Question

Using the valence bond approach, predict the shape, hybridisation and magnetic behaviour of [Ni(CO)₄]. (at. no. of Ni = 28)

Answer 1

1. The atomic number of nickel is 28, and its ground-state electronic configuration is [Ar] 3d⁸4s².
2. In the complex [Ni(CO)₄], nickel is in the zero oxidation state because CO is a neutral ligand.
3. The electronic configuration of Ni in this state is 3d⁸4s².
4. Carbon monoxide (CO) is a strong field ligand according to the spectrochemical series.
5. Being a strong field ligand, CO causes the pairing of electrons in the 3d orbitals of nickel.
6. As a result, the 3d orbitals become filled, while the 4s and 4p orbitals remain vacant.
7. The vacant orbitals (one 4s and three 4p) undergo hybridisation to form four equivalent sp³ hybrid orbitals.
8. Each of these hybrid orbitals accepts a lone pair of electrons from CO ligands through coordinate bonds.
9. Due to sp³ hybridisation, the geometry of the complex is tetrahedral.
10. All the 3d electrons are paired, the complex has no unpaired electrons.

Therefore, [Ni(CO)₄] is diamagnetic.

Answer 2

Since CO is a neutral ligand, the oxidation state of Ni in [Ni(CO)₄] is 0. Ni has an oxidation state of 0 in [Ni(CO)₄], and its configuration is 3d⁸4S². As the electrons pair together inside the orbital, making CO a strong field ligand. As a result, the ligand-CO’s incoming electron pairs fill the remaining ‘s’ and ‘p’ orbitals, filling all five d-orbitals (2 in ‘s’ and 6 in ‘p’). Since there are four CO atoms in the complex, -CO fills the one’s orbital and the three ‘p’ orbitals. Because there are no unpaired electrons, the hybridization is sp³, and the geometry is tetrahedral, making it diamagnetic.