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

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?

• Mass of phosphine is nowhere near 88. Not that it matters much, though. – Ivan Neretin Feb 5 at 6:08
• Damn it, I confused myself because it has a boiling point of -88C. It's PF3 that has a mass of 88 as well. So actually my question should be: why does phosphine have a higher boiling point than phosphorus trifluoride? – Rafael Feb 6 at 0:59

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.