What is the electron-domain geometry of XeF4?

Answer 1

Remember, the geometry of a molecule has to do with how many electron pairs AND lone pairs there are around the CENTRAL atom. Basically, count up the lines and the pairs of dots.

Notice that 'Xe' has 8 electrons in its valence shell, and 'F' has only 7.

When we put 4 'F' atoms around 1 'Xe' atom, each Flourine atom is going to want to 'share' one of Xenon's electrons so that it can have 8 electrons.

Also remember that when an atom shares electrons, it shares in pairs. Each shared bond has 2 electrons.

Now count up the number of electron PAIRS around the CENTRAL atom. We have 4 bonds (one with each Fluorine atom), and 2 lone pairs of of electrons bringing our count up to 12 electrons around Xenon, or 6 pairs total. (This violates the 'Octet Rule', but it's ok in this situation)

Here is a list of geometries according to the number of Electron Domains:

2 'ED's (electron domains) = Linear shape

3 'ED's = Trigonal Planar

4 'ED's = Tetrahedral

5 'ED's = Trigonal Bi-pyramid

6 'ED's = Octahedral

The electron-domain geometry of XeF4 would be Octahedral.

Hope this helped!