Question

When neutral or faintly alkaline $KMn{O}_4$ is treated with potassium iodide, iodide ion is converted into 'X'. 'X' is -

• A. $I_2$
• B. $I{0}_4^{-}$
• C. $I{0}_3^{-}$
• D. $I{0}^{-}$

Answer 1

Answer: (C)

$KMn{O}_4+I^{-}+O{H}^{-}\to \underset{(X)}{Mn{O}_2+I{O}_3^{-}+H_{2}O}$