Do filled Orbitals also hybridize?

Question

Recently I was watching a video on Valence Bond Theory, on the MIT Open-Course youtube channel. The teacher there said that

For hybridization to occur, electron promotion from fully filled orbitals to empty ones is a must.

But here in a book it is specifically mentioned that

Electron promotion, though common, but is not 'a must' for hybridization.

Also, it is clearly written that sometimes even the filled orbitals take part in hybridization.

I tried finding some examples of such a case on the web, but could not. It would be a great help if someone tells me even one example where the fully filled orbitals also take part in hybridization, with an explanation why it is so.

The link for the 41 min long video is here.

Answer 1

First off, hybridization is a concept chemists developed to help explain reality (their observations). Just like resonance theory and Huckel MO theory, it is often (but not always) a useful way to explain the world around us. A "rule" on hybridization: hybridization occurs in response to a bonding interaction. Further, hybridization involves the mixing of filled (or partially filled) atomic orbitals to produce new atomic orbitals that can be used to form "directed" bonds (molecular orbitals) that produce more stable molecular systems than an unhybridized atom might produce. Finally, energetically speaking, hybridization is a net neutral process. When low energy AOs are mixed with higher energy AOs, the resultant new AOs have the average energy of the initial AOs. So if some electrons are "promoted", then some are lowered in energy as well

For example, in the formation of methane, carbon mixes the electrons in the 2s and three 2p atomic orbitals to produce 4 $\ce{sp^3}$ hybridized atomic orbitals. The energy of these 4 equivalent $\ce{sp^3}$ orbitals is in between that of the starting 2s and 2p orbitals. In the case of methane this use of $\ce{sp^3}$ orbitals to produce a tetrahedrally shaped molecule, produces a molecule that is lower in energy than the alternative molecule formed from unhybridized s and p orbitals on carbon. In the case of phosphine ($\ce{PH_3}$) the situation is reversed. The $\ce{P-H}$ bonds in phosphine are directed along the x, y and z-axes. The central phosphorous atom remains unhybridized. In the case of phosphine, hybridization does not produce a lower energy molecule, so the central phosphorous atom in phosphine remains unhybridized.

Personally, I like the concept of hybridization and find it quite useful. But you have to be careful where to apply it. Hybridization is best when it is used to explain reality after the fact.

Answer 2

With hybridisation, you only hybridise the number of orbitals you need i.e. that are filled. So in methane, with four bonds, you have four electron pairs which need to fit somewhere. So you need four orbitals. Therefore we say that three 2p orbitals hybridise with the 2s orbital, making four sp³ hybrids (the superscripts denoting the number of each pure orbitals involved) each with 1 electron in them. Note that we had to invoke the idea of hybridisation here, since the electron configuration of carbon is 1s²2s²2p², and this would only give us two bonds. We know experimentally that this isn't true (in other words, methane exists!) and so have to conjure a trick to work around it.

Alternatively, let us look at the CH₃⁺ cation. This has three bonds, and so three electron pairs. This time we need to invoke the idea of hybridisation to give us three orbitals. So we say that two 2p orbitals hybridise with the 2s orbital, to give three sp² orbitals. This leaves us with one p-orbital that we didn't hybridise, which would be a non-bonding orbital (in fact, empty), and so this lies perpendicular to the plane of the molecule.

For another example, to clarify further, consider ammonia. Now ammonia has three bonds, but it also has a lone pair. We count the lone pairs in our hybridisation scheme. Therefore we also need to hybridise to obtain four orbitals, like in methane, despite only having three bonds. Thus ammonia is also sp³, mixing three 2p and one 2s orbital. Note however that in contrast to methane, three of these hybrid orbitals are singly occupied, and one is completely filled (the one which contains the lone pair).

Now, we can look at cases by which orbitals do not hybridise.

For heavy atoms such as lead and bismuth (the same groups as carbon and nitrogen respectively) hybridisation does not occur. This is referred to as the inert pair effect. In this case inert pair refers to the unwillingness of the 6s² orbital to undergo hybridisation. This is because heavy atom self-self bonds i.e. Pb-Pb are incredibly weak. We can imagine that some energy is expended in the hybridisation process because we are mixing s orbitals with p-orbitals of the same principal quantum number, of which the p are always higher in energy (said to be non-degenerate). So the hybrid orbitals are higher in energy than the pure s, and ever so slightly lower than the pure p (but not by much, as the p-character of an sp³ hybrid is 75%). So, the energy gained in forming a heavy atom bond such as Pb-Pb isn't enough to warrant hybridisation, since hybridisation requires energy.

Notice that I've used anthropogenic words to explain this i.e. (we hybridise as opposed to the atoms hybridise). This is deliberate. It's unclear whether hybridisation really occurs or not; indeed more sophisticated theories can be used to explain the phenomena. Hybridisation is more of a caveat used to explain something we couldn't readily explain with the previous level of theory. It's perhaps telling that we only invoke hybridisation when we need to. It's a characteristic of a bad theory when that happens. The idea is invoked readily in chemistry however, so you would do well to familiarize yourself with the terminology.If you're interested more in this topic, look up molecular orbital theory for a more nuanced explanation.

Answer 3

You mean like ammonia? If the filled orbital did not partake in hybridization, then the molecule would be flat.

Water would be linear if the filled orbitals did not partake in hybridization.