Question

What are the formal oxidation numbers of xenon, and fluorine, in `XeF_5^(-)`?

Answer 1

We got `XeF_5^(-)`....

Explanation:

The xenon atom has a formal negative charge.

The oxidation number of xenon in `XeF_5^-` is `Xe(+IV)`

For fluorine `Z=9`; and there are 6 lone pairs electrons, and 1 electron fro the `Xe-F` bond....hence the fluorines are neutral with 2 inner core electrons.

Around xenon there are 5 valence electrons from the `Xe-F` bonds, and 4 electrons from the two lone pairs....9 electrons, where neutral xenon has 8 valence electrons...hence `Xe^(-)`