 This is the last lecture of the business end of this course. We are going to talk about wave functions of butadiene, pi wave functions of butadiene. So, we are developing a Huckel-MOT treatment for butadiene. We have expressed the pi wave functions pi molecular orbitals as a linear sum of the Pz orbitals of the 4 participating carbon atoms and we have written the secular equation. In this secular equation, we have realized that since the carbon atoms are equivalent, Hii has to be equal to Hjj that has been equal to Coulomb integral and for all practical purposes all measurement will be with reference to this Coulomb integral. So, this is really our 0 for measurement. If you remember what we discussed in the last lecture, alpha is really the energy of a Pz orbital in the sigma framework of the molecule. So, whatever stabilization, destabilization takes place has to be done with from with respect to that. So, it will always be alpha plus something alpha minus something and overlap integral for our purpose we set it to 0. It is not set to 0 for all pairs of atoms in a little more advanced Huckel treatment, but today well we are at the end of the course we are not going to start a new discussion now. So, we are going to set it to be equal to 0 and the justification for setting s equal to 0 or nearly equal to 0 is that we are talking about pi interaction. So, overlap is poor overlap integral is going to be small. So, when we apply all these in this secular determinant it simplifies and we simplified further by substituting x equal to alpha minus e by beta. And we say that x then is essentially negative of energy in terms of beta in units of beta. We find out what x is and hence we have obtained what the energy levels are. The next thing we want to do is we are going to find well we are going to write down we will not find you know how to find it right we did it for ethylene we know that we can go back to the equations for the coefficients and substitute these values of alpha beta everywhere in the integrals and you can determine C1, C2, C3, C4 keeping in mind additionally that each of these molecular orbitals have to be normalized and each of the participating atomic orbitals are of course normalized. So, we will not do it I encourage you to do at least one by yourself but here I will just show you the results and I will write here E1, E2, E3, E4 to remind you that as you go higher up we are we are calling the lowest energy level to be E1 and do not forget that plus well 1.61804 beta really means a negative energy because beta itself is negative. So, these are the coefficients that one obtains if one does the treatment that we just discussed what we will do is we like to draw pictures do not we? So, we will try and draw pictures cartoons qualitative pictures well semi-quantitative pictures you can say of these molecular orbitals. So, what we see is that look at the first one there are two things to see first of all all signs are plus and second thing is the coefficients are not equal. So, there is a smaller contribution from chi 1 greater contribution from chi 2 equally greater contribution from chi 3 smaller contribution from chi 4. So, what will this M1 look like to do that what we will do is we are going to draw these p orbitals showing the signs of wave functions in each lobe. So, we are going to and I am not going to tell you which is which we are going to draw one lobe with solid color one lobe with white and you can decide which one is plus which one is minus as you know this plus minus sign of wave function is just relative. But in the first wave function first is that on top of this x y plane in direction of plus z all the lobes have the same wave function of same sign the lobes below have opposite sign to that but same sign among themselves and I have also drawn the heights of these orbitals proportional sort of proportional to the square of the coefficients or coefficients coefficients actually 0.3717 is smaller number than 0.6015 that is why chi 1 looks small here chi 2 looks bigger and coefficients of chi 2 and chi 3 are same. So, they are of the same height coefficient of chi 4 is the same as chi 1. So, chi 1 and chi 4 they are of the same height but smaller than chi 2 and chi 3. What will happen if I want to draw a similar diagram before that at the risk of sounding repetitive the solid lobe is it plus or minus I do not know and I do not care you decide can be plus can be minus as long as you stick to the same convention throughout. The crux of the matter is that top of the on top of the x y plane the wave functions are all wave functions at the same sign below the plane they also have the same sign but this sign is opposite to the sign of the wave functions above the x y plane. What about chi 2 0.6015 chi 1 0.3717 chi 2. So, now it is a turn of chi 1 to be larger chi 2 to be smaller beside there is a sign change between 2 and 3 minus 0.3717 chi 3 minus 0.6015 chi 4. So, this is the diagram I hope it is not very difficult to understand and we will come back to this sign change business what about chi 3 0.6015 chi 1 minus 0.3717 chi 2. So, sign change already minus 0.3717 chi 3. So, no sign change between 2 and 3 but another sign change between 3 and 4 and change in sign as well as size as well this is what chi 3 would look like. What is chi 4 plus 0.3717 minus 0.6015 plus 0.6015 minus 0.3717 this is what it will be like. So, now one thing that is very useful is to see where the nodes are of course in the first one except for that x y plane which is already there for p orbitals there is no new node that comes in as a result of formation of the MOs. For the second one chi 2 however, you see between 2 and 3 there is a sign change. So, you have a nodal plane right here I am drawing a dashed line where it is really a nodal plane. What about chi 3 between 1 and 2 there is a sign change between 2 and 3 there is no sign change between 3 and 4 there is a sign change in the opposite direction. So, one node here another node here what about chi 4 sign change between 1 and 2 and between 2 and 3 and between 3 and 4. So, how many nodes are there 1 2 3. So, if you remember the energies energy of chi 4 is highest energy of chi 1 is lowest. So, this is in line with something that we have said several times more the number of nodes higher is the energy right. So, that is what we learn first thing that we learn from the wave functions that it is in keeping with this nice model that we built. Now, we learn to determine 2 very important quantities 2 very important properties of the molecule first is charge distribution as you know charge distribution may or may not be uniform. If you take H 2 charge is distributed uniformly if you take H f that is not the case. So, now we learn a little more quantitative way of determining charge distribution from coefficients of atomic orbitals and this is something that is done at in higher level quantum chemical calculations as well using softwares like Gaussian or Gamma or something like that. So, first thing that I want to say here is sum over n equal to 1 by 4 C i n square is equal to 1 where i is a molecular orbital. So, I would be 1 2 3 and 4 and n is carbon atom number this also is 1 2 3 and 4. But do not get confused between this set of 1 2 3 4 and this set of 1 2 3 4 I should have perhaps written Roman numeral or something or Dev Nagari numeral to differentiate between the 2, but I have not. So, please bear with me all this treatment is available in Macquarie's quantum chemistry book as usual. So, C i n what is that C i n is the coefficient of the ith molecular orbital in the nth carbon atom. So, for example, this minus 0.3717 is the coefficient of molecular orbital number 3 very bad example in carbon atom number 3. So, it is C 3 3 let us take this one that is better. Minus 0.6015 is the coefficient of molecular orbital number 4 in carbon atom number 2. So, it is going to be C 4 2. I hope you have understood. Now, the sum over i for sum over n C i n square that has to be equal to 1 is not it because this sum of squares is like this over the atoms and if you remember that this chi 1 and chi 2 this chi 1 chi 1 is there is no chi 2 when you sum like this this chi 1 is normalized. So, what happens when you take the squares just square and add well you get something like this the sum of squares turns out to be 1 just do it yourself 2 into 0.3717 square plus 2 into 0.6015 square will turn out to be 0.999 something 1 great. So, what does that have to do with charge distribution? Well the total pi electronic charge on the nth carbon atom is q n is equal to sum over i equal to 1 to 4 n i C i n square. Now, what did I just say? Let us see n i is the number of electrons in the ith MO what is the number of electron in chi 1 2 what is the number of electrons in chi 2 2 what is the number of electrons in chi 3 and chi 4 0. So, this n i is 2 for chi 1 2 for chi 2 0 for chi 3 0 for chi 4. Now, you have to multiply that number by C i n square C i n square remember i ith the coefficient of ith molecular orbital in nth carbon atom or the other way around whatever you want. So, this is the if you take n i C i n square what is it? C i n square is the contribution population you can say and n i is the number of electrons. So, multiply them together you get the contribution of charge for that MO in this atom and you sum over all sum over all the electrons you get the total charge distribution on the nth atom. So, let us do that for one for all values of n take any value of n say take n equal to 2 for example second carbon atom what I am asking you to do is 2 multiplied by 6015 square plus 2 multiplied by 0.3717 square that turns out to be 1 do it for any carbon atom it will turn out to be 1. So, what that tells us is that there is uniform distribution of pi electrons over the 4 carbon atoms there is no polarization. There is a first thing that we had said from valence bond theory and from common sense but this is something that we arrive at using the charge distribution calculated from the coefficients of atomic orbitals. The second thing that we want to talk about is bond order see C i r multiplied by C i s remember C i r is a contribution of the ith MO in the rth atom C i s is the contribution of the ith MO in the sth atom when you take a product of this coefficients that gives us the pi electron charge in the ith MO between adjacent atoms r and s. So, if you want to find out pi bond order between adjacent atoms r and s they are adjacent right 1 and 2 2 and 3 3 and 4 that is all then this will write this bond order to be P r s pi equal to sum over i equal to 1 to 4 and summing over i that is summing over all the MOs n i multiplied by C i r C i s where n i is the number of electrons in the ith MO. So, if I want to find say P 1 2 pi what will it be what is C i r in that case what is the first C i r C 1 1 then what is C i s C 1 2 so that multiplied by 2 because we said that psi 1 has a population of 2 electrons are there in psi 1. So, 2 into C 1 1 plus 2 into C 1 2 then again psi 2 also has 2 electrons so 2 into C 2 1 into C 2 2 we do not go any further because there is no need of considering psi 3 and psi 4 because they are not even populated. So, what is C 1 1 0.3717 what is C 1 2 0.6015 what is C 2 1 0.6015 what is C 2 2 0.3717. So, you multiply them together multiplied by 2 and this is what you get P 1 2 pi turns out to be 0.8942. So, the pi bond order remember pi bond order between 1 and 2 is 0.8942 remember there is already a sigma bond. So, overall bond order will be 1.8942. So, what do you expect it to be in a valence bond picture you draw in some resonating structures you draw double bond between 1 and 2 in some resonating structures you draw double bond between 2 and 3. So, not in all the resonating structures do you have a double bond between 1 and 2 that is why the overall bond order is less than 2 or if you want to talk only about overall pi bond order that is less than 1 and that is something that comes out very nicely P 1 2 pi turns out to be 0.8942. Similarly, let us find out what is P 2 3 pi. P 2 3 pi would be C 1 2 C 1 3 we want to find the bond order between atom 1 and atom 2 but let us not forget that only psi 1 and psi 2 are populated. So, we have to worry about coefficients of psi 1 well we have coefficient we have to worry about the coefficient of the first and the second demos in these atoms. So, 2 into C 1 to C 1 3 plus 2 into C 2 to C 2 3 that turns out to be 0.4473 please work it out yourself. What does that mean? That means that pi bond order of well let us finish the story it is not very difficult to understand that P 3 4 pi has to be equal to P 1 2 pi because of symmetry right there are i integral if I just flip the molecule then 1 and 4 will change positions. So, and 2 and 3 will also change positions. So, what we see is that for these bonds 1 3 1 2 and 3 4 pi bond order is 0.8942 a little less than 1 and for 2 3 pi bond order is little less than half. Does this agree with the valence bond pictures that you drew please draw the resonating structures and convince yourself that this pi bond order is actually in agreement with what you get from valence bond theory using resonance as well. So, now we have discussed small molecules right we have discussed ethylene we have discussed butadiene. So, in McFarie's book Benzin is not worked out but the solution is provided I have not written it down because you might have instead of reading from the slides you read it from the book. But if you want to handle bigger molecules you do not want to stop you have such a beautiful easy theory semi empirical theory of course we do not want to stop at butadiene we want to go to big molecules we want to talk about say naphthalene the problem is in case of even naphthalene the secular determinant is daunting it is a 10 what is the order of the determinant 10. So, who is going to solve this it becomes a very very tedious problem that is where see remember we have talked about symmetry earlier we have said by symmetry coefficients must be either equal to well c 1 has to be equal to plus minus 2 we have said by symmetry this pi bond order of 1 3 has to be the same as that of 1 2 has to be the same as that of 3 4. So, symmetry actually has a big role to play in simplifying problems of quantum mechanics. So, one can use symmetry to factorize this determinant break this determinant down into smaller ones and from there very elegantly you can find out the expressions you see this is an expression for one of the molecular orbitals of naphthalene and one can work out all the energies from this equal approximation without having to use a computer actually using this symmetry factorization this is discussed nicely in cotton's book on chemical applications of wood theory among other books in fact all this whatever I have written here the diagram everything is copied from cotton's book. So, we need to learn symmetry before we can talk about larger molecules but we do not have time to do this I will give you some pointer in the concluding lecture which is to follow but for now this is the end of our discussion of a Huckel treatment of conjugated bimolecular systems.