 Now that we have talked about methane a polyatomic molecule with sigma bonds only and now that we have got an idea that we can generate molecular orbitals that are delocalized over the entire molecule but they can give you an idea about the electron density between two given atoms. The field is set for a discussion of molecules that have bi bonds and we will start with the simplest such molecule that one can think of and that is ethylene. And here we are going to use this kind of a Cartoon notation always. Here we are actually drawing these constant probability surfaces the way they are usually drawn in textbooks and our notation is going to be that we are going to denote the p orbitals the atomic p orbitals as chi 1, chi 2, chi 3 whatever is required and we are going to denote the molecular orbitals as size. So, before we do that let us discuss in very brief what pi electron approximation is. First of all the structure of the molecule is considered to be built by the sigma framework and while building this it does not matter whether one has used valence bond theory or molecular orbital theory whether hybrid orbitals have been used or not the structure of molecule is supposed to be in place. So, pi bonding comes into play only after a sigma bond is established that is the realm of this theory. We are not talking about molecules that have exclusive pi bonds only right. So, the pi framework has to be defined first. Sorry the sigma framework has to be defined first and pi electrons are considered to move in a fixed effective electrostatic potential involving sigma electrons not due to but involving sigma electrons what does it mean that means there are number of nuclei. So, if you take a pi electron a given pi electron they are considered to move under the joint field of this nucleus, but this field is shielded by the remaining electrons remember when we talked about particle in a box we are said that we can approximately describe conjugated polyenes by a particle in a box model free electron model there also the consideration was that there is shielding of the nuclear charge by the remaining electrons when we talk about any given electron. Here we are considering that the shielding is such that it is not complete the electron does not become free as long as it is in the molecule the potential energy is not equal to 0 necessarily, but there is shielding effective electrostatic potential is there and in this framework pi electrons are derogalized all over. The method to do this the simplest method simplest method is Huckel molecular orbital theory Huckel U has actually two dots on top of it and this is a semi empirical approximate method as you will see will not make any attempt to write down the Hamiltonian completely will not make any attempt for an exact solution because it cannot be achieved rather we will use certain experimental conditions experimental considerations to build our semi empirical approximate description of this molecules in question with this background let us discuss ethylene and remember when I say let us discuss ethylene I mean let us discuss the pi electrons of ethylene. So, this is the cartoon that we have shown you earlier the pi wave function will be written as psi phi is equal to c 1 chi 1 plus c 2 chi 2 where chi 1 and chi 2 as we said earlier are the two p orbitals of the two atoms here c 1 and c 2 are the coefficients before going further is there any reason why the magnitude of c 1 should be any different than the magnitude of c 2 actually there is not by symmetry I mean 1 and 2 are arbitrary levels we are using we can interchange them. So, there is no reason why either of these coefficients will have more or less magnitude than the other remember square of coefficient is a contribution of that p electron in the pi molecular orbital thing molecular orbital square of coefficient is the contribution of that p orbital in the molecular orbital in the pi molecular orbital. So, there is no reason really why mod c 1 should be anything other than mod c 2. So, we are going to work this out more formally, but I hope we understand this clearly. So, we might as well take c out and write c multiplied by chi 1 plus minus chi 2 that is the kind of wave function we are going to get that is something we can sort of guess even before going into any further. Now, let us write Schrodinger equation H psi equal to H psi very simple. In the next step let us expand the wave function Hamiltonian what is the Hamiltonian here Hamiltonian remember we are saying that the any electron that we talk about is considered to be moving in an effective potential provided by all the nuclei in question and the other electron. So, if I am talking about chi 1 well if I am talking about electron in on the first atom then I am talking about this electron moving in the field of nucleus 1 nucleus 2 and electron 2 that is what the Hamiltonian is if you want you can write down the expression will not do it because we are not going to make any attempt to actually solve it exactly. So, Hamiltonian operates on c 1 chi 1 plus c 2 chi 2 to give us E multiplied by c 1 chi 1 plus c 2 chi 2. Is chi 1 or chi 2 an eigenfunction of this Hamiltonian no they are not they are eigenfunctions of the hydrogen atom Hamiltonian did not coordinates of 1 or 2. Is c 1 chi 1 plus c 2 chi 2 an eigenfunction of the Hamiltonian H hat yes it is the combination is an eigenfunction that is why we have been able to write the Schrodinger equation in this form. The only trouble is that there is no direct way of solving this and finding E. So, we are going to try an indirect method that is used very commonly in quantum mechanical problems. What do we do we left multiply by chi 1 and integrate over all space. We are trying to generate 2 different equations why because we do not know what the mixing coefficients are c 1 and c 2 2 unknowns we need 2 equations and this is how we generate them. For the first equation left multiplied by chi 1 integrate over all space what do you get integral well c 1 multiplied by integral chi 1 H hat chi 1 plus c 2 multiplied by integral chi 1 H chi 2 is equal to E multiplied by c 1 integral chi 1 multiplied by chi 1 over all space plus c 2 multiplied by E integral chi 1 chi 2 over all space in bracket notation this is what we get. Please make sure that you understand how we get this expression how we get this equation c 1 integral chi 1 H chi 1 plus c 2 integral chi 1 H chi 2 equal to E multiplied by c 1 into chi 1 chi 1 integrated over all space plus E multiplied by c 2 integral chi 1 into chi 2 integrated over all space. Do we know how to find out these integrals we actually do not but we can at least give them some names. So, the first one integral chi 1 H chi 1 I call it H 1 1 in general integral chi i H chi i is going to be called H ii then integral chi 1 H chi 2 is given the name H 1 2 I hope you can see the pattern here the name of the integral is H from Hamiltonian subscript of the first wave function then subscript of the second wave function like that. On the right hand side we call integral chi 1 chi 1 over all space as S 11 of course I hope it is not very difficult for us to realize that this S 11 is going to be equal to 1 because chi 1 is a p orbital that is normalized and another thing here we are working with real p orbitals we are working with p x p y p z in this kind of a molecule it is more convenient if you talk about the molecule in x y plane and you consider chi 1 and chi 2 to be along z axis. So, of course that is going to be 1 but for some time let us retain the notation S 11 because then we get some nice good looking symmetric expressions. So, we will just keep it for now and integral chi 1 chi 2 is S 12 and that also you know right overlap integral as we have encountered in our discussion of H 2 plus the only difference is that for H 2 plus the overlap integral came from this kind of an overlap head on here we are talking about this kind of an overlap side on ok. So, let us write this equation in terms of these integrals and let us bring everything on the right hand side to the left hand side. So, that on the right hand side with the with 0 what do we get C 1 multiplied by H 11 minus E S 11 plus C 2 multiplied by H 12 minus E S 12 equal to 0. So, a linear equation in which the variables are C 1 and C 2 and the coefficients are H 11 minus E S 11 and H 12 minus E S 12 respectively right. So, that is the first equation that we sought. How do we get the second equation? Same thing just let us multiply by chi 2 and integrate over all space I encourage you to please do it yourself and convince yourself that the expression you get is this. And when you do that one thing that should come to your notice is that here I have written H 12 actually I should have written H 21 because I am left multiplying by chi 2. So, the p orbital in the bra vector is actually chi 2 and not chi 1. Why have I written H 12? Well without going into further detail believe me when I say that utilizing the properties of Hermitian operators H is a Hermitian operator remember H 12 is equal to H 21, it comes from something called turnover rule and S 12 of course, will be equal to S 21 because it is just integral of a product of two functions f 1 f 2 and f 2 f 1 give you the same thing. So, we got two equations and we have two unknowns C 1 and C 2. So, we should be able to in principle solve this and get expressions and find out what C 1 and C 2 are. Let us write this system of linear equations as a matrix equation like this square matrix containing the coefficients multiplies a column matrix containing the variables to give us a column matrix elements of H are 0 and 0. Now, see how do I solve this equation? One solution is that C 1 is equal to 0 and C 2 equal to 0, but if that is the case then there is no mixing yeah there is no wave function because psi pi is going to be 0 into chi 1 plus 0 into chi 2. So, there is no wave function that is a trivial solution. What is how can we hope to get a non-trivial solution a non-zero value of C 1 and C 2? Well, this can exist if this matrix cannot be inverted. The square matrix here if it is such that it cannot be inverted that is the only hope of getting non-zero solutions for C 1 and C 2. What is the condition for an inverse of a matrix to exist? The condition is that its determinant should not be equal to 0. Here we want the opposite thing we want the inverse to not exist. So, the condition is that the inverse of the square matrix should not exist. Therefore, this determinant of the square matrix should be equal to 0. When this determinant equal to 0 the inverse does not exist and that is the only hope of us getting C 1 and C 2 that are non-zero. So, this determinant is called the secular determinant and secular determinant equal to 0 is called a secular equation. So, for the rest of the discussion we are going to try to see how one can solve for this secular equation.