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(A) The number of oxidation states exhibited by transition elements depends on the number of electrons available in the $(n-1)d$ and $ns$ orbitals.For

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$3d^5\,4s^2$

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(A) The number of oxidation states exhibited by transition elements depends on the number of electrons available in the $(n-1)d$ and $ns$ orbitals.For $3d^5\,4s^2$ (Manganese,$Z=25$),the total number of electrons available for bonding is $5 + 2 = 7$. Thus,it shows oxidation states ranging from $+2$ to $+7$.For $3d^2\,4s^2$ (Titanium,$Z=22$),the total is $2 + 2 = 4$.For $3d^3\,4s^2$ (Vanadium,$Z=23$),the total is $3 + 2 = 5$.For $3d^5\,4s^1$ (Chromium,$Z=24$),the total is $5 + 1 = 6$.Therefore,the configuration $3d^5\,4s^2$ exhibits the maximum number of oxidation states.

Which of the following outer electronic configurations of elements exhibits the maximum number of oxidation states?

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