Question

Answer the following about the complexes \([ \text{FeF}_6]^{3-} \text{ and } [\text{Fe(CN)}_6]^{4-}\):
(i) Write the hybridization involved in each case.
(ii) Which of them is the outer orbital complex and which one is the inner orbital complex?
(iii) Compare their magnetic behaviour.
[Atomic number: Fe = 26]

Hint:

When studying complexes, remember that strong field ligands, such as CN\(^-\), lead to low-spin complexes with paired electrons, while weak field ligands, like F\(^-\), result in high-spin complexes with unpaired electrons.

Answer 1

(i) The hybridization of \([ \text{FeF}_6]^{3-}\) is \(d^2sp^3\), indicating an octahedral geometry, with iron in the +3 oxidation state. In the case of \([ \text{Fe(CN)}_6]^{4-}\), the hybridization is also \(d^2sp^3\), but the stronger field ligand CN\(^-\) leads to pairing of electrons in the \(d\)-orbitals.
(ii) \([ \text{FeF}_6]^{3-}\) is an outer orbital complex because it uses the \(4d\) orbitals of iron, whereas \([ \text{Fe(CN)}_6]^{4-}\) is an inner orbital complex, using the \(3d\) orbitals.
(iii) \([ \text{FeF}_6]^{3-}\) is paramagnetic due to the presence of unpaired electrons in the \(d\)-orbitals, whereas \([ \text{Fe(CN)}_6]^{4-}\) is diamagnetic because all electrons are paired in the low-spin configuration induced by the CN\(^-\) ligand.