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Report (any type)/Brief report
22/09/2021
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.
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I want to thank your team (Jeff) for finding a new writer when you were unable to reach my existing writer to revise my paper. I appreciate your professionalism and efforts. Thanks
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