OCR A-Level Chemistry AS Depth in chemistry (02), June 2018: Question 3

✅ Correct Answers

• (i) Ionic Equation: Br₂(aq) + 2I⁻(aq) ➔ I₂(aq) + 2Br⁻(aq) (Multiples allowed; state symbols not strictly penalized if omitted).
• (ii) Reactivity Trend: Iodine has a larger atomic radius, greater shielding (more inner electron shells), and weaker nuclear attraction on the incoming electron compared to bromine.

💡 Key Knowledge

• Halogen reactivity decreases down Group 7 as atoms get larger.
• Reactivity depends on how easily an atom can attract and capture an electron into its outer shell.

🧠 Exam Technique

• For trend explanations, always compare both elements clearly (e.g., "Iodine has..." rather than vague general statements).
• Use comparative language: atomic radius, shielding, and nuclear attraction are the required marking points.

❌ Common Errors

• Talking about atoms losing electrons instead of gaining them.
• Mentioning ionisation energy instead of electron attraction/nuclear attraction.
• Forgetting balancing numbers in ionic equations (missing the 2 in front of I⁻ and Br⁻ ).

✅ Correct Answers

• (i) Disproportionation explanation: Iodine is simultaneously oxidised and reduced in the same reaction. Oxidation state changes from 0 in I₂ to +1 in HIO (oxidised) and -1 in HI (reduced).
• (ii) Chlorine disadvantage: Toxic/poisonous, or forms halogenated hydrocarbons / carcinogens.

💡 Key Knowledge

• Disproportionation: A redox reaction in which the same element is both oxidised and reduced.
• Chlorine is added to drinking water to kill bacteria, but reacts with organic matter to form hazardous chlorinated by-products.

🧠 Exam Technique

• To score full marks on disproportionation, explicitly state: (1) it is disproportionation, (2) state the oxidation number of the element in the reactant, and (3) state the respective oxidation numbers in both products.

❌ Common Errors

• Stating chlorine causes cancer directly (examiners require reference to forming toxic compounds/carcinogens or reacting with hydrocarbons).
• Omitting signs on oxidation numbers (writing 1 instead of +1 ).

✅ Correct Answer

• Final Answer: +431.5 kJ mol⁻¹ (or 432 kJ mol⁻¹ to 3 sig fig). Award 2 marks.

📐 Step-by-Step Calculation

1. Recall formula: ΔH = Bonds broken - Bonds formed
2. Set up equation: -184 = (436 + 243) - 2(H-Cl bond enthalpy)
3. Rearrange for bonds formed: 2(H-Cl) = 436 + 243 + 184 = +863 kJ mol⁻¹
4. Divide by 2: H-Cl bond enthalpy = 863 / 2 = +431.5 kJ mol⁻¹

🧠 Exam Technique

• Always check the final answer line first. If the correct numerical value with the correct sign is present, full marks are awarded instantly.
• Remember that bond enthalpies are always positive values since breaking bonds requires energy (endothermic).

❌ Common Errors

• Dropping the minus sign on the final answer if calculating through standard energy cycle approaches ( -431.5 kJ mol⁻¹ gets 1 mark for wrong sign).
• Forgetting to divide by 2 at the end to find a single H-Cl bond enthalpy.

✅ Correct Answers

• (i) Equation: Br₂(l) ➔ Br₂(g) (Must include correct state symbols).
• (ii) Energetics: Endothermic, because energy is required to overcome intermolecular forces / London forces (induced dipole-dipole interactions) between molecules.

💡 Key Knowledge

• Enthalpy change of vaporisation refers strictly to changing a substance from liquid to gas.
• Boiling/vaporisation requires breaking intermolecular forces, not covalent bonds within the molecule.

🧠 Exam Technique

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• State symbols are vital when an equation specifically defines a phase change like vaporisation.
• Be precise with terminology: use London forces or induced dipole-dipole interactions. Avoid banned terms like "van der Waals' forces" if specified by strict marking guidelines.

❌ Common Errors

• Stating that covalent bonds ( Br-Br ) are broken during boiling (this is a very common misconception).
• Confusing endothermic and exothermic terminology.