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8873 / 2027
Polymers and Organic Chemistry

Lesson 3 of 12

Molecular shapes

Describe the local geometry around a carbon before the whole molecule.

A-Level 8873, revised syllabus (2026-2027)

  • I can explain organic molecular shapes through sigma and pi bonding.Syllabus

    Syllabus 8873, 9(f). Explain organic molecular shapes through sigma and pi bonding.

Read shape from sigma and pi bonding

Describe the local geometry around a carbon before the whole molecule.

Three structures that establish the pattern
MoleculeCarbon bonding and local shapeAngles and freedom of movement
Ethane, CH3CH3Each carbon forms four sigma bonds and has a tetrahedral arrangement. The C-C bond is one sigma bond from head-on overlap.About 109.5° around each carbon. Rotation about C-C is possible without breaking the sigma bond; the whole molecule is not planar.
Ethene, CH2=CH2Each carbon has three regions of sigma bonding in a trigonal planar arrangement. Parallel p orbitals overlap sideways above and below the plane, forming one pi bond.About 120° around each carbon. The molecule is planar; rotation about C=C is restricted.
Benzene, C6H6Each carbon has three sigma bonds in a planar hexagonal framework. The six parallel p orbitals overlap around the ring, giving delocalised pi electron density above and below it.About 120° around every carbon. All six C-C bonds are equivalent; benzene is not a ring of alternating independent single and double bonds.

Benzene: a planar ring with delocalised pi bonding

A hexagon with an internal circle is the preferred representation of benzene. Each corner represents a carbon attached to one hydrogen. The circle represents delocalised pi bonding, not an extra atom or separate small ring.

The ring and its six C-H sigma bonds lie in one plane. The delocalised pi electron density lies above and below that plane.

A multiple bond counts as one direction of electron density when you predict the shape around an atom. In ethene, each carbon has a double bond but only three directions, so it is trigonal planar. You do not need hybridisation labels here. Use the number and arrangement of sigma bonds, and say where sideways pi overlap can happen.

Worked exampleApply the pattern to an unfamiliar chain

Predict the approximate angles around each carbon in propene, CH2=CHCH3.

Try it first. Then show the worked stepsHide the worked steps
  1. The two double-bonded carbons each have three sigma-bond directions. Their local arrangements are trigonal planar with angles near 120°.
  2. The CH3 carbon has four sigma-bond directions, so it is tetrahedral with angles near 109.5°.
  3. The C=C consists of one sigma and one pi bond; the adjacent C-C single bond is a sigma bond. Do not assign 120° to every carbon just because the molecule contains one double bond.
Answer

The alkene region is locally planar; the methyl carbon is tetrahedral.

Check your understanding

What shape is expected around the carbonyl carbon in CH3CHO?

  1. Trigonal planar, about 120 degrees, with one sigma and one pi bond in C=O
  2. Tetrahedral, about 109.5 degrees, as the carbon forms four bonds
  3. Linear, 180 degrees, as the C=O double bond pulls the groups apart
Show the answer

Trigonal planar, about 120 degrees, with one sigma and one pi bond in C=O

The carbon has three regions of electron density: bonds to CH3, H and O. These spread out in a plane at about 120 degrees.