 In this lecture we are going to complete our discussion of molecular orbital theory of H2 plus dihydrogen cation. So far we have written down the Hamiltonian and the reason why you are working with H2 plus as we have said several times already is that it is a unique case of a 1 electron molecule. So here we have minus H cross square by 2 Me divided by square that is the kinetic energy term of this lone electron minus Q square by Ra minus Q square by RB these are the potential energy terms for the attraction of this electron by the 2 nuclei and Q square by capital R where capital R is the inter nuclear separation is the term in the Hamiltonian operator for inter nuclear repulsion. So we have formulated Schrodinger equation this way and we have said that we are going to construct the molecular orbital not by solving Schrodinger equation directly even though it is possible we are going to construct it by a rather unique technique by taking a linear combination of atomic orbitals and when we do that this is where we are we made a little bit of progress identified a something called an overlap integral integral 1 SA 1 SB overall space and we have got 2 wave functions psi 1 this is the bonding orbital 1 by root over 2 plus 2 s multiplied by 1 SA plus 1 SB and psi 2 1 by root over 2 minus 2 s 1 SA minus 1 SB we have plotted the electron distribution these are the contour diagrams and these are the profiles that you take through these 2 and from these we have said that it is quite obvious that in this case in the upper case that is where you have a build up of electron density between the 2 nuclei that is going to lead to stabilization and here you have a depletion of electron density if you take square of this between A and B that is going to increase that is going to contribute to increased inter nuclear repulsion so this will be a higher energetic state. So this here is called bonding situation bonding lowers energy and this here is called anti-bonding situation anti-bonding orbital why because there is an increase in energy compared to isolated atoms if there was no change in energy we would have called it non-bonding orbital right. So let me write the expression like this E 1 is equal to integral psi 1 H psi 1 overall space. So if you expand this and again the way to do this is once I write this please stop the video do the expansion yourself it is elementary and then when you reach a point where you get some strange looking quantities then you should turn the video back on okay. So this is what we have we have replaced psi 1 by 1 by root over 2 plus 2s multiplied by 1 SA plus 1 SB in the bra vector as well as the ket vector of course I can take the constants out and I will get 1 by 2 plus 2s outside so that is sorted inside we have this wave function H hat wave function integrated over all space expanded you are going to get 4 terms integral phi 1 SA H hat phi 1 SA plus phi 1 SB H hat phi 1 SB plus phi 1 SA H hat phi 1 SB plus phi 1 SB H hat phi 1 SA and now we have to expand these terms and write them in some way that we can understand. Similarly we can write an expression for E 2 the only difference between E 1 and E 2 is that for the last two terms in E 2 the signs are negative whereas the signs are positive for all terms for E 1 that is the only difference terms are actually all the same all right. So let us try to evaluate this one by one slowly this is where we are right now we have got the wave function we have got an expression for the expectation values of energy for bonding as well as anti-bonding situation. Let us work with the energy of the bonding orbital and let us try to expand this a little bit while doing that since we do not know what these integrals are we are going to define them we are going to give them some name that is the best we can do at the moment because see as I said 1 SA is not really again function of H hat so you cannot just write E 0 and bring it out that is not going to work. So what we say is whenever we have the same orbital before and after in the bra vector and ket vector we call that integral H ii or H jj both are actually the same it does not matter because both are 1s orbitals. When we have say 1 SA in the bra vector and 1 SB in the ket vector or the other way round we call it H ij. Now H ij is equal to H ji that comes from the property of Hermitian operators there is something called turnover rule where if you have something like integral phi 1 H phi 2 you might as well write integral phi 2 H phi 1 does not matter it is all the same for this course we are going to take it axiomatically and the last thing that we know already is the overlap integral that in any case is very easy to understand. So we are going to write the expression for energy in terms of these integrals now what is the first one integral phi 1 SA H hat phi 1 SA that is H a a right or H 11 or H ii whatever we want to call it what about the second one that is also going to be H ii because we have the same 1 SB in the bra vector and ket vector the last two terms are going to be H ij or H 12 or H a b whatever we want to call it because there we have if we have a in the bra vector we have b in the ket vector if you have b in the bra vector we have a in the ket vector. So this is what we get 2 in the numerator and 2 in the denominator cancel each other you are left with this expression H ii plus H ij divided by 1 plus S ij and I hope it is not very difficult to understand that for e2 we will get more or less the same expression except for the fact that instead of plus sign we are going to get minus sign okay please work it out yourself and satisfy yourself that there is a case what is the next task at hand we should try to evaluate H ii and H ij we should try to simplify their expressions in terms of some things that we know. So let us do that to evaluate it we will write down the expression for the Hamiltonian now when I do that I can take these first two terms in bracket isn't it essentially a one electron Hamiltonian so we are going to write it as H hat 1 electron so now our Hamiltonian H hat is equal to a one electron Hamiltonian minus q e square by rj plus q e square by capital R and remember by Born-Oppenheimer approximation this last term is essentially a constant okay. So this is what it is so H ii is equal to integral phi 1 S i H hat phi 1 S i we are going to put this expression for the Hamiltonian in the expression for H ii and obviously we will get three terms in that case we will get a sum of three integrals these are what they are let me just write it like this we will need some space to write more things as well H ii turns out to be integral 1 S i here instead of a and b I am just writing i and j more general coefficient more general notations 1 S i H hat 1 e 1 S i now this is something nice remember this phi 1 S is 1 S orbital and what is the one electron Hamiltonian the 1 S orbital is going to be an eigen function of one electron Hamiltonian so energy of 1 S orbital is actually going to come out so this first one is very simple to work out what about the second one what about the third one we will see even in the second one you can see that 1 by r is constant so that can come out third one that is more interesting okay so this is how we write it now first we take this constant out it becomes q e square divided by r integral phi S i phi S i now we know that that is equal to 1 we are working with normalized S orbitals that is great now what we do not know is the third one first one I already said I do not know why I have not written it yet but it will come also in the first one I think you can understand what will happen energy of 1 S orbital is going to come out inside you are going to be left with integral 1 S and 1 S which is going to be 1 and so this will simply become the first term will simply become energy of a 1 S orbital second term is q square multiplied by well capital S no just q square by capital R no capital S third term is minus q square multiplied by phi 1 S i phi 1 S j divided by R j what is this this is called a an integral j the name is Coulomb integral when you say Coulomb what do you think which field of physics would you think of if I took the name of Coulomb electrostatics right so this Coulomb integral has got to do with electrostatics as well see what do I have in the numerator I have 1 S square phi 1 S square what is phi 1 S square remember that is the energy density right so inside the integral I essentially have phi 1 S square I will write i also and here it is R j so if I draw it like this this is the 1 S orbital I am talking about if this is i and the distance I am talking about is this R j right so what is this if I just multiplied by electronic charge that gives me charge of this electron cloud okay this square is essentially your probability density probability density multiplied by charge gives you charge density so in the numerator if you just multiply by E which is just a constant that is a measure of that is your charge density so we have a charge density at a separation of R j from the second nucleus so this essentially then gives me well one term gives me for a particular value of R j the potential for electrostatic attraction between the second nucleus and the electron cloud around the first nucleus okay similar treatment is encountered if you want to talk about say electrolyte solutions in electrolyte solutions what we do is we take say a positive ion cation and we take the ionic atmosphere to be a delocalized negative charge to keep charge balance so similar treatment is there as well you consider the electrostatic attraction between this point positive charge and this negatively charged cloud ionic atmosphere same thing here and you are integrating over all space so that will give you the total potential energy for electrostatic attraction between one nucleus and the electron charge cloud electron cloud on the near the other nucleus alright so it has a physical meaning j Coulomb integral talks about an electrostatic interaction okay so let us write the expression for HII now the first term is E I E 1 S as we said second term is Q E square by R third term is minus Q E square j okay very nice we got HII already why do we want HII because HII appears in the expression for energies E1 and E2 what is the next integral HIJ HIJ is a little more interesting why because unlike HII you have one S orbital in the graph vector a different S orbital in the ket factor and you might as well start guessing what which quantity which integral we are going to encounter in a similar like that in a situation like this so this is your HIJ again the first one there is no problem we still get energy of 1 S orbital second one once again your 1 by R is a constant it comes out but when 1 by R goes out what do you have inside the integral integral phi 1 S I phi 1 S j I hope why now we all recognize this integral this integral is simply the overlap integral S I J as we are putting it here more often we just call it capital S so what is H I J first term we have no problem ah I do not know why I have written it in so many steps anyway second term is Q E square by R integral phi S I phi S j third term is this integral which is which looks somewhat similar to the Coulomb integral but not exactly the same we have 1 S orbital in the graph vector we have another S orbital in the ket vector remember here R J is not a constant okay so how you evaluate these integrals will come to that okay but let us write this expression the first one what happens is you take the energy of 1 S orbital out no problem with that but inside you are going to have integral phi 1 S I phi 1 S j that is again your overlap integral so first one will be energy of 1 S orbital multiplied by overlap integral that is the difference between H I I and H I J second one also is Q E square by capital R multiplied by S overlap integral ah shows up in H I J it did not show up in H I I or H I J now in the third term again we have an integral which at the moment we do not know what to do with we will give it a name it is since the earlier integral was called J we go alphabetically and we call this one capital K integral phi S I phi S J capital phi S I phi S J divided by R J and there are textbooks in which they try to make sense of it but I do not think it is a good idea this is a purely quantum mechanical quantity it is called an exchange integral or resonance integral so this is what your expression is it is very important to understand it is a purely quantum mechanical concept there is no classical analog so to say that when this electron is here the that charge cloud interacting with the other nucleus and then when they change places that gives extra energy all this is ah trying to extrapolate the classical logic too much into the quantum world you cannot do it beyond an extent so please understand that it is a purely quantum mechanical concept I nobody knows to be honest any classical analog for this so let us not even try okay so we have the expression for H I J we have the expression for H I I what will I do we will now just put the expressions in the expressions for E 1 and E 2 and here I am going to go fast because it is so easy I will just put everything in if you want you please pause and do it yourself please do it yourself then you will understand this is the expression for E 1 this is the expression for E 2 now look at the expression what do we get E 1 S is the energy of an isolated 1 S orbital what is Q square by capital R you see this plus Q is square by capital R appears in both E 1 and E 2 so that is essentially the inter nuclear repulsion between ah A and B so no matter what you do that inter nuclear repulsion term will be there and depending on capital R it will be large or small if it is very far away capital R is large the second term will be close to 0 if it is very close to each other it will be very large but it is going to increase so if I just consider the two terms the first two terms then I can draw like this this here is the energy of the 1 S orbitals both of these energies would increase to some value and this is going to be how much this difference will be Q is square divided by capital R as you understand that this extent of destabilization is going to ah change depending on what is the value of capital R now we have something interesting the third term has a negative sign in both the cases but here in the numerator we have J plus K and in the denominator also we have 1 plus S here in the numerator we are J minus K and in the denominator we have 1 minus S now J K S these are integrals we have already shown how S varies and S is definitely something that depends strongly on capital R J and K can be evaluated not analytically but numerically meaning for different values of capital R you can just put in all the values and by brute force you can calculate what will be with what will that integral be by doing summation it is ah not very difficult to do if you know a little bit of computer programming okay this is where computer programming comes in big time in chemistry all right so this is what you get for E plus or E 1 you have a function that goes through a minimum this is the energy remember of the bonding orbital and here this is the energy of the anti-bonding orbital which increases monotonically when you when you decrease capital R from infinity to a very small values this position position or inter nuclear distance where you get an energy minimum for E 1 that is the inter nuclear bond distance and you can look at these insets so here this is a situation where the two nuclear very far apart from each other see what the energy is and you can work out what S is when they are at inter nuclear separation this is what the wave function would look like there is strong reinforcement okay and in case of the minus combination of the orbitals this is the situation when they are very far away it is still 0 when they are very close say when they are at this equilibrium bond length now you have this destructive interference between the two wave functions and energy goes up to this value and another point to note is that at equilibrium bond length stabilization of the bonding orbital is actually less in magnitude compared to the destabilization of the anti-bonding orbital you can explain this qualitatively by talking about buildup of electron density and depletion of electron density between the two nuclei but only qualitatively alright so these are the energies of H 2 plus now the way we proceed now is that we can just fill in electrons if you have more electrons here that is how you handle H 2 and other diatomic molecules homonuclear diatomic molecules but before that this is an executive summary of what we have learned so far and another thing that I want to say here is that this is what happens when you talk about hydrogen if you want to talk about homonuclear homonuclear diatomics like C 2 N 2 O 2 and so on and so forth you might need to you will need to invoke the combination of not only 1s orbitals but also 2s orbitals 2p orbitals and so on and so forth you can do it in exactly the same way and you can generate orbitals bonding and anti-bonding in this way. So going very quickly through this because it is not all that difficult for you to figure out one thing that I would like to point out is that symmetry of these orbitals is actually they have a role to play later on in many different things so let us note the symmetry and let us note the symmetry using these linear combination of p orbitals when there is pi bonding side on overlap this is the kind of contour diagram that you generate see here we have plus sign and here we have minus sign right sign of wave function just right beside it plus and minus and here you have a node so if you start from any point of the wave function go through the center equal distance on the other side you get an a change in the sign of the wave function right so this is called anti-symmetric with respect to inversion and the term for it is ungerade so these orbitals these molecular orbitals come with the subscript u so this is called pi u orbital you can neglect this one for now why pi because it is a pi interaction if I go back to the orbitals that we have drawn molecular orbitals what does this orbital look like when we combine this 1s and 1s for the bonding situation it is like something like this the contour diagram would be something like this now see it is plus everywhere right so if you go from one any point through the center equal distance to the other side you get no inversion at all no change in sign of wave function so this is called a sigma g wave function g for gerade gerade and ungerade are german words well gerade means symmetric ungerade means symmetric as far as we are concerned now what about the anti-bonding orbital for the h2 plus there we take something like this we have depletion so you have a node in between right this is the node so here the wave function is plus here it is minus start from any point go through the center equal distance on the other side you have inversion so this is called sigma u well for now let us say that this is sigma because it is sigma interaction and this one here is pi because you have pi interaction there is more to it it comes from the symmetry notations will not get into that in this course but this is what we get for sigma interaction involving s orbitals this is what we get for pi interaction involving p orbitals so the bonding orbital here so for sigma interaction you involving 1s orbital the bonding orbital has gerade symmetry the anti-bonding orbital is ungerade symmetry for pi bonding with p orbitals the bonding orbital actually is ungerade and anti-bonding orbital you see this is plus this is minus this is plus this is minus so anti-bonding orbital is actually gerade pi g and another thing that we often do is that for anti-bonding orbitals we put star okay different books use different notations but this is the notation that is used most universally star for anti-bonding and gerade and ungerade to indicate whether they are symmetric or anti-symmetric with respect to inversion right so this is in a nutshell this symmetry of orbitals that we just discussed you can have different kinds of bonds sigma pi and delta sigma is head on pi is with p orbitals this is pi with d orbitals this is pi no with d orbitals this is pi of course I do not have enough fingers to show you the 2 lobes at the bottom as well for d orbitals these are the 2 lobes at the top this interaction is p interaction so once again s orbitals sigma interaction no pi interaction possible p orbitals sigma interaction pi interaction d orbitals well you can have sigma interaction also right d orbitals you can have this kind of sigma interaction or maybe I should show this kind of sigma interaction where is pi interaction like this and this is delta interaction phase on head on side on phase on so these are the different kinds of bonds we will not discuss the symmetries of orbitals and all these in this course but it is a good exercise for the students to work out by themselves. Now what we have done so far is that we have generated the alphabet by which we can discuss homonuclear diatomics in terms of molecular orbital theory we have generated the molecular orbitals using the single electron one electron molecule that is H2 plus next we are going to learn how we can fill in electrons into these same orbitals and how we can develop a molecular orbital theory of dihydrogen.