Hello, in this lecture, I'll be discussing electron configurations and the difference between poor and valence electrons. So, remember quantum theory gave us information about n and l, the principal quantum number, which tells us about how far from the nucleus is the electron. And the quantum number l which tells us about the shape of that orbital. An electron configuration for an atom is simple a list of the occupied sub-levels showing the number of electrons in each sub-level. So the notation looks like this, "nl#". We're going to have the n quantum number followed by the letter that represents the l sublevel. And then as an exponent, we'll have a number that tells us the number of electrons in that sublevel. Remember an s sublevel can hold two electrons, but the p sublevel can hold six electrons, because there's three different orientations of the p orbitals. I think the best way to learn this is through an example. So, let's write the electron configuration of business. Bismuth is one of the late metals. It's a beautiful, shiny, silver metal. Its oxidized surface is particularly attractive. It tends to have these bright colors. And you might see those if you visit a natural history museum. I've actually bought a sample of bismuth, because it's so beautiful. Let's write the full electron configuration of a bismuth atom. So, you see, bismuth is down here in period number six. While that six gives us the principle quantum number, n, of the highest energy electron of bismuth. But bismuth doesn't have just one electron. Bismuth is element number 83. That means it has 83 protons and 83 electrons when it's neutral. Each of those 83 electrons has a different set of quantum numbers. But remember we can only determine the quantum numbers n and l experimentally. So let's give a list of the electrons in bismuth, showing only the n and l sub-levels, and how many electrons are in each of these sub-levels. That's called the electron configuration. The closest electrons to this [UNKNOWN] nucleus are in the 1S sub-level. There are two electrons in that sub-level. So I can write 1 s2, to show those two electrons. The next highest energy electrons in the Bismuth Atom are in the 2s orbital. And there are two electrons in that orbital as well. As we increase the energy of the electrons, the next spot that's available is the 2p sub-level. There are six electrons in the 2p sub-level. Remember, there are three p orbitals at the 2p sub-level and each of those orbitals can have two electrons in it. Then we need to wrap back around the periodic table to the third period. There are two 3s electrons. And there are six 3p electrons. Do you see how I'm filling out this electron configuration as I move across the periodic table? Now I'm going to wrap around to the fourth period. There are two 4s electrons. Then we reach a part of the periodic table where it looks like there's a little bit of misregistration. Remember for a d orbital, l equals 2, which means the smallest value of n that's allowed is 3. So we have some 4s electrons. But the next set of electrons aren't going to be at n equals 4, they're going to be in the 3d sublevel. And there are 10 electrons there, because there are 5 d orbitals at the 3d sublevel. Then we go back to n equals 4. And we have some electrons in the 4p orbitals. And there are six electrons there, two in each of the orbitals. We can keep doing this process, wrapping around to see that we have two electrons in the 5s orbital, 10 electrons at the 4d sublevel. 6 electrons at the 5p sublevel. Then we get to the 6th period, and we have two electrons in the 6s orbital. And then, remember, if we have an f orbital, l equals Well, it was 0 for s, remember. It was 1 for p, it was 2 for d. So l equals 3 for f. n has to be greater than l. So if l equals 3, then n has to be at least 4. So this becomes the 4f sublevel. So, there are 14 electrons at the 4f sublevel. There are ten electrons if we, now, we see for the d. We're marching out for the 3, 4, 5, at the 5d sublevel. And, finally we get to the area where business is located. It's this third column over in the p block. So that means there are 1, 2, 3 p electrons in that highest orbital, highest sublevel. So we have 6p3. So this is the full electron configuration. If we added up all these exponents, 2 plus 2 plus 6 plus 2 plus 6 plus 2 plus 10, Let's say x plus 2 plus 10 plus 6 plus 2 plus 14 plus 10, plus the last 3. We would have 83 electrons. The spots that are below where bismuth are are not occupied. Those sub-levels do exist, but they do not contain electrons in the ground state. So remember, we talked about how, if we did something to excite the atom, the atom could promote electrons to higher sub-levels. And that would happen in the excited state of bismuth. We might get some electrons moving up into this seventh period here. But at the ground state, the highest energy electrons. Are here, at the 6S, 4F, 5D and 6P orbital levels. Wow. That was a lot of work writing all of that down, wasn't it? Because we had to basically write down the condition of all 83 electrons. So chemists like to have shortcuts. And one of the shortcuts we have is called the noble gas core configuration. Well, if we look on the periodic table, and we go to the row above where bismuth is, we find the noble gas in that row, and we see that it is Xenon. Xenon has the electron configuration shown here in this blue box. So if we wrote xenon in square brackets, that indicates all of the electrons that are drawn in that blue box. And then we have some electrons left over, which we would have to write out separately. The 6s, 4f, 5d, and 6p electrons we'd have to write out separately. So this is an abbreviated way of writing the electron configuration. It's called the noble gas core electron configuration. Okay, now it's your turn for practice. I want you to go ahead and use that shortcut we just talked about, the noble gas core, to write the electron configuration for Rutherfordium. Now, if you have an older periodic table, it might not have Rutherfordium named, but it is located right here in this box. [INAUDIBLE] So please go ahead and try to write out the electron configuration for Rutherfordium in its ground state. Thank you for working that problem. To use the noble gas core configuration, we have to go to the row above the atom of interest. So here we're going to go above Rutherfordium's row. And we see that above it there is Radon. So we put Radon in square brackets and that indicates all of the electrons that are contain in a Radon atom. But Rutherfordium has 104 electrons which is more electrons than Radon. So we have write out those highest energy electrons. It's got 7s2 electrons, 14f electrons and the 5f level. And, it's got 2 electrons at the 6d sublevel. So, this is the full electron configuration of Rutherfordium using the noble gas core abbreviation. Let's do little bit more practice. Here, I'm going to show you a diagram of a specific atom. This is an atom in the neutral state. And it has this electron configuration, 1s2, 2s2, 2p6, 3s2, 3p4. Which neutral atom has this electron configuration? There's the noble gas abbreviation but would this be for silicon, carbon, aluminum, or sulfur? Go ahead and put in your answer now. Thank you for answering that question. Hopefully you looked on your periodic table. And we're able to deduce that Sulfur is the atom that has that electron configuration. The last thing I'd like to cover relative to electron configuration is Hund's Rule. Hund's Rule talks about how the electrons prefer to fill up what are called the degenerate sublevels. Degenerate sublevels are orbitals that are at the same energy. For example, we have three 2P orbitals. They're at different orientation, but they're all at the same energy level, so we say that they're degenerate. According to Hund's Rule, the lowest energy arrangement of electrons in an unfilled sublevel, is first one that Maximizes the number of electrons that are unpaired. So if you have degenerate sub-levels, the electrons first prefer to have one in each of those orbitals. And then, if we have to pair some, we can. But the electrons that are in the degenerate orbitals that are unpaired prefer to have. Identical spin. Now I want to go back to Pauli's exclusion principle. We cannot violate Pauli's exclusion principle, we are not allowed to violate that rule. Hund's rule though we can violate that. If we violate Hund's rule that just gives us A higher energy electron configuration. We can't draw two electrons in the same orbital with the same spin. That would violate [UNKNOWN] exclusion principle. But we can have electrons in different orbitals that are unpaired, but which have different spins. And that's just a slightly higher energy arrangement of the electrons. So, here's an example for practice. Which of these arrangements of the d electrons has the lowest energy? So, this is an element that has 4 electrons in the d sub level The first thing we talked about is that the electrons prefer to be unpaired. Remember, electrons have like charges, and like charges repel each other. So the electrons don't want to be paired. So we can cross off a. The next thing we learned is that according to Hund's rule, the electrons like to align their magnetic moments so that they're all pointed in the same direction. The only one where they're all pointed in the same direction is c. So c is the lowest energy arrangement of the d electrons. Here's one for you to try. This is a different atom. And this case, there are 7 d electrons. Which of these arrangements of d electrons has the lowest energy? Thank you for answering. The first thing you probably realized is that one where there are more electrons paired than need to be can't be the lowest energy. So then we have three answer choices where they are two sets of paired electrons. And then we need to look at the electrons being with identical spin when they are in the unpaired state. So, in this one, I have one that's up, one that's down, and one that's up. That doesn't seem to fit Hund's rule. Here, I have 3 that are all unpaired, and they all have identical spins. So that looks pretty good. Let's just double-check by looking at d. Again, we see some with opposite spins that are in pairs. So that one's not good. So the best answer here is b. Whenever I show this example in class, there are always a few students who don't like the fact that I paired the electrons over here on the right instead of on the left, because In the United States, when we read and write, we go from left to right. So people like for things to go from left to right. But I point out to them that all five of these d orbitals are at the same energy level, they're all degenerate. So it doesn't matter which one we prefer to fill with two electrons first. So, b is the lowest energy arrangement here. And, it doesn't matter that the ones that have pairs are over on the right.