# Why do we describe the "ELECTRONIC" geometry of the water molecule as tetrahedral, but its molecular geometry as bent?

$\text{Well, there are 4 electron pairs around the central oxygen atom,}$ $\text{so..........}$
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 $\angle H - O - H$ should be ${109.5}^{\circ}$, which of course is the ideal tetrahedral angle - certainly this is the $\angle H - C - H$ bond angle we observe in methane.
However (and there is always a $\text{however}$), 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 $\angle H - O - H$ bond angle down from the tetrahedral angle to approx. ${104.5}^{\circ}$ by electrostatic repulsion of like charges. We may make the same argument for the ammonia molecule, $N {H}_{3}$, $\text{trigonal pyramidal}$ BUT tetrahedral to a first approximation. Capisce?