d
Oxidation State of \(Ti \) in the complex \(\left[\mathrm{Ti}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}\)

\(\mathrm{Ti}^{3+}=1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{1}\)

no. of unpaired electron in \(d\) orbital is one. Oxidation State of \(V\) in complex \(\left[\mathrm{V}(\mathrm{gly})_{2}(\mathrm{OH})_{2}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+}\)

\(=x+2 \times 0+2 \times(-1)+2 \times 0=+1\)

\(\therefore x=+3\)

\(\mathrm{V}^{3+}=1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{2}\)

no. of unpaired electron in \(d\) orbital is two.

Oxidation State of \(Fe\) in the given complex is \(+2\)

\(\therefore \quad \mathrm{Fe}^{2+}=1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{6}\)

no. of unpaired electron in \(d\) orbital is four.

Oxidation State of \(\mathrm{Co}\) in the given complex \(\left[\mathrm{Co}(ox)_{2}(\mathrm{OH})_{2}\right]^{-}\)

\(=x+2 \times(-2)+2 \times(-1)=-1=x-4-2=-1\)

\(\therefore x=+5\)

(not possible, common ox. no. of \(\mathrm{Co}=+2,+3,+4\) ) \(\mathrm{Co}^{5+}=1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{4}\)

It should be an inner orbital complex (\(d'sp\)  hybridisation) containing only one unpaired electron. So the complex having highest paramagnetism would be the complex of iron containing four unpaired electrons.