 Now, this is the topic today we are going to discuss spinel. Spinel are the ones which is let us say this nice looking gem looking thing or gem basically magnesium aluminum oxide. It is beautiful in color of course, color comes from you know where the color is coming from those transition. Now, definition is spinel is very simple. You can have either same metal or different metal ok. It could be let us say F3O4, CO3O4 or it could be Mg Al2O4. 3 metal should be there, 2 of them should be the same as you see at least 2 of them has to be the same will be the same usually speaking again or one is different Mg Al2O4 something like that. Now, AB2O4 is the general formula for spinel. The normal spinel are the one there are 2 type of spinel as I said normal there would be abnormal 2 spinel or inverse spinel. Now, we do not say abnormal inverse spinel. Normal spinel are the one where you have this electronic configuration. It is always going to be O4, 4 oxides are there that means, minus 8 each oxide minus 2 4 of the oxide will be minus 8. Now, minus 8 means you have almost one possibility in realistic possibility 3 plus 3 plus 2 of course, other possibilities are there those are less likely never happens for spinel. So, 3 plus 3 plus 2 since it is 3 plus in the normal behavior or normal spinel cases you would expect the octahedral geometry or octahedral geometry will be preferred by 3 plus and tetrahedral geometry will be preferred by 2 plus. Let us look at the you know over here magnesium, aluminium this is octahedral you can see 1, 2, 3, 4, 5, 6 each aluminium is octahedral. You look at magnesium it is having 4 ligands it is in a tetrahedral I think most likely 1, 2 maybe 3 and 4. Magnesium is in tetrahedral geometry, aluminium is in octahedral geometry ok. Yes. So, over here this is further showing you that 1, 2, 3, 4. So, one of them so, it is a little bit complicated looking thing, but if you go by unit cell if you break down the complex structure and look at the core of it. Each of the metal ions A B 2 O 4 both the B will be in octahedral geometry ok. Sometime it is just not clear by looking one glance at it ok. You have to just look it little bit carefully you will be able to understand where it is ok. And these are usually the crystal structure. So, that should not be any ambiguity you just have to look it carefully. So, both the octahedral will be in plus both the plus 3 oxidized metal will be in octahedral state and the one which is 2 plus will be in tetrahedral geometry and 4 oxides are there. If this is the case this is normal spinel. If other way around if one of these octahedral M 3 plus has to be pushed out from these octahedral side to tetrahedral side that will be called inverse spinel. So, all it depends on whether one of these M 3 plus will be in octahedral side or tetrahedral side that is it. Of course, always there will be one metal 3 plus which is in octahedral side this is common. The second metal 3 plus whether it is exchanging with this other metal that is what you need to see is it clear? Normal spinel normal behavior 3 plus should be octahedral right it is a privileged class 3 plus higher oxidation step octahedral should be right. It is a high charge everything should be ligand should be interacting more and so on, but that is like normal behavior. If it is other way around octahedral side is coming out and tetrahedral is going into the octahedral one or 2 plus is going into the octahedral one this is when it is called inverse spinel, inverse spinel normal spinel. Now, when what is the criteria? Simply you have to see metal this metal in 3 plus whether it prefers octahedral or metal 2 plus what it prefers let me show you. So, M 3 plus ion has when M 3 plus ion has a higher crystal field stabilization energy ok compared to M 2 plus ion then this is a normal spinel ok. If M 2 plus ion has a higher CFSE then it is inverse spinel. So, you do not have to really look at the tetrahedral you do not have to look at the tetrahedral you take the metal ion M 3 plus M 2 plus for a given metal ion you find out M 3 plus and M 2 plus they are CFSE separate CFSE. If M 3 plus is more then normal if M 3 plus is less having less CFSE compared to M 2 plus then it is inverse just read these two lines. So, expectation is M 3 plus in octahedral and this is usually going to be high spin should have high CFSE of course, opposite is true M 2 plus octahedral high spin should have low CFSE CFSE calculation we have learned how to do. If this is done normal opposite inverse normal spinel inverse spinel do not mix up with calculation of octahedral versus tetrahedral sometime people end up doing that they go into wrong direction ok. So, you have to just calculate octahedral site stabilization energy like how much stabilization is there. So, that is what it is written M 3 plus ion has a higher CFSE CFSE in an octahedral field compared to M 2 plus ion in normal spinel and so on do not calculate for the M 3 plus in tetrahedral ok. So, this is where I mean yes then how will you conclude right. So, you have to figure out the stability of M 3 plus versus M 2 plus right this other otherwise it will get confused what will be will be comparing with right CFSE for tetrahedral is usually low 4 nines of delta O what to calculate ok. So, let me give you an example it will be clearer yeah more negatives see any in terms of energy in chemistry physics everywhere energy means the one which is having low CFSE or low anything low repulsion is stable low energy I mean it is stable ok. Anything low would be stabilized it is like as I was I think I was trying to say it is like home feeling you want to feel home everybody wants to belong somewhere and that is where they they want to go right. So, always low ok stability should be maximum yeah no that is the splitting delta. So, that is the splitting we are comparing when we are not changing anything ligand keeping constant metal oxidation state varying metal with a higher oxidation state will have higher splitting ok that is somewhat true here, but I mean we are not saying that extend to which we are trying to tell here is M 3 plus M 2 plus yes you are partially right M 3 plus M 2 plus you just calculate the CFSE ok. Let me give you this example hopefully things will be clear ok M n 3 O 4 ok. M n 3 O 4 of course, here oxygen usually always I mean not always again there will be one example which is other way usually oxides for these cases are going to be the weak field ligand ok. It is a weak field ligand now unless the metal is very high you know very high oxidation state or so, oxide will be the weak field ligand ok. You do not have to worry about the oxide too much except in one question which will come. Manganese 2 plus D 5 since it is weak field ligand it is a high spin that is what I was saying high spin. Now, high spin is 0 CFSE Manganese 3 4 3 plus D 4 is T 2 G 3 E G 1. So, you are going to get net minus 0.6 delta 0 that is what you have calculated right should be minus 12 minus 6. So, sorry plus 6 minus 12 plus 6 is minus 6 D Q ok. Now, this is the normal spinel how about Fe 3 O 4 can you calculate Fe 3 O 4 whoever is trying to calculate try give yourself one minute Fe 3 O 4 Fe 3 plus and Fe 2 plus ok. Try whoever got it hold on for a minute inverse spinel yes that is the correct answer just I will in a moment I am looking for something. So, I hope you have got this right Fe 2 plus D 6 T 2 G 4 E G 2 high spin 3 and 2 cancels out 3 over here and 2 over here cancels out. So, minus 4 it is easy right over here 3 and 2 0 CFSE right. Now, Fe 3 as you can see is having less CFSE compared to Fe 2. So, that is it. So, inverse spinel do not only request is do not complicate too much you think, but for calculation do this because if you end up calculating tetrahedral Fe 3 plus case tetrahedral Fe 2 plus case tetrahedral it is going nowhere ok. It is a simple thing higher oxidation state should try to stay in the octahedral side. The moment higher oxidation state becomes less favorable compared to the lower oxidation state switch of the power happens. It is just the BJP and Congress you have to pick up one of them in the center not the local or democratic or republican you do not have any choice ok. Now, so, this is the special case, special case of D 8 octahedral. So, what is the special case? We are eluding before that these D G 2 orbital are the one getting repelled by 2 electrons. This is a D 8 electronic configuration as you can see T 2 G 6 E G 2 and this is the one repelled by 4 ligands that is unfair that is what happens is Z orbital try to stabilize to minimize the energy to gain more energy. So, you start with an octahedral situation, and you see D 8 you go blank. D 8 octahedral will tend to form square planar geometry ok. Because the Z you will see square planar complex is formed because Z will get destabilized sorry stabilized. Z is getting stabilized as you can see Z orbital will get destabilized. This is an unfavorable situation, this is a situation which cannot be tolerated. This is 2 ligands repelling 4 ligands repelling. This is the unique situation 1 1 unique situation same electron repelled by 1 electron repelled by 2 1 electron repelled by 4. In order to bring some sort of calmness to the system Z orbital will stabilize. If you want to stabilize X 2 Y 2 orbital then the problem becomes stabilization means repulsion is less that is the stabilization repulsion less is the stabilization. If you want to stabilize D X 2 Y 2 orbital then you have to elongate 4 of the ligands. Stabilization means you know reducing the repulsion. If you are stabilizing Z square then you just need to take out those axial ligands. Take it was here 2 axial ligands over here you pull it out. So, Z will be stabilized D Z square will be stabilized. Of course, further it will be stabilized. Any Z component there will get stabilized. So, D X Z 2 will be stabilized D Y Z will be stabilized. This is the scenario where square planar complex goes. D 8 octahedral usually will be I mean if it is possible always it will go to the square planar situation. That is also true D 8 tetrahedral or D 8 4 ligands are not tetrahedral D 8 configuration 4 ligands are going to be square planar ok. This is where you see iron cobalt nickel, nickel is nickel 2 plus D 8 is square planar all right. Now, this is one scenario. Other scenario is that easy orbital is unsymmetrically filled. See you can see that majority of the syllabus is biased towards octahedral ok. Tetrahedral not too much, but that is because octahedral is favored octahedral is historically more important and that is what most of the study has been done and it is a common geometry, more common geometry I would say. Now, this is the yantailer distortion. Of course, you can some people pronounce at differently in British in England you will go pronounce differently US it will pronounce differently. Yantailer some people tell some people tell a jantailer whatever it is it is the same thing jetty distortion. So, what we see if this one is unsymmetrically filled what are the situation when it is E g 3 or E g 1 that is the only two possibility. E g 2 will be symmetrically filled E g 4 will be symmetrically filled E g 1 or E g 3. If it is filled then again you are having some sort of a problem what is the problem that is what we are coming to let me go to the picture ok. This is what is yantailer distortion if of course, T 2 g is completely filled ok completely filled or symmetrically filled either T 2 g is T 2 g 3 E g 1 or T 2 g 6 E g 3 all these configuration. So, what are the configuration T 2 g 3 E g 1 T 2 g 6 E g 1 T 2 g 6 E g 3. So, wherever unsymmetrical situations are possible of course, T 2 g 3 E g 3 is not possible right T 2 g 3 E g 3 is not possible because it would be T 2 g 4 T 2 g 3 E g 4 is not possible. So, this sort of thing you should be able to recognize very quickly T 2 g 3 E g 4 actually should be T 2 g 4 E g 2 right. Why means it is a stabilization you look at very good question right. So, T 2 g 3 E g sorry E g this is did I say 3 everything I said this is correct 3 plus 3 6 4 plus 2 6. Now, T 2 g 3 E g 2 this is the T 2 g 3 E g 2 high spin if you are even saying. Now, how it can be going over there because it is giving more energy or destabilizing the system. This sixth electron should come here see the always filling rules anywhere I mean you read a Fuhn's principle you read whatever other principle system has to be stabilized. When two scenarios are there system goes for stabilization, but when high spin low spin situation is there because high spin is accessible it goes for high spin or it can go through low spin. Here after filling out 1 2 3 if it is high spin 4 5 6th electron will come back from here it is home right it is stable. So, T 2 g 3 E g 3 is not a valid configuration it is T 2 g 4 E g 2 it is invalid I mean literally it is possible for an excited state. If you excite T 2 g 4 E g 2 then it can come to T 2 g 3 E g 3 this is usually all we are talking about is ground state electronic configuration ok. Any confusion on that? No hopefully not. Now J T J Anteler distortion T 2 g either fully filled or half filled T 2 g 6 or T 2 g 3 here it has to be unsymmetrically filled means E g 2 and 4 is out of question E g 1 or E g 3 2 orbitals are there ok. What can happen? Two scenarios are there either d g 2 can be stabilized or d x 2 y 2 can be stabilized. Now stabilization again stabilization means what? Stabilization means repulsion is less ok. Stabilization means less repulsion less repulsion when it can happen when ligand is not coming. If Z square is stabilizing that means along that direction ligands are far. So, this will be too long this that will be. So, the left hand scenario this is a base for octahedral scenario we are drawing this is the equatorial plane axial was here. Now this is let us say normal scenario if I want to say that d z square orbital is stabilized that means this ligand is not this is the d z square orbital direction right. So, that means ligand is not coming close to this d z square orbital this is where the d z square orbital is right. So, it is not coming close that means it is going far and that is how d z square can be stabilized. That means two of them two of the ligands should be at a longer distance and four of them at a shorter distance or relatively shorter distance. The opposite is here when relatively speaking d x to y 2 is getting stabilized that means these are going out stabilized means repulsion is less ligands in the second scenario ligands will be further away and z will was over here. So, z in this is called z in z direction getting in this is called z out ok it is a very very simple thing. If you just try to understand very simply do not complicate things it is that very simple. Now as you can see z out this is the z out if you keep on moving more and more z out becomes your square planar right. So, here previous case what we were discussing if z becomes stabilized more more more and so that it is you are taking too far from the metal center octahedral is no longer octahedral it is becoming square planar ok. So, z in and z out yeah sorry no these are having same energy right no these are see these are degenerate you cannot dictate the term you cannot tell who is where you want it is just this is just for nomenclature sake I have written just to you know just to tell you or just to write you something I have written which one is d z 2 or which one is d x 2 you do not know it is same energy right degenerate means same energy 5 d orbitals before crystal field theory we used to think that same energy we do not discriminate right. So, this is just a representation sake not all the cases, but when it is unsymmetrically filled then it is becoming a problem. See three four scenarios are there E g orbital why E g orbital is getting part of why T 2 g not because those are the one you can see these are the outside if you see the splitting wide those are the one facing the maximum of the ligand repulsion right. In E g 4 configurations are there E g 1 E g 2 E g 3 E g 4 E g 2 scenario we have already discussed E g 2 scenario is the one where we are getting most most often we are getting that d 8 T 2 g 6 E g 2 E g 2 that is a d 8 electronic configuration it tends to go for a square planar one fine you remained with E g 1 or E g 3 E g 1 and E g 3 scenario we are trying to discuss here. So, when it is symmetrically or full field then usually we do not get bother it is you know other way or one way or the other it is just the same I mean you know it is completely full. So, nothing to really compare with, but in this scenario when you have unsymmetrical spilling then you have the chance of splitting. See similar split the extent of splitting you can see is not the same here again of course, usually speaking these are the one E g are the one which are going to face the ligand directly and thereby splitting starts over there and then of course, sometime it also get affected T 2 g gets affected that is why we have we were showing initially we did not split it. Initially we when we were discussing T 2 g 6 E g 2 we were showing that d z 2 is getting stabilized when it is getting further stabilized and this then it is going to get split further ok. So, look at the slide almost in every book it has been written really well if you have any confusion please come back or discuss with your friends. So, z in and z out this is z out z out means z you are pulling out stable stabilization happening ok. Now, this is a slide which is mostly left for you to digest little bit square planar geometry. So, if you want to memorize sometime things can be problematic we want to read you mainly let us say octahedral 3 2 octahedral where is tetrahedral tetrahedral is not given here anyway the geometry and thereby they are splitting how they are going to split this is not all of them is part of the syllabus again this is just a overview giving an idea how ligands are coming how d orbitals are getting perturbed which direction d orbitals are located and then from which direction ligands are coming of course, that is what going to determine what is their geometry. If it is a square planar where it is coming you know if it is a trigonal bipyramidal or TBP 3 of them in the equatorial pen triangle one from the top one from the below. So, which are the orbitals going to get affected into what extent and so on. So, these are you know these are these are in the book or may not necessarily you have to read it you are, but you are supposed to read octahedral tetrahedral square planar I think for fun you can look at that try to justify little bit ok. So, that is it for this chapter.