Oxidation number of Mn in K_2MnO_4 and in KMn | vedclass

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Question

(A) For $K_2MnO_4$: Let the oxidation number of $Mn$ be $x$. The sum of oxidation states is $2(+1) + x + 4(-2) = 0$. Solving this,$2 + x - 8 = 0$,so $

Vedclass · Accepted Answer

$+ 6$ and $+ 7$

Vedclass · Answer

(A) For $K_2MnO_4$: Let the oxidation number of $Mn$ be $x$. The sum of oxidation states is $2(+1) + x + 4(-2) = 0$. Solving this,$2 + x - 8 = 0$,so $x = +6$.For $KMnO_4$: Let the oxidation number of $Mn$ be $x$. The sum of oxidation states is $1(+1) + x + 4(-2) = 0$. Solving this,$1 + x - 8 = 0$,so $x = +7$.Therefore,the oxidation numbers are $+6$ and $+7$ respectively.

Oxidation number of $Mn$ in $K_2MnO_4$ and in $KMnO_4$ are respectively

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