 Hello and welcome to this new segment of CD spectroscopy and MOSBA spectroscopy for chemist. My name is Arnab Dutta and I am an associate professor in the Department of Chemistry at IIT Bombay. So, in this particular segment we are going to look into applications of MOSBA spectroscopy. So, so far we have covered the basic of MOSBA spectroscopy that it is actually an exchange of energy between nuclear state where the ground state of an atom preferably a 57 iron isotope atom which goes from i equal to half to i equal to 3 half excited state. And this change is a high energetic change where it actually requires 14.4 kilo electron volt of energy. And this particular energy is exchanged during this MOSBA spectroscopy and over there we get a signal and the signal is defined with two different parameters isomeric shift and quadrupole splitting. And now let us take a look how we can use this MOSBA spectroscopy and the associated parameters to find out what is the fate of a chemical reaction. So, this is the number 5 example of the application site. So, over here we are going to look into gamma radiolysis of an aqueous solution of iron 2 sulfate 7H2O salt. And what is gamma radiolysis? It is nothing but you are shooting that with gamma ray so that it can start a reaction by exchanging electrons. So, let us see how it actually happens. So, iron sulfate 7H2O when we put that in water it actually get dissolved very nicely and it forms this iron 2 H2O whole 6. So, over here this salt is loosing it is sulfate ion and binding with 6 different water molecules in an octahedral coordination geometry that is shown over here. All of them is binding with water molecule and iron plus charge is plus 2. So, overall charge is also plus 2. So, I should write in a Roman form that it is actually happening. Now, before going let us first look into it how the MOSBA spectra of this particular system will look like. We have covered that one of the earlier segment but just to recapitulate it over here. So, iron plus 2 system in octahedral geometry how it should look like. So, in the terms of electric field gradient or EFG it depends on two factors. One is the lattice that means the coordination geometry one is the balance or distribution of the electrons. So, in this particular molecule you can see that this molecule is bind in a very symmetric octahedral geometry. So, you do not expect any lattice electric field gradient because there is no asymmetry over there. However, the balance electric field gradient I do expect to see why because this is an octahedral geometric system. So, my D orbitals will split up in EG and T2G symmetry and as you have discussed also earlier the ligands can be explained with respect to sigma donor, pi donor, sigma donor only or sigma donor, pi acceptor and depending on that this energy gap between them is going to split which is known as the delta octahedral or crystal field splitting energy or ligand field splitting energy. So, this depends on the particular quality of the ligand and which particular atom is coordinating to the metal. Over here oxygen is binding is group 16. So, it is a sigma donor, pi donor which actually ensures that over there this EG and T2G gap is actually quite low that means it prefers a high spin system. Now iron II is a D6 system. So, this will go to 1, 2, 3, the fourth electron will go over here, fifth electron here, sixth electron comes over here. Now you can see EG is symmetric, T2G is asymmetric because of one extra electron on the lower spin. So, that is why balance electron exist lattice not in the beginning. So, what do we expect that there will be a quadruple splitting because of the presence of the balance interaction. So, what we expected is the following. So, I am drawing the MOSBUS spectroscopy this is the percentage of transmitter scale. So, from 100 to 0 how I am going to see and this is the x axis is the velocity the Doppler velocity which is actually representing the energy over here. So, we expect that first there will be one line and that will split up in two lines because of this balance interaction the balance asymmetry and the electric field gradient associated to that is the expect but in reality what we see is the following what we see is that following is this. So, over here if I take the average of this two line that is going to give me the delta value and this difference is the delta Eq this is in the reality whereas the average of the expected one is almost at the same position. So, the delta value or isomer shift almost at the same position but this splitting quadruple splitting is much more larger in the case of the reality compared to the expected. So, why so because there is one other phenomena happening over here and that is known as the Jan Taylor distortion because of the asymmetry present in the T2G level that is going to undergo further splitting and this particular energy gap is not going to help this one you are gaining some energy from this configuration that is going to happen. So, you will see Jan Taylor distortion Jan Taylor distortion and what is happening the way I have drawn this is the x square minus y square this is the z square orbital this is the x y orbital these are the y z and x z orbital. So, you can see the x y based orbitals are going higher in energy that means in the z axis it is actually going out the ligands are actually moving out. So, z based orbitals are going to be stabilized and this actual structure will be elongated on the z axis and it will be shrinked on the equatorial plane or x y plane and due to this interaction now your lattice is also having an asymmetry because from a perfect octahedral now it is moved to a D4H symmetry. So, it will also trigger lattice contribution for electric field gradient and valence interaction is already there and there is a both of them will come into the picture and because it has two components now to incorporating asymmetry around the metal. So, I am going to see a higher value of quadrupole splitting. So, that is why quadrupole splitting or delta Eq and that is shown over here. So, that is what I am going to see if I dissolve iron sulphate solution in water I am going to see this particular spectra. Now move to the next one. So, I have the solution present over here. So, this is the transmittance and this is the free millimeter per second. So, over here just to highlight you over here I am actually putting one spectra on top of others. So, it is just overlaid. So, this is 100 percent transmittance to 0 percent this is also start from 100 percent here and 0 percent here and so on and so forth you can put multiple of them aligning so that I can see in the same graph what is the change is actually happening. So, that is I am going to do it over here. So, this is the iron H2O whole 6 2 plus solution with channel or distorted geometry and I am seeing this particular system. Now, what I am going to show I will try to put the original data black and the new appearing system in the beginning. So, in this particular system we actually hit gamma ray. This gamma ray is a little bit different energy than the gamma ray we are using for MOSBUS spectroscopic, but this gamma ray is still high energy enough so that it can trigger some reaction in this chemistry. And what do you expect the gamma ray actually does? The gamma ray the first thing it does it actually reacts with water and produces OH dot and AH dot radicals. So, it actually breaks down one oxygen hydrogen bond homolytically. So, each of them get one electron oxygen get one hydrogen get one and they create the respective free radicals. This free radical is going to react with this iron H2O whole 6 because this water molecule is getting hydrolyzed like that it can be one of the water molecule which is already being bound with iron and one of them is creating this system. They will charge it still remain plus 2 and then what happens this OH dot radical is very reactive. And that is going to interact with this iron sample which is present in plus 2 oxygen state and tie 2 captures one electron from this iron 2 so that iron becomes iron 3 and then this OH dot taking one electron it becomes OH minus. So, that is what is actually happening over here. So, overall charge is still 2 plus because it is OH minus negative charge 3 plus charge so overall charge is still 2 plus. So, that is what is supposed to happen. If it is really happening let us take a look into MOSBUS spectroscopy to find it. So, when we begin in 0 minute this is the spectra we are seeing no other thing only iron H2O whole 6 widely splitted quadrupole splitting due to the gental distortion which actually creates an asymmetry not only from the lattice state but also from the valence state. But what happens when we hydrolyze it so over here we continue to see these bands as it is that means it is actually staying there. But additionally we started seeing another set of peaks appear which has a little bit lower delta Eq value and not only that the delta value for this new one is actually a little bit on the negative side compared to the delta value with the original iron 2 plus system. So, we continue to do this experiment further and slowly we see that the original graph the original values is actually shrinking down further and the red one is actually increase in size and at one point of time there is nothing left but only this. Now, the question is why so over here when you are talking about these two system this is nothing but it is representing the iron H2O whole 6 gental or distorted system. But when this reaction is happening I am actually generating another material H minus so that is I am generating and I am also creating an iron 3 plus system. Now, if I have an iron 3 plus system what will be the contribution of the electrons or distribution of the electrons over here. Now H2O and OH- are all sigma donating pi donating orbitals or ligands sigma donating pi donating ligands. So, over here is going to be high spin system iron 3 plus is a D5 system. So, all the 5 electrons will be stored like this. So, now you can see all of them are symmetrically oriented. So, there will be no valence contribution to the electric field gradient. So, it will be 0 value for the valence. But for the lattice one you can see lattice one it is asymmetry because 5 of them are water one of them is hydroxide. So, lattice EFG will be there non-zero. But overhaul compared to this one where you have both valence and lattice EFG is present when you are oxidizing it you are losing the valence one. So, your overall asymmetry is getting shrink down and that is why the delta Eq value also shrink down because now you are losing the valence EFG. Why the delta value is moving that is straight forward because I am moving from iron 2 plus to iron 3 plus and as you have discussed earlier iron 3 plus means less amount of D electron. Less amount of D electron means that this is going to D6 I should say D6 versus D5. So, let me just draw it over here D6 versus D5. So, that means more shielding effect on iron 2 more shielding effect means actually less electron density of the S electron coming to the system and that is actually affecting the overall isomer shift because we know S electron density psi 0 square value is going to multiply the delta R which is negative in nature. So, this psi 0 square will be higher in the case of D5 or iron 3 plus system because it has more chance to see the nuclear and this delta R value is negative. So, it is going to shift more negative side. So, that is why we are seeing this particular change over here that slowly it is shrinking down the delta Eq value and also shifting towards the negative direction and by that I can follow this reaction of the hydrolysis that it is actually happening over here. So, the MOSBA spectroscopy not only giving me an idea how this change is happening, but also what is actually happening over here. So, this is one of the nice examples of MOSBA spectroscopy to follow a chemical reaction. Now, we will take one more example of such change where we will talk about a ligand exchange reaction. So, again we will start with the same sample FeSO4 7H2O dissolve that in water and we will be creating this Fe hexahydrate sample plus 2 charge is developing over here. And then in this sample I am going to include 6 equivalent of cyanide iron in the form of potassium cyanide and try to find out what is actually happening. Is the cyanide is good enough to kick out some of the water molecules from the primary coordination of the iron or not. So, to look into this matter we did this experiment at a lower temperature to slow down the reaction so that we can follow that. So, the reaction was done at 5 degree centigrade and we continue to follow the MOSBA spectroscopy. So, I am going to show you multiple MOSBA spectroscopy the x axis is the Doppler velocity millimeter in second. This is the percentage of transmitters so I will show different blocks of different spectra. So, each block will cover 100 to 0 percent of transmitters and not only that in this particular direction I am also moving with time. So, in the beginning I see a very well separated doublet like that. So, why it is so? Because this is the iron H2O hole 6 which is showing you giant teller distortion. So, due to that we are having lattice and valence electric field gradient both of them are present and that is why I see a huge delta difference. This is happening at 0 minute. That means before we even add any cyanide to that only the iron sulphate is solubilized in water and it goes to its expected hexahydrates format showing some giant teller distortion. The next thing what we started seeing after almost 3 hours that the original peaks are present over there. But additionally we started getting another signal very close to it. So, now I have a 2 doublet, one doublet in the original position and there is another doublet coming over here. This is what is happening at 160 minute almost 2 hour 40 minutes later. Then we run that for almost 12 hours keeping at at 5 degree centigrade and the data we are getting is the following. The original position is very limited. So, obviously the original iron excited sample is going out. This data we are seeing earlier. So, let me draw that in a different color so that we can follow that also. So, that particular data actually start increasing in size. So, that was actually happening and very interestingly we also found there is an another signal generating on that side. So, let me put them in the green line. So, another signal start showing up over here but that is a singlet. So, what we are getting is one original position is singlet. There is a doublet and the original well-splitted doublet. So, original doublet that I know that is from this original structure. Now, why this singlet and doublet is coming up? So, this is what we observe at 720 minute. So, let us run the reaction a few more hours. So, then we go to 1500 minute more than 12 hours and what we got more than almost I would say 20 hours now. So, what we are getting this first peak is almost gone. Not only this first peak is almost gone that means primary iron excited sample is already exhausted. Not only that this red color doublet is also almost gone a little bit higher compared to the original one and this green color sample is actually becoming the major signature. This is actually what are you happening after more than 20 hours. Then we further look into it 6000 minute almost 100 hours later what is happening and over there we found the only signals present over there is this green signal. Only the green singlet is remaining the rest of them are gone. So, that is the MOSBUS spectra of this particular system. Now, our goal is to figure it out what is actually happening over here. So, to understand that we have to understand what is actually happening with this iron excited sample when it is reacting with cyanide ligates. So, what we have found that the reaction begin it reacts with 5 cyanide altogether and create this molecule. This is in iron 2 states overall charges plus 2 in the beginning now is still in plus 2 state but now 5 of the water molecule is kicked out of the cyanide and over here is kicked out I am going to get this particular molecule overall charge will be 3 minus because 5 negative charge from cyanide and plus 2 charge from iron plus 2 balancing is 3 minus this is the molecule I am preparing how this molecule will look like. So, cyanides all around 5 of them and only one water molecule present over there iron is in plus 2 state. Now, compared to these 2 what are the difference I expect? This is an iron pentacyanide system cyanides as we expected it do 14 base molecule carbon is actually coordinating. So, that is why it is will be a low spin system. So, the electronic configuration from this it will be low spin system. So, it will be paired up this energy gap is too high because of the presence of this strong sigma donor and pi acceptor system of cyanide that is a huge change when it is exchanging water versus cyanide water is a sigma donor pi donor it prefers high spin cyanide is a sigma donor pi acceptor it prefers a low spin system and in this case of low spin you have iron 2 d6 system like this all paired up. So, there will be no valence contribution to the e of g lattice contribution. So, it is no contribution. So, it will be 0 lattice contribution will be still there because you can see there is a difference in the coordination geometry is a cyanide and water. So, there is asymmetry present over there. So, it will continue to have the lattice electric field gradient but not the valence electric field gradient and that is why what we expect between these two this system over here it will shrink down on its quadrupole splitting. So, that means are we saying that this is the lower quadrupole splitting that we are seeing that is because of this particular sample over there and which is giving me this doublet there is a possibility of that and that is what is actually happening because of the cyanides is making the system a low spin system the quadrupole splitting is shrinking down because there is no valence contribution from there. Now, the question is why it is now moving on the negative side previously when you start from the beginning that is the delta value of the starting material follow the color this one shifted further negative why it is negative side. That answer is also found in the cyanides. Cyanides are as we just said is a pi acceptor ligand. So, it has pi star orbital which is going to interact with the metal d orbital and it is going to transfer electron density out of the metal to the ligand. So, this back bonding will happen electron density will move out and as we have discussed in the earlier segment as we are moving the d electrons out you are cutting down your d electron density cutting down your shielding effect for the d electrons and S electron has more chance to see the nuclei and if that is happening your psi 0 square value is increasing and that is going to affect the delta value which is dependent on this and the delta R value which is negative in nature. So, multiply that two together I am going to shift to a negative value if my d electrons are going out of my sample and that is exactly happening my d electrons are moving out and I am having a delta value for the shifted. So, what happens to the last one what is the green one then the green one is actually what is happening I am producing iron hexasinide sample and that is the system which is actually giving me this green color signal over here. It is singlet why it is singlet because you can see iron hexasinide is going to have a very similar electronic configuration like this one d6 system all paired up because exchange of one water with another cyanide molecule is not changing its property a lot with respect to high spin or low spin it is still remaining low spin that is why no valence contribution and also now we have all symmetric cyanide around it. So, the lattice contribution of EFG is also 0. So, there is no electrophil gradient parent in this iron 2 hexasinide sample and that is why it is showing us one singlet and why it is further negative delta value because one extra cyanide is coming exchanging on water further more pi electron movement from the metal d to the pi state of the orbital and that is actually triggering further electron density out of the metal system which is bringing down the electron density increasing the propensity of a electron density to be present in the nucleus which is multiplied with the negative term of delta r and I am moving toward that negative side. So, that is what is actually happening over their system. So, over here now you can see that whole system is showing me that there are three different components work there. One is the beginning one then there is five water molecule replaced by five cyanide molecule one and the final one all hexasinide sample. So, this is I can see it over time how it is behaving and I can see this five cyanide water molecule bound system this red one is intermediate one which slowly also transfer to the final product of the hexasinide one. So, over here it is showing that MOSWA spectra can show you this system. So, over here I am just summarizing the information that we found over here. So, it is going to have a jantoler distortion. So, that is what I am going to see a huge distribution of delta eq anywhere you are showing a graph you should show the axis and then five cyanide molecules comes and replace it. So, over here there is no more electric field gradient from the valence, but the lattice one still there and previously g was there and valence e f g was there. So, that is why over here I am expecting that this system is going to show small shifted value of quadrupole splitting. So, it will be much smaller and not only that the delta value that we say it is going to shift towards a more negative direction because now the cyanide with their all pi back bonding is going to move the delectron out of it. And finally, it moves one more cyanide replacing the final water molecule iron is only plus 2 oxidation state in all loops them this is a system. So, over here lattice e f g is also now gone and valence e f g is gone previously and it remains the same. So, but there are no contributions at the end what I am getting is actually a singlet system over here as a transmittance. So, that is what is actually happening and over here now if I want to draw a graph for kinetic following of this molecule. So, percentage of relative abundance of each of the material. So, first one is this Fe H 2 O whole 6 2 plus that is going to be 100% in the beginning and slowly it will move out and it is going out after a while. So, this is Fe H 2 O whole 6 2 plus next one is this one which actually start forming after a while and then it will shrink down and go down over time. So, this is this Fe H 2 O C N 5 overall charge 3 minus all iron in plus 2 and the final one is this green one which actually come into the picture much later and then it slowly try to be there and at one point of time it will cease and 100% is reached and this is the iron C N 6 4 minus charge over there. So, by that you can see you can follow the kinetics of the reaction very nicely with Mossberg spectroscopy and over there one question might come to your mind why I am seeing this molecule where 5 of them is actually already exchanged why cannot I see 1 2 3 4 cyanide exchange system those are actually happening but those reactions are happening much much faster rate even at 5 degree centigrade and it is happening before I can record a Mossberg spectroscopy. So, that is why that is remaining such a system that I am only seeing this particular sample one of the stable intermediate and that is the first thing I am able to see the previous one I cannot see if I go to very low temperature we might be able to capture those other intermediates also. So, with that I would like to conclude over here so this is another nice example how you can follow Mossberg spectroscopy to find out the kinetics of a molecule and figure it out how these molecules are behaving and undergoing different intermediates from a particular precursor to a final product. So, with that we would like to conclude over here and we look forward for the other application examples of Mossberg spectroscopy in the coming classes. Thank you. Thank you very much.