Final answer:
The sp³ and sp³d² hybridizations involve all orbitals forming sigma bonds or holding lone pairs, leaving no unhybridized p orbitals to form pi bonds. Thus, they cannot form pi bonds.
Step-by-step explanation:
Hybridized orbitals are a result of combining atomic orbitals to explain molecular geometry. In sp³ hybridization, one s orbital and three p orbitals combine to form four hybrid orbitals with equal energies, resulting in a tetrahedral geometry. These orbitals are used to form sigma bonds (σ-bonds), as each sp³ hybrid orbital can overlap with another sp³ hybrid orbital or an s orbital of a hydrogen atom end-to-end to form a σ-bond.
Similarly, in sp³d² hybridization, s, p, and d orbitals mix to form six hybrid orbitals arranged in an octahedral geometry, where again, all orbitals are used to form σ-bonds. In both sp³ and sp³d² hybridization, all hybrid orbitals are occupied either by lone pairs or used to form σ-bonds, leaving no unhybridized p orbitals available to form pi bonds (π-bonds). π-bonds require unhybridized p orbitals that can overlap side-by-side. As such configurations lack these, π-bond formation is not feasible in molecules where all the orbitals are sp³ or sp³d² hybridized.