# How is the valence electronic structure of the water molecule described?

There are 6 electrons from the oxygen, and 2 electrons from the hydrogens. There are 8 electrons in total ($4$ $e$ pairs) to distribute round the central oxygen atom; for which VSEPR makes a prediction.
4 electron pairs arranged around the oxygen atom would assume a tetrahedral geometry; bond angles are ${109.5}^{\circ}$. So the $H - O - H$ bond angle should (ideally) be ${109.5}^{\circ}$, as should the angle between the lone pairs. Because the lone pairs are larger and more diffuse than bonding pairs, they will tend to decrease the $\angle H - O - H$ to some extent. So water is a bent molecule with a $\angle H - O - H$ of about ${105}^{\circ}$ (actually ${104.5}^{\circ}$). (The shape of and the bond angle between the lone pairs are irrelevant to the structure of the molecule when it is determined.)
Note that this treatment predicts that all of the constituent atoms are neutral, i.e. 1 electron associated with each $H$, and 6 electrons (+2 inner core) associated with $O$. This treatment also predicts the structure of water's Group VI homologues, hydrogen sulfide, and hydrogen selenide, and hydrogen telluride. What would we predict the structure of the hydronium ion, ${H}_{3} {O}^{+}$ to be; where does the positive charge lie?