Question

When neutral or faintly alkaline KMnO4 is treated with potassium iodide, iodide ion is converted into 'X'. 'X' is -

• A. $I0^{-}_{4}$
• B. $I0^{-}_{3}$
• C. $I0^{-}$

Answer 1

The Correct Option is C

KMnO₄ is a well-known oxidizing agent, and its behavior changes with the pH of the solution in which it is used. Here, the reaction is taking place in a neutral or faintly alkaline medium.

Reaction in a Neutral or Faintly Alkaline Medium:

In a neutral or faintly alkaline medium, potassium permanganate (KMnO₄) has a moderate oxidizing property and can oxidize iodide ions (I^-) into iodate ions (IO₃^-). However, for the given question, the iodide ion (I^-) is eventually converted into hypoiodite ion (IO^-).

Explanation:

1. In an alkaline medium, the reduction of MnO₄^- occurs according to the half-equation: \text{MnO}_4^- + 2\text{H}_2\text{O} + 3e^- \rightarrow \text{MnO}_2 + 4\text{OH}^-
2. This reduction capability leads to the change of iodide ions (I^-) to hypoiodite ions (IO^-) when moderately alkaline conditions are applied, as described in the overall reaction: \text{I}^- + \text{H}_2\text{O} \rightarrow \text{IO}^- + 2\text{H}^+ + 2e^-
3. Ultimately, this reaction supports the conversion of iodide ions into hypoiodite ions in a neutral to faintly alkaline medium.

Therefore, when neutral or faintly alkaline KMnO₄ is treated with potassium iodide, the iodide ion is converted into the hypoiodite ion, IO^-.

Conclusion:

The correct answer is IO^-.