Question

\( [NiCl_4]^{2-} \) is paramagnetic while \( [Ni(CO)_4] \) is diamagnetic though both are tetrahedral. Why? [At. No. Ni=28]

Hint:

Weak field ligands usually produce high spin complexes; strong field ligands produce low spin complexes.

Answer 1

The difference in magnetic behaviour between \( [NiCl_4]^{2-} \) and \( [Ni(CO)_4] \) arises from the nature of the ligands and the resulting electronic configurations of nickel in both complexes.
1. For \( [NiCl_4]^{2-} \):
- The atomic number of nickel (Ni) is 28, so its ground state configuration is \( [Ar] 3d^8 4s^2 \). In the +2 oxidation state (as in \( [NiCl_4]^{2-} \)), Ni loses two electrons, and its configuration becomes \( 3d^8 \).
- In a tetrahedral crystal field, the d-orbitals split into two sets: the lower energy \( e \) set and the higher energy \( t_2 \) set. For \( [NiCl_4]^{2-} \), chloride (\( Cl^- \)) is a weak field ligand, so it does not cause significant splitting of the d-orbitals. As a result, the two electrons in the \( 3d^8 \) configuration remain unpaired, which results in the complex being paramagnetic.
2. For \( [Ni(CO)_4] \):
- In \( [Ni(CO)_4] \), Ni is also in the +2 oxidation state, so its configuration is \( 3d^8 \), similar to \( [NiCl_4]^{2-} \). However, CO is a strong field ligand, and it causes a larger splitting of the d-orbitals in the tetrahedral field.
- CO causes the pairing of electrons in the lower-energy \( t_2 \) orbitals, resulting in all the d-electrons being paired. As a result, \( [Ni(CO)_4] \) is diamagnetic because there are no unpaired electrons in the complex.
In summary, the difference in magnetic behaviour is due to the strength of the ligands. Chloride (\( Cl^- \)) is a weak field ligand, leading to unpaired electrons and paramagnetism in \( [NiCl_4]^{2-} \), while carbon monoxide (CO) is a strong field ligand, causing electron pairing and diamagnetism in \( [Ni(CO)_4] \).