Among the following options, identify the one which exhibits the greatest number of oxidation states.
Among the following options, identify the one which exhibits the greatest number of oxidation states.
$\mathrm{Fe}$
$\mathrm{Mn}$
$\mathrm{Cr}$
V
Solution
$\mathrm{Fe}(Z=26)=[\mathrm{Ar}] 4 s^2 3 d^6$
( $\therefore$ Highest oxidation state of $\mathrm{Fe}$ is +3 )
$\operatorname{Mn}(Z=25)=[\mathrm{Ar}] 4 s^2 3 d^5$ (5 unpaired electrons in $3 d$-orbital and $2 e^{-}$in $4 s$.)
Mn can show 6 oxidation states from MnO to $\mathrm{MnO}_4^{-}$.
$
\mathrm{Cr}(\mathrm{Z}=24)=[\mathrm{Ar}] 4 s^1 3 d^5
$
( $\therefore$ Highest oxidation state of $\mathrm{Cr}$ is +3 )
$
\mathrm{V}(Z=23)=[\mathrm{Ar}] 3 d^3 4 s^2
$
( $\therefore$ Highest oxidation state of $\mathrm{V}$ is +2 )
Hence, greatest number of oxidation state is shown by Mn