Question

In which of the following hybridizations does the orbital ${d_{{x^2} - {y^2}}}$ is involved?
A. $s{p^3}$
B. $s{p^3}{d^2}$
C. $s{p^3}{d^3}$
D. None of these

Answer 1

We have to understand and memorize the concept of basic hybridization and the orbitals that get involved in the respective hybridization. From it we can get our answer for the given orbital.

Complete answer:
At first we have to get to know the type of hybridization and the basic theory of the same.
1. $sp$ hybridization: $sp$ hybridization is observed when one s and one p orbitals of the same shell of an atom mix to form 2 equivalent orbital. The new orbitals formed are called $sp$ hybrid orbitals.
2. $s{p^2}$ hybridization: $s{p^2}$ hybridization is observed when one s and two p orbitals of the same shell of an atom mix to form 3 equivalent orbital. The new orbitals formed are called $s{p^2}$ hybrid orbitals.
3. $s{p^3}$ hybridization: When one ‘s’ orbital and 3 ‘p’ orbitals belonging to the same shell of an atom mix together to form four new equivalent orbital, the type of hybridization is called a tetrahedral hybridization or $s{p^3}$. The new orbitals formed are called $s{p^3}$ hybrid orbitals.
4. $s{p^3}d$ hybridization: $s{p^3}d$ hybridization involves the mixing of 1s orbital 3p orbitals and 1d orbital to form 5 $s{p^3}d$ hybridized orbitals of equal energy. They have trigonal bipyramidal geometry.
5. $s{p^3}{d^2}$ hybridization: $s{p^3}{d^2}$ hybridization has 1s, 3p and 2d orbitals, that undergo intermixing to form 6 identical $s{p^3}{d^2}$ hybrid orbitals.
6.$s{p^3}{d^3}$ hybridization: $s{p^3}{d^3}$ hybridization has 1s, 3p and 3d orbitals, that undergo intermixing to form 7 identical $s{p^3}{d^3}$ hybrid orbitals.
Hence, if we try to capture the above theory in single lines then it will be as below
$sp \rightarrow s{p_x}$
$s{p^2} \rightarrow s{p_x}{p_y}$
$s{p^3} \rightarrow s{p_x}{p_y}{p_z}$
$s{p^3}d$ or $ds{p^3}$ $\rightarrow$ $s{p_x}{p_y}{p_z}d_{z^2}$
$s{p^3}{d^{^2}} \rightarrow s{p_x}{p_y}{p_z}{d_{{x^2} - {y^2}}}d_{z^2}$
$s{p^3}{d^3} \rightarrow s{p_x}{p_y}{p_z}d_x d_y d_z$
Hence, from the above theory we can conclude that the orbital ${d_{{x^2} - {y^2}}}$ is involved in $s{p^3}{d^{^2}}$ hybridization.
Therefore, option B is the right answer.

Note:
Other than the above hybridization, $ds{p^2} \rightarrow {d_{{x^2} - {y^2}}}s{p_x}{p_y}$ is one more hybridization in which the given orbital is involved. But generally we considered the above hybridization only, still $ds{p^2}$ hybridization should also be kept at the back of our mind.