Question

Which of the following complexes exhibits the highest paramagnetic behaviour?
where, gly = glycine, en = ethylenediamine and bpy = bipyridyl moities)
(At. no. of $Ti=22,V=23,Fe=26,Co=27$)

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

Answer 1

Answer: (C)

The electronic configuration of
$V(23)=[Ar]4s^2,3d^3$
Let in ${\left[V(gly{)}_2(OH{)}_2{\left(N{H}_3\right)}_2\right]}^{+}$oxidation state of $V$ is $x$.
$x+(-1)\times 2+(-1)2+(0\times 2)=+1$
$x=+5$
$V^{5+}=[Ar]4s^0,3d^0$ (no unpaired electrons)
The electronic configuration of
$Fe(26)=[Ar]4s^2,3d^6$
Let the oxidation state of $Fe$ in
${\left[Fe(en)(ppy){\left(N{H}_3\right)}_2\right]}^{2+}\text{ is }x$
$[x+(0)+(0)+(0)\times 2]=+2$
$x=+2$
$F{e}^{2+}=[\text{ Ar }]3d^6$ ($\because 4$ unpaired electron)
but, bpy, en and $N{H}_3$ all are strong field ligands, so pairing occurs and thus, $F{e}^{2+}$ contains no unpaired electron.
The electronic configuration of
$Co(27)=[Ar]4s^2,3d^7$
Oxidation state of $Co$ in ${\left[Co(Ox{)}_2(OH{)}_2\right]}^{-}$
$x+(-2)\times 2+(-1)\times 2=-1$
$x=+5$
$C{o}^{5+}=[Ar],3d^4$ [4 unpaired electrons]
ox and $OH$ are weak field ligands.
The electronic configuration of
$Ti(22)=[Ar]4s^2,3d^2$
Oxidation state of $Ti$ in ${\left(Ti{\left(N{H}_3\right)}_6\right]}^{3+}$ is $3$ .
$T{i}^{3+}=[Ar]3d^1$ (one unpaired electron)
Hence, complex ${\left[Co(Ox{)}_2(OH{)}_2\right]}^{-}$has maximum number of unpaired electrons, thus show maximum paramagnetism.