Question

Hybridization of Ni in ([Ni(CN)$_4$]$^{2-}$): ____.

• A. sp³
• B. dsp²
• C. sp²
• D. d²sp³

Hint:

Contrast this with $[NiCl_4]^{2-}$. Since Cl⁻ is a weak field ligand, no pairing occurs, the hybridization is sp³, and the geometry is tetrahedral.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The hybridization and geometry of coordination complexes depend on the oxidation state of the central metal and the strength of the surrounding ligands (Crystal Field Theory).

Step 2: Detailed Explanation:
1. Oxidation State: In $[Ni(CN)_4]^{2-}$, Cyanide (CN⁻) is -1. So, $x + 4(-1) = -2 \Rightarrow x = +2$. Ni is in the +2 state. 2. Electronic Configuration: $Ni^{2+}$ is $3d^8 4s^0$. 3. Ligand Strength: CN⁻ is a strong field ligand. It forces the electrons in the $3d$ subshell to pair up. 4. Result: After pairing the 8 electrons in $3d$, one $3d$ orbital becomes vacant. This vacant $d$ orbital, along with one $4s$ and two $4p$ orbitals, hybridizes to form four dsp² hybrid orbitals. 5. This leads to a square planar geometry and a diamagnetic complex.

Step 3: Final Answer:
The hybridization of Ni in $[Ni(CN)_4]^{2-}$ is dsp².