molecular orbital theory - Hyperlithiated Carbon Species

Below is replicated Question 1 from the Final Qualifying Exam of the Australian Chemistry Olympiad, 2004B, here.

Question:

The theory of promotion-hybridization is quite successful at explaining why covalent molecules with a central atom from period 3 or higher (e.g. $\ce{SF6}$, $\ce{PF5}$, $\ce{I3}$) are stable even though they are hypervalent (i.e. they appear to "exceed the octet rule"). However, hybridization is inadequate in explaining the remarkable stability of molecules such as the so-called "hyperlithiated" carbon species such as $\ce{CLi6}$. Molecular orbital theory can be applied to $\ce{CLi6}$ (Reed and Weinhold, 1985) for an alternate explanation to when and why hypervalent molecules with central atom of period 2 can be stable.

Of all the different types of atomic orbitals (e.g. $s$, $p_x$, $d_{yz}$, etc) of a general central atom, six of these will participate in the forming of sigma bonds with the molecular orbitals of six ligands in an octahedral geometry. The valence $s$ orbital is one of them. Note that the ligands approach the central atom along the three coordinate axes.

(a) Which other five atomic orbitals can also do this?

Now, we shall consider the molecular orbitals (MO) of the 6 $\ce{Li}$ $2s$ orbitals (in an octahedral geometry). The lowest energy MO is given below in figure 1:

According to quantitative calculations, the MO in figure 1 is the only MO lower in energy than the uncombined lithium $2s$ orbital; all other MOs are slightly higher in energy than uncombined $\ce{Li}$ $2s$ and can thus be considered as antibonding. (Note that this is generally true for 2 to 8 Li atoms arranged symmetrically around the central atom).

The $s$ orbital will overlap with the MO given in figure 1. Exactly one of the other five MOs will have the right symmetry to overlap (and form a sigma bond) with each of the other five atomic orbitals of the central atom.

(b) Hence draw the other five MOs obtained by linear combinations of the 6 $\ce{Li}$ $2s$ orbitals. Indicate their relative energies on an energy diagram.

When the central atom is carbon, three of the six MOs (of the 6 $\ce{Li}$ $2s$ orbitals) will not combine with the corresponding atomic orbital of carbon. One of these is the one drawn in figure 1. Because carbon is much more electronegative than lithium, the $2s$ orbital of carbon is too low in energy to mix with the MO in figure 1.

(c) Which other two MOs will not combine when carbon is the central atom? Provide a rationale as to why this is so.

(d) Hence, draw the molecular orbital energy diagram of $\ce{CLi6}$. Keep in mind the relative energies of carbon AOs and the six Li MOs (as carbon is more electronegative than lithium).

(e) Using the energy diagram in part (d), explain the stability of $\ce{CLi6}$.

(f) Propose an alternative to the "octet rule" when predicting stability of hyperlithiated species of period 2 atoms. Using this rule, identify possible neutral (i.e., not cations or anions) hyperlithiated species of nitrogen and oxygen that can exist due to the same stability reason as that of $\ce{CLi6}$.

(g) Can $\ce{CH6}$ exist? If so, why so? If not, why not?

My response:

(a) These are the $p_x$, $p_y$, $p_z$, $d_{z^2}$, and $d_{x^2 - y^2}$ orbitals.

(b) Assuming analogy with octahedral complexes, Googling provides this, pg. 10 of which appears to answer the question. Given that $E_d$ < $E_s$ < $E_p$, the $d$ orbitals represented would be found at the bottom and $p$ at the top of the energy diagram.

• Is an octahedral complex analogous to $\ce{CLi6}$?


• What is a linear combination of atomic orbitals?

(c) By logical reasoning, these would be the two similar $d_{z^2}$, and $d_{x^2 - y^2}$ orbitals. I am uncertain regarding the rationale.

• Is it because it is unfeasible for carbon's electrons to be in $3d$ orbitals?

(d) I really have no idea.

• If three of the six possible MOs will not combine, does that mean that they are not present in the MO diagram?


• Since carbon is more electronegative than lithium, it should contribute more to the bonding orbitals, while lithium should contribute more to antibonding orbitals.


• How are the 10 electrons in $\ce{CLi6}$ distributed? In order to impart stability, there must be more electrons in bonding than in antibonding orbitals. Can we simply have three bonding MO orbitals with 2 electrons and three antibonding MO orbitals with 4 electrons between them?

(e) By logical reasoning, the larger number of electrons in bonding rather than in antibonding orbitals imparts stability to the molecule since bonding electrons are not entirely counteracted by antibonding electrons.

(f) I really have no idea.

• By guesswork, I would propose the decuplet rule, in which hyperlithiated species from period 2 are stable with 10 valence electrons, producing the species $\ce{NLi5}$ and $\ce{OLi4}$.

(g) By logical reasoning, yes: it satisfies the conditions of (f).

• My gut says no.

Given that my background in chemistry is limited to high school AP Chemistry, please curtail the material provided in your responses. Thanks for all your help in advance!

Answer

(a) Correct.

(b) Let me start with Linear Combination of Atomic Orbitals.

That's a quantum calculation method that assumes that all molecular orbitals can be derived from (basic) atomic orbitals by a method called linear combination. You might have heard of linear combination from vector geometry. It's basically just adding things up with different preceding factors. So 2x - 3y is a linear combination of x and y.

The thing about linearly combining atomic orbitals is that you must end up with the same amount of orbitals as you started out with (so you combine 6 atomic orbitals to make 6 molecular orbitals). There's also additional stuff you need to take care of (the squares of all the preceding factors of one AO need to sum up to give one), but that's only of marginal importance here.

An example of linear combination would be your average H–H sigma bond. You have two atomic orbitals (1s of each of the hydrogens) which you combine. I'll call the 1s orbital of the left hydrogen l and that of the right hydrogen r, for simplicity. The two simplest linear combinations are l + r and l - r. In the first one, you get positive interference, i.e. you add a positive phase to a positive phase – think of it as the positive half of a sine wave added onto another positive half of a sine wave: it'll get bigger. This type of interaction is favourable, and will thus give you a molecular orbital energy that is less than that of a separated atomic orbital.

But you also need to remember to create a second MO, the l - r. Here, you would be adding plus to minus, which gives zero. So you have a plane in between the two protons where the resulting wave function has zero value, is non-existant (a nodal plane). Because the nodal plane is between the atoms, the resulting molecular orbital is antibonding, or higher in energy. The basic rule is that the more nodal planes an orbital contains, the higher its energy level will be.

The question asks you to find the further five linear combinations that will lead to those group orbitals capable of overlapping withthe central carbon's orbitals. So you basically know what to look for; and yes: the CLi₆ is analogous enough to an octahedral complex, that you can just copy that image. If you didn't have either the picture or the hint, you would need to use group theory and reduction formulae, which I won't cover here (for those interested: the group orbitals will transform as a_1g, e_g and t_1u).

I don't understand your reasoning for putting the d-type orbitals lower in energy (i.e. more stable) than s- or p-type ones. Usually, it's s < p < d. But the better way to answer the question would be using nodal plane counting: The orbital that will mix with the carbon'S s orbital has no nodal plane, and will therefore be the lowest (a_1g, for those interested). The three that will mix with the carbon's p orbitals are second, because they contain one nodal plane each (t_1u). Finally, those that will mix with d orbitals are highest, because they contain two nodal planes (either parallel or perpendicular to each other. Their energy level will be the same, though, says group theory) (you guessed it, those are e_g).

(c) Yes, that can be regarded as the easiest answer. Carbon, being an element of the second period, only 'has' (term is incorrect, but you get the picture) 1s, 2s and 2p orbitals. 3s and 3p might be accessible by irradiation. 3d is not, as it is even higher in energy than 4s.

The same rationale applies to the higher order main group elements. Even though silicon, being an element of the third period, could be considered to access d orbitals to create anions like $\mathrm{SiF}_6^{2-}$, the d orbitals' energy is too far removed from that of the s and p orbitals, that it cannot be considered to take part in bonding.

Note that transition metals – the elements usually associated with octahedral surroundings and d-orbitals – usually use 4s and 3d (and sometimes 4p) orbitals. 3d is a lot closer to 4s and 4p than to 3p.

(d) Start off by thinking what the relative energy levels of the orbitals in question are. Carbon, as was pointed out, is more electronegative than lithium, so all of its orbitals will be lower in energy than the lithium orbitals (the difference is significant here).

That should give you a picture of carbon orbitals on one side (2s, rather far-ish down, three 2p's, still pretty far down) and the three sets of lithium group orbitals on the other side (the one that I called a_1g being the lowest, but probably higher than carbon's 2p's; the three I called t_1u next and the two e_g last; those inside a group at the same height, of course).

Next, you need to find out which orbitals can recombine favourably. If you attempted to combine the orbital shown in figure 1 with a carbon 2p orbital, that intereference would be non-bonding or zero (you have bonding interactions on one side cancelled out by exactly the same amount of antibonding interactions on the opposite side). Note that they helped you by saying that the figure 1 orbital (and thus also carbon's 2s) and the two lithium orbitals that would bond with d-type orbitals do not take part in favourable interactions; so only the p-type ones remain, each of them creating a bonding-antibonding pair of molecular orbitals.

Moving from the edges to the centre of the diagram, you would move the carbon's 2s plus the three lithium group orbitals not taking part horizontally into the middle. Carbon's 2p and the other three lithium group orbitals are then mixed, lowering the 2p's energy while raising the lithium orbitals' energy.

You should end up with an energy diagram of the following type:
C(2s) - 3 C(2p) -- 1st Li -- 3 2nd Li - 2 3rd Li. It's not clear (nor can it be deducted a priori) whether the 2p orbitals will end up lower than the 2s (or the 2nd Lithium group higher than the third), but it is to be assumed that they shouldn't change their places too much.

Finally, fill in the electrons in the way you're used to. Two go into C(2s), six into the three C(2p), and two into the first Li orbital (the one shown in figure 1). To answer your questions:

• non-combining MOs can or can not be present in the MO diagram. They should, if they are part of the valence orbitals. They should not if they are core orbitals (note we always omitted the seven 1s orbitals). Not combining however means, that they do not change height from where you put them initially.



• Yes, that is correct, but that is more of a consequence we can think of after mixing has been done.


• If we include the nonparticipating orbitals, the question is no longer a question. If they weren't present, though, you would do exactly what you proposed. You would fill in three electrons spin up into three antibonding MOs and one further electron spin down. The resulting system would be degenerated. It would not be stable, and would reduce symmetry to force the three orbitals apart, creating a discrete energy level for the single paired electron.

(e) Yeah, that was a pretty good guess into the wild, that would always be correct ;)

To use the molecular orbital diagram as a reasoning, though: Consider the state before bonding. The four carbon orbitals have four electrons between them (2 in 2s, 2 degenerated in 2p). The six lithium orbitals have six electrons between them (2 in the figure 1 orbital, 4 in the three orbitals that will take part in bonding; this state is also degenerated). After the bonding, not only have we moved four electrons out of relatively unstable orbitals into a lot more stable ones (those from the lithium orbitals into the carbon-centered ones), but we have also removed two degenerated states and turned them into clearly defined ones. Both are good for the system's stability.

(f) I think your guesswork is pretty much spot on. (The orbital diagrams would be totally different ones, though; as NLi₅ would be a trigonal bipyramid (symmetry group D_3h) and OLi₄ a tetrahedron (symmetry group T_d) as opposed to the carbon octahedron (O_h).

(g) I'm pretty sure this is a yes-but case. Yes, just looking at what we have for Li would suggest that H would do the same thing. But, hydrogen's atomic orbitals are lower than lithium's, as it is a lot more electronegative. That in turn means that there is significant mixing to be observed between C(2s) and the figure 1 type orbital. Therefore, the C(2s) orbital would again be stabilised, while the hydrogen group orbital would be destabilised – most likely so far that it reaches the energy level of the three higher group orbitals. While they were clearly empty in the lithium case, they would most likely be populated in the hydrogen case, destabilising the system.

Another thing to consider is the relative stability of CH₄ versus CLi₄. Again, the all equal phase group orbital of the lithiums would be too high to significantly mix with the C(2s) orbital in a tetrahedric case. The all equal phase group orbital of hydrogens does mix quite significantly, creating a much more stable CH₄ molecule than CLi₄. Thus, CH₆ would eliminate hydrogen to create CH₄, while CLi₆ is stable enough (and CLi₄ unstable enough) to not eliminate Li₂