 Hello everyone and welcome back to the course on actinide chemistry Yesterday, we have gone through some basic concept of PE And there we have seen that as we know that ph is one of the very crucial variable in controlling the Herbalysis behavior of actinide ion in the solution. Similarly, we thought of a variable that can be called as the PE maybe similar way that we call like ph that is nothing but ph is minus log of hydrogen and activity of concentration Similarly, suppose you can have a variable that is called PE that is again It can be very similar to the ph can be minus log of electron activity So we assume that suppose such a variable exists and somehow we can measure this for a system Then this will tell about the reducing or oxidizing Nature of the reducing or oxidizing capacity of the medium And then we have tried to see that how we can write different relations between PE PE node and The nature of or you can say the composition of the solution So here we have started with the the half cell reaction and we have gone through all these calculations So that we have put to this final equation And if you see this equation very carefully here we are relating our PE with the composition so you can say depending on your oxidized species And reduce the species so if you know this your concentrations of oxidized species and reduce species You can have some idea about PE in a very reverse way also You can say if you know the PE you can have some idea about the composition And then we have compared it with the noisest equation and we have drives of the relationship like this one For this one where you can see that In some of the slides you feel that I have given the term EH instead of purely E It is just to make you remember that In all these calculations rather I should say that it is Your E node is basically with respect to the standard hydrogen electrode So because of that we have given the notation as so many times I have used this H also So it is not very difficult to do you can just say that it just This was the relationship that with which respect you have measured your potential. So with this We have seen that we have a set of axial reactions Which standard of potential as we have measured with respect to SH3 They are known and if you know this Then from here you can just simply use this equation that is ln k is equal to minus E node f divided by RT and you will get the value of k And this is the k value And when you know this k value you can easily calculate what is P E node that is nothing but log of k divided by And so you can calculate this value and again you can fit it into this equation that we have derived And from there you can use these relations to draw your EHPS diagram and we have also seen that What are the conditions or rather I should say what are the boundary conditions that exist when you are using water as a medium This conditions arises mainly because the water as you can see Can oxidize suppose you have water It has both the limit it can oxidize to O2 And it can reduce to SH2 So these are the limited sort of conditions That water can undergo and because of that we should work In the conditions which limits these reactions We should not be having a condition where what is getting oxidized to oxygen or it is getting reduced to the hydrogen So based on this equation these are the reduction equations. We have derived two relationships P is equal to 20.75 minus pH and here we have seen that the slope is Minus one and here also we have derived the equation for the reducing limit of the diagram and there also we are getting slope of Minus one and here we can see this is our Conditions this is our oxidizing condition above this line the activity of oxygen is more than one Similarly below this line, this is the reducing limit of the media below this line The activity of hydrogen is one and all the reactions when we are considering the water medium Could be happening inside this room All the reactions should be inside this zone and then we have seen that When we use the equation that we have derived that P is equal to P0 plus 1 by N log of oxidized species divided by reduced species Then we can use this equation and we'll get a different lines in the pH PE diagram and again this is your oxidizing and reducing limits and then we have talked about two kind of line and then area The first thing you can see the lines can be horizontal The lines can be vertical or the lines can be small When we talk about the vertical lines, these are pure acid based reaction and they do not depend on the potential When we talk about the horizontal lines, they are show a pH independent reaction And these lines when you see the boundary lines, suppose you have seen this line The activity of these two Are same on this line. So you can say they mark a place where the two species exist in equilibrium Then we have also discussed about the area Larger the stability area or larger the area species will occupy more is the stability of the particular species and with this knowledge of pH PE diagram We can understand some redox chemistry of the actinide into the environmental media because if you see the pH The ranges are from 2 to 1 which is obviously your environment pH should come within this pH Similarly, your pH ranges are also in a range where the environmental age comes into picture So with this knowledge Can we draw because this I have shown you for an actual name, but Can we draw for other mental lines? So this is the list I've given you for different actinides starting from actinium to Novalium and you can see that for all these actinides I've given you the standard potentials And as I've told you in the previous slide that if you have information about this standard Reduction potential you can easily drive the LMK values And once we have this LMK value You can use these two equations you can drive From this LMK you can drive PE and that PE you can put it here and then from this PE node You can drive an equation between PE and depending on the Reactions suppose your reaction does not have any hydrogen in that then obviously the pH term will not come to the picture And similarly if your equation does not have any electron term into that Then PE term will not come to the picture. So suppose you take a reaction suppose this one I take In which both the things are there. You have a proton you have an electron So if for such kind of reactions you will get Equation which involves both PE as well as pH So you can draw you just know the equilibrium you can easily draw the lines for the php diagram using this set of equations Why these equations or why these lines are very important to us? So when we talk about the redox chemistry, you all know that this kind of series that I've given here It is very useful when you plan to have an experiment In which you are trying to oxidize or reduce a species Using some will depend on oxygen. For example, suppose I have a species that is NPO2 plus And I want to reduce it to N4 So for this direction To be feasible. I want a couple which can facilitate this reduction. So by looking the standard potentials By looking this standard potential You can choose The two couples and you can add them and you can see that how the reduction is taking place or in presence of which couple Which couple is more favorable? But we have to see that As we have seen that these are all reduction couple And you have to see the delta g values That is nothing but minus nf e naught And so the two reactions you have to see whose e naught Is more positive. So when e naught is positive, it means delta g is more and more negative So that reduction is more feasible So if you have a couple whose reduction is more feasible compared to other couple Then obviously the couple whose reduction is more feasible will go into the Reaction rather than the one whose reduction couple is not Completely not feasible. So based on that we can choose different couple and we can carry out this kind of production reactions They are very important and one of the example. I just want to discuss is that For neptunium again for neptunium 5 we use a fairer sulphamide And if you see this couple neptunium 5 to neptunium 4 In the presence of acid you have to take care of that also in the presence of acid. It is roughly 0.74 And if you see the iron couple It is from iron 3 plus That is what we have shown to iron couples This is 0.7 So this reduction using iron Of neptunium 5 to neptunium 4 is not looks very feasible In this as you can see from the couple this is Giving you more negative delta g value compared to this one So in those conditions, what we generally do As I have written we use the complexing agent that is called fairer sulphamide But the complexing agent will do it will Try to change the thermodynamics of the reactions What it will do during the reaction it will react with both iron 3 As well as neptunium 4 to facilitate the forward reaction So even in situations where the reduction is not black feasible We can try to do the reduction just by adding some kind of complexing agent So this complexing agents this protons everything that you are having in the medium has some effect on your enote And this can definitely change the mode of the reactions so with this We move to the The subject that is basically a current spectroscopy want to discuss Many I see spectroscopy the first thing that comes into the mind is the the color And as you can see from here the lanthanides and Ectanates also I will show in the next slides That they both are having way different colors very beautiful colors And here I have shown that For some of the lanthanides Which are colored or some are colorless will try to discuss that why they are colorless and why some of them are having color The question comes here is like why why we see the color first thing so as you all know that We have a spectrum that uv spectrum and There are certain peaks for different metal ion and suppose you are having peak or absorbance in certain rates according to that You have some spectral colors And what we see is not that spectral color, but we see is the complementary color to the spectral color But before going to the lanthanidic spectroscopy or the uv studies or the emission studies I hope that you have a fundamental understanding of the reasons in the electromagnetic spectrum that we are using for the uv studies or an ir studies So I assume that you know at least those basic things, you know at least lembut be at low that how your evidence is related to the Epsilon to the concentration and to the Path length of the cell you are using for the measurement So I hope you have the fundamental knowledge of all these things. So we can go directly to the lanthanides or actinides absorption spectroscopy And as I've shown you that we see different colors And this is also true for lanthanides and actinides depending on the evidence lines Or you can see the bands in the uv spectra But before going to that let us try to have some information about Term symbols. What is term symbol? In general, we all know about the electronic configuration. We have Some certain configuration that we know we have certain configurations But if you see the configuration, we are having a number of orbitals to field up with a given set of number of electrons And if you see that there are number of permutations and combinations are available that can be used for the feeling of electron into the orbital for example If you see you are having, let us say you are having a d orbital Which is having 10 possibilities And you have a new 2 electrons. So the number of electrons you can say is I am saying 10 possibilities because each orbital can have to add the max 2 electrons So we have 10 possibilities. So in these cases if you see the combination that are possible is 10c2 And from the basic mathematics, you know that if a relation having ncr it is nothing but n factorial divided by n minus r factorial into r factorial. So for this system like 10c2 What will be there if you calculate the number of combination or the number of These in which the electron can be filled in these orbitals Is very huge. Here if you see that is 10 factorial divided by 2 factorial into 8 factorial. If you solve it you will be getting 90 divided by 2 which is nothing but So here you can say for a simple system of d2 You are having almost 45 combinations that are possible So how do these possible combinations because electronic configuration can give you all this 45 configuration based on the ways in which the electron can be filled in the orbital, but Do all these configuration have the same energy But they are degenerate The answer is no Depending on your field They may not be degenerate So we know that there are possibilities of different kind of electronic configuration And each configuration will give rise to something called a term symbol one term symbol A term symbol is basically if you have an electronic configuration and you want to understand the energetic part of that configuration We have to talk about the term symbol because they will relate to the energetic part of the electronic configuration And as I have written here For case of european which has a configuration of 4f6 It's degeneracy it can be lifted either by partially or totally By server perturbation that is acting on this ion such as the electron-electron repulsion Or the spin orbit coupling or the crystal field perturbation and sometimes the zemen effect I hope you have a basic understandings of Spin orbit coupling that Will not talk about much of this spin orbit coupling, but just For your revision We can have two type of coupling then one is called the LS coupling or russians van der coupling First and the second you can talk about the JJ couplings What is the basic difference between the two coupling schemes is that The first coupling scheme that is the LS couplings scheme or the russians van der coupling scheme in which what we'll do Suppose we have a system for electronic configuration in which you are having Two or three electron and every electron suppose you are having three electron Then every electron has a spin as one as two as three and every electron has a individual And when you talk about the LS coupling what we do we sum up All the spins of the electrons we get a total value We sum up all the else to get a total value of 10 and this will Again couple weekly to give something called j. So here what we do with individually first couple the spins Then momentum and then we'll try to get coupling of snl to give the j value Whereas in the jj coupling, which is mostly true for the higher elements Here we start with the coupling of snl and here we are getting j for individual electron And these j couple weekly to give jj coupling. So mostly these are LS couplings are valid. Maybe up to atomic number of 30 But after that mostly it is the jj coupling that takes place But we will try to discuss mainly this LS coupling when we are using this term symbol because this is the Way we can always try to explain in a very simple way about the elaborating functions that are happening in Lengthenets and actinide and this is the enough for our understanding of actinide or lengthenide spectroscopy. So here I've given you some example that How you can calculate the number of configuration or rather we say the number of microstrates For a given number of orbiters and given number of electrons. So here I've taken example of f6 And as I've shown you for the d2 here also you can say that Since we are having f we can have 14 possibility of electron arrangement and then order number of electrons are 6. So the ways in which we can arrange them is nothing but 14, 3, 6. And if you solve this you'll be getting around 3003 ways You can just imagine you have almost 3003 ways of the arrangement of electron into The f orbiters when we are having simple and f6 configuration and all these configurations are not energetically same So we'll try to see that what configuration has what energy and what is the ground sugar configuration and this understanding comes from the term symbol Because the term symbol will tell you about the stability or you can say the energetics the energetic labeling of this electronic configuration How do we find out the term symbol? This is Not a very tedious process and I hope many of you might have gone through this in your BC or MSc classes, but just for the sake of completion I've just Give a very quick example of f6 which I have taken so Just you see the first line if you see If you just assume the electron electron repulsion you will be having 119 groups Of different terms in bulk and obviously if you sum those 119 group the total number should be 3003 And each term having a degeneracy of 2s plus 1 into 12 plus 1 Where s is again as I've told since we are talking about the ls coupling It is the total spin quantum number and l is the total orbital angular quantum number So if you have the idea about the total spin quantum number you can easily calculate Something that is called spin multiplicity which is nothing but 2s plus 1 and as you can see there is a One term here. So if we're having even number of electron We'll get an odd parity if we're having an odd number of electron you are getting even spin multiplicity So for 4 f6 It is even system right here in even number of electron 4 even number of electron we should get odd spin multiplicity So in this case we can get maximally 1 3 5 7 Etc. So, let us try To drive term symbol for this f6 configuration Which will be written form like this where we write on top this spin multiplicity L is the prolingual momentum that we have seen is sum of l1 l2 l3 that is the orbital momentum of all the electron And j is nothing but it talks about the coupling of l and s which goes from l plus s2 l minus s It goes from this so let us see what will happen when you have a system like the f6 if you have a system like f6 And if you try I'll just give you the ground states And try like one two three four five six So you should say like plus three plus two plus one zero minus one minus three And f6 electron fill up you can start like one two three four five six And if you see the total spin since you are having six electron and everybody is like in the spin of half So the total spinning is nothing but It's three so your total spinning is coming Three what about the total error? Since you can see Some of the else are positive some are adding negative means the direction of the orbital momentum is different so if you sum You'll find the total l value is coming positive three So when you are having s is equal to three L is equal to three what is the spin multiplicity is nothing but two s plus one It will be coming around When you are having l three, what does it mean that? Your angular momentum as I've written here also these terms are denoted by capital letter of the latent alphabet So when you're having l is equal to three you are here So what we are getting is f is equal to what we are getting is f So now as I've written The term symbol is two s plus one l j two s plus one in this case is seven L is three which is dealing me So you can say you have a spay group you have a term symbol that is seven What about the g's number as I've shown that they depend the coupling of Spin and the orbital angular momentum and the values can be from l plus h to l minus s Since again, we are having l is equal to three This is equal to three your l plus s is six and your l minus is Zero you can start from six and go up to the zero value here. I have just shown you the same in the diagram That seven f and then j value takes five four three two And zero how this ordering is coming into the picture that we will just see in the next slide that How the order comes into the picture? So we are having these six states and we'll see that how they are arranged So the ground state term symbol how to get the ground state term symbol The ground state term symbols can be easily done by the Moon's rule The first thing is the spin multiplicity should be on the largest scale If the spin multiplicity is same for the two term That you are having the same spin multiplicity then you have to look for the orbital angular momentum Rule three is that if the electronic shell is less than half build the ground state has the lowest possible day And if it is more than half build it should be having the highest possible day Since we are talking about an f6 system Which is less than half build So it is less than half build so we should be having a ground state with the low j value And since everybody is having the same multiplicity You can just arrange them on the basis of j the ground should should be having the lowest value that is seven f zero So if we see here the ground state is seven f zero and others are seven f ones seven f twos seven f three like that So with this idea of term symbol, let us try to understand the difference spectroscopy of lengthenites and actinites The basic difference between the spectroscopy or UVV spectra of lengthenites actinite compared to One of the d block elements is the involvement of different Orbitals if we see the d orbitals, they are more diffused when you see the f orbital They are more centered compared to the d orbital and or the electronic currents that are happening in the f orbitals Are quite deep buried inside So their intensity is not very much dependent on the external environment But because the four f shells are weakly effective They are weakly affected by the most ions or the ligands that represent outside because they are inside this You have a field orbital of five f two and five p six And the f orbitals are inside so they do not see much of the Outer environment and because of that the transitions are quite sharp when you compare them with respect to The d block elements But if you have internal comparison between lengthenite and actinite Then lengthenites use four f actinites use five f And here again we see if you see the spatial distribution of four f and five f These are more diffused So you can easily say that if four f is more diffused So if you compare the broadness of lengthenite and actinite in a given oxygen state You can see the five f may be a little more broader because of the more diffusion compared to the four f And they are also More interactive with the ligand field compared to the lengthenite We'll see some of the spectra of travel and lengthenite and actinites But before that we'll just see this setup. This is the I think very well known to you that how we measure this the UV visible adoption we have a covet In which we can put our analyte and this is the double beam UV spectrometer in which we use a monochromatic beam Which will split into two parts and then what is going through the reference that's nothing but everything except the metal ion or analyte that I want to use and then we take a reference of that and this You know the lambda to be at low and since you are getting a from this kind of absorbance you will get something like this In which your a is nothing but this And since you are getting a you know the concentration you can always Calculate what is the epsilon or the molar activity? This is I think everybody knows I will not spend much of time on this So what kind of transitions are generally possible in lengthenite and actinite? I have already told that from optical point of view they are Well a part of other elements because they are deep buried orbitals and They do not participate much as compared to the deep look elements into the bonding And what kind of transition that we can expect in lengthenite and actinites? These are three kinds of transition one is called Interconfiguration transition the second is interconfiguration transition and the charge transfer we'll discuss them one by one When I say the term interconfiguration Rather sometimes it is also called as FF transitions These are ideal apparatus for beta right because we know that for a transition to be allowed This should not be zero this should be plus one plus management Similarly When you talk about this and then to trick change in the spin multiple should be zero. So this rule is not followed here Because your transition is between F and F But still we see some transition we'll see why and then intraconfiguration transitions Here compared to this transition it is more favorable because it is like retail out And the third one is charge transfer we'll see in What kind of actinites we'll get this kind of transition And here I've just shown you some transition of uranium that we are getting And if you see Carefully then you see that just by changing the complexation conditions when you're going from one complex to another complex There is some change in the evolution pattern and that is generally used to identify which kind of species is there and we are also Using this kind of pattern to get the stability constant value or we are getting the information about the strength of complexation But we'll discuss some of these things in the next slide So yeah, the first thing is like intraconfiguration FF transitions as a rule they are Very forbidden, but still they are Seen in the spectrophotometer. Why because If you see the selection rule it is obviously not allowed But these states are not purely F in the nature when you see there is some admission of Of the opposite parity into this configuration You can say there is always some perturbation of this F state wave function with the other orbit it may be Deorbited so there is some perturbation or some mixing in this state and because of this mixing when we say The parity is same. It is not true There is a bit different parity and because of that it is slightly allowed. You can say the rule is Bit relaxing and because of that we get this kind of transitions as they are deep buried inside They are not affected by the ligand field. They told you that The outer electrons that are present there They will not allow the ligand field to come up to the orbital and because of their Leopardy forbidden reason or maybe partially allowed reason their intensities are not very high They are very low. So the molar activity of these kind of transitions are not very high They transitions are sharp as that shown in the is the previous reason that Since they are deep buried inside and they are getting shifted from the outermost shells, which are filled And some of the example we'll see again in the coming slides Then we come to the inter-configuration of transition that is every transitions as we can see here the del is L is Following the Leopardy rule So these are fairly allowed transition But again as we see the electron is going from F to D And this is shielded But this is not that shielded because of that It's broadness is called on a higher side compared to the Inter-configurational transitions Since it is allowed transition. So it is also Intense transition and some of the examples are like serium, typhositive, serium, prosycinium and When we talk about the the third transition that is called charge transfer transitions in what kind of scenario we can be through this kind of transition So to get these up to transition that is inter-configuration transition and inter-configuration transition The requirement is the F should have some electron and should not be zero It should have some electron because you are trying to excite electron even in your FF You are having some electron here and you are putting electron in this in your FD again You are having electron here and you're putting here But in certain cases when there is no F electron In those cases also if we see some kind of transition That are mainly arising from the charge transfer and one of the example is urinial ion that is O22 plus if you see this is urinium 6 And if you see the electronic configuration of urinium 6 you will get an x0 state So it is a zero But if you take over the spectrum you'll get a very good spectrum as I've shown you in the previous slide So for the x0 system Both the upper transition that is inter-configuration transition and inter-configuration transition are not possible And we are only Getting the spectrum that we get because of the charge transfer and why the charge transfer is happening and how it looks like We'll see that in the next slide so When you just compare these the actinide ion and the lanthanide ion is actinides having comparatively broader compared to the lanthanides because the pi of orbital is more difficult compared to the lanthanides And we'll try to see that how the spectrum will look like and We'll continue this in the next section. Thank you. Thank you very much