Question
Which of the following complexes exhibits the highest paramagnetic behaviour?
Where gly = glycine, en = ethylenediamine and bpy = bipyridyl moities
(At. no. ; Ti = 22, V = 23, Fe =26, Co = 27)
-
[V(gly)2 (OH)2 (NH3)2]+
-
[Fe(en)(bpy)(NH3)2]+
-
[CO(OX)2(OH)2]-
-
[Ti(NH3)6]3+
Solution
C.
[CO(OX)2(OH)2]-
Greater is the number of unpaired electrons, larger is the paramagnetism.
[V(gly)2(OH)2(NH3)2]+
V23 = [Ar] 4s2, 3d2
Oxidation state of V in [[V(gly)2(OH)2(NH3)2]+ is
x + (-1) x 2 + (-1) x 2 + (0) x 2 = +1
x=+5
V5+ = [Ar]3d0
[Fe(en)(bby)(NH3)2]2+
x + (0) + (0) + (0) x 2 = +2
x = +2
Fe2+ = [Ar] 3d6
But en , bby and NH3 all are strong field ligands, so pairing occurs, thus no unpaired electrons.
[Co(OX)2 (OH)2]- is
Oxidation state of Co in [Co(OX)2(OH)2]- is
x + (-2) x 2 + (-1) x 2 = -1
x-6 = -1
x = +5
Co5+ = [Ar]3d4
[Ti(NH3)6]3+ is +3, thus it contains 1 unpaired electron.
Hence, [Co(OX)2(OH)2]- has highest paramagnetic behaviour.
But en , bby and NH3 all are strong field ligands, so pairing occurs, thus no unpaired electrons.
[Co(OX)2 (OH)2]- is
Oxidation state of Co in [Co(OX)2(OH)2]- is
x + (-2) x 2 + (-1) x 2 = -1
x-6 = -1
x = +5
Co5+ = [Ar]3d4
[Ti(NH3)6]3+ is +3, thus it contains 1 unpaired electron.
Hence, [Co(OX)2(OH)2]- has highest paramagnetic behaviour.
