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

Aug 14, 2017

There's only one structure.

${\text{XeF}}_{5}^{+}$ has contributions from...

• $\text{Xe": " "" "" 8 valence electrons}$
• $\text{F": " "" } 5 \times 7$ $\text{valence electrons}$
• $\left(+\right) : \text{ } - 1$ $\text{valence electrons}$

and thus has $42$ of them to distribute.

1. Start with the skeletal structure and that uses $5 \times 2 = \underline{10}$ of them.
2. Each fluorine atom can hold three lone pairs, i.e. $3 \times 2 \times 5 = \underline{30}$ accounted for.
3. That leaves $\underline{2}$ electrons for one lone pair on the $\text{Xe}$.

The axial-equatorial bond angles are IDEALLY ${90}^{\circ}$, and the trans bond angles are IDEALLY ${180}^{\circ}$.

But of course, the lone pair on $\text{Xe}$ does not allow that... it crunches the equatorial fluorines like an umbrella. The true bond angles are less, by maybe $2 - {8}^{\circ}$.