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Topic: London Dispersion Forces in Polar Molecules  (Read 277 times)

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Offline Lamda

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London Dispersion Forces in Polar Molecules
« on: January 14, 2020, 05:13:01 AM »
Hello everyone,,
London dispersion forces in noble gases: they arise due to an accidental, uneven distribution of the electron cloud in the atom. This in turn creates a temporary dipole in the atom which induces the formation of a similar dipole in a neighboring atom (provided they're close enough). The result is a short-lived force between the 2 atoms.

The premise that an accidental uneven distribution of the electron cloud occurs in a noble gas atom makes sense since there is not a permanent dipole moment in a noble gas atom.

What I'm trying to find out though is how could I extend this concept of an accidental temporary distribution of the electron cloud over a polar molecule? A polar molecule exhibits a permanent dipole moment, i.e. there is always a 'preferred location' on the molecule at which the electron density is higher, namely the more electronegative element. I say 'always' because if it wasn't, the dipole moment also would not be called permanent. If this accidental distribution of electron density were to occur in a polar molecule, it would alter the permanent dipole. But then why would such accidental distribution of the electron cloud in a polar molecule happen in the first place? By simply looking at the electronegativities of the elements, we can tell how the electron density is distributed over the molecule. I just can't rationalize how or why such accidental distribution of the electron density would need to take place in a polar molecule.

Any thoughts?

Offline mjc123

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Re: London Dispersion Forces in Polar Molecules
« Reply #1 on: January 14, 2020, 06:23:52 AM »
Quote
If this accidental distribution of electron density were to occur in a polar molecule, it would alter the permanent dipole.

But not permanently. There is a small, fluctuating dipole moment superimposed on the "permanent" DM; the time-average DM of the molecule is the "permanent" DM.

Quote
I just can't rationalize how or why such accidental distribution of the electron density would need to take place in a polar molecule.

By the same token, why should it take place in a noble gas atom? Given the Coulomb potential of the nucleus, there is no reason (is there?) for the electron distribution to be anything other than spherically symmetrical. Yet it does fluctuate; if it deviates from the equilibrium state, there is a force pulling it back towards that state, but it is constantly fluctuating.

Offline Babcock_Hall

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Re: London Dispersion Forces in Polar Molecules
« Reply #2 on: January 14, 2020, 09:23:55 AM »
I am not sure what is meant by "accidental."

Offline Borek

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Re: London Dispersion Forces in Polar Molecules
« Reply #3 on: January 14, 2020, 09:33:05 AM »
I am not sure what is meant by "accidental."

Probably the same as in "random fluctuations". This is a common mistake in wording, especially for non native speakers.
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Offline Enthalpy

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Re: London Dispersion Forces in Polar Molecules
« Reply #4 on: January 15, 2020, 05:55:24 PM »
Any thoughts?

I disagree just with the time-dependent description, including "fluctuation", "short-lived" and so on. The interaction is static.

It's something a bit abstract in quantum mechanics. The proper description of two particles is a single wavefunction ψ(ra, rb, t) where ψ is a complex scalar function as usual but it depends on both particles, here on their positions ra and rb.

Generally, ψ(ra, rb, t) can't be written as ψa(ra, t)×ψb(rb, t). As the particles rearrange, for instance electrons in molecules, they find combinations of positions that are more favourable than if each particle had a distribution independent of the other ψa(ra, t)×ψb(rb, t). As two electrons repel an other, they are more likely far from an other than very near.

But QM does not need a change over time for that. If you consider a helium atom as a simpler example, its two electrons on 1s have a distribution perfectly independent of time. QM calls it a "stationary" solution, which means "immobile" more or less.

Then the wavefunction is written as a ψ(ra, rb)×ei2πEt/h where |ψ| is independent of time and |ei...| too, that is, the probability density is perfectly static. Nevertheless, ψ(ra, rb) can't be written as some ψa(ra)×ψb(rb). The probability density of finding both electrons very close to an other is weak because they repel an other. But the probability density of any electron has spherical symmetry nevertheless.

In other words, It's not a matter of "when" an electron is around that position, then the other has a consequent distribution of probability density. It's "if". This has no equivalent in the macroscopic world hence is abstract to me.

By the way, this is quantum entanglement. It is extremely common.

The interaction of two molecules does things similar to the helium atom, just more complicated because there are more electrons.

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Polar molecules do present the interactions of non-polar molecules too. But dipole interactions tend to be stronger, and then the others maybe neglected.

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