 So, so far we talk about the oxidation states and now we know exactly where it is coming for the higher oxidation state and where it is coming for the lower oxidation state. What happens if I now also want to have an idea about the spin state? What happen in a low spin state and a high spin state? And over here I am actually giving you an idea of the change of the spin state at the same oxidation state. Don't change oxidation state and spin state at the same time because there are two different things. Can you have an idea like which way it will go if it changes both the spin state and the oxygen state? Yes, you can have an idea. But if you really want to have a very good qualitative idea, change one at a time. Okay, so right now say I am keeping the oxygen state same, only the spin state is changing, okay? Oxygen state is constant, that is one of the assumptions I am taking so that we can understand what is actually happening over there. Now, a low spin state and high spin state, what are the factors defines whether it will be a low spin state and high spin state? Anyone? Rishabh, what are the factors controls whether my iron complex will be low spin state or high spin state? Lose one? Yes. What are the factors that controls my iron will be a low spin state or high spin state? The ligands which iron is coordinated with, if it is strong field ligand then it will be low spin state and weak field and high spin state. So that is very true. So it is about the ligand mostly controlled. Say if it is a strong field ligand, then you are going to have a low spin state. If it is a weak field ligand, you are going to have a high spin state. Now what is the difference between strong field and weak field ligand? Strong field ligand basically are generally sigma donor and pi acceptor ligands. Whereas on the other hand, the weak field ligand are typically sigma donor, pi donor and you can also have cases where you have only sigma donor ligands. For an example, group 151 like amine systems, that is only sigma donation possible, no pi interaction. Now, when you have a pi acceptor or pi donor, that actually changes the things a bit. So now say this is my metal, this is the d orbital and it is interacting with my ligand because it can have a pi acceptor property. That means some part of the ligand can actually take some electron density out of the system. That is what the pi acceptor means, acceptance with respect to the ligand. Some of the electron density is going out of the metal. So which particular electron density do you think it is going to be out, SPDF from the metal? Anyone, if I am talking about iron, which part of the electron density is going out? I am actually already shown over there, it is the d electron density. So the d electron density actually goes out. So the d electron density is actually going out. So it is going to have low d electron density. It is very much similar to the higher oxidation state. You have very low d electron density. So what will be the shielding effect? It is going to be low. What will be the s electron density inside the nucleus? Because there is not much of electron density of the d. So s electrons are having enough time with the nucleus. So this high zero square value, especially for the sample that we are talking about, it is going to be high because it is spending too much of time to the nucleus. But however we know the delta R by R value is negative like last time. So what will be my delta value? It will move towards the negative side. So low spin system is nothing but very similar in the logic compared to the higher oxidation state because I am losing my s electron density, a d electron density towards the ligand and the s electron density have very less shielding and it has enough time to interact with the nucleus, enough time to be inside the nucleus. That is that important term. Now coming back to the weak field, now it is quite straight forward to you guys. What is going to happen? So it is a sigma donor, pi donor. So it is going to have some pi donation coming from the ligand towards the metal. So the d electron density, it is going to be high because why it is happening? We talk them as a d pi, p pi interaction. So now the electron density is moving on the other way. Previously it is moving to the ligand, now it is coming from the ligand. Now you have more d electron density, high d electron density. So that means your shielding effect is also going to be very high. The d electrons will shield the s electrons from seeing the nucleus, psi 0 square value. It is going to be low because most of the s electrons are not even seeing the nucleus properly. However the delta r by r value is negative and all together the delta value will be on the positive side. So this system, the high spin system is nothing but very similar to high oxidation, sorry low oxidation state system where you have a lot of d electron density. So whenever you get a system and you are asked to talk about like whether the delta value will be on the positive side, delta value will be on the negative side, what you have to look for? You have to look for the d electron density and then connect it to the shielding effect. More d electron density, more shielding, less d electron density, less shielding. And from the d electron density through the shielding effect you are trying to have an idea how much s electron density is present there. And that means higher the s electron density with little shielding effect that is going to have an effect on the high zero square value. And with respect to that you can be either negative side or positive side just remember that delta r by r is a negative value. So that is what is going to happen. So you can see on the low spin state you are going to have mostly on the negative side, high spin state you are going to lie most on the positive side. Again the disclaimer over here I am going to give, it does not mean it has to be a negative value or positive value, it is a respective term, it is a comparative term. It will be comparatively negative side, it will be comparatively to the positive side. So that is going to happen for iron with respect to the different oxidation state and different spin state. Any questions or queries up to this point? And now I am going to show you a table. So this is the table taken from the Solomon liver book and over here they actually put all the different complexes mostly found and over there they have put the oxidation state in the parenthesis and spin state over here. So for an example the first two state what they are saying iron plus one oxidation state these two, s equal to 3 by 2 means it is high spin state, s equal to half means low spin state. So now you can see with the same oxidation number iron plus one the low spin state is lying on the negative side and high spin state is lying on the positive side. Now take a look you can take any either of them over here say iron plus three spin five by two iron plus three spin half you can say iron plus three spin five by two is most on the positive side high spin state and low spin state is lying on the negative side. So it is up to you how do you want to remember I do not remember them very well with acronyms so I generally try to follow the logic high spin state low spin state what is happening to the d electron density what is the effect on s electron density what will be the effect on the delta value because the equation will be with you all the time okay and over that delta r by r is the negative time. So that is how you can feel and over there as you can see iron plus one c start from a very positive value and iron plus six end up to be a very negative value so not only the actual value is negative but also they move up the direction as you can see iron plus one to iron plus two to iron plus three to iron plus four to iron plus six you can see it is slowly moving towards the negative side and that is what we found that higher oxidation state means you have less electron density more less electron density in the nucleus but multiply with the negative terms it will be coming on the negative side okay so that is what you guys have to look into. Generally if you have a question or query you generally try to take a look with respect to change in one parameter and try to find what the parameter is oxidation state or the spin state change and then try to logically find out which direction the delta value should go any questions or queries up to this point if not over there we haven't discussed about the source what is the source generally people use so typically the source use is metallic iron so that means that cobalt 57 if you remember it has been created a radioactive system which creates the iron 57 system it is created at a metallic form that is one of them and sometime this salt is also used you have learned this particular system in your probably class 11th year sodium nitro-prucide dihydrate sodium nitro-prucide solution you have used for the nitric group this ring test and all those things sodium iron CN5 NO 2H2 so either of these two compounds is actually created from this thing so if you remember cobalt 57 is also created with a bombardment with a deuterium and that is done on either a metallic iron or onto particular this particular complex one of them most of the experiment you will find they are done with a metallic iron because they actually make a metallic iron foil and on the top of that the bombard the deuterium created the cobalt 57 sample and sell them it has a half-life of 270 days and that is taken as the source okay so that is what is happening the first important parameter of the mosbar spectroscopy the isomer shift or the center shift or chemical isomer shift or the delta value that we are talking about so that gives you a very direct idea what is the oxidation state what is the spin state of the system but that is not the only parameter that can help you to find out what is happening in a molecule there is another effect comes into the picture and that is known as the quadrupole effect so what is the quadrupole effect so over here we are talking about the nucleus and electron interaction and over here we are taking how much the nucleus is getting affected by the electron density sitting around it and if you remember one of the assumption we have taken that the nucleus is spherical and that has the consequence that we are talking about only about a monopole and that is true if the nuclear state we are talking about has this nucleus state of equal to zero or equal to half in these two conditions your nucleus is actually spherical but if your nuclear state is equal to greater than half now you have a non-spherical nucleus now the system is going to have some effect because now say your nucleus look like this so previously once the electron density comes inside the nucleus because it is here all the different direction it is having similar interaction but now if you have a non-spherical nuclear so that is the nucleus if you have a non-spherical nucleus now it will depend where the electron density is inside the nucleus on the side over here or in the middle or where so in three dimension where it does exist because the charge distribution inside the nucleus is not homogeneous anymore because previously in a spherical nuclei the charge is positive all around the place it is homogeneously distributed but a distribution like this you might have extra positive charge on this side and this side has less positive charge so in the other way I can say this is a system where I have positive end on this one and negative end on this one and that is what is known as quadrupole and this quadrupole is created when you have a non-spherical nuclei so now the question is how do I count it the effect the effect is counted with a system called quadrupolar moment which is nothing but integration of the charge of the proton so this is the proton charge how it is distributed into the charge density into r square into 3 cos square theta minus 1 into dv I'm coming into that what are these things dv is a small volume that where you are trying to measure and then you have to do that over all the area of the nucleus but what is the r and theta r and theta are the spherical coordinates I'm trying to put down the electron density inside the nucleus and this is showing me exactly where the electron density is inside the nucleus now if you have a sphere this 3 cos square theta minus 1 value becomes 0 because theta value becomes 54 degree for a spherical interaction and that gives you this value 0 so quadrupole moment will be 0 for spherical moment but if it is a non-spherical system it will have some value and that is going to have some effect on the state of the interaction between the nucleus and electron and what is happening with the nuclear state before going further anyone have any idea why non-spherical nucleus should have a quadrupole moment if anyone has any question please go ahead and ask