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Author: Subject: Polarity and condensation physics
roXefeller
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[*] posted on 13-1-2021 at 16:16
Polarity and condensation physics


I was quizzing my daughter the other day on science and one question was draw a water molecule and label it. I made sure she understood the V-shape was important to get right in the drawing because almost all of the physical properties we know of water come from the polarity and van-der-waals forces. I drew a CO2 molecule as comparison, which has no polarity and is a gas at RT for example. It got me thinking about other nonpolar chemicals and why the distinction. For instance, propane, similar to ethanol but nonpolar and a gas at RT. Great. Why does this break down at higher molecular weights, say octane, which is liquid, albeit volatile, while something super jeavylike uranium hexafluoride is a gas. I know there are almost no van der waals forces on the UF6, but there shouldnt be many for octane either. A little explanation?



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Bedlasky
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[*] posted on 13-1-2021 at 19:59


Octane is longer than propane, have more C-H bonds, there is more space for van der Walls forces than in propane.

UF6 is clearly non-polar molecule. Bonds are highly polar, but sum of dipol moments is zero because of its structure. Forces keeping molecules together are small, this is reason why it is gas at RT. UF4 is solid, because it is a polar molecule.




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DraconicAcid
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[*] posted on 13-1-2021 at 20:04


UF4 will have a lot more ionic character, and that's why its a solid at RT. A tetrahedral UF4 would be just as non-polar as the octahedral UF6.

Don't disregard the strength of London dispersion forces. The higher the mass of the molecule, the more electrons there are, and the more polarizable the molecule is (and thus, the stronger the London forces).




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[*] posted on 13-1-2021 at 21:21


Tetrahedral UF4? No way. Lets consider that you have UF4 in gaseous state - - > divided to single molecules. Single UF4 molecule have pyramidal (and not tetrahedral) shape because of lone pair electrons (uranium have six valence electrons). So it's definitely polar.

[Edited on 14-1-2021 by Bedlasky]




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DraconicAcid
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[*] posted on 13-1-2021 at 21:23


Lone pairs? In a d or f subshell? Not a chance.



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[*] posted on 13-1-2021 at 22:19


Quote: Originally posted by DraconicAcid  
Lone pairs? In a d or f subshell? Not a chance.


Sorry, my fault.

https://sci-hub.se/https://www.sciencedirect.com/science/art...

So why it is UF4 more ionic than UF6? And why electrons in d and f shell can't be lone pairs?

[Edited on 14-1-2021 by Bedlasky]




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DraconicAcid
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[*] posted on 13-1-2021 at 22:40


I'm not sure I could say why d and f electrons don't act as lone pairs in the same way that p or sp hybrid electrons do. I just know that they don't. That's why [MX4] systems (such as FeCl4(-) or CoCl4(2-) ions) are tetrahedral, regardless of how many d electrons the metal has (unless they have 8, in which case they might be square planar, but that still doesn't work for VSEPR).

UF6 will be more covalent than UF4 because the uranium(VI) ion is smaller and more highly charged than the uranium(IV) ion- it is much more polarizing, so it can force the fluoride ion to share electrons (i.e., covalently bond). It's the same reason that SnCl2 acts like a typical ionic compound (water soluble solid, forms conducting solutions), and SnCl4 is a typical molecular compound (liquid, soluble in CCl4, forms non-conducting solutions).




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roXefeller
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[*] posted on 14-1-2021 at 13:08


I get why octane is liquid while propane isn't. Those are two separate comparisons. My question was why octane starts to develop polarity? It's heavier than propane obviously. But uf6 is way heavier than all of that.

Took a quick peek at the Wikipedia article for London dispersion forces. I think that answers my question best. The forces and the dipoles aren't static but can rearrange instantaneously in different modes to produce an overall condensed nature.

Quote:

The effects of London dispersion forces are most obvious in systems that are very non-polar (e.g., that lack ionic bonds), such as hydrocarbons and highly symmetric molecules like bromine (Br2, a liquid at room temperature) or iodine (I2, a solid at room temperature).
...
Larger and heavier atoms and molecules exhibit stronger dispersion forces than smaller and lighter ones.


[Edited on 14-1-2021 by roXefeller]




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[*] posted on 15-1-2021 at 10:06


Uranium hexafluoride is NOT a gas. It is a solid that sublimes at +56 Celsius.
The heaviest gas is actually tungsten hexafluoride - boils at +17 Celsius.

When a molecule has no polarity, the examples include noble gas atoms, and diatomic molecules.

The strength of London forces depends, for one, on the number of electrons. So look at noble gas series
He - 4 K
Ne - 27 K
Ar - 87 K
Kr - 121 K
Xe - 166 K
Rn - 211 K

When the number of electron is equal, consider how strongly the electrons are held. He and H2 have each 2 electrons, yet boiling points He - 4 K; H2 - 20 K. The one concentrated nucleus of He holds the electrons more tightly.
Now compare the hydro- and fluorocarbon series:
CH4 - -162 C - CF4 - -128 C
C2H6 - -88 C - C2F6 - -78 C
C3H8 - -42 C - C3F8 - -36 C
n-C4H10 - 0 C - n-C4F10 - -2 C
n-C5H12 - +36 C - n-C5F12 - +28 C
F has more electrons than H, but holds them more strongly. Going from methane to heavier fluorocarbons, for small molecules the smallness of H prevails, for bigger the low polarizability of F. Which is why the nonpolar fluorides are volatile.

The reason UF4 is less volatile than UF6 is the same "CF2" is less volatile than CF4. Since UF4 is "unsaturated" compared to UF6, it polymerizes (into crystals).
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