3.5 Shapes of molecules
- Syllabus
- 9701–2028–2029
- Topic
- 3.5
- Level
- AS
Learning objectives
3.5.1The shapes of, and bond angles in, molecules• State/explain: the shapes of, and bond angles in, molecules by using VSEPR theory, including as simple examples:- BF3 (trigonal planar, 120°)- CO2 (linear, 180°)- CH4 (tetrahedral, 109.5°)- NH3 (pyramidal, 107°)- H2O (non-linear, 104.5°)- SF6 (octahedral, 90°)- PF5 (trigonal bipyramidal, 120° and 90°)3.5.2The shapes of, and bond angles in, molecules• Predict the shapes of, and bond angles in, molecules and ions analogous to those specified in 3.5.1
VSEPR explains molecular shape by minimising electron-pair repulsion
Valence-shell electron pairs around a central atom repel one another and arrange as far apart as possible in three dimensions. Each single, double or triple bond counts as one electron domain, and each lone pair counts as one domain.
Lone pairs occupy domains but are not named as atoms in the molecular shape. Their electron density is concentrated closer to the central atom, so repulsion follows lone pair–lone pair > lone pair–bond pair > bond pair–bond pair and lone pairs compress neighbouring bond angles.
| Molecule | bonding domains | lone pairs on central atom | shape | bond angle(s) |
|---|---|---|---|---|
| BF₃ | 3 | 0 | trigonal planar | 120° |
| CO₂ | 2 | 0 | linear | 180° |
| CH₄ | 4 | 0 | tetrahedral | 109.5° |
| NH₃ | 3 | 1 | pyramidal | 107° |
| H₂O | 2 | 2 | non-linear | 104.5° |
| SF₆ | 6 | 0 | octahedral | 90° |
| PF₅ | 5 | 0 | trigonal bipyramidal | 120° and 90° |
CH₄, NH₃ and H₂O all have four electron domains, but replacing bonding pairs with lone pairs increases repulsion and reduces the observed angle from 109.5° to 107° and then 104.5°. By contrast, the two C=O double bonds in CO₂ count as two domains and point 180° apart.
Do not count the two electron pairs in a double bond as two separate domains, and do not name a shape from the molecular formula alone. Count domains around the central atom, then omit lone-pair positions when naming the molecular shape.
Transfer VSEPR by matching central-atom electron domains
For an unfamiliar molecule or ion: identify the central atom; count total valence electrons, adding electrons for negative charge or subtracting for positive charge; construct the bonding and lone pairs; count electron domains; then match the domain pattern to a reference VSEPR arrangement and state the bond angle.
| Species | central-atom domains | Analogy | Prediction |
|---|---|---|---|
| NO₃⁻ | 3 bonding, 0 lone | BF₃ | trigonal planar, 120° |
| NH₄⁺ | 4 bonding, 0 lone | CH₄ | tetrahedral, 109.5° |
| PCl₄⁺ | 4 bonding, 0 lone | CH₄ | tetrahedral, 109.5° |
| PF₆⁻ | 6 bonding, 0 lone | SF₆ | octahedral, 90° |
For NH₄⁺, nitrogen supplies five valence electrons, four hydrogens supply four, and the positive charge removes one: 5 + 4 − 1 = 8 electrons. All four pairs are N–H bonding pairs, so four equivalent domains adopt the tetrahedral arrangement and give 109.5°.
When an analogous species contains lone pairs, begin with the ideal electron-domain angle and use stronger lone-pair repulsion to justify compression. Transfer an exact numerical angle only when the species is directly analogous at the level expected by the question; real angles can vary with the atoms and bonding present.
Ionic charge changes the electron count, not the VSEPR rule. Multiple bonds remain one domain, lone pairs are omitted from the shape name, and molecular polarity is a separate conclusion taught in section 3.6.