Read the passage carefully and answer the questions.

Sodium dichromate $Na_2Cr_2O_7$ and Potassium dichromate $K_2Cr_2O_7$, are strong oxidizing agents; in acidic solution, its oxidising action can be represented as follows:

$Cr_2{O_7}^{2-}+ 14H^++ 6e^-→2Cr^{3+}+ 7H_2O$ $(E° = 1.33V)$

The chromates and dichromates are interconvertible in aqueous solution depending upon pH of the solution. The chromate ion is tetrahedral whereas the dichromate ion consists of two tetrahedra sharing one corner with Cr-O-Cr bond angle of 126°

Potassium permanganate is prepared by fusion of $MnO_2$ with an alkali metal hydroxide and an oxidising agent like $KNO_3$. This produces the dark green $K_2MnO_4$ which disproportionates in a neutral or acidic solution to give permanganate. It is a strong oxidizing agent. Hydrogen ion concentration of the solution plays an important part in the reduction of permanganate to manganate, manganese dioxide and manganese(II) salt.

Potassium permanganate forms dark purple (almost black) crystals which are isostructural with those of $KClO_4$. The salt is not very soluble in water (6.4 g/100 g of water at 293 K), but when heated it decomposes at 513 K.

Alkaline/neutral $KMnO_4$ oxidizes iodide ($I^-$) ion to

$IO_{3^-}$

The correct answer is Option (1) → $IO_{3^-}$

Core Concept

The oxidizing behavior of $\text{KMnO}_4$ depends strongly on the medium:

• Acidic medium $\rightarrow \text{MnO}_4^-$ reduces to $\text{Mn}^{2+}$ (strongest oxidizing action).
• Neutral / mildly alkaline medium $\rightarrow \text{MnO}_4^-$ reduces to $\text{MnO}_2$ (moderate oxidation).
• Strongly alkaline medium $\rightarrow \text{MnO}_4^-$ reduces to $\text{MnO}_4^{2-}$.

In neutral or alkaline solution, iodide ($\text{I}^-$) is oxidized to iodate ($\text{IO}_3^-$), not just iodine.

Option-wise Explanation

1. $\text{IO}_3^-$

In neutral/alkaline medium, permanganate oxidizes iodide further beyond $\text{I}_2$ to iodate. This is a higher oxidation state ($+5$ for iodine), consistent with the oxidizing strength of $\text{MnO}_4^-$ in this medium.

2. $\text{IO}_2^-$

This is an unstable and uncommon oxidation product for iodide under normal redox conditions with $\text{KMnO}_4$. It is not the typical product in aqueous chemistry.

3. $\text{I}_2$

Iodine ($\text{I}_2$) forms in milder oxidations (like with $\text{Fe}^{3+}$ or acidic conditions under controlled reaction), but $\text{KMnO}_4$ in alkaline/neutral medium oxidizes $\text{I}^-$ further than $\text{I}_2$.

4. $\text{I}_2\text{O}_4$

This is not a standard aqueous oxidation product of iodide in redox reactions. It is not relevant to typical permanganate chemistry.