 Hello everyone and welcome back to the series of lecture on actinical chemistry. In the last lecture we have studied about some spectroscopic aspects of proletariat and actinide and in continuation to that we will try to understand some of the utilities of the emission spectroscopy and here as we have also discussed in the last slide last lecture that how we can record this spectra that is the excitation spectra emission spectra and actin spectra and here the excitation source we are using is basically some electromagnetic wave that is basically UVVs generally we use we generally say it is like photo luminescence and in short form we also say it like PL studies so in this photo luminescence we excite with the source and then we measure the emission and we will try to extract some of the information that we are getting from this excitation spectra and emission spectra as well as on the lifetime spectra. So in brief I have told you that how to measure this excitation spectra in the excitation spectra or to measure the excitation spectra first thing you have to do you have to fix the emission wavelength you have to fix it and once you fix it then you just scan the excitation monochromator and then you will be getting some peak so suppose I am taking an example of europium 3 and this peak generally comes around 394 nanometer so this is my excitation in which my emission is fixed my emission is fixed when I talk about the emission spectra what I will do in emission spectra since I know this is the peak that is the giving the maximum excitation I will fix my excitation at this peak and then I record the emission spectra so by that we are having information about the excitation spectra we are having information about the emission spectra then what we will do we will take this maxima this maxima we will fix this to maxima and we will try to record the decay half life or the half life of the or the decay life of the complex that we are interested in so as I showed you that we can measure excitation spectra you can also measure the emission half lives we will talk about that emission half life later so what are the information that we can get from this excitation spectra I have shown you several excitation spectra that I have taken from Google so you can see there is a difference when you start from A to E you can say sometime you are getting and most of the time you are getting peak in this reason and that is the excitation spectra of europium with some complex so you are getting a peak at 394 but if you see there is a difference here and sometimes you get a peak here so what this suggests and marking of information we can draw on this kind of spectra let us assume that you are having a two complex then one complex I say is europium L and the other complex you can say like europium L2 let us say this is one and I am just arbitrarily assigning europium L1 to this spectra and the L2 to this spectra so just by looking at this spectra what I can say there is a difference in this but it says that this sharp transitions are in both this spectra so we know that the FF transitions are sharp but you can see this transition is not sharp this is a broad addition and it is not an FF transition rather it is a charge transfer band or charge transfer transition so what we can say from this charge transfer transition or charge transfer transition band is that out of these two complex that I have written to say L1 L2 this particular ligand that is L2 having a better overlap or more overlap to europium rather than L1 so what it is doing there is a charge transfer you are exciting the ligand and ligand is transferring to the europium and from there you are getting some emission so we can get this kind of information from the extension spectra even in certain complexes the charge transfer band is so large or you can say so intense because you take this spectra of europium with TTA I hope you must have heard this name TTA in your extraction classes if you take the spectra of this the charge transfer band is too huge you can say as these transitions are just like very very small as compared to the charge transfer band so the information of the metal ligand or overlap can be seen from this kind of spectra okay now you have some excitation spectra and then you are sitting on the excitation maxima let us say here or here whatever excitation maxima you choose and then you record an emission spectra so suppose I hit on some maxima and then I record the spectra and suppose again I am just arbitrarily assigning it L1 and L2 and you get some kind of spectra like this European L1 and L2 so what just by seeing this spectra one thing is very clear to me the spectra do not look very much similar so what I can say that the environment around europium in these two complexes are different what it is we will see later but it is different as we see this is different from this so this is the direct information that we can get from emission spectra but we should do it very closely then we can see there are several transitions that I have marked there and you know that we start with FD0, FD0, FD0 and we give transition to 7F0, 7F1, 7F2 and all the transitions are marked here so let us say the first transition that is 5D02, F1, F2, 0 many times we just say this is J0 and this is J0 so many times we just say it is like 0, 0 transition sometime I am also using the term like 0, 0 transition so if you see this 0, 0 transition then it is like 0, 1 this is 0, 2 like that you can name like that also so the information that is there in the 0, 0 transition let us see pure 0, 0 transition but this is my 0, 0 transition that happened to be found near 550 nanometer for europium we also know that the transitions that we say that are coming from 5D0 to 7, 0 to 7F1 we have seen that there is a splitting of J level there is a splitting of J level and this J level will be split into 2J plus 1 states that I have seen in the previous slide like this yeah you can see that this J level depending on the symmetry can split maximum into 2J plus 1 so but when we see about the 0, 0 transition what is the J value, J is 0 so you have 2J plus 1V, 1 so you can say even in symmetries of different environment this transition cannot split what is the use of this let us say let us say I have a spectra I have just recorded a spectra I am getting it like this and the reason that is maybe 577 to 579 or so I am getting these 2 peaks one thing I am sure this is not the splitting why because it is a 0, 0 transition so what information it gives to me that there is at least present of 2 europium species with 2 different dominant species of europium are present in the complex so this kind of information I can that figure from this 0, 0 peak let us say the other peak that is 0, 1 peak the 0, 1 peak can split into J plus 1 giving rise to 3 so if you are having this peak as you can say in this spectra there is some splitting similarly when you are having to 0, 2 this can give you 5 peaks so it can split into maximum 5 this all depends on the symmetry around the europium ion so lower is the symmetry more is the splitting so we can get information about the symmetry class if you know splitting of all the lines and then you have to just see the table and suppose you are having some system which is having XR on the symmetry and if you just see the first should be 1 then the second should be split into 2, third should be 3 likewise if you see if you are matching with splitting of your splitting of the pattern is matching with this table then you can safely say that that the symmetry around my europium ion in this particular complex is XR 1 so you can get that information also from this there is one more thing that we can know as asymmetry ratio so suppose they are having a transition 0, 1 and then 0, 2 and we have two transitions 0, 1 transition we have discussed it is called magnetic dipole whereas 0, 2 transition is called electric dipole we have also seen that this magnetic dipole transition do not get much affected by the ligand environment so if we take the ratio of the electric dipole transition to the magnetic dipole transition we are getting something that is called asymmetry ratio that is we are taking a ratio of the electric dipole transition with the magnetic dipole transition what it means that the 0, 2 transition divide the 0, 1 transition and in a typical spectrum of water if you take this ratio the water spectrum is like this one then this this is your 0, 1 this is your 0, 2 and if you take the ratio it will be somewhere between 0.6 so generally it so happens that when there is an increase in this ratio we say that the asymmetry around the europium ion is increasing so that asymmetry information also sometime we can derive just by using this asymmetry ratio so now we can derive some information okay we know this is the symmetry and this is the kind of complex that is there with europium now let's try to see how we can deal about the lifetimes I'll come back to this slide after sometime yeah so we also kind of record the lifetime in which we fix the excertion spectrum we fix the excertion wavelength we fix the emission wavelength we fix both that's certain wavelength and then we give an excertion pulse and then we monitor the decay and when we are getting this kind of decay we simply use this equation to fit this decay and depending on the nature of the complex we can get different lifetime but generally we start with the mono exponential we fit and then suppose we are not able to fit then we go for the bio exponential and try exponential but generally we restrict ourselves to bio exponential only but can fit it into different kind of exponential until once you are getting a good fit but not more than bio try because if you are using a very high exponential then it gives a good fit but it is logically not very valid so why the lifetime is getting decreased generally assume that for the lengthenite cases or the ignite cases the decrease in lifetime is mainly due to the water we can say it is the OH vibrations of the water molecule so most of the time we assume that the decay in the lifetime is mainly concerned with the vibration of the OH excertion of the water if water is the only responsible ion for the molecule for the decay then we can directly relate the number of water molecule in the primary sphere to this decay curve and it has been done and this is the relation this is the linear relation basically that directly relates the number of water in the primary adjacent sphere to the decay line so you have to just fit the curve you will get the decay constant and from the decay constant you are just putting into this equation and from this equation you can getting the number of water in the primary adjacent sphere and for european it so happens that the lifetime in water is coming around 110 microsecond which is giving you a life water molecule around 809 so for european we can get this 809 water molecule with lifetime of around 110 microsecond okay what is the utility of this let us assume that you are having european with let us say eight water molecules one two three four five six seven eight and then you add a ligand in the system and suppose your ligand is a bioenter system what will happen we assume that it will just go and bind to the mental ion so that now my ligand is here so you can say out of the eight position two positions are occupied by the ligand molecule now if I just measure the lifetime of this molecule and we will get the number of water molecule which we are getting eight in the case of european with the aqua ions or in the aqua adjacent sphere this should be reduced to six so if the hydration number is reduced to six we can say yes there is some complex formation similarly suppose you increase the ligand concentration or the pH depending on the nature of the ligand you may be getting minus two two complex assuming again a bioenter ligand then four water molecules should be deployed and then by putting this equation you should be able to get four water molecules only so that to exemplify that I've given you the example of a compensation with some polyamino carboxylate and this is for a curium and they again for the curium we can see this is the equation they use these are the various different polyamino carboxylate ligand and if you see the first one here you can see you start here which is the hydration number of around nine to ten and the movement you add this ligand at pH two there is a sharp difference in the number of hydration sphere what does it mean that at this pH there is a strong complexion that now the ligand is complex then most of your water molecule has been removed compared to that if you see this curve if you see this curve there the number of primary hydration sphere are still intact there are some water molecule in the primary hydration sphere you can say about the density of the ligand that will affect the by the density of the ligand here you can maybe having a one to one complex or here maybe you are having one or two complex but whether it is one of two one or one to two we cannot say directly by this because that will depend on the density suppose you are having a ligand which is having such as ADTA even with one is to one it can remove mostly all the water molecule from the your inner sphere of the metal ion this is for NTMA again you can see the two plateau at certain pH you are having one is to one if you increase the pH you are getting one is to two so just by recording the lifetime a different pH you can have some information about the number of water in the primary hydration sphere and you can also tell something about whether it is a one is to one complex or one is to two complex till now it is okay let us say we are having only one species the things are looking very easy till now but suppose in the system we are having more than one species suppose you are having let us say two species you are one of two one and fewer ones to two you are having a mixture in those cases what will happen that instead of a mono exponential curve you will get some curve which is it can easily be fitted into a bi exponential curve you will get a curve like this and when you fit it into the mono exponential you may not be getting very good fitting then you go to the bi exponential fitting and you are getting some good fitting so now we know that your system is having more than one based on the lifetime spectra but suppose I want to know that what is the emission or excitation spectra remember that this excitation and emission spectra that we have recorded that we have recorded in the steady state when I say steady state it means there is nothing to do with the time we have just excited and we have recorded but the decay lifetime says there are two species and suppose I am interested in getting the spectra of both the species that okay what is the spectra of the short lived or one of two one complex and what is the spectra of the long lived or must do complex so now as I was telling about that suppose you are having two species and you want to know that what is the contribution or rather what is the nature of the spectra of one of two one and one of two two for those kind of cases we cannot use the steady state emission profile why because there is no mining information but from the timing information that is the lifetime spectroscopy we know that one species is decaying faster than the other for simplicity I have assumed two species let us say you have a system which is having two species one is decaying with the lifetime worker 100 micro second and another with the micro second when I say lifetime I mean this is 100 micro second when we record the spectra if you look it carefully suppose you record a spectra which is giving like this these pieces with 100 micro second will decay first and that species with 500 micro second will be having a long tail why because the the key constant that we derived is just by fitting this equation if you see this equation these two equation then we can say okay 100 is decaying very fast compared to the 500 which is obviously giving you a tail and I want to record the spectra so initially with a steady state suppose I record a spectra which looks like this that I've given here and this is the steady state that's I have written t is equal to 0 and for simplicity I assume that these two peaks this peak and this shoulder is coming from the shorter component that is the 100 micro second component and the rest of the spectra is the from the 500 micro second component so what I want to record I want to record both I cannot do with the steady state so now what I will do I will do time residual fluorescence spectroscopy and this what I will do I will excite my complex with a wavelength that is say 394 in this case and then I'll wait for certain amount of time why to wait because suppose I excited this particular time and I'm waiting for this time this is my waiting time this is time I'm waiting and then I'm recording spectra what will happen let us say you have 100 and 500 and you are waiting for around the 500 micro seconds suppose this limit is around 500 micro second if you see the decay of the 100 micro second species in the 500 micro second species you'll find that if I am waiting up to 500 micro second you have given almost 5 life to the 100 micro second species whereas for the 500 micro second every one only 1 so if you see the decrease if you see that the intensity which is a combination of both 500 and 100 but if you see the decrease the decrease is this the intensity decrease is proportional to e to the power t by tau in this case is my t is my waiting time so my t is 500 and my rho is only 100 so here I'm getting a value of 5 it is the power 5 you can see my intensity at around 500 micro second is nothing but I know divided by it is power 5 which means that there is a decrease of almost 140 times so there is a decrease of around 148 times in the excitation state of the smaller component whereas if you see the longer component it has only one half life or one decay life so for this the decrease is just like 2.72 times or mainly the signal that is there is 63 percent from the L2 and hardly 1 percent from the L1 or you can see the shorter component so if we are giving a decay and you are missing after a decay the profile which you'll get is mainly composed of the long live component so I'm expecting that this peak should be not there the first peak then we are getting second peak and the shoulder should not be there and where should there be so if it is a spectra of a plus b then this should be the spectra of the b assuming that b is having a lifetime of 500 micro second so in this way you know a plus b you know b and you can you can subtract a plus b minus b this is not that direct we have to understand the quantum also but assuming that everything is the similar for both the complexes I can say that you can just subtract it to get the a so you can record the exception spectra as well as emission spectra using this time results in spectroscopy so now we have some idea and some application of for this spectroscopy now I just want to have a very brief discussion about the retrospective that we have already discussed but I just want to give you some of the examples that for the ectonites I have already discussed to you about this table that what are the redox potential of different ectonites and from this table you can easily see that some of the redox potentials are way on higher side so let us say for thorium it is high so it is very difficult to reduce thorium 4 to thorium 3 but some of the ectonites such as the plutonium having redox all of around one volt so there are a lot of chemistry a lot of redox chemistry can happen so why we require this redox chemistry because many a time it happens that we require this for the separation because most of the time our separation is based on the oxygen state of the metal ion and you must have heard and you must have listened this purest process about the purest process many a time we play with the oxygen state for their determination that is the generally we do for the pyronym estimation using a Davies Gray method and suppose you want to understand the chemistry of the retinal oxygen state even in those cases we want to have information about the redox potential so that we can play with the redox potential and with certain complexing agent or we can add some you know some different kind of redox couple that can either easily reduce or oxidize my metal ion so to exemplify suppose I've given a very simple example let us say that from our class book chemistry suppose you have here having a couple of iron that is getting reduced to iron 2 and you are having a couple that is between iodide and iodine gas so suppose I take these two and I will mix them and after that I'll just add a drop of starch what I found then the color of the solution is becoming blue why the color of solution is becoming blue because if you see this couple it so happens that the reduction potential of this is 0.77 the first one iron and reduction potential of this is low that is 0.54 so it so happens that iron will get reduced whereas this will go oxidize to iodine gas and when you are adding starch to the iodine gas it will make a complex that will be blue in color now to this blue color you add two drops of edda now what edda will do edda can complex iron 3 plus and iron 2 plus and the moment it will complex their redox behavior will change now the reduction of couple is no more 0.77 rather it decreases to something like 0.12 and since the redox potential has been changed so the course of reduction will change and you will get a colorless solution why because now there is no more iodine now what you are getting is now iodine is getting reduced to iodide and your iron is getting oxidized to iron 3 plus so this is the variation in the redox couple I just want to emphasize just by addition of some ligand how we can do it mathematically it is a very simple equation that I've taken from a paper you can just easily see that suppose we are having a redox couple that is m plus is going to mn minus 1 and you can always write delta g for that similarly similarly suppose there is a complexation for the complexation you can again write so this is the you have started with the oxidized one this is the product that I do use one again you can write for the complexation I am assuming that l is a neutral ligand and when you write this you can write the corresponding stability constants also that is k m and k mn and with simple mathematical juggling what you found that you can easily write this equation and if you closely look at this equation then now you are not talking about the pure meta line now you are talking about the complexes the ml complex is getting reduced to ml2 complex with the total gain of n electrons so now we are talking about the redox potential of this particular couple and if you see mathematically what we found if you write the equation for this you found that the e0 complex is nothing but e0 aqa minus rt upon nfl km upon kmn this is the important term why because now your potential is totally dependent on the stability of these two complexes as I have shown you that edt example that the complexes of edt with iron 3 and iron 2 their stability must be different so their stability constant must be different so which is more stable depending on that you will get a potential which is either left shifted or right shifted your potential may be more negative than the original or maybe more positive than the original one generally it so happens that when you are adding some kind of complexing agent some of the complexes get stabilized and you will get a reduction of the lower potential and this is again just I have shown you that the davis gray meta it is a very important method in the estimation of uroname in the nuclear industries here what people have done that they played cleverly with the product potential of iron complex iron couple and iron couple so if you see that the iron couple having a potential of around 660 millivolts in sulfuric acid but the moment you start adding phosphoric acid there is a decrease in the reduction potential whereas under similar condition if you see the couple of uranium that is almost unchanged and then rising so here if you see that the reduction couple of iron is higher side than the uranium whereas the reverse is true here what it means that here iron will try to reduce and uranium will get oxidized whereas here uranium will get reduced so now you are having uranium 6 that will get reduced to uranium 4 so now uranium 6 will get to uranium 4 uranium will get reduced and iron will get oxidized so just by changing the media we can see there is a huge effect of medium and the potential and just by with the playing the media we can play with the reduced potential and that is generally used to carry out different transformation that a lot of transformation even for the actinides but instead of metallic reducing or oxidizing agent we are mainly preffering something which is the kind of more organic so hydroxylamine hydrochloride or hydrazine or some time hydrogen peroxide they are mostly used in changing this couple of actinide and some of the reagent that are generally used for necronium in aqueous phase are given here and similarly for protonium also have given here so this is just to make you understand that in different kind of media the potential can shift here and there and this understanding of potential and this understanding of their shifting with different kind of reagents are generally used for their either selective reduction or selective oxidation and this process along with the acceptance that we use are generally used for the operation chemistry or understanding the chemistry of the lanthanides and actinides into the aqueous phase and with this I just want to end the today's lecture and thank you thank you very much