Chemical Bonding: BrO₃^- Lewis Structure

Drawing the Lewis Structure for BrO₃^-

Viewing Notes:

• The BrO₃^- Lewis structure has a total of 26 valence electrons. This includes the electron represented by the negative charge in BrO₃^-.
• You need to put brackets around the BrO₃^- Lewis structure as well as a negative charge to show that the structure is a negative ion.
• If you calculate the formal charges for the initial BrO₃^- Lewis structure you'll find that the Bromine (Br) has a +1 charge. You'll want to form double bonds with the central Bromine atom and two of the oxygen atoms to reduce the formal charges on the Bromine and Oxygen atoms. Doing so will give the entire structure a -1 charge (remember it's BrO₃^- so it should have a -1 charge).

Transcript: This is the BrO3- Lewis structure. Bromine has 7 valence electrons. Oxygen has 6, and we have 3 Oxygens, and then we need to add this extra valence electron here for a total of 26 valence electrons for BrO3-. Bromine is the least electronegative, we'll put that at the center, and then we'll put the Oxygens on the outside.

Since we have a total of 26 valence electrons, we'll put two between atoms to form chemical bonds. We've used 6. And we'll go around the Oxygens and fill their octets--8, 10, 24, and then back to the Bromine, 26. So we've used all the valence electrons. We can see that each of the atoms has 8 valence electrons so the octets are full for all the atoms.

At this point it looks like we have a pretty good Lewis structure; However, Bromine is below period 2, row 2 on the periodic table. It can have an expanded octet, meaning it can have more than 8 valence electrons. So we should look at our formal charges to make sure this is the most likely Lewis structure for BrO3-. For Bromine, we have 7 valence electrons from the periodic table; minus the nonbonding, these 2 right here, minus the bonding. We have 2, 4, and 6. Six divided by 2. That gives us a +2 charge for the Bromine.

For the Oxygen--and each of the Oxygens are the same, so we'll just do one--we have 6 valence electrons on the periodic table. Minus nonbonding, 6; minus the bonding, 2 divided by 2. So we have a -1 charge on each of the Oxygens. We'd like our formal charges to be as close to 0 as possible. Of course, we'll still have a -1 here, so we'll have to have a -1.

To do that, we can take and form double bonds with the Oxygens and the Bromine. Let's take these valence electrons here and share them to form a double bond. If we were to recalculate our formal charges, we'd see now the Oxygen has a formal charge of 0 (this Oxygen). And the Bromine, instead of a +2 will have a +1. So let's do one more Oxygen and form a double bond here.

Now the formal charge on this Oxygen is -0. The Bromine has a formal charge of 0, and this is still -1. That makes sense: the total charge--formal charge here is -1 and that matches with the -1 here. One last thing we need to do: this is a negative ion, the BrO3- ion--Bromate ion. We need to put brackets around the structure to show that it is a negative ion.

So that's the Lewis structure for BrO3-. This is Dr. B., and thanks for watching.