Please solve This Question sirCoordinaton compound question Once asked in aipmt also

[V(gly)2 (OH)2(NH3)2]+ V23 = [Ar] 4s2, 3d3 Oxidation state of V in [V(gly)2 (OH)2(NH3)2]+  is

x + (-l)x2 + (-l)x2 + (0)x2 = +1

x =+5

Vs+ =[Ar]3d° (No unpaired electron)

[Fe(en)(bpy)(NH3)2)2+  Fe26 =[Ar]4s2, 3d6

Oxidation state of Fe in [Fe(en) (bpy) (NH3)2]2+ is

x + (0) + (0) + (0) x 2 = +2

x=+2

Fe2+ = [Ar]3d6

But en, bpy and NH3 all are strong field ligands, so pairing occurs, thus no unpaired electrons. [Co(OX)2(OH)2]-

CO27 = [Ar]4s2, 3d7 

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 (4 unpaired electrons)

[Ti(NH3)6]3+

Ti22 =[Ar]4s2, 3d2

Oxidation state of Ti in [Ti(NH3)6]3+ is +3, thus it contains 1 unpaired electron.

Hence [Co(OX)2(OH)2]- has highest paramagnetic behaviour.