 Now, coming back to that one more time, iron 5726, 1s2, 2s2, 2p6, 3s2, 3p6, 3d8. That is the condition at iron 0 oxidation state in metallic form. So, how many S orbiters you can see over here? There is one, there is one, and there is one. There are three different S orbiters. Now, which one of them are getting affected? Now, over here, during the study of your chemistry, you learn this interesting term called core electron and valence electron. So, over here, this set of electrons are known as the core electrons, because they are remaining at an orbital which is very close to the nuclei, and rarely it is getting affected by what is happening outside during the reactivity of the molecule. During the reactivity of a molecule, the 1s and 2s are very minimally affected, whereas these electrons are known as valence electrons, because these are the maximum changes you are going to see. You can see an electron exchange, you are going to see that from this particular shell, number three for iron. So, that is why these are called the valence, these are called the core. Now, over here, what are the changes happening? The change is happening in the 3d orbital. Now, the 3d orbital, it is getting affected by both 1s and 2s orbiters and also 3s orbiters. Now, 1s and 2s, as I just said, it is the core electron, so their effect is very low, minimal. Whereas 3s orbiters, it is very lying almost at the same shell, so they have a better interaction between the 3s and 3d. So, their influence, the 3d orbital influence on 3s orbital is maximum. So, this makes minimal and maximum same with respect to ds interaction. So, that means if I change anything on the d orbital, which of the electrons are going to get affected, so on and so forth. So, that is what I am going to say over here. Any questions up to this point, because these are very important. Are we understanding why the d orbital is affecting my S electron density? And why do I care about S electron density? Because this is the S electron density that is the only thing that can go inside the nucleus and that is the psi 0 square value, which is very important. And this value is important because in the equation, this is the value is going to get affected. So, over here, I am trying to see if my d... Hello, sir. Yes. Sir, d and S orbital overlap is very low. So, is really d orbital affect the S orbital? Good. So, it is not really overlap, it is a different interaction. So, I am coming into that what the interaction is. So, for that, let me take an example. Okay. So, now I am going to take two examples, iron plus 3 and iron plus 2. And with these two systems, I am trying to explain what is actually happening and how the d and S electron is actually getting affected. And it is showing me that there is a bad network right now in IIT. So, I have very little to do. So, we will see how much you can go ahead. So, now iron plus 3 and iron plus 2. If I want to write what is the valence electron configuration for iron plus 3, it is 3d5 for iron plus 2, it is 3d6, right? Now, how does d electron and S electrons are connected? This is connected to something called shielding effect. Now, what is a shielding effect? Shielding effect can be learned in a different ways, but I really like that pictorial diagram, like exactly what is happening. And for that, I will go to the radial distribution function. So, it is the psi square, which is giving you how much electron density is distributed with respect to the distance from inter-nuclear distance. Oh, sorry. The distance from the nucleus. Now, what happens? Say, this is my S electron density. So, how do you see? If you remember, we draw that last time. So, this is my 3s electron density, how it looks like. And how the d orbital density looks like? d orbital density looks like something like this. Now, over here, you can see in this particular area, there is a possibility that you have d electron density and also have s electron density. And both of them are fine for the attention of the nucleus, that whether the nucleus is attracting that electron density or not. And over here, if I have too much of d electron density, what will happen? That it will fight with this electron density and ensure that the d electrons are getting affected with the nucleus more compared to the... s electron density. So, that means the interaction between the nucleus and s orbital can be affected by the 3d orbital. More the 3d orbital electron density, more will be that effect. And this effect is known as the shielding effect. Now, if I want to say that 3d5 versus 3d6, these are 5 electrons, there is 6 electron, which you are expecting to see more effect in cutting down the interaction between the nucleus and s electron density. Where I have more d electrons, that means 3d6. So, over here, I will write, over here, I have more shielding effect. And for this one, I have less shielding. Now, the number of s electrons it is constant. In both cases, there is 2 s electrons, 3s2, right? So, it is 3s2 is common and then it is 3d5 or 3d6. Now, the s electron density is common. So, over here, if it is less shielding, that means this s electrons of 3s2 have more possibility to interact with the nucleus. That means, if I want to find what is my psi 0 square value, this one is going to have a higher number. And in the case of 3d6, because I have more shielding, the s electrons are getting hampered compared to 3d5 or Fe plus 3. And that is why there will be a little bit lower contribution of s electron density inside the nucleus. I am going to take time over here because this is the most important part over here. If we are connecting the s electron density with respect to the oxidation state. The oxidation state of iron, when it is changing, it is changing the d electron density, nothing else. Now, how the d electron density is affecting the s electron through the shielding effect? Because they are trying to cover the similar area where both s electron and d electron is fine for the nuclear attraction. And if you have more interaction, that means you have more possibility that the s electron is going inside the nucleus. At r equal to 0. Now, if you have a lot of d electrons, that is actually taking most of the attention of the nucleus. So, s electron is not really having enough time with the nucleus, so you have less possibility that the electron density will be inside the nucleus. And that is given by the shy zero square. So, that is how the thing is getting affected through the shielding effect. Anyone has any problem in understanding that, please raise your voice, because that is very important. So, Orko, is it getting clear now how the s and d electrons are contributing with each other? Okay. So, now, if this is the case, I know this shy zero square value will be higher value in case of iron plus three and lower value in case of iron plus two. But again, these are relative terms with respect to when I am comparing between iron three and iron two. If I have a iron three system, if I have a iron two system, iron three is going to have higher S electron density inside the nucleus, iron plus two is going to have lower S electron density inside the nucleus. Now, if you want to put iron plus four over here, it will have much more higher value. If you want to put iron plus five, it will have much more higher value. If you want to put iron plus one, a much lower value, iron 0 probably the lowest value possible. So, that is how the oxidation state controls the a-selectron density inside the nucleus through the shielding effect. Now, this is we are sorted out that how much a-selectron density is going to be there. Now, look into that equation delta 4 by 5 pi z square, sorry I probably missed a pi value over here earlier, my bad. 4 by 5 pi z square r square delta r by r into psi 0 square of sample minus psi 0 square of source. Now, over here my source is going to be a constant value, that is the same source I am using and now say you are trying to differentiate between iron plus 3 and iron plus 2 that you are taking the same source. So, this is becoming a constant value. What you are changing is your sample whether it is a plus 3 versus plus 2. Now, looking back over here what we found psi 0 square is going to be a higher value for iron plus 3 compared to iron plus 2. So, this value is going to be higher for iron plus 3 compared to iron plus 2. So, this total value psi 0 square sample minus psi 0 square source is going to be higher value for iron plus 3 compared to the iron plus 2. Now, if this is fine now comes this particular term delta r by r which I already defined is nothing but the difference between the excited state radius minus ground state radius and what I have said earlier it is actually a negative value for iron 57 because iron the radius of the excited state is actually smaller compared to the ground state. Now, imagine if this is a negative term all the rest of them are constant values. This is a negative term I am multiplying with this particular term psi 0 square sample is the only variable over here source is a constant and now this is going to be a higher value for higher oxidation state and lower value for lower oxidation state. So, compared to that if I multiply that with a negative number what I can say this delta value will be on the negative side for iron plus 3 and positive side for iron plus 2. Take your time look into the equation and find out what is the variable and why the delta value is becoming more negative if I have a higher oxidation state think about it. This equation says the only thing that can affect the delta value that means the energy gap between the source and the sample is the S electron density. So, I am trying to connect the oxidation state with the S electron density they are connected by the shielding effect more that the electrons you have more is the shielding effect less has the chance that S electron density will be inside the nucleus for the logical check. Now, source is a constant and I am trying to differentiate different samples. Now this sample value of psi 0 square will be higher when I have a higher oxidation state that means lower d electrons that means high possibility of the dase electrons are inside the nucleus. However, I have to multiply that with this particular term delta r by r which is negative because the radius of excited state in iron is a negative term. So, that is why in iron higher oxidation state means you are going to move negative side and lower oxygen state means you are going to show in the positive side. Any doubt or question up to this point? Please show. Okay. If not, now the question is what do I mean by this positive and negative values? So, this delta value that I was talking about from so on, this delta value is nothing but the energy gap between the excited state and the ground state between the source and sample and that is what is I am actually going to see in the real MOSBAR spectroscopic data is the following. I am going to show the data as percentage of transmittance which can have a value from 100 to 0 and over here I am going to see a band like this. Why I am seeing a band like this that we have discussed earlier over here that I am going to match the source and sample energy gap. Now over here I am going to move this system towards the sample or against the sample and this is written by this Doppler velocity millimeter per second and this velocity is nothing but a function of the energy itself. So, it is actually a one of a hidden way I am writing that velocity of the Doppler effect and this is actually connecting my energy and this is going to have a particular value if I do not have to move anything at all if it is matching at zero and this side is positive that means I am going towards the sample this side is negative that is against the sample. This is going to be absolutely at zero only when your source and sample are exactly the same exactly nothing not a single parameter difference only then you are going to have v equal to zero value but if the energy gap of my sample so let me just draw it so say this is the ground state and excited state of my source and say my sample is such that right now it has higher difference between the ground state and excited state e sample one. So higher energy state means it has higher value than the source itself how I can match it that means I have to move forward with the help of this Doppler velocity only then I can match the energy and that is going to be somewhere around there say it is plus v1 so that will be written as this delta value energy gap difference between the source and the sample now say I have another sample where I have the energy gap such a way that it is actually a little bit lower compared to the ground of the source says e sample two where the energy gap is actually less than the source previously it was higher than the source so that was I have to move it towards the sample but now if it is less than the sample less than the source the sample energy gap is less than the source I have to slow it down or move on the other side so that means I have to move away from the sample say around this value is like minus v2 and that will be the delta value or the energy gap difference between the sample and source. So depending on the energy gap with respect to the source we can have a positive value we can have a negative value and this way we can have this particular value delta can tell you what is the energy difference between the excited state and the ground state of the nuclear state i equal to 3 half and i equal to half and this delta value over here is known as which is called isomer shift or sometime it is also called chemical isomer shift okay and sometime it is also called center shift centered with respect to the source. So if you now find a term isomer shift that means I am calling I am talking about the delta value which is nothing but the energy gap difference between the source and sample if it is as a positive value that means this energy gap difference is actually higher compared to the source if it has a negative value that means it has an energy gap lower compared to the source that's all and when we talk about earlier when you're talking about this iron plus three versus iron plus two you can have a negative side you can have a positive side that means what I'm talking about that if I have a sample like this for iron plus three where should the iron plus two should come now say iron plus two I'm showing with red this should come towards the anytime you're drawing a graph you should draw what are the accesses so it should come on the positive side or towards the negative side so as we just discussed iron plus three is there there is the delta value for it where should the delta value come for iron plus two positive side on negative side so that is what we have discussed earlier that iron plus two generally comes on the positive side with respect to the iron plus three iron plus two will be on the negative side doesn't always mean it has to be a negative value it is comparatively towards the negative side okay so that means iron plus two over here will be on the positive side so this positive side means so say this is the delta iron plus three past two and this is the delta of iron plus three so say the delta value of iron plus two I'm just giving you an example say minus point one but delta iron plus two will be negative to that minus point one so you can say delta iron plus two is positive side compared to the delta of iron plus three but doesn't always mean it has to be a positive value so this positive and negative are actually the relative term which side they are lying positive side or the negative side that's all so mole will be the positive oxidation state and negative and lower oxidation state so lower oxidation state will move towards the higher side and higher oxidation state will move towards the negative side why again connect that back to the s electron density shielding effect and that equation where the high zero square value is there but you have to remember there is this particular term of delta r by r which is negative and that is why it put everything on the other side okay so any questions are query up to here why a higher oxidation state should lie on the negative side by a lower oxidation state should lie on the positive side look back into the equation in the exam or assignment you will have the access to the equation you have to just look into it and then connect the dots like what is actually happening over there