 We are discussing Huckel MOT for ethylene. We have formulated the problem. We have identified these integrals H11, H12, S11, S12 and we have obtained this secular equation where this determinant is equal to 0. Now the next thing is to try and find solutions for this secular equation. Once again remember that Hij where i and j can be 1 and 2 in this case. This is equal to Hji and this is the form. Sij of course is equal to Sji and this is the form. Now let me remind you of this interesting aspect of Huckel MOT. We do not have to write the Hamiltonian. We try to formulate without doing it explicitly. Also we remember that for equivalent carbon atoms H11 must be equal to H22. Hii is the same. Hii equal to Hjj even though i and j can be different for equivalent hydrogen atom. Why? Because remember this is just the Hamiltonian is formulated considering an electron to move in the joint field of the nuclear and the other electrons and electrons are indistinguishable. We will come back to this later on. But let us say that this equivalent for equivalent carbon atoms H11, H22 we write this as alpha and alpha is called Coulomb integral. Why? Now we can perhaps continue the discussion. It is called Coulomb integral because it is all about electrostatic interaction of this electron in the joint field of nucleus and nuclei and the other electrons. So, that H11 and H22 in the secular equation we are going to write alpha here and we are going to write alpha here. We are left with H12 and S11, S12. Well H12 and H21 are written resonance integral beta. It is something that should remind us of our discussion of H2 plus problem. Here we have one orbital, one atomic orbital in the bra vector, another atomic orbital in the ket factor Hamiltonian in middle. So, this is a quantum mechanical quantity that is called resonance integral. So, H12 equal to H21 as we have said earlier we are going to call it beta. And beta is actually a negative quantity as we will discuss later on as well. So, how have we simplified this? This is alpha minus E multiplied by well S11 as we know is actually 1. We do not have to worry about it. So, this will become the first term will become alpha minus E. Second one H12 what is that beta, beta minus E multiplied by S. This one is H12 what is H12 that is beta, beta minus E S and this one is alpha minus E. So, this is the secular equation this is what it boils down to alpha minus E, beta minus ES, beta minus ES alpha nice symmetric determinant that equal to 0. If you expand the determinant what do you get? You get a quadratic equation and let me try to write that quadratic equation for you. What is the quadratic equation? Alpha minus E multiplied by alpha minus E I will write it as square minus beta minus ES multiplied by beta minus ES that is also beta minus ES square. Remember if I have a determinant like this a, b, c, d that is actually a, d minus b, c. I am sure you all know this but I thought I will write it just in case. Now what kind of an equation is this? x square minus y square and what is x square minus y square? It is x plus y into x minus y. Let me write it in this form. So, I get alpha minus E plus beta minus ES multiplied by alpha minus E minus beta plus ES is equal to 0. Considering the first term I write alpha minus E plus beta minus ES equal to 0. So, from here what do I get? I 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. See what just happened. I did not try to solve Schrodinger equation exactly. I have not written the Hamiltonian at all yet I have an expression for the energy E. In terms of integrals which I do not know but let us see whether we can figure out what they are. Meanwhile, let us try to work out the second integral also. This one, the second term also. When I equate that to 0 what do I get? Similarly, I get E multiplied by 1 but this time minus S is not it have minus E plus ES here that will be equal to alpha minus beta. So, in this case I get E equal to alpha minus beta divided by 1 minus S. We have two expressions two roots of energy and together we can write E turns out to be E alpha plus minus beta divided by 1 plus minus S. So, we have got the expression. Now, we will start worrying about what alpha beta S all these are. So, to start with let us just put back this expression into the first equation that we got. We will now formally determine what is C1 what is C2. So, put it in C1 multiplied by alpha minus E plus C2 multiplied by beta minus ES is equal to 0. This is what we can get easily right because H11 is alpha and H12 is beta S11 is equal to 1 S12 equal to S. Now, what we will do is we will put in these expressions for energy here as well as here and there are two expressions I am going to work out one you work out the other. I will work out this one E plus alpha plus beta divided by 1 plus S. When I put it there what do I get this equation I get C1 multiplied by alpha minus alpha plus beta divided by 1 plus S plus C2 multiplied by beta minus alpha plus beta multiplied by S divided by 1 plus S that equal to 0 expand it a little bit C1 multiplied by I will have 1 plus S in the denominator anyway in the numerator I will have alpha plus alpha S minus alpha minus beta for C2 I get beta plus beta S minus alpha S minus beta S very simple actually there is no need for me to work it out like this you can do it yourself. But what have I got from here C1 multiplied by alpha plus alpha S alpha and alpha go so I will just write it here because there is no space there. So, C1 and 1 plus in this in the denominator that also goes we do not care about it alpha S minus beta now I will write minus here C2 multiplied by what happens here beta S beta S goes to this beta minus alpha S. So, I will just write alpha S minus beta equal to 0 from here what do I get do not I get C1 equal to C2 and is in this the result that we had guessed already just from symmetry we just proved it a little more formally that is all. So, we get C1 equal to C2 and since we got this by taking the plus combination we call it C plus. So, your wave function becomes C plus multiplied by chi 1 plus chi 2 please work out the other one and C convince yourselves that is going to be C minus multiplied by chi 1 minus chi 2. When I normalize this that is very easy I will not work it out because we have done something similar for H2 plus we get C plus is equal to 1 by root over 2 into 1 plus S. This is what we have got psi plus is root over 1 by 2 into 1 plus S corresponding energy E plus is alpha plus beta divided by 1 plus S psi minus your homework is 1 by root over 2 multiplied by 1 minus S multiplied by chi 1 minus chi 2 E 1 is equal to alpha minus beta. Now, let us remind ourselves that alpha really is the energy of a px or by pz orbital in the sigma framework of ethylene considering ethylene to be in xy plane energy of pz orbital in alpha framework sigma framework of ethylene. So, whatever energy we calculate should be with respect to this right what is the stabilization what is the destabilization what is beta beta is integral chi 1 H chi 2 that is the interaction energy delocalization energy. So, we set alpha to 0 and we write beta to be 1. So, we write everything in terms of 1, but experimentally you can find out that the value is minus 75 kilojoule per mole. How can you find out experimentally using spectroscopy you see where the transitions are and you can work out easily. What is S? S is at best a small number as you will see in the simplest version of Huckel molecular orbital theory S is conveniently set to 0. But we will talk about that in a little more detail when we discuss the next problem when we add two more carbon atoms here. Before we conclude our discussion of ethylene let us try to think what kind of energy level diagram we are going to get E minus and E plus which one is lower and which one is higher and also let us try to draw the wave functions very simple this psi plus what will psi plus be it is chi 1 plus chi 2 that means something like this what is psi minus is going to be chi 1 minus chi 2. So, if I write it like this plus minus this will be minus plus who has said this is plus minus this is minus plus nobody has said if you want it in a different way you please do it yourself. So, this one really is the bonding orbital bonding pi orbital this one is the anti-bonding pi orbital and I would now like you to remind yourselves of these orbitals pi u and pi g pi u and pi star g that we had drawn in one of the earlier lectures and you can understand that there is a build up of electron density between the atoms in the bonding situation there is a depletion in anti-bonding situation. So, this energy is higher this energy is lower and also from the expressions you can see we said that beta is a negative quantity minus 75 kilo joule per mole. So, E plus is going to be lower energy E minus is going to be higher energy. So, if we set alpha to be equal to 0 if you draw like this I will write alpha equal to 0 then energy level diagram is such that this is the lower energy corresponding to psi plus bonding orbital is a higher energy corresponding to psi minus the anti-bonding orbital. What are the energy gaps? This one is beta this one is beta and that brings us to the question how do we know it is minus 75 kilo joule per mole well see how many pi electrons are there in ethylene there are 2 right where will they reside they will reside here if you can bring about a pi pi star transition you can see what energy is absorbed the energy of that photon h nu is going to be basically 2 beta and you can find out well 2 mod beta is what I want to write from there you can find out what beta is remember in this case this is actually minus beta beta is a negative quantity. So, minus beta is a positive quantity and this is how you can determine it experimentally this is how it has been determined that beta is minus 75 kilo joule per mole. We will discuss this in more detail when we go to the next problem that is Huckel theory of Butadiene