Benzene and resonance
About this app
Drag p orbitals together into the π cloud, compare bond lengths with a caliper and pull up the hydrogenation steps: why the double bonds in benzene do not jump and where the resonance energy of about 150 kJ/mol comes from. With Hückel’s rule and exam questions.
With explain mode: an AI tutor walks you through the app, points out what matters in the scene and answers your questions.
What it is about
All six carbon atoms in benzene are sp² hybridised. Their p orbitals stand perpendicular to the ring plane and overlap all the way round, so six π electrons belong to the whole ring. The two Kekulé formulas are resonance structures: two notations for one single state. The double bonds do not jump. All six bonds are 139 pm long, and benzene is about 150 kJ/mol more stable than an imagined ring with three isolated double bonds. This amount is called the resonance energy.
What the app shows
A benzene ring stands in 3D space with its atoms, σ framework, lines of the resonance structure, p orbitals and the π cloud above and below the ring plane. A caliper reads the bond lengths, and a scale places them next to ethene and ethane. A switch shows the imagined case of jumping bonds with changing values and a distorted hexagon. The hydrogenation diagram compares cyclohexene, hypothetical cyclohexatriene and benzene.
What you try out
You drag an orbital upwards: first they all align in parallel, then they flow together into the cloud. You switch between the resonance structures and see that the cloud stays the same. You move the caliper around the ring and pull up three hydrogenation steps in the diagram. Then you set a guess for benzene and compare it with the measured value. Finally you predict for five rings whether they are aromatic by Hückel’s rule, and test yourself with seven questions.
Frequently asked questions
- What is the resonance energy of benzene?
- It is the amount by which benzene is more stable than an imagined ring with three isolated double bonds. Hydrogenating cyclohexene releases about 120 kJ/mol, so three double bonds would give about 360 kJ/mol. For benzene about 208 kJ/mol is measured. The difference of 152 kJ/mol is usually rounded to about 150 kJ/mol.
- Do the double bonds in benzene jump back and forth?
- No. The Kekulé formulas are two notations for one single state. All six bonds are equally long, about 139 pm, and the six π electrons are delocalised over the whole ring. The resonance arrow ↔ links notations, whereas an equilibrium arrow ⇌ links two substances.
- What does Hückel’s rule say?
- A ring is aromatic if it is cyclic, planar and fully conjugated and has 4n + 2 π electrons, so 2, 6, 10 and so on. Benzene has six π electrons and fulfils the rule with n = 1. Cyclobutadiene with four π electrons does not.
- What can I do in the app myself?
- You let the p orbitals overlap, measure the bonds with a caliper, pull up the hydrogenation steps in the diagram and give your own guess before the measured value appears. An explain mode guides you through all steps.
Subject: Chemistry
More from Chemistry
- Binding equilibrium — kinetics, Kd and ΔG° (3D)
- Chemistry lab (3D)
- Chemistry lab: reactions
- Chemical equilibrium and Le Chatelier’s principle
- Chirality: molecule and mirror image
- Electrolysis of water (3D)
- Enzymes and substrates: lock and key
- Firework colours: flame tests and spectral lines
A VisuApp by heyprof: interactive, free in your browser, no sign-up. What is a VisuApp? · All apps