 I once again welcome you all to MSP lecture series on interpretative spectroscopy. So this is 25th lecture in the series and in my previous lecture I was telling about microstates and how to determine term symbols and how to write microstates for a given electronic configuration. So let me continue from where I had stopped to make you once again familiar with determining term symbols and also ground state term symbols and also microstates for different electronic configurations. So you can see here I have given for 2p1 and 3p2. So for example if you want to determine together for two electronic configurations say one electron is 2p1 state and one electron in 3p1 state then the combined number of microstates how to determine. It's much simpler you have to determine for both and then take the product of that one that would give you the combined number of microstates for two electronic configurations. So here you can see for example 2p1, 2p1 means you remember the formula n factorial over r factorial into n minus r factorial where n is the total electron capacity of the orbit we are considering or a sub-shell we are considering and r is number of electrons. So here if you take 6 factorial over 6 minus 1 factorial means you can write it 6 factorial over 5 factorial into 6. So and then it goes then you get 6 in both the cases you get n product is 6 into 636. So after this one I would be writing again to make you familiar with that one and similarly you go for 3p1 and 3d1 and in case 3d1 we will be having as many as 10 microstates and then 10 into 6 it would show 60. Similarly if you take 2s1 and 2p1 in case of 2s1 we will be having only two possible ways that you can yourself you can look into it and then we will be having 2 into 6 that will be 12. In similar way you can calculate for 3d1 and 4p1 that is also going to be giving 60 and then 3d1, 4d1 then you know that d1 system irrespective of which sub-shell it is will be having 10 microstates. So 10 into 10 would be 100. So this is how you can write. So now again let me go back to simple ones like d2 system. d2 system first what you should do is write 5 orbitals d orbitals you write azimuthal quantum number for each sub-shell then place two electrons here and then the sigma L, L equals sigma L and this is the L value or ML. So this will be 3, 3 means you know that L would take anywhere S, P, D, F, G, H, I it goes like that with values of L equals 0, 1, 2, 3, 4, 5, 6 and so on. Now we have 3 choose F. So F is the symbol now then we have to take the multiplicity S equals 1 therefore 2S plus 1 equals 3. So this 2S plus 1 will come here. Recall the term symbol we are using this J, J is plus or minus S and if the sub-shell is less than half field it is L minus S and if the sub-shell is more than half field it is L plus S. For example d2 is L minus S you have to consider and if it is d7 it should be L plus S. Now 2S plus 1 is 3 and now L we know the value L equals 3 and S also equals 3 we have so L minus S is 0. So what we get is 3 F naught this is the term symbol for a ground state and similarly we can go for F3 we can go for it F3 we put here 3 electrons all are unpaired in the ground state. Now we calculate the L value L equals now 6, 6 means I will be coming here and then we have 3 electrons are there 3 electrons S equals 3 by 2. So 2S plus 1 equals 4. So here multiplicity will be 2S plus 1 will be 4 and now L equals 6 and S equals 3 by 2 is there. So L plus X will be 6 minus 3 by 2 this is 9 by 2. So this will be the ground state term symbol. So this is how one should write and if you are curious let us look into P2 system also P2. P2 here S equals 1 therefore 2S plus 1 equals 3 and then L equals 1 therefore P this will be P P3 and then here less than half field L equals 1 and S equals 1. So L minus S equals 0 so this will be the ground state term symbol for P2. So in this fashion you should be able to calculate what should you should remember is 2S plus 1 L j, j can take any way L plus S or L minus S if it is L plus S is more than half field electronic configuration and L minus S is less than half field electronic configuration. Now let us look into for the same P2 system number of microstates. Number of microstates formula is n factorial over r into n minus r factorial. So here n is the total capacity of the sub shell we are considering here is P. So it should be 6 factorial and then r is number of electrons I can write here 6 factorial over 2 factorial and 6 minus 2 factorial. So this one I should write 4 factorial now it is 15. Number of microstates possible for P2 electronic configuration is 15 let us look into D2 as well. So D2 the total capacity of D orbit is 10 factorial. So it is 10 factorial over 2 factorial 10 minus 2 factorial this can be written something like this 2 factorial is 2 this is 8 factorial. So we have 45 states microstates. So this is how we should be able to determine without any problem let us look into one more F2 system here. F2 system it is a 14 factorial over 2 factorial 14 minus 2 factorial. So this can be written as 12 factorial this is 91. So for F2 electronic configuration the number of microstates is 91. So this is how one should be able to write. I hope you should be able to write for any given electronic configuration or any metal ion if any metal ion is given you have to determine the ground electronic configuration look into the actual state of the metal and arrive at the D system once you arrive at the D system calculate L and identify the term symbol SPDFGHI whatever and then find out multiplicity 2S plus 1 and then consider J value as L minus S if the electronic configuration shows the sub shell is less than half field and if it is more than half field consider J value of L plus S so that is very simple. So let us continue now. So as I mentioned just now we found out that there will be 15 microstates for P2 system then we can examine them you can see here I have written all possible states for the given electronic configuration of P2. So initially you can have something like this two electrons this is the one we are considering as ground term and next what you have can you can shift this one here this is another possibility and then you can have this are the possibilities then we can reverse the spin in this fashion and we get another three here and then we can have one upward spin and one downward spin we can have this three and another one is for example if it is downward spin this is this you can be downward spin you can go like this we have another three and then we can pair both the electrons and then we can arrange them in either in minus 1 or L equals 0 or L equals plus 1 value. So this is how we can write as possible microstates it is very lengthy and laborious if you want to write for 120 210 or 45 microstates are there but nevertheless if you are curious you can always start writing and check whether the determined value is correct or not experimentally. So why these arrangements have different energies that is the question the mean distance between the electrons may vary from one arrangement to another average inter electron repulsions will vary. So that means the arrangement and the mean distance between the electrons may vary because of inter electronic repulsions and hence these levels will be split into a number of levels of different inter electronic repulsions and hence they will be possessing different energies. So although p orbitals degenerate and have the same energy the electrons present in them interact with each other and result in the formation of a ground state and one or more excited states the first one we determine the ground state and then because of interaction or coupling what you can have is you can have several excited states. So besides electrostatic repulsion following factors influence each other what are those factors by interaction or coupling of magnetic fields produced by their spins by coupling of the magnetic fields produced by the orbital motion of the electrons that is called orbital angular momentum and when several electrons occupy a sub shell the energy states obtained depend on the result of the orbital angular quantum number of the electron. So these points should be remembered the resultant of all the L values is denoted by a new quantum number L that is called resultant angular quantum number this is L resulted angular quantum number it can take values from 0 to n and for 0 the term is s for 1p for 2d and 3f 4g 5h 6ik comes j is excluded because j we are taking for spin orbit coupling that is the reason the term is omitted while considering the term for resultant angular quantum number depending upon the values. For example L equals plus 1 0 minus 1 is there and you can have L equals 1 0 2 values this is how you can consider. So now let us come back again to p2 configuration what you should remember is angular momenta is quantized into packets of magnitude of h over 2 pi this is governed by quantum mechanical rules for p electron L equals 1 so orbital angular moment will be 1 into h over 2 pi that is shown by an arrow of unit length. So this will be shown by an arrow of unit length for 2p electrons the way in which the L value may interact with each other can be shown diagrammetrically in this fashion. For example all these three interactions are possible since the angular momentum is quantized. So that means whether we can arrange in different ways to have different resultant values now the permissible arrangements are those where the resultant is a whole number of quanta this we should remember. So the resultant the interaction is such that the resultant will be a whole number of quanta for p state the vector L must be at an angle to each other such that the resultant is a whole number. So this we should remember and that means for example let us take L equals 1 and L equals 1 we have 2. So that means the resultant will be L equals 2 when L equals 2 is there we have d state and then they can also interact in such a way that the resultant will be L equals 1 here in this one we get p state of course you can recall L equals 0 1 2 3 4 s p d f g it goes like that. So L equals 1 it is the p state when they interact in this fashion L can have value of 0 in that case what we get is s state. So this is how we can arrive at the different microstates by looking into the resultant all these three interactions are possible since the angular moment is quantized. So all these things are listening in whole number of quanta so that is the reason this is quite possible. So now what is spin orbit coupling when several electrons are present in a sub shell the overall effect of the individual orbital angular momentum L is given by the resultant angular quanta number capital L. Similarly the overall effect of the individual spins m s is given by the resultant spin quanta number s sigma m s. In an autumn the magnetic effects of well and s may interact or couple giving a new quanta number called j that is called total angular quanta number this is essentially the vectorial combination of L and s that is also called as resultant coupling or spin orbital coupling for example in case of p to case we know that L values will be 2 1 0 and corresponding terms will be d p and s and s can have values of 1 and 0. So now let us look into the coupling of spin orbit so that means coupling of spin angular momenta is m s equals plus half then they can couple with each other so that s equals 1 interaction and now they can oppose each other in this fashion so s equals 0. So this is how we get s equals 1 and s equals 0 value through coupling of spin angular momenta. So now let us look into p to case again to see in the spin orbit interactions so value equals we know the value 2 1 0 and corresponding terms will be d p s and then s equals we have 1 and 0 values are there. Now let us look into the interactions assuming they are vectors so we should remember this resultant should be having a whole quanta number quanta so L equals 2 and s equals 1 can interact in such a way that the net resultant will be j equals 3. In this case the 3 will come here 3 d 3 will be the state and then they can also interact in this fashion L 2 that is the reason we have taken the double the value of this one here and then L 2 and s can interact in such a way that resultant will be j equals 2 and then we have here 3 d 2 state and then they can also interact in this fashion now the j value will be 1 the resultant and then we will be having 3 d 1 state. So next we go for this is s equals 1 and L equals 1 now L equals 1 this is L equals 2 and s equals 1 we are considering here. Now let us consider L equals 1 and s equals 1 now the net resultant will be j equals 2 and this will be 3 p 2 state and now we can have interaction in this way so that j value equals 1 and here we get 3 p 1 state and now if you interact in this fashion the j equals 0 we get 3 p 0 state this is the ground state we call it as L minus s and we call here L plus s here. Similarly now we can consider L equals 2 and s equals 0 this is the last one and in L equals 2 and s equals 0 j equals 2 and L equals 1 and s equals 0 j equals 1 and L equals 0 and s equals j equals 0. So what we get is here 1 d 2 state and 1 p 1 state and then 1 s 0 state. So this is how you can also show diagrammetically by considering this as vectors and then you can arrive at the different ground state term symbols along with j values. j values you can clearly see how we are getting the values here. So each of these arrangements corresponds to an electronic arrangement spectroscopic state represented by a term symbol here this is a typical term symbol. So now let us come back to again electronic spectra of transcendental complexes. In transcendental complexes electrons may be promoted from one energy level to another one those two energy levels between which electronic transition occurs are essentially called as HOMO and LUMO highest occupied molecular orbital and lowest unoccupied molecular orbitals. Such electronic transitions you know require high energy and during electronic transitions low energy vibrational rotational transitions will also occur because each energy electronic state is composed of several vibration levels and which in turn composite of several rotational levels as a result what would happen we can see as possible vibration rotational excitation accompanying in electronic transition. So in electronic spectra such transitions are too close in energy to be resolved into separate absorption bands as a result this would cause broadening of absorption bands in DD spectra. So bandwidths for the same reason the bandwidths in electronic spectra are in the order of about 1000 to 3000 centimeter minus one. So in a free gaseous metal ion the orbits degenerate all the orbitals would be having the same energy as a result no DD transition are observed in a complex degeneracy is lost and splitting can be seen in case of octahedral complex T2G ground state and EG excited state and of course in case of tetrahedral complexes opposite of octahedral happens and we will be having E as ground state and T2 as the excited state and depending upon the ligand field what would happen we will be having different orbits coming into the picture as HOMO and LUMO. So between which we can see DD transition electronic transition that is called as DD transition. The magnitude of delta O depends on the nature of the ligands and affects the energy of electronic transitions and hence the frequency of absorption maxima. The spectrochemical series we know that in the spectrochemical series we know the position of each ligand by considering its donor and acceptor properties and then if the ligands are of weak field in nature then the HOMO-LUMO gap will be smaller, Christopher's debris energy will be smaller and delta would be smaller and requires very less energy for the promotion of the electrons or excitation of the electrons. So extent of splitting is related to ligand position in the spectrochemical series. So from that point of view we have to remember several factors and the one at most important factor is the spectrochemical series and the rank provided to a ligand in the spectrochemical series whether it is increasing order or decreasing order of crystal field, stabilization energy or crystal field strength. So let me continue in my next lecture more information about electronic spectrochemical complexes. Until then have an excellent time. Thank you.