3
$\begingroup$

I just studied atomic orbitals in a theoretical QM class, and I'm left with several questions, that are probably more questions in quantum chemistry:

  • Many orbitals seem to have a preferred axis - how is that axis "chosen"? Does it move constantly, or is it actually a fixed direction in space? Why is there an axis at all when the problem is spherically symmetric? Does the shape of the orbitals have any bearing on chemistry?
  • We can see lobes in preferred directions - do these actually play a role in chemical reactions, or maybe in crystallography for the arrangement of crystals?
  • If I understand correctly, electrons are actually represented by superpositions of spherical harmonics - with an arbitrary superposition, the probability density for the electron could be anything - can we actually find the coefficients of the superposition an electron actually is in?
  • Lastly, there is an orbital which has a high probability of the electron being at the center of the nucleus (I don't remember which one) - is that anything interesting?
| cite | improve this question | |
$\endgroup$
  • $\begingroup$ Orbitals do not have preferred axis. Without a symmetry breaking field the orientation of an atom in free space is undefined. In orbital calculations the results are determined by a random choice of phase, which is usually chosen along one of the coordinate axes for easier analysis and visualization. That's a mathematical trick, not a physical phenomenon. As soon as you have more than one atom in physics, of course, the symmetry of free space is broken and the orbitals will align to form the lowest energy state of the configuration. $\endgroup$ – CuriousOne Apr 29 '15 at 18:53
  • $\begingroup$ Perfect! Makes complete sense - needed to bridge the gap between the theory class and what really happens. The theory class couldn't care less apparently about the practical "details" :-) $\endgroup$ – Frank Apr 29 '15 at 20:32
  • $\begingroup$ Is it fair to say that the orbitals are undefined for free space? The orientation of the atom being undefined, the orbitals themselves are undefined, and if there is no potential, there is no interaction, so we are not observing anything anyway? We don't know where the electrons are for the free atom, and we can't even observe that anyway. $\endgroup$ – Frank Apr 29 '15 at 20:48
  • $\begingroup$ Atomic orbitals are defined perfectly, but if you throw a bunch of atoms with well defined atomic orbitals into free space, they will tumble around and the average orbital that one would measure with some methods (e.g. by scattering light) might look spherical. Hold the same atoms with the "tweezers" of an electric or magnetic field in an aligned ensemble, and then the orbitals will start showing up in the scattering data. Please note that it's actually not quite that simple... e.g. chiral molecules will show polarization and randomly aligned crystal powder has x-ray refraction peaks! $\endgroup$ – CuriousOne Apr 29 '15 at 21:15
  • $\begingroup$ Thanks - that's what I thought - in free space, pretty much you would get "spherical", there is no reason it shouldn't be symmetrical - not because there are no orbitals, but because the "preferred axis" in most of the orbitals is just random on average. $\endgroup$ – Frank Apr 29 '15 at 22:08
2
$\begingroup$

Regarding your first two points: The symmetry axis of an orbital is free for a free atom. If it's bound to some other atom through one of these one-dimensionally elongated orbitals, the orientation of one orbital is fixed.

If you take e.g. carbon, silicon or germanium, you have one s orbital and three p orbitals, which are oriented perpendicular to each other. Because they are so close in energy, they hybridize and therefore lead to a an sp3 hybrid. In a simple picture, the electrons repel each other as far as possible, which leads to bonding in a tetrahedral configuration with bond angles of 109.4 °. Therefore the atomic orbitals influence the crystal structure or symmetry.

Your third point is honestly unclear to me.

I'm not sure, what you mean by the fourth one. The "center of mass" has different spacings with respect to the core. But there should not be an orbital, which has its highest probability located at the core. This would mean that electrons overlap with the core, which is not energetically favorable.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks! My third point was about the spherical harmonics (sorry - just out of computing things with $Y_{lm}$, very theoretical, no background in chemistry) being only "pure states" from which superpositions are actually built, like I'm used to building wave functions $|\psi\rangle=a|+\rangle+b|-\rangle$, for example (just out of the theory class again! :-) $\endgroup$ – Frank Apr 29 '15 at 14:19
  • $\begingroup$ The fourth point is just from reading a snippet of Wikipedia on "Atomic Orbitals": "The s-orbitals for all n numbers are the only orbitals with an anti-node (a region of high wave function density) at the center of the nucleus. All other orbitals (p, d, f, etc.) have angular momentum, and thus avoid the nucleus (having a wave node at the nucleus)." $\endgroup$ – Frank Apr 29 '15 at 14:20
1
$\begingroup$

$$ \newcommand{\ket}[1]{| #1 \rangle} $$

I'll try to answer the last two.

with an arbitrary superposition, the probability density for the electron could be anything - can we actually find the coefficients of the superposition an electron actually is in?

I'm a little confused about what you mean here. If we are given $\ket{\psi}$ as a combination of, say, $\ket{n \ell m}$ states, we can use that to calculate which spherical harmonics are present, since we know (for hydrogenic atoms) how to decompose $\psi(\vec{x}) = R_{n \ell} (r) Y_{\ell m} ( \Omega)$. Sure, the probability density "could be" anything, but if you want something really weird you might have to engineer it. (Look into recent work on Rydberg atoms for some fun superpositions.) On the other hand, do you mean, given a real electron in an atom, is there a way to directly measure the probability distribution? That's a much harder experimental question.

Lastly, there is an orbital which has a high probability of the electron being at the center of the nucleus (I don't remember which one) - is that anything interesting?

Actually yes. It's the $s$ ($\ell = 0$) orbital, because any angular momentum will keep you pushed out of the nucleus. This becomes important in hyperfine splitting, where the electron interacts with the nuclear magnetic moment, and this is obviously much easier if you spend time close to the nucleus, so the effect is larger.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Yes - my question is the harder experimental one: the theory is all well and tidy, I can do exercises all day long where the superposition coefficients are given, but what does it mean in practice in the lab? But my fast-changing understanding of quantum chemistry is that at that point, the question becomes more to compute approximations for clouds of electrons around molecules? $\endgroup$ – Frank Apr 29 '15 at 15:36
  • $\begingroup$ Looks like there's some neat stuff done with field-emission microscopy, but not really my field--I'm just googling and passing a link along blogs.nature.com/news/2009/09/… $\endgroup$ – zeldredge Apr 29 '15 at 15:39
  • $\begingroup$ ...high probability of the electron being at... I've always wondered: In a diffraction experiment, we talk about the probability of detecting a free electron at some location, but what sense does it even make to talk about locating an electron that is bound to an atom. You can't detect it, so why talk about anything other than the shape of the wave? $\endgroup$ – Solomon Slow Apr 29 '15 at 16:05
  • $\begingroup$ @jameslarge You can't detect the electron, you can detect the effects of its interactions, e.g. the hyperfine splitting. A more dramatic effect is the annihilation of positronium. Positronium is a bound state of an electron and a positron. The orbitals are hydrogen-like, as the system is just two charges, like hydrogen. But the wave function of the electron has some amplitude at the position of the positron. The two can (and do) interact, annihilating each other. $\endgroup$ – garyp Apr 29 '15 at 16:13

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.