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

Step 1: Understanding the Concept:
The hybridisation of a central metal ion in a coordination complex is determined by its coordination number and the nature of the ligands (strong field or weak field) according to Valence Bond Theory.
A square planar geometry specifically corresponds to a certain type of hybridisation.
Step 2: Key Formula or Approach:
Determine the oxidation state and electron configuration of the central metal ion.
Apply Valence Bond Theory considering the strength of the ligand (strong field ligands cause pairing of electrons) to find the available orbitals for hybridisation.
Step 3: Detailed Explanation:
Let's analyze the complex ion \([\text{Ni}(\text{CN})_4]^{2-}\).
First, find the oxidation state of Nickel (\(\text{Ni}\)): Let it be \(x\). The cyanide ligand (\(\text{CN}^-\)) has a charge of \(-1\).
\[ x + 4(-1) = -2 \] \[ x = +2 \] The electronic configuration of neutral \(\text{Ni}\) (Atomic number 28) is \([\text{Ar}] 3\text{d}^8 4\text{s}^2\).
The electronic configuration of \(\text{Ni}^{2+}\) is \([\text{Ar}] 3\text{d}^8\).
The five 3d orbitals contain 8 electrons. According to Hund's rule, there are 3 fully filled orbitals and 2 half-filled orbitals.
Next, consider the nature of the ligand. Cyanide (\(\text{CN}^-\)) is a strong field ligand.
Strong field ligands cause the pairing of unpaired d-electrons against Hund's rule.
Therefore, the two unpaired electrons in the 3d orbitals pair up, emptying one 3d orbital (specifically the \(d_{x^2-y^2}\) orbital).
Now, the central metal ion needs four empty orbitals to accept electron pairs from the four \(\text{CN}^-\) ligands.
It utilizes the one emptied \(3\text{d}\) orbital, the empty \(4\text{s}\) orbital, and two empty \(4\text{p}\) orbitals.
These orbitals mix to form four equivalent hybrid orbitals.
The resulting hybridisation is one \(d\), one \(s\), and two \(p\) orbitals, which is written as \(\text{dsp}^2\).
The geometry associated with \(\text{dsp}^2\) hybridisation is square planar.
Step 4: Final Answer:
The hybridisation present is \(\text{dsp}^2\).