### Drawing the Lewis Structure for XeF2

Viewing Notes:

• The Lewis structure for XeF2 requires you to place more than 8 valence electrons on Xe.
• Hydrogen (H) only needs two valence electrons to have a full outer shell.
• Xenon (Xe) can have more than 8 valence electrons in your Lewis structure.
• You'll want to calculate the formal charges on each atom to make sure you have the best Lewis structure for XeF2.

Transcript: Hi, this is Dr. B. Let's do the XeF2 Lewis structure. Xenon, on the periodic table, has 8 valence electrons, plus Fluorine, 7, although we have two Fluorines so we'll multiply that by 2. That'll give us a total of 8 plus 14: 22 valence electrons. Xenon is the least electronegative. It'll go at the center, and we'll put the Fluorines on the outside. So we have 22 valence electrons. Let's form chemical bonds by putting two here, so that's 4; then around the Fluorines, 6, 8, 10, 12, 14, 16, and the center, 18, 20.

So we still have two valence electrons left over, we still have a pair left. So I could try to form a double bond with the Fluorine, but I know Fluorine's very electronegative. It doesn't normally form double bonds. I also know that Xenon can have more than 8 valence electrons, so I'm just going to put the extra two right here. What we'll do now is, we'll look at the formal charges to see if this is actually the structure that makes sense, or we maybe need that double bond.

Let's take a look at the formal charges, then. So on Fluorine right here, we have 7 valence electrons because it's in group 7 or 17 on the periodic table. We have nonbonding electrons, 6 of those. And then two bonding electrons which we divide by 2. Seven minus 6 minus 1 is zero. For the Xenon, it's in group 8 or 18, has eight valence electrons. We're looking at nonbonding, we have these two and then up here, these four, for a total of 6. And then bonding electrons, we have 2, 4; so 4 divided by 2. Eight minus 6 minus 2 is zero.

When I look at the Lewis structure for XeF2, and I look at the formal charges, I can now see that the formal charges are zero and that this is going to be the best Lewis structure for XeF2. This is Dr. B., and thanks for watching.

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