 So, in the previous class, we are discussing one of the important topic is how the molecular structure and molecular properties are interconnected. And for that, we find out that this originally Schrodinger equation H i equal to E psi can be used as one of the starting point where we can actually use it to find out how it is actually happening. H is represented by the Hamiltonian and which typically defines the surrounding of an electron that we are actually mostly looking into and try to find out its wave function, its energy and other properties. So, over there, when we talk about this H or Hamiltonian, we found that is connected to the surrounding and that actually has a huge role to play during the regulation of this molecular properties. So, if we want to understand how this surrounding is playing a huge role, we have to define it or understand it properly. And we found that we can understand it with the help of mathematics. With the help of mathematics, we can understand and we can use symmetry elements because the symmetry elements can define how we can actually project the electron in different orientation and how its relation with the other existing atoms and molecules. So, then we also find out there are different molecules, billions of them, but we can all define them with the help of only five symmetry elements, five different kinds of symmetry elements. First is proper axis of rotation, then there is the plane of reflection, then there is center of inversion or center of symmetry or center of reflection. Then we have S n, which is the improper axis of rotation where you do a C n followed by a sigma H operation. And at the end, the operation is the identity operator. The importance of the identity operator will come again when Professor Leela will teach you, but the system is that this identity operator is required to ensure these symmetry elements can have a particular relation in between them. We can actually club them or assemble them such that you actually form a group and that is which is known as the point group. So, this point group suggests that each of the molecule can be defined in respect of what are the different symmetry elements present. And if you have different symmetry elements, there is a particular relation also connected among them. For example, as we were saying earlier, if you have a C n axis of rotation where n is your principal axis, that means n is the highest number possible. At that time, if you have n number of C 2s perpendicular to that, you have to have n number of C 2s. You cannot have n minus 1 or n plus 1 or any other number. Either you have n number of C 2s or nothing perpendicular to it. If you have sigma v planes in a molecule where the principal axis of rotation is C n, you have to have n number of sigma v's or nothing. So, this kind of interesting relations exist between them because they are actually following the simple laws of mathematics. And when we combine them together, we find they can be termed as point groups and each of the point group can be defined in terms of called character table. And if I can find out what is the point group of a molecule, which is a pretty straightforward method that we have discussed in the earlier class, you just have to ask a few questions to the molecule with the answer of yes or no. And after a few questions, you can end up what will be the point group of the molecule. Then look into the character table of that particular point group. This is very similar to the periodic table, which is already available in the internet, which has been prepared well ahead by the mathematician. Just look into that and from there, you can define different characteristics of a molecule, whether it is spectroscopic interaction, so there is a bonding interaction, all these things you can connect back to the character table and the point group. So that thing in our mind, we figured it out that yes, the symmetry elements, the point group, the character table are the important factors. And if I want to define a molecule, I need to find out what is the point group. And from there, I can comment on different molecular properties. So one of the molecular properties that we are more interested in is known as the chirality or optical activity. And last class, we try to figure it out what is chirality or what is the optical activity. And if I remember correctly, we have five different definitions come out of it. So I am again writing them one by one. So chirality is such that a molecule which can rotate the plane polarized light will come into that a little bit later probably next class, what is a plane polarized light. So that is will be defined by chirality. If it is a chiral molecule, it will rotate. If it is a chiral molecule, then it is not. Second definition we have that take the mirror image of a molecule and see if it is superimposable or not. If it is non superimposable, then it is chiral molecule. If it is superimposable, that means it is a chiral. Superimposable means that is going to match exactly what your starting molecule is. The third definition we have found that your molecule cannot have any plane of reflection. It should not have anything. The fourth definition was that your molecule should not have a center of symmetry. And the last definition was your molecule should not contain any improper axis of rotation. So these are the five definition we found that can define whether my molecule is chiral or not. So one of the other way I actually try to define that all these particular systems, especially the last four are actually interconnected among them. To exemplify that, we actually took the example of this particular chiral molecule. This is nothing but carbon connected with four different groups. And what we found that first take the mirror image. So that means you are following the definition two to find out how it is actually done. And we found this molecule looks like this. And then we try to rotate it along this CH axis to ensure that it looks as close as similar to the original molecule where this chlorides and the CH bonds on the plate and on the right hand side of the molecule. And we did that and when I rotate, we found everything is same but the fluorine and bromine, these two systems actually exchange plates. So that's why it is not anymore superimposable. So these two molecules are non superimposable. So we said that this is nothing but a chiral molecule. And then we try to find out when you are taking the mirror image over here, this particular function, it is nothing but a sigma reflection, the plane of reflection. And then when you try to rotate it and try to see how it is superimposable or not because that is what we typically do even without understanding or even with a comprehending that we are actually nothing but doing a rotation. When you try to find the mirror image, it is superimposable or not. So over here, we are trying to do different CN operations. So we are doing a sigma operation, we are doing a CN operation. So basically you are doing a SN operation. And some of you might argue that this plane of reflection might not be in the perpendicular to the CN axis we are doing, but we learned that it doesn't matter a molecule can have only one particular reflection. So outside the molecule, if you want to put a plane of reflection and take a reflection plane, it doesn't matter where you put it, you are going to find the same reflection. It is kind of, you are looking at a mirror, doesn't matter how do you look into the mirror, you are going to find the reflection of yourself, no other person will show up on the other side. So similarly, no matter which particular way you put the mirror, if you are looking at a particular molecule, the reflection would be that particular molecule and it can be only one particular reflection. So in that respect, we can say over there what we are doing a CN and sigma age, we can say so we are doing a SN operation. And that means these two systems, if it is superimposable or not of the mirror image and the SN is nothing but the same. And similarly we can found that if my SN, the N is one, that means I am doing a C1, that means leave the molecule as it is and only doing a sigma plane. That means S1 is going to be equivalent to a sigma. So that defines corollary number three, that a molecule cannot have a sigma plane because basically it is having a S1. Similarly, if you do some mathematics, you can find the S2 is nothing but center of symmetry. So similarly, it defines the corollary number four. If you have a S2 system or a center of symmetry, you also cannot have a chiral molecule because the mirror image will be very similar to superimposable to the original molecule. So with that respect, we can figure it out. All these four corollaries, two, three, four, five, they are nothing but the same thing defined in different ways. So from there, we find out that what we need to find that whether the molecule having SN or not. Now finding an SN might not be that easy. So what we are going to do for that, we are going to find out the point group of the molecule. And as we know, from the point group, I can look into the character table. And in the character table, all the possible symmetry elements are already given. And over there, I have to just glance it and to see if there is any SN axis present here or not. And by SN, I mean also sigma and i because there are variations of S1 and S2. And what we found there are only two different point groups present in the world which doesn't have any of this SN symmetry elements. And those are CN and DN point groups. These are the two point groups have no SN axis. And by no SN axis, I also include S1 equivalent to sigma and S2 equivalent to i. So this is something we have to remember. So what we need to know is to find out the point group and figure it out if it belongs to CN or DN. If it belongs to CN and DN, your molecule will be guided. If it is not, it is accurate as simple as that. And again, we look into that how to find out the point group of the molecule very easy. Ask a few questions to the molecule and they will reveal what is the point group of the molecule. So that is how we are actually generally find out what is the point group of the molecule and whether it is a chiral or not. So as it is inorganic special cast and even the name is actually inorganic complexes. So I'm giving you an example. And this example, I'm giving a general example. Say metal is there, which is actually coordinated in an octahedral geometry. So now over here, there are two important things. Coordination geometry and the other term is symmetry. They are not the same thing. Coordination geometry means how many bonds it has, what is the geometry it is having. So if it has six bonds and this particular geometry is a, it is an octahedral coordination geometry, an octahedral coordination. But it doesn't mean that it is actually an octahedral symmetry because if you want to have octahedral symmetry, you have to have six equivalent ligands connect to it such a way that you go any direction through the center of the molecule, you're going to find a same thing. So this molecule, if you have six same ligands, say this l ligand is nothing but water molecule. Then yes, your molecule is not only octahedral geometry, but it's symmetry is also octahedral. So over here, the symmetry is also octahedral. But now say I took the same molecule with octahedral geometry. But instead of water, I have oxygens over there and the oxygens are nothing but coming from the carboxylet oxygen. So it is say as a oxalate group, two carboxylet group come to each other and over there now say they're connected. Now once they get connected like this, also the geometry is octahedral, coordination geometry. But the symmetry is not octahedral. Why? Because over here just see if I see at this seat at this particular point and go through this molecule, I don't find a linker over here. It's missing. Similarly over here, the linker is missing. Similarly over here, the linker is missing. So although the coordination geometry is octahedral, the symmetry over here is not an octahedral one. So be very careful when you define the difference between octahedral coordination geometry and octahedral point group symmetry. They are two different things. So now what is the point group of this particular molecule then? So let's take another deep look into it. So over here, what is the point group of this molecule? So again, it is not a junior level class. So I'm not wasting my time to find out exactly where the different symmetry elements are. So I'm just giving you the answer. So this molecule has the highest symmetric axis or rotation present over here is C3. And where is the C3 present? To look into that, what I am going to do, look into this ligand a little bit carefully. So this ligand, which is nothing but oxalate ligand, too negative, which is outside ligand and connected by a carbon-carbon linker. So this particular linker, when it is present over here, you can see that two particular bindings for the same ligand, what we call them as bidentated ligand. This term dented comes from the density. That means how many times it is biting the metal? It's biting two times. And over there, they're connected to that. And now see, take a look into each of the ligands. There are three such ligands. One over here, one over here, and one over here. And each of them has two pairs of metal oxygen bonding. And if you look carefully, one of them, you can say towards you, one of them is actually backwards to you. Take an example of this one. This system over here. So I'm writing O1 and O1. So it is O1A. It is O1B. You can say that O1A is actually towards you, because this wage bond means above the plane of the paper. And that dotted bond means it is behind the page of the paper. So that means O1A is towards you, O1B is backward. Similarly, over here, if I say it is 2A, it is 2B, you can say 2A is on the plane of the paper, 2B is on the backside. So that means again, 2A is towards you, 2B is on the backside. Over here, say it is 3A and 3B. You can say that 3A is above the plane of the paper, 3B is on the plane of the paper. So relatively, we can say that the 3A is towards you, 3B is backward. So each of them has two sets of metal oxygen bond. One of them is towards you, one of them is backward to you. Now connect all of them, which is above the plane of the paper. So this one, this one, and this one. And connect those, which are actually backside of the paper. So let me probably put a different color to differentiate it out. Now if I draw the molecule one more time, and now what I am doing, drawing the linker in a dotted line so that we can see the triangles properly. So one triangle is this one above the plane of the paper. One triangle is backwards. Now if you rotate the system through over here, a C3, 120 degree rotation, because you see it is a triangle over here, forming over here, other triangle forming over here. So there are two triangles oppositely oriented sitting there. So if you rotate that, you will see a similar structure, exactly similar structure you will find. And that is going to give you the same structure. So there the C3 is present. So for an example, if I rotate it, this particular system is going to go over here, this particular oxygen is going to go over there. And this connection is also moving in the same direction. Then you can figure it out. This is the C3 axis. Now if you have a C3 axis, obviously I am not going to, it is not linear, not special group. So if it is a C3 axis, the next question we ask to it, whether it belongs to D or C point groups. For having to, for the D or dihedral point groups, you have to have C2's perpendicular to the C3. And if you have C2, you have to have 3 or nothing. And there is present C2's over here through this oxygen-oxygen connector. If you rotate that 180 degree, you can see you are going to get the same thing. And as I said, if one is present, because the principal axis C3, the other two should be present and would be in a very similar position. So this will be over here going through this oxygen-oxygen box. So you have C3 and you have 3C2 perpendicular to C3. And there is no other symmetry elements present there other than obviously centrosymmetry. Sorry, the identity factor E, the rest of them are not there. So it is belong to a point group D3. And that is why this molecule is going to be chiral molecule. So what you now need to do is either say directly, yeah, it is going to be chiral, or you can also do that same operation that we actually learned from our beginners that take a mirror image and see, try to see if it is superimposable to its original structure or not. So let's try that. So here is this one, here is this one, here is this one. And then over here, I'm taking the sigma for the mirror image. So first draw the molecule, we'll draw the connector later because coordination geometry. So the coordination geometry is going to be octahedral. So over here, then draw the oxygen first. And then at the end, we'll draw the connector because that is the only thing that's changing. So over here, this is the connector. And I'm writing O1 O1 to understand it better. So they come over here. So that will be the connector. So this is O2 O2. So that will be over here and over here. So there will be this connector. And obviously the O3 O3 will be somewhere around here. So now this molecule, is it superimposable or not? So again, after the sigma orientation may not be that easy to understand. So we have to try to rotate in such a way that it actually matches to its original structure as much as possible. So what is the original structure? Metal in the center, that is it. But over here, see these two oxygens, which is one above the plane, one below the plane, is on the right hand side. Now it comes on the left hand side. So let's put this one on the right hand side. How we can do that? If we rotate that over here 180 degree. And if we do that, what you're going to get is the following. Again, first draw the oxygens. That's going to be the same. So if I rotate it over here 180 degree, these two oxygens system will come on the right hand side. That is my O101. Now if I rotate it 180 degree, the O2 on the top one is not going to change its position because that is sitting on the axis of rotation. But this O2, this O2 is going to change and come to this side 180 degree rotation. So it will come to the forward side. So this is the O2. So now their connection is such like this. And obviously the last of the O3s will be connected from the bottom. But just imagine how to do that. So this is the O3 over here. That is going to be same position. This O3, which is in the forward, move 180 degree and go to the backward side. So now you can see these two molecules, although the oxygens are almost in the similar positions, but their connectors are not. Because over here you can see this O2, O2 connector is kind of on the back side. Now it is come to the forward. Similarly, this O3, O3 is kind of in the forward side. Now it goes to the backward side. So that is why these molecules are not super impossible. Obviously this molecule is going to be chiral molecule. And that we have found out earlier when we said this is the point group of D3. Now the nomenclature of this kind of inorganic molecules, how to do that? It is again very simple. What we have to find out? Again the connectivity between the same ligand molecule, because the ligand molecules are bidented at this condition. We have to find it out. Each of the end of the same bidented molecule, how they are connected? Because each molecule has one top side and one bottom side. So top side I am putting a star and bottom side I am putting a double star. So this is the star double star for this particular ligand. For these two you can say this is the top and this is on the backward, relatively. And over here this is the top and this is on the backward side. With respect to on the plane of the paper you have drawn. This is wedge bond top. This is dotted bond bottom. This is solid bond. That means on the plane of the paper this is dotted bond. That means backward side. Similarly this is wedge and this is on the plane of the paper. So even if it is plane of the paper it is relatively backward side compared to this one. Now you look into them and try to find out which direction you are moving from going to top to bottom. So over here I am moving this direction. Over there I am moving this direction. Moving this direction. You can see all of them are moving in the similar direction. So if it is moving in the similar direction then try to find out what is the overall rotation. You can see it is a rotation on the anticlockwise or lever rotatory. And this particular system with anticlockwise rotation it is known as the lambda isomer. On the other hand side if you look into this particular molecule over here I can see this is the top. This is the bottom. This is the top. This is the bottom. This is the top. This is the bottom. And now which direction it is moving now. All of them you can see it is moving on the right hand direction. So it is a clockwise movement and clockwise movement and this is actually known as the delta isomer. So generally for this thing to understand which is lambda and which is delta and lambda. So try to put the molecule in such a way that will be easier for you. For an example if you get this particular orientation it is the same molecule but it may be a little bit difficult to get out the delta and lambda. So try to put in this particular orientation so that at least you can find which is top and which is bottom very easily and then try to figure it out. And with respect to that you can find out which is the lambda and which is the delta isomer. And over here this molecule is called delta and lambda isomer no matter whether the C2 axis is present there or not. So over there the C2 axis are present because both end of the ligand is same over here. They may not be the case if both ends are not same. For an example you are using one end amine and one end carboxylate. Then the molecule is the ligand is still bidented but not symmetry. And in that case if you put a molecule with this particular ligand three of them and then you can still find a molecule very simple similarly looking but you might be losing the C2 over there. So that molecule will belong to a C3 point group. So be very careful when you look into this kind of multi-dentate inorganic molecules which are actually optically active. Very careful which point group it belongs to and how to find out their nomenclature delta or lambda isomer. So any question till this time? Say anyone have any question? So over here so far we have gone through that optical activity is important and optical activity does exist even in an inorganic molecule. Even the coordination geometry is not a chiral geometry and if it has bidented ligands three of them if it's coordinated it can produce a optically active or chiral complex. And now we know how to do that, how to find its point group, how to find this the nomenclature of the different enantiomer. So this delta and lambda isomer obviously they are enantiomer very similar to DRL isomer that you find out from the organic chemistry.