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. $[V(gly)_2(OH)_2(NH_3)_2]^+$
• B. $[Fe(en)(py)(NH_3)_2]^{2+}$
• C. $[Co(ox)_2(OH)_2]^-$
• D. $[Ti(NH_3)_6]^{3+}$

Answer 1

Answer: (C)

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