Topic 2 of 6
Orbital overlap and molecular shape
Sigma/pi overlap, VSEPR and deductions for unfamiliar molecules.
A-Level 9476 (2026-2027)
End-on overlap gives sigma; sideways overlap gives pi
A double bond contains one sigma and one pi bond.
A sigma (σ) bond has electron density along the line joining the nuclei. It forms by end-on overlap: s-s in H2, s-p in a simple H-Cl description, or p-p. A pi (π) bond forms by sideways overlap of parallel p orbitals, producing electron density above and below that line. The two pi regions together constitute one bond.
The extra bond in ethene comes from sideways overlap
Two nuclei lie on the horizontal internuclear axis. A sigma bond occupies the region directly between them. Parallel vertical p orbitals on the two carbons overlap above and below that axis to form one pi bond.
| Connection | Sigma bonds | Pi bonds |
|---|---|---|
| Single bond | 1 | 0 |
| Double bond | 1 | 1 |
| Triple bond | 1 | 2 in mutually perpendicular planes |
| Whole ethene molecule | 5: four C-H and one C-C | 1: between the carbon atoms |
A carbon-carbon single bond can usually rotate without destroying its sigma overlap. Rotation about a double bond requires loss of the pi overlap and is restricted. Bond rotation is therefore not the same as rotating the entire molecule in space. Detailed d-orbital overlap is outside the overlap requirement here.
Count electron regions, then describe the atoms
Multiple bonds count as one region. Lone pairs affect shape even though they are not atoms.
VSEPR places regions of electron density around a central atom as far apart as possible. A single, double or triple bond counts as one region for this purpose. A lone pair is another region. The electron-region arrangement includes lone pairs; the molecular shape names the positions of the atoms only.
| Molecule | Bonding regions / lone pairs | Molecular shape | Typical bond angle |
|---|---|---|---|
| BF3 | 3 / 0 | Trigonal planar | 120 degrees |
| CO2 | 2 / 0 | Linear | 180 degrees |
| CH4 | 4 / 0 | Tetrahedral | 109.5 degrees |
| NH3 | 3 / 1 | Trigonal pyramidal | About 107 degrees |
| H2O | 2 / 2 | Bent | About 104.5 degrees |
| SF6 | 6 / 0 | Octahedral | 90 degrees between adjacent bonds; 180 between opposite bonds |
A lone pair is held by only one nucleus and occupies more space near the central atom than a bonding pair. The usual repulsion order is lone pair-lone pair > lone pair-bonding pair > bonding pair-bonding pair. The tetrahedral electron arrangement therefore gives progressively smaller bond angles from CH4 to NH3 to H2O. Exact angles also depend on the atoms and bonds: do not assign water's angle to every bent molecule.
Worked example
Deduce an unfamiliar ion
Predict the shape of H3O+.
- Count valence electrons: oxygen contributes six; three hydrogens add three; the positive charge subtracts one. Total = eight.
- Three O-H shared pairs use six electrons, leaving one lone pair on O.
- Four electron regions give a tetrahedral arrangement; three bonded atoms and one lone pair give a trigonal pyramidal molecular shape.
Trigonal pyramidal. A basic VSEPR comparison predicts an angle near the tetrahedral value, modified by its lone pair; it does not justify copying the exact NH3 angle.
Worked example
A double bond is one region
SO2 has two S-O bonding regions and one lone pair on S. What shape follows?
- There are three electron regions, so their reference arrangement is trigonal planar.
- Only two of those regions contain bonded atoms.
- The molecule is bent, with an O-S-O angle slightly below the 120-degree reference.
Bent, not linear. Counting each double bond as two directions would give the wrong model.