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

Hint: 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.