The electron configuration of

V (23) = [Ar], 4s 2 , 3d 3

Let in [V(gly) 2 (OH) 2 (NH 3 ) 2 ] + oxidation state of V is x.

x + (– 1) × 2 (– 1)2 + (0 × 2) = + 1

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

[Fe(en)(ppy)(NH 3 ) 2 ] 2+ is x.

[x + (0) + (0) + (0) × 2] = + 2

x = + 2

Fe 2+ = [Ar] 4s 0 , 3d 6 (  no 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] 4s 2 , 3d 7

Oxidation state of Co in [Co (OX) 2 (OH) 2 ] –

x + (– 2) × 2 + (– 1) × 2 = – 1

x = + 5

Co 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 [Ti(NH 3 ) 6 ] 3+ is 3.

Ti 3+ = [Ar], 3d 1 (one unpaired electron)

Hence, complex [Co(OX) 2 (OH) 2 ] – has maximum number of unpaired electrons, thus show maximum paramagnetic.