# Why does Phosphorus trifluoride molecule have dipole moment?

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?

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.

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

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.

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