Hello everyone, welcome back to the series of lectures on actinide chemistry and as we have discussed a bit about the aquatic chemistry of actinide elements, so today we will try to learn something about PHP diagram and before going to the presentation I just want to have a mention of the references that I have viewed for this presentation. The first is the chemistry of actinide and transactinide elements and second is the aquatic chemistry. I have also used some of the program for the presentation plot that I have taken from the KTS that is known as Medusa or Hydra. Some of the images are taken from the internet or web as per the convenience. I have modified them. So these are the few references that I have mainly used. Beside that, whatever references that I have used I have tried to quote them on the respective slides. Before going to the next lecture I would just like to have a mention of whatever we have read in the previous lecture and a recap of that. So there we have started with the electronic configuration and we have seen that the group oxidation state of lanthanides are plus 3. Whereas when you talk about the actinides, they can have a variable oxidation state starting from plus 2 to almost 7. So there we have seen that this kind of variation in the oxidation state of the actinides are mainly coming because here the filling happens in the pipe of orbital which is having close at proximity with the 6D and slowness of orbital and because of that we are getting such a huge variation in the oxidation state. We have also learned about the different ions that are possible in the actinide and we have seen that the divalent, trivalent, tetravalent. These ions are basically spherical. But when we talk about the other valency that is pentavalent ion and the hexavalent ion, they are no more spherical but are linear.  This is called transdioxo compounds. So here we can see that the charges are 2 plus. So you can see the total charge is 2 plus whereas we say it is like hexavalent. So uranium 6. But if you start above the properties of these ions, what you found that they are neither behaving like pure 2 plus or 6 plus but in between, somewhere in between. So lots of group has worked on this and finally they have come to the conclusion that the bonding or the chemical properties of this kind of, this kind of oxidation state that is actinide we say in general are dependent on the charge which is 2.2 units in the case of pentavalent and 3.3 units in the case of hexavalent. We have also tried to see that what is the hydrogen structure of different actinides in the water. And here we have just discussed that for the trivalent, they start with around 9 water molecules and they reach up to 8 water molecules. And there is a transition where we are having number of water molecules, some of them have having 8 and some of them are having 9. So it is the transition. This is about the first sphere and we have also seen in the second sphere or the total hydration number is increasing in the reverse order. And we have tried to rationalize this by size and total surface charge density. So with this basic background, I would like to go to the next slide that now we are going to discuss. So what are the different ionic radii? We have discussed about the lanthanide actinide contraction and here just I have given you some of the values that are there in the different oxidation state for the ionic radii. Let us start from trivalent, tetravalent, pentavalent and hexavalent. This we have already discussed in the lanthanide actinide contraction. We have shown you some graph for that. So I am not going into much of the detail of this. This is just for your reference that some of the values are given and as you can see there, having a decreasing current because of the lanthanide contractions or actinide contraction in the respective actinides. So once you are having a water, if you add water into the medium, what will happen that your water will be there in the primary conditions fail. So now the question arises that what is the coordination number of a given oxidation state? So here in general, the trivalent prefers plus 9, tetravalent prefers between 6 to 12, but generally 10, you can say and the pentavalent as I have shown you that these are not the spherical ion but are the linear ion. So they have a very different coordination number from 6 to 8. Exavalent is again 6 to 8 and tetravalent which is specifically exist in the alkaline medium only having a coordination number of around 6. But this is a very rough estimate. So I can give you some of the example where the ions which are in the trivalent state can have coordination number which are very different from whatever I have written. So for example, in general I have shown you this, this is the coordination number 9, this is the aqua molecule. You can see 8, 9 water molecules are surrounded the lanthanum ion. But there are cases in which the coordination number goes as blue as to 3. Why? Because in general the coordination number depend on two factors. One is the isoflalanthanide ion and the second is the ligand grounding. Or when I say ligand grounding there are two very important factors that is called first-order grounding and the second-order grounding. What I mean by this first-order grounding or second-order grounding is that let us talk about the first-order grounding. So when your ligands are very small molecule like this, a water molecule or having an iodine ions. When your ligands are having a small molecule and they are directly coordinating like this, you can say directly coordinating and the secondary sphere is having very small molecule like there you have an H. So in that cases this ordering or you can say the primary crowd ring is because of this oxygen but there is no or very few secondary crowd ring.  So beside the first one the second one is to be that they cause some sort of steric hindrance or some sort of crowding and because of this kind of crowding the coordination number generally drastically decreases. So here in this example as you can see the coordination number has decreased to 3. So such a decrease is mainly coming because of the secondary crowding. I have also shown you one example where the coordination number is again increased to 10. So saying the coordination number of almost 9 is okay for the solution chemistry but when we talk about the overall compounds or the other components of the solid state also then the coordination number varies to a very large extent. For example let us talk about the thorium. Thorium can have coordination number from 4 to as I as 15. So such a huge number of coordination number is possible in the case of thorium compared to the other metal ions. In general their trivalence are generally 9 coordinated. Tetravalence are generally you can say 10 to 12 coordinated. What about the pentavalent and hexavalent? Here have you shown you some example of the pentavalent. As you can see this is having a structure that we say that a linear structure that is this kind of molecule and you can see these positions are called axial position and there we say this is called equatorial position. So here you can easily see that the axial positions are already occupied and they are occupied by the oxygen. So now the ligand has only choice to come into the equatorial plane and in general the equatorial plane have almost 5 to 6 ligand 5 to 6 coordination number and here this is an example of the aqua complex where you can see 5 aqua ions are coming into the equatorial plane. So the total coordination number is 5 from equatorial which can be 6 or maybe 5 to 6 I can say in equatorial plus 2 you have in axial position which is given the total of 7 to 8 you can see. So in general these are having 7 to 8 coordination number. Again the same is true for the tetravalent same is true for the pentavalence also because these are known for uranium but suppose there is a case of neptunium again which is again a linear ion with a plus charge and here again we can have opposite water molecule in the primary coordination sphere or you can say in the equatorial plane given the total coordination number of around 7 to 8. So this is all about the coordination numbers of the different oxygen state of the actinides. So with the knowledge of hydration and the coordination number let's try to understand what is the thermodynamics of hydration. So as we have seen that when you put any gaseous ion or an ion in the gaseous state to the aqueous system there is stabilization or the overall delta G is becoming more and more negative. We have also seen that when you have an ion there is a gaseous sphere one is called primary one is called secondary gaseous sphere and we have also talked about the number of water molecule in the primary and the secondary gaseous sphere. So when the molecule in the gaseous state you are putting into the aqueous media what will happen there the water molecule will try to arrange around this metal ion and there is some interaction between the water molecule and this metal ion and because of that interaction there is a release of energy or you can see the process is exothermic in nature the overall delta is becoming negative. But at the same time when you see the entropy factor because we know that when you talk about the delta G there are two factors one is delta H and one is entropy. So there are two factors which control the total delta G value one is delta H which are shown that when there is interaction between the water molecules or the solvent molecule with the metal ion there is huge stabilization or the agrothermicity in the delta H. But what about the entropy factor? Entropy is basically as we know it is about the randomness. So if the randomness is increasing entropy is increasing but if the randomness is getting decreased the entropy will be decreased and here you can see when the water molecules are in the bulk they are more free or more having more entropy. But the moment these water molecules are coming in close proximity to the metal ions they are having some sort of structure. So now the entropy of the system is getting decreased. So in a way I can say with the total delta G that we want to derive for the hydration of these aqueous ions the delta H is favoring whereas the delta S is not favoring. But the amount of delta S that is there if you see from the table keep in mind that these values are in kilo joule for the delta H and for delta S it is in joule. So I have written specifically in joule because that is the joule we generally prefer. So if you use the delta H values they are of the order of 3,500 whereas if you use the delta S it is between 300 to you can say 400. So when you take total value that is the t delta S and you subtract from delta H to get delta G but we found that the total contribution of entropy term to this delta G value is not more than 3 to 5 percent. So this stabilization is mainly come on from the delta H. So it is anthropically not favor but anthropically very favor. We are giving you some plot. Here I have plotted all these three quantities that is delta S, delta H and delta G. You can see the delta H is keeps on decreasing or I can say is anthropically more favorable because we are getting a more and more negative value as we are moving and the delta S is also following almost the same trend. And when we take the difference according to the equation we are getting delta G which is again negative. But why this kind of trend? Because when we talk about the delta H, suppose we take delta H first, as we move in this series what will be there? There is a decrease in the size or you can say total increase in the z by r ratio. So when the z by r ratio or the anac potential is getting increased there is more and more stronger interaction between the metal ion, between the metal ion and water molecule. So this interaction is becoming stronger and stronger and because of that the energy released is more and more. So that's why we are getting more and more negative energy. Similarly, when there is more interaction, so suppose there is low interaction after this. So the only structure of water molecule is there in the primary sphere and the secondary is not very much influenced. But the moment you are getting to a very higher potential, a very high anac potential there is some sort of structuring in the secondary sphere also and because of that you can see there is a constant decrease in the entropy term. And then again we will just take care of entropy and tell you to find out del G value and what we found that the delta G values are again very much negative in these cases which is given the stabilization to the ions in the aquatic media. Once you have this ion in aquatic media, they are stabilized in some form. What else can happen to them? The next thing that can happen to them is the hydrolysis. So when you have ions in the media, they can act as a surrounded by some water molecule that is the some X number of water molecule. So the next thing that can be there is the hydrolysis because this metal ion has tendency to attract the oxygen from the H2O and slowly so happens that this kind of hydrolysis equilibrium can establish into the aquatic media. And as I have shown you that we are having different kind of ions that is spherical and linear. So their hydrolysis behavior is very much different from each other. And if you assume this kind of equilibrium for the hydrolysis, you can always write hydrolysis equations. And in second case suppose you are adding to some of the alkali to the media then obviously we can write this kind of equilibrium and these two equilibria are related with each other with the ionic product of water and you can get this. This is a very simple mathematical equation that you can always write and you can get information about different kind of equilibria that is there. But the most important thing that I want to discuss here is that at what point hydrolysis will occur and which are the metal ion which are more prone to the hydrolysis. In general as we have written that it is dependent on the oxidation state of the metal ion. And when we say oxidation state the trend I have given is this that it is most prominent for the tetravalent followed by the hexavalent, trivalent and the pentavalent. And keep in mind that when I say hexavalent I assume a charge of 3.3 and it is 3. So you can see there is a good decrease. You start with 4 plus and then you have a charge of 3.3 then 3 and then 2.2. So it is perfectly going with the charge of the central metal ion. Second thing is the the oxidation state which are having a different structure they are called actinides. Generally actually they actinize with the pentane which are having linear. So their hydrolysis is very very poor again because of the charge. If you see the value wise if you see the thorium hydrolysis constant with respect to others. I have given some of the values here. If you see thorium block a value for this hydrolysis reaction is around minus 2.5 whereas for uranium it is minus 5.25. These are the log k values. So if you try to write them in the form of k this will be like 10 to the power minus 2.5 and 10 to the power minus 5.25. What I mean by that if you just take the ratio you can say the thorium tendency of thorium to get hydrolyzed is almost 500 times more than that of the urinal ion. And when you talk about again uranium to emplissium it is again almost 100 times than uranium. And this is very weak as you can say it is going from minus 7 to minus 11. So it is almost 1000 times lower. So these hydrolysis is taking place depending on the ions. So when you have ions and they get hydrolyzed how their hydrolysis tendency is varying with the atomic number we can say in the series in the lanthanide series. So here in this particular graph I have shown you the onset of precipitation with respect to the atomic number. Here you can see as we are moving in the chain from the lanthanum to lute calcium the pH at which the hydrolysis will occur is decreasing. What does it mean that the ionic potential as we move the size is decreasing so the ionic potential is increasing. So now the ionic potential is becoming so higher that even at low pH you can get the hydrolysis. So this is for the trivalent basically I have shown you the trivalent. But what about tetravalent? I have shown you in the previous slide that tetravalents are almost 500 times more put into the hydrolysis compared to the trivalent. So when you see the tetravalent and you compare the thorium with another tetravalent that is plutonium if you see the hydrolysis constant these are again order of magnitude higher for the plutonium compared to the thorium. This is again you can explain because as we are moving from here to here there is a decrease in size and since the charge is almost same the ionic potential is on a very high side and then they form this is very rapidly compared to the thorium. So when we talk about this hydrolysis for the normal trivalent they are or maybe you can say pentavalent and hexavalent pentavalent mainly they are forming a kind of complex which we say the M OH M kind of complex which is you can say mononuclear or only with one metal ion. But when we talk about the tetravalent ion their hydrolysis is basically we say it is a polynuclear why polynuclear because it so happens many times that more than one metal ions are there in the hydrolysis. When you just add some LKD they will just make some mononuclear complex or a monohydroxy complex and the moment you add little more LKD the pH goes little on the higher side they come together all the molecules come together with a basic unit this and they try to make some amorphous hydroxy phases. Tetravalent this kind of phenomenon is very very prone and I must tell you that if you see an example if you see that let us assume that this end is one here and you are having some tetravalent and if I say that the number of proton really easy per metal ion so what I'm trying to say that with if you react with this with this and there is one proton release super metal ion one proton is released the moment this goes to two or beyond two you can see the number of proton release goes to beyond two the hydrolysis is so fast or so rapid that the number of units of thorium that come together is more than hundred or you can say the cluster is having more than hundred thorium atoms in this hydrolysis products. So the basic unit of this is this they start with this in the very freshly prepared and when there is a freshly prepared as I have shown you this equilibrium when they are really much freshly prepared with this equilibrium one can easily understand that if you increase this acidity you can get back so when they are freshly prepared and you are trying to increase the acidity of the medium we are able to dissolve them but what if they are getting age with time what will happen they make this kind of structure and if you put it even for the longer time they will make this kind of structure you can see some of the hydroxy groups are now removed and there are oxygrups oscillation is happening so the tendency of formation of this is very very slow but if it has found its solubility just by increasing the acid of the media is very very poor so this depolymerization this phenomenon is basically a polymerization so when I say depolymerization we are trying to get back from this to our metal ion so this depolymerization is making very very difficult when we are talking about the aging and this depolymerization tendency is basically dependent on the percentage of oxo and hydroxyl bridging if the percentage of oxo bridging is very high then this depolymerization is very very difficult whereas if the percentage of hydroxyl bridging is high we can have still some sort of back you can say depolymerization with this so now the question arises okay we know something about the hydrolysis we know the trend like yes tetravalent is having very high hydrolysis constant then compared to the other trivalent or divalent but suppose somebody asked me that can you tell me about that what is the fraction what is the fraction of hydrolyzed species that is formed when I am doing some reaction at a given pH so how we can calculate this value so for that I have given you some scheme you start with the assumption that yes you are having this kind of equilibrium which is having the certain log k value obviously and then the other in which I have started from mono hydroxy to dry hydroxy sorry it is misprinted it should be a dry hydroxy to the dry hydroxy and finally to z amount of OH groups so when we have this kind of equilibrium we can write the log k value and I assume that you are familiar with these two terms that is the k1 and beta 1 beta 2 sorry this is the stepwise constant and this is the cumulative when I say cumulative what I mean that this beta 2 is nothing but k1 into k2 so these are cumulative so with this knowledge of log beta values and log k value how can we arrive at the fraction or the percentage of hydrolyzed species at a given pH so what we can do we are having this knowledge and whatever metals and we are adding suppose we have added some amount the total m that ever whatever we have added now in the system at a given pH or at a given condition is present as m free plus different hydrolyzed species starting from mono diet rye also the species are present so your total m is always fixed and this fractions keep on changing depending on your pH suppose you can write this total m in terms of this k value beta values and proton so what I have done I have taken the relation between k1 beta 2 the relation between obviously between this quantity that is given by k1 and then between this quantity by beta 1 and I have tried to write all those terms with respect in terms of pre metal ion and the proton concentration so if you are able to write in this manner what we'll get we get a linear equation or we get a not linear I should say we get equation like this in which the m total is nothing but 3 plus 1 plus k1 by h1 b2 by h2 plus b3 by h3 so we can get this equation just for your simplicity I have assumed that suppose we have a metal ion which is just forming a mono complex mono hydrolyzed complex so in that kind of condition what will happen I assume that this is forming mono hydrolyzed so for simplicity I am just removing this other terms so the only thing I am left with is mn total that is nothing but mn free and 1 plus my first protonation constant divided by my proton constant concentration and I will try to find out what is the fraction of species or the mono hydroxy species that is formed in the different pH condition the first thing I have tried to find out the fraction of hydrolyzed species at pH 10 before that obviously I have to give some log k value so here I have assumed log k1 value of minus 11.3 with this log k1 value of minus 11.3 what will happen at pH 10 so the relation is very simple we know that that since we have assumed only a mono hydroxy species this is quite simple we have mn total is equal to mn free 1 plus k1 into h to the power minus 1 and since we are having a log k1 value of minus 11.3 this k will become 10 to the power minus 11.3 and pH is we are taking as 10 since pH is 10 so my proton concentration will be 10 to the power minus 10 and here if you see carefully this is h to the power minus 1 so when we write this is my h so when we write h to the power minus 1 that is minus 1 then this will be 10 to the power 10 I will display so when we are writing this equilibrium it is becoming 10 to the power 10 now we just put this value and the values are for the k1 10 to the power minus 11 but h to the power minus 1 10 to the power 10 we just solve this what we found that after solving and you just try to take the ratio of these two what we found that at this particular pH of 10 with the first hydrolysis constant of minus 11.3 the fraction of pre-metalline are more than 95 percent so you can say the fraction of hydrolyzed are less than 5 percent or in 25 percent we can try to find out that at a given pH if the log k values are known what is the fraction of hydrolyzed species that are present in the solution that is it with the pH to pH 11.3 now we just do again the very simple mathematics and what we found that at this particular pH 50 percent of the total metamine are present as pre what it means that since I have assumed the permission of only mono hydroxy species my total is nothing but some of these two mono hydroxy I am free I am saying is that 50 percent of the total 50 percent of the total is acting as a pre so this is 50 percent what is the rest? rest is only 50 percent what I can say that at this particular pH both of them are almost 50-50 percent and if you see carefully this pH is nothing but close to your log k value this is just the negative of your log k value so what you can say that if you have information about this log k value and if you know this simple calculation you can easily tell that at what pH what kind of what kind of species are forming and where the hydrolysis will start so this is the very simple case I have discussed with the mono hydroxy species but obviously in real systems we do not have this kind of species we generally having a combination of mono diatribe and in those cases you have to take care of all these equations all these equations all these terms in the equation and you have to solve it and after solving this what you will get is nothing but a plot in which the x axis is basically your pH and the y axis you can plot the fraction of the ion so you can say initially a hundred percent is your total or V free and then with the protein with the increasing pH there is a decrease and there is an increase in the hydroxy species this kind of diagrams are generally known as speciesion diagram or you can say the hydrolysis diagram where we see that how the hydrolysis of the metal ion is taking place as a function of the pH so here I have shown you some of the diagram for the hydrolysis of European and British and Nefunium here I have also shown you what run one trend that how this first hydrolysis constant is changing when you are going to the cities when you're going to the Latin cities here again you can see that the first hydrolysis constant keeps on increasing when you are going from the prosodium to lutecium why this because again here the charge is assumed to be plus 3 only and with the charge is plus 3 size is decreasing the Z by R ratio is again increasing so they are having more and more ionic potential and because of this more and more ionic potential their tendency to react with the oxide ion with the hydroxyl ion is increasing and you can say there is some change that is basically we called is a gadolinium break kind of things because here if the configuration is becoming F7 that is half field configuration and because of that there is some perturbation in this curve otherwise you can say this is a monotonically increasing graph over the first hydrolysis constant of the length and heights and this I have given to the log k value and the total concentration these are the two parameters that are required to plot this kind of graph because as I've shown you in this equation if you see this equation here in the bottom what we require is m total this is the unknown parameter and the pH at which we want to calculate the speciation these two things are required other thing k beta 2 all these betas are known so if you have the information of total concentration the log k values this is basically the cumulative ones so if you have the information of these two you can easily plot this kind of graph and I hope you will try this kind of graphs and you will try to plot them for different metal ions so here I've shown you for europium amyricium and nephronium so here I just want to discuss the case of nephronium because this is the one which I have shown you for minus 11.3 as I've shown you that at pH 11.3 50% is free and I say free it is free nephronium and 50% is complete and the same situation you can see here and this pH is nothing but around 11.3 because there is some secondary hydroxide also but very close to 11.3 so what we can say that before this pH this free nephronium will dominate after this pH the hydroxide species will dominate and here again I'm showing you for the amyricium and as you can see from the values of the first hydrolysis constant you can say that the amyricium is more prone to hydrolysis compared to europium the values are less negative so they are more prone to hydrolysis compared to the length of the amyricium. Here the trend again is the trivalent is giving more hydrolysis than the trivalent and then the lowest one we are assuming for the pentagalent so again I've given you some examples of plutonium one thing I just want to mention that in all this graph I have assumed that there is no solid phase formation is happening because if your metal concentration is on the higher side and there can be precipitation or formation of solid oxide or hydrolysis then this diagram may not be truly valid so with the assumption that there is no solid phase formation all these diagrams are drawn with this assumption because in case of plutonium if you do not assume this assumption that there is no solid phase formation you'll get lots of solid phases here. Again the example of pyrinium was shown here here what I want to show you that in case of uranium till now whatever we have shown you they are only mononuclear when I say mononuclear only one metal is involved with certain number of hydroxyl ions but when I say polynuclear or I can say more than one metal channel involved so plutonium is also going this kind of species but uranium is also prone to form this kind of species you can see even at a pH circumneutral pH less than 6 it is forming this kind of complexes in which more than one uranium moieties are coming together to form this kind of hydrolyzed complex. So with this now we have an idea that okay we are adding a metal ion we put into the solution and if it is going to get some kind of reaction or some kind of hydrolysis at what pH it will happen and you can easily plot or you can get some idea depending on the first constant that at what pH I should start my experiment and at what pH they will just start from the hydrolyzed species. So I can say I have information about the speciesian diagram that what species or what hydrolyzed species in particular will exist at a even pH but as we have read in the previous that when we talk about these actinides they are having very very oxygen state because even we are not talking much because they are in highly alkaline media. So this we are not talking much so we can for the time being we just assume that the oxygen state from even nobile we are not talking much so you can say from plus 3 to plus 6 and when we say plus 6 again it is in the form of linear isotope and plus or 2 plus bending whether it is a pentavirant or a pentavirant. So now we have information of their hydrolysis what is the trend of their hydrolysis but since they are having different oxygen state also how these oxygen states are varying in the solution or do we have any information about the oxygen state that can be present in water at a given condition at a given condition when I say I mean that certain PE value what is PE that we will try to understand that what the term PE and what I mean by the term PE so let us assume for very simplicity that you have a pH that we all know that what pH stands above is minus log of activity of proton so we all know that pH level is neutral and when we are increasing the pH to the higher side that is in the range from 7 to 14 we are having low proton density or we are saying that resolution is becoming less and less active and we are saying that we are decreasing the pH what we mean that okay the pH is going to be decreased it means we are having high proton density so similar to the pH scale which we are using for getting information about the hydrolyzed species can we have some sort of scale in which we talk about the electrons the availability of electron or the electron activity so you can say very similar to pH if there exists some scale which I can say PE which is nothing but minus log or you can say activity of electrons although the free electron do not exist in the media but assume that you have certain sources that somehow you are having some kind of products upon which are able to give some electron into the media we'll talk about this in the letter but assume that you have some electron and we can make this kind of scale that is called PE what will happen as the PE will increase what will happen that the number of electron get reduced or you can say there is less electron density what do I mean by less electron density that this condition is becoming oxidizing condition and when the PE is getting decreased what I mean that when your PE is is on the lower side I do not have given the values of the PE that what is the range we are talking about we will discuss it in the letter part but suppose you have certain range from X to Y your PE is getting decreased when your PE is getting decreased what will happen that the abundance of electron electron density will increase so here the conditions are more and more reducing so just to you can always compare in the pH whenever you are a bit concerned so then you can always compare when we are reducing the PE we are going to the reducing condition and when we are increasing the PE we are going to the oxidizing condition okay so if we can use this kind of variable similar to the pH we can always get information about the addux PE here so we'll continue for this pH and PE relationship and we'll see that how these variables will change the redox activity as well as the properties of the metal and mainly activate into the solution and thank you thank you for listening