 Hello and welcome to this new segment of CD and MOSBIR Spectroscopy for Chemistry. My name is Arnab Dutta and I am an Associate Professor in the Department of Chemistry at IIT Bombay. So, in the previous segment, we are discussing about MOSBIR spectroscopy. Today, we will revisit the basic and fundamentals of MOSBIR spectroscopy and then jump into different applications of MOSBIR spectroscopy that is happening all around us and we will discuss how MOSBIR spectroscopy is aiding us to understand the conditions, the oxidation state and this MOSBIR spectroscopy is going to help us to understand the oxidation state, spin state and the molecular structure of different complexes. So, let us start. So, we know in MOSBIR spectroscopy, this is actually a spectroscopy where we change the ground state of a nucleus. So, it is a nuclear state we are talking about and then we excite it to the its particular excited state. Now, as we are changing the nuclear state, the energy are very high for this particular transition. So, we need transition in the gamma ray region which is very energy dense and this particular energy we actually develop by using another set where excited state actually relax back to the ground state and this is coming from a source. So, the source is already present in the excited state which is relaxing to the ground state and in the meantime it is leaving the gamma ray which is being captured by my sample and the sample goes to the excited state and if there is a resonance then we will see a band over there. So, now when we are doing this experiment what do we expect that will be the unit of percentage of transmittance which we can also use as an absorbance unit if it is needed and over there there is a energy axis and if I am talking about the transmittance. So, first there will be all the things will be transparent because the gamma is not accepting at all but when this resonating condition is stuck I should get a line over here and typically the line is a little bit broad one and you get a signal like that. So, let me just redraw a little bit. So, we get a signal like that and that says us we are absorbing them but over here the systems of MOSBAR spectroscopy a little bit different than the other spectroscopy that we discussed like optical spectroscopy or NMR or EPR. Why because over here this gamma ray when it is coming out it is a huge amount of energy. So, that is why what happens this source and the sample get some momentum shift during this. So, when the gamma is coming out this is actually shifted on the back side because it is leaving the energy whereas on the ground state of the nuclear state of the sample when it is accepting the gamma ray it actually also move back a little bit because of this momentum. So, that is why unless this momentum energy is concluded and properly associated with this overall energy change it is very tough to get the resonance condition. So, that is why what happens the actual source is actually make a static system similar goes to the sample and this particular condition when I am putting in a static condition. So, how I am actually putting it in a static condition I am putting it in a form of a solid matrix or in a lattice condition and by that I ensure the momentum based recoil is not happening so it is coming to have a recoil less condition and at that particular condition I can expect that there is a very better chance or a very good chance to have a resonance over there. And on that top of that we also make this full system static for sample whereas the source we put it on a wheel which can be movable on either direction because what happens that over here we are using a source which is actually going to develop for an example I am taking an example of iron 57 that I am generating and my sample is also iron 57 and we have discussed earlier in detail why not all the possible isotopes cannot be MOSBAR active only a few of them are MOSBAR active and iron 57 is one of the most prominent one. So, over here let us say my source and sample both are iron 57 but their energy of this particular ground state and excited state of the nuclear state is going to vary depending on what is there around it and that is why we are doing MOSBAR spectroscopy so that we can see the effect of surrounding environment on the nucleus which will be reflected in the MOSBAR spectra. Now over here because the surrounding environment is not the same there will be slight difference and that is why the source and sample even if we put them static the resonating condition may differ because they are not actually experiencing the same environment. So, this slight difference I can cover with the help of momentum because previously momentum was actually ensuring that I am not getting resonating condition. Now momentum I am actually using strategically so that I can ensure that resonating condition is being achieved and for that we actually move this source either on the towards the sample or backwards to the sample and using Doppler effect to the system so that I can generate a good resonating condition. So, that is why in the actual MOSBAR spectra the unit of energy is giving by velocity and I am moving at millimeter per second speed so not a high speed only slight speed to ensure that we are seeing a resonating condition and that is how the MOSBAR spectroscopy look like and over there it depends on like when we actually going to match and by that particular velocity I can define the different conditions of the particular iron as an example of MOSBAR active isotope. Now over here we want to ensure one more thing when we talking about this particular system to be static and we are talking about some recoil less energy that we are talking about so when something is recoiling what are the different things it is going to happen one thing is going to happen the translation energy that means the molecule or the sample is moving either forward or backward depending on whether it is getting an gamma ray photon or leaving a gamma ray photon. But over here making this source and sample static I am getting that translation energy totally out but there is one more system the vibration one. So, this recoil can happen also the vibration through this particular molecule molecule is there with all these bonds and bond angles and it can vibrate and this vibration energy can also contain energy which is coming from a momentum based energy transfer and this particular energy of vibration it can be written in the form of n h cross i where n can be different value 0 1 2 3 depending on the different conditions of its particular vibration state and this is quantized and over there I wanted to always be at n equal to 0 then this system can be actually recoil less system. So, absolute recoil less system for that it has to be such a system that it is totally static and there is no vibration even exist over the system. Now over here this particular expression I draw this is actually showing a different vibrational state what is that is actually going to experience for vibration and over here this is actually nothing but an expression of a phonon based system which is connected to the temperature. So, as we hit that a system we see that it is actually going to be at high temperature it is actually vibrating more it is actually going to higher energy of phonons and over here we wanted to go to at lower energy of phonons as possible if it is possible phonon energy is close to 0 then we will be achieving this absolute recoil less system. So, that is why temperature plays a huge role during MOSBAR spectroscopy and we are going to get a better resolved signature in MOSBAR spectroscopy if I go to a lower temperature system. So, that is actually going to happen when you are doing the MOSBAR spectroscopy. So, over here we have established why temperature is a huge factor and how we are going to look into a MOSBAR spectra on the y axis is percentage of transmittance varies from 100 to 0 and on the x axis it is velocity at one point it will be 0 then it is the positive direction then it is the negative direction and this is the speed by which I am moving my source either towards that means the positive or far away that is the negative from the sample and we are creating a Doppler effect to ensure that we are having a resonating condition. Now, when you are talking about a perfect matching of a system there are two important factor comes say in MOSBAR spectra percentage of transmittance is there, velocity is there and we first get a signature signal and where I am getting the maximum we call them it is the delta which is isomer shift. Now, what is an isomer shift? So, during the transition as you are discussing and let us say I am talking about a 57 iron. So, over here the ground state is i equal to half system whereas excited state is i equal to 3 by 2 system. So, this is the transition I am trying to achieve over here. So, first I am going to get it from a excited source which is going to leave this gamma ray and which will be absorbed by my sample and go to this excited is still the same i equal to 3 by 2 and ground state is i equal to half system and over here it is actually coming down from here. So, this is actually happening now over here when you are talking about this the source and this is the absorber or sample is going to have subtle difference on its energy of this particular nuclei. So, how nuclei energy is getting affected by the electron density and we have discussed earlier there is a possibility finite possibility that the s electron density can reside in the nuclei. So, it can bridge the nucleus and can stay there. So, that is why that is why the s electron density we are worried about now in iron if you take a look into it it can have 1s electron, it can have 2s electron, it can have 3s electron. So, this electrons can be there when you are talking about a valence shell and among them this 3s electron density we are more worried about because 1 and 2s are the core electrons. So, they will be minimally affected by the other valence electrons whereas, the 3s electrons will be very much affected in the valence shell by the valence d orbitals electrons and over there one particular factor that comes in it is the shielding effect. So, it is nothing but a competition between s and d electron in the 3 shell that means the 3d and 3s electron who is going to capture the attention of the nuclei and over there 3s electron has a finite possibility that it can reside inside the nuclei and that is going to be controlling what will be happening on the energy of this ground state and excited state of this iron over here. Now, if I want to look a little bit look into much closely and try to find an equation. So, this isomer shift can be explained in the following delta is equal to 4 by 5 pi z is square r square delta r by r and then psi 0 square of the absorb from the absorber minus psi 0 square of the source over here absorber and source that means that iron centers we are talking about and psi 0 square means s electron density. So, this psi 0 square over here minus s electron density and over here this 4 pi by 5 z z is the atomic number of the iron we are talking about e is the electronic charge r is that nuclear radiate the atomic radiate of the system. On the other hand the dr by r it is a very important aspect over here what is delta r by r. So, delta r by r especially delta r is the difference of the nuclei when it is in the excited state minus when it is from the ground state and for iron system 57 iron this value of delta r is actually negative why because the excited state radius is actually lower than the ground state radius and when you talk about this excited state ground state I am talking about the nuclear spin value of half for the ground state nuclear spin value of 3 half for the excited state. So, talking about the nuclear states and over there a nuclear state at 3 by 2 is actually smaller it shrinks down when it is go to the excited state compared to the ground state I equal to half and that is why this delta r value is going to be a negative one. And then we talk about this electronic difference the s electron density difference and as we discussed we can see the absorber and the source is not going to have all the time the s electron density present in the system same and why it is different because it depends on the s electron density and s electron density on the other hand depend on the d electron density and that is going to vary from compound to compound and that is going to reflect in this particular value of psi 0 square absorber. And then I am going to take a particular source if I subtract that I am going to get a particular delta value. So, even if I take the same system as a sample and source only then I am going to get a 0 value or by some other features I am going to get a similar amount of s electron density in the nuclear I am going to achieve the same thing. So, that is what is actually happening over here and over there we have to remember this delta r value to be negative and what is the consequence of that that we are going to follow right now. So, over here I am taking examples of iron 3 plus and iron 2 plus and over here I am trying to rationalize how we can explain what would be the trend of the isomer shift which will be positive which will be negative. So, let us go through that first we find how much d electrons you have. So, iron plus 3 is a 3d 5 system and plus 2 is a 3d 6 system. Then the next question we found if these are the 3d 5 and 3d 6 electrons what about the shielding effect the shielding effect on s electron the 3s electron. Now, 3d 5 versus 3d 6 more the d electron more is the shielding. So, comparatively this is actually having low shielding effect this is actually going to have high shielding effect but over here I am saying it is comparative comparatively iron 2 plus has high shielding effect and iron 3 plus has low shielding effect and if that is the case then what is going to happen on psi 0 square value on the absorber lower is the shielding effect that means s electron has more chance to go into the nuclear and more chance to go into the nuclear that means more chance to find it over there. So, the psi 0 square which is actually the electron density inside the nucleus is going to be having a high value compared to the 3d 6 system and again it is a comparative system right. Now, we already know this full expression we had for the delta value I am writing it over one more time over here with a lot of constants and all those things. So, let me try that one more time 4 by 5 pi z is square r square into delta r by r into psi 0 square of the absorber minus psi 0 square of the sample or the source. So, that is the full expression and over here these are all constant delta r by r depends on the delta r value which we found negative delta is nothing again the difference of the radii of the nucleus in the excited and the ground state and that is why what will be the delta value trend it is a high value multiplied with a negative number. So, it will be on the negative side whereas this one is a low value multiplied the negative side. So, it will be on the positive side again comparatively and over here I want to mention this negative does not mean it has to be negative number it is a comparatively negative for an example it value might be 1 unit whereas r and 2 plus will be 2 units. So, we can say 1 unit is actually on the negative side compared to 2. So, that is what we mean by negative side similarly this positive side does not mean that it has to be have a absolute positive value it can be minus 0.1 and this minus 0.1 is sitting positive compared to minus 1. So, you can say yes is on the positive side we are not talking about the absolute value but actually the trend of the system. So, over here you can see iron plus 3 will be on the negative side iron 2 plus will be on the positive side and this is going to happen if I talk about the iron oxidation state and taking all the other factors remaining constant only difference is the iron oxidation states as we go from low to high oxidation state the delta value what is going to happen you can say on the negative side for the high oxidation state. So, that means delta value is going to negative side from the positive side. So, lower is the oxidation state it will be on the positive side higher is the oxidation state it will be on the negative side again it is not absolute negative or positive it is a relative negative or positive. So, with that now we will take some examples but before that one more particular factor remain over here and that is the quadrupolar splitting. So, this quadrupolar splitting factor what it says that typically we expected that this transmittance and this is the velocity and again this velocity it is reduced velocity but it is actually representing the energy axis. So, any particular spectroscopy it has two axis one is the absorbance or transmittance which says like how much system you have and an energy axis which says that exactly what do you have. So, over here starting from 0 to 100 we expect that we will get a signature signal like this but in certain times this signature actually split it up like this and why it is happening so this is actually happening. So, this is where is the delta value is isomer shift but this signal is split it and why does it split that is because split and why does it split because over here when you are talking about the excited state and ground state we talk about ground state is half and excited is 3 by 2. Now if there is a quadrupolar moment a quadrupole moment exist in the molecule that means it is not equal to 0 then this states will split up the half will remain because there is no other possibility the plus minus half will stay together whereas this excited state will split up plus minus half and plus minus 3 by 2 because excited is 3 by 2 it can split up in 3 half and half and 3 half and half are not the same system they have different distribution of the charge density and which is going to interact with the quadrupole moment quadrupole moment and non zero quadrupole moment can distinguish them so that way they are going to split up and that is why instead of one signature I am going to get two and they will be equally splitted that is why if I average it out I will say where it should be so that is the isomer shift and over here I am going to see a difference between them and that difference is known as quadrupolar splitting. I want to mention over here this quadrupolar splitting defense whether your molecule have a electrical field gradient or not which sometimes it is written as EFG so it has to be non zero fellow and when I can expect an electrical field gradient for two different systems first you can see that during the lattice contribution the lattice contribution is nothing but the coordination geometry. If your coordination geometry is symmetric you do not have an electric field gradient because everything is quite symmetric but if that is asymmetric then your electric field gradient is non zero so you will see a splitted most perspective scope. So what do I mean by coordination geometry is affecting the electric field gradient so now example you have a perfect octagonal geometry around a metal all of them are equally bound with a ligand and that ligands are remaining all same so that means electron density is well distributed all around the octagonal geometry and that is not going to create any electric field gradient but if you remove one of this particular ligand and change that to another one so then what will happen you would now disrupt the overall symmetry of the charge distribution and there you create an electric field gradient and that will showcase in the lattice contribution. And the next one is the valence contribution which is the same thing electronic distribution of the metal how it is going to be symmetric for example if you have a T2G3 EG0 system we will call it as a symmetric system and my EFG will be equal to 0 but now if you say T2G2 EG1 and that becomes asymmetric or EFG is not equal to 0 at that condition it will showcase the quadruple splitting whereas in the first case because it is symmetric it is not going to show up. So that is the other point of valence and lattice contribution which I have also discussed in the earlier class in details. So that is the backup of all the different topics that we are going to cover in the next applications. So this is actually a setup where we revisited our thought process and the important parameters that we are going to use in the upcoming segments when you are talking about the applications of MOSBUS spectroscopy. So again going back the important factors are isomer shift quadruple splitting and how they can tell me about the oxidation states, spin state and about the coordination geometry around the molecular complex. So with that we would like to conclude this segment over here. Thank you, thank you very much.