What type of intermolecular forces are exhibited in BF3?

[joerog]

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.

 

[DrDu]

Of course a molecule like BF3 has nonvanishing permanent higher dipole moments, the first being the quadrupole moment. However, the corresponding fields fall off very rapidly with distance, so that they are important only for short ranges.

 

[joerog]

1) By "BF3 has nonvanishing permanent higher dipole moments", are you implying that BF3 as a whole exhibits an identifiable electric pole?
2) Why do the fields fall off very rapidly?
3) When you say that these permanent higher dipole moments are important only for short ranges, how do their ranges compare to those involved in London dispersion forces? I know that London dispersion forces are also effective over short distances.

 

[DrDu]

1)Yes, the first is a quadrupole moment.
2+3) The electric field of a monopole falls of as $1/r^2$ (Coulombs law), that of a dipole as $1/r^3$, of a quadrupole as $1/r^4$. Consider your favourite textbook of electrodynamics for details and maybe you find this talk interesting:
http://crm2.univ-lorraine.fr/pages_perso/Angyan/Documents/IMF/pdf/imf-part1.pdf

 

[joerog]

Thanks a lot. That link was helpful. However, I read that London dispersion forces start "to decay" as R^-6, which is a shorter range than that of a quadrupole electric field. Given that BF3 is known to make LDF, doesn't that mean that over such range the electric field of the quadrupole should also be effective?

 

[DrDu]

LDF?

 

[joerog]

Excuse me, London dispersion forces.

 

[DrDu]

You have to be carefull which distance dependence you use. I would consider the thermally averaged quadrupole-quadrupole interaction, which i believe to falls of like $1/R^8$, i.e. more rapidly than London forces. Take also in mind that London or dispersion forces is often used in sensu lato, meaning all kinds of multipole interactions.

 

[joerog]

Thanks for the clarification on the distance. Regarding the term "London dispersion forces", I am referring to those generated by instantaneous temporary dipole moments due to the movement of the electrons about the molecule. I'd take it that a quadrupole is a permanent type of electric pole. That's why I was making a distinction between the nature of the electric field of a quadrupole and that involved in London dispersion forces.