 I once again welcome you all to MSP lecture series on interpretive spectroscopy. Let me continue discussion on EPR spectroscopy. In my last lecture I started discussion on hyperfine splitting, so let me continue. If you can see here to decide number of lines very similar to NMR, we use this rule 2 and I plus 1 rule, N is number of equivalent species nucleus or other species and then I is the spin value, so in all these things we are considering N equals 1 and so in absence of any a radical or an anion would show because of only presence of one electron it shows no interaction we will see only one line. When it interacts with one nucleus with the spin I equals half, so we see two lines, simple rule just in case of one I will show you here. So, we see two lines here and similarly when the nuclear spin I equals 1 is there we observe three lines and then in case of spin 3 by 2 we observe four lines. When we have two equivalent nuclei with the spin I equals half then it would be 2 into 2 we expect three lines here and similarly when we have three equivalent nuclei with I equals half we will see four lines and when we have four equivalent nuclei with I equals 1 we will see nine lines here. So, this is how we can understand hyperfine splitting simply applying this to a 9 plus 1 rule. So, how this hyperfine splitting happens is because of hyperfine interaction if an electron couples to several sets of nuclei then the overall pattern is determined by first applying the coupling to the nearest nuclei and then splitting each of the lines by the coupling to next nearest nuclei and so on. So, this is how it goes very similar to what we do while writing splitting tree or coupling interaction tree in case of NMR. The number of hyperfine lines is called NHFS hyperfine lines from a group of N equivalent nuclei of spin we use to N I plus 1 that is what I showed you in my previous slide. So, N total equals this is how it comes. So, I have some examples I will show you later. For example, if you have more than one type of nuclei how they are going to interact depending upon how far they are from the electron. So, now let us look into the relative intensities for I equals half again I equals half I equals 1 I equals 3 by 2 we were using Pascal's triangles in case of NMR. Very similarly we can also look into the relative intensity of the lines when we observe hyperfine splitting. For example, 0 we will see only one line when N equals 1 we will see two lines when N equals 2 we see three lines the intensities are 1 is to 2 is to 1 in case of 3 we have four lines 1 is to 3 is to 3 is to 1 when we have 4 we have 5 lines 1 is to 4 is to 6 is to 4 is to 1 and it continues like this. Now you can see in case of N equals 1 we will see two lines doublet 3 we will see triplet and when 4 are there we will see quadrate when 5 are there we will see quintet and when 6 are there we call sextet and when we have 7 we will see a septet line. So this relative intensity diagram whatever I have shown holds good for nuclear spin I equals half. Similarly, one can also write relative intensities for I equals 1 and again 0 one line will be there and 1 3 will be there for example, if you consider 14 N I equals 1 and then when you have 2 we will be having 5 lines 1 is to 2 is to 3 is to 2 is to 1 ratio and when we have 3 we will be having 7 lines 1 is to 3 is to 6 is to 7 is to 6 is to 3 is to 1 it goes and 4 and then 5 and 6 6 I can show you here we'll be having 13 lines 13 lines. So this how we can calculate and find out the relative intensities when they are interacting with a nuclear with I equals 1. So N equals 1 a triplet N equals 2 we saw that I showed you same thing I am repeating here again. Now let us look into some examples to begin with we shall consider a simple radical such as methyl radical here. Methyl radical we are talking about this electron here this would interact with the 3 protons of I equals half then if you see here if you update 2 N I plus 1 rule we expect 4 lines and how these 4 lines will come again it is very similar to NMR if you assume these 4 lines for example these 3 lines would be interacting in this fashion for example all of them the local magnetic field all of them can be aligned with respect to applied magnetic field or 1 can be down and they have 3 combination and 2 can be down and 1 up we have these combination and they are all degenerate and then we have all 3 opposing the applied magnetic field we will be having this one so we can arrive at 1 is to 3 is to 3 is to 1 this is exactly similar to NMR where we look into the coupling very similarly and then you can see the transitions this is 1 transition 2 3 4 4 transitions are there and we will see this how we will observe the EPR spectrum in case of methyl radical and if you want to assign this MI values this is plus 3 by 2 half minus half and minus 3 by 2 now let us look into this radical anion and here this electron would first couple with this 4 14 N that means if I use 2 N I plus 1 rule again we are expecting 9 lines I am not worried right now these things and also hydrogens we have here if you just leave and look into this one primarily we get 9 lines having this intensity ratio here so this is for radical anion so now let us look into examples of metal complexes coupled with non-equivalent nuclei so one is 15 N one is H hydrogen both of them are non- equivalent and they couple one after the other and now we can use either this way so if you recall how we were writing so first if you see here one doublet will be there this is either coupled with this one or this one both are having spin I equals half and then each one will be again so four lines will be there so that's what we see here we see doublet of doublets that can be seen when is coupled or interacting with nuclei which are non-equivalent then you can also use this fashion 2 N I plus 1 and 2 MI plus 1 you can use for two different one and multiply this also gives the same number here and if you want to look into the transitions to begin with we have something like this and then we have it is split like this first it will be split into say nitrogen and then if it is hydrogen we have and then if you see here this one two three four possible transitions can be seen and that is observed in this if you have spectrum of this metal complex only highlighted is one hydrogen and one 15 N here this is about 14 N here earlier we considered 15 N so I equals half now if you consider here first it is with hydrogen it is a doublet and then it would split into triplet of intensity 1 is to 1 is to 1 is to 1 so this is what is shown here and then of course here these are all the MI values for 14 N and this is for 1 H and then you can see this many lines six lines will be the six transitions are shown here one two three four five six and also we see here this is a very beautiful EPR spectrum with non-equivalent one one with I equals one and one I equals half now let us look into benzene radical anion benzene radical anion now this would couple with six equivalent hydrogen atoms seven lines will be there so if you look into a benzene radical anion so this is how we see EPR spectrum in the form of a septate with AH values 3.75 Ga or this one is 375 similarly one can also look into one hydroxy ethyl radical here in case of this one we get a quarter of doublets here so that means this one first couples with CH3 protons then it gives a quadrate and then each line in the quadrate will be split further because of hydrogen to doublet and we call it as quadrates of doublets we can see here this for one hydroxy ethyl radical let us look into acetamide radical here in acetamide radical we have here we have to consider this one and also we have to consider this one and also we have to consider this one so first let us say it couples with nitrogen it gives three lines and then each line will be split by two hydrogens again it will be a triplet and then these two will split further into triplet and you will get something like if you try to write here so each one this one will be three two plus one three and then this will be giving three one and this will be three so we get 27 lines we get in this one 27 lines so benzene radical if you see first it splits this one into a triplet three and then we have one two three four five five means six we will see 18 lines here so we can also tell how EPR spectrum of benzene radical would look like and then acetamide radical would look like this one okay now let us look into hexachromes manganese two plus and here electron spin is s equals half and nuclear spin is i equals 5 by 2 so since it is i equals 5 by 2 and nuclear spin it would split six lines here so we can see so that means if you record EPR spectrum for hexachromes manganese two plus this is how EPR spectrum would look like and also if you expand here you can write all possible states between which transition occurs or zema splitting happens can be seen here one two three four five six lines will be so this how you can show it is very similar to splitting or coupling tree we show in case of NMR spectroscopy so now let us look into one more metal complex vanadate ag-ag so here i equals 7 by 2 so this interaction with vanadium nucleus would be something like this here you can expect here since it is 7 by 2 one can expect eight lines and eight lines can be seen here very beautiful spectrum is here and then these corresponding transitions are shown here so this how one can analyze EPR spectra and then interpret first you find out equivalent nuclei that are going to interact and how they are going to split or how the hyper plane splitting looks like what are the relative intensities and then you can also show how the coupling or interaction happens in this fashion and then you can your job is done and here eight lines are a and here the g value is 2.0023 now let us look into of course i showed you benzene radical in case of benzene radical as i mentioned six hydrogen atoms are there and these six hydrogen atoms will interact with lone electron to give seven lines and here these seven lines are there again this is a beautiful spectrum EPR spectrum of benzene radical anion or sometime one can also write here this also shown like this here i think this is better to show something like this here now this is again interesting very nicely hyper plane splitting is shown in case of 13 butadiene anion radical here you can see here so here we have these if i say designate HA they are equivalent first it would couple with these four equivalent nuclei so that means basically five lines will be there you can see a quintet initially and then now these two will couple equally and each one will be a triplet here so there will be total of 15 lines so this is from one of the old journals of journal of chemical physics so initially it splits with four equivalent hydrogen atoms to give this quintet and then each one would be coupled with hb these two identical ones equivalent ones and a triplet and then we have triplets of quintet how you can say is triplets triplets of quintet now let us look into cyclopentadiene radical this is anion radical here and now this this is coupled with five equivalent hydrogen atoms so again use same 29 plus one rule it should show six lines here now let us look into naphthalene anion radical how it is made you take naphthalene in say one two dimethoxy ethane add potassium to that one immediately we can see the coloration blue coloration is coming that's essentially due to the formation of naphthalene anion radical so when you look into this radical in epr epr shows very nice hyperfine splitting you can see here initially it splits into one two three four five lines why that four lines are happening yes you should be able to see this and naphthalene and then naphthalene has eight hydrogen atoms yet hydrogen atoms can be designated something like this here we have this is one type of proton you can call it as beta proton and then we have this alpha protons are another four so that means in one is to one ratio we have two non-equivalent type of hydrogen atoms they would interact with this electron radical and split that in in this fashion so first it splits into a quintet and then each line is further split into a quintet to show this kind of multiple peaks that one can also see by just looking into as I mentioned 2ni plus one into 2ni plus one so it is first these things will be 25 lines you can expect here very nicely you can see 25 lines so one unpaid electron interacting with two sets alpha and beta of four equivalent protons so the ESS spectrum would thus show 2ni plus one 2ni 2 designation is given this is alpha and this is beta and then four plus one here four plus one five 25 lines by considering the pattern and coupling constant of A1 is 4.90 g and A2 equals 1.83 g of hyperpain splitting the species formed is consistent with naphthalene radical anion so this is a very beautiful one and this naphthalene anion is extensively used in organic chemistry and especially when sodium and potassium we want in stoichiometric amount in very small quantities it is very difficult to use from fresh sodium cutting and weighing because of its reaction with water to form and sodium hydroxide or sodium oxide something like that in that case what happens no matter how careful we are it is very difficult to weigh exact amount of sodium or potassium but on the other hand if we make something like this and if we standardize the solution we know the molarity and then we can use that solution as a source of sodium from that point of view this naphthalene sodium naphthalate or potassium naphthalate is quite useful in organic synthesis and also in organometallic synthesis so let me stop and continue discussion on more EPR problems in my next lecture until then have an excellent time thank you