 We were discussing Mosbauer spectroscopy and last day we have discussed two important parameters. One is the isomer shift and one is the quadrupole splitting. So I would like to go through that one more time just to ensure that all of you are following it properly. So don't need to remember anything or memorize anything just try to follow the logic how we are actually doing that. So first of all for Mosbauer spectroscopy we are changing the nuclear state. So taking 57 iron as an example we are coming from an excited state and coming to the ground state excited state is 3 by 2 ground state is i equal to half and over there we are releasing gamma ray and this energy will be captured by a sample. So this is the source and then there will be sample which will capture the energy. Now as we have discussed depending on the s electron density there can be directly go inside the nucleus for a finite time and can modify the energy of the nuclear state. So that's why in a sample and source the ground state and excited state may not be exactly at the same state. So when we are trying to go from the similar i equal to half to i equal to 3 half energy gap that energy gap may not be the same energy gap. It might be totally different energy gap so let me put in a different color. So that is can be happening or it can also happen that the energy gap is actually shrink down either of that can happen and at the end these are the samples and you can see the energy gap is not actually the same. While you are running this experiment we are doing something called recoil less condition or recoil less transitions which is actually the invention of Rudolf Mosberg from whom we actually got this name of Mosberg spectroscopy and this recoil less transfer ensures that we have enough overlap so that we can see the signals properly. However the energy is not matching properly and over there we can use the similar Doppler effect to ensure that the energies match and for that what happens we actually keep our sample and detector stagnant and move the source towards or far away from the sample and by that we actually match the energy gap and depending on that we can actually get the following different kind of spectral data so y axis is transmission percentage x axis is velocity by which I am actually moving my source towards or far away from the sample so towards is actually going to give me a positive number far away is going to be a negative number and at point time you will get zero so if this energy gap over here is perfectly matches with the energy gap with the sample what you are going to get a signal like this where it will be perfectly at zero but if it is higher in energy like in this case you have to move it to give it a positive Doppler effect so that the energy can match or if it is a lower energy like in this case you have to move it away from this sample so that the energy can match and depending on that this kind of signals you are going to get so over here you can see where the sample should be with respect to source it totally depends on the energy gap difference and this gap of the energy gap difference whatever the number you are getting we put it as delta or isomer shift so this isomer shift is nothing but the actual value on this velocity scale which is going to give you a relative energy with respect to the source so it is very much important and a lot of you have the question so how do I know why is the zero it's totally dependent on the what is the source you are using so source is very much important over here and we can have two different sources regularly use majorly it is used the metallic iron and sometime it is used sodium nitroprouside solution one of them is actually used metallic iron is the major news so whenever you are actually doing and recording a graph of mosba spectra you have to mention what is actually your source otherwise people might not know and people might use a totally different source and can have a totally different number so that is why you have to mention what is your source during the experiment and if somebody else is doing with the same source they should get the similar result but they use a totally different source they are going to get a totally different numbers of isomer shift the overall trend can be same but the numbers will be different okay and this delta value the isomer shift or chemical isomer shift it's actually can be expressed by this particular equation to psi zero square of sample minus psi zero square of the source okay so over there this is the full equation and over there there are few of the variables are there so from sample to sample if I'm using the same source so this is going to be a constant so that is why the source is very much important because if I change the source this particular value will be different so if I use the same source this value will be constant the rest of them are constant this is actually a value which shows what is the change in the radii from the excited state and the ground state so the excited state of iron is actually smaller compared to the ground state so delta r by r is actually a negative term it's a negative number okay so all the things now altogether the sample is going to be differed with respect to this value psi square zero psi zero square sample which is giving you the possibility of a electron density inside the nucleus for the sample so that is the only thing it is going to be changed so it can be a higher number it can be a lower number so this is the only thing it is going to change the rest of them are constant so what can happen so if your psi zero square sample which means the s electron density inside the nucleus it can increase or it can decrease so these are the two different possibilities right so case number one if the psi zero square value increases what is going to happen if it increases so this number over here is going to increase this is the constant this is a negative number so what will happen this is going to be subtracted from the source value and then multiply with a negative number and this is actually a very high value if it the psi zero square value increases that means the in the sample it loss of s electron density is present there and has a chance to go inside the nucleus so it is going to be a higher number which is multiplied with a negative number because delta r by r is a negative number so all together what is going to happen with the isomer shift or delta value it is going to be on the negative side doesn't mean to be a negative number all the time because it depends on what is the source value you are using but what is going to happen if I slowly increase my s electron density what I am going to see is that this value is going to increase the size your sample minus size your source multiply with a negative number so my number is going to shift towards a negative side okay and case two what happens if my size your square sample value that means a s electron density value decreases if this value now decrease what will happen if this value decrease and I'm multiplying with the negative number that is going to shift on the positive side okay so these are all relative terms negative and positive what I'm saying these are all relative terms it shows that which side I am going to move from sample to sample because in Mossberg spectroscopy generally compared with different samples so either I am looking in oxidation states spin state change so over there what my final goal is to find what is the change in the s electron density if it is increasing then this overall this whole value will increase and all together it is multiplied with a negative number so all the things will move towards the negative side okay and the other possibility is that my s electron density is decreasing and at that point of time this value is going to decrease and multiply the negative number so it will shift towards the positive side so that is the whole point we have to kind of follow to find out how the oxidation state or spin density or sorry the speed state is going to control my isomer shift or delta value so the next thing is that we have covered this thing in last class I'm going to do that one more time case number a say if you have an iron plus three system versus the iron plus two system first I'll find out what is the electronic configuration this is a d5 system this is a d6 system so this has much more higher d electron density it has a lower d electron density what will happen with respect to the shielding effect shielding means how the d electron is going to shield the s electron density from interacting with the nucleus so more the d electron more will be the shielding effect less the electron less will be the shielding effect so this less or more it is comparative term only between this iron three and iron two I'm considering okay so if I also give you iron one versus iron five you have to compare like which iron states I'm talking about and between them which is having more delay intensity so it is not actually absolute term but a related term okay so shielding effect says that it iron three plus has less shielding iron two plus has more shielding so what will be happening psi zero square sample value which is nothing but three s electron density because we have already discussed that it is one s two s and three s these three condition the electrons can go inside the nucleus for the iron and among them three s is actually most closely controlled by the 3d because they all belongs to the valence shield one s and two s in course so their effect is very minimal so this three s electron density which is going to have more chance to go inside the nucleus it is going to be the iron plus three because less shielding so the s electrons are free to interact with the nucleus so psi zero square is going to have a higher value and this is going to have a low value again these are all relative term between iron three and iron two if I change the oxidation state this relativity can change now I'm going to multiply delta r by r this is going to be a negative term for both of them because this is a constant value what is the change of the ionic radii for iron when it's excited state versus down state when the nuclear state is going to be i equal to 3 by 2 versus i equal to half a nuclear radii we are actually talking about so it's going to be negative so altogether what will be the isomer shift value then so you can see it is a negative value multiplied with a higher term so it is going to be a on the negative side on this hand it is negative value with a lower value so it is going to be on the positive side okay so higher is the oxidation state slowly the delta value isomer shift will go towards the negative side lower is the oxidation state it is going to move towards the positive side of the isomer shift okay that is going to happen as you are changing the oxidation state the next thing we have discussed is the spin state now say I have a low spin iron versus a high spin iron now over here we try to find out what will be the relative condition for the isomer shift which direction it will move so again we are going to look into the system regularly so it will the system differs between low spin and high spin with respect to the ligand generally for low spin it has to be a pi accepted ligand for high spin it is generally a pi donor or it can be also only sigma donor ligand is also possible no pi donation at all but we are actually not considering this at this moment so we are considering pi donor ligand and pi accepted ligand so that we are a very good strong field ligand in this case are very weak field ligand in this case now if this is happening what is the molecular picture so this is a metal present with a ligand and it is pi accepting ligand so that means a pi star orbital of the ligand is active and over there I am having a d pi p pi interaction but the electron density is going out from the d towards the ligand whereas in the pi donor ligand orbital generally a pi orbital comes over here what happens the electron density over there is going from this side in this way it comes to this side so this is also a d pi p pi interaction but over there the electron density come from the ligand to the metal side so electron density is moving this side now over here the rest of them is same d electron density what is going to happen which side is going to be higher d electron density it is going to be the highest because you are taking more d electron density out of the ligand over here consider at this moment the oxygen state are same so we start with the common d electron before we start with interacting with the ligand but once it's interacting with the ligand this pi donor is getting more electron density back towards the metal so the d electron density is going to be higher more different density in the d orbital whereas the pi accepting ligand it is going to remove some electron density towards the ligand so it is actually going to lose some d electron density then the rest of them are very same as the last time what is going to happen with respect to the shielding effect less d electron density so it will be low high d electron density it will be high what is going to happen with the psi 0 square sample that means it is nothing but the 3 s orbital interaction with respect to the nucleus so it is going to be actually a high value because it has more chance because it is having less hindrance from the d electron density before it can interact with the nucleus and this is going to be low number for this delta r by r it is always negative for iron 57 so what will happen to the delta value so this is a high value multiplied with the negative terms it will be on the negative side it's a low value multiplied with the negative term it is going to be on the positive side so if you consider the same oxidation state but you are changing only the ligand interaction high spin versus low spin low spin is always going to be on the negative side high spin is always going to be on the positive side again don't need to remember everything just logically think what has happened to the s electron density and how the s electron density is modulated by the d electron density with the shielding effect okay so this is what is going to happen with respect to the isomer shift with respect to the spin state and oxidation state the next thing we discussed about the quadrupole splitting we have discussed in details so today I'm just taking the gist of it so to see a quadrupole splitting what do you need to have two things first your nucleus have to have a quadrupolar moment active that means eq should be not equal to zero and it is only going to happen if your state is greater than half so it is only going to happen for iron 57 when you're talking about the excited state where the i equal to three by two when you're talking about the ground state i equal to half so over here eq is zero so only the excited state can actually vary and create a quadrupolar moment second condition you should have a electrical field gradient that means the electrical field around the nucleus should be asymmetric or inhomogeneous and this is known as electric field gradient or e f g and that should be equal to non-zero value and this actually comes out with two different condition first one is the lattice contribution which says that the ligands when it is surrounding the atom it should be in a symmetric manner for an example octahedral geometry square panel geometry they are all or spherical geometry they are all symmetric distribution so in these conditions you will see a lattice contribute you don't see a lattice contribution but if it breaks down that symmetry you will see a lattice contribution for example if you have a tetrahedral geometry you have a lattice contribution in an octahedral geometry if you change one of the ligands you go to c4 v symmetry you are going to see a lattice contribution the second one is called the valence contribution it depends on how the electrons are actually distributed around the system so if you start with a very simple octahedral geometry and say my symmetry is t2g3 eg0 so over there t2g3 is a symmetric management of the electrons around the nucleus so it is not going to be contributing to the valence shape however if it is t2g3 eg1 it is going to contribute now say if you have a t2g6 eg0 it is not going to contribute to valence contribution if you have a t2g6 eg2 it is not going to contribute t2g3 eg2 they are not going to contribute so these are very symmetric systems which are not going to contribute similarly you can think about also about that tetrahedral geometry e2 and t2g0 it is not going to contribute e2 t2g3 it is not going to contribute so these kind of systems are not going to contribute because they actually surrounds the electron around the molecule in a symmetric manner so the electric field which is going to be generated will be very symmetric so it cannot create a gradient but any deviation from this it will contribute to there and the valence contribution can keep down okay so two contribution lattice and valence contribution