Question

According to Valence Bond Theory, the hybridization of the central metal ion in [Ni(CN)$_4$]$^{2-}$ is:

• A. sp$^3$
• B. dsp$^2$
• C. d$^2$sp$^3$
• D. sp$^2$

Hint:

For complexes of coordination number 4:
If the metal is $\text{Ni}^{2+}$ ($d^8$ configuration) and the ligand is a strong-field ligand (like $\text{CN}^-$, CO), pairing occurs, resulting in $\text{dsp}^2$ (square planar, diamagnetic).
If the ligand is a weak-field ligand (like $\text{Cl}^-$, $\text{F}^-$), no pairing occurs, resulting in $\text{sp}^3$ (tetrahedral, paramagnetic).

Answer 1

The Correct Option is B

Step 1: Ni oxidation state.
Ni is +2, config $3d^8$. CN$^-$ is strong-field; forces pairing, leaving 2 empty 3d orbitals.
Step 2: Geometry.
Square planar + 4 ligands = $dsp^2$ hybridization.
Answer: $dsp^2$