 Solution

Ozone is bent, with two equivalent O–O bonds of approximate bond order $1.5$. Its useful localized orbital hybridization description is $s{p}^2$ at all three oxygens. Two hybrids at the central oxygen form sigma bonds and a third contains a lone pair; the three perpendicular p orbitals form a three-centre four-electron pi bond.

The three pi orbitals are bonding, nonbonding and antibonding, filled as ${\pi }_{\mathrm{b}}^2{\pi }_{\mathrm{nb}}^2{\pi }_{\ast }^0$. The terminal-centred nonbonding orbital is the Homo, conventionally $1a_2$; the delocalized antibonding orbital is the LUMO, $2b_1$. The two principal resonance structures, ${}^{-}\mathrm{O}-{\mathrm{O}}^{+}=\mathrm{O}$ and $\mathrm{O}={\mathrm{O}}^{+}-{\mathrm{O}}^{-}$, distribute the extra pi bond across the two links.

The 18 valence electrons comprise four sigma-bond electrons, ten in five sigma-framework lone pairs, and four in the pi system. This gives a paired ground-state model. More accurate treatments include electron correlation; the simple bond-order picture is not an exact charge-density calculation.^[ref]