### Drawing the Lewis Structure for IO4-

Viewing Notes:

• With IO4- be sure to add an additional valence electron to your total because of the negative sign.
• There are a total of 32 valence electrons in IO4-.
• 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 IO4-. This means the structure with the formal charges closest to zero as possible.
• Be sure to put brackets and a negative sign around the IO4- Lewis structure to show that it is an ion.

Transcript: This is the IO4- Lewis structure. IO4- has a total of 32 valence electrons, and that includes this extra valence electron up here. Iodine is the least electronegative, we'll put that at the center, and the Oxygens will go around the outside. We'll put 2 electrons between the atoms to form the chemical bonds. Then we'll go around the outside to fill the octets on the Oxygens until we use all 32 valence electrons. So we have 8, 10, 12, and 32.

At this point you can see that each of the Oxygens has 8 valence electrons, and the Iodine in the center, it also has 8. So the octets are fulfilled on each of the atoms, and we've used the 32 valence electrons, so this seems like a pretty good Lewis structure for the IO4- ion. The thing is, that Iodine can hold more than 8 valence electrons, so we really want to check the formal charges on Iodine to make sure that this is the most likely structure for the IO4- ion.

Looking at the Iodine atom, we can see Iodine is in group 7 on the periodic table so it has 7 valence electrons there. Nonbonding, all of these electrons there are involved in chemical bonds with the Oxygen, so that's 0. and then bonding, we have 2, 4, 6, 8, so we actually have 8 over 2. Seven minus 4 gives us a +3. So the formal charge on the Iodine is +3. If we look at the Oxygens, group 6 on the periodic table, six valence electrons; we have 6 nonbonding--and each of the Oxygens is the same, we only need to do one. And then bonding, we have 2; 2 divided by 2. Six minus 6 is 0, minus 1, gives us a minus 1. So the formal charge on all the Oxygen atoms is -1.

If we add all these formal charges up, we do get this negative 1 here, but we really want our formal charges to be as close to 0 as possible and that's not the case here. If we take and form a double bond with an Oxygen, that should lower the formal charge on the Oxygen to 0, and also bring the formal charge on the Iodine closer to 0 as well. Since we have a +3, let's do that to three of them. When we form these three double bonds with the Oxygens here, we can see that the Iodine now has a formal charge of 0, and these Oxygens with the double bonds, they also have a formal charge of 0. And the single bond stayed the same. We didn't change that. That has a formal charge of negative 1. That negative 1 matches the -1 charge on the IO4- ion. So with these formal charges closer to 0, this is a much better structure for IO4-. Finally, since it is an ion, we do need to put brackets around it and the negative sign there.

So this is the best Lewis structure for IO4-. This is Dr. B., and thanks for watching.

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