Why is argon a noble gas given that the 3d subshell is still empty?
More generally, why is it that the filling of a p sub-shell makes an element noble rather than s, d, or f sub-shells, or completed n-levels?
Let's start with the prior that in a hydrogen atom the principle quantum number determines the energy – so maybe full shells correspond to noble gases.
Now, I understand that in helium the n=1 level is full, so it is a noble gas.
Then, beryllium has a full 2s subshell, but empty 2p, so it is a metal.
Neon has the n=2 level full, and is a noble gas.
Magnesium has a full 3s subshell, but not a full n=3 level and is a metal – so far so good.
But, oops, argon has full 3s and 3p subshells but empty 3d, yet it is a noble gas.
At this point we look at the aufbau principle and say "ah, your prior is no good, the 4s is lower energy than 3d, and this is the first time it has kicked in – so there you go".
But, as we continue down the periodic table there are more noble gasses and they happen only with the filling of a p subshell. Why is krypton noble, but not zinc? Even palladium cleanly fills n=4 up through 4d (with only 4f empty) but it is still not a noble gas. Ytterbium fills an f subshell for the first time and it isn't noble. What makes the filling of a p subshell so specially inert compared to s, d, and f subshells?
The aufbau principle does not explain this. Draw the classic "diagonal aufbau chart" and circle where the noble gasses are. It doesn't add any insight. Noble gasses only happen when you fill a p subshell. Why?
Edited to add: When an electron is promoted out of a two electron s-shell into an empty p-shell the spin will flip and energy is saved because of the Pauli exchange (two same spin electrons can avoid each other). The one remaining s electron, and the one new p electron benefit from each having a sub-shell to themselves – kinda a push vs. the s to p promotion. This, to me, intuitively explains why full s shells aren't noble.
User4552 points out that brute force calculation shows that energy gaps after a p shell is filled are large. But it doesn't help my intuition to just hear "the computer calculations reveal that …". So can anyone help with intuition on why full p-shells, rather that full d-shells, are noble?
Best Answer
As a proxy for an atom's reluctance to form chemical bonds, we can use the energy required to lift one of its outer electrons to the next higher orbital, pretending as usual that the "orbital" concept is still at least approximately valid in a multi-electron atom. Then the question is why this energy gap tends to be significantly larger after filling a p-shell ($\ell=1$) than it is after filling s- or d- or f-shells ($\ell=0,2,3$).
As usual, let $n$ and $\ell$ denote the radial and angular quantum numbers, respectively. Let's accept that orbitals are filled in order of increasing $n+\ell$, and then in order of increasing $n$ whenever the first rule is neutral. (Reality is a little more complicated, but these rules mostly work pretty well.) The following list shows the filling order defined by these two rules, along with the number of nodes in the orbital wavefunction's radial factor: \begin{align} n+\ell &\hskip1cm (n,\ell) &\hskip1cm \text{radial nodes} & \\ \hline 1 &\hskip1cm (1,0) &\hskip1cm 0 &\hskip1cm \text{helium}\\ 2 &\hskip1cm (2,0) &\hskip1cm 1 \\ 3 &\hskip1cm (2,1) &\hskip1cm 0 &\hskip1cm \text{neon}\\ 3 &\hskip1cm (3,0) &\hskip1cm 2 \\ 4 &\hskip1cm (3,1) &\hskip1cm 1 &\hskip1cm \text{argon} \\ 4 &\hskip1cm (4,0) &\hskip1cm 3 \\ 5 &\hskip1cm (3,2) &\hskip1cm 0 \\ 5 &\hskip1cm (4,1) &\hskip1cm 2 &\hskip1cm \text{krypton} \\ 5 &\hskip1cm (5,0) &\hskip1cm 4 \\ 6 &\hskip1cm (4,2) &\hskip1cm 1 \\ 6 &\hskip1cm (5,1) &\hskip1cm 3 &\hskip1cm \text{xenon} \\ 6 &\hskip1cm (6,0) &\hskip1cm 5 \\ 7 &\hskip1cm (4,3) &\hskip1cm 0 \\ 7 &\hskip1cm (5,2) &\hskip1cm 2 \\ 7 &\hskip1cm (6,1) &\hskip1cm 4 &\hskip1cm \text{radon} \\ 7 &\hskip1cm (7,0) &\hskip1cm 6 \\ 8 &\hskip1cm (5,3) &\hskip1cm 1 \\ \end{align} In every noble-gas case, and only in these cases, the next available level is an s-orbital ($\ell=0$), which has more radial nodes than the levels immediately before or after it in the sequence. (This pattern works for helium, too, even though helium doesn't have a filled p-shell.) If we could understand intuitively why energy gap between the ground state and the first excited state is relatively large when the first excited state is an s-orbital, then we would have at least a partial answer to the question.
The order in which the shells are filled (the sequence shown above) indicates that radial nodes are more costly than "azimuthal" nodes, because for a given $n+\ell$, the cases with smaller $\ell$ (fewer "azimuthal" nodes) have higher energy. Accepting this trend as an axiom, we can focus our intuition on the radial part.
Intuitively, if we think of the radial nodes as "no-fly zones" for that electron, then having a larger number of radial nodes may correspond to having less freedom to re-arrange the multi-electron system to minimize the energy in the presence of electron-electron interactions. Here's the key idea: by analogy with a construction-induced traffic jam, the impact of each additional node (each additional no-fly zone) could be an increasing function of the number of nodes already present. This picture suggests that lifting an electron from a filled p-shell up to the next available s-shell should be more costly than, say, lifting an electron from a filled d-shell up to the next available p-shell, because the former lift requires increasing an already-larger number of radial nodes. Using the magnitude of the energy-gap as a proxy for an atom's reluctance to form bonds, this suggests that the noble gases should be less reactive (relatively), at least among atoms with the same value of $n+\ell$.
The key idea is that the energy-cost of each additional radial node is an increasing function of the number of radial nodes already present. This seems to be consistent with the information I've seen, but I have no real justification for anticipating it except for the dubious traffic-jam analogy. Even if the intuition is correct, it isn't quantitative enough to predict just how noble the noble gases are. It is suggestive at best, but at least it doesn't rely entirely on a computer. That's why I thought it was worth posting.