Hi, this is Dr. B. Let's do the XeF4 Lewis
structure.
Xenon has 8 valence electrons.
Fluorine has 7, but we have four of the Fluorines;
so that gives us 8 plus 28: 36 valence electrons.
We'll put Xenon in the center, it's the least
electronegative; and then Fluorines on the
outside, all four of them.
We'll start by putting two between atoms to
form chemical bonds, and then around the Fluorines.
We have 8, 10, 12, 14, 16, 18, 20, 22, 24,
26, 28, 30, and 32.
So we have 32, that gives us four more valence
electrons that we need to deal with.
Probably not going to be a double bond.
Fluorine's very electronegative, and really
doesn't form those double bonds.
But I know that Xenon can have more than eight
in its outer shell, so I'm going to take these
extra two pair of electrons right here, and
then put them right here, and that'll give
me a total of 36 valence electrons that I've
used.
Everything has octets; and Xenon, it has more
than an octet, but that's OK for Xenon.
If you wanted to check it, you could look
at the formal charges and you'd find out that
this is the best structure for XeF4, xenon
tetrafluoride.
This is Dr. B., and thanks for watching.
