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My book says that from drawing the Lewis structure for the molecule below, you can conclude that it has dipole moment. Can someone please explain why?

structure of PF3

I understand that the dipole moment expresses the polarity of the bond. And that the dipole moment of a molecule is the sum of the dipole moments for each bond in the molecule.

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  • $\begingroup$ So if net dipole moment is for polar compounds, that would make $\ce{PF3}$ a polar compound. And it is. $\endgroup$ – M.A.R. Feb 11 '15 at 13:10
  • $\begingroup$ @MARamezani Yes but how can you see it from the lewis structure? $\endgroup$ – Shmoopy Feb 11 '15 at 13:11
  • $\begingroup$ I can see that from the difference between the electronegativities of F and P, and the shape of the molecule. $\endgroup$ – M.A.R. Feb 11 '15 at 13:12
  • $\begingroup$ @MARamezani Then why does BF3 have a moment dipole of 0? $\endgroup$ – Shmoopy Feb 11 '15 at 13:14
  • $\begingroup$ Because of its shape. I'll come in with an answer. $\endgroup$ – M.A.R. Feb 11 '15 at 13:15
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Fluorine is more electronegative than phosphorous.

The phosphorous is at the apex of a pyramid, the base of the pyramid being an equilateral triangle with a fluorine atom at each vertex.

The F-P-F angles are 96 degrees.

Each F-P bond contributes to the net dipole moment, as a vector from the P to the F. The net dipole moment is the vector sum of the vectors along the three P-F bonds.

If all four atoms were in a plane (a trigonal planar geometry), there would be no net dipole moment. But because the geometry is pyramidal, the vector sum is not zero and there is a net dipole moment.

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  • $\begingroup$ So the reason that BF3 has dipole moment of 0 is because that B can have 3 bonds thus making BF3 a triangular plane? $\endgroup$ – Shmoopy Feb 11 '15 at 13:26
  • $\begingroup$ @Shmoopy BF3 is trigonal planar (all four atoms are in the same plane); therefore, BF3 has no net dipole moment. $\endgroup$ – DavePhD Feb 11 '15 at 13:35
  • $\begingroup$ Do not forget the lone pair in $\ce{PF3}$, it contributes significantly to the dipole moment. (In reverse direction, but it does contribute.) $\endgroup$ – Martin - マーチン Feb 11 '15 at 14:40
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    $\begingroup$ @Martin excellent point, but for P will the lone pair be mostly s-like, (rather than sp3 hybridized)? $\endgroup$ – DavePhD Feb 11 '15 at 14:46
  • $\begingroup$ Argh, of course, you are right! There won't be much contribution from the lone pair in a (mostly) s orbital. Somehow I was confusing it with the lighter analogon. $\endgroup$ – Martin - マーチン Feb 11 '15 at 14:51

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