Drawing the Lewis Structure for SO_{3}Viewing Notes:
Transcript: Hi, this is Dr. B. Let's do the SO3 Lewis structure. Sulfur has 6 valence electrons. Oxygen has 6, but we've got three Oxygens, for a total of; 6 plus 18; 24 valence electrons. Let's put Sulfur at the center and then the Oxygens around the outside, all three of them. Now we'll put two valence electrons between each atom to form a chemical bond. We've used 6. And then around the Oxygens on the outside. So we have 6, 8, 10, 12, 14, 16, 18, 20, 22, 24. And that's how many valence electrons we used. But you can see the Sulfur in the center only has 6 valence electrons, so we need to resolve that. If we share these two electrons, right here, between them, that'll give everything an octet. We're still only using 24 valence electrons. So this looks like a pretty good structure: the Sulfur has 8, and then each of the Oxygens has 8 as well. The problem is that Sulfur can often hold more than 8 because it's in the third period, or row, of the periodic table. So we'll want to calculate our formal charges to make sure that this is the best structure. You see here that they aren't zero. That means this is probably not going to be the best structure, that there might be a better structure we can use for the Lewis structure of SO3. One thing I can tell is, this plus 2 here makes me think that I need to move some of these valence electrons into the middle. In fact, two pair of them. So I've done that now, and you can see that we have three double bonds; that Sulfur has 12 valence electrons around it. Sulfur can hold up to 12. So let's recalculate the formal charges now that we've made these changes. So now, when we look at this Oxygen right here, which is the same as the other three, we can see on the periodic table Oxygen has 6 valence electrons. And we have nonbonding, these ones right here, they're not involved in chemical bonds. There are four of those. And then bonding, we have 4 of those, but we divide that by 2 to give us zero for all the Oxygens. So all the Oxygens have a formal charge of zero. When we look at the Sulfur, we know that Sulfur's in group 6 on the periodic table; 6 valence electrons. All of the electrons here on the Sulfur are involved in chemical bonds, so we have zero nonbonding; and a total of 2, 4, 6, 8, 10, 12 bonding valence electrons, divided by 2. So 6 minus zero minus 12 over 2; so 6 minus 6 equals zero. So we can write the formal charge for Sulfur as zero. So we have formal charges of zero for each of the atoms in SO3. That makes this the best Lewis structure for SO3. This is Dr. B., and thanks for watching. 
