Question

Which of the following complex exhibits the highest paramagnetic behaviour?

• A. $\left[V\right(gly{)}_2\left(OH{)}_2\right(N{H}_3{)}_3{]}^{+}$
• B. $\left[Fe\right(en\left)\right(bpy\left)\right(N{H}_3{)}_2{]}^{2+}$
• C. $\left[Co\right(OX{)}_2(OH{)}_2{]}^{2-}$
• D. $\left[Ti\right(N{H}_3{)}_6{]}^{3+}$

Answer 1

The correct option is C $\left[Co\right(OX{)}_2(OH{)}_2{]}^{2-}$

a) $V^{+2}$ Configuration $\left[Ar\right]3d^2$

2 unpaired electron

$u=\sqrt{2\left(4\right)}=\sqrt{8}BM$

b) $F{e}^{2+}$ Strong field ligand: $en,bpy,\left(N{H}_3\right)$ Configuration $\left[Ar\right]3d^2$

$\upharpoonleft \upharpoonright \upharpoonleft \upharpoonright \upharpoonleft \upharpoonright \square \square$

$0$ unpaired electron.

$\therefore u=0$

c) $C{o}^{2+}$ weak field ligand $Ox,OH$ Configuration $\left[Ar\right]3d^7$

$\upharpoonleft \upharpoonright \upharpoonleft \upharpoonright \upharpoonleft \upharpoonleft \upharpoonright$

$3$ unpaired electrons.

$u=\sqrt{3\times 5}=\sqrt{15}BM$

d) $\left[Ti\right(N{H}_3{)}_6{]}^{+}$ Configuration $\left[Ar\right]3d^1$

$1$ unpaired electron

$u=\sqrt{1\left(3\right)}=\sqrt{3}BM$

In complex C, there is more number of unpaired electron therefore, it exhibits the highest paramagnetic behaviour.

Hence, the correct option is C.