In neutral solution, thiosulphate is oxidized by potassium permanganate to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(SO_4^{2-}\)
Potassium permanganate is a strong oxidising agent whose product depends on the medium: in acid \( \ce{MnO4^{-}} \) is reduced to \( \ce{Mn^{2+}} \), while in neutral or faintly alkaline solution it is reduced to MnO2 (Mn from +7 to +4, 3 electrons per ion). In neutral solution it oxidises thiosulphate almost quantitatively to sulphate:
\( \ce{MnO4- + 2H2O + 3e- -> MnO2 + 4OH-} \) (× 8)
\( \ce{S2O3^2- + 10OH- -> 2SO4^2- + 5H2O + 8e-} \) (× 3)
Electron balance: 8 × 3 = 24 electrons gained and 3 × 8 = 24 electrons lost. Adding and cancelling water and hydroxide:
\( \ce{8MnO4- + 3S2O3^2- + H2O -> 8MnO2 + 6SO4^2- + 2OH-} \)
The average oxidation state of sulphur rises from +2 in \( \ce{S2O3^{2-}} \) to +6 in SO42−.
Why the other options are incorrect
Option (1) \(S^{2-}\) — Incorrect. Sulphide has S in the −2 state; going from +2 to −2 is a reduction, which an oxidising agent such as \( \ce{KMnO4} \) cannot bring about.
Option (2) \(SO_3^{2-}\) — Incorrect. Sulphite (S +4) is an intermediate; permanganate oxidises sulphite further to sulphate, so it is not the final product.
Option (4) \(S\) — Incorrect. Sulphur (oxidation state 0) is a reduction product of thiosulphate (it forms when thiosulphate is acidified), not an oxidation product.
NCERT (2026-27 edition):
In Neutral or Faintly Alkaline Solutions
