Solveeit Logo
Login

Question

Chemistry Question on d -and f -Block Elements

When neutral or faintly alkaline KMnO4 is treated with potassium iodide, iodide ion is converted into 'X'. 'X' is -

A

I04−I0^{-}_{4}

B

I03−I0^{-}_{3}

C

I0−I0^{-}

Answer

I0−I0^{-}

Explanation

Solution

KMnO4+I−+OH−−>MnO2+IO3−+H2O(X)KMnO_4 + I^- + OH^- {->} \underset{(X)}{MnO_2 + IO_3^- + H_2O}