Blow up a few long balloons and tie them together at their necks. They don't pile on top of each other — they swing apart until they're spread as evenly as possible around the knot. The groups of electrons around an atom do exactly the same thing. They're all negatively charged, so they repel, and they settle into whatever arrangement keeps them as far apart as they can get. That arrangement is the shape of the molecule. That single idea is the whole of VSEPR — Valence-Shell Electron-Pair Repulsion.
Look at the central atom and count two things: the number of atoms bonded to it, $N_X$, and the number of lone pairs sitting on it, $N_E$. Add them to get the steric number,
— the number of electron "groups" that have to share the space. Each value of SN has exactly one most-spread-out answer: 2 groups go to opposite sides ($180°$, linear); 3 groups make a flat triangle ($120°$); 4 groups point to the corners of a tetrahedron ($109.5°$). Methane, CH₄, has 4 bonds and no lone pairs, so $\text{SN}=4$ and its four H atoms sit at $109.5°$ — the classic tetrahedron.
A double or triple bond still counts as just one group, because all of its electrons point the same way — that's why CO₂ ($\text{SN}=2$) is linear, not bent. A lone pair counts as a group too, but it spreads over a wider angle than a bond, because no second nucleus is pulling it in. So lone pairs push the bonds harder than bonds push each other:
Start from the tetrahedral $109.5°$, swap one bond for a lone pair and ammonia (NH₃) drops to $107°$; swap two and water (H₂O) closes to $104.5°$. In the sim above, the X and E sliders set $N_X$ and $N_E$, and the lone-pair repulsion factor sets how much harder those lone pairs shove.
① Set X = 4, E = 0 (methane) and read $109.5°$; now raise E to 1, then 2, and watch the bond angle fall toward ammonia and then water. ② With water loaded, drag the lone-pair repulsion factor upward and watch the H–O–H angle compress even further. ③ Switch to the Dipole Moments view and compare CO₂ (symmetric — the bond arrows cancel, $\mu=0$) with H₂O (bent — the arrows add up to a net $1.85$ D, which is why water is such a good solvent).
The Valence Shell Electron Pair Repulsion (VSEPR) theory of Gillespie & Nyholm (1957) states that the geometry around a central atom is dictated by the requirement that bonded atoms (X) and lone pairs (E) arrange themselves to minimise mutual electrostatic and Pauli repulsion. The arrangement that minimises this repulsion is determined by the steric number $\text{SN} = N_X + N_E$.
| Symbol | Meaning | Unit / Range |
|---|---|---|
| $\text{SN}$ | Steric number = total electron domains | 2–7 (integer) |
| $N_X$ | Number of bonded atoms (σ-bonds) | integer |
| $N_E$ | Number of lone pairs on central atom | integer |
| $\theta_{ij}$ | Angle between domains $i$ and $j$ at the centre | degrees |
| $r_{ij}$ | Inter-domain distance on a unit sphere = $2\sin(\theta_{ij}/2)$ | dimensionless |
| $k_{ij}$ | Repulsion weight: lp–lp > lp–bp > bp–bp | dimensionless |
| $p$ | Repulsion exponent (typically 6–12 for Pauli core) | dimensionless |
| $\mu$ | Dipole moment of molecule | D (Debye) = 3.336×10⁻³⁰ C·m |
For H₂O: $N_X=2$, $N_E=2$, $\text{SN}=4$ → tetrahedral electron geometry, but bent molecular shape because we only "see" the two H atoms. With an idealised tetrahedral angle of 109.5° and the lp–bp repulsion factor pushing the H atoms together:
The 5° distortion from the ideal tetrahedral angle is exactly what gives water its bent shape, polarity, hydrogen bonding, and ultimately makes life possible.