Identify the oxidation states of Mn when MnO_ | vedclass

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Question

(D) The disproportionation reaction for $MnO_4^{2-}$ occurs as follows:$3 \stackrel{+6}{MnO_4^{2-}} + 4 H^{+} \longrightarrow \stackrel{+4}{MnO_2} + 2

Vedclass · Accepted Answer

$+7, +4$

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(D) The disproportionation reaction for $MnO_4^{2-}$ occurs as follows:$3 \stackrel{+6}{MnO_4^{2-}} + 4 H^{+} \longrightarrow \stackrel{+4}{MnO_2} + 2 \stackrel{+7}{MnO_4^{-}} + 2 H_2 O$Here,$MnO_4^{2-}$ (oxidation state $+6$) is oxidized to $MnO_4^{-}$ (oxidation state $+7$) and reduced to $MnO_2$ (oxidation state $+4$).Thus,the oxidation states of $Mn$ in the products are $+7$ and $+4$.Therefore,option $(D)$ is the correct answer.

Identify the oxidation states of $Mn$ when $MnO_4^{2-}$ ion undergoes disproportionation reaction under acidic medium.

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