Question

What type of hybridisation is present in the square planar geometry of the complex \( [\text{Ni}(\text{CN})_4]^{2-} \)?

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

Hint:

Strong field ligands like \(\text{CN}^-\) often produce low-spin complexes. Square planar geometry usually corresponds to \(\text{dsp}^2\) hybridisation.

Answer 1

The Correct Option is B

Concept:
Square planar complexes generally show \(\text{dsp}^2\) hybridisation. In \([\text{Ni}(\text{CN})_4]^{2-}\), cyanide ion is a strong field ligand and causes pairing of electrons.

Step 1: Find oxidation state of Ni.
Let oxidation state of Ni be \(x\). \[ x + 4(-1) = -2 \] \[ x - 4 = -2 \] \[ x = +2 \] So Ni is \(\text{Ni}^{2+}\).

Step 2: Write electronic configuration of \(\text{Ni}^{2+}\).
Nickel atom: \[ \text{Ni} = [Ar]\,3d^8 4s^2 \] So, \[ \text{Ni}^{2+} = [Ar]\,3d^8 \]

Step 3: Use ligand strength.
\(\text{CN}^-\) is a strong field ligand, so it causes pairing of electrons and gives square planar geometry. Square planar geometry corresponds to: \[ \text{dsp}^2 \] Hence, the correct answer is:
\[ \boxed{(B)\ \text{dsp}^2} \]