# Can orbitals in a molecule be occupied via excitation like they can in atomic orbitals?

I am facing a conundrum in the way I am viewing molecule orbitals due to the existence of excited states. What I wanted to know is, since atomic orbitals can exist in excited states (i.e. you get hydrogen accessing states outside the 1s orbital and such when a high-enough energy photon is used), is this the case with molecular orbitals, where 'technically' many more states exist than shown on a molecular orbital diagram? I mean, they tend to only show the valence and core electrons, but not past that, hence the question. For example, could methane, that only whose MO diagram only shows molecular orbitals composed of 1s, 2s, and 2p orbitals, have an electronic transition that occupies a 3s orbital? And if so, how would that occur? Would it occupy one particular atom or be 'spread out' like most electrons in MO theory? Sorry, long question, but something that is bugging me for some strange reason.

• Methane does not exactly have 1s, 2s, and 2p orbitals. Instead, it has molecular orbitals composed of those. And yes, any molecule has multiple excited states that can be reached. – Ivan Neretin Nov 13 '17 at 8:06
• Okay, fixed that. Tend to make mistakes in terminology when I write in a rush. – Matthew John Jackman Nov 13 '17 at 8:08

Most of what applies for atomic orbitals in atoms can be transferred to molecular orbitals in molecules. This includes that electrons can be excited to higher lying orbitals, creating excited configurations.

Also, just as for the hydrogen atom, there are in principle infinitely many orbitals available. They are simply truncated since the higher lying ones are less important (and for numeric reasons).

• Thanks for the answer. In the case of virtual orbitals, would the 'cut-off' for 'important' orbitals tend to be set really only to those that aren't outside your typical molecular orbital diagram for that reason? I mean, of course excluding the addition of things like polarization basis sets and such. That is actually why I got the question actually - was reading about post-HF methods and was wondering about what they consider as the "full CI" if there are technically infinite states... guessing it must just be the 'important ones' i.e. valance electrons, maybe some rare Rydberg ones, yes? – Matthew John Jackman Nov 13 '17 at 8:25
• Usually the cutoff includes all valence orbitals (or what you can form fram all valence AOs). Things like polarization functions are about the AO basis. But also influences the MOs, since you get as many MOs as AOs are supplied. Which also means FCI depends in the AO basis set. It can only consider as many MOs as are available. – Feodoran Nov 13 '17 at 8:33
• I guess I would like to ask one more addition to what we are talking about, if you don't mind - do we know the defined shapes of these molecular orbitals like we do for, say, the ones generally shown in MO diagrams? Or is it a bit more abstract since we are talking about more complex mixtures of AOs? – Matthew John Jackman Nov 13 '17 at 9:44
• If am not sure if I get your question right, but occupied orbitals can be quite intuitive, e.g. core orbitals which are al,pst identical to AOs, or bonding MOs where you can relate the number of nodes to the bonding character. These orbitals which are "generally shown in MO diagrams" do come somewhere from ;) However, virtual orbitals get more abstract and occasionally change with the AO basis set. Maybe this is worth a separate, new question ... – Feodoran Nov 13 '17 at 11:56

Customarily, LCAO molecular orbital schemes are drawn using the atomic orbitals of the valence shell, ignoring both core orbitals and most higher-shell orbitals. For example:

• the MO scheme of methane will contain the $\mathrm{1s}$ orbitals of the hydrogens and the $\mathrm{2s}$ and $\mathrm{2p}$ orbitals of carbon because these correspond to the valence shell

• the MO scheme of $\ce{CO2}$ will use the $\mathrm{2s}$ and $\mathrm{2p}$ orbitals of both carbon and oxygen and ignore the $\mathrm{1s}$ core orbitals as well as all higher orbitals

• the MO scheme of $\ce{HOCl}$ will use hydrogen’s $\mathrm{1s}$ orbital, oxygen’s $\mathrm{2s}$ and $\mathrm{2p}$ orbitals and chlorine’s $\mathrm{3s}$ and $\mathrm{3p}$ orbitals (see also the next example)

• when constructing the MO scheme of hexafluoridoferrate(III), the $\mathrm{2s}$ and $\mathrm{2p}$ orbitals of fluorine are used while the $\mathrm{3d, 4s}$ and $\mathrm{4p}$ orbitals are used — these together correspond to the valence shell.

For each atomic orbital entered into the equation the solution must include exactly one molecular orbital. So having two orbitals interact will always result in a more bonding orbital (the same phases overlap) and a more antibonding orbital (opposite phases overlap). Thus, we will end up with $8, 12, 9$ and $33$ three orbitals respectively for the examples above. This already provides us with ample orbitals to excite electrons into; in the methane case, these would correspond to four $\unicode{x3c3}^*$ orbitals, three of which are degenerate due to symmetry.

Nothing is stopping us from including the core orbitals in the calculation or additional unoccupied atomic orbitals. In fact, quantum chemistry programs often include these orbitals in the calculations. The core orbitals do no harm (you will end up with a set of MOs at very low energy that are basically confined to the nuclei; linearcombining these will lead to exactly the set of core orbitals you threw into the equation). The virtual orbitals are actually typically used as polarisation functions to provide a better final result.

Including the virtual atomic orbitals will merely create more virtual molecular orbitals at the top of the energy scale. These can indeed be accessed by excitation with photons of appropriate wavelengths — in theory. In practice, they are rather far removed and photons arriving with appropriate wavelengths will also be energetic enough to do much more damage, e.g. ionise electrons completely. As an example, the electrons in molecular orbitals of $\ce{[FeF6]^3-}$ corresponding approximately to the ligands’ p orbitals already require ultraviolet photons to even be excited to the $\mathrm{t_{2g}}$ orbitals. Imagine how much energy would be needed to excite a $\mathrm{t_{2g}}$ electron to levels formed with $\mathrm{3s}(\ce{F})$ or $\mathrm{5s}(\ce{Fe})$ orbitals!