\( \ce{MnO4^- + H2O + I^- -> MnO2 + OH^- + X} \)
Compound ‘X’ in the above reaction is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → IO3−
The product of permanganate depends on the medium. In neutral or faintly alkaline solution \( \ce{MnO4^{-}} \) takes 3 electrons and is reduced to MnO2 (Mn: +7 → +4). Under these conditions it oxidises iodide all the way to iodate (I: −1 → +5, 6 electrons):
\( \displaystyle \text{Reduction:}\ \ \ce{MnO4^- + 2H2O + 3e^- -> MnO2 + 4OH^-}\quad(\times 2) \)
\( \displaystyle \text{Oxidation:}\ \ \ce{I^- + 6OH^- -> IO3^- + 3H2O + 6e^-} \)
\( \displaystyle \ce{2MnO4^- + H2O + I^- -> 2MnO2 + 2OH^- + IO3^-} \)
So X is the iodate ion, IO3−.
Why the other options are incorrect
Option (1) I2 — Incorrect. Iodine is the product in acidic medium, where \( \ce{MnO4^{-}} \) goes to \( \ce{Mn^{2+}} \):
\( \displaystyle \ce{2MnO4^- + 10I^- + 16H^+ -> 2Mn^2+ + 5I2 + 8H2O} \)
Option (2) IO22− — Incorrect. There is no such stable ion of iodine; it is not a product of this reaction.
Option (3) IO4− — Incorrect. Periodate (I in +7) would need a stronger oxidation than permanganate gives in neutral solution.
NCERT (2026-27 edition):
Oxides and Oxoanions of Metals – “… 2. In neutral or faintly alkaline solutions:
(a) A notable reaction is the oxidation of iodide to iodate:
\( \ce{2MnO4^- + H2O + I^- -> 2MnO2 + 2OH^- + IO3^-} \)”