Question

In alkaline medium, \( \text{MnO}_4^- \) oxidizes \( \text{I}^- \) to:

• A. \( \text{IO}_4^- \)
• B. \( \text{IO}^- \)
• C. \( \text{I}_2 \)
• D. \( \text{IO}_3^- \)

Answer 1

The Correct Option is D

To understand this question, we need to look at the redox reaction between permanganate ion \( \text{MnO}_4^- \) and iodide ion \( \text{I}^- \) in an alkaline medium. 

The permanganate ion acts as a strong oxidizing agent. In an alkaline medium, it can oxidize iodide ions to iodate ions according to the balanced chemical equation:

\[2 \text{MnO}_4^- + \text{I}^- + \text{H}_2\text{O} \rightarrow 2 \text{MnO}_2 + \text{IO}_3^- + 2 \text{OH}^-\]

This reaction shows that in an alkaline medium, \( \text{I}^- \) is oxidized to \( \text{IO}_3^- \) (iodate ion) when reacted with permanganate ion. The presence of hydroxide ions from the alkaline medium helps stabilize the formation of manganese dioxide (\( \text{MnO}_2 \)).

Let's analyze the options:

1. \(\text{IO}_4^-\) - Periodate ion, not typically formed in alkaline conditions with this oxidation.
2. \(\text{IO}^-\) - Hypoiodite ion, not the product in this specific reaction.
3. \(\text{I}_2\) - Molecular iodine, not a product of this reaction in alkaline medium.
4. \(\text{IO}_3^-\) - Iodate ion, correct product of this reaction.

Thus, the correct answer is that in an alkaline medium, \( \text{MnO}_4^- \) oxidizes \( \text{I}^- \) to \( \text{IO}_3^- \).

Answer 2

The reaction of MnO₄^- in an alkaline medium with I^- produces IO₃^- as the product.
2MnO₄^- + H₂O + I^- → 2MnO₂ + 2OH^- + IO₃^-
Thus, the oxidation of I^- in the presence of MnO₄^- under alkaline conditions leads to the formation of IO₃^-.
So, the correct answer is : Option (4)