Question

Discuss the nature of bonding in the following coordination entities on the basis of valence bond theory:
(i) [Fe(CN)₆]^4-
(ii) [FeF₆]^3-
(iii) [CO(C₂O₄)₃]^3-
(iv) [COF₆]^3-

Answer 1

(i) [Fe(CN)₆]^4-
In the above coordination complex. Iron exists in +2 oxidation state.
Fe²⁺: electronic configuration is 3d⁶ 

As CN^- is a strong field ligand, it causes pairing of the unpaired 3d electrons.

Since there are six ligands around the central metal ion, the most feasible hybridisation is d²sp³ 6 electrons pairs from CN^- ions occupy the six hybrid d²sp³ orbitals.

Hence, the geometry of the complex is octahedral and the complex is diamagnetic (as there are no unpaired electrons).

(ii) [F_eF₆]^3-
Oxidation state of Fe is +3
Orbitals of Fe⁺³ ion.

There are 6F^- ions. Thus, it will undergo d²Sp³ or sp³d² hybridisation. As F^- is a weak ligand, it does not cause pairing of electrons in the 3d orbitals. Hence the most feasible hybridisation is Sp³d².

Hence the geometry of the complex is found to be octahedral.

(iii) [CO(C₂O₄)₃]^3-

Cobalt exists +3 oxidation state in the complex.

Oxalate is a weak field ligand. Therefore, it cannot cause pairing of the 3d orbital electrons.

As there are 6 ligands, hybridisation has to be either Sp³d² or d²Sp³ hybridization.

As there are no pairing of electrons Sp³d² is the most stable hybridisation.

Oxalate being a bidentate ligand donates 2 pairs of electrons each.

Hence, the geometry of the complex is found to be octahedral.

(iv) [COF₆]^3-
cobalt exists in the +3 oxidation state.

Again, fluoride ion is a weak ligand. It cannot cause the pairing of the 3d electrons. Sp³d² hybridised orbitals of CO³⁺ ion are

Hence, the geometry of the complex is octahedral and paramagnetic.