Question

What is the hybridisation of \([Ni(CN)_4]^{2-}\)?

• A. sp\(^3\)
• B. sp\(^2\)d
• C. dsp\(^2\)
• D. sp\(^3\)d

Hint:

Square planar complexes typically exhibit dsp\(^2\) hybridization and are commonly seen in metal complexes involving transition metals such as nickel, platinum, and palladium.

Answer 1

The Correct Option is C

To determine the hybridization of \([Ni(CN)_4]^{2-}\), we first consider the electron configuration of the central metal ion, nickel (Ni). The atomic number of Ni is 28, giving it the ground-state electron configuration of [Ar] 3d⁸ 4s². In this complex, Ni is in the +2 oxidation state, leading to the loss of 2 electrons from the 4s orbital: [Ar] 3d⁸.

We then analyze the ligand field created by the ligands. Cyanide (CN^-) is a strong field ligand. Strong field ligands cause pairing of the 3d electrons:

• The 3d subshell: (↑↓)(↑↓)(↑↓)()(↑)

Next, we evaluate the coordination number and hybridization:

1. With four CN^- ligands, the coordination number is 4.
2. Strong field ligands like CN^- result in dsp² hybridization due to the pairing and involvement of the inner 3d orbital.

The unpaired 3d electrons result in the hybridization involving the d, s, and two p orbitals, forming dsp² hybrid orbitals.

Therefore, the hybridization of \([Ni(CN)_4]^{2-}\) is dsp².

Option	Hybridization
sp3	Incorrect
sp2d	Incorrect
dsp2	Correct
sp3d	Incorrect