molecular geometry- XeF2?

[Mdude]

One Kaplan problem I did said that XeF2 has linear molecular geometry.

Another problem said that 2 lone electron pairs surround Xe in XeF4.

Can someone explain these, please? Thank you so much!

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[RPedigo]

Sure... so remember that xenon has access to the low-lying d-orbitals and so it can have what we call an expanded octet. That means that there can be more than eight electrons around the xenon. Sulfur and phosphorus are other common elements that exhibit this phenomenon.

So xenon has 8 valence electrons, and each of the two fluorines have 7 valence electrons for a total of 22 electrons, or 11 electron pairs. So if we place Xe in the center and the F on the outside, we need three lone pairs on each fluorine (fluorine likes to make one bond to keep its formal charge zero), so we've used up six electron pairs, because we have three lone pairs on each fluorine. So fluorine makes one bond each, so that's two bonds total, so that's two more electron pairs.

We have now used 8 electron pairs, and have three left. So let's put them on the xenon and see what happens. With three lone pairs on the xenon, we have three lone pairs and two bonds on xenon, which gives it a formal charge of zero. Let me know if you need help calculating formal charge. So since we have five things attached to the xenon (three lone pairs, two bonds) we adopt a trigonal bipyramidal geometry, but three of the spaces are occupied by lone pairs, which gives you a linear shape.

http://www.up.ac.za/academic/chem/mol_geom/xef2_2.gif

The concepts of shape are pretty tough and hard to explain over a message board, but please let me know what parts are confusing and I can further elaborate.

 

[tncekm]

RPedio nailed it, but I personally would have a hard time interpreting that if I wasn't familiar with AXE and VSEPR, so here is an alternate explanation; hopefully it helps.

I just think of it this way:

- You've got your "central atom", and that's Xe. Xe is a noble gas, right, but forget about that for a minute and just accept that its bound to Fl.
- Each Fl will donate an electron to Xe and Xe will reciprocate (Fl makes single bonds), so there are two bonds, and two of the 8 electrons from the Xe valence shell are "used up" and there are three lone pair remaining.
- First priority is realizing that electron repulsion is very strong, and the electrons will orient themselves in a fashion that will make each one as far apart as possible from the others. The only way to do this is when there are 3 electron pairs is to make a "triangle".
- Second priority is orienting the Fl. The electron pairs have made a "triangle" now the electron clouds from the Fl will be repelled by the lone pairs, and they will orient themselves as far away from each other and the lone pairs as possible. So, one goes above the triangle and the other goes below.

So, what you have is an Xe atom with three lone pairs oriented in a planar triangle shape (I visualize them horizontally) with one Fl atom below the triangle and another above. The result is a linear shape for the XeFl2 molecule, but if you include the electrons in the geometry, its technically trigonal bipyramidal.

 

[oohchki]

Another problem said that 2 lone electron pairs surround Xe in XeF4.

This one is very similar to the one already explained. Xenon has 8 electrons and each fluorine has 7, for a total of 8 + (7 x 4) = 36 electrons.

Since fluorine likes to form only one bond, each fluorine has a single bond (2 electrons) to Xenon, and is surrounded by 3 lone pairs (6 electrons), so each of the fluorines is using up 8 electrons, times 4 = 32 for all of them.

36 - 32 = 4, which are the electrons in the 2 lone pairs surrounding Xe. (geometry is octahedral and shape is square planar)

 

[tncekm]

Good explanation, but wouldn't the geometry be square bipyramidal? Edit: nm, I guess that's essentially the same shape. 😀

 

[Mdude]

thanks all- that makes perfect sense now. I appreciate your help!