Match List-I (reducing agent treated with potassium permanganate) wit…

The correct answer is Option (1) → A-IV, B-II, C-III, D-I

What a permanganate oxidation gives depends on the hydrogen ion concentration. In acid solution MnO₄⁻ is reduced to Mn²⁺; in neutral or faintly alkaline solution it is reduced to MnO₂. So the product formed from one reducing agent can differ with the medium, as it does for iodide.

(A) Iodide ion, in acid solution — This matches (IV) I₂. Iodine is liberated:

    \( \displaystyle \ce{10I- + 2MnO4- + 16H+ -> 2Mn^2+ + 8H2O + 5I2} \)

(B) Iodide ion, in neutral or faintly alkaline solution — This matches (II) IO₃⁻. Iodine goes from −1 all the way to +5 as iodate:

    \( \displaystyle \ce{2MnO4- + H2O + I- -> 2MnO2 + 2OH- + IO3-} \)

(C) Thiosulphate ion, in neutral or faintly alkaline solution — This matches (III) SO₄²⁻. Thiosulphate is oxidised almost quantitatively to sulphate:

    \( \displaystyle \ce{8MnO4- + 3S2O3^2- + H2O -> 8MnO2 + 6SO4^2- + 2OH-} \)

(D) Hydrogen sulphide, in acid solution — This matches (I) S. The sulphide is oxidised and sulphur is precipitated:

    \( \displaystyle \ce{H2S -> 2H+ + S^2-} \)

    \( \displaystyle \ce{5S^2- + 2MnO4- + 16H+ -> 2Mn^2+ + 8H2O + 5S} \)

NCERT (2026-27 edition):
4.4.1 Oxides and Oxoanions of Metals – “1. In acid solutions:”

“(a) Iodine is liberated from potassium iodide : 10I^– + 2MnO₄^– + 16H⁺ → 2Mn²⁺ + 8H₂O + 5I₂” … “(d) Hydrogen sulphide is oxidised, sulphur being precipitated”

“2. In neutral or faintly alkaline solutions:”

“(a) A notable reaction is the oxidation of iodide to iodate: 2MnO₄^– + H₂O + I^– → 2MnO₂ + 2OH^– + IO₃^–”

“(b) Thiosulphate is oxidised almost quantitatively to sulphate: 8MnO₄^– + 3S₂O₃^2– + H₂O → 8MnO₂ + 6SO₄^2– + 2OH^–”