 We are discussing hydrogen atom wave functions. We have shown you some beautiful pictures and we have learnt how to depict s orbitals 1s, 2s, 3s. We have learnt how nodes come. We have also learnt what happens when we can when we multiply them by this r square and generate the radial probability distribution function. Now, we try and show you some even more beautiful pictures like this. So, where does this come from? Let us just see. So, now we move on to p orbitals and the simplest p orbitals of course are 2pz orbitals. Here principal quantum number is 2. Azimuthal quantum number is 1. Magnetic quantum number for 2pz is equal to 0. Now, why is it called 2pz? We will learn very soon. Let us have a look at the functional form constant multiplied by now we have r, r to the power 1 multiplied by e to the power minus zr by 2a. So, see increasing function of r and this one is an exponential decay multiply them together. It is not a surprise that it goes through a maximum. So, what you see here is a radial part of the wave function starts with a 0 at r equal to 0 and of course, psi starts with a 0 at r equal to 0 as well goes up and comes down. And from the previous module, previous couple of modules, I think you are now familiar and you can make sense of this kind of 3 dimensional plots. Just do not forget please that the vertical axis is really psi. In fact, in this case, vertical axis is R. Now, we have to start worrying about the angular part. The angular part in case of 2pz orbital is cos theta. If you remember the relationship between the Cartesian coordinates and spherical polar coordinates, you will recall that cos theta is equal to z by r. So, I can simply write z by r here. So, the effect will be that that r in the denominator and this r is going to cancel each other. So, what we are left with is something in z. So, what will this orbital look like? Now, we bring in symmetry in a very qualitative manner. Will you agree with me if I say that this orbital will have exactly the same symmetry as the z axis? Or if you have problem with that, let me put it in an even simpler form. Wherever z is positive, this orbital is going to be positive. Remember, it is being multiplied by an all positive or 0 function. So, except for r equal to 0 and r equal to infinity where the radial part is 0, the function has to be positive when z is positive and negative when z is negative. So, where is the node? The node arises from z equal to 0 and z equal to 0 being a node, the function is going to change sin. So, the way we have drawn it here is I am not going through the exercise of trying to plot it in real time because you have seen enough of it. Now, you can do it yourself. So, we have plotted it is that the vertical axis is psi, this horizontal axis is z and the one perpendicular to the plane of the projection is x, I have neglected y for the time being. So, cos theta is along z. So, on this side where z is positive cos theta is also positive and what you get is you get a positive part of the wave function and wherever z is negative, you get a negative part of the wave function. And why does it go through a maximum? Why is it sort of like a cone? Because if you go back to the previous projection, see this is the r dependent part. So, r dependent part is multiplied by cos theta. Wherever cos theta is equal to 1, where is cos theta equal to 1? Where is cos theta equal to 1? I think you know that at theta equal to 0, that is where it has 1, that is where it has a value of 1 and you have purely the radial part in the total wave function. Then as cos theta decreases from 1 to 0, what happens is that this radial part keeps on getting smaller and smaller and smaller. The radius dependent does not change. What happens is that this maximum becomes a little smaller. That is why you get this kind of a 3 dimensional picture and when z becomes negative, of course, the entire thing becomes negative and what you see here is the contour lines. So, if I look from the top or from the bottom, what kind of projection do I get? What kind of a contour diagram do I get? I get something like this. Do you agree? Look down from the top. This is where the positive peak is. This is where the negative peak is. Check. This is the z axis. This is the x axis. So, whenever z becomes negative, the wave function becomes negative as well and go through any radius. It goes through a maximum, is not it? Starting from here, it goes through a maximum, then comes down. And if you remember what we had said, when we had shown you that example of contours of an island, the lines are close together where the gradient is deep. The lines are further apart from each other where the gradient is gentle. So, just looking by at it, what you see is that near the node, near the node, we have a steep gradient and then once it goes to the maximum, which is somewhere here, the falloff is much gentler. That is the case everywhere. You go from in this direction, this direction, this direction everywhere the same thing happens. It is just that the maximum value of psi it reaches is different depending on theta. Let me maybe demonstrate that if I can. Let us say you take this section. What will happen? Psi will go up up to this value. Remember, this is going up. The values are given here. It is going up from out to in. The inner point here is the maximum value. So, you get a function that looks something like this. Now, let us say I go along this radius. What will happen? Still it goes up and then comes down. But what I see is that I am not reaching this high value. The value is lower. So, you get an similar looking function, but the maximum value reached is not going to be much, is not going to be that much. What about here? When you go in this direction, this is what you get. What happens when you go? Let us say in this direction, similar function given my poor artistic skills, but maximum reached is lesser. So, it is 0 of course, wherever z equal to 0 and it is 0 at infinity as well, r equal to infinity as well. So, I just take you back and I will ask you to take a look at this 3D plot vertical axis being psi and the contour. And I will request you to make sure that you have completely understood that this and this are one and the same. There are just two ways of drawing it. Now, so here we get a node. This node is different from the node we had in say 2s or 3s orbital. Why? Because those nodes arose by equating the radial part of the wave function to 0. Here, I get the node by equating the angular part cos theta to be equal to 0. It is just that cos theta translates as z equal to 0 as well. So, this here is the node nodal plane you can say. And this is our familiar p orbital 2 lobes, one with plus sine of wave function, one with minus sine of wave function. What happens if I try to construct psi square? Well, we get 2 lobes once again, just that both become positive. Then what should I multiply by? I should multiply by sine square theta. Should I multiply by sine square theta? Remember, what is there in the volume element? R square sine theta. So, I should multiply it by R square sine theta and see what kind of shape we get. We will not deliberate upon this question right now in the next module or the module after that we are going to come back and we are going to briefly discuss this issue. For now, let us just go through the orbitals and have a look at p as well as d orbitals. Now, but the thing is you can construct the probability and the probability roughly follows qualitatively the shape of the contours that we had drawn but of course, they will not be exactly the same. So, this is the familiar p orbital that you get. Why is it that we like to plot constant probability surfaces? We like to do that because it is probability that tells you where the electron is. So, when you want to work out the things like bonding and all, you better know what probability is. Now, so this is another way of drawing the probability distribution as dots, we are not familiar with this. Now, we come to a peculiar problem. We are handled to be z. Now, the moment we try to go to the other 2 orbitals, orbitals for m equal to plus 1 and m equal to minus 1, we encounter the problem that we have e to the power i phi and e to the power minus i phi there. Remember, the phi part of the wave function is e to the power i m phi. Here, i m can take up values of plus 1 and minus 1. How am I supposed to draw it in a real surface? I cannot. So, what we do is we use a very convenient tool of quantum mechanics to try and make the situation such that we can draw something. More about this later, let me just state here that the way to handle these imaginary orbitals is that we remember that quantum mechanical orbitals are all linear. We had said it towards the beginning of this course, they are all linear. What is the meaning of quantum mechanical orbitals being linear? It means that when a operates on some wave function psi 1 to give some eigenvalue and it operates on some other wave function to give some eigenvalue, then a operates on a1 psi 1 to give a1 multiplied by eigenvalue of psi 1 multiplied by psi 1. This is the meaning of an operator being a linear. Also, a hat operating on say psi 1 plus psi 2 is equal to a hat operating on psi 1 plus a hat operating on psi 2. Since we have said it quite some time ago, why do not I just write it once? What I am saying is the meaning of quantum mechanical operators being linear is let us say I have some operator a hat operating on psi 1 to give some eigenvalue a1 a2 sorry a hat operating on psi 2 to give a2 sorry for my childish and writing. So, what I mean is that suppose a hat operates on say c1 psi 1 plus c2 psi 2 then I get c1 a hat psi 1 plus c2 a hat psi 2. This is the meaning of an operator being linear. We are going to use this and try to use some linear combination. So, what we are saying essentially is that we are if you have a linear combination of acceptable solutions then they should also be acceptable functions. Is it always the case or is it not the case always? Let us see. So, let us write this now. What I get is I get a1 c1 psi 1 plus a2 c2 psi 2 should actually included this in the slide itself. So, now see if this is to be an eigenvalue equation what is the condition? Eigenvalue equation means this c1 psi 1 plus c2 psi 2 should somehow be able to go into a bracket. That is only possible if you can take out a1 and a2 when will that happen? It will happen when you can write like this a multiplied by a hat sorry about that I do not need to write a hat here c1 psi 1 plus c2 psi 2. When will this happen? This will happen only if if and only if a1 equal to a2 equal to a or in other words the a linear combination of two acceptable wave functions is two acceptable eigenfunctions of an operator is also an eigenfunction provided the two wave functions have the same eigenvalue. So, here think of Hamiltonian operator what are we using orbitals for so far to find out energy. So, Hamiltonian operator is a total energy operator. What is the eigenvalue for 2 pz and 2 p plus 1 and 2 p minus 1 they are the same whatever is the eigenvalue for n equal to 2 is the eigenvalue energy eigenvalue for all these orbitals. If you remember l and m do not make any contribution to energy for a 1 electron system it is n all the way. So, these 2 psi plus 1 and psi minus 1 these p orbitals actually satisfy the condition that they have the same eigenvalues. So, if you take any linear combination you will generate another wave function which will be an eigenvalue of Hamiltonian with the same energy associated with it that is very important and so we do not care. See we have solved your Schrodinger equation in a particular way that is why we have got the solutions. If you solve it in some other way we could have got perfectly valid solutions that are different from this one that might sound a little confusing but that is how it is. So, we are always allowed to generate new functions new solutions knowing already existing acceptable solutions and what I have shown you is that if the if you have 2 functions whose eigenvalues of an operator are the same then the linear combinations no matter which linear combination I took C1 and C2 they are perfectly general. You take any linear combination of these wave functions now may I use the word degenerate since we are using energy here. If you have 2 degenerate wave functions same energy eigenvalue then any linear combination of these 2 wave functions is also going to be degenerate with the functions that we started with we will have the same eigenvalue. So, we are allowed to take such linear combinations. Now we know what happens when we add say e to the power i phi with e to the power minus i phi it becomes real or we subtract. So, we will use those relationships between trigonometric and exponential forms of imaginary numbers and this is what we will do. The first linear combination we take is 1 by root 2 multiplied by psi 2 3 plus 1 plus psi 2. So, 2 1 minus 1 and psi 2 1 plus 1 you just add them what do we get 1 by 8 root pi 1 by a 0 to the power 3 by 2 r by a 0 e to the power minus r by a 0 the radial part and the constant part is the same that is why energy is same. So, that goes outside the bracket well even sin theta goes outside the bracket and you are left with e to the power i phi plus e to the power minus i phi inside the bracket. What is e to the power i phi plus e to the power minus i phi it turns out to be cos phi. It turns out to be cos phi not exactly cos phi there is a factor of 2. So, you are left with sin theta cos phi 1 by root over 32 pi 1 by a 0 to the power 3 by 2 multiplied by r by a 0 e to the power minus r by 2 a 0 sin theta cos phi it is a real function. It is a real function moreover do you remember where sin theta cos phi arises C z equal to r cos theta what is x x is equal to sin theta cos phi is not it r sin theta cos phi. So, the sin theta cos phi is a function which lies along x has the same symmetry as x. So, this is your familiar psi 2 p x orbital. Similarly, if you take a if you subtract one of these orbiters from the other in that case we need an i in the denominator it is not root over i root over 2 into i. So, this if you have i in the denominator then we get something like this sin theta sin phi and if you remember sin theta sin phi multiplied by r is nothing but y. So, this is your 2 p y orbital. So, what we have learned here is something once again we miss when we study high secondary or BSC or even MSC. Very common question that we ask often in PHD admission interviews or in other places is what is the m value for 2 p x orbital. Once again many students do not know what to say many students arbitrarily say plus 1 many students arbitrarily say plus 2 sorry minus 1 many students say it can be plus 1 or minus 1 none of these answers are completely correct some are outright wrong the last one is not completely correct. See what we get here is that we obtain 2 p x orbital or 2 p y orbital for that matter by taking a linear combination of the orbitals for which m values are defined. So, m value is not defined for 2 p x or 2 p y. It is not associated with any particular m value if you now go back to the our initial discussions what is the meaning of m value? Meaning of m value is the z component of angular momentum right and the possible m values here are plus 1 and minus 1. So, the z component of angular momentum is going to be plus 1 or minus 1. So, if you take this e to the power i phi wave function and somehow measure the z component of angular momentum every time you measure you get plus 1. If you take this second one m equal to minus 1 e to the power minus i phi wave function every time you measure you get minus 1 l z value. If you take 2 p x and if you perform say 10 lakh measurements for 50 percent of the time 50,000 times well no 10 lakhs right. So, 5 lakh times you will get plus 1 for the remaining 5 lakh times you will get minus 1 it is not please remember 2 p x and 2 p y orbitals are not eigen functions of the l z operator. So, average value you see what the average value will be if it is plus 1 half the time and minus 1 half the time you cannot measure l z like that. So, do not forget that they are actually generated from m equal to plus 1 m equal to minus 1 orbital m is not defined z component of angular momentum is not defined is the total angular momentum defined actually it is because total angular momentum is root over l into l plus 1 and l here is plus 1 1. So, the length of the arrow is defined theta is not defined that is all I am saying if you measure many times half the time you will get one value the remaining half the time you will get another value. So, but that solves our problem we have generated this and now we can draw it similar figures just that instead of being along z they will be along x and y respectively. So, interestingly the Pz orbital arises automatically because there it the m value is 0 e to the power i phi is equal to 1 that is why you do not even have a phi part showing up here it is 1. So, when m equal to 0 the orbital is already a real orbital real wave function do not have to do anything if m is non-zero however then we in chemistry we always generate real wave functions by taking appropriate linear combinations why is that so because we do not want to do too much of math I mean you might roll your eyes and say that what are you doing in this course believe me it is not much you take a quantum physics course it will be much much more rigorous mathematically. So, students of physics are absolutely fine e to the power i phi e to the power minus i phi when they can understand it we understand better if we can draw some picture that is why without loss of much generality we prefer to take this appropriate linear combinations and generate orbitals that are real so that is how we generate Px and Py orbitals. Similarly for d orbitals what is the m value going to be it will be 0 plus minus 1 plus minus 2 you can understand for m equal to 0 we will get a real d orbital anyway however for other m values you are not going to get you are not you are going to get an imaginary value. So, what you do is you take linear combinations of m equal to plus 1 orbitals to generate something you take linear combinations of m equal to plus 2 orbitals what am I saying m equal to plus minus 2 orbitals generate something else I will say that again for d orbitals m equal to 0 once again yields a real orbital and we will see what it is in the next module I thought we will discuss P and d orbitals in the same module but time is running out so we will take a break and if you take m equal to plus minus 1 they are again then you will get exponential terms like this e to the power i phi e to the power minus i phi what we do is we take appropriate linear combinations. Then the other one will be e to the power 2 i phi e to the power minus 2 i phi for m equal to plus minus 2 we take linear combination of those 2 as well we do not take linear combination of e to the power i phi and to the power 2 i phi we always take linear combinations of wave functions where m values are same as far as modulus is concerned that is how we generate the d orbitals. So, like the orbitals even in those familiar 5 d orbitals that we have only one comes directly the other are generated from imaginary orbitals that we get from solution of Schrodinger equation so that is what I wanted to say about 2p orbitals let us move on to 3p orbitals this is your 3p z orbital and understandably we will only talk about z what you have here is constant multiplied by r multiplied by 6 minus z r by a so where is the node going to be r equal to 6 a divided by z if you write in terms of a and put z equal to 1 it is just going to be 6 e to the power minus z r by 3 a no problem with that multiplied by cos theta of course now what happens see when you equate this you get this radial node as we said already so the function will start from 0 why because r is there go through a maximum come down go through the node change sign and then become 0 asymptotically so this is your radial node what is the locus of the radial node sphere of course now without showing you that good looking picture to start with let us see how we can draw the wave function without actually plotting it completely actually working out the numbers how we can draw the contour plot what you do is first of all draw the radial node then draw the angular node cos theta equal to 0 that is z equal to 0 draw the angular node now what happens is whenever the wave function crosses any of these nodes it has to change function it has to change sign isn't it so let us draw some contour somewhere and we know very well that the contours contour lines come closer to each other towards one of the nodes and then they are further apart right when they come towards an angular node this was something can be plus can be minus does not matter just draw this so what happens is when you start drawing from here draw the line here it has to turn right because it cannot cross the node without changing sign and here we want to retain the same value so no change in sign so this is what it is now cross any of the nodes sign would have changed let us say we cross the angular node sign changes you get a similar looking loop but with opposite sign right now let us cross the radial node once again sign change will be there and you get a loop like this on the other side also you get a loop like this so every time we cross a node sign change is going to be there what I am saying is that if you are asked to plot an orbital or any function ever figure out the nodes draw the loop that occurs to you first and then remember that as you cross a node sign has to change that is the simple okay now this is your contour diagram in a different orientation and this is what we have been waiting to see the 3D plot with contours the point I want to emphasize here is that once again you see the inner lobes of the orbital are so much larger in fact it is not very easy to get these contours when you make the plot because they are very very shallow okay this is plus this is minus this is minus this is plus their maximum values are so small that you have to work a little bit to even draw the contours to get those values not so easy but when you incorporate the volume element then these become the major lobes because we start talking about probability so this here cos theta is a node so this is the surface of constant probability for 3pz orbital when we are showing it please do not forget the science when you do that and as usual this is another way of showing it using distribution as dots right that is what we wanted to say about p orbitals in the next module we will talk about d orbitals