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: Count Ligands:
Four chloride ligands means coordination number 4. That points to either $sp^3$ or $dsp^2$.

Step 2: Choose:
Cl is a weak ligand and does not pair up the 3d$^8$ electrons, so $dsp^2$ is not used. The complex is tetrahedral with $sp^3$.

Step 3: Rule Out:
Options (B) and (C) are octahedral types. So (A).

Final Answer:
Option (A). \[ \boxed{\text{(A) } sp^3} \]