Among the following outermost electronic conf | vedclass

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(C) The oxidation state of a transition metal depends on the number of electrons available in the $(n-1)d$ and $ns$ orbitals. For $3d^5 4s^2$ (Mangane

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$3d^5 4s^2$

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(C) The oxidation state of a transition metal depends on the number of electrons available in the $(n-1)d$ and $ns$ orbitals. For $3d^5 4s^2$ (Manganese,$Mn$),the total number of valence electrons is $5 + 2 = 7$. Therefore,it can exhibit a maximum oxidation state of $+7$. Comparing the given configurations: $A) 3d^3 4s^2$ (Vanadium,$V$) shows up to $+5$. $B) 3d^5 4s^1$ (Chromium,$Cr$) shows up to $+6$. $C) 3d^5 4s^2$ $(Mn)$ shows up to $+7$. $D) 3d^6 4s^2$ (Iron,$Fe$) shows up to $+6$. Thus,the configuration $3d^5 4s^2$ shows the highest oxidation state.

Among the following outermost electronic configurations of transition metals,which one shows the highest oxidation state?

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