[BLANK_AUDIO]. Hi there. So, we're going to move on today to talk about the angular wavefunctions that you can obtain from the solution of the Schrรถdinger equation for hydrogenic type atoms. So we have mentioned these before, when we said we can, you can separate the variables into a radial part and a, an angular part. And the notation that we use for the angular part is we have Yl, m sub l. These are the quantum numbers for the angular part. And then you have your spherical polar coordinate theta and phi. So, we also told you that the l quantum number cannot value 0, 1, 2, so on up to whatever N minus 1 is. And that the ml values can vary from minus l, minus l plus 1 minus l plus 2, zero and so forth up onto plus l. So these are now the, now the rules for the quantum numbers that come out of the, after solution. But that [UNKNOWN] for the, the radial solution, so I'm going to give you a, a general formula for them. So let's say this is a, a general formula. It's an approximate formula, but it helps to understand the values that come out. So let's do it here that you have Y, so it's l m sub l, you have theta phi and that's equal to, so we're going to go some constant times a sine theta and/or cosine theta function. And that's all the power of l. And then you have another bit here, e to the i, our imaginary number, the square root of minus 1. The ml quantum number phi. Okay, that's phi there. So let's try to apply that to some specific examples, as we did in the case of the radial functions, and see if we can make some sense out of the values that come out. So let's our, our first case is that you have l equal 0. And you're going to have ml is equal to 0. And you probably know from your general chemistry, that's a s-orbital. And the angular functions to find the shapes of the s-orbital should probably tell you that the, the shape of an s-orbital is spherical. But let's see where that, or what we're trying to do is let's see where that comes from. So we have Y00 and the solution Schroedinger equation for a hydrogen atom says that's a constant value. It's 1 over the square root of 2, 4 pi. And that is independent of the theta and phi values. Now why should that be from this formula here? And it's instructive to go back to our definition of spherical polar coordinates. So we had our let's put it well let's put it over here. We have our z axis, here we have our y axis, and this one coming out towards us is our x axis. And then we would find a point in this coordinate sphere. Draw a line to that point. And then that angle there, the angle that makes what the z axis would define as theta. So, z axis. And then the angle here along the coming from the x axis, we define as, as, phi. And theta compared from zero to pi so it can go all the way to plus zed to minus zed. And phi is going from, from, from zero to two pi. So if we tried to plot that for our on that. So let's redraw the axis system again. So here we have our z. Here we have our, our y axis. Here we have our, our x axis. So this fact, the fact that this comes out just to be a constant, means the, the value, lets do it in two bits. We're going to do the theta dependence first. And our theta dependence, remember is going from z equals zero to z equals pi. So that's, going to be a constant value all the way along. So if we take our first point, which would say, let's say, be, be up here. Then as we go, as we vary pi from zero to pi, then we're just going to get a constant value. It's going to describe a semi, a semicircle, which we'll draw approximately like that. And likewise for the phi dependance, the phi is going around this way, so you can imagine for, if you play in the z and the y, zy plane. We defined this here, now we're going to move that around 360 degrees or 2 pi radians, and what we're going to end up when we sweep that around is we're going to end up of course with a, let's put an arrow here, we're going to end up with a, a sphere. And this sphere is going to look something like I've, like I've put in here. So this is the typical spherical shape for an s-orbital that you see in your, in your general chemistry text. So let's move on now and talk about the next orbital shape. And the next orbital is well the quantum numbers your going to have now, move on to l equals 1. And now of course you have ml equals, it can equal 0 and it can equal plus and minus 1. So let's look at the the ml equals 0 case first. And the solution for that case is following, so it's square root of 3 over 4 pi, cosine of theta. So the difference here is now that you have a constant again. But now there is a dependence on the, for this function, on the theta term that's a cosine. It's a cosine theta, theta dependence. Again I didn't mention it in the, in the s-orbitals, but you can see. Let's look back to this general formula I wrote down here first. If you remember for Y zero zero that was zero, so this term here just goes to, didn't appear, because you have some surprise zero that's one. And also you have e to the i m l phi here. And that's also zero. So for the Y zero zero, you just have the constant. For the l now is 1, ml is 0, so this is zer, this is going to be 1 again. But now we have a, a sin or a cosine term coming in. In this case it's a cosine theta term, and it's to the power of l, which is 1. So that's just going back to our our general formula, just to make it clear. And so what we can do now as well is we can try to, try to plot this this orbital so we can see what it looks like. So again we have it Let's put it, let's just move it down a little bit here. So I'm going to have z axis here and I then I have the y here before and then the x coming out here. So before, remember if we, if we drew the I'm going to do this in red, if we drew the the theta dependence for the s-orbital, we just had constant. So it formed a semicircle. This is a constant value. But here it does depend on theta, so we need to know the cosine theta dependence. And the rest of that, because that goes d theta, when it's along the z axis, is going to be, is going to be 1, so at the value along this point here at theta equals 1, it's going to be this value here. So let's that's approximately there. And now we have to work out when we vary theta, it's going to vary as cosine theta. Well the maximum value of for the cosine theta term is when theta is equal to zero, it's going to be 1. And then it's going to, it's going to steadily decrease, and of course when cosine theta comes to 90, it's going to be zero. So what you do is if you plot that, it's called a polar plot, you might get something very approximate like, like that. So that's the, the cosine theta dependence, and these values will all be positive. After you 90 degrees, of course cosine theta goes negative, so you could generate another lobe, exactly identical, but on the bottom. So, this time, again, we'll just do this, which is very approximate. And that's the, and that the negative lobe. So, that would be staying, say, on the z, let's say that's staying on the z y plane. Then, of course you now have to do your phi dependence. And your phi of course is, is this angle here, so you sweep around to 360 degrees or two pi radians. And like before what you will do, you need to measure this. You will generate a dumbbell shape on bottom, a three dimensional dumbbell shape here on the bottom, and you would generate a three dimensional dumbbell shape on the bottom. So again, we can put one in what will look like shown over here. So here on the right we have our, our, our orbital generated from this rotation around, around the phi. And just to put it in just to make consistent, so here's our z axis. [BLANK_AUDIO]. And here's our y, and our, our x is coming out of the plane. So what you can see is we've generated dumbbells above and below this xy plane and we could, so this would be again. This is the positive region, and this is the negative, negative phase. So this, of course, is the shape that you all know as being due to the, to the p, p z, p z orbital. So you can see that these familiar shapes that you see in general chemistry, they all come out of the angular solutions for the Schroedinger equation for the, for the hydrogenic atom system. Now I'm not going to dwell too much longer on this, but I am going to show you the solutions for l equals one ml equals minus one. And again using our formula for that, you have y 1 minus 1. And that's going to be equal to 3 over square root of 3 over 8 pi, sine of theta, e to the i phi, so you remember i is our imaginary number, square root of minus one. And for l equals 1, m l equals plus 1, and that of course corresponds to Y at 1, plus 1 if you like. And that's minus the square root of 3 over 8 pi. And that's sine of theta e to the i phi. And I've noticed a mistake. This should be e to the minus, minus i phi here, because m l is equal to minus 1 again, going back to your, your general formula. That's where this comes from. So here for these ones you had k is constant k and then, you had a sine theta, or a cosine theta. We have sine theta term for these ones. L is 1 so it's the power of 1, and now of course we have m l is either plus 1 or minus 1, so you have for the minus 1, e to the minus i phi. For the plus 1, you have e to the i, e to the i phi. So let's go back down here. So you might say well the the Y one at zero one corresponds to p zed orbital you're all familiar with. And it's it's natural to say, well, do these correspond to p x and p y orbitals. And they do, but it's not, they're not simply derived from them. What you have to do is the p x and the p y are actually combinations of these two. The p x is the addition combination. So if you do y minus 1 1. So let's, let's, let's write that down. So you have the p x is Y 1 minus 1, plus Y 1 plus 1. And the the p y is the difference between them, so that would be the Y 1 minus 1 minus the Y 1 plus 1. And I am not going to go into it, but to show that and it's not that difficult to show it you need to be familiar, or you need to know, that this e to the i phi is this Euler's relationship. Let's put it down here. So Euler's relationship says that e, say to the plus or minus i phi, that's equal to cosine of phi and then plus or minus i sine of phi. So, if you expand them that way, you should be able, or you would be able to show that the p x actually corresponds to the addition, and the p y corresponds to the to the subtracting the two. So as I say, you can do that perhaps as an exercise. I'm not going to develop it here. But hopefully this little snippet has has shown you if you like, the origin of the shapes that you, you see for orbitals in general chemistry textbooks. And you can explain how the, how the s-orbital shape comes along, and you can also explain how the p z orbital shape comes along. And for some of you should also be able to probably work out the p x and the p y [BLANK_AUDIO]