2025 Edexcel IAS-Level Chemistry (XCH11) 模擬試題連答案詳解

1. Calculate moles of oxygen: \( n(\text{O}_2) = \frac{1.44}{24.0} = 0.0600\text{ mol} \). 2. Use the stoichiometric ratio to find moles of metal chlorate: \( n(\text{MClO}_3) = 0.0600 \times \frac{2}{3} = 0.0400\text{ mol} \). 3. Calculate the molar mass of \( \text{MClO}_3 \): \( M(\text{MClO}_3) = \frac{4.90}{0.0400} = 122.5\text{ g mol}^{-1} \). 4. Subtract the molar masses of chlorine and oxygen to find the molar mass of M: \( M(\text{M}) = 122.5 - 35.5 - 3(16.0) = 39.0\text{ g mol}^{-1} \). This atomic mass corresponds to potassium (K).

Award 1 mark for correct answer C. Correct calculation of moles of oxygen (0.0600 mol), moles of metal chlorate (0.0400 mol), and molar mass of metal (39.0 g/mol) leading to potassium.

The successive ionization energies show a large jump between the 3rd and 4th ionization energies (from 2745 to 11577 \( \text{kJ mol}^{-1} \)). This indicates that the 4th electron is removed from an inner shell, which is closer to the nucleus and experiences less shielding. Therefore, the element has 3 valence electrons and belongs to Group 13 (Group 3). Since it is in Period 3, the element is aluminium.

Award 1 mark for correct answer C. Identification of the jump between the 3rd and 4th ionization energies as indicating 3 valence electrons, matching aluminium in Period 3.

A: \( \text{BF}_3 \) has 3 bonding pairs and 0 lone pairs, giving a trigonal planar shape. B: \( \text{H}_3\text{O}^+ \) has 3 bonding pairs and 1 lone pair on the oxygen atom, resulting in a trigonal pyramidal shape. C: \( \text{NH}_4^+ \) has 4 bonding pairs and 0 lone pairs, giving a tetrahedral shape. D: \( \text{CO}_3^{2-} \) has 3 regions of electron density and 0 lone pairs, giving a trigonal planar shape.

Award 1 mark for correct answer B. Correctly identifying that \( \text{H}_3\text{O}^+ \) has 3 bonding pairs and 1 lone pair, resulting in a trigonal pyramidal shape.

The rate of hydrolysis of halogenoalkanes depends on the strength of the carbon-halogen bond and the mechanism of substitution. C-I bonds are weaker than C-Cl bonds, making iodoalkanes more reactive than chloroalkanes. Furthermore, tertiary halogenoalkanes react much faster than primary halogenoalkanes because they undergo nucleophilic substitution via the \( \text{S}_\text{N}1 \) mechanism, which involves a highly stable tertiary carbocation intermediate. Therefore, 2-iodo-2-methylpropane (a tertiary iodoalkane) reacts the fastest.

Award 1 mark for correct answer D. Identifying that the tertiary iodoalkane has both the weakest carbon-halogen bond and undergoes rapid \( \text{S}_\text{N}1 \) hydrolysis.

A: The solubility of Group 2 hydroxides increases down the group. B: The solubility of Group 2 sulfates decreases down the group. C: The thermal stability of Group 2 nitrates increases down the group because the cationic radius increases and charge density decreases down the group, leading to less polarization of the nitrate anion. D: The first ionization energy of the elements decreases down the group due to increased shielding and atomic radius.

Award 1 mark for correct answer C. Identifying that the thermal stability of nitrates increases down Group 2 as the cation polarising power decreases.

For an alkene to exhibit E/Z stereoisomerism, both carbon atoms of the double bond must be attached to two different groups. In 1,2-dichloroethene, each double-bonded carbon atom is bonded to one hydrogen atom and one chlorine atom, enabling cis-trans (E/Z) isomerism. In propene, 2-methylbut-2-ene, and 2-methylpropene, at least one carbon of the double bond is bonded to two identical groups.

Award 1 mark for correct answer C. Identifying 1,2-dichloroethene as the only alkene listed with different groups on each of the double-bonded carbons.

A termination step in a free-radical mechanism involves two radicals colliding and combining to form a stable molecule, reducing the concentration of radicals. The reaction in D shows a methyl radical and a chlorine radical combining to form chloromethane, which fits this definition. A is initiation, while B and C are propagation steps.

Award 1 mark for correct answer D. Correctly identifying the reaction between two radicals to form a stable covalent molecule as a termination step.

When the temperature is increased, molecules gain kinetic energy. The Maxwell-Boltzmann distribution curve flattens and shifts to the right. Therefore, the peak of the curve (representing the most probable energy) moves to the right and is lower in height. Because the total number of molecules in the system remains constant, the area under the curve remains the same.

Award 1 mark for correct answer B. Correctly identifying that the peak shifts to the right and becomes lower, while the total area remains constant because the number of molecules is conserved.

First, convert the pressure to Pa: \(p = 101 \times 10^3\text{ Pa}\). Convert volume to m\(^3\): \(V = 125 \times 10^{-6}\text{ m}^3\). Convert temperature to K: \(T = 120 + 273 = 393\text{ K}\). Use the ideal gas equation: \(pV = nRT \implies n = \frac{pV}{RT} = \frac{101 \times 10^3 \times 125 \times 10^{-6}}{8.31 \times 393} \approx 0.00387\text{ mol}\). Calculate the molar mass: \(M = \frac{\text{mass}}{n} = \frac{0.216}{0.00387} \approx 55.8\text{ g mol}^{-1}\). A molar mass of approximately 56 g mol\(^{-1}\) corresponds to the molecular formula C\(_4\)H\(_8\).

1 mark: Correct calculation of molar mass to determine C\(_4\)H\(_8\).

Hydrolysis rate of halogenoalkanes is determined by the strength of the C-X bond and the mechanism. Iodoalkanes have the weakest C-X bonds and thus undergo substitution faster than bromoalkanes or chloroalkanes. Furthermore, 2-iodo-2-methylpropane is a tertiary halogenoalkane which reacts rapidly via the stable tertiary carbocation in an \(\text{S}_\text{N}1\) mechanism.

1 mark: Correct selection of 2-iodo-2-methylpropane based on bond enthalpy and carbocation stability.

Under hot, concentrated conditions, chlorine gas reacts with NaOH according to the equation: \(3\text{Cl}_2 + 6\text{NaOH} \rightarrow 5\text{NaCl} + \text{NaClO}_3 + 3\text{H}_2\text{O}\). Chlorine is reduced to \(\text{Cl}^-\)\ (oxidation state -1 in NaCl) and oxidized to \(\text{ClO}_3^-\)\ (oxidation state +5 in \(\text{NaClO}_3\)).

1 mark: Correctly identifies both oxidation states as -1 and +5.

There is a very large increase between the 3rd and 4th ionization energies (from 2745 to 11578 kJ mol\(^{-1}\)), indicating that the fourth electron is removed from an inner quantum shell. Therefore, element X has three outer shell valence electrons (Group 13, Aluminum). It forms \(\text{X}^{3+}\) ions, resulting in an oxide with the formula \(\text{X}_2\text{O}_3\).

1 mark: Correct identification of the oxide formula using successive ionization energy trends.

The reaction proceeds via electrophilic addition. The secondary carbocation intermediate \(\text{CH}_3\text{C}^+\text{HCH}_3\) is more stable than the primary carbocation intermediate \(\text{CH}_3\text{CH}_2\text{C}^+\text{H}_2\) because of the electron-releasing inductive effect of the two adjacent methyl groups. This lowers the activation energy for its formation, making 2-bromopropane the major product.

1 mark: Correct explanation based on secondary carbocation stability and the inductive effect of alkyl groups.

The amide ion, \(\text{NH}_2^-\), has 2 bonding pairs of electrons and 2 lone pairs of electrons around the central nitrogen atom (isoelectronic with \(\text{H}_2\text{O}\)). According to electron pair repulsion theory, lone pair-lone pair repulsion is greater than lone pair-bonding pair repulsion, which is greater than bonding pair-bonding pair repulsion. This reduces the bond angle from the tetrahedral angle of 109.5\(^\circ\) to approximately 104.5\(^\circ\).

1 mark: Correct identification of NH\(_2^-\).

Pentane, 2-methylbutane, and 2,2-dimethylpropane are structural isomers. More branched isomers (such as 2,2-dimethylpropane) have a more spherical shape. This reduces the surface area of contact between adjacent molecules, leading to weaker London forces (intermolecular forces) which require less thermal energy to overcome, resulting in a lower boiling temperature.

1 mark: Correctly identifies 2,2-dimethylpropane based on branching and intermolecular forces.

A catalyst provides an alternative reaction pathway with a lower activation energy. It does not alter the average kinetic energy of the molecules, so the Maxwell-Boltzmann distribution curve remains completely unchanged. Because the activation energy barrier is lowered, a greater fraction of molecules have energy greater than or equal to this new activation energy, which increases the rate of reaction.

1 mark: Correctly identifies that the distribution curve is unchanged but a lower activation energy increases the fraction of successful collisions.

Mass of water lost = \(4.93\text{ g} - 2.41\text{ g} = 2.52\text{ g}\). Number of moles of anhydrous \(\text{MgSO}_4\) = \(\frac{2.41}{120.4} = 0.0200\text{ mol}\). Number of moles of \(\text{H}_2\text{O}\) = \(\frac{2.52}{18.0} = 0.140\text{ mol}\). Ratio of \(\text{H}_2\text{O} : \text{MgSO}_4\) = \(\frac{0.140}{0.0200} = 7.00\). Therefore, \(x = 7\).

Award 1 mark for the correct calculation showing x = 7 and selecting option C.

The rate of hydrolysis depends on both the halogen atom and the carbon skeleton of the halogenoalkane. Iodoalkanes hydrolyse faster than bromoalkanes and chloroalkanes because the \(\text{C}-\text{I}\) bond is weaker (has a lower bond enthalpy) than \(\text{C}-\text{Br}\) and \(\text{C}-\text{Cl}\) bonds. Furthermore, tertiary halogenoalkanes hydrolyse much faster than secondary or primary ones because they react via an \(\text{S}_\text{N}1\) mechanism, which involves the formation of a stable tertiary carbocation intermediate. Therefore, 2-iodo-2-methylpropane (a tertiary iodoalkane) reacts the fastest.

Award 1 mark for identifying 2-iodo-2-methylpropane as the correct answer and choosing option C.

As you go down Group 2, the size (ionic radius) of the metal cation increases, while its charge remains constant at \(2+\). Consequently, the charge density of the cation decreases, and its polarising power on the large, polarisable nitrate anion decreases. Since the nitrate anion is polarised to a lesser extent, the \(\text{N}-\text{O}\) bonds in the nitrate ion are weakened less, which makes the compound more thermally stable and requires more heat to decompose. Thus, thermal stability increases down the group.

Award 1 mark for identifying that increased ionic radius decreases the polarising power of the cation, leading to increased thermal stability.

Successive ionisation energies increase as more electrons are removed. A very large jump is observed between the third and fourth ionisation energies (from 2745 to 11577 \(\text{kJ mol}^{-1}\)). This huge increase shows that the fourth electron is being removed from an inner, fully closed electron shell which experiences a significantly greater effective nuclear charge. Therefore, the element has three valence electrons and is in Group 3 (Group 13).

Award 1 mark for identifying that the major jump in ionisation energies occurs between the 3rd and 4th ionisations, indicating Group 3.

All four options exhibit \(E\)-\(Z\) isomerism because both carbons of the double bond have two different groups attached. When reacting with \(\text{HBr}\), unsymmetrical alkenes react via Markovnikov\'s rule, where the major product is formed via the most stable (tertiary) carbocation intermediate. For 3-methylhex-2-ene (\(\text{CH}_3\text{CH}=\text{C}(\text{CH}_3)\text{CH}_2\text{CH}_2\text{CH}_3\)), adding \(\text{H}^+\) to C2 forms a stable tertiary carbocation at C3. Subsequent nucleophilic attack by \(\text{Br}^-\) produces 3-bromo-3-methylhexane. The C3 atom in this product is chiral as it is bonded to four different groups: \(-\text{CH}_3\), \(-\text{CH}_2\text{CH}_3\), \(-\text{CH}_2\text{CH}_2\text{CH}_3\), and \(-\text{Br}\). In contrast, the major product from 3-methylpent-2-ene is 3-bromo-3-methylpentane, which contains two identical ethyl groups and is not chiral.

Award 1 mark for selecting 3-methylhex-2-ene as the correct alkene that yields a chiral major product.

The amide ion, \(\text{NH}_2^-\), has 4 electron pairs around the central nitrogen atom (5 valence electrons from N, 2 from H, and 1 from the negative charge, giving 8 electrons). Two of these pairs are bonding pairs and two are lone pairs. The 4 electron pairs arrange themselves in a tetrahedral geometry to minimise repulsion. Because lone pair-lone pair repulsion is stronger than lone pair-bonding pair and bonding pair-bonding pair repulsion, the H-N-H bond angle is compressed from the tetrahedral angle of \(109.5^\circ\) by approximately \(5.0^\circ\) to \(104.5^\circ\), giving a bent shape.

Award 1 mark for identifying that the amide ion has a bent shape with a bond angle of 104.5 degrees.

A termination step always involves the reaction of two radicals to form a stable covalent molecule. The combination of two methyl radicals (\(\text{CH}_3^\bullet + \text{CH}_3^\bullet \rightarrow \text{C}_2\text{H}_6\)) stops the chain reaction and yields a small amount of ethane as an impurity. Options A and D represent propagation steps, and Option B is also a propagation step.

Award 1 mark for identifying the correct radical termination reaction that produces ethane.

A catalyst increases the reaction rate by providing an alternative reaction pathway with a lower activation energy, thus decreasing \(E_\text{a}\). Increasing the temperature increases the kinetic energy of the reacting particles, which increases the rate by raising the fraction of collisions with energy greater than or equal to the activation energy. Temperature does not alter the activation energy of the reaction pathway itself.

Award 1 mark for correctly distinguishing the physical effects of temperature and catalysts on activation energy.

From the chemical equation, 2 mol of \\text{Ca(NO}_3\\text{)}_2\\ produces 4 mol of \\text{NO}_2\\ and 1 mol of \\text{O}_2\\, giving a total of 5 mol of gas. Thus, the molar ratio of reactant to total gas produced is 2 : 5 (or 1 : 2.5). The moles of gas produced = \(0.100\text{ mol} \times 2.5 = 0.250\text{ mol}\). Total volume of gas = \(0.250\text{ mol} \times 24.0\text{ dm}^3\text{ mol}^{-1} = 6.00\text{ dm}^3\).

1 mark for the correct answer C.

Since alcohol \\mathbf{Y}\\ resists oxidation by acidified potassium dichromate(VI), it must be a tertiary alcohol. The precursor halogenoalkane \\mathbf{X}\\ must therefore be a tertiary halogenoalkane. Analyzing the options: 1-bromobutane is primary, 2-bromobutane is secondary, 1-bromo-2-methylpropane is primary, and 2-bromo-2-methylpropane is tertiary. Thus, \\mathbf{X}\\ is 2-bromo-2-methylpropane.

1 mark for the correct answer D.

Solubility of Group 2 sulfates decreases down the group. Both lattice energy and hydration enthalpy decrease (become less exothermic) because the cation radius increases. However, because the sulfate ion is very large, the change in lattice energy is relatively small. The hydration enthalpy of the cation decreases much more rapidly because the charge density of the cation decreases significantly. As a result, the enthalpy of solution becomes more endothermic (less exothermic) and the solubility decreases down the group.

1 mark for the correct answer C.

The successive ionisation energies show a very large jump between the 3rd (2745 kJ/mol) and the 4th (11578 kJ/mol) ionisation energies. This indicates that the 4th electron is removed from a shell closer to the nucleus, meaning there are 3 electrons in the outer shell. Therefore, the element is in Group 3 (Group 13).

1 mark for the correct answer B.

The electrophilic addition of HBr to 2-methylbut-2-ene involves the formation of carbocation intermediates. Addition of H+ to C3 forms a tertiary carbocation, while addition of H+ to C2 forms a secondary carbocation. Since a tertiary carbocation is more stable than a secondary carbocation due to the greater electron-releasing inductive effect of the three alkyl groups, the major product is formed via the more stable tertiary carbocation intermediate.

1 mark for the correct answer B.

Using VSEPR theory: NH3 and H3O+ have 3 bonding pairs and 1 lone pair, leading to a trigonal pyramidal shape with a bond angle of approximately 107 degrees. NH4+ has 4 bonding pairs and 0 lone pairs, which is tetrahedral with a bond angle of 109.5 degrees. NH2- has 2 bonding pairs and 2 lone pairs on the nitrogen atom (isoelectronic with H2O). The greater repulsion of the 2 lone pairs reduces the tetrahedral bond angle to 104.5 degrees.

1 mark for the correct answer D.

HF has the highest boiling temperature because of strong hydrogen bonding. For HCl, HBr, and HI, the main intermolecular forces are London forces, which increase in strength as the number of electrons in the molecules increases down Group 7 (HCl < HBr < HI). Therefore, the correct order is: HCl < HBr < HI < HF.

1 mark for the correct answer B.

Using Hess's Law: Enthalpy of formation = Sum of enthalpies of combustion of reactants - Sum of enthalpies of combustion of products. Reactants sum = \(-394 + 2 \times (-286) = -966\text{ kJ mol}^{-1}\). Products sum = \(-726\text{ kJ mol}^{-1}\). Enthalpy of formation = \(-966 - (-726) = -240\text{ kJ mol}^{-1}\).

1 mark for the correct answer A.

First, calculate the number of moles of the hydrocarbon using the formula: \(n = \text{Volume} / \text{Molar volume} = 1.04\text{ dm}^3 / 24.0\text{ dm}^3\text{ mol}^{-1} = 0.0433\text{ mol}\). Next, calculate the molar mass of the hydrocarbon: \(M_r = \text{Mass} / \text{Moles} = 1.30\text{ g} / 0.0433\text{ mol} = 30.0\text{ g mol}^{-1}\). The molar mass of ethane (\(\text{C}_2\text{H}_6\)) is \((2 \times 12.0) + (6 \times 1.0) = 30.0\text{ g mol}^{-1}\), which matches this calculated value.

1 mark for the correct option C.

There is a very large increase between the 2nd and 3rd ionization energies (from 1450 to 7730 \(\text{kJ mol}^{-1}\)). This indicates that the third electron is removed from a shell closer to the nucleus, meaning the element has 2 valence electrons and belongs to Group 2. Group 2 metals form ions with a \(2+\) charge (\(\text{X}^{2+}\)), which react with chloride ions (\(\text{Cl}^-\)) in a 1:2 ratio to form \(\text{XCl}_2\).

1 mark for the correct option B.

The shape is determined by the electron pair repulsion theory. \(\text{BF}_3\), \(\text{CO}_3^{2-}\), and \(\text{NO}_3^-\) all have central atoms with 3 bonding regions and 0 lone pairs, giving them a trigonal planar shape. In contrast, \(\text{NF}_3\) has 3 bonding pairs and 1 lone pair on the central nitrogen atom, resulting in a trigonal pyramidal shape.

1 mark for the correct option C.

Methylpropane and butane are alkanes with only London forces. Methylpropane is branched and has weaker London forces than straight-chain butane, giving it the lowest boiling temperature. Propanone is polar and experiences permanent dipole-dipole forces, which are stronger than London forces. Propan-1-ol exhibits hydrogen bonding due to its polar \(\text{-OH}\) group, which is the strongest intermolecular force among these compounds, giving it the highest boiling temperature.

1 mark for the correct option A.

Reducing ability of halides increases down the group. Iodide (\(\text{I}^-\)) is the strongest reducing agent of the halides. It reduces sulfur in concentrated sulfuric acid (oxidation state \(+6\)) to hydrogen sulfide (\(\text{H}_2\text{S}\)), where the oxidation state of sulfur is \(-2\). Bromide only reduces sulfur to sulfur dioxide (oxidation state \(+4\)), while chloride and fluoride are not strong enough to reduce concentrated sulfuric acid.

1 mark for the correct option D.

The chemical equation for the formation of propene is: \(3\text{C(s)} + 3\text{H}_2\text{(g)} \rightarrow \text{C}_3\text{H}_6\text{(g)}\). According to Hess's Law using combustion data: \(\Delta_f H^\theta = 3 \times \Delta_c H^\theta [\text{C(s)}] + 3 \times \Delta_c H^\theta [\text{H}_2\text{(g)}] - \Delta_c H^\theta [\text{C}_3\text{H}_6\text{(g)}] = 3(-394) + 3(-286) - (-2058) = -1182 - 858 + 2058 = +18\text{ kJ mol}^{-1}\).

1 mark for the correct option B.

The rate of hydrolysis is determined by two factors: the strength of the carbon-halogen bond and the mechanism. Bromides react much faster than chlorides because the \(\text{C-Br}\) bond is weaker than the \(\text{C-Cl}\) bond. Tertiary halogenoalkanes (such as 2-bromo-2-methylpropane) react much faster than secondary or primary ones because they undergo hydrolysis via the \(\text{S}_\text{N}1\) mechanism, which involves the formation of a highly stable tertiary carbocation intermediate.

1 mark for the correct option D.

For geometric (\(E\)/\(Z\)) isomerism to occur, each carbon atom of the \(\text{C}=\text{C}\) double bond must be attached to two different groups. In propene, but-1-ene, and 2-methylbut-2-ene, at least one of the double-bonded carbons is attached to two identical groups (two \(\text{-H}\) atoms in propene and but-1-ene, and two \(\text{-CH}_3\) groups in 2-methylbut-2-ene). Pent-2-ene (\(\text{CH}_3\text{CH}=\text{CHCH}_2\text{CH}_3\)) has two different groups on both double-bonded carbons (one has \(\text{-H}\) and \(\text{-CH}_3\), and the other has \(\text{-H}\) and \(\text{-CH}_2\text{CH}_3\)), so it exists as geometric isomers.

1 mark for the correct option D.

1. Calculate moles of NaOH: \(n(\text{NaOH}) = 0.0150 \times 0.200 = 3.00 \times 10^{-3} \text{ mol}\). 2. Moles of HCl in 25.0 \(\text{cm}^3\) aliquot: \(3.00 \times 10^{-3} \text{ mol}\) (due to 1:1 reaction ratio). 3. Total excess moles of HCl in 250.0 \(\text{cm}^3\): \(3.00 \times 10^{-3} \times 10 = 3.00 \times 10^{-2} \text{ mol}\). 4. Initial moles of HCl: \(n(\text{HCl})_{\text{initial}} = 0.0500 \times 1.00 = 0.0500 \text{ mol}\). 5. Moles of HCl reacted with calcium carbonate: \(0.0500 - 0.0300 = 0.0200 \text{ mol}\). 6. Moles of \(\text{CaCO}_3\) reacted: Since \(\text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{CO}_2 + \text{H}_2\text{O}\), the mole ratio is 1:2. Moles of \(\text{CaCO}_3 = 0.5 \times 0.0200 = 0.0100 \text{ mol}\). 7. Mass of \(\text{CaCO}_3 = 0.0100 \times 100.1 = 1.001 \text{ g}\). 8. Percentage purity: \((1.001 / 1.50) \times 100 = 66.7\%\).

[1 Mark] Calculates moles of NaOH used: \(3.00 \times 10^{-3} \text{ mol}\). [1 Mark] Deduces moles of HCl in aliquot and scales up to find excess HCl in 250 \(\text{cm}^3\): \(3.00 \times 10^{-2} \text{ mol}\). [1 Mark] Calculates reacted moles of HCl: \(0.0200 \text{ mol}\). [1 Mark] Uses 1:2 ratio to calculate moles of \(\text{CaCO}_3\): \(0.0100 \text{ mol}\). [1 Mark] Calculates mass of pure \(\text{CaCO}_3\): \(1.001 \text{ g}\). [1.5 Marks] Calculates percentage purity: \(66.7\%\) (accept 66.7% to 67.0% depending on exact molar mass of calcium carbonate used).

(a) The broad absorption at \(3350 \text{ cm}^{-1}\) is due to the \(\text{O-H}\) bond of an alcohol. (b)(i) The three secondary alcohol isomers are pentan-2-ol, pentan-3-ol, and 3-methylbutan-2-ol. (ii) Pentan-3-ol cannot produce a fragment at \(m/z = 45\) via single adjacent \(\text{C-C}\) cleavage. Pentan-3-ol is symmetrical: \(\text{CH}_3\text{CH}_2\text{CH(OH)}\text{CH}_2\text{CH}_3\). Cleavage on either side of the \(\text{CH-OH}\) group results in the loss of an ethyl radical (\(\cdot\text{C}_2\text{H}_5\)), leaving behind the fragment \([\text{CH}_3\text{CH}_2\text{CHOH}]^+\) at \(m/z = 59\). The other two isomers can lose a larger alkyl group to leave a \([\text{CH}_3\text{CHOH}]^+\) fragment at \(m/z = 45\).

[1.5 Marks] (a) Identifies O-H bond (1) and alcohol class (0.5). [3 Marks] (b)(i) Draws correct skeletal structures of pentan-2-ol (1), pentan-3-ol (1), and 3-methylbutan-2-ol (1). [2 Marks] (b)(ii) Identifies pentan-3-ol as the isomer (1) and explains that it only forms a fragment at \(m/z = 59\) because cleavage on either side yields an ethyl loss rather than a propyl/isopropyl loss (1).

(a) \(2\text{Mg(NO}_3)_2\text{(s)} \rightarrow 2\text{MgO(s)} + 4\text{NO}_2\text{(g)} + \text{O}_2\text{(g)}\). (b) Down the group: cationic radius increases while the charge remains \(2+\), so the charge density of the cation decreases. The cation has less polarizing power and polarizes/distorts the nitrate anion less. This weakens the \(\text{N-O}\) bond in the nitrate ion less, making the nitrate more thermally stable and requiring more energy to decompose. (c) Equation for calcium nitrate: \(2\text{Ca(NO}_3)_2\text{(s)} \rightarrow 2\text{CaO(s)} + 4\text{NO}_2\text{(g)} + \text{O}_2\text{(g)}\). Moles of \(\text{O}_2 = 1.50 / 24.0 = 0.0625 \text{ mol}\). Moles of \(\text{Ca(NO}_3)_2 = 2 \times 0.0625 = 0.125 \text{ mol}\). Mass = \(0.125 \times 164.1 = 20.5125 \text{ g}\) (rounds to \(20.5 \text{ g}\)).

[2 Marks] (a) Correct reactants and products (1), correct balancing and state symbols (1). [3 Marks] (b) Cationic radius increases / charge density decreases (1), polarizes/distorts nitrate anion less (1), weakens N-O bond less / more thermal energy needed to decompose (1). [1.5 Marks] (c) Moles of \(\text{O}_2 = 0.0625\) (0.5), moles of calcium nitrate = 0.125 and final mass = \(20.5 \text{ g}\) (1).

(a) Element **Z** belongs to Group 6 (or Group 16). Reason: There is a very large increase (jump) between the 6th and 7th ionization energies, indicating the 7th electron is removed from an inner electron shell closer to the nucleus. (b) Element **Z** is Sulfur (S). The electronic configuration of \(\text{S}^{2+}\) is \(\text{1s}^2\text{2s}^2\text{2p}^6\text{3s}^2\text{3p}^2\). (c) The element to the left of Sulfur is Phosphorus (P). Phosphorus has a HIGHER first ionization energy than sulfur. Phosphorus has the valence configuration \(\text{3s}^2\text{3p}^3\) with three singly occupied p-orbitals. Sulfur has the valence configuration \(\text{3s}^2\text{3p}^4\) with one paired set of electrons in a p-orbital. The mutual repulsion between the two paired electrons in sulfur's 3p orbital makes it easier to remove the first electron compared to phosphorus, despite sulfur's higher nuclear charge.

[2 Marks] (a) Group 6 (or 16) (1) and explanation of the jump between 6th and 7th IE (1). [1.5 Marks] (b) Sulfur / S (0.5) and configuration \(\text{1s}^2\text{2s}^2\text{2p}^6\text{3s}^2\text{3p}^2\) (1). [3 Marks] (c) States Phosphorus has a higher first IE than sulfur (1); explains that P has three unpaired electrons in 3p orbitals whereas S has a pair of electrons in one 3p orbital (1); explains that mutual repulsion between paired electrons in S makes its electron easier to remove (1).

(a) Electrophilic addition. (b) Major product: 2-bromo-2-methylbutane. Minor product: 2-bromo-3-methylbutane. (c) Mechanism: The double bond of 2-methylbut-2-ene attacks the hydrogen of \(\text{H}^{\delta+}-\text{Br}^{\delta-}\), breaking the \(\text{H-Br}\) bond (shown with curly arrows). This forms a tertiary carbocation intermediate: \(\text{CH}_3-\text{C}^+(\text{CH}_3)-\text{CH}_2-\text{CH}_3\), and a bromide ion (\(\text{Br}^-\)). A curly arrow from the lone pair on the bromide ion to the positive carbon of the carbocation forms the major product, 2-bromo-2-methylbutane. (d) For a molecule to exhibit \(E/Z\) isomerism, each carbon atom of the double bond must be attached to two different groups. In 2-methylbut-2-ene, one of the double-bonded carbons is attached to two identical methyl groups, so it cannot show \(E/Z\) isomerism.

[1 Mark] (a) Electrophilic addition. [1 Mark] (b) Correct structures of both 2-bromo-2-methylbutane and 2-bromo-3-methylbutane. [3.5 Marks] (c) Dipole on H-Br and curly arrow from C=C to H and H-Br bond to Br (1.5); correct structure of tertiary carbocation (1); curly arrow from bromide lone pair to the carbocation carbon (1). [1 Mark] (d) Explains that one C of the double bond has two identical methyl groups attached (1).

(a) The dot-and-cross diagram of \(\text{SF}_4\) should show the sulfur atom with 4 bonding pairs (one shared with each fluorine atom) and 1 lone pair (total of 10 valence electrons on S). Each fluorine atom has 1 bonding pair and 3 lone pairs (8 valence electrons). (b) Shape: Seesaw (or distorted tetrahedral). Explanation: Sulfur has 5 electron pairs in its valence shell (4 bonding pairs and 1 lone pair). These pairs arrange themselves in a trigonal bipyramidal orientation to minimize repulsion. The presence of the lone pair (which exerts greater repulsion than bonding pairs) results in a seesaw shape. (c) The bond angle in the ammonium ion is \(109.5^{\circ}\) (tetrahedral), which is larger than that in ammonia (\(107^{\circ}\), trigonal pyramidal). In \(\text{NH}_4^+\), there are 4 bonding pairs and no lone pairs, so all repulsions are equal. In \(\text{NH}_3\), there are 3 bonding pairs and 1 lone pair; since lone pair-bond pair repulsion is stronger than bond pair-bond pair repulsion, the N-H bonds are pushed closer together, reducing the angle.

[2 Marks] (a) Correct dot-and-cross diagram showing 4 bonding pairs and 1 lone pair on S (1), and correct outer electrons on F (1). [2.5 Marks] (b) Predicts seesaw shape (1); states S has 4 bonding pairs and 1 lone pair (1); explains trigonal bipyramidal baseline to minimize repulsion (0.5). [2 Marks] (c) Identifies bond angles: \(109.5^{\circ}\) for \(\text{NH}_4^+\) and \(107^{\circ}\) for \(\text{NH}_3\) (1); explains that lone pair-bond pair repulsion in \(\text{NH}_3\) is stronger than bond pair-bond pair repulsion, reducing the angle (1).

(a) \(\text{CH}_3\text{CH}_2\text{CH}_2\text{OH(l)} + 4.5\text{O}_2\text{(g)} \rightarrow 3\text{CO}_2\text{(g)} + 4\text{H}_2\text{O(l)}\). (b) \(\Delta_{\text{c}}H^{\ominus} = \sum \Delta_{\text{f}}H^{\ominus}(\text{products}) - \sum \Delta_{\text{f}}H^{\ominus}(\text{reactants})\). \(\Delta_{\text{c}}H^{\ominus} = [3 \times (-394) + 4 \times (-286)] - [-303 + 0] = [-1182 - 1144] + 303 = -2326 + 303 = -2023 \text{ kJ mol}^{-1}\). (c) Reasons: 1. Heat loss to the surrounding air or copper calorimeter. 2. Incomplete combustion of the propan-1-ol (forming carbon monoxide or carbon soot). 3. Evaporation of the alcohol from the wick.

[1.5 Marks] (a) Balanced equation (1), correct state symbols (0.5). [3 Marks] (b) Correct Hess's Law cycle or expression (1); correct substitution of values (1); correct calculation of final enthalpy value: \(-2023 \text{ kJ mol}^{-1}\) (1) (penalize missing negative sign or incorrect units). [2 Marks] (c) Any two correct reasons such as heat loss to surroundings (1), incomplete combustion (1), or evaporation of fuel (1).

(a) Increasing the total pressure shifts the equilibrium position to the left (towards reactants). This is because the reactant side has fewer moles of gas (2 moles) than the product side (4 moles). The system opposes the increase in pressure by moving in the direction that decreases the number of gas molecules. (b) When temperature is increased, the rate of both the forward and reverse reactions increases. However, because the forward reaction is endothermic, its rate increases by a greater factor than the rate of the exothermic reverse reaction. Consequently, the rate of the forward reaction is temporarily greater than the rate of the reverse reaction, causing the equilibrium position to shift to the right until a new equilibrium is established. (c) On a Maxwell-Boltzmann distribution, raising the temperature from \(T_1\) to \(T_2\) shifts the peak downwards and to the right. The activation energy (\(E_{\text{a}}\)) remains constant. At the higher temperature \(T_2\), a significantly larger fraction of the molecules possess energy greater than or equal to the activation energy (\(E \geq E_{\text{a}}\)), which results in a higher frequency of successful collisions and thus a faster rate of reaction.

[2 Marks] (a) Shift to left (1); explanation based on fewer moles of gas on the LHS (2 vs 4 moles) (1). [2.5 Marks] (b) Increasing temperature increases rates of both forward and reverse reactions (0.5); rate of the endothermic forward reaction increases more than the exothermic reverse reaction (1); therefore, the position of equilibrium shifts to the right (1). [2 Marks] (c) Explains that the distribution curve at higher temperature has a lower peak shifted to the right (0.5); explains that a larger fraction of molecules have energy \(E \geq E_{\text{a}}\), resulting in more frequent successful collisions (1.5).

(a) \(\text{CoCl}_2\cdot x\text{H}_2\text{O(s)} \rightarrow \text{CoCl}_2\text{(s)} + x\text{H}_2\text{O(g)}\)

(b) Mass of water lost = \(3.146\text{ g} - 1.716\text{ g} = 1.430\text{ g}\)
Moles of anhydrous \(\text{CoCl}_2\) = \(\frac{1.716}{58.9 + (2 \times 35.5)} = \frac{1.716}{129.9} = 0.01321\text{ mol}\)
Moles of \(\text{H}_2\text{O}\) lost = \(\frac{1.430}{18.0} = 0.07944\text{ mol}\)
Ratio \(x = \frac{0.07944}{0.01321} = 6.01 \approx 6\)
So, \(x = 6\).

(c) Further decomposition of the cobalt chloride would result in the loss of chlorine gas, which escapes as a gas. This makes the measured mass of the remaining anhydrous solid lower than it should be. Consequently, this is interpreted as a larger mass loss, which is incorrectly attributed to water. Thus, the calculated value of \(x\) would be higher than the true value.

(a)
- 1 Mark: Correct formulae and balanced equation: \(\text{CoCl}_2\cdot x\text{H}_2\text{O} \rightarrow \text{CoCl}_2 + x\text{H}_2\text{O}\)
- 0.5 Mark: Correct state symbols (s) for reactant and anhydrous salt, and (g) or (l) for water.

(b)
- 1 Mark: Calculates mass of water lost as \(1.430\text{ g}\) and moles of water as \(\frac{1.430}{18.0} = 0.07944\text{ mol}\).
- 1 Mark: Calculates molar mass of \(\text{CoCl}_2 = 129.9\text{ g mol}^{-1}\) and moles of \(\text{CoCl}_2 = \frac{1.716}{129.9} = 0.01321\text{ mol}\).
- 1 Mark: Finds the ratio \(\frac{0.07944}{0.01321} = 6.01\) and rounds to the nearest whole number, \(x = 6\).

(c)
- 1 Mark: Recognises that further decomposition causes chlorine gas to escape, resulting in a lower measured mass of anhydrous residue (or an apparently larger loss of mass).
- 1 Mark: Concludes that the apparent mass of water lost is too high, leading to an overestimation of \(x\) (the calculated value of \(x\) increases).

(a) Percentage of \(\text{X} = 100\% - (35.1\% + 6.6\%) = 58.3\%\).
Molar ratios:
\(\text{C} = \frac{35.1}{12.0} = 2.925\)
\(\text{H} = \frac{6.6}{1.0} = 6.60\)
\(\text{X} = \frac{58.3}{79.9} = 0.730\)

Dividing by the smallest value (0.730):
\(\text{C} = \frac{2.925}{0.730} = 4.01 \approx 4\)
\(\text{H} = \frac{6.60}{0.730} = 9.04 \approx 9\)
\(\text{X} = \frac{0.730}{0.730} = 1\)
Thus, the empirical formula is \(\text{C}_4\text{H}_9\text{X}\).

(b) The molecular ion peaks at \(m/z = 136\) and \(138\) in a 1:1 ratio are characteristic of a compound containing one bromine atom, as bromine has two isotopes, \(^{79}\text{Br}\) and \(^{81}\text{Br}\), in approximately equal abundance. Therefore, \(\text{X} = \text{Br}\). Since **Y** is a secondary halogenoalkane with 4 carbon atoms, its IUPAC name is 2-bromobutane.

(c) The reaction of 2-bromobutane with aqueous \(\text{NaOH}\) yields butan-2-ol. The skeletal structure of butan-2-ol is a 4-carbon chain with a hydroxyl (\(-\text{OH}\)) group attached to carbon 2.

(a)
- 1 Mark: Calculates correct percentage of halogen \(\text{X} = 58.3\%\).
- 1 Mark: Correctly calculates the molar ratios of \(\text{C}\), \(\text{H}\), and \(\text{X}\) by dividing percentages by relative atomic masses (\(2.925 : 6.60 : 0.730\)).
- 1 Mark: Obtains the simplest whole-number ratio of \(4 : 9 : 1\) and states the empirical formula as \(\text{C}_4\text{H}_9\text{X}\).

(b)
- 0.5 Mark: Identifies \(\text{X}\) as Bromine / \(\text{Br}\) based on 1:1 isotopic abundance of \(m/z = 136\) and \(138\).
- 1 Mark: Identifies **Y** as 2-bromobutane (must specify '2-' and 'butane' correctly).

(c)
- 1 Mark: Names the organic product correctly as butan-2-ol (allow 2-butanol).
- 1 Mark: Draws the correct skeletal structure of butan-2-ol showing a four-carbon chain with a hydroxyl group attached to the second carbon.

(a) Ionic equation:
\(\text{Cl}_2 + 2\text{OH}^- \rightarrow \text{Cl}^- + \text{ClO}^- + \text{H}_2\text{O}\)
Oxidation number of chlorine in \(\text{Cl}^-\): \(-1\)
Oxidation number of chlorine in \(\text{ClO}^-\): \(+1\)

(b) Balanced molecular equation:
\(3\text{Cl}_2 + 6\text{NaOH} \rightarrow 5\text{NaCl} + \text{NaClO}_3 + 3\text{H}_2\text{O}\)

(c) Disproportionation occurs because chlorine is simultaneously oxidised and reduced in the same reaction. In both reactions, the starting oxidation state of chlorine in \(\text{Cl}_2\) is 0.
- In cold NaOH, some chlorine atoms are reduced to \(-1\) (gaining 1 electron each) and some are oxidised to \(+1\) (losing 1 electron each).
- In hot NaOH, some chlorine atoms are reduced to \(-1\) (gaining 1 electron each) and some are oxidised to \(+5\) (losing 5 electrons each).

(a)
- 1 Mark: Correct balanced ionic equation: \(\text{Cl}_2 + 2\text{OH}^- \rightarrow \text{Cl}^- + \text{ClO}^- + \text{H}_2\text{O}\) (ignore state symbols).
- 1 Mark: Correct oxidation state of \(-1\) for \(\text{Cl}^-\).
- 1 Mark: Correct oxidation state of \(+1\) for \(\text{ClO}^-\).

(b)
- 1.5 Marks: Correctly balanced molecular equation: \(3\text{Cl}_2 + 6\text{NaOH} \rightarrow 5\text{NaCl} + \text{NaClO}_3 + 3\text{H}_2\text{O}\) (award 1 mark if species are correct but coefficients are incorrect).

(c)
- 1 Mark: Defines disproportionation as a reaction where the same element is simultaneously oxidised and reduced.
- 1 Mark: Explains that chlorine begins at oxidation state 0 and goes to both negative (\(-1\)) and positive (\(+1\) or \(+5\)) oxidation states, showing both reduction and oxidation.

(a) There is a large jump between the 4th and 5th ionisation energies (from \(4356\) to \(16091\text{ kJ mol}^{-1}\)), which indicates that the 5th electron is removed from an inner quantum shell. Therefore, **Z** has 4 valence electrons and belongs to Group 4 of Period 3, which is Silicon (\(\text{Si}\)).
The electronic configuration of the stable \(\text{Si}^{2+}\) ion is \(1s^2 2s^2 2p^6 3s^2\).

(b) The first four electrons are removed from the outer \(3s\) and \(3p\) subshells (\(n=3\) shell). The 5th electron is being removed from an inner main energy level (the \(n=2\) shell), which is closer to the nucleus and experiences significantly less shielding from inner shells. This results in a much stronger electrostatic attraction to the nucleus, requiring a much greater input of energy.

(c) Phosphorus has the outer electron configuration \(3s^2 3p^3\), where the three \(3p\) orbitals are each singly occupied. Sulfur has the configuration \(3s^2 3p^4\), where one of the \(3p\) orbitals contains a pair of electrons. The mutual repulsion between the two paired electrons in this \(3p\) orbital of sulfur makes it easier to remove one of them than from the singly occupied orbitals of phosphorus, overriding the effect of the higher nuclear charge of sulfur.

(a)
- 1 Mark: Identifies element **Z** as Silicon / \(\text{Si}\).
- 1 Mark: Writes correct electronic configuration for \(\text{Si}^{2+}\): \(1s^2 2s^2 2p^6 3s^2\) (allow \([\text{Ne}]3s^2\)).

(b)
- 1 Mark: Explains that the 5th electron is removed from a shell closer to the nucleus (or an inner shell / \(n=2\) shell).
- 1 Mark: States that this electron experiences less shielding, leading to a much stronger nuclear attraction.

(c)
- 1 Mark: States the outer electronic configurations or notes that phosphorus has singly occupied \(p\)-orbitals whereas sulfur has one doubly occupied \(p\)-orbital.
- 1 Mark: Explains that there is repulsion between the paired electrons in the \(3p\) orbital of sulfur.
- 0.5 Mark: Concludes that this repulsion makes it easier to remove the outer electron in sulfur, requiring less energy.

(a) The two structural isomers are 1-bromobutane and 2-bromobutane.
Skeletal structure of 1-bromobutane is a 4-carbon chain with a bromine atom at position 1. Skeletal structure of 2-bromobutane is a 4-carbon chain with a bromine atom at position 2.

(b) Mechanism:
1. The \(\pi\) electrons of the \(\text{C=C}\) double bond in but-1-ene (\(\text{CH}_2\text{=CH-CH}_2\text{-CH}_3\)) attack the hydrogen atom of \(\text{H-Br}\). A dipole is drawn on \(\text{H-Br}\) with \(\text{H}^{\delta+}\) and \(\text{Br}^{\delta-}\).
2. The \(\text{H-Br}\) bond breaks heterolytically, with the electron pair going to the bromine atom to form a bromide ion (\(\text{Br}^-\)).
3. This creates a secondary carbocation intermediate: \(\text{CH}_3\text{-CH}^+\text{-CH}_2\text{-CH}_3\).
4. A curly arrow is drawn from the lone pair on the bromide ion (\(\text{:Br}^-\)) to the positively charged carbon atom of the carbocation, yielding the final product, 2-bromobutane.

(c) The major product is 2-bromobutane. The reaction proceeds via a secondary carbocation intermediate (\(\text{CH}_3\text{CH}^+\text{CH}_2\text{CH}_3\)), which is more stable than the primary carbocation intermediate (\(\text{CH}_2^+\text{CH}_2\text{CH}_2\text{CH}_3\)) because it has two electron-releasing alkyl groups that stabilise the positive charge via the inductive effect.

(a)
- 1 Mark: Both names correct: 1-bromobutane and 2-bromobutane.
- 1 Mark: Both skeletal structures drawn correctly.

(b)
- 1 Mark: Curly arrow from the \(\text{C=C}\) double bond to the \(\text{H}\) atom of \(\text{H-Br}\), and correct dipole \(\text{H}^{\delta+}-\text{Br}^{\delta-}\) with arrow from bond to \(\text{Br}\).
- 1 Mark: Correct structure of the secondary carbocation intermediate showing positive charge on C2.
- 1 Mark: Curly arrow from lone pair of \(\text{Br}^-\) ion to the positively charged carbon.

(c)
- 0.5 Mark: Identifies 2-bromobutane as the major product.
- 1 Mark: Explains that the secondary carbocation intermediate is more stable than the primary carbocation because of the electron-releasing inductive effect of two alkyl groups.

(a) Phosphorus is in Group 5 and has 5 valence electrons. The positive charge means one electron has been lost, leaving 4 valence electrons. These 4 electrons form 4 bonding pairs with 4 chlorine atoms, with zero lone pairs around the central phosphorus atom. To minimise repulsion, these 4 bonding pairs arrange themselves as far apart as possible, resulting in a tetrahedral shape with a bond angle of \(109.5^\circ\).

(b) Sulfur hexafluoride (\(\text{SF}_6\)) has 6 bonding pairs and 0 lone pairs around the central sulfur atom. The shape is octahedral with bond angles of \(90^\circ\) (and \(180^\circ\)). The drawing should show a central S atom with 4 F atoms in a horizontal plane (using wedges and dashes to show depth) and 2 F atoms positioned vertically.

(c) In carbon dioxide (\(\text{CO}_2\)), the central carbon atom has two double bonds (two bonding regions) and no lone pairs. These two regions of high electron density repel each other to opposite sides, resulting in a linear shape with a bond angle of \(180^\circ\). In sulfur dioxide (\(\text{SO}_2\)), the central sulfur atom has two bonding regions and one lone pair. The presence of the lone pair causes extra repulsion, pushing the bonding regions closer together and resulting in a bent (non-linear) shape with a bond angle of approximately \(119^\circ\).

(a)
- 1 Mark: Correctly states that \(\text{PCl}_4^+\) has 4 bonding pairs and 0 lone pairs around the phosphorus atom.
- 1 Mark: Predicts tetrahedral shape and \(109.5^\circ\) bond angle.
- 1 Mark: Explains that electron pairs repel to get as far apart as possible to minimise repulsion.

(b)
- 1 Mark: Correct 3D representation showing octahedral geometry (wedges and dashes used correctly).
- 0.5 Mark: Correctly names the shape as octahedral and states bond angle as \(90^\circ\).

(c)
- 1 Mark: Explains that \(\text{CO}_2\) has 2 double bonds (or bonding regions) and no lone pairs on the carbon atom, leading to a linear shape.
- 1 Mark: Explains that \(\text{SO}_2\) has 2 bonding regions and 1 lone pair on the sulfur atom, which repels the bonding pairs to form a bent shape.

(a)
Initiation step:
\(\text{Cl}_2 \xrightarrow{UV} 2\text{Cl}^\bullet\)

Propagation step 1:
\(\text{CH}_3\text{CH}_2\text{CH}_3 + \text{Cl}^\bullet \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2^\bullet + \text{HCl}\)

Propagation 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) Hexane (\(\text{C}_6\text{H}_{14}\)) is formed during a termination step when two propyl radicals collide and combine.
Equation:
\(2\text{CH}_3\text{CH}_2\text{CH}_2^\bullet \rightarrow \text{C}_6\text{H}_{14}\) (or \(\text{CH}_3(\text{CH}_2)_4\text{CH}_3\))

(c)
1. Further substitution can occur, leading to di-, tri-, and poly-halogenated alkanes (such as dichloropropanes).
2. Isomeric mixtures can form because substitution can occur at different positions on the carbon chain (for example, forming 2-chloropropane alongside 1-chloropropane).

(a)
- 1 Mark: Correct equation for initiation showing homolytic fission of \(\text{Cl}_2\) into chlorine radicals under UV light.
- 1 Mark: Propagation step 1 showing propane reacting with \(\text{Cl}^\bullet\) to form a propyl radical and \(\text{HCl}\).
- 1 Mark: Propagation step 2 showing the propyl radical reacting with \(\text{Cl}_2\) to yield 1-chloropropane and regenerate the chlorine radical.

(b)
- 1 Mark: Explains that it occurs via a termination step where two propyl radicals combine.
- 0.5 Mark: Correct balanced equation for the termination step: \(2\text{C}_3\text{H}_7^\bullet \rightarrow \text{C}_6\text{H}_{14}\).

(c)
- 1 Mark: Identifies that further substitution occurs (forming multi-substituted products).
- 1 Mark: Identifies that structural isomers are formed (substitution can occur at different carbon positions).

(a) Sketch should include:
- y-axis labelled 'Number of molecules' and x-axis labelled 'Energy'.
- An asymmetric curve starting at the origin, peaking, and asymptotically approaching the x-axis at high energy.
- \(E_{mp}\) marked at the energy corresponding to the peak of the curve.
- \(E_a\) marked as a vertical line towards the higher energy end of the distribution.
- \(E_{cat}\) marked as a vertical line to the left of \(E_a\).

(b) A catalyst provides an alternative reaction pathway with a lower activation energy (\(E_{cat} < E_a\)). On the diagram, this shifts the activation energy line to the left. Consequently, a significantly larger fraction of molecules have energy greater than or equal to the new activation energy (represented by the area under the curve to the right of \(E_{cat}\)), leading to a higher frequency of successful collisions and thus a faster rate of reaction.

(c) The curve for \(T_2\) (higher temperature) should:
- Start at the origin and cross the first curve only once.
- Have a lower peak that is shifted to the right (higher energy).
- Have a flatter, broader tail at high energies, showing more molecules have high kinetic energies.

(a)
- 1 Mark: Correctly labels both axes (y: Number of molecules, x: Energy) and draws a curve starting at origin, rising to a peak, and asymptotically approaching the x-axis.
- 1 Mark: Correctly positions \(E_{mp}\) at the maximum peak of the curve.
- 1 Mark: Correctly labels \(E_{a}\) and positions \(E_{cat}\) to the left of \(E_{a}\).

(b)
- 1 Mark: Explains that the catalyst provides an alternative reaction pathway with a lower activation energy.
- 1 Mark: Refers to the diagram to explain that a larger fraction of molecules now have energy greater than or equal to the activation energy, leading to more frequent successful collisions.

(c)
- 1 Mark: Draws a correct curve for \(T_2\) with a lower peak, shifted to the right, crossing the first curve only once.
- 0.5 Mark: Mentions that the peak is flatter/lower and shifted to the right.

(a) The ionic equation is:
\(\text{CO}_3^{2-}(\text{aq}) + 2\text{H}^+(\text{aq}) \rightarrow \text{CO}_2(\text{g}) + \text{H}_2\text{O}(\text{l})\)

(b) Step-by-step calculation:
1. Calculate moles of \(\text{HCl}\) used in the titration:
\(n(\text{HCl}) = \frac{20.00}{1000} \times 0.125 = 2.50 \times 10^{-3}\text{ mol}\)

2. Determine the moles of \(\text{Na}_2\text{CO}_3\) in the \(25.0\text{ cm}^3\) sample:
According to the equation, \(1\text{ mol}\) of \(\text{Na}_2\text{CO}_3\) reacts with \(2\text{ mol}\) of \(\text{HCl}\).
\(n(\text{Na}_2\text{CO}_3) = \frac{2.50 \times 10^{-3}}{2} = 1.25 \times 10^{-3}\text{ mol}\)

3. Determine the moles of \(\text{Na}_2\text{CO}_3\) in the original \(250\text{ cm}^3\) solution:
\(n(\text{Na}_2\text{CO}_3)_{\text{total}} = 1.25 \times 10^{-3} \times 10 = 1.25 \times 10^{-2}\text{ mol}\)

4. Calculate the molar mass (\(M_r\)) of the hydrated salt, \(\text{Na}_2\text{CO}_3 \cdot x\text{H}_2\text{O}\):
\(M_r = \frac{\text{mass}}{\text{moles}} = \frac{3.575}{0.0125} = 286\text{ g mol}^{-1}\)

5. Calculate the mass of water and find \(x\):
\(M_r(\text{Na}_2\text{CO}_3) = (2 \times 23.0) + 12.0 + (3 \times 16.0) = 106\text{ g mol}^{-1}\)
Mass of water of crystallisation = \(286 - 106 = 180\text{ g mol}^{-1}\)
\(x = \frac{180}{18.0} = 10\)

Part (a) [Total: 2 marks]
- Correct ionic species and balancing [1 mark]: \(\text{CO}_3^{2-} + 2\text{H}^+ \rightarrow \text{CO}_2 + \text{H}_2\text{O}\)
- Correct state symbols [1 mark]: \((\text{aq})\) for reactants, \((\text{g})\) and \((\text{l})\) for products.

Part (b) [Total: 4.5 marks]
- Calculation of moles of \(\text{HCl}\) [1 mark]: \(2.50 \times 10^{-3}\text{ mol}\)
- Calculation of moles of \(\text{Na}_2\text{CO}_3\) in titration and scaling up to \(250\text{ cm}^3\) [1 mark]: \(1.25 \times 10^{-2}\text{ mol}\)
- Calculation of molar mass of hydrated sodium carbonate [1 mark]: \(286\text{ g mol}^{-1}\)
- Subtracting anhydrous molar mass (\(106\text{ g mol}^{-1}\)) to get \(180\text{ g mol}^{-1}\) [1 mark]
- Solving for \(x = 10\) [0.5 marks]

(a) \(\text{Cl}_2(\text{g}) + 2\text{NaOH}(\text{aq}) \rightarrow \text{NaCl}(\text{aq}) + \text{NaClO}(\text{aq}) + \text{H}_2\text{O}(\text{l})\)

(b) Disproportionation is a reaction in which the same element is simultaneously oxidised and reduced. In this reaction:
- Chlorine in \(\text{Cl}_2\) has an oxidation state of 0.
- It is reduced to an oxidation state of -1 in sodium chloride, \(\text{NaCl}\).
- It is oxidised to an oxidation state of +1 in sodium chlorate(I), \(\text{NaClO}\).

(c) When chlorine reacts with hot, concentrated sodium hydroxide, it is oxidised to a higher oxidation state, forming sodium chlorate(V) instead of sodium chlorate(I). The formula of the new salt is \(\text{NaClO}_3\).

Part (a) [Total: 1.5 marks]
- Correct species for reactants and products [1 mark]
- Correctly balanced equation [0.5 marks]

Part (b) [Total: 3 marks]
- Definition of disproportionation as simultaneous oxidation and reduction of the same element [1 mark]
- Stating the starting oxidation state of chlorine in \(\text{Cl}_2\) is 0 [1 mark]
- Stating that chlorine is reduced to -1 in \(\text{NaCl}\) AND oxidised to +1 in \(\text{NaClO}\) [1 mark]

Part (c) [Total: 2 marks]
- Stating that a different chlorine-containing salt containing chlorine in a higher oxidation state (+5) is formed [1 mark]
- Correct formula: \(\text{NaClO}_3\) [1 mark]

(a) Hess's Law Cycle:

Top row: \(\text{C}_4\text{H}_8(\text{g}) + \text{H}_2\text{O}(\text{l}) \xrightarrow{\Delta_r H^\theta} \text{C}_4\text{H}_9\text{OH}(\text{l})\)

Arrows pointing down from both the reactants and products to the common combustion products:
\(4\text{CO}_2(\text{g}) + 5\text{H}_2\text{O}(\text{l})\)

- Left downward arrow represents: \(\Delta_c H^\theta [\text{C}_4\text{H}_8(\text{g})] = -2718\text{ kJ mol}^{-1}\)
- Right downward arrow represents: \(\Delta_c H^\theta [\text{C}_4\text{H}_9\text{OH}(\text{l})] = -2676\text{ kJ mol}^{-1}\)

(b) According to Hess's Law:
\(\Delta_r H^\theta = \Delta_c H^\theta(\text{reactants}) - \Delta_c H^\theta(\text{products})\)
\(\Delta_r H^\theta = -2718 - (-2676) = -42\text{ kJ mol}^{-1}\)

(c) The reaction is too slow at standard conditions without a catalyst, or has a very high activation energy, making direct calorimetric measurement inaccurate because heat would be lost to the surroundings over a long period.

(d) The reaction is exothermic because the standard enthalpy change is negative (\(\Delta_r H^\theta < 0\)), meaning energy is released to the surroundings.

Part (a) [Total: 3 marks]
- Correctly showing reactants, products, and the shared combustion products: \(4\text{CO}_2(\text{g}) + 5\text{H}_2\text{O}(\text{l})\) with correct states [1 mark]
- Correct direction of both combustion arrows pointing downwards [1 mark]
- Correct labeling of arrows with the respective values/symbols [1 mark]

Part (b) [Total: 1.5 marks]
- Correct equation/expression based on Hess's Law [1 mark]: \(-2718 - (-2676)\)
- Correct final value with units: \(-42\text{ kJ mol}^{-1}\) [0.5 marks]

Part (c) [Total: 1 mark]
- The reaction is too slow / has a high activation energy [1 mark]

Part (d) [Total: 1 mark]
- Exothermic because the enthalpy change is negative [1 mark]

(a) Element X is Sulfur (S).
There is a very large increase (jump) between the 6th and the 7th ionisation energies (from \(8500\text{ kJ mol}^{-1}\) to \(27100\text{ kJ mol}^{-1}\)). This indicates that the 7th electron is being removed from an inner quantum shell, which is closer to the nucleus and experiences less shielding. Therefore, X has 6 electrons in its valence shell, placing it in Group 16 (Group 6). Since X is in Period 3, it must be sulfur.

(b) The third ionisation energy is the energy required to remove one mole of electrons from one mole of gaseous \(\text{S}^{2+}\) ions:
\(\text{S}^{2+}(\text{g}) \rightarrow \text{S}^{3+}(\text{g}) + \text{e}^-\)

(c) Phosphorus has the outer electron configuration \(3\text{s}^2 3\text{p}^3\), meaning each of the three \(3\text{p}\) orbitals contains a single, unpaired electron. Sulfur has the outer configuration \(3\text{s}^2 3\text{p}^4\). In sulfur, one of the \(3\text{p}\) orbitals contains a paired set of electrons. The mutual repulsion between these two paired electrons in the same orbital makes it easier to remove one of them, resulting in a lower first ionisation energy than phosphorus.

Part (a) [Total: 2.5 marks]
- Identifies X as sulfur / S [0.5 marks]
- Identifies the large jump between the 6th and 7th ionisation energies [1 mark]
- Concludes that X has 6 valence electrons and is in Period 3 [1 mark]

Part (b) [Total: 2 marks]
- Correct species in equation: \(\text{X}^{2+} \rightarrow \text{X}^{3+} + \text{e}^-\)\ or using S [1 mark]
- Correct gaseous state symbols: \((\text{g})\) on both ionic species [1 mark]

Part (c) [Total: 2 marks]
- Identifies the configurations: P has singly occupied \(3\text{p}\) orbitals, S has one paired orbital / \(3\text{p}^4\) [1 mark]
- Explains that spin-pair repulsion between paired electrons in S makes the electron easier to remove [1 mark]

(a) Sodium hydrogencarbonate is added to neutralise any remaining unreacted hydrochloric acid in the organic layer. The neutralisation reaction produces carbon dioxide gas, which increases pressure inside the funnel. Releasing the stopper periodically vents this gas to prevent pressure build-up. (b) Anhydrous calcium chloride acts as a drying agent to remove trace water from the organic product. The liquid is dry when it changes from cloudy to completely clear. (c) Error 1: The thermometer bulb is placed too low (it should be at the T-junction of the still head to measure the boiling temperature of the vapour being condensed). Error 2: Water enters the condenser from the top instead of the bottom (it must enter from the bottom to ensure the jacket is completely filled with water for efficient cooling). (d) Moles of 2-methylpropan-2-ol = 10.0 / 74.1 = 0.13495 mol. Since the reaction ratio is 1:1, theoretical moles of 2-chloro-2-methylpropane = 0.13495 mol. Theoretical mass of 2-chloro-2-methylpropane = 0.13495 * 92.5 = 12.483 g. Percentage yield = (7.50 / 12.483) * 100 = 60.1%.

Part (a): 1 mark for neutralising excess acid; 1 mark for stating carbon dioxide gas is produced; 1 mark for venting to prevent pressure build-up. Part (b): 1 mark for drying agent/removing water; 1 mark for observation of transition from cloudy to clear. Part (c): 1 mark for correcting thermometer position; 1 mark for correcting water inlet/outlet direction. Part (d): 1 mark for calculating moles of reactant (0.135 mol); 1 mark for calculating theoretical mass (12.48 g); 1 mark for final percentage yield to 3 significant figures (60.1%).

(a) The ionic equation is H+(aq) + OH-(aq) -> H2O(l). (b) Total volume of solution = 25.0 + 25.0 = 50.0 cm3. Mass of solution (m) = 50.0 g. Temperature change (\(\Delta T\)) = 26.8 - 20.2 = 6.6 K. Heat energy transferred (q) = m * c * \(\Delta T\) = 50.0 * 4.18 * 6.6 = 1379.4 J = 1.3794 kJ. Moles of acid reacting = volume * concentration = 0.0250 dm3 * 1.00 mol dm-3 = 0.0250 mol. Since the reaction stoichiometry is 1:1, 0.0250 mol of water is formed. \(\Delta H\) = -q / n = -1.3794 / 0.0250 = -55.176 kJ mol-1, which rounds to -55.2 kJ mol-1. (c) Polystyrene is an excellent thermal insulator compared to copper, reducing heat transfer to the surroundings. An improvement to minimize heat loss further is to place a lid on the polystyrene cup or place the cup inside a larger beaker filled with cotton wool insulation.

Part (a): 1 mark for correct formulas and balancing; 1 mark for correct state symbols. Part (b): 1 mark for mass of solution (50.0 g) and \(\Delta T\) (6.6 K); 1 mark for calculating heat energy q (1379.4 J or 1.38 kJ); 1 mark for calculating moles of acid/alkali (0.0250 mol); 1 mark for dividing q by moles; 1 mark for final answer of -55.2 kJ mol-1 with negative sign and correct units. Part (c): 1 mark for polystyrene being a better thermal insulator / copper is a good conductor; 1 mark for placing a lid; 1 mark for explanation of how it reduces heat loss.

(a) The gas is carbon dioxide (CO2). The ionic equation is CO32-(s/aq) + 2H+(aq) -> CO2(g) + H2O(l). (b) The Group 2 cation is magnesium (Mg2+). Group 2 hydroxides become more soluble down the group, so magnesium hydroxide has very low solubility and forms a white precipitate. Group 2 sulfates become less soluble down the group; magnesium sulfate is highly soluble, so adding sulfuric acid yields no precipitate, confirming Mg2+ over calcium, strontium, or barium. (c) Dissolve the solid in deionised water, add dilute nitric acid followed by silver nitrate solution. A white precipitate of silver chloride (AgCl) forms. To distinguish it from bromide, add dilute aqueous ammonia; the silver chloride precipitate will completely dissolve, whereas silver bromide only dissolves in concentrated ammonia.

Part (a): 1 mark for identifying carbon dioxide (CO2); 1 mark for correct reactants and products in the ionic equation; 1 mark for correct state symbols. Part (b): 1 mark for identifying magnesium (Mg2+); 1 mark for stating that Group 2 hydroxides become more soluble down the group (Mg(OH)2 is insoluble); 1 mark for stating that Group 2 sulfates become less soluble down the group (MgSO4 is soluble); 1 mark for linking both pieces of evidence to rule out Ca2+, Sr2+, and Ba2+. Part (c): 1 mark for adding nitric acid and silver nitrate; 1 mark for stating a white precipitate forms; 1 mark for stating that the precipitate dissolves in dilute ammonia.

(a) Heating to constant mass ensures that all the water of crystallisation has been completely driven off and the reaction is complete. (b) Mass of hydrated salt = 24.78 - 22.34 = 2.44 g. Mass of anhydrous BaCl2 = 24.42 - 22.34 = 2.08 g. Mass of water lost = 2.44 - 2.08 = 0.36 g. Molar mass of BaCl2 = 137.3 + (2 * 35.5) = 208.3 g mol-1. Moles of BaCl2 = 2.08 / 208.3 = 0.009986 mol. Molar mass of H2O = 18.0 g mol-1. Moles of H2O = 0.36 / 18.0 = 0.0200 mol. Mole ratio (x) = Moles of H2O / Moles of BaCl2 = 0.0200 / 0.009986 = 2.003, which rounds to the whole number 2. (c) The lid is placed loosely to prevent the solid hydrated salt from spitting out of the crucible during vigorous heating, which would decrease the final recorded mass of anhydrous salt, while still allowing the water vapour to escape freely.

Part (a): 1 mark for ensuring all water of crystallisation is removed; 1 mark for ensuring the reaction is complete/constant mass reached. Part (b): 1 mark for calculating mass of BaCl2 (2.08 g) and mass of H2O (0.36 g); 1 mark for calculating Mr of BaCl2 (208.3); 1 mark for moles of BaCl2 (0.0100 mol); 1 mark for moles of H2O (0.0200 mol); 1 mark for dividing moles to get x = 2. Part (c): 1 mark for preventing loss of solid by spitting/crackling; 1 mark for allowing steam/water vapour to escape; 1 mark for explaining how loss of solid would affect calculated mass accuracy.

(a) The cotton wool plug allows carbon dioxide gas to escape freely while preventing any acid spray/droplets from escaping, which would cause an overestimation of the mass loss. (b) Mass loss = mass of CO2 escaped = 0.44 g. Moles of CO2 produced = 0.44 / 44.0 = 0.010 mol. Since 1 mole of CO2 is produced for every 2 moles of H+ consumed, moles of H+ consumed = 0.010 * 2 = 0.020 mol. Volume of acid = 50.0 cm3 = 0.0500 dm3. Change in concentration of H+ = 0.020 / 0.0500 = 0.40 mol dm-3. Average rate of consumption of H+ = change in concentration / time = 0.40 / 40.0 = 0.010 mol dm-3 s-1. (c) A single large marble chip has a much smaller surface area than medium-sized chips. This reduces the frequency of collisions between H+ ions and the calcium carbonate surface, resulting in a slower initial rate of reaction (flatter curve slope), though the same final mass loss will eventually be reached.

Part (a): 1 mark for allowing CO2 to escape; 1 mark for preventing loss of acid spray/droplets. Part (b): 1 mark for moles of CO2 (0.010 mol); 1 mark for moles of H+ consumed (0.020 mol); 1 mark for volume conversion to dm3 (0.0500 dm3); 1 mark for calculating change in [H+] (0.40 mol dm-3); 1 mark for final rate of 0.010 mol dm-3 s-1. Part (c): 1 mark for stating that a single large chip has a smaller surface area; 1 mark for fewer successful collisions per unit time/lower collision frequency; 1 mark for stating the rate of mass loss decreases/slope is less steep.

(a) C6H11OH -> C6H10 + H2O. The reaction type is elimination (or dehydration). (b) Step 1: Protonation of the hydroxyl group in cyclohexanol by the acid catalyst to form a protonated alcohol (-OH2+). Step 2: Loss of a water molecule from the protonated alcohol to form a cyclohexyl carbocation. Step 3: Loss of a proton (H+) from an adjacent carbon atom to regenerate the acid catalyst and form the C=C double bond in cyclohexene. (c) Add the mixture to a separating funnel, add saturated sodium chloride solution, insert the stopper, shake, and invert to release pressure. Allow the phases to separate. Cyclohexene is insoluble in water and has a lower density (0.81 g cm-3) than the aqueous phase (1.20 g cm-3), so the organic layer containing cyclohexene will be the top layer. Run off the lower aqueous layer. (d) Add a small piece of clean sodium metal to a sample of the dried cyclohexene. If no cyclohexanol is present, there will be no effervescence/bubbles (since cyclohexanol would react with sodium to produce hydrogen gas).

Part (a): 1 mark for balanced molecular equation; 1 mark for identifying elimination/dehydration. Part (b): 1 mark for protonation of the alcohol oxygen; 1 mark for loss of water to form a carbocation intermediate; 1 mark for loss of a proton to form the double bond and regenerate the catalyst. Part (c): 1 mark for shaking and venting in a separating funnel; 1 mark for identifying that cyclohexene is in the top layer; 1 mark for explaining density comparison (0.81 < 1.20 g cm-3) determines layer position. Part (d): 1 mark for identifying a suitable test reagent (e.g. sodium metal, PCl5, or acidified potassium dichromate(VI)); 1 mark for correct observation showing absence of alcohol (e.g. no effervescence for Na/PCl5, or remains orange/no colour change for dichromate).