If KMnO_4 is reduced by oxalic acid in the acidic medium, then the ox…

The correct answer is Option (3) → +7 to +2

In acidic medium, permanganate oxidises oxalic acid (oxalate) to \( \ce{CO2} \) and is itself reduced to \( \ce{Mn^{2+}} \):

    \( \ce{5C2O4^{2-}} \) + \( \ce{2MnO4^{-}} \) + \( \ce{16H^{+}} \) → \( \ce{2Mn^{2+}} \) + \( \ce{8H2O} \) + \( \ce{10CO2} \)

In \( \ce{MnO4^{-}} \): x + 4(−2) = −1 ⇒ x = +7. In \( \ce{Mn^{2+}} \) the oxidation number is +2. So Mn changes from +7 to +2 (five electrons gained per Mn).

NCERT (2026-27 edition):
Oxides and Oxoanions of Metals – “… (c) Oxalate ion or oxalic acid is oxidised at 333 K:
5C₂O₄^2– + 2MnO₄^– + 16H⁺ → 2Mn²⁺ + 8H₂O + 10CO₂”