 Now we are going to discuss the wave functions of butadiene and we will see how we can arrive at some important information about the molecule from a knowledge of these wave functions. What you see here is an example of one of the wave functions that we are going to encounter in a few minutes. But before that this is where we are we have worked out the Huckel-MOT formulation for butadiene. We have expressed the pi molecular orbital as a linear sum of these 4 p orbitals on the 4 carbon atoms. And in the secular equation what we have done is we have said that equivalent carbon atoms have H i i equal to H j j equal to alpha and we have said that alpha to 0 and we make all the measurements from there because essentially that denotes the energy of a P z electron in the molecular framework. We have said that this H i j and H j i they are equal to resonance integral when only when i and j are adjacent to each other. So 1 and 2, 2 and 3, 3 and 4 no other combination. For all other combinations we set them to be equal to 0 and we have discussed why we are justified in doing that. And the overlap is set to 0 in all cases because we are talking about a pi overlap anyway which is not all that strong it is fairly weak. Alright and then we have simplified a little bit and we have written the determinant in terms of beta x is essentially energy in terms of beta setting alpha to be equal to 0 that is how we can read it and we have found some values of x from there we constructed the energy diagram. Once we know this energy diagram from our knowledge of ethylene we know that we can now plug this expression for energy back in the 4 linear equations that we had for the molecular orbitals and we can determine the coefficients. We can but we will not do here because it is a little tedious whoever is interested is welcome to give it a try. It will take a little time but you can do it now that you know ethylene anyway. I am just going to show you the results. These are the results these are the energy levels E1 E2 E3 E4 actually in hindsight I should have written them E Roman 1 Roman 2 Roman 3 Roman 4 because I have denoted the wave functions by Psi 1 Psi 2 Psi 3 Psi 4 where 1 2 3 4 are in Roman numerals. This is just to differentiate the label for the molecular orbitals from the label of the atomic orbitals that participate in the linear combination or you can think labels of the atoms themselves. So, these are the coefficients that come out and once you look at the coefficient you see this nice symmetry that was there in the secular determinant sort of reflected here as well. Mod of the coefficient coefficient is just 0.3717 or 0.6015 right and they just keep changing places you all always have 2.3717 and 2.6015 as coefficients no matter which molecular orbital you take is just that in the first one all of these coefficients have positive sign in the other three two of them have positive sign whereas two of them have negative sign very nice symmetric permutation combination kind of happy situation. So, now what we will do is we will draw the sort of cartoon represented what we have drawn here is also a cartoon representation a cartoon representation of these MOs. Here you see we have drawn chi 1 chi 2 chi 3 chi 4 as orbitals with the same height because they are all independent p orbitals. Now, we are going to multiply their height by the coefficient. So, now the heights are going to change since chi 1 for example the height of chi 1 will be little more than half of the height of chi 2 height of chi 2 and chi 3 will be same because both have coefficients 0.6015 height of chi 4 is going to be same as height of chi 1 and once again little more than half of the heights of chi 2 and chi 3 right. So, this is your chi 1 that is how we have drawn it of course this is all approximate cartoon depiction but we understand this very nicely. And also another thing that I have done here is since we have minus signs coming up later on I have drawn two lobes in two different colors one of them is plus one of them is minus which one is plus which one is minus I do not know and I do not care as long as we decide that solid ellipses are plus and empty hollow ellipses are minus or the other way round as long as we stick to one convention throughout we are good. Now, let us think of what will happen for chi 2. In chi 2 the magnitudes are going to just get reversed right because now chi 1 is multiplied by 0.6015 chi 2 is multiplied by 0.3717. So, obviously this chi 1 orbital will be bigger chi 2 orbital will be smaller remember when I say chi 1 orbital will be bigger or smaller I mean actually chi 1 multiplied by coefficient one more thing will happen and that is we have 2 minus signs here. So, if I have taken it like this for chi 3 and chi 4 I should have hollow lobes at the top and solid lobes at the bottom. So, this is your chi 2 I hope this is clear not difficult at all what about chi 3 I encourage you to work out chi 3 yourself before going to the next step. Well first of all we have alternate plus and minus signs and magnitude is more or less similar to chi 2, but signs are going to reverse not exactly alternately between 1 and 2 and between 3 and 4 there is no sign change between 2 and 3 this is chi 3 and this is chi 4 there is a sign change after every atom and when I say sign change of course we know that we are talking about nodes. So, let us try and draw some nodes one node is there already the molecular plane itself is a node that comes from the basic properties of the p orbitals anyway. In addition some nodes arise is there any node other than the molecular plane in chi 1 not really no node what about chi 2 do you have a node in chi 2 yes we do this is a node that is why the sign changes what about chi 3 we have 2 nodes here between 1 and 2 and here between 3 and 4 between 2 and 3 there is no node and if you are a little sorry that there is no node between 2 and 3 well our wishes fulfilled in the last one and we have a node between any pair of neighboring carbon atoms and that is why your signs keep on changing every time we have drawn the wave functions this is how it is drawn and once you understand this things like benzene should become cake work. Now, let us see can we get some idea about things like charge distribution can we get some idea about things like bond order from these coefficients from this orbitals. So, first thing to remember is that if I sum over the what am I summing over n is carbon atom right. So, I am summing from left to right if I sum from and here I have made a mistake because n should be from not Roman 1 to Roman 4 because Roman 1 to Roman 4 remember is actually the designator for MO and not atoms. So, this is 1 to 4. So, when we sum over all the atoms then I should get 1 when I sum this square of coefficients why that is the normalization condition is not it because remember is it safe to write here it is safe to write here I think this condition we have made sure that it is always full it is a psi 1 psi integral psi 1 psi 1 equal to 1 what does that mean it means that 0.3717 square I will just write it once integral chi 1 chi 1 before going any further what is this integral this is equal to 1 because chi 1 chi 2 these are all normalized by themselves plus what is the second term second term I can write something like this 0.3717 into 0.6015 multiplied by integral chi 1 chi 2 I am normalizing remember. So, I have to integrate over all space this what is this we have said that the overlap integral is equal to 0 in Huckel treatment. So, this is equal to 0. So, the only things have to that we have to worry about are integral chi 1 chi 1 integral chi 2 chi 2 integral chi 3 chi 3 integral chi 4 chi 4 right. So, that is what it is. So, and these integrals are one anyway. So, essentially what you get is 0.3717 square plus and right here 0.6015 square and again you have 6015 square and 3715 square. So, I can just write 2 multiplied by this this is equal to 1 and you can see that that is actually the case. So, we have written these in normalized form I keep this because this is something that is going to come handy right now. So, that is the first thing that comes from normalization condition ok. Now, let us think about charge distribution what is the meaning of charge distribution. If I think of a particular carbon atom the pi electronic charge on it is given by CIN square multiplied by Ni summed over i equal to 1 to 4 where Ni is the number of electrons in the ith MO and here again I should write i equal to Roman 1 to Roman 4 Ni Ci I N square. What does this actually mean? Well square of coefficient is the contribution of that orbital and Ni is the number of electrons in it. So, what is the number of electrons in psi 1 it is 2 what is the number of electrons in psi 2 it is 2 once again what is the number of electrons in psi 3 and psi 4 0 this is what we have drawn earlier. So, that is what we have to find that will give us the total pi electronic charge on nth atom we can work with the first atom. Let us work out what it is for the first atom yeah what will it be 0.3717 square multiplied by 2 see 0.3717 square multiplied by 2 plus 0.6015 square multiplied by 2 that we know already that that is equal to 1 you can it is just plain arithmetic you can do it. So, it does not matter which n you take does not matter which atom you take 1, 2, 3 or 4 when I say 1, 2, 3 or 4 I mean these these these 1, 2, 3, 4. So, it does not matter which atom you take pi electronic charge is always 1 which means that the charge the electrons are distributed uniformly across the molecule. So, this is a result that we know is correct. So, it is good that we have arrived at that. So, when we drew that valence bond theoretical picture we had some charge separation does that mean that the charge density on electron number carbon 1 number 1 or carbon number 4 is any different not really the average value still comes out to be same. But here we get very elegantly the idea that we have uniform distribution of pi electron a situation that is sort of analogous to dihydrogen H2 not HF. In HF you do not have a uniform distribution of electrons here we have a uniform distribution of pi electrons like the uniform distribution of sigma electrons in dihydrogen that is what we see. Another thing that we can work out is bond order from the coefficients. So, this CIR into CIS gives you the pi electron charge in the ith MO between adjacent atoms R and S. So, if I multiply 0.3717 by 0.6015 then I get an idea of pi electron charge density between 1 and 2. And the bond order is given by this this product multiplied by Ni where Ni is the number of electrons in the ith MO sum over i equal to 1 to 4. So, let us try to do that. What is P12 what do I multiply by 2 C11 C12 0.3717 0.6015 plus 2 C21 C12 what are these 2's these 2's are the number of electrons Ni and these are the coefficients for 1 and 2 that we get for from orbital 1 and orbital 2. So, essentially 0.3717 0.6015 0.6015 0.3717 and the product is added. So, basically that gets multiplied by 4 actually because 1 2 comes from here and there are 2 such terms the answer that you get is 0.8942 pi bond order between 1 and 2 carbon 1 and carbon 2 is 0.8942. In fact, when I did do the calculation I get 8943 what I have gone with the value given in the book. What about 2 and 3 similarly using the coefficients this time which coefficients will use 0.6015 0.6015 0.3717 minus 0.3717 please remember that it is not plus 0.3717 when you multiply a positive quantity by a negative quantity you get a negative quantity. So, you do that 2 C12 C13 plus 2 C22 C23 then you get 0.4473 what is P34 by symmetry it is the same as P12 0.8942. So, what is the total bond order that we get approximately 0.9 plus 0.9 is 1.8 plus 0.4 is a little more than 2 is not it? But that comes because we have done the calculation this way you will not get whole numbers here because remember the stabilization of bonding the stabilization of anti-bonding these are also not exactly the same. So, we get approximately total bond order that is similar to what we get from valence bond theory. But the more important picture that we get here is that the pi bond order between 1 and 2 and 3 and 4 is about double the pi bond order between 2 and 3 and that is what you expect from valence bond theory resonance picture as well because in order to get the double bond between 2 and 3 you need to have charge separation which is not such a happy situation. So, we get a similar picture than what we would have got from valence bond theory the good thing is we get it in a more mathematically rigorous way using Huckel approximation that can easily be extrapolated to bigger molecules like this naphthalene. So, naphthalene the problem is that it is too large a molecule the determinant is huge. So, what we do there is that we use symmetry to factorize the determinant and we get expressions like this this phi 1, phi 2, phi 3 these are the p orbitals and from there one can work out the energy levels and you can work out these are the wave functions anyway. What you see here B3g, Au so on and so forth these are let us for now just say that these are the symmetry levels of this orbitals that are involved there. So, that opens up an entirely new angle altogether what we learned from there is that using symmetry you can simplify quantum mechanical problems to a very great deal. Have we done that in a small way already actually we have right remember what we said we said that P34 is equal to P12 by symmetry. So, this is just the tip of the iceberg it is a hint that symmetry has an important role to play whenever we try to do a quantum mechanical treatment of these big molecules. But let that be the story for another day. Today we have learned that in butadiene using Huckel theory we can work out the energies we can work out the wave functions and from the wave functions we can work out the electron distribution and it turns out that electron is distributed uniformly over the molecule and also we can work out the bond order. When you go to bigger molecules like benzene and all similar treatment is extended.