dimethylpropane (a central carbon atom bonded to 4 surrounding methyl groups in a cross-like skeletal shape).

(a)(ii) 2,2-Dimethylpropane is highly branched and has a spherical shape, which minimizes its surface area. Pentane is a straight-chain molecule with a larger surface area. Pentane molecules can pack more closely together, resulting in more points of contact between adjacent molecules. Therefore, pentane has stronger London forces (instantaneous dipole-induced dipole forces) than 2,2-dimethylpropane, requiring more thermal energy to separate the molecules and boil.

(b)(i) Initiation step: \(\text{Cl}_2 \rightarrow 2\text{Cl}^\bullet\)
Type of bond fission: Homolytic fission (where each chlorine atom retains one electron from the shared pair in the covalent bond).

(b)(ii) Propagation steps:
Step 1: \(\text{CH}_3\text{CH}_2\text{CH}_3 + \text{Cl}^\bullet \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2^\bullet + \text{HCl}\)
Step 2: \(\text{CH}_3\text{CH}_2\text{CH}_2^\bullet + \text{Cl}_2 \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{Cl} + \text{Cl}^\bullet\)

(b)(iii) The UV light provides the activation energy / photon energy required to break the covalent bond in chlorine (\(\text{Cl}-\text{Cl}\)) to initiate the free radical substitution.

(b)(iv) Hexane is formed when two propyl radicals react together in a termination step:
\(2\text{CH}_3\text{CH}_2\text{CH}_2^\bullet \rightarrow \text{C}_6\text{H}_{14}\) (or \(2\text{C}_3\text{H}_7^\bullet \rightarrow \text{C}_6\text{H}_{14}\))

Maximum 12 marks:
(a)(i) 3 marks total:
- Pentane skeletal formula AND correct name [1]
- 2-methylbutane skeletal formula AND correct name [1]
- 2,2-dimethylpropane skeletal formula AND correct name [1]

(a)(ii) 3 marks total:
- 2,2-dimethylpropane is more branched / spherical / has less surface area of contact [1]
- Pentane has more points of contact between molecules [1]
- London forces are weaker in 2,2-dimethylpropane / stronger in pentane, requiring less energy to break [1]

(b)(i) 2 marks total:
- Correct initiation equation: \(\text{Cl}_2 \rightarrow 2\text{Cl}^\bullet\) (dots must be clear) [1]
- Homolytic fission [1]

(b)(ii) 2 marks total:
- Propagation 1: \(\text{CH}_3\text{CH}_2\text{CH}_3 + \text{Cl}^\bullet \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2^\bullet + \text{HCl}\) [1]
- Propagation 2: \(\text{CH}_3\text{CH}_2\text{CH}_2^\bullet + \text{Cl}_2 \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{Cl} + \text{Cl}^\bullet\) [1]
(Accept \(\text{C}_3\text{H}_7^\bullet\) for propyl radical)

(b)(iii) 1 mark:
- To break the \(\text{Cl}-\text{Cl}\) bond / provide energy for homolytic fission / generate chlorine free radicals [1]

(b)(iv) 1 mark:
- Correct termination equation: \(2\text{C}_3\text{H}_7^\bullet \rightarrow \text{C}_6\text{H}_{14}\) (structural or molecular formula) [1]

(a)(i) Stereoisomers are molecules with the same structural formula but a different spatial arrangement of atoms.

(a)(ii) In both molecules, the \(\text{C}=\text{C}\) double bond has restricted rotation due to the presence of the \(\pi\)-bond. However, for stereoisomerism to exist, each carbon atom of the double bond must be attached to two different groups. In but-2-ene, each of the double-bonded carbons is attached to a hydrogen (\(-\text{H}\)) and a methyl group (\(-\text{CH}_3\)), which are different. In but-1-ene, one of the carbon atoms in the double bond is bonded to two identical hydrogen atoms, preventing stereoisomerism.

(a)(iii)
- (E)-but-2-ene skeletal structure: a central \(\text{C}=\text{C}\) double bond with the two terminal methyl groups on opposite sides of the double bond (trans configuration).
- (Z)-but-2-ene skeletal structure: a central \(\text{C}=\text{C}\) double bond with the two terminal methyl groups on the same side of the double bond (cis configuration).

(b)(i) The mechanism of electrophilic addition:
1. A dipole is drawn on \(\text{HBr}\) as \(\delta+\) on \(\text{H}\) and \(\delta-\) on \(\text{Br}\).
2. A curly arrow is drawn from the double bond of but-1-ene to the \(\text{H}\) of \(\text{HBr}\).
3. A curly arrow is drawn from the \(\text{H}-\text{Br}\) bond to the \(\text{Br}\) atom.
4. This produces a secondary carbocation intermediate: \(\text{CH}_3\text{CH}_2\text{CH}^{+}\text{CH}_3\) and a bromide ion, \(\text{Br}^{-}\), which has a lone pair of electrons.
5. A curly arrow is drawn from the lone pair on the \(\text{Br}^{-}\) ion to the positively charged carbon of the carbocation to form the final product, 2-bromobutane.

(b)(ii) The electrophilic addition of \(\text{HBr}\) to but-1-ene can proceed via two different carbocation intermediates. The formation of the major product, 2-bromobutane, proceeds via a secondary carbocation (\(\text{CH}_3\text{CH}_2\text{CH}^{+}\text{CH}_3\)), whereas the minor product, 1-bromobutane, proceeds via a primary carbocation (\(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2^{+}\)). Secondary carbocations are more stable than primary carbocations because they have two electron-donating alkyl groups (instead of one). These alkyl groups release electron density towards the positive carbon atom, dispersing the positive charge and making the intermediate more stable and therefore faster/more likely to form.

Maximum 12 marks:
(a)(i) 1 mark:
- Same structural formula but different spatial arrangement / arrangement of atoms in space [1]

(a)(ii) 2 marks total:
- C=C double bond restricts rotation / does not allow free rotation [1]
- But-2-ene has two different groups on both carbons of double bond whereas but-1-ene has two identical groups (H atoms) on one carbon of the double bond [1]

(a)(iii) 2 marks total:
- Correct skeletal drawing of (E)-but-2-ene with methyls trans [1]
- Correct skeletal drawing of (Z)-but-2-ene with methyls cis [1]

(b)(i) 4 marks total:
- Curly arrow from C=C bond to H AND curly arrow from H–Br bond to Br [1]
- Correct dipoles: \(\delta+\text{ H}\) and \(\delta-\text{ Br}\) [1]
- Correct structure of secondary carbocation intermediate AND \(\text{Br}^-\)\ with a lone pair [1]
- Curly arrow from lone pair of \(\text{Br}^-\)\ to \(\text{C}^+\)\ of the secondary carbocation [1]

(b)(ii) 3 marks total:
- Major product goes via secondary carbocation and minor via primary carbocation [1]
- Secondary carbocations are more stable than primary carbocations [1]
- Due to the electron-donating inductive effect of two alkyl groups (releasing electron density to disperse the positive charge) compared to only one [1]

HF has the highest boiling temperature because it is the only hydrogen halide capable of forming intermolecular hydrogen bonds, which are significantly stronger than London forces and permanent dipole-dipole forces. For HCl, HBr, and HI, the boiling temperatures increase down the group because the number of electrons per molecule increases, which increases the strength of the London forces. Therefore, the order of decreasing boiling temperature is HF > HI > HBr > HCl.

1.05 marks: Correct option selected (A). 0 marks: Any other option selected or left blank.

The thermal stability of Group 2 carbonates increases down the group. The magnesium ion has a smaller ionic radius than the barium ion, resulting in a higher charge density. This allows the magnesium ion to polarise the carbonate ion's electron cloud more effectively, weakening one of the carbon-oxygen bonds within the carbonate ion and facilitating its decomposition into magnesium oxide and carbon dioxide at a lower temperature.

1.05 marks: Correct option selected (A). 0 marks: Any other option selected or left blank.

The rate of hydrolysis of halogenoalkanes depends on the strength of the carbon-halogen bond. Down Group 7, the C-X bond length increases and the bond enthalpy decreases, meaning the C-I bond is the weakest and breaks most easily. Thus, 1-iodobutane reacts the fastest to produce a yellow precipitate of silver iodide. The C-Cl bond is the strongest and breaks least easily, meaning 1-chlorobutane reacts the slowest. The order of rates from fastest to slowest is 1-iodobutane > 1-bromobutane > 1-chlorobutane.

1.05 marks: Correct option selected (A). 0 marks: Any other option selected or left blank.

The equation for the formation of propane is: \( 3\text{C(s)} + 4\text{H}_2\text{(g)} \rightarrow \text{C}_3\text{H}_8\text{(g)} \). Using Hess's Law and enthalpy of combustion data: \( \Delta_f H^\ominus = \sum \Delta_c H^\ominus(\text{reactants}) - \sum \Delta_c H^\ominus(\text{products}) \). This gives \( \Delta_f H^\ominus = [3 \times (-393.5) + 4 \times (-285.8)] - (-2219.2) = [-1180.5 - 1143.2] + 2219.2 = -104.5\text{ kJ mol}^{-1} \).

1.05 marks: Correct option selected (A). 0 marks: Any other option selected or left blank.

When temperature increases, the average kinetic energy of the molecules increases. On the Maxwell-Boltzmann distribution curve, this shifts the distribution to the right. Since the total number of molecules remains constant, the area under the curve must remain constant, which causes the peak of the curve to become lower and flatter. The activation energy is unaffected by temperature.

1.05 marks: Correct option selected (A). 0 marks: Any other option selected or left blank.

The molecular formula \( \text{C}_3\text{H}_6\text{O} \) has one degree of unsaturation. The strong, sharp IR band at \(1715\text{ cm}^{-1}\) indicates the presence of a carbonyl group (\(C=O\)). The lack of a broad band in the \(3200\text{ - }3600\text{ cm}^{-1}\) range rules out an alcohol (\(O-H\)) group. Propanone is a ketone (\( \text{CH}_3\text{COCH}_3 \)), which contains a \(C=O\) group and no \(O-H\) group, perfectly matching the IR data. Propanoic acid contains an \(O-H\) group and has the formula \( \text{C}_3\text{H}_6\text{O}_2 \). Both propan-1-ol and prop-2-en-1-ol contain \(O-H\) groups.

1.05 marks: Correct option selected (A). 0 marks: Any other option selected or left blank.

Iodide ions are powerful reducing agents and reduce sulfuric acid (where sulfur is in the +6 oxidation state) to sulfur dioxide (+4), sulfur (0), and hydrogen sulfide (-2). Therefore, \( \text{H}_2\text{S} \) is a reduction product. Hydrogen iodide is formed via an acid-base reaction, not redox. Iodine is an oxidation product of iodide. Potassium hydrogensulfate is formed in the initial acid-base reaction and sulfur is not reduced.

1.05 marks: Correct option selected (A). 0 marks: Any other option selected or left blank.

The value of the equilibrium constant, \( K_c \), is only affected by temperature. The forward reaction is exothermic. According to Le Chatelier's principle, decreasing the temperature shifts the equilibrium to the right to release heat. Since the equilibrium position shifts to the right, the concentration of products increases relative to the reactants, which increases the value of \( K_c \). Changing pressure or adding a catalyst does not alter the value of \( K_c \).

1.05 marks: Correct option selected (A). 0 marks: Any other option selected or left blank.

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

Using a Hess's Law cycle with enthalpy changes of combustion:
\(\Delta H_{\text{f}}^{\theta} = \sum \Delta H_{\text{c}}^{\theta}(\text{reactants}) - \sum \Delta H_{\text{c}}^{\theta}(\text{products})\)

\(\Delta H_{\text{f}}^{\theta} = [3 \times (-394) + 4 \times (-286)] - (-2021)\)
\(\Delta H_{\text{f}}^{\theta} = [-1182 - 1144] + 2021 = -2326 + 2021 = -305 \text{ kJ mol}^{-1}\)

[1 mark] - Correctly identifies the stoichiometry of the formation reaction (3C and 4H2) and applies Hess's Law to calculate \(-305 \text{ kJ mol}^{-1}\) (Option A).
[0 marks] - For incorrect calculations (e.g. B, C, or D).

Boiling point in halogenoalkanes is determined by the strength of intermolecular forces. Both molecules have London forces and permanent dipole-dipole forces. Although the C-Cl bond is more polar than the C-I bond, the iodine atom in 1-iodobutane has significantly more electrons than the chlorine atom in 1-chlorobutane. This results in much stronger London (dispersion) forces between 1-iodobutane molecules, which require more thermal energy to overcome.

[1 mark] - Correct choice is C.
[0 marks] - Incorrect options: A is chemically incorrect (C-Cl is stronger than C-I); B is incorrect as dipole-dipole forces are actually stronger in 1-chlorobutane; D is incorrect as no hydrogen bonding is present in these compounds.

During the reaction of solid sodium iodide with concentrated sulfuric acid, iodide is a strong enough reducing agent to reduce sulfur from its +6 oxidation state in \(\text{H}_2\text{SO}_4\) to various species: \(\text{SO}_2\) (S oxidation state +4), \(\text{S}\) (S oxidation state 0), and \(\text{H}_2\text{S}\) (S oxidation state -2). The lowest oxidation state among these is -2, corresponding to hydrogen sulfide, \(\text{H}_2\text{S}\).

[1 mark] - Correct choice is A.
[0 marks] - B, C are reduction products but not the lowest oxidation state; D (HI) is an acid-base product, not a reduction product of sulfur.

The broad band at \(3200\text{–}3600 \text{ cm}^{-1}\) indicates an O-H group (alcohol). The absorption at \(1650 \text{ cm}^{-1}\) indicates a C=C double bond. The absence of an absorption in the \(1700\text{–}1750 \text{ cm}^{-1}\) range confirms there is no carbonyl (C=O) group, ruling out propanal and propanone. An unsaturated alcohol with three carbons and one double bond is prop-2-en-1-ol.

[1 mark] - Correct choice is C.
[0 marks] - A and B contain a C=O group which would show a strong peak at \(1700\text{–}1750 \text{ cm}^{-1}\); D (cyclopropanol) is cyclic and lacks a C=C double bond, so it would not show a peak at \(1650 \text{ cm}^{-1}\).

As temperature increases, the particles gain kinetic energy, shifting the Maxwell-Boltzmann distribution curve to the right. To maintain the same total number of particles (which corresponds to the area under the curve), the peak must flatten and become lower. Therefore, the peak moves to the right and becomes lower, while the total area under the curve remains constant.

[1 mark] - Correct choice is B.
[0 marks] - Other options describe incorrect shifts or incorrect changes in the total area.

The rate of hydrolysis of halogenoalkanes is determined by the strength of the carbon-halogen bond (bond enthalpy), rather than the polarity of the bond. The C-I bond is the weakest bond among the three because iodine has the largest atomic radius, resulting in the longest and weakest shared pair of electrons. Therefore, 1-iodobutane reacts the fastest.

[1 mark] - Correct choice is B.
[0 marks] - A is incorrect because bond enthalpy, not polarity, is the dominating factor; C and D are chemically incorrect explanations.

Down Group 2, the size of the metal cation increases while its charge remains \(2+\). This decreases the charge density of the cation, reducing its ability to polarise (distort) the large carbonate anion. Because the carbonate anion is less distorted down the group, more thermal energy is required to break the C-O bond to form the metal oxide and carbon dioxide, making the carbonate more thermally stable.

[1 mark] - Correct choice is A.
[0 marks] - B describes a decrease in thermal stability; C and D attribute the trend to incorrect factors (lattice energy decrease actually opposes the stability trend, and ionisation energy of the metal is not the cause of thermal decomposition trends).

Hydrochloric acid is a strong acid and is fully dissociated in aqueous solution, whereas ethanoic acid is a weak acid and is only partially dissociated. For the neutralisation of ethanoic acid to proceed to completion, the undissociated ethanoic acid molecules must first dissociate into ions. This dissociation step is endothermic (requires energy to break the O-H bonds), which reduces the overall amount of heat released, making the overall process less exothermic.

[1 mark] - Correct choice is A.
[0 marks] - B, C, and D are incorrect chemical explanations.

Butan-1-ol (A) and 2-methylpropan-2-ol (B) can both form intermolecular hydrogen bonds due to their hydroxyl (\(-\text{OH}\)) groups, whereas ethoxyethane (C) and butanal (D) only experience weaker dipole-dipole interactions and London forces. Between the two alcohols, butan-1-ol has a straight carbon chain, allowing for a greater surface contact area between molecules compared to the branched structure of 2-methylpropan-2-ol. This results in stronger London forces in butan-1-ol, requiring more energy to overcome and leading to a higher boiling temperature.

Correct Option: A (1.05 marks)
- Award 1.05 marks for the correct selection of option A.
- Reject B: Branched isomers have a smaller surface contact area and weaker London forces, resulting in a lower boiling temperature.
- Reject C and D: These compounds do not contain O-H bonds and cannot form intermolecular hydrogen bonds with themselves, leading to significantly lower boiling temperatures.

When potassium bromide reacts with concentrated sulfuric acid, the bromide ions are oxidized to bromine (orange-brown vapour), and the sulfuric acid is reduced to sulfur dioxide (\(\text{SO}_2\)). \(\text{SO}_2\) is a reducing agent that reduces the orange dichromate(VI) ions to green chromium(III) ions, turning the paper green. Bromide is not a strong enough reducing agent to reduce the sulfur further to elemental sulfur (yellow solid) or hydrogen sulfide (gas with a bad-egg smell). In contrast, iodide ions are stronger reducing agents and reduce sulfuric acid to \(\text{SO}_2\), yellow sulfur, and hydrogen sulfide.

Correct Option: C (1.05 marks)
- Award 1.05 marks for the correct selection of option C.
- Reject A and B: Fluoride and chloride ions are weak reducing agents and do not reduce concentrated sulfuric acid to produce sulfur dioxide.
- Reject D: Iodide ions are very strong reducing agents and will reduce sulfuric acid all the way to elemental sulfur (yellow solid) and hydrogen sulfide (bad-egg smell).

In nucleophilic substitution reactions of halogenoalkanes, both bond polarity and bond enthalpy play a role. However, bond enthalpy is the dominant factor in determining the rate of reaction. The C–I bond is much weaker (has a lower bond enthalpy) than the C–Cl bond because iodine has a larger atomic radius, resulting in less effective orbital overlap with the carbon atom. Therefore, the C–I bond breaks much more easily during the reaction, which outweighs the effect of the C–Cl bond being more polar.

Correct Option: B (1.05 marks)
- Award 1.05 marks for the correct selection of option B.
- Reject A: The C–Cl bond is more polar than the C–I bond because chlorine is more electronegative than iodine; however, bond polarity does not determine the rate trend here.
- Reject C: The C–I bond has a lower bond enthalpy, not a higher one, compared to the C–Cl bond.
- Reject D: Chlorine is more electronegative than iodine, not less.

(a) The reducing ability of the halide ions increases down Group 7 from fluoride to iodide. As you go down the group, the ionic radius of the halide ion increases and there is more shielding from inner electron shells. Consequently, the attraction between the nucleus and the outermost electrons becomes weaker, allowing the outer electrons to be lost more easily.

(b) The balanced equation is: 8NaI + 9H2SO4 -> 8NaHSO4 + H2S + 4I2 + 4H2O. Alternatively, the ionic equation is: 8I- + 10H+ + SO42- -> H2S + 4I2 + 4H2O.

(c) Three distinct observations include: 1. Purple vapour (gas) due to the sublimation of iodine. 2. A dark grey or black solid forming on the sides of the tube due to iodine condensing. 3. A yellow solid (sulfur) precipitating. 4. A rotten-egg smell from the gaseous hydrogen sulfide (H2S).

(d) Chloride ions are weaker reducing agents than iodide ions because they are smaller and their outer electrons are more strongly held by the nucleus. Concentrated sulfuric acid is not a strong enough oxidizing agent to oxidize chloride ions to chlorine gas; hence, only an acid-base reaction occurs to yield hydrogen chloride gas.

(a) [3 Marks]
- 1 mark: Stating reducing ability increases down the group.
- 1 mark: Mentioning increased ionic radius and shielding.
- 1 mark: Explaining weaker attraction to outer electrons, meaning they are lost more easily.

(b) [2 Marks]
- 1 mark: Correct reactants and products.
- 1 mark: Correct balancing.

(c) [3 Marks]
- 1 mark each for any three valid observations: purple vapour/fumes, grey-black solid (iodine), yellow solid (sulfur), bad egg smell (hydrogen sulfide), or steamy/misty fumes (hydrogen iodide).

(d) [2.25 Marks]
- 1.25 marks: Explaining that chloride is a weak reducing agent (or less easily oxidized than iodide).
- 1 mark: Stating that concentrated sulfuric acid is not strong enough to oxidize chloride ions.

(a) The rate of hydrolysis increases in the order: 1-chlorobutane < 1-bromobutane < 1-iodobutane. The rate of reaction depends on the strength of the carbon-halogen bond. Down Group 7, the bond enthalpy of the C-X bond decreases (C-Cl is strongest, C-I is weakest) because the halogen atom gets larger and orbital overlap is less effective. Therefore, the C-I bond is broken most easily, leading to the fastest rate of hydrolysis. This bond enthalpy factor outweighs the bond polarity factor.

(b) Mix equal amounts of each halogenoalkane in separate test tubes with ethanol, which acts as a mutual solvent. Add equal volumes of aqueous silver nitrate solution to each tube. Heat the mixtures in a water bath at a constant temperature (e.g., 50 °C to 60 °C). Measure the time taken for a precipitate to appear in each test tube. The 1-iodobutane will form a yellow precipitate quickest, 1-bromobutane will form a cream precipitate at a moderate speed, and 1-chlorobutane will form a white precipitate slowest.

(c) The ionic equation is: CH3CH2CH2CH2Br + OH- -> CH3CH2CH2CH2OH + Br-.

(a) [4 Marks]
- 1 mark: Correct trend (rate increases: chloro < bromo < iodo).
- 1 mark: Stating C-X bond strength / bond enthalpy decreases down the group.
- 1 mark: Explaining that the weaker C-I bond breaks more easily (lower activation energy).
- 1 mark: Stating that bond enthalpy is the dominant factor over bond polarity.

(b) [4.25 Marks]
- 1 mark: Use of ethanol as a common solvent.
- 1 mark: Use of aqueous silver nitrate.
- 1 mark: Constant temperature water bath (fair test).
- 1.25 marks: Measuring the time taken for precipitate to appear (and linking fastest to yellow AgI, slowest to white AgCl).

(c) [2 Marks]
- 1 mark: Correct structures of reactants and products.
- 1 mark: Correct balancing and ionic charges.

(a) Step 1: Calculate the temperature change: dT = 48.7 - 20.2 = 28.5 °C.
Step 2: Calculate heat energy transferred to the water: Q = m * c * dT = 100.0 * 4.18 * 28.5 = 11913 J = 11.913 kJ.
Step 3: Calculate the amount of propan-1-ol burned: n = mass / molar mass = 0.750 / 60.0 = 0.0125 mol.
Step 4: Calculate the enthalpy change: dH = -Q / n = -11.913 / 0.0125 = -953 kJ/mol.

(b) Reasons for the value being less exothermic: 1. Heat loss to the surrounding air. 2. Heat absorbed by the copper calorimeter/beaker itself instead of the water. 3. Evaporation of the propan-1-ol from the wick after weighing but before ignition (or while cooling). To minimise heat loss, a draught screen (wind shield) can be placed around the apparatus, or the beaker can be insulated. To minimise evaporation, the cap should be placed back on the spirit burner immediately after extinguishing.

(c) The balanced equation is: C3H7OH(l) + 4.5O2(g) -> 3CO2(g) + 4H2O(l) (or 2C3H7OH + 9O2 -> 6CO2 + 8H2O).

(a) [5 Marks]
- 1 mark: Correct temperature difference of 28.5 °C.
- 1 mark: Correct heat calculation of 11913 J (or 11.9 kJ).
- 1 mark: Correct moles of propan-1-ol of 0.0125 mol.
- 1 mark: Evaluation of 953 (kJ/mol).
- 1 mark: Correct negative sign and unit of kJ/mol.

(b) [3.25 Marks]
- 1 mark: Stating heat loss to surroundings.
- 1 mark: Stating heat absorbed by calorimeter / evaporation of fuel.
- 1.25 marks: Stating a correct minimisation technique corresponding to one of the identified errors (e.g., draft shields, insulating the container, capping the burner).

(c) [2 Marks]
- 1 mark: Correct species.
- 1 mark: Correctly balanced equation (accept multiples).

(a) Compound X is butan-2-ol. Its skeletal structure consists of a four-carbon chain with a hydroxyl (-OH) group attached to carbon-2. Compound Y is butanone. Its skeletal structure consists of a four-carbon chain with a carbonyl oxygen double-bonded to carbon-2.

(b) Compound X (butan-2-ol) will show a broad absorption band in its IR spectrum between 3200-3750 cm-1, which is characteristic of the O-H (alcohol) bond. This peak will be completely absent in compound Y. Compound Y (butanone) will show a strong, sharp absorption band between 1675-1750 cm-1, which is characteristic of the C=O (carbonyl) bond. This peak will not be present in the spectrum of compound X.

(c) The equation for the reaction is: CH3CH(OH)CH2CH3 + [O] -> CH3COCH2CH3 + H2O. (Skeletal/molecular representation is also acceptable).

(a) [4 Marks]
- 1 mark: IUPAC name 'butan-2-ol'.
- 1 mark: Correct skeletal structure of butan-2-ol.
- 1 mark: IUPAC name 'butanone'.
- 1 mark: Correct skeletal structure of butanone.

(b) [3.25 Marks]
- 1 mark: Mentioning O-H alcohol stretch in X at 3200-3750 cm-1.
- 1 mark: Mentioning C=O stretch in Y at 1675-1750 cm-1.
- 1.25 marks: Explaining clearly that the O-H peak is absent in Y and the C=O peak is absent in X.

(c) [3 Marks]
- 1 mark: Correct reactant structures.
- 1 mark: [O] as oxidant and correct organic product.
- 1 mark: H2O as a product and balancing.

**(a)** The balanced equation is:
\(\text{(CH}_3)_3\text{COH} + \text{HCl} \rightarrow \text{(CH}_3)_3\text{CCl} + \text{H}_2\text{O}\)
2-methylpropan-2-ol is a tertiary alcohol. It easily undergoes nucleophilic substitution via an \(S_{\text{N}}1\) pathway because it forms a stable tertiary carbocation intermediate. Butan-1-ol is a primary alcohol and must react via an \(S_{\text{N}}2\) mechanism, which has a much higher activation energy and is extremely slow at room temperature.

**(b)(i)** To wash the mixture:
1. Place the mixture into a separating funnel, stopper it, and invert it to shake gently.
2. Release pressure regularly by opening the tap while the funnel is inverted (and pointed away from people).
3. Safety precaution: Shaking the mixture produces carbon dioxide gas from the reaction of unreacted hydrochloric acid with sodium hydrogencarbonate. Releasing the pressure prevents the stopper from flying off or the glassware from cracking.

**(b)(ii)** Anhydrous sodium sulfate acts as a drying agent to absorb any remaining water/traces of moisture from the organic layer. Once dry, the organic liquid will change from cloudy to clear.

**(c)** 2-methylpropan-2-ol contains a highly polar \(\text{O-H}\) bond, allowing it to form strong intermolecular hydrogen bonds. 2-chloro-2-methylpropane contains polar \(\text{C-Cl}\) bonds, so its molecules are held together by permanent dipole-dipole interactions and London forces. Hydrogen bonds are significantly stronger than permanent dipole-dipole interactions or London forces, requiring much more energy to overcome, resulting in a higher boiling point for the alcohol.

**(d)(i)** Step 1: The carbon-chlorine bond breaks heterolytically. Draw a curly arrow from the \(\text{C-Cl}\) bond to the chlorine atom. This produces a tertiary carbocation intermediate, \(\text{(CH}_3)_3\text{C}^+\), and a chloride ion, \(\text{Cl}^-\).
Step 2: A hydroxide ion acts as a nucleophile. Draw a curly arrow from a lone pair on the oxygen of the hydroxide ion, \(\text{OH}^-\), to the positively charged carbon atom of the carbocation. This forms the product, 2-methylpropan-2-ol.

**(d)(ii)** The reaction occurs in multiple steps. The first step (formation of the carbocation) is the slow, rate-determining step, which depends only on the concentration of the halogenoalkane. Since the hydroxide ion is only involved in a subsequent fast step, its concentration does not affect the rate, hence it is absent from the rate equation.

**(e)(i)** A yellow precipitate (silver iodide) forms much faster than a white precipitate (silver chloride) / a yellow precipitate is observed almost instantly, whereas the white precipitate takes significantly longer to appear.

**(e)(ii)** The rate of hydrolysis depends on the bond enthalpy of the carbon-halogen bond, not its polarity. The \(\text{C-I}\) bond enthalpy is much lower (weaker bond) than the \(\text{C-Cl}\) bond enthalpy (stronger bond). Consequently, the \(\text{C-I}\) bond is much easier to break, resulting in a much lower activation energy and a faster reaction rate.

**(a) [3 Marks]**
• MP1: Correct balanced chemical equation: \(\text{(CH}_3)_3\text{COH} + \text{HCl} \rightarrow \text{(CH}_3)_3\text{CCl} + \text{H}_2\text{O}\) (1)
• MP2: Identifies that 2-methylpropan-2-ol is a tertiary alcohol / forms a highly stable tertiary carbocation (1)
• MP3: Identifies that butan-1-ol is a primary alcohol / primary carbocations are highly unstable, meaning it must react via a slower \(S_{\text{N}}2\) pathway with high activation energy (1)

**(b)(i) [3 Marks]**
• MP1: Shakes the separating funnel with stopper in and inverts it (1)
• MP2: Opens the tap while inverted to release built-up gas/pressure (1)
• MP3: Explains that pressure builds up due to carbon dioxide gas (\(\text{CO}_2\)) produced from the acid-carbonate reaction, which must be released to prevent the stopper popping out/glass cracking (1)

**(b)(ii) [2 Marks]**
• MP1: Explains that anhydrous sodium sulfate acts as a drying agent / removes traces of water (1)
• MP2: States the visual change is from cloudy/turbid to clear (1)

**(c) [3 Marks]**
• MP1: States 2-methylpropan-2-ol has intermolecular hydrogen bonds (1)
• MP2: States 2-chloro-2-methylpropane has permanent dipole-dipole interactions and London forces (but no hydrogen bonds) (1)
• MP3: States that hydrogen bonds are stronger than dipole-dipole/London forces, requiring more energy to break (1)

**(d)(i) [4 Marks]**
• MP1: Curly arrow starting from the \(\text{C-Cl}\) bond to the chlorine atom (1)
• MP2: Correct structure of the tertiary carbocation intermediate: \(\text{(CH}_3)_3\text{C}^+\) (1)
• MP3: Correct structure of leaving chloride ion: \(\text{Cl}^-\) (1)
• MP4: Curly arrow from a lone pair on the oxygen of the hydroxide ion (\(\text{OH}^-\)) to the positive carbon atom of the carbocation (1)

**(d)(ii) [1 Mark]**
• MP1: Explains that the rate-determining step (slow step) only involves the halogenoalkane / hydroxide ions only react in a subsequent fast step (1)

**(e)(i) [1 Mark]**
• MP1: States that a yellow precipitate (silver iodide) forms much faster than a white precipitate (silver chloride) (1)

**(e)(ii) [2 Marks]**
• MP1: States that the \(\text{C-I}\) bond enthalpy is lower than the \(\text{C-Cl}\) bond enthalpy / \(\text{C-I}\) bond is weaker than the \(\text{C-Cl}\) bond (1)
• MP2: Concludes that less energy is required to break the \(\text{C-I}\) bond, leading to a faster rate of hydrolysis (accept that bond enthalpy, not polarity, is the dominant factor) (1)

**(a)** The balanced equation is:
\(\text{(CH}_3)_3\text{COH} + \text{HCl} \rightarrow \text{(CH}_3)_3\text{CCl} + \text{H}_2\text{O}\)
2-methylpropan-2-ol is a tertiary alcohol. It easily undergoes nucleophilic substitution via an \(S_{\text{N}}1\) pathway because it forms a stable tertiary carbocation intermediate. Butan-1-ol is a primary alcohol and must react via an \(S_{\text{N}}2\) mechanism, which has a much higher activation energy and is extremely slow at room temperature.

**(b)(i)** To wash the mixture:
1. Place the mixture into a separating funnel, stopper it, and invert it to shake gently.
2. Release pressure regularly by opening the tap while the funnel is inverted (and pointed away from people).
3. Safety precaution: Shaking the mixture produces carbon dioxide gas from the reaction of unreacted hydrochloric acid with sodium hydrogencarbonate. Releasing the pressure prevents the stopper from flying off or the glassware from cracking.

**(b)(ii)** Anhydrous sodium sulfate acts as a drying agent to absorb any remaining water/traces of moisture from the organic layer. Once dry, the organic liquid will change from cloudy to clear.

**(c)** 2-methylpropan-2-ol contains a highly polar \(\text{O-H}\) bond, allowing it to form strong intermolecular hydrogen bonds. 2-chloro-2-methylpropane contains polar \(\text{C-Cl}\) bonds, so its molecules are held together by permanent dipole-dipole interactions and London forces. Hydrogen bonds are significantly stronger than permanent dipole-dipole interactions or London forces, requiring much more energy to overcome, resulting in a higher boiling point for the alcohol.

**(d)(i)** Step 1: The carbon-chlorine bond breaks heterolytically. Draw a curly arrow from the \(\text{C-Cl}\) bond to the chlorine atom. This produces a tertiary carbocation intermediate, \(\text{(CH}_3)_3\text{C}^+\), and a chloride ion, \(\text{Cl}^-\).
Step 2: A hydroxide ion acts as a nucleophile. Draw a curly arrow from a lone pair on the oxygen of the hydroxide ion, \(\text{OH}^-\), to the positively charged carbon atom of the carbocation. This forms the product, 2-methylpropan-2-ol.

**(d)(ii)** The reaction occurs in multiple steps. The first step (formation of the carbocation) is the slow, rate-determining step, which depends only on the concentration of the halogenoalkane. Since the hydroxide ion is only involved in a subsequent fast step, its concentration does not affect the rate, hence it is absent from the rate equation.

**(e)(i)** A yellow precipitate (silver iodide) forms much faster than a white precipitate (silver chloride) / a yellow precipitate is observed almost instantly, whereas the white precipitate takes significantly longer to appear.

**(e)(ii)** The rate of hydrolysis depends on the bond enthalpy of the carbon-halogen bond, not its polarity. The \(\text{C-I}\) bond enthalpy is much lower (weaker bond) than the \(\text{C-Cl}\) bond enthalpy (stronger bond). Consequently, the \(\text{C-I}\) bond is much easier to break, resulting in a much lower activation energy and a faster reaction rate.

**(a) [3 Marks]**
• MP1: Correct balanced chemical equation: \(\text{(CH}_3)_3\text{COH} + \text{HCl} \rightarrow \text{(CH}_3)_3\text{CCl} + \text{H}_2\text{O}\) (1)
• MP2: Identifies that 2-methylpropan-2-ol is a tertiary alcohol / forms a highly stable tertiary carbocation (1)
• MP3: Identifies that butan-1-ol is a primary alcohol / primary carbocations are highly unstable, meaning it must react via a slower \(S_{\text{N}}2\) pathway with high activation energy (1)

**(b)(i) [3 Marks]**
• MP1: Shakes the separating funnel with stopper in and inverts it (1)
• MP2: Opens the tap while inverted to release built-up gas/pressure (1)
• MP3: Explains that pressure builds up due to carbon dioxide gas (\(\text{CO}_2\)) produced from the acid-carbonate reaction, which must be released to prevent the stopper popping out/glass cracking (1)

**(b)(ii) [2 Marks]**
• MP1: Explains that anhydrous sodium sulfate acts as a drying agent / removes traces of water (1)
• MP2: States the visual change is from cloudy/turbid to clear (1)

**(c) [3 Marks]**
• MP1: States 2-methylpropan-2-ol has intermolecular hydrogen bonds (1)
• MP2: States 2-chloro-2-methylpropane has permanent dipole-dipole interactions and London forces (but no hydrogen bonds) (1)
• MP3: States that hydrogen bonds are stronger than dipole-dipole/London forces, requiring more energy to break (1)

**(d)(i) [4 Marks]**
• MP1: Curly arrow starting from the \(\text{C-Cl}\) bond to the chlorine atom (1)
• MP2: Correct structure of the tertiary carbocation intermediate: \(\text{(CH}_3)_3\text{C}^+\) (1)
• MP3: Correct structure of leaving chloride ion: \(\text{Cl}^-\) (1)
• MP4: Curly arrow from a lone pair on the oxygen of the hydroxide ion (\(\text{OH}^-\)) to the positive carbon atom of the carbocation (1)

**(d)(ii) [1 Mark]**
• MP1: Explains that the rate-determining step (slow step) only involves the halogenoalkane / hydroxide ions only react in a subsequent fast step (1)

**(e)(i) [1 Mark]**
• MP1: States that a yellow precipitate (silver iodide) forms much faster than a white precipitate (silver chloride) (1)

**(e)(ii) [2 Marks]**
• MP1: States that the \(\text{C-I}\) bond enthalpy is lower than the \(\text{C-Cl}\) bond enthalpy / \(\text{C-I}\) bond is weaker than the \(\text{C-Cl}\) bond (1)
• MP2: Concludes that less energy is required to break the \(\text{C-I}\) bond, leading to a faster rate of hydrolysis (accept that bond enthalpy, not polarity, is the dominant factor) (1)

(a) A polystyrene cup is a highly effective thermal insulator, which significantly minimises heat transfer to and from the surroundings.

(b) Adding zinc in excess ensures that all of the copper(II) ions in the solution react completely, making copper(II) sulfate the limiting reactant and allowing the stoichiometric calculation of the enthalpy change based on the initial moles of copper(II) sulfate.

(c)(i) The student should plot temperature (y-axis) against time (x-axis). They would draw a cooling line of best fit through the points from minutes 5 to 10, and then extrapolate this cooling line back to minute 4 (the time of mixing). The difference between the extrapolated temperature at minute 4 and the initial temperature (20.2 °C) gives the corrected maximum temperature change (\(\Delta T\)). This graphical correction is necessary because heat loss to the surroundings occurs simultaneously with the reaction, making the measured maximum temperature lower than the theoretical maximum temperature rise under adiabatic conditions.

(c)(ii)
1. Calculate the temperature change:
\(\Delta T = 38.2^\circ\text{C} - 20.2^\circ\text{C} = 18.0^\circ\text{C}\)
2. Calculate the heat energy transferred (assuming density is \(1.00\text{ g cm}^{-3}\), so mass \(m = 50.0\text{ g}\)):
\(q = m \times c \times \Delta T = 50.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ }^\circ\text{C}^{-1} \times 18.0^\circ\text{C} = 3762\text{ J} = 3.762\text{ kJ}\)
3. Calculate the number of moles of limiting reactant, copper(II) sulfate:
\(n(\text{CuSO}_4) = 0.400\text{ mol dm}^{-3} \times \frac{50.0}{1000}\text{ dm}^3 = 0.0200\text{ mol}\)
4. Calculate the molar enthalpy change (adding a negative sign because the reaction is exothermic):
\(\Delta H = -\frac{3.762\text{ kJ}}{0.0200\text{ mol}} = -188.1\text{ kJ mol}^{-1}\)

(d) Sources of error (any two):
1. Incomplete reaction due to slow reaction rate or large zinc particle size.
2. The heat capacity of the polystyrene cup and thermometer itself is neglected.
3. Evaporation of water from the reaction cup.
4. Measurement uncertainties in the volume of solution (from using a measuring cylinder) and temperature readings (due to thermometer resolution).

Minimisation of one error:
- Use finely powdered zinc to increase the surface area and ensure the reaction completes as rapidly as possible.
- Use a volumetric pipette instead of a measuring cylinder to measure the volume of CuSO4 solution to decrease measurement uncertainty.

Part (a): [1 mark total]
- Award 1 mark for stating that polystyrene is an insulator or reduces heat loss to the surroundings.

Part (b): [1 mark total]
- Award 1 mark for stating it ensures complete reaction of copper(II) ions / CuSO4.

Part (c)(i): [3 marks total]
- Award 1 mark for plotting a temperature-time graph and extrapolating the cooling curve back to minute 4.
- Award 1 mark for explaining that the corrected temperature rise is the difference between the extrapolated temperature and the initial temperature.
- Award 1 mark for explaining that heat is lost during the reaction, meaning measured maximum temperature is lower than theoretical maximum.

Part (c)(ii): [4 marks total]
- Award 1 mark for calculating temperature change = 18.0 °C.
- Award 1 mark for calculating heat energy = 3.762 kJ (or 3762 J).
- Award 1 mark for calculating moles of CuSO4 = 0.0200 mol.
- Award 1 mark for final calculated value of -188.1 kJ mol^-1 (accept -188 kJ mol^-1) with a negative sign.

Part (d): [4 marks total]
- Award 1 mark each for identifying two distinct, valid practical errors (max 2).
- Award 1 mark for suggesting a correct modification matched to one of the listed errors.
- Award 1 mark for using precise and clear scientific communication throughout.

(a) The student should weigh by difference: first, weigh the weighing boat containing the sulfamic acid; transfer the solid into the dissolving beaker; then reweigh the weighing boat containing any remaining solid residues; finally, subtract the mass of the empty boat and residue from the initial mass of the boat and solid to find the precise mass transferred.

(b)
1. Completely dissolve the weighed sulfamic acid solid in a clean beaker using a volume of distilled water (e.g., approximately 100 cm³), stirring with a clean glass rod.
2. Quantitatively transfer the solution from the beaker into a 250.0 cm³ volumetric flask using a clean funnel.
3. Rinse the beaker, glass rod, and funnel several times with distilled water, transferring all rinse washings directly into the volumetric flask.
4. Carefully add distilled water until the level is just below the graduation mark, then add water dropwise with a teat pipette until the bottom of the meniscus is aligned exactly on the graduation line. Stopper the flask and invert it repeatedly to mix the solution thoroughly.

(c)(i)
- Moles of sulfamic acid:
\(n = \frac{1.942\text{ g}}{97.1\text{ g mol}^{-1}} = 0.0200\text{ mol}\)
- Concentration of standard solution in 250.0 cm³ (0.250 dm³):
\(\text{Concentration} = \frac{0.0200\text{ mol}}{0.250\text{ dm}^3} = 0.0800\text{ mol dm}^{-3}\)

(c)(ii)
- Moles of sulfamic acid in the 25.0 cm³ aliquot titrated:
\(n = 0.0800\text{ mol dm}^{-3} \times \frac{25.0}{1000}\text{ dm}^3 = 2.00 \times 10^{-3}\text{ mol}\)
- Since the molar ratio is 1:1, moles of NaOH present in the titre = \(2.00 \times 10^{-3}\text{ mol}\).
- Concentration of NaOH solution:
\(\text{Concentration} = \frac{2.00 \times 10^{-3}\text{ mol}}{0.02185\text{ dm}^3} = 0.09153\text{ mol dm}^{-3} \approx 0.0915\text{ mol dm}^{-3}\)

(d) A suitable indicator is phenolphthalein. The colour change at the end point in the conical flask (as NaOH is added from the burette into the acid) is from colourless to pale pink (permanent pink). (Alternatively, methyl orange can be used, changing from red to orange/yellow).

Part (a): [2 marks total]
- Award 1 mark for describing weighing of the container and solid, and reweighing the container after transfer.
- Award 1 mark for explaining that subtracting the residual mass yields the true mass of solid transferred.

Part (b): [4 marks total]
- Award 1 mark for dissolving the solid fully in a beaker with stirring prior to transferring.
- Award 1 mark for performing quantitative transfer including washing the beaker, funnel, and stirring rod, adding all washings to the volumetric flask.
- Award 1 mark for filling the volumetric flask until the bottom of the meniscus is on the 250 cm³ graduation mark.
- Award 1 mark for inverting the flask multiple times to ensure a homogeneous solution.

Part (c)(i): [2 marks total]
- Award 1 mark for calculating the moles of solid sulfamic acid = 0.0200 mol.
- Award 1 mark for calculating the concentration of the standard solution = 0.0800 mol dm^-3.

Part (c)(ii): [2 marks total]
- Award 1 mark for calculating the moles of acid in the 25.0 cm³ aliquot = 2.00 x 10^-3 mol.
- Award 1 mark for calculating the concentration of the NaOH solution = 0.0915 mol dm^-3 (allow 0.0915 to 0.0916 mol dm^-3).

Part (d): [2 marks total]
- Award 1 mark for selecting a correct indicator: Phenolphthalein (or Methyl orange).
- Award 1 mark for specifying the correct colour change at the end point: colourless to pale pink (or red to orange/yellow).

(a) Reflux allows a reaction mixture to be heated to its boiling point for a prolonged period without loss of volatile reactants, products, or intermediates, as their vapours condense back down into the reaction vessel.

(b)(i) The flask is fitted with a condenser mounted in a vertical position. The condenser jacket must have cold water entering through the lower connection and exiting through the upper connection to ensure the cooling jacket remains completely filled. The top of the condenser must remain unstoppered (open to the air) to prevent pressure build-up.

(b)(ii) Anti-bumping granules provide nucleation sites to promote smooth and even boiling, which prevents sudden violent boiling and dangerous liquid splashing.

(c)(i)
1. Periodically invert the separating funnel and carefully open the tap to release built-up gas pressure from carbon dioxide generation.
2. Ensure the stopper is held securely in place during shaking, and remove the stopper before running off the lower layer to prevent vacuum lock.

(c)(ii)
\(\text{H}^+\text{(aq)} + \text{HCO}_3^-\text{(aq)} \rightarrow \text{CO}_2\text{(g)} + \text{H}_2\text{O(l)}\)

(d) A suitable anhydrous drying agent is anhydrous calcium chloride, \(\text{CaCl}_2\) (or anhydrous sodium sulfate, \(\text{Na}_2\text{SO}_4\), or anhydrous magnesium sulfate, \(\text{MgSO}_4\)). On drying, the appearance of the organic liquid changes from cloudy to clear and transparent.

Part (a): [2 marks total]
- Award 1 mark for stating that it prevents loss of volatile substances / reactants / products.
- Award 1 mark for explaining that it allows prolonged heating / increases reaction rate.

Part (b)(i): [3 marks total]
- Award 1 mark for vertical condenser.
- Award 1 mark for water in at bottom and out at top.
- Award 1 mark for stating the system must be open at the top (no stopper on condenser).

Part (b)(ii): [1 mark total]
- Award 1 mark for smooth boiling / preventing bumping.

Part (c)(i): [2 marks total]
- Award 1 mark for releasing pressure periodically / opening the tap of inverted funnel.
- Award 1 mark for holding stopper/tap secure or removing stopper before running off layers.

Part (c)(ii): [2 marks total]
- Award 1 mark for the correct reactants and products: H+ + HCO3- -> CO2 + H2O.
- Award 1 mark for correct state symbols: (aq) for reactants, (g) and (l) for products.

Part (d): [3 marks total]
- Award 1 mark for naming a suitable drying agent: anhydrous calcium chloride (CaCl2), anhydrous sodium sulfate (Na2SO4) or magnesium sulfate (MgSO4). [Reject anhydrous cobalt chloride]
- Award 1 mark for stating the appearance changes from cloudy to clear.
- Award 1 mark for stating the drying agent does not dissolve but clumps/remains solid.

(a)(i) Dip a clean nichrome or platinum wire into concentrated hydrochloric acid and heat it in a hot Bunsen burner flame until it produces no colour. Dip the clean wire back into concentrated hydrochloric acid, touch a small amount of the solid sample **X**, and place the wire into the non-luminous (blue) Bunsen flame, observing the flame colour change.

(a)(ii) Calcium ion, \(\text{Ca}^{2+}\).

(b)(i) Bromide ion, \(\text{Br}^-\).

(b)(ii)
\(\text{Ag}^+\text{(aq)} + \text{Br}^-\text{(aq)} \rightarrow \text{AgBr(s)}\)

(c)(i)
\(\text{Moles of AgBr} = \frac{2.818\text{ g}}{187.8\text{ g mol}^{-1}} = 0.01501\text{ mol}\)

(c)(ii)
- Since **X** is a Group 2 metal halide containing bromide, its formula is \(\text{MBr}_2\). Therefore, 1 mole of \(\text{MBr}_2\) contains 2 moles of bromide ions, meaning:
\(\text{Moles of X} = \frac{0.01501\text{ mol}}{2} = 0.007505\text{ mol}\)
- Molar mass of **X**:
\(\text{Molar mass of X} = \frac{1.500\text{ g}}{0.007505\text{ mol}} = 199.9\text{ g mol}^{-1}\)
- Molar mass of the metal, \(\text{M}\):
\(\text{Molar mass of M} = 199.9 - (2 \times 79.9) = 40.1\text{ g mol}^{-1}\)
- This molar mass matches calcium (Ar = 40.1), which confirms the metal ion is indeed calcium, consistent with the brick-red flame test. Thus, the formula of **X** is \(\text{CaBr}_2\).

Part (a)(i): [3 marks total]
- Award 1 mark for cleaning a nichrome or platinum wire with concentrated hydrochloric acid and heating until clean.
- Award 1 mark for dipping the clean wire in acid and then the solid sample.
- Award 1 mark for heating the sample in a non-luminous (blue) Bunsen flame and observing the color.

Part (a)(ii): [1 mark total]
- Award 1 mark for identifying Calcium / Ca2+.

Part (b)(i): [1 mark total]
- Award 1 mark for identifying Bromide / Br-.

Part (b)(ii): [2 marks total]
- Award 1 mark for correct formulas of reactants and products: Ag+ + Br- -> AgBr.
- Award 1 mark for correct state symbols: (aq) + (aq) -> (s).

Part (c)(i): [1 mark total]
- Award 1 mark for calculating moles of AgBr = 0.01501 mol (or 0.0150 mol).

Part (c)(ii): [4 marks total]
- Award 1 mark for dividing moles of AgBr by 2 to determine moles of X = 0.007505 mol.
- Award 1 mark for calculating the molar mass of X = 199.9 g mol^-1 (accept 200 g mol^-1).
- Award 1 mark for subtracting the two bromide masses to get 40.1 g mol^-1 for the metal.
- Award 1 mark for identifying the metal as calcium and explicitly stating this is consistent with the brick-red flame test result.