Question

The hybridization involved in complex $ {[Ni(CN)4]^{2-}}$ is (At.No. Ni = 28)

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

Answer 1

The Correct Option is C

To determine the hybridization involved in the complex \([Ni(CN)_4]^{2-}\), we must follow a systematic approach considering the electronic configuration of the central metal atom and the ligand involved.

Step 1: Determine the oxidation state of Nickel:

• The complex is \([Ni(CN)_4]^{2-}\).
• Let the oxidation state of Nickel be \(x\).
• Each CN ligand carries a charge of \(-1\). Therefore, for four CN ligands, the total negative charge is \(-4\).
• The overall charge on the complex is \(-2\).
• The equation for oxidation state is: x + 4(-1) = -2 。
• Solving for \(x\), we get \(x = +2\).

Step 2: Determine the electronic configuration of Ni\(^{2+}\):

• Neutral Nickel (\(Ni\)) has an atomic number of 28, hence its electronic configuration is \text{[Ar]} 3d^{8} 4s^2.
• For \(Ni^{2+}\), we remove two electrons from the 4s orbital: \text{[Ar]} 3d^{8} 4s^{0}.

Step 3: Determine the hybridization in the presence of cyanide ligands:

• CN\(^-\) is a strong field ligand, which causes pairing of electrons in the \(\text{3d orbital}\).
• This pairing results in the configuration: \text{[Ar]} \, 3d^{10} 4s^0 4p^0 with all 3d electrons paired.
• To form four coordination bonds, the electron orbitals used are 4s, 4p, and one 3d orbital:
• The hybridization is thus \(dsp^2\).

Conclusion:

The hybridization of the complex \([Ni(CN)_4]^{2-}\) is dsp^2. This means the complex forms a square planar geometry.