Question

${\mathrm{KMnO}}_4$acts as an oxidizing agent in an alkaline medium. When alkaline ${\mathrm{KMnO}}_4$is treated with $\mathrm{KI}$, Iodide ion is oxidized to

• A. $\mathrm{IO}$
• B. ${\mathrm{IO}}^{-}$
• C. ${{\mathrm{IO}}_3}^{-}$
• D. ${\mathrm{I}}_2$

Answer 1

The correct option is C

${{\mathrm{IO}}_3}^{-}$

The explanation for the correct answer:

Option (C) ${\mathrm{IO}}^{-3}$

1. It is a crystalline solid that's odorless and ranges from purple to magenta in color.
2. Water, acetone, acetic acid, methanol, and pyridine are all soluble in it.
3. Potassium permanganate is a powerful oxidizing agent, it can be utilized as an oxidant in a wide range of chemical reactions.
4. When alkaline ${\mathrm{KMnO}}_4$ is reacted with$\mathrm{KI}$, Iodine is oxidized. $\underset{\mathrm{Potassium}\mathrm{permagnate}}{2{\mathrm{KMnO}}_4\left(\mathrm{s}\right)}+\underset{\mathrm{Potassium}\mathrm{Iodide}}{\mathrm{KI}\left(\mathrm{s}\right)}+\underset{\mathrm{Water}}{{\mathrm{H}}_2\mathrm{O}}\left(\mathrm{l}\right)\to \underset{\mathrm{Manganese}\mathrm{dioxide}}{2{\mathrm{MnO}}_2\left(\mathrm{s}\right)}+\underset{\mathrm{Potassium}\mathrm{Hydroxide}}{2\mathrm{KOH}\left(\mathrm{l}\right)}+\underset{\mathrm{Potassium}\mathrm{Iodate}}{{\mathrm{KIO}}_3\left(\mathrm{s}\right)}$
5. Oxidation state of iodine changes from $0$ to $+5$ i.e. increase in oxidation number. $\stackrel{0}{\mathrm{I}}\to \stackrel{+5}{{{\mathrm{IO}}_3}^{-}}$
6. Hence, it is the correct option.

Therefore, When alkaline ${\mathrm{KMnO}}_4$is treated with$\mathrm{KI}$, Iodide ion is oxidized to ${{\mathrm{IO}}_3}^{-}$