Question

[NiCl₄]²⁻ is paramagnetic, while [Ni(CO)₄] is diamagnetic, though both are tetrahedral. Why? (Atomic number of Ni = 28)

Why is [NiCl₄]^2– paramagnetic but [Ni(CO)₄] is diamagnetic ? (At. nos. : Cr = 24, Co = 27, Ni = 28)

Answer 1

In [NiCl₄]²⁻, Ni is in the +2 state. Cl⁻ is a ligand which is a weak field ligand which does not cause pairing of unpaired 3d electrons. Hence, it is paramagnetic. In [Ni(CO)₄], Ni has 0 oxidation state. CO is a strong field ligand, which causes pairing of unpaired 3d electrons. No unpaired electrons are present in this case. So, it is diamagnetic.

Answer 2

In [NiCl₄]²⁻, the oxidation state of Ni is +2. Chloride is a weak field ligand and does not cause pairing up of electrons against the Hund's rule of maximum multiplicity. As a result, two unpaired electrons are present in the valence d-orbitals of Ni which impart paramagnetic character to the complex. On the other hand, carbonyl is a strong field ligand and causes pairing up of electrons against the Hund's rule of maximum multiplicity. As a result, no unpaired electrons are present and hence, the complex is diamagnetic.

Answer 3

[NiCl₄]²⁻ and [Ni(CO)₄] both are tetrahedral. But their magnetic characters are different. This is due to difference in the nature of ligands.

Ni⁺² = [Ar] 4s⁰3d⁸

Ni⁺² has 2 unpaired electrons hence, this complex is paramagnetic.

In Ni(CO)₄, Ni is in the zero oxidation state i.e., it has a configuration of 3d⁸4s².

[Ni(CO)4]

Ni = [Ar] 4s² 3d⁸

But CO is a strong field ligand. Therefore, it causes the pairing of unpaired 3d electrons. Also, it causes the 4s electrons to shift to the 3d orbital, thereby giving rise to sp³ hybridization. Since, no unpaired electrons are present in this case, [Ni(CO)₄] is diamagnetic.