Drawing the Lewis Structure for XeF4

Viewing Notes:

  • In the XeF4 Lewis structure Xe is the least electronegative and goes at the center of the structure.
  • The Lewis structure for XeF4 requires you to place more than 8 valence electrons on Xe.
  • Xenon (Xe) can have more than 8 valence electrons in your Lewis structure.
  • Hydrogen (H) only needs two valence electrons to have a full outer shell.
  • You'll want to calculate the formal charges on each atom to make sure you have the best Lewis structure for XeF4.

Transcript: 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.