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 \(d^2sp^3\), signifying octahedral geometry. Iron is in the +3 oxidation state. For \([ \text{Fe(CN)}_6]^{4-}\), hybridization is also \(d^2sp^3\), but the stronger ligand CN\(^-\) causes electron pairing in the \(d\)-orbitals.
(ii) \([ \text{FeF}_6]^{3-}\) is classified as an outer orbital complex, utilizing iron's \(4d\) orbitals. Conversely, \([ \text{Fe(CN)}_6]^{4-}\) is an inner orbital complex, employing the \(3d\) orbitals.
(iii) Paramagnetism in \([ \text{FeF}_6]^{3-}\) stems from unpaired electrons in its \(d\)-orbitals. In contrast, \([ \text{Fe(CN)}_6]^{4-}\) is diamagnetic, as the CN\(^-\) ligand induces a low-spin configuration where all electrons are paired.