Understanding Electron Configurations

Hello, this is a lecture topic video discussing electron configurations. So far, we essentially know that from the Schrodinger equation, we can get a pretty good probability of where to find an electron around the nucleus of any particular atom. Again, this falls from the Schrodinger equation, and from that Schrodinger equation essentially comes out these quantum numbers, these four principal quantum numbers. So we have our n, which again is our principal quantum number, our l, which is our angular momentum quantum number, our m sub l, which is our magnetic quantum number, and then m sub s, which is our spin, okay, our electron spin. So now what we want electrons, and we want to start populating these different orbitals that are described by these four quantum numbers. So the one thing we need to really keep in mind all right, is something called the Pauli exclusion principle. This is essentially saying that in any given atom you cannot have two electrons that have the exact same set of four quantum numbers. Let's look at helium as an example. So remember we're in n equals one, all right, for both hydrogen and helium. Well what can we have, all right, for our angular momentum quantum number? So remember our rules, all right? L can be anywhere from zero up to n minus one. So in this case, L only can have one possible value, and that is zero. And remember that for each angular momentum quantum number that we have, we have a letter designation. So for zero, this is our S subshell. Again, this is a shell. this designates our sub-shell. And then from there, what can we have? We can have our our magnetic quantum number, and that can be anywhere from minus L through zero up through positive L. Well in this case, all right, our epsilon has to equal zero as well. Okay, this tells us essentially the orientation of the orbitals within a given subshell and it also tells us about the number okay of orbitals so for l equals zero all right we are in the s subshell and we can only have one magnetic quantum number all right and that's zero and that tells us that again we have one orbital within our s subshell so we have an s orbital within an s subshell and it only has one orientation remember these are spherical in nature, one of the easiest ones to kind of visualize. And so what do we do? We want to start populating this. Well, again, hydrogen only has a single electron, okay? And so if we're thinking about the fourth quantum number, which is our spin, all right, our spin can either be plus one-half or minus one-half. And it doesn't really matter which one it is. convention usually says we start to use our positive one-half spins first, all right, but these things can flip on each. So a positive one-half spin can end up being a minus one-half spin. They kind of flip back and forth, all right. Well, how would we represent that? All right, if we're doing with our positive one-half spin, all right, we would indicate it with this partial arrow pointing upward. So this particular electron has four quantum numbers. of one, zero, zero, plus one half. Well, helium shares essentially most of this being the same, right? Because now we have two protons, we have two electrons. So N equals one for both of them, or L, as we already saw for hydrogen, equals zero. for both of them, and our m sub l equals zero for them as well. So up until this point, all right, for, let's say this is electron number one, this is electron number two, you would say, well, they have the same quantum numbers. They do, but remember, For any given orbital, we can have two electrons, okay? We can have two electrons, only two electrons. And they have to have... opposite spins from each other. So we got to go one step further and say that this electron can have our plus one half spin. So we can draw that up here. But then our other electron for number two would have to have, because again, the Pauli exclusion principle, no two electrons in any given orbital can have the same four quantum numbers. And this is in fact, any electron within an atom, okay? And so this has to be equal to minus one half. And so we would populate it this way, and this should be like that, represent like that, okay? So again, this is a Pauli exclusion principle, all right? And so what we want to do is we want to start essentially taking these different orbitals, all right? We got our shells, we get our subshells. all right, and we get the orbitals within these given subshells, okay, and we want to start populating them with electrons, right. So what we can do is we can start to move on past the first period of the periodic table or our n minus one and we can move up to say our n minus two and how would that look? Well we already said, okay, this again is looking at our n minus one and so our hydrogen all right we would populate a 1s with a single electron okay so here this is what we would call our electron configurations over here and this is our kind of orbital and they should reflect one another okay so this would be our hydrogen and then when we get to helium what would we do we would finish up with that shell. Okay. Where within that shell, we have one S subshell and one S orbital within that subshell, which carries two electrons, again, opposite spins, meaning that they have four different quantum numbers. All right. Or the set of four quantum numbers is different between the two. All right. Well, let's move to N equals two. Well, when N equals two, what can our momentum be? 0 and 1. All right. That means now we have two different subshills. We have S and we have our P. Okay. And remember S only has a magnetic quantum number of 0, whereas P. Okay. So this is, so if we're looking at M sub L, this would be 0. Okay. And for P, it would be minus 1, 1. and zero. Okay, so for our S subshell, again we expect only one orbital, an S orbital, okay, with one orientation. Whereas with our P, you see three blocks. Well this explains why, okay, because we have three different P orbitals, each of them having a different orientation, okay. And again, each of these orbitals can hold two electrons with opposite spins. Again, uh no two electrons within an atom can have the same four quantum numbers describing it all right and so since we have uh these three different orbitals each can hold two electrons all right we can have six electrons populating uh the this 2p subshell all right so if we look at lithium all right what we want to do is now we're on the 2s and so we've added again an extra proton for lithium but we've also at our gained our extra electrons. So what we can do is we can start to populate our energy level diagram. The first two electrons for hydrogen and helium will go into our 1s, like we already discussed. And then the next electron that's introduced when we get up to lithium, all right, is going to go into our 2p orbital. Notice we're populating the lowest energy, all right, subshell first within any given shell, okay. And so again, remember that the M tells us about our our energy and our size and L tells us about our shape. But in addition to that, this also tells us about the types of subshells that we have. And generally speaking, N, our S, subshell, is lower energy than our P, is lower energy than our D, and lower energy than our S. So what we want to do is we want to populate the S subshell first when we're talking about lithium. And then after that's completely populated, we'll move to our p-orbital. All right, well, how about beryllium? Well, beryllium, again, we've now gained an extra proton and an extra neutron or electron, excuse me. So where are we going to put that? Well, now, okay, we're going to finish out our lowest energy subshell first. All right, the orbital within our lowest energy subshell. And so it's going to look like this. All right. And now what are we going to do? Let's go to boron. Well now in this case, we now have a 2p1 electron, all right, because again our 2s orbital is now full. Okay, so the question is where is this electron going to go for boron? Okay, well the fact of the matter is it doesn't really matter. Okay, is it going to be this guy here, this one, or this one? It doesn't matter because these are called degenerate orbitals, all right? All three of these p orbitals within this p subshell, all right, have the exact same energy. That's what degenerate means. So when we're talking about this extra electron that's introduced because we've moved now to boron, all right, it can go to any one of these p orbitals within this p subshell. So if we look, all right, By convention, we typically start with the one that's in the lower or the left-hand side, and we usually give it the up spin, all right, or the plus one half ms quantum number. So now we can start to fill in our orbital energy diagram. So there's our 1s completely full. Now we're moving up in energy because, again, the principle is we want to fill up our lower energy subshells first, and then what do we do? We fill up our... 2s and now we are finally into r2p okay all right so now that takes us to boron all right well what about the next element that's carbon okay so now the question is we've gained an extra all right 2p electron that's what the 2p2 again tell just what these superscripts mean that mean this is the number of electrons in that particular orbital, okay, or that particular subshell. So the question is, where is the second P electron going to go, all right? So what you can do, we already know that from boron, one of these is already populated, all right? So where's that second one going to go? Well, we can think of this like the bus, all right? You get into a bus or you enter into a bus and you have these three long seats. all right, one of them is occupied and two of them are not, all right. So what are you inclined to do, all right? You're inclined to probably go into one of the empty seats, all right, and that is in fact what happens in this case. We populate one of those extra orbitals, one of those empty orbitals, all right, and the reason we do this is something called Hun's Rule, all right, which is a pretty simple rule. All right, and what it says is that the most stable arrangement of electrons and orbitals, okay, of equal energy, which again we have three p orbitals, all have equal energy. So the most stable arrangement of electrons and orbitals of equal energy is the one in which the number of electrons with the same spin is maximized. Okay, and so in other words, the only way we can do that, right, is if we start to half fill each of these orbitals first. Okay, so let's move forward a little bit and fill in the energy level diagram. So that takes us through helium, this takes us through beryllium, and now we're boron and carbon. So what we want to do is maximize the number of electrons, okay, which have the same spin. Again, remember, this is representing our plus one half. So now, if we would have populated this initial orbital right here, we would have had to have had an up and a down spin, or a plus one half and a minus one half. Well, if we do that, We are in blatant disregard for Hund's rule, okay, which again states that we want to kind of fill in our orbitals, all right, where we maximize the number of electrons with the same spin, all right, in our given subshells, all right. And so again, that's Hund's rule. That tells us that we want to keep our electrons unpaired. Until each of those orbitals within a given subshell, all right, are all halfway filled. And when they're halfway filled, they all have to have the same spin. All right. All right. So that takes us through carbon. All right. Let's look at nitrogen. All right. Well, in this case, all right, we have to, again, consider Hund's rule. Now we have three P electrons. And you may say, well, how are we going to occupy? All right. Or where are we going to put that last, that third electron in that? in that p subshell? Well, again, we want to follow Hunt's rule where we maximize, all right, the number of electrons with the same spin, okay? And so we're going to half fill each of those orbitals within that 2p subshell. All right, so we can fill out our energy level diagram again. Now each of these has a plus one half spin, all right? And we can do oxygen, okay? Well, now what are we going to do with this now fourth electron in that P subshell? All right. Well, now what we're naturally going to have to do is we're going to have to start to pair electrons. OK, we're going to start pairing our electrons. And so what would this one look like? Now we have this guy is completely full of this. this 2p orbital within this p subshell. Both of these electrons have, again, different quantum numbers that describe them. Well, the first three are going to be the same, but that last quantum number, the spin quantum number, again, one is going to have a one-half and the other one is going to have a minus one-half spin. Okay. So we start to fill in, all right. After each of our orbitals are halfway filled, now we start to pair our electrons. And again, with this idea that any given orbital can only have two electrons in it. Again, what's different about them? They have different spin quantum numbers. And then we get to fluorine, right? And now what we see, we have this fifth one. So now we have, all right, two of these P orbitals within this P subshell that are filled. And they fill out our P. orbital diagram here, energy diagram, right, until finally we get to neon, where we have six p electrons to fill up, all right, these p orbitals within this p subshell. And now what we have, all of those orbitals, those p orbitals, are filled with electrons. Again, Pauli Exclusion Principle says that each of these electrons has to have their own unique set of quantum numbers. Okay. All right. So essentially, what are we doing here? All right. We're filling these orbitals, okay, according to what we call off-balls principle. Off-balls principle. And this is the principle that states that as protons are added, okay, so as you see, as we go from hydrogen all the way down, all right, to neon, All right, what have we done? All right, we've changed the identity of the element, and we've done that because we've added protons in each case, all right, as we proceed through the periodic table. So as protons are added one by one to the nucleus to build up that given element, all right, electrons are similarly added, all right, to these hydrogen-like orbitals. Okay, so that, again, is the off-ball principle, and what it's saying, again… is that essentially what we want to do is we want to start adding electrons, all right, from the lowest energy orbital to the highest energy orbital. And you can see that that's what we've done. What did we do? We go from the 1s, again, that's the lowest energy level, then followed by the 2s, all right, followed by the 2p. And then as we would proceed, all right, to our third period. where our quantum, our principal quantum number is n equals three, what would we do? We would start to fill the 3s. Again, notice here on our y-axis, essentially, all right, is energy, all right? So the 3s is the next lowest, followed by the 3p, and so on and so forth. We start to see some interesting things, all right, when we get to the 3d, and we'll talk about that in just a second. So the question is, how are you going to remember this off-ball principle, this filling order, if you will? Well, there's a couple of different ways that we can do this. One way is to use the periodic table. Many times, like... as shown here, you'll see that the periodic table is essentially kind of sectioned off into four different regions, all right, and each of these regions has its own unique color, all right. Well if we kind of, and this is these different colors are essentially representing our different kind of L, our angular quantum numbers, angular momentum quantum numbers, so our subscript. So again, L equals zero, one, two and three these are the only ones that we will ever deal with in this class okay where l is our s okay sub shells one is our l or excuse me p sub shell two is our d sub shell and three represents our f sub shell well you can see again that these are color coded accordingly all right so here's our essence All right. These are P's here. These are D's and these are F's. And so what you can see if we kind of zoom in on this a little bit is that they're ordered. OK, according to their energy levels, because, again, remember, S is lower energy than P, lower energy than D. and lower energy than our F. All right, so this is the order we want to fill these in. So again, if we look at N equals one, we only have an S subshell. So of course, we're gonna fill that first. But then as we go to N equals two, all right, remember L can equal zero or one. So we'll have an S and a P subshell, which we do. Here's our S, here's our P, all right? And so what we're going to do is we're going to populate the S first, and then we're going to move to populating our P orbitals within our P subshell. Then what happens when we get to N equals 3? Well now we see something interesting. N equals 3 means that L can equal to 0, 1, or 2. This is our S, E, and our D. All right, but in this case what we're going to do is we're going to populate the 3S, all right, like we have before. All right. It's always the lowest energy of our subshells. Then we're going to populate our 3P. OK, so 3S, then 3P. And then what are we going to do? Well, we're going to skip our 3Ds. All right. They're there because this principal quantum number of three. All right. But what's interesting here is that we're going to fill again, again, our four. We're going to fill our 4S first because N equals 4, L equals 0, 1. or three all right so again s p d and f all right so we're going to fill the 4s prior to the 3d okay and then we're going to fill in our four p's well the question is why are we filling in 4s before our 3ds because if we're looking at this um this kind of pecking order up here, you might think that you'd want to fill in your 3Ds first. But again, the periodic table is kind of organized, all right, according to these occupations or how the electrons occupy the different orbitals, all right? But the reason why we fill in our forest versus our 3D first, all right, is because of something called crossover, all right? When we start to increase the energy, all right, from one shell to another, sometimes we get this crossover, all right? So again, Remember, we're on our third principle quantum number, n equals 3. We can have a 3s, we can have a 3p, and a 3d, all right? But then when we get to n equals 4, we can have our 4s, our 4p, our 4d, and our 4f, all right? Well, what you can see, though, the way these are organized on this energy kind of diagram here, that, of course, our 1s is the lowest, then, of course, our 2s all right, is the next lowest followed by our 2p. And then finally, what do we see? All right, we see that we get our 3s and then our 3p. So just like we'd expect. And then what do we see about the 4s and the 3d? It is crossing over the 3d subshell. Actually is at a higher energy level, all right? than the 4s subshell. Okay, so what we do is we occupy the 4s subshell, that 4s orbital, before we occupy those 3d orbitals. All right, and again this is called crossover. All right, but the way the periodic table is organized, it kind of lends itself to you being able to see this pattern. Okay, the one thing you've got to remember is that don't mess up and actually say that this is the 4D subshell because it's not. It's the 3D. All right. Because again, remember this principal quantum number of three tells you that it is going to have these 3D orbitals associated with it. So what do we do? We proceed from the 4S to the 3D because of crossing over, and then we can follow our trend. All right. So 4S. to the 3d and then what are we going to do we're going to move up to our 4p okay which is right here and then the this kind of pattern kind of continues on for our fifth period all right and then finally all right you see in the sixth period we actually start to see the introduction of these 4f electrons because again when we get to four all right when we get to four that tells us that we should have this F subshell, which has these F orbitals, but we do not start to populate them until we actually get to the sixth period on the periodic table. So, again, this crossing over leads to some, I guess you would call them, unexpected filling of energy levels, but again, the periodic table kind of lends itself to allowing you to figure out which of these orbitals we're going to fill first when we're following this off-ball principle. Again, off-ball principle telling us that we want to start from our, start filling our lowest energy orbitals first and proceed to the higher energy orbitals. The other one, all right, is this kind of schematic here. This is another way to remember kind of the filling order that might jive with you. All right, so what you do, all right, is you kind of start with your 1s, and you label the different s orbitals going downward, all right? Every period is going to have that. Then, of course, remember if we're in n equals 1, we don't have any p orbitals. So what we do is we go down. This is essentially n equals 2, and n equals 2 tells you you're going to have an f subshell and a p subshell. n equals 3 is now going to have an s, a d, all right? And what we do, okay, so we can kind of skip this spot here, then start with our P's, go down in our P's, and then skip one here. All right, start with our D's, and then go down in our D's. And then finally, we're going to skip this and start with our F's, okay? And so how do, how is this going to help us? Well what we want to do is we want to kind of fill the diagonals with our electrons and then imagine that we're going to come up here and we're going to start to fill the next diagonal which would be R2s, right? And then again we can kind of go back and we can fill in this next diagonal, right? So on and so forth. And so then we go 3P, 4S, and finally 3D, 4P, 5S, so on and so forth. So you got to pick one of these two things, all right, that's going to help you remember kind of this... this ordering of these orbitals in terms of their energy. Because again, the off-ball principle states that we have to be able, or we have to fill, all right, these orbitals from lowest energy to the highest energy, okay? And so you want to pick one, all right, that again, that jives with you. All right. So as you can see though, all right, we left off with neon specifically, all right, with this particular electron configuration. And as you see, as we proceed through the elements, what happens? These, these Electron configurations with these elements with bigger atomic numbers, they start to get really, really long, quite repetitious and quite cumbersome. OK, so what we do is we have a shorthand and it's called a noble gas configuration. So as we increase in these energy levels, all right, we also increase in the size of our of our elements. OK, so remember, again, you can kind of think of. as we move through these principal quantum numbers, or as we move down the periodic table from the first period where n equals one, all the way down to the seventh period where n equals seven, all right? You can kind of think of these things as shells, okay? And as we proceed through, what do we do? We kind of fill up the shells, and then we start over, all right? Again, this gets very repetitious. So what do I mean by that? Well, here's neons, all right? Full electron configuration. We fill up the 1s2. then we fill up 2s2 and then 2p6, all right? Then as we move from neon to sodium, all right, it has the same, okay, initial electron configuration. This 1s2, this 2s2, this 2p6, all right? Well, what does that represent? Well, that represents neon, all right? But now we have this extra electron. Where's it going to go? Well, it's going to go into this 3s. sub-shell, this 3s orbital. Again, that's what that one tells us. We're going to put one electron there. But what can you see? This kind of neon configuration essentially remains the same throughout this whole period of the periodic table, from sodium to magnesium, aluminum, silicon, phosphorus, sulfur, chlorine, and argon. What do we see? All of these guys essentially share the same core electron configuration. So what do we do? We kind of shorten this, and we say that this is, all right, our core electron configuration. We have a core electron configuration that matches that of neon. And so these are what we'd be calling our core electrons, right? Those are all those electrons that kind of fill up that single shell. And now we can get this distinction between core electrons and what we call valence electrons. And these valence electrons, these are outer shell electrons, all right? These are the electrons that are occupying these orbitals and these outermost energy levels, these higher energy levels, all right? So we get these core electrons, and now we get our valence electrons, okay? And so what we can do is we can proceed through. from sodium to magnesium. Instead of saying for magnesium, we have 1s2, 2s2, 2p6, a 3s2 electron configuration. We can use our noble gas configuration and say we have essentially the core electrons, all right, from the neon, all right, and followed by these valence electrons, these 3s2 electrons, okay? So again, we get this concept of core electrons and valence electrons. We're kind of distinguishing or kind of categorizing two different types of electrons, all right? And again, we can proceed to aluminum, where again, we have this 1s2, 2s2, 2p6, 3s2, 3p1, all right? Again, this first part, the 1s2, 2s2, 2p6, that again is our core electrons followed by what we're showing here in our noble gas configuration, followed by our valence electrons, 3s2, 3p1. And what we can do is we can proceed all the way down through argon, where argon now has this neon core, these kind of core electrons, and then our valence electrons are 3s2, 3p6, all right? Or what can we do? We can say that this now is another set of core electrons that we filled. And so instead of using this kind of... configuration with neon. Now what are we going to be if we proceed past argon? We're going to start with this argon being in parentheses, this AR being the elemental symbol for argon being in parentheses. Again kind of distinguishing that we filled up a whole shell. We filled up all those sub shells, all right, all those orbitals within those sub shells within that particular period of the periodic table. All right so we got full electron configurations. And we also have noble gas configurations. And then finally, what are we not really talked about? Well, we haven't talked about the d orbitals as of yet, okay? The d's or the f's. So we'll talk about the f's last, all right? And so what you might imagine, all right, is that, again, here's argon. This is where we left off, okay? So what we can do is we can say, put our parentheses here, and that notation, this argon within parentheses, also is equivalent to this full electron configuration, all right? Again, because it's shorter, and again, these full electron configurations can be quite repetitious and cumbersome, what we can do is we can shorthand and use our noble gas configurations. Now we're at N equals 4. We're at potassium. As you might expect, we've talked about this already, because of crossover, We fill up the 4s electrons before we start to fill the 3d orbitals. Okay. Again, this is crossover. These, the 4s subshell, that 4s orbital. All right. And those 3d orbitals, again, there are five of those. All right. They're really close energy levels, but the 4s orbital is lower than those d orbitals so we fill that first and again this this falls out very naturally from the periodic table okay so we're going to fill those 4s all right um or uh orbital first all right and that's going to come from those electrons from potassium and calcium and then what are we going to do we're going to get into our again these are our transition our transition levels called the D block, all right? Because again, now we're going to start talking about our D subshell, all right? Our D orbitals, okay? And what do we get? We get 3D1, we're talking about scandium. And then again, as you expect, we would then get for titanium, we would get 4S2, then 3D2, vanadium, 4S2, 3D3. And then what do we do? We come to... one of our two inconsistencies that we see with the transition, all right, we see that chromium and copper have some very interesting, all right, electron configurations, all right, because what would we expect for chromium, all right, for chromium, we would expect to have a 4s, and then again, and slightly higher, energy, talking energy here, all right, we'd have our 3d orbitals, all right, and so what we'd expect if we're following the same sort of pattern is that the four acids would get filled, and then finally 3d, again, because of Hunt's rule, we're going to try and maximize the number of electrons with the same spin, all right, we would expect this, okay, we would expect four of those d orbitals to have a single electron. in them. All right. But if we look at chromium, is that actually what we see? No, we don't see that at all. Right. We see that we have 4s1 and then we have 3d5. So this is a very consistent, kind of a weird way of saying it, but a consistent inconsistency within the transition metals. All right. Well, why is it this 4s1 versus this 3d, 4s1, 3d5? versus 4s2 3d4. Well, what you can see here is that there's this kind of open seat at the table, all right? And what we do, instead of having this configuration, all right, this 4s2 3d4, is we can move one of those electrons from the s, okay, into this unoccupied 3d orbital, and that gives us 4s1 3d5. Now, when does this happen? Well, remember that these energy levels are pretty close to each other. You can go back here and take a look. All right. They're pretty darn close to each other. But through a lot of research, people have found that having half filled, one half filled d orbitals is stable. Thank So having half-filled d-orbitals is stabilizing. So in other words, because we have this 4s1 3d5 and those five d-orbitals are half-filled, that is actually more stabilizing than having this 4s2 3d4 kind of electron configuration. And so that is a very consistent trend that you see with this group that's kind of headed off chromium. Okay. So it proceeds downwards into the fifth and the sixth periods. All right. And so again, a consistent inconsistency. Well, now what we also see, all right, going from manganese to nickel is exactly what you would expect. All right. We would have our 4s2, 3d5, then 4s2, 3d6, and then 4s2, 3d7, until finally... We get to nickel, which is 4s2, 3d8, again, following the expected kind of pattern. But then we get to copper. All right, copper, this group headed by copper, also provides another example of a consistent inconsistency in the periodic table. Because, again, what would we expect for copper? We would expect, again, if we're going, trying to draw something fairly accurate here, all right, we would have 4s2. followed by our 3d's and energy here. We would expect to be have nine of these filled. Notice I'm filling them according to Hunt's rule. We're maximizing the number of electrons in these orbitals that have the same spin. We would expect something like this a 4s2 3d9 but in fact All right. Experimentation says that it's actually 4s1 3d10, right? Because what's happened again? Well, again, these orbitals are really close to each other. So what we can do is we can promote one of these 4s electrons in here to this final d orbital so that all these d orbitals are actually filled with electrons, right? And this is what they found, all right, for this whole family. Again, a consistent inconsistency in our... electron configurations for these transition metals. All right, well, why is that? Well, in addition to these half-filled d-orbitals, they have also found that completely filled d-orbitals are also stabilizing. And so again, that's why in this one case, what we can do is promote one of these 4s electrons into a d orbital to give us for this this copper family if you will this 4s1 3d10 electron configuration so again these are just some inconsistencies that are really important and it's fairly simple to explain when when you think about the fact that they've determined that half filled all right when all your d orbitals are halfway filled or all your d orbitals are completely filled that provides extra amount of stabilizing energy. Okay. Now, finally, it takes us to our F. All right. All right. Our F subshell or F orbitals. Again, this is an F subshell. Again, what are we talking about here? This is where L equals to three. And what do we have here? Magnetic quantum numbers can be minus 3, minus 2, minus 1, 0, plus 1, plus 2, and plus 3. So we have an F subshell that has seven different F orbitals, again, all having different orientations. And so what are we going to do? When we get down here, this is where we're on the sixth period. So. So n equals 6, all right, and n equals 7, all right? But again, we get these all the way up here in period 4 where n equals, all right, because n equals 4, again, just to review, l equals 0, 1, 2, 3, all right? So that tells you that our f orbitals, that's where they're coming from. So that's where the 4 comes from. all right it's from that principal quantum number all right so what do we do we got um lanthanum and actinem and what we're going to do we're going to fill those with one d electron before finally we start to take our electrons and filling up these seven, okay, these seven F orbitals within this F subshelf. Again, there are 14 of these, all right? That makes some sense because we, again, we have seven, seven orbitals. They can all hold two electrons each. Again, plus one half, minus one half, right? So we have 14 electrons that we can use to fill these F orbitals. All right, so again, there's one, two, three, da, da, da, da, all the way through 14 different elements for each of these periods, okay? So, lanthanides have 14 elements within them. Our actinides have also 14 elements within them. That means we are, you know, kind of gaining these 14 different electrons in both of these cases so that we can fill these seven F orbitals, okay? And so it makes a lot of sense now, but when we look at the electron configurations, all right, of these F subshells or these F orbitals within the F subshell, there are all kinds of inconsistencies. Now for an exam, all right, this is important. We will never ask you one of these about, or expect you to know these inconsistencies. And the inconsistencies are all represented here by these red arrows. Okay. So you don't need to know those. But what we do want you to know would be the ones where it follows the given pattern. All right. So, for example, you can see cesium here. It's a success to 5D1, which that comes from lanthanum right here. That's the 5D1. And then we get 4F1. So that follows the pattern. OK, that follows a pattern. That's something we could ask you on an exam. All right. This guy, number 64. All right. Also follows the pattern. So on and so forth for these ones that are not highlighted with this red number. All right. So, again, watch out for inconsistencies like with a textbook question. They could ask you those sorts of things. But on an exam, we would only ask you ones that actually follow the kind of pattern, this off ball principle pattern that we've been talking about. All right. So let's try a sample problem. So write out the full electron configuration. So the full one, which again, those can get to be cumbersome. All right. And then the double gas configuration for indium. All right. And so what do we need to do first? The first thing we need to do is we need to find indium on the periodic table. So let's go take a look. Where's indium at? All right. Let's see. And then it is number 49, okay? So it's number 49. It's in this P block, right? And all right, and it's in our one, two, three, four, N equals five sub-level. But it has, oops, it has 49 electrons that we need to work with, okay? And then what do we need to do? We need to start filling. orbitals of increasing energy. All right, so n equals 5. So let's start. So we want to do our full electron configuration first. So we're going to go 1s2 and 2s2, okay, because now... We're here. Okay, let's see a little bit as we proceed through. And then we're going to our P's. So 2, P, 6. That takes care of these guys. And then we're going to go 3S, 2, and then 3, P, 6. And now we're here. We've got 4S2. And now we're at 3D. 3d10 and then 4p65s2 and then this d shell this d sub shell all right again holds 10 electrons and you can see we're one past that So our 4D subshell is also completely 4. So 4D10 followed by one more electron. And we're now in the N equals 5, 1, 2, 3, 4, 5, 5P1. All right. And so this would be our full electron configuration. Now, what can we do? Well, we should check ourselves. we should have 49 electrons that are represented. Again, these superscripts tell us how many electrons are in each of these subshells. So we got 2, 4, we got 10, 12, 18, 20, there's 30, 36, 38, 48, and finally 49. So that is our full electron configuration for indium. Well, what would the noble gas configuration be? So what do we need to do? We need to proceed to, instead of n equals 5, let's look at the noble gas that is at n equals 4. We want the n equals 4 noble gas, and that will be our short angle, all right? And that was krypton, all right? So we can see krypton right there. And krypton would have, all right, this complete electron configuration. Okay, and so we can shorthand this to the electron configuration of Krypton. Again, these are our core electrons. We're going to have this distinguishing feature here, core electrons versus valence electrons. Those are our higher energy electrons occupying our higher energy orbitals. And so we can have this Krypton kind of abbreviation, followed by 5s2, 4d10. 5, 1. All right. So again, this is our full electron configuration here, and this would be our noble gas configuration here. And what do we want to do? We can also draw the orbital diagram for indium, or in this case, let's draw it for cobalt. Okay. So let's find cobalt on the periodic table, and it should be in that first, it's right here. Okay. cobalt number 27. Okay, so that's in again n equals four. All right, so it's a transition metal. All right, and let's draw its electron configuration or its orbital diagram. Excuse me. So what we want to do again, draw this little kind of arrow going up again representing our energy. All right, because we want to occupy our lowest orbitals first, our lowest energy orbitals first. according to Hofwald principle. So we can say that the one S is here. So let's fill that, right? Then what are we gonna do? We're gonna go up to our two S and then our two P, then our three S and then three P. and of course this goes up again with our 3d. Again remember when n equals 3, l can equal 0, 1, or 2. This is our s, this is our p, our d. All right so we got our 3d orbitals right here. Okay well now again because of crossover All right, we can also go ahead and fill in our 4s. Excuse me. Remember, our 4s is slightly lower in energy, all right, than our 3d. We can even draw these exaggerated a little bit more. All right, so let's start filling. All right, so we look at the first n equals 1. All right, we got two electrons there. Then n equals 2. We got two s electrons, and we got six 2p electrons. So let's fill those in. We're going to partially fill each of these with the same spin. Then we're going to pair them. That takes us to 3s, so right here. All right, so we can fill these. And then we got our 3p electrons. So we can fill these. Partially occupy each first. Same spin. Then fill them. All right, and finally, we get us to 4s. We're going to fill those first. There's a crossover. They're lower in energy than the 3d orbitals. And how many electrons do we have to fill in our 3d? One, two, three, four, five, six, seven. One, two, three, four, five, six, seven. So this is our orbital energy diagram. for cobalt. Now again, notice because of Hund's rule, I partially filled each of those d orbitals within that d3d subshell first before I started to pair them up. And again, there's no inconsistencies here, all right, because cobalt is not chromium and it is not the copper group. So always remember the half filled or completely filled the orbitals, remember that that is stabilizing, leading to those consistent inconsistencies in our electron configurations. All right, with that, here's a participation question. The question is, which elements do the following electron configurations, whether it be the full electron configurations or the noble gas electron configurations? Okay, those shorthands, which electron configurations do these things represent? Which elements do they represent? All right. With that, all right, I hope you have a great rest of your day. And if you guys have questions, don't hesitate to reach out. I'm happy to help.

So now what we want electrons, and we want to start populating these different orbitals that are described by these four quantum numbers. So the one thing we need to really keep in mind all right, is something called the Pauli exclusion principle. This is essentially saying that in any given atom you cannot have two electrons that have the exact same set of four quantum numbers. Let's look at helium as an example. So remember we're in n equals one, all right, for both hydrogen and helium.

Well what can we have, all right, for our angular momentum quantum number? So remember our rules, all right? L can be anywhere from zero up to n minus one. So in this case, L only can have one possible value, and that is zero. And remember that for each angular momentum quantum number that we have, we have a letter designation.

So for zero, this is our S subshell. Again, this is a shell. this designates our sub-shell.

And then from there, what can we have? We can have our our magnetic quantum number, and that can be anywhere from minus L through zero up through positive L. Well in this case, all right, our epsilon has to equal zero as well. Okay, this tells us essentially the orientation of the orbitals within a given subshell and it also tells us about the number okay of orbitals so for l equals zero all right we are in the s subshell and we can only have one magnetic quantum number all right and that's zero and that tells us that again we have one orbital within our s subshell so we have an s orbital within an s subshell and it only has one orientation remember these are spherical in nature, one of the easiest ones to kind of visualize.

And so what do we do? We want to start populating this. Well, again, hydrogen only has a single electron, okay? And so if we're thinking about the fourth quantum number, which is our spin, all right, our spin can either be plus one-half or minus one-half.

And it doesn't really matter which one it is. convention usually says we start to use our positive one-half spins first, all right, but these things can flip on each. So a positive one-half spin can end up being a minus one-half spin. They kind of flip back and forth, all right.

Well, how would we represent that? All right, if we're doing with our positive one-half spin, all right, we would indicate it with this partial arrow pointing upward. So this particular electron has four quantum numbers.

of one, zero, zero, plus one half. Well, helium shares essentially most of this being the same, right? Because now we have two protons, we have two electrons.

So N equals one for both of them, or L, as we already saw for hydrogen, equals zero. for both of them, and our m sub l equals zero for them as well. So up until this point, all right, for, let's say this is electron number one, this is electron number two, you would say, well, they have the same quantum numbers.

They do, but remember, For any given orbital, we can have two electrons, okay? We can have two electrons, only two electrons. And they have to have...

opposite spins from each other. So we got to go one step further and say that this electron can have our plus one half spin. So we can draw that up here.

But then our other electron for number two would have to have, because again, the Pauli exclusion principle, no two electrons in any given orbital can have the same four quantum numbers. And this is in fact, any electron within an atom, okay? And so this has to be equal to minus one half.

And so we would populate it this way, and this should be like that, represent like that, okay? So again, this is a Pauli exclusion principle, all right? And so what we want to do is we want to start essentially taking these different orbitals, all right? We got our shells, we get our subshells. all right, and we get the orbitals within these given subshells, okay, and we want to start populating them with electrons, right.

So what we can do is we can start to move on past the first period of the periodic table or our n minus one and we can move up to say our n minus two and how would that look? Well we already said, okay, this again is looking at our n minus one and so our hydrogen all right we would populate a 1s with a single electron okay so here this is what we would call our electron configurations over here and this is our kind of orbital and they should reflect one another okay so this would be our hydrogen and then when we get to helium what would we do we would finish up with that shell. Okay. Where within that shell, we have one S subshell and one S orbital within that subshell, which carries two electrons, again, opposite spins, meaning that they have four different quantum numbers. All right.

Or the set of four quantum numbers is different between the two. All right. Well, let's move to N equals two.

Well, when N equals two, what can our momentum be? 0 and 1. All right. That means now we have two different subshills. We have S and we have our P.

Okay. And remember S only has a magnetic quantum number of 0, whereas P. Okay.

So this is, so if we're looking at M sub L, this would be 0. Okay. And for P, it would be minus 1, 1. and zero. Okay, so for our S subshell, again we expect only one orbital, an S orbital, okay, with one orientation. Whereas with our P, you see three blocks. Well this explains why, okay, because we have three different P orbitals, each of them having a different orientation, okay.

And again, each of these orbitals can hold two electrons with opposite spins. Again, uh no two electrons within an atom can have the same four quantum numbers describing it all right and so since we have uh these three different orbitals each can hold two electrons all right we can have six electrons populating uh the this 2p subshell all right so if we look at lithium all right what we want to do is now we're on the 2s and so we've added again an extra proton for lithium but we've also at our gained our extra electrons. So what we can do is we can start to populate our energy level diagram. The first two electrons for hydrogen and helium will go into our 1s, like we already discussed.

And then the next electron that's introduced when we get up to lithium, all right, is going to go into our 2p orbital. Notice we're populating the lowest energy, all right, subshell first within any given shell, okay. And so again, remember that the M tells us about our our energy and our size and L tells us about our shape. But in addition to that, this also tells us about the types of subshells that we have.

And generally speaking, N, our S, subshell, is lower energy than our P, is lower energy than our D, and lower energy than our S. So what we want to do is we want to populate the S subshell first when we're talking about lithium. And then after that's completely populated, we'll move to our p-orbital.

All right, well, how about beryllium? Well, beryllium, again, we've now gained an extra proton and an extra neutron or electron, excuse me. So where are we going to put that? Well, now, okay, we're going to finish out our lowest energy subshell first.

All right, the orbital within our lowest energy subshell. And so it's going to look like this. All right. And now what are we going to do? Let's go to boron.

Well now in this case, we now have a 2p1 electron, all right, because again our 2s orbital is now full. Okay, so the question is where is this electron going to go for boron? Okay, well the fact of the matter is it doesn't really matter.

Okay, is it going to be this guy here, this one, or this one? It doesn't matter because these are called degenerate orbitals, all right? All three of these p orbitals within this p subshell, all right, have the exact same energy.

That's what degenerate means. So when we're talking about this extra electron that's introduced because we've moved now to boron, all right, it can go to any one of these p orbitals within this p subshell. So if we look, all right, By convention, we typically start with the one that's in the lower or the left-hand side, and we usually give it the up spin, all right, or the plus one half ms quantum number. So now we can start to fill in our orbital energy diagram. So there's our 1s completely full.

Now we're moving up in energy because, again, the principle is we want to fill up our lower energy subshells first, and then what do we do? We fill up our... 2s and now we are finally into r2p okay all right so now that takes us to boron all right well what about the next element that's carbon okay so now the question is we've gained an extra all right 2p electron that's what the 2p2 again tell just what these superscripts mean that mean this is the number of electrons in that particular orbital, okay, or that particular subshell.

So the question is, where is the second P electron going to go, all right? So what you can do, we already know that from boron, one of these is already populated, all right? So where's that second one going to go? Well, we can think of this like the bus, all right? You get into a bus or you enter into a bus and you have these three long seats.

all right, one of them is occupied and two of them are not, all right. So what are you inclined to do, all right? You're inclined to probably go into one of the empty seats, all right, and that is in fact what happens in this case.

We populate one of those extra orbitals, one of those empty orbitals, all right, and the reason we do this is something called Hun's Rule, all right, which is a pretty simple rule. All right, and what it says is that the most stable arrangement of electrons and orbitals, okay, of equal energy, which again we have three p orbitals, all have equal energy. So the most stable arrangement of electrons and orbitals of equal energy is the one in which the number of electrons with the same spin is maximized. Okay, and so in other words, the only way we can do that, right, is if we start to half fill each of these orbitals first.

Okay, so let's move forward a little bit and fill in the energy level diagram. So that takes us through helium, this takes us through beryllium, and now we're boron and carbon. So what we want to do is maximize the number of electrons, okay, which have the same spin. Again, remember, this is representing our plus one half.

So now, if we would have populated this initial orbital right here, we would have had to have had an up and a down spin, or a plus one half and a minus one half. Well, if we do that, We are in blatant disregard for Hund's rule, okay, which again states that we want to kind of fill in our orbitals, all right, where we maximize the number of electrons with the same spin, all right, in our given subshells, all right. And so again, that's Hund's rule.

That tells us that we want to keep our electrons unpaired. Until each of those orbitals within a given subshell, all right, are all halfway filled. And when they're halfway filled, they all have to have the same spin.

All right. All right. So that takes us through carbon.

All right. Let's look at nitrogen. All right.

Well, in this case, all right, we have to, again, consider Hund's rule. Now we have three P electrons. And you may say, well, how are we going to occupy?

All right. Or where are we going to put that last, that third electron in that? in that p subshell? Well, again, we want to follow Hunt's rule where we maximize, all right, the number of electrons with the same spin, okay?

And so we're going to half fill each of those orbitals within that 2p subshell. All right, so we can fill out our energy level diagram again. Now each of these has a plus one half spin, all right? And we can do oxygen, okay? Well, now what are we going to do with this now fourth electron in that P subshell?

All right. Well, now what we're naturally going to have to do is we're going to have to start to pair electrons. OK, we're going to start pairing our electrons. And so what would this one look like?

Now we have this guy is completely full of this. this 2p orbital within this p subshell. Both of these electrons have, again, different quantum numbers that describe them. Well, the first three are going to be the same, but that last quantum number, the spin quantum number, again, one is going to have a one-half and the other one is going to have a minus one-half spin. Okay.

So we start to fill in, all right. After each of our orbitals are halfway filled, now we start to pair our electrons. And again, with this idea that any given orbital can only have two electrons in it. Again, what's different about them? They have different spin quantum numbers.

And then we get to fluorine, right? And now what we see, we have this fifth one. So now we have, all right, two of these P orbitals within this P subshell that are filled.

And they fill out our P. orbital diagram here, energy diagram, right, until finally we get to neon, where we have six p electrons to fill up, all right, these p orbitals within this p subshell. And now what we have, all of those orbitals, those p orbitals, are filled with electrons. Again, Pauli Exclusion Principle says that each of these electrons has to have their own unique set of quantum numbers. Okay.

All right. So essentially, what are we doing here? All right.

We're filling these orbitals, okay, according to what we call off-balls principle. Off-balls principle. And this is the principle that states that as protons are added, okay, so as you see, as we go from hydrogen all the way down, all right, to neon, All right, what have we done? All right, we've changed the identity of the element, and we've done that because we've added protons in each case, all right, as we proceed through the periodic table. So as protons are added one by one to the nucleus to build up that given element, all right, electrons are similarly added, all right, to these hydrogen-like orbitals.

Okay, so that, again, is the off-ball principle, and what it's saying, again… is that essentially what we want to do is we want to start adding electrons, all right, from the lowest energy orbital to the highest energy orbital. And you can see that that's what we've done. What did we do?

We go from the 1s, again, that's the lowest energy level, then followed by the 2s, all right, followed by the 2p. And then as we would proceed, all right, to our third period. where our quantum, our principal quantum number is n equals three, what would we do?

We would start to fill the 3s. Again, notice here on our y-axis, essentially, all right, is energy, all right? So the 3s is the next lowest, followed by the 3p, and so on and so forth.

We start to see some interesting things, all right, when we get to the 3d, and we'll talk about that in just a second. So the question is, how are you going to remember this off-ball principle, this filling order, if you will? Well, there's a couple of different ways that we can do this. One way is to use the periodic table. Many times, like...

as shown here, you'll see that the periodic table is essentially kind of sectioned off into four different regions, all right, and each of these regions has its own unique color, all right. Well if we kind of, and this is these different colors are essentially representing our different kind of L, our angular quantum numbers, angular momentum quantum numbers, so our subscript. So again, L equals zero, one, two and three these are the only ones that we will ever deal with in this class okay where l is our s okay sub shells one is our l or excuse me p sub shell two is our d sub shell and three represents our f sub shell well you can see again that these are color coded accordingly all right so here's our essence All right.

These are P's here. These are D's and these are F's. And so what you can see if we kind of zoom in on this a little bit is that they're ordered. OK, according to their energy levels, because, again, remember, S is lower energy than P, lower energy than D.

and lower energy than our F. All right, so this is the order we want to fill these in. So again, if we look at N equals one, we only have an S subshell.

So of course, we're gonna fill that first. But then as we go to N equals two, all right, remember L can equal zero or one. So we'll have an S and a P subshell, which we do.

Here's our S, here's our P, all right? And so what we're going to do is we're going to populate the S first, and then we're going to move to populating our P orbitals within our P subshell. Then what happens when we get to N equals 3?

Well now we see something interesting. N equals 3 means that L can equal to 0, 1, or 2. This is our S, E, and our D. All right, but in this case what we're going to do is we're going to populate the 3S, all right, like we have before. All right. It's always the lowest energy of our subshells.

Then we're going to populate our 3P. OK, so 3S, then 3P. And then what are we going to do? Well, we're going to skip our 3Ds.

All right. They're there because this principal quantum number of three. All right.

But what's interesting here is that we're going to fill again, again, our four. We're going to fill our 4S first because N equals 4, L equals 0, 1. or three all right so again s p d and f all right so we're going to fill the 4s prior to the 3d okay and then we're going to fill in our four p's well the question is why are we filling in 4s before our 3ds because if we're looking at this um this kind of pecking order up here, you might think that you'd want to fill in your 3Ds first. But again, the periodic table is kind of organized, all right, according to these occupations or how the electrons occupy the different orbitals, all right? But the reason why we fill in our forest versus our 3D first, all right, is because of something called crossover, all right?

When we start to increase the energy, all right, from one shell to another, sometimes we get this crossover, all right? So again, Remember, we're on our third principle quantum number, n equals 3. We can have a 3s, we can have a 3p, and a 3d, all right? But then when we get to n equals 4, we can have our 4s, our 4p, our 4d, and our 4f, all right?

Well, what you can see, though, the way these are organized on this energy kind of diagram here, that, of course, our 1s is the lowest, then, of course, our 2s all right, is the next lowest followed by our 2p. And then finally, what do we see? All right, we see that we get our 3s and then our 3p. So just like we'd expect.

And then what do we see about the 4s and the 3d? It is crossing over the 3d subshell. Actually is at a higher energy level, all right? than the 4s subshell. Okay, so what we do is we occupy the 4s subshell, that 4s orbital, before we occupy those 3d orbitals.

All right, and again this is called crossover. All right, but the way the periodic table is organized, it kind of lends itself to you being able to see this pattern. Okay, the one thing you've got to remember is that don't mess up and actually say that this is the 4D subshell because it's not. It's the 3D. All right.

Because again, remember this principal quantum number of three tells you that it is going to have these 3D orbitals associated with it. So what do we do? We proceed from the 4S to the 3D because of crossing over, and then we can follow our trend. All right. So 4S.

to the 3d and then what are we going to do we're going to move up to our 4p okay which is right here and then the this kind of pattern kind of continues on for our fifth period all right and then finally all right you see in the sixth period we actually start to see the introduction of these 4f electrons because again when we get to four all right when we get to four that tells us that we should have this F subshell, which has these F orbitals, but we do not start to populate them until we actually get to the sixth period on the periodic table. So, again, this crossing over leads to some, I guess you would call them, unexpected filling of energy levels, but again, the periodic table kind of lends itself to allowing you to figure out which of these orbitals we're going to fill first when we're following this off-ball principle. Again, off-ball principle telling us that we want to start from our, start filling our lowest energy orbitals first and proceed to the higher energy orbitals. The other one, all right, is this kind of schematic here.

This is another way to remember kind of the filling order that might jive with you. All right, so what you do, all right, is you kind of start with your 1s, and you label the different s orbitals going downward, all right? Every period is going to have that. Then, of course, remember if we're in n equals 1, we don't have any p orbitals.

So what we do is we go down. This is essentially n equals 2, and n equals 2 tells you you're going to have an f subshell and a p subshell. n equals 3 is now going to have an s, a d, all right? And what we do, okay, so we can kind of skip this spot here, then start with our P's, go down in our P's, and then skip one here.

All right, start with our D's, and then go down in our D's. And then finally, we're going to skip this and start with our F's, okay? And so how do, how is this going to help us? Well what we want to do is we want to kind of fill the diagonals with our electrons and then imagine that we're going to come up here and we're going to start to fill the next diagonal which would be R2s, right?

And then again we can kind of go back and we can fill in this next diagonal, right? So on and so forth. And so then we go 3P, 4S, and finally 3D, 4P, 5S, so on and so forth. So you got to pick one of these two things, all right, that's going to help you remember kind of this...

this ordering of these orbitals in terms of their energy. Because again, the off-ball principle states that we have to be able, or we have to fill, all right, these orbitals from lowest energy to the highest energy, okay? And so you want to pick one, all right, that again, that jives with you. All right. So as you can see though, all right, we left off with neon specifically, all right, with this particular electron configuration.

And as you see, as we proceed through the elements, what happens? These, these Electron configurations with these elements with bigger atomic numbers, they start to get really, really long, quite repetitious and quite cumbersome. OK, so what we do is we have a shorthand and it's called a noble gas configuration.

So as we increase in these energy levels, all right, we also increase in the size of our of our elements. OK, so remember, again, you can kind of think of. as we move through these principal quantum numbers, or as we move down the periodic table from the first period where n equals one, all the way down to the seventh period where n equals seven, all right? You can kind of think of these things as shells, okay?

And as we proceed through, what do we do? We kind of fill up the shells, and then we start over, all right? Again, this gets very repetitious.

So what do I mean by that? Well, here's neons, all right? Full electron configuration. We fill up the 1s2. then we fill up 2s2 and then 2p6, all right?

Then as we move from neon to sodium, all right, it has the same, okay, initial electron configuration. This 1s2, this 2s2, this 2p6, all right? Well, what does that represent? Well, that represents neon, all right? But now we have this extra electron.

Where's it going to go? Well, it's going to go into this 3s. sub-shell, this 3s orbital.

Again, that's what that one tells us. We're going to put one electron there. But what can you see?

This kind of neon configuration essentially remains the same throughout this whole period of the periodic table, from sodium to magnesium, aluminum, silicon, phosphorus, sulfur, chlorine, and argon. What do we see? All of these guys essentially share the same core electron configuration. So what do we do? We kind of shorten this, and we say that this is, all right, our core electron configuration.

We have a core electron configuration that matches that of neon. And so these are what we'd be calling our core electrons, right? Those are all those electrons that kind of fill up that single shell.

And now we can get this distinction between core electrons and what we call valence electrons. And these valence electrons, these are outer shell electrons, all right? These are the electrons that are occupying these orbitals and these outermost energy levels, these higher energy levels, all right?

So we get these core electrons, and now we get our valence electrons, okay? And so what we can do is we can proceed through. from sodium to magnesium. Instead of saying for magnesium, we have 1s2, 2s2, 2p6, a 3s2 electron configuration.

We can use our noble gas configuration and say we have essentially the core electrons, all right, from the neon, all right, and followed by these valence electrons, these 3s2 electrons, okay? So again, we get this concept of core electrons and valence electrons. We're kind of distinguishing or kind of categorizing two different types of electrons, all right?

And again, we can proceed to aluminum, where again, we have this 1s2, 2s2, 2p6, 3s2, 3p1, all right? Again, this first part, the 1s2, 2s2, 2p6, that again is our core electrons followed by what we're showing here in our noble gas configuration, followed by our valence electrons, 3s2, 3p1. And what we can do is we can proceed all the way down through argon, where argon now has this neon core, these kind of core electrons, and then our valence electrons are 3s2, 3p6, all right?

Or what can we do? We can say that this now is another set of core electrons that we filled. And so instead of using this kind of...

configuration with neon. Now what are we going to be if we proceed past argon? We're going to start with this argon being in parentheses, this AR being the elemental symbol for argon being in parentheses. Again kind of distinguishing that we filled up a whole shell.

We filled up all those sub shells, all right, all those orbitals within those sub shells within that particular period of the periodic table. All right so we got full electron configurations. And we also have noble gas configurations.

And then finally, what are we not really talked about? Well, we haven't talked about the d orbitals as of yet, okay? The d's or the f's. So we'll talk about the f's last, all right? And so what you might imagine, all right, is that, again, here's argon.

This is where we left off, okay? So what we can do is we can say, put our parentheses here, and that notation, this argon within parentheses, also is equivalent to this full electron configuration, all right? Again, because it's shorter, and again, these full electron configurations can be quite repetitious and cumbersome, what we can do is we can shorthand and use our noble gas configurations.

Now we're at N equals 4. We're at potassium. As you might expect, we've talked about this already, because of crossover, We fill up the 4s electrons before we start to fill the 3d orbitals. Okay.

Again, this is crossover. These, the 4s subshell, that 4s orbital. All right. And those 3d orbitals, again, there are five of those.

All right. They're really close energy levels, but the 4s orbital is lower than those d orbitals so we fill that first and again this this falls out very naturally from the periodic table okay so we're going to fill those 4s all right um or uh orbital first all right and that's going to come from those electrons from potassium and calcium and then what are we going to do we're going to get into our again these are our transition our transition levels called the D block, all right? Because again, now we're going to start talking about our D subshell, all right?

Our D orbitals, okay? And what do we get? We get 3D1, we're talking about scandium. And then again, as you expect, we would then get for titanium, we would get 4S2, then 3D2, vanadium, 4S2, 3D3. And then what do we do?

We come to... one of our two inconsistencies that we see with the transition, all right, we see that chromium and copper have some very interesting, all right, electron configurations, all right, because what would we expect for chromium, all right, for chromium, we would expect to have a 4s, and then again, and slightly higher, energy, talking energy here, all right, we'd have our 3d orbitals, all right, and so what we'd expect if we're following the same sort of pattern is that the four acids would get filled, and then finally 3d, again, because of Hunt's rule, we're going to try and maximize the number of electrons with the same spin, all right, we would expect this, okay, we would expect four of those d orbitals to have a single electron. in them. All right.

But if we look at chromium, is that actually what we see? No, we don't see that at all. Right.

We see that we have 4s1 and then we have 3d5. So this is a very consistent, kind of a weird way of saying it, but a consistent inconsistency within the transition metals. All right.

Well, why is it this 4s1 versus this 3d, 4s1, 3d5? versus 4s2 3d4. Well, what you can see here is that there's this kind of open seat at the table, all right?

And what we do, instead of having this configuration, all right, this 4s2 3d4, is we can move one of those electrons from the s, okay, into this unoccupied 3d orbital, and that gives us 4s1 3d5. Now, when does this happen? Well, remember that these energy levels are pretty close to each other.

You can go back here and take a look. All right. They're pretty darn close to each other. But through a lot of research, people have found that having half filled, one half filled d orbitals is stable.

Thank So having half-filled d-orbitals is stabilizing. So in other words, because we have this 4s1 3d5 and those five d-orbitals are half-filled, that is actually more stabilizing than having this 4s2 3d4 kind of electron configuration. And so that is a very consistent trend that you see with this group that's kind of headed off chromium. Okay.

So it proceeds downwards into the fifth and the sixth periods. All right. And so again, a consistent inconsistency.

Well, now what we also see, all right, going from manganese to nickel is exactly what you would expect. All right. We would have our 4s2, 3d5, then 4s2, 3d6, and then 4s2, 3d7, until finally...

We get to nickel, which is 4s2, 3d8, again, following the expected kind of pattern. But then we get to copper. All right, copper, this group headed by copper, also provides another example of a consistent inconsistency in the periodic table. Because, again, what would we expect for copper? We would expect, again, if we're going, trying to draw something fairly accurate here, all right, we would have 4s2.

followed by our 3d's and energy here. We would expect to be have nine of these filled. Notice I'm filling them according to Hunt's rule. We're maximizing the number of electrons in these orbitals that have the same spin.

We would expect something like this a 4s2 3d9 but in fact All right. Experimentation says that it's actually 4s1 3d10, right? Because what's happened again? Well, again, these orbitals are really close to each other. So what we can do is we can promote one of these 4s electrons in here to this final d orbital so that all these d orbitals are actually filled with electrons, right?

And this is what they found, all right, for this whole family. Again, a consistent inconsistency in our... electron configurations for these transition metals. All right, well, why is that? Well, in addition to these half-filled d-orbitals, they have also found that completely filled d-orbitals are also stabilizing.

And so again, that's why in this one case, what we can do is promote one of these 4s electrons into a d orbital to give us for this this copper family if you will this 4s1 3d10 electron configuration so again these are just some inconsistencies that are really important and it's fairly simple to explain when when you think about the fact that they've determined that half filled all right when all your d orbitals are halfway filled or all your d orbitals are completely filled that provides extra amount of stabilizing energy. Okay. Now, finally, it takes us to our F.

All right. All right. Our F subshell or F orbitals.

Again, this is an F subshell. Again, what are we talking about here? This is where L equals to three.

And what do we have here? Magnetic quantum numbers can be minus 3, minus 2, minus 1, 0, plus 1, plus 2, and plus 3. So we have an F subshell that has seven different F orbitals, again, all having different orientations. And so what are we going to do?

When we get down here, this is where we're on the sixth period. So. So n equals 6, all right, and n equals 7, all right?

But again, we get these all the way up here in period 4 where n equals, all right, because n equals 4, again, just to review, l equals 0, 1, 2, 3, all right? So that tells you that our f orbitals, that's where they're coming from. So that's where the 4 comes from.

all right it's from that principal quantum number all right so what do we do we got um lanthanum and actinem and what we're going to do we're going to fill those with one d electron before finally we start to take our electrons and filling up these seven, okay, these seven F orbitals within this F subshelf. Again, there are 14 of these, all right? That makes some sense because we, again, we have seven, seven orbitals.

They can all hold two electrons each. Again, plus one half, minus one half, right? So we have 14 electrons that we can use to fill these F orbitals. All right, so again, there's one, two, three, da, da, da, da, all the way through 14 different elements for each of these periods, okay? So, lanthanides have 14 elements within them.

Our actinides have also 14 elements within them. That means we are, you know, kind of gaining these 14 different electrons in both of these cases so that we can fill these seven F orbitals, okay? And so it makes a lot of sense now, but when we look at the electron configurations, all right, of these F subshells or these F orbitals within the F subshell, there are all kinds of inconsistencies.

Now for an exam, all right, this is important. We will never ask you one of these about, or expect you to know these inconsistencies. And the inconsistencies are all represented here by these red arrows. Okay. So you don't need to know those.

But what we do want you to know would be the ones where it follows the given pattern. All right. So, for example, you can see cesium here. It's a success to 5D1, which that comes from lanthanum right here. That's the 5D1.

And then we get 4F1. So that follows the pattern. OK, that follows a pattern.

That's something we could ask you on an exam. All right. This guy, number 64. All right.

Also follows the pattern. So on and so forth for these ones that are not highlighted with this red number. All right. So, again, watch out for inconsistencies like with a textbook question. They could ask you those sorts of things.

But on an exam, we would only ask you ones that actually follow the kind of pattern, this off ball principle pattern that we've been talking about. All right. So let's try a sample problem. So write out the full electron configuration. So the full one, which again, those can get to be cumbersome.

All right. And then the double gas configuration for indium. All right. And so what do we need to do first? The first thing we need to do is we need to find indium on the periodic table.

So let's go take a look. Where's indium at? All right.

Let's see. And then it is number 49, okay? So it's number 49. It's in this P block, right?

And all right, and it's in our one, two, three, four, N equals five sub-level. But it has, oops, it has 49 electrons that we need to work with, okay? And then what do we need to do?

We need to start filling. orbitals of increasing energy. All right, so n equals 5. So let's start.

So we want to do our full electron configuration first. So we're going to go 1s2 and 2s2, okay, because now... We're here. Okay, let's see a little bit as we proceed through.

And then we're going to our P's. So 2, P, 6. That takes care of these guys. And then we're going to go 3S, 2, and then 3, P, 6. And now we're here. We've got 4S2. And now we're at 3D.

3d10 and then 4p65s2 and then this d shell this d sub shell all right again holds 10 electrons and you can see we're one past that So our 4D subshell is also completely 4. So 4D10 followed by one more electron. And we're now in the N equals 5, 1, 2, 3, 4, 5, 5P1. All right. And so this would be our full electron configuration.

Now, what can we do? Well, we should check ourselves. we should have 49 electrons that are represented. Again, these superscripts tell us how many electrons are in each of these subshells. So we got 2, 4, we got 10, 12, 18, 20, there's 30, 36, 38, 48, and finally 49. So that is our full electron configuration for indium.

Well, what would the noble gas configuration be? So what do we need to do? We need to proceed to, instead of n equals 5, let's look at the noble gas that is at n equals 4. We want the n equals 4 noble gas, and that will be our short angle, all right? And that was krypton, all right? So we can see krypton right there.

And krypton would have, all right, this complete electron configuration. Okay, and so we can shorthand this to the electron configuration of Krypton. Again, these are our core electrons.

We're going to have this distinguishing feature here, core electrons versus valence electrons. Those are our higher energy electrons occupying our higher energy orbitals. And so we can have this Krypton kind of abbreviation, followed by 5s2, 4d10.

5, 1. All right. So again, this is our full electron configuration here, and this would be our noble gas configuration here. And what do we want to do? We can also draw the orbital diagram for indium, or in this case, let's draw it for cobalt.

Okay. So let's find cobalt on the periodic table, and it should be in that first, it's right here. Okay. cobalt number 27. Okay, so that's in again n equals four.

All right, so it's a transition metal. All right, and let's draw its electron configuration or its orbital diagram. Excuse me. So what we want to do again, draw this little kind of arrow going up again representing our energy. All right, because we want to occupy our lowest orbitals first, our lowest energy orbitals first.

according to Hofwald principle. So we can say that the one S is here. So let's fill that, right? Then what are we gonna do? We're gonna go up to our two S and then our two P, then our three S and then three P. and of course this goes up again with our 3d.

Again remember when n equals 3, l can equal 0, 1, or 2. This is our s, this is our p, our d. All right so we got our 3d orbitals right here. Okay well now again because of crossover All right, we can also go ahead and fill in our 4s.

Excuse me. Remember, our 4s is slightly lower in energy, all right, than our 3d. We can even draw these exaggerated a little bit more. All right, so let's start filling.

All right, so we look at the first n equals 1. All right, we got two electrons there. Then n equals 2. We got two s electrons, and we got six 2p electrons. So let's fill those in. We're going to partially fill each of these with the same spin.

Then we're going to pair them. That takes us to 3s, so right here. All right, so we can fill these.

And then we got our 3p electrons. So we can fill these. Partially occupy each first. Same spin.

Then fill them. All right, and finally, we get us to 4s. We're going to fill those first.

There's a crossover. They're lower in energy than the 3d orbitals. And how many electrons do we have to fill in our 3d? One, two, three, four, five, six, seven.

One, two, three, four, five, six, seven. So this is our orbital energy diagram. for cobalt.

Now again, notice because of Hund's rule, I partially filled each of those d orbitals within that d3d subshell first before I started to pair them up. And again, there's no inconsistencies here, all right, because cobalt is not chromium and it is not the copper group. So always remember the half filled or completely filled the orbitals, remember that that is stabilizing, leading to those consistent inconsistencies in our electron configurations. All right, with that, here's a participation question. The question is, which elements do the following electron configurations, whether it be the full electron configurations or the noble gas electron configurations?

Okay, those shorthands, which electron configurations do these things represent? Which elements do they represent? All right. With that, all right, I hope you have a great rest of your day. And if you guys have questions, don't hesitate to reach out.

I'm happy to help.

Transcript for:Understanding Electron Configurations

Transcript for:
Understanding Electron Configurations