 We are slowly getting towards the climax of our course. We have talked about homonuclear diatomics, heteronuclear diatomics. We have discussed how MOT can nicely explain why carbon monoxide behaves as a strong sigma donor through carbon atom and a good pi acceptor through the same carbon atom. Now what we want to do is we want to venture into polyatomic molecules. So today and for the rest of this course we are going to discuss molecular orbital theory for polyatomic molecules and today we will discuss the case of only one molecule the favorite molecule of all chemists methane very nice tetrahedral molecule. We have already discussed the valence bond approach towards understanding a molecule like this. Let us see how different MOT is does it give us the same result? Does it give us some new information? What happens? So the motivation for doing this of course is that polyatomic molecules have a particular kind of geometry and this geometry is explained very nicely in valence bond theory by the use of hybrid orbitals. Now it is not as if you cannot set up molecular orbital theory for polyatomic molecules using hybrid orbitals. One might be tempted to do so because then the geometry is retained. The problem is hybrid orbitals are associated with localization. If you remember what we had discussed for carbon monoxide the whole reason why hybridization of carbon atom orbitals was invoked was that we would get very highly directional sigma hybrid orbitals H1 or H2 whatever we call it at that time. So if you use hybridization and use hybrid orbitals and you construct MO picture then you would get a an MO picture where electron clouds would be strongly localized. But then that gives up the biggest advantage molecular orbital theory has over valence bond theory and that advantage is that of delocalization. What I am trying to say is this. Suppose we are talking about methane. I will draw that box as usual. How do we draw it? I do not remember exactly how I have drawn it later on but I will just draw it like this put one dot here leave aside the next two vertices put another dot in the next again leave aside the other vertices. So where will I go? I will go here and here. So of course these are the 1S orbitals let us see PQRS these are hydrogen atoms or we can say that they are the 1S orbitals. What do I do? Where is carbon atom at the body center somewhere here? So I can take an SP3 hybrid orbital here say pointing towards P. Then when I draw the MO I will call it psi P that will be some C1 multiplied by psi 1S of P plus C2 of H1 I will call this H1. Similarly you can write psi Q will be equal to some C3 into psi 1S of Q plus C4 into H2 where H2 would be the hybrid orbital pointing towards your Q. So this way I can write 4 molecular orbitals that would be the localized MOT for methane. If that is the case what kind of energy diagram would I get? Sorry I should have actually made a slide where I had I would have shown all this for whatever reason I forgot but let me please bear with me this is the we can draw this energy diagram without much fuss. Let us say this is your carbon 2S and 2P and we have generated 4 equivalent SP3 hybrid orbitals so that is 3, 4 and let us say hydrogen atom 1S wave function has energy somewhere here. So what I am saying is if this is the case I should draw 4 1, 2, 3, 4 because there are 4 hydrogen atoms right PQRS. So the simple the local localized MOT would give me a picture like this I have 4 degenerate molecular orbitals well 4 degenerate bonding molecular orbitals 4 degenerate anti-bonding molecular orbitals the bonding ones would be all doubly occupied and this is the case if I look at ionization energy how if I look at ionization energy what kind of ionization energy should I get? Only one value of ionization energy. So now let me talk a little bit about photo ionization photo electron spectrum. What is photo electron spectrum? You give so much of energy is x-ray typically or sometimes very high energy you will write that an electron comes out and goes into the continuum right electron gets detached the molecule gets ionized. So typically what you do is you use very high fixed value of energy let us say 20 electron volts and let us say kinetic energy sorry let us say ionization energy of the orbital from which this is removed let us say that is 13 electron volt. So kinetic energy of the electron is going to be 7 electron volt 20 minus 13 simple. Let us say there is some other energy level at 11 electron volt then kinetic energy will be something like 20 minus 11 9 electron volt. So if I plot something like intensity against kinetic energy of electron or I can just write ionization energy I should get one band right single band in photo electron spectrum is what is predicted from localized molecular orbital theory which is nicely in accordance with our valence mode picture. However as I said this gives up on the biggest advantage and that biggest advantage is your delocalization. So what we will do is we will talk about delocalized MOT but before that I request you please remember this discussion. If I use localized molecular orbital theory then I am going to get one value of ionization energy one band in the photoelectron spectrum. That being said let us go ahead and discuss delocalized molecular orbital theory picture. What does that mean? I want delocalization so I will not use hybrid orbital and we will remember that atomic orbitals with proper symmetry can only be combined. So now see carbon atom central carbon atom has two orbitals that are relevant in our discussion 2s and 2p 2s and 2p right and this 2s and 2p can separately form linear combinations of hydrogen 1s orbitals with matching symmetry. So what we do is we first construct a linear combination of P, Q, R and S 1s orbitals and we make combinations of particular symmetries that is why these are called symmetry adapted linear combinations Salks in short or SLCs. If I am speaking Hebrew and Latin right now please bear with me for a few more minutes things will fall in their place. But remember symmetry adapted linear combinations are constructed using the atomic orbitals of the pendant atoms carbon is central atom and hydrogen atoms are sort of hanging or are suspended from carbon they are called the pendant atoms. You have to perform symmetry adapted linear combinations linear combinations of pendant atom wave functions orbitals that have the matching symmetry with 2s orbital and then later on 2p orbital what would the matching symmetry be in this case if we take P plus Q plus R plus S okay I am calling it Psi Hs Psi 1s of P plus Psi 1s of Q plus Psi 1s of R plus Psi 1s of S that has matching symmetry is not it. So if I take linear combinations like this C1 multiplied by Psi 2s plus minus C2 multiplied by this Psi Hs what do I get when the sign is plus then I get bonding orbital bonding molecular orbital when the sign is minus then what happens this one is plus sorry sorry this one is plus let us say I have plus here plus here minus here minus here then for this part of the MO you are going to get bonding for this part of the MO you are going to get anti-bonding so it is a non-bonding situation or if 1 is plus 3 or minus same thing will happen. So I hope now you understand what the meaning of symmetry matching is symmetry of the linear combination of atomic orbitals or symmetry of the salks have to be the same as the symmetry of the orbital atomic orbital of the central atom. So for 2s orbital the salk that has compatible symmetry is just plus plus Psi 1s P plus Psi 1s Q plus Psi 1s R plus Psi 1s S what about P orbital which linear combination of hydrogen orbitals will have matching symmetry with P orbital remember here this slope is plus and this slope is minus sign of wave function. So if I want to have matching symmetry then P should be plus Q should be plus R should be minus S should be minus is not it yeah that is what I have shown here filled circles means plus and empty circles means minus this has matching symmetry. So if I take a plus combination between these two I get bonding if I take minus combination then I get anti-bonding. So we will call this Psi HP is Psi 1s plus Psi 1s P plus Psi 1s Q Psi 1s R minus Psi 1s S you can now take linear combinations see what has happened. I am constructing molecular orbitals by taking linear combinations of a lesser number of central atom wave functions I do not have to take 2s and 2p together I will take either 2s or 2p and depending on which atomic orbital I take of the central atom I will choose a linear combination of compatible symmetry of the pendant atom orbitals symmetry becomes very very important that is point number 1. So how is it that symmetry of 2s and 2p are different well 2s has no node and this one has a node. So if you think of inversion for example this will be anti-symmetric with respect to inversion whereas your 2s will be symmetric with respect to actually everything. So 2s orbital is totally symmetric meaning no matter which symmetry operation you use its sign does not change it remains the same whereas by using some symmetry operations well there is no center of inversion actually in case of a tetrahedron but we do have C2s right. So now think of this C2 turned by 180 degrees what happens this goes here that goes here so this becomes negative. So what I am seeing is that this orbital here is anti-symmetric with respect to C2 there is another C2 here is anti-symmetric with that C2 also however with respect to this C2 it is symmetric similarly you can think of the planes this for example is a plane but I will show this plane this is better is this a plane plane of symmetry yes it is might as well it is the other one just to ensure that I do not confuse you there is everything else actually you see the plane so with respect to this plane what happens to this linear combination pqrs q goes to s and p and r remain in the same position so it changes. So there will be symmetry operations which will change the sign of this linear combination so it will be symmetric with respect to some it will be anti-symmetric with respect to some so behavior in response to symmetry operations is not the same for 2s and 2p orbitals they have different symmetries point number 1 point number 2 I have only drawn let us see the pz orbital if I draw px for example if I draw px like this this is x then which linear combination of pendant atom wave functions will have matching symmetry it will be well this is plus that is minus so p and r have to be have negative coefficients q and s have to have positive coefficients so that will be minus p plus q minus r plus s this kind of linear combination will be required similarly you can work out what it will be for pz now all these linear combinations are different from each other but they have the same symmetry px, py, pz they are in different orientations they have the same symmetry so we have two kinds of molecular orbitals now one arising out of 2s other involving 2p two kinds of molecular orbitals these and these they have different symmetries and remember this kind of orbitals is 3 fold there are 3 m o's of the same symmetry will they have the same energy actually no because there is no node here there is a node here so what we expect now is that we should have two groups of molecular orbitals very different from what we were expecting from localized mo t treatment we now expect that there will be two groups of molecular orbital there will be a group of one arising out of 2s there will be a group of three involving the px py pz orbitals of the central carbon atom will they have same energy no because remember more nodes translates into more energy. So now we have this interesting situation this one I have drawn so I do not have to draw by hand we have 2s orbital and 2p orbitals of carbon atom we have 4 1s orbitals of hydrogen atom they combine but this time we do not hybridize the carbon atom orbitals rather we take an appropriate linear combinations we have 4 linear combinations from here also right p plus q plus r plus s p minus q minus r minus s p plus q minus r minus s and I forgotten what I said so you work out what the fourth one will be. So I get 4 degenerates alks here instead of the 4 1s orbitals so the first one gives me the sigma bonding molecular orbital with 2s that we call it sigma 2s correspondingly there will be an anti-bonding sigma 2s star orbital and there will be 3 degenerate 2p orbitals 3 degenerate 2p star orbitals different picture is obtained from delocalized mo t approach to methane from what we got from the localized mo t approach first let us fill in the electrons 2 4 6 8 how many ionization energies will be there 2 right so we should have 2 kinds of ionization energy because ionization energy means let us see if I can draw a straight line now let us say this is the continuum this will be one ionization energy this will be another ionization energy which is more obviously this so sigma 2s orbital is associated with an ionization energy that is greater than ionization energy of sigma 2p point number 1 point number 2 is that the population of sigma 2p is thrice that of population of sigma 2s so in photoelectron spectrum the band corresponding to sigma 2p which means the lower energy lower ionization energy band that should be 3 times as intense as the band arising from sigma 2s that is the higher ionization energy band now is a moment of truth I show you the photoelectron spectrum and you can judge for yourself which picture is correct localized mo t picture where we expect one kind of ionization energy or delocalized mo t picture where we expect first of all 2 kinds of ionization energies and secondly the band corresponding to the lower ionization energy should be 3 times strong as that of the higher ionization energy the answer to that remember Max Planck experimental results are the only truth here is the truth you see lower ionization energy has a larger band higher ionization energy has a lower band well even before the saying that I should have said that there are 2 bands meaning 2 ionization energies so the delocalized mo t picture gives me the correct result localized mo t picture does not I better erase this so that it does not cover up what I have written later so we do have 2 kinds of MOs 2 kinds of energy levels and that should disturb us what is the meaning of 2 kinds of MOs I mean the bonds are all equivalent right otherwise how is it a tetrahedron bond lengths are all the same so is that a contradiction is there a conflict between this delocalized mo t that gives us 2 different kinds of MOs and our knowledge that the 4 bond lengths are equal actually there is no contradiction it is important to remember that MOs are not bonds MOs tell you about the wave function and in delocalized molecular orbitals every wave function is delocalized over the entire molecule go back and have a look at the wave functions each of the wave functions no matter whether it is sigma 2s or sigma 2p is delocalized over all the 4 hydrogen atoms and the carbon atom bond by definition refers to the electron density between 2 given atoms carbon atom and a particular hydrogen atom so what happens is that each of these MOs contributes equally to each bond because they are distributed homogeneously so every MO contributes the same amount to the electron density build up between a given pair of carbon and hydrogen atoms well between carbon and a given hydrogen atom so this is an important thing to understand MO and bond are not equivalent to each other they are interrelated but they are not equivalent so there is no contradiction whatever we get from valence bond picture equal bonds is what we can rationalize by MO picture as well and this is a good example of a very fundamental quantum mechanical principle that we had stated towards the beginning of our class properties of the system that you see depends on the experiment to perform eigen value that you get depends on which operator you apply take the same psi make Hamiltonian operator on it it will give you energy make px operate on it it will give you linear momentum along x similarly the experimental equivalent of using operators is to use different techniques you do photo ionization spectroscopy then you probe the actual energy levels the MOs photo ionization energy gives you an knowledge of MOs so you get to see the MOs you do rotational spectroscopy or you do extra crystallography if possible not about methane then you get an idea about the bond lengths so this is very much like the proverbial blind man trying to figure out what an elephant looks like the one who touches the foot says the elephant looks like a pillar the one who touches the trunk says the elephant looks like a hose pipe the one who touches the task says the elephant looks like a pointed sword a pointed rounded sword on a spear elephant does not look like any of these the holistic picture of the elephant is a linear sum of everything so you perform a different experiment you get to see different property like this proverbial blind man something different body parts of the elephant this is an important take home message that is sort of a side product of our discussion of molecular orbital theory for methane okay so we have learned something very important we have learned that it is better to use delocalized molecular orbital theory when we are talking about when we are talking about polyatomic molecules and we get to understand that we can actually figure out which of our theoretical models is correct by comparing with experimental results even then you can go wrong remember both theory gave us the correct value of read work constant still we had to discard it but at least your theory should match some experimental observation so that is what we wanted to say about methane so we are done with discussing sigma bonds now we go on to the last part of our discussion that is for pi bonds pi molecular systems right and that is where we will see that we use another kind of approximation called Huckel approximation that is what we are going to learn in the next two or three classes and that will take us to the end of this course.