The oxidation number of Mn in the product of | vedclass

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(A) The reaction for the alkaline oxidative fusion of manganese dioxide is:$2MnO_2 + 4KOH + O_2 \rightarrow 2K_2MnO_4 + 2H_2O$The product formed is po

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(A) The reaction for the alkaline oxidative fusion of manganese dioxide is:$2MnO_2 + 4KOH + O_2 ightarrow 2K_2MnO_4 + 2H_2O$The product formed is potassium manganate,$K_2MnO_4$,which contains the manganate ion,${MnO_4}^{2-}$.To find the oxidation state of $Mn$ in ${MnO_4}^{2-}$:Let the oxidation state of $Mn$ be $x$.$x + 4 	imes (-2) = -2$$x - 8 = -2$$x = +6$Thus,the oxidation number of $Mn$ in the product is $6$.

The oxidation number of $Mn$ in the product of alkaline oxidative fusion of $MnO_2$ is

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