 Hello there, last time we have seen how the EPR spectrum of any radical or other species can split into various hyperfine line due to the interaction of electron magnetic moment with the nuclear magnetic moment and we call them the hyperfine splitting. We saw lot of examples where the splitting comes from the spin half nuclei particularly from proton and we saw certain patterns that we can see in the spectrum when there are many equivalent spin half nuclei are there and we said that one can use the Pascal triangle to find out the number of lines and their relative intensities. So here we have this triangle again displayed here so when there is no nuclear spin we get one line that is at the middle then we have one spin half nuclei it is split into two so this much is the splitting due to the nuclear spin. So if I have got two nuclear spin then again it is split by the same amount here from this to this, this gap is same as this gap there but the intensity ratio becomes 1 is to 2 is to 1 that way we can build up the various possibilities for any number of nuclear spins. Today we are going to generalize this idea of many equivalent spin where the nuclear spin is more than half. Last time we have seen several examples of this kind one example is here that this free radical temporal here the unfair electron interacts with the nuclear spin of nitrogen and we saw that it gives APR spectrum this kind where three lines are of equal intensity. Let us recapitulate how we arrived at that first that electron spin in a magnetic field splits into two because of the spin half mx equal to minus half and plus half now this interacts with the nuclear spin of i equal to 1 or mi equal to plus 1, 0 and minus 1 these are the possible mi values. Similarly, the upper energy level is also split into three plus 1, 0 and minus 1 and the transition involves the change of the electron spin from this to this without changing the nuclear spin. So we got three transitions of this kind. So since all these three levels are almost equally populated we get the spectrum with intensity ratio of 1 is to 1 is to 1 which appears in this pattern 1 is to 1 is to 1. So this gap is same as this gap which is a measure of the interaction of the nuclear spin with the electron spin. We also saw copper complex which is of course a very complicated one if we see it now this spectrum comes from the naturally occurring copper complex now naturally occurring copper as two types of isotopes copper 63 and 65. So to understand the origin of this or rather to simplify the spectrum let us get the look at the spectrum where we have only one enriched isotope that is copper 63 and here is the spectrum of course spectrum has become simpler than earlier and what it shows is that there are four lines of different intensity by intensity if I mean the height of this one that this height is bigger than this which is bigger than this and this is the least height for same time if you notice carefully the widths are different. So one has to be careful in deciding what the intensity is. So for an absorption spectrum which looks like this type of thing this is a magnetic field we normally say intensity is area of this that is related to let us say if we take the full width and the half intensity of this and measure this one the height of this one so this I will call delta B half that is full width that half maximum of the intensity and this one this let us call it I then this area is proportional to I into that we all understand but the appear spectrum that appears here is in the form of derivative line. So what can you say about its area for a derivative line the shape is of this kind so if you simply add this area and this area this is going to give reactant will be equal to 0 so that is of no use we have to somehow therefore relate this intensity to this type of intensity. So what is done here is the first you convert the derivative signal to an absorptive signal of this kind by integration that is you start from this end to keep on adding this area various slabs are taken then keep on adding this little small strips then continue to do this all the way to the extreme right then it will give an absorption profile of this kind and then the area of this will be calculated in the similar fashion as this one or one can actually do numerical integration of this kind get add all the little strips then I will get the area of this one so further integration gives the area now one can go through this numerically with advantage of computer this can be done very easily but one can also get a very good approximation to this in the similar way that we have done it here by measuring the full width or half maxima and this with intensity here so he is similarly in a similar fashion we can define two quantities one is this intensity let us call this peak to peak intensity so that is from the top of this to bottom of this one I call it I peak to peak and analogous manner that the width that is used here we can define a width from this to maybe is a different color this to this extend this one this width from this peak position to this peak position and I call it delta B peak to peak so in this case the area of this after doing all the suitors in two times integration area that we find here will be this area will be proportional to this I peak to peak times the width that we have here delta B peak to peak square of that so it is not the product of this and this the way it comes here the reason is simple that we have to integrate it two times to get the area first integration gives this second integration of this one that is why the this comes in the form of product so because this area is now proportional to the width square even a small change in the width can really change the height of the signal quite appreciably now that is seen here that width of this line is slightly broad in a broader than this one this is broader than this one this is broader than this one so this is the narrowest line so consequently the height is maximum here so if we can integrate all these areas either by computer or in the absence of that we can even manually find out the height of all these lines and also measure the width of this peak to peak and if you calculate it will find that all four of them have exactly same intensity so that means that the apparent intensity difference is not of any consequence it is the integrally all four lines have the same intensity so why four lines coming now is easy to understand the nuclear spin of copper is 63 so the way we have seen here the 3 by 2 is the nuclear spin so this will split into four lines this will also split into four lines and then I will get four transitions of equal intensity and that is precisely seen here so why these lines have different width is something we will not try to understand now but we will understand later for the timing we notice that this is indeed a very special feature of this copper complex when you compare all the spectrum that you have seen for free radicals where the line widths are essentially same for all hyperfine lines now to understand the this spectrum now we try to see let us see look at this the various nuclear properties of these two isotopes where is the difference copper 63 has about 69 percent natural abundance 65 has 30 percent or 31 percent to be more accurate both of them have got the same nuclear spin 3 by 2 and 3 by 2 here so what I have got in this spectrum is actually two types of complex one has got copper 63 other copper 65 so both will therefore split into four lines because the spin of this is for copper 63 now copper 65 similarly will give similarly four line spectrum and because it is nuclear spin also 3 by 2 what we have is a superposition of these two but when you superpose them will they exactly overlap let us try to understand that first now center of the spectrum which is here for this for this here will they be same for these two species what decide the center of the spectrum that comes from this relationship that this value g factor or g value of this one so both the spectrum are recorded in this at the same time because sample had 63 as well as 65 so they experience the same frequency so the difference can come if they have got different g factor now can g factor be different for these two species the g factor is an electronic property of the molecule it depends on how the electronic structure that the molecule has now when I have got these two isotopes what is the difference difference only in the nuclear isotope now for and copper nuclear is reasonably heavy nuclear so for such heavy nuclei usually the electronic wave function does not depend on the nuclear mass only if the nuclear mass is very light for example hydrogen atom when you have got either it is proton or deuteron the wave function can slightly be affected but otherwise this difference is so the mass is so high compared to the mass of electron that this will have no effect on the electronic structure of this complex so the g factor will be same for both of them in other words center of the spectrum will be same for both the species then where else can the difference be is the splitting going to be different we do see that splitting is different here that if you see very carefully this side there is small hump is there so possible that it could come from overlap up to hyperbolic line similarly this could this is obviously coming from to hyperbolic line which are overlapping here though it is not very clearly seen only part of this line is seen here so it indicates that this splitting that you see here and here they are not quite the same but only evidence is that we see some sort of difference here but they are overlapping here so can the hyperbolic splitting constant be different further you go back to the table of the properties of copper nuclear you see that magnetic moment of copper 63 is 2.43 and for copper 65 is 2.54 indeed the magnetic moments are different copper 65 has slightly bigger magnetic moment than copper 63 that means the electron in the copper complex sees a little bigger magnetic field that is created by the 65 nucleus so this splitting will be just a little bit bigger than this one this will be slightly bigger than this one now how is it going to change so you know this gap will be bigger than this gap from here this gap will be still bigger than the gap that is here same way the gap that is here will be little bit less than the gap that is here and the gap between these to this will be much smaller than the gap that is here so now you can imagine that if we now overlap these two I will see spectrum which may look like some sort of splitting here and maybe little bit splitting here but more permanently here and the top of that the width of these are not same across this from right to left the widths are more here and least here so what happens in the process the resolution is such that is going to be such that this will be seen much more clearly than this one and when you take the derivative we get let us say here we get one line which looks like this this type of derivative signal and other line this type of thing so if there is a partial cancellation of this part of the spectrum with this part of the spectrum so net result is that we get a spectrum which looks like this type here the lines are broad so what we see is small hump and not so well is of line here what is more you see that copper 65 here as about 31 percent natural abundance and 63 has 70 percent or say 69 percent so this intensity is correspondingly going to be contribution of this will be somewhat smaller well exactly is the same ratio 30 percent intensity will come from here and 70 percent will come from there so we will see this we will have smaller signal than this one and that is exactly what we see here so we understand how this signal appears again the width changes gradually from the right most to the left most signal the other example is for us earlier is this dye vanadyl complex this complexes to vanadyl center here given here VO and VO there connected by this oxygen bridge at their other part of the ligand is not shown here it is this two bonds here these two and there are lot of things around this one which of which are of not much interest to us so all that is shows that these two vanadyl the way they are bound and form the complex they look completely equivalent so what sort of spectrum we expect from this two equivalent vanadyl complex this is the spectrum that appears here so we see that again the widths are not same unlike the copper complex where the width increased gradually from right to left here the narrow lines are in the middle and a broad lines are at the two sides and a broader here and also here so that again we will not try to understand the origin of this right now we will postpone it to some other lecture for right now we will see if the intensities are same or not here we both a similar argument that width and height together decide the intensity and for derivative line we use this formula we will see that intensities of these lines are not same intensity changes gradually from here become maximum here and goes down in this way so how does one account for this different intensities now so now having learned so many different types of nuclear now it should not be difficult for us to generalize say this to this particular case for vanadyl complex. For that we find out the magnetic properties of the vanadium nucleus see vanadyl complex is has the nucleus vanadium a vanadium it has two isotopes 50 V and 51 V so what here the natural abundance is only 0.25% for 50 V and 99.75% is 51 so we can as well assume that this is absent there and the spectrum that we see here is coming from this isotope and its nucleus spin is 7 by 2 so 7 by 2 means how many lines I will get from all the discuss that I have seen here it will give 8 lines of equal intensity for only one vanadium then the second vanadium is going to split all the lines further and I have shown it here. So if this is the line position in the absence of vanadium then when I got one such nucleus of vanadium whose nucleus spin is 7 by 2 see this splits into say 8 lines this is the center so this splits such as this gap is equal to this gap and this equal to this gap so the 8 lines are split are seen here now that is due to the first nucleus. When I have the second nucleus each of these lines are going to split further by the same amount and same number of lines so 8 line comes from here see the gap between this and this is same as gap between any of this line here similar this line is split into again 8 lines that way all the other lines are shown here. So what I see here therefore is that at a given position more than one contribution comes so here therefore count the number of lines that I can see here this indexed one there are two lines come there so the indexed two that if you count all of them there will be 15 lines with the indexed given here 1 2 3 4 5 6 7 8 7 6 5 4 3 2 1 and this is what is seen here though it is not very obvious one can actually integrate this various lines and come to this conclusion there. So this shows that van der eil complex when there are two van der eil there is perfectly understandable from the whatever we have learnt here the way the various half of a line that appear there and how equivalent nuclei can produce different type of line intensities. So can we generalize this Pascal triangle that we have seen for spin half system we are trying to do that now. So here the Pascal triangle is shown for spin one nucleus let us see how it is created it starts from the line intensity one where there is no nuclear half of an interaction and now there suppose number of such nucleus one which i is equal to 1. So then this will split into three lines it is the one is to one is to one this can be found out by adding three numbers comes from this three that is apparent the top of that one is directly above other is from the left other from the right. So here we can assume that there is zero is here zero there so this zero zero one gives one here i get this three lines from here here and zero is here then this gives one there similarly this one i get three numbers from here here and then I mean zero there then i get one zero zero that is fine one is to one is to one that is what we have seen in case of temporal now i equal to one but n equal to two. So if all the same argument now that to get a number here i try to find out the number which is directly on the left in this above and right. So i have got here zero one one gives me two but here then directly above is zero zero one this gives me one then if we what the intensity here i can expect i get in value one one one is three similarly here one one zero gives one one zero zero gives sorry this gives two so one two three two one is the relative intensity of spin one nucleus and where the two of them are present there that you can generalize this that every time i get number from here for example here i get add the numbers which is directly the left and above and right the two one zero gives three that is the numbers are created there. So this is not Pascal triangle but very much like the Pascal triangle we add three numbers one is one is the left right and right directly above now we have seen the copper spectrum where spin is three by two can you create a Pascal like triangle for three by two again procedure is very similar but not quite the same and the triangle is shown here when there is no hard performance splitting that is the intensity one is here when one spin three by two splits the line there will be four transitions. So to get this one is to one is to one is to one all i do is the i add now two numbers from the left and two numbers from the right now it is let us say how it is done there i will get four transitions here here here here so zero zero zero zero so to get the intensity here i add two numbers from the left and two numbers from the right so this and this together gives a value one here. Now for similarly here i add these two numbers from left and these two numbers from the right this gives me one then to get the line here i add two numbers from this left and this two numbers from the write case 1 and similarly here I get add this and this is this one. So, this looks simple 1 1 1 1, but let us say we understand the procedure correctly. So, for that we go down once more if we have two such nucleus of spin 3 by 2 then I will each of the lines will split into 4. So, there will be line here and line here and for this I would here here similarly, but here here. So, to get the intensity here I have this is 0 0 0 I add this and this to get a value 1 that get the intensity here I add this and this. So, then I get a value this to get the value here I add this to and this to. So, this gives me 3. So, then further this will give me this this and this this is 4 then this and this gives me 3 2. So, 1 2 3 4 3 2 1 that is the way the line numbers are created here. So, you can check if really you can figure out how this numbers are found out. So, we see that Pascal triangle is a triangle which works for spin half system, but we can generalize this thing for other type of nuclei as well we have generalized to i equal to 1 and i equal to 3 by 2. Now, we are going to generalize this pattern that we have seen here and summarize what we have seen for various nuclei. So, if there are let us say n nuclear spin and with a individual spin angle of momentum is that is a little i then the total angle of momentum becomes n times i. Here for example, whenever i equal to 1 here now with this little i we change the notation this small i for each nucleus. Now, when the 2 of them there n equal to 2 then my i becomes 2. Similarly, when i equal to 3 by 2, but n equal to 2 2 of them are there then total i becomes 3. So, then total nuclear spin term number becomes n times i and its component will be minus i to plus i changing in units of 1. So, total components equal to therefore, 2 i plus 1. So, each of this produces a local magnetic field that the electron sees. So, there will be number of hyperfine lines given by this number 2 times i plus 1 those many lines will be seen there. Now, if this total i happens to be an integer then there will be line at the center why because then 2 i plus 1 is an odd number. So, odd number will give a line at the exact center and if i is a half integer then 2 i plus 1 becomes an integer. So, there will not be line in the center here for example, here 3 by 2 was there. So, there is no line to the center here and the relative intensities will be given by the Pascal triangle for spin half nuclei and analogously for other nuclei we can generate a triangle of similar kind. So, summary is that all EPS spectra characters by the basically 3 things. One is the line position for any of the spectrum here is given by this relationship that is essentially given by the G value and the width which is also very important property we saw that for most of the organic free radicals the widths are very similar but not necessarily true for transraminal complex, copper complex or verandhyl complexes widths are different and of course, the hyperfine splitting and how many of them are there and their relative intensities these are the characteristics of EPS spectrum. Now, having learnt this much it is time to decide that whether EPR or ESR mean the same thing or not. I mentioned in the first lecture that we must discuss this and come to an agreement after some time it is time we did that. You see that spectrum comes from the married moment of the electron. Now, married moment can come either from the orbital motion of the electron or the spinning motion of the electron or it can come from both this is the relationship between the angular momentum and the magnetic moment. If the married moment comes from the spin angular momentum then this is the relation mu is the magnetic moment G beta S, S is the spin angular momentum. If it comes from the orbital motion of the electron then orbital angular momentum L and married moment due to L are related in a very similar manner. Only difference is the value of G that is used for spin angular momentum is about 2.00 something these numbers are not of consequence to us and the G value that is used for orbital motion is precisely called 1. So, that is what the difference is only difference in the quantity of and not quality qualitatively both are very similar most organic criticals have very little orbital angular momentum. So, in that case calling them ESR spectrum should be ok, but system for example, in which has both motion due to S and L that is married moment coming from the orbital motion at the spin motion what will you call them think about it. Now, it is also possible that system has no spin angular momentum, but it has an orbital angular momentum as a married moment comes from that motion and you can get a corresponding parametric resonance spectrum. Then calling it an ESR spectrum would be simply wrong because there is no spin motion present in the particular system. So, here is an example oxygen molecule we are all familiar with and it is a electronic configuration in the ground state is shown here is 1 S 2 S 2 P are the orbitals of the oxygen atom and there are 8 electrons there and these middle lines these middle horizontal lines are the molecular orbitals. So, this 8 electrons from one oxygen atom and 8 electrons from the other oxygen atom produces this set of electronic mechanism and you must have read in your MHC or may be undergraduate. Here important thing is that these 2 electrons are unpaired here they are occupying 2 different pi orbitals. So, that produces this triplet state of the oxygen molecule. So, oxygen molecule in the ground state is a parametric species and one can actually see its EPR spectrum. Now, if this oxygen molecule excited it produces a singular state and it produces 2 types of singular states and the way the differences you see here these 2 spins are parallel. One of the excited state is that these 2 become anti-parallel now, but since these 2 occupy different orbitals this can be a singular state and that singular state is has this set of electronic configuration this excited state, but it can have another excited state where these 2 electrons are forced to stay in a 1 molecular orbital then actually this spins are necessarily going to be paired. So, this is a singlet state, but this pi orbital gives us to a net orbital angle momentum of lambda which is equal to 2 give us to this delta state. So, this electronic state is 1 delta G has no spin angle momentum, but orbital angle momentum and one can see its magnetic resonance spectrum which is shown here. So, this singlet oxygen gives us to these 4 lines and this spectrum also other line which is present here which is very intense it actually comes from the ground state of oxygen triplet oxygen there. So, you see that one can see the EPR spectrum of a molecule which has no spin angle momentum, but only pure orbital angle momentum. So, this cannot be called an ESR spectrum by the way I must point out that this looks like hyperbolic hyperbolic line that we have seen so far, but in oxygen molecule which is this nucleus as 16 is also 16. So, this has no magnetic moment this also no magnetic moment. So, these lines that are appearing here though they look like hyperbolic line, but they are not this is interesting that something else is seen here this experiment was done in the gaseous phase. So, we take oxygen gas in the spectrometer and try to look at it its parametric resonance spectrum. So, here the oxygen molecule undergoes out of rotational motion and rotational motion are also angular momentum. So, they can also give rise to a coupling of angular momentum to the orbital motion of that and that can produce such lines which appears as if they are coming from certain hyperbolic interaction, but that is not so. Anyhow, coming back to this that this suddenly this cannot be called an ESR spectrum. So, electron parametric resonance is a very general term now which includes both parametric coming from orbital motion as well as spin motion. So, we will continue to use the term electron parametric resonance in these lectures. So, with this we come to an end of this discussion.