Chemistry Stack Exchange is a question and answer site for scientists, academics, teachers, and students in the field of chemistry. It only takes a minute to sign up.
Sign up to join this community
The electron configuration of calcium is 2, 8, 8, 2, where up to that point each shell, asides from the first shell counts up to 8 - why then does scandium have an electron configuration of 2, 8, 9, 2?
What causes the 3rd shell to start filling up, rather than the fourth?
The relative energies of the electronic subshells have been calculated for atoms in the vicinity of $Z=20$ (J. Chem. Educ., 1994, 71 (6), 469), and the result is surprising:
Looking at this graph, by all means the electronic configuration of scandium should in fact be $\ce{1s^2 2s^2 2p^6 3s^2 3p^6}$ $\color{blue}{\ce{3d^3}}$ in order to minimize orbital energies! The graph does not show the energy of the $\ce{4p}$ subshell, but it would lie somewhat above both curves shown. However, there is an important effect not being considered here, which is the destabilizing interelectronic repulsions. Electrons in a same subshell tend to repel each other more (intrasubshell repulsion) than electrons in different subshells (intersubshell repulsion). It turns out (and no one can really qualitatively explain why, see below) the repulsion is such that in an atom with $Z=21$, the configuration $\ce{[Ar] 4s^2 3d^1}$ is lower in energy than other potential candidates, such as $\ce{[Ar] 3d^3}$ or $\ce{[Ar] 4s^1 3d^2}$. Theoretically, the odd configurations $\ce{[Ar] 4s^2 4p^1}$ or $\ce{[Ar] 4s^1 3d^1 4p^1}$ would have lower interelectron repulsion energies than the observed ground state, but in these cases the lowered interelectron repulsions do not compensate the requirement of populating the higher-energy $\ce{4p}$ subshell. The comparison of several possible electron configurations and the joint minimization of orbital energy and electron repulsion energies does not seem to be something that can be done without resorting to heavy calculations or direct experimental validation.
In addition to some material I linked in comments, this article from Eric Scerri's blog (a chemist who focuses on aspects of periodicity, including electronic distribution) states, regarding the electronic configuration of scandium compared to calcium and to the $\ce{Sc^3+}$ ion (emphasis mine):
This amounts to saying that all three of the final electrons enter $\ce{3d}$ but two of them are repelled into an energetically less favourable orbital, the $\ce{4s}$, because the overall result is more advantageous for the atom as a whole. But this is not something that can be predicted. Why is it 2 electrons, rather than one or even none? In cases like chromium and copper just one electron is pushed into the $\ce{4s}$ orbital. In an analogous case from the second transition series, the palladium atom, the competition occurs between the $\ce{5s}$ and $\ce{4d}$ orbitals. In this case none of the electrons are pushed up into the $\ce{5s}$ orbital and the resulting configuration has an outer shell of $\ce{[Kr] 4d^10}$.
None of this can be predicted in simple terms from a rule of thumb and so it seems almost worth masking this fact by claiming that the overall configuration can be predicted, at least as far as the cases in which two electrons are pushed up into the relevant s orbital. To those who like to present a rather triumphal image of science it is too much to admit that we cannot make these predictions. The use of the sloppy Aufbau seems to avoid this problem since it gives the correct overall configuration and hardly anybody smells a rat.
So unfortunately, it seems that even though most of the effects which combine to result in the observed electronic configurations are known, there is no qualitative way to predict where the configurations are going to mismatch with the Aufbau principle or the energy levels of the orbitals. I have read the statement that the Aufbau principle is most decidedly wrong for practically every atom with respect to the placement of the orbital energy levels, but incredibly it happens to predict the configuration of the valence shell for most atoms.
The electron configuration for scandium is:
$$\ce{1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^1}$$
The $\ce{4s}$ and $\ce{3d}$ orbitals are relatively close in energy, so there are a few anomalies in the order of filling. A fair question might be, why isn't the scandium electron configuration $\ce{[Ar] 3d^3 4s^0}$. There is a special stability associated with completely filled orbitals, so the $\ce{3d^1 4s^2}$ configuration is of lower energy than the alternative $\ce{3d^3 4s^0}$.
I'm no expert but I have researched into this.
I'll cover the advanced answer first because it's important. But you might be able to follow the jist of it and then see the more basic one after.
Part of answering the question comes to order of energy levels, and that's an advanced question, in that it's currently debated among academics. And the answers once reached, will take some time to get from academic papers into undergraduate textbooks.
And the basic question(in the sense of question with the basic answer), is given a particular ordering of energy levels as per the aufbau principle, which is the assumption your books are working with, how does one explain what you asked, the electronic configuration of scandium.
You would need to know about subshells and the aufbau principle. But follow this and get the jist and then see the basic answer that follows.
Advanced answer
The standard answer in the books(re ordering of energy levels), is that 4s<3d and that would explain your question about the electronic configurations of neutral atoms very well. (But leaves some problems to explain the electronic configurations for ions). So your question, which is asking about the electronic configuration of neutral atoms, is easily easily addressed under that view.
Beyond that, there has been some debate. Eric Scerri has argued http://ericscerri.blogspot.com/2012/06/trouble-with-using-aufbau-to-find.html that actually for Scandium, 3d<4s (This explains the electronic configuration for ions), and he gives some explanations for how that happens, even though he thinks 3d<4s Scandium. And his explanation makes ions easier to explain. But he then has to provide some explanations (And does), to explain the electronic configuration of the neutral scandium atom. He says atoms go into 3d first but jump up to 4s because it's more stable. And he thinks this process books have of trying to predict electron configurations is thus a bit silly, because the reality of what's happening is complex.
There has though been a response to Eric Scerri, by Geoffrey Neusner. And he has stated that Neutral scandium has 4d<3s(as the standard view says). But that in Scandium *ions*, the order of energy levels of orbitals gets rearranged such that 3d<4s. See this article https://www.thinkib.net/chemistry/page/37492/the-electron-configuration-of-scandium and the paper associated with it https://tinyurl.com/2kvc96my or here
I think the ball is now in Eric Scerri's court to respond to Geoffrey Neuss's paper.
Orbitals fill up low energy first, then to higher energy. (at least I don't think that model is in dispute by Scerri or Neusner or anybody)
At a very high level (and not relevant to the energy of orbitals subject), they'd say electrons are not particles, some say they are waves, or something with properties of each. And also, at a high level see this ACS paper, https://pubs.acs.org/doi/pdf/10.1021/ed200673w they say electrons don't "occupy" orbitals, they might even be spread out. And orbitals are just visual representations mapping to some Mathematics, not direct representations showing shapes of regions of space. But for the purposes of considering electronic configurations, people are fine with the concept *as a model*, of orbitals as regions and of electrons in orbitals.
You are not asking about ions. So, you're in luck. The answer is fairly simple.
But either way. What one needs to know is. What is the order of the orbitals.
At a higher level of study, it might depend on the atom and the charge on the atom. And hopefully some people are working on that will be putting that information out there in the future.
The "standard answer" says the order of orbitals is famously as per aufbau and as is clear from the blocks in the periodic table.
Basic answer
Assuming orbital energy level ordering as per the Aufbau rule.
And this is even covered by Middle School Chemistry, if the material is of ludicrously high quality like this
https://www.middleschoolchemistry.com/lessonplans/chapter4/lesson3
"When the third energy level has 8 electrons, the next 2 electrons go into the fourth energy level."
and
"The third energy level can actually hold up to 18 electrons, so it is not really filled when it has 8 electrons in it. But when the third level contains 8 electrons, the next 2 electrons go into the fourth level. Then, believe it or not, 10 more electrons continue to fill up the rest of the third level."
And you see the electronic configurations in that format here https://ptable.com/?lang=en#Properties
At 16-19 you might learn about the Aufbau principle and order of energy levels and subshells. Shown in abbreviated form here https://sciencenotes.org/list-of-electron-configurations-of-elements/
Relevant at Undergraduate level too
The list of subshells can be determined looking at the blocks of the periodic table too. Here it is as mentioned in that article "1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, and 7p"
So you see the "third shell" is only partially complete, then the fourth fills a bit, then the third continues. Then the fourth does some more(but doesn't complete). An inner shell can have electrons added to it. And so that's where many books will put aside a simple view of just the shells, and from scandium onwards will tend to put away the format of electronic configurations of K,L,M,N like 2,8,9,2 And they'll go to showing electronic configurations as e.g. 1s2 2s2 2p6 3s2 4s1 e.t.c. And that model is in a way, simpler for elements from scandium onwards in the sense that it's clear which subshell is going to fill up next.
This question that was closed might have some information of interest as it shows there is uncertainty over this in academia. Why do 3d orbitals have lesser energy than 4s orbitals in transition metals?
