Question

The d-orbitals involved in $s{{p}^{3}}{{d}^{2}}$ or ${{d}^{2}}s{{p}^{3}}$ hybridization of the central metal ion are:
(this question has multiple correct options)
a.) ${{d}_{{{x}^{2}}-{{y}^{2}}}}$
b.) ${{d}_{xy}}$
c.) ${{d}_{yz}}$
d.) ${{d}_{{{z}^{2}}}}$

Answer 1

From the given hybridization we can say that the geometry of the compound has octahedral. So, to solve this question, consider the splitting of d-orbitals in an octahedral complex, when approached by a ligand.

Complete step by step solution:
When any negatively charged species or ligand approaches a metal, the energy of the orbitals increases. This leads to the splitting of orbitals as –

In octahedral complexes, the splitting of orbitals always takes place in this manner.
The electron first goes to the lower energy orbital, i.e. ${{d}_{xy}}$.${{d}_{yz}}$,${{d}_{{{z}^{2}}}}$and then to the higher energy orbital, i.e., ${{d}_{{{x}^{2}}-{{y}^{2}}}}$and ${{d}_{{{z}^{2}}}}$. Therefore, bonding electrons go to the high energy orbitals which decide the hybridization of the compound.
In case of $s{{p}^{3}}{{d}^{2}}$ hybridization, nd (outermost) orbitals are involved. The compound is therefore also known as outer-orbital complex.
In case of ${{d}^{2}}s{{p}^{3}}$ hybridization, (n-1)d (next to outermost) orbitals are involved. The compound is therefore also known as inner-orbital complex.
Therefore, the answer is – option (a) and option (d).

Additional information: In short, if inner d-orbital is used for bonding, the hybridization will be ${{d}^{2}}s{{p}^{3}}$. Similarly, if outer d-orbital is used, the hybridization will be $s{{p}^{3}}{{d}^{2}}$.

Note: Irrespective of the differences between $s{{p}^{3}}{{d}^{2}}$ and ${{d}^{2}}s{{p}^{3}}$, there are certain similarities, such as –
Both hybridization results in octahedral geometry.
Both have six hybrid orbitals.
There is an angle of 90 degrees between hybrid orbitals.