Question

Number of electrons transfer in each case when $KMn{O}_4$ acts as an oxidizing agent to give $Mn{O}_2,M{n}^{2+},Mn(OH{)}_3,andMn{O}_4^{2-}$ are respectively:

• A. 3, 5, 4, and 1
• B. 4, 3, 1, and 5
• C. 1, 3, 4, and 5
• D. 5, 4, 3, and 1

Answer 1

The correct option is A 3, 5, 4, and 1

$Mn$ exhibits its highest oxidation state of (+7) in $KMn{O}_4$. In $Mn{O}_2$ it is in the (+3) oxidation state. All this can be easily computed using the rules we apply for calculating oxidation numbers/states that we explored in the chapter Redox Reactions. $M{n}^{2+}$ is in (+2) oxidation state while $Mn(OH{)}_3$ is in the (+3) oxidation state. In the manganate anion - $Mn{O}_4^{2-}$ the central metal is in the (+6) oxidation state.