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Video transcript

- [Voiceover] We've already looked at the electron configurations for potassium and for calcium but let's do it again really quickly because it's going to affect how we think about the d orbitals and so we find potassium which is in the fourth period on the periodic table. If we do noble gas notation to save some time, we work backwards and the first noble gas we hit is argon, so we write argon in brackets. We know argon has 18 electrons and potassium has 19 electrons. Potassium has one more electron than argon and so we put that extra electron into a 4s orbital because for potassium the 4s orbital is lower energy than the 3d orbitals here. We have increasing energy and that electron goes into a 4s orbital so the complete electron configuration using noble gas notation for potassium is argon in brackets 4s 1. For calcium I should say. All right, we have one more electron then potassium and so that electron's going to go into the 4s orbital as well and so we pair our spins and we write the electron configuration for calcium as argon in brackets 4s 2. For the calcium two plus ion, so if you're thinking about forming an ion here, we're talking about the electron configurations for a neutral atom meaning equal numbers of protons and electrons. With the atomic number of 20, 20 protons and 20 electrons. If we lose two electrons, we have a net deposited two charge. We form the calcium to ion. The two electrons that we would lose to form the calcium two plus ion are these. These two electrons right here in the 4s orbital. The electron configuration for calcium two plus would be the same as the electron configuration for the noble gas argon here. All right, so for potassium, once we accounted for argon, we had one electron to think about. For calcium, once we counted for argon we had two electrons to think about. If we go to the next element on the periodic table, that's scandium. That's one more electron and calcium. We have three electrons to worry about once we put argon in here like that. This is where things get weird. Now we have to think about the d orbitals and once again things are very complicated once you hit scandium because the energies change. When you hit scandium even though these are very small energy differences, now the energy of the 4s orbital is actually higher than the energy of the 3d orbitals. We're talking about once again increasing energy and so that's pretty weird. All right, so if you think about these three electrons, where are we gonna put them? Well your first guess, if you understand these energy differences might be, okay, well I'm gonna follow Hund's rule. I'm gonna put those electrons in the lowest energy level possible here and I'm going to not pair my spins and so I'm going to write my electron configuration like that for scandium. You might think it would be argon 3d 3 but that's not what we observed for the electron configuration for scandium. Actually two of these electrons actually move up to the higher energy orbital so two of those electrons move up to the 4s orbital here like that. The electron configuration turns out to be 4s 2, 3d 1. It's actually 4s 2, 3d 1 or if you prefer 3d 1, 4s 2 once again with argon in front of it. Either one of these is acceptable. This is weird so like why did those electrons, why did those two electrons go to an orbital of higher energy? There's no simple explanation for this. All right, so even though it might be higher in energy for those two electrons, it must not be higher energy overall for the entire scandium atom. There are many other factors to consider so things like increasing nuclear charge. The scandium has an extra proton compared to calcium and then there are once again many more factors and far too much to get into in this video. Unfortunately there is no easy explanation for this but this is the observed electron configuration for scandium. How do we know this is true? How do we know that the 4s orbital is actually higher energy than the 3d orbitals? We know this from ionization experiments. For example if you form the scandium plus one ion, the electron configuration for the scandium plus one ion, so we're losing an electron from a neutral scandium atom. This turns out to be argon 4s 1, 3d 1 or once again you could write argon, 3d 1, 4s 1. Where did we lose that electron to form our ion? We lost that electron from the 4s orbital. We had 4s 2 here and here we have 4s 1. We lost this electron and that only makes sense if the 4s orbital is the highest in energy because when you lose an electron for ionization, you lose the electron that's highest in energy. That's the one that's easiest to remove to form the ion. The 4s orbital is actually higher in energy than the 3d orbitals. You don't see this a lot in text books and I think the main reason for that is because of the fact that if you're trying to think about just writing electron configurations. Your goal is to write, let's say you're taking a test and your goal is to write the electron configuration for scandium. The easiest way to do that ... Let me go ahead and use red here. The easiest way to do that if you want to write the electron configuration for scandium, you look at the periodic table and if you're doing noble gas notation, the noble gas that precedes it is of course argon right here. That takes care of the argon portion and then looking at the periodic table you would say this could be 4s 1, 4s 2, 3d 1. That gives you the correct electron configuration, argon 4s 2, 3d 1. But it's implying that the d orbitals, the 3d orbitals fill after the 4s orbital and is therefore a higher energy and that's not true actually. It does help you to just assume that's the case if you're writing an electron configuration but that's not what's happening in reality. We need to think about the other elements here. We just did scandium. Next let's move on to titanium. Thinking about titanium, so the next element in the periodic table if your question on the test was write the electron configuration for titanium, the easiest way to do it is just once again to think about argon. Put argon in brackets and then think to yourself, this would be 4s 1, this would be 4s 2, this would be 3d 1 and this would be 3d 2. You could write 4s 2 and then 3d 2 or once again you could switch 3d 2 and 4s 2. Once again this is implying the d orbitals fill after the 4s orbital which isn't true but it does get you the right answer. It's useful to think about it both ways. It's useful to think about the energy levels properly but the same time if your goal is to get the answer the fastest way possible, looking at the periodic table and running through the electron configuration might be the best way to do it on test. Let's look at some of these other elements here so we've just talked about scandium and titanium. All right, so let's go down here. Let's look at this little setup here. All right, so we just did scandium and titanium. All right, so scandium was argon 4s 2, 3d 1. We talked about two electrons in the 4s orbital, one electron in the 3d orbital. We just did titanium 4s 2, 3d 2 or once again you could switch any of these. When you're doing orbital notation, adding that second electron to a d orbital. Here's the electron that we added so we didn't pair up our spins. We're following Hund's rule here. Next element is vanadium so we do the same thing. One more electron, we add that electron to a d orbital but we add it to, we don't add it to one of the ones that we've already started the fill here, we add that electron to another d orbital, so once again following Hund's rule. Things get weird when you get to chromium. Let me use a different color here for chromium. If you're just thinking about what might happen for chromium, chromium one more electron to think about than vanadium. You might think, let's just add that one electron to a 3d orbital like that and then be done with it. If you think about it, you might guess 4s 2, 3d 4. Let's go ahead and write that. 4s 2, 3d 4, so question mark but that's not actually what we get. We get 4s 1, 3d 5. That electron, this electron here, let me go ahead and use red. So you could think about this electron. We expect it to be there, we expect it to be 4s 2, 3d 4. It's like that electron has moved over here to this empty orbital to give you this orbital notation. All right, so that's just an easy way of thinking about it and in reality that's not what's happening if you're building up the atom here because of the different energy levels. But just to make things easier when you're writing electron configurations, you can think about moving an electron from the 4s orbital over to the last empty d orbital here. Some people say that this half filled d subshell, let me go and circle it here. This half filled d subshell is extra stable and that might be true for the chromium atom but it's not always true so it's not really the best explanation. The real explanation is extremely complicated and actually just way too much to get into for a general chemistry course. Next element is manganese. All right, let me go ahead and stick with blue here. Manganese, one more electron than chromium here. Chromium we had six electrons here, and manganese we need to worry about seven electrons. This is kind of what we expect, just going across the periodic table. Let me go ahead and do this for manganese. Let me use green here. You might say okay, that's 4s 1, that's 4s 2 and then 3d 1, 3d 2, 3d 3, 3d 4, 3d 5. You might say to yourself 4s 2, 3d 5. This precedes how we would expect it to. All right, and the same thing with iron, so 4s 2, 3d 6. All right, so that takes care of iron and once again now you can start to pair up your spins. We've seen that in earlier electron configurations. Next cobalt, one more electron to worry about. All right, so 4s 2, 3d 7 makes sense and you can see here would be the electron that we added and we paired up our spins again. Nickel, same trends. We add one more electron, 3d 8. That makes sense, here's the electron that we added and once again we got a weird one. All right, so when we get to copper. So copper you might think ... Let me use red for copper so we know copper's red. We think about it, writing one more electron. If we took the electron configuration here for nickel, we added one more electron. You might guess that would be the orbital notation for copper but that's not what we see. We've taken this electron here and moved it over to here, like that. This gives us a filled d subshell here. Once again one explanational see for that is extremely stable for copper and that might be true for copper. All right, and that leaves us only one electron here in our 4s orbital. The electron configuration is 4s 1, 3d 10 but all these general chemistry explanations are just a little bit too simple for reality but if you're just starting out, they're pretty good way to think about it. Then finally zinc, zinc makes sense. We're adding one more, writing one more electrons. We just took care of copper. For zinc we have one more electron and so you could think about this being 4s 2 right here and then we have 3d 10, one, two, three four, five, six, seven, eight, nine, 10. 4s 2, 3d 10 or 3d 10, 4s 2 with argon in front of it gives you the complete electron configuration and you can see, you've now filled your 4s orbital and your 3d orbitals like that. Once again pretty complicated topic and hopefully this just gives you an idea about what's going on.
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