WEBVTT 00:00:01.000 --> 00:00:04.000 The following content is provided by MIT OpenCourseWare 00:00:04.000 --> 00:00:06.000 under a Creative Commons license. 00:00:06.000 --> 00:00:10.000 Additional information about our license and MIT 00:00:10.000 --> 00:00:15.000 OpenCourseWare in general is available at ocw.mit.edu. 00:00:15.000 --> 00:00:19.000 We are back here for some more stimulating discussion of 00:00:19.000 --> 00:00:23.000 chemistry. And I want to begin today by 00:00:23.000 --> 00:00:27.000 just saying that I find it particularly ironic that on the 00:00:27.000 --> 00:00:33.000 day that I am all prepared to speak to you about the molecular 00:00:33.000 --> 00:00:38.000 orbital energy levels of polyatomic molecules for which I 00:00:38.000 --> 00:00:41.000 am using methane as an interesting example, 00:00:41.000 --> 00:00:47.000 the boards are screwed up. The reason for that apparently 00:00:47.000 --> 00:00:52.000 is that this screen came down and will not go back up. 00:00:52.000 --> 00:00:57.000 And one of the reasons that I have designed my presentation 00:00:57.000 --> 00:01:02.000 technique for you this semester so heavily around the use of the 00:01:02.000 --> 00:01:06.000 blackboard is that last year, when I taught the class, 00:01:06.000 --> 00:01:11.000 I had a lot of feedback from students about how the one day 00:01:11.000 --> 00:01:17.000 when we had audiovisual problems and I couldn't use my PowerPoint 00:01:17.000 --> 00:01:22.000 and instead gave an impromptu chalk talk, that that was their 00:01:22.000 --> 00:01:25.000 favorite lecture. Maybe next year we will 00:01:25.000 --> 00:01:31.000 entirely be back to PowerPoint. I don't know. 00:01:31.000 --> 00:01:33.000 We will see. But, fortunately, 00:01:33.000 --> 00:01:38.000 I did write out my notes this morning in electronic format, 00:01:38.000 --> 00:01:42.000 so I can make use of them on the side boards. 00:01:42.000 --> 00:01:47.000 And, in addition, I want to display for you some 00:01:47.000 --> 00:01:50.000 molecular orbital shapes and properties. 00:01:50.000 --> 00:01:55.000 And I will do that by starting out -- 00:01:55.000 --> 00:01:56.000 Actually, I would like to start, if I could, 00:01:56.000 --> 00:01:58.000 with this one. I will go back to the document 00:01:58.000 --> 00:02:00.000 camera in a moment. 00:02:10.000 --> 00:02:13.000 Starting out right where we left off last time, 00:02:13.000 --> 00:02:18.000 you will recall that we were looking at diatomic molecules. 00:02:18.000 --> 00:02:22.000 We started out with a seven orbital problem two lectures 00:02:22.000 --> 00:02:24.000 ago. We looked at the planar BH 00:02:24.000 --> 00:02:28.000 three molecule. We had a molecule where the MOs 00:02:28.000 --> 00:02:33.000 were mixing in two dimensions. And then last time, 00:02:33.000 --> 00:02:37.000 when we looked at homonuclear and heteronuclear diatomic 00:02:37.000 --> 00:02:41.000 molecules the MOs were, in fact, mixing in just along 00:02:41.000 --> 00:02:45.000 an axis in one dimension. And so, when we left the 00:02:45.000 --> 00:02:48.000 situation last time, we had arrived at some kind of 00:02:48.000 --> 00:02:53.000 an understanding of the energy levels in the carbon monoxide 00:02:53.000 --> 00:02:56.000 molecule. And these were being developed, 00:02:56.000 --> 00:03:00.000 especially, with reference to CO being a poisonous molecule, 00:03:00.000 --> 00:03:03.000 -- -- whereas, N two, 00:03:03.000 --> 00:03:07.000 another ten valence electron diatomic molecule was very, 00:03:07.000 --> 00:03:09.000 very inert. Why was that? 00:03:09.000 --> 00:03:13.000 It turns out that you have the very electronegative oxygen atom 00:03:13.000 --> 00:03:17.000 on one side of the diagram with its very low energetically lying 00:03:17.000 --> 00:03:21.000 orbitals mixing with the carbon orbitals that are higher in 00:03:21.000 --> 00:03:24.000 energy. And the neat thing about this, 00:03:24.000 --> 00:03:28.000 for the carbon monoxide molecule, is that it results in 00:03:28.000 --> 00:03:32.000 an asymmetry of the molecular orbitals -- 00:03:32.000 --> 00:03:35.000 -- because the coefficients are not the same on the carbon side 00:03:35.000 --> 00:03:39.000 as they are on the oxygen side as those atomic orbitals blend 00:03:39.000 --> 00:03:43.000 together into the four sigma and four pi molecular orbitals that 00:03:43.000 --> 00:03:46.000 we have. We have a total of eight MOs, 00:03:46.000 --> 00:03:49.000 and they distribute energetically as four energy 00:03:49.000 --> 00:03:52.000 levels in ascending energy that are sigma in symmetry. 00:03:52.000 --> 00:03:56.000 That is, cylindrically symmetric about the internuclear 00:03:56.000 --> 00:03:58.000 axis. And then we had four that were 00:03:58.000 --> 00:03:59.000 pi. Two energies, 00:03:59.000 --> 00:04:02.000 a pi bonding with four electrons and a pi antibonding 00:04:02.000 --> 00:04:06.000 with just virtual empty orbitals. 00:04:06.000 --> 00:04:09.000 And because this highest occupied molecular orbital of 00:04:09.000 --> 00:04:14.000 carbon monoxide is so heavily concentrated on the carbon end 00:04:14.000 --> 00:04:17.000 of the molecule, you can essentially view this 00:04:17.000 --> 00:04:21.000 as a carbon-based lone pair. It has this big lobe here, 00:04:21.000 --> 00:04:25.000 through which the carbon can serve as a nucleophile or as a 00:04:25.000 --> 00:04:30.000 base and bind to something that has an empty orbital. 00:04:30.000 --> 00:04:34.000 And if that something that has an empty orbital also has a 00:04:34.000 --> 00:04:39.000 filled orbital that matches the symmetry of the LUMO of carbon 00:04:39.000 --> 00:04:44.000 monoxide, then you get the possibility for multiple bonding 00:04:44.000 --> 00:04:49.000 between the carbon of the CO molecule and that other entity. 00:04:49.000 --> 00:04:53.000 And we will see that that other entity may, in fact, 00:04:53.000 --> 00:04:58.000 be a d-block transition element, such as iron. 00:04:58.000 --> 00:05:03.000 And what I wanted to do was to go ahead and show you that the 00:05:03.000 --> 00:05:06.000 LUMO is an interesting and unsymmetrical one, 00:05:06.000 --> 00:05:10.000 as well. Let's take a look at that. 00:05:18.000 --> 00:05:21.000 And so, in this picture I have gone ahead and included a little 00:05:21.000 --> 00:05:24.000 ball and stick diagram underneath the representation of 00:05:24.000 --> 00:05:26.000 the orbital lobes. Here is the carbon, 00:05:26.000 --> 00:05:31.000 and the oxygen is colored red. And you can see that the lobes 00:05:31.000 --> 00:05:35.000 on the LUMO, the lowest unoccupied molecular orbital in 00:05:35.000 --> 00:05:39.000 this system, are also large on carbon relative to oxygen. 00:05:39.000 --> 00:05:43.000 The same as we had found for the highest occupied MO. 00:05:43.000 --> 00:05:47.000 What this means is that that carbon atom in the CO molecule 00:05:47.000 --> 00:05:50.000 is simultaneously a base in the sigma system. 00:05:50.000 --> 00:05:55.000 And because this is our lowest unoccupied molecular orbital, 00:05:55.000 --> 00:06:00.000 our carbon atom is actually an acid in the pi system. 00:06:00.000 --> 00:06:02.000 And that makes it electronically very 00:06:02.000 --> 00:06:06.000 complimentary to d-block elements, as we will see when we 00:06:06.000 --> 00:06:10.000 get into the chemistry of elements that do have valence 00:06:10.000 --> 00:06:14.000 d-orbitals in addition to s and p, coming up in a couple of 00:06:14.000 --> 00:06:17.000 lectures. And so what can happen is that, 00:06:17.000 --> 00:06:21.000 let's say that a carbon monoxide molecule somehow 00:06:21.000 --> 00:06:24.000 erroneously finds its way into your blood stream, 00:06:24.000 --> 00:06:29.000 you have there hemoglobin, the iron atom at the center of 00:06:29.000 --> 00:06:34.000 which acts reversibly to bind the dioxygen molecule. 00:06:34.000 --> 00:06:37.000 But it irreversibly binds to the carbon monoxide molecule 00:06:37.000 --> 00:06:41.000 because the iron center, which is sitting at the center 00:06:41.000 --> 00:06:45.000 of a heme unit that I mentioned, and we will talk about the 00:06:45.000 --> 00:06:49.000 structure of hemoglobin and the way that it functions a little 00:06:49.000 --> 00:06:51.000 bit later. But if CO gets in there, 00:06:51.000 --> 00:06:55.000 it binds to the iron through the HOMO, acting as a base to 00:06:55.000 --> 00:06:58.000 the metal acting as a sigma acid, and also through 00:06:58.000 --> 00:07:02.000 d-electrons being able to act as a pi base from the metal binding 00:07:02.000 --> 00:07:07.000 with this LUMO of the carbon monoxide molecule. 00:07:07.000 --> 00:07:09.000 And this LUMO is doubly degenerate. 00:07:09.000 --> 00:07:12.000 That means there are two of them at the same energy. 00:07:12.000 --> 00:07:16.000 And we will come back to this issue of orbital degeneracy 00:07:16.000 --> 00:07:19.000 later in today's lecture, but that electronic asymmetry 00:07:19.000 --> 00:07:24.000 of carbon monoxide and the fact that the HOMO and LUMO are both 00:07:24.000 --> 00:07:27.000 centered on the carbon end of the molecule are what make CO 00:07:27.000 --> 00:07:31.000 such an interesting molecule and such a reactive molecule and 00:07:31.000 --> 00:07:36.000 therefore a poison. And so now let us please go 00:07:36.000 --> 00:07:39.000 back to the document camera. 00:07:46.000 --> 00:07:49.000 What I have shown here are the five platonic solids. 00:07:49.000 --> 00:07:53.000 Actually, I spent a lot of time over the weekend practicing 00:07:53.000 --> 00:07:56.000 drawing these on the chalkboard so that I would be able to do 00:07:56.000 --> 00:08:00.000 that very nicely for you. And here it comes that we do 00:08:00.000 --> 00:08:05.000 not have that option today, so I will show them to you this 00:08:05.000 --> 00:08:07.000 way. And the platonic solids are the 00:08:07.000 --> 00:08:11.000 five regular polyhedra. And that means that they are 00:08:11.000 --> 00:08:13.000 polyhedra that have, as their faces, 00:08:13.000 --> 00:08:16.000 regular polygons that are all identical. 00:08:16.000 --> 00:08:20.000 And the reason I am bringing this up today is that we had 00:08:20.000 --> 00:08:24.000 talked about issues of symmetry with reference to molecular 00:08:24.000 --> 00:08:27.000 orbital theory. And these are all very highly 00:08:27.000 --> 00:08:29.000 symmetric shapes, as you can immediately 00:08:29.000 --> 00:08:34.000 recognize. The buckyball that we looked at 00:08:34.000 --> 00:08:38.000 earlier is not here because it has both five-membered rings and 00:08:38.000 --> 00:08:42.000 six-membered rings on the faces. And so it doesn't have all the 00:08:42.000 --> 00:08:46.000 faces identical. And those that do have all the 00:08:46.000 --> 00:08:49.000 faces identical are the platonic solids. 00:08:49.000 --> 00:08:52.000 And the complete list of those is shown here. 00:08:52.000 --> 00:08:55.000 With triangles, you can put four equilateral 00:08:55.000 --> 00:08:59.000 triangles together to make a tetrahedron. 00:08:59.000 --> 00:09:02.000 There is actually a journal called Tetrahedron. 00:09:02.000 --> 00:09:05.000 And there is another one called Tetrahedron Letters. 00:09:05.000 --> 00:09:08.000 And there is another one called Tetrahedron Reports. 00:09:08.000 --> 00:09:12.000 And the reason is that organic chemistry is carbon-based, 00:09:12.000 --> 00:09:16.000 and carbon so often gets involved in forming tetrahedra. 00:09:16.000 --> 00:09:19.000 And so the tetrahedron is a very important symbol of organic 00:09:19.000 --> 00:09:22.000 chemistry. And we will get back to that. 00:09:22.000 --> 00:09:25.000 And with triangles, if instead of putting four of 00:09:25.000 --> 00:09:28.000 them together, you put eight of them together 00:09:28.000 --> 00:09:32.000 this way, you get the octahedron. 00:09:32.000 --> 00:09:36.000 And the octahedron is equally important as a symbol for 00:09:36.000 --> 00:09:39.000 inorganic chemistry. And the reason for that is that 00:09:39.000 --> 00:09:44.000 when we have metal ions in solution, they often sit at the 00:09:44.000 --> 00:09:48.000 center of a coordination octahedron in which six ligands 00:09:48.000 --> 00:09:52.000 bind to that metal center. And we are going to get into 00:09:52.000 --> 00:09:56.000 how that was discovered pretty shortly by Alfred Werner, 00:09:56.000 --> 00:10:00.000 another Nobel laureate in chemistry. 00:10:00.000 --> 00:10:05.000 And with triangles you can also make the icosahedron shown here, 00:10:05.000 --> 00:10:08.000 and this has 20 faces. And the icosahedron is 00:10:08.000 --> 00:10:12.000 important in chemistry. And certain elements, 00:10:12.000 --> 00:10:16.000 for example boron, actually have structures in the 00:10:16.000 --> 00:10:20.000 solid state in which the boron atoms are arranged at the 00:10:20.000 --> 00:10:23.000 vertices of a regular icosahedron. 00:10:23.000 --> 00:10:28.000 So this is a chemically important shape as well. 00:10:28.000 --> 00:10:31.000 And if you go from triangles to squares you can, 00:10:31.000 --> 00:10:35.000 of course, make a cube. And probably the most familiar 00:10:35.000 --> 00:10:39.000 of a platonic solid to us. And then, if we go up to 00:10:39.000 --> 00:10:43.000 pentagons and put those together, as you can see here, 00:10:43.000 --> 00:10:47.000 we can put 12 pentagons together in a regular way, 00:10:47.000 --> 00:10:52.000 that gives rise to what we call the pentagonal dodecahedron. 00:10:52.000 --> 00:10:57.000 And you can imagine that each of these vertices could be a C-H 00:10:57.000 --> 00:11:02.000 unit, carbon-hydrogen. And then each carbon would have 00:11:02.000 --> 00:11:05.000 its valence of four satisfied. And there would be all single 00:11:05.000 --> 00:11:08.000 bonds at the parameter of this round molecule. 00:11:08.000 --> 00:11:11.000 And, in fact, that dodecahedron molecule was 00:11:11.000 --> 00:11:15.000 the objective of organic synthesis for quite a long time 00:11:15.000 --> 00:11:18.000 before it was finally successfully synthesized by Leo 00:11:18.000 --> 00:11:21.000 Paquette. You would be surprised with the 00:11:21.000 --> 00:11:24.000 kinds of challenges, maybe, sometimes that do drive 00:11:24.000 --> 00:11:28.000 chemists to do what they do. And that is the platonic 00:11:28.000 --> 00:11:32.000 solids. I am going to be focusing our 00:11:32.000 --> 00:11:36.000 attention on methane. And, if I could have the side 00:11:36.000 --> 00:11:39.000 board go back to the computer for a moment, 00:11:39.000 --> 00:11:43.000 that would be great. These are the notes for today's 00:11:43.000 --> 00:11:48.000 lecture that I wrote up this morning that you will be able to 00:11:48.000 --> 00:11:51.000 download later today, or maybe you already can. 00:11:51.000 --> 00:11:53.000 I don't know. In any event, 00:11:53.000 --> 00:11:57.000 when we approach the methane problem in molecular orbital 00:11:57.000 --> 00:12:02.000 theory, what we do is recognize that the tetrahedron can be 00:12:02.000 --> 00:12:06.000 conveniently inscribed into a cube by placing hydrogens at 00:12:06.000 --> 00:12:12.000 alternating corners of the cube, as shown here. 00:12:12.000 --> 00:12:16.000 And then, when we do that, the carbon atom of the methane 00:12:16.000 --> 00:12:19.000 molecule lies at the very center of this cube. 00:12:19.000 --> 00:12:24.000 And this way of arranging the system in space will help us to 00:12:24.000 --> 00:12:28.000 simplify the problem of the molecular orbital energy levels 00:12:28.000 --> 00:12:32.000 for the methane molecule. And we are going to be 00:12:32.000 --> 00:12:35.000 ultimately working toward a comparison of the molecular 00:12:35.000 --> 00:12:39.000 orbital description of the methane molecule with that 00:12:39.000 --> 00:12:43.000 afforded by valance bond theory. Now, some of you approached me 00:12:43.000 --> 00:12:46.000 after lecture last time and said, what happened to 00:12:46.000 --> 00:12:49.000 hybridization? And, if you wanted to talk 00:12:49.000 --> 00:12:52.000 about the methane molecule and the language of valence bond 00:12:52.000 --> 00:12:57.000 theory, you would say that this carbon atom is sp three 00:12:57.000 --> 00:13:00.000 hybridized in order that it may form simultaneously four two 00:13:00.000 --> 00:13:04.000 electron bonds between the central carbon and the four 00:13:04.000 --> 00:13:09.000 peripheral hydrogens. Well, where hybridization went 00:13:09.000 --> 00:13:12.000 is that it stayed with valence bond theory. 00:13:12.000 --> 00:13:15.000 Let me reemphasize the distinction between the 00:13:15.000 --> 00:13:20.000 molecular orbital treatment of molecule electronic structure as 00:13:20.000 --> 00:13:24.000 contrasted with the valence bond theory in which hybridization 00:13:24.000 --> 00:13:27.000 plays a role. It does not play a role in the 00:13:27.000 --> 00:13:31.000 language of MO theory, so please keep those quite 00:13:31.000 --> 00:13:35.000 distinct. Now, we need to choose a 00:13:35.000 --> 00:13:38.000 coordinate system for our problem. 00:13:38.000 --> 00:13:43.000 And I won't spend too much time discussing how we choose the 00:13:43.000 --> 00:13:47.000 coordinate system, but what we have done in the 00:13:47.000 --> 00:13:51.000 present case with methane is to choose the x, 00:13:51.000 --> 00:13:57.000 y, and z axes each to come out of the faces of the cube. 00:13:57.000 --> 00:14:02.000 And we do this because that choice of coordinate system will 00:14:02.000 --> 00:14:06.000 lead to the equivalence of the carbons' 2px, 00:14:06.000 --> 00:14:09.000 2py, and 2pz orbitals in 3D space. 00:14:09.000 --> 00:14:13.000 We could choose a different coordinate system, 00:14:13.000 --> 00:14:17.000 and that would make the problem more difficult. 00:14:17.000 --> 00:14:22.000 So, what we have done is chosen a convenient set of coordinates 00:14:22.000 --> 00:14:27.000 where the x, y, and z axes each come out of the 00:14:27.000 --> 00:14:31.000 faces of our cube. Now, we recognize that the 00:14:31.000 --> 00:14:34.000 methane problem, like the homonuclear and 00:14:34.000 --> 00:14:38.000 heteronuclear diatomic molecule problems we talked about last 00:14:38.000 --> 00:14:40.000 time, is an eight orbital problem. 00:14:40.000 --> 00:14:44.000 The diatomics N two and CO each had ten electrons to go 00:14:44.000 --> 00:14:47.000 into their eight molecular orbitals. 00:14:47.000 --> 00:14:50.000 And here, with methane, we have eight electrons, 00:14:50.000 --> 00:14:54.000 four coming in as the valence orbitals from the carbon atom, 00:14:54.000 --> 00:14:57.000 valence electrons from the carbon atom, rather, 00:14:57.000 --> 00:15:01.000 and four coming in as the four valence 1s electrons for the 00:15:01.000 --> 00:15:06.000 four hydrogens. And our eight orbitals are the 00:15:06.000 --> 00:15:08.000 carbons 2s, 2px, 2py, and 2pz orbitals. 00:15:08.000 --> 00:15:12.000 And we have four more orbitals, the 1s from each of the four 00:15:12.000 --> 00:15:15.000 hydrogens. We know how many electrons are 00:15:15.000 --> 00:15:18.000 going to be able to go into our MO energy level diagram. 00:15:18.000 --> 00:15:22.000 We know how many MOs we are going to have because we know 00:15:22.000 --> 00:15:26.000 how many valance AOs we have. And those numbers are equal at 00:15:26.000 --> 00:15:30.000 eight. The choice of this problem is 00:15:30.000 --> 00:15:35.000 predicated on my desire to show you how things progress as we go 00:15:35.000 --> 00:15:40.000 from a planar system to a linear system to now a system that has 00:15:40.000 --> 00:15:45.000 molecular orbitals that stretch out in three-dimensional space. 00:15:45.000 --> 00:15:49.000 Now, we are going to take the following approach. 00:15:49.000 --> 00:15:52.000 We are going to say, how can we generate linear 00:15:52.000 --> 00:15:56.000 combinations of the four hydrogen 1s orbitals that will 00:15:56.000 --> 00:16:01.000 be so constructed as to match the nodal properties of the 00:16:01.000 --> 00:16:06.000 carbon atomic orbitals that are the valence orbitals of that 00:16:06.000 --> 00:16:08.000 carbon? Okay. 00:16:08.000 --> 00:16:12.000 So that is our goal, and that is our strategy. 00:16:12.000 --> 00:16:17.000 And let me say also, that in this tetrahedral 00:16:17.000 --> 00:16:23.000 symmetry, you can recognize that you cannot distinguish any one 00:16:23.000 --> 00:16:28.000 of the four vertices of a tetrahedron from the other 00:16:28.000 --> 00:16:31.000 three. And that is important. 00:16:31.000 --> 00:16:36.000 And that is also true from the way that we have decided to 00:16:36.000 --> 00:16:40.000 situate the hydrogen nuclei on this cube relative to our 00:16:40.000 --> 00:16:43.000 coordinate system. In other words, 00:16:43.000 --> 00:16:47.000 none of these four hydrogens lie on any of the three 00:16:47.000 --> 00:16:51.000 Cartesian axes. And their relationship to the 00:16:51.000 --> 00:16:55.000 coordinate system and to the shape of the cube each is 00:16:55.000 --> 00:17:01.000 indistinguishable from that of the other three hydrogens. 00:17:01.000 --> 00:17:04.000 And that is an important aspect of the way that we set up the 00:17:04.000 --> 00:17:07.000 problem. And why do I mention that? 00:17:07.000 --> 00:17:10.000 Because I mentioned that when constructing MOs in general, 00:17:10.000 --> 00:17:15.000 one of the first things you can do, if you are going to be able 00:17:15.000 --> 00:17:19.000 to take advantage of symmetry to help you solve the problem of 00:17:19.000 --> 00:17:23.000 these MO energy levels and how they are constructed from linear 00:17:23.000 --> 00:17:27.000 combinations of atomic orbitals is to identify symmetry-related 00:17:27.000 --> 00:17:32.000 sets of atoms in orbitals. If you are keeping that in 00:17:32.000 --> 00:17:37.000 mind, you will realize that we are treating all four hydrogens 00:17:37.000 --> 00:17:41.000 together because they are indistinguishable from one 00:17:41.000 --> 00:17:46.000 another in the tetrahedral shape of the methane molecule. 00:17:46.000 --> 00:17:50.000 So, that is important. If we had a more complicated 00:17:50.000 --> 00:17:55.000 molecule, such as if we were looking at -- 00:17:55.000 --> 00:17:58.000 I get to use a little bit of board space today. 00:17:58.000 --> 00:18:02.000 Here is a slightly more complicated hydrocarbon 00:18:02.000 --> 00:18:04.000 molecule. Here is a slightly more 00:18:04.000 --> 00:18:07.000 complicated hydrocarbon molecule. 00:18:07.000 --> 00:18:11.000 And in a molecule like this, this is cyclohexane drawn in a 00:18:11.000 --> 00:18:15.000 chair confirmation. Sir Derek Barton actually was 00:18:15.000 --> 00:18:19.000 the Nobel laureate who got his prize for discussing 00:18:19.000 --> 00:18:22.000 conformational aspects of hydrocarbons. 00:18:22.000 --> 00:18:26.000 We would find that the hydrogens split up into two 00:18:26.000 --> 00:18:30.000 sets. They are not all equivalent in 00:18:30.000 --> 00:18:34.000 the cyclohexane molecule in this frozen out chair confirmation 00:18:34.000 --> 00:18:38.000 because we have one type that is equatorial and one type that is 00:18:38.000 --> 00:18:42.000 axial on each of these carbons as we go around this 00:18:42.000 --> 00:18:45.000 six-membered ring. And so there are different sets 00:18:45.000 --> 00:18:48.000 that you would have to treat separately. 00:18:48.000 --> 00:18:50.000 But methane is more symmetrical. 00:18:50.000 --> 00:18:54.000 And all four hydrogens in the methane molecule are identical, 00:18:54.000 --> 00:18:58.000 so we are going to treat them together in developing our 00:18:58.000 --> 00:19:02.000 linear combinations. And, coming back to this, 00:19:02.000 --> 00:19:06.000 what we are going to do, once again, to generate these 00:19:06.000 --> 00:19:10.000 things, is to imagine the central atom atomic orbital as 00:19:10.000 --> 00:19:13.000 growing out in space out to where the hydrogen nuclei are. 00:19:13.000 --> 00:19:17.000 And then we are going to take the sign of the wave function 00:19:17.000 --> 00:19:21.000 that we find out there and apply it to those hydrogens that are 00:19:21.000 --> 00:19:26.000 out there in space in order to see what our linear combination 00:19:26.000 --> 00:19:29.000 must look like. If you imagine the carbon 2s 00:19:29.000 --> 00:19:32.000 orbital, which has the same sign, positive, 00:19:32.000 --> 00:19:35.000 everywhere, and imagine it growing out to where the 00:19:35.000 --> 00:19:38.000 hydrogens are, and we are labeling the 00:19:38.000 --> 00:19:40.000 hydrogen A, B, C, and D, as shown here. 00:19:40.000 --> 00:19:44.000 And we will keep that same labeling scheme throughout the 00:19:44.000 --> 00:19:47.000 development of this problem. You would see that the carbon 00:19:47.000 --> 00:19:51.000 2s positive wave function, as we imagine it growing out to 00:19:51.000 --> 00:19:54.000 where the hydrogens are, would touch each of these 00:19:54.000 --> 00:19:58.000 hydrogens at the same time. And, thus, we confer the sign 00:19:58.000 --> 00:20:02.000 positive on all four of the contributors to this first of 00:20:02.000 --> 00:20:07.000 our linear combinations. And what that will represent is 00:20:07.000 --> 00:20:12.000 simultaneous bonding of the carbon 2s orbital with all four 00:20:12.000 --> 00:20:15.000 hydrogens and with the same sign everywhere. 00:20:15.000 --> 00:20:19.000 And so, we can write down one-half A plus B plus C plus D, 00:20:19.000 --> 00:20:23.000 where I am using A here to 00:20:23.000 --> 00:20:28.000 represent the hydrogen 1s orbital associated with hydrogen 00:20:28.000 --> 00:20:32.000 labeled A. So I have just abbreviated that 00:20:32.000 --> 00:20:33.000 in A. And, similarly, 00:20:33.000 --> 00:20:37.000 B is the hydrogen 1s orbital associated with hydrogen B and 00:20:37.000 --> 00:20:39.000 so on. So, this is a linear 00:20:39.000 --> 00:20:42.000 combination of these four hydrogen orbitals. 00:20:42.000 --> 00:20:44.000 And it is normalized with a half, here. 00:20:44.000 --> 00:20:48.000 And we have the same coefficient on each of the four 00:20:48.000 --> 00:20:52.000 hydrogens, so the size of the lobe that you would draw at each 00:20:52.000 --> 00:20:55.000 four of these positions would be the same. 00:20:55.000 --> 00:20:59.000 That is number one. We have built a linear 00:20:59.000 --> 00:21:03.000 combination based on the nodal properties of the carbon 2s 00:21:03.000 --> 00:21:06.000 atomic orbital. And now, let's do two more. 00:21:06.000 --> 00:21:11.000 And we are going to base these on the nodal properties of the 00:21:11.000 --> 00:21:15.000 central carbon 2px orbital and the central carbon 2py orbital. 00:21:15.000 --> 00:21:19.000 I am using the same coordinate system as I started with, 00:21:19.000 --> 00:21:24.000 namely, that the positive x-axis comes out of the face of 00:21:24.000 --> 00:21:28.000 the cube and comes toward us, and the positive y-axis comes 00:21:28.000 --> 00:21:33.000 out of this face of the cube and goes that way. 00:21:33.000 --> 00:21:35.000 And we will look at z in a moment. 00:21:35.000 --> 00:21:40.000 That one is going straight up out of the top face of the cube 00:21:40.000 --> 00:21:44.000 for positive z. What you can see is that if you 00:21:44.000 --> 00:21:48.000 imagine the carbon 2px atomic orbital as just growing out to 00:21:48.000 --> 00:21:53.000 where the four hydrogens are, we will see that the hydrogens 00:21:53.000 --> 00:21:57.000 in positions A and B will experience the positive lobe of 00:21:57.000 --> 00:22:02.000 the carbon 2px orbital at the same moment as the two hydrogens 00:22:02.000 --> 00:22:06.000 in back labeled C and D experience the negative lobe of 00:22:06.000 --> 00:22:12.000 the carbon 2px orbital. And so, we give positive phase 00:22:12.000 --> 00:22:16.000 to A and B and negative phase to C and D. 00:22:16.000 --> 00:22:20.000 And we give them the same coefficient everywhere such that 00:22:20.000 --> 00:22:25.000 this linear combination is normalized as one-half A plus B 00:22:25.000 --> 00:22:30.000 minus C minus D. 00:22:30.000 --> 00:22:33.000 That is our second linear combination of hydrogen 00:22:33.000 --> 00:22:36.000 orbitals. And this one has been so 00:22:36.000 --> 00:22:40.000 generated as to match the nodal property of the carbon 2px 00:22:40.000 --> 00:22:44.000 orbital, the key feature of which is that it is everywhere 00:22:44.000 --> 00:22:48.000 positive along positive x, and everywhere along negative x 00:22:48.000 --> 00:22:51.000 it is negative. And the 2py orbital of carbon 00:22:51.000 --> 00:22:56.000 has the property that everywhere in the plus y region of space, 00:22:56.000 --> 00:22:59.000 it is positive, so that C and B will be 00:22:59.000 --> 00:23:05.000 associated with a positive sign for this linear combination. 00:23:05.000 --> 00:23:09.000 And the 2py orbital is negative everywhere along negative y, 00:23:09.000 --> 00:23:12.000 back here. And that is where hydrogens A 00:23:12.000 --> 00:23:15.000 and D are located. So, A and D will carry a 00:23:15.000 --> 00:23:18.000 negative sign for this new linear combination, 00:23:18.000 --> 00:23:22.000 which is one-half of minus A plus B plus C minus D. 00:23:22.000 --> 00:23:26.000 This is just one way of writing down 00:23:26.000 --> 00:23:32.000 the picture that we see here. And there is only one left to 00:23:32.000 --> 00:23:37.000 be generated from the nodal properties of the carbon 2pz 00:23:37.000 --> 00:23:42.000 orbital, which has its positive node oriented along positive z 00:23:42.000 --> 00:23:46.000 and its negative lobe oriented along negative z, 00:23:46.000 --> 00:23:50.000 and it has the x,y-plane as a nodal surface. 00:23:50.000 --> 00:23:54.000 And so, therefore, hydrogens A and C will carry a 00:23:54.000 --> 00:24:00.000 positive sign because they are oriented and located in the plus 00:24:00.000 --> 00:24:05.000 z region of space. Whereas, hydrogens B and D are 00:24:05.000 --> 00:24:09.000 located in the negative z region of space, so they will carry a 00:24:09.000 --> 00:24:13.000 negative sign. And that one can be normalized 00:24:13.000 --> 00:24:16.000 also with a normalization coefficient of one-half. 00:24:16.000 --> 00:24:19.000 And so it is A minus B plus C minus D. 00:24:19.000 --> 00:24:23.000 And just don't forget that these letters 00:24:23.000 --> 00:24:27.000 refer to the 1s orbitals associated with the hydrogens 00:24:27.000 --> 00:24:32.000 that are so labeled. And we started with identifying 00:24:32.000 --> 00:24:37.000 the fact that the four hydrogens in this problem are all 00:24:37.000 --> 00:24:40.000 equivalent. And, thus, their 1s orbitals 00:24:40.000 --> 00:24:44.000 are all equivalent, indistinguishable from each 00:24:44.000 --> 00:24:47.000 other with respect to their spatial orientation. 00:24:47.000 --> 00:24:51.000 And, so we took four atomic orbitals, and we are 00:24:51.000 --> 00:24:56.000 constructing from them four linear combinations of those 00:24:56.000 --> 00:25:01.000 four atomic orbitals. I want to emphasize what I have 00:25:01.000 --> 00:25:04.000 written here, which is that with the 00:25:04.000 --> 00:25:08.000 coordinate system we have chosen and in this high tetrahedral 00:25:08.000 --> 00:25:13.000 symmetry, the carbons 2p x, y and z orbitals are identical. 00:25:13.000 --> 00:25:17.000 And, just like in a carbon atom floating free in space, 00:25:17.000 --> 00:25:21.000 they are degenerate. That means they have the same 00:25:21.000 --> 00:25:24.000 energy. And so we will talk about 00:25:24.000 --> 00:25:29.000 orbitals in atoms being degenerate in a moment. 00:25:29.000 --> 00:25:33.000 But we find that the 2p x, y and z orbitals on that carbon 00:25:33.000 --> 00:25:37.000 are going to have the same energy, and they are degenerate. 00:25:37.000 --> 00:25:41.000 And the molecular orbitals constructed from them in this 00:25:41.000 --> 00:25:44.000 tetrahedral system will have the same property. 00:25:44.000 --> 00:25:47.000 They will be degenerate. So, something that is 00:25:47.000 --> 00:25:51.000 occasionally associated with high symmetry is multiple MO 00:25:51.000 --> 00:25:55.000 degeneracy. Next, knowing what our carbon 00:25:55.000 --> 00:25:59.000 atom atomic orbitals are and knowing also what our four 00:25:59.000 --> 00:26:03.000 linear combinations are of the hydrogen 1s orbitals, 00:26:03.000 --> 00:26:07.000 we can now allow them to interact in ways that are either 00:26:07.000 --> 00:26:11.000 bonding or antibonding. And they interact, 00:26:11.000 --> 00:26:14.000 of course, according to their nodal properties. 00:26:14.000 --> 00:26:19.000 Those linear combinations that have the same nodal properties 00:26:19.000 --> 00:26:23.000 as a particular carbon atom atomic orbital will lead to a 00:26:23.000 --> 00:26:29.000 bonding interaction and an antibonding interaction. 00:26:29.000 --> 00:26:33.000 And we have an eight orbital problem, so we better see eight 00:26:33.000 --> 00:26:37.000 energy levels appearing in our molecular orbital energy level 00:26:37.000 --> 00:26:39.000 diagram for this methane molecule. 00:26:39.000 --> 00:26:43.000 And, when we go down here, starting at the lowest energy 00:26:43.000 --> 00:26:46.000 molecular orbital, we find that it can be 00:26:46.000 --> 00:26:49.000 constructed by adding to the carbon 2s orbital, 00:26:49.000 --> 00:26:53.000 which is, once again, spherically symmetric, 00:26:53.000 --> 00:26:57.000 this linear combination that is A plus B plus C plus D. 00:26:57.000 --> 00:27:00.000 And we can draw this with a 00:27:00.000 --> 00:27:03.000 surface like this, which envelopes all five of our 00:27:03.000 --> 00:27:08.000 nuclei which has the same sign everywhere. 00:27:08.000 --> 00:27:10.000 And so our lowest lying molecular orbital, 00:27:10.000 --> 00:27:14.000 our most bonding molecular orbital, our most stabilized MO, 00:27:14.000 --> 00:27:17.000 which can house a pair of electrons because it is singly 00:27:17.000 --> 00:27:21.000 degenerate, is one that will look something like this. 00:27:21.000 --> 00:27:23.000 It envelopes all five of our nuclei. 00:27:23.000 --> 00:27:26.000 And, effectively, the pair of electrons is 00:27:26.000 --> 00:27:30.000 associated simultaneously with all five nuclei. 00:27:30.000 --> 00:27:33.000 And the wave function has the same sign everywhere, 00:27:33.000 --> 00:27:38.000 just like the carbon 2s orbital from which this is partially 00:27:38.000 --> 00:27:41.000 constructed does. And then, next we have three 00:27:41.000 --> 00:27:45.000 molecular orbitals that are each built by forming bonding 00:27:45.000 --> 00:27:48.000 interactions with the carbon's p-orbitals, px, 00:27:48.000 --> 00:27:53.000 py, and pz with the linear combination that matches that 00:27:53.000 --> 00:27:56.000 carbon orbital in terms of the nodal properties. 00:27:56.000 --> 00:28:00.000 And we built them that way so that becomes particularly 00:28:00.000 --> 00:28:04.000 straightforward. This one here is coming out, 00:28:04.000 --> 00:28:07.000 the face that points to the right. 00:28:07.000 --> 00:28:11.000 And so you can see that this one is based on the carbon 2py 00:28:11.000 --> 00:28:13.000 orbital. And, if that carbon 2py orbital 00:28:13.000 --> 00:28:18.000 makes a bonding interaction with the linear combination denoted 00:28:18.000 --> 00:28:21.000 as minus A plus B plus C minus D, 00:28:21.000 --> 00:28:25.000 the bonding interaction here is signified by this plus sign. 00:28:25.000 --> 00:28:30.000 So, we have a plus sign here and a plus sign here. 00:28:30.000 --> 00:28:34.000 And, thus, we have a molecular orbital that has the same nodal 00:28:34.000 --> 00:28:37.000 properties as a carbon 2py orbital does, 00:28:37.000 --> 00:28:41.000 which means that the x,z-plane is a nodal surface for this 00:28:41.000 --> 00:28:45.000 molecular orbital. And what it looks like is kind 00:28:45.000 --> 00:28:49.000 of a two-bladed propeller. You have a plus phase that is 00:28:49.000 --> 00:28:54.000 distributed, enveloping the two hydrogens that we labeled B and 00:28:54.000 --> 00:29:00.000 C, along with that positive lobe of the carbon 2py orbital. 00:29:00.000 --> 00:29:04.000 So that you have that forming one blade of our two-bladed 00:29:04.000 --> 00:29:08.000 propeller. And the blade with the other 00:29:08.000 --> 00:29:12.000 sign is rotated relative to the first by 45 degrees. 00:29:12.000 --> 00:29:16.000 Sorry, by 90 degrees, in fact, and will involve the 00:29:16.000 --> 00:29:21.000 distribution of bonding character over carbon's negative 00:29:21.000 --> 00:29:26.000 2py lobe interacting with these two hydrogens back here, 00:29:26.000 --> 00:29:32.000 which are hydrogens A and D. And so we will visualize these 00:29:32.000 --> 00:29:34.000 in a moment to make that more clear. 00:29:34.000 --> 00:29:37.000 But these three bonds, one oriented along x, 00:29:37.000 --> 00:29:41.000 one oriented along y, and one oriented along z all 00:29:41.000 --> 00:29:44.000 look exactly like this. They are just rotated 90 00:29:44.000 --> 00:29:47.000 degrees to each other in space, just like the x, 00:29:47.000 --> 00:29:51.000 y, and z Cartesian axes are. I have only drawn one of them, 00:29:51.000 --> 00:29:55.000 but we have three of them because there is a three-fold 00:29:55.000 --> 00:30:00.000 orbital degeneracy in this molecular orbital. 00:30:00.000 --> 00:30:03.000 So, singly degenerate and then triply degenerate. 00:30:03.000 --> 00:30:07.000 And then, as we go up in energy, we will find that our 00:30:07.000 --> 00:30:10.000 lowest unoccupied molecular orbital is constructed by 00:30:10.000 --> 00:30:13.000 subtracting. It is the same as I have 00:30:13.000 --> 00:30:15.000 written here, but with a minus sign, 00:30:15.000 --> 00:30:19.000 the C 2s subtracting A plus B plus C plus D. 00:30:19.000 --> 00:30:21.000 A, B, C and D hydrogens all 00:30:21.000 --> 00:30:25.000 have the same sign to their wave function as each other, 00:30:25.000 --> 00:30:29.000 but they have the opposite sign as the contribution from the 00:30:29.000 --> 00:30:34.000 carbon 2s orbital to this molecular orbital. 00:30:34.000 --> 00:30:37.000 And what that results in, as I have tried to indicate 00:30:37.000 --> 00:30:41.000 with some red dashes here, is that you generate a nodal 00:30:41.000 --> 00:30:46.000 surface in between the carbon and hydrogen nuclei that will go 00:30:46.000 --> 00:30:50.000 completely around and will intersect perpendicular to each 00:30:50.000 --> 00:30:54.000 of these carbon-hydrogen inter-nuclear vectors. 00:30:54.000 --> 00:30:58.000 You have a change of sign as you are traversing the path from 00:30:58.000 --> 00:31:03.000 carbon to any one of the four hydrogen nuclei. 00:31:03.000 --> 00:31:06.000 And that is what we get by just reversing the sign relative to 00:31:06.000 --> 00:31:09.000 the bonding counterpart of that orbital. 00:31:09.000 --> 00:31:11.000 And that makes this one anti-bonding. 00:31:11.000 --> 00:31:15.000 We will give it an asterisk here to denote antibonding 00:31:15.000 --> 00:31:17.000 character to this molecular orbital. 00:31:17.000 --> 00:31:21.000 And now, ascending in energy, we will find that for each one 00:31:21.000 --> 00:31:25.000 of these orbitals that is bonded and oriented with respect to x, 00:31:25.000 --> 00:31:28.000 y or z in Cartesian space, we find that there is an 00:31:28.000 --> 00:31:32.000 antibonding counterpart. Again, what we are doing is 00:31:32.000 --> 00:31:36.000 taking, for example, carbons 2py and subtracting the 00:31:36.000 --> 00:31:40.000 linear combination that corresponds to carbon 2py in 00:31:40.000 --> 00:31:42.000 terms of its nodal symmetry properties. 00:31:42.000 --> 00:31:47.000 We are subtracting minus A plus B plus C minus D. 00:31:47.000 --> 00:31:49.000 The carbon 2py has the same 00:31:49.000 --> 00:31:53.000 plus, its positive lobe is oriented along positive y, 00:31:53.000 --> 00:31:57.000 just as it is down here, but now we have reversed the 00:31:57.000 --> 00:32:02.000 sign of the four hydrogens. And what this introduces will 00:32:02.000 --> 00:32:07.000 be antibonding nodal surfaces in between the carbon and the 00:32:07.000 --> 00:32:11.000 hydrogens. And they are strong antibonding 00:32:11.000 --> 00:32:16.000 interactions because of the good directional overlap of, 00:32:16.000 --> 00:32:20.000 say, the positive lobe of carbon's 2py with these 00:32:20.000 --> 00:32:26.000 s-orbital contributions in the negative from hydrogens C and B, 00:32:26.000 --> 00:32:28.000 here. And back here, 00:32:28.000 --> 00:32:33.000 A and D rotated, of course, by 90 degrees. 00:32:33.000 --> 00:32:37.000 One of the points that I was illustrating last time and 00:32:37.000 --> 00:32:41.000 really trying to emphasize in the context of MO theory is 00:32:41.000 --> 00:32:45.000 that, in general, we expect orbitals to be higher 00:32:45.000 --> 00:32:48.000 in energy if they have more internuclear nodes. 00:32:48.000 --> 00:32:53.000 And the energy of the orbitals is also affected by which atomic 00:32:53.000 --> 00:32:58.000 orbitals are close to the MO in energy and also by factors of 00:32:58.000 --> 00:33:01.000 overlap. And normally, 00:33:01.000 --> 00:33:04.000 for example, a sigma bond is typically 00:33:04.000 --> 00:33:09.000 associated with greater overlap than a pi bond because of the 00:33:09.000 --> 00:33:13.000 directionality of the overlap. And here, you have an 00:33:13.000 --> 00:33:17.000 interesting case, where this p-orbital is not 00:33:17.000 --> 00:33:23.000 directed right at the hydrogen. And it is not a pi bond either. 00:33:23.000 --> 00:33:27.000 It is something in between sort of sigma and a pi type of 00:33:27.000 --> 00:33:32.000 orientation. It is an oblique interaction of 00:33:32.000 --> 00:33:36.000 the hydrogen 1s wavefunction with that lobe of the p-orbital. 00:33:36.000 --> 00:33:40.000 But, still, it gives rise to very good overlap down here, 00:33:40.000 --> 00:33:44.000 and consequently a very strong antibonding character up here in 00:33:44.000 --> 00:33:47.000 counterpart. And, from calculation, 00:33:47.000 --> 00:33:51.000 this is the order of the energy levels that come out of that. 00:33:51.000 --> 00:33:55.000 And so we have only four energy levels, two of which are triply 00:33:55.000 --> 00:33:59.000 degenerate, so we indeed have eight wave functions that 00:33:59.000 --> 00:34:04.000 correspond to the molecular orbitals of this system. 00:34:04.000 --> 00:34:25.000 Question down here? Could you say that louder? 00:34:25.000 --> 00:34:26.000 Why did this not have the one-half? 00:34:26.000 --> 00:34:29.000 The one-half is missing there because I forgot to put it 00:34:29.000 --> 00:34:32.000 there. Yes, to be normalized it should 00:34:32.000 --> 00:34:35.000 have a one-half there. That is right. 00:34:35.000 --> 00:34:39.000 But then there is a further point that that brings up, 00:34:39.000 --> 00:34:43.000 which is the following. What I am implying here is that 00:34:43.000 --> 00:34:48.000 there are equal amounts of carbon 2s in the bonding orbital 00:34:48.000 --> 00:34:51.000 and as in the antibonding counterpart. 00:34:51.000 --> 00:34:55.000 But that need not be the case. It could be that this is 0.6 00:34:55.000 --> 00:35:01.000 here and that this 0.4 up here. And that is a greater level of 00:35:01.000 --> 00:35:06.000 detail than I am really interested in going into for the 00:35:06.000 --> 00:35:09.000 purposes of this problem. But your point, 00:35:09.000 --> 00:35:13.000 with respect to the coefficient, is well taken. 00:35:13.000 --> 00:35:16.000 Thanks. Christine, maybe we can fix 00:35:16.000 --> 00:35:20.000 that on the one that gets posted on the web. 00:35:20.000 --> 00:35:26.000 Let's now look at these things using our VMD program. 00:35:37.000 --> 00:35:41.000 Starting from the lowest energy and we will work our way up. 00:35:48.000 --> 00:35:52.000 Which orbital is this? This is a molecular orbital of 00:35:52.000 --> 00:35:54.000 the methane molecule. 00:36:00.000 --> 00:36:04.000 One way you can identify orbitals is where they are with 00:36:04.000 --> 00:36:06.000 reference to the HOMO or the LUMO. 00:36:06.000 --> 00:36:09.000 In this case, the HOMO is triply degenerate. 00:36:09.000 --> 00:36:13.000 And there is one orbital lower in energy than the HOMO. 00:36:13.000 --> 00:36:17.000 This would be then the HOMO minus one because it is one 00:36:17.000 --> 00:36:21.000 below the HOMO in energy. That is the one which is the 00:36:21.000 --> 00:36:24.000 carbon 2s, plus A plus B plus C plus D. 00:36:24.000 --> 00:36:27.000 That is that MO there. 00:36:33.000 --> 00:36:39.000 Actually, let's go ahead and take that one away. 00:36:49.000 --> 00:36:50.000 Sorry about that. 00:36:55.000 --> 00:37:00.000 Let's see if you can figure out which one this is. 00:37:14.000 --> 00:37:18.000 In order to figure out which one it is, you need to be able 00:37:18.000 --> 00:37:23.000 to realize where the nuclei are and where your internuclear 00:37:23.000 --> 00:37:26.000 nodes are, if any. And, if so, how many. 00:37:26.000 --> 00:37:30.000 This is one of our hydrogen contributions. 00:37:30.000 --> 00:37:33.000 Here is a hydrogen contribution with opposite sign. 00:37:33.000 --> 00:37:38.000 And there is a piece of our carbon 2p orbital in the center. 00:37:38.000 --> 00:37:43.000 Now, this is just one of three identical counterparts that are 00:37:43.000 --> 00:37:47.000 rotated 90 degrees with respect to each other in space, 00:37:47.000 --> 00:37:51.000 so I am not going to show you all three of them. 00:37:51.000 --> 00:37:55.000 They all look exactly the same. Is this one bonding or 00:37:55.000 --> 00:38:00.000 antibonding? Yeah, this one is antibonding. 00:38:00.000 --> 00:38:03.000 And it does not have the same sign on all the hydrogens, 00:38:03.000 --> 00:38:07.000 so it has to be part of our LUMO, which is triply degenerate 00:38:07.000 --> 00:38:09.000 and involves bonding or, in this case, 00:38:09.000 --> 00:38:13.000 the antibonding between a carbon 2p orbital with a linear 00:38:13.000 --> 00:38:17.000 combination of four hydrogens. You can see that when you have 00:38:17.000 --> 00:38:20.000 an antibonding orbital, you are depleting the electron 00:38:20.000 --> 00:38:25.000 density in the space between the nuclei rather than increasing it 00:38:25.000 --> 00:38:30.000 as you do in a bonding orbital. Let's go ahead and look at the 00:38:30.000 --> 00:38:34.000 bonding counterpart of this one. And we will do that. 00:38:34.000 --> 00:38:38.000 And let's delete it. One thing that you will get, 00:38:38.000 --> 00:38:43.000 if you haven't already gotten it, is you are going to get a 00:38:43.000 --> 00:38:48.000 link to be able to download all the files that I used to show 00:38:48.000 --> 00:38:51.000 you molecular orbitals with in class. 00:38:51.000 --> 00:38:55.000 Because this program I am using, this VMD program, 00:38:55.000 --> 00:39:01.000 is available on Athena. And you can also download it 00:39:01.000 --> 00:39:06.000 yourself for free onto your own computer if you want to. 00:39:06.000 --> 00:39:10.000 And so this is going to give you the opportunity to study 00:39:10.000 --> 00:39:16.000 these orbitals and rotate them around at your leisure on your 00:39:16.000 --> 00:39:19.000 desktop. And now, let's look at the 00:39:19.000 --> 00:39:24.000 bonding counterpart to that antibonding orbital that we just 00:39:24.000 --> 00:39:27.000 looked at. This one is the two-bladed 00:39:27.000 --> 00:39:33.000 propeller to which I referred. And you can see that this 00:39:33.000 --> 00:39:37.000 orbital has the same symmetry with respect to its nodal 00:39:37.000 --> 00:39:41.000 properties as a carbon 2p orbital because it has a single 00:39:41.000 --> 00:39:45.000 nodal plane. And one side all the signs are 00:39:45.000 --> 00:39:49.000 positive, and on the other side all the signs are reversed. 00:39:49.000 --> 00:39:54.000 And we have a nice piece of bonding going on there and there 00:39:54.000 --> 00:39:59.000 when we have this oblique interaction of one of the lobes 00:39:59.000 --> 00:40:04.000 of a p-orbital simultaneously with two of the hydrogens that 00:40:04.000 --> 00:40:08.000 it sits directly between in space. 00:40:08.000 --> 00:40:13.000 There is one of our triply degenerate components of our 00:40:13.000 --> 00:40:17.000 highest occupied molecular orbital. 00:40:17.000 --> 00:40:23.000 And then, there is just one left to visualize. 00:40:36.000 --> 00:40:40.000 All right. We will delete that one. 00:40:50.000 --> 00:40:53.000 We looked at our most symmetric and most bonding molecular 00:40:53.000 --> 00:40:55.000 orbital. And now, we are going to go 00:40:55.000 --> 00:40:59.000 ahead and look at the antibonding counterpart of that 00:40:59.000 --> 00:41:00.000 one. 00:41:13.000 --> 00:41:17.000 And so this is clearly a molecular orbital that is formed 00:41:17.000 --> 00:41:20.000 using some piece of the carbon 2s orbital. 00:41:20.000 --> 00:41:24.000 And you see that you get a change in sign as you go from 00:41:24.000 --> 00:41:28.000 carbon to each of the hydrogens, that accordingly the electron 00:41:28.000 --> 00:41:32.000 density is depleted in the region of space between the 00:41:32.000 --> 00:41:36.000 nuclei. And that is a characteristic of 00:41:36.000 --> 00:41:39.000 an antibonding orbital. And this one, 00:41:39.000 --> 00:41:42.000 according to our diagram, in fact, is the LUMO. 00:41:42.000 --> 00:41:47.000 So, as you will see over here, that was what we predicted our 00:41:47.000 --> 00:41:51.000 lowest unoccupied molecular orbital would look like. 00:41:51.000 --> 00:41:54.000 And then, up here, we had our triply degenerate 00:41:54.000 --> 00:42:00.000 LUMO plus one based on the carbon 2p orbital interactions. 00:42:00.000 --> 00:42:04.000 Now, you can see what these four types of molecular orbital 00:42:04.000 --> 00:42:06.000 would look like. And now, you wonder, 00:42:06.000 --> 00:42:10.000 just what kind of prediction does this make regarding 00:42:10.000 --> 00:42:13.000 experimental observables. And so let's look at this 00:42:13.000 --> 00:42:15.000 picture. 00:42:25.000 --> 00:42:30.000 This diagram contains a number of interesting features. 00:42:30.000 --> 00:42:34.000 It is a diagram that plots ionization energy here on the 00:42:34.000 --> 00:42:39.000 x-axis against a number of different molecules on the 00:42:39.000 --> 00:42:42.000 y-axis. At the top you have simply a 00:42:42.000 --> 00:42:46.000 neon atom. Neon is our noble gas atom that 00:42:46.000 --> 00:42:50.000 lies just immediately to the right of fluorine in the 00:42:50.000 --> 00:42:53.000 periodic table. If anything is more 00:42:53.000 --> 00:43:00.000 electronegative than fluorine, it would actually be neon. 00:43:00.000 --> 00:43:04.000 This inert gas that is extremely difficult to ionize 00:43:04.000 --> 00:43:09.000 because it holds so tightly onto all eight of its electrons. 00:43:09.000 --> 00:43:14.000 Does it hold equally tightly onto all eight of its electrons? 00:43:14.000 --> 00:43:19.000 The answer is no because, as you scroll across here in 00:43:19.000 --> 00:43:23.000 ionization energy, you see that its 2p electrons 00:43:23.000 --> 00:43:29.000 ionize here, and its 2s electrons ionize way over here. 00:43:29.000 --> 00:43:31.000 The type of diagram I showed you last time, 00:43:31.000 --> 00:43:35.000 where the 2s orbital, as you go across the periodic 00:43:35.000 --> 00:43:38.000 table, decreases in energy faster than does the set of 2p 00:43:38.000 --> 00:43:42.000 orbitals associated with an atom, is expressed in this 00:43:42.000 --> 00:43:44.000 diagram here because we have neon. 00:43:44.000 --> 00:43:47.000 And then we are going across the periodic table. 00:43:47.000 --> 00:43:51.000 This next molecule is HF, an eight electron species, 00:43:51.000 --> 00:43:54.000 and H two O, an eight electron species, 00:43:54.000 --> 00:43:57.000 and then NH three, ammonia, the other eight 00:43:57.000 --> 00:44:02.000 electron species. And then over here we get to CH 00:44:02.000 --> 00:44:05.000 four, the molecule that we have just 00:44:05.000 --> 00:44:08.000 discussed today. And what is very interesting 00:44:08.000 --> 00:44:12.000 here is that the ionization energies of the molecular 00:44:12.000 --> 00:44:16.000 orbitals, in the case of the molecules, and of the atom in 00:44:16.000 --> 00:44:18.000 the case of neon track very nicely. 00:44:18.000 --> 00:44:22.000 Here is the molecular orbital that has 2s character of HF, 00:44:22.000 --> 00:44:25.000 of H two O, of NH three, 00:44:25.000 --> 00:44:29.000 and of CH four. And it is rising in energy as 00:44:29.000 --> 00:44:34.000 we go from right to left across the periodic table. 00:44:34.000 --> 00:44:38.000 And that is because the carbon 2s orbital is much higher than 00:44:38.000 --> 00:44:41.000 the neon 2s orbital is, for example. 00:44:41.000 --> 00:44:46.000 And the carbon 2s orbital is contributing to that most 00:44:46.000 --> 00:44:50.000 bonding of molecular orbitals of the methane molecule. 00:44:50.000 --> 00:44:55.000 And then there is this manifold here, that corresponds to 00:44:55.000 --> 00:45:00.000 ionizing an electron out of the HOMO of methane. 00:45:00.000 --> 00:45:04.000 And it is a broadened peak distribution here because of the 00:45:04.000 --> 00:45:08.000 different vibrational states that the methane can be found in 00:45:08.000 --> 00:45:11.000 when, in fact, it experiences an incoming high 00:45:11.000 --> 00:45:15.000 energy photon that is capable of knocking the electron out of the 00:45:15.000 --> 00:45:17.000 methane molecule and ionizing it. 00:45:17.000 --> 00:45:20.000 This is called photoelectron spectroscopy. 00:45:20.000 --> 00:45:24.000 And you will be reading about it also in the context of your 00:45:24.000 --> 00:45:30.000 problem set because we have a problem that deals with this. 00:45:30.000 --> 00:45:33.000 And I point you to a piece of your text that covers 00:45:33.000 --> 00:45:37.000 photoelectron spectroscopy. But the key point here is that 00:45:37.000 --> 00:45:40.000 if you had all your eight electrons at one energy, 00:45:40.000 --> 00:45:44.000 as valance bond theory might suggest for a molecule like 00:45:44.000 --> 00:45:49.000 this, you might think you should only see one peak in the PES of 00:45:49.000 --> 00:45:51.000 methane. But you see two because 00:45:51.000 --> 00:45:55.000 methane, the molecule has two molecular orbital energy levels, 00:45:55.000 --> 00:45:57.000 one of them being triply degenerate. 00:45:57.907 --> 00:46:00.000 See you on Wednesday.