Question

What is the Lewis structure of `XeF_2O`? What should be its geometry?

Answer 1

Well we got `8_"Xe"+6_"O"+2xx7_"F"*"valence electrons"`...i.e. 28 electrons to distribute over four centres...

Explanation:

And so we got 14 electron pairs....and this is easier to represent on paper than it is on this editor.

We gots 3 bonding electron pairs, and two lone pairs around the central xenon atom. Electronic geometry is `"trigonal bipyramidal"`....but molecular geometry is `"T-shaped...."`

And for a Lewis structure of `O=Xe(-F)_2`...around the xenon and the oxygen atoms there are TWO lone pairs`"..."`around the fluorine atom there are THREE lone pairs...

Given the influence of the lone pairs....we would anticipate that `/_F-Xe-F>180^@`..note that lone pairs SHOULD be stereochemically active in that as lone pairs they are close to the parent atom, and exert a disproportionate stereochemical influence. You should look up the molecular geometry of this beastie.