Now early in your careers. Yeah, you've won many hats. Yes You have separately been in workplaces.  Perhaps Peter first to you if you've been in a situation in a in another life where you've been in a workplace And you knew that you were suffering Yeah, no how to handle it at the time Yes, and I was in that situation because I am in during the 90s I did have bipolar disorder Diagnosed and I and I took medication for it and suicide was one of those options that I contemplated and Until I reached a point in one day to say that's it no more and I took a hundred percent responsibility I didn't know how I was gonna do it, but I started crawling out of that hole What was the first step to taking that responsibility that as you said that's it to realize to realize That it no matter how loving and beautiful people are around me No one is going to care about my life or should care about my life more than I do. I Need to take care of my life. I need to find out my answer I'm an unique individual and what might work for you is not exactly what works for me But what did work for you? Well for me what I did was to go and research What had people that felt great about life did that I didn't do that's what worked for me. So You know Nelson Mandela He seems to have it together He was 27 years in prison if I'd been 27 years, I don't think I would have lasted to you know What did he do? How did he approach the world? What was it was his way of thinking so I read a biography on Nelson Mandela? I tried to soak up his personality his way of acting of understanding the world Gandhi was another one You know going from trying to fit in with society to becoming a disruptor and and through peace through peaceful men Through peaceful means to get rid of an empire in India. That's a that's a huge achievement So how did he approach people and and and it was incredibly similar But the Mandela's of the world and the Gandhi's of the world and the Branson's of the world or mother Teresa's of the world They have a way of operating in which they do not act as victim ever And that was the that was the lesson for me and and to to realize that there was a pity party going on in my brain You two work closely together Have you ever challenged Peter and said stop feeling sorry for yourself? Oh, absolutely We keep each other accountable to that all the time. It has to happen Absolutely, and and I think it's helpful to have someone like like that around you You know whether it's someone in the workplace whether it's someone at home And you know a friend a family member. Yeah Did you discover to yourself the effects of mental ill health in a workplace? Look, I don't think I've ever I mean we talk about where's that line where it crosses over into a mental health problem Something to be diagnosable or not. I wouldn't say that Cross that line, but definitely, you know our levels of mental well-being fluctuate and there are definitely periods at work for one reason or another where I can remember, you know having been more stressed and Usually it comes down to the dynamics of in in amongst people socially you had a HR boss HR boss, that was not very HR I've had a number of different HR Because that tends to be the area of a workplace where I gravitate to and so I've worked with HR a lot and I've had You know exactly what I was praised for in one from one manager was what was a problem with another manager and You know the good thing is I think because we adopt that philosophy of taking responsibility and taking that personal responsibility And not going into the victim space that kind of allowed me to say well What are my options here? How am I going to deal with this? Can I manage up? Is there any what can I change in this dynamic? How do I want to approach it or when is the time for me to move on to a new challenge and Take on something new because this situation simply isn't working for me And then they can be very difficult decisions for people to make You know do I stay do I go how do I should I change me? Should I try and change them? How do I tease all of that out? But for me that Making a decision and moving forward was what was most important not to stay stuck in Okay, great. Yeah


Hello and welcome to this new segment of CD spectroscopy and MOSBUS spectroscopy for chemist. My name is Arnab Datta and I am an associate professor in the department of chemistry at IIT Bombay. So, in the previous segments we are discovering the different symmetry elements and operations present in the nature and having a mathematical perspective of this particular systems. And from there we figure it out that there are different groups can be formed with the different symmetry elements present. And out of these 5 symmetry elements that is present in a molecule that could be present in a molecule are identity operator, single axis of rotation, plane of reflection, improper axis of rotation and center of inversion. And all of them can be combined in a particular way to differentiate them in different four segments of the groups. And those we have defined how we can figure it out and then we try to find out by asking rational question how I can easily figure it out what will be the point group of a molecule. So, in the last class we have done that, so today we will just recap a little bit and then go through different examples of molecules and try to do this exercise to find out the point group of a molecule. So let us begin. So, first of all again we are going to showcase that how to figure it out the point group of a molecule. So you can say it is point group determination. So what I am going to do we are going to ask some question the first question we are going to ask is the molecular structure that we are dealing with is it linear or not. And the question we will ask in such a way that will be binary either yes or no answer. If it is yes the next question I am going to ask do you have a center of symmetry or not and again two possible answers yes or no. If it is yes then it belongs to the point group of D infinite H, if it is no it belongs to the molecular point group of C infinite V. So these are the things when you have a linear molecule. If the answer is no, if the molecule is not linear the next question we ask whether they belongs to any particular cubic groups or not. And by this particular cubic group by mean only two particular groups that is mostly common in chemistry octahedral and tetrahedral. If the answer is yes the molecule is not linear but it belongs to particular point group which is easy to figure it out then it belongs to this particular cubic groups. And most of the time we will find the molecules that we are dealing with is not that simple it is not linear or neither it belongs to a particular cubic group it will be much complex than that and there we need to ask the next question it is not linear it is not cubic group we will ask do you have a axis of rotation or not. If the answer is no that means it belongs to a non-rotational group then the next question we ask whether you have a center of symmetry or not if the answer is yes it belongs to CI point group if the answer is no the next question we ask do you have a sigma or not. If the answer is yes this belongs to CS point group if it is no it is C1 the most simple of the molecule without any symmetry operation or element present other than the identity operator. And this three groups C1 CS and CI belongs to non-rotational group but if the molecule has a CN present then we can be either in the single axis rotation group or dihedral group and the factor that differentiate between them is this one do you have a number of C2 perpendicular to that principal axis CN or not if the answer is no that means I belong to the CN based point group and over there then I figure it out whether they have sigma h or not if yes that will be CNH point group if the answer is no then the next question we ask whether you have n number of sigma v's or not along with you if the answer is yes they belong to CN v point group if the answer is no then it belongs to CN point group. So it does not have n number of C2 perpendicular to CN but it can have sigma h so it is CNH it can have sigma v's it will be CN v or nothing it will be only CN and all this thing belongs to single axis rotation groups and if it does have n number of C2 perpendicular to CN then the molecule belongs to DN point group. So we go to do you have a sigma h the similar line of questioning that we have done just over here we just going to repeat it sigma h if the answer is yes the molecule belongs to DNH point group if the answer is no then we ask the question do you have n number of sigma d's with you over here there are sigma d's because this is the presence of C2 is such that if the sigma v's are present over there they will be bisecting those C2's so that means they will be recognized as sigma d and if it does have those sigma d's that will belong to DN d point group and if it does not have anything other than CN and in number of C2 perpendicular to CN that will come to a DN point group. So with that we actually configure that these are the dihedral groups so that we are going in went into in detail in the last segment now we are going to use them how to do this questionnaire and figure it out how we can figure it out the point group of a particular molecule. So we will take some examples so the first example we are taking is the BF3 molecule we have looked into this molecule earlier it is actually planar molecule the plane of the molecule is drawn in such a way that it is perpendicular to the plane of the board I am drawing right now and what is the point group of this molecule? So what we are going to do start asking questions to this molecule the first question you ask to the molecule are you linear and you can see obviously the answer is no then the question is do you belong to any particular cubic groups that means I am asking whether you have octahedral or tetrahedral geometry and obviously from this geometry you can see it is no then comes do you have a principal axis and the answer is yes there is a principal axis over here where you can rotate 120 degree and get to this molecule. So if I draw this molecule from top view motion so this is the side view and this is the top view of the molecule and you can see this molecule is on such a way that over here through this boron if I wrote it 120 degree I am going to get an indistinguishable and superimposable structure so that means there is a C3 present over there which is drawn over here in this particular side view molecule. So the answer is yes and it has a C3 now once you have a C3 now it belongs to either single axis rotation or dihedral group so figure it out whether it is dihedral or not next question I will be asking do you have 3C2 perpendicular to your C3 so it is n number of C2 perpendicular to your CN so n is 3 over here so I am asking do you have 3C2 which is perpendicular to C3. So you can see over here along with each BF bond I have a C2 I can rotate 180 degree and I am going to get a similar structure indistinguishable and super imposable because this BF bond is remaining same the other Florence exchange places see in this structure from the top view over here and it is not only over here it can belong to any of the C2 belongs to the BF bond so as I have told earlier if a molecule have a perpendicular C2 compared to its principal axis either you will have n number of it n is the CN axis or this principal axis or none so once you find one of them obviously you see in a symmetric way you can find the other also so I found one C2 so there will be other 2C2 so you can have 3C2 perpendicular to C3 so the answer is yes. The next question we ask whether it has a sigma H or not sigma H means principal axis is this C3 a plane perpendicular to it and you can see all the atoms belongs to that plane so that means it will be a mirror image on this particular plane reflection which will be exactly sitting on the top of the original molecule so that means it is a sigma and it is a sigma H. So the next question we ask whether you have a sigma H or not answer is yes and this is nothing but from the top view molecule is the plane of the board and in this particular system is the plane perpendicular to the board and you can see answer is yes so the molecular point group for this molecule will be D3H that we can figure it out from here. So that is how we actually figure it out all these particular molecules. The next question let us do ammonia this is the structure of ammonia the lone pairs so it is a trigonal pyramidal structure and we try to find out what is the point group again same line of questioning is this molecule linear obviously not is this molecule belong to any cubic group. So some of you might need to say yes might think about saying yes because it is having a tetrahedral structure yes it may have a tetrahedral geometry around it but it does not have all the same ligands all places because it is a lone pair over here so that actually breaks the tetrahedral symmetry and it is not a tetrahedral geometry so it is not belong to any cubic group. Next question is what is the CN and the CN is actually this nitrogen where you are having a C3 axis of rotation you can see nitrogen remain as it is all the other hydrogen changes places so yes CN is present and it is C3 axis and the next question is do you have sigma H? See if sigma H has to be there just to be through the center molecule and perpendicular to it and you can see it is not a plane of reflection because an nitrogen remain as it is but these 3 hydrogen will come on the top of this plane which is not the superimposable and indistinguishable structure. So that is why this molecule does not belong a sigma H and the next question is do you have 3 sigma Vs so just drawing the structure one more time so you can see this particular plane the plane of the board containing this NH bond is going to remain same this hydrogen will replace with each other so superimposable indistinguishable structure and that is going to happen in 3 different planes belong to each of the NH bond so that will be sigma V plane because that actually contains each of them contains the principal axis C3 so that is why they are all C sigma Vs and yes it is present so this point group of this molecule will be C3V so that is the 2 examples we have gone so far. Now we are going to move a little bit more after BF3 and ammonia we go to the next set of example. The next set of example I am going to do is N2F2 what is the structure of the molecule and that is the interesting thing this molecule can have 2 different structures this is the nitrogenate and double bond and then you can have 2 different orientations of the fluorine either they can be in the same side so that is known as the cis geometry. So let us take it is a cis so this is the cis geometry structure and what is the other structure possible so it is possible that this molecule is such that these fluorines are in opposite inside so that is the trans molecule. So we will try to find out the point group of each molecule one by one first let us go to the cis again questionnaire is very much similar are you linear question is answer is no next question is do you belong to any cubic group answer is no what is the Cn present in this molecule just when you look into it you can find there is a C2 present over here if you wrote at 180 degree the nitrogen and nitrogen bonds exchange places the fluorine exchange places but you are going to get a super imposible and indistinguishable form so that means a C2 is present is yes it is C2 next question do you have 2 C2 perpendicular to that C2 so that means somewhere around here or somewhere around here and those are not having any C2 because if you rotate over here you can see the nitrogen remaining same that the fluorines come on the top side from the bottom so that is not going to match the original configuration so there is no that means it does not belong to a dihedral group next question do you have a sigma H so if it has a sigma H it has to be in this particular plane perpendicular to the C2 and the answer is no it does not have a sigma H because these two fluorines again going to change its place if we do this operation next question is do you have 2 sigma V's so now over here you can see this plane perpendicular to this plane of the drawing and the plane of the molecule itself the plane of the molecule contains all the atoms so the reflection will be exactly on the same place so that is a sigma V for sure because it contains the C2 also and this one perpendicular to is actually bisects each another and all of them goes to their respective places so that is also sigma V so these are the two sigma V's present over there answer is yes so this molecule belongs to C2 V point group for this C's molecule now the trans molecule now the trans molecule we are asking to do the same question whether you are linear or not the answer is no QB group answer is no do you have a CN now previously our CN was somewhere around here now if you rotate that the fluorine will come to this side this one go to this side so they are not really matching to this original position with 180 rotation but what happens if I rotate over here this is the plane of the molecule perpendicular to it and if you rotate 180 degree the nitrogen remain in a similar position but this nitrogen comes from the left to right and right to left and over there the nutrients are remaining in the same place similar thing happens to the fluorine with 180 degree rotation through this axis over here which is perpendicular to the plane of this molecule what you are going to get this fluorine is going to come over here and this particular fluorine is going to go over there so that means it remain as it is so that means there is actually a C2 present over here so answer is yes and it is a C2 our next question do you have two C2 perpendicular to that C2 that means somewhere around here or somewhere around here so none of them rotate 180 degree and give you a same structure this rotation over here the fluorine will come to left to right to left not on the top of each other same thing over here if you rotate from there bottom to the top top to the bottom but not really on top of each other so the answer is no next question do you have a sigma H now sigma H is now the plane of the molecule because your principal axis of C2 is now perpendicular to it see if I draw this molecule like this this is the plane of the molecular thinking and this is where the C2 is so you can see C2 is actually perpendicular to the plane of the molecule and which is actually containing all the atoms over there so it is a sigma plane but because it is perpendicular to it is sigma H so answer is yes it has a sigma H and point group will be C2 H so please take a look into it one more time at your favorable condition and find out whether it is actually making sense to you C is C2 V trans C2 H now the next one we are going to do with XCF4 molecule so how does this XCF4 looks like so again we are going to draw two different orientation one is from the side view this one and one is from the top view so this is the plane of the molecule so now when we look into that we start asking the same question are you linear or not answer is no QB group answer is no you have a CN the answer is yes it is actually having a C4 directly go through xenon if you noted 90 degree all the fluorine exchange places that goes to a similar looking orientation where the xenon is not even changing its position so it actually have a C4 so over here C4 so these are the 90 degree angles so once that sorted out that I have a C4 now the next question is whether it belongs to deep hydro group or not so do you have four C2 is perpendicular to C4 so over here you can see we have gone through that earlier also two sets of C2s will be there which will be going through the fluorine xenon fluorine bond whereas the other two will be going through the xenon fluorine xenon bond fluorine xenon fluorine bond there is other C2 and there is a other C2 so all together you have 1 2 3 and 4 4 C2 present as it is yes belongs to dihedral group then it is a sigma H or not sigma H will be a plane perpendicular to the principal axis which is the molecular plane in the molecular plane you already have four fluorines and xenon sitting over there so even if you reflect them each and every of them is reflecting on to each other so it is symmetric molecule and it is going to give you a similar configuration from the original bond so sigma H answer is yes so the point group of the molecule is D4H so that is the point group of this molecule the next one we are going to do is the following is a cobalt molecule with 5 ammonia 1 chloride and the overall structure charges plus 1 because it is a cobalt 2 system now looking into that you can see there are 5 ammonia 1 chloride ligands so the geometry is going to be octahedral because that is the geometry the metals preferred in biology so now I am putting ammonias over here all around the place and one chlorine over there and over here what you can see that this molecule obviously not linear but does it belong to a cubic group because this molecule is an octahedral symmetric structure it has but over here you have to understand this although it looks like octahedral coordination geometry the symmetry is not octahedral because to have a symmetry octahedral you have to have all the six ligands connected to the central atom same so over there there are 5 ammonias but one of them is chlorine which is dissimilar and that actually breaks the overall symmetry the coordination geometry is octahedral no problem with that but the symmetry is not octahedral so you have to understand the difference between coordination geometry and absolute symmetry coordination geometry is like how they are actually interacting with the all ligands and how the bonding direction coming from the central atom but the symmetry defines what are the identity even for the atoms connected to the central atom and with respect to that it is not an octahedral so it is not belong to any cubic group then comes what is the principal axis of this molecule so principal axis of this molecule belongs to this when it is going to chloride cobalt ammonia system and it is having a C4 axis over the 90 degree rotation if I try to take a look from the top view of this molecule I am going to see a cobalt over here and all ammonia sitting what it is called the equatorial plane and this axial position has chloride also has an ammonia at the bottom but they all belong to this one particular line so now if I rotate C4 along with this line that means over here the chloride cobalt and ammonia does not change the position the ammonia in the equatorial plane change the position and that is actually making sure that I have a C4 axis next question is do I have 4 C2 perpendicular to the C4 or not if it has to have 4 C2's it will be somewhere around here and here you can see if you rotate it the ammonia and cobalt ammonia present in the equatorial plane is remaining as it is the cobalt is remaining as it is but this chloride and ammonia present in the axial position they will be changing its position and that will be not superimposable and indistinguishable from the original structure so that is why we can say it does not have 4 C2 perpendicular to this C4 then the next question does it have any sigma h sigma h again will be the equatorial plane because that is the only plane perpendicular to that C4 all the ammonia in the equatorial and cobalt is going to be as it is if I do a sigma operation around this particular plane but this chloride and ammonia will exchange places and again it is breaking the symmetry so you can say there is no sigma h then the next question comes do you have 4 sigma v's then and that answer is yes because those are actually present along with this equatorial ammonia cobalt ammonia bond so for an example like this particular plane is one of them and it actually contains ammonia chloride and ammonia in the axial position onto it the C4 and these 2 ammonia is going to reflect it same thing happens when I am going through this particular way so there is one sigma v this is a sigma v we can have sigma v's going between the ammonia cobalt ammonia bond so altogether 4 so for an example over here the plane of the board I draw in this side view it is actually going to be one of the sigma v planes so I have 4 sigma v planes 2 of them goes through the bonds ammonia cobalt ammonia in the equatorial plane and 2 of them will go between those 2 bonds so as it is yes I have 4 sigma v's and this will be C4 v molecule in the point rule now I go to the next one which is I am drawing a alkene type molecule hydrogen hydrogen this is position and the rest of them is a chloride and a bromide so what is the point group of this particular molecule so this ethylene derivative as the same question are you linear answer is obviously no do you belong to any cubic group answer is no and the next question come do you have an axis of rotation Cn and here is this thing if it is acceleration it can be either in this particular system if you rotate you can see bromine and hydrogen changing places over here hydrogens are remaining as it is but the chlorine and bromine is 6 spaces so no acceleration over there one is perpendicular to this board again the hydrogen and hydrogen is actually changing places because this hydrogen will come to opposite trans this hydrogen goes to the opposite bromide so that is also not working so altogether this molecule does not have any Cn axis of rotation then comes does it have a center of symmetry center symmetry has to be in the center of molecule and you can see this chlorine versus hydrogen chlorine versus bromide so obviously no i there is a sigma plane so that sigma plane is the plane of the molecule because this is a planar molecule alkene molecule and this molecule is actually planar and that contains all the atom on its place so if you do a mirror image all of them actually mirroring on each other's original position so that is why it contains a sigma plane and over there as we asked there is no Cn axis present over there I cannot differentiate the sigma as a sigma h sigma v or sigma d because there is no Cn axis present so that is why it remains as only sigma and the point group of this molecule is Cs there is a point group of this molecule we will go to the next one number 8 which we are actually doing an alkene kind of molecule there I have a chlorine I have a fluorine on the back hydrogen on the front and then hydrogen on the back fluorine over here and chlorine over here so this is a alkene based molecule over here and we try to find out which is the point group of that so one of the other way to look into this molecule better is the Newman projection you know that we do we draw say I am looking from this particular direction to this molecule and over there I am going to see one dot is the front carbon that is having a chloride on the top fluoride on the left hydrogen on the right hand side if I am looking from this particular direction and on the back carbon I will draw as a circle and over there chlorine is over here fluorine is over here hydrogen is over here so there is a Newman position of the same molecule so these are the same molecule now over here I am going to ask the same questions are you linear answer is no next question do you belong to a QB group the answer is still no do you have a CN axis of rotation so over here you can try to find different axis of rotation from here or here or even perpendicular to that you can see whatever the rotation you are going to do you are not going to get any new configuration which matches to the original so that means this molecule does not have an axis of rotation present in this molecule next question does it have a sigma plane so you can see the sigma plane we can have it along this plane of the board and you can say like it is nothing but this plane over here in the Newman projection you see chlorine carbon carbon chlorine remains same but hydrogen and fluorine will be replacing space similar case over here that is actually not going to work and the similar question is that plane of the board perpendicular to it if I take a sigma plane so that is also not working so any particular plane is actually not going to be working so there is no sigma now the question does it have a center of inversion it has to be in the middle over here if it is there so this hydrogen passes through that is it reaching to a hydrogen so it is easy to see from this Newman projection hydrogen going through that center of symmetry the midpoint between the carbon carbon bond you can reach a hydrogen fluorine reaches a fluorine chlorine reaches a chlorine so that is how it is going through and over there it retains is center of inversion and carbon carbon obviously is doing the reflection so that means yes it is at the center of inversion and the point group will be C i so I would suggest you to take a look into this molecule on its own and try to visualize it and then you can do it properly that yes there is a center of inversion at the midpoint between C C bond so with that we go to our next example which is actually a ethane molecule present in its eclipsed configuration so as we know the ethane molecule can be present in this kind of configuration this is known as the eclipsed configuration which is understood better if I draw the Newman projection of it so the front carbon hydrogen on the top hydrogen on the left hydrogen on the right the back carbon as a circle this is the hydrogen over there almost on top of each other we are drawing a little bit moved over there so that we can see it properly so that is how it is happening so almost on top of each other what is the symmetry of point group of this molecule the question you are going to ask are you linear answer is no are you cubic group tetrahedral or octahedral answer is no then comes the question what is the C n axis of this molecule and the C n axis is actually present over here along with the C c axis if you rotate 120 degree so over here 120 degree this hydrogen comes over there this go there this go there so similar and same thing happens on the backside carbon also so that is why he has a C n axis is present it is a C 3 next question do you have 3 C 2 perpendicular to the C 3 so that we can differentiate between dihedral and single axis relation group so does it have any C 2 perpendicular to C 3 so for that what we need to take a look into it is that where it is possible so the C 2 is actually present over here where so it is actually going through the C c bond and through this over here one over here and the other one over here so basically one one example over here going through this hydrogen carbon hydrogen middle of that over here if you rotate 180 degree this hydrogen comes over here this hydrogen comes over here and the rest of them also changing its place because all of them are hydrogen so they are remaining as it is carbon carbon also position changes but the overall final configuration you are going to see will be similar to the original configuration so there is a C 2 over here and the similar C 2 present also along with the other hydrogen hydrogen bonds the other one from there which is seen better if we look through this particular structure between the hydrogen carbon hydrogen bond and through this carbon carbon bond there is a 3 C 2 present and obviously all of them you can see is perpendicular to that original principal axis C 3 so now the next question is do you have a sigma h sigma h has to be the perpendicular to the principal axis in this molecule is a plane on the top Newman projection is the plane and over there it is the perpendicular plane to the board of the board and over here you can say yes the sigma h is present because all of them are reflecting on its own hydrogen to hydrogen carbon to carbon so it is present the molecular point above this molecule will be d3h because it has a C 3 it has 3 C 2 because the C 3 and also a sigma h now the last example of this system is cyclohexane and you are going to draw the chair form of it and you try to find out what is the point group of this molecule that is how the chair from looks like and these are all the position of where the hydrogen should come into so this is the molecular structure we have to find out the point group of it so first question is is it linear answer is no next question whether it belongs to any particular cubic group answer is still no then the question is what is the principal axis of C n that is actually going through this molecular structure through the middle and it has a C 3 it is not very easy to see it from here so what I am going to do I am going to draw a structure of a cyclohexane in this kind of way top view if I look it from the top over here you can see three of them is actually above the plane and three of them is bottom the plane so the top ones I am actually drawing as positive it is not a positive charge or anything just to showcase that it is above the plane and three of them as negative again not a charge but to showcase they are below the plane so but here positive means above the plane and negative means below the plane and now if you rotate it from here 120 degree so this is 60 this is 120 this 60 120 60 120 you can see the above the plane carbons reaches to the above the plane condition and similarly this is also going to get to the below the plane system so they are remaining as it is so there means there is a C 3 present over here so yes this is present you have a C 3 next question do you have three C 2 perpendicular to that C 3 that is the question we have so that means where it can have so it is possible to have it if it is going through here so it is better to see it from here see if you rotate 180 degree over here you can see 180 degree this bottom one present over here 180 degree rotation it will go to the top the top will come to the bottom so this negative will go to the positive positive will come to the negative this above the plane will go to the below the plane below the plane will come above the plane similar over here so there is a C 2 present over here which is bisecting the C C bond and there are three possibilities because as we said if you find one of them other two will be also there because either it will have three of them or none and it will be similar position going through the C 3 perpendicular to the C 3 axis the principal axis and going through the C C bond so yes this molecule is having the three C 2s next question does it have a sigma h sigma h has to be the plane of this because it will be the perpendicular plane and over here when you do that you can see the plane is actually perpendicular to this principal axis but this above one will go to the down down will come to the bottom so this plus and minus signature will be changing because they are not on the plane of this sigma this is above or below plus belongs to above negative belongs to below and they will change their position so that is why there is actually no sigma h present then the question do you have sigma v is present so for the sigma v is what we can see over here there is a sigma v plane present so let me just draw that in a different color to showcase it properly so here is the sigma v plane present why does the sigma v plane because over here you can see the negative and the positive that I shown over here which belongs to this plane the above and below one will return to the plane this positive side will be reflecting this negative side will be reflecting so that top remains on the top bottom remains on the bottom everything remains same so the sigma v plane is actually going through this carbon carbon atom not through the bond as c2 that means there will be another one coming over here sigma v and another one going through this one and you can see each sigma v is bisecting the c2's right this particular c2 and this particular c2 you can see is bisected by this sigma v similarly this c2 and this c2 and this c2 is bisecting this and this one and this one is bisecting this one so yes it is present and it is nothing but sigma d's so that means it will be a d and d or d3d point group this structure so that is how we differentiate the different molecules with respect to the point group and that is how we figure it out the point group of different molecules so with respect to that we would like to conclude over here and we hope now we have a very good exercise how to find a point group and how to ask this question and figure it out so we would like to request you to follow the question that will be coming in the assignment and do it and if you have any issues on the visualization try to draw that in different orientation that fits your imagination better and try to resolve what is the point group of the molecule with that we would like to conclude this particular exercise session over here and we look forward to discuss further on the chirality and its connection to the point group thank you thank you very much


Let us move on to the next system. This system is not as exciting or as dramatic as airships because it is just a balloon which is tethered to the ground. But even though it is tethered to the ground, it can be made use for very interesting and far reaching applications. So this is a photograph of a small portable airship which is launched from the back of a pickup truck. So what is it? Basically an aerodynamically shaped tethered LTA platform. So it is not a vehicle, it is not an aircraft, it is a platform because it is expected and desired to remain stationary. So one huge degree of freedom we have removed. The payload that can be mounted on this is again limited by our imagination. Most people use it for either communication equipment or for surveillance equipment. Now what are the desired features of a good aerostat? For a given size, we should be able to carry the largest amount of payload, look like just like any other aeronautical system, we should have high payload fraction. It should be able to maintain a specified altitude on its own without any great intervention. And once it maintains its position, it should remain stable. That means as the wind direction changes or disturbances occur, it should on its own align itself and not demonstrate a continuous motion which will basically distract it from its application. Now all of these are affected by the envelope shape and hence the sizing of the envelope, the shaping of the envelope and very accurate calculation of the size, location and type and shape of the fins that you see on the back is an important exercise. And we have to learn all these things as part of this course. So this particular aerostat as an example can carry around 10 to 15 pounds of payload at a height of 700 to 2000 feet and it is not very long, it will probably easily fit into this room but it can remain up for around 48 hours. Now my question to you is should it not remain endlessly, why should there be 48 hour limit or so for this particular system? So what do you think is the reason due to which 48, yes, I would also encourage people to speak their names because we are still new to each other. Swami, from which department are you? Manu Technology. So might be because of the leakage of the gas contained inside over a period of time, that is precisely the reason. If we are able to make an envelope which has got little or no leakage, then actually there is no limit to endurance but typically fabrics have a particular leak rate and that introduces the limit to how many hours you can keep it without popping up. In this case it is around 48 hours which is good enough if you want to do, if you want to have a system for around two days to do surveillance, just one deployment and it is there. Interestingly it can withstand winds of around 45 knots which becomes almost 90 kilometers per hour approximately and it just needs one crew member who is standing below you can see. So it is a single man operation or a single woman operation and it can be easily deployed and the same person can even drive the vehicle and take it around. Alright, let us look at what are the key parts of the aerostat. So the first and the most important part obviously is the envelope. Here we say the helium chamber because we are assuming that helium is being used but it could also contain hydrogen or any other lifting gas. So it is basically the one that contains the lifting gas so it is nothing but a container with nothing but an envelope to carry a given volume of gas to control the pressure inside the envelope and to ensure that the envelope does not reach breaking point due to stretching because of the change in temperature or because of any other reason which causes stress in the envelope. We have a relieving system called as the air balloon ne. Please note the T in this word is silent. So it is called as air balloon ne. It is actually a small air bag inside the gas bag and by controlling intentionally the volume of this air bag that means by taking out the air or taking in the air in this gas bag you can control the volume available for the helium or hydrogen or LTA gas inside to expand or contract. So we use it essentially for buoyancy control and stress relieving the envelope. Then we have a payload platform now in this sketch we have shown the payload directly attached to the envelope and we have shown it below the balloon ne but it could be actually anywhere on the body of the aerostat. And in many cases such as the picture which I showed you ahead the payload is mounted not on the envelope but at a particular place called as a confluence point below the aerostat. So the payload or the equipment that you want to carry could be directly on the balloon if the local loads can be handled by the envelope material or else you can delink that by mounting it on a point called as a confluence point. And then the connection between the envelope and the ground is this tether. In this case we have shown some kind of a catenary profile because we are expecting it to be under high wind conditions but there is one serious mistake in this figure. I do not know whether you can make out what do you think is not correct at least conceptually in this particular figure. At the ground yes the tether will be connected to something on the ground in this case they have shown a winching. Yeah so this is the winching and mooring system not necessary I mean you one can have various kinds of systems the sketch shows a system which has a small you know cup like thing to hold the nose etc but one can mount it anywhere so it is a conceptual sketch what is it just a minute I mean you are saying something that can happen if the wind is very high when you fly a kite for example in high wind conditions this is the kind of profile you see for even the kite. So the tether can remain in high winds in this particular shape yes you had you had a point yes your name name Ashish yes why should it face opposite yeah so it is actually facing opposite the wing that is right the wind if the wind is from left to right then that is the right direction in which it is facing why should it face opposite to the wind it should not it is remaining stationary so the best way to do remain stationary is to face the wind and use the fins on the back because as the wind direction changes the fin will bring you back this is called as weather caulking so the direction is right yes Jagdish that is just a conceptual sketch as I said that tether is connected to the envelope it may not be there may be some other I know this is a conceptual sketch I agree there is something more fundamental which is wrong yes Sunira yes can be the payload can be on the balloon as I said it can be on the balloon or it can be below anything else okay fine so basically what we wanted to show was a three tail configuration with the third one hidden because it is behind and this one is actually the top fin it may it may seem to you as something inclined towards you but actually it is meant to be vertical okay I tell you the problem is with the angle that the aerostat envelope is shown actually if there is wind it will trim to some angle of attack it will not remain horizontal the balloon will actually acquire some equilibrium angle okay so I will correct this figure I have the correct figure I do not know how this wrong figure has come here this was made by one of our M. Tech students and I had told him that this has to be corrected but it is my mistake I put the wrong figure in the presentation okay yes it could be but in that case the tether will not be so much catenary when there is a catenary position we normally expected to trim at some angle but you are right I could alter the center of gravity and create a situation where the trimmed angle is kind of 0 but normally that is not the case okay so let us go ahead the tail fins that are mounted behind are essentially for static and dynamic stability okay so now there are two words here static stability and dynamic stability so the aerospace engineers among you I am sure you will appreciate the difference between the meaning of static and dynamic stability but for the benefit of others who may not know or for those who have forgotten let us have a brief idea about what is meant by static stability what is meant by dynamic stability and what is the difference between them so any volunteers to explain the difference between these two terms then worry about aerostats talk about aircraft in general we have the same things apply here what is meant by static stability of an aircraft or a system so Nina we want to try yes so what you said is if a force is applied on the system it returns back to the original position no this is not static stability this is of course if this happens if this happens the system is statically stable but something less also is acceptable okay so this is not the requirement this is more than the requirement what do you think is the minimum requirement for a system to be statically stable okay so we do not have to really use the many new terms to explain it basically we should say if a system is in equilibrium and if some unbalanced force acts on it or an external force acts on it what is the tendency of the system under the action of this external force if the response of the system natural response without any intervention is to tend towards the original position rather than going away then it is statically stable okay it may not lead to stability so what may happen is if I disturb a system if I take it let us say away it should tend to come back it may exceed and go on the other side it is still statically stable okay but then as Pratik mentioned over a period of time normally we expect the oscillations to slowly reduce and it should actually come back to the position that is dynamic stability so dynamic stability is what happens in the end does it actually come back static stability is does it tend to come back or not so we should have both in aerostats we should have static stability that means we should have a tendency of it to automatically react in such a way that it tends to come back and not only that over a period of time or a reasonable period of time we want it to actually also come back not keep on oscillating for whole life so the pair since help in both and their size from both the considerations the tether is basically a cable which attaches the envelope to the ground and the payload is what you as a user mount onto the system either on the envelope or at a point below it and the ground station is something that helps in two things in winching and in mooring so what is the meaning of winching what is the meaning of mooring and how are they different can someone try to explain what comes to your mind winching and mooring yes your name Kiran the winching is a process of holding in the place and what is mooring okay so it is not correct actually mooring is the holding in a particular place and winching is basically in raising or lowering it to the desired location so a winch is a system which pulls or pushes to intentionally change the location and a mooring system is the one where you attach it when you do not want it to move so you have to have a winching system for up lowering and raising and lowering and a mooring system to hold it when you do not want for example if you want to leave the aerostat on the ground for some reason for a couple of hours you do not want it to keep moving around you want it to be moored on the ground okay this is a moored aerostat developed in India by a DRDO laboratory called as aerial delivery R&D establishment in Agra so pioneering institute or lab in the country which has done work in the area of LTA systems mostly they have worked in aerostats but now I believe they are also starting to work on airships so this is one of their first aerostats so you can see that there is this hull or envelope now hull and envelope are interchangeable words they come from the aerospace lineage and which comes from the ship lineage or naval lineage hull is basically the body so hull is as same as anonymous with the envelope you have fins on the back and notice the fins are quite large in size in aerostats and then you have this mooring mast and here you have a system for winching it up and down and then you also see something called as confluence lines which are basically lines which are then attached together I am sure when you fly kites you will remember when you when you flew kites that we attach a single tether to the kite with two ropes towards the end which are properly sized okay the angle between the two ropes is actually done how much is the angle between the two ropes as children when you flew kites did someone teach you how to do kani so we hold the knot it should be it should hit the top portion at the top attachment point and bottom attachment point so the two of them should be the same length which means what if I degree angle that gives the best characteristics but those of you who are kite enthusiasts if I want to have a highly maneuverable kite what should I do I was taught by my seniors that there should be one inch or one finger width more on the top that gives you a more maneuverable kite if the wind conditions are very poor and you are going to basically fly by giving it a constant jerk then the bottom should be longer than the top so these are the things you learn similarly there are confluence lines which are used to attach the envelope to the tether which and the length of these lines and their orientations helps in deciding the stability characteristics we will study this when we come to the stability of aerostats this is one of the recent aerostats developed by the same laboratory and this photograph was taken at the Aero India show so you can see that there is a very neat winching and mooring system developed incidentally by LNT in Poway and then you have this envelope and here you can see the payload is mounted in this case directly on to the envelope okay okay so let's see and okay now this particular aerostat 17 crore rupees that they mentioned please note the bulk of the money has gone in the radar development or the cost of the radar is much more than the aerostat okay my guess would be that the aerostat system everything together will be not more than 4 to 5 crores and the remaining money will be for the radar okay but the whole system is what is important not just the balloon okay so now we have NDTV's confirmation that aerostats are available and made in India


Let's get over to I'm am Mr. Basil Chapman as we do each and every Tuesday at 20 past the hour and don't forget folks Basil doesn't outstanding show here every trading day 10 to 11 Eastern Time as a great newsletter the opening call now It's very easy to get the opening call folks come over to our website at TFN You go into newsletters you go on the left hand side. You can see the opening call You can get the opening call for one month for a hundred and forty nine dollars six months for six ninety five, which is the savings of a Hundred ninety nine dollars at twenty two percent in one full year for eleven ninety five Which is the savings of five hundred ninety three dollars or thirty three percent now They all come with a thirty-day money back guarantee folks Basil has a huge amount of archives that are out there Also, when you get the newsletter you're gonna get all the archives You can understand how Basil looks at the market every day and how he uses the tools that are on All of those archives each and every day Basil Chapman. It's a little chilly in Boston right now It's going to be warming up very soon Well, you know, it's funny Caitlyn's down here and she's telling me it's 28. That's just 28 man We're sitting outside having dead, you know lunch at 85 and it It's 37 in Boston. It feels like 28 folks. Okay, so yeah That's because you get used to whatever it is, but this this has been a pretty darn good winter. I must say No, no, I know morning. Yeah global warming says in this particular instance very little snow So yes, so talking about things falling down. We're looking at the down which is okay Hi, no 39,000 889 in this daily chart here. We've pulled back from this So these are the techniques I discussed in my webinar the last webinar I did a week ago Tuesday I discussed the how you use the inside track how this little mini channel right here can become right there Can become a repellent zone when it comes at the bottom. It could be a propellant zone So those are techniques that I talk about and we try to use in the positions that we have now at this particular time The week each art is still very strong. So we've got called long positions, but we have started a game We've got a short-term trading position short in the Dow My anticipation is that there might be a rally, but I think it's gonna create this pattern I call the H but dreaded H pattern where it comes down sharpening that makes a little arch formation and comes back and retest The left side loan takes it out. That's what I'm looking for, but When the nine period moving average, you know, I've spoken to you about this for some time and I did this in the webinar as well I was showing how so this is I'm going to talk about that in a moment. That's Draftings, but right now I'm talking about the Dow So look, this is the daily chart on the left You've got this the chart with this bar chart with the green on the way up and red on the way down The nine period moving averages is like green line the black lines of 14 period moving average This is the on balance volume. So on the right. It's just three three things This is the gray is the actual price is the current price daily chart And then you've got the green the black and you can see that every pullback has still seen the green line hold But I'm I'm thinking that the next pullback if if there is a sharp pullback over the next few days going to say the beginning of April then we will go underneath that green line will go underneath the black line and that will start It'll change to pink and that'll be negative. So I'm anticipating that it hasn't happened yet But there are a couple of things that I've started to take in some profits Just little small bits off long positions for instance We've been long for quite a while because a couple of years actually the DB Agricultural fund and I kept one of the grains were not doing well and I just see no I Don't know why this is going up with the chart looks fantastic And then I realized the other day I had spoken about this on my show But I just didn't put the two things together if you look at CC, which is cocoa. Yeah, right You won't believe this. I mean talking about rocket. This is a monthly chart. Yeah, and from the moment in September of 22 2022 September we are now in March of 2024 and this has been there has not been a peak All the way up That is still a leg a to the upside In the this is cocoa. Yeah, so we've gone for get out there by chocolate folks Yeah, don't go find stores that don't know that cocoa is going at ten thousand bucks a ton You know, it's wild too about that DBA Basel is that that's the largest waiting in that whole DBA, you know It's 12. I didn't realize it's 12.4 percent Correct higher than that is just government bonds. So right and I showed that to subscribers this morning I sent out the the chart of the components and I said I didn't realize that it got and that was really a very big influence now I think that Coco is getting a little toppy here and I'm anticipating There is going to be a pullback and I'm watching it very closely So we took a little bit off our DBA. We've got about 77 percent gain in this since we've had it We took a little bit off, but I'm watching it closely now something I thought I would mention because I talk about the different techniques that I show subscribers to my opening call and we We talked about them I have a video every weekend for about an hour and I discuss these different patterns and the charts and the stocks that we have so we've got Draft kings from about January around about the 37 38 level And it was a pattern when it hit this 45 point. This is the daily chart on the left It hit this pattern. It made a cup formation and then pulled back And I actually forgotten that I typed this in I actually Enlarged this where it says chapwave cup and handle cup and ladle to leg d the difference between a cup and handle For those of you look at classic formations Um The cup and handle makes the cup formation and then it makes a little bit It looks like a teacup and then it has a little bit of a handle and then it breaks to the upside Investors business daddy has a whole thing for decades about how they use it, but you've got to time it perfectly My problem with it is that invariably if you even if you time it correctly It pops over the left side lip and then it comes back into the handle and just stays there So I developed a long time ago when I discovered that if in the chapwave methodology You break out about above the left side lip. That's 45 point 62 in this case that was back in uh in February If you can take that out in leg c It says you should go to at least a d and then you've got to be careful And then you can come back and test the lip that would be 45 62 So we took a little bit off today Very nice gain in this but what's interesting is it's just gone to that leg d And it broke out in the pattern that we were talking about it's in a leg d in the weekly And in both cases the technicals are still very strong And it's in a leg e in the monthly chart the all-time high was 74.38 back in march of 2021 So I I like this very much personally. I'm actually not for sports betting. I just I've now I grew up in the amateur win of the olympic games was amateur So I guess I fight just I'm not used to that but in the meantime back in the Back in the weekly chart you can see that this breakout had I'll I'll open this up and I'll show it In a little bit more detail. Look at this It's really like a bowl formation. You can call it I call it a cup But basically it's a bowl and it broke out And it's walking the nine period moving average You can see from the last time that it crossed in the weekly chart positive was in february the week of the third It was in the 17 area And it's just walked this green line all the way and this is a weekly chart So this is looking very good to me. It should pull back a little bit soon But I I think that we're looking at stocks under the radar that could Go against any major market trend if there is a sudden turn down and folks It's very easy to get battle news that I come over to our website at tfnn Going to newsletters you're going to see the opening call on the right hand side Basil, you have a great one safe one. We look forward to show tomorrow. Thank you very much Tom you too. Stay right there folks come right back