Lesson 3 of 34
Hybrid orbitals form the sigma framework; unhybridised p orbitals form pi bonding
The local bonding pattern predicts tetrahedral, planar or linear geometry.
A-Level 9476 (2026-2027)
I can describe carbon hybridisation in the reference molecules.Syllabus
Syllabus 9476, 11.1(b). Describe carbon hybridisation in the reference molecules.
I can explain the reference molecular shapes and bond angles.Syllabus
Syllabus 9476, 11.1(c). Explain the reference molecular shapes and bond angles.
I can predict shape and angles in analogous molecules.Syllabus
Syllabus 9476, 11.1(d). Predict shape and angles in analogous molecules.
Hybrid orbitals form the sigma framework; unhybridised p orbitals form pi bonding
The local bonding pattern predicts tetrahedral, planar or linear geometry.
| Molecule | Carbon hybridisation and shape | Carbon-carbon bonding |
|---|---|---|
| Ethane | Each C uses four sp3 orbitals; tetrahedral, approximately 109.5°. | One sigma bond from end-on overlap. Rotation is possible without breaking the sigma bond. |
| Ethene | Each C uses three coplanar sp2 orbitals; trigonal planar, approximately 120°. | One sigma bond plus one pi bond from sideways overlap of parallel unhybridised p orbitals. |
| Benzene | Each C is sp2 and the ring is planar, approximately 120°. | A sigma framework and a delocalised pi system from all six parallel p orbitals; all C-C bonds are equivalent. |
| Ethyne | Each C uses two sp orbitals; linear, 180°. | One sigma bond and two pi bonds formed by two perpendicular sets of p orbitals. |
Hybridisation is a model of bonding. One s and three p orbitals combine to make four sp3 orbitals. One s and two p orbitals make three sp2 orbitals. One s and one p orbital make two sp orbitals. Any p orbitals left over can overlap sideways to form π bonds. To decide the shape, count the regions of σ bonding around an atom. A double bond still counts as one direction.
Worked exampleTransfer the model to an unfamiliar molecule
Predict local geometry in CH3CH=CHCN.
Try it first. Then show the worked stepsHide the worked steps
- The CH3 carbon has four sigma bonds and is approximately tetrahedral.
- Each C=C carbon has three sigma-bond directions and is approximately trigonal planar.
- The nitrile carbon has two sigma-bond directions, one to carbon and one within C≡N; it is approximately linear.
- The alkene pi orbitals must remain parallel, explaining restricted rotation about C=C.
Approximately 109.5° around CH3, 120° around the alkene carbons and 180° at the nitrile carbon. Geometry is local; a molecule need not have one hybridisation throughout.
Check your understanding
What is the hybridisation and shape around each carbon in ethyne, HC≡CH?
Show the answer
sp, linear at 180 degrees, with one sigma and two pi bonds in C≡C
Each carbon has two regions of sigma bonding, so it uses two sp orbitals and is linear. The two unhybridised p orbitals on each carbon overlap sideways to give two pi bonds.