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?

  • $\begingroup$ Mass of phosphine is nowhere near 88. Not that it matters much, though. $\endgroup$ – Ivan Neretin Feb 5 '19 at 6:08
  • $\begingroup$ 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? $\endgroup$ – Rafael Feb 6 '19 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.

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  • $\begingroup$ Thank you. Actually I screwed up my question completely. I meant to ask: why does PH3 have a higher b.p. than PF3 when it is less massive and has virtually non-polar bonds, whereas PF3 has polar bonds and is asymmetrical? They both have the same shape. $\endgroup$ – Rafael Feb 6 '19 at 1:02

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