Question

Find the oxidation number of elements in each case :

$Mn$ in $K_{2}Mn{O}_4,K_{2}Mn{O}_3,M{n}_3{O}_4,MnS{O}_4$ and $K_{3}Mn{F}_6$.

• A. $6,4,\frac{8}{3},2,3$
• B. $10,5,\frac{8}{3},3,3$
• C. $6,5,8,2,3$
• D. None of the above

Answer 1

The correct option is A $6,4,\frac{8}{3},2,3$

Let $x$ be the oxidation number of $Mn$ in $K_{2}Mn{O}_4$.
Since the overall charge on the complex is $0$, the sum of oxidation states of all elements in it should be equal to $0$.
Therefore, $2(+1)+x+4(-2)=0$
or, $x=+6$
Smilarly,
Let $x$ be the oxidation number of $Mn$ in $K_{2}Mn{O}_3$.
$2(+1)+x+3(-2)=0$
$x=+4$
Let $x$ be the oxidation number of $Mn$ in $M{n}_3{O}_4$.
$3x+4(-2)=0$
$x=+\frac{8}{3}$
Let $x$ be the oxidation number of $Mn$ in $MnS{O}_4$.
$x+(-2)=0$
$x=+2$
Let $x$ be the oxidation number of $Mn$ in $K_{3}Mn{F}_6$.
$3(+1)+x+6(-1)=0$
$x=+3$
Thus, the oxidation number of $Mn$ in $K_{2}Mn{O}_4,K_{2}Mn{O}_3,M{n}_3{O}_4,MnS{O}_4$ and $K_{3}Mn{F}_6$ are $6,4,\frac{8}{3},2$ and $3$ respectively.