J.R. S. answered 11/30/19
Ph.D. University Professor with 10+ years Tutoring Experience
There will not be any hydrogen bonding in the first one, CH3-O-CH3 because to have H-bonding you need to have a H atoms attached to either a F, O or N atom. Not the case here. But this molecule will exhibit weak dipole-dipole because of the somewhat negative O atom.
In SO2, you have mainly London dispersion because it is non polar
In CH3NH2 you have hydrogen bonding because a H atom is bonded to a N atom
In PF5 you will have London dispersion because there is no H-bonding and the molecule is non polar so there won't be any dipole-dipole.
In XeCl2 you will have London dispersion because no H-bonding and again, the molecule is non polar so no dipole-dipole interactions.