Question

On the basis of crystal field theory explain why $Co\left(III\right)$ forms paramagnetic octahedral complex with weak field ligands whereas it forms diamagnetic octahedral complex with strong field ligands.

Answer 1

Electronic configuration of $Co=\left[Ar\right]3d^{7}4s^2$

and electronic configuration of $C{o}^{3+}=\left[Ar\right]3d^{6}4s^0$

With weak field ligand ${\Delta }_0<p$ (pairing energy) so, the electronic configuration of $Co\left(III\right)$ will be $t_{2g}^4{e}_g^2$ i.e., it will have four unpaired electrons hence will be paramagnetic.

Octahedral splitting in presence of strong field ligand

With strong field ligand ${\Delta }_0<p$ so, pairing of electrons occur thus, the electronic configuration will be $t_{2g}^6{e}_g^0$ i.e., it will have no unpaired electron hence will be diamagnetic.