Question

In ${\mathrm{sp}}^3$ hybridization how many atomic orbitals are involved?

Answer 1

Hybridization:

• Hybridization is the process of combining two atomic orbitals of nearly equal energy to produce the same number of hybrid orbitals having the same shape, equal energy, and orientation such that there is minimum repulsion between these hybridized orbitals.

${\mathbf{sp}}^{\mathbf{3}}$ hybridization:-

• ${\mathrm{sp}}^3$ hybridization represents the tetrahedral structure of molecules. In which one $\mathrm{s}$ orbital and all three of the $\mathrm{p}$ orbitals hybridize to form four ${\mathrm{sp}}^3$ orbitals, each consisting of $75\%$ $\mathrm{p}$ character and $25\%$ $\mathrm{s}$ character.
• Hybridization of a ‘$\mathrm{s}$’ orbital with all three ‘$\mathrm{s}$’ orbitals$({\mathrm{p}}_{\mathrm{x}},{\mathrm{p}}_{\mathrm{y}}\mathrm{and}{\mathrm{p}}_{\mathrm{z}})$ results in four ${\mathrm{sp}}^3$ hybrid orbitals.
• ${\mathrm{sp}}^3$ hybrid orbitals are oriented at a bond angle of $109.5°$ from each other. This $109.5°$ arrangement gives tetrahedral geometry which can be shown as;

Hence,In ${\mathrm{sp}}^3$ hybridization total of $4$ atomic orbitals are involved given as; one $\mathrm{s}$ orbital and three $\mathrm{p}$ orbitals$({\mathrm{p}}_{\mathrm{x}},{\mathrm{p}}_{\mathrm{y}}\mathrm{and}{\mathrm{p}}_{\mathrm{z}})$.