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

Answer 1

Answer: (D)

+ 4 to + 6

Answer 2

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,
$x+(2\times -2)=0$ $x-4=0$ $x=+\text{ }4$
In ${{K}_{2}}Mn{{O}_{4}}$ oxidation state of $Mn$
is, $2+x+(4x-2)=0$
$2+x+(-8)=0$
$x-6=0$
$x=+\text{ }6$
It means that oxidation state of $Mn$ changes from +4 to+6.