Drawing the Lewis Structure for XeO2F2

Viewing Notes:

  • The Lewis structure for XeO2F2 requires you to place more than 8 valence electrons on Xe.
  • 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 XeO2F2.
  • There a total of 34 valence electrons in XeO2F2.

Transcript: This is Dr. B. Let's do the XeO2F2 Lewis structure. For XeO2F2, we have a total of 34 valence electrons. We'll put the Xe in the center and then we'll put an Oxygen here and here, and a Fluorine right there and there. We'll put a chemical bond right here between the atoms. Each bond represents two electrons, so we've used 2, 4, 6, 8 valence electrons. And then we'll fill the octets of the outer shell. So we have 8, 10, 12, 32, and remember we had 34. So we've used 32 and we'll put those remaining two right here on the central Xenon atom.

When you check the formal charges, you'll find out that it looks just like this. So while all the charges add up to zero, overall we'd like to have everything, each of the atoms, at zero itself. So let's try another structure and see how that looks. So to get rid of that +2 formal charge on the Xenon, I'm going to move two electrons from this Oxygen to form a double bond here. That should get rid of part of it. And then this here should get rid of the rest of it. So I think that will make the formal charges for the entire molecule zero, and each of the atoms, as well.

When we recalculate the formal charges, we can see they're now all zero. So forming those double bonds with the Oxygen solved our problem. We're still using 34 valence electrons, so that's OK. So that makes this the best structure for XeO2F2.

This is Dr. B. with the Lewis structure for XeO2F2, and thanks for watching.