Question

Hybridisation and geometry of [Ni(CN)$_4$]$^{2-}$ are
 

• A. sp$^3$ and tetrahedral
• B. sp$^3$ and square planar
• C. sp$^3$ and tetrahedral
• D. dsp$^2$ and square planar

Hint:

Strong field ligands cause electron pairing and favor square planar geometries in transition metal complexes.

Answer 1

The Correct Option is D

Step 1: Identify the Metal Ion
Nickel in [Ni(CN)$_4$]$^{2-}$ has an oxidation state of \( +2 \).
Its electronic configuration: \( 3d^8 4s^0 \).
Step 2: Effect of Ligand
CN$^-$ is a strong field ligand, which causes pairing of electrons in the \( d \)-orbitals.
This leads to dsp$^2$ hybridization, forming a square planar geometry.
Final Answer: [Ni(CN)$_4$]$^{2-}$ undergoes dsp$^2$ hybridization, leading to a square planar shape.

Answer 2

The complex [Ni(CN)₄]^2- is analyzed to determine its hybridization and geometry. Nickel (Ni) in this complex is in the +2 oxidation state.

In the formation of this complex, the electronic configuration of Ni in its ground state is [Ar] 3d⁸ 4s². For Ni²⁺, it becomes [Ar] 3d⁸ as it loses two electrons from the 4s orbital.

Considering the strong field ligand cyanide (CN^-), it causes pairing of the electrons in the 3d orbitals, resulting in an electron configuration for Ni²⁺ where the 3d orbitals are fully filled except one which is unoccupied.

Here, hybridization involves the 3d, 4s, and two 4p orbitals of nickel, resulting in dsp² hybridization. Four CN^- ligands form coordinate bonds by donating electron pairs to the hybridized orbitals.

This dsp² hybridization leads to a square planar geometry for the complex.

Hybridization	Geometry
dsp2	Square Planar