In alkaline medium, \( \text{MnO}_4^- \) oxidizes \( \text{I}^- \) to:
- \( \text{IO}_4^- \)
- \( \text{IO}^- \)
- \( \text{I}_2 \)
- \( \text{IO}_3^- \)
The Correct Option is D
Approach Solution - 1
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:
- \(\text{IO}_4^-\) - Periodate ion, not typically formed in alkaline conditions with this oxidation.
- \(\text{IO}^-\) - Hypoiodite ion, not the product in this specific reaction.
- \(\text{I}_2\) - Molecular iodine, not a product of this reaction in alkaline medium.
- \(\text{IO}_3^-\) - Iodate ion, correct product of this reaction.
Thus, the correct answer is that in an alkaline medium, \( \text{MnO}_4^- \) oxidizes \( \text{I}^- \) to \( \text{IO}_3^- \).
Approach Solution -2
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)
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