What is the proper orbital hybridization O and N in $\ce{H_2O}$ and $\ce{NH_3}$? The rules say that they are sp and sp2 hybridized, respectively, but some sources say both are sp3. I have problems with both, and I'll try to explain them.
1) In the first case: According to elementary rules, the total number of hybridized orbitals should be equal to the number of sigma bonds. Thus, oxygen should be sp hybridization and then nitrogen should have sp2 hybridization (in the above molecules). However, this possibility does not relate well to the VSEPR model: in water, the hydrogen atoms would be striaght and perpendicular to the two perpendicular unhybridized p-orbitals.
2) If they have sp3 hybridization, (and why would they?), then there are some contradictions with electron delocalization/conjugation. In phenol, the delocalization in the benzene group is supposed to combine with an nonbonding orbital in oxygen. However, there's a 109.5 degree angle around the O atom, so then both nonbonding domains would be in the wrong orientation for conjugation. (I reference this page: http://chemguide.co.uk/analysis/uvvisible/bonding.html).
As a side question, does the conjugation/delocalization in phenol change oxygen's formal charge?