How many structures are there for #"XeF"_5^(+)#?

1 Answer
Truong-Son N.
Aug 14, 2017

There's only one structure.

#"XeF"_5^(+)# has contributions from...

  • #"Xe": " "" "" 8 valence electrons"#
  • #"F": " "" "5 xx 7# #"valence electrons"#
  • #(+): " "-1# #"valence electrons"#

and thus has #42# of them to distribute.

  1. Start with the skeletal structure and that uses #5 xx 2 = ul10# of them.
  2. Each fluorine atom can hold three lone pairs, i.e. #3 xx 2 xx 5 = ul30# accounted for.
  3. That leaves #ul2# electrons for one lone pair on the #"Xe"#.

The axial-equatorial bond angles are IDEALLY #90^@#, and the trans bond angles are IDEALLY #180^@#.

But of course, the lone pair on #"Xe"# does not allow that... it crunches the equatorial fluorines like an umbrella. The true bond angles are less, by maybe #2 - 8^@#.

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