When KMnO_4 acts as an oxidising agent and ul | vedclass

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(C) In $KMnO_4$,the oxidation state of $Mn$ is $+7$. The number of electrons transferred is equal to the change in the oxidation state of $Mn$. $1$. F

Vedclass · Accepted Answer

$1, 3, 4, 5$

Vedclass · Answer

(C) In $KMnO_4$,the oxidation state of $Mn$ is $+7$. The number of electrons transferred is equal to the change in the oxidation state of $Mn$. $1$. For $[MnO_4]^{2-}$,$Mn$ is $+6$. Change $= |7 - 6| = 1$. $2$. For $MnO_2$,$Mn$ is $+4$. Change $= |7 - 4| = 3$. $3$. For $Mn_2O_3$,$Mn$ is $+3$. Change per $Mn$ atom $= |7 - 3| = 4$. $4$. For $Mn^{2+}$,$Mn$ is $+2$. Change $= |7 - 2| = 5$. Thus,the number of electrons transferred is $1, 3, 4, 5$.

When $KMnO_4$ acts as an oxidising agent and ultimately forms $[MnO_4]^{2-}$,$MnO_2$,$Mn_2O_3$,and $Mn^{2+}$,the number of electrons transferred in each case respectively is:

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