Lewis Structure of XeO3?

[Curtis Ngo_4E]

In Xenon trifluoride, which Lewis structure would you use? One in which there are 3 double bonds, one with 3 single bonds, or one with 2 single bonds and 1 double bond? Is formal charge the deciding factor in this scenario?

[Kevin Wright]

For the Lewis Structure of Xenon trioxide, it would be drawn with 3 double bonds and a lone pair of electrons on Xenon. Yes, the formal charge is the deciding factor for this. If they were all single bonds, it would fulfill the octet rule, but would result in a formal charge of 3+ for Xenon and 1- for the oxygen atoms. Xenon has unfilled d orbitals and therefore can have an expanded octet. So Xenon double bonds with the oxygens which results in a formal charge of 0 for Xenon and 0 for oxygen

[Victor Qiu 1C]

I think Kevin Wright’s answer is correct. Just to add on it, Xe forms 3 double bonds with 3 O, respectively, and has a lone pair electrons. Also, each O has 2 pairs of lone pair electrons.

[Tania Peymany 1A]

Yes, both Kevin and Victor are right. I would also like to point out that the molecular shape is also very important when drawing out the Lewis structure. In this case, the molecular shape would be triagonal pyramidal. But this is for XeO3, and I noticed that Curtis put Xenon Trifluoride, in which the Fluoride would have a -1 charge. This would lead to 30 valence electrons. Then 24 valence electrons would be used to fulfill the octets for the 3 Fluoride atoms, leaving a remaining 6. These remains 6 would then be applied to the Xenon as 3 lone pairs. And as Kevin mentioned, Xenon can have an extended octet, allowing for these extra valence electrons to surround the Xenon atom.