 Okay, shall we start guys? Yes. Yes, sir. Yeah, so tetrahedral complex we have this Kind of structure right elegance are present at the corner of the tetrahedral We are going to see that splitting in orbitals how the orbital is fixed in tetrahedral Okay, so obviously you see the bond angle is 298 minutes along any of these lines. We cannot assume any access right But if axis is there it will be something like this it is present 90 degree like this suppose this is the axis we have x y just a rough Diagram I am trying to draw right so obviously what we can say the ligands are approaching the metal It is not along the axis but it is in between the axis right in between the answers you see obviously the non Exial orbital that you have in this case the non axial orbital experience more repulsion and hence it goes to the Higher energy level so if you look at the splitting in this orbital the splitting is like this see It's just the ulta reverse of octahedral complex All the you know the excess is same Y axis and x axis the t2g orbital goes to the higher energy level and EG comes down to the lower than the energy level. Okay, everything is same Excess and all everything is same and in this again if you if you have to distribute the electron all those things You have to consider you build a strong build one soon and soon, right? Energy gap everything but those things are not there. Okay, you don't have to think about this if it is there You have to consider like that Now the splitting here is given as delta T. You see T stands for tetrahedral This is delta T. Okay, and there is a relation of delta O and delta T And this relation is established Experiment and the relation is delta T for tetrahedral is equals to four by nine times of Delta four by nine times of Delta This is experimentally determined no proof nothing many times they have asked this question Right, so you must have to keep this from this only you can understand easily that delta T is of less than Delta This is also So yes, tell me So this relation is for a particular pair of calm for the particular complex. Yeah So, but then how will one complex of octahedral and tetrahedral? see if it forms like See we can have Six forms a for a PCL six forms like there's two different types of arrangement if it is there Then we have this kind of relation tetrahedral complex and and if you know the different Legans obviously if the ligands are so different the two different molecules then we cannot Compare these two right because the different molecules are there, but yes, if you have some compounds like For example, if I take F e F4 we can have different charge on this no on the complex and we can have different charge on this With this or F e This charge may be different on this and then it can show the you know the complex Okay So this is tetrahedral complex right every other thing is same you don't have to think much about it like Because we have discussed a lot of things in octahedral complex those things we can apply here also, but that is not So just let it be now if you have a square cleaner complex then how the orbitals Did you draw this all of you? What is that on the top non-axid orbiter? there Non-axidor experience more repulsion Okay, not a square planar in a square planar obviously you see the Structure is this and Since it is square planar so all the ligands are along this line, right? There is no Ligands along DZ square orbital middle we have here So one thing is sure that the orbital DX square Y square experience the maximum repulsion, right? Like in octahedral the repulsion of DX square Y square and DZ square is same But here DZ square won't have similar repulsion like DX square Y square So the T2g orbital also splits into two parts So a eg orbital also is splits into two parts and to T2g also splits into two parts because of the different repulsion in different So let me show you the splitting here. Then you will understand it how it happens. So a square pyramidal the Splitting is like this. See what happens here? Since the ligands is approaching the metal along this this two orbital DX square Y squared So if this one obviously have the maximum repulsion, so this is the maximum energy Right in between this two if you compare DX Y and DZ square, right? The DX Y orbital is in XY plane and ligands are approaching along XY plane only That's why the DX square DX Y will have more repulsion than DZ square DZ square and DX Z Y Z you see obviously it is an XZ plane and it is in YZ plane right perpendicular to XY plane But DZ square, you know this orbital has an electron cloud in the XY plane. Yes or no? Yes, this is DZ square It has the electron cloud in the XY plane and that's why this Orbital experience slightly more repulsion than DX Z and D Y Z Hence the splitting is this They haven't asked question on all this, but yes, you should know the at least the order of this Which orbital should have the maximum energy and will have the maximum energy, which has the minimum energy The ligands along this line, we do not have ligands right in comparison to the octahedral complex. Yes Okay Now the first part of this chapter that is coordination compound is almost done We are left with only last, you know part of it the first part then after this will start with Isomerism in coordination compound the last part of this chapter So we are left with only organometallic compound Organometallic compound if you remember we have discussed it in organic chemistry also when we are doing Grignard reagent I said that this is the this is an organometallic compound Organometallic compounds are those compounds in which we have the bond between carbon and metal Okay, so there are three types of organometallic compound Right RMGX is one that we have a specific name Name for that tell you what So first of all it I don't it is just a theory You just need to know a few structures and two names that is it nothing much you have to keep in mind Right on these are the compounds in which These are the compounds in which Carbon metal bond is present Bounds in which Carbon metal bond is present. This is classified into three categories The bond is present classified into three categories. The first type we have Sigma bonded complex Sigma bonded complex okay Simple in this type of complex that there's a sigma bond between the carbon and the metal Okay, so carbon and metal joined with the sigma bond carbon Metal and hence we call it a sigma bonded complex the example we have RMGX Grignard reagent We can have a L2 CH3 Six, this is also sigma bonded complex Right electron deficient LCS3 Dimerizes this is the example Next exam next type we have pi bonded complex in pi bonded complex obviously pi electrons are involved with metal Okay, pi electrons are involved with metal. These are the complex These are the complex in which in which the metal ion Metal ion is bonded with bronze and We can have pi electrons of alkene, benzene, alkyne and any other molecules Okay, pi electron clouds By electron clouds for example, you see these are structures that I'm going to draw these structures are important Okay, they have asked these things like that. Okay You see this we have a five-membered ring five membered ring metal is iron And this side we have another five-membered ring this pi electron cloud it donates to this metal This is the bonding we have this compound We call it as ferrocene all these name you have to memorize ferrocene we also represent this as metal is iron So we write here this one see this Right here fe Then a Greek letter ETA ETA Right ETA dash For one molecule we have C5H5 the structure of the ligand. This is the ligand we have C5H5 we have two times the whole two and We'll write a number on the top of it ETA that number is the number of carbon atom in the ligand Okay, that is five carbon so five we have Once in need they have asked this question They have given the structure and ETA to the power x they have written x ETA to the power x and the question was what is the value of x? Okay, x here or the number here is the number of carbon atom in the ligand One atom you can see or number of atoms in the ETA to the power 5 the name is ferrocene representation One more example we have in fact to we have Five electrons of benzene you see this chromium and this attached with this we call it as dibenzene chromium benzene chromium and Representation of this is CR for chromium ETA to the power 6 C6 H6 Okay Another example you see we have a cyclic compound platinum here. We have chlorine chlorine chlorine metal is this and this metal The Legand the another the fourth ligand is this one Carbon with two hydrogen CH2. This is the carbon with two hydrogen This is a complex part and The name of this compound is Jesus salt G salt That representation is the cation first Metal is platinum cl3 3-chlorine atom and this is ETA C2 H4 So the pi bonded complex that you have are the ligands which pi electrons contains here You see CLT 3 ligands have written simply but the pi electron ligands we used to represent with this third type of organometallic compound we have and We call it as Sigma and pi bonded complex Sigma and pi bonded come right on these are the complex these are the complex in which These are the complex in which gone between and carbonyl compound carbonyl group in the metal and carbonyl group Is present basically the metal carbonyl compound in that only we have sigma and pi bonded complex Right the oxidation is state of the metal is zero right this one is also important on this also They have asked question many times. Okay, right on the oxidation is state of the metal is zero for example You see and I see a four and I see a four Since the ligand is neutral So oxidation is state is zero CR CO6 CR CO6 in this type of compound only the Sigma and pi bonded complex we call it. Okay, the important point here is I'll give you one note here on This note they have asked question many times in the exam They generally asked which a state point is correct regarding this kind of complex like this, okay Then this note everything is there and you can understand it. I'll explain the note also later on once you finish it right down first In this type of compound metal and carbon atom in this type of compound Metal and carbon atom Both sigma and pi character both sigma and pi Sigma bond is formed sigma bond is formed by bond is formed by The donation of lone pair of electron donation of lone pair of electrons nation of lone pair of electrons of Carbon lone pair of electrons of carbon the vacant orbital bone pair of carbon to the vacant orbital of metal To the vacant orbital of metal pi bond forms the pi bond forms due to the back bonding bond forms Due to the back bonding of a fully filled D orbital fully filled D orbital the vacant anti-bonding orbital carbonyl group to the vacant Anti-bonding by orbital to the vacant Pi orbital the carbonyl group excuse me, sir Sir if a pond is being formed by the donation of electrons from Carbon, then how is it a sigma bond? I mean is it a Did you copy this? Yes, sir. Okay, so you see how it happens First of all, you see this carbonyl group co. How many electrons it has? Plus six We have 14 electrons right and for 14 electrons if you draw the configuration according to MOT molecular orbital This would be Sigma 1s sigma star 1s Sigma 2s Sigma star 2s pi 2p y pi 2p z Sigma 2p x Then pi star 2p y pi star 2p z Then sigma Star 2p x. This is for n Less than equal to 14 if it is greater than 14 14 to 20 Then we have to interchange this This will come over here and this two will come this way. There's two we need to interchange that is only Difference we have. Okay. Now since you have 14 electrons if you fill this 14 electrons two four six eight nine ten 11 12 14 You see this pi First of all it starts these are anti-bonding orbital site anti-bonding pi, which is vacant anti-bonding pi orbitals The metal that donates electron back in back bonding it donates electron into the pi star orbital Anti-bonding orbital. So if you look at the metal, you know, which has the five electron Suppose I'm just rough diagram. I'm drawing Suppose one orbital is this another orbital is this and This is metal right One electron pair of metal which involves in back bonding and carbon monoxide is this Sorry carbon is look carbon monoxide is C triple bond O Oxygen has one lone pair and the carbon has one lone pair in its orbital One lone pair of carbon and this is the pi star orbital anti-bonding So the sigma bond forms for that Carbon donates its lone pair of electron into the vacant orbital of the metal This forms a sigma bond and pi bond forms since it is a pi star anti-bonding orbital So from the metal the D electron Donates its pair of electron into the vacant anti-bonding pi Orbitals of this carbonyl group. This forms the pi bond. So pi bond forms by the back bonding of pair of electron of metal into with the anti-bonding pi orbitals of carbon Carbonyl group and sigma bond forms by the donation of lone pair of electrons of carbon to the vacant orbital of metal This is we have in carbonyl compound Yes, you tell me now. Yes, sir So what are actually coordinate bonds of D? Yes coordinate bond are like no It's a type of covalent bond only Yes, right Why they have done this change is here because the most of the compounds which involves coordinate and covalent bond They have the similar properties That's why they merge the two Yes Got it. So this is the note that I have given you on that they have asked question many times in J. E. O. In which statement is correct or which statement is you know incorrect like that? So all these whatever we have done There are so many questions possible Like small small points when it is high spin complex when it is low spin complex or low spin complex What is the nature of the ligand big ligand or a strong ligand all these are kind of questions They have asked many times Okay, so this is the first part of this in our chapter The second part which is the last part of this chapter we have which is that is nothing but the isomerism in coordination Okay, so coordination compound shows both kind of isomerism is structural in a studio Studio part is the only which is important Right, but they have asked question from a structural also in a studio We have geometrical as well as optical isomerism possible, right? So we'll start with this Write down isomerism in coordination So I had one small So why can't it be why can't the oxidation number not be zero? Which one where for the metal carbonyl bond Yes, sir, I'm asking why can't it be zero? I mean why can it be? Why does it come to zero? Why does no, huh? See and I see I have given you the example and I see you the code This is a complex Metal carbonyl compound we have CRCO6. There's no charge. You see This ligand is neutral There's no charge on the complex that it is stable also nickel. How many electrons? What is atomic number of nickel? 20 28 28. Yes, eight full ligands, right? 4 into 2 8 electrons it provides 28 plus it 36 It is a stable. Why why you know without any charge only to the stable and it is not So, but what if it's something else like you had given examples that there was like VCO6 That is different. We have metal carbonyl compounds plus some other ligands also present Yes, so that is a different case here. We have metal and carbonyl compound Okay, that's why you see I have given the examples of this tool. Yes. Okay Yeah, so isomerism in coordination compound two types of isomerism I'm just not going into detail of all this because you know what is isomerism and all the two types of isomerism It shows the structural and stereo, right? So first we'll see is structural isomerism You'll also we have certain, you know definitions and then examples of it Stereo isomerism we have to see how to draw the different isomers and how to find out the given is the drawn Instructure is different from the other There we have to understand of it, but here just we have definitions and then examples of that like we had in the In the isomerism chapter. Okay So the first types of structural isomerism we have we call it as ionization isomerism Definition write down very simple definition on ionization. It gives different ions are called ionization isomers Basic thing is same like the structural formula must be same Without that we cannot even imagine of isomerism, right? So the formula must be same the molecular formula of the compound, right? If they if on ionization they give different ions then they call I just an isomerism, okay Write down these isomers have same molecular formula. These isomers have same molecular formula and produces when the ligands and the outer atoms means outside the coordination sphere The ligands and the outer atoms are interchanged. For example, we have CO NH 3 for CL 2 BR Okay On ionization. It gives BR minus and this plus if I write down the another Molecule of this which is CH 3 and it's CO and it's 3 4 cl br And one chlorine outside So both has the same molecular formula, but on ionization they give different ions and hence these are ionization isomers Okay, second type is hydrate isomer is obviously this phenomenon is isomerism Hydrate or we also call it a solvent Isomerism down. This is a special type of ionization isomerism in which the water molecules are involved in which water molecules are involved CR H2O 6 cl 3 outside present and CR H2O 5 cl Cl 3 5 cl Cl 2 So what a molecule is present outside. Okay, so this is hydrate isomer is Solvent isomer is Sir, could you go back just a minute? Yes, sir. Thank you, sir. Next is linkage isomer is Or what this is possible with embedded ligand simply the linkage is different. Okay, right down this type of isomer is this type of isomerism is theoretically possible is theoretically possible possible in any coordination compound in any coordination compound containing an Embedanted ligand containing an embedded ligand example very simple one C O NH 3 5 NO 2 2 plus and C O NH 3 5 O N O linkage Oxygen is a donor atom nitrogen is the fourth one coordination isomeries coordination isomeris write down The compound which has the compound with had which has we dares both cate ionic and an ionic complex ions which has Both cate ionic and any complex ions the compound which has both cate ionic and anionic complex ions shows this type of isomerism, shows this type of isomerism and this is obtained by and these isomers are obtained by and these isomers are obtained by by the interchange of ligand in both ions. For example, we have CuNH3 for PTCL4, this is the complex. Could you tell me the charge on this complex and this complex? Plus 2, plus 2, minus 2. Plus 2 and minus 2 because the total charge is 4 will divide equally plus 2 minus 2. Now if you write down the coordination isomers of this, what you do is exchange the ions here, the ligands, sorry. CL here and PTCL3 NH3, this is another molecule here. Here we have plus 1 charge and here we have minus 1 charge. Total charge will be plus 2 and minus 2 always. Further if you write down CuNH3, 2Cl2, PTCL2, NH3, CuNH3, Cl3, PTCL, NH3, whole price. Here the charge would be 0 and 0, 0, 0. Further here the charge would be minus 1 and here it would be plus 1. Here it is minus 2 and plus 2. All these possible combinations we can draw but out of this combination, out of this 5 combinations, this one is not possible. This one is not possible. Why this one is not possible? I will tell you. Because the charge on the ions is 0 and there is no attraction between the two part of it. If the charge is 0, this is also independent, this is also independent. There is no interaction. This as a whole is not a compound now because there is no interaction between the two part here. That's why this we cannot consider. This is not the possible isomers. The answer for this one is 1, 2, 3, 4 isomers are possible. We will write down one more last one. Sir, in the actual compound, will they exist as mixtures? No, we are just drawing it. The actual compound is this only. Whatever is written, that will be the compound. It won't exchange. It is not like isomers, it is not like resonance that things get changed. Shifting off ligands is not possible on its own. But if you write down the structure like this, it is possible. We are talking about isomers. We can write down the... Yes, sir. So one last type here. We will finish it. Write down next one. That is polymerization isomerism. This type of isomerism exists in compounds having the same stoichiometric composition. Same stoichiometric composition, but different molecular composition. Same stoichiometric composition, but different molecular composition. Same stoichiometric composition, but different molecular composition. PTNH3Cl2. And you see this one. PTNH34 complex part. And PTCl4 complex part. This too, if you see the ratio of platinum NH3 and Cl in the two compounds. It is 1 is to 2 is to 2. And here it is 2 is to 4 is to 4, which eventually comes out to be 1 is to 2 is to 2. So this kind of ratio, if it is same, different molecular formula, but stoichiometric composition is same. So this compound is polymerization isomerism. Did you get it? Yes, sir. So this is it for structural isomerism. Here we have only definitions and then examples of them. Next class is chapter will do studio isomerism. That is very important. We'll finish this and we'll start another chapter. So module, center module you can solve. I'll share tomorrow. I'll share one PDF also on this, which you can try. Okay. Yeah. Anything else guys? Okay. Thank you guys. Thank you, sir. Thank you, sir. Thank you.