 So, now the actual factor that is going to affect over here is known as electrical field gradient. What is an electrical field gradient? Very rough term or very overview term what we can say it is nothing but the in homogeneity in the electrical field. Electrical field coming from here from the S electron density penetrating the nucleus and that one is going to interact with your nucleus. And if you have a quadrupolar moment and if you have an inhomogeneous electrical field only then you are going to see the effect of the non-spherical field. So, let us come into that what are the pictures I can have first say my field is homogeneous anywhere any direction you go your field is homogeneous that is how it looks like does not matter where you put your system you are going to see the similar field. So, over there the electrical field gradient or the difference in the electrical field with respect to the direction or the position is not present. So, we said EFG short from for electrical gradient is going to be 0. So, if you have electrical field gradient 0 then if you put a spherical field or if you put a non-spherical nucleus it does not really matter because your electrical field gradient is already 0. So, even if you have a quadrupolar moment active system which is given by EQ non-equal 0 a non-spherical nucleus and I am putting into a very homogeneous electrical field it is not going to be affected because you are not having the electrical field change with respect to the difference of the presence of the quadrupolar moment. And over here the quadrupolar moment is already 0 electric field gradient is 0. So, nothing matters. So, in either two cases you are not going to see any difference. Now, what happens? I filled this inhomogeneous. So, now you can see over here you have a larger electron density over here it is lower electron density it is the inhomogeneous field. So, my EFG is not equal to 0. Okay, I am drawing two fields like this and now you can imagine what are the different things you can change you can put a spherical nucleus EQ equal to 0. Although the electrical field gradient is non-zero because the nucleus is not creating any quadrupolar moment you are not going to see any effect on the nuclear states in this condition. However, if now your nucleus is a non-spherical one and you have a non-zero value of quadrupolar moment your electric field gradient is also there it is inhomogeneous. In this particular condition you are going to see some effect of it in the transition. Of the energy state transition of the system only when your system is giving some quadrupolar moment that means you have to have a non-spherical nuclei non-spherical nuclei is only possible if i is greater than half only then your quadrupolar moment will be active that is the condition and your electric field gradient should be also active that means you should have an electron density around the nucleus which is inhomogeneous and that is very easy to find how let us take a look into that. But before going there let us try to find out what is the effect of that if I have a system which is having an electric field gradient positive non-zero and quadrupolar moment non-zero system. So first iron 57 system I have a ground state of i equal to half an excited state of i equal to 3 half now say in this condition this ground state can it be active for quadrupolar moment the answer is no because i equal to 0 and i equal to half their spherical system so it cannot create eq this one can create a quadrupolar moment but now say i am putting it in a homogeneous field homogeneous electric field that means efg is equal to 0 so then what i'm going to see only one band in the terms of mosbar spectroscopy i'm going to see a band like this this is the percentage of transmittance this is v equal to millimeter per second wherever it is cutting that is the delta value isomer sheet that is how it looks like that is what we studied so now say the same system i am talking about but now say i am taking a system where the electric field gradient does exist it is a inhomogeneous field so if you have a system where you have a quadrupolar moment that will be very affected so i equal to half it is equal to 0 so there is no effect of that so it doesn't matter so this is electric field gradient non-zero but your quadrupole moment is 0 so it is a classic case of this one so you're not going to see any effect however i equal to 3 by 2 it is i greater than half so it is a non-spherical system and you also have an electric field gradient present so the system i want to look into over here what will happen that will split up in two different system one will be plus minus half one will be plus minus three half and what will be the difference it is that known as delta e cube or quadrupolar splitting okay this is known as quadrupolar splitting because that is generated from quadrupolar moment and previously you are going to have only one signal now what you're going to see two signals and you can very easily see the energy gap between these two will be nothing but an average of the previous value so previously we are seeing a value like this right now what i'm going to see is the following two signals one from the lower energy one for the higher energy lower energy means lower side v higher energy means higher side v and if you take an average of that that will be the original value without any splitting so i'm not exactly done in the middle of it so that is how it is going to look like so this blue line is the one you are going to see and the difference you are going to get it at two different positions if you take an average of that whatever the average value would be that will be the delta value or isomer sheet and the difference between these two it will be the delta eq or quadrupolar splitting okay so that is the two terms you are always going to find in a mosbar spectroscopy the quadrupolar splitting and isomer sheet isomer sheet will be always there quadrupolar sheet depends on the condition only if you have a electric field gradient non-zero only then it will be value and for i equal to half and i equal to three half the ground state of the iron 57 cannot split the excited state can split that will depend whether you have an electric field gradient present or not any questions up to here if not we'll go to the next part how do i find out my electric field gradient is non-zero or zero because that is going to be the only important factor we'll controlling them and this is given by only the coordination side by two different system one is called the lattice contribution which is nothing but with respect to the coordination geometry and second thing is known as the valence contribution which is coming from the electronic distribution around the middle center so these are the things which are going to say whether you are going to have a electric field gradient present or not it is always better to understand with an example so let's look into an example so the example i'm making is the following iron with six water molecules iron is in plus two oxidation state okay so i'm taking this iron it has six coordination and all of them are water molecules so the first thing we are going to look into about the coordination geometry the coordination geometry in this system you can see it is perfectly octahedral system right so if it is a perfect octahedral system do you expect that just because of the coordination it is going to create any difference in the electronic field around the iron all are six water molecules so all should be same so it shouldn't be showing any electronic distribution difference what i'm going to say if instead of this one of the water molecules over here if i put a bromide now i create a asymmetry in the system now it will create a electronic distribution in homogeneity and at that condition it will be contributing which is known as the lattice contribution however at this moment if you don't have that bromide if you have all six water molecules you shouldn't see any effect right however what is the electronic distribution it is an iron plus two system so that means 6d electrons it is octahedral system so it is going to have t2g and eg the electronic configuration i'm drawing 6d electrons so first three goes over there where the fourth one goes that depends on whether it is a high spin or low spin now water molecule group 16 sigma donor pi donor it is going to be a weak field ligand that means it's a high spin system so the fourth electron will go there fifth here sixth one here so now the electronic distribution is it symmetric symmetric electronic distribution means if you have a t2g3 eg0 it is a symmetric electronic distribution if you have t2g3 eg2 it is symmetric electronic distribution if you have t2g6 eg2 symmetric electronic distribution if you have t2g6 eg0 for low spin complexes that is symmetric electronic distribution, but this one is not. EG is fine, but T2G case not all the orbitals have similar electron density. So that means it has some inhomogeneity. Such a kind of electronic inhomogeneity coming from the electronic distribution, it is known as the valence contribution. So over here in this case of iron, water molecule, what do you expect? There should be valence contribution and there shouldn't be any lattice contribution. For the first check, that is what we believe. So that means there should be an electric field gradient present, because as we said, electric field gradient has two contribution, valence or lattice. Lattice means simply look into the coordination side, see if it is symmetric or not. If it is square geometry, four of them same, symmetric. Octaheddles, all six of them are same, symmetric. Linear, two of the ligands are same, symmetric. But if it is not, if you have different ligands on either end, then it is a symmetric. Valence electron contribution is coming from the electron density distribution. Either the T2G or EG should have same number of electrons, only then you can have a valence electron contribution. For tetrahedral also you can think about the similar way. Now, in this particular case of iron system, the system is not run over here, because what I said it is a T2G4 EG2 system. And then with this particular electronic configuration, one thing can happen which is known as jam-teller distortion. So what will happen in jam-teller distortion? These orbitals will split up. So the valence cell is still fine, it is still inhomogeneous. So valence contribution will be still there. So nothing change over there after the jam-teller distortion. But what is happening in the coordination side? Previously, all of them has six water molecules and all of them are like this. But now two of them will move out, that is the jam-teller distortion. So say these two, the red ones will have now different bond distance compared to these ones. So altogether previously you have a perfect tetrahedral. Now after jam-teller distortion you do not have it. So these two are different than compared to the equatorial one, the axial bonds are different. So now your coordination side is also not totally symmetric. So now you are also going to have some lattice contribution due to the jam-teller distortion. And both those things will happen in the same time. So it will have a very strong electric field gradient present, a non-zero electric field gradient. And that is why what will happen for this iron H2O whole 6 2 plus, what we say it has going to have a delta value and it is going to have a non-zero water polar splitting value. And what you are going to see is the following. Like these two values, plot it in velocity, y axis is percentage of transmittance. These values experimentally found at minus 0.5 millimeter per second and 2.9 millimeter per second. So what should be the delta value average of them? Average it out, it will come out at 1.2 millimeter per second. And what is the delta EQ difference between these two? That is nothing but 3.4 millimeter per second. So that is how you figure it out. What is the isomeric shift and what is the quadrupolar splitting? So I take two minutes extra today, so I will stop over here.