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QUESTION

# How can structure of XeF3+, XeF5+ , XeOF4 , XeF4 are possible according to VSEPR theory?And also how we can determined which atom or molecule have how many bond and lone pairs?

First you must draw the Lewis structures of these .

Then you use 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 .

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

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

Now we use 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 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.