 they are very easy to monitor so that we are going to follow again with some examples. So one of the examples I discussed last class so I'm not going to go into the details of it so I'm just giving you the examples. It is iron in an octahedral geometry with six water molecules. So what is going to happen over there? So first it is octahedral geometry so the first thing you are going to do is to draw the octahedral geometry electronic distribution. So it is iron plus two to start with so six electrons and it is going to be a high spin system because H2 is a sigma donor tides on a ligand. So six electrons are going to contribute one two three the fourth and fifth electron goes over there high spin system. Sixth electron comes over here so now over here you can see just looking into that first thing we are probably going to say there will be no lattice contribution and there will be valence contribution right because the valence cell is in homogeneously filled up. However this kind of electronic configuration also triggers the jantailer distortion. So jantailer distortion will be there and what will happen these orbitals will split up. Similarly these orbitals will split up and you are going to have this contribution and once you have jantailer distortion this is going to be stabilized this particular configuration and what will happen? These two axial ligands generally have been going to have different bond lengths compared to this square panel geometry. So there will be six outer molecules bind to the iron but two of them the axial ones are different compared to the equatorial ones. So that means previously what we say is the lattice contribution is not present that is wrong because due to this jantailer distortion now it is going to be present. So it will having both valence contribution both lattice contribution and at the end what we are going to have is a very strongly splitted line. So generally once the results is shown it is given both the values the delta value the isomer shift say it is given 1.2 millimeter per second. I am giving you the data from the original literature and the splitting the quadrupole splitting it is written as delta aq and isomer shift is written as delta this is the quadrupole splitting and this is the isomer shift. So quadrupole splitting is 3.4 millimeter per second now if I ask you to draw the MOSBUS spectroscopy how it is going to look like. So what you already know said in reality I am going to see two splitted lines like this. Now the question is what is the position of these lines on the x scale of velocity millimeter per second and this is the partial meter transmitters and over here what we are going to find out so let's say this is x this is y and we want to find x and y. So this is already given over there what is x minus y the difference between them that is delta aq value right this is given over there 3.4 millimeter per second and what is the average of x and y that means what will be the peak before it's splitted that is going to be the value of the original isomer shift so x plus y divided by 2 that will be 1.2 millimeter per second that will be the original delta value which is 1.2 millimeter per second and delta aq is 3.4 millimeter per second. So now if you solve this equation you can find x and y so I'm not doing the full mass x is minus 0.5 millimeter per second y is 2.9 millimeter per second so you can say this is minus 0.5 this is 2.9 so these are the values are getting and that is how you actually find out where the lines would be okay and before going further down the question is why the quadruple splitting happens this is because basically when you look into the excited state of i equal to 3 by 2 and the ground state of i equal to half we generally talk about this is the only signal I'm going to see so we show a line like this but in reality what happens due to this quadrupolar splitting it is for this quadrupolar moment it is going to affect only the non-spherical one that means i equal to greater than half states so over there ground state is i equal to half so it is going to remain the same i equal to half but this state is going to split up it is going to split up in plus minus half and i equal to plus minus 3 half so there are total four state possible in between 3 half plus 3 by 2 plus half minus half minus 3 by 2 and among them what happens because of this quadrupolar moment this system can differentiate between plus minus half and plus minus 3 half so plus 3 half and minus 3 half come together over here and plus half and minus half come together over here which are all generating from i equal to 3 half so over here now you can have two interactions okay so instead of one line you are going to see two lines and what is this difference this difference is the delta eq which is shown over here and what would happen if the splitting is not there this will be exactly in the middle so that is why their average value what is actually supposed to be the isomer shift okay isomer shift is before you consider the quadrupolar split where the line should be cutting the y-axis and quadrupolar splitting is after you consider the quadrupolar effect and how where the lines that is how the signals can be distributed any questions or query up to this point please go ahead if not we will go ahead and over here we are going to talk about some of the other examples and over there we will discuss the answers of the quiz tree so now take the example of an iron system bound with six ions and this is an iron plus two state so as you know cyanide carbon is actually from group 14 so it is going to be a pi acceptor sigma donor system so it is going to be a strong fill ligand so no spin complex so what will be the splitting is going to be and easy so now if you consider it is a iron two system what is going to happen we are going to see all the six electron is going to be present in the t2g level right so in this particular condition do you expect any gentler distortion the answer is no no gentler distortion because this configuration if you if you do the splitting you are not going to gain any energy so no gentler distortion so all the six ligands around this system going to be as same as it is there so there will be no lattice contribution because all the bonds will be exactly same point what about balance contribution second balance contribution is also going to be zero because it is a very symmetric t2g six easy zero system so what that t2g is already well fulfilled no asymmetry is no electrons there so also totally symmetric so all together t2g six is zero is totally symmetric system so there will be also no balance contribution so what is be the efg electric field gradient is going to be zero so that means you don't expect any quadrupolar splitting over here because your electric field around the metal ion is not creating any asymmetry and that is what actually observed when we measured the system the delta value the isomer shift for this iron system is actually minus point one three millimeter per second with delta aq value zero that means no quadrupolar splitting and you actually see a single bank like this for the fecn six system with iron at plus two of sinister so you're taking this fecn six four minus system so what I'll charge is four minus to show that there is the plus two charge of the iron and over there where it is actually cutting it is cutting at minus point one three millimeter per second compare this system with respect to the iron with four water molecules six water molecules earlier the delta value as one point two and with cyanides it's moved to minus one point three so which side it is moving from water to cyanide it is moving towards the negative side one point two two minus point one three why it is moving to negative side oxidation state remaining same only changes the spin state high spin to low spin low spin means it is moving some delay condensate out of it and that means it is actually creating more chance of the ace electron density especially the select on density to interact with the nuclear because of this shielding effect and that is increasing the high zero square value and it is multiplied with the delta r by r which is a negative term and that is why I am going to a more negative side so this is exactly the argument we have given over there so consider that weekend of the plexiglass cyanide pi donor is water keeping iron plastic so cyanide should be in the negative side water should be in the positive side and exactly that is what we observe water with one point two cyanide to minus one point three okay so that is what we are gaining now say instead of same system but I have a three minus two because now the iron is iron plus three state iron plus three step over there now if I draw it the electronic distribution so it is a d5 system it is still a low spin system because all cyanides around here one two three four five so that is how the electronic distribution is there t2g5 eg0 now you can have a gentler distortion the eg orbital there is no point of drawing the gentler distortion because there is no electrons are together so that is how it is going to look like and you are going to see some gentler distortion so what is going to happen two axial things are going to move out so you are going to see some lattice contribution all the ligand metal bonding is not going to be same so that is contribution will be non-zero or you're going to see that is contribution what about balance contribution now it is actually a t2g5 system the t2g5 eg0 system t2g5 means all the t2g orbitals are not symmetrically filled up so there is asymmetric present so balance contribution will be there so what do you expect over here you are expecting a quadrupolar splitting and in reality what we see it has a delta value of minus 0.03 millimeter per second and delta eq value close to 0.5 millimeter per second so you can draw this thing so this is going to be a very closely line two lines and that is what is going to show it over there so now the first question is where is the delta value so what are you going to do take the average of these two lines and draw in the middle and this is the value of the delta value minus 0.03 millimeter per second look into the value compare that with the previous one iron 2 it was minus 0.13 so iron 3 iron correctly yeah iron 2 is there iron 3 is in minus 0.03 so over here it is actually moved towards a little bit on the positive side compared to the iron 2 over here which is on the negative side so what is the expected thing if you have a low oxygen state that should be lie on the positive side and iron 3 which is a higher oxidation state it should lie on the more negative side but it is showing the opposite trend why it is so that is because the measurement also depends on the temperature so sometime the temperature also has a huge role to play and over here the temperature where the iron 3 is measured it's not at the same temperature as the iron 2 so that is why it is showing the data in the more positive side but if you can able to measure that at the similar temperature then the value starts shifting and at the low temperature it starts shifting close to 0.5 millimeter per second and then it shows that it is actually going to be a little bit on the further negative side which is expected for higher oxidation state so it is going to move to minus 0.5 millimeter per second so that is why you have to be very careful when you are discussing and comparing different oxidation state of a similar system like iron you have to ensure all the other experimental conditions remaining as same as possible otherwise you cannot compare them directly and over here what do you expect with respect to the splitting of the line the splitting of the line will be more sharper at lower oxidation state compared to the sorry lower temperature compared to the higher temperature okay why it is so it is again very similar to what you expect at lower temperature the electric quick gradient will be much sharper because the asymmetry will be holding it up and at higher temperature because the electron has more space to move and you are giving some thermal energy to move the electron so whatever the asymmetry you are creating around the nucleus they can get destroyed and that is why you are going to see some broad peaks at higher temperature so as you are going to lower temperature you are going to see sharper peaks so that what do you expect for iron plus 2 to iron plus 3 system in the cyanide you are expected to see the shift of the delta value keeping all the other experimental values same and the secondly is the splitting in iron 2 you are not expecting any splitting in iron 3 you expect the splitting to be there because both lattice and balance contribution will be there one more thing you're probably going to see over here in iron cyanide 3 plus system versus iron 2 H2O whole 6 system here the splitting is way too large compared to the splitting in the iron 3 why that is because if the splitting is happening with the asymmetry the asymmetry of the electrons generating from the EG level is actually affects more because the EG level shows the shifts in the DX square minus and DZ square orbital which actually creates more asymmetry around with the movement of this axial limit whereas a change in the T2G level DX Y DYZ DX they create some asymmetry but it is not as strong as it is happening in the EG level and that is also shown in the Mosba spectroscopy with respect to the quadrupole splitting quadrupole splitting is going to be much higher when the gentler distortion is happening to the EG level also if the easy level is not happening it is only happening to the T2G level your quadrupole splitting is going to be smaller okay next example now say I am taking an iron complex of the Geometry over here five of them are cyanide this is actually a nitrocyl group iron oxidation state is 2 so what is going to happen so again we are going to look into that we're going to find this is the T2G this is the EG there is a nitrocyl group is actually replaced a cyanide group now cyanide and nitrocyl both are pyrocycling even so it is going to be staying a low spin complex it is actually iron 2 D6 low spin complex so it is going to be like this now over here do you expect any gentler distortion answer is no it is a totally T2G 6 EG 0 system so do you expect any balance contribution the answer is no because it is all very symmetrically distributed T2G 6 EG 2 is easy what about lattice contribution now over here a lattice contribution will be present because all the equatorial bonds are same but one bond in the axial is cyanide one bond in axial is nitrocyl group so they are different and that difference will be shown up in the system and over here but we are going to see is the following the delta value is going to be minus 0.27 millimeter per second and delta EQ value is going to be 1.85 millimeter per second which shows this delta EQ value what you are seeing is coming from lattice contribution there is no balance contribution but lattice contribution is there and look into this value 1.85 compared to that water molecule bound system when both lattice and balance contribution was there it is 3.4 now one of them is cutting off the balance contribution is going out in the case of this complex and that is why the delta EQ value is going down you can see it is kind of additive more contribution you are getting from both lattice both balance and it is going to be contributing over there now this value of delta equal to minus 0.27 it is measured at the same condition where iron CN 6 4 minus is measured at delta equal to minus 0.13 millimeter per second so iron with all cyanides is minus 0.13 iron with five cyanides and one nitrocyl it is minus 0.27 so what you can comment on the pi acceptance property pi acceptor property comparison between nitrocyl and cyanide if we consider this delta values so you can see the delta value over here is more negative of the nitrocyl group containing system so that means why it is more negative go back to your logic that means it is going to have higher value of S electron density in the nucleus the size 0 square value is higher that means it has more S electron interacting with the nucleus that means it has less shielding that means it has less D electron so how it can happen it can happen if nitrocyl is pulling out more S D electron compared to cyanide so over there by comparing that keeping all the other systems same pi donor property wise nitrocyl is actually a little bit better than cyanide in this particular scenario and that is why the delta value is moving towards a more negative side now if you put it another particular group over here X where you are actually trying to look into how the pi acceptor property is compared to cyanides and you just have to follow that isomer shift if it is going to more positive side that means the pi acceptance property is poor if it is moving more negative side that means pi acceptor property is bad so that is how most words spectroscopy can give you such nice information about the molecular property what is happening at the molecular level of the molecule where the pi acceptance property even with one particular ligand can be distinguished okay any questions up to this point