XeF4, Chemistry?

Question

1. What is the electron geometry of xenon tetrafluoride XeF4?

2. What are the two possible locations for the lone pairs in xenon tetrafluoride?

3. what is the lewis structure for xenon tetrafluoride?

4. which of the above figurations is most favorable?

5. what is the molecular geometry of xenon tetrafluoride?

Answer 1

All answers in my link

Answer 2

I think if you systematically go through the steps in drawing a lewis structure, you can answer this question. First, you take all of the valence electrons from Xe and the 4 F atoms and add them together (36). Draw out a preliminary structure with the 4 F atoms bonded to Xe with one bond each. subtract 2 electrons for each bond you used (8), which gives you 28 electrons to distribute around the compound. Assign electrons to the most electronegative atoms, in this case F. You will have 6 electrons on each fluorine, for a total of 24, leaving you with 4 more to put on Xe. You'll notice that Xe now has 4 bonds and 2 lone pairs, which means you need 6 molecular orbitals. In order to get 6 MO's, you need to mix in 6 atomic orbitals- 1 s orbital, 3 p orbitals, and 2 d orbitals, to give you sp3d2 hybridized orbitals.
For an atom with 6 MO's around it, VSEPR says you should have a square bipyramidal structure. so with 4 F atoms and 2 lone pairs, you can have 1 of 3 structures, 2 of which are the same. You can have both lone pairs in equatorial positions, or you can have 1 equatorial, 1 axial, or you can have both axial. Realize that the axial positions are 180 degrees apart while the equatorial positions are 90 degress apart, and the axial-equatorial angle is also 90 degrees apart. If I was a lone pair, I'd want to be as far away from another lone pair as possible, so I'd want to be in an axial position. So, if both lone pairs occupy axial positions, then the F's are all equatorial, forming a square planar structure.