Question

In ${\text{s}}{{\text{p}}^{\text{3}}}{{\text{d}}^{\text{3}}}$ hybridization, which orbitals are involved?
A.${{\text{d}}_{{{\text{x}}^{\text{2}}}{\text{ - }}{{\text{y}}^{\text{2}}}}}{\text{, }}{{\text{d}}_{{{\text{z}}^{\text{2}}}}}{\text{, }}{{\text{d}}_{{\text{xy}}}}$
B.${{\text{d}}_{xy}}{\text{, }}{{\text{d}}_{yz}}{\text{, }}{{\text{d}}_{zx}}$
C.${{\text{d}}_{{{\text{x}}^{\text{2}}}{\text{ - }}{{\text{y}}^{\text{2}}}}}{\text{, }}{{\text{d}}_{xy}}{\text{, }}{{\text{d}}_{xz}}$
D.${{\text{d}}_{{z^2}}}{\text{, }}{{\text{d}}_{yz}}{\text{, }}{{\text{d}}_{zx}}$

Answer 1

Here we have to choose the orbitals are involved on ${\text{s}}{{\text{p}}^{\text{3}}}{{\text{d}}^{\text{3}}}$ hybridization. Hybridized orbitals are the ones which are formed when mixing of atomic orbital happens. The new hybrid orbitals are suitable for forming bonds between the two atoms. The hybrid ${\text{d}}$ orbital comprises of 5 orbitals which are ${{\text{d}}_{{{\text{x}}^{\text{2}}}{\text{ - }}{{\text{y}}^{\text{2}}}}}{\text{, }}{{\text{d}}_{{{\text{z}}^{\text{2}}}}}{\text{, }}{{\text{d}}_{{\text{xy}}}},{\text{ }}{{\text{d}}_{yz}}$ and ${{\text{d}}_{zx}}$.

Complete step by step answer:
Let’s start with discussing hybridized orbitals for better understanding of the question. Hybridized orbitals are the ones which are formed when mixing of atomic orbital happens. The new hybrid orbitals are suitable for forming bonds between the two atoms. The hybridization of ${\text{d}}$ orbitals produces less stable orbitals as compared with hybridized ${\text{s}}$ and ${\text{p}}$ orbitals. In the formation of molecules having the hybridization of ${\text{s}}{{\text{p}}^{\text{3}}}{\text{d}}$ to ${\text{s}}{{\text{p}}^{\text{3}}}{{\text{d}}^{\text{3}}}$, hybridized ${\text{d}}$ orbitals are used. The hybrid ${\text{d}}$ orbital comprises of 5 orbitals which are ${{\text{d}}_{{{\text{x}}^{\text{2}}}{\text{ - }}{{\text{y}}^{\text{2}}}}}{\text{, }}{{\text{d}}_{{{\text{z}}^{\text{2}}}}}{\text{, }}{{\text{d}}_{{\text{xy}}}},{\text{ }}{{\text{d}}_{yz}}$ and ${{\text{d}}_{zx}}$.
We are asked about the orbitals which are used in ${\text{s}}{{\text{p}}^{\text{3}}}{{\text{d}}^{\text{3}}}$ hybridization. The orbitals which are used are as follows, ${\text{s}}$ orbital, ${{\text{p}}_{\text{x}}}{\text{, }}{{\text{p}}_{\text{y}}}{\text{, }}{{\text{p}}_{\text{z}}}$ orbitals and ${{\text{d}}_{{{\text{x}}^{\text{2}}}{\text{ - }}{{\text{y}}^{\text{2}}}}}{\text{, }}{{\text{d}}_{{{\text{z}}^{\text{2}}}}}{\text{, }}{{\text{d}}_{{\text{xy}}}}$ orbitals.
According to the options given we need to mark only ${\text{d}}$ orbitals and hence the answer to this question will be option A. ${{\text{d}}_{{{\text{x}}^{\text{2}}}{\text{ - }}{{\text{y}}^{\text{2}}}}}{\text{, }}{{\text{d}}_{{{\text{z}}^{\text{2}}}}}{\text{, }}{{\text{d}}_{{\text{xy}}}}$.

Note:
The hybridized ${\text{s}}$ orbital has the maximum screening effect and hence is the most stable hybrid orbital. The $3{\text{p}}$ hybrid orbitals have less screening effect and hence are little less stable when compared with the ${\text{s}}$ hybrid orbital. The lowest screening effect is shown by ${\text{d}}$ hybrid orbitals and hence they are the least stable.