In the laboratory, a manganese(II) salt in solution is oxidised to pe…

The correct answer is Option (4) → Peroxodisulphate ion, \( \ce{S2O8^2-} \)

Turning Mn²⁺ into permanganate raises manganese from +2 to +7, so a strong oxidising agent is needed. In the laboratory this is done with peroxodisulphate, which is itself reduced to sulphate:

    \( \displaystyle \ce{2Mn^2+ + 5S2O8^2- + 8H2O -> 2MnO4- + 10SO4^2- + 16H+} \)

Each Mn loses 5 electrons (10 for two Mn), and each S₂O₈²⁻ takes up 2 electrons (10 for five ions).

Why the other options are incorrect

Option (1) Potassium nitrate — Incorrect. KNO₃ is the oxidising agent used when MnO₂ is fused with an alkali metal hydroxide; that fusion gives green manganate, not permanganate from a Mn(II) salt.

Option (2) Zinc sulphate — Incorrect. Zinc sulphate is not an oxidant. It only catalyses the oxidation of a manganous salt to MnO₂ by permanganate in neutral solution.

Option (3) Dilute sulphuric acid — Incorrect. Acid makes manganate (+6) disproportionate into permanganate and MnO₂; it cannot oxidise Mn²⁺ to +7.

NCERT (2026-27 edition):
4.4.1 Oxides and Oxoanions of Metals – “In the laboratory, a manganese (II) ion salt is oxidised by peroxodisulphate to permanganate. 2Mn²⁺ + 5S₂O₈^2– + 8H₂O → 2MnO₄^– + 10SO₄^2– + 16H⁺”