Which of the following complexes exhibits the highest paramagnetic...

Key Idea : Greater is the number of unpaired electrons, larger is the paramagnetism. $$\begin{aligned} & {\left[\mathrm{V}(\mathrm{gly})_2(\mathrm{OH})_2\left(\mathrm{NH}_3\right)_2\right]^{+}} \\ & \mathrm{V}_{23}=[\mathrm{Ar}] 4 s^2, 3 d^3 \end{aligned}$$ Oxidation state of $\mathrm{V}$ in \left[\mathrm{V}(\mathrm{gly})_2(\mathrm{OH})_2\left(\mathrm{NH}_3\right)_2\right]^{+} is $\begin{gathered} x+(-1) \times 2+(-1) \times 2+(0) \times 2=+1 \\ x=+5 \\ \mathrm{~V}^{5+}=[\mathrm{Ar}] 3 d^0 \end{gathered}$ (No unpaired electron in option (A)) So, option (B), $$\begin{aligned} & {\left[\mathrm{Fe}\left(\text{ en) }(\text{ bpy })\left(\mathrm{NH}_3\right)_2\right]^{2+}\right.} \\ & \mathrm{Fe}_{26}=[\mathrm{Ar}] 4 s^2, 3 d^6 \end{aligned}$$ Oxidation state of $\mathrm{Fe}$ in \left[\mathrm{Fe}(\mathrm{en}) \text{ (bpy) }\left(\mathrm{NH}_3\right)_2\right]^{2+} is $\begin{gathered} x+(0)+(0)+(0) \times 2=+2 \\ x=+2 \\ \mathrm{Fe}^{2+}=[\mathrm{Ar}] 3 d^6 \end{gathered}$ But en, bpy and $\mathrm{NH}_3$ all are strong field ligands so pairing occurs, thus no unpaired electrons. So, option (C), $\begin{gathered} {\left[\mathrm{Co}(\mathrm{OX})_2(\mathrm{OH}]_2\right]^{-}} \\ \mathrm{Co}_{27}=[\mathrm{Ar}] 4 s^2, 3 d^7 \end{gathered}$ Oxidation state of $\mathrm{Co}$ in \left[\mathrm{Co}(\mathrm{OX})_2(\mathrm{OH})_2\right]^{-}$is$x+(-2) \times 2+(-1) \times 2=-1 $$\begin{aligned} & x-6=-1 \\ & x=+5 \\ & \mathrm{Co}^{5+}=[\mathrm{Ar}] 3 d^4 \\ & \text{(4 paired electrons)} \\ & \text{Now, option (D),} \\ & {\left[\mathrm{Ti}\left(\mathrm{NH}_3\right)_6\right]^{3+}} \\ & \mathrm{Ti}_{22}=[\mathrm{Ar}] 4 s^2, 3 d^2 \end{aligned}$$ \mathrm{Ti}_{22}=[\mathrm{Ar}] 4 s^2, 3 d^2$Oxidation state of Ti in$\left[\mathrm{Ti}\left(\mathrm{NH}_3\right)_6\right]^{3+} is +3, thus it contains 1 unpaired electron. Hence, $\quad\left[\mathrm{Co}(\mathrm{OX})_2(\mathrm{OH})_2\right]^{-}$has highest paramagnetic behaviour.