Question

What is the oxidation number of $Mn$ in $KMn{O}_4,K_{2}Mn{O}_4,MnS{O}_4,Mn{O}_2$ and ${Mn}_3{O}_4$ respectively?

• A. $+7,+6,+2,+4,$ $+\frac{8}{3}$
• B. $+4,+6,+2,+4,$ $+\frac{8}{3}$
• C. $+7,+6,+2,+3,$ $+\frac{8}{3}$
• D. None of these

Answer 1

Answer: (A)

$+7,+6,+2,+4,$ $+\frac{8}{3}$

Let $x$ be the oxidation number of $Mn$.
Since, the overall charge on the complex is $0$, the sum of oxidation states of all elements in it should be equal to $0$.
$K_{2}Mn{O}_4$; $2+x+4(-2)=0;x=+6$
$MnS{O}_4$; $x-2=0;x=+2$
$Mn{O}_2$; $x+2(-2)=0;x=+4$
${Mn}_3{O}_4$; $3x+4(-2)=0;x=\frac{8}{3}$
The oxidation number of $Mn$ in $KMn{O}_4,K_{2}Mn{O}_4,MnS{O}_4,Mn{O}_2$ and ${Mn}_3{O}_4$ is $+7,+6,+2,+4$ and $+\frac{8}{3}$ respectively.