Drawing the Lewis Structure for BBr3

Viewing Notes:

  • The BBr3 Lewis structure is similar to BF3 and BCl3 since F and Cl are in Group 7 and have 7 valence electrons.
  • Boron (B) doesn't need 8 valence electrons to have an octet (Boron often only needs 6).
  • If you're not sure you have the best Lewis structure for BBr3 you can calculate the formal charges. You'll find the B in BBr3 only has 6 valence electrons.
  • For the BBr3 Lewis structure there are a total of 24 valence electrons available.

Transcript: Hi, this is Dr. B. Let's do the Lewis structure for BBr3. Boron, on the periodic table, is in group 3, it has 3 valence electrons; Bromine has 7, but we have three Bromines, for a total of 24 valence electrons. We'll put Boron in the center, it's the least electronegative. Then we have three Bromines around the outside there, and we have 24 electrons. Start by forming a chemical bond between each of the atoms there that's 6; and then on the outside, 8, 10, 12, 14, 16, 18, 20, 22, 24, and we've used up all the valence electrons.

You can probably see right away that Bromine, they're all full, but the Boron in the center, that only has 6. What we could do is share some electrons and make a double bond. The thing is, I know that Boron actually does OK with 6 valence electrons. It doesn't have to have 8. So I'd like to check and see if the formal charges tell me that this is the appropriate structure before drawing the other one. I'm skeptical about a double bond to an atom like Bromine, which is really electronegative.

So let's check the formal charges. We'll do this Bromine first right here. So Bromine has 7 valence electrons, we look on the periodic table; nonbonding, these ones right here that aren't forming bonds, there are 6 of those; and then bonding, these 2 right here, so we'll put 2 over 2. Seven minus 6 minus 1 equals 0. So the formal charge for this Bromine is 0, and since all these Bromines are the same, the formal charge will be 0 for all of them.

Next, let's do the Boron. Boron, on the periodic table, 3; minus nonbonding, they're all in bonds, so 0; minus bonding, we said we have 6. Six over 2 equals...3-3, 0. So the formal charge on Boron is 0. Because all of the formal charges are 0, I'm comfortable that this is the correct structure. This is the one that we want to draw. So I'm not going to worry about the double bonds.

That's the Lewis structure for BBr3. This is Dr. B., and thanks for watching.