Transcript - Lecture 9

Well, last time we had seen that when you write this electron configuration that that electron configuration is a shorthand way of writing the wave functions for the electrons in a multi-electron atom within this one electron wave approximation.

But we also noted that at most we gave, or we assigned a hydrogen-like wave function to two electrons. And we noted that that was because of something you already knew about called spin. And spin last time we saw was a quantum mechanical phenomenon.

Spin is the intrinsic angular momentum build into the particle. And we saw that that spin is designated by a fourth quantum number. That fourth quantum number here is m sub s and it had two allowed values, and those allowed values were one half and minus one half.

And we said that those two allowed values or those two states represented a spin up and spin down. Two different polarities, in a sense, of this quantum mechanical phenomenon called spin. How was spin actually experimentally observed? Well, really thanks to these two gentlemen.

Here is Uhlenbeck and Goudsmit. These were very young scientists. I think post-docs at the time. And what they were doing was looking at the emission spectra from sodium atoms. And this is 1925, and so they knew enough already to know at what frequency they should expect to see some emission. And so they looked at the emission, disbursed it, looked at that frequency.

And instead of seeing emission at that particular frequency where they expected, they saw, so if this is the expected, what they really saw was some emission at a little bit lower frequency and some emission at a little bit higher frequency. Not much but a little bit lower and a little bit higher. And in spectroscopy, this kind of emission is actually called the doublet.

Well, they were quite excited about and thought much about the reason why there might be a doublet. And they reasoned that they could explain that doublet if the electron were existing in one of two states, one of two spin states. This was revolutionary.

And so they took their results excitedly to the resident established scientist that was closest who was Wolfgang Pauli. And Wolfgang Pauli was not a nice man, but they told Wolfgang Pauli about what they saw. And Wolfgang said, rubbish, you publish that and you will destroy your scientific careers. And so they left dejectedly.

No sooner had the door shut than Wolfgang sat down and wrote a paper on the presence of a fourth quantum number, mS. One of the travesty, really, in the history of science. That is well known now. But it really take, though, Dirac another three years later to actually show where this mS came from, this fourth quantum number came from. He had to use the relativistic Shrodenger equation, which we didn't look at. He solved that. And when he solved that out drops this fourth quantum number, mS. But Pauli did make a contribution. He worked on this problem and worked on the theoretical basis of it. And in the end came out with something called the Pauli exclusion principle.

And the Pauli exclusion principle here simply says that no two electrons in the same atom can have the same electron wave function and the same spin. Or, another way to state that is that no two electrons in the same atom can have the same set of four quantum numbers.

For example, in our electron configuration here for neon, we've got two electrons in this 1s state. Well, this electron here has the quantum numbers 1 0 0 ??. This electron also in that 1s state has the quantum numbers 1 0 0 -?? . So the n, l and m quantum numbers are the same for these two electrons.

But what is different is the spin quantum number. One is ?? and the other is -?? according to the Pauli exclusion principle that says you cannot have two electrons in the same atom with the same set of four quantum numbers. And likewise, for example, this electron right here, quantum numbers 2 1 -1 +??. But this electron, quantum numbers 2 1 -1 -??.

No two electrons have the same set of four quantum numbers. So that's why when we were actually writing this electron configurations, which is the shorthand representation for the wave functions in the one electron approximation for these multi-electron atoms, that's why we didn't give the ten electrons in neon, for example, we didn't all give that a 1s hydrogen-like wave function.

OK. Now what we want to do is we want to take a look at what the wave functions for the individual electrons in a multi-electron atom actually look like. And to do that we're going to look at the radial probability distribution function because that tells us a little bit about what the wave functions look like.

Because the radial probability distribution function, as a function of r, is the probability of finding the electron a distance between r and r plus dr. That's what the radial probability distribution function tells us. And so what I did here is to plot the radial probability distribution function for each one of the electrons in the argon atom.

Each one of the electrons in the different states for the argon atom. And the first thing I want you to notice is that these radial probability distribution functions for each one of these electrons look very similar in shape to the radial probability distribution functions for the electron in a hydrogen atom.

In fact, they have the same shape. If you look here at the 1s wave function, which is in green here, you see it starts out as zero, as all radial probabilities do. It increases, goes to a maximum, and then drops of exponentially. That's exactly what the 1s wave function did in a hydrogen atom. If you look at the 2s wave function, it starts out at zero, it goes to maximum, and then there is a node.

There is a node right here. And then it goes to another maximum and exponentially decays. That's exactly what the 2s wave function looked like in the hydrogen atom. For the multi-electron atoms, these wave functions have the same basic shape and they have the same nodal structure. They have the same number of radial nodes, same number of angular nodes. The big difference between these wave functions and the hydrogen atom is the distances from the nucleus.

For example, if you look at the 1s radial probability distribution for argon, you find that that most probable value of r is a tenth of an a knot, a tenth of a Bohr radius. What was it for a hydrogen atom in the ground state?

A knot, right? So this 1s electron here is ten times closer, on the average, into the nucleus than the electron in a hydrogen atom. And the same is the case for the 2s wave function, 2p, 3s, 3p. It's all a lot closer in if you compared those actual values. And the reason it is is because of the Coulomb attraction between the electron and the nucleus.

As you go to argon, which has now a plus 18 times e charge on the nucleus, that's a greater attractive interaction. The Coulomb interaction is the charge on the electron times the charge on the nucleus. It is that greater electron attraction that pulls the electrons in closer to the nucleus.

That is why that 1s electron, the most probable value of r is at a tenth of a knot instead of one a knot. Yes?

Well, they don't collapse in a sense. They don't collapse into the nucleus. Certainly the probability of finding the electron closer to the nucleus is greater.

But it is true that as you get larger and larger atoms you have more and more electrons. And so they do generally get bigger. And we will see that when we actually look at the size and look at the trends in the radius as we go across and down the periodic table. We will look at that a little bit more in detail and we will see that, in general, if you add electrons they get bigger.

Because you do, in fact, have more electrons and you put them in these outer shells. But if you just do a one-on-one comparison of 1s to 1s in hydrogen, this is closer in, this is closer in, this is closer in. Make sense? OK. But now also, since I have this diagram up here on the board, I want to just convey an idea that you, I think, are already familiar with.

And that is this idea of shells. When you say a shell you think of well, the n=3 shell, the n=2 shell, the n=1 shell, right? But the word shell also connotes some spatial information. And you can see that spatial information here in this radial probability distribution function. What I want to point out is that the most probable value of r for the n=3 shell, well, those most probable values aren't identical for 3p and 3s.

But they are very similar. They are similar compared to the most probable value for any electron in the n=2 shell. And those are similar than the most probable value for the electron in the 1s shell. So here you can see that spatial information.

You can see that, on the average, the n=3 shell is further out from the nucleus than the n=2 shell than the n=1 shell. I just wanted to illustrate the concept of shells because we had that radial probability distribution function up here. And it shows you, kind of nicely, how those shells are indeed separated in space on the average. And that is important, as we're going to see in a moment.

OK, so that's a quick description of what the wave functions look like for the electrons in a multi-electron atom. They have the similar structure as the hydrogen atom wave functions. They're just all closer in because of the charge on the nucleus. But now what we've got to do is have to look at the energies of the different states for these electrons.

So we looked at the wave functions. And now we've got to talk about the energies. Here is that energy level diagram for the hydrogen atom that we drew a few days ago. You see that the n=1 state here, which is the 1s state, is lowest in energy. The n=2 state here is higher in energy.

n=3. n=4. And you see that we have four degenerate states at n=2. That is the 2s state is the same energy as all the 2p states. And at n=3 we have 9 degenerate states. And at n=4 we've got 16 degenerate states. But in a multi-electron atom, that means starting with helium, things change.

And the first big change is that all of these energies here are lower compared to the hydrogen atom. The energy of the 1s state is lower than it is in the hydrogen atom. The 2s state is lower. The 2p state is lower. The 3s state is lower. All of them are lower. Why do you think that is the case? Anybody?

Pardon? It's closer to the nucleus. Yeah, it's closer to the nucleus because the Coulomb interaction is greater. So that Coulomb attraction is greater, that energy is greater, the binding energy is greater. It's more negative, the same reason as the wave functions were in closer to the nucleus.

So these energies are lower because of that larger attraction because that nucleus has a greater charge on it, greater positive charge. But the other thing that you see is that the 2s state has no longer the same energy as the 2p state. The 2s state is no longer degenerate with the 2p state. And the 3s state is no longer degenerate with the 3p state.

And the 3p state is no longer degenerate with the 3d state like it is in a hydrogen atom. How are we going to understand that? Why is this degeneracy broken or lifted as it is sometimes called? Well, to explain that what I am going to have to do here is to talk about a concept called shielding.

And shielding then is what creates what we call an effective charge that we're going to talk about. Now, to do so I want to point out this. That all of these binding energies right here, these binding energies e sub n, but now it's e sub nl. It depends on the angular momentum quantum number.

Those binding energies are minus the ionization energy of the electron in that nl state. That's the physical significance of that. But what we're going to do here is that we're going to set that binding energy equal to this, this effective z squared r sub h n squared.

And now I'm going to switch to the board and we will talk about that a little bit and explain this effective charge and the shielding so that we can understand why the 2s state is lower in energy than the 2p state. I just said these states e sub nl, physically significant is that they are minus the ionization energy.

And I'm going to approximate that binding energy as a hydrogen atom binding energy in the sense that I'm going to set that equal to that Rydberg constant, r sub h over n2, and this is minus.

But out here I'm going to put a z. This z is not going to be the charge on the nucleus but rather it's going to be something called the effective charge. So z, e, f, f. And it's going to depend on both n and l quantum numbers. This I'm calling the effective charge.

And what I'm going to do eventually here is to use the experimentally measured ionization energies to back out an effective charge for each electron in each n sub l state. That's what I'm going to do. But now what is this effective charge? Well, suppose I take a helium atom and we have a nucleus here.

And the charge on that nucleus is plus two. And, of course, helium has two electrons. And so I have, say, an electron number one over here and I have electron number two over here. But the bottom line is that for this electron one right over here, this electron one does not experience the full nuclear charge.

Why? Because, on the average, there's another electron in between it and the nucleus. And so the nuclear charge that is experienced by this electron is actually lower than the full nuclear charge because this electron here is negatively charged.

And it, in effect, cancels part of this positive charge on the nucleus. And so this electron doesn't experience the full nuclear charge. It experiences something smaller than the full nuclear charge. And that charge, at something smaller than the full nuclear charge, is that we call the effective charge.

How are we going to determine that effective charge? Well, we're going to use some experimental measurements to do that. We are going to look up the ionization energies, or measure the ionization energies. And we'll see how we do that a little bit later, but you can guess how we do that. We're going to look them up and we're going to back out the effective charge. So let's do that for helium.

If you go and you look up the ionization energy for a helium atom, you find that it is 3.94 times 10 to the minus 18 joules. What I can do is I can rearrange that equation for the effective charge, Zeff. And it is n=1, l=0. And that will be n squared times that ionization energy over r sub h all to the one half.

The ionization energy, when we talk about ionization energy it's usually from the ground state, so n is going to be equal to one here. The ionization energy I just said was 3.94 times 10 to the minus 18 joules. Rydberg's constant in joules is 2.180 times 10 to the minus 18 joules.

If I solve that I find the effective charge is 1.34 plus. So the charge on the helium nucleus is plus two, but the effective charge for those electrons in the helium is less than that. It's 1.34. Why? Because on the average there is another electron in between that electron and the nucleus. And so the full nuclear charge isn't experienced.

That is the effective charge. Now this effective charge and the physical concept here of shielding, that's what's so important in why the energy of the 2s state is lower than the energy of the 2p state. So the energy of 2s --

An electron in the 2s state is more strongly bound than in the 2p state. That energy is lower than the 2p state. Why is that the case? Well, let's think of the lithium atom. You know that the electron configuration of lithium 1s2 s21. The electron configuration of lithium is not 1s2 2p1, right?

Because that's a higher energy. But why? Well, to answer that we've got to look at a radial probability distribution again. And I'm going to draw it for the 2s state. So here it is for the 2s state. And I'm going to draw it for the 2p state. Here it is for the 2p state.

So this is 2p and this is 2s. Now, we have to do a thought experiment here. And that is this. If you look at the 2s radial probability distribution function there is a part of this distribution function that is really close to the nucleus. And, for the sake of this argument, let me say that for this part of the radial probability distribution function the effective nuclear charge is plus three.

Now, that is, of course, not the case. But bear with me for a moment. We're saying for this part of the radial probability distribution function the effective charge is plus three. Well, that is the nuclear charge. But we're saying it is close enough. We'll just call it plus three.

However, for this part of the radial probability distribution function here for the 2s, we're going to say the effective charge Zeff is plus 1e and plus 3e here. Why? Well, because in lithium there are these 1s electrons that are really close in here.

And we're going to say that those 2s electrons scream completely the nuclear charge. So you've got a z equal plus three and you've got two electrons here, electron one and electron two, and you've got this electron number three out here. Since this is a minus one and this is a minus one, the effective charge here for this way out electron is going to be plus one.

But now what about the 2p state, if we had an electron in that 2p state? Well, what you can see is that because the most probable value of r of the 2p state is in about the same place as it is for the 2s state, it's not exactly the same. It's actually a little closer in, but it's about the same place. We're going to say that for an electron in the 2p state, the effective nuclear charge is going to be one.

But now, in order to calculate or think about the total effective nuclear charge, what we're going to have to do is to average that effective charge over these probability distributions. So if the 2s probability distribution has a part of it where the effective charge is plus three and part of it where it is plus one.

If I average over that, well, the effective charge is going to be, I don't know, plus two or something. But for the 2p wave function where the effective charge is all plus one, it doesn't have this lobe that is really close to the nucleus where the effective charge is large. Well, that effective charge for the 2p is going to be smaller than it is for the 2s.

On the average here, the effective charge Zeff, for the s2 is going to be greater than Zeff for the 2p electron. Since that effective charge is greater, well, we showed right up here that the binding energy, therefore, is going to be more negative because that z is larger.

And so that's why the 2s state is lower in energy than the 2p and so on and so forth. 3p is lower than 3d, etc. Make sense? It's because of this phenomena of shielding and because the 2s wave function always has this probability of the electrons being really close to the nucleus where the effective nuclear charge is larger.

If you average the nuclear charge over the probability distributions, well, the effective charge for that 2s wave function is always going to be larger than the 2p wave function. Therefore, the 2s is lower in energy than the 3p.

OK. That is an important point. Now --

How do we actually write the electron configurations for all of the atoms in the periodic table?

Well, I think that you already know that we use something called the Aufbau principle. And the Aufbau principle is the building up principle. And what that says is that you make a list here of the states and you order the states according to their energy, the lowest energy at the bottom here and then keep on going.

And we will talk about the energies of these states in a moment. But then, in order to write the electron configuration, you start putting in electrons into these states, one electron at a time starting with the lowest energy state. So for hydrogen, well, that left electron configurations 1s. For helium, well, that electron configuration is going to be 1s squared.

But you already know that we have to put in that second electron obeying the Pauli exclusion principle. So we're going to fill up these states, one electron at a time, lowest energy first. And then keep going to the next higher energy level. And we're going to do so heeding the Pauli exclusion principle as we go.

So that's helium. Here's lithium. Now we've got lithium. We're going to do beryllium. Well, we have to heed the Pauli exclusion principle. That electron has to have the opposite spin of this electron. And now we come to the 2p state. We put an electron in the 2p state here. So we're at nitrogen. And it doesn't matter whether you put it in 2px, 2py or 2pz.

Because they have all the same energy. But now we have got to add another electron. We have got to get to carbon. And when we get to carbon here we're going to have to follow something else called the Hund's rule. And Hund's rule says the following. It says that when electrons are added to the states of the same energy, which is what we've got right here.

A single electron enters each state before a second one enters any state and the electrons, the spins remain parallel. So where do we put carbon's electron? Do we put it here? No, because we violate something called Hund's rule.

It means that we have to put it in 2py or 2pz. But do we put it in here? No. It's upside down. According to Hund's rule, the spins have to remain parallel in order to have the lowest energy configuration. And so we've got to put it in here and with the spin being parallel to the spin of the other electron.

So now we get to oxygen. And we put the electron, again, in one of the other 2p states. The spin has to be parallel. And then we keep going. Now we can start to double up on the 2p states. That electron goes in the other spin. That one goes in the other spin. That one goes in the other spin.

Now we're finished with the 2p states. And now we're going to put the electron in 3s state. And the next one. Now we're ready for the 3p states again. So the first one goes into one of the 3p states, the second one goes into one of the other 3p states and the spins remain parallel. And we keep going. So that's the Aufbau principle.

That's how you write the electron configurations here then for all of the atoms in the periodic table, heeding those two rules. Let's take a look at this a little more in detail. We're going to start here just looking at the third period, or the third row in the periodic table. So we're starting here with sodium and going across.

And here is the electron configuration of sodium. And you see I put two electrons in 2px, two in 2py, two in 2pz. Well, if I'm just writing it like this, I don't mind that you just write them as 2p6.

Because you cannot really tell the difference here anyway. So you can just write it as 2p6. Now, just a few reminders of things that you probably already know. And that is that these electrons, 1s, 2s and the 2p6, there are four electrons. They make up the inner gas configuration. So electrons that make up an inner gas configuration we call our core electrons. And this inner gas configuration is that of neon.

And so, actually, if you were to write the sodium electron configuration it's fine to just write it as neon and then with a 3s1 next to it. The 3s1, of course, is the valence electron. Valence electrons are those electrons that are beyond the closest electron gas configuration.

So if we went across this third row, we would just fill up then the 3p states. And then we get to the configuration, the inner gas configuration here of argon. Let's look at the fourth period here starting with potassium and going across.

For potassium, the electron configuration is that of argon plus one electron. And now what I want you to notice is that that extra electron went into this 4s state. It did not go into the 3d state. It turns out for potassium that the 4s state is actually lower in energy than the 3d state. The same thing here for calcium, the next electron, it goes into the 4s state.

So the configuration is argon 4s2. It's not argon 3d2. 4s is lower in energy than 3d. I will show you a pneumonic in a moment to try to remember that. Next scandium, another electron. Well, now the 3d states start filling up. 4s2 3d1. Titanium 4s2 3d2. Vanadium 4s2 3d3. Chromium we have a problem.

Chromium we've got an exception here. You might expect it to be 4s2 3d4 because we're just filling up the d states. It turns out that's not the case. It turns out that the chromium configuration is 4s2 3d5. How do you know this? You don't know this. There was no way to really predict this.

This is an experimental observation. It turns out that having a half filled 4s shell and a half filled 3d shell is more stable than having a filled 4s shell and four electrons in the 3d states. This is an exception. This is an exception you have to know. Yes?

I am going to say that in a moment. It is an exception also for molybdenum. And we'll talk about it. It is not the exception for tungsten or below. So as we continue to go across, manganese, now we're going back to our pattern. 4s2 3d5. Iron 4s2 3d6. Cobalt 4s2 3d7.

Nickel 4s2 3d8. And here we've got another exception, copper. Copper, if you followed our pattern, you might expect to be 4s2 3d9. It's not. It is 4s1 3d10. Again, it turns out that the overall atom is more stable if that 4s state is half filled and the 3d state is totally filled.

Would you be able to predict that? No, you're not going to be able to predict that. This is an experimental observation and this is another exception that you do have to know. Next, zinc. Well, we're back to our pattern here. This is 4s2 3d10.

And then once the 3d10 states are filled, or the 3d states are filled, then we start filling the 4p states as we go across here, this fourth row. And we get to the krypton inner gas configuration. And these all follow a pattern. Just fill the 4p states up. So chromium and copper are two exceptions you have to know.

Now, what happens as we go across the fifth period here, right here? We're starting with rubidium. Well, again, the electron goes into the 5s state and not the 4d state. 5s is lower than the 4d. And for strontium, the electron goes into the 5s state, not into the 4d state.

So you fill up those 5s states before you start filling up the 4d states here. And just like copper and chromium are exceptions in filling the 3d states, molybdenum and silver are exceptions in filling the 4d states. These four exceptions, chromium, copper, molybdenum and silver are exceptions that you do have to know.

Along this fifth period here there are other exceptions. I don't expect you to know that, but there are four exceptions that you do need to know. Yes?

Right. I actually don't care which way you write the electron configuration. I will care if I specifically ask you which state is lower in energy. But if you just write it, I actually don't care how you write it. OK? OK. Now, how do you remember what the ordering is of these states?

The energy ordering. Well, here's a pneumonic for you. Maybe you've seen this before. I'm going to start out writing 1s like this, and then I'm going to write right underneath it 2s and then over one 2p, then 3s, 3p, 3d, then 4s, 4p, 4d, 4f and then 5s, 5p, 5d, 5f, 6s, 6p, 6d, 7s, 7p.

And that's all that's going to be important right now. And now how do I use this pattern to generate the energy ordering? Well, of course, 1s is lowest. But now I'm going to draw diagonals through this. So, for example, 1s is lowest. And then the next higher energy state is 2s. And then the next higher energy state is 2p.

And that state after that, that's higher in energy is 3s. And then I draw another diagonal so that the next higher energy state is 3p and 4s. And draw another diagonal 3d, 4p, 5s. Draw another diagonal 4d, 5p, 6s. Another diagonal 4f, 5d, 6p, 7s and another diagonal.

So that's how you remember the relative energy ordering of these states. You should be able to write the electron configuration of all atoms in the periodic table, including the four exceptions. There are other exceptions, but I won't expect you to write those. Now, I have to tell you about something that is a little confusing.

And that is when we write down the electron configuration of ions. What I'm about to say has nothing to or doesn't change in any way what I just told you about writing the electron configuration of neutrals, but it does have something to do in writing the configuration of positive ions.

And here is the story. If I actually look at the energies of the individual 4s states and 3d states as we go across the fourth row, potassium, calcium, scandium, titanium, well, the 4s states in potassium and calcium are actually lower in energy than the 3d states. And, indeed, when we wrote the electron configuration of the neutral, we put that electron, that extra electron making the potassium in the 4s state and that extra electron in the calcium making a 4s state. But it turns out that right at this element, scandium, z=21, the energy of that 3d state drops below the energy of the 4s state.

Does that have a consequence for how you write the electron configuration of scandium or titanium, for example? No. The electron configuration for titanium is still what I said, argon, 4s2 3d2. The fact that this 3d state is now lower in energy does not effect writing the electron configuration of the neutral.

Now, you might ask if the 3d states here of titanium are lower in energy, why don't the 4s hop over into the 3d? They don't because this configuration, again, minimizes the electron repulsion. If you put all four electrons in the 3d state there would be a greater electron repulsion and then the overall energy would actually be higher. What I'm plotting here is just the energy of 3d and just the energy of 4s.

But when you're looking at the electron configuration of an ion you have to look at all the interactions and all of the energies. So for the neutral it's still what I said. However, now if I form an ion, if I take titanium and I make titanium two plus, you know the electrons that come out? The electrons that come out are the electrons in the 4s state because they're the higher energy state.

And so the electron configuration of titanium is the argon 3d2. That is important. And, similarly, for that fifth row, for rubidium, strontium 5s is lower than 4d. But right here at yttrium 4d drops below 5s.

That does not effect the configuration that we wrote for the neutrals. So, for example, here is silver. It's a 5s1 4d10. But if we make a silver ion, silver plus, it's that 5s electron that leaves because it's the higher energy electron. And so the configuration is krypton 4d10. OK. See you on Friday.

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