Molecular Geometry and Polarity

Shape from Repulsion

VSEPR theory rests on one idea: electron domains around a central atom push each other as far apart as possible. A domain is any single bond, any multiple bond (counted once), or any lone pair.

The molecular shape is what you see when only the atoms are drawn, so lone pairs change the name without changing the domain count. Four domains give a tetrahedral arrangement, but methane (0 lone pairs) is tetrahedral, ammonia (1) is trigonal pyramidal, and water (2) is bent.

Lone pairs also squeeze the angles. A lone pair is held by one nucleus only, so it spreads wider than a bonding pair and pushes harder. The tetrahedral $109.5^\circ$ therefore falls to $107^\circ$ in ammonia and $104.5^\circ$ in water. Repulsion strength ranks: lone-lone > lone-bond > bond-bond.

Molecular polarity needs two things: polar bonds, and a geometry that fails to cancel them. Bond dipoles are vectors, so a symmetric arrangement of identical bonds sums to zero. $\mathrm{CO_2}$ has two strongly polar C=O bonds pointing in exactly opposite directions and is nonpolar; $\mathrm{H_2O}$ has two similar bonds at $104.5^\circ$ that add to a large net dipole. $\mathrm{CCl_4}$ is nonpolar for the same reason as $\mathrm{CO_2}$, while $\mathrm{CHCl_3}$ is polar because one substituent differs and the cancellation fails.

Common pitfall: deciding polarity from the bonds alone. Polar bonds are necessary but not sufficient, the geometry decides whether they cancel. Carbon dioxide is the standard counterexample: very polar bonds, zero net dipole.