Question

Which compound amongst the following has the highest oxidation number for $Mn$? $KMn{O}_4,K_{2}Mn{O}_4,Mn{O}_2$ and $M{n}_2{O}_3$.

Answer 1

$KMn{O}_4$	$+1+x-8=0$	Ox.no. of $Mn$
	$x=+7$	$+7$
$K_{2}Mn{O}_4$	$+2+x-8=0$
	$x=+6$	$+6$
$Mn{O}_2$	$x-4=0$
	$x=+4$	$4$
$M{n}_2{O}_3$	$2x-6=0$
	$x=+3$	$+3$

Thus, the highest oxidation number for $Mn$ is in $KMn{O}_4$.
Sometimes, oxidation numbers have such values which at first sight appear strange. For example, the oxidation number of carbon in cane sugar $\left(C_{12}{H}_{22}{O}_{11}\right)$, glucose $\left(C_6{H}_{12}{O}_6\right)$, dichloromethane, etc., is zero.
Cane sugar $\left(C_{12}{H}_{22}{O}_{11}\right)$
$12\times x+22\times 1+11(-2)=0$
$12x+22-22=0$
so, $x=0$
Glucose $\left(C_6{H}_{12}{O}_6\right)$
$6\times x+12\times 1+6(-2)=0$
$6x+12-12=0$
So, $x=0$
Dichloromethane$\left(C{H}_{2}C{l}_2\right)$
$x+2\times 1+2(-1)=0$
$x+2-2=0$
So, $x=0$.