Question

Can ${\mathrm{sp}}^3$ form $\mathrm{\pi }$ bonds?

Answer 1

${\mathrm{sp}}^3$ hybridization:

1. ${\mathrm{sp}}^3$ hybridization is the mixing character of one $2\mathrm{s}$ orbital and three $2\mathrm{p}$ orbitals to form four hybrid orbitals of the same energy.
2. Now, the compounds having this type of hybridization must have the four orbitals bonded to each other.
3. For four orbitals to bond together only four single bonds are required and the single bonds are $\mathrm{\sigma }$-bonds not $\mathrm{\pi }$-bonds.
4. Hence, ${\mathrm{sp}}^3$ do not form $\mathrm{\pi }$-bonds.