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 the +II oxidation state.

Fe²⁺ : Electronic configuration is 3d⁶

Orbitals of Fe²⁺ ion:

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

Since there are six ligands around the central metal ion, the most feasible hybridization is d²sp³.

d²sp³ hybridized orbitals of Fe²⁺ are:

6 electron pairs from CN⁻ ions occupy the six hybrid d²sp³orbitals.

Then,

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

(ii) [FeF₆]³⁻

In this complex, the oxidation state of Fe is +3.

Orbitals of Fe⁺³ ion:

There are 6 F⁻ ions. Thus, it will undergo d²sp³ or sp³d² hybridization. As F⁻ is a weak field ligand, it does not cause the pairing of the electrons in the 3d orbital. Hence, the most feasible hybridization is sp³d².

sp³d² hybridized orbitals of Fe are:

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

(iii) [Co(C₂O₄)₃]³⁻

Cobalt exists in the +3 oxidation state in the given complex.

Orbitals of Co³⁺ ion:

Oxalate is a weak field ligand. Therefore, it cannot cause the pairing of the 3d orbital electrons. As there are 6 ligands, hybridization has to be either sp³d² or d²sp³ hybridization.

sp³d² hybridization of Co^3+:

The 6 electron pairs from the 3 oxalate ions (oxalate anion is a bidentate ligand) occupy these sp³d² orbitals.

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

(iv) [CoF₆]³⁻

Cobalt exists in the +3 oxidation state.

Orbitals of Co³⁺ ion:

Again, fluoride ion is a weak field ligand. It cannot cause the pairing of the 3d electrons. As a result, the Co³⁺ ion will undergo sp³d² hybridization.

sp³d² hybridized orbitals of Co³⁺ ion are:

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