Next question is from Eric Summer Hayes. I see a lot of professional athletes, mostly baseball players, that are overweight. Would those players be able to increase their athletic abilities by dropping some body fat, or is there actually an advantage to packing on extra weight? This is a cool question. Yeah. Justin must have picked this question already. Yeah, it's near and dear to my heart. Yeah, yeah. No, I like fatty athletes out there. I like this because I've heard both sides argued really well. And I can see we're, you know, let's take someone like a Pablo Sanibal, right? Who's got like a lot of extra weight on him. Super athletic, though. Yeah, right, and then you can't deny it. And then you have to ask someone like that, the way he crushes a baseball, if some of that extra weight is used into his benefit for momentum. Now, would he be a little bit more agile at third base that if he was 20 or 30 pounds lighter? Sure. You can't help but think, yeah. Right, sure. I could see that. But then he also might not hit the ball as far if he was smaller. So there's, I think there's places, and then same thing with like linemen, right? If you had a lineman who weighs 300 and something pounds, but you made him lose 50 to 60 pounds, he would be much more agile. He'll be faster, right? But then would he be less of a movable force, and would he be as dominant as a lineman? So. Yeah, it's an interesting idea because I've seen, I've seen like both kind of perspectives. Like, so I've seen linemen that actually, like you think they have to just be big. They just have to be massive and, you know, immovable. Where I've seen some linemen that might have like smaller stature and, you know, a smaller body, but they're very powerful. And the way that they can snap and move and the athleticism has knocked these big ogres on their back. And it just, it depends on like how you are grounded and how much snap that you have. And so for me, it's like, it's all based on power output. Doug, will you pull up a picture of Aaron Donald? I want to see what his physique looks like. So I think there's a, I think you're right. I think there's a point, right? So I think there's a point where it's too much. How do we judge that? As viewers, it's what's really tough. I'm pulling up like arguably one of the best right now, right? So look at that. Cause if you change too, if you change their body weight in there, it's going to change their mechanics. It's all going to get effective. So they're going to have to relearn a lot of these skills, you know, in their smaller frame.  So there's the best lineman in the game right there. Yeah. So he don't look, he definitely don't look fat. No, no, he looks powerful as hell though. You know what? It's way more complex than the black and white, this person's lean, this person's fat, therefore it's way more complex than that. Like for example, when you look at like weight class based sports like wrestling or mixed martial arts, for example, you may see a heavy weight, you may see two heavy weights. One of them kind of looks a little soft, big guy, right? The other one is shredded and muscular.  Okay, now let's pretend that they have equal skills. Like the skills are identical. Not necessarily, maybe the really buffed lean guy had to like force and build his body to be a heavy weight. And the other guy, it's kind of soft. He lives at that body weight, he's super comfortable with it. He's not stressing and pushing his body to be that heavy. He's just a naturally big dude, in which case he's got the advantage. He's totally got the, you know, Fedor Emilienko was like that. He looked kind of doughy. And he would just- He didn't think he'd have the gas tank he did, but he could go forever. No, so it's way more complex than that. What's his face? That's local, that's- Can't even ask questions. No, no, no, no, no. Well, him too. Yeah, him too, but I'm thinking of, why can't I think of his name? AKA guy right now. Just did his last fight. I mean, he has one of the best gas tanks in the game. Heavy weight, both light- Mark Hunt? No, come on, bro. I don't know. Jesus. It starts with a C. Why can't I not think of his name right now? I can't believe I'm drawing- I'm a brain fart, it's contagious. I see that, dude. These are terrible, Doug, please.  Thank you, Daniel Carmié. Thank you, thank you, Jesus. That's a perfect example, though. French names are hard. Look at what he looks like, dude. He does not look like he- Well, as I'm saying, it's way more complex than that.  You lose weight and change the shape of your body. You've changed the body, now you're not as good at moving and using that body like you were with the other one.  And so you have to get bigger in order to be able to stand your ground at this point. And I believed them. And I thought that I had to gain at least like 30, 40 pounds or I was gonna get messed up. And I did, and I did not play anywhere near as good as I did 30 pounds lighter than that. And I should have just listened to myself because really it was about getting stronger. It was about getting more powerful, having more snap, being there first, like predicting the outcome before they even could see it. So that to me is way more advantageous. And you know, I think intuitively you know what body weight you're most efficient at. And I think a lot of people kind of find that themselves. Yes, and you know, we tend to confuse extreme aesthetics with health and performance. There's a wide range of body fat percentage that is healthy, right? So for a guy could be 8%, he could be as lean as 8% or as high as 18%. By the way, if you saw a picture of a guy in 18%, totally different, right? One, he looks dad boddish, maybe even a little heavier than that. The other way he looks, you got a six pack. And you could say, well, the 8% guy is healthier version. Not necessarily that range right there doesn't make that big of a difference. There's a lot of other factors that determine health. And within that range right there, you could be very healthy or unhealthy. Go outside of that, then you start to run into some problems. Same thing with athletic performance. You can't necessarily look at a guy or girl and say that they're gonna be really good or bad. I mean, you can in extreme cases, like I could look at someone like, they're probably not gonna run very fast.  But you could, but outside of that, it's hard to tell. Now there's certain things that I might look for because I'm a trainer. I might look at like someone's hip development, upper back development that usually means that they have some good power. Is it a guarantee? No, I've seen far too many times somebody who looked like they should not be powerful and fast and completely. It's like the steroid talk. Yes.  Like there's, you know, there's, I've been wrong on both sides. So it's like, it's too hard for an outsider to speculate. Oh, he's 20 pounds or 30 pounds overweight. Now that being said, I think there's also, there's extremes versions of this. There's definitely athletes that come in, they sign a big contract. And they're locked in. They get lazy.  Start playing first base. Yeah. Yeah, right, right. So they come in, so they're, so it's not saying that, that, you know, everybody the size that they're all coming in at professional sports is probably the best size for them. That's not true either. There's definitely people that, you know, like I said, sign contracts. They know they've got a deal for the next five to 10 years and they get lazy.  And had they not put on a few extra pounds, they'd be a better athlete. So they're examples of both sides. Yeah.


Hello and welcome to the session. In this session we discussed the following question which says, a wire is bent in the shape of a circular ring and then it is re-bent in the shape of an equilateral triangle each of whose size measures 8.8 centimeters find the diameter of the ring. Before moving on to the solution let's recall the formula for the perimeter of an equilateral triangle this is equal to 3a where this a is the length of each side of the equilateral triangle then we have the formula for the circumference of the circle this is equal to 2 pi r where r is the radius of the circle this is the key idea to be used in this question let's move on to the solution now we are given that each side of the equilateral triangle is of measure 8.8 centimeters so the perimeter of the equilateral triangle would be equal to 3 into 8.8 centimeters and this is equal to 26.4 centimeters now we take let r be the radius of the circular ring formed now in the question as it is given that the wire is bent in the shape of a circular ring then it is re-bent in the shape of an equilateral triangle so the perimeter of the equilateral triangle is equal to the circumference of the circle the perimeter of the equilateral triangle is 26.4 centimeters and this is equal to 2 pi r which is the circumference of the circle from here we get r is equal to 26.4 upon 2 into pi now we put the value for pi that is 22 upon 7 so r is equal to 26.4 upon 2 into 22 upon 7 from here we get r is equal to 26.4 into 7 upon 2 into 22 now 22 1.2 times as 26.4 so we get r is equal to 1.2 into 7 upon 2 then 2 0.6 times as 1.2 so r is equal to 0.6 into 7 which is equal to 4.2 centimeters thus we have the radius of the circular ring is equal to 4.2 centimeters and therefore the diameter of the circular ring is equal to 2 into the radius that is 4.2 centimeters and so this is equal to 8.4 centimeters is the diameter of the circular ring so 8.4 centimeters is our final answer this completes this session hope you have understood the solution for this question


Hydrogen-powered aircraft, there's two approaches here. There's like a direct hydrogen combustion using just a burning hydrogen as if it was a jet fuel, for example. However, the approach we're taking for now would be a hydrogen fuel cells to provide power. So if we have a hydrogen fuel, it gets plumbed into the fuel cells. The fuel cells reverse electrolysis reaction happens. So hydrogen reacting with oxygen in the ambient air, it produces electricity. And that electricity gets sent into electric motors, and those electric motors then turn on a propeller, for example, thus providing the thrust that needed to power the aircraft.


In the second video, in my course on more advanced topics in linear algebra, I talk about arithmetic with complex numbers. It is very simple to do and is very much like working with real numbers. When we work with complex vectors and complex matrices, we will use the simple arithmetic that is covered in this lecture. So grab your pencil and paper and let's do some complex arithmetic. So what we are going to do now is just do some complex arithmetic. And the first thing we are going to do is scalar complex number multiplication. And so let's start with this scalar complex number, but we are going to just constrain ourselves a little bit. And we are going to say let this scalar and let's make this scalar k, let that be an element for now only of the real numbers. Later on we will look at what happens when we have a complex number so that we multiply two complex numbers with each other. So let's have this number z, this complex number z, that is going to be a plus b i. And we now know real number, real number, imaginary unit i and multiply b by i, this is our complex number.  And all we are going to have here, we are just going to have the normal distribution that we learned at school. So nothing complicated there, so we are going to have k times a and we are going to have k times b times i. And so let's have a little example of that, well let's first of all define that. So that is going to be our definition, we multiply a real number by a complex number. And so there is our definition, so let's just have a little example. And we are just going to say let's do three times four, let's do that five times i. Pretty simple, three times four and three times five, that is what we are going to have. So that is three times four plus here we have three times five times i. And that is going to be twelve plus fifteen i. So really nothing difficult there whatsoever. Let's go to the addition of complex numbers, let's look at the addition of complex numbers. And now we are going to have two complex numbers, let's make this one z. We are going to make this be a plus b times i. And let's make a second one w is another common variable that we have when we do talk about complex numbers. And let's make this c plus di. And now do please remember the fact that c and d, they are both real numbers. So if we are going to define this addition z plus w, that is going to be a plus b times i plus c plus d plus i. d times i I should say. And now I am just going to group all my real parts and my imaginary parts. So the a and the c, they are both the real parts. So I am going to put them together, that is my real part. And I am going to take out i as a common factor there so that we have our imaginary part b plus d times i. And there is our definition for complex number addition. And you can see very easily there it is just adding the two real parts and adding the two imaginary parts. There is really nothing to it. Shall we do one of these? Let's say 3 plus 4i. And I am going to add to that 7 plus 3i for instance. And that is just going to equal these two real parts, go together. 3 plus 7 is 10. Plus it is 4i plus 3i. So remember it is 4i plus 3i, that is what I am talking about there. And it is the same as you would say 4x plus 3x. We know how to do that or we can take out this common factor i. And then we just left with 4 plus 3. In other words that is a 7i. So plus 7i, very simple. Which reminds me of something because we have always been dealing just with these plus and plus and plus and plus. It is always a positive. But what if we want to do negative? And so one little other thing that we can add here. Let's do let k equal negative 1. And so if I have negative 1 times, let's do 3 plus 4i. And I am just going to get negative 3 negative 4i. So certainly we can have negatives in there. That is not an issue for us at all. Because what that allows us is to do the subtraction of complex numbers. So let's do the subtraction of complex numbers. Let's have this definition again. We are going to have z equals a plus bi. Now remember that b is a real number. So it might be negative 3. So I can have a plus negative 3i. That would be negative 3i. I think you get what I am talking about. Not particularly difficult. So let's have c plus di. And all I want to calculate now is z minus w. I am subtracting w from z. So I am going to have a plus bi. And I am going to subtract from that the following. c plus di. And really nothing difficult here. You can clearly see where this is going. 3 plus b times i. I am going to distribute this negative 1 throughout. So that is going to become negative c. And that is going to become negative d times i. And all I am going to do is group together the real and imaginary parts. So I am going to have a minus c. That is going to be my new real part.  So let's say that we have 3 plus 4i. And I am going to subtract from that 7 plus 7i. What am I going to get here? I am going to get 3 plus 4 times i. Minus 7 minus 7 times i. Just group these together. It is going to be 3 minus 7. And I add to that 4 minus 7i. 4 minus 7 times i. And I am going to get 3 minus 7 is negative 4. And 4 minus 7 is negative 3. Negative 3i. As simple as that. And what we have to find here is the subtraction of complex numbers. So where shall we put it? Let's put it right there. We have to find the subtraction of complex numbers. So let's do something that is slightly more fun. Although we do get to add complex numbers. That is a real thing. And when we are going to talk about vectors and matrices. That is of course what we are going to do. So let's talk about the product of complex numbers. Here we go to the product of complex numbers. So I am going to have two complex numbers. Z and W. Let's make this A plus B i. Let's make this one C plus D i. Where all these things A and B and C and D. Remember they are all members of this set of real numbers. Real part, imaginary part, real part, imaginary part. Two complex variables. And I want to multiply them. I want to say Z times W. And it is going to be nothing other than just the expansion of what you already know. So I am going to have A plus B i. And I am going to multiply that by C plus D i. And guess what? We are just going to do this same distribution. So I am going to have A times C and then A times D i. And I am going to have B i times C and B i times D i. Simple, simple stuff. A times C is A C. A times D i is positive A D i. I am going to have... I am just going to put these things in order because they do commute B i C. So I am going to say that is B times C times i. And now I just have to be careful. I have got B times D. So that is B times D. This is my B times D. But I am also doing i times i. And remember what i times i is? Well that is i squared. And remember that is how we defined our complex, our imaginary number. I should say that is negative one. So it is B times D i times i is negative one. So that becomes negative B D. And now I am just going to group together the real and imaginary parts. So that is going to be A C minus B D. Plus I am just going to group those two. Take at i as a common factor. B D plus B C times i. So you have just got to watch out for that i times i. Which is going to give you a negative one. So that you end up with your negative B D. So let us write that down there. That is an important definition for us. The product of complex numbers. Let us do a little bit of an example. Let us just have 3 plus 4 i. And I am going to be slightly more difficult. Let us multiply that by 3 minus 3 i. We are going to have 3 times 3 that equals 9. I am going to have 3 negative 3 that gives me negative 9 i. I am going to have 4 i times 3 that is positive 12 i. And now I have got to be careful. 4 times negative 3 is negative 12. But i times i is negative one. So that gives me a positive 12. So I am going to end up with 9 plus 12 for my real part. And I can add to that. Let us do, it is a negative 9 and a 12. So let us make 12 minus 9 times i. 9 and 12 is 21. And 12 minus 9 that is going to be 3. So plus 3 i. So just check the arithmetic there. I am famous for making arithmetical errors. And that is why we have computer languages and calculators. Because who wants to do arithmetic? So check on that one for me. But I think you can see clearly what we do when we do the product of complex numbers. So let us move on. And I want to talk to you about something very special. And that is called the complex conjugate. And the way that we define that, if I have a complex number z, there is my variable z. And that is a plus bi, a real part, an imaginary part. I define the complex conjugate. Let us put it there as z with a little bar on top. So just a little bar. That is going to be a plus. And I am just going to multiply this b part by negative 1. So that would just mean z bar in this instance is going to be a minus bi. So in practice what we usually just do is just swap the sign of that imaginary part. So a plus 3 i, a plus bi I should say it is complex conjugate is a negative bi. And so that is a very important thing. So let us do a little example. Let us have that z equals 3 plus 3 i. And that means z bar, this complex conjugate. Well I am going to multiply that positive 3 by negative 3 and that gives me 3 minus 3 i. That is great. Let us do another one. Let us have w equals 4 minus 4 i. So what is w conjugate going to be? Well I have got to just multiply negative 4 by negative 1. That gives me a positive. So that is 4 plus 4 i. Can you see that we always just swap that sign in practice? That is all we are going to do. So if we think about the argn plane. Let us just put a little sticker there. All these things that are so important that I want you to remember. That is how we define the complex conjugate. Right there, complex conjugate. And so if we think about the argn plane. It just really means if I display or think about a complex number as a vector. And so if I have my vector there. Remember that is going to be my a part component. That is my b component. It just literally means we are just reflecting this along this real axis line. So I am just going to still have the positive a. But now I have the negative a positive a. But from the positive b I am going to go to the negative b. Or from negative b I go to the positive b. So the fact is if this was my complex number z. This one right here. That is going to be z bar. So that is kind of cool. Complex conjugate. Where can I use the complex conjugate? Well let us go back to something we have already done. And we looked at the norm of a complex number. So let us just remind ourselves very quickly what the norm of the complex number was. So we had this idea if z was equal to a plus b i. I had this norm of z. And remember that was the two bars. And that was just this real number. That I take the positive square root of the real part squared. Plus the imaginary part squared. Now let us think about something. Let us do something very special here. I am just going to write it out. So you can see where this is going. I am going to say let us do the norm of z. But we are going to square the norm. Let us square the norm. Well if I square the norm what am I going to get? a squared plus b squared isn't it? I am squaring the square root. So I am going to get a squared plus b squared. Now let us do something crazy. Let us do something crazy. Let us multiply z by z bar. That is crazy isn't it? z times z bar. So what is this going to be? a plus b i times what is its complex conjugate. Well it is going to be a minus b i. And if I do that multiplication what do I get? a times a is a squared. And I have a times negative b i. That is negative a b i. Positive b i times a. That is plus positive b i. And negative a b i. Positive a b i. Those two can's allow. And now I am left with b times negative b. That is negative b squared. But I have i times i which is another negative one. So that becomes positive b squared. So just do that so that you are comfortable with that. And hey look at that. We have the fact that. Let us get a little geniart. We have this nice little thing there. Let us put it down there. Because z bar is nothing other than, let us put it out here, that is the norm of this complex number squared. The norm of the complex number squared is equal to the complex number times its complex conjugate. These things commute. So I can have them the other way around. It doesn't really matter. And then if I take the square root of this. So if I take the square root of the complex number times its complex conjugate, then I am just going to get the norm, the length. If I want to think of it as a vector, I am just going to get the length. So this is kind of an important and quite interesting if you think about it. And so that would be one use of us thinking about this idea of a complex conjugate is because I can do these wonderful things. Instead of calculating this Pythagorean theorem as far as the norm of a complex number is concerned, thinking about the length there of that vector from the origin. For any point, and remember any point is a point, a comma b, any point on this plane defines for me the whole set of complex numbers. But I can think of the square, at least of the norm, as being the product of a complex number and its complex conjugate. As I say, all the other way around. Now that we've revisited the norm, we can also revisit the argument. Remember the principle argument, this angle that we make? Now there's a slightly different way to think about it. I'm not going to discuss it here, but if you do want to watch the extra videos on using Matlab to do these calculations, we'll just look at another way of thinking about this argument where we're going to have a positive argument on this side. So from zero to pi, we'll call that a positive angle. But as soon as we go beyond pi, as far as the radiance, this angle is concerned, we're going to rather go clockwise. And so on this open interval from zero to pi, and we're going to call that a negative angle. And that's it. I think you know enough of complex numbers so that we can start thinking about complex vectors, which is not this. Remember this is a complex number turned into a vector. I'm talking about a whole vector of complex numbers and even better, a matrix full of complex numbers. That'll be coming up next in this lecture series.


Hey guys shipper1000 here got monkey behind me so about these fairies this is we're on the garage this is monitor of the cameras one two three and four now four is over here it's pointing at the boat and the front door which the porch light is always on it comes on at night it goes off in daytime this is the camera that i'm catching these flying things in this is the only camera i'm catching them in i'm not catching them in here that's what's weird i will get like some you know bugs or a total you know you can actually tell it's a moth once in a while but this is where they usually are in this one right here and they'll be active for you know 10 20 minutes and then they'll just go away as you can see there are a couple little bugs flying past here and there but they were real active and we went outside earlier and when we came back in well when we went outside there was no bugs flying around so why are they not attracted to the porch light or this camera they're only attracted to this camera now i think they probably are moths like one of them it's called the the luna moth and it could be that so here's here's the porch light it's like i said it's always on i'll show you the cameras how close they are let me lighten this up for you here some if i can there we go so there's the there's our cameras the one on the left of course is pointing that way that's the one don't pick up anything the one on the right that one there is the one that you can see pointing out this way so that's what's weird now honestly i do think it is the type of moth or maybe even a galley nipper but how come they are not attracted to this light sometimes we'll come out here and there'll be all kinds of bugs here but when i'm catching them in the number three camera which is the one over there on the right that's pointing that way when i'm catching them in that camera there's no moths around here and there's no moths in front of that camera right there which is very strange why not you see how close they are together what a foot apart that's the only weird thing that i you know that i find weird you can see how close they are i would say eight to twelve inches apart so when you're seeing them fly over off off right inside the screen which would be toward this camera how come they don't show up on that camera really weird that's what's bugging me i i do think they're a moth or a galley nipper probably a moth what do you think monk probably think they're moths yeah i like the fairy yeah they do look like fairies yeah i mean if fairies exist right but as an investigator i have to investigate this i just can't say okay i think they're moths so that's what they are right because somebody might have proof that they're not moths that's right i never know you know so but fairies are supposed to be mythical i've never seen a fairy me either i've seen a lot of moths but usually when i see a moth in one of them cameras i mean it looks bigger than life you know it looks huge all right i've seen moths in there and i could you know explicitly say that's a moth you know i can say that's what it is we got those great big ones like bigger than the palm of our hands i've seen them come up and actually land on the camera and you can see their wings but their biggest life these aren't that big all right but yeah i'm guessing moths but still yeah you like the fairy idea yeah me too but like i said i've got to keep an open mind i can't say they are fairies i can't say they're not how come they don't go to the light i don't know i can't say they're moths and i can't say they're not you know 100 sure i mean i can say if you know or maybe i you know but yeah as far as yeah i don't understand why they're like like we'll see well you know come out here tonight we'll see a bunch of mosquitoes and stuff flying around that light and when i go in and look at the monitor that camera will pick them up over here which would be the right part of the camera on the screen you can see and you can tell what they are but whatever these things are whether they're moths or or fairies they're definitely only showing themselves on that right camera and high which is weird it is so what do you guys think i mean it's weird you know that's why i never give a definitive that's what it is because i can't prove it's fairies and i can't prove it's a moth maybe somebody can prove it's a moth right which would be cool but it's also cool if they can't do because of mystery you know weird homework yeah so yeah all right well i had her come out with me because i'm scared of the dark yeah right all right he keeps a safe don't he oh yeah okay guys anyway thanks for watching i appreciate it again let me know in the comments if if you can prove that's a moth that'd be great but if you can't that'd be great too you know yeah but see there's no bugs flying around here and there wasn't earlier the last video you saw was that i just seen the light fly oh the last the the last video if you watched the last video the last clip on that video was filmed tonight okay and i came out here and there was some bugs flying around there but when i walked over there i couldn't see it not not a damn bug not one bug so pretty cool all right guys anyway there you go yep kind of an update but anyway shea bearer met the man alleged we're gone for now i can sit yourself for now bye y'all and we'll see you in the next one bye bye guys take care


So, today we will discuss metal olefin complexes. For the first time in this series, we are going to look at compounds where the metal is bonded to two carbon atoms simultaneously. In organometallic terms, one often indicates the formation of an olefin complex by writing them as m coordinated to two carbons by drawing a line in between the two carbons. In the nomenclature, one indicates this as an eta 2 complex, meaning both carbon atoms are bonded at not necessarily equidistant, but almost in a similar fashion. So, that both carbons are interacting with the metal simultaneously. So, in these systems, the olefin which is usually, if you consider the four atoms which are bonded to the olefin, I will label them as A, B, C and D. If A, B, C and D and the two carbons are in a plane, then the metal is in a position below that plane. So, the metal itself is in a plane below the plane which is containing the four atoms A, B, C, D and the two carbons which are interacting with the metal. So, let us take a look at some of these complexes. It is interesting that there is a very similar bonding pattern between olefins and metals and carbon monoxide and metals. The given take of electrons that we discussed with metal carbonyl complexes exists with metal olefin complexes also. So, the type of synthesis that we encounter in metal olefin complexes are very similar to what we encounter in metal carbonyls. So, let us take a look at a few of the synthetic methods. Then, we will examine a few structures and the bonding pattern in these molecules today. So, synthesis of olefin complexes in general involves a simple ligand substitution. So, a carbon monoxide ligand can be substituted by an olefin. One can also reduce a metal salt in the presence of an olefin and this is exactly what we did for carbon monoxide. We used to reduce the metal salt in the presence of carbon monoxide sometimes at high pressure and that led to the formation of a metal carbonyl complex. If you want to make a metal olefin complex, we can do the same trick namely take the metal in a higher oxidation state, reduce it with a suitable reducing agent which will not reduce the olefin and under those circumstances, we can prepare a metal olefin complex. There is also a specialized method called the metal atom synthesis. In the metal atom synthesis, one generates a vapor of the metal which will co-condense with the olefin or any ligand of interest. If we can do that, then a metal olefin complex can be generated without interference from any other ligand. Finally, one can in some instances take an unsaturated complex and add an olefin and that will lead to a metal olefin complex. Let us take a look at some very simple methods of generating metal olefin complexes. The group 10 elements which are usually the ones having D 10, it is actually group 11 which has a D 10 complement of electrons. So, they are oxidized to the plus 1 oxidation state. So, silver 1, gold 1 and copper 1 are suitable examples. So, you can very easily make these olefin complexes especially the copper and the silver metal complexes by simply taking the metal salt and reacting them with an olefin. All you have to do is to keep a noncoordinating anion and if you take a nitrate for example, it is quite efficient as a noncoordinating anion and you can treat it with an olefin, a suitable olefin, a weak complex is formed. Now, you might be wondering why is it a weak complex? You remember that the type of interaction that is involved in this give and take of electrons between the metal and the olefin. You need the metal to be present in a state which is ready to give electrons to the anti-bonding orbitals of the olefin in this instance. The carbon monoxide case was also similar to form a carbon monoxide silver plus complex or a copper 1 complex. It was quite difficult because the back bonding is poor and so, necessarily you form weak complexes. Nevertheless, these complexes are extremely useful because the weak interaction is convenient for establishing an equilibrium and any application that requires only a weak interaction will benefit from such a situation. So, the very common application of silver complexes is that one can quote chromatographic columns with silver ions and then this weak interaction is sufficient to distinguish between various olefin complexes. So, in other words the stability constant of two different olefins if they are different, if you can have a the cis and the trans isomers of two butene they can be very easily separated utilizing the strict because the stability of the trans butene is less than the cis butene complex and this is something which we will consider later in detail, but suffice it to say that this stability difference is able to differentiate between these two olefin complexes in terms of the stability constant and hence the rate at which the silver coated column will discriminate between the trans and the cis butene. So, a similar situation occurs with the copper complex, if you take a copper 2 plus ion you can reduce it with copper metal in the presence of a solvent like acetonitrile CH3CN. The key to this reaction is the fact that you have a non-coordinating anion or a weakly coordinating anion the perchlorate anion and also the fact that you have a stabilizing solvent. Copper 1 is stabilized in the presence of acetonitrile because acetonitrile is a pi accepting ligand and so it stabilizes the lower oxidation state of copper in this case it is copper 1. So, it is possible to make copper perchlorate which is in the plus 1 oxidation state which I have put within quotes because this itself is an unstable system. This is unstable and it is only stable if it is present in the if acetonitrile is also present in the medium. So, one can use other non-coordinating anions very conveniently the other ligand which is very commonly used is the triflate counterion. So, copper 2 triflate is taken and the reduction is carried out with copper metal which is the most convenient because no other interfering metal is present in the medium. So, you take copper 2 plus reduce it with copper metal and in the presence of triflate perchlorate or nitrate and then you can reduce it then you can reduce it to the copper 1 state and if an olefin is present you can form the metal olefin complex fairly readily. But as I mentioned the copper and the silver salts are reasonably stable, but they cannot be used extensively because they tend to decompose. The reduction can be carried out electro chemically this is sometimes convenient if you use two copper electrodes and try to pass current in the presence of acetonitrile and an olefin. The olefin complex will be formed when the copper 2 plus in solution the copper sulphate solution can be taken and the reduction can be electrolysis can be carried out with copper electrodes. It is convenient to do so then the anode and the cathode get coated with the copper 1 compound especially if the solvent is sufficiently stabilizing one can get very good yields of the copper 1 complex. Now, it is not always necessary to reduce the metal from the higher oxidation state to the lower oxidation state using a metal. Here is an example where ethanol functions as a reducing agent ethanol gets oxidized to a standard height. So, ethanol gets oxidized to C H 3 C H O. In other words, two hydrogens have been removed from ethanol and those are the reducing elements they reduce the rhodium chloride to rhodium 1, rhodium 3 chloride here it is present in rhodium 3 and here it is present in the rhodium 1 oxidation state. And you can see that because you have two rhodium with two chlorins now in the presence of acetylene if acetylene if in the presence of ethylene in the presence of ethylene you will form a rhodium 1 complex. And in this case it is two rhodium atoms coordinating to four ethylene moieties and this complex is quite stable. And because ethylene is quite volatile this can also be used as a precursor for making other olefin complexes. So, this is yet another way of reducing the metal salt and generating the metal olefin complex. So, let us take a look now at some other few more complicated methods. In this instance you have a bis cyclopentadienyl cobalt complex and as you know ferrocene is the only bis cyclopentadienyl complex that has got an 18 electron structure in the plus 2 oxidation state. So, cobalt obviously has got more number of electrons than necessary to form an 18 electron system. Since cobalt has got 9 electrons it has 1 electron more and if you pump in one more electron using say potassium is the reducing agent which will become k plus and it will add one electron to the medium. And this one electron will add on to the cobalt and so cobalt if it becomes cobalt minus 1 this turns out to be a 20 electron complex and it is quite unstable. What it does is that it liberates cyclopentadienyl anion with potassium as the counter ion and during this process cobalt from the plus 2 oxidation state its present in the plus 2 oxidation state it has got reduced to cobalt 1 and the cobalt 1 is now coordinated to 2 ethylene molecules which are in the reaction medium. So, this complex can be drawn as follows you can have 2 ethylene moieties which are bonded in a symmetrical fashion to ethylene to the cobalt 1 species which is present in the center. Although we have not studied cyclopentadienyl complexes up to now we can understand that because we have 5 carbons coordinated to the cobalt 1 they will in the neutral method give 5 electrons. So, the neutral method we will have 5 electrons from cobalt and 9 electrons 9 electrons from cobalt 5 electrons from the C p and 2 electrons each from these olefins and so we have 4 electrons from ethylene. So, we have a total of 18 electron. So, 18 valence electron complexes formed very conveniently and this turns out to be a stable system. So, this y t is in fact known for cobalt rhodium and the rhodium complex is extremely interesting and if time permits we will look at the structure of this complex and the type of interactions that can be there in this system. Now, the most famous olefin complex that is present in the literature that is available in the literature is what is called the Zeiss complex. The Zeiss complex has in fact a very strange complex that was formed and it was accidentally discovered by Zeiss when he boiled potassium tetrachloroplatonate with ethanol and it was formed in the and finally it was found out that it was an ethylene complex that was present in the product. So, there was one ethylene moiety which is coordinated to the platinum and there are 3 chlorines because platinum is present in the plus 2 oxidation state. It has got a negative charge and this negative charge is now balanced by a potassium ion. So, we started with a platinum 2 complex. We have not carried out a reduction here, but what we have done is we have generated ethylene and we have used ethylene in the reaction medium and coordinated the ethylene to the platinum which kicks out 1 chloride ion and forms an ethylene complex. So, this is an example of a simple substitution there is no reduction involved. So, this is an example of a simple substitution reaction a chloride ion is replaced by an olefin and because the olefin is reasonably strongly bound to platinum 2 in spite of the fact that the oxidation state is plus 2 platinum because it is a lower group element. It has got sufficiently large orbitals it interacts very conveniently or very readily with ethylene and it forms a nice olefin complex. Now, a few words about this complex is pertinent here one of them is a fact that the complex is found in the presence of water. This whole reaction was done with dilute hydrochloric acid and notice that the structure of the molecule is also containing the lattice the crystal structure of the molecule is also containing a molecular water. So, this is very strange because one often thinks that organometallic compounds cannot be synthesized in the presence of water and the very first olefin complex that was synthesized and there was way before the discovery of ferrocene it had a molecular water in the lattice to begin with and it also was made by boiling it in dilute hydrochloric acid with ethylene. So, simple substitution reactions are possible with several elements especially with elements which have either a d 8 configuration and they have only 4 ligands around the coordination sphere. It is very easy to add a ligand or to substitute a ligand. So, here is an example here is an example with rhenium and notice that the molecule rhenium is again interacting with ethylene molecule and in this instance what has happened is AlCl3 in the reaction medium AlCl3 has grabbed this chloride ion. So, this chloride ion has been grabbed by the aluminum in the process and ReCO5 plus moiety has been formed and ReCO5 moiety will be co-ordinatively unsaturated and it grabs the ethylene and forms an ethylene complex and the counter ion is now AlCl4 minus. Another example is the generation of a vacant coordination site in the complex by using a silver salt by using the principle that silver one ion is halophilic. It will tend to react very readily with any halide ion in the coordination sphere of a metal. Here we have taken iron complex which has got an iodide and the iodide is removed by the silver. So, the iodide and silver react together and form silver iodide and this silver iodide is now removed and that leaves a vacant coordination sphere on iron which leads to the formation of an olefin complex which we have pictured here. So, in all these examples either there is an assisted removal of a ligand from the coordination sphere of a metal complex and that vacant coordination sphere is occupied by an olefin. In all these examples we have used simple ethylene. In fact, ethylene forms a very strong complex with metals and substitution in fact destabilizes the system except under some special circumstances which we will look at in a few minutes. In coordination chemistry it is possible to have more stable complexes formed when you have two places of attachment for the ligand. So, in other words if I have a metal complex where two NH 2 groups are present on the ligand then the coordination of both NH 2 groups to the metal turns out to be a more stable situation than when you have a single point of attachment. So, CH 3 NH 2 coordinated to a metal versus CH 2 versus ethylene diamine coordinating to a metal. These two systems are differentiated by the so called entropic effect the chelate effect which leads to more stable complexes. A similar situation can happen in the case of olefins and here we have pictured cyclo-octadiene which is written in a convoluted form purely because it has to bend in that fashion in order to bind to the metal. So, that both olefin faces can be facing the palladium atom. So, here you have a weakly coordinating benzonitrile pHC N that leaves the coordination sphere of palladium and the cyclo-octadiene is a chelating ligand. It is a olefin complex, but it is a chelating ligand and it can now conveniently form a metal olefin complex. Now, it is also possible for us to replace a carbon monoxide as we had mentioned when we discussed substitution reactions. Carbon monoxide in metal complexes can be readily displaced especially in the transposition if one can use a ligand which will not compete with carbon monoxide for pi accepting property. Although I mentioned in the beginning of this lecture that ethylene can have pi accepting property it is not as good a pi acceptor as carbon monoxide. Carbon monoxide as I mentioned to you earlier is a ligand par excellence. It is the only ligand which has got extremely good pi accepting property and it is very convenient to use it in a variety of situations, but when you want to do a substitution reaction with an olefin the trans carbon monoxide gives way. So, that one carbon monoxide will have better pi accepting property from the metal and a weakly coordinating ligand can be present in the coordination sphere. Here we have iron in the zero oxidation state. It is a 18 electron complex where carbon monoxide has been displaced. So, that the other two carbon monoxides which are present in the same plane will have better pi bonding characteristics from the iron. So, you are replacing one pi accepting ligand by a weaker pi accepting ligand. This again is a general principle and we have managed to do this in this instance with this di ethyl malleate which is a good ligand for replacing the carbon monoxide. Now, I have mentioned that ethylene is a good ligand for metal in zero oxidation state, but very often it is not possible to prepare homo elliptic complexes. By homo elliptic complexes I mean only those complexes, only those ligands which are present around the metal are of the same type where in this case an olefin. So, it is possible in a few instances to prepare homo elliptic complexes with olefins and one example is when you can reduce nickel dichloride with tri alkyl aluminum. When you do that in the presence of ethylene you can prepare a tris ethylene complex of nickel. Unfortunately, this complex is not very stable. This complex is not extremely stable and is stable with only up to a temperature of 0 degrees centigrade. So, this is useful, but nevertheless it is not a stable system. A slightly more stable system is a bis cyclo octadiene complex of nickel. Here because of the chelate effect the stability is increased nickel is again in the oxidation state of 0. So, you have a D 10 system. So, you have a D 10 system and this D 10 system now has got 4 olefins coordinated to it in such a fashion that it is almost like a tetrahedral complex of ethylene. Here you have 2 cyclo octadiene and it is quite much more stable than the ethylene complex which is a 16 electron complex. So, the nickel complex with 3 ethylene is a 16 electron system with 2 cyclo octadanes it is an 18 electron system and this is 18 electron system is much more stable and this can be isolated and characterized very easily. Now, I want to talk about displacement of 1 olefin by another olefin. Now, this is not the method of choice for making olefin complexes, but in a few instances ethyliens can be replaced by another ethylene or an olefin complex. Now, this has to if it has to be thermodynamically viable the complex that is formed must be much more stable the complex that is formed must be must be much more stable than the complex that was present initially. And when you replace ethylene with a cyano ethylene or acrylonitrile then the stability increases this must be an electronic effect because anytime you have a substitution on ethylene the stability usually goes down, but in this instance it has increased and it can be seen immediately that if you put an electron withdrawing substituent on the ethylene this leads to greater stabilization because the pi star orbital on acrylonitrile the pi star orbital is at a lower energy level and from the nickel the electron density can flow into the pi star orbital more readily and so the back bonding is greater in this instance. So, the greater degree of back bonding the greater degree of back bonding in these complexes leads to greater stability. Now, let us move on there is one unusual method of synthesizing metal olefin complex and this is by removing a hydride ion. If you take the tritile cation the tritile cation is the one which is a cation which is stabilized by the presence of three phenyl groups and this can abstract a hydrogen atom. And if it abstracts a hydrogen atom from an alkyl complex which we have already learnt about if this hydride is abstracted by the tritile cation then one ends up with the ethylene complex in this particular case it is a propene complex which is formed and this propene complex is quite easy to conveniently isolated using this method. So, alkyl complexes can be converted to olefin complexes in a few instances by the use of this tritile cation. Now, we have discussed a few methods of making metal olefin complexes and we have discussed in the during this process the reasons for their stability, but let us just systematize them and collect our thoughts together. First of all we said that carbon monoxide can be replaced by olefins in the transposition and this is possible because carbon monoxide is a stronger pi accepting ligand and it will displace or it will displace the carbon monoxide in the transposition and replace it with an olefin and this will lead to greater stability because the olefin itself is a poorer pi acceptor. So, the metal is able to donate more electron density to the carbon monoxide and the m c o bond becomes stronger and this leads to greater stability. So, a second example that I told you about is a replacement of an ethylene with acetonitrile acrylonitrile and acrylonitrile because of its greater pi accepting property it also can replace a simple ethylene. Here again we are talking about better pi accepting character of the olefin, but in this instance the pi accepting property of the olefin that is coming in is much better and so the nickel zero complex that we had in the on the reaction side on the reactant side this complex turns out to be less stable than the acrylonitrile complex purely because on the right side we have better pi accepting property. So, the pi accepting property goes up in this instance and so you have greater stability. So, apart from this electronic effects we also have very significant steric effects in the case of metal olefin complexes and usually it is seen that the trans complex is less stable than the cis complex if there are two substituents and the mono substituted complex is always more stable than the disubstituted complex and of course, the ethylene was the best ligand when you have no electronic effects to talk about then the ethylene complexes in fact the best system to study. So, the steric effects play an important role and we will see in a few minutes why this is the case. Let us take a look at the complexes in greater detail the structural aspects you have eta 2 bonding which means both carbons are equally bound and all the hydrogens if they are ethylene, if it is an ethylene there are 4 hydrogens all 4 hydrogens are equidistant from the metal. The back bonding that we have referred to which means the metal gives electron density into the pi star orbitals lengthens the carbon carbon bond. So, you have a carbon carbon bond which can lie anywhere between a C C double bond to a C single bond C and the if there is such a back bonding interaction we will see in a minute that there is a barrier to rotation of the olefin with respect to the metal carbon metal olefin axis. So, if this is the olefin that we are talking about then in this axis you can rotate the olefin or the metal complex with respect to the olefin, but this rotation has a barrier and this is because of pi interactions between the metal and the olefin all these factors have to be explained and we will be able to explain each one of these factors each one of these factors using some of our structural features. Here I have indicated that the carbon carbon bond length in ethylene is 134 picometers or if you prefer the angstrom unit to be connected it is 1.34 angstroms and if one forms the complex then the carbon carbon bond length increases. Similarly, the frequency of the carbon carbon double bond stretch the stretching frequency in ethylene itself is 1623 centimeter minus 1 and when it is coordinated to platinum as in Zeiss salt it is reduced remarkably it goes down by about 100 centimeter minus 1 and this is exactly the ballpark figure we obtained for carbon monoxide complexes also when you interact the carbon monoxide with the metal then if the metal is the zero oxidation state the metal is capable of putting electron density into the pi star orbital and similarly, here in this instance it is putting electron density into the pi star orbital of the olefin and we will picture that in a few minutes. So, here is the initial interaction and we have already discussed the people who have discovered this type of bonding do or chat and done concern do or chat and done concern or the three people who develop this mode of bonding. So, the primary interaction in this mode of bonding is the donation of electron density from the ligand. So, this is the ligand that we are talking about and here is the metal and usually the ligand donates electron density to the metal and we have pictured that here it is giving electron density to the metal from the filled pi orbitals of the olefin. So, the filled pi orbitals are present and they do not have a node between the two carbons. So, here are the two carbons and the two carbons have a pi cloud on either side of the carbon and one side is capable of donating electron density very significantly into an empty orbital on a metal. So, this is the sigma bond people have estimated that the sigma interaction involves transfer of about 0.24 electrons 0.24 electrons from the ligand to the metal. So, although we call it two electron donation the two electrons are present in the olefin, but approximately 0.24 electrons are transferred or at least 24 percent of the electrons are about 12 percent of the pi electron density is transferred completely to the metal. So, this is the sigma interaction we call it sigma interaction because one can rotate the one can rotate the olefin about this axis and you will not lose any overlap as you rotate the olefin. So, let us take a look now at the pi electron pi interaction again this is part of the Dewar chat Duncanson model of bonding and what we have is a pi star orbital of the ethylene. Now, interacting with a filled orbital on the metal. So, here is the metal and here is the olefin the two carbons are present here and the pi star interaction involves an anti phase overlap of the two p orbitals on the carbon the ones that are not hybridized to interact with the hydrogens and. So, this pi star orbital now accepts electron density from the metal into the empty orbitals on the olefin. So, the situation is very similar to what we encountered in carbon monoxide electron density from the metal is going into an anti bonding orbital and the symmetry of this interaction is a pi interaction because you will notice that if I rotate the metal or the olefin with respect to one another around this axis of interaction then the bonding is immediately disrupted and there is a breakage of this bond. So, this must involve pi symmetry. So, these are pi interactions and people have estimated that in this pi interaction about 0.22 electrons are transferred from the metal to the olefin. So, almost equal amounts of electron density are transferred from the ligand to the metal and from the metal to the ligand. Now, if one can believe these calculations then one can say that there is a slight excess of electron density transferred from the ligand to the metal. So, the sigma bonding is a little bit more important than the pi interaction, but you can see that it has to be a synergistic interaction. If you donate more electron density to the metal then the metal in turn can donate more electron density to the pi star orbital. So, this doer chat Duncan's and model promotes or suggests that there is a synergistic interaction between the metal and the olefin. So, here I have summarized both pictured both in the same slide and you can see how the two interactions complement one another and it is a completely synergistic interaction. One of course has sigma symmetry and the other has got pi symmetry. So, rotation does not break this bond, whereas this bond is completely broken the pi bond is broken when you rotate it. So, that is why you have a small barrier for the rotation of an olefin. This barrier can be pictured here in the Zeiss complex I have illustrated it for you saying that the if the three chlorins around the platinum or in the x y plane. If you think that this is the x y plane then the ethylene is parallel to the z axis. The ethylene carbon carbon axis is parallel to the z axis. So, that is how we want to talk about this interaction then the d x z orbital. If this is the x axis then the d x z orbital is the one which would be suitable for interacting with the pi star orbital on the olefin. So, d x z orbital is pictured here and the d x squared minus y square orbital will be the orbital which primarily accepts electron density from the pi from pi of the olefin electron density flows into this. From the d x z it flows into the pi star orbital of the ethylene. So, I hope you will be able to appreciate this dual interaction that stabilizes metal olefin complexes very significantly. And as I mentioned earlier you do have a barrier to rotation this is the original geometry that we talked about. If you rotate it you will end up with the four carbons the four hydrogens on the carbon the same plane as the p t c l 3 moiety. They will all be in this particular plane and when you do that then you end up with some steric interactions which destabilize this situation. But you will notice that the d x y orbital on the platinum and the pi star of the olefin can interact just as much as the d x z was able to interact with the pi star. So, here d x in this orientation d x y can pump in electron density into the pi star of the olefin. So, both ways you can have pi interactions the sigma is not disrupted at all, but the pi star can also be there in this orientation, but because of steric influence you do not have a very stable situation. So, the steric interaction is pictured here in general it is more important when you have a bulky group in this position then it interacts with the chlorine significantly, but even with an ethylene the molecule prefers this orientation much better. So, that is the more stable crystallographically characterized system. So, let us move on let us take a look at a few of the characteristics that are significant. One is the fact that you have slight re hybridization of the carbon the carbon in the ethylene in simple ethylene it is hybridized in a sp2 fashion. So, the carbon has got an sp2 hybridization. Now, when it is bonded it seems to have a slight re hybridization at the carbon center and that leads to a bending of the hydrogens away from the metal. So, if the metal is bonded in this position then the hydrogens are moved away from the metal and this moving away leads to a slight re hybridization of the carbon and various explanations have been given for this particular explanations. We just talked about the mild steric interaction that can be present. There is also an electronic effect that makes the carbon bend away in such a fashion that there can be better overlap between the metal and the carbon orbitals. So, initially carbon carbon bond distance is around 136 or 134 picometers. Here it is 1.34 angstroms and it can hold in the complex they go all the way from 146 picometers or 1.46 angstroms to even 1.54 angstroms and this is the distance that you have for a carbon carbon single bond and surprisingly the back bonding can be more and more significant in some instances especially if you have electron withdrawing groups and then the bond distance can increase all the way to 1.54 angstroms. Now, if you have a cumulene where you have two double bonds consecutively then the central carbon we are talking about systems like this then the central carbon is sp hybridized. So, these are allenes and these allenes get significantly bent when you are binding them or interacting them with a metal atom. Here I have pictured a platinum atom which is interacting with an allene moiety and the angle in the free allene is 180 degrees and there is no difficulty in understanding that because the central carbon is hybridized within sp hybridization, but the moment you have interacted with an olefin then the strain of the central carbon is released and it goes moves towards the 120 degree angle that you would end up with in a sp 2 hybridization and it becomes close to 142 degrees here. The carbon carbon bond distance is around 148 picometers and it is close to the carbon carbon double bond distance. Now, let us take a look at the frequencies of metal olefin complexes because this is also interesting. As I mentioned the free olefin is 1.34 angstroms or 134 picometers when it is bonded to a metal then the bond distance increases and the frequency decreases. So, you can see about 100 or less than 100 centimeter minus 1 decrease in the metal olefin. You have a 100 centimeter minus 1 decrease in the carbon carbon stretching frequency. So, this frequency is talking about the carbon carbon stretching frequency here, nu c c and this is the c double bond c distance that we are talking about here. You will notice that when you have a 4 D or a 5 D element then the bond distance is different. There can be greater or more pi interaction between the metal and the olefin and if you have a metal in a 0 oxidation state then the pi interaction the bonding between the 2 carbons is decreased even further. So, if you have iron in 0 oxidation state this is a complex with iron in 0 oxidation state the bond distance is 1.46 angstroms it is long. Whereas, in the case of rhodium 1 this is a plus 1 oxidation state and so back bonding is less and so the olefin has got shorter bond distance. So, this is less pi interaction in this case. So, the discussions that we had with carbon monoxide with respect to greater pi back bonding and greater pi bond bonding resulting in frequencies that are smaller or greater reduction in stretching frequencies hold good in this instance also. If there is less back bonding then the bond is stronger and the frequencies are higher. So, there is inverse correlation between the 2 of them. So, how do we know how much the metal is interested in pumping electron density back into the olefin. One way to measure this is to look at the ionization potential for the metals and here I have listed a few of the ionization potentials and you will notice that that as you go down the group you tend to have a remarkable change in the ionization potentials. Nickel is easiest to ionize it is easy to remove the electrons from nickel 0 that is also a D 10 system. Platinum 0 is also a D 10 system, but it requires almost 2 times the amount of energy to remove the electron density from platinum, but the platinum complexes with olefins are extremely good purely because in spite of this higher energy of ionization higher energy required for ionization it is possible to have good overlap between the platinum and the olefins. In the case of nickel one realizes that it is easy because of the fact that you have a 3 D element and this also results in very good in very good bonding between the 2 systems. Pelladium 0 should not by what we are seeing here in terms of ionization potential nickel and platinum should form better complexes than palladium and that is in general true. So, let us take a look at the rhodium and the iridium complexes once again the rhodium complex is easier to ionize and the iridium complex has got a higher ionization potential and so rhodium forms very good complexes very very good complexes very stable complexes we have seen a few of them in the last few transparencies. Similarly, copper and silver because they are positively charged they are already oxidized they have very high ionization potentials they form weak complexes. So, these are systems where they have weak complexes and you have no complexes with zinc 2 plus. So, let us proceed further time that we have let us take a look at the donor acceptor properties can we correlate the donor acceptor properties with the pi accepting property and the stability of the complex. In general the stability of the complex if it is plotted on the y axis if it keeps increasing if you have very good if you have very good electron accepting groups electron accepting groups on the olefin then they form stable complexes. So, you have trans disino compounds. So, you have C n and C n then it forms a much more stable complex with the metal because you can have better pi accepting property. Whereas if you have no substituent then you have poorer pi accepting property and that is pictured around around here. And then you have systems which have got substituents which prevent the close approach of the olefin to the metal. So, if you have cis it is more stable than trans we have already explained this because trans cannot twist in such a way that it can approach the metal in a closer fashion. Whereas cis by bending the substituents on the carbon carbon double bond in such a way it can move it away from the metal and so it can have better interactions. And so this is the mono substituted systems and then this is your ethylene itself simple ethylene and these are with weaker electron accepting groups. The systems which are pictured here have weaker electron accepting group. So, this is a general tendency they are very poor or in unstable complexes when you have tetra substituted systems and they are not pictured here in this graph at all. So, here again if you have this electron accepting property is quantified using the hammock parameter which you encounter in organic chemistry. If the hammock parameter is plotted on the x axis then you will notice that as a hammock constant may becomes larger and larger electron accepting property increases and then the binding constant is increasing. So, there is a positive correlation between the two of them. So, the barrier to rotation is proportional to the extent of pi interaction. In general you have better barriers to rotation when you have 0 oxidation state metal complexes. The rhodium complex that we are going to talk about has got a barrier of 15 kilo calories per mole, but this barrier to rotation need not correlate with the frequency decrease. Frequency decrease independently correlates with the back bonding, but because there is a combination of steric effect and pi effect these two need not correlate one with the other. So, here I have pictured for you what happens to the carbon-carbon bond distance when you have an electron withdrawing substituent. Here I have tetra four fluorines attached to the carbon the olefins and here I have simple ethylene. This distance is hardly increased from 1.34 angstroms it has gone only to 1.36 angstroms. So, this is hardly a 0 2 angstrom increase in the case of the ethylene complex whereas the carbon-carbon bond distance in the case of tetra fluoro ethylene is almost similar to what you have for a carbon-carbon bond distance in diamond. So, carbon-carbon bond distance in diamond would be close to 1.54 angstroms and that is pretty much what you have in the case of the tetra fluoro. So, there is a lot of electron density going in from the nickel to the pi star orbital and this results in a very strong weakening population of the pi star and weakening of the carbon-carbon bond. So, there are some similarities between carbon monoxide and ethylene complexes. It is not just the case where you have eta 1 and eta 2 coordination modes between the ligand and the metal, eta 1 in the case of carbon monoxide and eta 2 in the case of ethylene. That is not the only difference or that is not the only similarity in terms of electron density donation from the HOMO probably ethylene is a little better than carbon monoxide HOMO in ethylene is bonding. So, when you remove electron density from the HOMO it weakens the carbon carbon bond. So, both carbon monoxide and ethylene accept electron density into the pi star orbital. So, this weakens the ethylene carbon-carbon double bond and so the CO on the other hand it has got a slight difference because the HOMO is non-bonding if not slightly anti-bonding. And as I explained when we are discussing the carbon monoxide complexes this is a controversial discussion that has been there in the literature, but it is understood that the carbon monoxide is not a bonding orbital. The highest occupied molecule orbital is not a bonding orbital and so removal of electron density does not affect carbon monoxide significantly in terms of the CO stretch. So, a complete discussion of the metal olefin bonding and then what we call as a Dewart Duncan's in chat model is given in a recent paper which is pictured which is reference is given here. And what we have understood from all these aspects is a fact that it can olefin complexes can be readily synthesized, structurally characterized and the bonding differences between carbon monoxide and olefins allow one to synthesize them by displacing carbon monoxide. And it is possible to have the significant reduction in the stretching frequency of olefin complexes due to the pi accepting nature of the olefins. And you can also use these metal olefin complexes to generate more stable olefin complexes by displacing them based on the pi accepting nature.


So just imagine as you're opening up that the ball is elastic and your hands are stuck to it and you're pulling this elastic ball bigger and bigger so you can feel two forces happening at the same time. As you're opening up the ball is trying to pull your hands together and then as you're moving them in together you feel the ball pushing back because it's a rubber ball and it's got buoyancy. It's got elastic. Expanding, pulling, pulling that chi, chi taffy. You can imagine this chi taffy and you're pulling that. And when you're contracting, same thing. You're squeezing this energy ball. Okay?


Yes! He doesn't stop. We are going at it. We went to okay. Episode five. The dogs are joining me today. Say hi, kiddos. Say hi. Hi, Fulverton. She thinks we're playing, I think. Hi, Fubi. Hi. What? Luna's looking at me like I did so. I don't know what Luna's up to right now. She's on some demon time. We're down bad. We're down horrendous. I will say that. We are coming off two straight losses. Taking us to two and two. I had this monstrous peen. I had this huge bratwurst laid out on the table when I was two and L. And progressively, it's just shrunken, shrunken to the shell. Now I'm looking at two inches hard max. It's not good. And if you're new to the channel and you just heard me say that, I don't have an explanation for you. I don't know why I said it either. We need a W. I don't know what it's going to take. I feel like a 99 Dark Matter Center is what it's going to take. The Andre Aiden was close to that, but the guy we played was just so much better than me that I didn't have a chance anyway. I feel like if I play a comparable opponent, like we're equally matched, at least pretty close. If I have a Dark Matter Center, I think I got a shot. So that's what we're going for today. See if we can get in this wheel spin. All right. The Wheel of 2K is spinning. If I can get a Dark Matter somehow out of this, I'd love to take a store pack. So good. Really good too. Eastern Conference. Playoff jackpot. Boom. Let's go. That's exactly what I needed. So, okay. Number one. That's like literally one of the best spins almost. It's not a full jackpot, but it's close. We can get any player that was from a playoff team. I kind of want to go with one of the finals team. So either Clippers, Sons, Bucks or Hawks. I can't believe the Hawks are a final. That makes no sense. I'm going to start with Dark Matter and let's look at each team. So the Dark Matter Sons players, there's the Devin Booker. I would love that. Well, it is a jackpot. So technically I could get Galaxy Opals too. Hawks. There's Dominique Wilkins and Clint Capella. Now that's just Chris Bosh all over again. Even if he's a Dark Matter, there is a Yanis. It's absurdly expensive. It doesn't look like it's going to happen. Thonmaker would be pretty cool. There's a Dark Matter Blake Griffin, but I don't need a power for it. He's too small to be my center. So I feel a little bit biased to the Sons right now because I tried Deandre Eight and didn't work and now we're going Devin Booker, but 58,000 MT for this absolute demon Devin Booker. I'm very excited. All right, Devin Booker, let's look at the stats here. Oh, he looks like... Mans is clapping Kylie Jenner too, right? That's the one he's clapping. Or is it Kendall? I don't really know. He's clapping a really hot moment. Good for you. 98-3, 98-mid, 96-close. Doggies. Come here. Yeah, guys in the absolute stud. A plus is all around the board other than post-defense, rebounding, and basketball IQ. 94 overall team right now. That is exactly the player I need. I'm not going outside like this, bro.  I'll show you the cheat sheet right now. We are two and two. So there's a lot still possible on the cheat sheet. Without further ado, ladies and gentlemen, let's see Devin Booker in action. Let's go. Look at it. Oh yeah. They literally play chase, bro. They play tag.   Is this Jason Kidd? Who's bullying me here? Damn, he's going to scrap. Oh, okay. Good start, boys.   That's Devin Booker. Oh, I thought he was going to throw that down. Where's my fault though? I should have set my feet with Devin Booker. I didn't set my feet. Dirk? Dirk wide open? Somebody's going to hit it. Let's go. Ah, ah, ah. Good D. Good D, Moran. Let's go make him earn it. You got to earn that bucket. All right, you earned it. Oh, Moran, wide open. Let's go. Six to two. Good start, boys. Damn, he just saws me up all the way, honest. Ooh, good step through. Oh, that was such a good step through, honestly, but Booker is a... Oh, I get the leaner. Devin Booker is over two on the... I mean, they're slightly covered, but I don't know, man. Good D. That oddball magic all the way, all the way. Ooh, Christmas. Vintage. Oh, and that's a rip. And that's a layup. And that's 12 to four. Let's go. Ooh, just saws me up. He is literally giving me the shot, and we're going to green it. Let's go. That was weird. 15 to eight. This is definitely one of my best starts ever. And he's going to give me a freebie with Jeremy Lin. That was a really bad release. But as we saw last episode, Jeremy Lin can strap.  Good defense. Oh, get there. Why? Why do they do that? There's a guy right in front of you. Just pass it to the guy. They literally just throw that pass like they've never thrown a pass in their life. I'm going to go with a pump fake. We're going to rip it over to Devin Booker, who's now over three on wide open buckets. And he just chucked up a half-core shot. Damn, Jeremy Lin covered. I shot that so late. He just closed the gap so quick. Devin Booker though. Wow, he's just going to leave Jeremy Lin. That's another green for Jeremy Lin. I don't know what he's doing on defense. Bit of a mismatch down here. Ooh, look at the glance from Chris Bosch. Back to Booker. We got to get Booker a two or something. Yeah. He's just got to see the ball hit the bottom of the net. He goes through the, ooh, and nails it. Well played. Oh, Jeremy Lin with the cross off the way. Oh, okay, I got fouled. Thank God. Jeremy Lin's free throw is eh. I'm really bad at free throws. I've noticed that in this game for sure. Ooh, take it back. 100%. Can we rip this? Can we rip this? Oh my God, oh my God. He's just jumping, jumping, jumping. Little pump fake. Yup. Oh, we have such a mismatch here. It's Trey Young on O'Kong Woo. Get in there, baby. Let's go. That's how you get a bucket. 17-26. Yup, back to, pump fake? Yup. Oh, I didn't realize how much I was on the clock. Shit. One of my best first quarters ever. We are locking in this Devin Booker. Even if he's not playing that one right now, we're locking him in. Look at Fobi. Fobbers. I missed you, baby. I missed my child so much. Did I just get a penalty? I don't care. It's my child. Oh, whoa, there's nobody on this dude. There's multiple people unguarded and he didn't find any of them. And he hits, he can hit that. All right, respect. Dude, I don't know. This guy might be lagging or something. O'Kong Woo is wide open for an ugly slow duck, but it's a dunk on the less. But I don't care, bro. I'm taking my dump. However, I can get him a rant. Oh, I take it back. Horrible defense. Hey, there's 24 seconds on a shot clock. You gotta play 24 seconds to get defense. Not just five. I understand. Ooh, rant. Yup, pump fake. Oh, that's what I was saying. That was a wide open D-Wade right there. Abuse too. Lean, ooh. Oh, stop. I'm supposed to be leaning too. Dude, did he just grief me? Wait, so he just, I could take the W right now. Cause he paused too many times. I could have taken the free W, but you're kidding. Now he's not gonna make it. I wanna win like a man, bro. I'm not to, ooh. No, no, no, no, no, you're supposed to dunk that. I'm so sad right now. You have to cut to the rim. All right.   We're maintaining our lead. If I lose now, after I had a chance to take a free win, we're just gonna clamp up and not let that happen. Oh, Marant. Marant! Marant does not wanna dunk. Just like in real life. He just wants to almost dunk every time. Yo, we got a 21 on this guy. Yup, fake, fake it. Indy honest for the one who just got so many layups today. Oh, that's Caruso. He's got Caruso at the top of the key. Is that a dark matter, Caruso? Who made that? What a dish. Is it like a fan favorites promo? Maybe that's why Devin Booker can't score.  That was a 2% covered. No way. Good box out.  And then he's gonna hit that? Got him jumping. Devin Booker with a leaner. Hey, he hit it. That's supposed to be a dunk, but whatever. Get in there, get in there, get in there. Ooh, good spacing. 31, 37. Damn, it's a little bit closer game than I would like. Let me be honest. Oh, he bricked it. Follow line. Yup, good spacing, Booker. He knows he's a shooter.  Let's go. Wow, he really wants this, huh? That ain't gonna be there. Yup, Devin Booker. Devin Booker on the court, baby. Nope. What is that? Are you shitting me?  That was one of the dumbest. No, you didn't just hit that.  It's unbelievable.  It stood up somehow. I have to take it two stops. And then he hits a Lamarcus, all just three, fully guarded. Glad I'm a five point lead. I have to put you on a tie out.  I'm trying to train them on the electric fence right now and she's not taking it. I can't right now, Phobie. When this game is over. He's so cute. Look at how cute Phobie is.  Don't chew on me for that court. 35, 42. Oh my God, Caruso with a nasty cross. Good Lord.   I actually can't believe this right now, bro. He's just fumbled the bag on some wide open threes, a wide open dunk. This is literally about to be a tie ball game. I'm not losing this game. I'll tell you that. He is, oh, excellent screen there.  Don't you dare.  Wide open, Dirk Navitzky, perfect release.  You have to be joking. I literally don't know what to say, bro. I literally don't know what to say, bro. All right, good defense there. We're gonna need a free bucket. Stop spacing to a three.  Are you gonna do it again? Everyone stop going for the highlight reel.  Devin Booker, Devin Booker redemption trial right here, man. Great defense, great defense Devin Booker. All right, Devin Booker. Hey, no Fubi, no Fubin. At one point, I think we had almost a 10 point lead, but I will shoot that. Let's go, 4% cover. Don't worry, I'll miss. LaMarcus Aldridge, green.  Let's go. D-Wade wide open, he's gonna break it. Oh my God, did you really? He hit it. Tied up in the third. Jeremy. I should probably save to last shot unless I see something absurdly wide open like my best third point shooter who doesn't break like Devin Booker. I'm over Devin Booker, bro. Oh, damn, nice move. That's fine though, I got 6.7 to score now. Oh, yup. That should be easy bucket. Let's go. Damn, we scored, but I scored a little too fast. He didn't even shoot one. Alrighty. That was a good way to close that quarter out. Oh my God, I'm sweating, boys. We gotta get out of this with a dub. We gotta make it out with a dub. Up by five, heading into the fourth quarter. Alex Caruso's a menace. I see a positive attitude, bro. I'd be bitching too much in 2K. Okay, not the world's greatest pass and Thiples run in the court. Please, I wish all of them could just do a layup or something so base. I don't know if there's a way to do that. I'm sure there is. I just don't know how to do it. Okay, throws an hoop when he had a layup. Why are you standing on a bounce? That is not open at all. All right, good knee, boys. Jay Lin, Jay Lin got it to the right. Foul in wants his starting spot back.  Yeah, Jeremy Lin might be going to start a rep to today, bro. To 10 point ball game. Look at the defense, look at the defense. Uh-oh, uh-oh, not the pass to the century. Now we're pulling away, baby. Finally, unshokable, unshokable, unshokable. Come on, Dirk, that was a great quick wheel. Start to fourth. Oh, he's not. I thought for sure, bro. Look at this run. It's about to be a 10-0 run after Dirk straps this through. 10-0 run to start the fourth. This is some elite hoopin' right now, damn. I got to see his shot meter. I got to watch him green on me. Yannis with the pump. Up to Devin Booker, who wants redemption? All right, Devin Booker's much better in the second half than the first so far. I'll say that much. Gidea, Gidea? Oh, stay up on it. That's fine. He used the whole clock for a dunk. Why are you pause-spanning me? You died by 13, bro. Did this slow? Ooh, Yannis. Damn it. Damn it. I seriously, I need to find a way to just get like a basic layup. 13 point game, though, so I feel good. Jay Lin clamped. Oh, is that Dreymon? Shit. 62.73. Ooh, they lost him. That's a green. Devin Booker just comes alive in the second half, I guess. No, you're not getting that bucket on the go. All right, dude, this is a massacre now. Wow, he's just gonna let me have it. I was expecting a lot. Didn't get it. Wow, all his luck completely ran out in the fourth. Damn, I can strap that. I'm hitting the leaner, because I'm rude. Oh my God, that was sexy. Oh, we got that takeover. I don't know what takeover this is. No, no.  He just hit a half court. He just hit a half court shot.  I didn't actually know that that was possible. No, not again. Not again. Oh, that was insane. I shouldn't have hit it anyway. Devin Booker with a rip, pass it up to Jay Lin, who's gonna slam it. Ay. Dude, Jeremy Lin caught a body today, too. 20-point game. It was 50 to 50. Or 55 to 55, right? This game was so close. Feeded to Devin Booker, who actually gets the shot off him. 85-60, Devin, we needed a win like that, literally so bad for my ego, partially. And so that we could lock in another solid player, Devin Booker.  Oh my God. I can't believe that game was close, dude.  All right, so Devin Booker had, he was my leading score of 24 points, and 40% from three. He really didn't even shoot that good a percentage. Jeremy Lin, five for eight. One for four from three, and Morant was five for 10, two for three. But that, ladies and gentlemen, is gonna be episode number five in the books. Devin Booker, bro. Three and two. So looking at the cheat sheet, the haircut is out of the question. Technically, I could still do a 24 hour stream, but hopefully I get at least one more win. All right, boys, I love ya. I will see you in the next episode. Peace out.


Hey guys, it's me Logan and we're back here again with another K-pop game video! And as you guys can tell by the total video, we're gonna be getting K-pop songs by the last 5 seconds and I reckon this is gonna be an easy peasy piece of cake of a game video. I don't know what I'm saying anymore.  You know, I got like 50% or something like that and this time we're gonna be back at it again by being cocky and we're gonna see how many we get right and I'm gonna actually remember my score because every other time I always freak and forget. So if you guys want to join in with me, comment down below what your score is and also what ones you got wrong or right because I'm probably gonna get like most of them right, not wrong because I am gonna be confident. But yeah, other than this Australian goodness, I'm freaking excited. I hope you guys are too. And let's do this. Oh, oh, J.K., don't cook. We're just kidding. Oh, we've had up, had up, had up. Oh, this is a remix. This is pretty dope. But I still like the original better. Oh, they wish me good luck. Thank you. Oh. Hey, mama, this is by EXO, ba, ba, ba. One, one, one, one, one.  Round one.  Nakna. Two for two. They're confident. They're confident. We're doing it. This is like my score counter, man. This is my score counter. This is just so I don't forget. Hard carry yet? Hard carry yet? But God 7. Oh, yeah, let's do this. We are free. I freaking love this song. So good. Probably one of my favorite God 7 songs. Hard carry, carry. OK, four. Let's go through the song, right? But I'm not sure. I've got it right. I've got it right. That we're on four for four. We're doing good. We're doing good because guess the song. So I'll guess the groups. We're on four for four. We're doing this. Wham. Oh, be they. Oh, dance is good. Five, man. We're five. Give me a five. Bam. Oh, it's the, no, thank you. No, jail for it. Why am I getting? I just saw this the other day. And I, oh my God, I'm five for six.  Oh, get down. Get down.  We're on six. I did say get down.  OK, we're on six. We are on six people. We are on six. We're doing it today. We got one wrong so far. Oh, I don't know. Iowa, what a man. Oh, yeah, let's do this. And let's do this. Seven. I love that song. I love that music video. Oh, Monster X. No. I knew it was hero. But I see it at the same time as the other thing. So I'm not going to count that. We're on seven. We are still on seven. Number 10. Here we go. Oh, Blackpink. Not the song wrong.  I knew it was playing with fire as well. Uh-oh, uh-oh. It's just like so many of these songs just popped in my head. Just like a movie by BTOB. You know what that means? You know what that means? We're on eight. We're on eight. We're doing this. I can't believe it. No, don't even know. Don't even know.  But we're on eight, guys. We're on eight. If we get two more, I'm going to be happy. 13. No, no, no. God damn it. I knew this song. We're still on eight, 14. Oh, a Russian roulette by Red Velvet. Gary is my bias for Red Velvet. Just kidding, you guys know. I love Gary. I've watched so many fun, funny moments. Oh, BTS. Ah, danger. Don't. Why did I say danger? We're on nine, though. We're on nine. I forgot to say we're on nine from the last one, right? That's not bad. We've got a nine. Bonus, bonus. Oh, I love this song. I just forgot that the ending was like that. We've got nine out of 16. That's not bad. We've got more than half. Oh, my god. If that is somewhat surprising, I'd be just some of you guys as surprised as well. I got freaking nine out of, well, I'd say it's 15, but there was a bonus, so nine out of 16. I got freaking nine out of 16. That's actually freaking good for me. I'm actually improving so much with these K-pop games. Like, I am going to be a god at it. Next time, I'm going to get 100% to, OK? I'm going to do it. Trust me, trust me. And if I don't get 100% to, I'm probably going to edit this video in. And it's going to be awkward. Yep. But you guys are supposed to be all for this video. And if you guys enjoyed it, please show some support by leaving a like and also comment down below what you got right and also what your score is and also what you got wrong. Actually, no, just tell me what your score is. Because, you know, saying what you got right and wrong, that's just too much work. I'm going to make you guys do a little work as possible. So just comment down your score. See if you guys did better than me. And if you guys did, then please teach me. Please teach me the ways. Please teach me the ways. And also, subscribe to guys on YouTube channel because there will be more reactions and more K-pop games coming to you guys in the near future. That's basically it. So I hope you guys have a great day. And just keep on smiling. And adios, amigos.