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Consider an atom (carbon) with a ground state valence electron configuration of:

$$ \underset{3s}{[\uparrow \downarrow]} \underset{3p}{[\uparrow \vert \uparrow \vert \; \; ]} $$

In some molecules such as carbon tetrafluoride, carbon can "expand" to a more energetic electron configuration similar to:

$$ \underset{3s}{[\uparrow]} \underset{3p}{[\uparrow \vert \uparrow \vert \uparrow ]} $$

Although these orbitals would usually be mixed together and also paired together with the opposite atoms they are attached to.

Are there any cases where the reverse can happen?

For example could phosphorous "contract" from

$$ \underset{3s}{[\uparrow \downarrow]} \underset{3p}{[\uparrow \vert \uparrow \vert \uparrow ] }$$

to

$$ \underset{3s}{[\uparrow \downarrow]} \underset{3p}{[\uparrow \downarrow \vert \uparrow \vert \; \; ] }$$

Looking at the geometry of p orbitals and taking into account molecular orbital theory I think this is always of non benefit unless possibly the materials are put under external strains such as high pressure or heat.

Also, in the cases of atoms like chromium which are "expanded" by default I'd imagine contraction to the typical state could happen.

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    $\begingroup$ No. The electrons in carbon do not 'expand' when carbon forms bonds. The electrons reside in molecular orbitals which are wholly different to the atomic orbitals. $\endgroup$ – bon Jul 13 '16 at 19:07
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@bon is right in that you are talking about molecular orbital theory. Speaking broadly, the redistribution of electrons the way you mention happens if the orbitals actually become energetically non-equivalent. This can happen if an atom is surrounded by ligands in certain way. Because orbitals have certain orientations, certain complexes (f.e, tetrahedral) will energetically be more favorable than others (f.e, octahedral).

A simple example is the Jahn-Teller effect: have a look here and here. Essentially, for low-spin d$^7$ configurations you will have something resembling your example, although those are d orbitals; I actually do not know if this happens with p orbitals that you use in your example.

This Wikipedia article on crystal field theory also provides a nice overview of the different ways the orbital energy levels can distort in. However, for better understanding, I suggest a good textbook (I can give a few suggestions if you're interested).

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There are a few things worth clarifying here.

Hybrid orbitals do not exist in physical reality. I think I have said this enough recently for it to be clear. Hybridisation is a mathematical tool to produce nicer 2-centre-2-electron descriptions of bonding. When carbon (or any other atom) forms bonds it does not actually promote electrons into hybrid orbitals. The electrons in a molecule occupy molecular orbitals which are delocalised over the whole molecule (although in some cases there may be very little contribution from some atoms). These molecular orbitals are filled by electrons so as to produce the lowest energy state, in just the same way as atomic orbitals are filled to produce the lowest energy state.

Species (atoms, molecules, ions or whatever) always adopt the lowest energy state unless excited. For carbon the lowest energy state of an atom is: $$\underset{2s}{[\uparrow \downarrow]} \underset{2p}{[\uparrow \vert \uparrow \vert \; \; ]}$$

For chromium it is (see here for an explanation): $$\underset{4s}{[\uparrow]} \underset{3d}{[\uparrow \vert \uparrow \vert \uparrow \vert \uparrow \vert \uparrow ]}$$

Unpairing electrons in carbon or pairing them in chromium requires energy and so these atoms will not exhibit this electron configuration under normal circumstances. It might be possible to generate these configurations by exciting the atom with light of the right frequency or by producing excited states through some decay process. Singlet oxygen (an excited state of $\ce{O2}$) can be produced in both of these ways.

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