Question

c). [CoF6] 3 is paragnetic whereas [ Co(NH3)6] 3+ is diamagnetic explain on the basis of V.B theory

Answer 1

In [CoF6]3- configuration of Co(III) is 3d6.
It has 4 unpaired electrons in its d-orbitals.

Though every ligand prefers d2sp3 hybridisation over sp3d2 because of stability reasons, but F- is a weak ligand and cannot force pairing of unpaired electrons in the d-orbitals of Co(III). So no vacant d-orbitals are available for d2sp3 and it’s only option is to go with sp3d2 hybridisation. As you can see, the 4 unpaired electrons remain as it is and give [CoF6]3- a ‘PARAMAGNETIC’ nature.



Next, in [Co(NH3)6]3+ also Co(III) has 3d6 configuration, meaning 4 unpaired electrons.

But, here NH3 is a strong ligand and is capable of forcing the unpaired electrons to pair up. So two vacant 3d orbitals become available and resulting hybridisation is d2sp3. As you can see due to pairing up NO unpaired electrons remain in the d-orbitals, and hence [Co(NH3)6]3+ is ‘DIAMAGNETIC’.

Answer 2

Fluorine is a weak field ligands but ammonia is a strong field ligands.

• In $[CoF_6]^{+3}$ Co has an oxidation state +3. So, Co has 6 valance electrons.
• As F is a weak field ligands so, it can not make a pairing. So, it has one unpaired electron and is paramagnetic.
• In $[Co(NH_3)_6]^{+3}$ Co has an oxidation state +3. So, Co has 6 valance electrons.
• As ammonia is a strong field ligands so, it can make a pairing. So, it has no unpaired electron and is diamagnetic.