 We have discussed the valence bond theoretical approach for dihydrogen molecule. Then we have gone on to discuss hybridization which gives us access to polyatomic molecules and their shapes. Once again let me remind you that hybridization is invoked to explain the shape it is not the cause. Methane is tetrahedral not because it uses SP3 hybrid orbitals but because of VACPR you need SP3 hybridization to describe the system that is all. Now that being done it is time for us to move over to molecular orbital theory. We have already given you a very very small executive summary of a comparison between valence bond theory and molecular orbital theory. The advantage of molecular orbital theory over valence bond theory is that delocalization can be handled very easily it is a general theory and excited states are accessible without any hassle. The problem is and this is something that we will encounter not today but in the next class we are going to see that sometimes it becomes a little too general it over does things. The ionic structure of H2 for example is over emphasized in a molecular orbital treatment. But nevertheless MOT is the most popular way of handling electronic structure at this level at least later on you might want to go a little further but I mean this is really the beginning. So we need to discuss and understand molecular orbital theory most of the use application of quantum chemistry quantum mechanics in chemistry involves molecular orbital calculations you can use different levels it is a different issue altogether. But then when we do that let us not forget our definition of orbitals an orbital is a one electron system. So when we talk about molecular orbital what we mean is one electron wave function for a molecular system right there cannot be any other electron. So what is the simplest molecule that has only one electron and that therefore cannot be handled by BBT in the first place that molecule would be H2 plus. And already a few classes ago a few lectures ago we have shown you the Hamiltonian for H2 plus as well as H2. There are more terms as you go from hydrogen atom to H2 plus to H2 the number of terms keeps increasing as number of as the as different interactions come in. Now what we are going to do is we are going to use our good old Born-Oppenheimer approximation and say that nuclear stationary with respect to electrons. So we are going to hold this capital R inter nuclear separation to be constant for the calculation and we are going to royally ignore the kinetic energy terms for the two nuclei. So the Hamiltonian we are going to use really is minus h cross square by 2 Me del square minus Q e square by Ra minus Q e square by RB plus Q square by capital R. I am not using atomic units here we should be equally comfortable with both. So for a change let me use this in the next discussion we have Huckel approximation perhaps we will use atomic units once again. So this is our Hamiltonian and it is a one electron system the difference is that unlike hydrogen atom it is not a central field problem there are two nuclei right two foci. So you cannot use spherical polar coordinates anymore you can use electrical polar coordinates and you can actually solve actually get an exact solution for Schrodinger equation for H2 plus but we will not do it it has been done I mean this published literature it is there in some textbooks but you find that they are there in exotic textbooks you will not find them in the textbooks that we use McPurray or Pillar or even Zabo because there is no need because you put in so much of trouble and use electrical polar coordinates and get some solution you cannot use them you do not know what to do when you bring in one more electron and talk about H2 which is the smallest real molecule. So the thing is this it is better to start using approximation right away since you cannot avoid it one step down the line. So the approximation that we are going to use is something we have learned while talking about variation method remember we had said that since in any case variation method allows us to change parameters and try and minimize energy with respect to these parameters how does it matter what kind of functions we use you might remember that elephant on the opening slide of a couple of the lectures that we had had some time ago those elephants were drawn using some mathematical function. So any wave function whatever is a real wave function even though we do not want to do it exactly we can use some of Gaussian or Lorenzian or exponential polynomial sums and products of these functions and we can actually simulate these actual the correct shape and that is good enough. So if we take an appropriate set of orthonormal functions orthonormality using orthonormal set brings in some advantages as we have learned in our discussion of many electron atoms and here the biggest advantage is that we already have a set of relevant orthonormal functions at our disposal and they are the atomic orbitals. So what we do is we generate the molecular orbitals by linear combination of atomic orbitals this is the first approximation of course later on atomic orbitals are not used and molecular orbitals are used by using just what I said products and sums of different kinds of functions but for this course we will stick to linear combination of atomic orbitals because they give us a nice insight of what is going on and it can give us fairly good estimates of energies. So LCAO as it is called is what is going to be used. For H2 plus the only atomic orbitals that we need to consider are the 1s orbitals because where is the electron? The electron will be in the lowest energy state and there is an energy gap between 1s and 2s. So this is something where implicitly we have used an important principle that is going to come back and get used many times in our discussion and that principle is that only orbitals of appropriate energies are going to participate in the linear combination and later on we are going to say only orbitals that have the correct symmetries will participate in linear combination correct symmetry and comparable energies without that they do not participate in this linear combination a little bit of difference in energy is okay but here the difference would be 1s and 2s so no cannot cannot makes no sense in bringing in 2s you might get a little bit of improvement if you do but at our level in this course we are good with this. So the LCAO MO we use is C1 1s a plus C2 1s b and I hope this reminds you of the wave function we had in the valence bond approach. There also we had used the 2 1s orbitals okay but the difference there was that we started with filled orbitals right we started with phi 1s a 1 multiplied by phi 1s b 2 then to that we added phi 1s b 1 phi 1s b 2 and then we proceeded here we do not bother about the electron to start with we are more interested in constructing the wave function and then we will fill in the electron when it is appropriate time right but let us work with the orbit with this molecular orbital to start with C1 1s a plus C2 1s b so it takes square of that this will give you probability density that is C1 square phi 1s a square plus C2 square phi 1s b square plus 2 C1 C2 phi 1s a phi 1s b. Now see is there any reason why in these 2 terms C1 square should be different from C2 square we had used exactly similar argument for valence bond theory also there is no reason why any of these 2 orbitals will make a different contribution so it makes sense to say that C1 square equal to C2 square or C1 equal to plus minus C2 okay so that we can write when C1 square equal to C2 square when C1 equal to C2 we write C1 equal to C2 equal to C a and when C1 is equal to minus C2 then we equate it to C b of course this is a bad choice well I should not say of course actually this is a bad choice because it is better to write b here and a here for reasons you will see very soon all right so what is the wave function then the wave function the first wave function where we use a linear combination with plus sign is C a multiplied by phi 1s a plus phi 1s b so in the subsequent discussion most of the time we have not written phi we have simply written 1s a plus 1s b what about psi 2 psi 2 would be C b into 1s a minus 1s b and I think most of you would know already that this plus combination is the bonding orbital this minus combination is the anti-bonding orbital why bonding why anti-bonding will come to it in its time but now this is what we have generally we like to draw pictures instead of writing algebraic expressions and this is a picture that we generally draw we draw two circles for the two s orbitals and we would put two plus signs to denote that the combination is plus and for anti-bonding we put a minus sign in one of them to denote that the combination is minus and in fact it is better for us since we know about contour diagrams and all we can look at the contours and understand better these are the overlapping contours of a and b orbitals here the only difference is that you have to put in the plus and minus sign and you can draw this psi square like this as dots remember we put a dot wherever there is some probability so we can generate the diagram of psi psi star psi square it will be like this no node here here there will be a node because in this region there is going to be a destructive interference so if you look at the profiles take a cross section of this orbital you will get two peaks for psi plus the bonding orbital here however 1s orbital points up the other 1s orbital points down this plus is minus remember so you are going to get a node a zero crossing point exactly between the two nuclei and if you take this psi square then psi square looks like a circus tent in case of the bonding orbital and it looks like two circus tents right beside each other for the anti-bonding orbital so generally what the argument that is given in many textbooks even though it is not it is a little bit of hand waving argument that there is a build up of electron density because psi square is more psi square dr will also be more here build up electron density between the two nuclei a and b and that is why the electron sort of shields the two nuclei from each other and act as a cement for them here actually there is a decrease in electron cloud if you just do two 1s orbitals and if there was no bonding at all then the picture would have been something like this I will draw normalized picture it would have been something like this so in this region you would have this much of electron density more than what you have for the anti-bonding orbital so generally even in class 1112 books this is the explanation that is given so we will not worry about that explanation for the moment just go ahead and see how it pans out to start with let us normalize this wave function we know that this is what we usually like to do first so integral psi 1 square dr in this case will be equal to well will it be dr it will be dr 1 dr 2 right there are is a double integral again that will be equal to 1 so you write this I am going very fast here because this is more or less what we have done in the valence bond theory approach as well so expand this you get 4 terms and when you look at the 4 terms they are familiar integral phi 1s a phi 1s a dr a is equal to 1 because 1s a is normalized same is true for 1sp and what is this integral 1s a 1s b d tau integral 1s b 1s a d tau you might remember from our valence bond discussion that these are overlap integrals remember overlap integrals these are simply overlap integrals so what do you have you have 1 equal to ca square multiplied by 2 plus 2s so ca turns out to be 1 by root over 2 plus 2s 2 plus 2s similarly you can find the expression for cb that is equal to 1 by root over 2 minus 2s I am going at breakneck speed because it is so simple please work this out yourself then only you understand fully good now you might remember that overlap integral is actually a function of inter nuclear separation we have done at least one of these this one we have done in fact for that we had given you an expression and we had shown you the plot it is not very difficult to see that when these 2 orbitals are infinitely far away overlap integral is of course 0 it comes closer overlap will go up up up up and finally it will be something like this what about this is a very curious case overlap integral when this r equal to infinity is of course 0 now even in this case what is the overlap integral is it 0 or is it something else yeah do not forget what overlap integral is you have to do a point by point multiplication of the 2 wave functions and adds so this is 0 so this is really an example of non-bonding interaction here when you have sigma interaction between an s and a p orbital then what will the overlap integral be like 0 here and then when I bring these close together say this is your p orbital minus plus change color and this is your s orbital plus now what happens is it 1 is it 0 what is it for this part is going to be plus for this part will be minus and the magnitude is same so it is going to be 0 at r equal to 0 0 at r equal to infinity somewhere in the middle it is going to peak where will it peak it will peak in the situation where this s orbital is here that is where it is going to peak so for inter nuclear separation here to let us say here okay so in our tutorial we have some problems in which you have to work this out I think you have you might have done it already in our assignment great so this is what we have we have an we have expressions for the bonding and anti-bonding orbitals that is great and now our job is to find the expression for energy expectation value of energy by using these wave functions for bonding and anti-bonding orbitals we will call the energy of the bonding orbital E 1 bonding MO E 1 and we will call the energy of the anti-bonding MO E 2 well it is perfectly okay to call the MO s orbitals they are orbitals but I prefer to call them MO s because we are also using atomic orbitals and by by now we have become used to referring to atomic orbitals as just orbitals okay now let us see what the expression for energy of bonding MO would be and once again I am going fast because this is now cakewalk for us right what is this integral first thing to do is plug in the value of psi 1 and take out the product of the normalization constant this is what we get E 1 is 1 by 2 plus 2 s multiplied by in bra vector we have phi 1 s a plus phi 1 s b in the ket vector we have h hat 1 s a plus 1 s b of course you should expand and you are going to get four terms and the four terms I hope do not look absolutely unfamiliar to us similarly we can get an expression for E 2 and point to note is that if you look at E 1 plus plus plus well of course this is also plus for E 2 the integrals remain the same only in these two cases plus becomes minus there is no other difference okay now let us take a closer look at these energies do we know these integrals have we encountered integrals like this integral phi s a h hat phi s a well phi 1 s a h hat phi 1 s a integral psi i h psi i what is it isn't it h i i integral phi 1 s b h hat phi 1 s b again that is h i i you can call it h j j but they are the same does not matter integral phi 1 s a h hat phi 1 s b what is that that is s i j and integral phi 1 s b h hat phi 1 s a is s j i but then s i and s j are equal because if you remember h hat is a Hermitian and so we can use turnover rule so these are familiar integrals for us we might as well write E 1 in terms of h i i h i j and for now we will write s i j and whenever we are bored of writing i j we are going to go back to s does not matter okay so E 1 turns out to be h i i plus h i j divided by 1 plus s i j you might as well write h 11 plus h 12 does not matter okay perhaps is better to write h 11 plus h 12 what about E 2 in E 2 if you remember the only difference was that the last two terms had minus negative coefficients negative signs so plug that in you get E 2 to be h i i minus h i j divided by 1 minus s i j both become minus okay so now to go further we need to evaluate h i i we need to evaluate s i j we already know what s i j is in fact we already know what h i i h i j are as well we are just revising so this is your Hamiltonian this time not in atomic units minus h cross square by 2 me del square minus q r square by r i you can recognize that these two terms constitute a one electron wave function minus q e square by r j plus q e square by r this is your this is for the potential energy for attraction between electron and nucleus sorry sorry sorry what did I say q e square by capital R is for inter nuclear repulsion nucleus nucleus repulsion so we will club these two together and we are going to write Hamiltonian of one electron so now our Hamiltonian has three terms so h i i is integral phi s i i left multiplying Hamiltonian operating on phi 1 s i okay so we will write like this same expression so no hamdan h i i equal to sum of these three terms now look at this 1 by r is a constant at a fixed inter nuclear distance right we are we are working under the ambit of Born-Oppenheimer approximation here I might digress a little bit and I might say that the Born-Oppenheimer approximation does not hold in all cases but that comes under the beyond approximation theory limit that is a little more complicated phenomenon for our purpose one open hammer approximation is actually very good approximation we can work without its ambit without losing general like generality okay so this is a constant at fixed inter nuclear distance fine so we can bring it out what did I do sorry so I can write like this integral phi 1 s i h 1 electron phi 1 s i we will come back to this plus q e square integral phi 1 s i phi 1 s i minus q a square q e square integral phi 1 s i 1 by r j phi 1 s j again I hope that these terms do not look unfamiliar to you because first of all phi 1 s i phi 1 s i integral of that what is that that is simply one it is normalized integral of phi 1 s i into 1 by r j phi 1 s i what is that is I hope you remember that it is the Coulomb integral j right remember Coulomb integral Coulomb integral denotes the electrostatic interaction right we had worked this out when we talked about dihydrogen molecule in your valence bond theory and I hope it is not very difficult to see that integral phi 1 s i h hat 1 electron phi 1 s a what will that be h hat that will be the expectation value of energy of 1 s orbital it does not matter whether I use a or b this energy of 1 s orbital is still the same so energy of 1 s orbital plus q e square by r minus q e square into j okay now let us evaluate the integral h i j there is not enough to evaluate h i i right I want energy finally so h i j similarly we can proceed and we can again use this 1 by r to be constant h i j will be first of all this is the value of energy for your 1 s orbital then q square by capital R as we know that this what is this integral of 1 s i 1 s j what is that going to be that is going to be s right overlap integral it will not be 1 and this integral minus q e square integral phi s i 1 by r j phi s j what is that we know that also right so the first term is e 1 s into s let me that let me work that out because I think I made a careless statement just few minutes ago I am working this out what is it h hat 1 electron operating on phi 1 s j gives me e 1 s so if I write it in bracket notation we have in bra vector phi 1 s i so what will I get this becomes e 1 s multiplied by phi 1 s j of course so you take this and plug it in and moreover you take this e 1 s out you are left with e 1 s multiplied by integral phi 1 s i phi 1 s j that is your s e 1 s s next one will be just q e square by r multiplied by s same s is there then if you look at the third integral it is k isn't it exchange integral or I did not tell you this name earlier it is also called resonance integral because here you are considering 1 s i and 1 s j in the same integral right so exchange so please remember we will come back to that later but let us write like this minus q e square into k and remember that k is a purely quantum mechanical concept there is absolutely no classical counterpart for k fine so we know h i i we know h i j we of course know s i j it is equal to s so I have written s i j equal to s here now let me plug in the values of h i i this is a h i i I hope you recognize e 1 s plus I have taken q e square common q e square multiplied by 1 by capital R minus j and h i j is e 1 s plus q e square multiplied by s by r minus k so e 1 will then be equal to if I now collect the coefficients of e 1 s q e square by r and minus q e square so what will be the coefficient of e 1 s e 1 s into 1 plus s isn't it 1 from here s from here what will be the coefficient of q square by capital R q e square by capital R so 1 from here and s from here so 1 by s minus q e square what will be the coefficient you take minus outside from this term you get j from this term you get k minus q square multiplied by j plus k that is e 1 so e 1 then turns out to be just simplify divide throughout by 1 plus s you get e 1 s plus q e square by capital R minus q e square multiplied by j plus k by 1 plus capital S right similarly you can work out an expression for e 2 and that turns out to be e 1 s plus q e square by capital R minus q e square into j minus k divided by 1 minus s so see e 1 s plus q e square what is that that is simply energy of 1 s orbital plus the nucleus repulsion term q e square by r is the potential energy for nucleus repulsion so this is the energy that we get if there is nothing else happening so this is basically destabilization this q e square by r is capital R so when you bring in the third term q square into j plus k divided by 1 plus s or minus q square into j minus k divided by 1 minus s this third term decides whether you get net stabilization or net destabilization so that depends on the integrals and remember we are constructing these integrals using 1 s orbital to start with but that will obviously not give me very good result so what is done is as we have discussed hydrogen type or hydrogen like orbitals stos gto's in 1 s orbital expression that z is kept as a variational parameter because here also you have two nuclei right we do not have a second electron yet but that z instead of z you write zeta variational parameter so that is varied and you can also vary n and using variation method you can find the upper bound for the energies that is how you find the expressions for e 1 and e 2 for given values of capital R another thing to remember is that j k s these are also integrals that depend on the value of capital R okay there is a very strong dependence of the energies and this is what you get for e plus you get a minimum and then it goes up for e minus it goes up all the way so this is how the energy of anti of bonding orbital varies as a function of R this is how the energy of bonding orbital varies as of bonding anti bonding orbital varies as a function of inter nuclear separation at equilibrium bond length you can see that stabilization the amount of stabilization that you have of the bonding orbital is actually less than the amount of destabilization that you have in the anti bonding orbital so if you have one electron in bonding orbital one electron in anti bonding orbital since energy of the system will not be 0 it will be more than 0 right and that is often explained by this hand waving arguments using this kind of orbital picture okay so we have already studied sigma bonding with 1 s orbitals I leave it to you to work out the at least shapes of bonding and anti bonding orbitals using 2 p orbitals this is from one of the textbooks I have forgotten which one but this one is wrong please do not look at that and this is what you get for pi bonding using 2 p orbitals okay just side on overlap and one thing that you should remember or know or understand is that symmetry of orbitals is often used as very important parameters for the hydrogen molecule ion for bonding orbitals it is symmetric with respect to inversion so it is called sigma g g for j ready the anti bonding orbital is anti symmetric with respect to inversion that is called sigma u do not think that always bonding orbitals are j ready and anti bonding orbitals are j ready that is not the case if you look at the bonding pi orbital used by side on overlap of 2 p orbitals you see that this bonding orbital for pi is 1 j ready and anti bonding orbital is actually j ready okay so we are going to use this in the nomenclature of m o s when we talk about actual molecules and using this I think you already know what are the types of bonds sigma pi delta okay these are examples of what kind of bonds are formed using different orbitals again I leave this to you as an assignment in case of any difficulty any question please feel free to contact us or ask in the open session this completes our discussion of h 2 plus little longish module sorry very fast but it is a reputation of what we have done earlier I am sure you will be able to handle it with this knowledge we next discuss h 2 molecule we bring in another electron and use the same orbital perhaps with modification for things like shielding and we see how that compares with the earlier valence bond approach and after that we go on to heteronuclear homonuclear diatomics of second row then heteronuclear diatomics polyatomic molecules finally we will talk about pi molecular systems