Originally Posted by
dihydrogenoxide
Also consider how electronegative Fluorine is, creating stronger van der waals forces, or london dispersion forces. Both compounds are straight line structure. However, Fluorine is more electronegative than iodine. So in NaF, the side with the Fluorine is slightly negatively charged, which creates a higher attraction to the Na side of the other NaF molecules. Therefore, it is harder to seperate the molecules, giving it a higher boiling point. This phenomenon can also be seen in NaI, but because Iodine is less electronegative than Fluorine, the london dispersion forces aren't as strong, so it has a lower boiling point than NaF.