Question

What is the percentage of $\mathrm{s}$ character in ${\mathrm{sp}}^3$ hybridized orbital?

Answer 1

${\mathbf{sp}}^{\mathbf{3}}$

• ${\mathrm{sp}}^3$ hybridization consists of one $\mathrm{s}$orbital and three $\mathrm{p}$orbitals to create a hybrid orbital

• It has a bond angle $109.5°$ and has a geometry of tetrahedral shown in the figure above.
• Since one part is $\mathrm{s}$and three parts are $\mathrm{p}$orbital, the percentage of$\mathrm{s}$ is $25\%$ and of $\mathrm{p}$ is$75\%$.
• The more the$\mathrm{s}$ character, the more will be the electronegativity percentage in that hybridization.

Hence, the percentage of $\mathbf{s}\mathbf{}\mathbf{in}\mathbf{}{\mathbf{sp}}^{\mathbf{3}}\mathbf{}\mathbf{is}\mathbf{}\mathbf{25}\mathbf{\%}$.