If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

Electron configurations of the 3d transition metals

AP.Chem:
SAP‑1 (EU)
,
SAP‑1.A (LO)
,
SAP‑1.A.3 (EK)

The Aufbau principle predicts that the 4s orbital is always filled before the 3d orbitals, but this is actually not true for most elements! From Sc on, the 3d orbitals are actually lower in energy than the 4s orbital, which means that electrons enter the 3d orbitals first. In this video, we’ll discuss this in more depth and walk through all of the electron configurations for the 3d transition metals.

.
Created by Jay.

Want to join the conversation?

  • piceratops tree style avatar for user Michael
    At , Jay says that the 4s orbital is the "highest in energy." However Sal in a previous video said that while the 4s orbital would be the "furthest away," in fact the 3d orbital is the "highest energy." He said this at in the video "electron configuration for d block element" while doing the electron configuration for Fe. Which is correct?
    (90 votes)
    Default Khan Academy avatar avatar for user
    • spunky sam blue style avatar for user Ernest Zinck
      4s is higher in energy than 3d until you get to Ca. Then the relative energies of 4s and 3d switch.
      As you go from left to right across the Periodic Table, you are adding extra protons to the nucleus and extra electrons around the nucleus.
      The various attractions and repulsions in the atoms change as you do this. It is these attractions and repulsions that govern the energies of the various orbitals.
      For rather complicated reasons, once you get to scandium, the energy of the 3d orbitals becomes slightly less than that of the 4s.
      The "furthest away" argument is suspect. Nobody really knows how big a 4s or a 3d orbital is, except in an H atom where there is only one electron.
      (40 votes)
  • duskpin ultimate style avatar for user Utkarsh Sharma
    Why do Chromium and Copper behave so weirdly ?
    Moreover will their EC's be Cr:{Ar}4s1 3d5 only if we write it in actual ?
    Cu:{Ar}4s1 3d10 only if we write it in actual ?
    Moreover if this does not happen in real what is the point of showing and understanding that weird Orbital Notation ?
    (9 votes)
    Default Khan Academy avatar avatar for user
    • orange juice squid orange style avatar for user Luke Yakielashek
      Cr and Cu are the two exceptions of electron configuration of atoms up to Kr. this is because a 1/2 or completely full D block has extra stability, therefore in the case of Chromium one electron will shift from the 4s block to fill the 3d block to exactly one half
      Thus:
      Cr: [Ar] 4s2 3d4 actually exists as: Cr: [Ar] 4s1 3d5 and;
      Cu: [Ar] 4s2 3d9 actually exists as: Cu: [Ar] 4s1 3d10

      This idea holds true for Mo and Ag in period 5 and so on
      (32 votes)
  • blobby green style avatar for user Sharan
    How can we write the electronic configuration for an element if the periodic table is not given to us ?
    (5 votes)
    Default Khan Academy avatar avatar for user
    • piceratops ultimate style avatar for user Just Keith
      You must know the atomic number of the element. Without that, you cannot determine the electron configuration.

      Given the atomic number, there is a set of rules that allow you to determine the electron configuration. However, in the simplified version of these rules that is understandable to a general chemistry student, there are about 19 elements that are exceptions. These you just have to learn.

      What you do is you start assigning electrons to the subshells using the following pattern (you completely fill up one subshell before moving onto the next higher) and you keep going until the total number of electrons assigned is equal to the atomic number:
      1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p
      There are no known elements that, in their ground state, have electrons in a subshell beyond 7p. This pattern will give you the correct configuration for all but about 19 elements.

      The actual filling order is more complicated. For example, the 4s and 3d have nearly the same energy level and so, in a way that varies according to which element it is, either of the two subshells could be the first to fill and there is no certainty that one will be completely filled before the other actually starts filling. There are similar irregularities in the filling of 5s and 4d as well as some other pairs of subshells.

      So, rather than going through the far more difficult to understand rules that do correctly predict the electron configurations for all elements, it is easier just to learn the elements that this simplified method doesn't work correctly for. Here they are:

      In the following elements, there is 1 electron, not the predicted 2, in the outermost s subshell, the "missing" electron is instead located in the outermost d subshell:
      Cu, Cr, Nb, Mo, Ru, Rh, Ag, Pt, Au

      In Pd (palladium) the outermost s subshell, predicted to have 2 electrons, is actually empty. Both of these electrons are instead in the outermost d subshell. Thus, the actual configuration of the Pd is [Kr] 4d¹⁰ (the 5s is empty).

      All the other exceptions are in the f block and just have to be learned on a case-by-case basis. However, it is unlikely you would be asked to know these in a general chemistry course. Here are the f-block exceptions:
      La, Ce, Gd, Ac, Th, Pa, U, Np, Cm, Lr
      Note: in one of his videos, Sal does the configuration of La, but he gets it wrong. He did not know that La is an exception to the rule.

      There is a simple chart that makes learning the filling order easy. Here is a link to it:
      http://www.mpcfaculty.net/mark_bishop/memory_aid_e_config.jpg

      Note: Nickel has two electron configurations that are commonly encountered over which there is a dispute amongst chemists as to which should be considered the ground state. I have assumed that the configuration that follows the rules is the ground state, since that is what is usually taught in general chemistry. However, this is not certain.
      (34 votes)
  • aqualine ultimate style avatar for user Srilakshmi Ajith
    At , Jay said that there are many other factors to consider, what are the other factors that need to be considered?
    (4 votes)
    Default Khan Academy avatar avatar for user
    • orange juice squid orange style avatar for user Maya Pawlikowski
      Maybe bonding with other elements, where being shared is actually is lower energy, (octet rule), or pressure, where the atoms want to either take up more or less space (Gases at high or low pressure respectively).

      Also charges, but he mentioned that.

      Hope this helped! :D
      -Maya
      (3 votes)
  • leaf green style avatar for user Assaf Shomer
    You keep saying that 4s orbital electrons have higher energy than 3d orbital electrons (for scandium).
    In addition, I thought that electrons want to fill the positions with the least energy before they fill positions with higher energy.
    Combining those two things I do not understand why would the higher-energy 4s orbital fill with electrons before the lower-energy 3d orbital.
    Is it because the repulsion of the electrons filling the 3rd shell makes it "harder" for the electrons to occupy that state rather than the 4s state, despite the last is of higher energy state?
    Anyway, this remained unclear to me throughout the video.
    Thanks in advance
    (2 votes)
    Default Khan Academy avatar avatar for user
    • piceratops ultimate style avatar for user Just Keith
      The 4s and 3d subshells have nearly the same energy level. How much of a difference, and which subshell is lower in energy, varies by element.

      In scandium and the other d block members of period 4, the 3d is lower in energy than the 4s. However, because the 3d orbitals are smaller and more closely packed than the 4s, the electron-electron repulsion that occurs in the 3d subshell is enough to make it possible for 1 or 2 electrons to have a lower energy by occupying the 4s. So, we have 3d and 4s co-filling rather than filling independently.

      So, with scandium, the first electron after completing the previous noble gas configuration goes into 3d. The difference in energy levels between 4s and 3d is small enough for the next two electrons to enter the 4s, because the electron-electron repulsion they feel if they enter 3d is enough to knock them into the 4s.

      The same process happens throughout the d-block. Sometimes the difference in energy levels between the s and d allows one electron to enter the s, sometimes it the difference allows 2 electrons to enter the s. And in palladium alone, the difference between d and s is so great that none of the electron are in the s.

      This leaves us with why you have s filling in Groups 1 and 2 instead of the d. That is simple. For the first few members to of Groups 1 and 2 there is no d to be filled. But with 4s-3d, 5s-4d, etc. there is the d to take into consideration. The explanation is simple: in Groups 1 and 2, but not in the d block, the s is lower in energy than the d. In the d block it is the other way around.

      Because the difference in energy levels between 4s-3d, 5s-4d, etc. varies by element, we have to look at each element on a case-by-case basis to see how these will fill.

      Similar irregularities appear in the f block, only now there are s, d and f subshells all with similar energy levels. So, those also have to be learned on a case-by-case basis.
      (6 votes)
  • starky tree style avatar for user Shreet Dave
    Why should Scandium have Energy level of 3d lower than 4s? As much as I know this thing called chemistry, every exception made in chemistry has a reason, an explanation. So why such an exception?
    (0 votes)
    Default Khan Academy avatar avatar for user
    • piceratops ultimate style avatar for user Just Keith
      3d and 4s have nearly the same energy level. The difference between the two varies from element to element. So, if the two levels are close enough on a particular element, one or two electrons can get bumped up to 4s due to electron-electron repulsion being greater than the difference between the two energy levels. A similar situation happens in period 5 with 5s and 4d. While most of the elements of the d block have the relevant s and d of close enough energy for 2 electrons to get bumped up the the s, there are some in which the difference is not small enough and only one gets bumped up. In Pd none of the d electrons get bumped up to the s -- thus it is the only element which contains no electrons in the shell to whose period it belongs.

      Note: In Groups 1 and 2, 4s is lower in energy than 3d, but starting with Sc, 3d is slightly lower than 4s.
      (10 votes)
  • female robot grace style avatar for user Debangee Das
    what exactly is the Hund's rule? I did not get it.
    (1 vote)
    Default Khan Academy avatar avatar for user
    • starky sapling style avatar for user ASK2.0
      Electrons have the same charge, thus they repel each other. Every subshell of an electron can hold two electrons but it will first try to "spread out" the electrons (Like people in a waiting room they will first gravitate toward a section where no one already is). Thus an one electron will go to each sub shell in an orbital before each gets a second.
      (6 votes)
  • duskpin ultimate style avatar for user Lily Martin
    Jay says that the 4s orbital fills before the 3d orbital, but i am a bit confused about this. He shows the electrons as being added to 3d after 4s is filled. Is it just an abstract idea? Or is there a way to understand it more clearly?
    (2 votes)
    Default Khan Academy avatar avatar for user
    • piceratops seed style avatar for user RogerP
      If you look at this webpage, there is a chart showing the relative energy levels of the different orbitals - http://www.chemguide.co.uk/atoms/properties/3d4sproblem.html.
      You'll see that the 3d orbitals are of slightly higher energy that the 4s orbital and this is why 4s is orbital is filled first.
      It can certainly be confusing! To add further to the confusion, when the 3d orbitals becomes populated with electrons, then 4s orbital has a slightly higher energy level than the 3d orbitals. This means that when such an atom loses electrons to make a cation, the electrons are first lost from the 4s orbital.
      (3 votes)
  • female robot ada style avatar for user Lydia Norris
    In this video, Jay said that at scandium the 4s and 3d orbitals switched back to 3d having a lower energy than 4s (while in the 3rd period 4s was slightly lower). How and why did the energies of the orbitals change?
    (2 votes)
    Default Khan Academy avatar avatar for user
    • hopper cool style avatar for user Iron Programming
      Unfortunately there is conflicting views on this topic, due to the oversimplification of the Aufbau principle.

      One answer I saw said that there was overlap of the subshells 3d and 4s due to the shape of their orbits. Once 3d subshells are filled in, they repel the 4s orbit pushing it back further from the nucleus making it have a higher energy level than the 3d.

      Now, why the order of filling up 3d vs 4s is quite complicated, and to be honest I am quite tired of hearing people say that myself.

      One thing I saw was that the 3d subshell is a bit more compact, making the 4s preferred in certain cases.

      As you might suspect it depends on the atomic number (aka the # of protons). As we go down the periodic table the nuclear strong force increases, and somehow this effects the choice to choose 3d over 4s or vice versa.

      I'm just sharing some of my thoughts & research but I am still looking into this so if anyone else sees this please update both of us on how it works exactly.

      Hope this helps!
      - Convenient Colleague
      (2 votes)
  • male robot hal style avatar for user Krish
    For all transition metals, do the energy levels of the 4s orbital become higher than the 3d orbitals?
    (2 votes)
    Default Khan Academy avatar avatar for user
    • leaf red style avatar for user Richard
      Yes the same affect is experienced by the rest of the transition metals. When the 3d orbitals are unfilled, the 4s orbitals are lower in energy and are filled first according to the aufbau principal. But once 3d electrons begin filling for transition metals they begin to push the 4s electrons farther from the nucleus making the 4s electrons now higher in energy than the 3d ones.

      Hope that helps.
      (2 votes)

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.