Welcome to module 35 of points anthropology part one, we shall continue our study of connectedness. So far things were not all that difficult, but now we are trying to take you a little deeper into connectivity. Here is a result that can be used to prove connectivity of a big class of interesting spaces. So, let us formally make a definition here. Once again, this is for namesake only. Take any set theoretic function f from x to y. For every y inside y, look at all points which are coming to y, all points of x is coming to y, namely f inverse of y. That will be called the fiber of f over y. What are the fibers of f? They are all inverse images of some point in y. For example, if you take a point y, there is no point which is going to that point f inverse of y will be empty. So, then I will say fiber of y at that point is empty. So, that is also allowed. It is purely set theoretic notion. There is no other condition on f. The word fiber is used in many higher mathematics. I am used to that. That is why I am putting that terminology here. So, everywhere fiber occurs, you can just write f inverse of y inside the core domain. Then you do not have to use this fiber word. Let f from x to y be any quotient map now, in particular is subjective. Suppose y is connected and all the fibers of f are also connected, namely I am writing it just for your sake, f inverse of y for every y is connected. They are all connected. Then conclusion is x is connected. You see, what we have seen earlier is that if x is connected, then f y is connected. Therefore, in a quotient map f of x is whole of y. So, if x is connected, f of x is connected, f x equal to y. So, y is connected. We are looking at a converse here. y is connected. You do not immediately conclude x is connected. But if the fibers are all connected, there is a extra condition, then you can do the other way around x is connected. So, it is quite a profound result, but the proof is very easy here. All that I do is take a separation instead of a, b I have called a 1, a 2, does not matter. Take a separation of x. Look at all points b i, all points of y below. So, there are points of y such that the fiber over that point is contained inside a i. b i, I am just defining it. That is my b i. This b i is subset of y. So, this I do for both i equal to 1 and 2. As such, b i could be empty, they could be anything. But here I am using two things. Namely, f is surjective and then it is actually a quotient map and the fibers are all connected. So, together these things will imply first of all, all these b i's are non-empty. b 1 union and b 2 is the whole space x that comes from surjectivity. And sorry, before that each f inverse of each point, namely the fiber being connected must be inside a 1 or a 2. Therefore, a 1 union a 2 will be the whole of x.  So, union of all f inverse of y is the whole of x. Therefore, y will be union of b 1 and a 2. What we want to show is this is, you know, we started with y as a connected space. This is a partition. This is a separation. That will be contradiction.  Similarly, f of a point in a 2 will have to be inside b 2. So, b 1 and b 2 are non-empty.  And to begin with, we start a 1 and a 2 are closed subsets. Inverse image of a set is closed implies the corresponding set is closed in the quotient topology, in this quotient map. Therefore, both b 1 and b 2 are closed. Clearly, a fiber cannot be both b 1 and b 2. It has to be in one of them, right? So, b 1 intersection b 2 has to be empty. So, all this show that b 1, b 2 is a separation for y and that is a contradiction. Why a contradiction? Because we started with a wrong assumption that x has a separation, okay? As an immediate corollary, we get a big theorem here. Take any finitely many connected spaces and then take their product that is connected. How do we do this? Do it for 2 at a time, then by induction it will follow for any finite product, right? For 2 at a time, how do you do? Look at x connected, y connected. Look at x cross y to y, the projection map or you can take x cross y to x also, no problem. That projection map is an open map. Any open surjective map is a quotient map. So, you can apply the previous theorem to this quotient map. Y is connected. What are the fibers of y? Fix a y, all points x comma y coming to y, nothing but x cross singleton y, which are all homeomorphic to x. Therefore, each fiber is connected because I start with the assumption that both x and y are connected. So, theorem says that the product is connected. Once you have proved it for 2 by induction, it will be true for all finitely many product. The above theorem, together with this corollary, it is an easy corollary. Sometimes you have to directly use corollary. Sometimes you may have to go back to the original theorem. That will be very useful in proving connectivity of a large number of tucaloid spaces. So, that is what we will do now, namely what are called as the general linear groups, the orthogonal groups, the unitary groups and so on. They are all inside n cross n matrices. So, let us make a formal introduction to these things which are very important in mathematics, in the central mathematics. Look at, of course, my k is any field you will be to begin with, but finally when you talk about biology and so on, I am taking k as either real numbers or complex numbers. Remember that. So, this notation m n cross m k denote the set of all n cross m matrices with entries in k. Clearly, under entry-wise addition and multiplication, it is a vector space of dimension n m over k. This is part of your linear algebra. Therefore, by choosing an appropriate and convenient basis, we can identify this vector space with k power n m, the Cartesian coordinate space because dimension is n m. There are many different ways of doing it. For example, you may write the rows of A as, see there are n rows and m columns. So, each n row, the one single row is a vector inside c power m, k power m. Write it the next row to the, you know, on the side of, on the right of that and so on, instead of writing the below. So, write them side by side so that it will look like a n m order to pull. For example, if you have 2 by 2, then you will get a 4 vector, okay, a 2 vector followed by a 2 vector. So, you get a 4 vector and so on. So, that will give you an identification of this vector space with k power n m, okay. It just depends upon which way you want to write. Any of them will be as good as any other one. So, you better choose according to your convenience. That is why convenient and appropriate. Appropriate to the context. But you should keep in mind that your original space has a matrix structure for many other purposes. So, in order to rely, in order to remind you that instead of writing k m n here, k n m here, n m m n, once you write 5 into 3 as 15, it is lost. 5 into 15 could be 1 into 15 also. So, you write it as n cross m. So, this is just a reminder that it is an n cross m matrix, okay. This notation is used. It is a very clever notation. It is not mine. It is there in the literature.  As a Euclidean space, they will be both same. As a vector space, they are the same, okay. As matrices, this will be different than k 1 cross n m or n m cross 1 and many other possibilities. Various subsets of m n k are of interest in higher mathematics. Now, if this notation m n k stands for n cross n, okay. So, instead of writing 2 of them, I am just writing n. That means they are square matrices. Here are a few of them. I am giving you definition of that. Look at all square matrices of size n such that the determinant is not equal to 0. It is the same thing as all those which are invertible. The S o n is inside now R n, okay. R n cross R n cross R n and so on. R n cross n. N cross n matrices with real entries such that A A transposes identity and determinant of A is 1. If you do not put this condition determinant of A, then this is called O n orthogonal group. So, this is a special orthogonal group, okay.  Here also you can have another one S u n wherein you put one extra condition determinant of A equal to 1 here, okay. I am trying to give you only a few of them here, but I just told you two more also here. Note that the first one g l n k is an open subset of k n cross n or all the matrices. Why? It is given by determinant of A not equal to 0. It just means that inverse image of the function determinant which is a polynomial function, okay. So, therefore, all those points which are you know determinant is a function into k, right. k is either c or r. So, not equal to 0 means that is an open set. So, it is the inverse image of that open set, okay. So, this is an open set whereas for similar reasons S o n, this is an equation, that is an equation. It is not an equation, okay. S o n and u n, they are closed subsets of m n, m n k either c or r. This notation is A transpose and this is A star is conjugate transpose, okay. So, I claimed that our previous theorem along with the corollary about finite products can be used to prove that all these three here at least and then many more are all connected spaces. So, let me present a proof of one of them, the first one, okay. When k is c, I am taking the gl and c, okay. If you take gl and r, then you have to be dangerous, then you have to put determinant of A equal to 1 or minus 1. That will have two different components, okay. Okay. The space gl and c, what is the space? It is now subspace, open subspace of the Euclidean space is connected for all n greater than equal to 1, gl runs, it starts with. So, just a temporary short notation g n depends upon n, right. So, I am going to prove this inductively. So, let us first prove c, g1, g1 is just c star. What is determinant of a 1 cross 1 matrix? It is just the element itself and that should not be 0. So, it is just c star which is c minus 0, okay. We know that c minus 0 is connected. Dropping out a finitely many points from r2, r3, rn, they are all connected that we have seen already, okay. So, now assume that the result is true for n minus 1 and n is the determinant of 2. That means, g n minus 1 is connected we are assuming, okay. Now, let us use this symbol En to this column, this is a column vector 0001 inside c power n. So, to write a column vector I will 0001 and put a transpose here that is all. Consider the map phi from n cross n matrices to c power n given by phi of a equal to a operating upon En. What does it give you? It would give you the last column of this matrix A, okay.  It is one of the coordinate functions. So, therefore phi is open surjective mapping, right. This is an open surjective mapping but we are not interested in the whole space here. We are interested in only an open subset of that g l and c, okay. So, all this I have told you that it is just a quotient map here because it is a projection map to the last n columns, okay. In particular, phi is an open mapping. Now, g l and c is an open subset of that map. So, it follows that phi restricted to g n is also an open mapping. But what is the image? If you take a non-zero, it is an invertible matrix that is determinant of A is not equal to 0. Each column is a non-zero vector that is a minimum thing. Therefore, the last column which is A of n, it is a non-zero vector. Therefore, the value is taking inside c minus c n minus 0 0 0. So, I will write it as c n star. What we want to say is that this is also surjective namely from g l and c namely g n to c n star. So, this is surjective is what I want to say. Why? Take a non-zero vector inside c n.  A basis. How many elements will be there in the basis? Exactly n elements. The first vector you write it as a last one. V is equal to the last vector. Then write the other n minus 1 vectors, column vectors they are all in 3 to the minus column vectors. You will get an n cross n matrix. That matrix because each column together they span the whole thing. So, that must be a basis. This is what we have assumed. So, the matrix will be invertible. Determinant is not equal to 0, which is the same thing as a matrix invertible. So, what we have? What is an element A inside g n whose nth column is v? It just means that if you take phi of A, it is v. Therefore, phi restricted to g n to c n star is surjective. We are exactly in the situation of theorem 3.34 that we have just proved. It is a surjective open mapping. So, it is a quotient map. c n star, what is it? c n cross c n cross c n. So, sorry, c cross c cross c minus 0 0 0, which we know is connected. What are the fibers of this phi? If we show they are connected, then you are in a good shape. You can immediately conclude that g n is connected by this theorem. Okay. Therefore, what we need to show is that each fiber here, phi inverse of v, this is connected. By definition, what is phi inverse of v? All those n cross n matrices inside g n, such that when evaluated and last on a n, which is the last column will be equal to v. This v is fixed. First thing I want to show is that all the fibers are homomorphic to each other. What is the idea? Finally, I want to prove that they are connected. So, if I prove one of them is connected, all others will be connected. So, let me prove that they are homomorphic to each other, which is much easier to prove. So, what you have to do? Start with any non-zero vector. Choose a in g n. This we have done earlier, such that a of e n equal to v because we know it is surjective now. Okay. It then follows that the left multiplication by a defines a homomorphism. Take an element here, multiply it by a on the left. What is the inverse map? Multiply by a inverse. So, this is a homomorphism of the whole space to itself, g n to g n to g n. But I want to say that this fiber goes to that fiber. What is an element of this fiber? All those b such that b of e n equal to e n. What is this one? All those say b prime of e n is equal to v. So, if you take an element such here b, a times b will have this property and go on us. Okay. So, homomorphism is very easier to verify here. Now comes, I have to just prove that one of the fibers namely phi inverse of e n, this is connected. You can see why I have done this one in the next step. Why the choice of e n is done here? Namely, this allows us to apply the induction, the induction hypothesis. Okay. So, the claim is that phi inverse of e n is connected. So, but what is it? I am just repeating it here. What is it? All those b in g n, b of e n is e n. This is the definition. I want to say that this thing is homeomorphic to g n minus 1 cross c n minus 1. Suppose I prove this one, then the theorem is over. Okay. This was only thing that was needed. Once one fiber is connected, all the fibers are connected. So, I can apply the previous theorem to conclude that g n is connected along with the induction hypothesis, it follows that g n is connected. So, all that I have to prove is that this set, this space is homeomorphic to this space. Look at what happens. b is in phi inverse of e n, means the last column is 0 0 0 0 1. Okay. So, this b has the last column. This is a big 0. This is a block matrix, by the way. This is n minus 1 cross n minus 1. This is n minus 1 cross 1. This is 1 cross n minus 1. And this is 1 cross 1. It is a block matrix. Okay. So, this whole column is just 0 0 0 1. That is the meaning of that it is e n. Right? This is some block b 1, which is an n minus 1 cross n minus 1 matrix.  Okay. So, every b which has this form, namely inside phi inverse of e n has this form. But determinant of this one is nothing but determinant of b 1 times 1. Therefore, this must be non-zero. The moment this is non-zero, it is element of g n minus 1. And what is this one? This is just a element of c n minus 1. Projection map here is to first n minus 1 coordinates here, n minus 1 cross n minus 1 coordinates. That is continuous. Projection to these coordinates also continuous. So, breaking b into two parts like b 1 comma u, that is a continuous function. This is what I am putting here. The compliment, I mean the inverse is also continuous from the whole six to this one. Therefore, this becomes a homeomorphism. This proves that g n c is connected. So, here little bit of many interesting topology is involved here. Okay. So, the method employed here is educative again. Not only the information that you have got now that g l n c is connected, okay. If you have learnt this proof nicely, it will help you in long way for many other things. Now I will have to go to another important thing here, okay. This time I may take a little more time. The topologist signed a card. Remember, I wanted you to, I want to, I promise that I will show you why the closure of a path connected that may not be path connected. That is what we are coming to here. This example will serve that purpose but pay attention to this example because it will serve as a many, many counter examples for us, okay. For us as well as for other people also. So, this thing, something you have to study properly. What you said? What is the sine curve going to do? It is starting with the graph of a sine function. Only thing is instead of sine x, we are taking sine pi by x. The x is inverted here. Therefore, the domain should exclude 0. At 0, it is not defined. So, I do not want to take 0 to infinity. I just take 0 to 1. That is enough for me. You can take 0 to infinity for other purpose. 0 excluded, 1 closed, no problem. Take the function sine pi by x and look at its graph. The graph is x comma sine pi by x, point like that, which will be subset for R2. Okay. So, the whole topology sine curve is going to be subset of R2. Okay. You can just look at this curve, but now we want to do something more here. Namely, we want to put the y axis part between minus and plus 1. So, 0 comma y between minus 1 and plus 1. This is the lines, closed line segment, okay, lying on the y axis, 0 comma y. So, you take the union of this sine curve, namely the graph of sine pi by x, union with b. That is my x. Okay. So, this is by definition the topologist's sine curve. Okay. Being the graph of a function, continuous function defined on an open or on some interval which is connected. Okay. So, the graph A is connected. Actually, it is path connected. The whole thing is a path. It is also path connected. All right. So, A is connected. A bar is the whole of x. So, this is what one has to see here. I will show you the diagram here, then it will be very clear to you. So, x is connected. A is actually path connected, but I am not going to use that. Just connectivity is enough to conclude that x is connected. Let us look at this curve. So, this is the sine curve. This is 1. That is 0. At 0, it is not defined. But what happens to this? This is 1 by 2. Okay. This is not 1 by 2. This is what? Yeah. 2, 1 by 2, see pi x by pi. So, yeah. So, you see for, let me see, this pi by x is there. Put x equal to 1 by 2, then sine 2 pi you get. That is 0. Okay. Like this. So, at all these 1 by n or 2 by n, 4 by n and so on, it will take either this point or this point or this point or this point or this point. It depends upon even and odd. So, it will keep going like this. This part is b. This line segment on the y axis. This is the y axis from minus 1 to plus 1. All the, the entire curve is between this y equal to minus 1 to y equal to plus 1 in that strip because sine x does not take values other than minus 1 to plus 1 in the interval minus 1 plus 1. So, this is like. So, now look at any point here. There are all these points coming closer and closer to this point. So, every point on the y axis from minus 1 to plus 1 is in the closure of this a. This is a. Therefore, a bar is the whole of x. Is that clear? So, every point on the y axis from minus 1 to plus 1 is a closure point of this a. Therefore, our theorem says that this space is connected. A itself is path connected because it is actually, itself is a path. Except that in the definition of path you must have a closed interval. So, we do not have a closed interval. So, what you do? Suppose you take a point here, from any other point here to any other point here, you can take the restriction of the function sine pi by x that will give you a path here. So, this is path connected. So, since a bar is x follows that a is not closed. Of course, it is easily seen that b being a line segment that is closed. We claim that both a and b are path components of x. That will show that x is not connected, x is not path connected. So, this is the final thing we have improved. That a and b are both path components. They are path connected because one is a closed interval, homeomorphic closed interval. Other one is the image of of a continuous function defined on an interval, open one side open as a cyclone. No matter, it is an interval. So, what we have to show is that take any point on the b part and any point on the a part. Two points to take show that there is no path from one to the other. That will show that these two are what path components. See, if you have taken a, what you have to do bigger than a, you have to take a point of b. So, you must be able to join a point of b to a point of a. So, I say there is no such path. So, if I prove this, what the claim that a and b are path components follows. Therefore, x is not path connected. They are exactly two path components. Suppose you have a path. Path means now what? Some continuous function defined on a closed interval. I can change the closed interval to any 01. There is no problem by re-parameterization. So, you can assume that there is a path from 01 to x such that omega 0 is inside b and omega 1 is inside a. So, remember x is subspace of r, r2. Look at the projection map r2 to r, the x axis, x projection, first projection x, y going to x. Then, pi composite omega 0, this is what? This is the x coordinate of omega 0. Omega 0 is inside b. So, the x coordinate will be 0 because the whole of b is just 0 cross minus 1 plus 1. So, pi of omega 0 is 0. Pi of omega 1 will be something on the x axis between open interval 0 to 1. So, it is positive. Of course, it is less than 1, less than or equal to 1, but it is not equal to 1. It is some positive number. By intermediate value theorem, see this map is from where? From x to r. So, the codomain is r. So, intermediate value theorem for this map 01 to r, not from x to r, but I have taken 01 to x and then composite with pi. So, 01 to r, it follows that for every k inside n, there exists a tk belong to 0 comma 1 by k. Why I want to take this one? Because I want to make this tk convert to 0 as k tends to infinity. That is why I have made this one. But these points are where there is some point here. So, what is the statement? Statement is look at omega 0, that pi composite omega 0 and pi composite omega 1 by k. Between these two, there must be some point, take that point conveniently, tk will go to that point, that is the intermediate theorem. So, such that pi of omega tk is equal to this point, 2 divided by sum 2 nk plus 1. All that I want to say is look at omega 0 is 0 and omega 1 by k instead of omega 1 here, sorry pi of omega 1 by k is some positive number. Between them, there will be a point like this. So, that is what I want to say. So, it can be that both of them are 0 and this I have to choose is that is not possible, that is the point. So, 2 divided by 2 n plus k, this is some integer, this is r integer is what I want, 2 n plus 1, 2 divided by 2 n plus 1 for some nk and such n. Once you have chosen this way, what happens is omega tk will be what? It is x coordinate is this point. So, I have to write the x coordinate here. Once x coordinate is written, this omega tk is some point inside the graph. So, it will look like this. See, what I am trying to do is if the point is already on the x y axis, I throw them out. After all, the curve starts from omega 0 and omega 1. So, between them, you have lots of parts of this path a, this is the graph of a. So, it must be somewhere in this graph, that is what I want. In other words, all these points, see this point is omega 1, right, pi kappa 3 omega 1 or something here does not matter.  This must be omega of something. This must be omega of something because they are all, once I have chosen this one, the point must be between these two. Point must be between these two like this. I keep choosing them closer and closer, that is the whole idea. So, once you have chosen that way, this will be sign of pi divided by this number, pi divided by this number 2 nk plus 1 by 2. So, what is sign of this one? It will be plus minus 1 and x coordinate is this one. So, that is precisely what I have been pointing out. x coordinate will be somewhere here or here or between these two points and y coordinate will be precisely that. x coordinate will be somewhere here, the y coordinate will be that. This, that, this, that does not matter, all of them may be this point. But I will always choose something after that here, one here which I can always choose them alternately. That is why I have taken it as odd numbers. All these things must be there. All that happens is by continuity, if you take the limit of this one, it must converge to some point in B. What is that point? Actually, it must be omega 0, this 0 it must be, right? Omega 0 is 0, 0, 0. Omega 0 is 0, it must come to that one because this goes to 0. But this point is plus minus 1, plus minus 1. This is not a convergent sequence. So, this contradicts the continuity of omega. So, it proves that there is no path, all right? So, this proves that this theorem 3.29 namely the closure of a path connected space is connected, that is not true. Closure of a connected space is connected. If you want to put path connectivity, it is not true. It is not valid for path connected space, okay? Let us show that this same example gives you a counter example for this also for path connectivity, okay? So, it does not take much time. So, we show that one also. Now put x equal to the topology sine curve and y equal to the closure interval is 0, 1. And f from x to y, the surjection projection map to the x coordinate, which is a quotient map which we know because it is surjective open mapping to begin with, restricted to x also it is surjective, all that I have to say that is all, okay? We know that x is not path connected, right? But y is path connected, y is just 0, 1, okay? What happens to the fibres? The fibres of f0 take any point between f inverse of 0 is nothing but sorry, f inverse of 0 is nothing but the b, the component b, 0 cross minus 1 plus 1. So, this is path connected. What is f inverse of any other point x where x is not equal to 0? Look at this picture. There the only part is the sine curve, which is a graph. The graph means for each point is only one point here. Each point is only one point. What is it? x comma sine pi by x. Therefore, the inverse image under the projection is precisely one point here. The fibres of each point between 0 open to 1, there are only one points. So, they are also path connected. So, the hypothesis is satisfied. Everything is satisfied yet the conclusion is wrong, okay? Conclusion says that this is path connected but that is wrong. Therefore, the theorem is not valid for path connectivity, okay? Okay, so this example, this example, you know, has a very peculiar property. You take a small neighborhood of this point on the y axis here, okay? A small neighborhood. Then look at the intersection of this entire thing there. It will have lots of segments of this curve, infinitely many of them. They are all disjoint with each other. Every point on the y axis, every neighborhood, no matter how small the neighborhood is, has this property. It is union of all segments. Disjoint, first segment will be on the x axis, that is a straight line. But otherwise, all other segments are part of this sine curve, right? So, this phenomenon we want to study and make it into a property namely locally disconnectivity or locally connectivity. That will be the next topic. We will do it next time. Thank you.


Ever heard of the term toxic positivity? Sounds oxymoronic. Yet it's there, it happens, and no, you're not being ungrateful. These seemingly harmless phrases probably come from a place of good intention, but no matter how someone intends it, they still tend to come off as indifference or condescension. Not only that, but they're also usually said when you're already feeling kind of crappy. This is not good. So whether you're worried about being the one saying it, or you're wondering why you feel bad about someone else saying these to you, let's look at these seemingly harmless yet toxic phrases. Number one, it's not the worst thing that can happen to you. Wow, that's dumb. This phrase carries the implication that you're ungrateful or weak. It indirectly adds shame and guilt to the person who is already suffering. Pain is subjective and relative to each person, so comparing it or trying to top it is not helpful in the least. Even though what someone is going through could seem minor to you, that same issue could be a colossal obstacle for someone else. Number two, I told you so. We all know it can be tempting, but we also know fully how painful it is to deal with someone who's gloating over us with I told you so. Especially when we've gone through the painful step of admitting we're wrong and are now going through the equally painful task of writing that wrong. No matter how much insistence there is that the phrase is being said to educate for the future or to show you care, it only sends a message that you can't be trusted or confided in, as a mistake means humiliation and condescension. Number three, it's nothing, don't worry about it. Similar to the first phrase, saying something is nothing is not helpful. We understand that it's meant to give a wider perspective, as in in the grand scheme of things, what has happened is not going to derail everything, so you're okay and can move past this. However, since it's such a short and abrupt phrase, what comes through is what you do and what you feel are unimportant. You don't matter. Ouch. Number four, you're strong, you can handle it. Well, it depends on the context, but if this phrase is given when a person has expressed a vulnerability or a pain point like grieving, they're looking to you for help, support, maybe even a solution. They feel weak and making the decision to ask you was already difficult. Simply telling them the opposite, you're strong, you can handle it, is the equivalent of saying, I reject this weakness you're showing, I refuse your request for help, deal with it on your own. This rejection of them, when they've opened up, carries the message that being vulnerable is wrong and they must hide themselves. Number five, you're in trouble. This is toxic when it's said when the person already knows that they're in for some bad times, like after crashing their friend's car or missing an important deadline. In other words, thanks Captain Obvious, you're not helping. A phrase like this during a highly stressful event can also sound like you're taking sadistic glee and thermos fortune. If this trouble was told in confidence to you, meaning the person needs at least a sympathetic ear, this phrase sounds almost like a betrayal of trust and they may think twice about opening up to you in the future about anything. Number six, time heals everything.  It's right up there with this too shall pass or time heals all wounds. It sounds like something someone says so they don't have to deal with the person with the problems anymore. When someone is in the deepest part of grief or in the depths of intense emotion, this phrase sounds flaky and unreal. It also sounds like a lie. Time isn't Wolverine's healing factor. Time is what allows a person to adapt, process, and cool down from the peak intensity of emotions. That's it. Profound things will always leave a mark. Barring amnesia. Generally, we know we're just trying to help and we say these things because we feel the need to contribute and support somehow. Caring is good and necessary. To make those efforts more effective, we need to be aware of how our words come across. Our vocabulary can be improved by being rooted in realism and the now, instead of an idealized future. When we better utilize our words to comfort others, we also do so for ourselves. Thanks for watching and good on you for wanting to be better for others. Did any of these phrases tweak something in you? Feel free to add, share, and discuss in the comments. See you soon.


Ottawa and Edmonton all tied up at four, though Edmonton winning the shots. Good job. The puck drops in the third period and this breakdown is brought to you by DraftKingsportsbook. Who wins the draw? Well, it's Norris. And that's the guy we're talking about in this whole little breakdown because what happens to him starts out pretty dangerous. Here he is. Eyes on number nine and stick to the helmet. That's a penalty, sir. I saw it. It wasn't cool. Ref's going to call it. He's going to skate away. He says, here, take your stick with you. Get out of here. Helmet just sitting on the ice. I'll say, uh-oh. Wow. Wow. Double rude. Rips it off his head, then hits it. So the stick, it goes under the visor and then he just kind of like does an extra pull to rip it off his head. Let's watch it again because everyone is super into it. And, you know, he's trying to lift his stick up, but Norris kind of moves his stick on his own, so then there's no resistance and it goes up and just under the visor.  Norris, meanwhile, has to get a new helmet because it's broken a little bit. He gives his helmet to the helmet doctor on the bench, takes number 16, tries to put it on and says, he's too small and then he just throws it away. Instead of giving it back to the guy, tries it on and says, uh, fix that one while you're at it too. So he gives a toothless laugh and now this guy, he's like, all right, well, guess I'll give you my helmet, Norris, if that's what you really want. And he says, yeah, that is what I want. Thanks. This feels great. We have the same size head. We should try to share shoes as well. Look how happy he is to be wearing his buddy's helmet. So silly. And they have the power play. So he's going to hang out on the power play for a while, wait till his moment, just chill, wait till his moment gets the puck, passes the puck, and then he sets up for the shot after making one more pass, bam, sets up, gets the puck back, not wearing his helmet. Uh, goal right there, far side, bam, in, easy, celebrates, does a little flying motion here on it. So that's, he's excited. He's not wearing his helmet. First time he scored a goal, not wearing his helmet. Does he give the helmet back to the friend? Kinda just kind of says, done with it, yours now. Doesn't even pass it to the right guy and it's on the ground and he's feeling good. Might wear that helmet all the time now, but he gets his back and they do end up winning the game. And that's probably the first time in his career, he scored a goal in the wrong helmet. If you bet on that, you probably made out well. Reminder, this breakdown was brought to you by the DraftKings Sportsbook, and I thank them for allowing me to cover such important moments in the great game.


Hello, guys. Good evening. Can you hear me? Yes. So today, guys, we are going to start a new chapter coordination program from organic chemistry, a very important topic. You're going to have one definite question from this. Okay. There are so many types, different different types of questions, forms from this chapter. This chapter requires a bit of information of chemical bonding, valence bond theory, hybridization, plus concepts of isomerism is also applicable here. We'll see how the coordination compound shows isomerism, geometrical and optical. All those things are important. In fact, on isomerism, optical isomerism, in fact, stereo isomerism, they ask questions every year in the exam. Okay. Is this fine whiteboard? Or do you want me to go for the blackboard? Blackboard. Black. So coordination compound. See, coordination compound. We also call this as, as complex compound, coordination compound or complex compound. It differs from the normal compound in many aspects, we discussed all those things one by one. But first of all, we'll see what is the use of this compound and why we are studying it. Obviously, they ask questions in the exam. But what is the use we have in data realignment? You must have heard about chlorophyll, right? Chlorophyll. Chlorophyll is the complex of complex of magnesium, right? It is a complex compound of magnesium. Okay. Hemoglobin, right? It is also complex compound. But it is a complex of iron. So with these two examples, you must have understood this, how this particular type of compound, coordination compound is important for our day-to-day life or for human being. That's one thing. Okay.  Why? We'll discuss this generally. Coordination compound, like the basic difference between this and the normal compound we have, that we'll try to understand. Let us take two examples here. One I am taking NaCl, which is a normal compound, simple compound. And it is a simple compound. And other one I'm taking that is K4FeCN6. It is a complex compound or coordination compound, complex compound or coordination compound. Okay. In NaCl, what happens, if you dissolve this in water, if you put this into water, it dissociates into, in H2O. It dissociates into its ions, that is Na plus aqueous and Cl minus aqueous. So it completely converts into its ions. Okay. Here, if you put this into water, and this also dissociates, it dissociates into 4K plus, plus FeCN6, 4 minus. So for this one, this part, the complex part here, the, you know, the portion here in the square bracket, this part, we call it as complex part. And this makes the entire difference, difference in the two examples we have. It is a complex part. So complex part, when you dissolve this complex compound into water, this complex part still maintains its identity here. It would dissociate into water. This, what we can say, this retains its identity in the solution. But this won't happen in case of simple compound. It completely converts into this. There is no retention in the properties. This molecule converts into ions. But here, we have the retention. So this is the basic difference between complex compound and the simple compound. Okay. That the complex compound retains its identity in the solution itself. Okay. This part won't dissociate. This is one major difference we have. Another difference is what? In complex compound, we have two types of valencies. Okay. Primary and secondary valencies. There is also a different, you know, kind of concept we have here. All these things, you know, we'll discuss one by one. Okay. I'm just trying to compare the two, you know, compounds here, complex and simple compound, so that you can understand what is the, I don't know, difference we have here in these compounds. The basic understanding you must have. Right. So we'll discuss all these things in detail.  Okay. Overlapping of orbital hybridization. Everything's are there. Okay. And this thing that I have said, it is, you know, it is very, very important. As important as the isomerism in coordination compound. Okay. So one by one, we'll discuss all this. So before going into this, that what is a complex compound, what are terms associated into it? Basically, they are different, different terms to use for coordination compound. But before going into those terms, let us first understand the classification of compound and where this complex compounds are placed in the classification of compounds. Okay. So next page you see heading is classification of compounds. Compounds mainly are classified into two categories. First of all, one is additional compound, additional compound also known as molecular compound. And then other one is simple compound. Simple compounds are simple molecules. We have examples here. NaCl, NaOH, MgCl2, HCl, many things. Okay. HCl, etc. All these are simple compounds. Further, this additional or molecular compounds again classified into two categories. The first one is double salt. And the second one is coordination compound. Coordination compound. So this is our topic today, coordination compound. It comes under additional or molecular compound. Okay. So one by one, we'll see the definition of additional compound, then double salt, and then coordination compound. Okay. So all of you write down the first thing here, that is additional or additional or molecular compounds. Write down when two or more, more simple, stable compounds joins together, joins together their forms, their forms additional or molecular compounds. Example is we have a KCl.MgCl2.6H2, carnalite, FeSO4.NH4.2SO4.6H2. This compound we call it as Mohr's salt. The name of the compound is, and this compound is carnalite. These are additional compounds because KCl, MgCl2, H2O, these are the simple compounds. When they join together, they form additional compounds. Now we know this additional compound is further classified into two categories, double salt and coordination compound. Let us write down the definition of double salt first. This chapter is very conceptual, logical chapter. It's not like the usual organic chemistry where you have to mug up things randomly. It's not like that. I would say 80, 85% things are there to understand. If you understand the concept, you can do it easily. But yes, a little bit of information is required. So that comes under 10 to 15%. That few things, if you memorize, if you know that, you can do it easily. So what all things are there? That's not very difficult to memorize. It's very basic. When you read some books, if you go through ones, you will understand those things and it will be there in your mind. A few things you know already. We will discuss that and you will understand this. Okay. So double salts are what? These are the compounds. These are the compounds which are stable in solid state, in solid state. But when dissolved in water, but when dissolved in water, it breaks down into individual constituent ions, constituent ion. For example, you see, if you take the example of gun light, KCl.MgCl2.6H2. If you dissolve this in water, it converts into K plus plus Mg2 plus plus 3Cl minus plus H2O will be H2O only because the solvent is water itself. Right? So this is the meaning of double salt. Could you tell me the n factor of this compound, not required here for this particular chapter, but yes, randomly I am asking. What is the n factor for gun light? Yes. Yes, it's three. How do we find out n factor for a salt? You can count total positive or negative charges. Here the positive charges is three and negative is also three and hence it is the answer. Yes, correct. That is the couple. Now the important one is we have coordination compound. Have you done this chapter in school? Yes, sir. Okay. Write it down. These are the compounds in which some of the constituents, some of the constituents, ions or molecules, some of the constituents, ions or molecules loses their identity, loses their identity. When dissolved in aqua solution, do not break up completely into ions. These are the compounds in which some of the constituents, ions or molecules loses their identity when dissolved in aqua solution, but do not break up completely into its individual constituents, ions. Okay, complete breakup is not there. Like example, I have given you K4 FeCN6. FeCN6 does not dissociate into its ions. Potassium loses its identity completely, but not the complex part, that is FeCN6 4 miles. So here you see some more examples. The first example I have already given you, that is K4 FeCN6 and this convert into 4 K plus plus FeCN6 4 minus. Okay. This one does not lose its identity. Whatever it is written in the square bracket, that is the complex part we have written in the square bracket. If you talk about this CO NH3 6 and CL3, this converts into CO, CO is cobalt here. CO NH3 6 3 plus plus 3 Cl minus. This compound, the complex part is the negative ion is the complex part. So we call it as complex anion, complex anion because anion is complex here. This one is complex cation, this particular complex cation. Similarly, we can have complex anion and complex cation, both possible because both part cation and anion could be complex also. Okay. Now, before going into this, let us try and understand what are the different different terms we have that we'll use in this chapter. The terms involved, we'll take one example and with respect to that already we'll discuss. Suppose the simplest one I'm taking K4 FeCN6 and this is present in the square bracket. So first of all, this complex part is written in the square bracket. This we call it as coordination sphere. Coordination sphere. Okay. This one is the central metal atom, central metal atom. That is, in short, we write it as CMA, central metal atom. We use this abbreviation for this central metal atom. Apart from central metal atom, whatever is written in the square bracket, we call it as ligands. What are ligands? K plus is what? Apart from this square bracket, this coordination sphere. The other part we call it as ionization sphere. Or we also call it as counter ion. Ionization sphere or counter ion, both are same thing. Okay. So this one is complex anion because a negative part is complex. Ligands are what? Ligands are electron pair donors. Electron pair donors. It donates electron pair, not electron, but electron pair to the central metal atom. That is ion here in this case. So these are the terms we are going to use in this chapter. Ligands coordination sphere, central metal atoms, in short, CMA and other things. Terminology. So first we'll see what are ligands? Next page. Ligands, it's classification. Ligands and it's classification. So write down. Ligands are atoms, ions or molecules which donates which donates electron pair to the central metal atom or ion, central metal atom or ion. It gives a pair of electron, right? Electron pair it is. This one is electron pair. One second. Just one second. So ligands are electron pair donors. Hence the bond between metal atoms, ligands, if you see, it is a coordinate bond, right? So metal atoms are attached, are attached with a coordinate bond with ligands. Okay. Since ligands are electron pair donors and metals are electron pair acceptors, hence we can say that ligands are, ligands behaves as the Lewis space since it's dorates electron pair and metal atom accepts electron pair. Hence metal atoms are Lewis acid, behaving as Lewis acid in the complex. Okay. Two, three things about ligands you should know. First of all, it behaves as Lewis space electron pair donors, right? Next thing it can be neutral, positively charged, negatively charged. Any three things possible negatively charged, positively charged or negatively charged. Okay. They have different density. What is density? How many pair of electrons a ligand can donate? That becomes the density of that ligand. In detail we'll discuss this. Okay. Then it attached with or in short I'll write down coordinate bond, coordinate bond or central metal atom CMA, central metal atom CMA. If you talk about the classification of ligand, we have two types of classification, types of classification. Another thing that ligands donates electron to the metal. So metal must have the capacity to accept electron and for that we need vacant orbital in the in the metal atom, right? And that is the reason mostly transition metal because transition metal has vacant the orbital. Hence mostly transition metals can easily forms complex compound. Okay. Can easily forms complex compound. Ligands are what? Ligands are electron pair donors. All those molecules which has electrons, electron pair, available electron pair, like lone pair or pi electron cloud, like in benzene, all those molecules can behave as the ligand. Okay. So must have electron pair present on it. Okay. So this is Louis space we have done. One point here you write down one note that central metal atom CMA, central metal atom CMA accepts a pair of electron from the ligand, accepts a pair of electron from the ligand. Hence behaves as Louis acid, central metal atom behaves as Louis acid accepts electron pair and hence behaves as Louis acid. Finish. Now the classification of ligand, we have two types of classification. One is classification based on charge. All these things I'm discussing here for ligands. Right. So classification based on charge, based on three types of, you know, obviously ligands we have. One is neutral. Other one is positively charged and other one is negatively charged. So neutral, then negatively charged ligand. Third one is positively charged. Okay. Neutral ligands, you should know the examples. Okay. It's important. I'll tell you why and how. Example. Neutral ligands, we have many examples. We have H2O, there's no charge, and it's, we have lone pair or nitrogen, no charge. We have NO, nitrosyl group, no charge, benzene we can take, pi electron cloud, no charge. Okay. And we have carbon monoxide also. All these are neutral. Yes. That also you can write ethylene diamine. That is also a neutral this thing. Neutral ligand, ethylene diamine. Okay. We'll discuss those examples also. Negatively charged ligands are, we have halide ion, nitrite ion, sulfate ion. Okay. Cyanide ion, hydroxide ion, positively charged ligands are very few, right. NH2, NH3, NH2, NH3. We have positive charge. Donor atom is this nitrogen, electron. And we also have nitrosonium ion NO plus. Nitrosonium, NiUm ion. Nitrosonium, metronium. Yes. That's also fine. Why charge are important here? Because, because you know when you calculate the oxidation state of metal, so for that you need charge on the ligand. Okay. We'll have the nomenclature of coordination components also. Right. The rules are very similar. Then you know with the nomenclature we have done very similar rules. But to find out the oxidation state of metal, because that is to mention in the name of the Rangam. Right. So to mention the, you know, oxidation state, we need to find out it. And for that we require the charge on ligand. And in the question, the charge on ligand is not mentioned. That you should know. Okay. That's why this charge is important. Very important. Okay. Now, the next classification of this thing of ligands are based on, no, it's dendicity. Write down next classification based on dendicity. Classification based on dendicity. So first of all, we need to understand what is dendicity of the ligand. Write down the definition of dendicity. Tell me, what is dendicity? Anybody? The number of atoms which can donate electron pair. The number of atoms which can donate electron pair than that ligand. Okay. Number of atoms which can donate electron pair that is dendicity. Okay. Anything else? Number of lone pairs that can be donated to the CMA. Number of atoms attached to the central metal atom. Yeah, that's correct. Number of atoms attached to the central metal atom is correct. Yes, right. Sana. Number of electron pair that is donated. That is also correct. Siddharth, when you say number of lone pairs that can be donated to CMA, that is not a correct definition. There's a thin line, the difference between the two. You will understand this. Let me discuss this. You'll get it. Okay. Yes. Number of ligands binded to the central atom. Number of ligands when they're to certain extent it's correct. Basically, if you say the number of atoms of ligand that is attached to the central metal atom is the dendicity of that ligand. Okay. If you say it is a number of electron pairs that can be donated by a ligand to the central metal atom, that is not correct because the dendicity can be anything 4, 5, 6, but it is not necessary that a ligand with dendicity 6 will show 6, so 6 coordination number in the complex always. Dendicity, the maximum capacity is 6, but it can show 5 also or 4 also. That's why we always take the number of atoms of ligand which is attached to the central metal atom in the complex. Okay. A thin difference, like I said, a thin line we have within the two, you'll get it like once I discuss it. So first of all, you write down what is dendicity, definition. It is a number of electron pairs, it is a number of electron pairs accepted by, number of electron pairs accepted by a central metal atom or iron, number of electron pairs accepted by a central metal atom or iron from a particular ligand, a particular ligand known as the dendicity of that ligand in that complex. Again, I am repeating this, ligand, you can donate 4, 5, 6, any number of, any pair of electron, but it is not necessary that a tridentate ligand or pentadentate ligand will always have dendicity 5 or 3 in the complex. It can be less than that also, depends upon the structure and oxidation state of the metal. Okay. So dendicity is the number of electron pair accepted by a central metal atom from which a particular ligand is, from a particular ligand is known as the dendicity of that ligand. Or in the other way, you can say in the bonding state, you just need to check the number of atoms of ligand that is attached to the central metal atom, that is the dendicity. Fine, anyways. So dendicity, based on dendicity, the ligands are classified into six categories, okay, into six categories. And you should know here the examples of each category plus the structure. Because a structure is required, sometimes they'll ask questions on the donor atoms, which one is the donor atom in a given ligand, right? For that, you should know the structures. The first classification, like I said, six classifications there. The first one is monodendric ligands. Monodendric ligands are those ligands whose dendicity is one, right? Dendicity is one. The example we have, okay, the example of monodendric ligand, just write down dendicity one and then write down example. We have H2O, oxysyl is the donor atom, NH3, nitrogen is the donor atom, Cn minus, Cn minus, so monodendric ligand. Then we have X minus monodendric ligand, CONO2 minus OH minus NNO. See, NO here, it's important, okay, we'll discuss this later. I'm just mentioning it here. NO is, as we say, it is very, you know, the property of this, whether it is, obviously it is monodendric, but whether it is, you know, neutral or positively charged, it depends upon the complex, okay? So it behaves in a different way in the, you know, complex. Important also we'll discuss this later because when you see the complex, you see here, if I write down this example, K4, Fp, Cn6, here the charge on cyanide is minus one. Once you know this, Cn minus is the charge, right? But in the complex, the charge is not mentioned, right? So we have NO is a ligand, we have NO plus is the another ligand, right? But for both, it is NO only that is mentioned in the, in the question, in the complex. So whether this NO is neutral or positively charged, that is a bit, you know, confusing to understand in the question. Based on this, they have asked question in JEE, okay? So we'll discuss this. So please take care of this NO thing, nitrosil, whenever it comes into the paper, whether it is neutral or positive, positively charged, you have to take care of that, okay? We'll discuss this when they, you know, go a bit ahead in this particular chapter, but I've just given you this information that NO, you have to be a bit careful in case of NO. As far as the charge on the ligand is concerned. Okay. So second type of classification we have, mono-dentate, then we have bidentate. By or di-dentate both are simply. By or di-dentate whose dendicity is 2. By or di-dentate whose dendicity is 2. The example we have, ethylene-diamine, ethylene-diamine, okay? In short, we write it as EN and the structure is CH2, CH2, N, N, H, and this is the donor atom, 2-nitrogen. Nitrogen are the donor atom. Dendicity is 2. By dendate or di-dentate? Done. Next. Next one is oxalate. Oxalate is C2O4, 2 minus, and its structure is CO minus, then we have double bond O this side, double bond O this side. Both oxygens are the donor atoms. Oxalate, O, X is the symbol here. Next one is glycinate. These are the donor atoms, nitrogen and oxygens. Glycinate, GLY is the symbol. The other one is carbonate. You all know the structure of carbonate ion. CO3, 2 minus, C double bond O, O, O negative, negative, and this is the donor atom. Popping this down. The next example is dimethylglyoxine. The structure is C double bond N, O minus, C double bond N, OH, CH3, and CH3. The donor atoms are this 2-nitrogen atom. Another one is dipyridine. Dipyridine is this. Present in the ring, the nitrogen. Both nitrogen atoms are the donor atoms. Oxygen can also donate, but since it is more electronegative, so nitrogen donor atom. COH also you can write here. Glyoxine. Yes sir. Yeah, dipyridine is this. Next one is tridentically. Post-tenticity is 3. The examples are diethylene, triamine, diethylene, triamine. Easily you can draw this structure. See CH2 whole twice is diethylene. Triamine is this. 1 NH2 here, 1 NH2 here. Diethylene and triamine. One hydrogen we have with this nitrogen. This is donor atom. This is donor atom and this is donor atom. The three donor atoms here. And the symbol for this one is diene. Next is terpyridine. Terpyridine. T-E-R-P-Y. And the structure here is copy this down. I'll go to the next page. Next slide down. Tetradentate, then we have penta, and then we have hexa. Three more. Tentradentate ligand. Density is 4. Okay. And the first one is nitriloacidate ion. For example, we have n, the structure is n. We have CH2, C double bond O, O minus. Same thing, CH2, C O, O minus. Next one, we have CH2, C double bond O, O minus. And all these oxygen atoms are donor atoms. Okay. Fourth atom is this nitrogen. Four donor atoms we have here. Another example is triethylene. Triethylene tetraamine. Triethylene tetraamine is this. Three ethylene groups we have. One is this. Tetraamine is this one. N, H. Here also we have NH. And this is attached to CH2 whole twice. Here also we have CH2 whole twice. One, two, three ethylene group, two amine group, and two more we have here, NH2, NH2. So triethylene tetraamine. Donor atoms are the four nitrogen atoms. One, two, three, and four nitrogen atoms are the donor atoms. Tetra-dentate we are done. Next one is pentadentate. Penta-dentate ligand. Penta-dentate ligand. Density is five. Example we have EDTA 3 minus. What is this? Ethylene diamine, dry acetate, dry acetate high. So ethylene diamine, right? So ethylene is this CH2 whole twice. Diamine is two nitrogen on both side of it. But dry acetate we have. Okay. So diamine dry acetate means one side we have CH2, CO minus double bond O. Here also we have CH2, CO minus double bond O. Here also we have CH2, CO minus double bond O, and one hydrogen. So donor atoms here you see. The donor atoms are one, then we have two, then we have three, and then we have five, one, two, three, four, five. Penta-dentate ligand. Two nitrogen and three oxygen atoms are the donor atoms. Exat-dentate ligand, the last one. Exat-dentate ligand. One second. Density is six. And the example is EDTA, EDTA 4 minus. Ethylene diamine, tetra acetate high. Could you draw the structure of it? Ethylene diamine, tetra acetate high. What is the structure? You can remove that fourth hydrogen or nitrogen atom and add one acetate high in here. Means this hydrogen you remove and add CH2, CO minus oxygen is the donor atom. Yeah, that is the structure of the galaxy. The structure I have given because donor atoms you must know. Sir, does it matter where the donor atoms are placed in the molecule? Sorry, what? Sir, does it matter where the donor atoms are placed in the molecule? Like if they're far from each other or something? No, distance and all there is no effect. Okay, sir. Distance, because it is arranged in the space, three-dimensional space and because of that only it can show a stereo isomerism which we'll discuss later. Okay, sir. Right, so when it donates electron to the metal atom then it arrange itself in such a way so that that donation and then bonding is possible. Yes, sir. Yeah, so these are the few things. Now the thing is why these coordination compounds are forming and why metals are accepting electron from this thing, what do you say, ligands? Okay, why metals are, like in normal compounds we say that metal and like atoms form bond to gain octane, to gain stability. This kind of theory is there. So here also metals accept electron from the ligands in order to gain the next higher noble gas configuration. Okay, suppose you have an example of, same, K46 atomic number of ion is 26. So what is the next higher noble gas configuration? What is atomic number of that? 26. 26 and the next higher noble gas is 36 that is scripted. Correct? So metal that takes electron, suppose ion is the metal, Fp has 26 electron and it takes electron from the ligand and it has tendency to make it 36, the total number of electrons, the net number of electrons. That is the tendency of the metal atom. This kind of behavior we observe in some compounds but it is not like it is always there. It is something like the bonding theory that we discussed in chemical bonding. There are different theories. We have valence bond theory, we have VSEPR theory for instructors and all and then we have molecular orbital theory, hybridization is another kind of theory of bonding. No, so there are different, different compounds which follow different kinds of bonding theory. So here also we have different, different theories. We have valence bond theory we have here also. We have Wernher's coordination theory we have here also. Apart from that, we have this rule also that all them like there are metals which has tendency to accept electron from the and gains the next higher noble gas organization. But it is not true for all the metal atoms. So this rule we call it as Siegwick EAN rule, effective atomic number rule. So what is this? So first of all write down the heading effective atomic number, effective atomic number in short we write EAN. We also call it as Siegwick, the name of the scientist is this EAN rule, Siegwick EAN rule. This rule is no, it is observed to be true in case of, true in case of metal carbonyl compound. Means metal and carbonyl compound is there, CO group is there. Then the metal in the rule is correct mostly and you know in some book it is written also Siegwick EAN rule is valid only when the metal has carbonyl you know ligands, CO is the ligand over there. So what is the effective atomic number? The effective atomic number for any metal in the bonding state if you try to find out then what you'll do. It's a very basic thing you can understand easily that effective atomic number means net number of electrons you're trying to find out after the bonding. That would be equals to atomic number which was initially there right atomic sorry this atomic number and in that way it it loses some electrons also. So we'll write oxidation state minus oxidation state with the charge with the sine. Means if it is plus three then we'll write plus three here minus three we'll write minus three oxidation state with the sine plus plus the number of electrons just a second okay. So it is the number of electrons gained from ligands. So basically if you think on this particular equation plus and minus it is the total number of electrons we are trying to find out on a metal in the bonding state okay. So for that what all things you need to know that you see here. Suppose the first example I am taking K4 FE CN6. We need to find out the effective atomic number for this complex that is the question effective atomic number for this complex okay. So we'll find out. Excuse me sir what is that on the right sir number of electrons gained from the ligands. I should write here ligand okay. So K4 FE CN6. So if you try to find out the effective atomic number for this thing iron here what we'll do with this formula or if you do not know this formula effective atomic number means that total number of electron the metal has in this bonding state. So what we'll do we'll write down the atomic number first of all 26 because iron has 26 electrons and if you try to find out the oxidation state what is the oxidation state of iron here could you tell me. Plus two plus two right means it has it has lost plus two electron right. So minus plus two oxidation state plus how many electrons it accepts from the ligand. There are six ligands and one ligand can give two electrons so six into two the number of electrons and when you solve this you'll get 26 plus 12 that is 38 38 minus 2 is 36 got it. 36 is the atomic number of krypton so next higher noble gas configuration. When you calculate this for CR CO6 CR CO6 calculated effective atomic number here and the next one is FE CO5 FE CO5 CO in here is the carbonyl ligand okay don't get confused with cobalt CO5 if CO is not the you know central metal atom it is ligand right if it is not ligand then it is cobalt calculate the effective atomic number for this too what are you getting EAN value here and EAN value here 38 you're getting in both I don't think so 36 in both okay the first thing you need to find out yes the first thing you need to find out is the oxidation state of the metal all these are neutral ligands zero is the charge right zero is the charge it has 26 plus 10 36 24 plus 32 36 so for this one the value is 24 minus 0 plus 6 into 2 and this is equals to 36 for this one is 26 minus 0 plus 5 into 2 and this becomes 36 electrons okay 36 electrons both are both complex has hence it follows syzvic EAN rule right but for some examples like this one CU NH3 whole 4 then SO4 we have here for this one you find out the effective atomic number tell me you are getting 35 right its effective atomic number you see NH3 is a neutral ligand sulfate is minus 2 so on this complex we have plus 2 charge and hence the oxidation state of copper is plus 2 is it right so the value of EAN yeah 35 is correct is copper is 29 oxidation state is plus 2 plus 4 into 2 and this becomes 4 into 2 and this becomes 30 they see the effective atomic number for this one is 35 it is not 36 that's why the syzvic EAN rule is not valid for all of complex okay but we have few application outfit which is very important we'll discuss that one more thing you try to understand we are we are looking out the oxidation state of metal and for that you should know what is the charge on the complex and ligands that is there because without that idea the charge on ligands or you know this thing um the charge on ligands you are not be able to find out the oxidation state of the metal you must have seen that in nomenclature of coordination compound also we have to mention the oxidation state of the metal just after the name of it that's why the charge on ligand you must know okay so this is one thing now you see this example what is the effective atomic number one question i'll show you here see this the question is the question is why mnco5 mnco5 the complex we have exist in exist in in its dimer form in its dimer form that is mn2 mn2co10 tell me this basically mnco5 does not exist mn2co10 exists mnco5 exists in its dimer form not because mnco5 will have 35 EAN so when it forms dimer that will form a bond between mn and mn so that will make it 36 mnco5 you said it has 35 EAN value right yes so what happens in the dimer form so when the bond is formed between two manganese atoms um EAN will increase by one because electron is shared so it becomes 36 which is clicked on EAN okay that's right no chillation is not there chillation is ring complex it forms right wait wait wait see if you find out for mnco5 you must have done this the effective atomic number for mnco5 and we are getting 35 here EAN value is 35 but if I ask you to find out the effective atomic number for mn2co10 what is this number what is this number mn2co10 for each manganese atom yes tell me did you calculate this Anjali sir you got it 36 yes sir how sir because 35 is in the mnco5 form and in the dimer form since there's a bond between the two manganese atoms it's going to become 36 you add one electron yeah correct correct so basically uh if you look at the rough structure if I draw here we have a manganese manganese bond right m and mn bond and five co is already there one two three four and five all these are co right co co co and co and co okay so this contains uh like you leave this electron here this one you let it be apart from this from this electron an entire thing has 35 plus one more electron we have here so 36 electron this manganese has and similarly we have 36 on this manganese also one two three four and five right co co co co and co okay so if you count the number of electrons manganese has 25 plus 10 35 plus one electron of this manganese in this one so both manganese atom has 36 electron which has noble gas configuration and hence mnco5 dimer is into mn2 co 10 which is the more stable form one note you write down here metal or metal iron metal or metal iron in a complex metal or metal iron in a complex tends to acquire metal or metal iron in a complex tends to acquire noble gas configuration noble gas configuration by accepting electron by accepting electron from the ligands however this rule this rule fails in many cases however this rule fails in many cases and and works best for metals in low oxidation state and works best for metals in low oxidation state low oxidation state and with metal carbonyl compound low oxidation state but low oxidization again they don't defile so it is basically works well for metal carbonyl compound okay now application of this rule is what that we can understand you know we can find out the coordination number first of all and we can also find out the oxidizing or reducing behavior of the complex we have application of it and they ask question on this also write down the heading next application of the first one is to find coordination number coordination number we haven't done now I'll tell you what is coordination number to find coordination number we haven't done few things okay I'll do it to find out the coordination number coordination number is what it is the number of uh you know the same thing like we did denticity coordination number is also the number of atoms of ligands that are directly attached to the um you know central metal atom so we can find out coordination number with the help of the EMEAN rule it's very simple I'll show you how suppose one uh you know a complex is given and the complex is this FE COX and they'll ask us to find out the coordination number means how many atoms of this ligand is attached to this eye so we'll find out EAN and we know according to this rule the effective atomic number should be equals to 36 that's what we need to do so EAN already has 26 electron and there's no charge on you know on the metal because it is zero it is neutral ligand plus suppose its coordination number is you know x I am assuming x into 2 is equals to 36 and you can solve this equation for x x value you'll get five here is the coordination number is five five ligands are required here carbon ligands FCO6 could you find out for Ni what is the coordination number for this one Ni COX is it four yeah x value is four we are getting when you solve this Ni is 28 okay coordination number is four this is the first application second one we can find out oxidizing and reducing behavior and reducing behavior let's see this example we have this mn CO6 and we have b vanadium CO6 if you find out EAN for this magnet is 25 minus zero plus 6 into 2 so we are getting 37 electrons more than 36 okay that's why what happens this mn CO6 has tendency to mn CO6 has tendency to lose electron and converts into mn CO5 mn CO6 plus what do you write mn CO6 plus plus one electron so this one has 37 electron so this one will have 36 electron and hence this is behaving as it is getting oxidized it is behaving as a reducing agent so really could you find out for this one b CO6 what behavior is this sir what is vanadium atomic number scandium titanium vanadium 21 22 23 23 is atomic number how many electrons this has EAN value 35 35 right so it has tendency to accept one electron right so this accepts one this is again this has tendency to accept one electron this plus one electron and this converts into b CO6 minus so it is getting reduced this is behaving as this is behaving as oxidizing agent so we can find out the oxidizing reducing property and the coordination number of the uh of the ligand so apart from this we have other few more different types of ligand we have seen the ligands based on their density that is more dented diented by dented triadentra pentarexa they have seen that few more we have you see this another type of ligand we got a flexi dented ligand flexi dented ligand right down the definition of this right down a poly dented ligand again one more thing poly dented ligands are those ligands whose density is two or more right so except mono dented ligand all ligands you can consider as poly dented ligand right except mono dented all ligands are considered as poly dented ligand correct so write down a poly dented ligand is found to have is found to have different density uh density hello am i audible yes yes yeah so what did you write a poly dented ligand is found to have different density in different compound and hence and hence they called flexi dented ligand their density is flexible depending upon the metal uh oxidation number plus structures and other things and what other ligands are present okay i'll show you some examples here this means what uh next slide right down it is not necessary it is not necessary that all donor atoms necessary that all donor atoms present in poly dented ligands all donor atoms present in poly dented ligands forms a bond with the central metal atoms forms a bond with central metal atoms or ions this means what if you talk about edta 4 minus what is the density for edta 4 minus 6 density is 6 right but it is not necessary that in all of the bonds its density is 6 only it can be 5 4 anything depending upon the molecule right means what if the density is 2 or more then it can change its density depending upon the requirement of the complex and hence it it is known as flexi dented ligand i'll show you this example you see since we have already understood norsedric eam rule so i want you to uh you know find out the density of density of carbonyl ligand that is c o in this complex c o n s 3 4 c o 3 this is second c o 3 carbonate uh ligand it is and we have b r outside of it another one example is c o n s 3 5 c o 3 carbonate and then we have b r tell me the density of carbonate in this compound and in this compound these two ligands and its density flexi dented is the properties of poly dented ligands all poly dented ligands can can show different different density in different different compounds so yes it is for all poly dented ligands first one you are getting two sana something doesn't depend upon the oxidation number of the that is what i'm asking you you have to calculate that oxidation number and then uh you know uh the other things yes okay i got the answer we'll we'll see we'll discuss it sana this is so you need to find out the oxidation number of the metal and then you see what is its requirement means how many electrons it requires to gain the next higher noble gas configuration accordingly you should you know think of that how many elect how many atoms are attached with the central metal what is zero there pratham what is the oxidation state of cobalt in both molecule tell me pratham tell me the oxidation state sir three plus three in both sir the first one is plus three and the second one second one is also same same as n s 3 is neutral only right that's why you that's why you see it is important to know the charge whether the ligand is neutral or any charge is present or not okay it's important so this we know this is carbonate iron so it is minus two charge on it this is zero and bromine is minus one so on the complex the entire complex we have plus one you assume this as x so when you find out the x value you'll get plus three yes or no and plus three this is the first thing so you should be very comfortable with it and how to find out the oxidation state and for that you should know the charge i have given you all those examples that is more than enough okay fine now you tell me what is the atomic number of cobalt 27 what 27 so cobalt is 27 and its electronic configuration is argon 4s 2 3d 7 so for co plus 3 the electronic configuration is argon why i'm doing this not required basically electronic one was fine let it be so basically we have 24 electron not don't not required to write down this configuration here we have 24 electrons right okay so this wants to have 36 electron the next higher configuration right so for 36 electron it needs 12 more electron right 12 more electrons means its coordination number is six here right it's going to have 12 electrons means there are six atoms which are donating electrons to the metal which is cobalt here out of six you see there are four NH3 molecules which are donating electrons to this cobalt we know NH3 is mono dented ligand so we can say all NH3 molecules are attached with cobalt out of six four coordination number has been fulfilled by NH3 now two are left so both the atoms both the donor atoms in co3 2 minus co3 2 minus carbonate ion is a bi-dented ligand but we have to see in the molecule in the complex how many atoms are donating electron according to that we can say that whether the density of carbonate ion is one or two right maximum capacity is two it is bi-dented ligand two pair of electron it can donate but what is the density in this compound we need to check okay so four electrons eight electrons are given by this four NH3 molecule it requires four more so all the two donor atoms in in carbonate ion here will donate electron to this cobalt hence here in this complex co3 2 minus is a bi-dented ligand it's behaving as a bi-dented ligand what sir why are we like according to EA and student fee we're taking at 27 as the atomic number why are we taking 24 okay does make a difference okay yes sir yes sir i'm taking 27 24 okay yes those examples of solution is zero yes sir understood yeah so here you see now here's the same thing it requires 12 electrons but out of 12 five NH3 is providing electron here and this cobalt needs only one pair so both carbon atom cannot both oxygen atom in carbonate ion cannot donate electron to this cobalt and cobalt does not require that much electron so cobalt will say okay only one oxygen atom i need only one pair so give me only one pair of electron okay so out of two only one oxygen donates electron to this cobalt hence here in this molecule this co3 is behaving as a monodentate ligand so it is bi-dented in the first one and monodentate in the second one so your answer is correct son okay one and two here you can think of so why not four ns3 donates and two donates from this side that is not possible because ns3 is monodentate ligand and if it is written here it means it is bonded with the cobalt so we have to consider all monodentate first then we'll move into the polydentate ligands so basically these two molecules if i draw a rough you know diagram of it the first one you see when we have four ns3 molecule we have cobalt here present so cobalt somewhere here and four ns3 like this one ns3 here one then two then three and then four four ns3 here and carbonate ion is this we have C double bond O minus O minus and both the carbonate ions are donating electron to the cobalt so this is the structure we have density is one two three four five six six coordination number density is two for this right and one for all the ammonia now what happens in the second one second one we have the same thing right but there are five ns3 so like i said all monodentate you have to consider first ns3 then we have ns3 here and then we have ns3 here and here we have one oxygen atom O minus carbon double bond O minus like this out of it capacity is two it can donate two pair but cobalt does not require two pair because already ns3 has given 10 electrons it requires only one only one oxygen donates so here it is monodentate ligand here it is did you understand this so what happens to the other oxygen it will be there as this so then what will the charge won't be neutralized sir sorry the charge won't be neutralized no sir no overall it will be neutral because there are electron it donates electron no overall it will be neutral because there will be some charge on cobalt also okay right overall if you calculate yeah so this is how the thing is right how the complex forms and you see i have given you the rough you know a rough idea of it how this is drawn if i go a bit ahead which we'll discuss a bit later you see this here if you look at this diagram this is structure this structures looks like a square bipyramidal yes or no yes yeah hence the structure of this molecule is square bipyramidal that's what i was talking about when it has to donate electron to the central metal atom it will arrange itself in the uh in the three dimensional space right and hence it these are capable of showing uh you know stereo isomerism if all the conditions are there correct so this is the coordination number six compound which is the most important one we'll discuss this in a stereo isomerism and it has octahed it is the octahedral complex okay you can see the structure four square corner and one on the top one on the bottom octahedral complex same thing we have here okay anyways so this is flexi dented ligand okay next one you write down chelating ligand second one is also square bipyramidal sir uh both are the octahedral square bipyramidal the same thing okay write down chelating chelating ligand see here i'll i'll just show you this structure chelation is a cyclic is a ring complex okay like you see you consider this one one two three it's it forms a ring right forms a ring so this kind of ligand which are capable of forming a ring like this right are called chelating ligand did you understand this so for chelation what kind of ligand be required poly dented or monodentated anybody poly dented poly dented right because if it is monodentated in case of phase three chelation is not possible ring structure won't form right that's why all poly dented ligands are also considered as chelating ligands okay because if they are capable of forming a ring cyclic ring that we call it as chelates the structure or the complex that that we get by the poly dented ligands in which a cyclic structure is there is called chelates chelates are generally more stable than the normal complex because of cyclic structure with only one exception that exception i'll show you first of all you write down the definition in this chelating ligand write down all poly dented ligands all poly dented ligands are chelating ligands are chelating ligands if on coordination if on coordination it results in the formation of if on coordination it results in the formation of a closed cyclic ring the formation of a closed cyclic ring next thing the complex forms in this way is called are called chelates the complexes forms in this way are called chelates and these chelates are more stable than the ordinary complex and these chelates are more stable than the ordinary complex right chelates are more stable than the ordinary complex with only one exception the exception is hydrogen if the ligands is hydrogen and h2 and h2 and both nitrogen donates electron to the metal it forms a three membered ring and hence it is lesser stable yes yes it's a second argument all poly dented ligands are chelating ligands if on coordination it results in the formation of a closed or cyclic ring the formation of a closed cyclic ring the complexes forms in this way are called chelates and these chelates are comparatively and these chelates are comparatively more stable than the ordinary complex but sir are there poly dented ligands that form a three-membered ring are stable sir no no three-membered ring is not stable i'm just giving one example right it's not like because of hydrogen it is not stable if three-membered rings are forming then it is not stable in the normal the previous example also is not stable this is not three-membered in the previous one this one four okay but yeah if three-membered ring is forming then it is not stable right i've just given you one example okay this is one exception now the next one is ambi dented ligand ambi dented ligand you must have you know i must have discussed this in case of cyanide the reaction of cyanide when we were doing in uh organic chemistry sir yeah so how is the ring more stable see ring structure is generally more stable like the overall the surface area decreases which enhances the stability of the molecule because of the ring structure we don't have any you know the theory for this we can say the surface area is less and then you know see actually what happens these are complex compounds complex in the term i'm using here in terms of you know it's a structure and orientation in this space correct so when the molecules atoms are arranged in this space right so they always have to you know they always wants to maintain the distance so that the overall repulsion has been minimized it happens on its own to minimize the repulsion only like they have atoms pair of electrons are there bond pair electrons are there so we have some random steric hindrance uh lone pair lone pair repulsion bond pair bond pair repulsion so to minimize all this the atoms molecules are arranged themselves in the space in that way right so when the structure is a ring structure it is forming so in that ring is structured the overall strain is very less right the distance is there is maintained overall strain is less we have a certain angle where the strain has been decreased a lot that's why we say that when cycling structure forms it has more so there are multiple things it's not like only one thing can talk about repulsion you can talk about you know bond angle and other things right so all these factors overall you know makes the structure ring structure more stable than the other on our open chain complex that we get right so in three member ring the angle strain is you know dominates on all other things the angle is strained between the two bond pair of like are so high and hence it is not the basic thing that we have discussed so many times in organic chemistry right so embedded ligands are what right down the ligands which has more than one donor atom or sites the ligands which has more than one donor atom or sites more than one donor atoms or sites available while forming complex while forming complex only one donor atoms only one donor atoms only one donor atom is attached to the metal attached to so could you repeat the last or your voice is breaking says attached to to metal I'll repeat again wait other ligands which has more than one donor atom or sites available but while forming complex only one donor atom is attached to the metal at a time it's not like all the donor atoms will attach so like for example you see cyanide once again guys yes so yes so we were talking about ambient it so cn minus you see the ligand is cn minus and when cn minus cn minus is this a carbon triple bond with nitrogen one lone pair on nitrogen carbon and a negative charge okay two donor sites carbon and nitrogen if carbon donates then it is known as cyanome if nitrogen donates it is isosynol right and cyanide isosynide forms isonates in the minor product right same thing we have if you take this one I'll go to the next page wait n double bond oh oh minus nitrogen just a second yeah and double bond oh oh minus correct so when nitrogen donates electron nitrogen and oxygen both has the capacity to do it because it has nitrito and because nitrogen is donating electron and double bond oh oh minus it is nitrito okay ambient we can talk about this scn minus this is a thiosynide this is scn minus when nitrogen donates it is isosynide basically the first two are the most important one okay one more I will show you in instead of sulfur we have oxygen ocn minus so it is a two more lone pair we have here here it also it is it is thiosynide it is cyanate oh oh because oxygen is donating if it is ocn nitrogen is donating it's lone pair then it is cyanate oh and so all these are ambient dendritic gas more than one donor site so why can't these form polydentate ligands so why can't these form see actually carbon see carbon nitrogen has triple bond right you you will get it on your own for for polydentate ligand you the bond must be like this no there must be some angle between the bond then only it can rotate and right but since it is triple bond here sp so 180 bond angle must be maintained right yes yes that's why it is not possible little bit you think about it you'll get it yes okay so that's the thing so no uh these are ambient dendritic okay now the next thing is okay next you write down coordination number coordination number coordination number right now it is a number of atoms of ligand it is a number of atoms of ligand that are directly bonded to number of atoms of ligand that are directly bonded to the central metal atom or iron by coordinate bond okay i'm repeating number of atoms of ligand that are directly bonded to the central metal atom or iron by coordination bond is known as the coordination number of that metal atom. Coordination number is for metal atom. Coordination number of that metal atom. With example, you will understand this. Suppose if I take CO and it's 34, 2 plus, right? So we have four nitrogen atom bonded with it. Coordination number is 4 here. If you look at this example, if I take CO, EN3 and 3 plus. What is the coordination number here? 6. It is 6 because EN is bidentated ligand and there are three ligands present. So it is 6. So if you write down the formula for coordination number, it is, you know, the number of ligands into its density. But you have to again cross check that how many atoms are bonded with. Because the one that we have done, the bidentated and moradentated ligand, the carbonate, for example, if you use this form, you'll get it wrong. So that, those kind of understanding you must have while you are doing this kind of questions. So this is the, in general, we use this formula. This formula we use. If you look at the structure here, this one, how it is bonded, we have a cobalt and we have EN, right? So we have two atoms present here. I'll just write down, you know, this and this and this. Two atoms will be like this, which is donating electron to the middle. This one ligand is this. Another ligand is this. And another ligand is this. Suppose these are the donor atoms here. It looks like fan, right? But actually it is, you know, it is the, at the corner of the square bipyramidal, just like that. Like this it is. Don't draw it, just I'm just telling you. It is at, on the top, on the top bonded like this and this. So if you, if you try to imagine this, imagine a square, right? Imagine a square, flat, not vertical, flat square you imagine. In the center we have metal. And just to draw a line like this, straight line like this. Here at the top, we have one atom and, and this corner, it is bonded like this. You have to imagine this a bit. Then you can understand. And this kind of imagination is required when we do studio isomerism over here, right? Center, you draw a line, flat square, right? Horizontal square, you draw a line, vertical line. Here you have one donor atom of the ligand. And another donor atom is at the corner of this square of the ligand. The line here, this, this point, this one donor atom of one ligand and another atom of the same ligand is here. And this is the ligand connected with each other. And these are donor atoms placed like this. Okay. This is CO, EDN3. We'll discuss this also later on when we do stereo chemistry in coordination compound. Mostly that part is the most important part we have here. Next, one more thing here is coordination is fear, coordination is fear. Write down the definition. This we also call it as coordination entity. Wait a second, let me just close the door. Okay, write down the central metal atoms or ions, the central metal atoms or ions or ions and the ligands, central metal atoms or ions and ligands are written within a square bracket, written within a square bracket, which is called coordination is fear or entity. It is a single unit. It is a single unit won't dissociate in the form of ions won't dissociate in the form of ions won't dissociate in the form of ions when dissolved in aqueous solution and retains its identity retains its identity. So guys, we'll take a break now. This is done with it. We'll start with nomenclature in coordination compound after this. Okay, so we'll resume the session at 645. Nomenclature, the basic rules are same. You just need to focus a little bit on the nomenclature of ligands. How do we write down the name of ligands when it is neutral, acidic or neutral, negatively charged or positively charged. Okay, in that way, how do we write down the name? But the basic rules are same here. Okay, so we'll resume at 645. Take a break.


Manure City factory footage this week. We're gonna do something a little different. Yeah So on Friday of last week it was Manure City Like how we're in it, but we're not in it. Yeah, it was Manure City week and Adafruit is I think one of the only electronic manufacturers in Manhattan. I think so. I think so and then it was also Manufacturing day. Yeah, so all in the same day. So we said, okay fine. We'll do a factory tour So this is a slight speed up. This isn't exactly Because I wanted to be around a little bit over a minute. So this is where we ship stuff One of my desk I have a desk on each floor here at Adafruit but this is where we do a lot of the shipping and storing of things all of this is Velocity storage. So the things that we ship the most of are closest to the shipping stations and you could see a whole bunch of Adafruit products around here Things are a little different since we've had COVID Happening in our lives. So obviously see people wearing masks. We changed some of the layout and some different things And we're also continually trying to maximize our space But I wanted to do a little video. So I asked, you know, who works here in manufacturing to run down and make a quick video because it's really hard to do multi-floor live broadcast and so this is the other Part of the tour that we did and then this side. This is the other side of One of our floors. This is where we have a whole bunch of storage. Yay boxes of stuff stuff and stuff and stuff All labeled and counted and and accounted for and tracked and yeah This is where all the back when we were an apartment company our storage was in the living room and Shipping was a closet.  Now, you know, we have a lot so it definitely has that like, you know, Costco feel a little bit and You can see more of that area then We also did a little bit of filming Of the other floor and this is where we're at right now. So this is where we do a lot of manufacturing Into the pick-and-place machines here, so two pick-and-place Samsung machines inspection space station oven testers for the upcoming products nitrogen generators The rework and testing benches a lot of these are we've actually spread out a lot of people. There's me. Hi. Hello And then there's Phil over there. Hi, and then we've got the hack sign and Let's see what else we got all the real storage and component storage Old oven that we keep around in case we ever need it and then a The TARDIS TARDIS. Well, we kind of saw that a little bit Okay, so that's samples and then that's where we film. So this is where we film right here. This is where you see us Hello, yeah, yeah, that's where we're sitting right now. That's where we're sitting now. Huh, that's cool I didn't realize it ended like that. Okay, and then I'm going to show the rest of the um Regular old factory footage that we do each week. So here we go And it wouldn't be a fruit factory footage unless uh You could see the Disney marvel, whatever Complex, I think I figured it out We are the watcher I guess we're watching what if this building is never completed. I won't interfere. I don't know man. Except for once There's going pretty fast


The Navy honored the remains of Hospital Corpsman 3rd Class Michael B. Judd, July 15th at Arlington National Cemetery. Judd was aboard a CH-46A C-9 helicopter that was shot down by enemy fire June 30th, 1967 in Vietnam. I joined the Navy when I was 17. He was my kid brother. He came in after me. I went into the submarine service and he went into the hospital corpsman service. He went with the Marines, served here in the Washington area and then went to Vietnam. The Department of Defense POW, Missing Personnel Office, used forensic identification tools and circumstantial evidence to identify Judd's remains. From the Defense Media Activity, I'm Petty Officer Andrew Johnson.