Discussion Forum

Identify the oxidation  states of Mn, when $MnO_{4}^{2-}$ ion undergoes disproportionation reaction under acidic medium

Answer:

Explanation:

 The disproportionation reaction for $MnO_{4}^{2-}$ occurs as follows:

 $3 Mn^{+6}O_{4}^{2-}+4H^{+}\rightarrow Mn^{+4}O_{2}+2Mn^{+7}O_{4}^{-}+2H_{2}O$

 Hence, $MnO_{4}^{2-}$ is oxidised to $MnO_{4}^{-}$ and reduced to  $MnO_{2}$ . Thus  , oxidation  state of Mn when $MnO_{4}^{2-}$  undergoes disproportionation reaction under acidic medium are +7 and +4 

$\therefore$   Hence option (d)  is the correct answer.