Final answer:
Delocalization of electrons occurs in molecules represented by resonance hybrids because it allows the electrons to be spread over more than two nuclei, resulting in more bonding. The resonance hybrid is always more stable than any of its contributing structures.
Step-by-step explanation:
Delocalization of π- or nonbonding electrons results in more bonding, i.e., delocalized electrons are spread over more than two nuclei. It results in lowering the potential energy, called resonance stabilization. The resonance hybrid is always more stable than the predicted stability of any of its contributing structures.
For example, the molecule benzene has two resonance forms. We can use either of these forms to determine that each of the carbon atoms is bonded to three other atoms with no lone pairs, so the correct hybridization is sp². The electrons in the unhybridized p orbitals form bonds. Neither resonance structure completely describes the electrons in the bonds. They are not located in one position or the other, but in reality, are delocalized throughout the ring. Valence bond theory does not easily address delocalization. Bonding in molecules with resonance forms is better described by molecular orbital theory. (See the next module.)