CHEMISTRY COMMUNITY

Why is it that AsF3 has a higher boiling point than AsF5, despite AsF5 being a bigger molecule? I understand that AsF3 is polar and has dipole-dipole IMF, but do we ignore the molecular weight comparison?

In this case, the molecular weight comparison is not as important. Even though AsF5 has stronger London dispersion forces than AsF3, the dipole-dipole interactions that AsF3 has are much stronger than any London dispersion forces. In general, you only need to focus on London dispersion forces and compare molecular weights when it's dealing with nonpolar molecules. If other types of intermolecular forces are present, you focus on those.

You would consider the weight if both molecules are not polar which is not the case here. AsF3 displays dipole-dipole interactions which indicates it's polar while AsF5 only displays LDF forces so it's not polar. Dipole-dipole is stronger than any LDF so AsF3 would have the higher boiling point.

AsF5 is non-polar while AsF3 is so it has more attractive dipole dipole forces, therefore requiring more energy to overcome IMFs

This question has to do with polarity, not size. Since AsF3 is polar, it has stronger dipole-dipole interaction than AsF5, a nonpolar molecule.

Dipole- Dipole is a stronger IMF than LDF so that is why AsF3 has the higher boiling point. Yes AsF5 has higher LDF but it does not matter due to the Dipole-Dipole.