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(B) To find the oxidation state of $Mn$ in each compound:$1$. In $K_2MnO_4$: $2(+1) + x + 4(-2) = 0 \implies x = +6$.$2$. In $KMnO_4$: $1(+1) + x + 4(

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$KMnO_4$

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(B) To find the oxidation state of $Mn$ in each compound:$1$. In $K_2MnO_4$: $2(+1) + x + 4(-2) = 0 \implies x = +6$.$2$. In $KMnO_4$: $1(+1) + x + 4(-2) = 0 \implies x = +7$.$3$. In $MnO_2$: $x + 2(-2) = 0 \implies x = +4$.$4$. In $Mn_2O_3$: $2x + 3(-2) = 0 \implies x = +3$.Therefore,the highest oxidation state of $Mn$ is $+7$,which is shown by $KMnO_4$.

The highest oxidation state of $Mn$ is shown by

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