Chemical Bonding: IO₃^- Lewis Structure

Drawing the Lewis Structure for IO₃^- (Iodate Ion)

Viewing Notes:

• With IO₃^- be sure to add an additional valence electron to your total because of the negative sign.
• There are a total of 26 valence electrons in IO₃^-.
• Iodine is in Period 5 on the Periodic table. Therefore it can hold more than 8 valence electrons. You'll want to check the formal charges to make sure that you have the best structure for IO₃^-. This means the structure with the formal charges closest to zero as possible.
• Be sure to put brackets and a negative sign around the IO₃^- Lewis structure to show that it is an ion.

Transcript: This is the IO3- Lewis structure: the Iodate ion. For IO3- we have a total of 26 valence electrons. Iodine is the least electronegative. We'll put that at the center, and then we'll put the Oxygens around the outside. We have 26 valence electrons for IO3-. We'll put 2 between atoms to form chemical bonds. We've used 6. Then we'll go around and fill the octets for the Oxygens. We have 6, 8, 10, 24; and then back to the central Iodine, 26. So it looks like we're done: each of the atoms has 8 valence electrons, so the octets are satisfied, and we've used 26 valence electrons.

However, when we look at the structure, Iodine is below period 2, row 2 on the periodic table. That means Iodine can have an expanded octet--can have more than 8 valence electrons--so we really need to look at the formal charges here to see if this is the most likely structure for IO3-. For the Iodine--Iodine has 7 valence electrons on the periodic table. Nonbonding, we have 2 right here that are not involved in a chemical bond. Minus bonding, we have 2, 4, 6, which we'll divide by 2. That gives us a +2 formal charge for the Iodine.

For the Oxygens, they're all the same so we'll just do one. Oxygen has 6 valence electrons from the periodic table, minus nonbonding; we have 6 nonbonding, minus bonding--2 divided by 2. That gives us a -1 formal charge for each of the Oxygens. We really want our formal charges to be as close to zero as possible. We'll still need to retain the -1, of course. To do that, we can form a double bond with the Oxygen by moving these valence electrons and share them with the Iodine. That'll result in a formal charge of 0 on this Oxygen and a +1 on the Iodine.

So we're getting close. Let's form another double bond between this Oxygen and the Iodine. When we do that, this Oxygen has a formal charge of 0, and now the Iodine has a formal charge of 0. We're still using 26 valence electrons, and each of the Oxygen atoms have octets. The Iodine has more than 8, but that's OK because it can have an expanded octet. Our formal charge for this entire structure here shows up to be -1, which makes sense. This is a negative ion. Because the formal charges are close to zero with this structure, that makes this the more likely Lewis structure for IO3-.

One last thing: we do need to put brackets and a negative sign around the Lewis structure to show that it's a negative ion.

And that's the Lewis structure for IO3-. This is Dr. B., and thanks for watching.