Monday, May 22, 2006

Lecture 038: Exam Review 2

Lecture 038: Exam Review 2

View Lecture


Jean-Claude Bradley: OK, so for the review session, I didn't receive any emails so I'm just going to do the problems we're getting in class here today. The other thing I want to point out is that for chapter nine, the exam will cover chapter one through eight. I was looking at the material that we covered, and the quiz that I gave you, and I didn't quite cover enough ground to be able to do that. But if you're taking 242, anything you did learn in chapter nine, that's where you are going to start, so one to eight for the exam.


[Student1 suggests a problem.]

Jean-Claude Bradley: This one right here? Ok, the SO. All right, so I'm starting off with HN3.


What is the charge of the central atom in HN3?


So available electrons? We're going to have nitrogen to contribute five. Five times three plus one. So, the hydrogen will have 16 and we're going to need eight for each nitrogen plus two for the hydrogen. So that's 26. We're going to share 26 minus 16, that's 10. So we have five bonds.


We have three nitrogen's and one hydrogen's position. Again we have different choices for what we can do. We can consider making triangles or other structures that are cyclic, but we want to avoid that. So if I have three nitrogens and I can't make a triangle out of them, we're going to make a straight line. That's the first thing we're deciding to do.


Now, we have another three bonds to put in a hydrogen so if I notice where else I can put the next two bonds, I pretty much have to do that. Therefore the only place I could logically put the hydrogen would be on the ends, because I already have four bonds on the central nitrogen. So that's the only place that the H could go.


Now I put my five bonds. I have to complete the octets. I'm missing four electrons on the left nitrogen. I'm missing no electrons on the central and one lone pair on the last.


Finally, the charges. So the nitrogen on the left has one, two, three, four, five, six electrons for charges. Nitrogen only has five so I have an excess, minus 1. The nitrogen in the middle: one, two, three, four electrons. Nitrogen has five so I'm going to have a positive charge; it's missing an electron. And the one on the right, one, two, three, four, five is neutral, so the question is what is the charge on the central atom? The central atom is nitrogen with the plus, so the answer here would be plus one. Does that make sense? In fact, this problem is extremely similar to azide, N3-. This is hydrazoic acid. It is just azide with a proton on one end, basically.


I have another question here on charges.


The Lewis Structure of SO, the sulfur has how many lone pairs? So for the available electrons we'll have six plus six equals 12. We'll need 16. It tells us that we have two bonds. With only two atoms, there's not much choice. We have to put the two bonds in the middle. Then we complete the octets. If we look for charges, six electrons with charges on each. Both sulfur and oxygen have six so there's no charges. So the question is, how many lone pairs? It would be two. So the answer you drew on the paper is two. Did you get that wrong?

Student1: Yeah.

Jean-Claude Bradley: Yeah, that's definitely the answer; two lone pairs.


And the Lewis Structure of SO4 two minus, the sulfur atom has what charge?


Ok, so available it would have six for the sulfur, six for each oxygen, so that would be 30. But we also have to add the two electrons from the two minus. We're going to need eight times five. So again, like all sulfur-oxygen compounds that we looked at, we're going to put the sulfur in the middle, surrounded by four oxygen. That uses up our four bonds. I can see here that's where you made your mistake. Is it because you didn't count the two negative?

Student1: Yeah.

Jean-Claude Bradley: So, basically if you count the two negative, you'd end up with only four bonds and then complete the octets. Those will all be negative one. The central sulfur then would be plus two.


Student1 comments on the procedure outlined.

Jean-Claude Bradley: Yes, if you have a plus, you remove that, because you are counting electrons. So the more negative, the higher the number. All right, so this is the charge on the central sulfur so the answer was plus 2.


SO2 is polar, has an average charge of negative one on each oxygen, has a plus two charge on the sulfur, has Van Der Waals as a dominant molecular force or none of the above?


So let's work out the Lewis Structure for SO2. So available we're going to have six times three equals 18. We're going to need 24. Share six. We've got three bonds. It's almost tempting to make a triangle; we know not to do that! We're going to put the sulfur in the middle with the oxygens around it. That uses up only two bonds. We have to use up a third one so we have to create a double bond on one of those oxygen. Complete the octets.


We look at charges. We have a negative on the right oxygen, a plus on the sulfur. Now, that's not the whole picture because that's only one of the resonance forms, so we have to draw the other resonance hybrid.


Ok, now is this molecule polar? SO2? It comes down to the hybridization of the sulfur. The hybridization of the sulfur is sp2. It's got three groups of electrons around it so that means I have 120 degree angles between the two oxygens. So it's bent like this. When we work out the dipoles, remember when you have charges, the charges will be more important than anything else, and you'll go from plus to minus. So I'll have one dipole moment in that direction for that hybrid. I'll have one dipole moment in that direction. Those two arrows can only cancel each other out if they are at 180 degree to each other. They are not, they're at 120, so the dipole moments don't cancel out and therefore, yes, the molecule is polar.


The average charge on each oxygen would be what? How do you figure that out? Negative one half, right. You look at all the resonance hybrids and you average them out. The charge on this oxygen is zero on the left, minus one on the right. Minus one plus zero dided by two is negative one-half, so central sulfur is plus one on the left, plus one on the right. It's average charge will be plus one.


We already know 'A' is the answer, but let's work these out to shed some light here. Is Van Der Waals the dominant intramolecular force? Well, no. It has a dipole moment so therefore will be dipole-dipole so 'A' will be the correct answer. The other questions are on chapter nine, and like I said, chapter nine will not be on the exam. So anything else in chapters one through eight? You guys exhausted everything on Wednesday?

Student2: The last question?

Jean-Claude Bradley: The last question? Yeah, I'm going to ask you to draw out how far you got on it and then come and see me after. OK, anything else, any general questions? Are we all good? So, for some reason I set the exam to start at two o'clock this afternoon, but there's no reason for that. It will start at 10 o'clock if you wanted to take it early. All right. Well, good luck and hopefully I'll see a few of you next term.



Transcription by CastingWords