## Wednesday, August 12, 2015 ... /////

### $4s$ gets filled before $3d$: this is no myth

(in almost all important atoms and ions, but the details are difficult to calculate)

Some chemists would like to make the teaching of orbitals even stupider than it is.

Yesterday, I was making the Google News search for "quantum mechanics" – one of the searches I do relatively often – and I was led to the "Education in Chemistry Blog". Eric Scerri, a chemistry lecturer at UCLA, is fighting against various "wrong claims" in the education of chemistry. The article he posted yesterday was titled

Five ideas in chemical education that must die - part three
and dedicated to the ordering of $3d$ and $4s$ orbitals. He believes that it's wrong for most chemistry textbooks to say that $4s$ gets occupied before $3d$. And he also criticizes writers and teachers who admit that many of these questions behave irregularly because of complexities of quantum mechanics that can't be calculated and settled by chemistry students.

Except that Dr Scerri is wrong. In the most important cases, $4s$ gets filled before $3d$, there is no contradiction because the occupation numbers in the shell depend both on the number of electrons and the number of protons, and the reliable answer for each atom or ion can indeed be extracted only from a quantum mechanical calculation that is both unsolvable analytically and inaccessible to chemistry students (and chemistry teachers, and most physicists, too).

At the basic school and in the first two years of the high school, I didn't really understand quantum mechanics. If you had asked me what was going on inside the atoms, I would have probably told you something about the Keplerian classical orbits. But I wasn't sure. Needless to say, this is an extremely bad starting point to understand the rules for the orbitals etc. – which are phenomenological simplifications of quantum mechanics (simpler than the full-fledged quantum mechanics but harder than the Keplerian classical orbits).

I've had some problems with our 300-pound chemistry teacher, bad grades because of that, and so on, and reduced but nonzero problems even with her successor that was hired after we (the temporarily omnipotent kids) have fired the gorilla in the wake of the Velvet Revolution.

I still believe that the education of chemistry is highly imbalanced, illogical, and terribly ineffective. Even though quantum mechanics is hard, I actually do think that schoolkids should be exposed to some quantum mechanics before they are taught advanced things about orbitals. In other words, I tend to agree with the Physics First paradigm. To say it modestly, I think that it should be tried at least at some schools.

Equally importantly, chemistry classes are full of memorization of technicalities that are simply not important in the grand scheme of things. I find it crazy when kids are forced to memorize chemical formulae for all amino acids – but the same students end up believing creationism or something like that, anyway. Such a combination is a proof of an imbalanced education system.

If you asked me to design new chemistry textbooks and similar things, the changes would be profound. But I want to talk about something that the chemistry books are basically describing correctly. The ordering of orbitals.

Orbitals of atoms and ions

Here is my sketch of this body of basic knowledge. To describe an atom or an ion, you need quantum mechanics. The zeroth approximation describes the non-relativistic motion of a nucleus of charge $+Ze$ ($Z$ is the number of protons and the number of neutrons doesn't affect the motion of the electrons almost at all because the neutron-electron forces, e.g. the gravitational ones, are negligible) and especially a bunch of $N_e$ electrons. We have $Z$ electrons for the atoms (which are electrically neutral); less than $Z$ electrons for positively charged ions; and more than $Z$ electrons for negative charged ions.

We may go to the center-of-mass system. Up to some small changes of the electron masses (reduced mass etc.), it means that we fix the location of the nucleus at $\vec r =0$. The Hamiltonian becomes$\eq{ H &= \sum_{j=1}^{N_e} \frac{|\vec p_j|^2}{2m_e} + V_{\rm pot},\\ V_{\rm pot} & = -\sum_{j=1}^{N_e} \frac{Ze^2}{4\pi\epsilon_0 |\vec r_j|} + \sum_{j\lt k=1}^{N_e} \frac{e^2}{4\pi\epsilon_0|\vec r_j-\vec r_k|} }$ Great. In other words, the energy contains the kinetic energy for $N_e$ electrons; the attractive (i.e. negative) potential energy between the nucleus of charge $Q=+Ze$ and the electrons of charge $Q=-e$; and the repulsive (i.e. positive) electron-electron potential energies.

This quantum mechanical problem is analytically unsolvable for $N_e\gt 1$, just like the three-body Keplerian problem which is already unsolvable in classical physics. However, what's funny is that the problem is easily solvable if we erase the electron-electron interaction term, the last term with $\sum_{j,k}$.

When we do so, the Hamiltonian becomes a simple sum $\sum_j$ of $N_e$ one-electron Hamiltonians that commute with each other and have nothing else to do with each other! And for each electron, the Hamiltonian is that of the hydrogen atom (electron's kinetic energy plus $V(r)\sim 1/r$), except for the extra factor of $Z$ in the electron-nucleus potential energy term.

The one-electron hydrogen-like problem may be easily solved, e.g. by using the $SO(4)$ symmetry. The energy eigenstates are either those in the continuous spectrum at $E\gt 0$, describing the ionized hydrogen atom (the eigenstates are slightly perturbed plane waves, if you wish); or, more importantly, the discrete spectrum describing the actual atoms (bound states).

The bound energy eigenstate may be described by the quantum number $n=0,1,2,\dots$, by $\ell= 0,1,2,\dots n-1$, and by $m=-\ell,-\ell+1,\dots, +\ell$. And the electron carries the internal spin, a pretty invisible qubit for one electron, $s_z=\pm 1/2$. The energy of the one-electron state equals $-E_0/n^2$ where $E_0=13.6\eV$. It only depends on the principal quantum number $n$. There are $n^2$ (or $2n^2$ if the spin is included) states in the shell with a given $n$ – the usual degeneracy.

Hydrogen-like ions which have $N_e=1$ electron and $Z$ protons in the nucleus happen to be "isomorphic" to the Hydrogen. They are just $Z$ times smaller and the binding energy is $Z^2$ times greater than for the corresponding state of the hydrogen atom (one $Z$ comes the explicit $Z$ in the potential term, another $Z$ because $1/r$ is $Z$ times larger than for the hydrogen).

We are still ignoring the electron-electron interactions, OK?

So right now, we may easily solve this truncated atom. Each of the electrons occupies an independent state. The total wave function has to be antisymmetric which imposes the Pauli exclusion principle. The result is that to get low-lying states of the atom, we are inserting the electrons to one-electron electron eigenstates, starting from the most bound ones – i.e. from $n=1$.

The chemists' convention for the values of $\ell$ is to replace the values $\ell=0,1,2,3,4,5,\dots$ by the letters $s,p,d,f,g,h,\dots$ which stand for spherical, principal, diffuse, fundamental, gargantuan, holy, etc. – bizarre historical adjectives invented in spectroscopy that have nothing to do with the shape of the orbitals.

The shells with $n=1,2,3,4,\dots$ have $2n^2 = 2,8,18,32,\dots$ electrons in total and these one-electron states are being gradually filled as we add additional electrons. So if we neglected the electron-electron interactions, we would fill the states with the indicated values of $(n,l)$:$1s \quad |\quad 2s, 2p \quad |\quad 3s, 3p, 3d \quad |\quad 4s, \dots$ The idea that we simply add electrons to the shells, one-by-one, in one order is known as the Aufbau principle (a German word for "construction"). We couldn't say whether $2s$ gets filled before $2p$: in the hydrogen atom, they have the same energy (up to small relativistic corrections that are easily beaten by the electron-electron interaction effects as soon as the latter are taken into account).

The very stable atoms would be those in which we fill the full shells. The full $1s$ atom has $2$ electrons (one is spinning up, one is spinning down) and it's the $Z=2$ helium atom. The next one has the full $1s$ and $2s,2p$ and $2+8=10$ electrons in total. It's the $Z=10$ neon atom. The next one would naively have $Z=2+8+18=18=28$ because it would fill $1s;2s,2p; 3s,3p,3d$. But the next inert gas, argon, actually has $Z=18$ only because it only fills the states$1s \quad | \quad 2s , 2p \quad | \quad 3s, 3p\quad {\rm (argon)}.$ The $2(2\ell+1) = 10$ electrons in the $3d$ i.e. $(n,\ell)=(3,2)$ states are already omitted. Why doesn't argon, the third simplest inert gas, want to fill all the $n=3$ shells including $3d$?

Well, in the mathematical problem neglecting the electron-electron interactions, the $3d$ one-electron states would be demonstrably energetically equivalent to those in $3s,3p$, so a very stable inert gas would be expected to fill $3s,3p,3d$, all of them. Because in the real world, this doesn't occur, the separate treatment of $3d$ has to have something to do with the electron-electron interactions.

And be sure that the electron-electron interactions must be blamed.

The problem is that the electrons in the $3d$ shell repel each other – and other electrons in the atom. A special problem with $3d$ is its highest possible value of $\ell$ allowed for $n=3$, namely $\ell=2$. A higher value of $\ell$ means that the wave function (and therefore the charge distribution) depends on the angular variable $\theta,\phi$ (describing the two-sphere) more intensely than it does for lower values of $\ell$.

Roughly speaking, this fact implies that the electrons are "more likely to be concentrated" and "more likely to be really close to other electrons". If they're very close, the term $e^2/4\pi\epsilon_0 r_{jk}$ gets (increasingly) large and positive. It gets even worse if you add many $3d$ electrons.

At a high enough value of $N_e$, the electron-electron interactions simply cannot be neglected, not even for the qualitative ordering of the states of the atom (i.e. the order in which we fill the distorted one-electron states). And some electrons are actually added to $4s$ before any electrons are added to $3d$. Even though $4$ is larger than $3$, some one-electron states in $4s$ get filled before the states in $3d$.

This "lower priority" for one-electron states with higher values of $\ell$ occurs well before argon. The very fact that $2s$ is written before $2p$ etc., even though the one-electron states in the hydrogen atom have exactly the same energy, is a manifestation of the fact that higher values of $\ell$ generally produce a higher positive contribution to the energy relatively to the $\ell$-independent hydrogen result, and are therefore discouraged.

The precise point at which the new $\ell$-dependent contribution from the electron-electron interactions becomes large enough to change even the order of the occupation of the shells is clearly a difficult computational issue. Orbitals with a lower value of $\ell$ are almost universally preferred – have a lower energy, mainly due to the more modest contributions from the electron-electron electrostatic repulsive (positive) terms.

But it just happens that for $3d$ and $4s$, the new contribution to the energy that increases with $\ell$ is large enough so that $3d$ is not only segregated from its former friends, $3s$ and $3p$; but the $4s$ state actually gets preferred.

If you look at the results – either experimentally or by a more complicated (numerical? perturbative?) calculation of the atomic orbitals – you will see that the atoms filling shells up to $3p$ are working (qualitatively) in the naive, hydrogen-based way – with the extra preference for lower values of $\ell$ among the formerly degenerate states; and the further shells are filled as shown by these electronic structures:$\eq{ 1s &\quad|\quad 2s , 2p \quad |\quad 3s, 3p\quad ({\rm argon}, Z=18),\\ [{\rm Ar}] &\,4s^1 \quad ({\rm potassium}, Z=19),\\ [{\rm Ar}] &\,4s^2 \quad ({\rm calcium}, Z=20),\\ [{\rm Ar}] &\,4s^2 3d^1 \quad ({\rm scandium}, Z=21),\\ [{\rm Ar}] &\,4s^2 3d^2 \quad ({\rm titanium}, Z=22),\\ [{\rm Ar}] &\,4s^2 3d^3 \quad ({\rm vanadium}, Z=23),\\ [{\rm Ar}] &\,{\bf 4s^1 3d^5} \quad ({\rm chromium}, Z=24),\\ [{\rm Ar}] &\,4s^2 3d^5 \quad ({\rm manganese}, Z=25),\\ [{\rm Ar}] &\,4s^2 3d^6 \quad ({\rm iron}, Z=26),\\ [{\rm Ar}] &\,4s^2 3d^7 \quad ({\rm cobalt}, Z=27),\\ [{\rm Ar}] &\,4s^2 3d^8 \,{\rm or}\, \,{\bf 4s^1 3d^9} \quad ({\rm nickel}, Z=28),\\ [{\rm Ar}] &\,{\bf 4s^1 3d^{10}} \quad ({\rm copper}, Z=29),\\ [{\rm Ar}] &\,4s^2 3d^{10} \quad ({\rm zinc}, Z=30).\\ }$ Starting from $Z=31$ gallium, both $4s$ and $3d$ are already fully occupied and additional states, starting from $4p$, begin to get filled rather regularly (for a while).

When you look at the table of elements between $Z=18$ and $Z=30$ above, you may decide that it's pretty complicated. But it's not so bad. In general, the electrons are being pumped into $4s$ before they are added to $3d$. The $4s$ shell can hold up to $2$ electrons and $3d$ may harbor at most $10$.

But there are exceptions – indicated by a different font. Chromium at $Z=24$ only has $4s^1$ – and the electron is one of those in $3d^5$ instead – even though almost all the neighbors have $4s^2$. A similar exception is the $Z=29$ copper with $4s^1 3d^9$. And to make things more complicated, the $Z=28$ nickel allows both states, either with $4s^1$ or $4s^2$.

I have mentioned a semi-heuristic argument that tends to "postpone" the shells with higher values of $\ell$. Similar semi-heuristic, partly technical yet qualitative arguments exist that explain the Hund's rules. (Normally, I would describe them here, in the partly personal Lumoesque way, but I don't want to distract you by many rules.)

But the ordering of the shells depend on "quantitative discrepancies growing sufficiently to change the picture qualitatively". You have virtually no chance to calculate, on the top of your head, whether the $\ell$-dependence caused by the electron-electron repulsion is enough to fill both electrons in $4s$ before any electron in $3d$, and so on. The electron-electron repulsion effectively adds $X$ electronvolts to one state relatively to another and the question is whether it's enough to beat $Y$ electronvolts from the hydrogen problem – which is needed to change the order. You clearly need a difficult calculation; the result simply can't be extracted from any quantative thought.

Dr Scerri complained that
Textbook authors and instructors react to this apparent conundrum by appealing to all kinds of contorted arguments that are intended to pacify students and lead them to think that the solution lies in the higher reaches of quantum mechanics, which are beyond their level of understanding.
But at least when it comes to this point, the textbook authors and instructors are absolutely right. The detailed arrangement – the "instinct" of Mother Nature who almost always wants to fill $4s$ before any electron is placed in $3d$, but sometimes makes an exception – cannot be extracted from any simple rule or verbal argument. It boils down to a complicated quantum mechanical calculation, one that cannot be solved analytically and one that is prohibitively difficult for chemists, students, and similar people even if one tries to address it approximately or numerically.

The actual energy eigenstates of the atoms cannot be factorized to the (antisymmetrization of) products of wave functions of the individual electrons. In other words, the electrons aren't really occupying the shells independently of each other, even though this independence is effectively what most chemists assume (and what the chemistry students are taught to assume, too). The independence is a wrong assumption exactly because of the electron-electron repulsion term. This is what makes the problem so complicated, this is what makes the fate of the electrons correlated with the fate of the other electrons (i.e. what makes the problem "irreducible" to the problem for individual electrons).

Ions

One more related point: if you want to study the electronic structure of ions with similar values of $Z$, you have to solve different problems than the problems of the atoms. There is no isomorphism here. If you have a positive ion, the relative importance of the electron-nucleus attractive interaction is higher than it is in a neutral atom with the same number of electrons!

So the positive ions are generally "more hydrogen-like" in the sense that it's a "slightly smaller error" if you neglect the electron-electron repulsion.

In practice, the electronic structure of the most important ions here is simple – because the most important ions have the electronic structure of argon, with no "loose" electrons added at all! These are particularly good configurations that save as much energy per electron as possible.

So the most important ions with $Z=19,20,21$ i.e. of potassium, calcium, scandium have the charge $Q=+1e$, $Q=+2e$, and $Q=+3e$, respectively: all the "loose" or "disputable" electrons – either from $4s$ or $3d$ – are completely removed. The ionization energy of these ions is "similarly large" as it is in inert gases – the electronic structures are the same. They are very "stable" ions.

Nevertheless, you may consider ions that are less ionized than these nice ions. My warning is still the same: you can't be sure about the electronic structures of either of these ions just because you were sure about the structures of the atoms or other ions. The problem for each ion or atom is inequivalent from others.

For example, consider the $Sc^{2+}$ ion instead of the "simple and nice" $Sc^{3+}$ ion. This $Z=21$ ion has $N_e=21-2=19$ electrons, the same number as the $Z=19$ potassium atom. So you could simply think that the electronic structure of the $Sc^{2+}$ ion is the same as the electronic structure of neutral $K$ atom. The $K$ atom has the electronic structure of argon, plus an extra $4s^1$ electron.

Because the number of electrons is the same, you could think that $Sc^{2+}$ has the argon plus $4s^1$, too. However, it is not the case: $Sc^{2+}$ has argon plus one $3d^1$ electron. Heuristically, it's because $Sc^{2+}$ is a more hydrogen-like problem – the electron-electron interactions are less important (relatively to the attraction to the more strongly charged nucleus) than they are in the $K$ atom. That's why the ordering of the shells ends up being more naive, hydrogen-like, and the first electron is actually added to $3d$ rather than $4s$.

Similar differences between ions and atoms with the same $N_e$ exist for other values of $Z,N_e$, too. Sometimes the electronic structures end up the same. Sometimes they are different. Which of them has a lower energy depends on the result of a difficult, analytically unsolvable quantum mechanics problem. There is no "contradiction" in the slight irregularity of the electronic structures. The energy contributed by a $3d$ or a $4s$ electron are very close to each other and every detail in this electron's interaction with the other electrons may be enough to declare $4s$ as the winner (most of the time), or the $3d$ as the winner.

In almost all the cases of atoms and ions that matter, it's a good rule to remember that $4s$ gets occupied before $3d$ does. Chromium and copper are half-exceptions because only one electron is placed in $4s$ and the rest goes to $3d$. Nickel is a quarter-exception because there is also a state just with $4s^1$, like for chromium and copper, but there's also the regular ground state with $4s^2$ which is almost exactly energetically degenerate.

And the ions that are exceptions from the rule that "$4s$ gets filled before $3d$" are not the most important (most energy-efficient) atoms that appear in important chemistry problems. So the rule that "$4s$ gets filled before $3d$" isn't true in some "totally rigorous, universal sense" but it is still vastly more often true than the opposite claim that "$3d$ gets filled before $4s$".

Dr Scerri finds the irregularities – and dependence on difficult quantum mechanical calculations – too inconvenient so he would prefer the textbooks and instructors to be completely silent about the fact that $4s$ states are filled before $3d$. But his "alternative" wisdom doesn't make any sense. For example, he wrote:
In the case of the scandium atom, the relevant ordering is that the energy of the $3d$ orbitals falls below that of the $4s$. Consequently, it is the $3d$ orbitals that are preferentially occupied, contrary to what is stated in 99% of textbooks.
No, 99% textbooks are totally correct in the case of the scandium atom because the scandium atom starts with the electronic structure of argon, completely fills $4s^2$, and then partially fills $3d$, namely by $3d^1$. The scandium atom is not an exception to the rule that $4s$ gets filled before $3d$. The exceptions with the "less than maximally" filled $4s$ are only some of the less important positive ions of scandium.

(Also note that he misleadingly wrote about the "energy of the $3d$ orbitals". But the energy of the atom simply cannot be written as a simple sum of the energy of orbitals. The electron-electron repulsion is what kills the independence of the electrons from each other.)

The true exceptions among atoms are chromium, copper, and partly nickel atoms that only have $4s^1$ even though they could have $4s^2$ – in the case of nickel, it's only one of two degenerate electronic structures that is "exceptional". But the "exception" is only partial because $4s$ could have stayed completely empty if $3d$ were "totally" preferred, but they always have at least $4s^1$.

Dr Scerri seems to half-realize that his comments make no sense and he has offered an amusing handwaving argument to obfuscate the contradictions of his picture:
There are some plausible reasons why this view persists. One factor is that in the case of potassium and calcium atoms, the $4s$ orbitals are genuinely preferentially occupied. But for these two atoms, the question of the relative occupation and ionisation of $4s$ and $3d$ orbitals does not even arise for the simple reason that there are no $3d$ electrons present.
Oh I see, so for potassium and calcium, the competition between $4s$ and $3d$ "does not even arise" because there are no $3d$ electrons present. That's very funny. It's exactly like saying "one can't decide whether Edward Witten is a faster sprinter than the Olympic winner because Witten hasn't even participated at the Olympics".

Cute. But Witten hasn't participated at the Olympics because he is a slower sprinter than the Olympic winner. And exactly the fact that potassium and calcium atoms have no $3d$ electrons is the most important proof or example of the fact that $3d$ states get filled after $4s$. Potassium and calcium atoms are the most important atoms (or ions) for which the competition between $4s$ and $3d$ matters. And $4s$ wins which is why no electrons at all are placed to $3d$ even though $4s$ is already getting filled.

The claim that "the competition is ill-defined" would only be valid if both $3d$ and $4s$ states remained empty (or if both Witten and the sprinter were absent from the Olympics), like in the case of argon and the "nice ions". But that's not the case. $4s$ is partly or fully filled in the potassium and calcium atoms, the sprinter participated, and therefore the contest has a clear result. The sprinter and $4s$ win, Witten and $3d$ lose.

$3d$ electrons can't be removed from the potassium and calcium atoms – because there are none to start with, as he says – but this claim strengthens the picture that in this environment, the $3d$ electrons are the more loosely bound ones and would get ionized first in similar atoms or ions. Sometimes, it may be the other way around – and $4s$ electrons get removed first. This is the case of the ionization of $Sc$ to $Sc^{2+}$ in which two $4s$ electrons are removed while one $3d^1$ electron stays there – much like you expect from the hydrogen-like ordering. This ionization follows the hydrogen-like rules because the positive ions are "closer" to the hydrogen problem because the nucleus-electron interactions are relatively stronger and the electron-electron repulsion is slightly less important in comparison.

You may also interpret this ionization example by saying that $4s$ gets removed first – even though it got filled first, too. This surprising (for some) conclusion follows from the complicated electronic structures of atoms and ions – in most of them, $4s$ electrons are more bound than $3d$ electrons but in some of them, exceptions of atoms and especially sufficiently positive ions, it's the other way around and $3d$ is more bound than $4s$.

Can Dr Scerri be really misunderstanding this simple point?

This kind of stuff (electronic structures) is surely being taught at high schools – and sometimes perhaps basic schools. But you may be pretty much sure that the education is useless at many places if even a media-savvy chemistry lecturer at UCLA – and together with him, undoubtedly, thousands of chemistry teachers who don't have any blogs – believes that something as simple as the claim that "$4s$ gets (almost always) filled before $3d$ but $4s$ may also be the first to go in the case of ionization" – must be wrong.

The statement is totally correct and there is no contradiction between the two parts of the statement.