 So with that, we come to the end of the city spectroscopy discussion. So in the city spectroscopy discussion, we have gone through this following points. First of all, why chirality is crucial? Because it connects molecular structure with molecular property. Then we look into the system why it is important because biology actually plays with chirality. So chirality is an important factor over there for controlling the molecular recognition, controlling the reaction rate. Even you can say chirality is a signature of biology. If you have an enantiomeric excess of a sample, you can say it is probably something biological is influencing over there. So we look into especially the amino acids. So study the amino acid structure. There are one letter and three letter codes and how it is happening. So we actually gone through them. And then we looked into the molecular origin why a system can be chiral. It is coming chirality from not only the molecule, but also the light, what is the LCP and RCP, how it is behaving. Then we found some of the parameters that we can use CD and ORD, which is coming from optical rotation and the ellipticity that it is coming. We found CD is actually a better parameter to find out the chirality of the molecule. And at the end, we have gone through some of the examples of chirality and how it can help us to not only know the structure of the molecule, but their stability, whether it is binding a metal or not, and all this particular information. So these are the take home messages from the CD spectroscopy. Please go ahead and ask any question if you have any before we go for the next topics. So please go ahead and ask any question. If you haven't, or any comment that you want to repeat some of the topics one more time, or you're not very happy of the expansion of a few topics you want to go ahead with one more time. That previously yesterday means that in the previous class we have told that about the directionality. So what is the means X polarized light, Y polarized light and Z polarized, and how we will distinguish them. Okay, so that is a very important question that is mostly going to be covered in Professor Leela's class. So I'm just going to be just a hint of it what is X and Y is equal as light. So when we actually do look into a system, we try to look into a system in a Cartesian axis right so we say okay so this will be my z axis this will be my y axis this will be my x axis. However, the molecule is always tumbling. Right. So there are two different accesses you can think about one is actually a molecular Cartesian system. And the other one will be your experimental Cartesian system. And this all difference between them is how you defined your XYZ ancestors, whose direction what is the angle and all this things. Now in the molecular Cartesian axis is a bit difficult to define because the molecule is not remaining at the same place all together right, unless you go to zero Kelvin, and even then some vibration will be there the molecule is not going to be stable. It is going different motion, hydration and rotation translation, all together you can say it is going to brownie and motions. So you cannot define a molecular axis is all the time very clear. So that is why what we do in experimental system, we actually specifically define this is my XYZ axis. Depending on that I actually measure all the different samples in that experiment. And whenever we define an XYZ axis, it means I am defining it with respect to experimental Cartesian system. Now over here in an experimental Cartesian systems, let's say this is your source of light. And this is actually a crystal that you put to create your linearly polarized light. And here is your sample. Now in the sample, if you look forward there are different molecules and they can have different orientations. All of them are going to be same. And how I'm going to define it. So I'm going to define it with respect to the XYZ axis of my experimental system. And over there I'm saying that I'm going to put my crystal such that I am going to get only this axis, or this axis, or this axis. So with respect to that I'm going to say I have a X polarized light Y polarized light and Z polarized light. And over there, the molecular motion is quite fast. And the interaction and time I'm giving to this molecule for this optical spectrum, it is happening in the millisecond to even second scale. That is my time region and within that time, a molecule will be very similar to the XYZ axis of my experimental section for a Y. And in that time if it is there, it will absorb the light depending on whether it is Z, Y or X polarized light active system. Okay, now how I can define that which particular transition is XYZ LI for that system again you have to look into the transition moment integral. Look into your excited state operator and ground state. And what you need to look into what is their symmetry representation. If you look at it, you can say like it is a pi to pi star these two Sigma star all this interaction, each orbital you can find out with respect to the symmetry element present in that particular point group. What is that symmetry representation, say it is a one say it is a two. Now a two and a one or together if you want to have it to be non zero, whatever is the symmetry presentation of me all these things should come together and give a non zero system and it is only possible in one particular condition, it has to be the a one that totally symmetric representation for that particular point only then it can be non zero. That means you have to find out the X such a way that it will be multiplied with a two and a one and give you a one, and that is very easy, because from the point group again you like learn all those things in the later part of the class. It says, if you want to have a one it is only possible if you multiply the same system to a one. That means a two and a x should be a one together and only then a one into a one, give you a one. If you multiply anything else b one b two a two with anything with the totally symmetric representation a one, you are going to get back that particular representation. So that means a two and a x multiplication should give you a one. And the other corollary of the point group are capable is if you want to have a one by multiplied by two different groups, it is only possible if you multiply the same system. A two into a two give you a one b one into b one give you a one b two into b two give you a one. Okay, so that means the X has to be a two symmetric, and then you look into the character table find out along with the a two, what is there. So for an example, let me take a graph of that. Let's say this point group of C to the point we're looking about. So over there, say I find out a one is a totally symmetric representation. And this is what I want after that. If you look at EMI it should be a one and say it is nothing but a shy excited state, my operator and ground state say my ground state is a one excited state is a two. Now I have to find new search that whatever it is, multiply that and give me a one. That should give me a one together, because you can see it is all one one one. So you have to multiply that thing with a one only then you will get one one one back for the symmetric representation. So a two and X should give you a one that means it should be an a two symmetric, because only a two into a two can give you a one. That means it should be a two symmetric and look into there over here, you see there is no XYZ so no matter what polarization you give this will be a transition that you will be never be seen. Now say my excited state is actually be one and ground state is still a one. So then it should be a be one again so that we can multiply all these things together and get a one. If it's a be one, then it should be a X polarized only then it should be happening. Similarly, for be to active system it should be Y polarized a one active system it should be Z polarized. And by that you can differentiate which particular operation is happening and if your molecular structure that you are imagining to be hold a particular point group it is true or not. Does it answer your question or go. Yes, sir. Okay, any more question before we close it up.