Why does phosphine have a dipole moment and a higher boiling point than carbon tetrafluoride?

Question

Phosphine, PH₃, and carbon tetrafluoride, CF₄, are small molecules of a similar size and the same mass of 88 au. CF₄ has a dipole moment of 0, which is unsurprising given its tetrahedral shape. However, I would expect P-H bonds to be non-polar (P and H have electronegativities of 2.19 and 2.2) and therefore the molecule to also have a dipole moment of about 0, despite being trigonal pyramidal and therefore asymmetrical. So I would expect it to exhibit similar intermolecular forces and a similar boiling point to CF₄. Yet PH₃'s dipole moment is quoted as 0.58 D. And PH₃ has a boiling point of -88 °C and CF₄'s is -128 °C.

Why does phosphine have such a higher dipole moment and boiling point than CF₄ despite it having non-polar bonds?

Answer 1

The dipole moment of $PH_3$ can be attributed to the lone pair on P, which is directed away from all three P-H bonds. You can imagine that there is a higher electron density near the lone pair, while the electron density is more or less uniform near the nonpolar P-H bonds. This does not happen for $CF_4$, so it has no dipole moment.