1
$\begingroup$

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 px, py and pz, 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?


Update: I found this question and this answer which both state that non-hybridized carbon has the configuration 1s2 2s2 2px1 2py1.

Why is this the case, and not, for example, 1s2 2s2 2py1 2pz1?

$\endgroup$
5
  • $\begingroup$ The hybridization happens when and only when a Carbon atom ¡s linked up to another atom. $\endgroup$ – Maurice Dec 1 '20 at 15:33
  • $\begingroup$ chemistry.stackexchange.com/questions/105045/… $\endgroup$ – Mithoron Dec 1 '20 at 21:16
  • 2
    $\begingroup$ The choice of coordinate system is arbitrary, you cannot distinguish the axes and you can relabel then as they fit your choice. $\endgroup$ – Martin - マーチン Dec 1 '20 at 21:39
  • $\begingroup$ @Martin-マーチン Thank you for your comment. I think it does answer my question; you might consider posting it as such. I than assume that saying that the $p_x$ and $p_y$ orbitals are filled first is just convention? $\endgroup$ – Jonas Dec 1 '20 at 21:43
  • 1
    $\begingroup$ Unfortunately this is maybe an approximation to an answer. This actually requires a deeper dive into quantum mechanics. There is a lot to unpack, and I don't have the time for it to write it down. But someone else might. In the meantime I'm happy my comment helped a bit. $\endgroup$ – Martin - マーチン Dec 1 '20 at 21:47
3
$\begingroup$

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.


* 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 $\hat{L}_z$).

$\endgroup$
1
  • 1
    $\begingroup$ That's a good answer, thanks for writing it up. I'm afraid I can't add anything else useful to it. So here's a fun fact: We often use the z-axis as the bond axis, which may lead to some confusion (see chemistry.stackexchange.com/q/57125/4945). Weirdly we also like to use the z-axis for empty orbitals (and lone pairs), like in carbocations (or anions), which may lead to even more confusing situations. Drag and drop, and mix and match. And boy, these things may become weirder ... As Ivan would say: so it goes. $\endgroup$ – Martin - マーチン Dec 2 '20 at 22:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.