 Finally we start talking about molecules and in whatever time is left in this course we will only talk about molecules. We start with a very simple approach valence bond theory which I think most of us would know at least qualitatively and then we end with a very simple approach for pi electron systems that is Huckel treatment. So whatever we discuss now is actually much simpler than what we have developed so far while discussing atoms and it seems that way because while discussing atoms we have built all the tools that we need to discuss molecules that is why it is going to sound much simpler. So here goes we start with valence bond theory of homonuclear dioramics and we get to know how is it that these atoms come close together and are very happy to be with each other even though we have this big fat positively charged nuclei which repel each other strongly. What is the reason for stabilization when these nuclei come together enveloped by the electrons and how are bonds formed, how are molecules formed as a result and that is what leads to everything in this universe. So there are two approaches to talk about this one is valence bond theory and the other is molecular orbital theory. We start with valence bond theory this essentially is an extension of Lewis electron dot model that everybody has studied in high school. Remember we used to put dots for electrons and then we used to share and we used to say that there has to be this stable octet and so on and so forth valence bond theory translates this very simplistic model of Lewis J and Lewis in language of quantum mechanics that is all. So it is very simple to understand and it is it relies completely on overlap of atomic orbitals and sharing of electron pairs that as I said it is just a translation of Lewis theory into quantum mechanical language. So atomic orbitals are retained with the modification we are talking about many electron systems well today we will talk only about H2 but then later on we want to talk about many electron systems. So there you have to take care of things like shielding and all in fact as we will see even for H2 you have to take care of shielding. So atomic orbitals will be modified accordingly. So actually it works fine for many systems and the beauty of it is its simplicity. The problem of it is also its simplicity because it is limited to 2 center 2 electron bonds as long as you limit yourself to ground state of 2 center 2 electron bonds it works beautifully. Problem is the moment you go beyond 2 centers or you have 1 electron or 3 electrons or more electrons then it does not work. So when you have more than 1 more than 2 centers you need to account for delocalization which is actually not provided for in valence bond theory. So you have to invoke this afterthought that is called resonance. Now resonance is something that is absolutely favorite with students of chemistry we know very well we love pushing arrows and we love talking about how a particular resonance structure is stable than the more stable than the others but resonance really is not something that is built into VBT it has been it is an add-on to extend the scope of the theory to delocalize systems. And even if you invoke resonance there is something you cannot do and that is you cannot access excited states as we might have mentioned in the passing earlier there is a lot of chemistry in excited states as well. All the photochemistry that you see well the very fact that you see the process of vision all that is something or the other to do with excited states. And valence bond theory is definitely limited to ground state only nothing else. So that is a very very severe problem with valence bond theory. So the way out is to use molecular orbital theory in which you consider the electrons to move in the joint field of nuclear and you set up the Hamiltonian well we will set up the Hamiltonian for valence bond theory as well today. So but then in molecular orbital theory you set up the Hamiltonian you can solve Schrodinger equation with the Hamiltonian exactly for H2 plus and we will not even do it. But the moment you add one more electron you encounter the usual complications that we got with many electron atoms and we get more because now you do not only have more than one electrons but you have more than one nuclei as well. So what we do is we generate molecular orbitals and I am sure you have come across this already we generate molecular orbitals as linear combination of atomic orbitals LCAO. So if you remember what we had talked about earlier we are actually synthesizing the wave function by using this orthonormal basis set of atomic orbitals. So you can use variation theorem there remember the good thing about molecular orbital theory is that it is a general theory. It can handle delocalization just like that you do not have to do anything special it gives you excited states just like that you do not have to do anything special it is general theory that is its advantage and that is its disadvantage also perhaps I might seem like I am talking in riddles because I said the same thing for BBT what is advantage is also the disadvantage I am saying the same thing for MOT what is advantage is also the disadvantage but it is true that is how it works your advantage sometimes unchecked advantage can become a disadvantage as we will see that sometimes is a bit too general it ionic structure of H2 for example is grossly over interpreted if you use molecular orbital theory but that will be the story for another day. Now you see let us write the Hamiltonian and to write the Hamiltonian we will start with a hydrogen atom then even though your molecular orbital even though valence bond theory it has no scope for discussing H2 plus ion will write the Hamiltonian for H2 plus ion just because it has fewer terms and then we will go on to H2 molecule so this is something that is very familiar to us ignore the picture for the moment just look at the expression for Hamiltonian Hamiltonian for hydrogen atom something is wrong hydrogen atom Hamiltonian now you can now in the picture is correct sorry about goofing up there so hydrogen atom you remember nucleus and electron okay so the height the Hamiltonian would be minus h cross square by 2ma del h square minus h cross square by 2me del e square where a stands for the nucleus or the atom as a whole e stands for the electron in it and minus q e square by r a gives you the potential energy for attraction of the electron experience the attraction experience by the electron because of the nucleus so this potential energy the first time here is kinetic energy of the atom as a whole largely kinetic energy of the nucleus this is the kinetic energy of the electron and very soon we are going to write all this in atomic units okay so now what we will do is we will go in steps we will add one more nucleus one more proton actually and we will discuss the case of H2 plus ion now when we do that all these three terms will remain of course some more terms will coming which more terms will come first of all we will have a kinetic energy term for this nucleus B we will have electrostatic attraction between this electron and the nucleus B so what we will have to write is we have to write something like minus q e square by r b for the potential energy term and for the additional kinetic energy term of the nucleus we have to write something like minus h cross square by 2 m b del s del b square of course do not forget m a and m b are one and the same they are just protons an additional term will come and that is due to nucleus nucleus proton proton repulsion that will be plus q e square divided by capital R where capital R is the inter nucleus separation right so this is the kinetic energy term for the second nucleus attraction potential energy for attraction between the electron and the second nucleus and this is the term for nucleus nucleus repulsion now what happens if you want to go to H2 molecule because remember balance bond theory has no scope for H2 plus we have to talk about H2 so we add one more electron right at the moment we add one more electron we add many terms how many more terms do we add whatever is there will remain and moreover since there is second electron we have to use a little different nomenclature so what we should do is that instead of del e square we have to write del 1 square where 1 is the well index of the first electron and there will be another term minus h cross square by 2 m e del 2 square that is for the second electron then the second electron will also feel attraction with h a and with h b so now the distances will also have to be rewritten so this separation between the first electron and nucleus a will write it as r 1 a separation between first electron and nucleus b will be r 1 b between second nucleus second electron and nucleus a will be r 2 a and that between the second electron and nucleus b will be r 2 b and accordingly we will get how many more terms well we will get something like minus q e square divided by r 2 a minus q e square divided by r 2 b this is due to attraction of this electron by these two nuclei nucleus nucleus repulsion is accounted for and the other term that you have to bring in is well of course kinetic energy of the second electron minus h cross square by 2 m e del 2 square and finally the additional term is electron-electron repulsion electron-electron separation is r 1 2 so that will be plus q e square by r 1 2 so I have shown you the additional terms that come in with respect to h 2 plus Hamiltonian in highlights great. Now of course this is a very complicated situation and we need some kind of an approximation before we can go ahead and fortunately we have in our hands something called Born-Oppenheimer approximation which essentially says that the total energy is a sum of all these energies so wave functions are all products of Hamiltonians and everything is separable. So the statement of Born-Oppenheimer approximation or the consequence of Born-Oppenheimer approximation that we will use here is this when we talk about the electron there is no need to worry about the nucleus nucleus is much more bulky it takes much longer its movement is in a different time scale altogether. So when we do electronic calculations we might as well take the nucleus to be stationary this is extremely important so it is like the something that is very light that moves fast something that is extremely heavy moves slower when we talk about the motion of the light particle we do not have to worry about the motion of the heavy particle that is all. So when we use this consequence of Born-Oppenheimer approximation how does this expression for the Hamiltonian simplify and to keep things very simple I am working with the Hamiltonian for H2 plus for H2 we will just add those additional terms. So first of all this del A square will you agree with me that this becomes 0 will you agree with me that del B square term becomes 0 because we are considering the nucleus to be stationary so there is no question of their kinetic energy. So these two we can ignore under Born-Oppenheimer approximation what else if these are stationary then this inter nuclear separation will remain constant for our calculation. So R can be taken to be a constant and we can perform the electronic calculation for a particular value of R see what the energy is then go back and do it for another particular value of R see what the energy is and we can plot energy as a function of R and see whether there is a minimum if there is a minimum then that is the happiest situation and that separation R is your equilibrium Born length. So it becomes very very simple all you have to work with now for H2 plus is this minus h cross square by 2 Me del A square minus Q square by R A minus Q square by R B. So life is simpler than what it is and in fact you can using elliptical polar coordinates you can actually solve it and get wave functions but we will not do it because again they will become useless the moment you go to H2. So what we do is we actually use linear combination of atomic orbitals as we will discuss later but today let us think how we handle this H2 problem hydrogen molecule dihydrogen molecule problem by this valence bond theory. So we start with the Hamiltonian for H2 molecule more terms as we have said earlier I am going a little fast here because see it is not all the difficult and we have discussed this many times if there is a problem please go back and just do iterations I think you will be fine. So this is our Hamiltonian and we have said what these different kinetic energy and potential energy terms are everything is accounted for kinetic energy of the electrons. We ignore the kinetic energy of the nuclei by bond open hammer approximation we consider the electron electron repulsion we consider electron nucleus attractions 4 of them and we consider electron nucleus repulsion keeping in mind that we can keep R to be a constant. So this is the Hamiltonian and if you write Schrodinger equation using this Hamiltonian this is what we get if you write in shortcut. Our purpose is to get an idea of wave function or purpose is to make an estimate of the energy and energy of course as we said a little while earlier the energy is going to be a function of R inter nuclear separation remember that. So the complicating factor here is that this R12 inter nuclear separation is not a constant you might remember from our discussion of many electron atom that whenever you have more than one electron in the system they move deviously they like to avoid each other but it is not as if their separation is always the same it is not. So we cannot really solve this directly so we need some simple some approximate solutions and we know by now how to use approximation methods. So we will start with a trial solution and see whether it makes sense. So let us think what kind of wave functions we can think of for H2 and while doing that we are going to use the atomic orbitals that we are so familiar with to try and construct the wave functions of the molecule molecular wave function. So and of course we will do that using common sense by and large so let us think. What happens when R equal to infinity now when I say infinity I do not mean infinity infinity a certain angstrom is sufficient to qualify as infinity for this atomic and molecular systems because if you think of the potential energy between these two atoms that becomes 0 asymptotically but becomes close to 0 within say 10 angstroms of inter nucleus separation. So at such separations which we can say roughly is infinity for all practical purposes electron number 1 moves in the field of nucleus a. Electron number 2 moves in the field of nucleus b. So electron number 1 would reside in the 1s orbital of nucleus a. Electron number 2 would reside in the 1s orbital of electron b. So what would the wave function be? Wave function would be the product of psi a 1 and psi b 2 psi a 1 means electron number 1 is in the 1s orbital of electron a sorry of nuclear of atom a psi b 2 means electron number 2 is in the 1s orbital of atom b. So a and b are the names of atoms 1 and 2 are the indices for electrons. So this is what it is when they are very far apart electron number 1 is happily blissfully ignorant about the presence of this nucleus b and electron number 2 is happily ignorant of the presence of nucleus a. Now what happens when they come close together? Let us say when the inter nucleus separation is close to equilibrium bond length does not even have to be exactly equal to equilibrium bond length close enough so that this electron experienced that nucleus and the other way around. Then it is not sufficient to write the wave function as psi a 1 psi b 2 that is only one of the components of the wave function we write it as psi 1. The other possibility is that electron number 2 might reside on in the 1s orbital of nucleus of atom b electron number 1 might reside in the 1s orbital of atom a. So that is the second possibility and that would be the second term psi 2 that would contribute to the wave function psi a 2 psi b 1. What is the wave function then? The wave function would be capital psi is equal to c 1 psi 1 plus c 2 psi 2. We are going to evaluate c 1 and c 2 maybe in the next class but we can actually make a sort of a guess that what c 1 and c 2 could be if you do not worry about normalization of psi. Can you think what I expect the ratio of c 1 and c 2 to b or rather what do I expect the ratio of the magnitude of c 1 and magnitude of c 2 to b? Please think about it we are going to come to that eventually right. So this is my wave function to start with. This wave function was proposed by Heitler and London and so this is called Heitler London wave function. Using this wave function we can write down Schrodinger equation H psi equal to E psi and then the problem is to evaluate the coefficients and evaluate the energy. I hope you remember what we do when we have problems like that. Let me write Schrodinger equation I just write it in terms of H for now. H operates on c 1 psi 1 plus c 2 psi 2 to give me something I will just call it E for now c 1 psi 1 plus c 2 psi 2. Well we have done similar things earlier. So what we do is we left multiply by this psi 1 and integrate over all space. So what do I get then integral psi 1 H psi 1 well and c 1 comes out plus c 2 integral psi 1 H psi 2 is equal to c 1 E integral psi 1 psi 2 plus c 2 E. I am taking E to be a constant on the right hand side because E is the eigenvalue even though we cannot solve it and find it well it is there. So we just take it out integral. Oh what did I do sorry sorry first one is wrong distracted a little bit. So c 1 E psi 1 integral psi 1 psi 1 over all space remember I am left multiplying by psi 1 and integrating over all space all are in this case because we are talking about 1 is orbitals c 2 E psi 1 psi 2. Now first thing is this is not equal to 0. This is very important to understand why is this not equal to 0 because see first of all your these are products of wave functions. And secondly we are talking about 1 1 s orbital of this atom and 1 1 s orbital of this atom. These 2 1 s orbitals are not mutually orthogonal. This is a concept where sometimes we get confused 2 orbitals or 2 wave functions of the same system if they are not the same of course they have to be orthogonal to each other but of different systems they are not orthogonal. In fact as we will see we are going to get some new quantity out of all this. So this is what you get and similarly if you left in left multiply by psi 2 what do you get? You get something like c 1 integral psi 2 H psi 1 plus c 2 integral psi 2 H psi 2 equal to c 1 E integral I just write psi 1 psi 2 because the order does not matter plus c 2 E integral psi 2 you are going to evaluate all these. For now what we do is we call this H 1 1 we call this H 1 2. We have encountered these earlier remember we call this H 2 1 and we call this H 2 2. And with this we can rearrange the equation also we can write something like c 1 we will write here multiplied by H 1 1 minus E s 1 1 I will call this s 1 1 we have encountered these do not this is s 1 2 plus c 2 multiplied by H 1 2 minus E s 1 2 and this way we get another equation also in c 1 and c 2. So system of linear equations we will get the secular determinant and for a non-trivial solution that secular determinant has to be equal to 0. So now let me show you that secular determinant here is the secular equation. You remember what we had written a little while ago we got this H 1 1 minus E s 1 as a coefficient of c 1 H 1 2 minus E s 2 as coefficient of c 2 in one the first equation in the second one this was a coefficient of c 1 this was a coefficient of c 2. So of course this is your secular equation. Now we have to evaluate these integrals one by one to do that first thing we will remember is that we are using normalized atomic orbitals and since we are using normalized atomic orbitals s 1 1 equal to s 2 2 has to be equal to 1 is that correct that is right what is s 1 1 s 1 1 is equal to integral psi 1 multiplied by psi 1 over all values of r. So that will turn out to be integral psi a 1 psi v 2 and the same thing. Now I hope with all our previous discussions of atoms it is not very difficult for us to see that this is really a double integral 1 in coordinates of electron number 1 1 in terms of coordinates of electron number 2 and I can write it conveniently as a product of 2 single integrals 1 in terms of 1 1 in terms of 2. So I get psi a 1 psi a 1 multiplied by psi v 2 psi v 2 very simple. And as these 1 s orbitals themselves are normalized we get that these are this is equal to 1 this is also equal to 1 not very difficult to understand in fact I will just erase this. So one thing is out of the way s 1 1 and s 2 2 are both conveniently equal to 1. What is s 1 2? S 1 2 is integral psi 1 psi 2 over all values of r. So I write down the expressions for psi 1 and psi 2 psi a 1 psi v 2 multiplied by psi a 2 psi v 1 integrated over all values of r and remember this also has to be a double integral because there are two kinds of coordinates 1 for 1 1 for 2. This again factorizes into an integral in electron number 1 integral in electron number 2 psi a 1 psi v 1 multiplied by psi a 2 psi v 2. Now what is the meaning of psi a 1? Psi a 1 is psi a 1 essentially means electron number 1 in the 1 s orbital of atom a psi a psi v 1 means same electron number 1 but in a different 1 s orbital different atom b. So this integral is not equal to 0 psi a and psi b are not orthogonal to each other because they are orbitals of two different atoms a and b this is something that we have discussed earlier. So what we say is and of course there is no difference between the two the indices are different but that is all. So we call this s we call this s and we call this s square and it has a name s is called an overlap integral. Why is it called an overlap integral? Because if I just draw these 1 s orbitals for you hopefully I do not have anything here so it is okay to draw here. Let us say this is where a is its 1 s orbital is an exponential decay and since we have drawn 3 d plots in the past I hope it is okay if I draw on this side also sorry that it has gone below let us see if I can correct this. So this is your psi a and let us say your a nucleus b is here for that we will use a different color this is the 1 s orbital 1 s orbital is an exponential decay remember. Now think what is the value of this integral? The value of the integral at this point is of course 0 because it is a product of this wave function and this wave function. Here also it is practically 0 the only region in which this integral has non-zero values is this the region in which both one psi a and psi b have reasonably non-zero values well values are actually if you want to be very very rigorous they are non-zero everywhere because they become 0 only asymptotically. But here what we are saying is practically I mean if it is 0.0001 that we take that to be 0. So only in this region where both the orbitals have more or less good non-zero values only in that region this integral will have non-zero value. So this is a region of overlap that is why the name is overlap integral okay and it is not very difficult to see from what we discussed just now that overlap integral is going to have dependence on r is going to be 0 at very large separation and for 1 s orbital it is going to be maximum at separation of 0 capital R equal to 0 right. So using elliptical coordinates one can work out will not work out it is given as a problem in Macquarie's book you can work out an expression and the expression turns out to be an exponential decay in R, R here is inter-nuclear separation multiplied by a polynomial in R okay and when we plot this this is the kind of plot that we get and this is what we had expected qualitatively also becomes 0 asymptotically asymptotically but as you come closer and closer and closer it becomes maximum okay this is the variation of overlap integral as a function of inter-nuclear separation in the assignments we are going to have several such exercises where we will give you different kinds of orbitals and ask you to work out how overlap integral varies as a function of inter-nuclear separation and you will see you will not get this kind of a variation always okay but that you work out in the assignments. Now we are back to what we are talking about we have simplified our secular equation a little bit this is how it was remember h11 minus es11 h12 minus es12 and so on and so forth we have sort of taken care of the ss we have said this s11 equal to s22 equal to 1. So this term will become h11 minus e this term will become h22 minus e and we have said that s12 is equal to s square so this will become h12 minus es square this one will become h12 minus es12 sorry es square again now here h12 and h21 are the same as we have said earlier remember you can use turnover rule so we are writing h12 and not h21 okay so here it is we have got this concept overlap integral we have introduced it and we have got this secular equation which is a little simpler than what it was earlier okay we have made some advance. Next we want to evaluate the h11 integral this is where we will start from in the next class.