Answer:
Option D
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.
