Neutral or weakly alkaline \(KMnO_4\) solution will oxidize iodide ion into:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → iodate ion
What permanganate forms depends on the medium. In acidic solution \( \ce{MnO4-} \) (Mn +7) is reduced to \( \ce{Mn^2+} \) and oxidises iodide only to iodine. In neutral or faintly alkaline solution it is reduced to \( \ce{MnO2} \) (Mn +4) and oxidises iodide all the way to iodate, iodine going from −1 in \( \ce{I-} \) to +5 in \( \ce{IO3-} \).
Each Mn gains 3 electrons and each I loses 6, so two \( \ce{MnO4-} \) are needed for one \( \ce{I-} \):
\( \displaystyle \text{Reduction: }\ce{MnO4^- + 2H2O + 3e^- -> MnO2 + 4OH^-}\quad(\times 2) \)
\( \displaystyle \text{Oxidation: }\ce{I^- + 6OH^- -> IO3^- + 3H2O + 6e^-} \)
\( \displaystyle \text{Net: }\ce{2MnO4^- + H2O + I^- -> 2MnO2 + 2OH^- + IO3^-} \)
Option (2) periodate ion — Incorrect. IO4– (iodine +7) is not formed; the product of this reaction is IO3–.
Option (3) iodite ion — Incorrect. IO2– (iodine +3) is not the product of permanganate oxidation.
Option (4) hypoiodite ion — Incorrect. IO– (iodine +1) forms from \( \ce{I2} \) and alkali, not from iodide and permanganate.
NCERT (2026-27 edition):
In Neutral or Faintly Alkaline Solutions
