Drawing the Lewis Structure for ClO_{4}^{}Viewing Notes:
Transcript: Let's do the ClO4 Lewis structure. Chlorine has 7 valence electrons; Oxygen has 6, we've got 4 Oxygens and we need to add this up here. This negative means we have an extra valence electron. Gives us a total of 32 valence electrons for this Lewis structure. Put the Chlorine in the center, it's the least electronegative; and Oxygens around the outside, like this. We'll form bonds between atoms by putting two electrons there We've used 2, 4, 6, 8, and then around the outside atoms: 10, 12, and 32. At this point, everything has octets, but since Chlorine's in period 3 of the periodic table, we should check our formal charges because it can hold more than 8 valence electrons. When we calculate the formal charges, we can see that the Chlorine has a +3 and the Oxygens have a 1. If we add that all up it does equal the negative 1, but we want our formal charges to be as close to 0 as possible. We need to resolve that. If we remove electrons from the outside of the Oxygens, that should resolve the formal charge of the +3 on the Chlorine. So we move those in to create double bonds. So now we can see that the charge on the Chlorine, after creating these three double bonds with the Oxygens, is zero. Each of the Oxygens has a formal charge of 0 as well, and this Oxygen over here, the one with the single bond, has a formal charge of negative 1. That makes sense because we have this 1 here. So we've used all of the valence electrons that we started out with, we have our formal charges as close to 0 as possible, and the charge of this ion here adds up to a minus 1 just like in our formula. There's one last thing we need to do: We'll put brackets around our structure to show that it is a negative ion, and then put our 1 charge on the outside. So this is the best Lewis structure for ClO4. This is Dr. B., and thanks for watching. 
