Drawing the Lewis Structure for BF3

Viewing Notes:

  • The BF3 Lewis structure is similar to BCl3 and BBr3 since Cl and Br 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 BF3 you can calculate the formal charges. You'll find the B in BF3 only has 6 valence electrons.
  • For the BF3 Lewis structure there are a total of 24 valence electrons available.

Transcript: This is the BF3 Lewis structure: Boron trifluoride. Boron has 3 valence electrons and Fluorine has 7, but we have 3 Fluorines so we multiply those together, and when we add it all up we have 24 total valence electrons for the BF3 Lewis structure. We'll put the Boron in the center, it's the least electronegative, and the Fluorines around it.

We have a total of 24 valence electrons. We'll put 2 between atoms to form chemical bonds; we've used 6. Then we'll go around the outside to satisfy the octets on the Fluorine or until we run out of valence electrons. So we have 6, 8, 10, and 24. So we've used all 24 valence electrons. We can see that each of the Fluorines has an octet; it has 8 valence electrons. But the Boron in the center only has 6 valence electrons.

It turns out that Boron is a bit of an exception. It can be OK with only 6 valence electrons. If you were to check the formal charges for Boron and then for the Fluorines, you'd find out that all the formal charges are 0 for the BF3 Lewis structure.

So that tells us that this is the most likely, or plausible, Lewis structure for BF3. This is Dr. B., and thanks for watching.