Chemical Bonding: XeH₄ Lewis Structure

Drawing the Lewis Structure for XeH₄

Viewing Notes:

• The Lewis structure for XeH₄ 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 XeH₄.

Transcript: Hi, this is Dr. B. Let's do the XeH4 Lewis structure. Xenon has 8 valence electrons. Hydrogen has 1, but we have four Hydrogens, for a total of 8 plus 4: 12 valence electrons. We'll put Xenon in the center and then Hydrogens on the outside. Now we'll take and put two electrons between the atoms to form chemical bonds. We've used 8, and we have Hydrogens with two each, so their outer shells are full. And Xenon also has eight. So we've used 8 valence electrons, but we have 12 total to work with.

So we still have 4 valence electrons. The thing is, Xenon can have more than 8 valence electrons. So we could add another four valence electrons to Xenon to solve our problem. Let's put two right here, and then two right here. So now we've used all 12 of the valence electrons. The Hydrogens, they're OK with two each, and the Xenon, it's OK also with 12.

So that's the Lewis structure for XeH4. You could check the formal charges to make sure that they were zero, and that this was the best structure. If you did, you'd find out that it is the best structure for XeH4. This is Dr. B., and thanks for watching.