The disproportionation of MnO_{4}^{2-} in aci | vedclass

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(B) The disproportionation reaction of manganate ion $(MnO_{4}^{2-})$ in acidic medium is given by: $3MnO_{4}^{2-} + 4H^{+} \rightarrow 2MnO_{4}^{-} +

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$4$

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(B) The disproportionation reaction of manganate ion $(MnO_{4}^{2-})$ in acidic medium is given by: $3MnO_{4}^{2-} + 4H^{+} \rightarrow 2MnO_{4}^{-} + MnO_{2} + 2H_{2}O$ Here,the manganese compounds formed are $MnO_{4}^{-}$ (where $Mn$ is in $+7$ oxidation state) and $MnO_{2}$ (where $Mn$ is in $+4$ oxidation state). Given that the oxidation state of $Mn$ in $B$ is smaller than that of $A$,we identify $A$ as $MnO_{4}^{-}$ $(+7)$ and $B$ as $MnO_{2}$ $(+4)$. In $MnO_{2}$,$Mn$ is in the $+4$ oxidation state,which corresponds to the electronic configuration $[Ar] 3d^{3}$. The number of unpaired electrons $(n)$ in $Mn^{4+}$ is $3$. The spin-only magnetic moment $(\mu)$ is calculated as: $\mu = \sqrt{n(n+2)} = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \ BM$. The nearest integer value is $4$.

The disproportionation of $MnO_{4}^{2-}$ in acidic medium results in the formation of two manganese compounds $A$ and $B$. If the oxidation state of $Mn$ in $B$ is smaller than that of $A$,then the spin-only magnetic moment $(\mu)$ value of $B$ in $BM$ is $.........$ (Nearest integer).

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