We know that the molecule of H₂O is V-shaped. This is what makes it a dipole.
But why is that? I mean, if the hydrogens have a partial positive charge, then they should try to get away from each other, until they are diametrically opposite to the oxygen molecule. But that doesn't happen. Why is this?
The oxygen atom fills its octet rule by forming two bonds. It shares 1 electron in a covalent bond with each hydrogen and has 4 remaining valence electrons. It is sp$^3$ hybridized and has the 4 non-bonding electrons in two lone pairs. An sp$^3$ hybridized atom has four attachment points spaced approximately 109$^\circ$ apart and has the shape of a pyramid with a triangular base. See this image for an example. You don't typically see the water molecule drawn in 3 dimensional space with the lone pairs of electrons, all you see is a planar molecule with H-O-H bond angle of ~109$^\circ$.
It is not the + charges on hydrogens, but the negative charges on the O-H bonds, and on the O it self that is dominant in determining the molecule's shape. There are four non-bonding electrons on O -two lone pairs- on top of the two O-H bonds.
So you should think of four groups around O, not two.
I mean, if the hydrogens have a partial positive charge, then they should try to get away from each other.
Well, this is not a complete answer to your question, but this could be a factor I guess.
Bond pair – lone pair repulsion is always greater than bond pair – bond pair repulsion
So what I want to say is that the repulsion between the O---H bond and the lone pair is so heavy such that it is really hard for two hydrogen atoms to get away each other as you are expecting to happen.
In the bonding of Oxygen with Hydrogen, Oxygen ends up with 2 extra electrons, giving it what is called the "bent" shape. This is because the attachments and the electrons all repel each other but the electrons are not drawn except in a Lewis Dot structure such as the one here: http://en.wikipedia.org/wiki/File:Water-2D-flat.png
