When neutral or faintly alkaline $\mathrm{KMnO}_4$ is treated with potassium iodide, iodide ion is converted into ' $\mathrm{X}$ '. ' $\mathrm{X}$ ' is

• A

  $\mathrm{I}_2$
• B

  $\mathrm{IO}_4^{-}$
• Correct answer
  C

  $\mathrm{IO}_3^{-}$
• D

  $\mathrm{IO}^{-}$

Solution

When neutral or faintly alkaline $\mathrm{KMnO}_4$ is treated with potassium iodide, $\mathrm{KMnO}_4$ is converted into $\mathrm{MnO}_2$ while iodide $\left(\mathrm{I}^{-}\right)$ion is converted inot $\mathrm{IO}_3^{-}$(iodate). The reaction takes place as follows:

Hence, the correct answer is option: 'C'.