 Welcome to this final segment of CD spectroscopy and MOSBUS spectroscopy for chemist. My name is Arnab Dutta and I am an associate professor in the department of chemistry IIT Bombay. So, over here over the past few segments we are actually covering the CD and MOSBUS spectroscopy in detail. So, in this final segment we are going to cover the main points of MOSBUS spectroscopy. So, in MOSBUS spectroscopy like any other spectroscopy we actually have covered a particular change in the state and over here the change we do from ground state to an excited state is the nuclear state. And over here I am taking the example of iron 57 our most favorite example which changes is ground state nuclear state of I half it goes to I 3 half. And because we are changing and nuclear state which are very stable so changing them is very difficult so that is why we need energy of very high magnitude and over there we need to use gamma ray energy one of the higher form of the electromagnetic radiation we can think of and the energy for this particular iron 57 is 14.4 kilo electron volt a huge energy. So, how we generate that energy that kind of energy is also generated from a iron 57 system which is already staying in an excited state and it is coming to its ground state of I equal to half and that is going to release that energy of gamma ray which will be absorbed by this sample system. So, this is my sample and this is my source which is also 57 iron and this source of 57 iron is generated from a cobalt system which actually is a meter stable state give us a 57 iron and which undergoes different change in the state of nucleus and develop this I equal to 3 by 2 state which is stable long enough so that we can perform this experiments and at the end we get a resonating condition and we see a signal that it is actually happening. Now, one of them will be percent of transmittance one of them should be energy now what kind of energy axis we actually use for that we have to know how we actually do this experiment. So, first we have a nucleus or an atom of a source sample so this is the source and over here we put the source in a solid matrix or in a lattice why we will coming into little bit later and this is my sample we put that in analogous lattice or solid matrix and over here when we actually release that gamma ray what happens gamma is a very highly energetic system when it is coming out this molecule the atom will like to go back as a backlash similarly this molecule will like to move to this direction and this is actually going to create problem during our resonance because anywhere we are going to change in the nucleus state energy which is actually happening in a very small scale and to have a resonating condition is very tough to get and over there if you are losing energy for this recoil it is going to be very difficult that is why we put them in lattice or solid matrix so that we can control any of this translation recoil. So, this is known as a recoil less condition and over here we make the translation almost to 0 but over there we still have the vibration energy the phonons which can be non-zero and that is why temperature would have an effect of the MOSBAR spectroscopy how it looks like and over here we actually make the sample static but the source we actually put in a dynamic system where we can on a wheel we can move it towards or backward from the sample and over here when I am moving the source towards or against the sample the nucleus looks very similar but it is surrounded by the electrons and that is by the valence electron different compounds different environment so that is why this energy of this ground state and etcetera will be slightly different and that difference we want to match properly during a resonating condition by moving this back and forth by using the Doppler effect. So, that is why the energy scale is actually given by a Doppler velocity so that is actually mentioned where I am going to get this resonating condition and this transmittance varied from 100 to 0 so that is what we have covered in the MOSBAR spectroscopy in the beginning there is the basics of the MOSBAR spectroscopy how we actually achieve the resonating condition. The next is what are the important factors the two important factors are this is the basic is the isomer shift and quadrupole splitting so what is isomer shift so isomer shift is a value exactly where my ground state and excited state energy matches between the source and the sample that means exactly where it matches and whatever the value I am getting over there in the terms of millimeter per second which is coming in lieu of an energy so this value will be my delta value or isomer shift and this isomer shift value is dependent on certain factors what are those so this is dependent on 4 by 5 z square r square where z is the charge for example if r57 I have a particular z value e is the electronic charge r is the radius of the nuclei and it is multiplied by delta r by r where delta r is nothing but change or the difference of the radius of the nucleus in the excited state minus ground state and in case of 57 iron this is actually negative value because the excited state is actually smaller compared to the ground state so it actually shrinks down when it goes to the excited state so that is delta r factor and this is also multiplied with the electron density present on the nucleus which is nothing but the s electron density because s electron density is the only electron density which has finite possibility to be present inside the nucleus which is non-zero at the nucleus and this is the value we get from the sample and the source and over there this is the s electron density on the nucleus and this is we can vary for 1s 2s and 3s in case of the iron 57 and over here 1s and 2s at the core electron so they have minimal effect and this is the valence electron and this valence electron density is affected by the environment more so this is going to create the difference between different oxidation states spin state and all those things. Now how does it vary we find out that it varies with respect to the oxidation states I am taking an example of Fe 3 versus Fe 2 how we can differentiate it is a d 5 system it is a d 6 system more d electron and less d electron over here in iron 3 more d electron in the Fe 2. Now d electron density affects the shielding it shields s electron from interacting with the nucleus it is a more of a like a competition who is going to take the charge from the nucleus and stabilizes them d or s if more d electron s is facing a huge problem. So the shielding effect is lower shielding effect is lower in case of iron plus 3 because it is not having that much of d electron and this will be higher comparatively compared to iron 3 and that means psi 0 square the value that we found over here it is going to be high in case of iron 3 plus because less d electron so it is having more chance of s electron going to the nucleus and that is psi 0 square is obviously is going to be the low in the case of iron 2 and that means this value of psi 0 square over here it is getting be much higher for this particular expression but it is going to multiply this in negative number so it will be on the negative side and this will be on the positive side. So just by looking into the delta value I can see which is having more d electrons or less d electron which is having the more d electron density or less d electron density by following this rational and we can also say how my delta value is changing from there we can say what is the oxidation state change is happening in that particular molecule or what is the electron density change happening inside the molecule and we have covered multiple examples of that to follow it up. The next thing is coming is the quadrupolar splitting which is given by this term delta eq and previously I want to mention this isomer shift is given in the value of millimeter per second that same Doppler velocity unit we are using also for the unit as the isomer shift. In the quadrupolar splitting also we are going to use the same system so first of all why we see a quadrupolar splitting so previously we said that is a ground state of i equal to half also have an excited state of i equal to 3 by 2. Now if my molecule is like that I am going to see only one band like this I am going to expect but what happens if I have a quadrupolar moment present then what happens quadrupole means that you have charge like this if it is present the ground state does not have any effect it remains as it is but the excited state it actually splits up in i equal to plus minus half and i equal to plus minus 3 half why because if it is a quadrupolar moment it can differentiate a half spin and more than half spin if it is a half spin that means it is having no quadrupolar moment so it will detect this i3 half differently than a non quadrupolar moment best i equal to half system so they split up so that means now you can have the possibility of two transition and your line will be split it and over here the splitting you get over here that will be your quadrupolar splitting and if you take the average of it that is where it should be before it have a splitting that will be your delta value one example similar over here that is the delta value over here and now you have two differences the difference between them is delta iq now when I am going to gain this part of quadrupolar splitting when you have a quadrupolar moment how I can have a quadrupolar moment so you will get a quadrupolar moment when you have an electrical field gradient so this is the parameter controls whether you have a quadrupolar moment or not so to have a quadrupolar moment you have to have a electric field gradient non zero so there are three different conditions that we have talked about we can think about first your system is spherical non quadrupolar is i equal to half system so in this condition if you put that in a field which is actually not asymmetric so over here your electric field gradient will be zero so it is a symmetric field you can see and your system is also symmetric i equal to half system so you are going to get electric gradient zero so over here you are not expected to see in quadrupolar splitting case number b you can have your system non spherical if you have a non spherical system that means it is i equal to greater than half that means now you can have some electric field gradient but at the same time you are putting that in a electric field which is symmetric so even in this case your electric field gradient will be zero because this system is can have quadrupolar moment but this cannot separate it because you put that in a totally symmetrical field so in this case also you are not going to get any quadrupolar moment case number c there is a possibility that now you put that in a field which is now asymmetric and you put your system but the system is i equal to half now the system is not quadrupolar it is not going to make effect even I have a asymmetric field asymmetric electrical field so over here if the steel remains zero I am not going to see any quadrupolar splitting the only way to get a quadrupolar splitting is you have to have a system which is having quadrupolar moment and put that in a field which is also asymmetric in nature so in this case you will have a non-zero electrical field gradient and you will see some quadrupolar splitting so that is the only condition you can have now at which particular condition I can have quadrupolar splitting where my electrical field gradient will be non-zero so there are two aspects one is the lattice contribution so you have to have a ligation configuration such that it is not symmetric so if you have all ligands same it is not going to show but if some of them are different and it is not symmetric so now I put a different one over here now you will say that yes I have an electrical field gradient over here so this is the lattice contribution your arrangement of the ligands should be asymmetric the second one is the valence contribution which says the electrons present in that system should be asymmetrically oriented so for an example if you have a system which is octahedral in nature so we disperse our d electron density in two different orientations it is under t2g and eg and say you have a symmetry like that a d1 symmetry sorry I should write in the t2g level t2g1 t2g2 they are going to be efg variant because over here it is asymmetrically but once you put the t2g3 it is not because all of them are symmetrically varied similarly we can go to other all the possible cases you can think of t2g3 eg1 it is active t2g3 eg2 it is not active t2g3 t2g4 eg2 it is active because now t2g4 the electron coming over there and that electron we are considering t2g5 eg2 it is also being active t2g6 eg2 again it is not active t2g6 eg3 active t2g6 eg4 again it is not active and you can go for and go on forth this is the high spin system I talk about you can also look into the low spin system and figure it out so a asymmetric electronic distribution is needed to trigger the valence contribution part and once either of them or both of them are active you are going to see a quadrupole aspect and you have also looked into we can by that figure it out whether it is a high spin or low spin system because they will have different contributions from the valence or lattice and we can figure it out that also very nicely from the MOSBus spectroscopy and the last portion we have covered I think I mean number 4 yeah is the mixed valence system and we have covered multiple examples of that where we have figured it out in mixed valence system you can have metals in different conditions 2 or 3 totally localized so over here you expect to see MOSBus spectra for each of them separately depending on what is their electric field gradient you can have quadrupole splitting and you can have particular delta value so you can figure it out whether it is iron plus 2 and iron plus 3 or not then it is possibility that they are actually interacting between them and becoming 2.5 plus 2.5 plus so over there you will get only one set of signals and that shows that it is now somewhere in between iron 2 and iron 3 so it is a total mixed valence system so the delocalization you can actually followed by MOSBus spectroscopy and with that system you can also do the temperature effect you can go to lower temperature from positive and negative direction positive negative it will be confusing say from room temperature to very low 5 Kelvin temperature the experiments you are doing and over there you are seeing these peaks how they are changing in their new peaks coming up and at one point of time you will see they are very well separated signals so you now see iron 2 plus and 3 plus back in a much more localized condition when you go to lower temperature so that you can figure it out and over there by looking into the splitting and all this thing you can see how the metal and the ligands are actually interacting in between them so that you can also figure it out so it is a metal part versus a ligand part that you can find out whether it is a pi donating or it is a metal part and a ligand part it is pi accepting so it is a movement of the delectron towards the metal or against the metal and by that we can figure out what is the delectron density is going to be present there and how it is going to affect the psi 0 square value and how it is going to measure the delta value so isomer shift will be also seeing some changes so typically when you have a lot of pi back bonding it is actually moving the delectron density out so it increases the propensity of a delectron density on the nuclei which I will see the shift of the delta value towards the negative direction and this one on the other side delta value on the positive direction and that is how we can find out the mix balance how it is actually happening at different conditions is there are the intermediates present that we can also unravel by using the MOSBA spectroscopy so over here in this full course we have given you multiple examples of MOSBA spectroscopy coming from biological samples coming from material samples and how even you can use MOSBA spectroscopy for some missions in Mars and try to find out thousand miles or thousand million miles from here in a different planet or different star system is there iron present over there what are the iron condition that we can figure it out from MOSBA spectroscopy so that is how MOSBA spectroscopy found multiple applications all over the world and its applications are increasing over time so that is we figure it out MOSBA spectroscopy is going to be a very unique spectroscopy in the coming days to understand iron specifically and also the other elements are coming like tin, europium also coming into the picture for this MOSBA spectroscopy so these are the segments we have been following for this MOSBA spectroscopy for this full multiple weeks of this particular course and I hope that now you have a very good idea about the basics of MOSBA spectroscopy how we can use it for different application to figure it out the different oxidation state spin state and also about the molecular geometry around the central atoms and I hope this is going to be very helpful towards you and I hope that you enjoyed this particular course. Thank you, thank you very much.