How many structures are there for "XeF"_5^(+)XeF+5?

1 Answer
Truong-Son N.
Aug 14, 2017

There's only one structure.

"XeF"_5^(+)XeF+5 has contributions from...

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

and thus has 4242 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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