Among the following options, identify the one which exhibits the greatest number of oxidation states.

$\mathrm{Fe}(Z=26)=[\mathrm{Ar}] 4 s^2 3 d^6$ ( $\therefore$ 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.