9
$\begingroup$

Do orbitals exist even when they are not occupied?

For example: $\ce{Cr^{+3}}$ has the configuration $\ce{[Ar]}\mathrm{3d^3}$ with the other two $\mathrm{3d}$ orbitals empty. We know the other two orbitals exist, since the metal "uses its empty $\mathrm{d}$ orbitals" to form complexes.

But, for $\ce{Na+}$ with configuration $\ce{[Ne]}$, do the $\mathrm{3s,3p}$ and $\mathrm{3d}$ orbitals exist too?

$\endgroup$
  • $\begingroup$ click on the edited $x$ ago to see that I did not touch anything that was not MathJax, thank you. (Brian is the culprit.) $\endgroup$ – Jan Nov 26 '15 at 13:56
  • $\begingroup$ @Jan Apologies. $\endgroup$ – yasir Nov 26 '15 at 14:03
7
$\begingroup$

Do orbitals exist even when they are occupied? Orbitals are mathematical constructs and as such they exist in a mathematical sense regardless of whether they are occupied or not. Whether you can actually say that orbitals exist in reality is a bit dubious. They certainly provide a good description of the observed electron density and behaviour of atoms but they are still only a model. Like everything in science, we develop a model to describe what we observe but that model is not perfect and does not necessarily represent what actually exists. Particularly on a quantum scale, the notion of well defined objects that exist in space starts to become a bit fuzzy.

$\endgroup$
  • $\begingroup$ In that case every atom can be thought of having empty orbitals.Then why do we argue this happens because this particular element has empty orbitals to accommodate the electron pair? Empty orbitals always exist as per your answer. $\endgroup$ – yasir Nov 26 '15 at 12:38
  • 3
    $\begingroup$ @yasir Yes, they do; but what is meant is usually accessable empty orbitals. Example: phosphorus and sulphur have unaccessable $\mathrm{3d}$ orbitals (they are high in energy and cannot really participate in hybridisation). A carbon-bromine bond has an accessable $\unicode[Times]{x3c3}^*$ orbital which can be attacked by the lone pair of a nucleophile because it is relatively low in energy. $\endgroup$ – Jan Nov 26 '15 at 13:08
  • $\begingroup$ @yasir By calculating (or measuring) energies or by qualified a priori guesses. $\endgroup$ – Jan Nov 26 '15 at 14:35
  • $\begingroup$ So for $\ce{Hg^{+2}}$ the $\ce{6d}$ orbitals are accessible?Actually it has something to do with oxymercuration demercuration mechanism, so bear with me. $\endgroup$ – yasir Nov 26 '15 at 14:58
  • $\begingroup$ @yasir - There is a finite probability that the electrons are in any number of infinite electronic states - which includes those higher energy "orbitals" - but the probability is exceptionally low in the ground state. That probability can change - higher or lower - depending on the environment (applied electric field, for instance). I'd say all the "orbitals" are accessible but some of them (the ones we see for ground state electronic configurations) are way more likely to be populated than the ones we'd expect to see populated in an excited state. $\endgroup$ – Todd Minehardt Nov 26 '15 at 16:19

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.