JEE Main 2022 · 27 July · evening · Single correct · +4 / −1
In neutral or alkaline solution, MnO₄⁻ oxidises thiosulphate to :
- (a) S₂ O₇²⁻
- (b) S₂ O₈²⁻
- (c) SO₃²⁻
- (d) SO₄²⁻
Answer and solution
Answer (d) SO₄²⁻
Solution
Chemistry ma’am · 4 steps
Permanganate is a strong oxidant, so it pushes the sulphur all the way to +6, which is sulphate.
- Permanganate's change. In neutral or alkaline solution, MnO₄⁻ is reduced only to MnO₂. Mn goes from +7 to +4, gaining 3 electrons per ion.
- Thiosulphate's change. In S₂O₃²⁻ the average oxidation state of S is +2. A strong oxidant like permanganate takes sulphur to its highest common state, +6, in the stable aqueous ion SO₄²⁻.
8MnO₄⁻ + 3S₂O₃²⁻ + H₂O → 8MnO₂ + 6SO₄²⁻ + 2OH⁻
- Check electrons. Each S rises by 4, so 6 S atoms lose 6 × 4 = 24 electrons. Eight Mn atoms gain 8 × 3 = 24 electrons. The charges also match: -14 on each side.
- So. Thiosulphate is oxidised to sulphate, SO₄²⁻, which is option (d).
The idea to keep. Strong oxidants like MnO₄⁻ or Cl₂ take thiosulphate all the way to sulphate. A mild oxidant like I₂ stops at tetrathionate, S₄O₆²⁻.
Answer from the NTA final answer key. Solution by PrepWiser.