Question

What type of hybridization is found in \([\text{Ni}(\text{Cl})_4]^{2-}\)?

• A. \(sp^3\)
• B. \(sp^3d^2\)
• C. \(d^2sp^3\)
• D. \(dsp^2\)

Hint:

Chloride is a weak field ligand, so Ni(II) with Cl forms a tetrahedral sp3 complex.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The hybridisation of a complex depends on the metal electron configuration and the strength of the ligand field.

Step 2: Find the Oxidation State:
In \([\text{NiCl}_4]^{2-}\), \(x - 4 = -2\), so Ni is \(+2\) and has the configuration \(3d^8\).

Step 3: Ligand Field:
Chloride is a weak field ligand. It does not force the d electrons to pair, so the 3d orbitals stay occupied and the 4s and 4p orbitals are used.

Step 4: Hybridisation:
One 4s and three 4p orbitals mix to give four \(sp^3\) orbitals, so the shape is tetrahedral and the complex is paramagnetic with two unpaired electrons.
A \(dsp^2\) square planar shape would need strong field ligands such as \(\text{CN}^-\). The \(sp^3d^2\) and \(d^2sp^3\) types give six-coordinate octahedral shapes, which need six ligands.

Final Answer:
The complex is tetrahedral with \(sp^3\) hybridisation, option (A). \[ \boxed{\text{(A) } sp^3} \]