 Now, once you know how the electronic configurations are and then you have to ask whether it is going to have any orbital angular momentum or not. Those questions obviously, when you read you will have you can verify it by this table. You do not have to verify right now you can sit down I hope the table is not wrong none of the data is wrong. So, you should be able to justify the data by itself this is like a self-check ok whether you have understood orbital angular momentum and T 2 G E G electronic configuration or not ok. Same is true in here for tetrahedral complex once again it is the same thing E and T 2 right E and T 2. So, E configuration usually will not give you any unsymmetrical feeling in E is not going to give you orbital angular momentum, but T 2 can right. So, based on that because E is d x 2 y 2 and d z 2 right. So, for example, d 1 d 1 for octahedral is T 2 G 1 right that means, orbital angular momentum is possible d 1 for tetrahedral is E 1 right it is splitting between E and T 2 is that clear any of you did not understand that I will be happy to discuss again. No this is the spin only value 1 unpaired electron 1.73 yes usually say usually what also happens is simply the splitting in the tetrahedral see you cannot directly compare octahedral versus tetrahedral you have to also understand the trend in the sense the splitting between E and T 2 is less in tetrahedral this is octahedral yes this is the observed value and this is the calculated value that is the maximum you get ok. Yes expected is you should get little bit higher than that, but it does not because some quenching is there it is more see we are not going to trick you that observe is less than the calculated that can happen sometime we will see more there could be some other contribution quenching by spin exchange which I will be discussing ok. What I am mainly trying to say is spin only values should be good enough, but when you see the value is more than the spin only value then orbital angular moment perhaps is there how to calculate it simply by figuring out whether you are having orbital angular moment or not. If you experimentally see the value is little less or exactly similar or very close that means, that some other way of quenching would be there which will be one at least one instant we will be discussing today ok. It is you know this is like too simplified in terms of teaching ok. In reality you can have lot of other practical components which maybe we are not discussing, but once again for this class we are saying that only thing you want to worry about is the spin only value. After that if you find that the value is little bit more sometime 0.1 more 0.2 more those are originating because of orbital angular momentum ok. Because you know orbital angular momentum will always add up it is it is it depends on the directionality right. So, if you are imagining let us say more of a spin up and then it is it is another vector is there to direction whatever this vector is having and I mean vector addition is usually you are going to take the modular value the absolute value. That is a good question again. We usually usually never actually worried for the orbital orbital angular momentum versus spin usually we deal separately and then take the pure value of it do the vector addition ok. Now, if you want to do the see it is it is orbital angular momentum actually does not depends on the spin right. It is the motion around the orbital right. So, spin only value is dependent on the spin. So, therefore, up and down is coming orbital angular momentum there is no spin component. Now, that is that is the motion simple the motion is we are discussing right simple motion it is not the up and down motion we are up and down motion is the spin I think that perhaps is the reason. So, let us say some spin are up and some spin are down ok. So, orbital contribution for all those will be taken as just the orbital contribution right up spin and down spin what is the orbital contribution total we are we are seeing, but we are not mixing up that with the spin value spin only value is separate and orbital value is separate ok. Let us calculate where between the orbital and the spin for that spin only value n into n plus 2 mu l plus s ok the main equation the first equation that is assume that it is 90 degree. I think that is that is that is independent of the angle why why we need an angle for that? Well the spin I do not I do not think spin and the orbit are of course, spin orbit coupling can be there those both the coupling can be there, but if you are individually taking how one is affecting the other if direction does not matter actually sorry yeah that is mainly so, that that that is when we have yeah we usually say that when it is less than half field that is l minus s we take when more than half field then we take l plus s value I think it is independent of the direction right ok. I will try to find out if it is dependent if the equation is dependent on the direction ok. Let me proceed further ok we have d z 2 and d x 2 y 2 which is not inter convertible and we are started saying that for d 2 these are the value and then we are saying that these are the orbital contribution whether possible or not yes or no answer. Now, same way it is true for tetrahedral complexes ok you can find out whether it is there or not if there is any discrepancy then then just let me know I think this should be this this is from the book directly. Other reasons for orbital contribution so, usually we are pretty much focused on the ground state electronic configuration. Sometime what happens is let us say T 2 G 6 E G 1, T 2 G 6 E G 1 means there should not be any orbital contribution, but still if you can excite T 2 G 6 E G 1 to T 2 G 5 E G 2 one of the electron from T 2 G level to E G level if you can promote then you can promote or then you can induce orbital angular momentum ok. That is where we are talking that excited state can sometime also contribute ok. These excited state contribution will not be as high as the ground state contribution ok. So, for example, over here T 2 G 6 E G 2 no orbital no orbital angular momentum should be there, but in excited state it should be T 2 G 5 E G 3 now orbital angular momentum will be there. Of course, the absolute value will be less compared to if some other compound is giving from its ground state ok. So, the orbital angular momentum not necessarily coming just from the ground state it can come from also excited state. So, therefore, it this leaves a option or this gives a option to set a question saying that look the spin only value should be this much or you calculate the spin only value which is let us say 1.73 or you know 2. or 4.89 whatever it is, but the observed 5.6 you do the electron distribution and see there is no way you can have orbital contribution orbital angular momentum, but still the value is high. Then of course, you have to think perhaps ground state is not contributing there is a chance may be in the excited state ok. So, under the normal condition if you cannot explain what is going on then you are open to include the excited state. Same is over here E 4 T 2 3 should not have any orbital contribution orbital angular momentum. Now, in excited state E 3 T 2 4 you can have it ok. In all these cases you see that mu observed is greater than mu spin only for both octahedral nickel 2 plus and tetrahedral cobalt 2 plus because these are the electronic configuration for octahedral nickel 2 plus that is there tetrahedral cobalt 2 plus that is there. So, at least these two terms should be very familiar octahedral nickel 2 plus tetrahedral cobalt 2 plus these are the one can give you excited state. Other than that if any question is given in the exam where you cannot once again cannot explain by normal understanding of orbital angular momentum like unsymmetrical filling in T 2 G it is symmetrically filled still the value is high you can try to explain that by contribution from excited state is it clear ok. You read it it should be. Now, this is how it is. So, T 2 G 6 is E 2. Now, one of the electron if you can convert if you put let us say light if you put energy or the distance between these two is not very high in excited state you can excited state you can achieve therefore, from T 2 G 6 E G 2 you can have T 2 G 5 E G 3. Now, in this configuration you can have the orbital angular momentum it is clear both ground state and excited state you need to consider. Now, this is another configuration this is for T 2 G 5 E G 3 I guess this is the one of course, in here you can have three different orientation that is what we are trying to say. So, not from ground state if you go to excited state excited state can have three different configuration right which is the origin for orbital angular momentum ok. No need to have this point just this is good enough. Ground state there is no orbital angular momentum excited state it should be there how it is coming I think you have already learned because the relative positioning of the electron can vary ok. Now, so what we are trying to tell you here is this is also a D 8 this is also a D 8 configuration nickel 2 plus is D 8 nickel 2 plus D 8 octahedral case this is the scenario you have nickel 2 plus D 8 tetrahedral case what is the electronic configuration nickel 2 plus octahedral and tetrahedral nickel 2 plus D 8 octahedral will be this is the electronic configuration ok octahedral with not so strong field ligand strong field ligand should have given you the square planar geometry this is nickel tetrachlora or nickel hexacoa or whatever weak field ligand should give you the octahedral species ok. Whenever you see D 8 be little bit causes D 8 can have octahedral and tetrahedral sorry octahedral and square planar of course, if it is given clearly that is no worry. D 8 with D 8 electronic configuration with strong field ligand like cyanide or carbon monoxide can give you square planar geometry sometime for these cases you do not have to worry, but in other cases you have to worry little now nickel 2 plus octahedral electronic configuration is given nickel 2 plus tetrahedral electronic configuration is D 4 T 2 4 E 4 T 2 4 right. So, E 4 T 2 4 right now the question simply is which one will have higher orbital or higher magnetic moment value both of them are having 2 unpaired electron here you have 2 unpaired electron here you have 2 unpaired electrons right octahedral tetrahedral both of them are having 2 unpaired electron experimentally which one you think will give you higher magnetic moment value. Tetrahedral why is that simply because this contribution for octahedral this orbital angular momentum contribution coming from the excited state in the ground state itself there is no orbital angular momentum in the excited state there is, but in tetrahedral case you can have orbital angular momentum due to the unsymmetrical filling of the T 2 G orbital right that is the reason and actually it can go up to for 2 unpaired electron it can go up to 4 Bohr magneton ok, B m is the unit for magnetic moment ok it is ok it is given. Now, so far we have discussed about the D block elements no that also depends on the ligand strong field if the ligand in strong field usually then only you will see ok. So, if it is a weak field ligand then you do not see too much of that ok alright. Now, weak field means fluoride chloride water these are the weak field ligand you have that you know electro electro electro electro electro electro chemical series right it is not electro electro electro chemical series. Now, so far we tried to discuss the magnetism of D block elements we have simply learned spin only is good enough some few special cases we have say that orbital angular momentum is essential and thereby values can be increased. If once again I think that is a valid point that if you see little bit decrease do not worry about it because there may be other reason we are not worried about less value or lesser value of an experimentally observed magnetic moment we are mainly worried about the higher value lesser value can come from some a lot of other factor. Now, we are now trying to discuss the magnetism or magnetic behavior of lanthanides or actinides you will not get into actinides just simply lanthanides. So, simply speaking lanthanides are of little bit different class of compounds ok. It has F orbital you know that F orbital is buried inside it is not the real outside electrons are the S even you know wherever if you see lanthanide cerium, prasidinium, yeah whatever em or whatever that the fortin are there cpr, em, sm, samarium, gadolinium whatever it is there all of those cases F electrons are buried inside. What essentially that tells you is ligand will have very little effect ligand cannot influence the magnetic moment value because the F electrons are buried inside in it ligand cannot affect the F electron too much or almost nothing. So, D orbitals or D block elements we were seeing that orbital angular momentum value are almost gone right that mu L component is very little only when unsymmetrical feeling is there then we are seeing, but over here you do not have to worry about anything you have to do both L and S component. Orbital and the spin component because orbital component cannot be restricted by the ligand. So, as if you are dealing with a free metal species, although metal complexes are there lanthanide complexes are there lanthanide ion is in the middle ligands are surrounded, but ligands cannot affect the magnetic moment value. But you have seen that for octahedral complex or tetrahedral complex or D block elements ligand can influence there is a splitting in the D orbital. F orbital we do not see such splitting or you can you can take it almost that there is no splitting in the F orbital because ligand and F orbital cannot interact too much that is fine. Therefore, you have to bring back your previous equation which has magnetic moment values by considering both orbital and the angular component ok. Now, the facts that is the fact the F electrons in lanthanides are buried in the n minus 2 cell. So, you are in for calculations what is the calculations? Let me show you this is the calculation this is something I guess you have to know if not remember. I think a lot of you know this already start learning. Now, the so, number of unpaired electrons number of unpaired electrons that is the can that component can give you the S summation of number of spin of the unpaired electron that is S you know L this L you know from this ML value right let me. So, term symbol you have heard of. So, I will discuss little bit 2 S plus 1 L j j equals L plus S or L minus S absolute value of these. Now, S equals number of summation of spin right or whatever way they that is written summation of spin let us say number of unpaired electron. Let us say you have 3 unpaired electron 3 unpaired electron that means, spin will be S will be half plus half plus half right. Now, if you have 3 unpaired electrons. So, this is your F orbitals 1 2 3 4 5 6 7. So, ML values are plus 3 plus 2 plus 1 0 minus 1 minus 2 minus 3 right 3 unpaired electrons simply means that right ML maximum 3 plus 2 plus 1. So, L equals 6 3 2 1 ok. Now, therefore, for that case 3 unpaired electron case you have 6 this L value at 6 6 I will come back to that you know S P term symbol S P D F G H I and so on. So, S L if L value is 0 then that is S if you have not studied before you just go for this Russell Sander's term. S P is 1 D is 2 F is 3 4 5 and so on 1 2 3 4 5. Now, this capital S you have learnt 3 by 2 3 by 2 right 3 by 2 it is going to be 4. So, 4 6 is going to be I I will come back and explain in a moment and J. J is L minus S for the less than half field configuration. So, J in this case will be 6 minus 3 by 2 that is going to be how much 9 by 2. So, this is going to be your term symbol ok. Although you do not need to really know this one like this J term for the class, but I think some of you at least 30 percent of you know or 40 percent of you know I do not know ok. So, what we are trying to tell you is this is the Russell Sander's term symbol you know kind of derivation. What it tells you? The total spin let us say you have F 3 electron 3 F electrons F 3 configuration ok C P R E M. So, what is that 3 3 plus will be C P R E M E M is what is the Einstein E M no E M E M is Einstein E M maybe. Anyways, prosidium is 2 plus is going to be F 2 I think E M is going to be 3 plus 3 sorry 3 unpaired electron. Anyway 3 unpaired electron if you have that 3 unpaired electron will give you spin value S capital S is going to be 3 by 2. You have to calculate the L value. L value L equals summation of M L 3 to 1. So, that is going to be 6 S you know L you know J J equals L plus S or L minus S. L plus S when it is more than half field more than F 7 ok and L minus S when it is less than half field. In this case 3 unpaired electron less than 7 F 7. So, L minus S L minus X that is becoming 9 by 2 ok. Now you left up with just this one 2 S plus 1 L J 2 S plus 1 S equals 3 by 2 you plug that in. So, that is becoming 4 ok. L L is your 6 that means, that 6 means if L equals 0 that is S, if L equals 1 that is P, if L equals 2 capital L equals 2 then D F G H I and so on. So, from there you get I and then J in this case we have calculated L minus S. Some of you who have who knows this that is fine no problem for you those who those of you who did not hear it before just google it you will be able to get it clear. I will show you one over here with preceding 3 plus sorry L plus S yes because less than half field. Less than half. F 7 yeah ok. Now let us look at here ok whoever knows can you please calculate by yourself preceding 3 plus I am saying that electronic configuration is 4 F 2 F 2 ok. F 2 can you calculate whoever knows it calculate whoever does not hear this before stay with me I will explain ok. Now F 2 that means, 2 unpaired electrons right 2 unpaired electrons means spin equals half plus half total S equals 1 are you following me fine if you get it right the answer is given here definitely. Now L L equals so, 2 unpaired electron it should be plus 3 and plus 2 L should be the maximum also right it is not like you will put at 0 and minus 1 or anything. So, that is L equals summation of M L M L is what plus 3 plus 2 plus 1 0 minus 1 minus 2 minus 3 7 of them 7 7 are of F for D it was plus 2 plus 1 0 minus 1 minus 2 if it is degenerate ok and so on for P it will be plus 1 0 minus 1 if it is degenerate anyway. So, this is the L value summation of M L 2 unpaired electron 3 and 2 5. Now J will be in this case since it is un half field less than half field less than 7 electrons F 14 F can have 14 electron 2 is less than half less than 7. So, L minus S L minus S 5 minus 1 4 ok. Now if you have that here 3 plus 3 plus 3 plus is 3 H 4 I think you have calculated right 3 H 4 how now you plug it in this equation 2 S plus 1 L J 2 S plus 1 means 2 plus 1 3 L L is going to be 5 that is going to be H. So, 3 H is it getting clear let me just some of you may not be seeing it ok. It is going to be 2 S plus 1 L J. So, 2 minutes break you can take you can go for whatever 2 plus 2 S plus 1 S equals 1 plus 1 L equals L equals 5 and J equals 4. So, these 5 is going to be 3 and 0 S p d F S p S p d F G H good and 3 H 4 4 is 2 J plus 1 sorry L minus S. Is it correct? Let me let me start I think this math part is way too simple for you guys it is say these 2 equations are there you calculate G J first G J with S and L value and J value and then plug this into this equation that is all. For your verification that is the table you should be able to calculate 2 3 of them by yourself to get confidence nothing else. The values are given equations are given and pair electrons are given and pair electrons are given and pair electrons are given and pair electrons are given and pair electrons are given. So, calculate 2 3 of them by yourself to get confidence nothing else. The values are given equations are given and pair electrons are given L S J value you should be calculating by yourself then mu value calculated value do not worry about the observed value too much just calculate the value you should be able to get exactly same value unless anything is given by mistake any wrong data are given I do not think any wrong data is given ok. So, once again we are talking about the lanthanides F orbitals are not perturbed by ligand therefore, both L and S component orbital and spin component of the magnetic moment you have to consider. Simply you have to know these 2 equations what are those 2 equations these are the 2 equations what is S what is L what is J that you know. Once you have that you have G J calculated you put that G J value and the J value you have already calculated from there you put all these mu value you should be able to get. Those mu values are calculated and shown in here for given electronic configurations for different electronic configuration it is given. So, you sorry that is that is mistake just delete that mu b ok. It is the calculation is shown for prosidium P R 3 plus you can see from for just clarity one of the example is shown here prosidium 3 plus which is F 2 electronic configuration you can calculate this one first yourself and then go on to calculate any of these ok. This is you do not really have to worry about this bottom term you can just calculate the just 1 S 2 F 3 H and so on for the class purpose, but as you know this is very simple 2 S plus 1 L J this J you do not really have to worry J equals L plus S or L minus S, but still I would say you should do it for the class purpose you may not need to worry. Less than half field electronic configuration L minus S what is the half field half field 7 L plus S will be for the more than half field and from there you can get this term symbol Russell Sanders term symbol it is called ok. Now, let me move on. So, this is the experimentally calculated value sorry theoretically calculated value this is the experimentally observed value it is quite close that is good enough ok. Here you do not have to worry about any orbital angular momentum because you are already taking care of it you are calculating based on that ok.