Draw the LCAO model of ethylene in the bonding and antibonding orbital. Distinguish the ground state of ethylene from its excited state. Distinguish \(\pi^{2}\) from \(\pi^{*} \pi\).

The LCAO model of ethylene can be constructed by combining the 2pz orbitals of the carbon atoms. In the bonding orbital (\(\pi\)), the wavefunctions of 2pz orbitals are added, resulting in constructive interference and electron density above and below the molecular plane. In the antibonding orbital (\(\pi^*\)), the wavefunctions of 2pz orbitals are subtracted, resulting in destructive interference and a nodal plane in the middle. In the ground state, ethylene has a double bond between carbon atoms and its molecular orbital configuration is \((\sigma_{1s})^2(\sigma_{1s^*})^2(\sigma_{2s})^2(\sigma_{2s^*})^2(\sigma_{2p})^4(\pi_{2p})^2(\pi_{2p^*})^0\). In the excited state, one electron is promoted from the \(\pi\) orbital to the \(\pi^*\) orbital, weakening the double bond, and the configuration becomes \((\sigma_{1s})^2(\sigma_{1s^*})^2(\sigma_{2s})^2(\sigma_{2s^*})^2(\sigma_{2p})^4(\pi_{2p})^1(\pi_{2p^*})^1\). The \(\pi^2\) symbol corresponds to the ground state, while \(\pi^*\pi\) corresponds to the excited state of ethylene.