Correction to Lecture Series #6Hybridization of Atomic Orbitals and its Influence on Molecular Structurep-σ∗p-π∗p-πp-σs-σ∗s-σnpatom 1npatom 2nsatom 1nsatom 2Figure 12. The MO diagram for homonucleardiatomic molecules from the first 3 rows of theperiodic table. Occasionally the π−π orbitalsand the π-σ orbitals will switch places. Seep-σ∗p-π∗p-πp-σ2pN2pOFigure 15. Incorrect MO diagram for the NOmolecule.Correction to Lecture Series #6In the previous lecture series, I presented the MO diagram for a heteronuclear diatomic molecule,NO. The two figures of relevance to this discussion are given below. Figure 12 is theappropriate figure for any homonucleardiatomic molecule from the first 3 rows (nod-electrons). However, figure 15 is wrongfor two reasons. One is relatively trivial:For heteronuclear diatomics, such as NO, theenergy of the p-σ and the p-π orbitals switchpositions so that the p-π orbitals are lowestin energy. Why does this happen? To tellthis truth, I can’t really give a good reason,but it is what is observed experimentally.The other mistake was that I didn’t adjust theatomic orbital energy levels of the N and Oatoms to reflect their ionization potentials.This was a moreserious error, and Iexplicitly point outthis difference in thecorrected Figure 15.p-σ∗p-π∗p-πp-σ2pN2pODifference in the IP between O atom and N atom. O atomIP is higher, so its energyLevels are lower. p-σ∗p-π∗p-πp-σ2pN2pO2pODifference in the IP between O atom and N atom. O atomIP is higher, so its energyLevels are lower. Corrected Figure 15. The p-orbital MO diagram for the NOmolecule and for other heteronuclear diatomic molecules in the firstthree rows of the periodic table.(?)HHHHCFigure 2. The assembly of CH4, based onthe shapes of the atomic orbitals on the Catom, which is located at the origin of thecoordinate system I have drawn.C1C2H1Figure 1. An incorrect bonding picture of the moleculeHCCH. The molecule is linear, and so we have both C2and H1 bonded to C1 through the pxorbital.Lecture Series 7Hybridization of Atomic Orbitals and its Influence on Molecular StructureAt the beginning of Lecture Series 6, I wrote the following: “Many of the structures that wehave been covering – especially the organic-molecule structures, should have raised somequestions. For example, when we learned about the atomic orbitals that are used by a carbonatom, we had a 2s orbital, and 3 2p orbitals. Each 2p orbital is aligned with an axis, so we had2px, 2py, and 2pz. The 2s orbital was spherically symmetric. If these are the atomic orbitals thatwe will need to form molecular orbitals, then shouldn’t the molecular shapes somehow resemblethe spatial distribution of the atomic orbitals? In other words, since all of the p-orbitals are atright angles to each other, shouldn’t carbon be characterized by bond angles that are 90o, insteadof 180o (AB2), 120o (AB3), and 109.5o (AB4)? Think about this question. It is important.“Let’s consider a second, but related anomaly. Some of the molecules that we havementioned do not appear to be in contradiction to the shapes of our molecular orbitals. Forexample, HC≡CH is linear both in fact, and by VSEPR theory. Consider just the carbon on theright side of this molecule. Itis bonded to a hydrogen andanother carbon, as shown inFigure 1. This can’t happen!What we are doing is taking 1atomic orbital, the 2px orbitalon C1, and generating twobonding MO’s from it. Wecan only get one bonding MO,and one antibonding MO. Look at it this way. C12px + C22px will give us one bonding MO, andone antibonding MO. C12px and H11s will give us one bonding MO, and one antibonding MO.Thus, we have used 3 atomic orbitals to generate 4 MO’s. This is not possible! We can onlygenerate as many MO’s as we have atomic orbitals. Therefore, even the simple structure ofacetylene (HCCH) can’t be explained by ourcurrent view of atomic orbitals and MO theory.Either the atomic orbitals or MO theory iswrong.Let’s consider another molecule --methane, or CH4. The Lewis dot structure,combined with VSEPR theory, states that thismolecule is tetrahedral. This is confirmed byexperiment. However, based on the atomicorbitals that we have at hand, we might startassembling this molecule as is shown in Figure1. In this figure, we have brought in 3hydrogen atoms, and located them so that theyare s/p-σ bonding with the px, py, and pzorbitals. Based on the above discussion ofacetylene, we know that we can’t bring in thefourth hydrogen atom and form a bond with one=+=2s2pxFigure 4. Adding and subtracting 2s and 2p orbitals fromeach other.90o109.5oFig. 3. At left is the structure of CH4 predicted bysimply considering the shapes of the p-orbitals. Thebinding sites are denoted by the oval pads. A 4th H isapparently swimming around the structure. At right isthe structure of CH4 as it actually exists. The HCHbond angles are noted.of the ‘unused’ lobes of a p-orbital.Those lobes only appear ‘unused’in the cartoon. They are actuallyused up in the already formedcarbon-hydrogen bonds. Thecarbon atom does have one otherorbital available for binding to thehydrogen atom, and that is the 2sorbital. However, the 2s orbital isspherically symmetric. Thus, whatwe might expect to form is amethane molecule with 3 hydrogenatoms located at right angles to oneanother, and a fourth hydrogen sortof swimming around the molecule,bound to the 2s orbital on thecarbon. This is in contradiction tothe actual tetrahedral structure.We compare these two structuresin Figure 3.It turns out that the answer to these apparent contradictions is that the atomic orbitals we knowand love actually change when we begin to consider bonding. They hybridize. What does thismean? All it means is that the orbitals ‘mix’ with each other to form new orbitals. Let’s lookand see how this happens by mixing a single s orbital with a single p-orbital. When we mixorbitals, we add or subtract them from one another. There are two ways to do this. The first isthe knucklehead way, and it entails looking up the arithmetic functions in chapter 13 thatdescribe the atomic orbitals. Then we simply add or subtract those functions from one another,and we plot the result. These would be the hybridized orbitals. The second way is much smarter(because it is simpler), and we show it in Figure 4. Here, we just take the graphicalrepresentations of the orbitals and add