Electron configuration of non-hybridized carbon

Question

I'm currently learning about orbital hybridization in carbon.

I see that carbon has an electron configuration of 1s² 2s² 2p².

However, there are actually three p-orbitals in the second shell, namely p_x, p_y and p_z, which could each hold two electrons. However, there are only two electrons to be distributed on these three orbitals. Which orbitals do they occupy? Since their energy levels are identical, is it random?

I also learned that the three p orbitals can be combined with the 2s orbital (sp³ hybridization) to form four new hybrid orbitals. It then makes sense that the four electrons available each occupy one of the four new orbitals (so that each of them holds one electron), because this way, they can have the greatest spatial distance to each other (I assume they "want" this as they repel each other).

However, how are the second shell electrons distributed in a non-hybridized carbon atom? Does such an atom even naturally occur or does it always hybridize?

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Update: I found this question and this answer which both state that non-hybridized carbon has the configuration 1s² 2s² 2p_x¹ 2p_y¹.

Why is this the case, and not, for example, 1s² 2s² 2p_y¹ 2p_z¹?

Answer 1

The answer comes in several layers, as Martin has alluded to. I'll try and give a short summary: each point goes slightly deeper than the previous one.

1. For an isolated atom, the labelling of the orbitals is arbitrary. That is to say, the $p_x$, $p_y$, and $p_z$ orbitals are all interchangeable. More formally, this reflects the isotropy of the system, which means that all directions are equal. There is nothing that can physically differentiate the $x$-axis from the $y$-axis or the $z$-axis.

2. You can still arbitrarily assign $x$-, $y$-, and $z$-labels to three perpendicular axes. If you kept these coordinate axes consistent between different atoms, such that every $p_x$ orbital pointed in the same direction, you should not expect to find in all molecules that the $p_x$ and $p_y$ orbitals are always occupied. Instead, on average, you would have one-third of them with the configuration $(p_x)^1(p_y)^1$, one-third with $p_xp_z$, and one-third with $p_yp_z$. Again, this symmetry stems from the fact that there is no physical factor that favours any of the three coordinate axes over each other.

3. The previous point assumes that you have made a measurement of the electrons, such that you can definitively say that they are in a specific set of orbitals.* But prior to this measurement, there is no reason that they should already be in that state. It's better to posit that an isolated carbon atom is in a superposition of those three states, i.e. $p_xp_y$, $p_xp_z$, and $p_yp_z$. My old question: What constraints are imposed on a wavefunction by the symmetry of the system? is somewhat related, except that it deals with boron (which has the same issue).

Having gone through this, there is still the question of why we write $p_xp_y$ when it's not really accurate. The answer is that it's far easier than dealing with a superposition of states (in (3)), and since space is isotropic, $xy$ is the same as $xz$ and $yz$ as long as you relabel the axes, which means that (2) is more of a technicality rather than a real "gotcha". Also, it's perhaps not worth overthinking this: once you form a bond, all of this ceases to have any meaning. In a molecule, electrons don't remember "where" they come from, so it becomes pointless to quibble over whether it was in the $p_x$ or $p_y$ orbital.

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* You can't measure individual electrons in an atom, you can only measure their collective properties. Also, I'm skeptical about whether you can measure a state of $p_xp_z$ given that $p_x$ and $p_z$ are eigenstates of non-commuting operators, but let's leave this aside for now (or assume, perhaps, that we're dealing with $p_{-1}$, $p_0$, and $p_{+1}$, which are all eigenstates of $\hat{L}_z$).