In the p-orbitals, there are 3 dumbbell-shaped lobes, each of which can contain 2 electrons. In the diagrams of the d- and f-orbitals, there are two or more lobes (or orbital shapes) or each value of m. In all of these cases, which lobes/shapes do the electrons occupy for any given element?


The question implies a misunderstanding about orbitals and their representation as dumbbell-shaped lobes. These plots show the relative orientation of orbitals to each other by showing regions that represent some threshold probability (say, 90%) of finding an electron contained therein. The underlying probability distribution has the same symmetry as the orbital itself because it defines the orbital. Any (spatial) orbital can hold up to two electrons$^1$, regardless of its shape. It thus makes little sense to ask in which lobe (or, further, which half of the dumbbell, since there is a nodal plane in between the halves) an electron is.

Moreover, any single atom is spherically symmetric in the absence of external fields, meaning that the shapes of the orbitals are not actually important to understand electron distribution in a free atom. The shapes are useful tools to understand certain bonding concepts such as crystal/ligand field theory nevertheless.

$^1$ a spin orbital holds one.

| improve this answer | |
  • $\begingroup$ doesn't answer the question, and kindly avoid insulting comments. $\endgroup$ – charlie K3 Jan 21 '18 at 0:13
  • $\begingroup$ to clarify: every question posed on every stackexchange site, and every lecture hall, implies some level of misunderstanding by the questioner. Otherwise no need to ask a question. Perhaps the answer you're looking for, albeit not necessarily correct, would be: "We cannot be certain in which of several possible orbitals any given electron resides, because of the XYZ Principle. $\endgroup$ – charlie K3 Jan 21 '18 at 0:27

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.