Question

Question #57594

Answer 1

First you must draw the Lewis structures of these compounds.

Explanation:

Then you use VSEPR theory to predict their shapes.

The number of lone pairs and bond pairs comes automatically from this procedure.

`bb"XeF"_3^+`

1. `"Xe"` is the least electronegative atom, so it is the central atom..

2. Your skeleton structure will have an `"Xe"` atom with 3 `"Xe-F"` bonds.

3. In your trial structure where every atom has an octet, there will be 26 valence electrons.

4. `"XeF"_3^+` has 8 + (3×7)-1 = 28 valence electrons — 2 more than in your trial structure.

5. Insert these electrons as a lone pair on the `"Xe"`. The `"Xe"` atom will have 3 bonding pairs and 2 lone pairs.

Now we use VSEPR Theory.

6. This is an `"AX"_3"E"_2` molecule. The electron geometry is trigonal bipyramidal, with two lone pairs in the equatorial positions. The molecular geometry is T-shaped.

`bb"XeF"_5^+`

Using the above procedure, you find that your trial structure has 40 valence electrons, but you actually have 42 electrons available.

Your Lewis structure will have 5 `"Xe-F"` bonds and a lone pair on the `"Xe"`.

This is an `"AX"_5"E"` molecule. The electron geometry is octahedral, with a lone pair at the bottom apex.

The molecular geometry is square pyramidal.

Now, can you do the rest? If you have difficulties, just post another question.