Lecture 21. Second Midterm Examination Review
Martin, Rachel
 Principles of quantum mechanics with application to the elements of atomic structure and energy levels, diatomic molecular spectroscopy and structure determination, and chemical bonding in simple molecules. 
 Good morning everybody. Today we're going to talk about the exam. Anybody have any questions before we do? What's the mean? I don't know. You know, again, it takes you guys an hour to take the exam. Multiply by 207 or whatever. There are four of us grading it. Okay, it doesn't take as long to grade it as it does to take it, but it's not going to be all graded and ready to go by the next class period. Sorry, we just, there's only four of us, we can't do that. So it's almost done being graded. We'll get that finished today. Then it gets sent to wherever it goes for scanning. And then as soon as I get the PDFs back, I will give them to you. But I have zero control over how long it takes to scan stuff. So I know you're anxious to see your score, but it just takes a little while. It's a big class. Okay, so let's talk about the exam. There's a question in the back? Yeah, what you've seen so far, how have people done it? It looks like people have done okay. It looks, looks all right. People ran out of time. I expected that. That's fine. It's normal. But overall, seems pretty reasonable. Okay, so let's get started going over this. All right, so first question. I are spectrum of carbon monoxide. Why is there a gap in the center? Yeah, there's no delta J equals zero transition. So anything mentioning that that transition is forbidden or that there's a, that the specific selection rule prohibits that, anything like that got full credit. Okay, so then the next question is what is the energy in wave numbers of this transition? And, you know, this is really straightforward. You just read it off the spectrum, right? Because since there's no delta J equals zero transition, that spot in the middle where there's no line is, that's where the new equals zero to new equals one transition is. And so if you just read it off the spectrum, you get something like this. It's hard to read from that little plot. So if you've got anything close, that counted. Okay, so the next question is using the spectrum to estimate the force constant for the CO bond. And I'm not going to work all of this out in the interest of time, but basically you need to use that transition frequency. So that's, that's what we do. Of course, this is the reduced mass and K is in newtons per meter. And the actual value, if you look it up for carbon monoxide, is about 1860. You know, again, it's hard to read the transition frequency off that little plot and that sort of determines the answer you get. So if it was consistent and you got something close, you got full points. So that was really straightforward. It's just kind of remembering which equation to use and plugging stuff in. As always with such things, there are opportunities to make mistakes, getting the units wrong and, you know, forgetting to convert things. But otherwise, that one was pretty straightforward. Okay. Next question, is carbon monoxide a perfect rigid rotor? And it's not and a lot of people wrote because life's not perfect, which is funny and probably reasonable. But it doesn't count as a good justification. So, you know, I guess John Mark said that at the review session. But, well, you're also about to learn that life's not fair because that doesn't get you the full credit. So, in order to get full credit, you had to write that there's centrifugal distortion and you can tell because the spacings on the spectrum are not exactly equal. It's kind of squished on one side and stretched on the other. Yes? Yes. I didn't grade that one. You can hope your TAs are nicer than I am. So, I think that's like, you know, when you go to iTunes and you buy a song and they charge you $99 instead of 99 cents, is that close enough or is it completely wrong? I know what I think. So, you can hope the TAs are nicer than I am. I'm not sure. That's like you gave the baby a thousand times too much medicine and he died. It's bad. Well, it is. Those order of magnitude errors cause huge problems in real life. Your Mars rover crashes into the surface of the planet. You don't get partial credit. You get fired. You know, there's it like the, you know, yeah, I'm making fun but like the, you know, the things where you have a small error in the last couple decimal places or whatever because of round off problems, that's not really a huge deal but, yeah, orders of magnitude matter. Okay. All right. So, how would the Raman spectrum of CO look different? So, if you're looking at a Raman spectrum instead of an IR spectrum, how would you be able to tell that? Okay. So, you had to give at least two differences and here are the ones that I could think of. You would have a Rayleigh line in the center of the spectrum which is not called a Q branch. It's a Rayleigh line. So, the Q branch is if you have, you know, something like a radical where there's electron orbital momentum and so the central transition isn't forbidden, then you see that. For a Raman spectrum, it's the Rayleigh line. Other things could be the spacing is four times the rotational constant or the line intensity is smaller. And so, any two of those were fine. Question in the back? You definitely got something for that. I don't remember exactly how we did that one. Yes? I was wondering, I was taking the exam, and I was thinking about a note saying that there's a Q branch and a P branch and an O branch in that spectroscopy. How would that, how does that factor into the whole thing? Okay. If you had labeled all three branches there, it doesn't do it because that's, you know, that's like, we can talk about it later if you want, but it seems like not worth spending a lot of time on. But yeah, if you drew it and labeled all three branches, then that's something different. But yes? Isn't there a difference in the number of peaks as well? What do you mean a difference in the number of peaks? The number of peaks that would show up on the IR versus the Raman spectrum? It would be pretty hard to tell. Because I thought that's what we were using our point groups before, the number of peaks. Sure, but it's the same molecule and it's a linear molecule. So if you had a, you know, if you had a complex polyatomic molecule, that would definitely be true. You would see different vibrational modes. But this thing only has one, it's a diatomic molecule so that it's only got one vibrational mode, right? So that wouldn't really be applicable here. But you're right, if you had a complex polyatomic molecule, that would be totally true. Okay, let's go on to the next one. All right, so for this one, it's sort of a, it's a standard Burge-Sponer plot but with a little bit of a twist which is that there's a break in the middle of it. So what that tells you in practical terms is that it has a weird potential. So this was from a paper in science a few years ago on beryllium dimer and looking at beryllium dimer forming. And at first the, you know, the dissociation energy of beryllium dimer was estimated incorrectly because it has this weird shape in the experimental potential. It deviates really strongly from a normal, worse potential. So why does it have this weird shape? Well, if you think about the electronic structure of beryllium dimer, you can kind of tell it's not going to form a very strong bond which is true. And, you know, since it has this weak bond, it's got a strange potential. And this was something that was a relatively recent result. It was in science a few years ago. Okay, so what does this mean for doing the problem? It means you can't really look at the, you know, the value for new max. So you're used to the X intercept telling you something about the convergence limit. And, you know, here that's kind of hard to get because, you know, we have these two breaks in the plot. But so anyway, the hint was to use the part of the data that's consistent with the Morse potential, which was the first part which is shown by this line here. And so you could do that. And so as always for these things, the slope tells you something about the potential. And so then we know that the Y intercept, which is normal in this plot, and DE is just, all right. So, and that was about 625. And so that's it, pretty straightforward. So the only part, the only part that had information content here was just knowing which equations to use and being able to get this off the plot. Okay, so then moving on to, yeah. I think I got a negative. Well, does that make sense? I mean, so it depends on, you know, you're thinking about, it's an absolute value thing. I mean, you're thinking about putting in, you have to put energy into the molecule to get it to dissociate, right. So that's, so. So if your X is negative, and your BE is positive, then your XE is going to be negative. Right. So if your XE is negative, and your BE is positive, then your DE is negative. But take the absolute value. But that's just, that's one of those things where it's, you know, the convention is sort of, you know, it could be either way depending on whether you're a chemist or an engineer and what you're doing. So, you know, as far as whether it's the dissociation or the energy is positive or negative. Yes. What? For the question, is that for the question? It is for XE, yeah. But. You will lose points if it's a negative? Like a point. It's not, it's not a big deal. OK. So now the next one. Looking at Frank Condon factors. OK. So the question is, write an expression for Frank Condon factor for a transition between these two states. And I wanted you to put in the states, but which is why I wrote down the functions. But so you got almost all the points for doing it in direct notation. So we have these two states. And the Frank Condon factor is. So that got almost all the points. People were really worried during the exam about which one's the initial and which one's the final state. It doesn't matter, right? So we're talking about a transition between them. And that amplitude isn't going to depend on which one is initial and which one is final. So, you know, it doesn't really matter. Yes. Because in your lecture notes, the Frank Condon factor, how do you think vibrational states in the form that's that OK for all the words vibrational states? Oh yeah, that's true. That's, so here's one of these cases where direct notation is ambiguous. And so that's why it's best to write in the states. So yeah, what you're saying is these quantum numbers aren't relevant to anything. Yeah, true. But so what I wanted you to actually do is, you know, write out like the integral. And so I said, new prime equals zero. So that's one. And then H2 is, right? And do you have to evaluate it? No. I just wanted to write the expression. So doing it in direct notation got you most of the points, but I wanted you to make the connection and write down the functions. OK, so the next question is we want to sketch this transition on an energy level diagram. All right, so we've got some electronic states. And, you know, it doesn't really matter what they look like as long as they vaguely look like a Morse potential. And then we have some vibrational states in there. And you should know that, you know, there's some zero point energy. So that would be new prime, double prime, sorry. And so that's basically it. And then, of course, we needed to label the transition. And all right, the way I drew it, it is difficult to do that. And so there's not very much overlap the way I drew it, but there you go. The next question is going to be if you drew the error going down, is that fine too? Yes, because we didn't specify which one's the initial and which one's the final state. So as long as you're consistent, good enough. I just wanted to see if you can make the connection between, you know, writing down these expressions sort of in quantum terms, plugging in actual functions, you know, not evaluating it but putting it in. And, you know, being able to go from that to looking at a potential diagram. Okay, next one, term symbols. So for those who spent endless hours in office hours writing out all the possible excited states of O2, it was time well spent. Okay, so the O2 molecule is in its ground state. And here, you know, there was some potential for confusion because the book sort of doesn't distinguish between the various electron configurations that you can get with the ground state electron configuration. But of course there is an actual ground state. It's, you know, you have to use Hoon's rule to get that. So it's the one where you have the most unpaired electrons. So let's look at this in terms of the molecular orbital diagram. So I'm just drawing the P orbitals. So I've got sigma g. All right, that's my, so that's my molecular orbital diagram from general chemistry of O2. And I'm ignoring the S orbitals because they're filled and they don't give me anything. And so I write out the electron configuration for this molecule, which gives me this. And again, since it's the ground state, we use Hoon's rule and put in the configuration with both of these electrons unpaired and in separate orbitals. Okay, everybody with me so far? Cool. Okay, so then the next question is what's lambda? And it is going to be zero. So we get a sigma term. And then if we, since it's a, so we have to decide whether it's g or u, and so we've got these two electrons in the, they're both in the pi star g orbital. So that's g times g. So this term is even. And they're in different orbitals. So it's minus. And what's its spin multiplicity? Oops. Nope. Just kidding. It's a triplet state because we have two unpaired electrons. I was getting ahead of myself in thinking of the next one. Okay, so that's the ground state. And, you know, there were a couple of, you know, of course you can put those electrons in different places and write a couple more term symbols. There's a delta term symbol, et cetera. Since I just asked for the lowest energy one, that was essentially a waste of time if you did it on this question. Okay, so the next one is we have one electron promoted from the homo to the lumo. And so, you know, again, general chemistry, highest occupied molecular orbital and lowest unoccupied molecular orbital. Some people asked about that. I think people just get nervous and forget stuff sometimes. Okay, so you have, these guys are all the same. And then there are a couple choices here. I'm not labeling the orbitals, but obviously they didn't change. And so, these things give you, there are two states. And if you only got one of them, so you had to get both of them to get full credit. If you only got one of them, we took off a point. So, what do you think? There are various side conversations, but it seems like everybody more or less gets it. Is that true? Okay, good. All right, so for, you know, the endless hours that we spent redrawing this picture paid off. Okay, so the next question has to do with NMR. All right, so this was, this question was about writing the matrix elements for I plus, the raising operator for a spin one half. And so, it's full definition was here. And you want to write this out as, so you know that I plus operated on alpha is zero. This is zero. And then, so I'll write the full thing. There we go. So, all right, so this gives us, there we go. That's more like it. So, one times alpha alpha. And so, the answer is, and if you just wrote the matrix because you had it written down on your sheet sheet, that didn't get full points. You needed to show some understanding of how to do it to get full credit. So, if you wrote out the full matrix or just wrote it out symbolically or, you know, put down the, how the matrix elements work, that all counted, as long as it showed that you knew how to do it. This collection of constants in front in the definition of the raising and lowering operators, I'm just letting you know for purposes of taking the final. You know, it comes out to one in this case, but if it wasn't to spin one half, it wouldn't, as you remember if we, you know, from doing some of the examples at office hours. Okay, so now let's look at NMR spectra. So, this was one of these things where exhortations to read the directions before doing it were really, really important. So, people did sad time wasting things like, there were people who didn't just draw the spectrum for the protons that were labeled. They, like some people actually crossed out my labels and relabeled the whole thing and drew all the protons. And it was sad because it must have taken forever and, you know, then people didn't have time for other things. People tried to draw a proton for the one that's labeled with the star, which of course has four bonds to that carbon. So, you know, I'm not being mean. I'm just pointing out, like, read the directions. There are things, you know, there are things in there that are meant to save you time and, you know, like not having to draw the spectrum for all the protons. Okay, so let's look at what we have to do first to get this. Okay, so we have a chemical shift table. And so one of the first things to do here is identify your various functional groups. So, we have A here. So we have our two protons here. And some people made the mistake of saying that this thing was a carbonyl. So that would be, and, you know, they, so they put it in this chemical shift range. And that's not what it is, right? It's got, you know, it's between these two oxygens. So it's an ether. So it would be toward the, you know, higher chemical shift end of the ether because it's got two of them. But so it should be somewhere around, you know, four or five ppm somewhere in there. Four and a half something. Okay, what else have we got? So D here is an aromatic. It has a proton as its neighbor. C is a methyl group. B has a proton there. And it's got, so B is going to be a complicated, a, multiple, right? Because it's got two neighbors on one side and three on the other and they're not equivalent. But so, okay, we figured out what chemical shift everything should be. So we can start drawing the spectrum. So you get points for labeling the axis. And let's start on this end. Okay, so D, our aromatic here is going to be, you know, between seven and nine ppm. So let's, you know, put it down here. And it has one neighbor. So it'll be a doublet. And then our ether is going to be the next one along in the chemical shift scale. And it just has two oxygens next to it, which we can probably assume are, you know, O16. And so it's just going to be a singlet. And then our methyl group here is going to be furthest toward zero. And it's going to be a doublet because it only has one neighbor. And then we're left with what is B. So it's got two sets of neighbors that are inequivalent. And, you know, there's two protons on one side and three on the other. And I didn't tell you which J coupling is larger. So it is either a triplet of quartets or a quartet of triplets. And if you wrote something like that and drew, you know, this, that's fine. Some people really carefully drew the picture out. You know, again, I want to know if you understand it, not your ability to draw these things beautifully because as you can see, I'm not so great at it. Yes? For the numbers, do we have to like B exact or do we just draw them like in that order? For the chemical shift range, you got points if they were in the right order relative to each other and the chemical shift scale was something reasonable. But, you know, thinking that, that this thing is a carbonyl and putting it at 12 ppm was not reasonable. Yes? It doesn't because it's exchangeable. It means it's popping on and off all the time in, you know, exchanges with the solvent. Although, you know, people who, who mixed in the nitrogen proton and the splitting, I think I just took off a point. It was, you know, if you understood the rest of it, that's not a huge deal. Okay, so that is, that is that one. And then, we said a new spectrum is collected with the carbon decoupling turned off. And we want to see the signal for the proton labeled A. Okay, so the carbon decoupling is turned off. So that means in our, so in our previous spectrum, we didn't see any coupling to C13. Now, the decoupling is turned off and we're going to. C13 is a spin one-half. And proton A, remember that was the one that was an ether and it didn't have any neighbors. So, it was a singlet before. And so, if it's not decoupled, it's a doublet. And that, that one was either right or wrong. And so then, we had an inversion recovery experiment is performed on the C13's in this molecule. So remember, this is the experiment that we do to measure the longitudinal relaxation time. So we put everything, we flip all the spins of the signal and then wait for it to relax and measure how long it takes to relax back to equilibrium. And remember what causes relaxation to equilibrium. It's fluctuating magnetic fields local to that nucleus. So something that has direct dipole-dipole couplings with protons, so if it's directly bonded to a proton, that's going to relax faster than something that doesn't have that. Remember, another important source of T1 relaxation is methyl rotation and also just molecular motion. And so, in all of these respects, you know, here's our molecule again. The one that's labeled with a star is non-protonated. It's aromatic, so it's probably going to be rigid conformationally. And C is a methyl. So the one that returns to equilibrium more quickly is C, the methyl. And basically, any one of these explanations worked. So if you got any reason why it's going to have a quicker relaxation time, that was good. Yes? So what do you mean by steric hindrance? I don't see how that has anything to do with it. But, you know, win me over. What does that have to do with it? Well, it's going to be a more than the A. It's not the argument about rigidity, it's kind of like a ball of the mind. Hmm, kind of, but it's not that it's starically hindered. I mean, if you had like a T-butyl group, that would still relax pretty fast because those metals can still rotate. So, sorry, no. You had to have something about conformational rigidity. Yes? What if you said like it was connected to a lot more atoms and so those atoms affect that carbon, so it's harder, but it takes it longer to come back to equilibrium? Well, but can, so which one is connected to more atoms? I think I said the star one. Well, but it's not, right? They're both connected to three, to four other atoms. It depends, it matters what those atoms are. So, you know, if you, if you, if you point if I said that one took a longer time. We can talk about it later. Wait, so you're asking if you just, if you got, which one took a longer time, right? And not the explanation, did that get partial credit? Yes, it did. Okay, that's it. That's the exam. So, again, the grading is almost done. We'll get it done as fast as we can today and then send it to get scanned. And next time we're going to start talking about statistical mechanics.
 Good morning everybody.  Today we're going to talk about the exam.  Anybody have any questions before we do?  What's the mean?  I don't know.  You know, again, it takes you guys an hour to take the exam.  Multiply by 207 or whatever.  There are four of us grading it.  Okay, it doesn't take as long to grade it as it does to take it,  but it's not going to be all graded and ready to go  by the next class period.  Sorry, we just, there's only four of us, we can't do that.  So it's almost done being graded.  We'll get that finished today.  Then it gets sent to wherever it goes for scanning.  And then as soon as I get the PDFs back, I will give them to you.  But I have zero control over how long it takes to scan stuff.  So I know you're anxious to see your score,  but it just takes a little while.  It's a big class.  Okay, so let's talk about the exam.  There's a question in the back?  Yeah, what you've seen so far, how have people done it?  It looks like people have done okay.  It looks, looks all right.  People ran out of time.  I expected that.  That's fine.  It's normal.  But overall, seems pretty reasonable.  Okay, so let's get started going over this.  All right, so first question.  I are spectrum of carbon monoxide.  Why is there a gap in the center?  Yeah, there's no delta J equals zero transition.  So anything mentioning that that transition is forbidden  or that there's a, that the specific selection rule prohibits  that, anything like that got full credit.  Okay, so then the next question is what is the energy  in wave numbers of this transition?  And, you know, this is really straightforward.  You just read it off the spectrum, right?  Because since there's no delta J equals zero transition,  that spot in the middle where there's no line is,  that's where the new equals zero  to new equals one transition is.  And so if you just read it off the spectrum,  you get something like this.  It's hard to read from that little plot.  So if you've got anything close, that counted.  Okay, so the next question is using the spectrum  to estimate the force constant for the CO bond.  And I'm not going to work all of this out in the interest  of time, but basically you need to use that transition frequency.  So that's, that's what we do.  Of course, this is the reduced mass and K is in newtons per meter.  And the actual value, if you look it up for carbon monoxide,  is about 1860.  You know, again, it's hard to read the transition frequency  off that little plot and that sort of determines the answer you get.  So if it was consistent and you got something close,  you got full points.  So that was really straightforward.  It's just kind of remembering which equation to use  and plugging stuff in.  As always with such things, there are opportunities  to make mistakes, getting the units wrong and, you know,  forgetting to convert things.  But otherwise, that one was pretty straightforward.  Okay. Next question, is carbon monoxide a perfect rigid rotor?  And it's not and a lot of people wrote  because life's not perfect, which is funny and probably reasonable.  But it doesn't count as a good justification.  So, you know, I guess John Mark said that at the review session.  But, well, you're also about to learn that life's not fair  because that doesn't get you the full credit.  So, in order to get full credit, you had to write  that there's centrifugal distortion and you can tell  because the spacings on the spectrum are not exactly equal.  It's kind of squished on one side and stretched on the other.  Yes?  Yes.  I didn't grade that one.  You can hope your TAs are nicer than I am.  So, I think that's like, you know, when you go to iTunes  and you buy a song and they charge you $99 instead of 99 cents,  is that close enough or is it completely wrong?  I know what I think.  So, you can hope the TAs are nicer than I am.  I'm not sure.  That's like you gave the baby a thousand times too much medicine  and he died.  It's bad.  Well, it is.  Those order of magnitude errors cause huge problems  in real life.  Your Mars rover crashes into the surface of the planet.  You don't get partial credit.  You get fired.  You know, there's it like the, you know,  yeah, I'm making fun but like the, you know, the things  where you have a small error in the last couple decimal places  or whatever because of round off problems,  that's not really a huge deal but, yeah,  orders of magnitude matter.  Okay. All right.  So, how would the Raman spectrum of CO look different?  So, if you're looking at a Raman spectrum instead  of an IR spectrum, how would you be able to tell that?  Okay. So, you had to give at least two differences  and here are the ones that I could think of.  You would have a Rayleigh line in the center of the spectrum  which is not called a Q branch.  It's a Rayleigh line.  So, the Q branch is if you have, you know, something like a  radical where there's electron orbital momentum  and so the central transition isn't forbidden,  then you see that.  For a Raman spectrum, it's the Rayleigh line.  Other things could be the spacing is four times the  rotational constant or the line intensity is smaller.  And so, any two of those were fine.  Question in the back?  You definitely got something for that.  I don't remember exactly how we did that one.  Yes?  I was wondering, I was taking the exam,  and I was thinking about a note saying that there's a Q branch  and a P branch and an O branch in that spectroscopy.  How would that, how does that factor into the whole thing?  Okay.  If you had labeled all three branches there,  it doesn't do it because that's, you know, that's like,  we can talk about it later if you want, but it seems  like not worth spending a lot of time on.  But yeah, if you drew it and labeled all three branches,  then that's something different.  But yes?  Isn't there a difference in the number of peaks as well?  What do you mean a difference in the number of peaks?  The number of peaks that would show up on the IR  versus the Raman spectrum?  It would be pretty hard to tell.  Because I thought that's what we were using our point groups  before, the number of peaks.  Sure, but it's the same molecule and it's a linear molecule.  So if you had a, you know,  if you had a complex polyatomic molecule,  that would definitely be true.  You would see different vibrational modes.  But this thing only has one, it's a diatomic molecule  so that it's only got one vibrational mode, right?  So that wouldn't really be applicable here.  But you're right, if you had a complex polyatomic molecule,  that would be totally true.  Okay, let's go on to the next one.  All right, so for this one,  it's sort of a, it's a standard Burge-Sponer plot  but with a little bit of a twist which is  that there's a break in the middle of it.  So what that tells you in practical terms is  that it has a weird potential.  So this was from a paper in science a few years ago  on beryllium dimer and looking at beryllium dimer forming.  And at first the, you know, the dissociation energy  of beryllium dimer was estimated incorrectly  because it has this weird shape  in the experimental potential.  It deviates really strongly from a normal, worse potential.  So why does it have this weird shape?  Well, if you think about the electronic structure  of beryllium dimer, you can kind of tell it's not going  to form a very strong bond which is true.  And, you know, since it has this weak bond,  it's got a strange potential.  And this was something that was a relatively recent result.  It was in science a few years ago.  Okay, so what does this mean for doing the problem?  It means you can't really look at the, you know,  the value for new max.  So you're used to the X intercept telling you something  about the convergence limit.  And, you know, here that's kind of hard to get  because, you know, we have these two breaks in the plot.  But so anyway, the hint was to use the part of the data  that's consistent with the Morse potential,  which was the first part which is shown by this line here.  And so you could do that.  And so as always for these things, the slope tells you something  about the potential.  And so then we know that the Y intercept,  which is normal in this plot, and DE is just, all right.  So, and that was about 625.  And so that's it, pretty straightforward.  So the only part, the only part that had information content  here was just knowing which equations to use and being able  to get this off the plot.  Okay, so then moving on to, yeah.  I think I got a negative.  Well, does that make sense?  I mean, so it depends on, you know, you're thinking about,  it's an absolute value thing.  I mean, you're thinking about putting in, you have to put energy  into the molecule to get it to dissociate, right.  So that's, so.  So if your X is negative, and your BE is positive,  then your XE is going to be negative.  Right.  So if your XE is negative, and your BE is positive,  then your DE is negative.  But take the absolute value.  But that's just, that's one of those things where it's,  you know, the convention is sort of, you know,  it could be either way depending on whether you're a chemist  or an engineer and what you're doing.  So, you know, as far as whether it's the dissociation  or the energy is positive or negative.  Yes.  What?  For the question, is that for the question?  It is for XE, yeah.  But.  You will lose points if it's a negative?  Like a point.  It's not, it's not a big deal.  OK.  So now the next one.  Looking at Frank Condon factors.  OK.  So the question is, write an expression  for Frank Condon factor for a transition  between these two states.  And I wanted you to put in the states,  but which is why I wrote down the functions.  But so you got almost all the points for doing it  in direct notation.  So we have these two states.  And the Frank Condon factor is.  So that got almost all the points.  People were really worried during the exam about which one's  the initial and which one's the final state.  It doesn't matter, right?  So we're talking about a transition between them.  And that amplitude isn't going to depend on which one is initial  and which one is final.  So, you know, it doesn't really matter.  Yes.  Because in your lecture notes, the Frank Condon factor,  how do you think vibrational states in the form that's  that OK for all the words vibrational states?  Oh yeah, that's true.  That's, so here's one of these cases where direct notation is  ambiguous.  And so that's why it's best to write in the states.  So yeah, what you're saying is these quantum numbers aren't  relevant to anything.  Yeah, true.  But so what I wanted you to actually do is, you know,  write out like the integral.  And so I said, new prime equals zero.  So that's one.  And then H2 is, right?  And do you have to evaluate it?  No. I just wanted to write the expression.  So doing it in direct notation got you most of the points,  but I wanted you to make the connection  and write down the functions.  OK, so the next question is we want  to sketch this transition on an energy level diagram.  All right, so we've got some electronic states.  And, you know, it doesn't really matter what they look like as  long as they vaguely look like a Morse potential.  And then we have some vibrational states in there.  And you should know that, you know, there's some zero point energy.  So that would be new prime, double prime, sorry.  And so that's basically it.  And then, of course, we needed to label the transition.  And all right, the way I drew it, it is difficult to do that.  And so there's not very much overlap the way I drew it,  but there you go.  The next question is going to be if you drew the error going  down, is that fine too?  Yes, because we didn't specify which one's the initial  and which one's the final state.  So as long as you're consistent, good enough.  I just wanted to see if you can make the connection between,  you know, writing down these expressions sort of in quantum  terms, plugging in actual functions, you know,  not evaluating it but putting it in.  And, you know, being able to go from that to looking  at a potential diagram.  Okay, next one, term symbols.  So for those who spent endless hours in office hours writing  out all the possible excited states of O2,  it was time well spent.  Okay, so the O2 molecule is in its ground state.  And here, you know, there was some potential for confusion  because the book sort of doesn't distinguish  between the various electron configurations that you can get  with the ground state electron configuration.  But of course there is an actual ground state.  It's, you know, you have to use Hoon's rule to get that.  So it's the one where you have the most unpaired electrons.  So let's look at this in terms  of the molecular orbital diagram.  So I'm just drawing the P orbitals.  So I've got sigma g. All right, that's my,  so that's my molecular orbital diagram  from general chemistry of O2.  And I'm ignoring the S orbitals because they're filled  and they don't give me anything.  And so I write out the electron configuration  for this molecule, which gives me this.  And again, since it's the ground state, we use Hoon's rule  and put in the configuration with both  of these electrons unpaired and in separate orbitals.  Okay, everybody with me so far?  Cool. Okay, so then the next question is what's lambda?  And it is going to be zero.  So we get a sigma term.  And then if we, since it's a, so we have to decide whether it's g  or u, and so we've got these two electrons in the,  they're both in the pi star g orbital.  So that's g times g. So this term is even.  And they're in different orbitals.  So it's minus.  And what's its spin multiplicity?  Oops. Nope.  Just kidding.  It's a triplet state because we have two unpaired electrons.  I was getting ahead of myself in thinking of the next one.  Okay, so that's the ground state.  And, you know, there were a couple of, you know,  of course you can put those electrons in different places  and write a couple more term symbols.  There's a delta term symbol, et cetera.  Since I just asked for the lowest energy one,  that was essentially a waste of time if you did it  on this question.  Okay, so the next one is we have one electron promoted  from the homo to the lumo.  And so, you know, again, general chemistry,  highest occupied molecular orbital  and lowest unoccupied molecular orbital.  Some people asked about that.  I think people just get nervous  and forget stuff sometimes.  Okay, so you have, these guys are all the same.  And then there are a couple choices here.  I'm not labeling the orbitals,  but obviously they didn't change.  And so, these things give you, there are two states.  And if you only got one of them, so you had to get both  of them to get full credit.  If you only got one of them, we took off a point.  So, what do you think?  There are various side conversations,  but it seems like everybody more or less gets it.  Is that true?  Okay, good.  All right, so for, you know, the endless hours  that we spent redrawing this picture paid off.  Okay, so the next question has to do with NMR.  All right, so this was, this question was  about writing the matrix elements for I plus,  the raising operator for a spin one half.  And so, it's full definition was here.  And you want to write this out as,  so you know that I plus operated on alpha is zero.  This is zero.  And then, so I'll write the full thing.  There we go.  So, all right, so this gives us, there we go.  That's more like it.  So, one times alpha alpha.  And so, the answer is, and if you just wrote the matrix  because you had it written down on your sheet sheet,  that didn't get full points.  You needed to show some understanding of how to do it  to get full credit.  So, if you wrote out the full matrix or just wrote it  out symbolically or, you know, put down the,  how the matrix elements work, that all counted,  as long as it showed that you knew how to do it.  This collection of constants in front in the definition  of the raising and lowering operators, I'm just letting you  know for purposes of taking the final.  You know, it comes out to one in this case,  but if it wasn't to spin one half, it wouldn't,  as you remember if we, you know,  from doing some of the examples at office hours.  Okay, so now let's look at NMR spectra.  So, this was one of these things where exhortations  to read the directions before doing it were really,  really important.  So, people did sad time wasting things like,  there were people who didn't just draw the spectrum  for the protons that were labeled.  They, like some people actually crossed out my labels  and relabeled the whole thing and drew all the protons.  And it was sad because it must have taken forever  and, you know, then people didn't have time for other things.  People tried to draw a proton for the one that's labeled  with the star, which of course has four bonds to that carbon.  So, you know, I'm not being mean.  I'm just pointing out, like, read the directions.  There are things, you know, there are things in there  that are meant to save you time and, you know,  like not having to draw the spectrum for all the protons.  Okay, so let's look at what we have to do  first to get this.  Okay, so we have a chemical shift table.  And so one of the first things to do here is identify your  various functional groups.  So, we have A here.  So we have our two protons here.  And some people made the mistake of saying  that this thing was a carbonyl.  So that would be, and, you know, they,  so they put it in this chemical shift range.  And that's not what it is, right?  It's got, you know, it's between these two oxygens.  So it's an ether.  So it would be toward the, you know, higher chemical shift end  of the ether because it's got two of them.  But so it should be somewhere around, you know,  four or five ppm somewhere in there.  Four and a half something.  Okay, what else have we got?  So D here is an aromatic.  It has a proton as its neighbor.  C is a methyl group.  B has a proton there.  And it's got, so B is going to be a complicated,  a, multiple, right?  Because it's got two neighbors on one side and three  on the other and they're not equivalent.  But so, okay, we figured out what chemical shift everything  should be.  So we can start drawing the spectrum.  So you get points for labeling the axis.  And let's start on this end.  Okay, so D, our aromatic here is going to be, you know,  between seven and nine ppm.  So let's, you know, put it down here.  And it has one neighbor.  So it'll be a doublet.  And then our ether is going to be the next one along  in the chemical shift scale.  And it just has two oxygens next to it,  which we can probably assume are, you know, O16.  And so it's just going to be a singlet.  And then our methyl group here is going to be furthest  toward zero.  And it's going to be a doublet  because it only has one neighbor.  And then we're left with what is B. So it's got two sets  of neighbors that are inequivalent.  And, you know, there's two protons on one side  and three on the other.  And I didn't tell you which J coupling is larger.  So it is either a triplet of quartets or a quartet of triplets.  And if you wrote something like that and drew, you know,  this, that's fine.  Some people really carefully drew the picture out.  You know, again, I want to know if you understand it,  not your ability to draw these things beautifully  because as you can see, I'm not so great at it.  Yes?  For the numbers, do we have to like B exact or do we just draw  them like in that order?  For the chemical shift range, you got points if they were  in the right order relative to each other  and the chemical shift scale was something reasonable.  But, you know, thinking that, that this thing is a carbonyl  and putting it at 12 ppm was not reasonable.  Yes?  It doesn't because it's exchangeable.  It means it's popping on and off all the time in, you know,  exchanges with the solvent.  Although, you know, people who, who mixed in the nitrogen proton  and the splitting, I think I just took off a point.  It was, you know, if you understood the rest of it,  that's not a huge deal.  Okay, so that is, that is that one.  And then, we said a new spectrum is collected  with the carbon decoupling turned off.  And we want to see the signal for the proton labeled A. Okay,  so the carbon decoupling is turned off.  So that means in our, so in our previous spectrum,  we didn't see any coupling to C13.  Now, the decoupling is turned off and we're going to.  C13 is a spin one-half.  And proton A, remember that was the one that was an ether  and it didn't have any neighbors.  So, it was a singlet before.  And so, if it's not decoupled, it's a doublet.  And that, that one was either right or wrong.  And so then, we had an inversion recovery experiment is  performed on the C13's in this molecule.  So remember, this is the experiment that we do to measure  the longitudinal relaxation time.  So we put everything, we flip all the spins of the signal  and then wait for it to relax and measure how long it takes  to relax back to equilibrium.  And remember what causes relaxation to equilibrium.  It's fluctuating magnetic fields local to that nucleus.  So something that has direct dipole-dipole couplings  with protons, so if it's directly bonded to a proton,  that's going to relax faster than something  that doesn't have that.  Remember, another important source  of T1 relaxation is methyl rotation  and also just molecular motion.  And so, in all of these respects, you know,  here's our molecule again.  The one that's labeled with a star is non-protonated.  It's aromatic, so it's probably going  to be rigid conformationally.  And C is a methyl.  So the one that returns  to equilibrium more quickly is C, the methyl.  And basically, any one of these explanations worked.  So if you got any reason why it's going  to have a quicker relaxation time, that was good.  Yes?  So what do you mean by steric hindrance?  I don't see how that has anything to do with it.  But, you know, win me over.  What does that have to do with it?  Well, it's going to be a more than the A. It's not the argument  about rigidity, it's kind of like a ball of the mind.  Hmm, kind of, but it's not that it's starically hindered.  I mean, if you had like a T-butyl group,  that would still relax pretty fast  because those metals can still rotate.  So, sorry, no.  You had to have something about conformational rigidity.  Yes?  What if you said like it was connected to a lot more atoms  and so those atoms affect that carbon, so it's harder,  but it takes it longer to come back to equilibrium?  Well, but can, so which one is connected to more atoms?  I think I said the star one.  Well, but it's not, right?  They're both connected to three, to four other atoms.  It depends, it matters what those atoms are.  So, you know, if you, if you,  if you point if I said that one took a longer time.  We can talk about it later.  Wait, so you're asking if you just, if you got,  which one took a longer time, right?  And not the explanation, did that get partial credit?  Yes, it did.  Okay, that's it.  That's the exam.  So, again, the grading is almost done.  We'll get it done as fast as we can today  and then send it to get scanned.  And next time we're going to start talking  about statistical mechanics.
Molecular_Structure___Statistical_Mechanics_131B__Lecture_21__Second_Midterm_Examination_Review_0_299.png Molecular_Structure___Statistical_Mechanics_131B__Lecture_21__Second_Midterm_Examination_Review_300_947.png Molecular_Structure___Statistical_Mechanics_131B__Lecture_21__Second_Midterm_Examination_Review_948_995.png Molecular_Structure___Statistical_Mechanics_131B__Lecture_21__Second_Midterm_Examination_Review_996_1295.png Molecular_Structure___Statistical_Mechanics_131B__Lecture_21__Second_Midterm_Examination_Review_1296_1391.png Molecular_Structure___Statistical_Mechanics_131B__Lecture_21__Second_Midterm_Examination_Review_1392_1643.png Molecular_Structure___Statistical_Mechanics_131B__Lecture_21__Second_Midterm_Examination_Review_1644_2279.png Molecular_Structure___Statistical_Mechanics_131B__Lecture_21__Second_Midterm_Examination_Review_2280_2339.png Molecular_Structure___Statistical_Mechanics_131B__Lecture_21__Second_Midterm_Examination_Review_2340_7031.png Molecular_Structure___Statistical_Mechanics_131B__Lecture_21__Second_Midterm_Examination_Review_7032_11279.png Molecular_Structure___Statistical_Mechanics_131B__Lecture_21__Second_Midterm_Examination_Review_11280_15731.png Molecular_Structure___Statistical_Mechanics_131B__Lecture_21__Second_Midterm_Examination_Review_15732_15827.png Molecular_Structure___Statistical_Mechanics_131B__Lecture_21__Second_Midterm_Examination_Review_15828_19571.png Molecular_Structure___Statistical_Mechanics_131B__Lecture_21__Second_Midterm_Examination_Review_19572_22931.png Molecular_Structure___Statistical_Mechanics_131B__Lecture_21__Second_Midterm_Examination_Review_22932_22979.png Molecular_Structure___Statistical_Mechanics_131B__Lecture_21__Second_Midterm_Examination_Review_22980_23075.png Molecular_Structure___Statistical_Mechanics_131B__Lecture_21__Second_Midterm_Examination_Review_23076_24047.png Molecular_Structure___Statistical_Mechanics_131B__Lecture_21__Second_Midterm_Examination_Review_24048_25679.png Molecular_Structure___Statistical_Mechanics_131B__Lecture_21__Second_Midterm_Examination_Review_25680_29267.png Molecular_Structure___Statistical_Mechanics_131B__Lecture_21__Second_Midterm_Examination_Review_29268_30527.png Molecular_Structure___Statistical_Mechanics_131B__Lecture_21__Second_Midterm_Examination_Review_30528_30731.png Molecular_Structure___Statistical_Mechanics_131B__Lecture_21__Second_Midterm_Examination_Review_30732_32279.png Molecular_Structure___Statistical_Mechanics_131B__Lecture_21__Second_Midterm_Examination_Review_32280_32399.png Molecular_Structure___Statistical_Mechanics_131B__Lecture_21__Second_Midterm_Examination_Review_32400_32615.png Molecular_Structure___Statistical_Mechanics_131B__Lecture_21__Second_Midterm_Examination_Review_32616_37679.png Molecular_Structure___Statistical_Mechanics_131B__Lecture_21__Second_Midterm_Examination_Review_37680_38351.png Molecular_Structure___Statistical_Mechanics_131B__Lecture_21__Second_Midterm_Examination_Review_38352_38411.png Molecular_Structure___Statistical_Mechanics_131B__Lecture_21__Second_Midterm_Examination_Review_38412_38567.png Molecular_Structure___Statistical_Mechanics_131B__Lecture_21__Second_Midterm_Examination_Review_38568_38975.png Molecular_Structure___Statistical_Mechanics_131B__Lecture_21__Second_Midterm_Examination_Review_38976_39323.png Molecular_Structure___Statistical_Mechanics_131B__Lecture_21__Second_Midterm_Examination_Review_39324_40043.png Molecular_Structure___Statistical_Mechanics_131B__Lecture_21__Second_Midterm_Examination_Review_40044_43199.png Molecular_Structure___Statistical_Mechanics_131B__Lecture_21__Second_Midterm_Examination_Review_43200_43799.png Molecular_Structure___Statistical_Mechanics_131B__Lecture_21__Second_Midterm_Examination_Review_43800_45359.png Molecular_Structure___Statistical_Mechanics_131B__Lecture_21__Second_Midterm_Examination_Review_45360_46571.png Molecular_Structure___Statistical_Mechanics_131B__Lecture_21__Second_Midterm_Examination_Review_46572_47339.png Molecular_Structure___Statistical_Mechanics_131B__Lecture_21__Second_Midterm_Examination_Review_47340_50027.png Molecular_Structure___Statistical_Mechanics_131B__Lecture_21__Second_Midterm_Examination_Review_50028_50087.png Molecular_Structure___Statistical_Mechanics_131B__Lecture_21__Second_Midterm_Examination_Review_50088_51035.png Molecular_Structure___Statistical_Mechanics_131B__Lecture_21__Second_Midterm_Examination_Review_51036_51107.png Molecular_Structure___Statistical_Mechanics_131B__Lecture_21__Second_Midterm_Examination_Review_51108_51335.png Molecular_Structure___Statistical_Mechanics_131B__Lecture_21__Second_Midterm_Examination_Review_51336_53147.png Molecular_Structure___Statistical_Mechanics_131B__Lecture_21__Second_Midterm_Examination_Review_53148_53471.png Molecular_Structure___Statistical_Mechanics_131B__Lecture_21__Second_Midterm_Examination_Review_53472_53963.png Molecular_Structure___Statistical_Mechanics_131B__Lecture_21__Second_Midterm_Examination_Review_53964_56891.png Molecular_Structure___Statistical_Mechanics_131B__Lecture_21__Second_Midterm_Examination_Review_56892_57179.png Molecular_Structure___Statistical_Mechanics_131B__Lecture_21__Second_Midterm_Examination_Review_57180_57227.png Molecular_Structure___Statistical_Mechanics_131B__Lecture_21__Second_Midterm_Examination_Review_57228_57467.png Molecular_Structure___Statistical_Mechanics_131B__Lecture_21__Second_Midterm_Examination_Review_57468_57496.png
Good morning everybody. Today we're going to talk about the exam. Anybody have any questions before we do? What's the mean? I don't know. You know, again, it takes you guys an hour to take the exam. Multiply by 207 or whatever. There are four of us grading it. Okay, it doesn't take as long to grade it as it does to take it, but it's not going to be all graded and ready to go by the next class period. Sorry, we just, there's only four of us, we can't do that. So it's almost done being graded. We'll get that finished today. Then it gets sent to wherever it goes for scanning. And then as soon as I get the PDFs back, I will give them to you. But I have zero control over how long it takes to scan stuff. So I know you're anxious to see your score, but it just takes a little while. It's a big class. Okay, so let's talk about the exam. There's a question in the back? Yeah, what you've seen so far, how have people done it? It looks like people have done okay. It looks, looks all right. People ran out of time. I expected that. That's fine. It's normal. But overall, seems pretty reasonable. Okay, so let's get started going over this. All right, so first question. I are spectrum of carbon monoxide. Why is there a gap in the center? Yeah, there's no delta J equals zero transition. So anything mentioning that that transition is forbidden or that there's a, that the specific selection rule prohibits that, anything like that got full credit. Okay, so then the next question is what is the energy in wave numbers of this transition? And, you know, this is really straightforward. You just read it off the spectrum, right? Because since there's no delta J equals zero transition, that spot in the middle where there's no line is, that's where the new equals zero to new equals one transition is. And so if you just read it off the spectrum, you get something like this. It's hard to read from that little plot. So if you've got anything close, that counted. Okay, so the next question is using the spectrum to estimate the force constant for the CO bond. And I'm not going to work all of this out in the interest of time, but basically you need to use that transition frequency. So that's, that's what we do. Of course, this is the reduced mass and K is in newtons per meter. And the actual value, if you look it up for carbon monoxide, is about 1860. You know, again, it's hard to read the transition frequency off that little plot and that sort of determines the answer you get. So if it was consistent and you got something close, you got full points. So that was really straightforward. It's just kind of remembering which equation to use and plugging stuff in. As always with such things, there are opportunities to make mistakes, getting the units wrong and, you know, forgetting to convert things. But otherwise, that one was pretty straightforward. Okay. Next question, is carbon monoxide a perfect rigid rotor? And it's not and a lot of people wrote because life's not perfect, which is funny and probably reasonable. But it doesn't count as a good justification. So, you know, I guess John Mark said that at the review session. But, well, you're also about to learn that life's not fair because that doesn't get you the full credit. So, in order to get full credit, you had to write that there's centrifugal distortion and you can tell because the spacings on the spectrum are not exactly equal. It's kind of squished on one side and stretched on the other. Yes? Yes. I didn't grade that one. You can hope your TAs are nicer than I am. So, I think that's like, you know, when you go to iTunes and you buy a song and they charge you $99 instead of 99 cents, is that close enough or is it completely wrong? I know what I think. So, you can hope the TAs are nicer than I am. I'm not sure. That's like you gave the baby a thousand times too much medicine and he died. It's bad. Well, it is. Those order of magnitude errors cause huge problems in real life. Your Mars rover crashes into the surface of the planet. You don't get partial credit. You get fired. You know, there's it like the, you know, yeah, I'm making fun but like the, you know, the things where you have a small error in the last couple decimal places or whatever because of round off problems, that's not really a huge deal but, yeah, orders of magnitude matter. Okay. All right. So, how would the Raman spectrum of CO look different? So, if you're looking at a Raman spectrum instead of an IR spectrum, how would you be able to tell that? Okay. So, you had to give at least two differences and here are the ones that I could think of. You would have a Rayleigh line in the center of the spectrum which is not called a Q branch. It's a Rayleigh line. So, the Q branch is if you have, you know, something like a radical where there's electron orbital momentum and so the central transition isn't forbidden, then you see that. For a Raman spectrum, it's the Rayleigh line. Other things could be the spacing is four times the rotational constant or the line intensity is smaller. And so, any two of those were fine. Question in the back? You definitely got something for that. I don't remember exactly how we did that one. Yes? I was wondering, I was taking the exam, and I was thinking about a note saying that there's a Q branch and a P branch and an O branch in that spectroscopy. How would that, how does that factor into the whole thing? Okay. If you had labeled all three branches there, it doesn't do it because that's, you know, that's like, we can talk about it later if you want, but it seems like not worth spending a lot of time on. But yeah, if you drew it and labeled all three branches, then that's something different. But yes? Isn't there a difference in the number of peaks as well? What do you mean a difference in the number of peaks? The number of peaks that would show up on the IR versus the Raman spectrum? It would be pretty hard to tell. Because I thought that's what we were using our point groups before, the number of peaks. Sure, but it's the same molecule and it's a linear molecule. So if you had a, you know, if you had a complex polyatomic molecule, that would definitely be true. You would see different vibrational modes. But this thing only has one, it's a diatomic molecule so that it's only got one vibrational mode, right? So that wouldn't really be applicable here. But you're right, if you had a complex polyatomic molecule, that would be totally true. Okay, let's go on to the next one. All right, so for this one, it's sort of a, it's a standard Burge-Sponer plot but with a little bit of a twist which is that there's a break in the middle of it. So what that tells you in practical terms is that it has a weird potential. So this was from a paper in science a few years ago on beryllium dimer and looking at beryllium dimer forming. And at first the, you know, the dissociation energy of beryllium dimer was estimated incorrectly because it has this weird shape in the experimental potential. It deviates really strongly from a normal, worse potential. So why does it have this weird shape? Well, if you think about the electronic structure of beryllium dimer, you can kind of tell it's not going to form a very strong bond which is true. And, you know, since it has this weak bond, it's got a strange potential. And this was something that was a relatively recent result. It was in science a few years ago. Okay, so what does this mean for doing the problem? It means you can't really look at the, you know, the value for new max. So you're used to the X intercept telling you something about the convergence limit. And, you know, here that's kind of hard to get because, you know, we have these two breaks in the plot. But so anyway, the hint was to use the part of the data that's consistent with the Morse potential, which was the first part which is shown by this line here. And so you could do that. And so as always for these things, the slope tells you something about the potential. And so then we know that the Y intercept, which is normal in this plot, and DE is just, all right. So, and that was about 625. And so that's it, pretty straightforward. So the only part, the only part that had information content here was just knowing which equations to use and being able to get this off the plot. Okay, so then moving on to, yeah. I think I got a negative. Well, does that make sense? I mean, so it depends on, you know, you're thinking about, it's an absolute value thing. I mean, you're thinking about putting in, you have to put energy into the molecule to get it to dissociate, right. So that's, so. So if your X is negative, and your BE is positive, then your XE is going to be negative. Right. So if your XE is negative, and your BE is positive, then your DE is negative. But take the absolute value. But that's just, that's one of those things where it's, you know, the convention is sort of, you know, it could be either way depending on whether you're a chemist or an engineer and what you're doing. So, you know, as far as whether it's the dissociation or the energy is positive or negative. Yes. What? For the question, is that for the question? It is for XE, yeah. But. You will lose points if it's a negative? Like a point. It's not, it's not a big deal. OK. So now the next one. Looking at Frank Condon factors. OK. So the question is, write an expression for Frank Condon factor for a transition between these two states. And I wanted you to put in the states, but which is why I wrote down the functions. But so you got almost all the points for doing it in direct notation. So we have these two states. And the Frank Condon factor is. So that got almost all the points. People were really worried during the exam about which one's the initial and which one's the final state. It doesn't matter, right? So we're talking about a transition between them. And that amplitude isn't going to depend on which one is initial and which one is final. So, you know, it doesn't really matter. Yes. Because in your lecture notes, the Frank Condon factor, how do you think vibrational states in the form that's that OK for all the words vibrational states? Oh yeah, that's true. That's, so here's one of these cases where direct notation is ambiguous. And so that's why it's best to write in the states. So yeah, what you're saying is these quantum numbers aren't relevant to anything. Yeah, true. But so what I wanted you to actually do is, you know, write out like the integral. And so I said, new prime equals zero. So that's one. And then H2 is, right? And do you have to evaluate it? No. I just wanted to write the expression. So doing it in direct notation got you most of the points, but I wanted you to make the connection and write down the functions. OK, so the next question is we want to sketch this transition on an energy level diagram. All right, so we've got some electronic states. And, you know, it doesn't really matter what they look like as long as they vaguely look like a Morse potential. And then we have some vibrational states in there. And you should know that, you know, there's some zero point energy. So that would be new prime, double prime, sorry. And so that's basically it. And then, of course, we needed to label the transition. And all right, the way I drew it, it is difficult to do that. And so there's not very much overlap the way I drew it, but there you go. The next question is going to be if you drew the error going down, is that fine too? Yes, because we didn't specify which one's the initial and which one's the final state. So as long as you're consistent, good enough. I just wanted to see if you can make the connection between, you know, writing down these expressions sort of in quantum terms, plugging in actual functions, you know, not evaluating it but putting it in. And, you know, being able to go from that to looking at a potential diagram. Okay, next one, term symbols. So for those who spent endless hours in office hours writing out all the possible excited states of O2, it was time well spent. Okay, so the O2 molecule is in its ground state. And here, you know, there was some potential for confusion because the book sort of doesn't distinguish between the various electron configurations that you can get with the ground state electron configuration. But of course there is an actual ground state. It's, you know, you have to use Hoon's rule to get that. So it's the one where you have the most unpaired electrons. So let's look at this in terms of the molecular orbital diagram. So I'm just drawing the P orbitals. So I've got sigma g. All right, that's my, so that's my molecular orbital diagram from general chemistry of O2. And I'm ignoring the S orbitals because they're filled and they don't give me anything. And so I write out the electron configuration for this molecule, which gives me this. And again, since it's the ground state, we use Hoon's rule and put in the configuration with both of these electrons unpaired and in separate orbitals. Okay, everybody with me so far? Cool. Okay, so then the next question is what's lambda? And it is going to be zero. So we get a sigma term. And then if we, since it's a, so we have to decide whether it's g or u, and so we've got these two electrons in the, they're both in the pi star g orbital. So that's g times g. So this term is even. And they're in different orbitals. So it's minus. And what's its spin multiplicity? Oops. Nope. Just kidding. It's a triplet state because we have two unpaired electrons. I was getting ahead of myself in thinking of the next one. Okay, so that's the ground state. And, you know, there were a couple of, you know, of course you can put those electrons in different places and write a couple more term symbols. There's a delta term symbol, et cetera. Since I just asked for the lowest energy one, that was essentially a waste of time if you did it on this question. Okay, so the next one is we have one electron promoted from the homo to the lumo. And so, you know, again, general chemistry, highest occupied molecular orbital and lowest unoccupied molecular orbital. Some people asked about that. I think people just get nervous and forget stuff sometimes. Okay, so you have, these guys are all the same. And then there are a couple choices here. I'm not labeling the orbitals, but obviously they didn't change. And so, these things give you, there are two states. And if you only got one of them, so you had to get both of them to get full credit. If you only got one of them, we took off a point. So, what do you think? There are various side conversations, but it seems like everybody more or less gets it. Is that true? Okay, good. All right, so for, you know, the endless hours that we spent redrawing this picture paid off. Okay, so the next question has to do with NMR. All right, so this was, this question was about writing the matrix elements for I plus, the raising operator for a spin one half. And so, it's full definition was here. And you want to write this out as, so you know that I plus operated on alpha is zero. This is zero. And then, so I'll write the full thing. There we go. So, all right, so this gives us, there we go. That's more like it. So, one times alpha alpha. And so, the answer is, and if you just wrote the matrix because you had it written down on your sheet sheet, that didn't get full points. You needed to show some understanding of how to do it to get full credit. So, if you wrote out the full matrix or just wrote it out symbolically or, you know, put down the, how the matrix elements work, that all counted, as long as it showed that you knew how to do it. This collection of constants in front in the definition of the raising and lowering operators, I'm just letting you know for purposes of taking the final. You know, it comes out to one in this case, but if it wasn't to spin one half, it wouldn't, as you remember if we, you know, from doing some of the examples at office hours. Okay, so now let's look at NMR spectra. So, this was one of these things where exhortations to read the directions before doing it were really, really important. So, people did sad time wasting things like, there were people who didn't just draw the spectrum for the protons that were labeled. They, like some people actually crossed out my labels and relabeled the whole thing and drew all the protons. And it was sad because it must have taken forever and, you know, then people didn't have time for other things. People tried to draw a proton for the one that's labeled with the star, which of course has four bonds to that carbon. So, you know, I'm not being mean. I'm just pointing out, like, read the directions. There are things, you know, there are things in there that are meant to save you time and, you know, like not having to draw the spectrum for all the protons. Okay, so let's look at what we have to do first to get this. Okay, so we have a chemical shift table. And so one of the first things to do here is identify your various functional groups. So, we have A here. So we have our two protons here. And some people made the mistake of saying that this thing was a carbonyl. So that would be, and, you know, they, so they put it in this chemical shift range. And that's not what it is, right? It's got, you know, it's between these two oxygens. So it's an ether. So it would be toward the, you know, higher chemical shift end of the ether because it's got two of them. But so it should be somewhere around, you know, four or five ppm somewhere in there. Four and a half something. Okay, what else have we got? So D here is an aromatic. It has a proton as its neighbor. C is a methyl group. B has a proton there. And it's got, so B is going to be a complicated, a, multiple, right? Because it's got two neighbors on one side and three on the other and they're not equivalent. But so, okay, we figured out what chemical shift everything should be. So we can start drawing the spectrum. So you get points for labeling the axis. And let's start on this end. Okay, so D, our aromatic here is going to be, you know, between seven and nine ppm. So let's, you know, put it down here. And it has one neighbor. So it'll be a doublet. And then our ether is going to be the next one along in the chemical shift scale. And it just has two oxygens next to it, which we can probably assume are, you know, O16. And so it's just going to be a singlet. And then our methyl group here is going to be furthest toward zero. And it's going to be a doublet because it only has one neighbor. And then we're left with what is B. So it's got two sets of neighbors that are inequivalent. And, you know, there's two protons on one side and three on the other. And I didn't tell you which J coupling is larger. So it is either a triplet of quartets or a quartet of triplets. And if you wrote something like that and drew, you know, this, that's fine. Some people really carefully drew the picture out. You know, again, I want to know if you understand it, not your ability to draw these things beautifully because as you can see, I'm not so great at it. Yes? For the numbers, do we have to like B exact or do we just draw them like in that order? For the chemical shift range, you got points if they were in the right order relative to each other and the chemical shift scale was something reasonable. But, you know, thinking that, that this thing is a carbonyl and putting it at 12 ppm was not reasonable. Yes? It doesn't because it's exchangeable. It means it's popping on and off all the time in, you know, exchanges with the solvent. Although, you know, people who, who mixed in the nitrogen proton and the splitting, I think I just took off a point. It was, you know, if you understood the rest of it, that's not a huge deal. Okay, so that is, that is that one. And then, we said a new spectrum is collected with the carbon decoupling turned off. And we want to see the signal for the proton labeled A. Okay, so the carbon decoupling is turned off. So that means in our, so in our previous spectrum, we didn't see any coupling to C13. Now, the decoupling is turned off and we're going to. C13 is a spin one-half. And proton A, remember that was the one that was an ether and it didn't have any neighbors. So, it was a singlet before. And so, if it's not decoupled, it's a doublet. And that, that one was either right or wrong. And so then, we had an inversion recovery experiment is performed on the C13's in this molecule. So remember, this is the experiment that we do to measure the longitudinal relaxation time. So we put everything, we flip all the spins of the signal and then wait for it to relax and measure how long it takes to relax back to equilibrium. And remember what causes relaxation to equilibrium. It's fluctuating magnetic fields local to that nucleus. So something that has direct dipole-dipole couplings with protons, so if it's directly bonded to a proton, that's going to relax faster than something that doesn't have that. Remember, another important source of T1 relaxation is methyl rotation and also just molecular motion. And so, in all of these respects, you know, here's our molecule again. The one that's labeled with a star is non-protonated. It's aromatic, so it's probably going to be rigid conformationally. And C is a methyl. So the one that returns to equilibrium more quickly is C, the methyl. And basically, any one of these explanations worked. So if you got any reason why it's going to have a quicker relaxation time, that was good. Yes? So what do you mean by steric hindrance? I don't see how that has anything to do with it. But, you know, win me over. What does that have to do with it? Well, it's going to be a more than the A. It's not the argument about rigidity, it's kind of like a ball of the mind. Hmm, kind of, but it's not that it's starically hindered. I mean, if you had like a T-butyl group, that would still relax pretty fast because those metals can still rotate. So, sorry, no. You had to have something about conformational rigidity. Yes? What if you said like it was connected to a lot more atoms and so those atoms affect that carbon, so it's harder, but it takes it longer to come back to equilibrium? Well, but can, so which one is connected to more atoms? I think I said the star one. Well, but it's not, right? They're both connected to three, to four other atoms. It depends, it matters what those atoms are. So, you know, if you, if you, if you point if I said that one took a longer time. We can talk about it later. Wait, so you're asking if you just, if you got, which one took a longer time, right? And not the explanation, did that get partial credit? Yes, it did. Okay, that's it. That's the exam. So, again, the grading is almost done. We'll get it done as fast as we can today and then send it to get scanned. And next time we're going to start talking about statistical mechanics.
know, the dissociation energy of beryllium dimer expressions sort of in quantum terms, plugging in homo to the lumo. And so, you know, again, orbital and lowest unoccupied molecular orbital. all the time in, you know, exchanges with the