Assuming the central atom has no lone pairs of electrons in its outer shell, here are the bond angles:
|
Number of ligands bound to central atom |
Bond angle(s) | Molecules with these angles |
| 2 | $180^{o}$ | BeH$_{2}$ |
| 3 | $120^{o}$ | AlCl$_{3}$ |
| 4 | $109.5^{o}$ | NH$_{4}^{+}$, CH$_{4}$ |
| 5 | $90^{o}$, $120^{o}$ | PCl$_{5}$ |
| 6 | $90^{o}$, $180^{o}$ | SF$_{6}$ |
The angle of $109.5^{o}$ is quite tricky to derive, it's known as the 'tetrahedral angle'. There's a cunning derivation on the Ask NRICH forum here.
If we didn't know anything about the water molecule, we'd expect it to be like BeH$_{2}$. However, oxygen has six electrons in its outer shell. Two of these form covalent bonds with hydrogen atoms, leaving two "lone pairs". These repel against each other and also against the electrons in the covalent bonds, and force the bonds to be closer than we might initially guess.
It seems the lone pairs repel other electrons more strongly than electrons in covalent bonds. This could then explain the different shapes of NH$_{4}^{+}$ and NH$_{3}$ as the lone pair forces the covalent bonds in NH$_{3}$ to be closer together than in NH$_{4}^{+}$ (which is symmetric).
In water, the two lone pairs force the O - H bonds even closer together than the N - H bonds in NH$_{3}.$