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0:06 ChemFact Education, if you want to create a molecular orbital diagram for any polyatomic molecule, there are some basics you need to understand. There will be many new things in this lecture , so listen to the entire lecture with patience, and you will become capable of creating and understanding the MO diagram of any polyatomic molecule. Okay? So, let's talk about the first thing; if we have any polyatomic molecule with more than two atoms, because we have already done diatomic ones. So, the simplest case we are left with is one that has three atoms. For example, if we take BeH2 or H2O, or there are many examples, we will learn them all. This is our first molecule for which we need to make an MO diagram . So, we have to keep in mind that if there were only two atoms, then on one
0:51 side there would be beryllium and on the other hydrogen; but here there are side atoms too, there are two hydrogens , so how will it be formed? So, after this, remember that in any polyatomic molecule, you know there is one central atom, which I have represented with 'a' , and if this is a triatomic molecule, let's suppose 'b' represents the side atoms; that is, I am explaining with a general case, then you can apply this to BeH2 or any MO diagram, and you will understand everything. So, whenever you want to create an MO diagram, you must remember that the MO diagram will be formed in this way. What will be on this side? Whose orbitals will be there ? The question will come to mind because if there are more than two atoms, keep this in mind: on this side are the orbitals of the central atom.
1:36 That is, whatever the central atom is in that molecule, its atomic orbitals will be there. Like, who is it in this case? Beryllium. So, if beryllium is the central atom, then what orbitals are in its valence shell? 2s and what is 2p. Although all three p-orbitals will be of equal energy, I have just shown them one above the other. So that if this is px, let this be py, and this one is pz, they will be equal in energy . But with what will px interact? With what will py interact? With what will pz interact? To keep that clear, I can draw them in this way as well. But in terms of energy, what will they be? They will be equal. This much is understood, sir. This means, sir, that on the left-hand side of the MO diagram for any polyatomic molecule, there will always be the central atom and its
2:22 atomic orbitals. Because we are going to apply this same concept to coordination compounds later. Where there is a metal in the center, surrounded by ligands. So, when we study Ligand Field Theory, the LFT diagrams we study are, in a way, what? They are molecular orbital diagrams, aren't they? So on this side, you understand that if I ask you to make an MO diagram for a complex, you will write the metal on this side and list the orbitals the metal has here. What will be on the other side? We are just about to start learning that. Because the main story is all on this side. Now let's talk about the second point: suppose I say we need to make an MO diagram for BH3, you know, sir, we will draw two sides like this. Who will come here? Boron's atomic orbitals. If we want to make an MO diagram for CH4, carbon's will be here, and if we want to make one for NH3, then nitrogen's
3:07 orbitals will be here. So we have understood one thing very clearly, sir, that if you want to draw the MO diagram for any polyatomic molecule, you first place the central atom on the left-hand side, and since I mentioned repeatedly that we will apply this concept in coordination too, we can say that metal is on one side. Now let's see what is on the other side. So whose did we explain just now? Let's suppose which molecule were we talking about? AB2, this is a general molecule. On the left-hand side, we have atomic orbitals of A, and what do we have on the right-hand side? On the right-hand side , we will have Ligand Group Orbitals, which are known as LGOs. I know a completely new word has come up, and it's on your mind: "Sir, what is this?"
3:55 So, what is the belief of this theory? Let's understand that. This theory believes that while the central atom's atomic orbitals are here, for overlapping with the central atom's atomic orbitals, the side atoms—and I am using the word "ligand" for whom? For the side atoms. So, this theory believes that for polyatomic molecules, the side atoms prepare various new combinations of their orbitals. They bring those combinations, and then what do they do with this central atom? They perform mixing of orbitals. So, you just realized that from now on, when making MO diagrams for polyatomic molecules, we need to consider the combinations produced by the side atoms
4:41 'atomic orbitals; I know there will be many questions in your mind. Sir, what will the combinations be? How many will there be? How will they be? How will we create them? We have to learn all of this. But we are proceeding based on the assumptions of this theory. So, we will understand every point carefully. That is, on this side, there will be those combinations that these ligands have produced. So that they can do what with the central atom's atomic orbitals ? They can combine. That is why what is their name? Ligand Group Orbitals. Meaning, ligands will first form a group on one side. Let me explain this in simple words: there is a central atom and side atoms. What will the side atoms do? They know that the central atom's atomic orbitals are arranged on this axis in terms of energy and symmetry. So now, all these side atoms
5:29 have a meeting somewhere else, and in that meeting, they decide that they need to overlap with them. So we must create such combinations that allow for overlap with them. This means there are two steps in creating an MO diagram. What is step number one? In step number one, the side atoms produce some combinations, which we call LGOs. You will always write in polyatomic molecules that all your side atoms, let's say two of them, their atomic orbitals will do what? They will combine. I am writing it clearly above the arrow. Combine. I did not say they will mix their orbitals. They will not mix. They will only produce what?
6:16 Combinations. Every single word will be clear. What did I tell you? Just keep listening very carefully. By combining among themselves, our ligands will produce different groups of orbitals— the combinations we have just named Ligand Group Orbitals. Suppose they have done that. Second point, step number two: these LGOs they have prepared will then be mixed with them. Meaning, the central atom's atomic orbitals and the LGOs created by the side atoms will mix this time; understand the difference between both words, as when LGOs are formed, orbital mixing does not happen, only different
7:01 combinations are formed. How many combinations, and how? We will learn why all those things are the way they are. We have to study LGOs in detail. So, you understood one thing: the side atoms will create them. They will bring the LGOs. Then we will perform mixing with the central atom's atomic orbitals . So the molecule, for instance, AB2. Then the molecular orbitals of the AB2 molecule will be formed. So this means that from today onwards, you have understood this. If I have to draw the MO diagram of BH2, you have understood this point that sir, on this side there will be Beryllium and on this side there will be Hydrogen's LGOs. So when I write H2, it is not a hydrogen molecule. These are the LGOs of two
7:48 hydrogen atoms. Who has created the LGOs? Two hydrogens. Exactly the same concept. If I ask you, tell me child, we have to draw the MO diagram of BH3. Has it entered your minds? Sir, on one side there will be Boron's atomic orbitals and on the other side, the LGOs formed by three hydrogens, which will mix to form molecular orbitals. If I ask, tell me child, how will the MO diagram of CH4 be formed? You will say sir, on one side Carbon, on the other side four hydrogen LGOs; in Ammonia, Nitrogen and on the other side, three hydrogens. Apart from this, if I talk about a complex like octahedral ML6. On this side the metal, here your six ligands' LGOs, so a small idea has been formed about what kind of molecular orbital diagram we will see. Now let's
8:33 move on to how, sir, will those LGOs be made? What kind will they be? So we know this thing, child. Because we are studying a molecule like BH2. So while drawing the MO diagram, we know Beryllium has one 2s orbital, whose symmetry we know, what kind is it? It is spherical. Right? Similarly, it also has p orbitals. 1, 2, 3, all three have the same energy. X, Y, and Z, in this way, there is 2px, 2py, and 2pz. You can write it out completely as well. Or else you must have understood already. You are smart enough to know these are 2px, y, and z orbitals. On the other side, now what will we bring? LGOs of whom? Hydrogen. Actually, when we
9:19 prepare LGOs, we have to keep in mind what the side atoms do. They produce such different combinations from their atomic orbitals so that they can adopt a symmetry that allows them to mix with the central atom's atomic orbitals, because I mentioned that for mixing, two things are very essential. One is energy and the other is symmetry. If their symmetry does not match theirs, then mixing will never happen. So it means these two hydrogens, if these two separately, keep this in mind, before drawing the MO diagram of anyone, you must have an idea of the shape of that molecule. If you don't have an idea of the shape, if you don't know where the
10:04 two hydrogens are coming from, you will never be able to prepare LGOs. Meaning, I know very well what kind of molecule BeH2 is. It is a linear molecule. This means I have an idea that the two hydrogens will be seen doing what to each other from 180 °? They will be seen approaching. So, we have now removed the beryllium. We are only talking about the H fragment. Only about how these two produce combinations. So that mixing can take place with its atomic orbitals. So we know that this hydrogen also has an orbital, and this hydrogen also has an orbital. So, keep this in mind, the first point worth noting in your
10:49 notebook, which you must have already written before, is that as many atomic orbitals as the ligands have, that many Ligand Group Orbitals will be generated . For example, if one hydrogen here has a 1s orbital and the other has a 1s, then there are two atomic orbitals, so how many Ligand Group Orbitals can be produced? Two can be produced. Because when we are making Ligand Group Orbitals, you should understand Ligand Group Orbitals as a form of combination of atomic orbitals. So, when we are producing these different combinations, producing LGOs, what will the energy of the LGO be equal to? It will be exactly the same as the energy of the atomic orbitals of the two hydrogen atoms. Why ? Because until mixing occurs, there will be no change in the orbital energy . Because bonding and antibonding were only forming upon mixing. When LGOs are formed, mixing does not take place.
11:36 Then we are only producing what? Different combinations. And when we are producing combinations, what is always in our minds? Because we will make combinations based on the fact that with whom do we have to cause the overlapping? With these. If we can create some symmetry from them that overlaps with these, then overlapping will occur. Otherwise, it will not. For example, the very first thing these two hydrogens will think is, "Bro, in what range does the energy of our LGOs fall? " Can they interact with 2S? Or can we interact with 2P? Or do we interact with both? So, the energy of hydrogen's 1S, as we have studied before, will come approximately here. So this is the LGO energy, and why did I make two? Because there will be two LGOs and the energy of both will be equal. Now keep
12:22 this in mind. Energy-wise, both hydrogens are thinking that we can interact with both 2s and 2p. Because the energy is compatible with both. Now they ask, with whom should we overlap? What kind of combination should we produce? So they say, since we see a lower energy 2s orbital, we think we want to overlap with Beryllium's 2s orbital. If we want to overlap with Beryllium's 2s orbital, what should our symmetry be? So if they know this is Beryllium's 2s orbital. I will show it like this. So if Beryllium's 2s is in a positive phase, then both of them will say, let's do one thing, we will also proceed in the same phase. If it is negative, both can proceed as negative.
13:08 So the first combination they must have produced, because they won't have Beryllium yet. So remove this Beryllium . So the first combination in their mind is this one. This hydrogen will also come in a plus phase. This one will also come in a plus phase. This is the first LGO that we have produced. And in this first LGO, I am naming it LGO one in-phase. Meaning both are in the same phase. With what can this overlap? If Beryllium's 2s comes in between them, a constructive interference will occur. If its phase is negative, then what will happen? Destructive. So, did you get a little
13:55 idea about LGO? That side atoms produce such combinations so that they can mix with their atomic orbitals. Now they have prepared one LGO for the 2s. How many more LGOs can they prepare? They could only do two in total. So one is done, they can do one more. Now they think, with which one among 2px, 2py, or 2pz should we overlap? So which internuclear axis do we always assume? Z, so they know that the axis from which we are approaching Beryllium is which one? Z. So the px and py orbitals are perpendicular to us. Because both of us are coming from here. So px will be here. And where will py be? It will be here. Whereas the s are coming from where? From here. Suppose two s are coming from here and p is from here. Can there be any interaction? No. So they said, we won't be able to interact
14:42 with px and py anyway. Why? Because the symmetry does not match. So, what can we match the symmetry with? With PZ. But what is Beryllium's pz like? pz is like this. Is it clear? So both of them think that if we have to bond with the pz orbital, because one lobe of its p orbital is positive. Then how will the other one always be? Negative. Or if this one is positive, then what will this one be? It will be negative. So, this second combination, what are they preparing for—that if it is Beryllium's 2pz, then one of us should be positive and the other one should be how? Negative. If we get this combination ready, then with whom will we be able to interact? With 2pz. I have named this LGO2. I have named this
15:28 out of phase. Or it is in a different phase. So, this is LGO1 and LGO2; what is the energy of both? It is equal. Now look carefully at these two hydrogen LGO combinations. Now tell me, do you understand that if there are ever polyatomic molecules, on one side are the central atom's orbitals. On the other side are some combinations made from the atomic orbitals of the side atoms, and those combinations will be of such a type that they can combine with these atomic orbitals. A little bit is being understood because this is just the first molecule. It will only be understood a little bit. Gradually, everything will become clear. Now look,
16:14 when both these hydrogens came with it. Then Beryllium will say, "Bro, tell me, do you have any LGO of such symmetry that is capable of interacting with my 2S?" They said, "Yes, our LGO1 is absolutely perfect for your 2S." So he will say, "Bring it here, son." So one will be formed, we have named it psi () one molecular orbital. If the phases get reversed, son, an anti-bonding one can also be formed. Beryllium will say, "I have 2px, 2py, 2pz, what combination have you come up with? With whom will you be able to overlap?" They said, " Since our molecule's shape is linear and if we consider the internuclear axis as Z, then we have brought a combination for PZ's combination, with which bonding and anti-bonding interaction can occur." So, with this PZ one, there is a bonding LGO2 one.
17:02 And there will be an anti-bonding interaction. In this way. Which two were left over? X and Y. Both of these are of the same energy. Whether I show this above or below, or together, it won't make any difference. But you should know that both are of the same energy. Let's just make them of the same energy. In this way, these are non-bonding. Is it clear? So this is the MO diagram of BeH2. If we name their molecular orbitals as wave functions, this is psi one. Let's suppose this is two. This is psi three, psi four, psi five, and psi six. This is their increasing order of energy. Okay? Alright, now did this point make sense? I will remove this from here. Because we will understand all concepts slowly from here. Now I will also tell you about their shapes and give you some idea about them. We know that
17:49 beryllium has two electrons in its valence shell. Okay? And hydrogen, one hydrogen has one, so two hydrogens have two, making it four electrons total. If you fill those four electrons, one, two , three, and four. Then you will see that two bonding ones are filled. The non-bonding and anti-bonding ones are what? They are completely empty. Now you have to look at each thing, child. What will be the shape of psi 1, what type of molecular orbital will it be? So you know in one, this was beryllium because it is formed by bonding interaction and these were the hydrogens; if they have had this bonding interaction, then psi 1 must be forming something like this. Are you getting the point? And if it is forming like this. This is the shape of psi 1. Three here. So this is the shape of psi
18:38 1. Now you tell me, is this gerade or ungerade? I can say, is the lowest energy psi 1 of BeH2 gerade or ungerade ? So what is the answer? Psi 1 will be gerade. Because everything is what? Either it will be a positive phase. Everything will be what? Negative. Because this interaction is a sigma interaction. Pi interactions are the ones like this, right? These interactions, whether you count this one or that one. How are they? They are on the sigma internuclear axis. That is why psi 1 is forming a type of sigma bond and it will be what? Gerade. Is it clear? Now let's talk about psi 2. What would psi 2 look like? So let's talk about psi 2. Also, look at this too. Does psi 1 have any nodes? The answer is no. What is the number of nodes? I will write it in brackets. Number of nodes is zero. Let's talk about psi two . What is psi two made of? Its PZ and
19:25 its S orbital. So, if its PZ is like this, plus and negative. And the two hydrogens have their orbitals as plus and negative because a bonding interaction occurred. So, look carefully; when the interaction happened, what is this entire phase? Positive, like this, positive. In the middle are these nuclei, like this. And all the phase on that side will be what ? Negative. Is that understandable? So, this is also a sigma interaction because they all happened on the nuclear axis. And I mentioned that sigma orbitals are what with respect to the internuclear axis? They are symmetrical. This means psi two is also sigma symmetric. Okay? Secondly, does it have a node? The answer is yes. How
20:12 many nodes are in it? One node is present in it. One nodal plane is present. Secondly, is it gerade or ungerade? So the answer is, if you go this way it is positive, if you go that way it is negative. So this psi two has what kind of symmetry? It has ungerade symmetry. You can be asked for the shape of psi two. Just like that, the shape of psi two is appearing. Clear? Now I will talk about how psi three and psi four will be. Because psi three and psi four, child, are just Beryllium's ...px and py orbitals, and we already know what the shape of px and py orbitals is like. So, there is no need for us to explain this. Let's talk about psi five. Okay? That psi five is formed in a way through anti-bonding interaction. What will its shape be like? So, let's get an idea. Look, in psi five, it must have happened that if
20:57 Beryllium's orbital was negative and these poor hydrogens 'LGOs were positive, then the interaction that occurred would be...It would be an anti-bonding interaction. Anti-bonding interaction means its representation would be that if these are three nuclei —1, 2, and 3—then we will represent it like this. One here, let's assume this represents the positive one. Sorry , these two should be positive, child, okay. This one; in the middle, the electron density will become something like this. And since this is the positive phase, if this is our internuclear axis, then it is like this . So, in this way, the electron density
21:44 is not in a circle; the electron density will be something like this. So , the bottom line is that you will see there is a nodal plane here, and what about the other nodal plane? It is here . So, how many nodes are there in psi five? There are two nodes. So, you can see how many nodes it has in total. It has become two. In the exact same way, let's talk about psi six. Well, tell me , is it gerade or ungerade? The answer would be, sir, this is plus. This is negative. This is plus. But if we rotate the internuclear axis, the circles will rotate just like that, so there is no effect on the phase. With respect to the internuclear axis, what is sigma? It is symmetric. So, it can be called sigma and it is gerade. Is that clear? Now, let's talk about the last one, six. How must six have been
22:30 formed? So, in that, we have had an antibonding interaction. Between what? One p orbital, plus-negative, and this negative-plus; that is it. Look at it carefully. If this is your internuclear axis, then look at the plus and negative; the hydrogen is here like this. This is the center of the beryllium period, and this is the other hydrogen. So, what was formed by the interaction, because an antibonding interaction occurred here. So, it must have become a bit like this over here. This lobe has become a little smaller. So, we can represent it in this way. It is slightly over here. Represent it however you want, you just have to show the electron density. After that, this lobe is right here. In this way. After
23:18 this, between these plus-negatives, what happened again? Antibonding. So, that is why it is like this, regardless of how I drew it. My point is that a nodal plane must have appeared here. It is clearly understandable. There is a nodal plane here, and what is here? One . How many nodal planes are there? Three. So, in psi six, there are three nodal planes. And if you look closely, whether they are of sigma symmetry or pi. Because even if you assume this is minus. This is plus. This is minus, this is plus. But if you rotate it along this internuclear axis, then with respect to it, what is it? It is symmetric. There will be no phase changes. So, this is also of which symmetry? It is of sigma symmetry. Is it gerade or ungerade? So, what type will the answer be? It will be of the ungerade type. While this one, how is it? Purely gerade. Why ungerade? Just
24:03 look, brother. This is plus, this is minus, this is minus, this is plus, so if you go from the center, the negative and the phase don't match at all. So, this means it is ungerade. So, you have this much information. And you have also understood that Beryllium has two sigma bonds. Look, does this thing repeat information? We have already made this. I am teaching you everything in complete detail. I know that very easy questions will come from this. From this MO diagram, we also learned that Beryllium has formed two sigma bonds with two hydrogens. This is what represents a sigma and another sigma. Okay? Suppose this is sigma gerade and this is sigma ungerade. Meaning, both types of bonds are not the same. They can have two different energies. So, in
24:48 IR, you can connect asymmetric stretching, etc., with this. So, this MO diagram, look, its belief and the Valence Bond Theory's belief are both different. So, can both their results also come out? Different. So, what did you learn while studying the MO diagram of BeH2? Understand very carefully. Because from today onwards, we will discuss more difficult cases. So, we have to keep in mind that whenever we make an MO diagram for any polyatomic molecule from today on, we will keep in mind, sir, on one side will be the central atom, which I am representing as metal. On the other side, now you might be starting to understand a little. On the other side, the ligands must have produced different combinations. They must have produced combinations of such a type. So that the central atom, with its atomic
25:34 orbitals, they can......be able to mix. Those different combinations that have been produced, what do we call them? LGOs, so that means these LGOs of these side atoms mix with the central atom. Depending on the symmetries that can be formed, those types of bonding and anti-bonding interactions can be formed . So, not very difficult questions come from MO diagrams. As much as I have told in the class. Keep listening very carefully. I will tell all the MO diagrams and I explain this because I have explained it to you in a very simple way. Moving forward, in the H2O MO diagram I will teach next, I will incorporate both quantum mechanics and group theory to show how LGOs are actually produced. Because I have already taught this method. I have taught this on the app as well. I have
26:22 mostly shown methods on how you can prepare LGOs. But what is the reality, brother? You can make LGOs for molecules like BH2 or H2, but the moment I ask you to make them for SF6, you will have to think a lot. So what is the actual, correct way to make LGOs , how do we find out, and until now, I haven't even written whether the interaction between them will form an A1 symmetry or B1. I mean, you know, if you have studied LFT, you know very well that you also assign names to the metal orbitals here, like the metal has A1, B1, and then the ligands will come from here. You will say, sir, the ligand has A1, B1. Then you say, sir, the ligand has B1, B2, there is no B1 at all. So how do we find out? How will we know, sir, which symmetry products
27:07 the ligand can form as LGOs? And for the metal orbitals, sir, how did we state that this is A1 or B1? So remember, didn't I teach you A1 and B1 character tables in group theory? What? But the molecule symbols there were capital letters. A1, A2, but when you use them for orbitals, just make them lowercase. Everything else regarding symmetry properties? Will remain exactly the same. Meaning, only the student who has studied group theory will be able to understand the H2O MO diagram, because I will teach it in full detail. Along with that, I will definitely keep teaching those tricks. I showed here that when I drew the MO diagram for BH2, I didn't use any symmetry at all anywhere. I will definitely do it in the H2O one, and we will study all the subsequent ones in this manner. We will learn the actual chemistry, the real chemistry. Because what problems did we face while just joking around? The problem is, son,
27:54 that in lower classes, the teacher didn't even tell you about the sigma-pi crossover. What is called a sigma-pi crossover? Going from N2 to O2. Have you been taught diatomic molecules or not? When we go from N2 to O2, we know this very well. For N2, the sigmas are at a lower energy. But for O2, the pi orbitals are at a lower energy. Man, I taught you the MO diagram, right? Why do you do such things? I have taught the MO diagrams. Only in N2 to O2, due to SP mixing, the energy of sigma and pi gets exchanged. I have taught it in complete detail. So, that is what is called the sigma-pi crossover. So, when our teachers taught us N2, O2, etc., in lower classes. They didn't give much of an idea about SP mixing. Even if they did, they would just give an idea up to heteronuclear molecules and leave it. Now, if we talk about LGO, most people
28:39 just read it in a way that, "these are formed, so these are LGOs." But I will teach you the way LGOs are actually formed. So that in the future, you can read any example from any book. Why are we unable to understand by reading books today, like why we don't understand how they drew this BF3 MO diagram? Everyone has the book. We all tried, but we didn't understand. Why? Because the teacher just made tricks out of the examples given, which is a good thing. The trick came in the exam, and we solved it. But we couldn't understand every case. And now, see in the H2O diagram how dangerously the MO diagram is constructed. How symmetries are derived. If you want to understand everything, you have to listen to every single detail. And if you want to draw the MO diagram directly, I have already explained on the app how to directly make the H2O MO diagram; it can be done
29:25 . But I will teach every single thing here in detail. You all keep studying along with me. Whatever doubts you have , you can comment. Best of luck. See you in the next lecture. Subscribe to our YouTube channel and download the app now.
