 Good evening. Can you hear me? Yeah, so I think they had discussed the structure of isomerism. Yes. Okay. So, so the last part of this chapter is stereo isomerism, stereo isomerism in coordination compound. I don't know the definition. The studio isomerism is because of the relative position of atoms of books in this space. Right. So, we know two types of isomerism. This has one is geometrical geometrical isomerism. The one is optical. Now to understand the geometrical and optical isomerism, you should understand the geometry also. So, the geometrical isomerism will take few assumption here and assumption is central metal atom is represented by represented by M. This is one thing monodendric ligands ligands are presented by monodendric ligands are represented by small letters a, b, c, d and so on by small letters a, b, c, d, e, f. Then we have symmetrical bidented ligand. Symmetrical means both atoms are same. Symmetrical is not because of the structure, but it is because of the donor. Then it is symmetrical bidented ligand. That's why we assume it as a, a, b, b, capital letter. And unsymmetrical bidented ligand is represented by a, b, c, d. This is the notation we are going to use. We'll see that in the last. Sure, sir. Okay. So, these are the notations we are going to use. Correct. Now, we are going to see the geometrical isomers of compounds with different coordination number. The most important one we have coordination number six, coordination number six. So the first type of, you know, complex we can have with coordination number six. And I'm considering complex with all monodendric ligand. Right. The first type is complex with all monodendric. The first example is we have MA6 type. What is MA6 type? What do you mean by this? Sir, all these surroundings are the same monodendric ligands. And same kind of, same ligand is being surrounded in that. Yeah. Yes. Six monodendric ligands. And all ligands are similar. We have six monodendric ligands. So obviously the coordination number six we have, its geometry is what? Its geometry is octahedral. Right. So how these six ligands are arranged around the metal M that we need to understand here. And by changing the position of the ligand, do we get different structures? If we get different structures, then the isomers, isomer possible. Otherwise it does not show isomerism. For example, you see this, how do we represent this MA6 type molecule? It has octahedral geometry. And we have these metal here at the center. And six ligands are present at the corners of octahedral, which is nothing but this. Since all ligands are the same, anywhere if you change the position of A, this AU place here, this AU place here. This does not make any difference. Hence, in this case, we have no geometrical isomerism possible. No GI for this. Did you understand this? You see MA5B type, five monodendric ligands of identical monodendric ligands, five. And one is different from the other. Okay. So how do we place this? Again, the geometry is octahedral. Metal is present here at the center. And wherever you want, you can place this B and the five A atoms. For example, I am placing five A atoms here at the corner of this square. One A on the top and B here. Wherever you want, you can place it. That's not a problem. Okay. One molecule is this. Can we draw one more molecule here? M here, A, A. I'll place B here, A here. Tell me, is this possible? Yes, sir. That's the same one. Both are the same one. And why it is same? For all the foreign angles and all of this thing. Yes. All the six positions are equivalent position here. Right. That's why A or B, whatever structure you write first or second, both are the same structure we have. Right. Because the relative position is same. You see, this B will have, at a certain distance, we have A, A all this A present. Here also we have same kind of thing. Right. Hence both molecules are identical, same molecules. And we do not have geometrical isomerism possible in this kind of. Tell me, what is the angle this one? So this angle is 90 degrees. You see, the square is in the, imagine the, you know, the screen of the computer or laptop, whatever. Right. The screen is the square. So this A is coming towards you, this A, one of the A. And the other one is going away from you. Like that you can imagine tetrahedral or octahedral geometry. Is it right? Are you able to imagine this? Yes, sir. Right. So one, this A is coming towards you. And this one is going behind the screen away from you. This is octahedral geometry. So obviously this A, this line and the plane of the square is 90 degrees. What is this angle here? This angle is also 90 degrees. What is this angle? 180 degrees. Yes. It's 180 degrees. Actually, you must understand how this octahedral looks like. Okay. Similarly, if you take third type of, you know, complex, which is M A4 B2 type. Could you tell me how many GIs possible, if possible then? Three. Last one, sir, please. No, Swami is getting two. Others tell me, okay. Shankar is getting two. Trippan is getting two. Venkat, what about you? Okay, Pratham is getting two. Doing so one minute. Amira, Niyati. Advaita is there. Yes. Sir, I got three. Okay, Amira is getting two. Niyati is getting two. Guys, you see, whenever you get this kind of question, suppose, you know, if I, any example you have, then best way is what? Write down in this form, in this form actually, A4 B2 type, so that you can easily, you know, draw the structure. Because you cannot memorize this, there are so many possibilities. And we are going to introduce all the possibilities here. But in the exam, you should know how to draw the structures. Right? Uttana is getting two. So first of all, what you do, what you have to do here, you have to draw the various possible structures. Most of you are getting three. So I'm drawing three, the square here. When you draw this three square. And obviously metal we have at the center of the square, in all the structures, metal will have at the center. Now randomly you place all the ligands around this metal at the corner of the octahydrate. Whatever you want, wherever you want, you can place. For example, suppose, I am placing all the A at the corner of the square. This is one possibility. And the two B, like this. This is one possible structure you can draw. Now what do you do? This A and B are like exchange. A, A, B, A. This is B and this is A. This is another possibility. What is the another possibility we have? Can we change the position of B here in this square? Or I will change the position of this B. I place this B here, A here, B here, A here. Alternate. Isn't it the same as the first one? Same as the first one. Okay. What is the, is there any other possibility? For the second one, if we interchange the top B with the right A. This one. A, B, this is what you are saying. Any other possibility? Can we draw anything else here? Sir, isn't second and fourth equivalent? Any other possibility you see that we can draw? First of all in the exam, obviously if you do some practice, you don't have to draw all the structure. You could identify this easily that how many structures we can draw. But suppose if you do not know, you just draw like this all the possible structure whatever you feel like, like the possible structure we have. Now you have to then analyze that whether the drawn structure that you have drawn, whether the structure you have drawn are different structures. Because the number of different structures are the geometrical isomers here. So you need to first identify also after drawing the structure that they are same or different. How to identify the given molecules are same or different. Okay. So for this what we do, we just take the reference structure. Like suppose the first one, this structure I take as a reference one. One, two, three and first structure, we take it as a reference one and the first structure at the corner of the square, we have all A ligand present. So if other structure also will try to draw a square and the corner of the square contains all A ligand. Like suppose this one. In this one, could you think of this particular structure? This is square you see, this is square. So when you rotate the molecule like this, it becomes the same as the first molecule. So what we can say first and third are equivalent structure. Is it clear? Similarly, first and second are different structure. We cannot compare first and second. Wait here, this is second. Second one you see, we have three A atoms at the corner and one B. So if you want to compare first and fourth, these two are different structures. If you want to compare second and fourth, so three A at the corner and one B. So we can easily draw this square hypothetically. We can think of this square, which contains three A at the corner and one B here at the corner. It means this second and fourth are also the similar structure. This is one way, one of the ways to identify whether the given structure is identical or different structures. Did you understand this? Yes, sir. Yes, sir. Got it? Okay. This is one of the ways, right? One also thing you can think of. We can draw trans isomer in this. We can define cis and trans isomer here. And cis and trans isomer, we define by the relative coefficient of the atom. Like you see, if the 2B atom present at 180 degree, the 2B atom present at 180 degree, this is trans isomer. The B present at 180 degree. Okay. Here you see the 2B atom, the angle between this B and this B, this bond. It is 90 degree, right? So when the bond angle is 90 degree here, then it is cis isomer. So this one is cis, this one is trans. And if you look at the bond angle here between B and B, it is 180. Here also it is 180 if you look at this. And that's why first and third are the same thing. Second and fourth, you see the bond angle is 90. These two are also same. Yes, sir. You need to change the relative position. If it is not changing, then the molecule is same, identical only. Here if you compare, another thing also here if you compare, the 2B is at 90 degree from all the A ligand, right? Here also we have the same thing. This B, if you see, it is at 90 degree from all the A ligand. You see, this angle is 90, this angle is 90, this angle is 90, this angle is also 90. All the angles are 90 here. Hence the molecules are identical. How many different way you can think of this structure, right? So you don't have to memorize this once you know. Once you finish this, you will easily identify that what are the possible structures we can do. Okay. Now another one you see, fourth type of molecule, M A3 B2 C type molecule. Tell me, if GI possible, then how many GI should be there? Okay. Done. How many isomers you are getting? Venkat, wait. The structures are 1, 2, 3A and 1C. This is one structure. One is this. Another one is this. Next we can draw B and C, which are B we have. You can also think of this structure, B, C, M. Like this you can draw many structures, but again you have to, you know, compare whether that drawn structure is same or different. Right? So here you see, we have 3A at the corner. With respect to B, we can say B is at 180. Here B is at 90. These two structures are different, definitely. Okay. This one I have written 4A here. I'll just change this. Suppose BL place over here. BL place over here. So this two, the B are at 90 degree to B atom. Here also B at 90 degree. Right? Here B at 180 degree. Here B at 180 degree. And here are the two B atom. Whereas we have written 4A. This is C. This is E for example. So here the two B atom are at 90 degree. So let us try and draw a square which contains 3A and 1C. Okay. 3A and 1C. We cannot draw over here, but here we can draw 3A and 1C. Can we do that? No, here also we cannot do that. Here we can draw it here. So first, second, fourth and fifth. So we can say first and fourth are identical structure. 3A and 1C. We cannot draw it here. Right. First and fourth are same. If you look at this one, 2A, 1B and 1C. Okay. So we can have 2A, 1B and 1C. If you look at this one and this one here, this structure, 4, second one and fifth one, are they similar structure or different structures? They are different. Which one? Second one and fifth one. They are different, sir. Different. So first, second and fifth are different. What about second and third? Second and third are also different, but third and fifth are the same structure. Third and fifth are the same structure. Because you see, we can have 2A, B and C, a square like this. So third and fifth are the same structure. So the three isomers we have, we have first one as an isomer. We have second one and third one. These three isomers we get in this morning. Sir, I didn't understand how second and fifth are different. Second and fifth. Yes, sir. You see, in the second one, we have 2A, B and C. Here also we have 2A, B and C. But the position of A you see here, it is at 90 degree. And here the 2A at 180. Oh, right. Yes, sir. B and C are at 90 degree here. B and C are at 180 degree. Okay. Yes, sir. Right. Yes, sir. So with respect to A, we can say this molecule is cis isomer. This one is with respect to A. Identical atoms we have to see. With respect to B and C, we cannot say. With respect to A, it is cis isomer. And this one is trans isomer. B or C we can also take here. cis, this one and trans. Okay, so one second and third are the different structures. Fourth and fifth you want you can draw. But these are the similar structures that we have already drawn. Now the problem is that this kind of question, if you try to solve by this method, it takes a lot of time in the example. Okay, so what is the best way you need to do here? And this is the method you are going to use in the exam. You don't have to draw these structures. It takes a lot of time. Just to consider A3 B2 C we have here. The molecule we have is m A3 B2 C. What do you need to do? You need to, you know, write down the ligands in a pair, right? Two ligands in a pair, three groups you have to form, three different possibilities, three groups you have to form. Like you see this A3 B2 AC, what, how you can, you know, form three groups in a pair. We can write both A together. Then one A we can write with another B and another B we can write with C. This is one group we get. Another one what we can write, we exchange A and B here. A, A, B, A and B, C. So we can write A, B. This A, B you let it be, as it is A, B and AC we can write another one. And B, B we had, we did not write together. So they write down A, A together. Then here we write down B, B together. And then A, C together. These are the three different, you know, possibility you can write. Any other possibility you can try, you won't get. Yes sir. So this actually, if you want to draw the structure after this, you see how do we draw the structure? See here, this pair actually represents the ligand which is present at the opposite corner of a diagonal. This pair, this pair and this pair. A, A is present A, A, A, B, A, C. This is the second A, A, A, B, A, C. A, A, A, B, C. You see this molecule here. The two ligands present at the opposite corners are A, A, bracket close. Another two, we have B, C, B, C, bracket close. Another two we have A, B, bracket close. Whether you write this B, C here or A, B here, that won't make any difference, right? But this molecule, this, you know, you know, arrangement, we can draw like this. Did you get my point? Similarly, here you see, for this one it would be two A, A, like two A at the opposite corner, A, C at the opposite corner and here we have what? Here we have B, B. This one you see, here we have A, A, B, B, A, C. This is the third one we have. This one it would be A, B, then A, C, then A, B again, second one. So like this you can frame different pairs of two ligands at a time and three groups you need to form here since it is octahedral. Did you understand this? It gives you the number of isomer that you get over here. See, if you go back and do it for the other molecules, that also fine, you can do it. You see for this one, A4 and B2. Consider this. A4 and B2. So what we can write, we can write A, A. Again, we can write A, A. We can write B. A, A, A, B, A, B. This is another isomer. So these are the two isomers possible here. You can see here, we won't get any third possibility like this. Clear understood? Yes, sir. If they ask you the number of isomers, you have to draw like this. You don't have to draw these structures in the exam. It takes a lot of time. But yes, you should have the understanding that when you have written the ligands like this in a pair, then how the molecule looks like, how can we draw the structure with this particular input? All these pairs are present at the opposite corner of a diagonal. So where exactly are these pairs drawn again? Diagonal opposite. Okay. Suppose you don't have anything. You just have these pairs. So how do we draw it? A, A, A. And then we have a metal here in the center. A, A here. So we'll place A, A here like this. Opposite. A again. So A and A. And we have B, B. This is one of the structure. And this one, if you write like this, A, A, B, B, A, A. This one and this one are the same thing. Right? Yes, sir. All of you understood? Yes, sir. So just a few time in the exam, you must take care of this how to draw the structure. Now, on the basis of this only, I am not going to draw that octahedral structure now. Just I am writing down the complex. You tell me how many isomers possible for this, if possible then. Okay, like you see the another one, I think it is a fourth one. Fourth or fifth one? Fourth one, I guess, right? No, fifth one. Yeah, fifth one. Okay, how many isomers we have here? M, A, 3, B, C, D type. You see all these are monodendritic ants. How many isomers we have here? In M, A, 2, B, 2, and C, 2 type. Done? So first one we are getting four and the second one we are getting. Okay, so the arrangement, the possible arrangement here are, we can write down A, A, then we can write down A, B, and C, D. Then we will have A, A, then A, C, B, D. Again we will write down A, D together, so A, A, A, D, B, C, and one more we have A, B, A, C. There are the four isomers possible for the first one. Okay, second one we can have A, A, B, B, C, C, A, B, A, B, C, and then we have A, A, B, C, B, C, A, C, B, B, A, C, and A, B, A, C, B, C. Five GI possible for this. Now one very important point here it is what? Out of this five geometrical isomers, this one, the last one that I have written here, this one is optically active also. You must take care of it. There is no plane of symmetry in this. Similarly you can draw for another molecules like we have M, A2, B2, C, D type and in this we have six GI possible, M, A2, B, C, D, A type. In this we have nine GI possible and last one we have M, A, B, C, D, EF type. In this we have 15 GI possible. Okay, and here we get this one, this has been asked in need exam, the last one. Fourth one, which one? This one, the fifth one. No sir, the M, A, 3, B, C, D. You are talking about the fourth one. Case five, the fourth sub example. A, B, A, C, D, yes sir. This also we can say. Sir, can we also say if there are no identical elements in the same ordered pair, they will be optically active. No identical? Like if all the elements in the ordered pair are different, if there's no A, A, A, B or something like A, A, B, B and R, if there's A, B, B, C, A, C and R, then it will be optically active. Is it right to say that? Optically active, we just check the plane of symmetry. Okay, we will discuss that. Optically active, optical isomers we will discuss after this geometrical isomers we will see that. So we will check the plane of symmetry over there, nothing much. Okay, yeah, M, A, B, C, D, E, F, type all are optically active. No plane of symmetry there. You can say that. One important example here, the last one is this. We have M, A, B, B, 3 type and this one, they've asked many times in the exam, J. For example, we have CO, NO2, 3, NH3. So whenever you get this kind of question, you represent it this way, then it will be easier for you to analyze. Okay, molecule is this, and this is the representation we have. So we can represent two structure for this and in this case we do not, you know, write C or trans term here. So we have metal here. One possibility where all the A present at 90 degree, all the A present at, we can place it this way. One on the top and the bottom will have B, one and then two and then three. This is one way. Another way is what? We'll have metal here. A will place over here. B here, A again here. So this molecule you see, so this looks like, this looks like the face of a triangle here. See this? This also looks like the face of a triangle. This particular representation, we call it as facial isomer. This one is, this one is facial isomer. We call it as FAC. In this one, the position of ligands are on the meridian of an octahedron, this position and this position, meridian of an octahedron. And this we call it as meridional isomer. Just this term you have to memorize that in short. So always remember, NA3B3 type, we represent it this way and we do not, you know, write down C or trans isomer here but we write down facial, that is FAC or mer isomer. So but here, NA3B3 type ligand. NA2 is whether you write here or not. NA2 has the tendency of to behave as an ambient ligand but here nitrogen is the donor act. Okay. If it is oxygen, then it would be written over there, ONO, not NA2. Okay sir. Sir, how do we write sister trans for the others before this? For which one? For any of the others, like NA3B2C or something. Sir, I will just copy this. Sister trans, we just see the identical ligand and their position. I will explain you one or two examples. Okay sir. See the simplest one is MA4B2 type. Okay. So in that case, you consider the position of B over there. In one of the isomers, you consider the two B ligand present at 90 degree and in another one, the two B ligand present at 180 degree. Yes. This one. Yes sir, that one you explained sir. Huh. Sir, after that, the next one also you explained after that like the MA3B CDS. Suppose you are talking about this one. Right. Sister trans. Okay. So this also you are told sir and then try to respect. Which one you are talking about? Next one sir, like supposing there are four ligands. MA3BCD or MA2B2C2. So in that case, usually we do not write anyone you can take it. See, suppose we have MA3, according to you, MA3BCD type. Correct. Yes. Okay. So what we can write, you draw the structure, we can write A, A here, A, then B, CD, we can write like this. Okay. Why with respect to A, you cannot define because the two A here, present at 180 as well as 90. Correct. So if you change the position here, like for example, we have metal here and what change we can do? We can do one thing. We can have this A place here, B on the top, B and C. Correct. So here you see all A are at 90. Correct. But if you look at the two structure here, you can take B and D with respect to B and D. You can differentiate the two. It is trans here and it is cis here. Any two ligands you can take if the structure is given. But we cannot define, it depends upon the structure what is given. With respect to A, you cannot define because A are at 180 as well as 90 degree also. So this we cannot take. Any two ligands out of the three B, CD we can take for this. Like in this one, this is one possible isomer. This is another possible isomer. Do we have another possibility in this? We can have another possibility also. B and C are at 180 degree. Many structures you can draw. B and C here, D you are ready to be as it is. A, A and A. Okay. Okay sir, I understood. This also you can take. So basically there is no proper definition of cis and trans. They just said that based on the angle we can see. But depending upon the structure you have to choose. Okay. With respect to this is cis or this is trans. Okay sir. Fine. So this is we have done for all the possibility we have done for octahedral complex with only merinded ligand. It is also possible that in the complex we have one bidented ligand and other are merinded ligand. Or more than one bidented ligand. That is also possible. Write down the next type here. Sir can you go to number 6 please. What? The board number 6 sir. Board number 6 you said. This one. Yes sir. Thank you sir. Okay so in all these molecules the one which is the most important one and which they have a question is this one. M A 3 B 3 type. That is facial and meridond. How do we write down facial and meridond? Facial is all the A identical ligand present at 90 degree. You see the position of this, this and this. All these are at 90 degree. When one of this you place at 180 you exchange the position of this to you get meridonal which is this one. Right. So here you see the 2A one of the A present at 90 degree from the other A ligand and 180 degree from the other A ligand. This is present at 90 as well as 180 and same thing we have for B also. 180 with respect to this 90 with respect to this. So basically if you interchange the position here you'll get meridonal. Why we call it as meridonal? That is not required but it is present. The ligands are present at the meridonal of the octahedron. Hence it is meridonal. Okay. But you should know how to write down the facial or meridonal isomer that is what I explained. All at 90 degree 2 at 90, 1 at 90. Okay. Now the next one is compounds containing, second type write down compounds containing bidented ligand. So for example the first one we have MAA 4 types. So 1 is bidented 4 is monodentated, so 6. 1 is identical bidented ligand and other 1 and 4 ligands are monodentated. Identical monodentated ligand. Coordination number is 6 for this complex. When coordination number is 6 the geometry is octahedral and to draw the structure draw a square. What you have to keep in mind here whenever you have bidented ligand whether it is symmetrical or unsymmetrical, a 2-donor atom in this case we have ANA 2-donor atom must present at 90 degree. 180 degree cannot must present at 90 degree. Right. For example you see ethylene diamine ethylene diamine I am just trying to make you understand this. We have nitrogen is a donor atom and this nitrogen is connected with carbon CH2-CH2 another nitrogen enter like this and it donates electron and forms a complex like this. So this 2-donor atom are connected by this bond. This bond or this connection we represent by like this this means AA atoms are joined together the bonds of the ligand that we have here. So maybe if you present this AA at 180 the bond length is not enough or if it is enough also then there will be more strain and hence we do not place this 2 ligand this no 2 atom at 180 degree always keep that in mind for bidented ligand the position occupied by the donor atoms of bidented ligand is always at 90 degree. So this is the the position of the 2 bidented ligands and all A present like this How many GIs are possible in this complex? One Only one structure no GIs. Only one so no GIs. So we always write in a pair so one structure cannot give you a pair no GIs possible because all these positions are equivalent it does not matter whether you place this A over here or this A over here or this A over here so all there is no GIs for this one if you look at the second example we have MAA and A3D type tell me how many GIs so one thing we can write here similarly just to save the time in the exam we can write AA then we can write again AA only one possible how many GIs is possible in this unless you have geometrical isomers possible with the relative position no you change the position of A and check whether you are getting the structure or not so the possible structure here is AA then we will write AA and then we will write AB see this is not the method it is not given in the book but to understand or to save time to count the number of GIs it is always better than to draw the structure in the exam another one is what AA we write as it is A1B and then AA and we do not have any other possibilities so two GIs are possible for this one sir in the previous one you will get AA 2 pairs of that and then okay so it is always at 90 degrees so you will not have the pair of AA okay next one you see third type MAA A2B2 tell me the number of GIs MAA ABCD AA 3 so it would be AA AA and BB AB AA AB CD AC AB AD AD AB CD similarly we can draw three more isomer here first and third are the same first and third is it okay CD I have written here AD AD it should be AC here AC is it fine sir I think you can have we have 6 GI possible AC AD AC AD AD AD and AB 6 GI possible like this and again I am telling you this thing you do not write in your school exam or in board exam okay this obviously gives you an idea that how many GI possible and you won't get wrong answer over here if you could find the number of GI like this okay this portion is deleted sorry? this portion is deleted your voice is not coming properly sir she said that this portion is deleted for our school it's not there fine then it is fine they won't ask you this but like every year I used to say this in the exam just for the information I have to count the number of GI's without drawing the structure okay yeah that's fine AA3 type I will draw the structure for this one then only you will understand how the structure possible for this one you see we have a square by pyramidal metal is here then AA all are bidented ligands must present at 90 degree so we can say these two are connected these two are connected and these two are connected one possibility is this okay another one is what metal present here AA this two oh sorry we can write next what tell me do you have any GI possible can we draw any structure here if I take this two here for example right then if you take this one we will have the same structure if you take this one sorry this one then it is not possible because these two are at 180 degree are you getting me one pair will always be at 180 degree yes that's why in this we do not have GI no GI possible right but why I have done this if you look at this molecule so this is not possible let me just remove it it is not possible but if you look at this molecule here if I take unsymmetrical ligand which is nothing but this mab3 type in this is it possible previous one case 5 we will have to write A capital A capital A no even if you want to get one structure we have to write capital A capital A so that means one pair of A's are at 180 degree with each other yes sir if we follow that coordinate system order pair what I said the two atoms donor atoms which belongs to the same molecule same ligand should not be there at 180 degree okay sir means you have multiple bioentery EN has suppose ethylene diamine so one ethylene diamine placed here another one placed here okay sir but then if you write the order pair A A and all you have to write A A A A A A A you have no choice what I said this is also not a genuine method we don't write like this this is just I have given you to count the number of GIs okay sir here there is no sense of writing it down A A A A A A does not make any sense yes sir it is not possible so that's how it is okay how to use it possible last one there is no GIE don't think sir why no sir they get me too two I don't know I am not getting your voice properly today for all of you some I think some network problem anyways then we have suppose I place A here and B here then I can place A here and B here and next is A here and B here right so we can say these two are connected these are connected and these are connected another possibility I think one more should be there one two then we can write A here and A here B here this one possible is it it's not same see these two GIs possible for this one so how are they different sir could you explain yeah I will tell you no it's not possible GIE mirror image is not possible here I will tell you in this one because you are talking about geometrical isomalism so mirror image we are not considering see here it's very simple and clear you can understand this look at the relative position of all A here we have these two at 90 this two at 90 but this is at 180 right in this one you see all A plays at 90 degree 90 these two are different relative position is different here two molecules are different but in the previous one it was A only everywhere right A so if you draw this one also the relative position is same so no GIE in the previous one right it looks like similar but the relative position of A is different to molecule yes so in fact this one is what we can say A, A and A here yeah that's right so these are two structures we have the second one here if you compare this with the facial and meridional one this one all A present at 90 so this one is fact and this one is model it looks like different but all A are at 90 here also and in facial also all A was at 90 got it yeah so this is it for coordination number 6 here the coordination number 6 is the most important one okay most important one they ask mainly from coordination number 6 only optical isomerism will discuss geometry okay so next you write down sorry so one minute I was copying that sir just a second sir done sir thank you so next is write down coordination number coordination number 4 coordination number 4 if you have complex contains contains all mono dented ligands all mono dented ligands for example the first one is MA4 type do we have any GI in this no GI present yes if you write down MA3B type what is the number of GI MA2B2 MA2BC MAB CB number of GI is in this so in this one we have no GI in this one we have cis and trans 2 GI possible if you want to write down just AA and BB in this one we have AABC so we have only 2 GI that is AABC and we have AB AC type in this and here we have one we can write ABCD together or we can write AC BD together or we can write ADBC any other possibility we have only 3 GI possible AB will be also possible I have written here too ok so since we have coordination number 4 so we will get 2 pair here 3rd pair won't be there if we have bidented ligand in this then what are the possibilities bidented ligand we have MAA A2 type MAA AB AB A2 type MAA MAA BB type MAA CD type AB CD type MAB AB type so how do we know if it's square, planar, tetrahedra done and what about the last one is square planar or tetrahedral we can understand only if you know the geometry sorry hybridization if you are getting it as SP3 it is tetrahedral otherwise SP2 gives square planar so hybridization gives you this information yes so we need to know that to know the no in the complex when you use that VBT you know that whatever hybridization you are getting because in this case that is there in case of trigonal bipyramidal where we have where we have you can talk about this thing what is the dipole moment you can talk about whether dipole moment is 0 or not or you can talk about number of bond lengths are same or different which gives you a square pyramidal geometry also but in case of coordination number 4 you will get whatever the geometry whatever the hybridization we have according to that square planar you will get DSP2 hybridization yes ok now in this one you see all these molecules from number 1 to number 6 the number of GIs are 0 no GIs possible in all these molecules you can think of by drawing the structure you will get no geometrical isomerism from 1 to 6 ok one thing you can keep in mind if you if coordination number is 4 and the ligand is symmetrical by dented coordination number 4 ligand is symmetrical by dented then GI is not possible that you can keep in mind coordination number 4 symmetrical by dented no GI possible here we have 2 GI and here we have again 2 GI the possibility is AA BB and we can also write AB BA here also similar way you can write down we have AC BD we have AD BC so these are the few examples we have done for coordination number 4 and 6 now we will see the optical isomerism we don't have much in this only few examples we need to discuss in optical isomerism right on the heading optical isomerism so you know the logic behind optical isomerism it is the property of a molecule towards plain polarized light we are going to give you the basic understanding what is optical isomerism and all that we already know what you need to know here first of all all linear geometry whatever linear geometry we have they do not show optical isomerism so write down the first point linear geometry does not show optical activity or optical isomerism because suppose we have square planar so in the square planar geometry the plane in which the molecule is present along that plane only we have a plane of symmetry we can divide the molecule into 2 equal half along that plane and hence we have a plane of symmetry so there is no optical isomerism possible for the square planar geometry or any planar geometry like linear, trigonal planar square planar optical isomerism is not possible ok next slide down tetrahedral geometry shows optical isomerism only if all the four ligands are different then only then we do not have any plane of symmetry there we are going to repeat that Yeah, tetrahedral geometry shows optical isomerism, tetrahedral geometry shows optical isomerism if all the four ligands are different. If any two are same, optical isomerism is not possible. Because when four ligands are different then only we do not have any planar symmetry. Yes sir, thank you. Okay, next slide down. What if it's a bidented ligand? Ah, wait, I'll do, wait. Right down next. The molecule of the complex which shows optical isomerism, complexes which shows optical isomerism must contain, or not must contain, which shows optical isomerism may contain at least one symmetrical bidented ligand, at least one symmetrical bidented ligand. Next one. Complex contains unsymmetrical bidented ligand, unsymmetrical bidented ligand, then also optical isomerism possible, may possible. If all these things are there then you can think of optical isomerism. Unsymmetrical bidented ligand and optical isomerism possible. Last point. Trans isomers, trans isomers contains plane of symmetry, trans isomers contains plane of symmetry, hence does not show optical isomerism. Hence does not show optical isomerism. There are few complexes which shows optical isomerism. I'll write down the general formula here of the complex. This type of complex, it shows geometrical, optical isomerism. MA2 B2 C2 type, we have discussed this. One of the combination shows optical activity. MAB CDEF type, optically active. If you have MAA, this is also optically active. MA2 type, this is also optically active. AA2 type, AB, this is also optically active. MAB3 type, this is also optically active. One minute I'm done, copy this? Yes sir. Now this too we have already seen. This complex, you see the structure of this. MAA3 type. How do we draw the structure here? Only one possibility we have, one structure we can draw. All the A atoms present like this. Do you have any symmetry here? All A atoms are there. Along the plane of the square, do you have symmetry? Sir, could you please repeat? Yeah, tell me. No sir, can you please repeat that sir, I didn't hear you. No sir, can you please repeat that sir, I didn't hear you. No sir. Some of you are saying yes, some of you are saying no. We do not have, why we do not have? If you take a plane here, along the plane of the square, this, we do not have mirror image over here. It's like, right? The position of the bident. See, fine. In this molecule, obviously I have given over there that this molecule shows optical isolation. If you do not have plane of symmetry here, no POS in this. There is a plane of symmetry, optical activity is not possible. Yes. So actually what happens here, this, you may get mirror image of this A over here. Because it is out of the plane, and it is below the plane we have. So this A bond is coming above this diagonal, right? This MA bond, it is just above it. And this bond is below this diagonal. Can you imagine this? Yes, tell me. That's why this thing, plane of symmetry is not possible in this one. The same logic you can apply to these, you know, complex also. MAA2, A2 diagonal. The only difference is what? This AA2 is fine. We can have two monodendrons. Yes, sir. You lost the connection. Yes, can you hear me? Yes, sir. So this is the, you know, plane of symmetry here. All the examples, if you write down A2 type here, or if you write down MAA2 AB type here, all these are optically active because of this one. There's one, you know, example here, which is that facial and meridional. MA3B3 type. MA3B3, this you have to memorize. Okay. Because one of the molecules you can easily understand there's plane of symmetry, but other one it is not visible in the complex, because it has alternative axis of symmetry, which is actually you need to change the position, angle, and then you have to imagine it, right? That's a very difficult thing to visualize in the complex. So you have to memorize this one. First of all, I'll draw the structure. The two possible structure here is one is facial, other one is meridional. So two square, metal at the center, and facial is all A are at 90 degrees. So one, two, three, all A are at 90 degrees. Meridional is, I'll draw it, right? B, B, and D. This is facial. All A are at 90 degrees. Meridional is one A here. Basically, these two position you have to interchange. You'll get meridional. So we have A, A. Here we get B. Here we have A, B, and D. This is meridional. In this one, it is very, along this plane will have the plane of symmetry. This A and this A will have the mirror image. Along this plane we have the plane of symmetry, right? So in this, we can say we have POS, and hence no optical isomerism. This molecule shows, okay? In this molecule, it is difficult to visualize, but this molecule contains optical, contains alternative axis of symmetry, right? So just keep this in mind. Sometimes they ask this question because we cannot see here, like if you try to imagine, you won't get here any kind of plane of symmetry because this thing you cannot imagine. It's very difficult to visualize. No, you have to draw the structure three times, two, three times, and then you have to see the plane of symmetry. This something is very difficult to visualize. So what I'm telling you here, ME3B3 type, first of all, the molecule is optically inactive. There's no second thought about it, optically inactive, right? If they ask you this question also, then you keep this in mind. This molecule, however, you know, plane of symmetry, we cannot visualize in this, but this molecule, you know, it contains alternative axis of symmetry, right? The plane of symmetry is there, which makes the molecule optically inactive. No optical iso-method. Okay. Apart from this, I've given you what all different formulas we have, general formulas, which shows optical activity. You need to just check the plane of symmetry by drawing the structure if you do not memorize. That's the method we have. Otherwise, you can keep this one. Excuse me, sir. What did you say AOS is? Alternating axis of... Could you just explain that, sir? What it means? We can do that. But, you know, usually what happens, you have to keep this in mind because in any molecule, you cannot, you know, identify this particular axis of symmetry. That's why mostly we don't talk about it. It is an element of symmetry that we generally discuss in this thing. Solid state. Okay, center of symmetry, plane of symmetry, alternative axis of symmetry. This rarely exists, but it is there. So what happens, you have to change the position into this and then you can find out the alternative axis of symmetry. Okay. Just a second. Do one thing. Let it be here. We'll discuss solid state in that only. I'll tell you how to do this. Because you need to draw the structure and then change the position. You'll get one particular structure here. Again, you change the position. Another structure you get. And then you can compare the first and third like that. You can say there is an alternative axis of symmetry. Okay. So let it be. We'll discuss a little bit this in solid state. Okay, we'll do that. Not much important, but here you need to understand that it needs to be three type. Okay. It contains, first of all, it does not show optical activity. Right. And one of the molecules is the left side. I have drawn. Okay. In that molecule, we have alternative axis of symmetry present. Right. And that's why it is optically inactive, which actually you could not see because it does not look like, you know, any kind of symmetry present in this. Sir, where exactly is the POS for the B case? POS for the B case. This one. Yes, sir. Along the plane of this one. Oh, the mirror image of A is A itself. Right. It's not by the end. It's not connected to the corner. Oh, okay, sir. If it is connected, then it would be difficult. Right. But here it is there because it's not by the end. Okay. Understood, sir. I'll show you one figure here. You'll understand the alternative axis of symmetry. Hold on, guys. See, in this way, we generally understand optically alternative axis of symmetry. Any molecule if it is given like this. Okay. If you rotate this by 90, 90 degree, right? Like this. Two comes over here. This three comes over here. Four comes over. We'll get this. Okay. Once again, I'm getting. Okay. If you rotate this by 90 degree or if you keep on rotating, okay, you will get a state where the first molecule, this is structure and the other one completely overlaps. Okay. It's looked like identical to each other. And that particular thing, we call it as alternative exercise. Here also the complete structure is not drawn. Okay. Four four axis of symmetry we have here. You need to draw it. You know, rotate it four times like this. Okay. So this kind of symmetry, that's what I was telling you. Even if you understand this, you cannot imagine in this, in the molecule, right? It's very difficult to imagine. That's why I'm a three B three. Why I have given you this example, because they have asked this question in the exam. Alternative axis of symmetry. We cannot visualize in the molecule. So when you look at the structure here, it looks like there is no symmetry present here. Right. Because it has alternative axis of symmetry. That's why I said, you have to memorize this. You need to rotate the molecule. And then if you find one state where the two molecules looks like completely identical, then we consider alternative axis of symmetry. This comes under elements of symmetry where we have center of symmetry, plane of symmetry, two-fold axis of symmetry, four-fold, six-fold axis of symmetry. I mean things comes into this. Okay. Even in solid state also, we won't discuss this in detail. Won't discuss this in detail. Mainly we'll see center of symmetry and plane of symmetry. So that is only thing we need to take care of. So this is it for this chapter. Okay. Coordination compound. We have done this thing in detail. Isomerism part, especially coordination number six optical isomerism is the most important one. They ask questions on hybridization related also, VBT, CFT also they ask question based on that. So there are many different kinds of questions that forms here. We have seen that on the basis of isomerism, they have asked question many times and VBT, they have asked question many times. One last thing I will tell you, you will also see some color based question in this chapter. Okay. What color this complex looks. So color of the complex is because of the DD transition. Transition between D and D orbital. And if D sub shell contains any unpaired electron, then only it shows color. If there is no electron in the D sub shell, it is, it does not show any color. Right. So DD transition. The orbital gap that we have the energy gap that we have, this chorus, this energy, it corresponds to one particular wavelength and corresponds to that only we have the color of the complex. So obviously this color of the complex, we cannot find out logically. We have the way to find out the color, because suppose if you solve this and you get the lambda, lambda value you got, then how do you memorize that wavelength of each color you cannot memorize. So you can find out the mathematical value of it, but still you cannot say that this wavelength corresponds to which color. It is impossible to memorize the wavelength of each color. That's why the color related complex that you get, what you do matters with, in a given particular oxidation state, it shows a different, it shows a color. You make a note of it. Okay. You will hardly have eight to 10 different types of, you know, ions with metals to make a note of it and try to keep this in mind before the exam, you can give it a revision of all these, right? These kind of things you will revise in last days, last week of the exam actually. So that you can do. Finally guys, so we are done with this coordination compound. Okay. We are going to start with a new chapter and that is physical chemistry. Solution we are going to start with.