Question

What is the oxidation number of $Mn$ in $KMn{O}_4$ and of $S$ in $N{a}_2{S}_2{O}_3$?

Answer 1

Let the Ox.no. of Mn in $KMn{O}_4$ be x.
We know that, OX.no. of $K=+1$
Ox.no. of $O=-2$
So, Ox. no. K$+$ Ox. no. Mn $+$ $4(Ox.no.O)=0$
or $+1+x+4(-2)=0$
or $+1+x-8=0$
or $x=+8-1=+7$
Hence, Ox.no. of Mn in $KMn{O}_4$ is $+7$.
Similarly, for S in $N{a}_2{S}_2{O}_3$,
$2(Ox.no.Na)+2(Ox.no.S)+3(OX.no.O)=0$
$2\times (+1)+2x+3(-2)=0$
$x=+2$
Hence, Ox. no. of S in $N{a}_2{S}_2{O}_3=+2$.