Is d2sp hybridisation possible like in [Ni(CN)4]?
The correct answer and explanation is :
Yes, d²sp³ hybridization is possible in complexes like $[Ni(CN)_4]^{2-}$. To explain this, let’s break down the concept and the electronic structure involved:
Hybridization in $[Ni(CN)_4]^{2-}$:
- Nickel’s oxidation state in $[Ni(CN)_4]^{2-}$ is +2, meaning the metal ion has lost two electrons. The electron configuration for neutral nickel (Ni) is $[Ar] 3d^8 4s^2$. After losing two electrons, the configuration for $[Ni]^{2+}$ is $[Ar] 3d^8$, with two electrons in the 4s orbital and six electrons in the 3d orbitals.
- Cyanide (CN⁻) is a strong field ligand, meaning it causes significant splitting of the metal’s d orbitals. In the case of $[Ni(CN)4]^{2-}$, the cyanide ligands induce enough splitting to force the pairing of electrons in the lower-energy d orbitals (specifically, the $3d{x^2 – y^2}$ and $3d_{z^2}$ orbitals), leaving a set of empty orbitals available for bonding.
d²sp³ Hybridization:
- To form six bonds (one per cyanide ligand), nickel needs six hybrid orbitals. For this to happen, nickel undergoes d²sp³ hybridization:
- The two electrons in the $3d$ orbitals pair up.
- One $4s$ orbital is also involved in hybridization.
- Three $3p$ orbitals combine with the two $3d$ orbitals to form six hybrid orbitals (d²sp³).
- These six hybrid orbitals form a octahedral geometry, which is typical for a complex with six ligands. However, due to the strong ligand field (from cyanide), the geometry is actually square planar, as observed in $[Ni(CN)_4]^{2-}$, where the electronic arrangement dictates this specific shape.
Conclusion:
While d²sp³ hybridization is generally associated with octahedral complexes, it can also be applied to square planar complexes like $[Ni(CN)_4]^{2-}$ due to the influence of the ligands and the specific electronic configuration of nickel in the +2 oxidation state.