Question about Bonding, is XeF2 non-polar? Think I got it, make sure I'm right!!

[TheRyuu]

This came up in my AP Chem (HS Class) class today and I want to know the truth . So I come to the all knowing AT'ers and ask them.

The compound XeF2 <-with the 2 being a subscript , is that compound considered polar or nonpolar? It has a symectrical bonding so it can be considered nonpolar just on that, yet it has unshared pairs of electrons which means in could be Polar as well right?

->.. ..
F - Xe - F
->>..

Symetrical, yet it has unshared electrons on the Xe? This AP Chem stuff is so confusing

(Arrows to move the dots over)

EDIT:
I think I got it.

XeF4 - is non-polar because the two pairs of unshared electrons cancel each other out.

XeF2 - non-polar, Possibly the same reason, and it apparently has nice symmatry.

Right?

 

[BrownTown]

non-polar

 

[Nathelion]

Definitely non-polar. Since the complexes (atoms) on the "sides" are identical, the 3D layout would be symmetrical.

 

[CycloWizard]

It has three pairs of unbonded electrons and two very electronegative elements. This tells me it's probably a trigonal bipyrimidal 3-D geometry (though general chemistry was years ago now, so not 100% sure, but I think that's what it's called), with the fluorines at opposing ends. Since the fluorines are antipodes, the molecule would be non-polar.

 

[TheRyuu]

My AP Chem Teacher has a PhD and even she doesn't know, so I don't really think it's as simple as saying the F atoms are on the "sides" and the 3D Layout would be symmetrical.

The F's may be antipodes, but with the 3 pairs of unshared electrons, thats a classic sign of a polar molecule.

 

[TheRyuu]

I think I got it.

XeF4 - is non-polar because the two pairs of unshared electrons cancel each other out.

XeF2 - non-polar, Possibly the same reason, and it apparently has nice symmatry.

Right?

 

[CycloWizard]

Originally posted by: wizboy11
My AP Chem Teacher has a PhD and even she doesn't know, so I don't really think it's as simple as saying the F atoms are on the "sides" and the 3D Layout would be symmetrical.

The F's may be antipodes, but with the 3 pairs of unshared electrons, thats a classic sign of a polar molecule.

The way I'm looking at it, the three unshared electrons will be centered around the axis (picture a triangle). Then, at the poles directly above and below the Xe atom are the F atoms. This would be the state of lowest energy, so it only makes sense that it would be the natural configuration of the molecule.

Originally posted by: wizboy11
I think I got it.

XeF4 - is non-polar because the two pairs of unshared electrons cancel each other out.

XeF2 - non-polar, Possibly the same reason, and it apparently has nice symmatry.

Right?

I would think XeF4 is very slightly polar. There is no symmetric configuration that it could maintain, which means it would generally have a shape like water. The free electron pair would definitely be drawn towards the four fluorines, since they are so electronegative. However, the fluorines will tend to repel each other and will not get as close as the hydrogens in water, which is why I think the polarity would be very slight.

 

[TheRyuu]

So, it could be both? I google it and a lot said it was non-polar.

This is confusing me SOO much (not that it really matters though).

Maby I can "show off" by explaing why (it's polar/non-polar)

 

[CycloWizard]

I didn't trust myself, so I Google'd 'xenon tetrafluoride polarity' and got this link, which says the following:

Xenon tetrafluoride has two lone pairs and four bonds around the central atom. That is a total of six groups. Therefore the arrangement of electrons is an octahedral. However, the lone pairs do not show up in the molecular geometry. Because lone pairs are particularly repulsive, they will get as far from each other as possible. In this example that is on opposite sides of the octahedral. Therefore the overall molecular geometry is square planar.

Because the lone pairs are exactly opposite of each other, they cancel. The bond dipoles of the xenon-fluorine bond are also exactly opposite each other. All bond dipoles cancel; the molecule is nonpolar.

Xenon difluoride (XeF2) wasn't so easy to find. I found this, which verifies that it does, indeed, have three pairs of unbonded electrons (page 5). This leads me to believe that I was right and it is nonpolar.

 

[BrownTown]

To be honest this really shouldn't be complicated at all. Xenon has 8 valence electrons, flourine has 1. So XeF2 has 10 valence electorns = 5 pairs. 2 are in the bonds and 3 are lone pairs. The three lone airs are all in the same plane with the 2 flourins in a line orthogonal to the plane. With XeF4 there are 12 electrons, so 4 bonds and 2 lone pairs. The four flourins are in the same plane at the four corners of a square with the 2 lone pairs in a line orthogonal to the plane. The only part that shoudl be confusing is why you are allowed to have 10 or 12 valence electrons when classically only 8 are allowed.

OK, actually flourine has 7 valence electrons, but only 1 of those is ever actually used in bonding normally, so the idea is the same... you just have 22 valence electrons, 4 are in the bonds, 6 around each flouring leave 6 around the Xenon which means 3 lone pairs.