 So far we have discussed diatomic molecules that is a good beginning and that is the only way to begin because unless we talk about H2 plus we would not get access to the molecular orbitals. But then having done that we cannot stop there can we because if we have a chemistry course in which we only discuss diatomic molecules then it is no fun we do not want to stop there we want to go beyond and we want to if possible talk about molecules like this. In this course we will not reach here this molecule incidentally is called calphostin C I had worked on this molecule we did some experiment and then our collaborators did a quantum chemical calculation in those days more than 20 years ago in the best supercomputer that was available in that university it took one year to get this optimized structure of calphostin C definitely way beyond diatomic molecule we will not go this far but we will at least try to talk about molecules that have some atoms few atoms and we will try to give you an idea of how one can start developing a molecular orbital theoretical treatment of molecules like this. Again when we do that we have to worry about sigma bonds and pi bonds we actually talked about pi bonds a little bit when we discussed carbon monoxide but we did not dwell upon it for too too long. So, what we will do is to start with we are going to discuss a system which has exclusively sigma bonds and then we will forget about sigma bonds completely and discuss a theory that is exclusively for pi bonds. So, here goes if I talk about polyatomic molecules the first thing that comes to my mind is that polyatomic molecules have certain shapes diatomic molecules can only be linear there are 2 atoms after all what happens if you have more than 2 atoms there if you have 3 atoms well the structure is determined by VACPR valence shell electron pair repulsion theory and essentially this is something that makes sure that the steric requirement of this so called bond pairs and lone pairs are met as we might have studied in class 11 and 12 lone pair lone pair repulsion is maximum followed by lone pair bond pair repulsion followed by bond pair bond pair repulsion and that is why if you have 2 electron pairs only 2 bond pairs you get a linear molecule like you do in BH2 or BCL3 if you have 3 like in BF3 it is a trigonal planar molecule if you have 4 like in methane we are going to talk about methane today you are going to get a tetrahedral molecule remember tetrahedral not square planar if it is just BH4 and if you have more if you have PA5 kind of molecule then the geometry is trigonal bipyramidal TBP for SF6 molecule we have an octahedral geometry and all this is because the repulsion between pairs of electrons has to be minimized. In valence bond theory hybridization is used as I said to generate appropriately oriented orbitals hybridization provides a directionality and that is its strength. Remember in molecular orbital theory the strength is delocalization and delocalization and directionality do not go hand in hand. So, in molecular orbital theory it is possible to use hybrid orbitals but that would give you a localized MOT which we do not like so much rather we want to build a delocalized MO picture of the polyatomic molecules that we are going to discuss and the molecule that we want to discuss today is methane. Delocalized molecular orbital theory picture of methane to do that we will consider this picture that AO's atomic orbitals with proper symmetry can only be combined and what I mean by that will become clear as we go ahead in the discussion. First let us think the central carbon atom which valence orbitals are there 2s and 2p this is 2s and this is 2p. So, we will try to do it like this for the hydrogen atoms that are there the orbitals that will participate in bonding they are the 1s orbitals. So, we will denote the hydrogen atoms as p, q, r and s all right and we are going to take appropriate linear combinations of 1s orbitals of p, q, r and s very often I am just going to call these 1s orbitals p, q, r and s and we will try to generate linear combinations which have the right symmetry to give a bonding and anti-bonding combination with the central 2s or 2p orbital of carbon atom. So, this idea of using linear combination of hydrogen 1s orbitals with matching symmetry generates what is called symmetry adapted linear combinations or slcs or salks as many people pronounce them. There is a more formal way of generating salks that is by using something called projection operators I want something some combination of p, q, r and s that I will be able to add to or subtract from the 2s orbital. So, what is the matching symmetry that I am looking for? See 2s orbital is just plus right there is no node here. So, if it is all plus then if I have plus sign of wave function of p and q and r and s that is good right let us say all the 4 1s orbitals on the hydrogen atoms have plus sign on the wave function. So, now I can write them like this psi hs is psi 1s of p plus psi 1s of q plus psi 1s of r plus psi 1s of r. Could you write all minus? Yes you could does not matter but the convention is to use plus wherever possible. Now, if I take linear combination of this combination this combination here is the salk symmetry adapted linear combination and this is the salk of orbitals of the pendant atoms. See the molecule is such that there is a central atom carbon and there are pendant atoms pendant means something that is hanging or projecting. So, this hanging atoms here are hydrogen atom. So, we take symmetry adapted linear combination of orbitals of the pendant atoms p, q, r and s in this case. So, if I now take a linear combination of this salk with the 2s orbital I can combine in 2 ways plus and minus. So, I can write like this c 1 a coefficient multiplied by psi 2s of c plus minus c 2 multiplied by this linear combination that is one of the molecular orbitals that I get involving the 2s orbital and a particular salk of the hydrogen atom 1s wave functions. Where is this wave function? I hope we all see that this wave function that I have written here psi m os is actually delocalized over the entire molecule. It is not only between carbon and p it is not only between carbon and q it is delocalized all over. So, the orbital is all over recognizing the fact that when an electron moves in the joint field it can be anywhere in the molecule. It does not have to be in one particular place, it cannot be in one particular place probability of finding it is distributed all over the molecule and that is what automatically takes us to electron density distributed over different centers and that takes us to sort of multi-center bonds you can call it. But is this a bond? Is a molecular orbital a bond? We will come to that discussion later on. Now, that being said let us think of the 2p orbital. The 2p orbital the two lobes have different signs. So, if I take p plus q plus r plus then what will happen? Then that does not have the right symmetry right because p plus q will give you a constructive interference you can say with the upper lobe of the 2p orbital r plus s will actually give you a destructive interference. So, I can say p and q will give bonding interaction r plus s will give you anti-bonding interaction overall you have a non-bonding interaction that means the MO is not formed. So, p plus q plus r plus s is not the right salc for forming MO with the 2p orbital of carbon. Then what would be an appropriate geometry? If I take p plus q minus r minus s that works does not it? p plus q minus r minus s we can write this this is a different salc from the one that I written earlier. But this salc has a right symmetry to give bonding and anti-bonding combinations with the central 2p atom on carbon sorry on the central 2p orbital on carbon atom on the 2p orbital of the central carbon atom. So, we have to use different salcs of the pendant atom for different orbitals of the central atom. I hope it is not difficult to understand that here let us say I had drawn the pz orbital I can generate a similar salc for px orbital where is px orbital here. So, px orbital would be let us say here this is plus this is minus if this is the case then what will be the right salc for the px orbital the right salc would be see this has to be plus q s has to be plus p has to be minus r has to be minus. So, I can write like this minus p plus q plus no minus r plus s. So, this gives us the salc that can give you bonding and anti-bonding MOs with the px orbital of your carbon atom. What about py I can draw like this let us say this thing is pointing towards us let us say that is minus and the one that is behind us that is plus this is your py for py what is the right combination that is required p and s must have plus sign q and r minus must have minus sign. So, plus p minus q minus r plus s this is the salc that gives us the right symmetry for forming the bonding and anti-bonding MOs with py orbital of the central carbon atom three such linear combinations are there great. So, now see we have two kinds of MOs one that involves the 2s orbital and the other that involves the 2p orbital what is the difference between the two one is that involves the 2s orbital there is only one possible MO and there are three MOs that are possible involving the 2px, 2py, 2pz atoms atomic orbitals that is point number one. Point number two is there is no node in the first MO there is a node in the second MO this node is the one that is already there in that particular p orbital. So, we have two kinds of MOs should they have the same energy from our elementary knowledge of chemistry we can expect that they will have different energy because the first one has no node second one has one node and generally energy goes up with the number of nodes. But then what is the experimental evidence to tell us that there are indeed two kinds of MO orbitals the experimental evidence comes in the form of photo ionization what is photoelectron spectroscopy we are familiar with ionization energy I think ionization energy would mean suppose this is an atomic orbital or a molecular orbital provide ionization energy the electron is promoted from here to an infinitely high level infinitely high level means it does not feel the attraction of the nucleus anymore it is not in the orbital you can think that is your ionization energy. Now let us say I have two different MOs in some atom or some molecule I can expect two different first ionization energies I am talking about first ionization energy here electron can go from the lower level electron can go from the higher level I will call this epsilon 1 I will call this epsilon 2 even though I actually drew them in the diverse order does not really matter now see so this is v equal to 0 kind of limit. Suppose I have excited this molecule with some radiation x typically which has a frequency nu what is the energy of a photon h nu and let us say nu is very high something like 20 electron volt or something like that 20 kilo electron volt or something like that right so what will happen the molecule is going to the electron is going to promoted beyond ionization limit this here is the ionization limit so if it goes beyond ionization limit what will what does that mean the energy that you provide h nu it will be too much actually it will be in excess so you can write it as h nu equal to i e plus kinetic energy of the free electron. So if I now use the same h nu so typically what you would have is you have your sample that would be irradiated with this very high energy photons with energy h nu and your detector is going to record what is the kinetic energy of the electrons that are set free. So see let us put some values let us say this nu here is 20 kilo electron volt let us say epsilon 1 is something like 1 and epsilon 2 is something like 1 well 1 will be the same thing something like 3 so what will be the kinetic energy in the first case ke 1 is going to be 20 minus 1 is equal to 19 and ke 2 is equal to 20 minus 3 is equal to 17 so you get two values so if I try to plot the spectrum of kinetic energies then I will get one line at 17 one line at 19. So actually I should write like this x axis is kinetic energy and what is y axis y axis intensity in the way I have drawn it here I should get equal intensity for both the lines and so I should get two lines like this let us say for whatever reason this one is doubly degenerate that means there are two MOs with the same energy same energy of one now what will happen there will be actually twice the probability of a transition occurring from here than from here yeah so then and these have to be all occupied energy levels remember otherwise you will not see anything so now what will happen is that this ke 1 this line is going to be twice as intense why because number of transitions from there is double we are assuming that both are all levels that we are talking about are occupied population of electrons in this higher energy level is going to be double the population of the lower energy level that is why that corresponding kinetic energy will be twice as much of course I do not have to plot kinetic energy in the x axis I might as well plot ionization energy then what will the plot look like now it will be just the other way round so for ionization energy of one the line will be twice as intense as the that for ionization energy 3 this is photoelectron spectrum in a nutshell for you and this is what is going to be used to establish that indeed there are two different kinds of MOs so let us draw the energy level diagram now so here briefly discussed hf and also we know that depending on which atom you are talking about energies of 1s orbitals are going to be different from each other so this is roughly to scale where the energies of the 1s orbitals of hydrogen atom and I have drawn four of them because four hydrogen atoms are participating in bonding there is somewhere between the energies of 2s and 2p orbitals of the central carbon atom now what happens we said that there are two kinds of MOs one involves 2s orbital and that is associated with lower energy let us put it in this is the sigma bonding orbital sigma 2s and you are going to have an anti-bonding orbital involving 2s also and for the remaining 3p orbitals they will participate to form sigma 2p and sigma star 2p as well and these are going to be triply degenerate as we have discussed next let us fill in the electrons how many electrons are there 2 4 6 8 4 from hydrogen 4 from the valence shell of carbon so you see both the bonding MOs actually are occupied so now if I do photo ionization then what kind of spectrum do I expect right I expect a spectrum with two bands isn't it or two lines actually you never get lines all spectra have some finite line width for different reasons so what kind of ionization energies do I expect I expect two kinds of ionization energies and this one is lower so the higher ionization energy should have should be associated with a smaller band in the photo electron spectrum the lower ionization energy band should be thrice as intense as the one with the higher ionization energy band and this is the spectrum for you taken from the net is drawn in the opposite direction but as you see for lower ionization energy the band is much more intense compared to the band for higher ionization energy and from here you can experimentally determine the what is the separation between these levels also okay so our model stands validated. Also often we ask we get the question what is the meaning of all this structure there is some structure here there is some structure is it 1 or is it 2 actually all this is Vibronic structure molecules vibrate as well using Born-Ock and Hammer approximation you can think that they are separated from electronic transitions they are not always they can couple so this is something called Vibronic structure again something we are not going to go into in any detail in this course at least what is what matters to us is the area under the bands the area under this band is thrice the area under this one so this band is associated with a lower ionization energy this is associated with a higher ionization energy and this is exactly in line with the MO energy levels energy level picture that we had drawn okay the model stands indicated what have we learned we learned that you can have two kinds of MOs in molecules like this and they are going to be associated with two energy levels however we know that the bond lengths are equal otherwise the model itself would have not been sustainable so do we have contradiction here we are saying there are two kinds of MOs and we are saying that all bonds are equivalent so is there a problem actually there isn't because it is important to realize that MOs are not bonds MOs are wave functions using wave function you can figure out the electron density you can figure out the probability distribution of electrons over the molecule and actually each MO contributes equally to each bond that is why you in end of the day you get a uniform distribution of electrons all over the molecule it is completely delocalized so if you think of the valence bond picture that most of us would know using sp3 hybrid orbitals there is no contradiction to be honest valence bond picture focuses on the bond a bond means electron density between two atoms molecule orbital talks about the wave function delocalized over the atom from there and this is something that we are going to learn in a little more detail when we talk about pi electronic systems we can actually figure out what is the electron density between two particular atoms also the pictures are correlated same story being told in two different ways there is no contradiction and it also brings us to this very important realization that which property of the molecule you see is determined by which experiment you do remember wave function collapse before you make a measurement the molecule is in an entangled state make a measurement there is a wave function collapse and you get to see some property of the molecule how do you bring about that wave function collapse in real life you perform some experiment so you perform photoelectron spectroscopy the idea you get is about molecular orbitals you do rotational spectroscopy we will talk about rotational spectroscopy later you can get an idea about the bond length okay so you do some other kind of experiment if it is possible to crystallize from crystals you can tell what the bond lengths are and what is the shape of the molecule so which experiment you perform will reveal different aspects of the molecule to you that so from this discussion today we have learned so many different important points about polyatomic molecules and now that we have discussed the molecule with sigma bonds only we will go over to other molecules where pi bonds are involved.