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Q.
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:
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.