[joeeykn245]

    I am a bit confused as to how to estimate melting points of molecules. For instance:

Which has a higher BP/MP,   CaO  or LiF?

I thought it was LiF because the smaller atoms result in stronger intermolecular forces because of increased charge density versus the much larger CaO.

Then again, the larger CaO would have stronger dispersion forces because the electrons are polarized more easily due to the larger size.

Which one of these is the deciding factor in estimating BP/MP?

[formaldehyde23]

Both of your compounds are ionic. That means you use lattice energy to determine which compound has a higher melting point.
The lattice energy is the amount of energy you need to exert in order to break an ionic solid into its aqueous counterparts.
CaO has a 2+ cation and a 2- anion. That means that the compound is much more strongly bonded to each other; the lattice energy is greater.
LiF has a 1+ cation and a 1- anion. LiF will therefore not have as high of a lattice energy than CaO.
So, since CaO has a higher lattice energy than LiF, it takes more energy to convert the solid into liquid, meaning that CaO has  higher melting point.

Your first idea is the distance between the centers of the atoms. The distance does not take priority in determining lattice energy as the charge of the cation/anion.

[joeeykn245]

Thanks for the detailed response formaldehyde!

Disregarding if the compounds are ionic however, what is the general trend with atom/molecule size and the strength of inter-molecular forces?

[formaldehyde23]

Strength of intermolecular forces depend on the type of atom that the attraction is connecting to. You can look at hydrogen bonds first: Hydrogen can only "hydrogen bond" with three distinct elements: Nitrogen, Oxygen, and Fluorine. If you have a larger atom such as Cl, it is much more difficult for hydrogen to "H bond" with the chlorine.
You can even look at ion-dipole interactions. In order for this intermolecular force to occur, the ion needs to be very small and highly charged so that it can be hydrated in solution.
So, the smaller the atom, generally, the greater the strength of the inter-molecular forces.

[lokifenrir96]

Actually, in general, as the size of the molecule increases, the strength of Van Der Waal's forces of attraction between the molecules increases. This is because of an increase in the surface area of contact between molecules, hence the ease of polarisation increases, and magnitude of dipoles increases, and so does intermolecular forces of attraction. This is, of course, assuming you're talking about instantaneous-induced dipoles, so the molecule must be non-polar.

Same holds true for permanent-dipole - induced dipole.

I'm not sure for permanent-dipole - permanent-dipole molecules, though. Anyone want to clarify?

For H-bonds, the strength of the H-bond would depend on the atom the H is covalently bonded to, as well as the atom the H is H-bonded to. So if it's Flourine, then the H-bond is likely to be stronger than if it were Chlorine, it would be weaker. Like formaldehyde mentioned, the more electronegative said atom is, the stronger the H-bond, and generally, the smaller the atom is, the more electronegative it is.

For ion-dipole interactions, yes the higher the charge density of the ions, the easier it is for water molecules to interact with them and cause them to dissolve.