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Confusion about the electron sub shells.

  1. Sep 24, 2013 #1
    Apologies for the long post this is going to be, but at the moment I am doing A-level chemistry and physics, and I am learning about electron orbitals and quantum physics at the same time, so I have reached a state of confusion. (At a higher level than that required for the A-level; I'm not having any difficulty accepting fact as fact, it's just that I want to try and get a more complete understanding.) I'm only 16, so please try and keep any explanations as simple as possible; certainly pre-university! (This description of events is going to be time shifted greatly, but I'm going to try and recount it in the most logical order I can.)

    So firstly about the 3d and 4s sub shells. My chemistry teacher has told me rather simply that the 4s sub shell has a lower energy level than the 3d sub shell, primarily, so that the 4s sub shell is filled before the 3d sub shell. However once the 4s sub shell is filled, the 4s and 3d sub shells 'swap' in terms of energy levels, so that the 4s sub shell has more energy than the 3d sub shell. So my first question is why exactly is this?

    I asked my physics teacher why this is, and his immediate answer was 'because their wave functions change' or something along those lines (I presume he was talking about the electrons in those sub shells, but I really have no idea).

    My chemistry teacher then described the example of the vanadium atom, where you have a full 4s sub shell, and 3 electrons in the the 3d sub shell. She told us that when you create a V+ ion, the electron is taken from the 4s sub shell because it is the sub shell with the highest energy. So I asked then about what happens when you add another electron to recreate an atom. Does it go back into the 4s shell because that's how it was before, or does it go into the 3d sub shell because then it will be in a lower energy state? She didn't know the answer, so that's my second question, along with why whatever happens actually happens. Also what happens to the 4s sub shell when it only has one electron in it? Does it drop back below the 3d sub shell, or stay where it is? Also what happens when you have a V2+ ion?

    Today, our physics teacher decided to make the lesson one big aside, in order to try and explain what we're learning about in chemistry at the moment. He explained the Pauli exclusion principle so that it made total sense, and it explained most of what I'd learnt in chemistry, but didn't quite explain the questions I had in chemistry.

    So anyway, I can't quite remember what I asked then, but I'm stuck with the memory of an explanation about how when an electron is added to a sub shell, that sub shell gains energy due to 'interaction energies' between the electrons in the shell, which made sense. However in particular I'm stuck with the idea that this interaction energy is responsible for the 4s sub shell rising above the 3d sub shell in terms of energy level when full. But this goes against firstly what my teacher said initially about changing wave functions, but also against my understanding that 3d sub shell remains below the 4s sub shell in terms of energy, even when the the 3d sub shell is filled. I mean surely (if the interaction energies thing is relevant) the energy of the 3d sub shell with 10 electrons will be more than the energy of the 4s shell with 2 electrons? So basically this is the same question as my first one, but also where do these 'interaction energies' come into things?

    My physics teacher then threw another spanner in the works (don't get me wrong, he's a brilliant teacher! :D ) by saying that electrons like to go around in pairs. Which seemed to go against everything I'd learnt about electrons; they're both negatively charged; they're both fermions, so they try to avoid being in the same states. Also in chemistry we'd been told that when filling out the diagrams
    ( http://www.google.co.uk/imgres?imgu...a=X&ei=nt1BUpOPMMXbtAamv4H4Ag&ved=0CEcQ9QEwCQ )
    using arrows representing electron spin, that we should make sure all the boxes in one shell are filled with one arrow, before we start filling the boxes with the second arrows. I also read an article that said the electrons have lower energies when not in pairs due to electromagnetic repulsion, so prefer not to be in pairs. So what on earth was my teacher getting at when he said this??!!!

    The reason I've put this in the physics section rather than the chemistry section is that my physics teacher appeared to have more of an understanding about the changing energy levels, which is the main crux of my confusion.

    As you can clearly see all my knowledge of this stuff is incredibly disjointed, and I feel like my head will explode if I don't get it cleared up! I may up remembering/coming up with more questions, but I think that's all of them.

    Thank you so so much for reading all this, I will greatly appreciate if you answer even one of my questions :) Thank you so much for any answers in advance.
  2. jcsd
  3. Sep 24, 2013 #2


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    I don't think there is an intuitive explanation for the details of orbitals at transition metals. You can calculate them (as solutions of the Schrödinger equation), and you can measure them. All those electrons influence each other, so the energy of orbitals levels depend on the number of electrons in other orbitals.

    I don't know how "like to go around in pairs" is meant here. They can have different spin states, so they are not in the same state. "Same orbital" does not mean "same position", if you measure their positions you will frequently find them on opposite sides, their repulsion is not stronger than the repulsion between random other electrons.
  4. Sep 24, 2013 #3


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    I'm not sure I can answer your question directly, but let me explain the origin of some of these phenomena, in the hope that it will clear some things out.

    Orbitals, and their designation by the quantum numbers n and l, come from the theory for hydrogen and "hydrogen-like" atoms, meaning atoms consisting of a nucleus and one electron. This system can be solved exactly (in the absence of relativistic effects) using the Schrödinger equation. The solutions come out to be nearly identical whatever the charge of the nucleus, which only changes the scaling of the orbital. From these calculations, we find that the energy of the orbital depends only on the quantum number n, such that, for instance, 2s and 2p orbitals have the same energy!

    Things get more difficult when you try to add a second electron. The equation now describes what is called a three-body problem, for which we can't have an exact mathematical solution, so we have to make approximations (or use computers to get numerical answers). Fortunately, we can still figure out many things from the equations.

    For one, the energy now also depends on l. And while you could imagine that the effect is small, such that for instance E(3d) < E(4s), as you now know it is the other way around, E(3d) > E(4s). A simple explanation for this come from an effect called screening. Consider lithium, which has three electrons, 1s22s1. The 2s electron will feel the presence of the two 1s electrons (like charges repelling), and this has the effect of reducing the charge of the nucleus for the 2s electron. While the 1s electrons feel the full 3+ charge of the nucleus, the 2s electron behaves as if the nucleus had a charge slightly less than 3+. (Screening is not perfect, so the charges of the two 1s electrons do not completely cancel out two of the positive charges of the nucleus.)

    I guess that you know that the electrons are smeared out around the nucleus, and do not "orbit" at a given distance from the nucleus. If you look at the radial distribution of the orbitals, you will see that s electrons come closer to the nucleus than p electrons, and themselves come closer than d electrons. Because of this, a 3s electron will feel a greater effective charge of the nucleus than a 3p electron, so the 3s will be more attracted to the nucleus and its energy will be lower. It turns out that, if the orbitals 1s, 2s, 2p, 3s, and 3p are all filled, a 4s electron will see a higher effective charge of the nucleus than a 3d electron, hence E(3d) > E(4s).

    That is the simple answer, but it is not the entire picture, as other things can happen. For quantum mechanical reasons, half-filled sub-shells have lower energies. This is why, for instance, while the electron configuration of V is 3d34s2, the next element Cr is 3d54s1.

    Replies to your other questions to come.
  5. Sep 24, 2013 #4


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    I don't know how to answer the question of what happens dynamically (i.e., when electrons are ejected or recaptured). But V is even weirder than that. The ground state is 3d34s2, and upon ionization, it is one 4s electron that goes away, resulting in 3d34s1. But the ground state of the ion V+ is actually 3d4! So again, adding or removing a single electron can affect the energy of the other electrons in a strange way.

    I don't know what he was talking about. Maybe Cooper pairs in superconductivity? Otherwise, you are right in thinking that electrons avoid each other as much as possible in atoms, filling different orbitals (for a given n,l) before two electrons will occupy the same orbital. This spreads out the electrons, it reduces the average repulsion between them. There is also a quantum phenomenon, spin correlation, that makes it more energetically favourable for unpaired electrons in distinct orbitals to have the same spin, so all those arrows you talk about should be pointing in the same direction.
  6. Sep 25, 2013 #5
    Thank you so much for you answers, they're very helpful :)

    "It turns out that, if the orbitals 1s, 2s, 2p, 3s, and 3p are all filled, a 4s electron will see a higher effective charge of the nucleus than a 3d electron, hence E(3d) > E(4s)."

    This is a convincing explanation as to why the 4s sub shell is filled first, but why then do they swap around in energy level once the 4s is full? Is it due to the energy you add to the shell with an electron, or is something else? I'm inclined to think that it is something else, unless the 3d shell does in fact rise above the 4s once again once it is full. Is this correct?

    "But the ground state of the ion V+ is actually 3d4!"

    So what you're saying here is that the remaining electron in the 4s shell moves into the 3d shell? I suppose the main source of my confusion here is with the swapping of the 3p and 4s shells; is the movement continuous, or instant. i.e. does the adding of one electron cause a small rise in energy, and the adding of the second electron cause a second rise? Or is there just a jump when 4s is filled?

    "This is why, for instance, while the electron configuration of V is 3d34s2, the next element Cr is 3d54s1."

    Same explanation for copper being 3d104s1?

    Again many thanks for taking the time to reply :)
  7. Sep 25, 2013 #6


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    This is a bit where the simple pictures break down. The 3d and 4s do not swap positions. You have to remember also that you are talking about different atoms. You fill in 4s before 3d in the sense that K is 4s1, Ca is 4s2, and Sc is 3d14s2, but they are three different atoms!

    For example, if you take a nuclear charge of ##Z=+22## and add 22 electrons, the valence electrons will be in the configuration 3d24s2 (that's Ti). But if you take a nuclear charge of ##Z=+23## and add 22 electrons, the valence electrons will be in the configuration 3d4 (that's V+).

    So take the rules you learn just as simple tools to get quick answers, which may be incorrect in some cases: when filling orbitals of neutral atoms, fill 4s before 3d, but when considering ionization of neutral atoms, the 4s electron requires less energy to remove.

    I hesitate into going into the dynamics of the process, because things get even more complicated. What is certain is that a 4s electron will not simply relax to a 3d orbital because l can only change by one unit at a time (4s could relax to 3p, if all such orbitals were not already filled). You would need other processes to take place, like a collision with another atom, for the 4s electron to relax.

  8. Sep 25, 2013 #7
    Thanks a lot for all your answers :) They've definitely cleared a lot of things up for me :)
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