0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Mass of phosphine is nowhere near 88. Not that it matters much, though. $\endgroup$ – Ivan Neretin Feb 5 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 at 0:59
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 at 1:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.