 Hello everyone, once again I welcome you all to MSB lecture series on transformative chemistry. This lecture is 55th in the series after this another 4 will be left, so let me continue discussion on UV visible spectroscopy with emphasis for coordination compounds. In my last lecture I did mention about microstates and also term symbols, how to write microstates for a given electronic configuration and what are the things you should remember and also how to determine term symbols for ground state and other excited states. So let look into this table here, I have given spin multiplicity that is spin angular contour number that is 2s plus 1, this is small s 2s plus 1 and this is s here and when we have 0 electron 2s plus 1 will become 0, so this s is this small s is 0 and then we have 1 electron half, so 2s plus 1 becomes 2 and this is called doublet and when we have no unpaired electrons it is called singlet because 2s plus 1 gives rise to one value and when we have 2 unpaired electrons this will be 3, this is called triplet, when we have 3 unpaired electrons this will be 3 by 2 or 1 and half, we get 2s plus 1 multiplicity 4 is quadrate in the same when you have 4 we get 2 for multiplicity will be 5 this called quintet, so you should remember these things it is easy to do it s is sigma s and then you have to go for 2s plus 1 multiplicity. So now for example if electronic configuration is given to write all possible L values for the electrons in this electronic configuration we have to see here we have 4f orbital has one electron and 5d orbital has one electron, so we should be able to tell L resulted angular contour number that is sigma L and spin resultant spin contour number that is 2s plus 1 we should be able to write. So how to do that one it is very simple of course only one electron is there and L value first we have to find out L value here if you see I have already given here you can see we have an L equals 3 and L equals in this case 2 we have in case of D how? So plus 3 plus 2 plus 1 0 minus 1 minus 2 minus 3 this is for f orbital, so here one electron will be here so your L value will be 3 similarly so plus 2 plus 1 0 minus 1 minus 2 this is for d orbital, so here one electron so L equals 2 so that means we have L equals if we take sigma L that is 3 plus 2 together that is equal to 5 and then what are the possible L values now we have to see interaction of those things and such that we should get whole number that means if you take like this one I showed you previously you take 2 unit length and 3 unit length this is for L equals 2 and this is L equals 3 and if you arrange all possible orientation such that the resultant could be whole number in that case what we get is L equals we will get 5, 4, 3, 2, 1 so these are the possible values for this electronic configuration so this is the answer for this one this is how one can answer the question. So now let us look into once again total angular momentum quantum number J so that I already mentioned if a given electronic configuration is more than half field then we have to consider L plus S if it is less than half field we have to consider L minus S so here total orbital angular momentum of an autumn that is L equals sigma L and the total spin angular momentum of a autumn S equals 2S plus 1 combined to form total angular momentum a number that is quantized by the number J L and S do not necessarily have to be pointing in the same direction therefore J can range from L plus S to L minus S that means L plus or minus S so this diagram tells you this J is shown in purple and orbital L is shown in blue this one and spin is shown in green so that how they interact to generate this one can be seen from this diagram. Now I have just listed all possible excited state electronic configuration for P2 system P2 system we calculated microstates as 15 now 15 possible ways we can arrange including the very basic one ground one that normally we write for all molecules in this fashion and then in excited states they need not have to follow up principle and all those things you can keep on writing like that when you write it I have also given here sigma MS spin and also sigma L values I have given here accordingly the term symbols are also given here for each one term symbol one can write so that is what I have done here for each electronic configuration of in the excited state I have given all so that means we should have 15 different term symbols so all the terms derived with LS coupling occur for an excited state of P2 for carbon here S2 P2 this we have considered P2 means carbon so however in the ground state of the atom 1 S2 2 S2 2 P2 that is the carbon the number of states is limited by Polish exclusion principle this restriction reduces the number of terms from 3D, 3P, 3S etc to 1D, 3P and 3S so we will end up with only these 3 states now so that means when we look into electronic transitions with P2 system or carbon we have to focus our attention to only these 3 energy levels. Now before we determine the ground state term we have to follow certain rules what are those the terms are placed in order depending on their multiplicities the multiplicities means 2S plus 1S value and when we are writing term symbols what we should remember is most stable state has highest S value and stability decreases S value decreases that means if there is a possibility of writing several term symbols we have to consider one having maximum multiplicity that means the maximum 2S plus 1 value so the ground state possesses the most unpaired electrons gives a minimum repulsion the why we have to choose is when you keep more unpaired electrons your 2S plus 1 value will be very higher in that case what happens repulsion will be minimum in the ground in that particular electronic configuration what so it would be more stable and obviously there is ground state. So that is the reason since more number of unpaired electrons will minimize repulsion as a result that has to be considered as ground state that means highest 2S plus 1 value for a given value of S multiplicity state with highest L is the most stable next you have to consider for example multiplicity is same and then we have to look for the highest L value so that means if there is a ambiguity for a given value of S and L again for example we have 2 or 3 states spin multiplicity is same and L value is also same in that case we have to see the smallest J value is the most stable if the shell is less than half filled then it is L minus S and the biggest J value is stable if the sub shell is more than half filled that is the L plus S these wounds rule have to be followed while determining the ground state term for an electronic configuration. For example let us consider now as I mentioned in my previous slide that P2 now has only 3 energy levels terms between which electronic transition can occur so now among them which is the ground state for example if you see 3P automatically the choice should be this one we should not worry about L value the moment you see highest possible one so 3P has a triplet state maximum it is automatically it becomes ground state term and then among these remaining these 2 D is the more because spin multiplicity is same because L value is more here so this is the second one so L equals 2 as the same and now no option that this one will be least in energy and in this case now 3P is the most stable means the lowest energy and now it can have 3 terms 3P 0, 3P 1, 3P 2 that I showed you if you refer to my previous lecture and here again when we have these 3 the smallest value of J is the most stable so among them if it is split into 3 levels the lowest energy level is this one it goes in the energy increases in this order because this is less than half filled so 0 will be the lowest one lowest value is the most stable one because it is less than half filled P2 so this is how one can write so all these things are consolidated in this figure here you can see this is the ground term and then again this is split and again this is the lowest energy 1, 3P 0 then 3P 1, 3P 2 and corresponding energy is also shown here in terms of kilojoules per mole and also centimeter inverse and also conversion is 83.5 centimeter inverse equals 1 kilojoules per mole and 1D2 and then 1SO so this is how one should be able to write or write the splitting of terms in carbon ground state for example now I have given try for oxygen 1S2, 2S2, 2P4 same exercise you follow to write all possible splitting terms in case of oxygen having S2P4 electronic configuration. So now let us discuss about whole formulation what is this whole formulation is all about when a sub shell is more than half filled it is convenient to work out the terms by considering the holes that is the vacancies in the various orbitals instead of considering number of vacancies rather than the large number of electrons so that means basically instead of counting number of more number of electrons if it is more than half filled it is very easy to say the number of vacancies that vacancies are called as holes for example consider oxygen it has 4 electrons that means it has 2 holes considering 2 holes instead of considering 4 electrons is similar to considering 2 electrons in case of carbon that has 2 electrons that means now we are considering 2 holes you can correlate this one with P2 electronic configuration where we have 2 electrons if this is only for simplification of all electronic configurations to explain and interpret spectral data for example now in case of carbon we have these ground states these terms are there among them this is the ground state you know and now in this case also we have same way and now in this case what happens smallest value of j is more stable P2 is less than half filled whereas in case of P4 the highest value is more stable so as a result what happens 3P2 becomes the least energetic and this is the ground term and then 3P1 so just order is changed here so that means why this kind of whole formulation simplifies so that you can compare P2 electronic configuration with P4 of course here it is same thing happens in case of D electronic configuration also although we have 10 different electronic configuration using this kind of simple formula whole formulation and number of electrons we can simplify and thus understanding and interpretation would be that rather easier. So now I have given for different electronic configurations of Pn P6-n Dn D10-n Fn F14-n which give identical terms you can see here P1 and P5 1 electron 1 hole so 2P ground term and now 2 electrons and 2 hole carbon and oxygen so 3P is the ground state other terms are 1s 1d so allowed ones now P3 P3 same again so no question of anything else so 4s will be the one okay 3 electrons are there and then 2P and 2d P6 again 1 and again if you take D1 D9 system whole formulation comes 1 electron is there and 1 hole is there so 2d you can check you know ground term is going to be the same if you examine D1 electronic configuration and D9 it would be same similarly D2 D8 would have 2 electrons and 2 holes 3f is the ground term and then other terms are these things D3 D7 4f other terms D4 D6 similar D5 is unique and D10 is unique so of course here no DD transition at all only ground term is there everything is 0 so 0 means L equals 0 and 1 multiplicity is only 1 so 1s you can write without even any calculation now electronic spectra of transmittal complexes so spectra arise because electrons may be promoted from one energy level to another one obviously so we see electronic spectrum when one electron is promoted from one state to another one such electronic transition require high energy during electronic transitions low energy vibrational and rotational transition will always occur because unless electrons are starting from V0 or 0 and then E1 or something that is not going to happen always molecule will be under vibrational rotational activity as a result what happens always during this electronic transitions this low energy vibrational resistance will always occur so in electronic spectra such transitions are too close in energy to be resolved into separate absorption bands as a result what happens electronic spectra look broader for example 1 even to E2 transition has several transition within vibrational from E1 to E2 and within vibrational rotational again so as a result what happens it is very difficult to resolve in electronic spectra as a result what happens that results in broadening of absorption bands as a result what happens most of the dd transitions dd spectra if you look into it it looks very broad for the same reason and also the bandwidths are in the order of 1000 to 3000 centimeter minus 1 spectrum of a colored solution may be measured quite easily using a spectrophotometer and monochromatic beam of light is passed through a solution onto your photoelectric cell amount of light observed at any particular frequency can read off or a whole frequency range can be scanned instead of using monochromatic beam. So now this absorbance is plotted the absorbance A is called optical density which is given by A equals log I0 over I I am sure you are all familiar with this term and this equation where I0 is intensity of the original beam of light and I equals intensity after passing through solution epsilon the term we are introducing now the molar absorption coefficient or molar absorptivity coefficient calculated from A in this equation A over Cl where C is concentration of the solution in moles and also 1 equals path length in centimeter usually it is 1 centimeter. So this is what exactly we call it as Beer Lambert's law what it says the amount of energy absorbed or transmitted by a solution is proportional to molar absorptivity of solution and the concentration of solute. So now let us look into selection rules that we have to follow before we start discussion on electronic spectra when we wrote all possible electronic transitions then your spectrum would really look complicated. So hundreds of transition absorptions will be there and then it is going to look like a mess and you may not be able to assess and interpret corresponding to each transition. So now we have a selection rule what happens it try to simplify the spectrum how it does that means not all the theoretically possible electronic transitions are actually observed that is good selection rules are there to distinguish between allowed and forbidden transitions that means some of them are allowed and some of them are forbidden why they are forbidden let us see allowed transitions occur forbidden do occur but much less common and they have always low intensity. So now let us talk about the first rule that is called Laporte selection rule or orbital selection rule what it says is transition which involve a change in the subsidiary quantum number delta L equals plus or minus 1 or Laporte allowed and have high absorbance that means in order to see a transition between two energy levels the delta L value should change that should be plus or minus 1 if not it is Laporte forbidden. For example if we consider calcium we do flame test for alkali metals and alkaline earth metals and they have very distinct colors they give for flame test and why that happens for example let us look into calcium calcium has s2 electronic configuration s2 p0 electronic configuration is there when you supply energy or heat it then what happens the electrons are excited from s2 one of the electron is promoted to p1 so we have s1 p1 electronic configuration. So now i changes by plus 1 so now that means here basically what happens this is Laporte allowed because you know it is going from s2 p so Laporte allowed L is changing as a result what happens it is very intense that information also come from epsilon value you can see 5000 to 10000 liters per more per centimeter. So that means it is a measure of whether it is Laporte allowed or not strictly speaking dd transitions are forbidden delta L equals 0 that is the reason if you look into dd transitions epsilon value will be anywhere between 5 to 10 liters per more per centimeter. So this is about dd that means all dd transitions are forbidden then the question is why do we see dd transition at all I will come to that one after talking about spin selection role. So during the electronic transition electron does not change its spin the opposite to that one here it should be delta L equals plus or minus 1 here delta L equals 0 that means a electron going with upward spin should go like upward spin only. So during transition this one cannot change like this so if it goes with upward it should go to the exact state in the same way and it should not change its spin that means delta S equals 0 whereas this one is and delta L equals plus or minus 1 these two are very very important. So now the question in hexa aquam manganese 2 plus dd transitions are spin forbidden why try to think about that one spin forbidden why yes let me write electronic configuration for this one we have something like this this is high spin. So now if this electron is going here so if this electron goes here then we will be having something like this this electron let us say it is gone so then where it should go it should go here or it should go here but it should go like this upward spin only it cannot change so that means it is not allowed that is the reason dd transition is forbidden if you have another electron so this only allowed to go here or here. So now with d5 electronic configuration all have the spin like this so in case of hexa aqua manganese spin forbidden that is the reason we see very very weak transitions in this one intensity is very weak in case of u is the spectrum of hexa aqua manganese 2 plus. So dd transitions are forbidden that means leopard forbidden because delta L equals 0 in dd transition and low absorbance values are also there I mentioned about that so that means because of a slight relaxation in leopard rule dd transitions are observed so how we are relaxing leopard rule for example when a transitive metal forms a complex metal is surrounded by ligands and mixing of d and p orbits may occur as a result they are no longer true dd in nature of course even if you take as simple as valence bond theory in an octahedral complex we are using d2 sp3 hybrid orbitals that means before a metal to ligand bond is established to form an octahedral complex metal utilizes its 6 orbitals d2 sp3 to mix them together to generate a new set of orbitals they have different orientation they do not really look like the original orbits from which they are constituted but they have the properties of all in the same ratio that means now in a metal complex compare the nature of d orbitals that is very different from the nature of d orbitals they are present in free gaseous metal ion as a result what happens in gaseous metal ion dd transitions are strictly forbidden whereas in case of complex as a result of mixing d orbitals lose their identity they are no longer dd orbitals as a result what happens dd transition occurs so this is how leopard relaxation comes into the picture and we see dd transitions mixing is common when complexes lack centrosymmetry mixing is even more facile when complexes lack centrosymmetry that is the reason if you look into manganese having d5 electronic configuration tetrahedral complexes they are little bit more colored and intense compared to hexa aqua or 6 coordinated manganese complexes having homolyptic nature and centrosymmetry and intensity of absorption increases if the mixing is more effective so that is also a measure so this can also tell us so now as that is what I mentioned here tetra bromo magnet if you see it is tetrahedral and if you look into pentamine chloro cobalt 2 plus is octahedral with unsymmetrical distribution this is colored but if you take hexamine or hexa aqua they are less intense in color for example here you have centrosymmetry and no mixing of p and d orbits are really prano sir and they are not colored however metal ligand bond vibrates so that ligand spends an appreciable amount of time out of the centrosymmetric equilibrium position as a result small amount of mixing occurs and low intensity transistors are observed again if the question is there why it is because always metal to ligand bonds are not static they will be vibrating with respect to the mean position and often during this kind of vibrations they come out of centrosymmetric equilibrium position and hence the mixing occurs and hence they show low intensity transitions despite their leopard forbidden. So now let us look into MN 2 plus or off white or pale flesh colored and the intensity is only 100th of that of spin allowed transition spin forbidden transistors are always weak and can be ignored thus analysis of spectra of transcendental complexes are greatly simplified by ignoring spin forbidden transition for all practical purposes spin forbidden transition you can just ignore. Thus for d 2 only terms that are needed to be considered are ground state 3f and the excise state 3p. So now let us focus our attention to only 2 states 3f ground state and then 3p state why 3f because l value is more in case of f that is the reason. So now splitting of electronic energy levels and spectroscopy states we can see here s are symmetrically spherical and unaffected in an octahedral field and p orbits are directional and are affected by an octahedral field all are equally affected and no splitting they become t1u. So now t2g and eg is there and eg is doubly degenerate and this is triply degenerate and this value is tq we know that. So now I have given here so the allowed transition and leopard forbidden all those things. So if you take charge transfer transitions they are leopard allowed and spin allowed that is the reason we see very high epsilon value example hexachloro titanate 2 minus and partly leopard allowed but spin allowed is all dd transition epsilon value is very low example I have taken here tetrahedral complexes of cobalt having bromides or chlorides and leopard forbidden and spin allowed is dd transition again we can come across in this case again epsilon value is low hexachloro titanium and hexachloro vanadium where we have centrosymmetry so they are forbidden but spin allowed we can see very weak and partly leopard allowed but spin forbidden in this case because of d5 electronic configuration and spin forbidden leopard forbidden again dd transition very very weak this is in case of hexachloro vanadium d5 electronic configuration we have and all high spin complex and it is both leopard forbidden spin forbidden and this we see only because of it is jumping out of centrosymmetric equilibrium position. So let me stop here continue in my next lecture more discussion on electronic spectroscopy of coordination compounds.