 I once again welcome you all to MSP lecture series on interpretive spectroscopy. In my previous lecture, I elaborated on different type of couplings, LS coupling or Russell Sanders coupling and how they are related to each other and what would happen and what is term symbol. I started talking about term symbol. So let me continue from where I had stopped. So term symbol is very important in electronic spectroscopy. A term symbol or a spectroscopic term essentially represents the energy level of microstates. What is microstate? Again I would elaborate and I would calculate microstates for different electronic configuration. So with the same energy of a given electronic configuration. That means ground state also you can have a term and also you have several possible exaggerated states and each transition should be identified that is the reason we can have different term. This is how the term symbols have been introduced. Term symbol is nothing but represented something like this L, J and J can take any value between plus or minus half. As I mentioned earlier for a sub shell which is less than half field L minus SC is the most stable one that is considered as a ground state and if for a sub shell with more than half field electronic configuration L plus SC is considered. Now L how to calculate L and what is 2S plus 1 and then what is J? We shall calculate for few electronic configurations that I would do later after little bit explaining about those things. L can have anywhere 0 to N and when we have L equals 0 and the term symbol we are giving is S all our capital and when N it is P and when it is 2 it is D, when it is 3 it is F, when the L value is 4 it is G and when L value is 5 it is H and 6 it is I. So of course you know that J is total angular momentum quantum number it can be L plus S or L minus S, 2S plus 1 is spin multiplicity. When you have no unpaid electron S is 0 2S plus 1 it is a singlet, when you have one electron it is half. So 2S plus 1 will be 2 it is called doublet, when we have unpaid electrons 2 S will be 1 plus 1 sigma S. So that is 1 and 2S plus 1 value will be 3 we call it as a triplet and then when we have 3 electrons the sigma S is 3 into half. So it is 1 and half and then 2S plus 1 will be obviously 4 and it is called quadrate and then when we have unpaid electron 4 S equals 2 and 2S plus 1 equals 5 is called quintet. So this is how we can identify and give the name for a state depending upon 2S plus 1 value as singlet, doublet, triplet, quartet, quintet and so on. Now to make you familiar let us find out the term symbol for different electronic configuration. Let us consider a simple P2 system, P2 system first what we have to do is that means 2 electrons are there in P orbital and write electronic configuration first you put 1 0 minus 1 this is L value, azimuthal quantum number and then we have 1 electron here, 1 electron is here. So now L can have anywhere between L can have 0, 1, 2, 3, 4, 5. So here L is 1 so 1 is there L so that means here we know that S, P, D, F, G, H, I it goes 6. So that means now we have 1 S equals 1 this is 0 so 1 means P we have to consider P here and then we have to consider 2S plus 1 value 2S plus 1 is 1 it will be 3, 3 and then here what we should consider is less than L minus S. So L equals 1, S equals 1 so L minus S equals 0 so this is 3P0. So this will be ground term so easily you can calculate here. Then let us look into D2 system, D2 system of course you have to write like this and then we have plus 2, plus 1, 0, minus 1, minus 2 or azimuthal quantum numbers and then we place electron here so then sigma L ok this is 2 plus 1, 2 plus 1 equals 3 so L equals 3 here L equals 3 means F will come here ok F. Now S is S equals sigma S equals half plus half equals 1 then 2S plus 1 equals 2 into 1 plus 1 equals 3 so that is 2S plus 1 is 3 and then it is less than half it L minus S 3 minus 1 equals 2 so this is the term symbol for D2 electron ground state. So this is how you should be able to do it. If you have still doubt let me go ahead and do 1 for F orbital so now let us take F3 you have to write 1, 2, plus 3, plus 2, plus 1, 0, minus 1, minus 2, minus 3, 3 electrons are there 1, 2 and 3. So value will be 6, 6 means basically here L equals 0, 1, 2, 3, 4, 5, 6 so you should remember which one it is so here S, P, D, F, G, H, I so I should consider I here ok and then 2S plus 1 is 1 and half into 3 by 2 plus 1 is 4, 4 will be 2S this is 2S plus 1 value 3 by 2, 4 here and then L, L equals 6 and S equals 4 so L minus S equals, S equals 3 by 2 here, S equals 3 by 2 so that means 3 by 2, 6 minus 3 by 2 will give you L minus S value so that means 12 minus 3 by 2 equals 9 by 2 this is here L minus S which is equal 9 by 2. So for F3 the ground term symbol is 4I 9 by 2 so this how you should be able to calculate I hope we have learned if not let me go for one more D 7 electronic complex more than half filled now let us consider 1, 2, plus 2, plus 1, 0, minus 1, minus 2, 1 electron here 2, 3, 4, 5, 6, 7 so now the value 4 plus 2, 6, 6 are there and minus 3 equals 3, 3 means you have to consider F so here term is F and then 2S plus 1, 2S plus 1 is this is 2 into 3 by 2 plus 1, 4 this is 4, this is 2S plus 1 equals 4 and then L equals 3. So now J is since it is more than half filled it is L plus S this is L plus S is 3 plus 3 by 2 3 plus 3 by 2 this will be 9 by 2 here 4F 9 by 2 will be the ground term you got it I hope you have understood. So here I will tell you again here 7 electrons are there and then here we have 4 plus 2, 6 are there out of 6 minus L value if you subtract we get L equals 3, 3 means it is F and then 2S plus 1 is we have 3 electron, 3 electrons are there 3 by 2 or 1.53 by 2 3 by 2 is the S value sigma S and then 2S plus 1 will be 4 so that 4 is 2S plus 1 we are putting here and then since it is more than half filled the J will be having L plus 1 S value will be as the lowest energy or least energetic one. So here 3 plus 3 by 2 will be 9 by 2 so this is the ground term so this how you should be able to calculate the term symbols. So now fine we can calculate but how to determine the ground state term that is very very important how I wrote I will show you so the terms are placed in order depending on their multiplicities multiplicity we are referring to 2S plus 1 value here. The most stable state has highest S value and stability decreases as S value increases the ground state possess the most unpaid electrons gives minimum repulsion why we are considering 2 plus 1 highest value because when you have 2S plus 1 highest value we have maximum number of unpaid electrons are there so that means you can anticipate it has to be ground state and it gives most stable low energy because of minimum repulsion as they are all singly occupied. For a given value of S the state with highest L is the most stable one for example 2 terms have the same spin multiplicity 2S plus 1 value in that case we have to go to the L value the highest L value 1 is the most stable one if there is ambiguity for a given value of S and L that due to some reason 2S plus 1 is also same then L is also same then we have to consider the J value the smallest J value is the most stable if the sub shell is less than half field in that case J be equal to L minus S and the largest J value is the most stable one if the sub shell is more than half field in that case J equals L plus S so we have to consider these terms here P2 system will be 1D how it is 1D see P2 system 1D 3P and 1S are there and 3P has a triplet state 3P has a triplet state maximum S is 2S plus 1 ground state term among 1D and 1S so we have this one this is a ground state then we have ambiguity between 1D and 1S because both of them have the same spin multiplicity but D has a larger L value so the D is the most stable so then D will be the first excited state you can consider and now the 3P has three terms 3P has three terms 3P 0 3P 1 3P 2 since the value J value for a less than half field will be L minus S as a result 3P 0 will be most stable one least energetic one this is what I showed you 3P 0 while calculating I directly calculated the ground state term 3P 0 here because L minus S will be 1 minus 1 0 will come and then the next one is 3P 1 and 3P 2 so this is how you should be able to determine the ground state for any electronic configuration and as I mentioned here what is important is some similarities are there between the different electronic configuration having the same ground term that means Pn and P6 minus n 6 is the maximum capacity and Dn and D10 minus n and Fn and F14 minus n give identical terms for example P1 and P5 one electron was less than completely feed 2P term is there and P2 P4 again two electrons are two less than completely feed electronic configuration 3P so there is some correlation is there that we should be able to identify P3 is unique we have 4S 4S is there because here 3 electrons are there unpaid electrons 2S plus 1 will be 4 and P6 we have 1S and D1 and D9 again one electron one less than completely feed we have 2D ground term and then D2 and D8 two electrons and two less than completely feed we have 3F D3 and D7 two electron less than half field two electrons more than half field 4F term we have and D4 one less than half field one more than half field D4 and D6 they have 5D and D5 is unique we have 6S and then D10 we have 1S so this how you can see the correlation when you see the same ground state term you can always bring some similarities as I mentioned here D1 D9 D2 DI D3 D7 or D4 D6 or P1 P5 and P2 P4 and we can understand and then it is easy also to remember there is no need to calculate again so now the term symbols for non-equivalent electrons for example it is considered 2P1 and 3P1 in that case let us consider L1 equals 1 here L2 equals 1 here and same thing we have L plus L1 plus L2 to L1 minus L2 value will be there and then we get L equals 2 1 and 0 values and similarly S1 equals half and S2 equals half then if you consider in the same way we can have S plus S1 plus S2 until S1 minus S2 so we have 1 and 0 values here that means when we have L equals 2 1 0 or 0 1 2 we have DPS and then S equals 2 S plus 1 then 0 1 and we have 1 we have 3 values are there so that means the corresponding spectroscopic term can also be written in this 3D we have 15 and 3P we have 9 and 3S we have 3 and 1D we have 5 and 1P we have 3 and 1S we have 1 so total of 36 is there that means 3D we have 15 what we call these are all called microstates number of microstates for a given term one can calculate that means now we have to calculate the microstates for different electronic configuration to know how many states are there and all states one can expect transitions from ground state to all these states but due to the selection rule we eliminate and make it very simple in fact I should tell you we come across only a two type of electronic transition as per the DD is concerned apart from D0 D5 and D10 they are unique D0 do not show anything D5 also very special why I would tell you later and D10 there is no DD transition because completely filled that means leave these things then we have D1 D2 D3 D4 D7 D8 D9 again similarities are there we can classify into simply two categories and then one category would show only one DD transition the other one would show three transitions so that means what are those things I would tell you maybe if possible today's lecture or maybe in my next lecture so now before that I shall I show you how to calculate microstates for example here total number of microstates now for these terms are given as 3D 3P 3S 1D 1P 1S for this combination non-equivalent electrons then the total number of microstates is 6 into 6 36 we have calculated here 36 how we arrived at that one also let's look into it now again before we proceed to calculate microstates for different electronic configuration whether it's a P or DRF the way we determine the term symbols for a ground state let's look into it first L1 equals 1 and L2 equals 2 here. Now again here if you see L1 plus L2 to L1 minus L2 it goes and we have L equals 3 to 1 values are there when you have L equals 3 to 1 values are there we can look into the term symbols PD and F and then here in two electrons are there and one electron is there when the two electrons are there S equals 1 it will be 3 so we have to consider 2S plus 1 value of 1 and 3 for this system we have one electron in 3P 1 and 3D 1 also we have one electron similarly S equals S1 equals half and S2 equals half and then S will be 1 and 0 so the total number of microstates possible here is 3 of 3D 3P and 1F 1D and 1P and this multiplicity is 2S plus 1 is 3 whereas here multiplicity is and because of 0 electrons so here 0 spin that means no unpaid electrons so that means total number of microstates will be 6 into 10 60 so that is shown here 3F we have 3 into 2 into 3 plus 1 21 and then here 3D we have 15 and 3P we have 9 and 1F we have 7 and 1D we have 5 and 1P we have 3 it's basically LS coupling we have shown here and then we get a total of 60 here so then how we got these things we can go back to microstates and look into it so what is microstate so here the number of arrangements of electrons in a given sub shell for a given electronic configuration so that means in the ground state and we can write only one electronic configuration such a way that maximum number of unpaid electrons are there and if it exceeds we will start paving but when you go to excited states they need not have to obey our principle and Hund's rule you can have anything in that case what happens several possible excited states one can think of and the sum of all possible excited states along with ground state are nothing but the microstates and to determine the microstate for a given electronic configuration we use this formula that is n factorial over r factorial into n minus r factorial n factorial is nothing but the total electron capacity of an orbital if it is a d2 it's a 10 okay if it is F it is 14 if it is G it is 18 if it is P it is 6 electrons so this how we go and L is azimuthal quantum number and r is number of electrons in the sub shell d2 if you take it r equals 2 and n equals 10 this is how we determine that one so some values are shown here so number of microstates for P2 6 factorial the P orbital capacity is 6 electrons then 2 is the r is 2 factorial and then this is n minus r 3 because 6 minus 2 so it comes around 15 how it comes 15 also I will show you and d2 similarly 10 factorial or 2 factorial into 10 minus 2 factorial it comes around 45 and here for d5 system 10 factorial and then here we have 5 factorial and then 10 in minus 2 factorial so we have 252 comes here I will calculate again showing you in more detail d1 we have 10 factorial and P1 we have 6 and then P3 we have 20 d1 we have 10 microstate and P1 we have 6 and P3 we have 20 if it is not clear here let me do it here let's consider simple P2 here P2 if you consider the total number of electrons r equals 2 and then n equals 6 so what we will do is the formula is doing n factorial 4 r factorial into n minus r factorial so here 6 factorial and then 2 factorial and then 6 minus 2 is 4 factorial so this one I can write 4 factorial 5 into 6 and 4 factorial into 2 goes we have 15 so the microstates for P2 is for P2 is 15 let's look into d2 here so d2 here r factorial equals 2 again and then n equals 10 the orbit total capacity is 10 so here if you use the same 10 factorial over 2 factorial 10 minus 2 factorial so this one can be written as 8 factorial into 9 into 10 and here 2 factorial to 8 factorial so it goes 45 so this is how you can calculate just look into it I would come up with few more electronic configurations both P and D and also with the FR but to make you familiar with calculating microstates and also determine the ground term again in my next lecture I will show you few ground term state determination and also microstates for several other electronic configuration until then so go through this course and then try to make familiar yourself to get some confidence about interpretation and utilization of the structures thank you