# Why is zinc not considered to have a noble gas configuration?

I am wondering why we say that, for example, zinc does not have a noble gas configuration?

I would like to say beforehand that this question may has its origin in a confusion about the word "shell" or "outer shell". Wikipedia states that:

The properties of the noble gases can be well explained by modern theories of atomic structure: their outer shell of valence electrons is considered to be full.

Ok, so argon is a noble gas and its configuration is: $\ce{1s^2 2s^2 2p^6 3s^2 3p^6}$. When we speak of a shell we mean the principle quantum number $n$. In this case, the highest principal quantum number is $n=3$. However, for $n=3$ there is also the $\ce{3d}$ subshell available, which is not filled by argon.

So I conclude that "filled shell" does not mean that all available subshells for a given principal quantum number (here $n=3$) have to be filled. Because the $\ce{3d}$ subshell is empty here. Instead, I conclude that "filled shell" means, that whenever we start to fill a subshell (here $\ce{3s}$ or $\ce{3p}$), they must be filled. If there is another subshell with $n=3$ available, it has to be completely empty (like $\ce{3d}$).

But when we look at zinc, we find the configuration: $\ce{[Ar] 3d^{10} 4s^2}$.

Again, for the highest principal quantum number (here $n=4$) we have a filled subshell (here $\ce{4s}$). Additionally, all shells of the $n=3$ are also filled. There is no subshell (neither for $n=3$ nor for $n=4$) which is partially occupied. So, according to the definition from Wikipedia, what is missing for zinc to be in a noble gas configuration?

Again, let me stress, that I am not asking for what a noble gas is (I know that zinc is neither a gas nor a noble gas). I am asking what I am doing wrong, when I conclude that zinc obeys the definition of a noble gas from Wikipedia?

• Maybe I'm misunderstanding your point, but wouldn't by this definition every alkaline earth element (beryllium, magnesium, etc) also have a noble gas configuration since they are filling there $2s$ subshell?
– Tyberius
Mar 30, 2018 at 17:46
• Yes, exactly. So what is wrong with this argument? Mar 30, 2018 at 17:47

The difference hinges on the word "valence" in "their outer shell of valence electrons is considered to be full".

That's the difference between a $\ce{1s^2 2s^2 2p^6 3s^2 3p^6}$ configuration and a $\ce{1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^2}$ one. It's not just filled principle quantum numbers or filled subshells, but (un)filled valence subshells.

The definition of valence is a bit circular here. A valence subshell is one which can react to form a bond. For argon, all the valence subshells are filled because there's a substantial stability gap in adding that last $\ce{3p}$ electron and the first $\ce{4s}$ or $\ce{3d}$ electron. (If you think about potassium, the $\ce{3d}$ orbital is even less easily filled than the $\ce{4s}$ one.) That's why the $\ce{3d}$ orbital doesn't count as a "valence" orbital for Argon -- it takes too much energy to fill. Even though it has a principle quantum number of 3, it (re)acts like it has one of 4.

Contrariwise for zinc, there's a relatively small energy gap between the $\ce{4s}$, $\ce{3d}$ and $\ce{4p}$ orbitals. You probably already know this for $\ce{4s}$ and $\ce{3d}$, as they trade back and forth during the transition metals, but $\ce{4s}$ and $\ce{4d}$ are also relatively close in energy. This is part of the reason why atoms like carbon can "borrow" a $\ce{2s}$ electron to make four one-electron-donated bonds, rather than being forced to make two one-electron-donated bonds and one two-electrons-borrowed one.

So the definition of valence orbitals doesn't depend on their quantum numbers, but on the energy required to fill them. That's why zinc is not a noble gas - the $\ce{4p}$ orbitals count as valence (reactive) orbitals for zinc even while the $\ce{4d}$ don't.

• Thanks for this great answer. So, in a pictorial way, am I correct when I say, that we have a noble gas, whenever all orbitals in these blue bubbles (see link) are completely filled? i.imgur.com/NaNnScV.png The blue bubbles could be interpreted as the valence orbitals of an element, whenever at least one electron enters a bubble. One can also see why this is the case, since the energy levels of the orbitals inside a blue bubble are relatively close to each other, why the jump from one bubble to another costs more energy. Apr 3, 2018 at 14:45
• @thyme Yes, that's about right.
– R.M.
Apr 3, 2018 at 15:06

Albeit few excellent answers are already provided, I’d like to add a few additional points to the argument for the benefit of the readers:

Each shell in an element can be referred to a collection of subshells (e.g, $\mathrm{s}$, $\mathrm{p}$, $\mathrm{d}$, and $\mathrm{f}$) with the same principal quantum number ($\mathrm{n}$). Filling up of electrons in subshells takes place according to Aufbau Principle and the Madelung Energy Ordering Rule. The Aufbau Principle states that in the ground state of an atom or ion, electrons fill atomic orbitals of the lowest available energy levels before occupying higher levels. The Madelung Energy Ordering Rule is based on the total number of nodes in the atomic orbital, $\mathrm{n + ℓ}$, which is related to the energy (here $\mathrm{n}$ is the principal quantum number representing shells and $\mathrm{ℓ}$ is the azimuthal quantum number, which give us information about subshells). Accordingly, orbitals with a lower $\mathrm{n + ℓ}$ value are filled before those with higher $\mathrm{n + ℓ}$ values. In the case of equal $\mathrm{n + ℓ}$ values, the orbital with a lower $\mathrm{n}$ value is filled first (https://en.m.wikipedia.org/wiki/Aufbau_principle). Basically, two subshells with the same principal quantum number essentially have totally different energy levels. For example, for $\mathrm{2s}$ subsell: $\mathrm{n = 2}$ and $\mathrm{ℓ = 0}$, therefore $\mathrm{n + ℓ = 2 + 0 = 2}$, while for $\mathrm{2p}$ subsells: $\mathrm{n =2}$ and $\mathrm{ℓ = 1}$, therefore $\mathrm{n + ℓ = 2 + 1 = 3}$. That means electrons will occupy in $\mathrm{2s}$ first before $\mathrm{2p}$.

According to the Madelung Energy Ordering Rule, the $\mathrm{4s}$ orbital $\mathrm{n + ℓ = 4 + 0 = 4}$ is occupied before the $\mathrm{3d}$ orbital $\mathrm{n + ℓ = 3 + 2 = 5}$, because the lowest value of orbitals is filled first. According to the Aufbau principle, the $\mathrm{4s}$ sublevel is filled before the $\mathrm{3d}$ sublevel because the $\mathrm{4s}$ has lower in energy. The $\mathrm{4s}$ sublevel is only lower in energy if there are no electrons in the $\mathrm{3d}$ sublevel. As the $\mathrm{3d}$ sublevel becomes populated with electrons, the relative energies of the $\mathrm{4s}$ and $\mathrm{3d}$ fluctuate relative to each other, and at the end, $\mathrm{4s}$ would have higher in energy as the $\mathrm{3d}$ sublevel fills. This is evident in transition metal elements when electrons are lost from their orbitals, they are lost from the $\mathrm{4s}$ first because it is higher in energy.

In above account, it is clear that why ‘the $\mathrm{4p}$ orbitals count as valence (reactive) orbitals for zinc even while the $\mathrm{4d}$ don't’ (the point made by @R.M.). According to the Madelung Energy Ordering Rule and Aufbau Principle, the $\mathrm{4p}$ orbital $\mathrm{n + ℓ = 4 + 2 = 6}$ is occupied before the $\mathrm{4d}$ orbital $\mathrm{n + ℓ = 4 + 3 = 7}$, based on the lowest value of energy. As a matter of fact, electrons go into $\mathrm{5s}$ even before $\mathrm{4d}$.

Historical Remarks about noble gases:
$\ce {He}$ (1895), $\ce {Ne}$ (1898), $\ce {Ar}$ (1894), $\ce {Kr}$ (1898), and $\ce {Xe}$ (1898) were discovered in Earth by an English group led by Sir William Ramsay during their pioneering work (1894-1898), while $\ce {Rn}$ was discovered separately by German scientist, Friedrich Ernst Dorn in 1900. Naturally, these gases were given zero oxidation number because of their inertness with compare to other elements. That concept changed completely in 1962 when the first nobel gas compound, $\ce{XePtF6}$ was synthesized.