Final answer:
Pi electron energy is lower in an aromatic ring compared to that in an open-chain compound, indicating greater stability due to aromatic delocalization. Breaking a pi bond in aromatics requires significantly more energy than in non-aromatics, reinforcing the inherent stability offered by the aromatic electron system. Molecular orbital theory further confirms the lower energy and increased stability of pi electrons in aromatic rings.
Step-by-step explanation:
When comparing pi electron energy in an aromatic ring to that in an open-chain compound, it's important to consider the concept of conjugation and the stability provided by the aromatic system. In the case of 1,3-butadiene, itself a conjugated system, the HOMO-LUMO gap is narrower than in isolated pi-bond systems like ethene, indicating more stability and lower energy of the pi electrons in the conjugated system. However, with aromatic rings, such as benzene, the stability is even greater due to aromatic delocalization; this results in a unique situation where the aromatic ring's pi electrons are at a lower energy than those in a non-aromatic conjugated system.
The aromatic stabilization is quite substantial; breaking a pi bond in an alkene costs about 260 kJ/mol, but disrupting the pi system in an aromatic compound by breaking a pi bond requires an additional 208 kJ/mol. This higher requirement for energy illustrates the lower energy and therefore greater stability of pi electrons in the aromatic system compared to an open-chain or isolated pi-bond system.
Furthermore, the effects of molecular orbital (MO) theory demonstrate that aromatic systems have some degenerate orbitals which are filled in a way that minimizes the energy of the system. The molecular orbitals in aromatic systems are lower in energy than those in open-chain compounds, again highlighting the relative stability of the pi electrons in aromatic systems