Drawing the Lewis Structure for SO_{4}^{2}
Transcript: Hi, this is Dr. B. Let's do the SO4 2 Lewis structure, for the sulfate ion. On the periodic table: Sulfur, 6 valence electrons; Oxygen also has 6, we have 4 Oxygens, multiply by 4; and these 2 valence electrons up here, we need to add those, as well. That gives us a total of 32 valence electrons. We'll put the Sulfur in the center, and then the four Oxygens will go on the outside. Next, we'll draw bonds between the Sulfur and the Oxygens, so there we have four bonds and we've used eight valence electrons. Let's go around the outer atoms and make sure they have octets. So we've used 8, 10, 12, and 32. Looking at the structure here, we see that each of the Oxygens has 8 valence electrons; 2, 4, 6, 8; as does the Sulfur here, 2, 4, 6, 8. But we're not quite done yet. Sulfur is in the third period of the periodic table. That means it can hold more than 8 valence electrons. So we really do need to check our formal charges. So to calculate the formal charge on the Sulfur: we see that Sulfur, on the periodic table, group 16 or 6, has 6 valence electrons. Up here, all of the electrons, all of them are involved in bonds, so that's going to be zero. And the bonding electrons, 2, 4, 6, 8; we've used 8 of those, and we'll divide that by 2. Six minus zero minus 4 gives us a +2 formal charge for the Sulfur. For the Oxygen, it's also in group 6 or 16, so it has 6 valence electrons. Up here, nonbonding, we have 6; and then bonding, we have 2. And all these Oxygens are the same, so we only need to do one. Six minus 6 minus 2 gives us a minus 1 formal charge for each Oxygen. If we add up all the formal charges, the 1, 1, 1, 1 and +2, we do get a total charge of negative 2. That does make sense, but with formal charges, we want them to be as close to zero as possible for the atoms. So let's see if we might be able to do another Lewis structure that has more zeroes for the formal charges. When I see this +2 charge here on the Sulfur, I know that I can move electrons from the outer valence electrons of the Oxygen into the middle. If I do that twice, if I move these two into the middle to form double bonds, and get rid of them, I think that'll get rid of the positive 2 charge. Let's try that and then recalculate our formal charges. So I've moved electrons from the outside of these two green Oxygens into the middle to form double bonds. Let's see how that changes the formal charges. So for Sulfur, 6 minus zero, there are no nonbonding; and now we have 2, 4, 6, 8, 10, 12 total bonding electrons. Six minus 6, that gives us zero. So the formal charge on the Sulfur is zero. If we look at the green Oxygens, you can see that we have 6 minus 4 of the nonbonding, and then 4 bonding; we divide by 2. Six minus 4 minus 2 is zero. Finally, looking at the blue Oxygens, we have 6 minus 6 nonbonding, and then 2 bonding divided by 2. Six minus minus 6 minus 1 is minus 1. So at this point, we see that we have mostly zeroes. But if you look, you have a negative 1 and a negative 1 here. That works well with this. Since more of the formal charges are zero, this is a better structure for SO4 2. Since it's an ion, there's one last thing we need to do. We need to put brackets around it to show that it's an ion and the charge of the ion. We add our 2 minus right there. And that is the Lewis structure for SO4 2. It was a bit of work, but we have the best structure here. Our formal charges are in good shape. We've used all the valence electrons. So that's it. This is Dr. B., and thanks for watching. 
