AQA A-Level Chemistry Paper 1, June 2024: Question 5

✅ Correct Answers

05.1 Chlorine with water:

Accepted: → instead of ⇌ , multiples, or the sunlight decomposition equation: 2Cl₂ + 2H₂O → 4HCl + O₂ .

05.2 Chlorine with cold, dilute NaOH:

Accepted: multiples; writing NaClO in place of NaOCl .

💡 Key Knowledge

• Both reactions are classic examples of disproportionation: chlorine is simultaneously oxidised (from 0 to +1 in HOCl / NaOCl) and reduced (from 0 to -1 in HCl / NaCl).
• HOCl (chloric(I) acid / hypochlorous acid) acts as an antibacterial agent and bleach.
• NaOCl (sodium chlorate(I)) is the active bleaching agent in household bleach.

❌ Common Errors

• Confusing the cold dilute NaOH reaction with the hot concentrated reaction (which forms NaClO₃ instead of NaOCl).
• Writing incorrect formulas for chloric(I) acid or sodium chlorate(I), such as writing HClO₃ or NaClO₃ .

✅ Final Marking Points

Oxidation: 2Cl⁻ → Cl₂ + 2e⁻ [1 mark]

Reduction: 2ClO₃⁻ + 12H⁺ + 10e⁻ → Cl₂ + 6H₂O [1 mark]

Overall: 2ClO₃⁻ + 12H⁺ + 10Cl⁻ → 6Cl₂ + 6H₂O (or simplified) [1 mark]

🧠 Exam Technique

• Always balance oxygen using H₂O and hydrogen using H⁺ in acidic redox questions.
• Remember that electrons must never appear in the overall balanced equation. Double-check that total charges on both sides match:
  Left side: 2(-1) + 12(+1) + 10(-1) = 0.
  Right side: 0. Perfectly balanced!

✅ Correct Answer

Equation:

Also accepted: 2NaCl + H₂SO₄ → 2HCl + Na₂SO₄

Role of chloride ions:

Base / Proton acceptor

💡 Key Knowledge

• Chloride ions (Cl⁻) are too weak a reducing agent to reduce concentrated H₂SO₄ (unlike Br⁻ and I⁻).
• Therefore, this is solely an acid-base reaction, not a redox reaction.
• Cl⁻ removes a proton (H⁺) from H₂SO₄ to form misty fumes of HCl(g).

❌ Common Errors

• Saying "reducing agent": This is an automatic 0 marks for the role. Chloride does NOT reduce sulfuric acid (no SO₂, S, or H₂S is formed).
• Writing an ionic equation when the question specifies solid sodium chloride. Always include the sodium species as given.

📐 Electron Pair Count for Central Cl

• Group 7 central atom = 7 valence electrons
• Negative charge (-1) = add 1 electron → 8 electrons
• Bonded to 2 chlorine atoms = 2 shared bonding pairs
• Remaining electrons: 8 - 2 = 6 non-bonding electrons = 3 lone pairs
• Total electron pairs: 2 bond pairs + 3 lone pairs = 5 pairs (electron geometry based on trigonal bipyramidal).
• To minimise lp-lp repulsion, all 3 lone pairs sit in equatorial positions (at 120°).
• The two Cl atoms occupy axial positions, giving a linear molecular shape.

✅ Required Diagram

✅ Correct Answer & Marking Points

Bond angle: 180° [1 mark]

Explanation: (2) bond pairs repel to be as far apart as possible [1 mark]

💡 Deducing the Structure

• Thallium (Tl) is in Group 3 → 3 outer shell electrons.
• A single positive charge (+) means it loses 1 electron: 3 - 1 = 2 valence electrons.
• Forms 2 single covalent bonds to chlorine atoms: uses both valence electrons.
• Result: 2 bonding pairs, 0 lone pairs.
• Two bonding pairs repel equally to achieve maximum separation, giving a linear arrangement with a 180° bond angle.

🧠 Exam Technique: Securing VSEPR Marks

• Always state the number of bond pairs and number of lone pairs explicitly.
• Use the magic phrase: "pairs of electrons repel to be as far apart as possible" or "repel equally".
• Merely stating that the molecule is "linear" will not gain the explanation mark! You must reference electron pair repulsion.