Correction to Lecture Series 6 In 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 the appropriate figure for any homonuclear diatomic molecule from the first 3 rows no p p p np atom 1 p np atom 2 p p 2p N s p p 2p O Figure 15 Incorrect MO diagram for the NO molecule ns atom 2 ns atom 1 s Figure 12 The MO diagram for homonuclear diatomic molecules from the first 3 rows of the periodic table Occasionally the orbitals and the orbitals will switch places See p p 2p N p p d electrons However figure 15 is wrong for two reasons One is relatively trivial For heteronuclear diatomics such as NO the energy of the p and the p orbitals switch positions so that the p orbitals are lowest in energy Why does this happen To tell this 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 the atomic orbital energy levels of the N and O atoms to reflect their ionization potentials This was a more serious error and I explicitly point out this difference in the corrected Figure 15 2p O Difference in the IP between O atom and N atom O atom IP is higher so its energy Levels are lower Corrected Figure 15 The p orbital MO diagram for the NO molecule and for other heteronuclear diatomic molecules in the first three rows of the periodic table Lecture Series 7 Hybridization of Atomic Orbitals and its Influence on Molecular Structure At the beginning of Lecture Series 6 I wrote the following Many of the structures that we have been covering especially the organic molecule structures should have raised some questions For example when we learned about the atomic orbitals that are used by a carbon atom we had a 2s orbital and 3 2p orbitals Each 2p orbital is aligned with an axis so we had 2px 2py and 2pz The 2s orbital was spherically symmetric If these are the atomic orbitals that we will need to form molecular orbitals then shouldn t the molecular shapes somehow resemble the spatial distribution of the atomic orbitals In other words since all of the p orbitals are at right angles to each other shouldn t carbon be characterized by bond angles that are 90o instead of 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 have mentioned do not appear to be in contradiction to the shapes of our molecular orbitals For example HC CH is linear both in fact and by VSEPR theory Consider just the carbon on the right side of this molecule It is bonded to a hydrogen and another carbon as shown in H1 C1 C2 Figure 1 This can t happen What we are doing is taking 1 atomic orbital the 2px orbital Figure 1 An incorrect bonding picture of the molecule on C1 and generating two HCCH The molecule is linear and so we have both C2 bonding MO s from it We and H1 bonded to C1 through the px orbital can only get one bonding MO and one antibonding MO Look at it this way C12px C22px will give us one bonding MO and one 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 only generate as many MO s as we have atomic orbitals Therefore even the simple structure of acetylene HCCH can t be explained by our current view of atomic orbitals and MO theory H Either the atomic orbitals or MO theory is C wrong Let s consider another molecule methane or CH4 The Lewis dot structure H H combined with VSEPR theory states that this molecule is tetrahedral This is confirmed by experiment However based on the atomic orbitals that we have at hand we might start assembling this molecule as is shown in Figure H 1 In this figure we have brought in 3 Figure 2 The assembly of CH4 based on hydrogen atoms and located them so that they the shapes of the atomic orbitals on the C are s p bonding with the px py and pz atom which is located at the origin of the orbitals Based on the above discussion of coordinate system I have drawn acetylene we know that we can t bring in the fourth hydrogen atom and form a bond with one 90o 109 5o Fig 3 At left is the structure of CH4 predicted by simply considering the shapes of the p orbitals The binding sites are denoted by the oval pads A 4th H is apparently swimming around the structure At right is the structure of CH4 as it actually exists The HCH bond angles are noted of the unused lobes of a p orbital Those lobes only appear unused in the cartoon They are actually used up in the already formed carbon hydrogen bonds The carbon atom does have one other orbital available for binding to the hydrogen atom and that is the 2s orbital However the 2s orbital is spherically symmetric Thus what we might expect to form is a methane molecule with 3 hydrogen atoms located at right angles to one another and a fourth hydrogen sort of swimming around the molecule bound to the 2s orbital on the carbon This is in contradiction to the actual tetrahedral structure We compare these two structures in Figure 3 It turns out that the answer to these apparent contradictions is that the atomic orbitals we know and love actually change when we begin to consider bonding They hybridize What does this mean All it means is that the orbitals mix with each other to form new orbitals Let s look and see how this happens by mixing a single s orbital with a single p orbital When we mix orbitals we add or subtract them from one another There are two ways to do this The first is the knucklehead way and it entails looking up the arithmetic functions in chapter 13 that describe 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 graphical representations of the orbitals and add or subtract the graphs Look at the left side of Figure 4 Here we are adding 2s 2px together Note that when the positive 2s orbital is added to the negative lobe of the 2px orbital we get a very small 2s lobe Of course we do We are adding a positive and a negative number The answer is going to be smaller 2px than the magnitude of either of …