Question

Based on valence bond theory, predict the magnetic behaviour of the [Ni(NH₃)₄]²⁺ complex.

Answer 1

1. In the [Ni(NH₃)₄]²⁺ complex, nickel is in the +2 oxidation state. This gives the electron configuration of Ni²⁺ as [Ar] 3d⁸.
2. Ammonia (NH₃) is a strong field ligand, meaning it can cause pairing of the d-electrons of the metal ion.
3. The coordination number of nickel in [Ni(NH₃)₄]²⁺ is 4, which suggests the complex adopts a square planar geometry.
4. Since ammonia (NH₃) is a neutral ligand and a strong field ligand (causing electron pairing), we expect that the hybridisation of the Ni²⁺ ion will involve the d, s, and p orbitals.
5. The hybridisation is d²sp^2, where two d-orbitals, one s-orbital, and two p-orbitals are involved in the bonding with the four NH₃ ligands.
6. This results in a square planar geometry.
7. The electron configuration of Ni²⁺ is 3d⁸. In the presence of strong field ligands (NH₃), the d-orbitals undergo splitting.
8. In a square planar geometry, the d-orbitals split into two sets:
   t_2g​ (lower energy, three orbitals)
   e_g​ (higher energy, two orbitals).
9. The strong field nature of ammonia (NH₃) will cause pairing of the electrons in the lower-energy t_2g orbitals. In this case, the Ni²⁺ ion has 8 electrons in the 3d-orbitals, and all of them will be paired.
10. Since all the 3d-electrons are paired, the complex has no unpaired electrons.

Therefore, [Ni(NH₃)₄]²⁺ is diamagnetic because there are no unpaired electrons in the complex.