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Question 1

Topic: 11.1

(a) The elements of Group 17 are called halogens. Complete Table 1.1.

(b) State the trend in volatility of the halogens chlorine, bromine and iodine. Explain your answer.

(c) Iodine is made by reacting bromine with sodium iodide.
(i) Construct an ionic equation for the reaction of bromine with sodium iodide.
(ii) State the role of bromine in the reaction. Explain your answer.

(d) Concentrated sulfuric acid is added to separate samples containing equal amounts of NaCl, NaBr and NaI. All three samples initially react to produce the hydrogen halide.
(i) Write an equation to describe the acid–base reaction that occurs when concentrated sulfuric acid reacts with NaBr.
(ii) Deduce which sodium halide, NaCl, NaBr or NaI, produces the largest percentage yield of hydrogen halide when concentrated sulfuric acid is added. Explain your answer by considering the relative reactivity of the halide ions as reducing agents.

▶️ Answer/Explanation

Solution

(a)

Explanation: The table is completed by identifying the state of each halogen at room temperature. Fluorine and chlorine are gases, bromine is a liquid, and iodine is a solid. This trend is due to increasing molecular size and stronger intermolecular forces down the group.

(b) The volatility decreases from chlorine to iodine.

Explanation: As we move down the group, the number of electrons increases, leading to stronger instantaneous dipole-induced dipole forces. This makes it harder for the molecules to escape into the gas phase, reducing volatility.

(c)(i) \(\text{Br}_2 + 2\text{I}^- \rightarrow 2\text{Br}^- + \text{I}_2\)

Explanation: Bromine oxidizes iodide ions to iodine, while itself being reduced to bromide ions. This redox reaction is represented by the ionic equation above.

(c)(ii) Bromine acts as an oxidizing agent.

Explanation: Bromine gains electrons from iodide ions, reducing itself to bromide while oxidizing iodide to iodine. This electron transfer confirms its role as an oxidizing agent.

(d)(i) \(\text{NaBr} + \text{H}_2\text{SO}_4 \rightarrow \text{HBr} + \text{NaHSO}_4\)

Explanation: This acid-base reaction produces hydrogen bromide and sodium hydrogen sulfate. The equation shows the initial step in the reaction between NaBr and concentrated sulfuric acid.

(d)(ii) Sodium chloride produces the largest percentage yield of hydrogen halide.

Explanation: Chloride ions are the weakest reducing agents among the halides. They do not further reduce sulfuric acid, unlike bromide and iodide ions, which react further, reducing the yield of HBr and HI.

Question 2

Topic: 10.1

(a) Sulfur chloride, SCl₂, is a liquid at room temperature. When SCl₂ is added to water, misty fumes are seen and a solution is made that turns universal indicator red.

(i) Identify the type of reaction that occurs when SCl₂ is added to water.
(ii) Name a chloride of a different Period 3 element that is also a liquid at room temperature and produces misty fumes when added to water.
(b) A molecule of SCl₂ contains two S–Cl covalent bonds.
(i) Complete the dot-and-cross diagram in Fig. 2.1 to show the arrangement of the outer electrons in a molecule of SCl₂.
Use × to show electrons from the chlorine atoms.
Use ● to show electrons from the sulfur atom.

(ii) Predict the shape of, and bond angle in, a molecule of SCl₂ by using VSEPR theory.
(c) Solid magnesium nitride, Mg₃N₂, is a crystalline solid.
(i) Deduce the oxidation numbers of magnesium and nitrogen in magnesium nitride to complete Table 2.1.

(ii) Magnesium nitride reacts with an excess of water to produce ammonia and magnesium hydroxide only. Construct an equation to describe this reaction.
(iii) Explain why the solution produced in the reaction in (c)(ii) has a pH greater than 7. Refer to the products of the reaction in your answer.
(d) Boron nitride is a white solid that melts above 2900°C. Fig. 2.2 shows part of the lattice structure of a crystal of boron nitride.

(i) Use Fig. 2.2 to deduce the empirical formula of boron nitride.
(ii) Suggest the identity of another crystalline solid that has atoms arranged in layers similar to that of solid boron nitride.

▶️ Answer/Explanation

Solution

(a)(i) Hydrolysis.

Explanation: When SCl₂ is added to water, it undergoes hydrolysis, a reaction where water breaks the bonds in SCl₂, producing misty fumes (HCl gas) and an acidic solution (turning universal indicator red).

(a)(ii) Silicon(IV) chloride / phosphorus(III) chloride.

Explanation: Both SiCl₄ (silicon tetrachloride) and PCl₃ (phosphorus trichloride) are liquids at room temperature and react with water to produce misty fumes of HCl.

(b)(i)

Explanation: The dot-and-cross diagram shows two shared pairs (●×) between S and Cl, with 6 non-bonding electrons (×) on each Cl and 4 non-bonding electrons (●) on S.

(b)(ii) Non-linear shape, bond angle 103–105°.

Explanation: SCl₂ has two bonding pairs and two lone pairs on sulfur, resulting in a bent (non-linear) shape with a bond angle slightly less than 109.5° due to lone pair repulsion.

(c)(i)

Explanation: In Mg₃N₂, magnesium has an oxidation number of +2 (group 2 metal), and nitrogen has −3 (as it gains 3 electrons to achieve a stable octet).

(c)(ii) Mg₃N₂ + 6H₂O → 3Mg(OH)₂ + 2NH₃.

Explanation: Magnesium nitride reacts with water to form magnesium hydroxide (a weak base) and ammonia (a weak base), balancing the equation ensures stoichiometric correctness.

(c)(iii) Due to OH⁻(aq) from Mg(OH)₂ and NH₃.

Explanation: The solution is alkaline (pH > 7) because Mg(OH)₂ partially dissociates to release OH⁻ ions, and NH₃ reacts with water to produce OH⁻ ions.

(d)(i) BN.

Explanation: The empirical formula is deduced from the 1:1 ratio of boron to nitrogen atoms in the lattice structure.

(d)(ii) Graphite.

Explanation: Graphite has a similar layered structure with carbon atoms arranged in hexagonal rings, analogous to boron nitride’s lattice.

Question 3

Topic: 6.1

(a) Define Le Chatelier’s principle.

(b) Reaction 1 describes the reversible reaction between yellow Fe³⁺(aq) and colourless SCN⁻(aq) to produce red FeSCN²⁺(aq).
reaction 1 Fe³⁺(aq) + SCN⁻(aq) \( \rightleftharpoons \)FeSCN²⁺(aq) ∆H = –x kJmol⁻¹
yellow colourless red
A mixture of Fe³⁺(aq), SCN⁻(aq), and FeSCN²⁺(aq) is at equilibrium at 20°C. The temperature of this mixture is then increased to 50°C and allowed to reach equilibrium. Deduce the changes that occur, if any, in the equilibrium mixture at 50°C compared to the equilibrium mixture at 20°C.
• change in appearance
• change in relative concentration of FeSCN²⁺(aq)
• change in value of the equilibrium constant, \(K_c\)

(c) In another experiment, equimolar amounts of Fe³⁺(aq) and SCN⁻(aq) are mixed together and allowed to reach equilibrium. The total volume of the mixture is 25.0 cm³.
reaction 1 Fe³⁺(aq) + SCN⁻(aq) \( \rightleftharpoons \) FeSCN²⁺(aq)
At equilibrium the mixture contains:
• [SCN⁻] = \(1.30 × 10^{–3}moldm^{–3}\).
• [FeSCN²⁺] = \(0.300 × 10^{–3}moldm^{–3}\).
(i) Calculate the initial amount, in mol, of Fe³⁺(aq) added to SCN⁻(aq) to produce this mixture.

(ii) Calculate \(K_c\) for reaction 1 and state its units. Show your working.

▶️ Answer/Explanation

Solution

(a) Le Chatelier’s principle states that if a system at equilibrium is subjected to a change in conditions, the equilibrium will shift to counteract the change and restore equilibrium.

Explanation: This principle helps predict the direction of shift in equilibrium when factors like concentration, temperature, or pressure are altered.

(b)

• Change in appearance: The mixture becomes paler red/more yellow as the equilibrium shifts left (endothermic direction).
• Change in [FeSCN²⁺(aq)]: The concentration of FeSCN²⁺ decreases because the reverse reaction is favored at higher temperatures.
• Change in \(K_c\): The equilibrium constant decreases because the reaction is exothermic (\(\Delta H = -x\)), and increasing temperature favors the reactants.

Explanation: Since the reaction is exothermic, increasing the temperature shifts the equilibrium to the left, reducing FeSCN²⁺ concentration and \(K_c\).

(c)(i) Initial amount of Fe³⁺(aq): \(1.60 \times 10^{-5} \, \text{mol}\).

Explanation: At equilibrium, [SCN⁻] = \(1.30 \times 10^{-3} \, \text{moldm}^{-3}\) and [FeSCN²⁺] = \(0.300 \times 10^{-3} \, \text{moldm}^{-3}\). The initial [SCN⁻] = equilibrium [SCN⁻] + [FeSCN²⁺] = \(1.60 \times 10^{-3} \, \text{moldm}^{-3}\). For 25.0 cm³, the amount is \(1.60 \times 10^{-3} \times 0.025 = 1.60 \times 10^{-5} \, \text{mol}\).

(c)(ii) \(K_c = 178 \, \text{mol}^{-1} \, \text{dm}^3\).

Explanation: Using \(K_c = \frac{[\text{FeSCN}^{2+}]}{[\text{Fe}^{3+}][\text{SCN}^-]}\), and substituting equilibrium concentrations: \(K_c = \frac{0.300 \times 10^{-3}}{(1.30 \times 10^{-3})(1.30 \times 10^{-3})} = 178 \, \text{mol}^{-1} \, \text{dm}^3\).

Calculation for Kc

Question 4

Topic: 4.2

(a) Define enthalpy change of formation.
(b) Iron is made when iron(III) oxide is heated with carbon monoxide, as shown by reaction 2.
reaction 2 Fe₂O₃ + 3CO → 2 Fe + 3 CO₂
Table 4.1 shows enthalpy change of formation data measured at 298K and 101kPa.

(i) Complete Table 4.1 by adding equations with relevant state symbols to represent:
• standard enthalpy change of formation for Fe₂O₃.
• standard enthalpy change of formation for CO.
(ii) Use the data in Table 4.1 to calculate the enthalpy change of reaction, \(∆H_r\), in \(kJmol^{–1}\), for reaction 2. Show your working

▶️ Answer/Explanation

Solution

(a) The enthalpy change of formation is the energy change when one mole of a compound is formed from its elements in their standard states under standard conditions (298K and 101kPa).

(b)(i)

Explanation: The standard enthalpy change of formation for Fe₂O₃ is represented as \(2Fe(s) + \frac{3}{2}O_2(g) \rightarrow Fe_2O_3(s)\), and for CO as \(C(s) + \frac{1}{2}O_2(g) \rightarrow CO(g)\).

(b)(ii) The enthalpy change of reaction, \(∆H_r\), is calculated using the formula:

\[ ∆H_r = \sum ∆H_f(\text{products}) – \sum ∆H_f(\text{reactants}) \]

Substituting the values from Table 4.1:

\[ ∆H_r = [2(0) + 3(-393.5)] – [1(-824.2) + 3(-110.5)] = -1180.5 + 824.2 + 331.5 = -24.8 \, kJmol^{-1} \]

Final Answer: The enthalpy change of reaction is \(\boxed{-24.8 \, kJmol^{-1}}\).

Question 5

Topic: 15.1

Hydrocarbon molecules contain covalent bonds.
(a) Define covalent bond.
(b) A C=C bond in an alkene is made from a σ bond and a π bond.
(i) Identify the hybridisation of the carbon atoms in a C=C bond in an alkene.
(ii) Draw labelled diagrams to show, in terms of orbital overlap, how the σ and π bonds are made in a C=C bond.
σ bond
π bond

(c) In electrophilic reactions involving alkenes the π bond of C=C is broken.
(i) Suggest one difference between σ and π bonds that explains why the π bond is broken in electrophilic addition reactions involving alkenes.
(ii) Complete Fig. 5.1 to show the mechanism for the electrophilic addition of hydrogen bromide to 2-methylpropene to produce the major organic product. Include charges, dipoles, lone pairs of electrons and curly arrows, as appropriate.

▶️ Answer/Explanation

Solution

(a) (electrostatic) attraction between nuclei of two atoms and shared pair of electrons

Explanation: A covalent bond is formed when two atoms share a pair of electrons, resulting in an electrostatic attraction between the positively charged nuclei and the negatively charged shared electrons.

(b)(i) \(sp^2\)

Explanation: In a C=C bond, each carbon atom forms three σ bonds (two with hydrogen and one with the other carbon) using \(sp^2\) hybrid orbitals, leaving one unhybridized p orbital for the π bond.

(b)(ii)

Explanation: The σ bond is formed by head-on overlap of \(sp^2\) hybrid orbitals, while the π bond is formed by sideways overlap of the remaining p orbitals perpendicular to the plane of the σ bonds.

(c)(i) EITHER (pair of) electrons in 𝜋 bond are further away from the nuclei so weaker attraction OR (pair of) electrons in 𝜎 bond are closer to the two nuclei so stronger attraction

Explanation: The π bond is weaker because its electron density is distributed above and below the internuclear axis, making it more accessible to electrophiles compared to the σ bond which has direct overlap along the axis.

(c)(ii)

Explanation: The mechanism shows the π electrons attacking HBr, forming a carbocation intermediate on the more substituted carbon (Markovnikov’s rule), followed by bromide ion attack to form 2-bromo-2-methylpropane.

Question 6

Topic: 22.2

(a) V shows stereoisomerism.

(i) Explain what is meant by stereoisomerism.
(ii) Deduce the number of stereoisomers of V. Explain your reasoning.
(iii) Deduce the molecular formula of V.
(iv) Name all the functional groups present in V.

(b) Fig. 6.2 shows two reactions involving V.

(i) Identify the role of reagent T for each functional group that reacts in reaction 1.
(ii) Suggest the identity of reagent U in reaction 2.

(c) Both functional groups in one molecule of Y react with an inorganic reagent to form one molecule of Q and one molecule of methanol, CH₃OH, as shown in Fig. 6.3.

(i) Part of the mass spectrum for Q is shown in Fig. 6.4. Only peaks with m/e greater than 198 are shown.

Calculate the relative abundance, x, of the peak at m/e = 201. Show your working.

(ii) Q contains only hydroxyl functional groups. Complete Table 6.1 to show the observations that occur when 2,4-dinitrophenylhydrazine (2,4-DNPH reagent) is added to separate samples of Y and Q.

(iii) Under certain conditions, 0.0020mol of Q reacts with an excess of sodium to produce a total of 44.8 cm³ of gas at s.t.p. Calculate the number of hydroxyl groups present in a molecule of Q. Show your working.

(iv) Use Table 6.2 to describe and explain two differences between the infrared spectrum of Y and Q in the region above 1500 cm⁻¹.

▶️ Answer/Explanation

Solution

(a)(i) Stereoisomerism refers to molecules with the same structural and molecular formula but different spatial arrangements of atoms/groups.

Explanation: Stereoisomers differ only in the 3D orientation of their atoms, such as in geometric (cis-trans) or optical isomers.

(a)(ii) Number of stereoisomers = 8.

Explanation: V has 2 chiral centers (each contributing 2 stereoisomers) and 1 C=C bond (contributing 2 geometric isomers), so total = \(2^3 = 8\).

(a)(iii) Molecular formula: \(C_{13}H_{20}O_3\).

Explanation: Counting all C, H, and O atoms in the structure gives the formula.

(a)(iv) Functional groups: ester, carbonyl (ketone), and C=C (alkene).

Explanation: The structure contains an ester (–COO–), a ketone (C=O), and a double bond (C=C).

(b)(i) Reagent T acts as a reducing agent for both C=O (carbonyl) and C=C (alkene) groups.

Explanation: T (likely \(H_2\)/Ni) reduces the ketone to an alcohol and the alkene to an alkane.

(b)(ii) Reagent U is sodium borohydride (\(NaBH_4\)).

Explanation: \(NaBH_4\) selectively reduces the carbonyl group without affecting the C=C bond.

(c)(i) Relative abundance \(x = 13.2\).

Explanation: From the mass spectrum, the ratio of peaks at m/e = 201 and 200 is 12:1.1. Thus, \(x = (12 \times 1.1) = 13.2\).

(c)(ii)

Explanation: Y (with C=O) forms an orange precipitate with 2,4-DNPH, while Q (no C=O) shows no reaction.

(c)(iii) Number of hydroxyl groups = 2.

Explanation: 0.002 mol Q produces 0.002 mol \(H_2\) (44.8 cm³ at STP). Since 2 OH groups produce 1 \(H_2\), Q must have 2 OH groups.

(c)(iv)

1. Y shows a C=O peak (1670–1740 cm⁻¹); Q does not.
2. Q shows an O-H peak (3200–3600 cm⁻¹); Y does not.

Explanation: Y has a ketone (C=O), while Q has alcohols (O-H) after reduction.

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