Of the following outer electronic configurati | vedclass

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(A) The maximum oxidation state of a transition element is determined by the total number of electrons in the $(n-1)d$ and $ns$ orbitals.For option $A

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$(n-1)d^5 ns^2$

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(A) The maximum oxidation state of a transition element is determined by the total number of electrons in the $(n-1)d$ and $ns$ orbitals.For option $A$: $(n-1)d^5 ns^2$,total electrons = $5 + 2 = 7$.For option $B$: $(n-1)d^8 ns^2$,total electrons = $8 + 2 = 10$ (but the maximum stable oxidation state is usually limited by the number of unpaired electrons and stability of the configuration).For option $C$: $(n-1)d^5 ns^1$,total electrons = $5 + 1 = 6$.For option $D$: $(n-1)d^3 ns^2$,total electrons = $3 + 2 = 5$.Among the given configurations,the configuration $(n-1)d^5 ns^2$ (e.g.,$Mn$ in $3d^5 4s^2$) allows for a maximum oxidation state of $+7$. While $d^8 s^2$ has more electrons,it does not exhibit an oxidation state of $+10$; transition metals typically show a maximum oxidation state corresponding to the sum of $ns$ and $(n-1)d$ electrons up to $Mn$ $(+7)$. Thus,$(n-1)d^5 ns^2$ is the correct choice for the highest common oxidation state.

Of the following outer electronic configurations of atoms,which one achieves the highest oxidation state?

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