 Hello. Hello. Yeah, shall we start? Yes, sir. Okay. So the next topic we are going to start is nomenclature coordination compound. So nomenclature, like, you know, we did in organic chemistry, here we have some rules. So we'll follow those rules to write down the name of these compounds, right? So, like, if you are going to write down the rules one by one, it will take a lot of time. I'll just show you for the reference for now. Okay. And that PDF I'll share with you. So you can copy those notes later on those points later on. Okay, we'll continue with it. Right. So let me just know show you just I'm stopped. I mean, just stop sharing my screen for a while and then we'll come back to this. I will see this. These are the rules we have of nomenclature. We'll just go through it quickly. And this PDF, this PDF is for the entire chapter. Okay, the entire chapter is given here. And there are some solved questions also, which you can practice later on when once we finish the chapter. So I'll share after the session the entire PDF with you. Okay, here we'll focus just on the rules which is given here a, b, c, d, all these rules. See all these rules if I'm going to dictate you, it takes a lot of time. Okay, that's why I'm skipping this. But these rules, you must write or keep these notes properly if you don't want to write the rules, because you have to revise it at times. Okay, see the first of all, the first rule is very simple. We'll name the compounds like we named the simple solve positive part first, and then the anand in the last like sodium chloride, any plus will write it first, sodium and then chloride. Here also, if you have positive part is this then potassium and then the name of this complex, FECN6. Okay, that is what written over here naming with the complex is such a potassium. Here, the complex cation it is. So we'll write down the name of cation first, that is complex part, and then we'll write in the last with the space chloride. Okay, the name of this complex is starting with the name of complex ion which is this, because whatever positive ion is there will write down the name first. K4 FECN6 here will write down the name of, name of this first, what we say, name of ligands first. Okay, first we'll write down the name of ligands, then we metal, then we write down the name of the metal. And just after the name of the metal, we'll write down the oxidation state of the metal in parenthesis. I'll show you some examples, we'll get it. Okay, if you have multiple ligands, more than one ligands present, then we'll write down the name of ligands in alphabetical sequence. Right, that's the basic rule we are following here. But how do we write down the name of ligands? So to write down the name of ligands, we have certain rules here you see, the ligands can be neutral, anionic or cationic. So if it is neutral, we have one particular rule type of name, anionic, one particular type of name, cationic, one particular name. So this you must go through once properly. Okay, see it is written here, neutral ligands are named as the molecules. Like you see C5H5, when it is pyridine, we'll write on pyridine. C5H5 whole phrase, try phenylphosphate, no change in this. This one, ethylene diamine, no change in this. Neutral ligands which are not named as the molecules are COS carbonyl, NOS nitrosyl, H2O as Yacqua, NH3 is amine, amine double amine, must take care of it. Okay, this is one point, the first point is very important. In fact, this C, the point number C, you have to go through properly, it's very important. Right, the neutral ligands named as the molecule, but these are some different names we have, COS carbonyl, NO is nitrosyl, H2O is Yacqua, NH3 is amine, A double amine, okay, not single amine, double amine. Anionic ligands ending with I, are named by replacing I'd with O, means I'd becomes I'd, I'd becomes I'd, I'd becomes I'd, I'd becomes I'd. So suffix O, we have to add for anionic ligands, okay, for anionic ligands, right, anionic ligands, I'd, O suffix you need to add. For example, you see, Cl- is chloride or chloro, BR- is bromide or bromido, Cn- is cyano, O2-oxo, OH-hydroxo, N3-nitrido, O2-peroxo, O2H-perhydroxo, S2-apiodol, like this, it is given, you can go through once, okay. So next we'll see, okay, OH-hydroxo and S2-imidol. Ligands with the name N in I'd or I'd becomes I'd by replacing ending E with O as follows, carbonate, carbonato, oxalate, oxalato, sulfate, sulfato, like that, acetate, acetato, okay. Then for positive ligands, you see we have third point, positive ligands ending with EM suffix, so hydrazineum, nitronium, okay, nitrosonium NO plus, okay. Now, this is the name of the ligand point number C, which is different from that we, from what we did in the nomenclature previously. So this point number C, you must take care of. Now, if ligands are present more than one, then repetition is indicated by prefix dietriotetra, et cetera. However, when the name of the ligand includes a number diapiridil, ethylene, diamine, bistris, tetrakis are used in place of dietriotetra. I'll discuss this example just a second. With example, you will understand, here you will get confused, okay. If ligand is already contains prefix like ethylene, diamine, or if it is polydentate ligand, then prefix bistris, oxalate, the same thing will discuss it. More than one type of ligand, alphabetical sequence you need to follow. After naming the ligands, the central metal and to be named immediately followed by this oxidation state in Roman under the brackets as per IUP, okay, in parenthesis, okay. NS2, NS4, 2CUCL4. NS4 will write down first because always get any part first, ammonium, okay. 4-chlorine, tetra-chloro, then cuprate. If it is negative, if the metal is in negative part, negative anion, then 8 will write down, cuprate, and then oxidation state of copper 2. After naming the central metal atom, iron, which is in the outer sphere, is to be named the naming of the sum of the complexes done as follows as per IUP. Okay, fine. Now, with example you will get, see this one, K4-efficient 6. So, positive part is potassium. So, potassium hexa-sino ferrate, negative ion, so ferrate. What is the oxidation state? Plus 2, so we will write down plus 2 in the Roman. Potassium hexa-sino ferrate, you see plus 2 here. Number 2, K2-PTCLC, positive part, potassium hexa-chloro, platinum, plus 4 is oxidation state. Here, we'll write down the positive part first. This part will name first. And in the complex part, we'll write down the name of ligand first and then metal. The ligand is hexa-amine, A double m, take care, hexa-amine, A double m, cobalt-3 chloride. One more point was there. I did not ethylene diamine on the point where we have single M in the spelling. Okay, I'll see that. Now you'll see, CR-H2O4Cl2Cl. This will write down first two ligands we have here. This H2O, the name is aqua, it is chloro. First to write alphabetically, aqua and then chloro. So tetra, aqua, dichloro, chromium, oxidation state of chromium is 3, tetra, aqua, dichloro, chromium 3, chloride. This one, it is amine-chloro. So di-amine double m, tetra-chloro, platinum, oxidation state is 4. Tri-amine, trichloro, cobalt-3. Okay, like this will name. Did you get it? Any doubt in this? A little bit of idea you have. I'll give you some examples now. Okay, we won't go like this. But did you get it this 4-5 point that I've discussed? Yes, sir. Yes, sir. We'll discuss some examples because if you have an example, you won't get it. Okay, let's discuss some examples. And I'll share this PDF. Okay, don't worry. Okay, now suppose the compound is this, which you have to write down the name. And the compound is this one, CO, NH3, 4, then H2O, CL, bracket close, CL. This is the first one. Second one is CU, EN2, SO4, Ni, CN4, 2 minus. I'll do the first one. So positive part first. The first we'll write down the complex part in the first one. That's a way. So we have this thing, 1, 2, 3, 3 ligands here. So this one is amine, this one is aqua, and this one is clodo. So first we'll write down A, Avala ligand, right? So A and A, amine, aqua, first you'll write amine, right? Alphabetically. The name of this compound is, this compound is tetra amine, T-E-T-R-A, and A double M, not single M, tetra amine, then aqua or water, then cloridlo, you can write or you can write clodo, cobalt. What is the oxidation state of cobalt? It's minus 1. So this one is plus 1. This is minus 1. This is 0. This is 0. So it is plus 2. Yes or no? Cobalt, then in the bracket, the oxidation state, chloride. This one I'll do in the last, wait for it. This one is tetra-sino, tetra-sino, nickelite, nickelite. Oxidation state of nickel is plus 2 here. So nickelite 2 ion, because it is not a molecule, it is an ion. So nickelite 2 ion. Okay. Now, this one you see the bis-triswala rule we'll use over here. The name of this ligand is what? Ethylene diamine or ethane 1,2 diamine? Okay, it is ethane 1,2 diamine. So the name of the ligand already contains di term with it, like ethane 1,2, di amine. So when the name of the ligand contains the term like di, tri, tetra or like that, then for these numbers, we do not use again di, tri, tetra 1. But we use bis-tris, tetra case like that. Okay. So if it is 2, so we'll run for this ligand, we'll write on bis bracket open ethane 1,2 diamine, ethane 1,2 diamine, bracket close copper. Oxidation state of copper is 2, copper 2 sulfate. Did you understand this? One more point that you have to take care of here, that when it is a mine, means more oriented ligand, then we are using double ampere. But when it is this oriented ligand, then we're using single ampere. If the complex part is positive ions and neutrals, then there's no special ending. We'll write down the name as it is. But when the complex is negative ions, we'll write down eight suffix like I have written here, other examples also I have shown you, it is eight suffix we have added. Got it? Some more examples I'll show you now. This one you do, COEN3, COONO, NH3SO4, ALH4, try these compounds. Yeah, I'll do it. This one you see, this one you doubt, this one. The name of this compound is ethylene diamine. For this three, we'll write down tris, we don't write try here because the name itself contains dietero. So we'll don't write try, we'll write on tris. Open bracket ethylene 1,2 diamine with a single amp, covalent. What is the oxidation state of covalent here? It is two, then space and sulphate. This one is the oxidation state if you see, it is zero and NO2, what is the oxidation state here? NO2, what is the charge? Minus one. So covalent will be plus three because it is minus two and on the complex we have plus two then. So this one is, it is pentamine and this one is nitrito O. ONO it is written, it means oxygen is the donor atom here. So nitrito O will write, if NO2 simply it is written, then nitrogen is the donor atom. Okay, so name of this compound is pentamine, pentamine, A double M, I, B, N, E, pentamine, then the other ligand is nitrito O, nitrito O, covalent, there's no space here. Okay, covalent three, so the ALH4 is lithium, what we should write the positive part first, lithium is fine but tetrahydride is correct, yes and lithium and then space tetrah, because it is hydride ion, right? So it is negative ion, so hydride ion becomes Ido, tetrahydride O, aluminate because of the negative ion. Oxidation state is, what is the oxidation state? This is three. It is, we have Li plus and ALH4 minus, it is a negative ion. So this, like this, we light on the name. Okay, I'll share that video, you'll see how some more questions will have the idea of it, how to do it. Just you need to follow the rules, okay? Sir, for organic reactions, this gives hydride ion, right? Sir, but the complex doesn't dissociate, so how does it give the hydride ion? In organic reaction, we have different conditions. We are talking, it's not like you cannot break this, it is not like you cannot break this. What we are talking about, we are talking about that when you dissolve this in water in aqua solution, then this part retains its identity. So I said what? That the complex part retains its identity in solution, right? What is organic reaction we have? We have elevated temperature, right? High temperature we are using. We are using this reagent in order to provide hydride ion, correct? So for that purpose, we use the necessary condition which is required. Yes, got it. It's just we are talking about, in solution it retains its identity. Yes, yes. In this nomenclature, you see, in some of the questions, they'll give you both, you know, part, ket ion and ion part complex. So in that case, you have to, you know, understand how to write down the name. Okay. Suppose we have this, not much important, but we'll see just one example. CRMH3 and the another one we have COCN6. So the name of this compound, I'll write down first. It is hexa amine because always we'll write down the positive part first. So hexa amine, chromium, chromium, oxidation state we let it be now, then some space, hexa amine, hexa-sino, cobalt, cobaltate, oxidation state we'll see. Okay. So oxidation is said, how do you assign? That's the thing, right? What do you do? You just find out the total charge. What is the total charge we have? NS3 we have zero, no charge on NS3, and CN has minus one. So total charge is six, like the magnitude six. So this charge you have to distribute equally on the two metal. So we'll write down three here and three here. This is the name of this compound. Try this one, one more. I'll just give you PT, NS3, 4, CUCL4. So it is tetra amine, platinum, 2, tetra clodo, cuprate 2. Is it? Yeah, that's right. The name is tetra amine, tetra amine, platinum, tetra clodo. If you're doing this, you know, these kinds of questions in the exam, then you don't have to name this like this. Okay. I would suggest, first of all, you find out the oxidation state of the metal, whether it is both complex or the normal one that we were doing before this, right? Find out the oxidation state first. Okay, especially when you look at the option, and if you see the oxidation state differs in all the options, then you just find out the oxidation state. According to that, you can remove the options easily, eliminate the options easily. Okay. So nomenclature questions, basically, it's not, you know, a good idea to solve this, this way, like, you know, you find out the name and all these things and like them the name and then compare with options. You can easily eliminate options there. Okay. Now, next thing is, sometimes what we have, we have bridging group ligands, okay. Bridge ligands are present. I'll show you how it is. For example, you see this one. Just we are looking at one example, because these things are not at all important. Okay. H2O4, then we have FE. And there is a ligand which acts as a bridge here, OH, OH. And it is connected to FE here, H2O4 times SO4, like this, SO4, so how do we write down the name of this? You can write down this as, you know, like, separately this part and this part, you can write down. Or you can also write down because you have two bridges like this. So we'll use bridges two. So we'll just die for that. Die. And we'll use a Greek word, mu, to represent the bridging ligand here. Mu means it is a bridge group ligand. Okay. We have, these things are written in the rules. Okay. It is not given in that particular PDF, because it is not important. Okay. But I'm giving you one, just one two example, I'm discussing so that you can do it. I have seen this kind of question. They have asked an example. Okay. Die, mu, mu represents the bridging group ligand. Okay. Mu. And then two hydroxy we have here. Right. So die means two hydroxy, mu means bridge. Then we'll write hydroxo, hydroxo. Then we'll write down, we have x204, x204 twice. So we'll write down bis, bis, the name of this is tetra aqua, tetra aqua, iron. Oxidation state of iron is, you see, minus one here, minus one here, minus two into two, minus four. So plus four here, plus four, plus two, plus six. We'll divide this plus six equally on the two metal. That is three. Okay. Bracket close, tetra aqua. I'll write down one bracket here also because this is for the tetra aqua. No, just a second. No, we have this one. One second. Die, mu, hydroxo we have done with this. And then this part we'll name it as bis tetra aqua. Okay. Bis tetra aqua, iron, C and then sulphate. It's the name of this. Mu, we use a Greek letter just to represent the bridging group ligand. Okay. I'll give you one point here, one note you write down. In this one note you write down. Right on the bridging ligand, the bridging ligand is expressed, is expressed by bridging ligand, is expressed by Greek letter, mu, repeating, I'm repeating myself, bridging ligand is expressed by a Greek letter, mu, which is added immediately before the name of the ligand and separated by a hyphen. Okay. No, die sulphate we don't put. Die sulphate we don't put. Simply we'll write down sulphate. Because when you have read bis tetra aqua, iron is there are two iron atom present. Right. To balance that we have, we should require two sulphates. So from there you got the idea that we'll have two sulphates and so that is not required. Sir, could you just repeat the last part of what you said? The point that in the note, right? Yes, sir. Yeah, I'm repeating myself. The bridging ligand is expressed by Greek letter mu, which is added immediately before the name of the ligand and separated by a hyphen. Okay. One more example on this will do and then we'll move on. You can have two different bridging ligand also. Here we have OHOH. So I've written di new hydroxo. If you have two different names, then you can write down the name of the two. Like you see this one NH3, NH3, whole four. Then we have obald, NH2, NH2, NH4, NH3, whole and then we have NH2. So what is NH2? NH2 is amido, you can write. NH2 is nitrito. Sir, the bridge is like three centred to electron bond kind of thing. Bridge one. Yes, sir. So because it's not a diidentated ligand, sir. Yes. Yeah, it's like three centred to electron, because NH2 already has only one lone pair now. So it is distributed among the three nucleus and COOCO. Okay. See, here we have two other things or two different bridging group ligand, right? One is amido, NH2, other one is nitrito, NO2. So we'll write down mu two times here. So the name of this compound you see, it is nitrito, it is amido. Alphabetically, we'll write this. So we'll write down mu amido. This means the NH2 group is present as a bridging bridge here between the two metal. Then one more, we have some mu, nitrito. Here we have a small n here, right? Small n. Then we have two times here, tetra amine. So we'll write down this open bracket, tetra amine, cobalt, what is the oxidation state? What is the charge? Nitrate. Tell me the charge here. NO3 minus, NO3 minus. So four minus. So four plus on this, right? Minus one, minus one. Plus six divided by two. Plus three we'll have. So we should write nitrito n also. Nitrito n. I know why we'll write, because it's a molecule, no? So no, sir. Nitrito n I meant for the bridging ligand. Nitrito. Oh no. See, you can write, that's not wrong. But when nitrito you have written, it means it is understood that nitrogen is already in it. Yes. So these are some bridging group ligands. Okay, not important. Okay, but I have given you a bit of test of it, like if you get, you can easily do this with the help of auctions. But I haven't seen the past five years they have asked questions. Okay, fine. So next is we'll start the bonding in coordination compound. Okay. How the bonding takes place in coordination compound. We have half an hour, so we can go with it. Fine. So right down next, bonding in coordination compound. There are different, different theories for this. Like we have in normal compounds, I told you, VBT, hybridization, molecular orbital theory. So here also we have different, different theory. Okay. We have, we have balanced bond theory here also, Werner coordination theory we have here. Like we have crystal field theory, splitting of orbitals. There are many things. Okay. That we are going to discuss here. So bonding in coordination compound. So the very first attempt to explain the bonding in coordination compound is done by Werner's Werner and we call it as Werner's coordination theory. Write down into this. It explains the property and structure. It explains the property and structure of various coordination compound, various coordination compound. Write down the main postulates of this theory are the first point you write down. According to Werner's coordination theory, the first point in this first postulates every element, metal means every element, two types of valencies, two types of valencies. That is, we'll have primary valency, primary valency. First you write down two points into this primary valency. And we have, we also have one more. You can write down three, four lines after three, four, five lines. We can write down secondary valency. These are the two valencies we have in coordination complex, primary and secondary. Write down primary valency. It corresponds to, it corresponds to the oxidation state of the central atom. CMA write down, central metal atom. It corresponds to the oxidation state of the central metal atom. And it is shown by dotted line. And it is shown by dotted line. Next point in this you write down. The primary valency is satisfied by, is satisfied by the neutral mon, sorry, satisfied by the negative ions, negative ions such as CL minus SO4 minus sulphate ion, etc. Sir negative? The primary valencies are satisfied by the negative ions, CL minus sulphate ions, all those things. So in the complex, if the negative ion is present, that those negative ions, only negative ions, okay, those negative ions satisfies the primary valency. And we represent the negative ions with the dotted line, because I have given you that it is represented by the dotted line, okay. So negative ion is represented by dotted. I'll draw the structure also, okay. Secondary valency write down. It corresponds to the coordination number. It corresponds to the coordination number, the central metal atom or ion, pointwise write down, okay. First point I've given you, second point. It satisfies by, it satisfies by either neutral molecules or negative ions, either neutral molecules or negative ions. What do you, like, what is the meaning of satisfies by, I'll discuss that in the last, first you write down the nodes, satisfies by the neutral molecules or negative ions. Till now, you see the negative ions satisfies both kind of valencies, primary as well as secondary, okay. Next one. Sir, can you repeat the first point? First point, it corresponds to the coordination number of the central metal atom or ion. So this one is coordination number. See this, this one is, this one is coordination number, but this one is the oxidation state of the metal, right. Second point I've said satisfies by either negative ion or neutral molecules, it is satisfied by only negative ions, okay. It is represented by solid line, like this. This one is represented by dotted line, like this, write down. It is represented by thick lines or solid lines, you can write. Next point. Secondary valencies are directional, which I did not give you this point there, okay, fine. Write it down. This is directional, you have to place this in a fixed direction, okay, with a fixed angle. Secondary valency, valencies are directional. Primary valency is non-directional, non-directional. This one is directional and because of this only, the isomerism is possible in coordination compound, okay, because you have to, you have to place the atoms, ligands around the central metal atom in a fixed direction at a fixed angle. That's why it shows isomerism. One more point here is this primary valency, it is because of the ionic bond. It is the ionic bond. The primary valency means the ions which are involved, we have ionic bond here in primary valency. But in this one, we have coordinate bond. Secondary valency, we have coordinate bond. Yes, right, non-ionical also. Sir, you said that primary valencies are satisfied only by negative ions, no sir? Yes, right. Sir, within in complexes like K4, FECN6 or something, K potassium, which is positive ion is satisfying the primary valency, no sir? No, it's not like that. I'll take one example, two minutes, okay, I'll take one example. Okay. Next, see, because of this directional in nature, so those, you know, ligands which satisfies the secondary valency, you have to place this in a different direction, in a fixed direction. And that's why we'll have a coordination polyhedral. You have a shape of it that we call it as tetrahedral geometry or square plane geometry and all those. That comes because of this directional nature of secondary valency. Okay. Now, these are the few properties you should keep in mind that primary is ionic bond, oxidation state, non-directional dot line, secondary valency is coordination number, coordinate bond, thick line, and directional in nature. Now, we can understand this with one example. I'll just write down this example here. Suppose I'm taking an example of CO NH3 6Cl3, this one. Could you tell me the primary valency for this one? What is the primary valency? What is the secondary valency? What is primary valency? We have to check oxidation state. Check the oxidation state. Primary is 3 and secondary is 6. Sorry? Maybe 3. Primary is 3 and secondary is 3. Primary is 3 because oxidation state is 3, right? So primary is 3 and secondary valency is 6 because the coordination number is 6. Like this, you'll get the primary valency and secondary valency. Now, if you want to draw the structure of it, the structure you can draw this way. You have cobalt in the center and we have 6 NH3 and this satisfies the neutral molecule and this satisfies the secondary valency because secondary valency is 6 molecules and this satisfies this. So we have 6 NH3 which is represented by the thick line 1 coordinate bond. This is 6 NH3 by the thick line. This is 3 chlorine which is present outside and ionizable. This is because we cannot break this complex. This is ionizable. So it is represented by the dotted line which is this one. This is this molecule. Now, what I'll do? I'll do one change in this. Suppose I am taking this one. CO NH3 CO NH3 5 CL and then CL2. Number of in this molecule, number of ionizable chlorine atom is what is 3 which is outside the in this one. Again, if you count the primary valency, primary valency is the oxidation state. That is 3 only. Secondary valency is again 6. So it is same here. This chlorine is a negative ion. It has a dual role here. It satisfies primary valencies as well as the secondary valency. It has a dual role and number of ionizable chlorine atom is 2. The ion which is performing dual behavior, that is not ionizable because it is inside the square bracket. We cannot ionize the chlorine atom which is present inside the square bracket. Number of ionizable chlorine atom is 2. Fine. Did you get it? Now, when you draw the structure of it, so how do we draw it? You see we'll take a cobalt here in the center. I'll just write down a line here like this. NH3 consider this as the coordinate only. We have only 5. So we'll have this and NH3 here. Then 4 and 5 and here we have suppose one chlorine. So dotted line is for primary valency chlorine and since the chlorine atom satisfies both valencies, so there is one thick line also. Means it satisfies both valencies and then we have one primary valency here and one primary valency. This one satisfies both valencies. Right down here. Satisfies both. Primary valency and secondary valency. Excuse me sir. Yeah. Sir in the first structure while writing the chlorine atoms, does it like show how should we know like between which ammonia molecule should we write the chlorine and all? Anywhere you place. It is non-directed. It doesn't matter. Okay. Yes sir. Wherever you feel like, it is non-directional. So there's no way. Like but for this it is important and because of this only it has octahedral geometry. Right. Square and then one top and one the bottom. Right. You know structure we are trying to draw but yes the dotted line you can place anywhere. Not a problem. Understood this one. Now I'll take one more example in this same one. This one you see. CO, NH3, 4, Cl2 and one chlorine atom outside. So if you count here the primary valency is again 3 and secondary valency is again 6 here and the structure is 1, 2, 3, 4, 4 NH3 and we have 2 chlorine atoms which has the dual role. Right. So chlorine and then one chlorine will have. Number of ionizable chlorine is only this one. Right. Reds say I will circle it. I will take one more. The same thing will place all the chlorine atom inside. CO, NH3, Cl3 like this. So here also we have the same thing. Primary valency is 3 oxidation state. Secondary valency is coordination number. When you draw the structure here, the structure is this CO, NH3, NH3 and chlorine is here. 1, 2, 3 dual behavior. So both times you draw. Chlorine, chlorine and chlorine and none of the chlorine atom is ionizable. Number of ionizable chlorine atom is 0. Number of ionizable chlorine atom is only 1. The chlorine and NH3 have to be alternate. Sorry. So the chlorine and NH3 have to be alternate? No, no, no. See, it's don't try. See, just you try to understand that how do we represent secondary and primary valency? Structure will do once we finish the valence on it. What is the hybridization we are getting? And according to that, what should be the geometry and what is the placement of the ligands? Okay. Coordination number is 6 and coordination number 6's geometry is octahedral. That I'm telling you. Okay. We'll discuss that. Okay. So octahedrally you have to place. That is it. Yes. Okay. So this is the structure. Now K4F is CN6. We have 6 cyanide ion. Right. So 3, the oxidation state is 2. Yes, sir. So those cyanide ions will be dual behavior. Okay, sir. Understood. Got it? Yes, sir. Okay. Now on this, one last thing we'll discuss. Okay. This way, a question like they asked about the molar conductivity of these ions. They'll give you 4 or 5 options and they'll ask you to arrange them in molar conductivity. Molar conductivity we know it depends upon what? It depends upon the number of ions. Correct. Number of ions in solution. Right. So we'll see how many ionizable protein atom is there. Correct. The first one, how many ionizable protein atom, how many ions we'll get in the first case? This one. Okay. After this. How many ions we'll get in the first, in this case, you see, in this case? Four ions. We'll get four ions. Correct. Because we have three Cl minus and one this one complex. Correct. This positive and this three Cl. Four ions we'll get here. Here we'll get three. And then we can count the others also. Right. So the molar conductivity will be maximum for this one and this one. And then we can go for the other. Did you get my point? Right. So the number of ions you count, according to that molar conductivity, you can compare. They can also give you like this. Like this one, it reacts with A, G and O3. So in which of this precipitate forms or where we get the maximum precipitation. So here we get three modes of agCl precipitation. Here we get only two agCl, then three agCl. In the last one, we have no precipitation. So precipitation based question also they ask in this. You just need to keep in mind the ions which are present, atoms which are present outside the square bracket are ionizable. And according to that only, we consider the reaction or the number of ions that we get in the solution. Understood my point. So this is the coordination theory. He said what that there are, you know, two types of valencies and a neutral ions. He talks about oxidation state, coordination number and many things. But there are two drawbacks also in this theory. He could not explain why the secondary valency is directional. Right. The optical properties of molecules he could not explain. That's the one drawback. Okay, it's right on the drawback of Werner's coordination theory. The drawback of Werner coordination theory. He could not explain, he could not explain the magnetic and optical properties of the complex. The magnetic and optical properties of the complex. Main drawback is this only. One more drawback was there that he again could not explain why only certain elements shows remarkable properties towards forming the complex, the coordination complex. Okay. But the main thing is this only could not explain the optical properties, means why the secondary valencies are directional. He could not explain that and the magnetic properties means obviously there are a number of electrons, unbeared electrons were there. Right. So now to overcome this, these problems, this drawbacks, we got another thing that is we have after this valence bond theory, then we have crystal field theory, CFT, ligand field theory we have. Okay. Molecular orbital theory also we have into this one. There are many theories after this. Okay. But our syllabus is we have to do only VBT and CFT that is valence bond theory and crystal field theory. Right. So after valence bond theory, you will be able to understand that structure of the molecule, the hybridization kyao raya and I got into that what could be the structure of the molecule and how the elements, you know, the ligands are placed around the central metal atom in order to in the structure. Okay. So this VBT we'll discuss in the next class. Right. We don't have time now. I can't start it. Right. I would request all of you that chemical bonding I have explained the valence bond theory that you must go through once before the next session. Understood guys? Yes sir. Yes sir. Yes sir. Okay. Fine. So we'll wind up the session here. See you in the next class guys. Thank you. Thank you sir. Thank you sir. Thank you sir. Thank you sir.