13
$\begingroup$

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.

Improve this question
$\endgroup$
  • 3
    $\begingroup$ I tried to watch that... $\ce{PH3}$ cannot be described in terms of VSEPR or hybridisation. The bond angle is about $93^\circ$. I could not watch any more after that. $\endgroup$ – Martin - マーチン♦ Jul 24 '14 at 7:36
  • $\begingroup$ @Martin I never said anything about PH3 and she DID NOT explain PH3 by VSEPR theory. If you had watched the video full you would have got my point. I think since you did not watch the video and neither you bothered even reading my question , so you should not comment anything that is not related to it. $\endgroup$ – user2619 Jul 24 '14 at 9:40
  • 2
    $\begingroup$ @NicolauSakerNeto I am not opposing hybridisation. Just the concept in which it is usually treated, as a rigid theory that is required to form certain bonds. Since the bonding angle is $93^\circ$ it would be wrong to neglect the influence of the s orbital. || Dear Hardik, after studying five years of chemistry, a doctorate and currently a postdoc in computational/ theoretical chemistry, I think I would be able to answer your question. Maybe my comment was a bit impulsive, and I apologize that I have offended you with it, I just wanted to point out an obvious flaw of the lecture. $\endgroup$ – Martin - マーチン♦ Jul 24 '14 at 17:16
  • 1
    $\begingroup$ Whatever you were trying to say, I clearly understood it wrong, I apologize for any inconvenience. I would like to add one last thing, your book mentioned electron promotion, this concept is outdated by at least twenty years now. I highly recommend to switch to a different book, e.g. Chemistry of the Elements, by N. N. Greenwood, A. Earnshaw; Inorganic Chemistry: Principles of Structure and Reactivity, by James E. Huheey et. al. ... $\endgroup$ – Martin - マーチン♦ Jul 25 '14 at 6:48
  • 2
    $\begingroup$ @Martin Yes you are right, this concept is outdated. The book although mentioned it as a separate small note. And the teacher in the video too, thats why i got confused. I will sure try to read the book you recommended. $\endgroup$ – user2619 Jul 25 '14 at 17:51
4
$\begingroup$

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.

Improve this answer
$\endgroup$
  • 1
    $\begingroup$ Actually, one of the lone pairs is unhybridized (basically pure $\ce{p}$) and the other lone pair is roughly $\ce{sp}$ hybridized. The two $\ce{O-H}$ bonds are roughly $\ce{sp^3}$ hybridized. $\endgroup$ – ron Jul 25 '14 at 1:19
  • $\begingroup$ @ron If that was the case, then the bond angle would be different fro what you measure. $\endgroup$ – LDC3 Jul 25 '14 at 1:49
  • $\begingroup$ How so? The O-H bonds are $\ce{sp^3} hybridized so the expected H-O-H angle would be around 109 degrees, close to the experimentally observed value. $\endgroup$ – ron Jul 25 '14 at 1:51
  • 1
    $\begingroup$ Sure you can. 2 sp3 orbitals require half of an s orbital and 1.5 p orbitals, leaving 1 p orbital for one lone pair, and half an s orbital and half a p orbital for the sp orbital the other lone pair to exist in. This hybridization scheme is consistent with the Photoelectron spectrum data for water. $\endgroup$ – ron Jul 25 '14 at 2:07
  • 1
    $\begingroup$ see isites.harvard.edu/fs/docs/icb.topic776365.files/… water is discussed on p. 18. Let me know what you think. $\endgroup$ – ron Jul 25 '14 at 2:42
11
$\begingroup$

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.

Improve this answer
$\endgroup$
  • $\begingroup$ Not sure what you're getting at, but I thought my answer might help the OP have a better understanding of the subject. If you feel the answer isn't helpful, you are free to down-vote and\or constructively criticize it. $\endgroup$ – ron Jul 25 '14 at 2:45
  • 1
    $\begingroup$ It does help, I was just pointing out that you could also include an example that the OP requests. It's easy to veer from the requested answer when there is so much to explain. $\endgroup$ – LDC3 Jul 25 '14 at 2:52
  • $\begingroup$ I agree with almost everything you say, but not the part where the new resultant AOs have an average energy. It is just not possible to assign energy values to hybrid orbitals, as these are no eigenfunction of the Hamilton operator (They do not produce the correct quantum numbers). $\endgroup$ – Martin - マーチン♦ Jul 27 '14 at 17:45
  • $\begingroup$ @Martin 2 questions, 1) if I theoretically mix 50% carbon 2s AO and 50% carbon 2p AO to create an sp orbital, why wouldn't its theoretical or estimated energy be 0.5*E(s) + 0.5*E(p); 2) if I mix two carbon sp AOs to form an sp-sp MO I can experimentally determine its energy (PES spectrum of acetylene) and see if it is halfway between the 2s and 2p energy levels for atomic carbon, right? $\endgroup$ – ron Jul 27 '14 at 17:56
  • $\begingroup$ 1) Theoretically you could do that, but the value you obtain would be a number without a physical meaning. One has to be very careful in using this number even as an estimate, since it is not an observable state. Hybrid orbitals are directionalized (localised) orbitals, that would violate the spherical symmetry of an atom and also the quantum numbers. The concept of hybridisation is very useful, in describing a general bonding pattern. It will give you wrong results for (excitation) energies, since if you distort the molecule (in any way) bonding and therefore also hybridisation changes. $\endgroup$ – Martin - マーチン♦ Jul 28 '14 at 2:41
6
$\begingroup$

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.

Improve this answer
$\endgroup$
  • $\begingroup$ Molecular orbital theory is not a competing interpretation. In MO terms, hybridisation is the mixing of different kinds of atomic orbitals of the same atom into a molecular orbital. It is still invoked as a mathematical concept and only treated as such. There can also by hybridisation in virtual (unoccupied) orbitals. The theory that is employed (by organic chemists mostly) is a very crude approximation. Hybridisation is always a result of bonding and therefore hybrid orbitals cannot be assigned an energy value. $\endgroup$ – Martin - マーチン♦ Jul 26 '14 at 9:18
  • $\begingroup$ @Martin, I'm not sure I completely agree. In inorganic chemistry, for example, you use the idea of a symmetry adapted linear combination of orbitals (SALCS) to rationalise the bonding in, say, octahedral complexes, instead of saying that they are sp3d2. Nowhere in the MO diagram of such complexes does it mention hybrid orbitals of this kind. Unless of course the process of taking SALCS is considered hybridising, which is something I'm not familiar with? $\endgroup$ – user7232 Jul 26 '14 at 12:28
  • $\begingroup$ The symmetry adapted linear combination is somewhat closer to the truth - meaning a less crude approximation - but it is still an approximation. Even everything we know so far about orbitals is an approximation, since knowing the exact wave function is not possible, we can still consider different atomic orbitals contributing to a molecular orbital. Any linear combination of ao towards an mo of the same atom is therefore a hybridisation. You cannot neglect the influence of this when working within the framework of linear combinations. $\endgroup$ – Martin - マーチン♦ Jul 26 '14 at 16:19
  • $\begingroup$ I see, thanks Martin, I understand your point. I'll edit my contribution accordingly to reflect this. $\endgroup$ – user7232 Jul 26 '14 at 23:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy