 Can you hear me? Yes, so we'll wait for a few more minutes. Okay. For others and then we'll start. Okay. Five minutes we have. Semester exam or something. Is it? Okay. When it is? Day after. With subject. Okay. Chemistry. Where is chemistry guys? What is the portion for that? Sleapers for chemistry exam. Still chemical bonding. Is it? Which are just some matters there. Okay. So I think almost half of the 11th grade chemistry is there. Yes. Everything in physical chemistry except the equilibrium chapter. Right. So guys, let's start with the session. I think last class we could not have the last class. Okay. So I'll plan to have a compensatory class for that. Okay. We'll do that. So last class we stopped at hybridization. Right. We discussed how the VBD valence bond theory could not explain the bonding of a CH code. Right. And then I said that we got a new theory and that is hybridization. So today in the session, we are going to start what is hybridization and how to find out hybridization. Okay. So let's start this. I still, I think there are many more. So let's try. We'll wait guys two more minutes. Okay. Because I can see there are only 28 participants till now. Okay. So let me drop a message. Okay. Fine. We'll start now. We'll start with hybridization. So what is hybridization? Hybridization is the intermixing of orbitals. Okay. So first of all, you're right. So what is hybridization? Hybridization is the intermixing of orbitals. Okay. So first of all, you write down hybridization is the intermixing of orbitals, intermixing of orbitals, orbitals of same atom, orbitals of same. It is also the redistribution of energy redistribution of energy. So what happens in this in some of the molecules before the bonding, the atomic orbitals of one particular actor. Okay. Combines together and forms hybrid orbitals. Okay. That's why we are saying it as intermixing of orbitals. The orbitals intermix combines together. Okay. The overlap combines together and forms a new set of orbitals. This new set of orbitals is called hybrid orbitals and this entire phenomenon is hybridization. Right. So by mixing of orbitals, there is redistribution of energy. Right. And intermixing of orbitals also the same. Okay. So write down hybridization is the hybridization is the, is the process of hybridization is the process of intermixing of atomic orbitals of the central atom. I'll write down here. It is the process of intermixing of orbitals of central atom, central atom. Write down through hybridization. The energy is redistributed energy is redistributed and new orbitals form orbital forms, which is called, which is called hybrid orbitals. Right. Which is called hybrid orbitals. Okay. All hybrid orbitals, same shape or similar shape and energy. All hybrid orbitals must have the same energy. One more thing here, the number of atomic orbitals combines atomic orbitals combines equal number of, equal number of hybrid orbital forms, equal number of hybrid forms. Copy this stuff. Suppose two atomic orbitals combines the two hybrid orbital forms, three, then three hybrid orbitals, four atomic orbitals, then four hybrid orbitals and so on. Okay. Let's discuss one more concept in this chapter that is molecular orbital theory. Right. There is a basic difference between molecular orbital and hybrid orbital. Hybrid orbital forms. When, when the orbitals of same combines. Okay. Or intermix. Orbitals of same atom. But in molecular orbital. We have orbitals of different data. That's the basic difference. At this point, did you copy this down? All of you? Yes. Next slide down. One side you write down orbital mixed. The stable you draw. This is hybrid orbit. Okay. So what happens suppose when S and P orbital mix, obviously of the same right forms to SP hybridized hybrid orbitals. Okay. Okay. When S and two P mix three atomic orbitals are mixing. So we'll get three SP two hybridized hybrid orbitals. When X, sorry, S is mixed with three P orbitals. Then we get four SP three hybridized hybrid orbitals. When X mix with three P and one D, then the hybrid orbital forms number is five and the hybrid orbital is SP three D. If it is one S, three P and two D, then the hybrid orbital forms are six SP three P. So this is the hybridization we have based on the number of atomic orbitals combines. You copy this down. Don't leave this. Yes. DSP two is also possible. That is a square cleaner geometry in this chapter. You don't, you don't have use of that. If you want to write, you can write it down. One D orbital, right? One S orbital and two P orbital. This gives you two, three, four DSP two hybridized hybrid orbital. Even D two SP three is also possible. Right. This kind of geometry and hybridization you will get in grade 12 when we study coordination compound. Okay. So this is the thing. Now there are few characteristics of hybrid orbital. You must remember. Okay. A few points I have given you already. You can continue in this also. Uh, next write down. I have given you already all hybrid orbitals must have the same energy. Right. Next write down like atomic orbitals, like atomic orbitals, hybrid orbital can also accommodate maximum drugs. Means in one hybrid orbitals, we can have maximum two like we had in atomic orbitals. Now this point is very important. This one. Like hybrid orbitals, hybrid orbitals forms only Sigma bond may contain lone pair. If you remember in organic chemistry, right? We had discussed a bit of hybridization and I told you that we'll see this in chemical bonding chapter. If you look at this structure and if I find out the hybridization of each carbon atom here, right? If you remember, I have told you that we just need to count one more thing I'll do to one lone pair on nitrogen. So if I have to find out the hybridize this one I have done already in organic chemistry. If you remember, I told you that you just count the number of Sigma bond and lone pair. Okay. Like you see for all these carbon atom, we have one one hydrogen attached, right? One hydrogen here, one hydrogen here and there is no lone pair on these carbon atom. Could you tell me how many Sigma bond we have here? This one we have done, right? How many Sigma bond we have this carbon atom? Could you tell me how many Sigma bond? See Sigma bond. Do we have any lone pair on the carbon atom? Do we have any lone pair on the carbon atom? No, right? That's why the hybridization is what? This carbon is sp2 hybridized because it's three Sigma bond. Have you done this kind of question in organic chemistry? All please respond. We have done this, right? So here you can understand the reason that why we are counting only Sigma bond and lone pair. Okay. Yeah. You can understand the reason here. Why we are counting only Sigma bond and lone pair. So reason is what? Because the hybrid orbital that forms this hybrid orbital forms only Sigma bond. It can never form pi bond here and it may contain lone pair. This is the property of hybrid orbit. Okay. We'll discuss more examples into this one. But for just for this reference, if I ask you to count the, to find out the hybridization of each carbon atom here, all the carbon atom in the ring is sp2 hybridized. Yes or no? Is sp2. This carbon is also sp2 hybridized. Okay. If you look at this carbon, this carbon is sp3. This carbon is sp2 because of three Sigma bond. This carbon has three Sigma bond, sp2. This carbon has three Sigma bond, sp2. This nitrogen has three Sigma bond and one lone pair. So it is sp3. Any doubt in this? You can tell me. All of you. Any doubt in this? Actually we haven't done anything new over here. We have already discussed all these things. Okay. That example just I've given you here so that we can understand that why we were counting only Sigma bond and lone pair if the structure is given in order to find out the hybridization of each atom. Okay. Remember, it is hybridization for one atom. Is this carbon atom or this carbon atom or this carbon atom? Hybridization is not defined for molecule. Okay. Keep that in mind. Okay. Now the next, you know, property here you write down. If there are, if there are five bonds to be formed, then equal number of atomic orbitals, atomic orbitals must be left, must be left unhybridized. Okay. Next one is the last one. Geometry of the molecule. Molecule can be determined, can be determined by knowing the hybridization. It's very simple to find out. Hybridization. It's very simple to find out hybridization. But if you want to understand the actual thing, you have to give a little bit of time. Okay. I can, you know, finish this hybridization in two minutes, two more minutes. I'll be able to finish this. Okay. But I'm not giving you just a small short trick we have here, which we use in solving questions, but I want you to understand the actual concept and then we'll see the trick. Okay. So first we'll discuss the actual concept. See, here what happens, according to the hybridization, we can find out the geometry. If hybridization is SP, geometry is linear. If hybridization is SP2, geometry is trigonal, trigonal planar. SP3, geometry is tetrahedral. Right? SP3D. Then the geometry is trigonal bipyramidal, TBB. SP3D2. Geometry is octahedral. And octahedral is nothing but same as square bipyramidal. Now you'll see the shape of the hybrid orbital. You don't have to, you know, memorize this, because if SNP combines or if PNP also combines atomic orbitals, the hybrid orbital that we get is this one. This is the hybrid orbitals we get when the atomic orbitals combines. Yeah, pentagonal bipyramidal, if you ask me, it is SP3D3. The geometry is pentagonal bipyramidal. Not very important, but you'll get there and it is SP3D3. Okay? So this is hybrid orbitals and these are atomic orbitals of same atom, obviously. Atomic orbitals of same atom. Now we'll try to understand how this hybridization explains the geometry, sorry, the bonding of CH4. Because of that only, we have this particular theory of hybridization. Geometry of DSP2 is a square planar, Rujita. DSP2, geometry if you want to write down, DSP2 is a square planar. Okay? You won't get this particular thing in this particular chapter. But yes, if you want to write, you can write it down. DSP2 is a square planar. Okay? Now, look at this example of CH4. Because of this only, we got this new concept, right? Of hybridization. CH4 molecule. You see, carbon has six electron and when you draw the electronic configuration, 1S2, 2S2, 2P2, if you look at this orbital diagram, 2S orbital has two electron and 2P also has two electron. So two electron here and one here. This is the distribution of energy. In ground state. This is in ground state. But since this carbon is attached with four hydrogen atoms, it means we must require four unpaired electron. So in order to get four unpaired electron, this two electron of this orbital, it jumps into the vacant orbital of T and we get here one electron and three electrons we get here. One, two. Okay? Now this has one S orbital and 3P orbital here, right? Before it make a bond with hydrogen, this four orbitals, one S and 3P, goes under hybridization. All these four atomic orbitals mixed together and forms a new set of orbitals. So how many orbitals we get here? Could you tell me? How many orbitals we get here? Four. Yes. Because four atomic orbitals combined, so we get four hybrid orbitals. Each hybrid orbitals has one electron because here also we had one electron present. Okay? This is SP3 hybridized hybrid orbitals. Then what happens? The four S orbital of hydrogen makes a bond with this SP3 hybridized orbital and we get the CH4 molecule. So if you look at this structure here, we have four SP3 hybridized orbital. So the structure is this. SP hybridized orbital is this. One, two, three and four. Four SP3 hybridized orbital of this thing, carbon, which has one electron each as it is written here, one electron each. Now with this one electron, the S orbital of hydrogen overlaps and makes a bond. This is the S orbital of hydrogen. All these are S orbitals. One, two, three, oh, this is again. This is hydrogen. This is hydrogen. This is hydrogen and this is hydrogen. S orbital. What is this overlap? Could you tell me? This overlap is SP3 S overlap. And in fact, we have four such overlaps possible. Present here, right? Everywhere we have SP3 S overlap. Okay? That is how the bond forms in CH4 molecule. What happens here, you see these four bonds are formed by the overlapping of SP3 and S orbitals. All four bonds are identical and same. And that is how it explains the property of CH4 and hydrogen. Did you get it? So this is the theory of hybridization. Now, if you look at this, one more example. All of you have done that? Suppose we have BECL2. Hybridization we need to find. Okay? So beryllium, it is 1S2, 2S2, right? And the 2P orbital is vacant here. There's no electron present in this 2P. If I draw the diagram here, this is 2S and this is 2P. 2S has two electron and there's no electron in 2P. But it has to bond with two chlorine atoms. This two unpaired electron for beryllium is required. So in excited state, what happens? One of the electron from 2S of beryllium, it jumps into 2P. And in excited state, the structure is this. 2S has one electron and 2P has only one electron. Before bonding with chlorine, this S and this P orbital, which has unpaired electron, this two intermix and forms a new set of orbitals, which we call it as hybrid orbitals. This one, right? And these two orbitals will be left unhybridized. Okay? 2P orbitals right down here, 2P orbitals, 2P orbitals left unhybridized. And these two orbitals, the hybrid orbitals SP, because SP and SNP combines, has one electron each as it was here. Now, when you draw the structure of this, obviously, BCL2, the hybrid orbital is this. And chlorine, the electron is present. This one has one electron, this one has one electron. And the chlorine, the electron is present in P sub-shell. So P sub-shell is this for chlorine. P orbital, in fact, is this for chlorine. Which has one electron each. Okay? This overlaps here and forms a bond along this line. So it's a sigma bond, right? This sigma overlap and this sigma overlap. Okay? So hybridization for B is this SP in this molecule. Yeah, if possible, then they can form pi bonds properly. Here it is not forming, but in other cases, if it is possible to form, then they will form pi bonds. Tell me, did you get it? All of you? Now, when you take one more example, you will have a better clarity of this. Let's take an example of CO2 molecule. Could you tell me the hybridization of carbon in this? So I got your answer. Others, tell me the answer. No, Gayatri, that's not correct. Rohan, that's not correct. No, Stuti, it's not correct. Is it done in the class? Like the school class? Yes, Prakul, right. Rujitha, that's correct. Yes, Rohan, that's correct. Say it's very simple. Carbon has 6 electrons. So it is 1S2, 2S2, 2P2. 2 electron, 2S. And 1 electron, 1 electron here. But in excited state, what happens? 1 electron from 2S, it jumps into 2P. And we get 4 unpaired electron. So this one is ground state. And this is excited state. Okay. You know the structure of CO2? The structure of CO2 is O double bond, C double bond. So if somebody asks you what is the hybridization of CO2, you can draw the structure and you can find the number of sigma bond and lone pair from the central atom. That is only one sigma here and one sigma here, 2 sigma and 0 pi. So it is sp hybridized carbon atom. That is what you can say directly, right? But why I have drawn this structure? Because there are 2 pi bonds. Yes or no? Can hybrid orbital forms pi bond? No, right? But the central atom, carbon has 2 pi bonds. Okay. It means out of this 4 orbitals here, 1, 2, 3, 4 of carbon, out of this 4 orbitals, 2 orbitals must be left unhybridized, right? In order to form a pi bond. So what we say that this S and this P in 1P goes under hybridization and it forms 2 electron and the 2 atomic orbital is this. So this one here you see, these are the hybrid orbitals, right? Hybrid orbitals and these 2 are the atomic orbitals. Atomic orbitals, right? 2 pi bond we need to form here. The 2 atomic orbitals must be there unhybridized for the carbon atom. Now, we will draw the orbital diagram. But before that, we will see the electronic configuration of oxygen also. It has 8 electron, 1S2, 2S2 and 2P4. 1S2, 2S2, 2P4, 2S2 electron and 2P4 has 4 electron. So 1, 2, 3 and 4. So what point I am trying to make is for oxygen there are 2 atomic orbitals which has 1, 1 electron. This is what you need to keep in mind. This is 2S, here also 2S, here also 2P. This is also 2S, 2S and this is 2P. So 2S and 2P orbitals have 2 electrons unpaired for oxygen atom. This is for carbon. Now let us draw the orbital diagram. For carbon, there are 2 SP hybridized carbon atom. See here. SP hybridized carbon atom here and 2 atomic orbitals with 1 unpaired electron. Look at this. This is SP hybridized carbon atom of carbon. It has 2P orbital also, atomic orbital. So one of the atomic orbitals is this and other atomic orbital is this. This is for carbon. All these orbitals has 1, 1 electron. So here we have 1, here we have 1, here we have 1 and here we have 1. Now oxygen has 2P orbitals which has 1, 1 electron each. So one of the P orbital is this and another P orbital is this for them. One P orbital is this. The other one is this for one oxygen atom and this is another one for another oxygen atom. So this P orbital has 1 electron. This P orbital is 1 electron. So all these orbitals has equal electron, equal electron like this, forms sigma here. Now when this forms here, one electron this side, one electron this side, when this forms a pi bond between, we have a sigma bond here and this atomic orbital. These are, if you look at the hybridization here, SP and P overlap we have this side, to form a sigma bond. SP and P overlap to form a sigma bond. Now to form a pi bond between carbon and oxygen, these two orbitals overlaps. It forms a pi bond, these two overlaps and here also these two orbitals overlaps and these two orbitals also overlaps, right? So what happens here you see, this middle one, this middle one is carbon. This is oxygen with a single bond. This is a single bond we have and we have a double bond here, which is this. This carbon has a single bond this side and a double bond here. So this is a structure. And for oxygen you see it has two lone pairs here. If you look at this, this 2s and 2p electrons are the lone pair on oxygen atom and that you can place here. Did you understand this? You see the pi bond that is forming here, it is because of the overlap of atomic orbitals, not the hybrid orbitals. So I have started this thing of hybridization. Till now you have to watch the video, you won't understand it now, right? But what I'm going to discuss after this that you will understand. Guys, anyone has doubt in this? How sigma and pi bonds are formed? Please respond to all of you guys quickly. Another thing is what you see, this kind of question if you're going to solve in the exam, this type, that it takes a lot of time, right? You have to draw a lot of a structure and it takes a lot of time. So in the exam, we won't do questions like this, especially in the Cayman test or any comparative exam. If they ask you to find out the hybridization, they will either give you the structure like I've given you the first one, okay, benzene and other things are there, like that, or they will give you the molecule which we have here, right? So if the structure is given, fine, you can count the number of lone pair and a sigma bond, right? But if molecule is given like this, CO2 and other molecules are there, then you have to draw the orbital diagram like this and then you'll see which orbitals are taking part in the reaction which are not taking part in the reaction and then you can choose the orbitals which goes into overlaps, intermixing and forms hybrid orbital here. This is what you need to do. Actual method is this, by which you can find out the hybridization. But like I said, this, we won't follow in the test, right? Then what we should do? We are going to use the steric number rule in order to find out the hybridization. You see this? What is a steric number rule? It is a trick basically, steric number rule. What is steric number? Steric number is equals to what? In VSEPR, I have discussed what is steric number? It is a sum of bond pair plus lone pair. Yes or no? Sum of bond pair and lone pair. How to find out bond pair and lone pair? We calculate the valence electron. We'll divide it by 8. The quotient gives you bond pair and the remainder gives you number of electrons in the lone pair. That's why this is equals to Q plus R by 2. The same thing we have done in VSEPR if you remember that. Now based on the steric number, you can say the hybridization of the molecule. The best method is the steric number and hybridization. If the steric number is 2, hybridization is sp. If it is 3, it is sp2. If it is 4, then it is sp3. If it is 5, then it is sp3d. This is the hybridization. We have understood this. All of you have copied. If in the question, in the command test, they'll ask you to find out hybridization, we are going to use this method steric number. But actual concept is this. Orbital overlap you need to check and then you can find out the number of orbiters that take parts in the hybridization and then we can say the hybridization. The steric number rule we are going to use find out the hybridization of hybridization of I3- NO3- X3O2F2 CLF3 also you can write IO2F2- Try all these. Tell me the hybridization and geometry. That all of you. For I3- where is electron, and where is electron is 22 7 x 3- because one negative charge. Now if you divide this 22 by 8, we have 2 bond pair and d-loan pair. So steric number is 5. Hybridization is sp3d. Geometry is TPP. Shape is linear. Okay, shape is linear, geometry is TVP, NO3 minus, valence electron, is it 24? Yeah, so then 24 divided by 8, so we have 3 bond pair and 0 lone pair, so it is 3 SP3D, sorry, it is 3, so it is SP2, geometry is trigonal planar, yes. Geometry is trigonal planar, next. XeO2F2, valence electron is, is it 34? So 34 by 8 gives you 4 bond pair and 2 lone pair, sorry, 1 lone pair, 32, 1 lone pair, and then we have 5 steric number, it is SP3D, okay, TVP. The geometry here, this one is, valence electron is 28, 28 divided by 8 is 3 BP and 2 LP, 5 SP3D. For this one, 34, right, so 34 divided by 8 is 4 BP and 1 LP, 5 SP3, any doubt in this? Yes, Rujita, one second. Rujita, do you have the notes for VSEPR? Have you written that table? Right, I want you to go back and look at this combination, for electron pair 5, this combination 2 and 3, what is the shape over there? Yeah, same only, means with this particular thing, you can find out hybridization, you can find out geometry, you can find out shape, so everything VSEPR, hybridization, VBT, it ends up over here, any given molecule if you have, okay, find out valence electron divided by 8, you'll get bond pair and lone pair, that gives you the steric number, right, with the steric number, you can write down hybridization, hybridization has a unique geometry, right, depending upon the bond pair and lone pair, you have the idea of shape of the molecule also. Second one, NO3 minus it is, so nitrogen has 5 electron, oxygen has 6, right, 6 into 3, 18 plus 5, 23, and since negative charge is there, so 23 plus 124, 24 divided by 8 is 3 bond pair and 0 lone pair, steric number is 3, hybridization is SP2, yes, no valence electron we need to take, so it should be 5, any doubt guys in this, are you finding it difficult or easy or moderate, yeah, these are not much stuff, okay, once you practice a bit more, you will have the, you know, idea of it, how to do it, okay, now after this what happens, if you look at the, sorry, if you look at the structure for oxygen molecule, oxygen molecule in VBT also, we have discussed this, that we have OO double bond and 2 lone pair on each oxygen atom, this is the structure of O2, so this is structure of O2 according to the valence bond theory, it suggests that oxygen molecule is diamagnetic in nature because all electrons are paired, okay, when all electrons are paired the molecule is said to be diamagnetic in nature, right, but experimentally this observation was wrong, okay, experimentally it has been observed that oxygen molecule is paramagnetic, okay, so this thing was found out to be wrong, it's not correct, actual thing is what, oxygen molecule is paramagnetic in nature, this is correct, once again guys, okay, so paramagnetic, right, so experimentally this is found to be paramagnetic in nature and you know, but VBT suggests that O2 is diamagnetic, so again there's a drawback in this here, right, because valence bond theory, yeah once again, once again it's regarding, right, valence bond theory suggests, see, first of all paramagnetic and diamagnetic, it is just a property of a molecule, okay, you know, paramagnetic molecules are those molecules which has unpaired electron, okay, which has unpaired electron, for example, if I write down NO2, how many electrons this has 23 odd number of electrons, it must has one unpaired electron, so when unpaired electron present, the molecule is said to be paramagnetic, when all electrons are paired, for example H2, H2 you see, there are only two electrons and these electrons are paired, right, okay, it's paired, so it is diamagnetic molecule, this is what the simple definition of paramagnetic and diamagnetic one, property is what paramagnetic molecule has unpaired electron and when it is placed in the magnetic field, it is weakly attracted towards the magnetic field, okay, so it experience some force in presence of magnetic field or electric field, force can be of attractive in nature or repulsive in nature, right, but if the molecule is diamagnetic, correct, then the molecule is diamagnetic, then there is no effect on this, whether you place this in magnetic field or in electric field, it won't experience any force, it won't get deflected from its path, okay, so this is what the definition of paramagnetic and diamagnetic molecule is, so what happens according to VBT oxygen molecule is said to be, is observed to be diamagnetic, but experimentally it has been proved that it is not diamagnetic, it is paramagnetic, because in presence of magnetic field it get attracted towards the magnetic field, are you understanding this, yes, correct, so obviously again this is the drawback or the flaw in VBT, valence bond theory, now to explain the magnetic property of oxygen, we got a new theory and that we call it as molecular orbital theory, okay, so heading all of you write down, you should know why we have, why we need this particular theory, you should know why we need the theory of hybridization, okay, so there was some drawback in VBT which explains, which gives us hybridization and again this molecular orbital, molecular orbital theory, did you understand, yes this, the background of it, did you understand why we need this theory, yeah, okay, now we see what is molecular orbital theory and how molecular orbital theory explains the magnetic behavior of oxygen, okay, so what is molecular orbital theory, first of all, you see like in hybridization atomic orbital combines and forms, and forms hybrid orbital, so here also in molecular orbital theory, orbitals of atoms combines and forms molecular orbital, let me write down these two points then probably you'll understand then what is the difference in the two things, one basic difference we have here, write down atomic orbitals, when I say atomic orbitals, orbitals combines and it forms hybrid orbital, hybrid orbital, again if I write down atomic orbitals, orbital combines and it forms molecular orbital, okay, then you'll ask me so what is the difference in the two, both you are saying atomic orbital combines and forms, then what is the difference between molecular orbital and hybrid orbital, yes, okay, so the basic difference is what when hybrid orbital forms, then the atomic orbital must belongs to the same atom, this atomic orbitals when it forms hybrid orbitals, this atomic orbitals belongs to the same atom, but when molecular orbital forms, atomic orbitals of different atoms combines, are you getting me, yes, now you go back and see this, the examples of CH4 I have discussed once again, yeah, the examples of CH4 I have discussed, you see this, VCl2 CH4, right, this what I said, this four orbitals goes under hybridization, right, all these four orbitals of carbon, we don't have here orbitals of hydrogen, is it, yes, no, the atomic orbitals of carbon combines and forms hybrid orbital, okay, only one atom is involved in this, if you look at this one, VCl2, the atomic orbitals of beryllium combines and forms hybrid orbital, okay, another one, atomic orbitals of carbon, you see this, carbon orbital, right, so atomic orbitals of carbons combines and forms hybrid orbital, no, you see, we have here, no, hybrid orbitals is the, it forms when different orbitals of same elements combines, you see this, that's what I'm telling you, different orbitals of same atoms combines, forms hybrid orbital, molecular orbital, when we have orbitals of different atoms involved, understood, right, so all these examples that I have given you, you can easily understand that this atomic orbitals of same orbitals takes part and it forms hybrid orbitals, different atoms of same elements you can take, you can talk about isotopes here, right, rarely it's, we cannot say it's not possible, right, it's possible, right, yes, got it, understood, so first point is the difference between molecular orbital and hybrid orbital is, in molecular orbital we have different atoms involved, right, now two, three points to write down and then I'll see, I'll discuss the other things here, write down the first point next year, in a molecule electrons are considered, electrons are considered to be associated with, in a molecule, electrons are considered to be associated with all the nuclei in the molecule, all the nuclei in the molecule, next write down as electron in atoms, as electron in atoms occupy atomic orbitals, similarly electrons in molecules, similarly electrons in molecules, yeah, as electrons in atoms occupy atomic orbitals, similarly electrons, electrons in molecules, in molecules are present in molecular orbitals, so the basic difference you must take care of, other things are very much similar to that of hybrid orbitals, okay, like the next point to write down, the number of molecular orbitals forms, the number of molecular orbital forms equals to the number of atomic orbital combines, the number of molecular orbitals forms equals to the number of atomic orbital combines, next, like atomic orbitals, like atomic orbitals, atomic orbitals, molecular orbitals can also occupy, sorry, molecular orbital orbitals can also accommodate two electrons, maximum of two electrons, molecular orbitals can also accommodate maximum of two electrons, so we are talking about the combination of atomic orbitals, so there are two different ways for combining of atomic orbitals, so what I'll write down here, molecular orbitals write down, molecular orbitals forms, forms by two different ways, the first one is linear combination, atomic, linear combination of atomic orbital and the second one is united atom method, in this too, this is not in our syllabus, only we have to study linear combination of atomic orbital, linear combination of atomic orbital is what, heading right down, linear combination of atomic orbital, according to this what happens, when two atoms combines, when two atoms combines, forms a molecule, these atoms combines in two different ways, one possibility is what, either they'll have the same phase, psi a and psi b are the wave function of a and b and psi a b is the wave function of a b, so a and b atoms, either they will have same phase or opposite phase, same or opposite phase, we are talking about the combination of orbitals of two different atoms, orbitals are nothing but the path of electrons, so it must have some wave characteristics, that's why we are talking about wave function here, this part is not important, they won't ask you any questions on this in J also, but since it is there, we are just discussing it, we have to keep in mind, when a and b combines with the same phase, then the molecular orbital that forms, we call it as bonding molecular orbital, when the phases are same, bonding molecular orbital, in short we read it as b m, when the phases are opposite different, then it is anti-bonding, anti-bonding molecular orbital, that is a b m, there is a formula we have of this combination, psi a b is given as n is the normalizing factor, c a is a constant, psi a plus c b into psi b, just write it down, you won't get any questions on this, okay, it is there in the linear combination of atomic orbital, not much important, n is the normalizing factor, n is the normalizing factor, it just you know ensures that the probability of finding an electron in the universe is 1, right, normalizing factor, c a and c b are the factors to minimize energy, c a and c b are the factors to minimize energy, next slide down one note here, yes normalizing factor is the you know factor which ensures that the probability of finding an electron in the universe is 1, okay, that is normalizing factor, c a and c b are the factors to minimize energy, c a and c b are the factors to minimize energy, now you see this how this molecular orbital forms in different combinations, okay, shape I am giving you here, okay just write it down, hardly you will get questions on this, I have seen only one questions on the shape, that's why I am giving you, but not much important, when you have SS combination, like I said there are two possible combinations, one is same phase, other one is opposite phase, suppose we have two s orbital, okay this combines with this and both has the same phase, plus minus here, plus combines with plus and minus, so this is when same phase combines, this forms BMO, bonding molecular orbital, which has a shape like this, elliptical we can say, where we have positive charge here, slightly negative charge here, negative charge here, it's sigma, it is sigma bonded forms, because along the inter nuclear axis the combination is taking place, along this axis, similarly if this is plus, this is minus, combines with minus and plus, then it forms anti-bonding, ABMO, ABMO the shape is this, here we have minus, here we have plus, and in between these two we'll have a nodal plane, important one is for PP combination, I'll do that just to finish it first, do you have any exam on Saturday, no it's not static, the charges are not static problem, that's why there is no no repulsion, it continuously moves and it is not the charge first of all, okay these are the phases of the orbital, plus and minus are not the charge here, take care of that, okay no problem Shraddha, we'll discuss that later, let it be, anyways, okay Friday then the class is not possible, let it be, if you have PP combination on PP combination they have asked one question that I have seen, that's why I'm giving, you just need to know the shape here, three possibility we have, because we have three orbitals here, this combines with this plus, combines with plus and it gives bonding molecular orbital which is this, minus minus plus and this is BMO, along the inter-nuclear axis, sigma BMO, bonding molecular orbital, but again if you have this one, minus plus minus plus then it is, a nodal plane we'll have here, okay this is sigma on innovative, this is sigma, so it will be like this, this is minus, this is plus and this is also minus and this is plus, okay, it is ABM, if you have pi combination here, plus plus minus minus, same phase plus combines with plus so we'll have this and we have electron cloud like this, negative something, okay, this is BMO again, plus plus combines with plus, when plus combines with negative plus minus minus plus, this is the nodal plane, only one kind of question is possible in this structure, that which of this combination gives sigma molecular orbital or pi molecular orbital, one is this, other type which of this combination gives bonding molecular orbital or gives anti-bonding molecule, that is it, you just need to know the shape here, see one more thing here, when it is bonding molecular orbital, right, bonding molecular orbital is represented by sigma and when it is anti-bonding then it is sigma star, means this star represents anti-bonding star, when there's no star it is bonding, okay, why sigma? Because it is along the inter-nuclear axis, right, that is why it is sigma, so sigma is BMO, sigma star is ABM, anti-bonding, okay, here also the similar thing here, this one is sigma BMO, this one is again along the inter-nuclear axis, ABMO is sigma star, right, this is pi because inter-nuclear axis is this, so it is pi BMO, but this one is pi star anti-bonding, now to conclude this what we can write, that when the atomic orbitals of two atoms combines, they combines in both ways, we cannot neglect the possibility of anyone, okay, atomic orbitals combines in both ways, means either they will form bonding molecular orbital or they will form, means they will form both bonding plus anti-bonding molecular orbital, we cannot ignore the possibility of forming any one of the kind of orbitals, right, both orbital forms, right, because the phase can be same or can be different also, so if suppose we have 1s combines with 1s, look at this, if 1s orbital combines with 1s, so s always gives you sigma, so we will get sigma 1s and we get anti-bonding also, sigma star 1s, star means anti-bonding, if 2s combines with 2s, we again get sigma 2s plus sigma star 2s, anti-bonding, if it is 2px, 2pz I will take first, 2pz combines with 2pz, it is sigma 2pz plus sigma star 2pz, then we have 2px plus 2py, one of the p orbitals already form sigma, this will form pi, 2px, 2px from pi, so pi 2px and pi star 2px, if you have 2pz and 2pz, then it is pi 2pz plus pi star 2pz, right, this forms sigma, we usually can, anyone you can consider as sigma, x and y also you can consider as sigma, yes that is what I am talking about Prabhu, okay, anyone you can consider as sigma, I am sorry the last one is y, wait, this is y, one second Prabhu, okay, so anyone you can consider as pi and others are sigma and others are pi, generally we consider inter-nuclear axis along z axis, that's why I have considered pz forms sigma here, right Prabhu, this is just the assumption we have, okay, but if you write suppose pi 2pz plus pi star 2pz, you won't get the wrong answer, okay, it is the assumption that the inter-nuclear axis is along z axis, that's why the pz orbital forms sigma, right, usually what happens, the energy of bonding molecular orbital is lesser than the energy of anti-bonding molecular orbital, right, so obviously you see if the orbital combines and forms all these orbitals, right, all these orbitals, in the molecule we have all these orbitals present, now when you distribute the electron in the orbitals like we do in electronic configuration, similarly here also you can distribute the electrons of the molecules in the molecular orbital, so since we have all these orbitals present, so here also we'll distribute the electrons according to the energy of the molecular orbital, right, so we always start with bonding molecular orbital and then we'll go through anti-bonding molecular orbital, the energy order is experimental that you should memorize, okay, but in general what we say that when you have this thing, the energy profile if you see for bonding and anti-bonding, suppose this is the energy axis we have y axis and suppose 1s combines with 1s, so 1s when combines with 1s it forms abmo and this also forms bmo, right, since bmo has lower energy, so I'll write down bmo down and abmo down, similarly you can also write down 2s combines with 2s, so it forms abmo, bmo and other orbitals also you can write, okay, so this is the energy profile we have, like I said that orbitals forms both anti-bonding and bonding, so you should know the overall energy of the orbital, the order you should know and based on that you can distribute the electron from low to high energy, okay, so if there are two possibilities here, if the total number of electrons equals to n, case one when n is less than equal to 14, 14 electrons, less than equal to 14 electrons, then this energy order you should know, orbital energy order and you will get questions from this only, the one that I am discussing now, this is only the most important part we have here, other parts are there, that's why we have discussed but not important, okay, so what is the energy order you see, sigma 1s is minimum, then we have sigma star 1s, then we have sigma 2s, sigma star 2s, after this we have pi 2py, pi 2py, then we have pi 2p x, sigma 2p z, sigma 2p z and then we have pi star 2py, pi star 2p x, sigma star 2p z, this is the orbital energy order, this orbital which I have written like this are degenerate orbitals, these two and this means equal energy order, and when you distribute electrons in this, you have to follow unscrew, while distributing the electron you have to follow, is that fine, then nice, yeah wait, wait, no n2 is this only, less than equal to 14, no, nitrogen has 14 electrons, so for n2 this is the order we have, the second case we have for distribution of electron, when the number of electron n is greater than 14 but less than equal to 20 in this case, only one change you have to do here, sigma 1s is minimum, sigma star 1s, sigma 2s, sigma star 2s, then we have sigma 2p z, pi 2py, pi 2p x, then we have pi star 2py, pi star 2p x, and sigma star 2p z, means this position we have interchanged, that is it, here also we have same thing, these two are degenerate, these two are degenerate equal energy orbital, right, we follow Hansel here, based on this first of all, we will try to find out the magnetic property of oxygen, O2 molecule, because that is why we have this, you know, the entire thing, that is molecular orbital carry, it has 16 electron, so when you distribute 2, 4 and write down separately this one, okay, 6, 8, 10, 11, 12, Hansel, right, first one on electron and then pairing, 11, 12, then we have 13, 14, 15 and 16, you see since we have unpaired electron present in the orbital, that's why oxygen is paramagnetic, it's 2 unpaired electron magnetic with 2 unpaired electron, okay, that is how the magnetic behavior of oxygen has been explained, okay, now the next thing and all these things are important, okay, for exam point of view, we can also calculate the bond order with respect to the molecule, with the help of the molecular orbital here, okay, by the way, what is bond order, how do we define it, what is bond order, I have done this, tell me guys, it is a number of, number of bond between the two atoms in a molecule, okay, that is how we define number of bonds, it is always for the two atoms in a molecule, it is not for the entire molecule, number of bonds present between, in the molecule orbital theory is not valid for more than 20, okay, it is not given there, if I take this molecule which is CH3, CH double bond, CH single bond CS3, suppose this is carbon, suppose this is the first carbon, the second, third and fourth, if I ask you, what is the bond order of C1, C2 carbon, what is the answer, bond order of C1, C2, one, because there is only one bond present, C2, C3, 2 and C3, C4, it is again one, right, why this is one, because the number of bonds are one between the two carbon atoms, the first point is bond order we define between the two atoms in a molecule, okay, I have also told you that this bond order may have the fractional value, right, this may have fractional value when resonance is possible, what I said, we can find out bond order with respect to the molecular orbital theory, okay, let's see this, what is the bond order of nitrogen molecule and n bond in nitrogen molecule, tell me the bond order, is it 3? Yes, 3, could you cross verify, we can also cross verify this with molecular orbital theory, with the help of molecular orbital theory, we can find out bond order with this formula, the bond order for, you know, for this molecular orbital theory is number of electrons BMO, bonding molecular orbital minus number of electrons in ABMO, anti-bonding molecular orbital and this is divided by, divided by, this is the formula we have for bond order, according to molecular, bonding, sorry molecular orbital theory, now what I want, I want you to use this formula and find out the bond order of N2, nitrogen molecule, how many electrons nitrogen has, the number of electrons for nitrogen equals to, no it's 14, how it is 28 or 7, 7 electrons in 1 nitrogen atom, 7 electrons in 1 nitrogen atom, 14 is atomic mass, right, so 14 is the total number of electrons for nitrogen, follow the order of orbitals given, it is the first order it follows, okay and then find out bond order, what you are getting, are you getting S3, yes it should be 3, right, so I'll write down the order here, the order is sigma 1s, sigma 2s, sorry, sigma 1s, sigma star 1s, then sigma 2s, sigma star 2s, pi 2p, x, pi 2py, sigma 2pz, then pi star 2py, pi star 2px, sigma star 2pz, if it is more than 14, then this 2, you need to interchange, that is it, it has 14 electrons, so if I distribute 2, 4, 6, 8, 9, 10, 11, 12, 14, okay, so bond order is the number of electrons in bonding molecular orbital, which is this, that is 6 plus 4, 10 minus 4 by 2, that is 3, that's why the bond order is 3, you can find out the bond order of hydrogen also, what you are getting from bond order of hydrogen, that's 2 electrons, so it is simply sigma 1s 2 and there is no electron, right, so bond order is equals to 2 minus 0 by 2, that is 1, what is the bond order you are getting for he2, it has 4 electrons, then we have sigma 1s 2, sigma star 1 is 2, so bond order is what, 2 minus 2 by 2, that is 0, 0 means what, there is no bond present, right, 0 means what, bond order is nothing but the number of bonds, so when the number of bonds is 0, then there is no bond between the helium atom, that's why you see helium is exist as helium atom only, it does not form he2 like hydrogen, this is not possible, are you getting it guys, this is not possible, we can conclude this, it does not exist, because there is no bond, it does not exist, right, so the molecule for which the bond order is 0, right, then the molecule does not exist, yes, yes, we can say that, we can generalize this, if the bond order is 0, then the molecule does not exist, correct Arshad, okay, even on this they have asked question in the exam, they will give you 4 options and they will ask you which one of these molecules does not exist, to find out bond order, if it is coming out to be 0, that is your answer, any doubt till here, all of you please respond, is it clear, the point is first 10 minutes whatever we discussed in molecular orbital theory, okay, is not important, but when we start discussing this, the energy order, sigma 1 is sigma star 1 is and all and then this bond order, mostly 99% of the cases, the last question from this portion, okay, must take care of that, now one more relation you see here, we know this relation we have discussed last class, last class, last class, that bond order is directly proportional to bond strength and it is inversely proportional to bond length, so once you know bond order, right, you can also compare bond strength and bond length, clear, one very important question and most probably you are going to have this questionnaire school exam and the question is compare the bond order, bond strength and bond length of O2 2 minus O2 minus O2 and O2 plus, for these four molecules are ions, so I will do this for first one, see O2 2 minus has 18 electrons, so order is sigma 1s, sigma star 1s, sigma 2s, sigma star 2s, sigma 2pz, pi 2px, pi 2py, pi star 2px, pi star 2py and sigma star 2pc, we have 18 electrons, 2, 4, 6, 8, 10, 11, 12, 13, 14, 15, 16, 17 and 18, so bond order for O2 2 minus is all these are bonding molecular orbital, right, 1, then 2, 3, 3, 4 and 5, 5 into 2, we have 10 electrons in bonding molecular orbital minus 2, 4 and 4, 8 divided by 2, that is 1, bond order is 1 for O2 2 minus, okay, similarly you can do if it is O2 minus, means it has one electron less, right, it means it has 17 electrons, so one electron that comes out from the anti-bonding orbital, right, then it would be 10 minus 7 divided by 2, that is 1.5, again we have O2, it has 16 electrons, so bond order is 16 electrons, bond order is we have 10 minus 6 divided by 2, that is 2 and O2 plus has 15 electrons, so bond order is 10 minus 5 divided by 2, that is 2 point, so if I write down the order which discussion they have asked many times in need in others exam, okay, so this is important here, so if I write down the order of bond order here, we have O2 2 minus, O2 minus, O2 and then O2 plus, if I go left to right, if I go left to right, O2 2 plus the bond order increases, bond order increases means bond strength also increases, bond strength increases means bond length decreases and hence the bond length increases from O2 plus to O2 minus, this is the order we have to go, yes O2 2 plus is 14 electrons, so we will consider the first one, right, so this example is very important, the oxygen line example, many times they have asked this question in the exam, in school exam also, in comparative exam also, okay, we must take care of all the three properties here and when the bond order is zero, then the molecule does not exist, that is also very important, okay, so this is it for molecular orbital theory, any doubt in this, any specific doubt? Next slide down, what other types of molecule? I know too we'll discuss in the last, it's a odd electron molecule we'll discuss in the last video, there's two types of bond we have, like three electron bond and one electron bond, odd electron molecule, we have just two three examples we'll discuss in the last video, now the next type of bonding we have, we call it as hydrogen bonding, right down, it is the electrostatic force of attraction, it is the electrostatic force of attraction, it is the electrostatic force of attraction, between the hydrogen, between the hydrogen or between hydrogen comma, attached to a more electronegative elements, in hydrogen attached to more electronegative element like chlorine, nitrogen, oxygen, etc., yeah, it is the electrostatic force of attraction, electrostatic force of attraction between hydrogen, between hydrogen comma, attached to more electronegative elements such as oxygen, nitrogen, chlorine, etc., the another electronegative element, done, did you write this? Okay, now what do you mean by this? First of all you see in this example, what I said, when hydrogen is attached to an electronegative element, suppose we have m, m is highly electronegative, highly electronegative, mainly we have it chlorine, we can have it chlorine or nitrogen or oxygen or chlorine, etc., but mainly we define for chlorine, oxygen and nitrogen. Okay, because of the high electronegativity of these atoms here, this attracts the bond pair of electron towards the side and after some time the bond becomes fuller, this atom will have delta negative charge and this atom will have delta positive charge, it is very much similar to that high effect, if you remember, inductive effect. Okay, this happens. Now, suppose if you have another electronegative element, suppose you have oxygen, hydrogen and hydrogen, delta positive, delta positive and delta negative, oxygen is also a highly electronegative element. Now, when these two atoms come closer, this electronegative atom and this hydrogen, then there will be an electrostatic force of attraction because of this delta positive and delta negative charge. There will be an electrostatic force of attraction, this electrostatic force of attraction is hydrogen bonding, copy this down. Yeah, okay, so hydrogen bonding, first of all, hydrogen involved, but hydrogen must be attached to the electronegative element, this must be electronegative. So, it is a bond electrostatic force of attraction between hydrogen and an electronegative element, first of all, but this hydrogen which is involved in hydrogen bonding must attached with an electronegative, understood this. So, this hydrogen bonding is possible in a molecule like H2O, in water it is possible, in energy it is possible, in alcohol it is possible, HF hydrogen chloride it is possible, so there are many examples for that. This happens because of the high electronegativity of the atoms attached to the hydrogen. Okay, if you look at the hydrogen bonding in water, all these molecules you see, this has delta positive, delta negative, delta positive charge, delta positive, delta negative, delta positive charge and we have delta negative, delta positive, delta positive, positive and positive, delta negative, positive and positive. Okay, this kind of charge separation is there. Now, when these atoms comes closer, this oxygen and hydrogen will have in force of attraction, electrostatic force of attraction, this will have in force of attraction, this will have in force of attraction, this will have in force of attraction. All these force of attraction are hydrogen, so water molecules are associated like this with other molecules and hence, this we also call it as an associated molecule, because all the molecules are connected with a, with hydrogen bonding, hence it is associated. It is possible with two different molecules also, for example, you see we have H2O and NH3, so nitrogen is delta negative, delta positive and this is also delta positive, delta negative, delta positive, delta positive, positive and positive. So these two will attract each other. This is the hydrogen bonding, electrostatic attraction, right? So it is possible between two different molecules. More, more charge density on the atom, more will be the electrostatic attraction and stronger the hydrogen bonding is, right? So right now as electronegativity increases, as electronegativity increases, charge separation increases, charge separation increases and the strength of hydrogen bonding also increases. Now hydration is something else, hydration is something else. Hydration is like if you dissolve sugar crystal into water, that is also hydration. There is no hydrogen bonding between sugar crystal and this thing, what we say, water molecule. Hence you see one thing you can understand here, obviously we cannot compare this with hydration because it's a different process, but there will be some kind of attraction. Hydration process is different, means we have one concept here that we say the molecule which can form hydrogen bonding with water can dissolve in water, that we can say, okay? But hydration process is different, hydrogen bonding is different. It is possible that the molecule which shows hydrogen bonding with water get hydrated in water, that we can say, okay? So this is the hydrogen bonding, the basic understanding you must have, more charge, more will be the electrostatic attraction and hence that's stronger the hydrogen bonding, okay? Now this hydrogen bonding are of two different types, right? Next slide down, types of hydrogen bonding. Two types of hydrogen bonding we have, two types. The first one is intermolecular hydrogen bonding, right? If you see the name itself suggests that this is the hydrogen bonding between the two molecules of same and different type like H2O and H3O example, right? That is intermolecular hydrogen bonding, okay? Write down, this type of bonding occurs, this type of hydrogen bonding occurs when the hydrogen bond accepted, when the hydrogen bond accepted and the hydrogen bond donor are present in two different molecules. Or write down this way, wait, are present in two molecules of same or different type, two molecules of same or different type. See hydrogen atom accepted is what? The molecule which donates hydrogen for hydrogen bonding means from which the hydrogen is involved, like you see this one. In this particular example, this molecule is H-atom donor because its hydrogen is involved in hydrogen bonding and this molecule is H-atom acceptor. So when we have two different hydrogen, atom, donor and acceptor molecule are different or same also, in H2O also it is possible. This is also intermolecular hydrogen bonding, right? That is intermolecular hydrogen bonding, write down between H2O and H3O example, hydrogen bonding occurs. The second type we have intramolecular hydrogen bonding, write down. In this type of bonding, in this type of bonding, the hydrogen atom donor and acceptor, the hydrogen atom donor or acceptor are present in the same molecule, are present in the same molecule. Look at this example. We have CH3CH double bond C HC double bond O. This is the bond we have within the molecule you see. This bond is H bond. Since it is within the molecule, so it is intramolecular, not intermolecular. It is intramolecular. Here also you can place any other atoms. This type of bonding is intramolecular bonding. Okay, this example you see if you talk about, you know ortho-paran metaposition, right? This is a benzene ring and any group is attached. We can have OH, NH2, CHO, any group. With respect to this group, this position is ortho position. This position is metaposition and this position is para position, right? This is also ortho. This is also meta, meta, right? So in this particular, you know, compound, any molecule of benzene, we have two ortho, two meta and one para position. Okay, two ortho, two meta and one para position. If you take the example of ortho-nitro phenol, right? Ortho-nitro phenol. So first of all, phenol is this and ortho-nitro means what? At ortho position, we have NO2 attached, right? So at ortho, we have NO2 present, positive negative. There is ortho-nitro phenol, right? When you talk about the para-nitro phenol, para-nitro phenol is this, OH and here we have nitro group N11O, N11O and here we have a single bond and positive bond. This molecule is O-nitro phenol. O stands for ortho. This is P-nitro phenol. P stands for para here. So in ortho-nitro phenol, what happens? We have intramolecular-hydrogen bonding possible. This intramolecular-h bonding. But in para-nitro phenol, we don't have intramolecular-hydrogen bonding. But here we have intermolecular-hydrogen bonding possible. Look at this because this cannot make a bond with hydrogen because the distance is more over here. But if you have another molecule, similar molecule, para-nitro phenol, if I place like this O and H and we have NO2 here, NO- this positive double bond O and these two attracts each other because the charge separation delta positive and delta negative. So this bond is again a hydrogen bonding but it is intermolecular hydrogen bonding because it happens between the two molecules. The donor in the acceptor atom is present in two different molecules. Where I have taken O and N, Ruchita, Achaib, oh my god, is my bad guy is this. You corrected it. This you corrected my bad guys. It should be like this. H, right side. Okay. So here we have delta negative, delta positive, delta negative, delta positive, delta positive and delta positive and delta positive. All of you are done, finished. Now because of intermolecular-hydrogen bonding, the boiling point increases. This is the proper T here, right? That's why the boiling point order if I write down for para-nitro phenol, the boiling point is more than that of ortho-nitro phenol. This you must remember. Right on the heading, properties, this is again, we have three more points in this. We'll finish it and then we'll take a break and take max to max five minutes. Okay? Just max to max five minutes. Right on properties affected by, affected by hydrogen bonding. There's three points in this. That is it. First one, boiling point. Just now I said, write down because of intermolecular hydrogen bonding, hydrogen bondings are the weak bonds. Okay? We don't talk about the stability because of hydrogen bonding. Right? It won't affect the stability much because of intermolecular hydrogen bonding. The intermolecular force increases, hence boiling point increases and hence the boiling point increases. This is the reason that H2O, the boiling point of H2O is more than that of H2S because in this we have intermolecular hydrogen bonding, H2S. Intermolecular hydrogen bonding, more boiling point. Same reason we have for NH3 also, more boiling point than PS3. In fact, this H2O because of more boiling point exists as a liquid but this H2S exists in the gaseous form and this question they asked in the exam also that H2O exists as a liquid but H2S as a gas. What is the reason? Okay? Second point, solubility in water. Right now the compounds which can form hydrogen bonding with water is solubility in water. The compounds which can form hydrogen bond with water are solubility in water. Like you see CH4, CH4, there's no electronegativity difference can't form hydrogen bonding is insolubility in water but if you talk about alcohol CS3OH this is soluble in water. Heard in the last point here is viscosity. Due to hydrogen bonding viscosity increases. Due to hydrogen bonding viscosity increases. This is it for hydrogen bonding. We'll have few more, two, three more points. You know topics we have to discuss small, small topics are there. Two examples we have that we'll discuss after the break. Okay? Take a break now. We'll resume at 6.50. Okay? Yeah, shall we start guys? See there are few small, small topics are left in this chapter. Okay? Miscellaneous topics and we have here. I guess it does not have connection with the previous thing that we have done. So we'll consider this under miscellaneous topics of chemical bonding. Okay? The first one we have bridge bond. Okay? Heading you right down all of you. Bridge bond. Bridge bond we have two types. You just need to keep the examples in mind. That is it. Okay? When you get that question, you'll understand that what you need to do here. Bridge bond is just a factual question you'll get. Two types of bond. The first one is three center, three center, two electron bond. And the other one is three center, four electron. This we call it as a bridge bond. Okay? These two types of bond. What is three center, two electron bond? Center means three nucleus. Center means here nucleus. The first example you see of three center, two electron bond. And why we are calling it as three center, two electron bond that you will understand with this example. Keep this in mind that center means nucleus we have here. Take an example of BEH2. BEH2 goes under dimerization. Okay? It dimerizes and forms and forms BE2H4. Dimerization means what? Two molecules combines together. That is dimerization. Okay? Polymerization, when we have more than two molecules combines, then it is polymerization. Okay? Now in BEH2 what happens you see on dimerization we get a stable structure and that's why the dimerization possible. Obviously logically you cannot understand that whether the dimerization is possible or not. That's why I said that it is a factual thing. You have to keep this in mind that in this case the dimerization is possible and in other case it is not possible. Like this example you see BEH2. Okay? So beryllium has vacant P orbital available in this bonding. If you look at the bond here BEHH and another molecule is this BEH and H. If you look at the electronic configuration of beryllium, beryllium has four electrons. So it has 1S2, 2S2 and 2P0. There's no electron in P sub-shell. Right? So in the excited state what we can say 2S has one and another electron jumps into the 2P shell of this. So it is vacant P orbital beryllium. And these two goes under hybridization forms SP hybridized orbital and these two are the vacant orbital. So the point I'm trying to make is along with this bond beryllium has vacant orbital P orbital present here vacant P orbital present. So here what happens this electron pair you see the electron pair which is present in this bond and electron pair which is present in this bond. This electron pair is attracted towards this beryllium goes into this vacant orbital and this is attracted towards this beryllium and that is how the dimerization takes place. So this converts into this molecule. Beryllium at the two corner this hydrogen beryllium one beryllium one hydrogen this side one hydrogen this side and here also we have H and V. This is the dimerized form of B2S. Okay I see this forms of bridge here. Right? This is a bridge between the two molecule. That's why it is bridge bond. Why we are calling it as two three center two electron bond? I have given you two bonds but the number of electrons are two only. It means that these two electron is you know associated with the nucleus of one two two beryllium atom and one hydrogen atom. This two electron is associated with this beryllium this hydrogen and this beryllium. So we have three nucleus one two and three and two electrons. So it is three center and two electron bond. Understood? Right? So this bond is the bridge bond. We have three center three center means three nucleus two four beryllium one for hydrogen. Right? This bond is the terminal bond. Right? And this bond here this we got it as three center two electron. Yes done guys three center because we have three nucleus. Right? And the two electron which was present here between this beryllium and hydrogen now in this dimerized form this two electron is associated with the two beryllium atom and one hydrogen atom this three atom. That's why we have three center. Three center means three nucleus and two electrons. So you have to memorize this that in BEH2 we have three center two electron bond. Yeah, I said no Shraddha see what happens. Beryllium BEH2 molecule is this. Now in this bonding state with two hydrogen atom beryllium has vacant orbital available. Hence it can accept the electron. Right? Correct? So this pair of electron in order to get dimerized this pair of electron is now associated with like we can say it donates to this orbital or it is you know associated with this nucleus this hydrogen and this nucleus three nucleus we have and we get a bridge bond like this it acts as a bridge between the two molecules. That's why the bond is bridge bond and it has three nucleus one two and three three nucleus so three center and since this two electron is you know distributed among the three atoms hence it is three center two electron one. Even this bond is also three center two electron bond. Yes you're right guys. This bond is also three center two electron one. You have to keep this in mind that in BEH2H4 we have a bridge bond that is three centered two electron. Another important example for this one and in fact this one is more important is for BEH3. BEH3 get dimerizes and converts into B2H6. Right? B2H6 di borane. B2H6 is di borane. Right? If you look at the bonding in BEH3 electronic configuration of boron has five electron one is two two is two two p one so in excited state you see it must have three unpaired electron in order to make a bond with hydrogen so excited state you see one orbital is vacant here so this one boron has one vacant orbital in fact you can keep this in mind boron whenever bonded with three bonds it has one vacant orbital and that's why this electron pair you see which is present here it is associated with this boron atom and this electron pair also associated with this boron atom and hence this two dimerizes and it converts into D H D H D hydrogen and hydrogen this bond is three center two electron bond. Okay so we have two three center two electric this is bridge bond and this is terminal bond terminal bond we also call it as this one we also call it as banana bond because this one it looks like banana that's why it is nothing else right and we have similar thing this side also right so it looks like banana hence we call it as banana bond there are few properties on which they ask question in exam and that you should remember the first property is what of this that is the bridge the bond length if you see if you take the bridge bond boron hydrogen the bond length of bridge bond is more than to that of terminal B H bond right if you talk about bond strength usually bond length and bond strength are opposite but here still the bond strength is more for bridge bond however its length is more than terminal B H there's no mistake here these two orders are correct bond length and bond strength usually it is inversely proportional but here the same order we do not have because the two electrons are you know distributed among three centers three nucleus that's why we have greater attraction and hence the bond strength of bridge bond is more than to that of terminal no it's not that we have two electrons only where where we have the four electrons is three center two electron okay so these two properties related to the bond length and bond strength you must remember then guys tell me yes so another type of bridge bond we have the second type that is three center four electron bond we have two examples in this the first one is BeCl2 in BeCl2 if you look at the structure beryllium attached with one chlorine again beryllium has vacant orbital i'm not going to draw the electronic configuration again we have discussed it in BeH2 similar you know configuration we have here chlorine has three lone pairs on it and we know beryllium has vacant p orbital that we have discussed already so what happens this chlorine donates its lone pair into the vacant orbital of this beryllium and hence we get a bridge bond here which is this BeClBeClCl and Cl so one bond pair will have here which is present as it is another one is of chlorine and chlorine has two lone pair here another one is of chlorine this one you see one two three center two plus two four electron that's why this bridge bond is three center four electron all these diversification processes are a spontaneous process okay the molecule will get dimerized on its own in order to gain the stability okay done one more example in this we have iCl3 which dimerizes as i2Cl6 okay why this happens is that you see here when you look at this iCl3 you can find out its geometry and you know hybridization you'll get sp3d this is tbt trigonal bipyramidal it has three bond pair and two lone pair the hybridization is sp3d for iCl3 now because of its hybridization the structure of iCl3 is this chlorine here chlorine here chlorine here at this corner we have one lone pair at this corner we have another lone pair so this is the bonding we have trigonal bipyramidal okay this is the lone pair of it lone pair and lone pair okay so here we have lone pair lone pair repulsion at this you know equatorial coalition now when it dimerizes shows three centered uh you know four electron bond it dimerizes and the structure will go like this iCl i and the two lone pair it is present over here now you see the two lone pair are placed at 180 degree to each other and hence the repulsion reduces and hence the stability increases in fact this all these chlorine are going into the plane all these are dotted line into the plane and these chlorine atoms are coming out of the plane right this is the structure we have and most importantly is what that these two lone pair now at 180 degree maximum distance and hence the stability increases and that's why it get dimerizes right iCl3 converts into i2Cl6 BCl2 converts into B2Cl4 three centered four electron bond now the next one is back bonding see the back bonding we have two types of back bonding the first one is p-pi p-pi back bonding you write it on this p-pi p-pi back bonding i'll explain why it is called p-pi p-pi back bonding you'll understand it okay first we'll see this example here p-pi p-pi back bonding is what suppose we have a molecule we have a huge application of this right BF3 molecule suppose we have so BF3 molecule you see boron is bonded with three fluorine trigonal planar structure we have and in this bonding state the boron has one vacant orbital that we know already and each fluorine atom has three lone pair on boron has only six electron you see this is the six electron has boron has okay so because of this electron deficient atom boron is so what happens any fluorine because all are the fluorine atoms only any of this fluorine atom donates its lone pair of electron to the vacant orbital of boron what is the benefit of it the benefit is boron has six electron only in this state when this when this fluorine atom donates its electron it has eight electrons so it's octet complete right so this is what the back bonding is so what is happening here you see the boron and fluorine boron has vacant p orbital fluorine has lone pair present in p orbital and this side we have again two fluorine atom present it is possible with any one of the fluorine atom because all has the same you know size and same shape so this lone pair it donates into the vacant orbital of boron and hence the octet of boron is complete right so what happens here this is p orbital this is p orbital so it is p p overlap we can say p p overlap the bond that forms here this bond has pi characteristics because this already has a sigma bond like inter-nuclear axis this is lateral overlapping so pi characteristics hence we call it as p pi p pi why back bonding because sigma bond has already been formed and then this bond is forming right that's why it is p pi p pi back bond understood tell me guys it's clear so p pi p pi back bonding is possible when the two atoms belongs to the second period this also belongs to the second period this also belongs to the second period mostly mostly it happens in this case back bonding it calls because one bond has already been formed one sigma bond then fluorine is forming a bond again after formation of a sigma bond fluorine is donating the molecule has been formed and then the fluorine is donating its electron back to boron means when you just draw the Lewis order structure will give electrons to fluorine fluorine donates its electron to boron that's why it is back bonding it forms after the formation of a sigma bond after the molecule has been formed that's why it is back on clear right so in this what happens you see the electron deficiency right on this point the electron deficiency of boron atom decreases due to back bonding the electron deficiency of boron atom decreases due to back bonding next line as back bonding tendency increases electron deficiency decreases right so if you look at this molecule one question they ask here that compared the due to back bonding the electron deficiency of boron decreases as the back bonding tendency increases yeah this one line I said one second due to back bonding the electron deficiency of boron decreases more back bonding or more tendency of back bonding lesser will be the electron deficiency more tendency of back bonding Lessor will be the tendency of electron deficiency, less of a tendency of less of the electron deficiency. Okay. Compare the Lewis acidity, very important question. Compare the Lewis acidity of BF3, BCL3, BBR3 and BIT3. See Lewis acidity is what? Lewis acidity is the tendency to accept electron. Lewis acids are those compounds, those compounds which has tendency to accept a pair of electron, pair of electron. Now all these boron are electron deficient. Okay. So because of back bonding what happens, electron deficiency decreases, electron deficiency decreases then the molecule has less tendency to accept electron and its Lewis acidity decreases. Correct. So if you look at this three molecule here. Okay. In BF3, the boron has 2P orbital, fluorine has 2P orbital. This 2P, 2P overlap we have here. Here we have 2P, 3P overlap, 2P for boron, 4P for bromine, 2P for boron and 5P for iodine. What do you mean by this? See when the orbital size is comparable then the overlap is easy and the extent of overlapping is maximum here. But when the orbital size is not comparable suppose just 2P, 2P overlap here. Suppose you have 2P, 3P overlap is something like this. Right. So here the overlapping tendency is less because of size difference. 2P, 4P. Again the size difference is large. Right. So it is 2P, 3P. It is 2P, 4P. So as the size difference increases, right, then the overlapping tendency decreases. Hence in BF3 molecule we have maximum tendency to overlap and forms the back bonding. Right. So back bonding tendency is maximum in BF3 and minimum in BI3. That's why the Lewis acidic behavior for these molecules is nothing but the tendency to accept electron. BI3 is the most electron deficient molecule hence the order is this. Lewis acidic behavior. Yes, did you understand this? More, you know, back bonding. See I'll explain this again, one second. See actually all these molecules have similar kind of bonding. For example, BF3 I'm writing it down. BF3 is this. And boron has vacant orbital. When it has vacant orbital, so it behaves as Lewis acid. It behaves as Lewis acid. Why? Because since the vacant orbital is present, so it has tendency to accept electron. If you donate electron pair to this, it can accommodate the electron pair into this vacant orbital. That's why it is Lewis acid. Okay. So if it is electron deficient, it has more tendency to accept electron. But the electron deficiency decreases because of back bonding. Okay. So fluorine what it does? Fluorine has the pair of electron lone pair, right. So since the orbital size is also comparable. So it donates its electron to the vacant orbital of boron. Since it accepts electron from fluorine. So boron has less tendency to accept electron from any other molecule. Means it's Lewis acidity. It's Lewis acidity decreases because of back bonding. Can we say that all of you respond because of back bonding? Can you say that tell me and what kind of back bonding is this? P pi, P pi back bonding. Right. Yes. So more the tendency to form a back bond lesser, you know, the behavior of this molecule. Right. The Lewis acidic behavior is lesser. We can see. Right. So if you look at BCL3, BCL3 is this. This also has similar kind of bonding. But the only difference is the P orbital of boron and chlorine. We have a difference in size because this is 2p and this is 3p. So more the size difference, lesser will be the tendency for. Hence what we can say the back bonding tendency, P pi, P pi back bonding tendency is maximum in BF3 and minimum in BAI3. What do you agree with me on this? P pi, P pi back bonding tendency is maximum in BF3 and minimum in BAI3. Yes. Is it clear? respond guys. All of you. Right. So maximum back bonding means minimum tendency to accept electron. It's lowest acidity decreases. That's why I said that BF3 shows maximum back bonding. So it's lowest acidity is minimum and BAI3 shows minimum back bonding tendency is lowest acidity is maximum. BF3, then we have BCL3, then we have BBR3. And then we have Nikaity. This is the lowest acidic nature for the trihalides of boron. Clear? Yes. Yes. One important thing. This order I have given according to this P pi, P pi back bonding must take care of that. Right. If you simply think of electronegativity, then you can say, sir, floating is the most electronegative atom. So it has the maximum tendency to accept electron. But that is not the case here. We cannot take the factor as electronegativity. Electronegativity is not the factor here. Not the major factor. The major factor is P pi, P pi back bonding. So according to electronegativity, if you write down the order, the order will be exactly opposite. Right. According to P pi, P pi, the order is this. So which factor dominates when according to that only will write down the answer. So here you have to keep in mind in case of BF3, BCL3 means boron trihalide molecule, P pi, P pi back bonding is dominating all other factors. And hence the Lewis acidic behavior is this. In and out. Correct. So fine guys. So we have another type of back bonding here, which we call it as P pi D pi back bonding. Okay. And then we have odd electron molecules. Yes, yes, yes, acidity, Lewis acidity, not acidity. I'm talking about Lewis acidity here. Okay. Because the definition of Lewis acid is what the molecule which has the tendency to accept a pair of electron is called Lewis acid. Right. The BI3 is the most strong Lewis acid. BF3 is the least strong Lewis acidity. Or the weakest Lewis acidity is BF3. Strongest one is BI3. Clear. Any doubt? Yeah. Okay. Next we have to start with P pi D pi and then odd electron bond. Okay. I won't start it today. We don't have time. Otherwise I have to leave it in between. Okay. So we'll do this in next class. Half an hour mostly I'll take. And then we'll finish off, finish it off. And then we'll start with gases acidity. I have gone a bit fast today. Okay. So I request all of you to revise the entire session once. Okay. Because we had to cover a lot since last class we missed. Right. That's why I've gone a bit fast. If you revise your notes, if you want I will share the videos also with you later on. Okay. Anything else? Okay guys. Thank you so much. Take care. Bye.