Intermolecular Forces in BF3

[habbababba]

Hi!

Boron trifluoride is a nonpolar molecule due to its high symmetry even though the covalent bonds within the molecule are polar. As a result, the only type of intermolecular forces in BF3 would be the London dispersion forces. I understand that these forces are exhibited by nonpolar molecules because of the correlated movements of the electrons in interacting molecules. Because the electrons in adjacent molecules "flee" as they repel each other, electron density in a molecule becomes redistributed in proximity to another molecule and this results in the creation of a temporary dipole moment in the molecule.

However, the boron atom in BF3 is still a partially positive center and the fluorine atoms represent partially negative ends. Wouldn't it make sense that this partially positive center of one BF3 molecule would attract at least one of the partially negative fluorine ends in another BF3 molecule? If yes, then this type of intermolecular forces would resemble (but not identical, by definition, to) the permanent dipole-dipole forces in water. In other words, the cause of these intermolecular forces is not an induced temporary dipole in neighboring molecules.

Any contribution is appreciated.

[Enthalpy]

Hi Habbababba, welcome here!

 

I'm very pleased with your suggestion. Though, other people here may understand the topic better than I do.

 

Trying to compare : BF₃ boils at 173K while CF₄ boils at 145K despite having more atoms. Whether carbon is easier to polarize than boron is unclear to me, but the four fluorines making the carbon inaccessible do favour the gas. This goes in the direction of BF₃ interacting by sub-molecular permanent polarization.

 

On the other hand, SF₆ boils at 209K, but is bigger, so it gives no clear hint to any explanation.

 

As an orientation for the BF₃ molecules in a solid, and for a few molecules in a liquid, with each rim facing the centre of a neighbour molecule, you have the example of solid benzene

http://en.wikipedia.org/wiki/Benzene(picture top right)

though it doesn't have to result from a permanent polarization in benzene.

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I dislike some wording details in Wiki's article

http://en.wikipedia.org/wiki/London_dispersion_force

because in a stationary state, electrons don't move nor flee in a molecule: they're immobile, in the sense that the amplitude (not the phase) of their wave function is static.

 

So it's not a matter of "instantaneous" position. The proper formulation is abstract, alas: one has to write one single wave function for the electrons that interact, as psi(r₁, r₂)*phase(t) for two electrons, and this psi - which is stationary - contains the electron interaction in that it differs from any psi(r₁)*psi(r₂).

 

At any instant, you may find an electron anywhere according to psi(r₁, r₂). It's the probability or finding a second electron at one place that depends on where you have located a first electron.

Edited February 13, 2015 by Enthalpy