 In this series of lectures we have seen many EPR spectra, we have analyzed some of them and measured the isotropic hyperbolic coupling constant for some of the spectra. We found the values are different for different nuclei. So, today we are going to ask this question why the different nuclei give different hyperbolic coupling constants. So, what is the theoretical basis for isotropic hyperbolic coupling constants? Of course, the simple answer will be that this type of interaction that is contact interaction which gives us to hyperbolic coupling constant. So it is here the wave function decides the hyperbolic coupling constant. Then can we predict the hyperbolic coupling constant from wave function or if we have the measured value of coupling constant can we test if the theoretically calculated wave function gives similar values. So, today we are going to discuss these issues in reference to organic free radicals. Let us take a few examples of some known coupling constants. This benzene anion radical this we know that it gives several line spectra and the hyperbolic coupling constant is let us call AH is equal to 3.75 Ga. If you take naphthalene anion radical this is got two sets of equivalent nuclei. So, the coupling constants are let us call AH1 is 5.01 Ga and AH2 equal to 1.86 Ga. Compared to that if we take the case of nap methyl radical which is 3 equivalent nuclei this coupling constant is equal to 2310 Ga. So, you see the values are quite different here this two coupling constant corresponds to of course two different sets of equivalent nuclei. So, if you mark them this color chalks this one group of equivalent nuclei and these are other. So, from the experiment we get these values but there is no way we can assign the value that whether this belongs to one set or this belongs to other one. So, can theoretical calculation help us assigning this. So, we will that is the aim of this today's lecture. Try to connect the observed value to some theoretical model of wave function which can either predict this or help us assign this or we can check if this is the value how good the wave functions are if it is able to make this sort of prediction. Immediate major problem if we take the stand that the hydrophilic coupling constant that we have been saying again and again comes from this sort of contact interaction. So, the immediate trouble is that if we take this aromatic hydrocarbon for this the this protons which are coupled they in fact lie in a plane where the molecular orbitals the one which actually suppose to hold the unpaired electron that molecular orbital has exactly a nodal plane where these protons are lying. So, in the simple minded molecular orbital these protons cannot give any splitting but we do see splitting. So, you see the admittedly that there is a problem that the wave function which gives this sort of electron distribution cannot predict this high-profile splitting constant. So, does it mean that we have to do very sophisticated quantum and calculation to arrive at that. Luckily there are very simple molecular orbital concepts which can be used to make very good predictions of the coupling constant that this sort of nuclei can give. We need to understand a few simple key concepts that is here. The concepts are electron density, spin density and spin population. Let us see what they are. So, if we have an N electron system they are given by a wave function of this kind which will have coordinates of all the electrons and this coordinate x includes the space coordinate and spin coordinates. By spin coordinate I mean whether it is alpha or beta spin and this should have all the information that we can think of about the system. So, a function for an N electron system can be written in this fashion. Now, here the xi are the full coordinates. So, for this the meaning of this psi star psi is that psi star psi with all the coordinates here is that if I have a volume element given by dx1 dx2 dxn this gives the probability that electron number 1 will appear in the volume x1 to x1 plus dx1 electron 2 will appear in the volume x2 to x2 plus dx2 and so on. Again this xs are the full coordinate consisting of space and spin. So, that is the meaning of this psi star psi. Now, if I want to find out the probability of finding electron 1 in a space x1 to x1 plus dx1 irrespective of where the other electrons are then what I should do is to integrate this function with respect to all the rest of the coordinates leaving this aside that is written here. That is probability of finding electron 1 in a volume element x1 and x plus dx1 is given by this integral where the integration is done with respect to all the coordinates other than the electron number 1 alright. Now, from this I can ask the next question is what is the probability in this volume now not the electron number 1 but any electron can come since there are n electrons then any of the electron 1 to n can appear there. So, that probability of finding any one electron in a volume element x1 and x1 plus dx1 n times this value and we call this one particle density function written as rho 1 x1 dx1. So, this gives the density function for any electron to be available in the volume element x1 to x plus dx1. As I said that x has space coordinate here x has a space and spin coordinate. So, if I integrate this rho with respect to only the spin coordinate then I will get a function which will be a function of only the let us call it R this is the space coordinates and that is nothing but the charge density where I have integrated with respect to the spin coordinate here. So, this is called electron density function p1 R1 integrated with respect to the space. So, this gives the probability of finding any electron in this range of volume given by R1 to R plus dr1. Naturally if I now sum over all possible space that will give me the number of electrons that is the normalization condition. So, to understand this abstract concepts let us take a simple example of 3 electron system 3 electrons such that 2 are paired 1 is unpaired. The way function for a many electron systems are written as a determinant form called slater determinant in which for each electron we use a spin orbital which is a function of again function of space and spin and the determinant form is written in terms of this if there are 3 electrons we need 3 spin orbitals. So, here we use this a, a bar and b and electron number 1, 2, 3. So, these are the spin orbitals and where the difference between a and a bar is that this has alpha spin function this has beta spin function this also has alpha spin function. So, that is shown in this slide if we expand this determinant then you get the full form of the function. We also use the space orbital a and b they are orthogonal and normalized to 1. So, you see that because of this same space part is used for 1 and 2 this electrons are paired and only 1 electron unpaired which is occupying the spin orbital b or the same thing differently space orbital b with a and alpha spin function. So, here now we can go through this sort of calculation that to find the 1 particle density function and then charge density function or electron density function we expand it and multiply psi star psi type of form here and integrate with respect to the other 2 electrons and so what is the 1 particle density function we can find out from this 3 electron system. If we do that 1 particle density function turns out to be a x, a x plus that is why it comes out to be. So, in a sense it should be implicitly clear that that is how it should be that electron 1 stays in spin orbital a, electron 2 stays in spin orbital a bar, electron 3 stays in spin orbital b. So, 1 particle density function has exactly square of this, square of this and square of this one, but that will come out naturally if we simply multiply the function with this conjugate and integrate with respect to coordinate 2 and 3. So, if we integrate now this with respect to the spin part we get the charge density or electron density function a, r, a, r plus this is also understandable that this 1 integrate with respect to electron spin this alpha spin will integrate to 1 this beta spin will also integrate to 1 and this will integrate to 1. So, this though it comes out to be this is the charge density or electron density function which looks like this. Now, this is same as you see that this comes from the alpha spin this also comes from alpha spin this comes from the beta spin. So, you can write it as the electron density function current coming from the alpha electron plus again 1 electron density function coming from the beta electron. So, with this in mind that the electron density function ultimately comes to be a sum of densities that comes from the alpha electron and the density that come from the beta electron we define spin density is defined as p u alpha r minus p beta r that is the difference of the density of alpha and beta that is the spin density function. So, what we have here now for this this 3 electron system paired 1 electron here the spin density function this is written as q 1 r q 1 r becomes. So, this is difference between alpha and beta spin state. So, alpha comes from this this and this is the beta. So, if you subtract alpha and beta this comes out to be beta beta this is the spin density. So, the value that I can get for such type of expression you see it will be always a square of some number. So, this is always going to be greater than or equal to 0. So, you cannot get a negative spin density for such type of wave function. We will see later that spin density the way it is actually going to influence the observed hyperbolic coupling constant can be negative. So, this picture therefore is not sufficiently broad to predict negative spin density because it is going to call as a square number positive therefore. So, we can immediately think of improving this wave function a 1 a this is prime bar and b 3 what is the difference earlier wave function was this here the alpha and beta this paired electron have the same spatial part only difference in the spin part by putting a prime here I use a different spatial part for the alpha spin and beta spin. So, this is a method which is called unrestricted Hartree-Fock method where the alpha and beta electrons which are paired need not occupy the same spatial orbital. So, now with this one if we do the same calculation now p 1 r will be and then the q 1 r will become again the same definition the difference between the alpha and beta population that will be here. So, in this condition as the spatial part for alpha and beta different this need not be same as that. So, this can be both positive and negative this is one way of having the possibility of generating negative spin density other way would be that instead of having a single determinant to represent the wave function one can get more than one determinant with and some of that which is called the configuration interaction type of wave function that can also predict a negative spin density. So, if the spin density the way it is defined now that q if we integrate respect to all space that is total spin density that will give me just the number 1 pi electrons in this particular case only 1 is there so this is only 1 so that way total spin density must sum to 1. Now for calculating the hyperbolic coupling constant how do we go about the Hamiltonian for that is g e beta e g n beta n psi 0 square s dot i this was for a single electron as a probability find the electron at the nucleus. Now here we are going to integrate over the space so we should immediately make some modification here that when integrate whenever it is necessary this will give the same result. So, this could be written as g e beta e g n beta n delta r some delta function with respect to space coordinate r s dot i. So, this will have essentially same result of that because delta function gives non zero contribution only when the r itself is 0 again this is for one electron case. If I have many electron then I must do same thing for all the electrons then for that I can write I will write it here I basically sum over all possible electrons which are going to interact with one nucleus what does it mean it is a simple generalization of this that this j is corresponds to all the electrons and this is the electron density at the nucleus for the jth electron that is the meaning of this and this is the spin angle momentum operator of the jth electron. So, what I have done here in going from here to here is that I put a s z i z here and that is possible if the interaction is not very strong so that the first order calculation is good enough then this same thing could be written as s z i z first order calculation. So, this give rise to the expression which is here. So, if you calculate the energy of this Hamiltonian that energy will be beta n and what it will give it will give me this this will give me this spin density one particle spin density at the position of the nucleus. So, here therefore we can associate this to our hyperbolic coupling constant a therefore can be equated to this part of this expression but this is the one we are trying to arrive at. This connects now the observed hyperbolic coupling constant to the calculated spin density at the nucleus. Now this calculation will depend on the type of wave function we can use here it is either this or this even better than that. So, the connection between the theory and experiment is this that this is experimentally measured this is theoretically calculated and see how well they match. So, I can use a theoretically calculated value to assign it is a various coupling constant to various different nuclei in a molecule or having measured the value and assigned it I can see how well my calculations are are they able to predict such measured values. So, theory and experiments are bridged in this relationship from this few ideas that is spin density electron charge density we need to extend this concept just a little bit more where we deal with the wave function which is of course written as a determinant form. But often these determinant forms are made of this molecular orbitals and molecular orbitals are in turn written in terms of atomic orbitals. So, we want to relate this concepts to ultimately the spin population in an atomic orbital. Now let us look at this slide now. So, this is let us say total wave function for a 2n plus 1 electron system here psi 1 psi 1 bar with a prime psi 2 psi 2 bar with a prime each of them have a paired electron only one is unpaired at the end here. So, this electron is unpaired all the other electrons are paired we have put the beta electron molecular orbitals with a prime we will see in a moment why it is convenient to distinguish them not just by bar but also with a prime. So, this molecular orbitals are written now in terms of the atomic orbitals this phi is at the atomic orbital. So, these are corresponding coefficients for a given molecular orbital here. So, with this or population of the mu atomic orbital orbital given as the coefficients of this here see the mu in the psi i molecular orbital the phi mu has a coefficient c i mu. So, here see that our wave function has 2n plus 1 electron and all this n electron 1 up to this n they have got alpha spin plus 1 more extra also alpha spin and other n electron with the psi 1 prime psi 2 prime they have got beta electron. So, this population of the mu atomic orbital is actually sum of the alpha electron and the beta electron and square of this is a coefficient of the atomic orbital there. So, as all the orbitals are normalized. So, we can get this is equal to 2n plus 1. So, with this type of atomic orbital population I can write now the spin population which is the difference between alpha and beta spin state spin population. So, exactly same thing we did earlier when we took the difference between alpha and spin state to call them spin density. Now, it is spin population on a particular atomic orbital mu here that is the way it is. What we have in mind is that if we know the atomic orbital and the corresponding coefficients we can calculate the spin density or and spin population on a particular atom. Now, in Huckel-Molkler orbital all the paired electrons they occupy the same space orbital. So, for them the C i mu and C prime they are same. So, this will all cancel each other and only thing that will be left here is the single electron which is unpaired. So, the rho mu will be for Cn plus 1 mu and Cn plus 1 mu. So, with this now let us make some connection to Huckel-Molkler orbital and see how much it can predict the observed splitting constants. Benz we have seen earlier methyl gives a hydroponically costly constant equal to 23 gauss. And benzene gives the a1 equal to 3.75 gauss. Now, methyl radical the electron is basically staying in one sp3 hybrid orbital and here this one electron is occupying a pi orbital. So, six such protons are there. So, it will spend one sixth of its time around this one proton. So, from this relationship that the one sixth of this happens to be this you see is very nearly equal to 3.8. So, value is very similar. So, it somehow shows therefore that the spin density near the carbon atom is correlating with the observed hydroponic coupling constant. Is it true for other organic radical or not naphthalene which we saw earlier the a1 H was 4.195 gauss and a2 H was 186 gauss. If we take the ratio of this ratio which is 2.66 for naphthalene that orbital which occupies the unpaired electron here that is what this sort of contribution from various atomic orbitals. The ratio of this spin density will be square of this square of this. So, we have come to 1.181 by 0.69 which is 2.62 and you see how well these are matching. So, this also shows that the spin density the carbon atom governs the observed hydroponic coupling constant and here because all 4 have a same coefficient they have also similar coupling constant. So, I can associate therefore that these 4 have this coupling constant and this 4 have this coupling constant. So, from this and this has been found for many such organic aromatic hydrocarbon radical. So, I can write a of H this observed hydroponic coupling constant is some sort of constant times the carbon hydrogen parameter some Q parameter times the spin population on the carbon atom. This is the this stands for the proton hyperbolic coupling constant this is the spin population on the carbon atom. This was found empirically true. So, though it is empirically true the how does it work because our fundamental problem still remains that the molecular orbitals as a nodal plane where the hydrogen atom is sitting there. So, it just cannot give rise to such splitting. So, the way function has to further improve to take that into account and the way it is done is now shown here. Let us take consider only the fragment of this benzene ring type of the fragment where only one electron is sitting in this p orbital which is part of the pigment orbital and this sigma bond as the CH group which gives the hyperbolic splitting. So, the way it goes is that I as such because this is plane a nodal plane here this cannot give any splitting because of this. So, let us see whether I can get some spin density here. So, for that this electron is here and this both the electrons are paired but I can think of a situation that this one one electron here other electron there and other is that this electron down this electron up and here. So, both are same essentially nothing different but if there is no difference in energy between these two of course they will contribute equally to the resultant way function that I can write here which is the linear combination of let us say psi 1 and psi 2 and total psi commutant as C 1 psi 1 plus C 2 psi 2 and if they are equally energetic then this will be same as this one and there will be no resultant spin density here. But it so happened that this one is just a little bit more stable than this one so that C 1 is just a little bit more than C 2 why is that simple qualitative argument is that you see these two electrons are parallel here these are anti-parallel the two electrons are parallel give rise to more stability that concept is used here when two electrons are parallel give more stability there is a Hohn's rule of fact. So, even though these two I thought is looking very similar this little possibility that these two parallel being slightly more stable give rise to this. So, that means I can get a net spin population appearing here these are not cancelling now so that way one can generate a net spin population here which can give rise to Hohn's splitting. So, it is possible now this has to be now implemented in the wave functions we refine the wave function by not just sticking to the ground state configuration that we have but we can mix excited state and get a better quality wave function which will give this sort of spin population which is appearing here what is more this also shows that if the overall spin system has a alpha spin state this produces a better spin state here. So, the spin density here is negative for such system that also has been seen experimentally. So, here this lecture we have seen how the wave function can be used to predict the spin densities or the calculated values can be used to assign the various observed values to different nuclei and the wave function can be improved if there is no matching between the observed value and the calculated value by doing improved calculation. Now with this we come to the end of this lecture and this it is taken from this textbook.