Benzene: Electrons in a Ring

The puzzle of benzene

Benzene (C₆H₆) consists of six carbon atoms in a flat hexagon, each carrying one hydrogen atom. On paper one draws alternating single and double bonds. Yet when you measure the bond lengths, all six are the same: 139.7 pm. That is exactly between a single bond (154 pm) and a double bond (134 pm).

The orbitals provide the explanation. The six electrons that would be responsible for the double bonds belong to no particular bond. They spread over the whole ring. This is called delocalization.

Orbitals are building blocks, not variants

The six orbitals in the picture are not six different versions of benzene. They are six patterns in which the electrons in the ring are possible, much as a drum can vibrate in several patterns at once. The six π electrons occupy the three lowest at the same time, and the other three stay empty. The actual picture is the sum of the occupied orbitals. You find it under “Total density”.

How the π orbitals arise

Each carbon atom has a 2p orbital standing perpendicular to the ring plane, like a dumbbell pointing up and down. Six such orbitals can be combined in six ways. The principle is the same as for any molecular bond: equal signs reinforce, opposite signs cancel. The more sign changes around the ring, the higher the energy.

• π₁ has no change and is the most strongly bonding. A ring of density above and below the molecule.
• π₂ and π₃ have one nodal plane through the ring axis. They have exactly the same energy.
• π₄* and π₅* have two nodal planes. They are antibonding and empty.
• π₆* has a sign change at every atom and is the most strongly antibonding.

The six π electrons fill the three lowest orbitals. The sum of their densities is a perfectly even double ring. That is why all bonds have the same length.

Aromaticity and Hückel’s rule

Benzene is surprisingly stable. Compared with three separate double bonds it gains about 150 kJ/mol, the resonance energy. This is because the electrons have more room in the lowest orbitals. Hückel’s rule says when this happens: a flat ring with 4n + 2 π electrons (2, 6, 10, …) is aromatic. Benzene with 6 electrons fulfils it. Cyclobutadiene with 4 electrons does not; it is unstable and reactive.

Why Kekulé is only a model

August Kekulé proposed the ring with alternating double bonds in 1865. In the Kekulé model of the view you see three localized π bonds. Because the ring can be drawn in two equivalent ways, we now assume that the molecule is a hybrid of both. The orbitals show it better: there are no individual double bonds, only a shared cloud.

Why this matters

• Aromatic rings are found in dyes, in the bases of DNA and in many drugs.
• The number of delocalized electrons decides which light a molecule absorbs. Benzene absorbs in the ultraviolet at about 255 nm. The larger the delocalized system, the further the absorption moves into the visible. This explains many colours.

Think about it

How many nodal planes through the ring axis does π₆* have?

Three. The sign changes at each of the six atoms, and the nodal planes run between them.

Why do π₂ and π₃ have the same energy?

Both are built the same way, just rotated by 90°. The hexagonal symmetry of the ring allows this rotation without changing the energy. This is called degeneracy.

Cyclobutadiene (four carbon atoms) is not aromatic. Why?

Four π electrons do not fit 4n + 2. The last electron pair would have to enter a high-lying orbital or stay unpaired. The molecule loses stability instead of gaining it.

Back to the basics: hydrogen atom and water.

A question or an idea?

Something unclear, an error you spotted, or a chapter you would like to see? Write to me. Good questions and their answers will appear here later as an FAQ.

Write a message

Last updated: 2026-10-03