(B) In $KMnO_4$,let the oxidation number of $Mn$ be $x$. The sum of oxidation numbers is $1 + x + 4(-2) = 0$,so $x - 7 = 0$,which gives $x = +7$. In $

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(B) In $KMnO_4$,let the oxidation number of $Mn$ be $x$. The sum of oxidation numbers is $1 + x + 4(-2) = 0$,so $x - 7 = 0$,which gives $x = +7$. In $MnO_2$,let the oxidation number of $Mn$ be $y$. The sum of oxidation numbers is $y + 2(-2) = 0$,so $y - 4 = 0$,which gives $y = +4$. The difference in oxidation number is $|7 - 4| = 3$.