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Hybridized AOs: Where is the lone pair?

  1. Jun 16, 2015 #1
    can unhybridized atomic orbitals hold sigma bond electron pairs?
    To my knowledge, hybridized AOs can hold both lone pair e- as well as sigma bond e- pairs.
    But I'm not sure if unhybridized AOs can do the same.

    Thanks in advance!:smile::wink:
     
  2. jcsd
  3. Jun 16, 2015 #2

    DrDu

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    Of course they can! The bonds in most compounds can be described in terms of unhybridized atomic orbitals, tetravalent carbon being the mayor exception.
    E.g. in H2O, you can accommodate the lone pairs in one s and one p orbital, and use two p orbitals to bind the hydrogens.
     
  4. Jun 18, 2015 #3
    But doesn't Oxygen atom possesses sp3 hybridized orbitals so as to give a tetrahedral geometry of molecule? Please correct me if I am wrong :D
    AAAVJKN0.JPG
     
  5. Jun 18, 2015 #4

    DrDu

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    Atoms don't "possess" hybridization. Hybridization is solely a theoretical model. Whether it is appropriate or better than an unhybridized model is often difficult to sort out. In the case of oxygen, the s and p orbitals have vastly different energies which makes hybridization unfavorable. It is unusual to include the lone pairs in the definition of the observed geometry, which refers to the nuclear arrangement, only. At best, we can measure the bond angle of about 104 degrees which is slightly larger than the 90 degrees between p-orbitals. This can be explained with some repulsion between the hydrogen atoms.
     
  6. Jul 8, 2015 #5
    The answers of Dr. Du are wrong. Tetravalent carbon and the oxygen in the water molecule are both sp3 hybridized. An easy example of an unhybridized atom in bonded form is the hydrogen molecule or hydrogen in normal hydrocarbons. And the s- and p-orbitals in oxygen don't have wastly different energies. The energy level of an orbital dependends on the electron occupation. 8 electrons, 4 electron pairs symmetric distributed around an oxygen atom. If there were no hybridization we would have three equal bonds from the p-orbitals and one different bond from s-orbital. But this is unstable in comparison to the symmetric form with 4 equal bonds. And with the electron occupation in oxygen plus two electrons from hydrogen atoms the oxygen fulfills the 8 valence electron rule. And again 8 electrons distributed over 4 orbitals in a molecule with tetrahedral geometry, this is only possible with sp3 hybridization. And from the statement to put one lone pair in s-orbital and one in a p-orbital in combination with the statement that they have wastly different energies we could conclude that there is difference in the lone pairs, but this is not what we observe in the reality. The energetic equality of the sp3 orbitals is responsible for the important acid-base chemistry of water. Without the energetic equality there would be no chance for autoprotolysis and so on.
     
    Last edited: Jul 8, 2015
  7. Jul 8, 2015 #6

    cgk

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    FoxOne, you are wrong in your assessment. DrDu's explanations are 100% correct; they are based on a quantum mechanical understanding of the wave function and its description. You seem to think that hybridization is somehow "real" and determines properties of a molecule, as your comments on the chemistry of water imply. Let me guarantee you that hybridization is not real; the form of hybridization you chose (or even, if you chose any at all) only has an influence on how you visualize the bonding in the molecule, but it has no implications whatsoever for the bonding itself. Real first principles quantum chemistry programs almost never use hybrid orbitals in electronic structure calculations, and they compute wave functions which are even quantitatively correct!

    I understand your confusion, as most introductory textbooks in organic, general, and physical chemistry have a horrible description of what orbitals actually are, and this lead to very wide spread and highly counter-productive erroneous believes. However, the meaning of orbitals is not something which lacks understanding in theoretical and quantum chemistry, what DrDu's explanations are based on.

    And, just for the record: I attached a bunch of pictures which visualize an exact representation of an accurate wave function for the H2O molecule. As you can see, sp3 hybridization is not necessary, a description with two inequivalent lone pairs (one of them being simply an oxygen p) works fine. In this representation, the two lone pairs have a difference in orbital energy of about 0.2 Hartree energy units (~200 kcal/mol), just as explained before.
     

    Attached Files:

  8. Jul 8, 2015 #7
    1. It can't be right. The hybridorbital (sp3) is a linear combination of the wavefunctions of the atomic orbitals. Easy quantummechanical explanation: An electron with a defined probability to find it in a room. What happens with the room of highest probability to find the electron when you bring a second s-orbital occupied with 1 electron close to this system? Maybe it will deform due to the different potential fields? I know that water is not sp3 hybridized, but it is hybridized because this happens when you make a linear combination of AO's from hydrogen and oxygen and this is an QM explanation. The real hybridization can be found here http://www.sciencedirect.com/science/article/pii/S0009261409004503.
    And I think you have an imagination problem. Because you try to tell me that the one-electron orbital wave functions of oxygen (s- and p-atomic orbitals) remains constant after bringing the hydrogen s-atomic orbitals close to it. This can't be.

    2. Your pictures directly show the hybridization mentioned in the literature in point 1. These are sp2 b1 hybrid bonds.

    3. I have a basic understanding of quantum mechanics, you are not the only one who studied, and i know what a orbital is. But you try to explain molecular bondings with atomic orbitals and this is completely wrong. Can you tell me why there are numerous groups working on Hartree-Fock methods to calculate molecular ORBITALS?

    4. The only new method which is working without orbitals is the so called multi-configurational self-consistent field method. This method is working with the linear combination of state functions because sometimes Hartree-Fock or density functional theory methods bring wrong or unrealistic results.

    5. Your argument with the computer programs is very funny. Orbital means one-electron orbital wave function. So when a program use wave functions it uses orbitals and when the program made a linear combination of this atomic orbitals we will get another orbital. Hope i can show you that you the confused one. And with your point that hybridization is nothing we can experimental proof i disagree. Because electron density measurements on the sulfate ion show directly the electron distribution of an sp3 hybridization. It can be found here http://rruff.info/doclib/zk/vol164/ZK164_219.pdf.
     
    Last edited: Jul 8, 2015
  9. Jul 8, 2015 #8

    cgk

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    Good... you are making the first step to understanding what we are talking about. And now for the interesting question: If hybrid orbitals are linear combinations of atomic orbitals, and molecular orbitals are linear combinations of hybrid orbitals... what is stopping you from expressing one and the same set of final molecular orbitals with DIFFERENT sets of intermediate hybrid orbitals? And what is stopping you from skipping the hybrid orbitals altogether and expressing the molecular orbitals directly as linear combinations of atomic orbitals? Or as linear combination of functions which are not even atomic orbitals themselves, but simply allow expressing the AOs/MOs efficiently?

    No one said anything to that effect. What was said is that you can express bonding situation by invoking the atomic orbitals directly, rather than hybrid orbitals (or choosing various different sets of different hybrid orbitals: here sp2 and sp3, for example). And this is obviously true, as the MOs can be expressed over AOs rather than hybrids.

    By the way, what you described is polarization, not hybridization. And, of course, strictly speaking, atomic orbitals in molecules are not real (i.e., physically observable) either...


    That is one way of seeing it. It shows that, just as DrDu said earlier, the two lone pairs need not be considered equivalent and that the O need not be described as sp3 hybridized. It thus contradicts your statements from post #5, where you incorrectly claimed that this cannot be true.

    By the way: I attached another set of orbitals to the current post. The set of orbitals attached here and the set in #6 are physically equivalent[1] and describe the same many-body wave function. Any physically observable quantity computed from either set of orbitals will come out equal to within numerical precision (i.e., 10-14 decimals digits in double precision arithmetic). Where are your sp3 vs sp2 hybrids now? Muhahaha!! (<- evil laugh). Mind blown? 8).

    h2o-boys-bond1.png h2o-boys-bond2.png h2o-boys-lp1.png h2o-boys-lp2.png

    And before you tell me that I am wrong and this cannot be true: No, I am not wrong, and yes, this is very much true.

    [1] Both of them represent the same DF-RKS/PBE/def2-TZVP wave function. The ones in #6 are Intrinsic Bond Orbitals (IBOs) made in IboView (which can be downloaded at www.iboview.org if you want to see it for yourself), the ones attached here are Boys orbitals. Since you mentioned Hartree-Fock: HF orbitals would be quantitatively somewhat different, but would look visually indistinguishable from the DFT orbitals, and they could be localized in the same way.

    Pro tip: This is a semi-anonymous internet forum. Some people take the time to answer 100s of highly technical questions about very particular topics in science and research. Ever wondered what kind of people might do this? What kind of day job those people might have?...

    If I guarantee you that I know all the gory details about electronic structure theory, and I know exactly how it is implemented in computer programs---what do you think? How could I be in a position where I can guarantee (hint, hint) that I know exactly (hint, hint) how it works?

    1. MCSCF is not a new method.
    2. MCSCF does use orbitals. Orbitals are what the CSFs or determinants are made of.
    3. There are *many* electronic structure methods which are not based on orbitals (e.g., various kinds of real-space quantum Monte-Carlo methods, various first-quantized variational wave functions, "explicitly correlated" wave functions) or which invoke orbitals, but only for computational reasons, not to express wave functions (e.g., various geminal methods like the anti-symmetrized geminal power).
    4. None of those have *anything* to do with the topic we had been discussing. The qualitative electronic structure of most molecules, especially of H2O, is well described by a single determinant, as used in Hartree-Fock and Kohn-Sham methods. These are perfectly adequate methods to base discussions of hybridization on.

    8).

    No, but seriously: You need to work on your attitude. Both in science and real life it is *very* important that one has not only a good understanding of how things work, but also to which degree what one thinks is "knowledge" can be trusted. You just happened to say a number of very, very wrong things in this thread, coupled with implying that your conversation partners are the ones who do not know what they are talking about (pro tip: they do know). Now, as mentioned, this is a semi-anonymous internet forum so this particular thread will most likely not come back to haunt you in the future. But I would *strongly* advise not to pull off anything like this in your day job or other real-life situations. It can backfire. Hard.

    One thing I like about physical facts is that they do not go away just because someone does not agree with them.
     
  10. Jul 9, 2015 #9

    DrDu

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    I am not going to feed a troll, here.
    With the OP in mind I just want to add something about the basis of Valence Bond theory.
    Valence Bond theory, or the Heitler-London method, as it was first called, starts from the idea that bonding in molecules can be described in terms of a simple product of the atomic orbitals of the atoms involved multiplied with an appropriate spin function. It turns out that already this simple approximations gives a very good description of bonding especially in compounds of non-metals, i.e. composed of atoms with a high electronegativity. This includes molecules like water, HF, F2 etc.
    Taking into account the change of the atomic orbitals, which may involve various degrees of hybridization, leads to a further improvement of the wavefunction but usually does not change the basic picture.
    For compounds including metals or semi-metals like boron or carbon the valence bond wavefunctions turn out not to be such a good starting point as it becomes necessary to also include lots of ionic structures or excited atomic states. Molecular orbital theory does usually a much better job here.
    Nobel prize winner Linus Pauling tried also to describe these metallic compounds in terms of valence bond theory, propagating especially the use of all kinds of a priori hybridizations, but many of his ideas had to be dropped when these models could be tested numerically with the advent of computer programs in the 1960's.
    Nevertheless his ideas still appear in most of the introductory chemistry books.
    As cgk certainly can confirm a considerable part of teaching theoretical chemistry consists in convincing chemists to be that a considerable part of what they have learned in school and introductory chemistry courses is outdated or even wrong.
     
  11. Jul 9, 2015 #10

    Borek

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    I can't speak for cgk, but I agree. I see it all the time, and it is not limited to theoretical chemistry.
     
  12. Jul 9, 2015 #11
    At cgk,

    1. My starting problem was this statement:
    This statement is wrong because the lone pairs of water distributed in one s- (spherical shape) and one p-orbital (dumbbell shape) would never gave an electron density distribution you gave with the simulated pictures. And when you start to accept that linear combination of atomic orbitals on an atom is hybridization our problem is solved.
    http://goldbook.iupac.org/H02874.html

    2. I gave you literature where you can read:
    "In molecular orbital (MO) theory, based on the linear combination of an atomic orbital basis, the hybridization concept does not appear directly, and is usually implicitly inferred from Mulliken population analysis [3] in terms of the gross charges on the heavy atom. Nonetheless, if the series of first-row hydrides, XHn, is studied by MO theory at an elementary Hückel level, sp3 hybridization on C is seen to stem directly from the high Td symmetry of CH4 [4], but the situation is more difficult when dealing with hydrides of lower symmetry. H2O, a triatomic bent molecule of C2v symmetry having an interbond angle 2Theta = 105° [5,6], seems particularly well suited for investigating the effect on its geometrical structure of sp hybridization on the central oxygen atom. If only p orbitals on O were used in bonding, the interbond angle would be 90°, giving
    O–H bonds strongly bent outwards and very far from the principle of maximum overlap [2]. Admitting sp hybridization within A1 symmetry gives a 3 x3 Hückel secular equation which is not easy to analyze in non-numerical terms."

    From: V. Magnasco, Chem. Phys. Lett. 2009, 212-216.

    3. You talk about IBOVIEW. Now you start to being very funny and i'm the one who is evil laughing at you. Please read following literature from the programmer of IBOVIEW.
    http://pubs.acs.org/doi/abs/10.1021/ct400687b G. Knizia J. Chem. Theory Comput., 2013,4834–4843.
    In page 4836 figure 1 a you will find the intrinsic bond orbitals of acrylic acid mentioned as sp-hybrid. Just because you're calculating IBO's it doesn't mean that they can't represent hybrids.
    In page 4837 equation 5 is used for the calculation of ionization energy shift. In it you can find a "(hybridization-dependent) proportionality constant."
    So i hope i can show you that hybridization is still a model discussed for the characterisation of chemical bonds and geometries and until you can't show me some scientific references which are proving the opposite. I imagine the molecular world with molecular and hybrid orbitals. And in fact we calculate electron density distributions. How you call these spaces of electron density is a thing of nomenclature rules, but the mixing of for example s-orbital character and p-orbital character is called hybridization and this mixing is a very good model for the describing of reactivity series and it's confirmed by the IBO calculations. Page 4838 part Cyclopropane in G. Knizia J. Chem. Theory Comput., 2013,4834–4843.

    4. Just because simulations use simplified mathmatically models (maybe with or without hybridization) it doesn't mean that the simplification represents the right way in describing the reality.

    5.
    I gave you a reference which reinforced my arguments. And this was a measurment of electron density not a simulation so please have another try to discredit me.

    6.
    I never say that molecular or hybrid orbitals are real, but the electron density distribution is observable and when this is in good agreement with our theories about hybridization i can imagine the world with that. And the electron density distribution is determining the chemical properties of a molecule and without energetic equalization of the electrons in water the manner of reaction is not explainable. Maybe in the future we are able to solve n-body problems analytically or find solutions of the schrödinger equation for whole molecules and with this we maybe find new nomenclature rules for the mixing of electronic states but until then i accept the actual definitions and discuss about it after the guidelines for research integrity and good scientific practice.
     
    Last edited: Jul 9, 2015
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