KMnO4 acts as an oxidising agent in acidic medium. When acidic KMnO4 reacts with KI, then I- gets converted to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → I2
In acidic medium, \( \ce{KMnO4} \) is a strong oxidising agent: Mn changes from +7 in \( \ce{MnO4^{-}} \) to +2 in \( \ce{Mn^{2+}} \), taking up 5 electrons. Iodide ions are oxidised to iodine:
\( \displaystyle \ce{MnO4- + 8H+ + 5e- -> Mn^2+ + 4H2O} \quad (\times 2) \)
\( \displaystyle \ce{2I- -> I2 + 2e-} \quad (\times 5) \)
\( \displaystyle \ce{10I- + 2MnO4- + 16H+ -> 2Mn^2+ + 8H2O + 5I2} \)
Why the other options are incorrect
Option (1) IO3− — Incorrect. Iodate is formed when \( \ce{KMnO4} \) oxidises iodide in neutral or faintly alkaline solution, not in acid:
\( \displaystyle \ce{2MnO4- + H2O + I- -> 2MnO2 + 2OH- + IO3-} \)
Option (2) IO− — Incorrect. Hypoiodite is not formed in this reaction.
Option (4) IO4− — Incorrect. Periodate (I +7) is not formed by \( \ce{KMnO4} \).
NCERT (2026-27 edition):
Oxides and Oxoanions of Metals – …
