 We have seen the octahedral complexes, we have seen the tetrahedral complexes. These are the two most important classes of compounds that are present over the let us say whatever chemistry known so far. There are other complexes as you have seen it could be you know pentacoordinated, the metal complex can be pentacoordinated. Two different types of geometry is possible for pentacoordination. Out of that trigonal bipyramidal is one of the preferred geometry. So, there are lot of things that we can definitely go on and discuss about it, but it is it is literally impossible to discuss each and every structure in detail, each and every metal complex kind of different geometry in detail and how they are you know ligand field will affect the d orbital splitting. So, geometry changes d orbital splitting pattern changes. I think just by learning octahedral and tetrahedral you have got that sense. Now it has all to do with the way the ligand is approaching. Sometimes ligand is approaching directly towards the atomic orbital of the of each of those metal or d orbital of the metals and thereby you can see the extent to which it gets stabilized or destabilized is differing. So, if you take technically speaking if you take a trigonal bipyramidal perhaps you would be able to at least understand why the splitting is such that ok. In the syllabus we for you guys we do not have any other geometry to discuss octahedral and tetrahedral. If possible you can look at little bit at trigonal bipyramidal TBP ok. Now square pyramidal or square planar is something which will come shown square planar is basically nothing, but octahedral geometry you are having you take out 2 z axis. You have octahedral scenario you take out 2 z axis that is become square planar. So, we will discuss that. So, in the last class once again we were discussing Warner coordination theory 18 electron rule balance bond theory and how good it is and then how bad it is and how good the crystal field theory is. Really we will stop there how good the crystal field theory is and not too much we will get into the MO approach molecular orbital approach ok. There are I mean no end to learning I think for this course purpose we will stop in there not too much afterwards ok. So, the major objective for this syllabus or for this chapter and the next chapter is definitely giving you an idea about these high spin low spin complexes. The spectrochemical series I hope you have come across this term in the last class. Crystal field stabilization energy, yarn tailored distortion and spinel these are the last 2 topics we will discuss today. I will briefly give you an overview what we have discussed in the last class. So, this is an octahedral complex metal center is there at the middle of it. So, that is at the center of the geometry. It is a metal center metal and ligand are almost having electrostatic interaction. It is a positive charge and negative charge are the ligand which are interacting with the metal center. Remember in the valence bond theory we were mainly assuming it is a covalent structure ok. As you see this is axial position that is also axial position. This is the direction where d z 2 orbital is d z square or d z 2. Now these are the direction these 4 1 2 3 4. These 4 are the ones where we have d x 2 d y 2 orbital directly it is not indirect. This is where is actually d x 2 y 2. So, therefore, from this you can understand why d z 2 and d x 2 y 2 orbital is most destabilized for octahedral geometry. Because they are the one which facing the music basically they are they are the one which is getting rippled most because ligand electrons and the d orbital electrons are rippling each other very simple. Now, all other orbitals d x y d y z d x z they are not facing these ligands directly. They are in between somewhere here here here here. So, in between. So, thereby they are not going to get rippled too much they are going to get rippled, but not too much relative to d x 2 y 2 and d z 2 they are going to stabilize ok. So, just a quick look at the orbitals what we have discussed in the last class d z 2 orbital d x 2 y 2 orbital ok. Now these are the other orbitals with respect to these three other orbitals you can see where the ligands are. Therefore, you can understand the stabilization or the destabilization of these. Yeah that is a good question. Usually that is where we will come to Janteler distortion ok. That I guess that you will be able to clearly understand when we are discussing the Janteler distortion. That is the actually the origin of Janteler distortion and J elongation and J din ok we will come back. Now, this is the same thing in a different bottle I think whatever fits you you just look at it. It gives you a very clear idea these are the ligands the black balls are the ligands and the and the metal orbitals or d orbitals are shown in here clearly ok no confusion right. Now, as we were showing in the last class so, these are d orbitals and this is let us say ligand electrons. Ligand electrons are coming for these orbitals to overlap right. Now first instance we will discuss the d z 2 interaction with the ligand. Ligand electrons comes they will be repelling each other therefore, d z 2 will be destabilized d z 2 is facing the ligand once again directly. This is the d x 2 y 2 1 ok. Now, here you see that the 4 ligands are coming along all these axes. To answer to the previous question one of the way you can see is these are 4 lobes, 4 lobes divided by 4 ligands and 2 lobes divided by 2 ligands kind of that is how perhaps it is you know you do not have to worry usually, but finer details when it is unsymmetrically filled right d z 2 and d x 2 although they are of same energy, but if they are unsymmetrically filled let us say one is there in the easy orbital or 3 is there in the easy orbital then the problem comes ok. Now so, you have seen the you have seen how they are coming and thereby they are also getting these orbitals are getting destabilized ok. All other d orbital like these 3 are like d x y, d x z and d y z they are really not facing the orbitals directly as you can see they are sitting right in between and thereby you can see. So, for example, over here we have shown d x z same is true for d x y and d y z thereby these orbitals are relatively mind you relatively stabilized with respect to d z 2 and d x 2 y 2. If you compare the free metal ion with respect to free metal ion everything is destabilized because free metal ion has no ligand thereby no repulsion. The moment ligand comes repulsion starts so, the systems energy goes up ok. Now overall therefore, you have 2 of 3 down ok, but net stabilization with respect to that bary center the center in the middle the net stabilization or destabilization for a completely filled d orbital let us say d 10 or d 5 in a high spin situation which we have discussed should be the 0. Stabilization if it is completely filled total stabilization and total destabilization has to be the same ok. Now this is where again once again this is kind of a clear picture this is a octahedral field we were showing 6 ligands coming at these box. So, if you are assuming this is a box 6 ligands are facing the way it is shown 1 2 3 4 5 6. Now this is the free metal ion this is the metal complex 3 of the t 2 g got stabilized 2 of the e g got destabilized ok. Now this is e g not e e will be for tetrahedral t 2 g not t 2 now of course, I think I have discussed this information already. Now since this is 3 of them this is 2 of them this distance between this bary center and the stabilization for t 2 g yeah ligand may be different yeah. Here we are thinking see we have to ideal deal with the idealistic one at the beginning m l 6 ligands are same m l 6 we are not thinking like 3 of the chloride 3 of the fluoride that is a mixed situation ok for just to discuss of course, those situation comes often comes, but this is if we do not understand I mean very simple situation how can we go complex ok. So, of course, let us say if you have 3 chloride and 3 cyanide what will be the situation right that is a special you know topic on this how finer details you can get in that d orbital splitting. Initially it may split and then they may further split if you have all of them let us say weak field like 3 fluoride and let us say 3 chloride they may be having behaving same, but the moment you have 1 weak field 1 strong field ligand things will be little bit complicated ok. So, overall how much ligand I think overall you have to see whether it is a strong field ligand or weak field ligand whether it is a first point whether it is a octahedral geometry or tetrahedral geometry and thereby go for it for the exam purpose for this course I do not think we will be tricking into anything which is mixed little bit more complicated than that. For exam purpose or for this course purpose you have to just know octahedral field splitting and tetrahedral and we will be discussing briefly about the square planers because that is kind of comes automatically. Furthermore if you want to learn maximum go to trigonal bipyramidal ok we will give those splitting, but you know as again it is a more complex than we would like to think we are just dealing idealistically ok that is a good question. So, over here we see that this stabilization should be 0.4 2 by 5 delta 0 0.4 delta 0 this destabilization would be 0.6 delta 0 we have discussed it or 4 d q 6 d q if it is delta 0 or 10 d q we have discussed right. Now, the first advantage of this is a crystal field theory is you can explain the magnetic properties you can expect to explain the you know spectroscopic behavior let us say UVB study why certain peaks are coming. So, this is where we were telling that if you have an unpaired electron if you are moving from T 2 G to E G the spectra you get will be something like this absorption maximum of course, we are not getting into finer details of the spectra, but this is the major origin where from electrons are moving and where it is going. So, from T 2 G this is a d 1 configuration 1 electron is there. So, T 2 G 1 that means, over here T 2 G 1 1 electron was over here you all these 3 orbitals are degenerate same energy only way it can go is up to here. Now, if you think if E G is splitted further these 2 orbitals are d x 2 y 2 and d z 2 if it is splitted further. So, the electron can go to 2 different levels right. Therefore, you can expect 2 different peaks whether intensity will be high or low that is of course, then you have to think about the symmetry and lot of other things which we will not be discussing, but you can sense where the spectra is coming from and where the electron is you know moving from I mean which orbital to where it is going ok. So, this is fine. So, you get the spectra. Now, we have also seen crystal field stabilization energy. So, you calculate you are given I think you have to really master this you should be able to do it in your dream ok. D 3 electronic configuration what is the crystal field stabilization energy if octahedral. D 5 what is if it is octahedral how many scenarios are there 2 scenarios how high spin and low spin right. So, what will be the crystal field stabilization energy. Any question in specially an exam if you see before even thinking too much I would say just imagine the scenario only 3 scenarios you have octahedral 2 scenario high spin low spin and tetrahedral only high spin scenario 99 percent case. We will discuss one case today which is otherwise, but that is it octahedral 2 scenario and tetrahedral 1 scenario done. So, once you have that I think the answer should come out almost like why something is preferred why something is not preferred. At least I would say out of 25 mark question 5 marks to 7 marks will be based on that indirectly I mean of course, it will not be given perhaps on a platter definitely you will be able to figure out about that. So, please do familiar you should not be fearing about calculating and electronic configuration should be always correct there should not be any mistake ok. Now, so D 5 system for example, one case we were giving. So, this is the D 5 high spin that means, 3 stabilization 2 destabilization CFSE should be 0 right that is what I was trying to say. So, net stabilization and destabilization should be 0 if it is symmetrically filled ok. So, this is 0 you do not have to even calculate if it is D 10 it is again 0 D 10 means all of them are full. So, stabilization is equals to destabilization. So, that is where let us say you are given D 7 and this is the configuration 5 6 7. So, 5 take out 5 because 5 3 plus 2 has stabilized destabilized and cancel each other out. Now, you just deal with 2 2 will be D 7 2 will be over here and here right. So, that will be minus 8 D Q or 0.8 delta 0 or delta octahedral. You should be able to do it really really quick without calculating and going through the simple math. Now, of course, D 7 you can have also high spin and low spin configuration. For example, this one what would be the CFSE minus. So, of course, minus that is out of question it is stabilization each of them are 0.4. So, 5 of them 20 minus 20 or minus 2 sorry yeah minus 2 delta 0 that is it delta 0 means delta octahedral delta O I mean different people pronounce it different way. So, this is it I mean if you do it really simply I am sure 5 to 7 month question will be there I mean invariably whatever it is the question is based on that it is going to be based on that. Now, of course, we have discussed it once again two scenarios high spin because the spin is maximum high spin low spin means spin is minimum you see one unpaired spin. Now, if you look at the I will come back to the magnetic behavior magnetic behavior is nothing, but due to the unpaired electron right. So, the moment you have unpaired electrons magnetic behavior comes because the parallel spin will cancel each other you want to spin on that direction and I want to spin on this direction. So, it will be cancel up canceling. So, this is the one will have low magnetic value magnetic moment value this is the one will have high magnetic value. This is the origin for molecular magnet a lot of molecules are magnet and these are the ones you see the application almost everywhere literally everywhere you take any electronic gadgets you take almost anything which is electronic in nature which has some fancy application it is the material which are having magnetic properties and that is why they are used how expensive it is based on that you know you will you will get your material. Of course, in lot of other cases you see the use of these things I will show one or two case today ok. So, dependence of delta 0 like how much splitting is going on that depends on the nature of the ligands as I was trying to tell you whether it is a weak field ligand or strong field ligand. Strong field means the splitting will be very high weak field means the splitting between T 2 G and E G will be small and therefore, for weak field ones will always see high spin means spin will not be pairing up ok. And the charge on the metal of course, if it is a high higher charge you will have higher separation if it is 5 D you will have higher separation compared to 3 D right. So, this is the spectro electrochemical series with respect to different ligands these are the stronger one stronger ligands these will these are the one which almost always will give you the low spin sorry high which one low spin complex because the splitting will be high strong field ligand splitting will be high. So, the spin pairing will happen it cannot go from T 2 G to E G ok. These are the ones which are likely to give you low spin complexes is it getting clear if it is you are getting I mean sometime it is little bit confusing it is either yes or no type of answer question I mean the understanding is either yes or no high spin or low spin right. So, just get it clear now this is the trend we have shown for tetrahedral case we were dealing with a completely different scenario which can be clearer from this picture. If you remember the previous picture for octahedral case which was nothing, but direct confrontation here it is like more of a political approach ok you know diplomatic approach you do not go direct you just talk with ok. So, they are talking they are not directly confronting thereby the scenarios are completely different since it is not interacting directly overlapping directly with let us say those E G orbitals previously we have seen for octahedral cases D x 2 y 2 and D z 2 D z 2 is over here actually and see the ligand where it is ligand is here D z 2 is here actually those orbitals are the ones those E G orbital for octahedral case are the ones which are farthest from the ligand which are farthest and thereby they are the one which will get stabilized the other 3 orbitals D x y D y z D x z those are the ones which are nearer not directly overlapping, but closer and thereby they are destabilized ok. Now, this difference is called delta T and this delta T is going to be 4 ninth of the delta O or delta 0 2 third coming from the number of ligands 6 to 4 octahedral to tetrahedral another 2 third coming from indirect approach not from direct approach. If it was directly approaching that is the case of octahedral case, but here indirect approach basically you can calculate based on the angle which angle it is coming ok. So, roughly these are like rough calculation it is becoming 2 third times 2 third 4 9 of delta 0. So, delta T is always less therefore, you never see almost never ever see low spin case for tetrahedral since the splitting is very less all ways you end up getting high spin case. So, never ever calculate tetrahedral for high spin sorry low spin. Now, this is the electronic configuration and their respective stability if you are comparing delta octahedral and delta tetrahedral directly. So, what happens how much stability is there if it is D 1 for octahedral case D 1 for tetrahedral case and we are comparing apple versus apple that means, high spin versus high spin high spin of octahedral and high spin of tetrahedral tetrahedral cannot have low spin. In tetrahedral case we have to normalize the value with the 4 9 of delta 0 these 2 you do it you will get this should be you should be able to get it ok.