In neutral or faintly alkaline medium, KMnO₄ being a powerful oxidant can oxidize, thiosulphate almost quantitatively, to sulphate. In this reaction overall change in oxidation state of manganese will be :

Solution

Chemistry ma’am · 3 steps

Medium decides the product: in neutral or faintly alkaline solution, permanganate stops at MnO₂.

1. Start at permanganate. In MnO₄⁻, oxygen is -2 each, so x + 4(-2) = -1 gives manganese an oxidation state of +7.
2. Product depends on medium. How far MnO₄⁻ is reduced depends on the medium. In neutral or faintly alkaline solution it goes only as far as MnO₂, where Mn is +4.

   8MnO₄⁻ + 3S₂O₃²⁻ + H₂O → 8MnO₂ + 6SO₄²⁻ + 2OH⁻

   Medium	Mn product	Mn oxidation state	Change from +7
   Acidic	Mn²⁺	+2	5
   Neutral or faintly alkaline	MnO₂	+4	3
   Strongly alkaline	MnO₄²⁻	+6	1

3. So. Manganese goes from +7 to +4, an overall change of 3, which is option (d). As a check, 8 Mn gain 8 × 3 = 24 electrons and 6 S lose 6 × 4 = 24, going from +2 to +6.

The idea to keep. For KMnO₄, always read the medium first: acidic gives Mn²⁺ (5 electrons), neutral or faintly alkaline gives MnO₂ (3), strongly alkaline gives MnO₄²⁻ (1).

Answer from the NTA final answer key. Solution by PrepWiser.