The most stable oxidation state of Mn is | vedclass

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Question

(A) The electronic configuration of $Mn$ $(Z=25)$ is $[Ar] 3d^5 4s^2$.When $Mn$ loses two electrons from the $4s$ orbital,it forms $Mn^{2+}$,which has

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$+2$

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(A) The electronic configuration of $Mn$ $(Z=25)$ is $[Ar] 3d^5 4s^2$.When $Mn$ loses two electrons from the $4s$ orbital,it forms $Mn^{2+}$,which has the configuration $[Ar] 3d^5$.The $3d^5$ configuration represents a half-filled $d$-subshell,which provides extra stability due to exchange energy and symmetry.Therefore,the $+2$ oxidation state is the most stable for $Mn$.

The most stable oxidation state of $Mn$ is

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