 We are about to begin the discussion of the very last topic of this course of hours and that is Huckel molecular orbital theory for conjugated molecules. We are all familiar with this kind of depiction of benzene for example in which these p orbitals participate in linear combination to form molecular orbitals the lowest energy of which looks like the one that is depicted here. This is very common place as you see I have taken this picture from Wikipedia. So, this kind of a system where we have conjugation alternate single and double bonds by valence bond theory description. The most common way of dealing with it is to use the pi electron approximation or to generate what is called the sigma pi picture. In this approximation what we say is that the structure of the molecule is determined by the sigma framework common sense makes perfect sense as we said the structure is given by how the sigma bonds are disposed with respect to each other. So, the sigma framework besides the structure of the molecule and the pi electrons are supposed to move in a fixed effective electrostatic potential due to sigma electrons. Now, what is the meaning of effective electrostatic potential that takes us back to what we have learned from many electron atoms and diatomic molecules and so on and so forth. So, here what we have is we have a number of nuclei it is a poly nuclear system anyway if you think of benzene or even ethylene the one that we are going to discuss in some detail. Now, so first of all we have said that the sigma framework is in place and then only we start discussion about the pi electrons. So, the pi electrons when we start talking about them would experience a joint field of all the nuclei but that field will be shielded by the sigma electrons that are already present there. So, the potential that these pi electrons field is going to be a fixed effective electrostatic potential that is what we mean by this sentence. The electrons are supposed to be delocalized over this framework. In our earlier discussion of ammonia for example, we have said that delocalization is the biggest strong point of molecular orbital theory. So, there is no reason why we should give up on this unique advantage that this theory offers us. And even in the valence bond picture extended valence bond picture when we try to draw the structure of conjugated systems we end up drawing resonance structures in which the double bond in one structure it is between say 1 and 2 carbon in the other structure it is between 2 and 3 and so on and so forth. So, delocalization over the framework is a very logical common sense kind of point to start from. With all these considerations the simplest approach to pi electron systems is provided by Huckel molecular orbital theory. So, remember Huckel theory is all about pi electrons we do not even talk about sigma electrons here. So, for all you know you might have treated the sigma electrons using valence bond theory using sp to hybrid orbitals something like that it does not matter because we are starting with the starting from the point when the sigma framework is already there. But Huckel treatment is essentially a molecular orbital treatment which is all about pi electrons delocalized pi electrons in conjugated systems. So, the first such system the easiest one that we can think of is ethylene and you might notice here that what I have really drawn a cartoon we have said time and again that this kind of orbital and all that this is not orbital this is something else this is the surface that contains a certain amount of probability and so on and so forth. But we are still using it because that is what is done in all pictorial depiction of Huckel MOT and so you better be familiarized that as well and it also brings out the fact that this is an approximate theory it is a semi empirical theory as we are going to mention once again unless I forget about it. So, this is your ethylene carbon 1 carbon 2 we are considering the sigma framework to be in x y plane. So, if you talk about sp2 hybrid orbitals then the sp2 hybrid orbitals are made up of s, px and p y orbitals. If you want to talk about MOT to describe the sigma bonding also then 2s, 2 px and 2 p y will be the orbitals involved in sigma bonding. The z orbitals in the way I have drawn this molecule the pz orbitals chi 1 and chi 2 for atom number 1 and atom number 2 are available for pi bonding. So, we are going to work only with chi 1 and chi 2 remember sigma framework is already there we are right now worried only about the pi molecular system. So, what is the first step write the MO how do you write the MO as a linear combination of the participating p orbitals participating atomic orbitals in this case pz orbitals. So, it is very straightforward psi pi is equal to c 1 chi 1 plus c 2 chi 2 nothing to explain it is very very simple here. What do I do next? Next what I should do is I should write Schrodinger equation. I have not written it here but I will write it and I will erase again. So, essentially I write Schrodinger equation h c 1 chi 1 plus c 2 chi 2 is equal to e c 1 chi 1 plus c 2 chi 2 then what do we do first we left multiply by chi 1 remember. So, what do you get integral I am so used to writing psi that even when I want to write chi I end up writing another psi. So, it will be something like c 1 square chi 1 h chi 1 plus c 1 c 2 integral chi 1 h chi 2 is equal to e multiplied by c 1 integral chi 1 chi 1. And I do not need to say chi 1 star here because I know pz orbital is a real one plus c 2 integral chi 1 chi 2 over all space. So, this is essentially what I get this is c 1 square and then I go ahead and I develop this treatment of linear I develop this system of linear equations and from there we get our by now familiar to us the secular equation right secular determinant is equal to 0. And whatever integrals I wrote here they all get written once again in an abbreviated form. So, this is what we have the secular equation h 1 1 minus e s 1 h 1 2 minus e s 2 h 1 2 minus e s 1 2 h 2 2 2 minus e s 2 I will read that again h 1 1 minus e s 1 1 h 1 2 minus e s 1 2 h 1 2 minus e s 1 2 h 2 2 minus e s 2 2 that determinant is equal to 0 and we know that this comes from the requirement that we have this non non trivial roots non trivial roots means roots in what coefficients c 1 and c 2. So, what we have written here is something that we have already shown you when I wrote this thing by hand h i j is integral chi i h chi j is equal to h j i s i j is integral chi i chi j equal to s j i the second one is very easy to understand it is a product of two functions integrated over all space sequence does not matter. First one would be very easy to understand if this Hamiltonian of was for that particular atom but it is you can be confused a little bit because chi 1 and chi i and chi j are the atomic orbitals and atomic orbitals belong to different atoms do not forget the Hamiltonian is of the entire molecule and interestingly knowledge of Hamiltonian is not required in Huckel theory. So far we have been the first thing we have been trying to do is to write the Hamiltonian in its analytical form we do not have to do it we do not need it in Huckel theory because Huckel theory is a semi empirical theory Hamiltonian is required to find energy is determined from known experimental results. So I do not need to know what the Hamiltonian looks like that is what makes Huckel theory very advantageous but it is also little dangerous because starting this point on one might start forgetting all that we have learned right because then we get some integrals even if you do not know anything before we will actually be able to get the right answers. So that is the danger of things that are easy one needs to be careful about that right. Now H11 has to be equal to H22 because we are talking about two equivalent carbon atoms. So there is no reason why the integrals will be different if I write chi 1 H chi 1 and chi 1 H or chi 2 H chi 2. So H11 and H22 are the same H11 and H22 these are one and the same these are called Coulomb integral alpha and what is the meaning of Coulomb integral we can say it now but let us wait a little bit and we will say it in a little while. Next thing is H12 and H21 H12 and H21 are equal to each other something that we have already discussed this is called resonance integral beta again we are going to discuss the physical meaning of beta what does beta stand for and why it is important. S12 equal to S21 not very difficult to understand that we will just call S. So how does this simplify instead of H11 and H22 I will write alpha instead of H12 in these two places and we write beta S11 is equal to 1 is not it S11 and sorry this is S22 not S21 perill of copy paste as usual this is S22 S11 and S22 they are equal to 1 is not it because the PZ orbitals are normalized after all. So what will I get in instead of the first one H11 is equal to alpha minus E into 1. So alpha minus E second element here 1 2 element will be beta minus ES same thing here and this 2 2 element will be alpha minus E right alpha minus E beta minus ES beta minus ES alpha minus E that determinant is equal to 0 and the rest is very simple you expand it and you get some expression for E and just in order to ensure that we all know how to expand determinants even though we have actually handle determinants many determinants already I just expand this one what does this mean it means alpha minus E whole square minus beta minus ES I do not know why I started writing so low whole square is equal to 0. So this is x square minus y square kind of situation on the left hand side so I know we know what it means it means x plus y into x minus y equal to 0 setting the first factor to be equal to 0 what do you get you get E multiplied by 1 plus s is equal to alpha plus beta or E is equal to alpha plus beta divided by 1 plus s and if you set this factor to be equal to 0 then what do you get you get E into 1 minus s is equal to alpha minus beta so E is equal to alpha minus beta divided by 1 minus s you write them together you get E is equal to alpha plus minus beta divided by 1 plus minus s and so these are the two energies that you get see how many linear combinations will be there there are two atomic orbitals right I should be able to combine them in two ways and even before going into the coefficients I know that 1 will be plus 1 will be minus so one will have higher energy than the other so which one has higher energy which one has lower energy alpha plus minus beta divided by 1 plus minus s right that is what we will see but before that let us find the coefficient so to do that let us go back to one of the equations that we got in the first place c 1 multiplied by h 11 minus E s 11 plus c 2 multiplied by h 12 minus E s 1 1 2 is equal to 0 okay this is what we have so c 1 multiplied by h 11 is alpha s 11 is 1 h 12 is beta and s 12 is s so you write like this c 1 multiplied by alpha minus E plus c 2 multiplied by beta minus E s is equal to 0 so if you put the value of E in both the terms where it occurs and well see your choice right you can put either the plus combination or you can put the minus combination so let us start with the plus combination E plus we will call it E plus since we are using the plus combination alpha plus beta by 1 plus s you substitute that that is mundane so you do it yourself it turns out that c plus is equal to c 2 sorry c 1 equal to c 2 we call both to be c plus since we got them by this plus combination so the wave function we call psi plus equal to c plus into chi 1 plus chi 2 and sorry about the typo here I wrote capital C here in the next one I wrote small c so but they are actually the same please do not get confused by normalizing this wave function psi plus knowing that chi 1 and chi 2 are actually normalized you get a value of c plus to be 1 divided by root over 2 into 1 plus s remember what you got for h 2 plus and all see the similarity 1 divided by root over 2 into 1 plus s okay so you have got an expression of the wave function we have got the corresponding energy of course you have another value of energy where you have minus sign between alpha and beta minus sign between 1 and s and you substitute that in this same equation by the way there is another equation remember you would get the same result if you substitute in that equation also so you get the wave function corresponding to E minus we call that psi minus is 1 by root over 2 into 1 minus s chi 1 minus chi 2 okay so one combination is plus one combination is minus equal contribution from both the atomic orbitals here and since I forgot to draw this diagram here I will draw it so I will draw it by hand this psi plus what would it be not very difficult to understand I can draw like this if you draw the first orbital this way then the first second orbital will be like this that is plus and for psi minus if you draw the first orbital like this second orbital has to point downwards okay because it is minus so here the only node that is there is the node that already existed in p orbital right x y plane here however you have an additional node between these two so this here is the bonding orbital this is the anti-bonding orbital and energy of the second one is higher than the energy of the first okay energy of the second one is higher than the energy of the first when is that going to happen when beta is negative right we said E minus is a higher value than E plus so obviously beta has to be negative and it makes perfect sense for beta to be negative we will come to that first let us get done with alpha it is important to realize that alpha essentially is the energy of a pz orbital in the sigma framework of ethylene short simple sentence but quite profound what is alpha integral chi 1 h chi 1 what is chi 1 this p orbital and in integral the same chi 1 appears in bra as well as ket factors what is Hamiltonian Hamiltonian for the entire pi molecular system okay so what terms would it contain even though we are not going to bother to write it it is going to contain terms in inter nuclear repulsion it is going to term contain terms in for attraction of this electron with electro nucleus 1 as well as nucleus 2 and so on and so forth so the Hamiltonian is really the Hamiltonian of the entire molecule so this this energy alpha that you get is the energy of pz orbital in sigma framework of ethylene even if there is one pz orbital that would be the energy so if you remember what we studied in h2 plus the three terms the first two terms said energy of 1s orbital and nucleus nucleus repulsion so it is sort of something like that alpha so that is the base value in this framework okay even if there is no bonding that alpha energy will always be there if you place this p electron 1 p electron in this sigma network so that is our starting point base value that is why okay we will see what it is said to very soon and what is beta beta is delocalization energy chi 1 h chi 2 integrated over all space right that accounts for delocalization of each electron over the entire molecule okay on all the pi m os so it would better be negative because if that energy is not negative then why will pi bond form negative energy means stabilization isn't it so what we do is we set alpha to be equal to 0 because all measurements are with respect to alpha and delocalization energy we can set to 1 very often we write the energies in terms of beta we do not even bother writing beta in well today we are going to write beta anyway but this is a measure of stabilization it is important to understand that it is a negative quantity the value of beta is minus 75 kilo joule per mole how do I know that it is minus 75 kilo joule per mole I know it from experiment because as you will see we know these we know the expressions right E plus and E minus now if I know what is the energy of transition between homo and lumo of your ethylene then I can figure out what the value of beta is okay this alpha in any case will set to 0 we will see what we will do to s and then whatever is the difference that turns out to be 2 beta okay from there you see that it turns out to be minus 75 kilo joule per mole and it is the beta is used as a known constant quantity in Huckel theory this is the approximation and this is the semi empirical aspect now what is s s actually is well we when we talk about butadiene we will see how we handle s but s is a very very small number let us just say that this is really a very small number so we do not bother about s most of the time it is okay if you set s to 0 if you set s to 0 then what will happen E plus will be beta E minus will be minus beta so you draw the energy level diagram like this I will draw it here I will draw it here let us say this is your energy I cannot draw it here because I have written this first I will draw it somewhere here energy goes up this is the energy level corresponding to psi plus this is the energy level corresponding to psi minus this energy turns out to be what beta and this energy turns out to be minus beta it is important to understand that minus beta is higher energy and if you want to be a little more accurate you might as well write that this energy is alpha all measurements are with respect to alpha anyway okay so we are setting s to 0 here we have shown you the expression in which s is not set to 0 but we are saying that s will be equal to 0 because it is pi interaction anyway we will come back to that when we talk about butadiene okay let us quickly complete the formulation of the problem for butadiene and then we will come back and talk a little more about it this is the picture for butadiene okay sorry for this kind of a situation this is the same secular determinant and that is equal to 0 this is a bigger secular determinants if you want to work it out it will require a little more effort obviously but the basic formulation is the same we are calling the pz orbitals of atoms 1 2 3 and 4 carbon atoms 1 2 3 and 4 chi 1 chi 2 chi 3 chi 4 so psi pi will of course be c 1 chi 1 plus c 2 chi 2 plus c 3 chi 3 plus c 4 chi 4 we are going to learn in the next module how to find out c 1 c 2 c 3 c 4 well we will not find out how you know already know how to do it from ethylene we just show you the values but what we can do here is that from our previous knowledge we know that this h i i h j j is Coulomb integral alpha so in all these diagonal elements instead of h i i i can write alpha h i j and h j i for ethylene we said the resonance integral beta but now what we will say is that it is equal to beta only for adjacent atoms so what we are saying is that for energy calculation we consider like pair wise potentials there is no need to consider interaction between 1 and 3 1 and 2 2 and 3 and 4 these interactions are enough now of course the electrons are delocalized over everything but it is an approximation right the contribution from distant terms is going to be small enough so in Huckel approximation they are neglected so what we will do is we will set h i j equal to h j i to be beta only for adjacent atoms and we will set them to be 0 for all others and similarly we will set s i j to be s i j to be equal to s j i to be equal to s only for adjacent atoms and in fact what we will do in this simple version of Huckel theory that we are discussing in this course is that we will set everything to 0 why is it justified to set s to be equal to 0 is that see already the structure is determined by the sigma network right change due to pi bonding is sort of like a correction term so the overlap in sigma is much more than in pi so overlap integral is negligibly small compared to sigma that is the motivation for making this approximation that we can set s to be equal to 0 for all in the simplest version or little more advanced version of Huckel theory you cannot do that you still have to consider the overlap integral for adjacent atoms but that is good enough so let us see where we have to write beta alpha we already know h 11 h 22 h 33 h 34 h 44 these are alpha and we set them to be equal to 0 what about beta see one and two are adjacent to each other right we set the s to 0 fine now for beta one and two are adjacent to each other so instead of h 12 we will write beta in the 1 2 element as well as in the 2 1 element then 1 and 3 are not adjacent so we do not bother 1 and 4 we do not bother 2 and 3 are adjacent to each other so h 23 is going to be beta in the 3 2 as well as 2 3 elements and then 3 4 are adjacent to each other so h 34 is going to be beta in these 2 elements right so knowing this is a substitute alpha beta this is what you get I hope that was not too fast wherever we have to write alpha wherever we have to write beta we write that and we set all the overlap integrals to be equal to 0 that gives us this kind of a nice simple matrix well determinant I am very sorry that this vertical line on the left hand side is not showing up something wrong in the compatibility so this is what it is now what will you do you have to expand this determinant and how do you expand the determinant it is a little long alpha in minus c multiplied by this 3 by 3 determinant minus beta multiplied by the 3 by 3 determinant in which the columns are beta 0 0 beta alpha minus c beta 0 beta alpha minus c t d as per doable once you do it but even before you do it what you do is you set x equal to alpha minus c by beta that will make it even simpler you can write it in terms of x and 1 and then you get a an equation like this x to the power 4 minus t x square plus 1 equal to 0 do not be daunted seeing x to the power 4 this equation is really a quadratic equation not in x but in x square so you know how to solve quadratic equations minus b plus minus b square minus root over b square minus 4 ac by 2 a do that and you get this value for x square 3 plus minus root over 5 minus b divided by 2 and now you know what x is just x square root in these 2 cases you get 4 answers you get plus minus 1.61804 and plus minus 0.61804 okay well there is 1 and 0 the digits after decimal point are all the same what is x if you set alpha to be equal to 0 then x really is minus e divided by beta is not it so it is x is an energy term in terms of beta but in terms of negative beta also let us not forget that now so put in the values and this is what you will get you get this energy levels not to forget that beta is negative so the lowest one is 1.61804 beta followed by 0.61804 beta minus 0.61804 beta minus 1.61804 beta and the barycenter is at alpha so measurements are with respect to alpha okay how many electrons are there 4 electrons is not it 4 pi electrons are there in butadiene so if I just fill in the electrons then I can draw like this if I draw in the convention and sense so these turn out to be the bonding orbitals occupied orbitals those are the anti-bonding unoccupied molecular orbitals so the homo-lumo energy gap how much is homo-lumo energy gap that turns out to be I will just consider 2 decimal places 1.22 beta that is the energy gap that we get between homo and lume okay so if you know the energy gap you can find out beta if you know beta you can find out the energy gap so these are the energy levels next we have to worry about wave functions what do the wave functions look like and we more importantly we learn what kind of information about the molecule we can get from these wave functions.