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[BLANK_AUDIO].

Â Hi there.

Â So, we're going to move on today to talk about the angular wavefunctions that

Â you can obtain from the solution of

Â the Schrà¸£à¸–dinger equation for hydrogenic type atoms.

Â So we have mentioned these before, when we said we can, you

Â can separate the variables into a radial part and a, an angular part.

Â And the notation that we use for the angular part is we have Yl, m sub l.

Â These are the quantum numbers for the angular part.

Â And then you have your spherical polar coordinate theta and phi.

Â So, we also told you that the l quantum number cannot

Â value 0, 1, 2, so on up to whatever N minus 1 is.

Â And that the ml values can vary from minus l, minus l plus 1

Â minus l plus 2, zero and so forth up onto plus l.

Â So these are now the, now the rules for

Â the quantum numbers that come out of the, after solution.

Â But that [UNKNOWN] for the, the radial solution, so I'm

Â going to give you a, a general formula for them.

Â So let's say this is a, a general formula.

Â It's an approximate formula, but it helps to understand the values that come out.

Â So let's do it here that you

Â have Y, so it's l m sub l,

Â you have theta phi and that's

Â equal to, so we're going to go some

Â constant times a sine theta and/or

Â cosine theta function.

Â And that's all the power of l.

Â And then you have another bit here, e to the

Â i, our imaginary number, the square root of minus 1.

Â The ml quantum number phi.

Â 2:27

Okay, that's phi there.

Â So let's try to apply that to some

Â specific examples, as we did in the case of

Â the radial functions, and see if we can make

Â some sense out of the values that come out.

Â So let's our, our first case is that you have l equal 0.

Â 2:44

And you're going to have ml is equal to 0.

Â And you probably know from your general chemistry, that's a s-orbital.

Â And the angular functions to find the shapes of the s-orbital should

Â probably tell you that the, the shape of an s-orbital is spherical.

Â But let's see where that, or what we're trying

Â to do is let's see where that comes from.

Â 3:07

So we have Y00 and the solution Schroedinger equation

Â for a hydrogen atom says that's a constant value.

Â It's 1 over the square root of 2, 4 pi.

Â And that is independent of the theta and phi values.

Â 3:29

Now why should that be from this formula here?

Â And it's instructive to go back to our definition of spherical polar coordinates.

Â So we had our let's put it well let's put it over here.

Â We have our z axis, here we have our y axis, and

Â this one coming out towards us is our x axis.

Â And then we would find a point in this coordinate sphere.

Â Draw a line to that point.

Â And then that angle there, the angle that

Â makes what the z axis would define as theta.

Â So, z axis.

Â 4:12

And then the angle here along the coming from the x axis, we define as, as, phi.

Â And theta compared from zero to pi so it can

Â go all the way to plus zed to minus zed.

Â And phi is going from, from, from zero to two pi.

Â 4:32

So if we tried to plot that for our on that.

Â So let's redraw the axis system again.

Â So here we have our z.

Â Here we have our, our y axis.

Â Here we have our, our x axis.

Â 4:52

So this fact, the fact that this comes out just to be

Â a constant, means the, the value, lets do it in two bits.

Â We're going to do the theta dependence first.

Â And our theta dependence, remember is going from z equals zero to z equals pi.

Â So that's, going to be a constant value all the way along.

Â So if we take our first point, which would say, let's say, be, be up here.

Â 5:19

Then as we go, as we vary pi from zero

Â to pi, then we're just going to get a constant value.

Â It's going to describe a semi, a

Â semicircle, which we'll draw approximately like that.

Â 5:34

And likewise for the phi dependance, the phi is going around this way, so

Â you can imagine for, if you play in the z and the y, zy plane.

Â We defined this here, now we're going to move that around 360 degrees

Â or 2 pi radians, and what we're going to end up when we

Â sweep that around is we're going to end up of course with a,

Â let's put an arrow here, we're going to end up with a, a sphere.

Â And this sphere is going to look something like I've, like I've put in here.

Â So this is the typical spherical shape for an s-orbital

Â that you see in your, in your general chemistry text.

Â 6:17

So let's move on now and talk about the next orbital shape.

Â And the next orbital is well the quantum numbers

Â your going to have now, move on to l equals 1.

Â And now of course you have ml equals, it can

Â equal 0 and it can equal plus and minus 1.

Â So let's look at the the ml equals 0 case first.

Â 7:02

But now there is a dependence on the, for

Â this function, on the theta term that's a cosine.

Â It's a cosine theta, theta dependence.

Â Again I didn't mention it in the, in the s-orbitals, but you can see.

Â Let's look back to this general formula I wrote down here first.

Â If you remember for Y zero zero that was zero, so this term

Â here just goes to, didn't appear, because you have some surprise zero that's one.

Â And also you have e to the i m l phi here.

Â And that's also zero.

Â So for the Y zero zero, you just have the constant.

Â 7:36

For the l now is 1, ml is 0, so this is zer, this is going to be 1 again.

Â But now we have a, a sin or a cosine term coming in.

Â In this case it's a cosine theta term, and it's to the power of l, which is 1.

Â So that's just going back to our our general formula, just to make it clear.

Â And so what we can do now as well is we can try to,

Â try to plot this this orbital so we can see what it looks like.

Â So again we have it Let's put it, let's just move it down a little bit here.

Â So I'm going to have z axis here and I then I have

Â the y here before and then the x coming out here.

Â 8:26

So before, remember if we, if we drew the I'm going to do this in red, if we

Â drew the the theta dependence for the s-orbital, we just had constant.

Â So it formed a semicircle.

Â This is a constant value.

Â But here it does depend on theta, so we need to know the cosine theta dependence.

Â 8:43

And the rest of that, because that goes d theta,

Â when it's along the z axis, is going to be, is

Â going to be 1, so at the value along this point

Â here at theta equals 1, it's going to be this value here.

Â So let's that's approximately there.

Â And now we have to work out when we vary theta, it's going to vary as cosine theta.

Â Well the maximum value of for the cosine theta term

Â is when theta is equal to zero, it's going to be 1.

Â And then it's going to, it's going to steadily decrease, and of

Â course when cosine theta comes to 90, it's going to be zero.

Â So what you do is if you plot that, it's called

Â a polar plot, you might get something very approximate like, like that.

Â So that's the, the cosine theta dependence,

Â and these values will all be positive.

Â 9:34

After you 90 degrees, of course cosine theta goes negative, so

Â you could generate another lobe, exactly identical, but on the bottom.

Â So, this time, again, we'll just do this, which is very approximate.

Â And that's the, and that the negative lobe.

Â So, that would be staying, say, on the z, let's say that's staying on the z y plane.

Â Then, of course you now have to do your phi dependence.

Â And your phi of course is, is this angle here,

Â so you sweep around to 360 degrees or two pi radians.

Â 10:08

And like before what you will do, you need to measure this.

Â You will generate a dumbbell shape on

Â bottom, a three dimensional dumbbell shape here

Â on the bottom, and you would generate

Â a three dimensional dumbbell shape on the bottom.

Â So again, we can put one in what will look like shown over here.

Â 10:29

So here on the right we have our, our,

Â our orbital generated from this rotation around, around the phi.

Â And just to put it in just to make consistent, so here's our z axis.

Â [BLANK_AUDIO].

Â And here's our y, and our, our x is coming out of the plane.

Â 10:54

So what you can see is we've generated dumbbells above and

Â below this xy plane and we could, so this would be again.

Â This is the positive region, and this is the negative, negative phase.

Â So this, of course, is the shape that you all know as

Â being due to the, to the p, p z, p z orbital.

Â So you can see that these familiar shapes

Â that you see in general chemistry, they all

Â come out of the angular solutions for the

Â Schroedinger equation for the, for the hydrogenic atom system.

Â 11:26

Now I'm not going to dwell too much longer on this, but I am going to

Â show you the solutions for l equals one ml equals minus one.

Â And again using our formula for that, you have y 1 minus 1.

Â And that's going to be equal to 3 over square root of 3

Â over 8 pi, sine of theta, e to the i phi, so you remember

Â i is our imaginary number, square root of minus one.

Â And for l equals 1, m l equals plus 1, and that of

Â course corresponds to Y at 1, plus 1 if you like.

Â And that's minus the square root of 3 over 8 pi.

Â And that's sine of theta e to the i phi.

Â And I've noticed a mistake.

Â This should be e to the minus, minus i phi here, because m

Â l is equal to minus 1 again, going back to your, your general formula.

Â That's where this comes from.

Â 12:45

So here for these ones you had k is constant k

Â and then, you had a sine theta, or a cosine theta.

Â We have sine theta term for these ones.

Â L is 1 so it's the power of 1, and now of course we have m l is either

Â plus 1 or minus 1, so you have for the minus 1, e to the minus i phi.

Â For the plus 1, you have e to the i, e to the i phi.

Â 13:13

So let's go back down here.

Â So you might say well the the Y one at

Â zero one corresponds to p zed orbital you're all familiar with.

Â And it's it's natural to say, well, do these correspond to p x and p y orbitals.

Â And they do, but it's not, they're not simply derived from them.

Â What you have to do is the p x and the p y are actually combinations of these two.

Â The p x is the addition combination.

Â So if you do y minus 1 1.

Â So let's, let's, let's write that down.

Â So you have the p x is Y 1 minus

Â 1, plus Y 1 plus 1.

Â And the the p y is the difference between them, so

Â that would be the Y 1 minus 1 minus the Y 1 plus 1.

Â And I am not going to go into it, but

Â to show that and it's not that difficult to show it

Â you need to be familiar, or you need to know,

Â that this e to the i phi is this Euler's relationship.

Â Let's put it down here.

Â So Euler's relationship says that e,

Â say to the plus or minus i phi, that's equal to

Â cosine of phi and then plus or minus i sine of phi.

Â So, if you expand them that way, you should be able, or you would

Â be able to show that the p x actually corresponds to the addition,

Â and the p y corresponds to the to the subtracting the two.

Â So as I say, you can do that perhaps as an exercise.

Â I'm not going to develop it here.

Â But hopefully this little snippet has has shown you if you like, the origin

Â of the shapes that you, you see for orbitals in general chemistry textbooks.

Â And you can explain how the, how the s-orbital shape comes along,

Â and you can also explain how the p z orbital shape comes along.

Â And for some of you should also be able to probably work out the p x and the p y

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Â