There are 4 valence electron pairs in water: 2 bonding, the O−H bonds; and 2 non-bonding, the O lone pairs. VESPER dictates that the most stable geometry of these bonding and lone pairs is tetrahedral.
And thus to a first approximation, the ∠H−O−H should be 109.5∘, which of course is the ideal tetrahedral angle - certainly this is the ∠H−C−H bond angle we observe in methane.
However (and there is always a however), because TWO of the electron pairs around oxygen are LONE pairs, these tend to lie closer to the oxygen atom. And these lone pairs tend to compress the ∠H−O−H bond angle down from the tetrahedral angle to approx. 104.5∘ by electrostatic repulsion of like charges. We may make the same argument for the ammonia molecule, NH3, trigonal pyramidal BUT tetrahedral to a first approximation. Capisce?
