To find the number of valence electron you get in one molecule of xenon oxytetrafluoride, add the number of valence electrons of each individual atom that makes up the molecule.
So, xenon oxytetrafluoride will have a total of
#8#valence electrons are coming from the xenon atom
#6#valence electrons are coming from the oxygen atom
#7#valence electrons from each of the four fluorine atoms
Now, xenon will be the central atom. It will bond with the oxygen atom via a double bond. This will ensure that oxygen has a complete octet.
The xenon atom will bond with the four fluorine atoms via four single bonds, which will ensure that each of the four fluorine atoms gets a complete octet.
These bonds will account for
#overbrace(1 xx 4)^(color(red)("one double bond")) + overbrace(4 xx 2)^(color(purple)("four single bonds")) = "12 valence e"^(-)#
#2#lone pairs on the oxygen atom
#3#lone pairs on each of the four fluorine atoms
#1#lone pair on the xenon atom
Don't worry about the fact that the xenon atom gets more than eight valence electrons here, it is more than capable of expanding its octet to accommodate these extra electrons.
The molecule's Lewis structure will look like this
Regions of electron density are bonds to other atoms (a single, double, or triple bond counts as one region of electron density) and lone pairs of electrons.
In your case, the xenon atom is surrounded by six regions of electron density
- a total of five bonds to other atoms
- one lone pair of electrons
According to VSEPR Theory, a molecule that has a central atom with a steric number equal to six and one lone pair of electrons falls into the
Indeed, a xenon oxytetrafluoride molecule looks like this