Question

Based on valence bond theory, predict the magnetic behaviour of the [Fe(H₂O)₆]²⁺ complex.

Answer 1

1. The complex [Fe(H₂O)₆]²⁺ contains water (H₂O) as a neutral ligand. Since the complex has an overall charge of +2, the oxidation state of iron in this complex is +2.
2. Iron has an atomic number of 26, so the electron configuration of neutral iron (Fe) is:
   Fe : [Ar] 3d⁶ 4s²
3. For Fe²⁺, it loses two electrons, both from the 4s orbital:
   Fe²⁺ : [Ar] 3d⁶
4. Fe has 6 electrons in the 3d-orbitals.
5. In the complex [Fe(H₂O)₆]²⁺, the coordination number of Fe²⁺ is 6, as it is surrounded by six water molecules.
6. A coordination number of 6 typically results in an octahedral geometry for the complex.
7. Since water (H₂O) is a neutral ligand and a weak field ligand, it does not cause significant electron pairing.
8. The hybridisation of Fe²⁺ in this case is d²sp³, involving two d-orbitals, one s-orbital, and three p-orbitals, forming six hybrid orbitals.
9. These six hybrid orbitals form bonds with the six water ligands, resulting in an octahedral geometry.
10. The water is a weak field ligand, Fe²⁺ will exhibit a high-spin configuration, where electrons are more likely to remain unpaired to maximize the number of unpaired electrons.
11. The complex is paramagnetic because it has 4 unpaired electrons in the Fe²⁺ ion.

Thus, [Fe(H₂O)₆]²⁺ is paramagnetic because it has unpaired electrons.