Drawing the Lewis Structure for N2O4

Viewing Notes:

  • There are a total of 34 valence electrons in N2O4
  • Nitrogen (N) is the least electronegative so the two Nitrogen atoms go at the center of the Lewis structure.
  • With N2O4 you'll need to form two double bonds between the two Nitrogen atoms and two Oxygen atoms to fill the octets with the available 34 valence electrons.

Transcript: This is the N2O4 Lewis structure. For N2O4, we have a total of 34 valence electrons. N is the least electronegative. We'll put the N's in the center, and then we'll put the Oxygens on the outside. We have 4 Oxygens. We'll put valence electrons between atoms to form chemical bonds and then we'll go around the outside and complete the octets on the Oxygens until we use all 34 valence electrons. So we have 10, 12, 14, and 34.

So we've used all 34 valence electrons that we started with, and we have the octets filled on the Oxygens; however, each Nitrogen only has 6 valence electrons, so we've not filled the octets for the Nitrogen atoms. We can take 2 valence electrons from up here and share them to form a double bond. Let's do that for both of these Oxygens here. And that should complete the octets for the Nitrogens.

You can see that the Oxygens have 8 valence electrons, still. Now the Nitrogens have 8 valence electrons, as well. And we're still using 34 valence electrons. So in the Lewis structure for N2O4, we're able to use all the valence electrons and complete the octets on each atom by forming those double bonds with the Oxygens and the Nitrogens here.

This is Dr. B. with the Lewis structure for N2O4, and thanks for watching.