Question

Using the valence bond approach, predict the shape and magnetic character of [Fe(CN)6]³⁻ ion.

Answer 1

1. The cyanide (CN⁻) ligand is a monodentate ligand with a charge of −1.
2. The overall charge of the complex is 3−, so the oxidation state of iron in the [Fe(CN)₆]³⁻ complex is +3.
3. Iron has an atomic number of 26, so the electron configuration of neutral Fe is:
   Fe : [Ar] 3d⁶ 4s²
4. For Fe³⁺, it loses three electrons. The electrons are removed first from the 4s-orbitals and then from the 3d-orbitals:
   Fe³⁺ : [Ar] 3d⁶
5. In the complex [Fe(CN)₆]³⁻, the coordination number of Fe³⁺ is 6, as it is surrounded by six cyanide (CN⁻) ligands.
6. A coordination number of 6 suggests that the complex will adopt an octahedral geometry.
   
7. Cyanide (CN⁻) is a strong field ligand, meaning it will cause pairing of electrons in the metal ion’s d-orbitals.
8. To form the bonds with the six cyanide ligands, Fe³⁺ undergoes d²sp³ hybridisation, which involves the mixing of two d-orbitals, one s-orbital, and three p-orbitals from the Fe³⁺ ion to form six hybrid orbitals.
9. The cyanide (CN⁻) is a strong field ligand, it causes pairing of the electrons in the lower-energy t_2g​ orbitals. In this case, the five electrons in the Fe³⁺ ion will all be paired in the lower-energy t_2g​ orbitals.
10. With all the electrons paired, there are no unpaired electrons in the Fe³⁺ ion.

Thus, the complex [Fe(CN)₆]³⁻ is diamagnetic because there are no unpaired electrons.