Question

Hybridisation of \(\mathrm{Ni}\) in \(\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-\) is:}

• A. \( s p^3 \)
• B. \( d s p^2 \)
• C. \( s p \)
• D. \( s p^3 d \)

Hint:

To determine hybridization in coordination compounds:
- Find the oxidation state and electron configuration of the metal.
- Check the coordination number and ligand strength.
- Use magnetic properties (paramagnetic vs. diamagnetic) to confirm geometry.
- Square planar (\(dsp^2\)) is common for \(d^8\) metals with strong-field ligands.

Answer 1

The Correct Option is B

To determine the hybridization of nickel in \(\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}\), we need to find the oxidation state of nickel, the number of ligands, the geometry of the complex, and the electronic configuration.
-
Step 1: Oxidation state of Ni
The complex is \(\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}\). Cyanide (\(\mathrm{CN}^-\)) has a charge of \(-1\). With four \(\mathrm{CN}^-\) ligands, their total charge is \(4 \times (-1) = -4\). The overall charge of the complex is \(-2\). Let the oxidation state of Ni be \( x \): \[ x + (-4) = -2 \quad \Rightarrow \quad x - 4 = -2 \quad \Rightarrow \quad x = +2 \] So, Ni is in the \(+2\) oxidation state (\(\mathrm{Ni}^{2+}\)). -
Step 2: Electronic configuration of \(\mathrm{Ni}^{2+}\)
Nickel’s atomic number is 28, with a ground state configuration of \([Ar] 3d^8 4s^2\). For \(\mathrm{Ni}^{2+}\), the two electrons are removed from the \(4s\) orbital: \[ \mathrm{Ni}^{2+}: [Ar] 3d^8 \] This gives eight electrons in the \(3d\) orbitals. -
Step 3: Ligand and coordination number
The complex has four \(\mathrm{CN}^-\) ligands, so the coordination number is 4. \(\mathrm{CN}^-\) is a strong-field ligand due to its position in the spectrochemical series, which causes significant splitting of the \(d\) orbitals.
-
Step 4: Geometry and hybridization
For coordination number 4, possible geometries are tetrahedral (\(sp^3\)) or square planar (\(dsp^2\)). To determine the geometry, we consider the magnetic properties and ligand strength: - \(\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}\) is known to be diamagnetic (no unpaired electrons), as confirmed by experimental data. - For \(\mathrm{Ni}^{2+}\) (\(d^8\)), in a tetrahedral field (weak-field, \(sp^3\)), the electrons occupy the \(e\) and \(t_2\) orbitals. For \(d^8\), the configuration is typically \(e^4 t_2^4\), which has two unpaired electrons, making it paramagnetic. Since the complex is diamagnetic, tetrahedral geometry (\(sp^3\)) is ruled out.
- In a square planar field (strong-field, \(dsp^2\)), the \(d\) orbitals split into \(d_{x^2-y^2}\), \(d_{z^2}\), \(d_{xy}\), and \(d_{xz}/d_{yz}\). For a strong-field ligand like \(\mathrm{CN}^-\), the \(d^8\) electrons fill the lower-energy orbitals (e.g., \(d_{xz}\), \(d_{yz}\), \(d_{xy}\), \(d_{z^2}\)) completely, leaving no unpaired electrons, consistent with diamagnetism. The square planar geometry uses one \(d\) orbital (usually \(3d_{x^2-y^2}\)), one \(4s\), and two \(4p\) orbitals, giving \(dsp^2\) hybridization.
-
Step 5: Confirm hybridization
The options are:
- \(sp^3\): Tetrahedral, paramagnetic for \(d^8\), incorrect.
- \(dsp^2\): Square planar, diamagnetic with strong-field ligands, correct.
- \(sp\): Linear geometry, not possible for coordination number 4.
- \(sp^3d\): Not typical for coordination number 4; often for higher coordination numbers.
Thus, the hybridization is \(dsp^2\), corresponding to a square planar geometry.