What is the product formed on oxidation of I− by MnO4− in a faintly alkaline medium?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → IO3−
In a neutral or faintly alkaline medium, permanganate is reduced to MnO2 and oxidises iodide (I = −1) to iodate (I = +5):
\( \displaystyle \ce{2MnO4^- + H2O + I^- -> 2MnO2 + 2OH^- + IO3^-} \)
Why the other options are incorrect
Option (1) IO4− — Incorrect. Periodate (I = +7) is not formed; in this medium the oxidation stops at iodate.
Option (2) I2 — Incorrect. Iodine is liberated from iodide by permanganate in acidic solution, where MnO4− is reduced to Mn2+.
Option (3) IO− — Incorrect. Hypoiodite (I = +1) is not the product; iodide is oxidised further, to iodate.
NCERT (2026-27 edition):
4.4.1 Oxides and Oxoanions of Metals – “2. In neutral or faintly alkaline solutions:
(a) A notable reaction is the oxidation of iodide to iodate: 2MnO4– + H2O + I– → 2MnO2 + 2OH– + IO3–”