The outer shell electrons in the central oxygen are not completely evenly distributed. There's 4 'pairs', two of which are entirely the oxygens while the other two pairs the oxygen has to share with the two hydrogens it's bonded to (at least the model is kinda like that).
As such there's a slight redistributing of the charges around the oxygen with the two unbounded pairs being a bit higher in electron charge density than the areas where the hydrogens connect to the oxygen. Hence the bonds are pushed together ever so slightly by this extra concentration of negative charge and you get a decrease in the bond angle of the hydrogens compared to say methane which is an evenly distributed tetrahedron of electrons.
I know that's very much 'high school chemistry' take on the electron arrangements etc but I doubt the answer requires a more 'quantum' approach than that.
According to Valence Shell Electron Pair Repulsion (VSEPR), a pair of electrons takes more "space" than the usual bonding pair of electrons.
As for a very simplistic answer, this is very correct.
Why when there are in the bond pair also 2 electrons and the lone pair 2 electrons. What is the problem? Also the sp3 hybrid orbitals are on same energy level so, they are all similar by size.
When unshared electron (electrons in lone pairs) are present, VSEPR does "predict" slight distortion of ideal bond angels. It is assumed that an outer atom "pulls out" in bonding electron pairs and makes them require less space about the central atom than the unshared electrons.
(1) The bond angle tends to distort to give more room ("space") to unshared electron pairs.
(2) in geometry with positions that are not equivalent, the unshared electron pairs will occupy the more spacious positions.
In other words, if there is one only one unshared pair electrons, that pair electron will not affect the geometry of the compound. For example, SnCl2 and BF3 have the same geometry, the angle between the atoms are 120 degree.
However, if there is more than 1 unshared pair of electrons, these pairs will try to locate as fas as possible from each others. That explained why H2O and CF4 are different from geometry.
I agree that H2O and CF4, for example, are sp3 hybridization. But the difference in shape (geometry) has nothing to do with the energy level when you comparing these 2 compounds. But if you compare the geometry of H2O when the bond angles are all 109.5 degree with the one that has less than 109.5 degree, then we will say that the one that has the bond angle less than 109.5 is more stable which means it's in the lower energy.
Also, VSEPR is a theory, so at least, we should not try to prove that it wrong! Some days, when you become a chemist, you can work on that and give another theory which shows that unshared pair electrons take less space than bonding pair electrons. But up until now, we agree with this theory and it helped us explain not only the stuff about geometry of compound, but also some mechanism in organic chemistry.
Have fun studying!
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