The electronic configuration of V(23)=[Ar]4s2,3d3 Let in [V(gly)2(OH)2(NH3)2]+oxidation state of V is x. x+(−1)×2+(−1)2+(0×2)=+1 x=+5 V5+=[Ar]4s0,3d0 (no unpaired electrons)
The electronic configuration of Fe(26)=[Ar]4s2,3d6 Let the oxidation state of [Fe(en)(ppy)(NH3)2]2+ is x. [x+(0)+(0)+(0)×2]=+2 x=+2 Fe2+=[Ar],3d6(∵4 unpaired electron) but, bpy, en and NH3 all are strong field ligands, so pairing occurs and thus, Fe2+ contains no unpaired electron
The electronic configuration of Co(27)=[Ar]4s2,3d7 Let the oxidation state Co in [Co(ox)2(OH)2]−is x x+(−2)×2+(−1)×2=−1 x=+5 Co5+=[Ar].3d4[4 unpaired electrons] ox and OH are weak field ligands, thus pairing of electron units does not occur.
The electronic configuration of Ti(22)=[Ar]4s2,3d2 Oxidation state of Ti in [Ti(NH3)6]3+ is 3 . Ti3+=[Ar]3d1 (one unpaired electron)
Hence, complex [Co(ox)2(OH)2]−has maximum number of unpaired electrons, thus show maximum paramagnetism