Evidence for sub-shells

Right. Okay. Electrostatic interaction. Are you ready? Not a big help. Yes. Go get it. Forgive me if I start. Okay. Electrostatic interaction. We talked about this in terms of the force that exists between charges. Okay, now there are two charges, positive charges and negative charges. In terms of particles, we know the positive charge resides in the nucleus of the atoms. Negative charges are on the electrons in the nucleus of the atoms. There may be more than one. In fact, there usually are more than one positive charge embedded. But the fundamental effect we're always looking at is some kind of either attraction between opposite charges or repulsion between similar charges. Now, we're not going to have much repulsion between the protons in a nucleus that's at all effective because they're already glued together by a strong nuclear attraction. So the only repulsive forces we're really interested in are electron electron forces. First of all, let's look at the force of attraction between these two particles or these two electrostatic charges. They will be apart by a distance we can define as R. The magnitude of a charge can be given by the symbol Z. So the charge magnitude. Z. Now, for the positive charge, we put it as Z plus. For the negative charge, Z minus. And so the force that pulls these two together is called the electrostatic force, electrostatic force, which in this case is an attraction. So it holds the two together. It's equal to. Sorry, it's proportional to. It's not equal to. It's proportional to. There is a proportionality constant which we don't need to go into. In fact, what I could do is I could write the proportionality constant as K some constant Z plus Z minus over R squared. So all we need to understand, to understand the factors that influence the strength of this electrostatic force is the magnitude of z plus the magnitude of Z minus and the distance between them. Just on a very simplistic level. Now, if we look at a couple of examples, if we have a positive charge of three plus and a negative charge of one minus, and they're separated by a distance of R, then the force is proportional to 3 times 1 over r squared. If we Compare that to 4 plus times attracting 1 minus, then the force is proportional to 4 times 1 over r squared. And if R is the same in both cases, well, this force must be larger. Do we understand that? Do you see that the force of attraction is larger here? Now, within atoms there are also repulsive forces, because although we have a nucleus in the centre. Just choose 12 positives. No, let's just make it simple. Two positive charges, a Helium nucleus. We've got electrons around the outside. These electrons are moving, of course, around the outside. So they will be in all different positions. We can't really average out the positions, but you should be able to see that there will be several forces involved here. There will not only be forces of attraction, there will be forces of repulsion. Now, on average, this electron is going to be further away from this other electron than the electron is going to be from the nucleus. And so the force of attraction, so the force of attraction, Which is proportional to 2 times 1 over R. I'm going to put Ra, the Ra squared, and the force of repulsion. Is going to be proportional to 1 times 1. In this case, 1 times 1 over re squared. This force of attraction, you can see that's w 2 over r where r is small, whereas this is 1 over r squared, where R is large. So I should have said R squared where R is small. This R is smaller than this R. So in all cases, both the top is bigger and the bottom is smaller. So the force of attraction is much larger than the force of repulsion. Now, we don't need to consider all of the individual attractions and repulsions between the electrons within atoms, but we do need to remember that there will be a negative component. There will be a repulsive component from electrons within atoms in the same way as that there is a strong attractive component from electrons attracted to protons. Now, let's just look at another example. Let's take lithium. Lithium has got a nucleus of three positive. We know that there are two regions of space orbitals. The first orbital has two electrons in, and the second orbital has one electron in. Now, the force of attraction of this four between these two, this is R. We'll call this one R2 because it's attracted to the outer, the second shell. And this one's R1. We should be able to see that even though the magnitude of the numerator is the same. 3 times 1, 3 times 1. In this case, the R is much larger here than it is here. So the denominator R squared is much larger for the attraction on the outer electron. So we've got our two forces, we'll call it force one is proportional to three times one over R2 squared. And this is force two proportional to three times one over R1 squared. If R1 is much smaller than R2, then this will be much smaller than this. So this will be much larger than this. So the force of attraction here is much larger than the force of attraction here.

Yeah, which is why it's easier to.

Use an outer electron, which is why it's much easier to remove an outer electron because it's held by less force. Alright, so are we all happy with this concept? Don't forget here as well. This electron will experience some repulsion from these two inner electrons. So the total force acting on this electron is not just the attraction, it's the attraction minus the repulsions. But remember, the repulsions are less important.

How would you take the repulsion into account if you were to.

Well, you can't really, because there are so many different repulsions. You just have to say there are some repulsive forces. We just say that this is attenuated. The value of this is changed a little bit. It's reduced a little bit by the repulsive forces of electrons between this and the nucleus. It's an attenuating factor. Aurora mobile phone, please. Not important now. All right. Okay, so let's put this into perspective with what we talked about last lesson. We talked about successive ionization. Now this is our successive ionizations graph for potassium set of ionizations. So what we're doing, what we're looking at here is the energy required to remove the first electron, then the next electron, then the next electron and the next electron. First thing I must point out is that this is a logarithmic scale down the side. The amount of energy does increase considerably. But when we take it in a logarithmic scale, we do get this nice pattern. So the very first point here, this represents this process. You see here we have a complete potassium atom with two electrons. Eight electrons. Eight electrons and one electron. So I've drawn the configuration similar to the configuration that you would expect down lower down in the school. There are 19 protons in the nucleus. So we're removing this outer electron. So we're moving to this situation where the outer electrons being taken away and you see what's left behind. Now the next electron to be taken away is going to be much closer to the nucleus. These circles, these energy shells are not to scale, of course. So if we move up to the next one, the next electron requires a lot more energy because it's now a lot closer. We're going from the eight electrons in the third shell to seven electrons in the third shell. We're removing an electron. However, the next electron to be removed requires just a little bit more energy. Well, why would it require a little bit more energy if it's the same distance?

Because there's one less electron. So there's less.

Very good. There's one less electron, and so now there's less repulsion. So we get this slow increase in energy because as we move remove electrons, there are less electrons repelling that outer electron. And so it's held slightly more tightly. And this continues all the way along, removing the six, six, the fifth, the fourth, the third, the second. Only one electron left here in the second shell to remove. Sorry, third shell to remove. And so the next electron is going to be removed from the second shell up to here. And there it is. And this electron is that much closer to the nucleus and so requires that much more energy. And so what happens again, we get the same pattern all the way through to this one with the last electron to be removed from the second shell. And once this is removed, we have to then remove one electron from the shell that's closest to the nucleus. So this pattern then shows us very clearly the effect of energy shells. I've chosen to show the changes for potassium because it illustrates very well the four shells of potassium. But this same pattern would appear for any element, but just starting at a different place. So if we started with, for example, carbon, we would start here. Carbon. Well, here. Sorry, start here for carbon. Carbon, we would be able to remove the four outer electrons and then we get a jump. If we started with, say, magnesium, it would be here. The first electron to be removed would be this one. One electron, then that one, and then we get this pattern of magnesium. In other words, if we're given a series of datum or series of data, we should be able to say whereabouts in the periodic table the element is just by considering the shape of the plot. If our plot looks like this, like this, two electrons and then a jump, and then eight electrons, we know we're in group two. If it looks like this one electron and a jump, we know we're in group one. If it looks like this four electrons, then a jump. We know we're in group four. So we should be able to predict which element or which group we're talking about by looking at the change in ionization or change in energy required for the successive ionizations. Okay, so so far this has given us an idea about the energy shells. But what it hasn't told us is it hasn't given us any information about the subshells. How do we get information to give us. How do we get evidence rather to support our idea for the Aufbau principle for the location of electrons in orbitals? Well, the answer to that is that this comes from also from ionization energies. But in this Case, not successive ionization energies, but individual first ionization energies. This is a graph which looks random. This is the graph of the first ionization energy of each successive element going up the periodic table. So we start with hydrogen. The amount of energy required to remove the outer electron for hydrogen is here. Let's see how much it is. Float over there. We say it's 1310. In fact, we calculated this last lesson, didn't we? From the convergence limit. 1310 kilojoules per mole. What holds that electron to the hydrogen nucleus? Well, the answer is the force of electrostatic attraction. It's in the first energy shell. Force of electrostatic attraction? Yes.

Isn't that the hardest one to lose.

Energy? Well, obviously not, because the next one's up here.

Yeah, but isn't it the one which is closest to the.

Well, obviously not, because the next one's up here. What's the next element? Helium. The amount of energy required to remove an electron from helium is 2,370. Alright, keep in mind the value for hydrogen was 1310. This is now 2370. Would someone like to comment on that value? Compare it to hydrogen. What would you expect it to be. Now? What would you expect the value to be knowing that Hydrogen's value is 1310? How much more, Lucas? The question is. The question is here, I'll repeat it. Hydrogen. The amount of energy required to remove that electron is 1310 kilojoules. This electron is in the first energy shell. It is held by the force of attraction between one positive charge and the electron charge and the distance between them. This electron is also in the same shell, but now it's held by an attraction between a double positive charge and the electron. What would you expect the value to be? Double the value for hydrogen, Less the repulsion, less the inter electron repulsion. Well, double the value for hydrogen would have taken it to 2,620. Is this 2620? It's a little bit less. So the difference in the first ionization between hydrogen and helium adequately demonstrates our electrostatic attractions instead of being one. Well, I'll show you it here. This is hydrogen. The hydrogen atom is the simplest case. The electron is held to the nucleus by a force of attraction between 1 electron and 1 proton only. So f is proportional to z1 z2 over r squared and z1 times z2 is 1. So the energy required to remove the electron is proportional to one over. Well, I should say that's the energy required. The energy is 1 over R. The force is R squared. The X. The energy is R. So the energy required is proportional to 1 over r. And there's the distance R between the nucleus and the electron. But let's look at helium. Now, helium. Well, what z1 times z2 is now 2 because the distance is the same. R is the same or more or less. There may be some shrinkage of the orbital due to the strong double plus charge. I don't want to go into that at all. We can do that. We can do this. We can appreciate this without thinking about that. If R is exactly the same, then we would expect this to be held by double the strength, therefore the energy to be two times the amount. However, there is also some inter electron repulsion. So to be taken into account, this reduces the amount of energy needed to ionize the helium, but only a bit. Do you get where I'm coming from? Do you get the principles I'm using? Yeah. Alright. So now let's remove this and look at lithium. It's right the way down here. So what's changed? Well, the nucleus has got bigger. Lithium has now got a three plus charge. But that must be the explanation. We have to now say, well, okay, we know the lithium is three positive charge. You would expect the energy to be much greater, but it's much less. So the answer must be that R must be a lot bigger. That's the only answer. R must be a lot bigger. And it is. R is now the extent or the distance between the nucleus and the outer shell. The outer shell is further away than the first shell. And so that explains lithium very well. Our explanation fits very well into what's going on. We wouldn't expect it to be. Well, it's difficult to expect it to be anything, isn't it? Because we don't know what R is. We just know R is a lot bigger. So the actual energy required is a lot less. It's a lot less. All right, so to go up to beryllium, then the next one, let's look at Lithium's value. Lithium's value 519. Beryllium is about 900. Beryllium, we're taking an electron away, the second electron away from the same shell. We would expect it to be more. The nucleus has now grown from three to four. And so we would expect the energy required to be maybe, you know, a quarter more than lithium. 4 plus attracting the outer electron. Here it's expected to be 4/3 times that of lithium, maybe less. A little. Because of inter electron repulsion, the value was 900. I can't remember what lithium was 590. It's there or thereabouts. Okay, so working on this theory, working on this principle. Aurora, can you put your phone away, please? Working on this principle, what would we expect the next electron to be greater or less in energy? We expect it to be greater, wouldn't we? Because we've got a third electron in the same shell with a five plus charge. We would expect it to be greater, but look, it's not, it's less. So we have to explain this somehow. Well, we can't explain this in terms of. Of attraction to the nucleus because this is now a five plus charge attracting the electron. There is only one way it can be less, and that is if. If what?

If the repulsion is greater than the delegated energy.

Well, we haven't seen that as a pattern, have we? We haven't seen the pattern of the repulsion doing this, otherwise it would have happened here. So there must be another reason. And the only thing that we can change is what? The distance. The only thing that can change is the distance. So this is telling me that this electron is located further away from the nucleus than these two electrons. It's the only answer. So this is evidence that there is something going on within the same energy level that we don't know about and that this electron is further away. Let's have a look at it. And this is evidence for the existence of P orbitals. The electron is now on average further away from the nucleus, even though it's in the same shell. Lucas.

But isn't it also statistically possible that you don't find that electron that far outside?

Statistically possible? Yes, but we have to talk about averages. And so that's why this electron, which I've labeled 2px, you see it's flashing there just to let you know which one we're talking about. The 2px electron is further away. There's evidence from the, from the drop we have here. And then we're going up, as we would expect, until we get to nitrogen. Something horrible is happening. So oxygen. Something horrible happens between nitrogen and oxygen. Well, in this case this is due to inter electron repulsion. But it's only because the electrons are pairing up in a P orbital. It doesn't affect electrons very much when they pair up in an S orbital. Why not?

Because they can arrange themselves at the opposite side.

Yeah, because they could be the opposite side. An S orbital is a big sphere. So it's possible for the electrons to be paired up but still be a long way away. But that's not the Case with a P orbital, the P orbitals, the P orbitals are here, they're much smaller. And so inter electron repulsion is significant in a P orbital. And this is what we've got now, even though I've put the two electrons in the opposite sides of the P orbital, there they still have a considerable repulsion. And that repulsion is enough to allow less or to make that less energy is needed to lose an electron. And we see the pattern then continues up to this electron, neon, when guess what happens? It falls all the way back to sodium because sodium is now in the 3s or the third shell, starting with the 3s orbital, and the pattern repeats itself. We go up, down to a P orbital, parallel electrons in P orbitals down, because we've got a pair in a P orbital, and then up to the end to the noble gas, which then drops all the way back down to the transition metal. And so our first. Our first ionization energy graph shows two things. Well, shows three things, let's be honest. The first thing it shows us is that we do get a general trend across a period. We get a general trend upwards across a period. But within that general trend there are two anomalies. So this is our general trend upwards. This anomaly is due to the first electron going into a P orbital. Electron in a P orbital. The second anomaly is due to pairing a P orbital or pairing up in a P orbital. The general trend upwards, of course, is due to increased nuclear charge, increased nuclear charge. Acting on the same energy level. So this is what we've got. And this trend is repeated here, exactly the same there, with two anomalies, a trend upwards across a period with two anomalous jumps or inflections, whatever you want to call it, two changes of direction which are for the same reasons. This one of course happens between beryllium and boron, this one between nitrogen and oxygen, whereas this one occurs between magnesium and aluminium and between phosphorus and sulfur. Any questions so far? There's something else the graph shows us as well. There is another pattern that we can see in the graph. I don't know if you can spot it. Can anyone spot another pattern in this graph? Aurora. It's also going down. In what way? Tell me. Explain to me.

So after each jump, it takes less energy every time.

Yeah, you're right, but you've not really expressed it in a way that we can all understand what you mean. Mariana.

The more you move towards the right, the.

Again, you're saying the same thing as aurora, but we're not being clear. I Need to be more clear about this, Lucas.

The further down you go at the period for the noble gases on the periodic table, the more. The less ionization energy is needed to.

Right, okay, so what you've done is you identified a particular group. If you take the elements from the same group, it doesn't matter which group. First of all, let's choose the noble gases. You'll see the three noble gases. There is a decrease moving down the periodic table. Why? Well, because.

The electrons are further away from the bonds at.

Yeah, but the nucleus is bigger.

Yeah, there's more electrons.

Well, so what does that mean?

So that means the number is bigger.

Yeah, but the. The distance is much bigger as well. It's that, Lucas.

There is more electron repulsion.

It is to do with electron repulsion. It's to do with. We call it electron shielding. As we move down the periodic table, whichever group we choose to look at, there are more and more shells of electrons between the outer electrons and the nucleus. And so we get an increase in shielding, increasing shielding. Of the nucleus. By inner electrons, inner electron shells, And this occurs. I chose the noble gases as an example, but it didn't have to be the noble gases. We could have looked at any group. Let's say. Well, this is this group. Which group's this? Which group's that, Mariana? Yeah, the alkali metals. Lithium, sodium, potassium. So what is this saying to us? It's saying that as we go down the periodic table, down a group in the periodic table, it becomes easier to ionize the elements. Does this have any consequences in terms of reactivity?

Yes, that's what we saw.

Do we see it with the noble gas, the alkali metals? Yeah, we see the alkali metals get more reactive as we go down the group. And this certainly contributes to it, the ionization energy. Like I said, it doesn't matter which particular group you choose. Always. We've got this increased shielding due to an increased number of inner electron shells. So let's just quickly summarize it. Where are we then? Well, we're back to the Aufbauer principle, aren't we? This directly shows us the Alfbau principle in its operation. It gives us evidence for our Aufbauer principle. We start off here. This hydrogen is 1s1. We're taking away an electron from the S shell. Helium 1s2. This electron is much easier to remove because it's in the 2s level. 1s2 2s1 1s2 2s2. But now 1s2 2s2 2p1. There's a drop. We've gone to a P orbital which is more diffuse. It's further away from the nucleus. This is 2p2, 2p3. But here, this is 2p4. So if we were to draw that there's element electrons in a box, we've got the first paired electron in a P orbital. So there is repulsion within this P orbital, which makes the energy required slightly less than in P3. Here we've got two P5, two P6. And this element, of course, is neon. Neon. So this one has got the configuration ne square brackets 3s1. Ne square brackets 3s2. There's a drop now, because this is ne in square brackets 3s2, three P1 in a P orbital, a little bit more diffuse. And then 3p2, 3p3. And this, of course, is 3p4. 3p4. Let's just draw it out over here. It's like this. The PX PY P Z. And we've got our first pair. So there is repulsion in the three level, and that's why less energy is needed. Then we've got 3P5, 3P6. And down here, what have we got? We've got AR4S1. This is calcium AR, square brackets 4S2. And then we've stopped. I've stopped. What comes after 4S2, 3P1? Nope. 3P1 was down here. Oh. What comes after 4S2? Mariana. Tell me. Mariana. 3D1, in fact. Now the D electrons start to fill up, the pattern changes for the ionization levels. We'll be having a look at that pattern in a separate lesson. Okay, what happens with the D orbitals?