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

In neutral or faintly alkaline medium, ${\text{KMnO}}_4$ acts as an oxidizing agent and is reduced to ${\text{MnO}}_2$. It oxidizes iodide (${\text{I}}^{-}$) to iodate (${\text{IO}}_3^{-}$).Reaction: $2{\text{MnO}}_4^{-}+{\text{H}}_2\text{O}+{\text{I}}^{-}\to 2{\text{MnO}}_2+2{\text{OH}}^{-}+{\text{IO}}_3^{-}$ Note: In acidic medium, iodide is oxidized to ${\text{I}}_2$.