$KMnO _4$ oxidises $I ^{-}$in acidic and neutral/faintly alkaline solution, respectively, to

Question

Statement-I : Consider the following pairs of ions $(\text{Sc}^{3+}, \text{Ti}^{3+}), (\text{Ti}^{4+}, \text{Ni}^{2+}), (\text{Cu}^{2+}, \text{Zn}^{2+})$ and $(\text{Cr}^{3+}, \text{Mn}^{3+})$. Out of these pairs three pairs consist of ions that are both coloured :
Statement-II : Among the lanthanide ions $\text{Eu}^{2+}$, $\text{Gd}^{3+}$, $\text{Ce}^{4+}$ and $\text{Tb}^{4+}$, the ion $\text{Tb}^{4+}$ is the strongest reducing agent.
Choose the correct option.

Answer 1

To answer the question about how $KMnO_4$ oxidizes $I^{{-}}$ in different solutions, we need to consider the chemical reactions involved.

Oxidation in Acidic Solution:

When $KMnO_4$ is used as an oxidizing agent in an acidic medium, it reduces to $Mn^{2+}$ ions and oxidizes iodide ions $(I^{-})$ to iodine $(I_2)$. The reaction can be expressed as:

$2KMnO_4 + 10I^{-} + 16H^{+} \rightarrow 2Mn^{2+} + 5I_2 + 8H_2O + 2K^+$

Thus, in acidic solution, iodide is oxidized to $I_2$ (iodine).

Oxidation in Neutral or Faintly Alkaline Solution:

In neutral or faintly alkaline conditions, $KMnO_4$ oxidizes $I^{-}$ to iodate, $IO_3^{-}$:

$2KMnO_4 + I^{-} + H_2O \rightarrow 2MnO_2 + IO_3^{-} + 2K^+ + 2OH^{-}$

Here, $I^{-}$ is oxidized to $IO_3^{-}$ in neutral or faintly alkaline medium.

Thus, $KMnO_4$ oxidizes $I^{-}$ to different products in different solutions:

In acidic solution to $I_2$
In neutral or faintly alkaline solution to $IO_3^{-}$

Based on the explanation above, the correct answer is $I_2$ & $IO_3^{-}$, which corresponds to the option "I₂ & IO₃^-".

$KMnO _4$ oxidises $I ^{-}$in acidic and neutral/faintly alkaline solution, respectively, to

The Correct Option is B

Solution and Explanation

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