Question

In the preparation of potassium permanganate, pyrolusite $(Mn{O}_2)$ is first converted to potassium manganate $(K_{2}Mn{O}_4)$ . In this conversion, the oxidation state of manganese changes from:

• A. + 1 to + 3
• B. + 2 to + 4
• C. + 3 to + 5
• D. + 4 to + 6
• E. + 5 to + 7

Answer 1

Answer: (D)

In the proportion of potassium permanganate
$Mn{O}_2$ converted to $K_{2}Mn{O}_4$
In $Mn{O}_2$ oxidation state of $Mn$ is,
$x+(2\times -2)=0$ $x-4=0$ $x=+4$
In $K_{2}Mn{O}_4$ oxidation state of $Mn$
is, $2+x+(4x-2)=0$
$2+x+(-8)=0$
$x-6=0$
$x=+6$
It means that oxidation state of $Mn$ changes from +4 to+6.