OCR A-Level Chemistry AS Breadth in chemistry (01), June 2022: Question 24

✅ Correct Answer

Oxidation and reduction of the same element.

💡 Key Knowledge

• A disproportionation reaction is one where a single element is simultaneously oxidized and reduced.
• You must use the word element (or specific name like 'chlorine' / 'Cl').

❌ Common Errors

• Using the word "atom" instead of "element" (this is strictly penalized by examiners).
• Using vague terms like "species".

✅ Correct Answer

Equation: Cl₂ + 2NaOH → NaClO + NaCl + H₂O

Explanation: Cl is oxidised from 0 (in Cl₂) to +1 (in NaClO), AND Cl is reduced from 0 (in Cl₂) to -1 (in NaCl).

🧠 Exam Technique

• Clearly state the initial oxidation number (0 in Cl₂) and both final oxidation numbers (+1 and -1) to secure both explanation marks.
• If you make a mistake in the equation (e.g., writing HCl instead of <>NaCl</> and H₂O ), examiners apply Error Carried Forward (ECF) to your oxidation number changes if consistent.

❌ Common Errors

• Omitting one of the products (like forgetting water or sodium chloride).
• Failing to state both directions of change (oxidation and reduction must both be explicitly linked to their respective products).

✅ Correct Answer

Reagent: Add aqueous silver nitrate ( AgNO₃ ) / Ag⁺ ions.

Observations: Precipitate colours must be linked. (e.g., Chloride gives white ppt; Bromide gives cream ppt; Iodide gives yellow ppt).

Equations: Ag⁺ + Cl⁻ → AgCl (or equivalent for Br⁻ / I⁻).

Identifications: B is NaBr (Sodium bromide) and C is CaCl₂ (Calcium chloride).

📐 Calculations & Deducing Formulae (Step-by-Step)

1. Test Identification: State the addition of dilute nitric acid followed by silver nitrate (or just silver nitrate as per mark scheme). Note observation of precipitate colours.
2. Deduce Halide in B (Group 1): Group 1 has a +1 charge (e.g., NaX). Molar mass range is 100–115 g mol⁻¹. If the halide was chloride (Ar = 35.5), mass of metal would be 64.5 (not Group 1). If bromide (Ar = 79.9), mass of metal = 100 - 79.9 = 20.1 (too heavy for Li/Na/K alone, wait—let's check: NaBr molar mass = 23.0 + 79.9 = 102.9 g mol⁻¹, which falls cleanly inside 100–115!). Therefore, B = NaBr.
3. Deduce Halide in C (Group 2): Group 2 has a +2 charge (e.g., MX₂). Molar mass range is 100–115 g mol⁻¹. For calcium chloride ( CaCl₂ ): Ca (40.1) + 2 × Cl (35.5) = 40.1 + 71.0 = 111.1 g mol⁻¹, which fits the 100–115 range. Therefore, C = CaCl₂.

🧠 Exam Technique & Guidance

• The test for halides requires the correct reagent ( AgNO₃ ) and correct observation of precipitates.
• Make sure you show clear working or logic for how you subtracted the atomic masses of the halides from the given molar mass range (100–115 g mol⁻¹) to verify the Group 1 and Group 2 metal ions.