Which one of the elements with the following outer orbital...

Key Idea Number of oxidation states exhibited by d-block elements is the sum of number of electrons (unpaired) in d-orbitals and number of electrons in s-orbital. (a) $3 \mathrm{~d}^3, 4 \mathrm{~s}^2 \Rightarrow \mathrm{O} . \mathrm{S}=3+2=5$ (b) $3 \mathrm{~d}^5, 4 \mathrm{~s}^1 \Rightarrow \mathrm{O} . \mathrm{S}=5+1=6$ (c) $3 \mathrm{~d}^5, 4 \mathrm{~s}^2 \Rightarrow \mathrm{O} . \mathrm{S}=5+2=7$ (d) $3 \mathrm{~d}^2, 4 \mathrm{~s}^2 \Rightarrow \mathrm{O} . \mathrm{S}=2+2=4$ Hence, element with $3 \mathrm{~d}^5, 4 \mathrm{~s}^2$ configuration exhibits largest number of oxidation states.