When neutral or faintly alkaline is treated with potassium iodide, iodide ion is converted into 'X'. 'X' is [2019]

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Solution

Q.
When neutral or faintly alkaline ${KMnO}_{4}$ is treated with potassium iodide, iodide ion is converted into 'X'. 'X' is [2019]

1 $I_{2}$

2 $I O_{4}^{-}$

3 $I O_{3}^{-}$

4 $I O^{-}$

Ans.
(3)

In neutral or faintly alkaline solutions:

$2 {MnO}_{4}^{-} + H_{2} O + I^{-} ⟶ 2 {MnO}_{2} + 2 {OH}^{-} + \underset{' X'}{{IO}_{3}^{-}}$

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