Consider the following reaction:MnO_2 + KOH + | vedclass

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(B) The reaction is: $2MnO_2 + 4KOH + O_2 \rightarrow 2K_2MnO_4 + 2H_2O$. Thus,$A$ is $K_2MnO_4$.In neutral or acidic medium,$K_2MnO_4$ disproportiona

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$4$

Vedclass · Answer

(B) The reaction is: $2MnO_2 + 4KOH + O_2 \rightarrow 2K_2MnO_4 + 2H_2O$. Thus,$A$ is $K_2MnO_4$.In neutral or acidic medium,$K_2MnO_4$ disproportionates: $3MnO_4^{2-} + 4H^+ \rightarrow 2MnO_4^- + MnO_2 + 2H_2O$.Here,$B$ is $MnO_4^-$ ($Mn^{+7}$,$d^0$ configuration,$n=0$,$\mu = 0 \ BM$) and $C$ is $MnO_2$ ($Mn^{+4}$,$d^3$ configuration,$n=3$,$\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \ BM$).The sum of magnetic moments is $0 + 3.87 = 3.87 \ BM$.The nearest integer is $4$.

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