Which of the following electronic configurati | vedclass

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(B) The maximum oxidation state of a transition metal is determined by the total number of electrons in the $(n-1)d$ and $ns$ orbitals.For $3d^34s^2$,

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$3d^54s^2$

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(B) The maximum oxidation state of a transition metal is determined by the total number of electrons in the $(n-1)d$ and $ns$ orbitals.For $3d^34s^2$,total electrons = $3 + 2 = 5$.For $3d^54s^2$,total electrons = $5 + 2 = 7$.For $3d^54s^1$,total electrons = $5 + 1 = 6$.For $3d^64s^2$,total electrons = $6 + 2 = 8$.Among the given options,$3d^54s^2$ (Manganese,$Mn$) can show a maximum oxidation state of $+7$,but the question asks for the configuration representing the highest oxidation state potential. Comparing the total valence electrons available for bonding,$3d^54s^2$ is the standard configuration for $Mn$ which exhibits $+7$. However,if we look at the options provided,$3d^64s^2$ (Iron,$Fe$) has $8$ electrons,but $Fe$ typically shows $+2, +3$. The configuration $3d^54s^2$ corresponds to $Mn$,which is well known for its highest common oxidation state of $+7$ in the $3d$ series.

Which of the following electronic configurations of a metal atom represents the highest oxidation state?

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