 Welcome you all to MSB lecture series on interpretive spectroscopy. In my last lecture, after discussing about hyperfine splitting, I showed a lot of very interesting examples with very clean EPR spectra. So let me continue again showing more examples to make you familiar with interpretation. This is about naphthalene and radical. It's very easy to prepare this one. You take naphthalene in 1, 2-dimethoxyethane, very dry freshly distilled 1, 2-dimethoxyethane and add sodium or potassium. You can immediately generate potassium or sodium salts of naphthalate. So that is essentially blue color that is due to the formation of radical. So we call it as naphthalene and as radical. And it's very interesting to record its EPR spectrum. And EPR spectrum would look like here. You can see here, initially there are 1, 2, 3, 4, 5 lines are there. And then these 5 lines are further split into 5 lines here, further split into 5 lines. So how that happens? And that we can understand if you look into the naphthalene. Naphthalene what we have is, you can see here, we have two different type of hydrogen atoms are there here. And they are called alpha hydrogen atoms. We have 1, 2, 3, 4 and we have beta hydrogen atoms are there. And then they are coupled non-equally. So first let us say they are coupled with alpha to give a 5 lines to an A plus 1 rule, quintet. And then each line is further split by this 4 beta equally to form another each line further splitting into quintet. Then we get 25 lines. So the 25 lines also very simply in a simple manner we can calculate. Take this one set. We have 4 of them. We have another set here. Another 4. So we can use simple this formula. So alpha 4 are there. Beta are also 4. This is how you can calculate very nicely the number of hyperfine lines in a given spectrum here. So another way is you can also start writing using a branch splitting tree or we used to call in NMR. So first it split into 5 lines and each line will be further split into 5 lines. So if you take count this one we have 5 into 5. This also gives 25 lines. Either this way or simply you can use this equation and you can work out. If we have 3 or 4 sets of non-equalant nuclear are there. This can continue something like this. So on it can continue like that. The product of this one would show you number of lines in the EPR spectrum. You can see here a beautiful spectrum of the same I have shown here. The splitting also I have shown here. In this one what happens some of the lines are merging. As a result what would happen? What we see is less number of lines here. This one is another interesting copper complex here. And copper 2 is there and we know that copper i equals 3 by 2 is. This is 63 enriched product we have only 63 copper we have here. And then we have next to that one we have 2 nitrogen 14n i equals 1. And then we have 2 different type of hydrogen atoms are there. This on hydrogen and 2 nitrogen atoms are equivalent. And then what we have is next to that one we have these 2 are identical. That means we will be seeing coupling of this one 3 by 2 first splits into 4 lines. And then each line splits further by 2 nitrogen and then 2 hydrogen and then 2 hydrogen. So, it gives a very complex system. For example, if you consider simply one can write something like this. Initially it gives if you consider 2n i plus 1 here it would be 4 lines. 4 you can see here 4 are there here 1 2 3 4. And then next this will be 2n i plus 1 here nitrogen it will be 5 lines. One of them I will take to show. So, next we have 2 hydrogen atoms are there it will be split into triplet. And then this will be further split into triplet. So, you can calculate the number of lines. But what happens because of overlapping of some of them very closely spaced what we see is we get 4 multiplets constituting each one 11 lines here in this fashion. So, we get almost 44 lines instead of more lines we expected. Because of overlapping we see something like this happening here. Again this is also taken from one of the oldest article published in Journal of Chemical Physics in 1958. Now let us look into Nitrobenzoite-Dianon radical. Here we have one side CO2 one side NO2 is there. And here if you just consider first this interacts with nitrogen it will give a triplet 2n i plus 1. And then it interacts with two different type of hydrogen atoms are there if you see Arto and Meta. That means first it gives a triplet because of coupling with nitrogen i equals 1. And then each line of this triplet will be split by these two one identical Arto hydrogen atoms triplet. And then again it will be split by these two identical ones to another triplet. So, what we get is triplet of triplets of triplet. What we get is triplets of triplets of triplets of triplets. If you want to express this one triplet of so this is the initial triplet of triplets of triplets. What we get is we have 27 lines. This 27 lines one can also calculate in this fashion. So, I have designated Arto hydrogen Meta nitrogen here. So, here we get 2n i plus 1 will be 3. And here what we get is two of them are there 3. And then here we get 3. So, you can get 27 lines. So, easily this way also you can calculate and you can account how many number of lines will be there in a given EPR spectrum. So, very nicely this para nitrobenzite shows hyper plane splitting in this fashion. Having 27 lines triplet of triplets of triplets. Now, let us look into benzo semi quinone radical anion here we have. So, this one is coupled equally to 4 hydrogen atoms i equals half. So, you can see a quintet here. And also the corresponding transitions also shown in this one. So, this is for benzo semi quinone radical anion. Now, let us look into more examples of hyper plane interactions. Now, if you consider this pyrazine anion here we have two nitrogen atoms are there and four equivalent hydrogen atoms are there. So, that means initially what happens? This will be first split into a quintet because of two nitrogens having i equals 1. So, here we have. So, we have 1 is to 2 is to 3 is to 1 ratio a quintet because of coupling with two equivalent nitrogen 14 nitrogens. And then each one will be further split into quintet because of 4 equivalent hydrogen atoms. So, each line of a quintet is further split into quintet. So, we have this ratio would be one will be like this 1 is to 4 is to 6 is to 1 1 is to 4 is to 6 is to 4 is to 1 it continues like this. And we get the spectrum something like this. This is the same spectrum is shown here you can see here hyperfine due to two equivalent nitrogen and then super hyperfine due to 4H equivalent nuclei. This is called super hyperfine the coupling of the coupling it continues first we call hyperfine and then further continues with more number of non-equivalent we call super hyperfine. After that one no matter how many are there they are all referred to as super hyperfine interactions or super hyperfine splitings we call. So, this another beautiful one we will see here again 5 into 5. So, 5 into 5 again you can use the same rule here 5 into 5. So, 25 lines will be there either you can calculate this way or one can also go in a stepwise manner the spectrum should be quintet with intensities 1 is to 3 is to 3 is to 1. And each of those lines will be further split into quintets with relative intensities of 1 is to 4 is to 6 is to 4 is to 1. This is another again a beautiful EPR spectrum showing hyperfine as well as super hyperfine splitting. So, now EPR G and isotropylers look into it. We all know that H nu equals G beta H or sometime the magnetic field is represented by H or B most commonly B is used nevertheless one should understand here B equals H. So, B or H are essentially same and one can also calculate G by considering this equation. So, for example, here Planck's constant is given in arc cities minus 27 6.62517 and then this is value is given here for a given frequency 9.114 into 10 to the power of 9 cycles. And then beta is given here and the field strength magnetic field is given in Gauss and then G can otherwise be calculated is using this radical as a reference. The relation between J the resultant vector of L and S J equals L plus S we are considering here whereas, in case of electronics spectroscopy we consider J equals L plus R minus S I mentioned already we are considering L equals J equals L plus S when is the sub shell is more than half field when the sub shell is less than half field we considering J equals L minus S. So, two values comes here but whereas, in EPR we are considering the resultant vectors L and S coupling that is L plus S if L is 0 then J will be equivalent to S. So, then we use this equation this is a standard known equation G equals 1 plus G into J plus 1 plus S into S plus 1 minus L into L plus 1 over 2 J into J plus 1 for free electron S equals half and L equals 0. So, here what happens J becomes S. So, now if you use the same analogy here we get G equals 2. So, this is how we can calculate G value. However, actual value of G of free electron is 2.0023 if you recall I mentioned in my earlier lecture this is due to the relativistic correction free radical G value is 2.0023 as unpaid electron is not confined to any localized orbital most freely over the orbitals. So, value is 2.0023 is considered here. So, in the transcendental complexes unpaid electron remains localized in a particular orbital due to loss of orbital degeneracy or spin orbital coupling and hence G value is different from 2.0023. So, if it is localized then the value would change in that case G value depends on both lambda and dq crystal field. So, one can use this expression to calculate G where the electron is localized it happens in case of metal complexes. So, you can use this formula here the K is extent of metal ligand delocalization under the banner of metal ligand covalency as per ligand field theory and lambda equals spin orbit coupling constant 10 dq equals crystal field splitting and then alpha depends on the ground state term whether it is 2 for e g and a for a to g a state and if lambda and 10 dq from u v data you can get or known now then K can be calculated using experimented G value obtained from EPR. So, K can be easily calculated and G can be obtained from EPR spectrum. G is very similar to coupling constant I had mentioned already in case of NMR. So, you can find out the G value and once you use that one this term can be calculated as other most of the terms are known now. So, now let us look into another example for a radical the magnetic field is 3810 Gauss the frequency of the microwave is 9600 megahertz what is the value of its g factor. So, we know now just I had discussed in one of the earlier slides that h nu equals this we are using g beta and b naught b naught is the field strength b is known and then what happened this can be simplified by considering h over b naught h over b naught is simplified. So, that is considered here for example, this equation now we can write something like this g equals h nu over something like this. Now, this term one can simplify after simplifying what we get is we get value of 4484 and then simply use here this is given megahertz this gives approximately 1.8 g value. So, this is how we can calculate g value here. So, let me stop here I am concluding now EPR and if any more interesting problems are there I will come back again while looking into the problems. As I mentioned lost 10 lectures 8 to 10 lectures are devoted to problems in the problems I am including problems from all these spectroscopic methods u v n m r mass EPR and also mass bar and I might take a couple of lectures on mass bar spectroscopy and later I begin discussion on solving more and more problems to make you familiar with the type of problems and then how to interpret data having data getting from more than one type of spectroscopic methods. So, it is essentially to make you familiar with interpretation and elucidation and understanding the reactions that are carried out and the product obtained. So, see you in my next class.