Drag any molecule to rotate it in 3D. The dashed amber arc marks the bond angle the AP exam expects; memorize the pattern: lone pairs compress angles below the ideal value.
CH₄: Tetrahedral
4 bonding pairs, 0 lone pairs → sp³, 109.5°.
| Molecule | Electron geometry | Molecular shape | Bond angle | Hybridization |
|---|---|---|---|---|
| CH₄ | Tetrahedral | Tetrahedral | 109.5° | sp³ |
| NH₃ | Tetrahedral | Trigonal pyramidal | 107° | sp³ |
| H₂O | Tetrahedral | Bent | 104.5° | sp³ |
| BF₃ | Trigonal planar | Trigonal planar | 120° | sp² |
| SO₂ | Trigonal planar | Bent | ≈119° | sp² |
| CO₂ | Linear | Linear | 180° | sp |
VSEPR ranks repulsion: lone pair–lone pair > lone pair–bonding pair > bonding pair–bonding pair. Each lone pair on NH₃ pushes the three N–H bonds together, dropping the angle from the tetrahedral ideal 109.5° to 107°. Water's two lone pairs squeeze the angle further, to 104.5°. Molecules with no lone pairs on the central atom (CH₄, BF₃, CO₂) keep their ideal angles.
When an FRQ asks you to "justify the bond angle," name both the electron-pair geometry and the molecular shape. Saying "water is bent because of its two lone pairs" earns partial credit; the full chain is: 4 electron domains → tetrahedral electron geometry → 2 lone pairs compress the H–O–H angle to 104.5°. Resonance (SO₂, CO₃²⁻, NO₃⁻) delocalizes electron density but does not change the VSEPR electron-domain count.