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Why molecules of small atoms have hybridization and the big ones don't

  1. Dec 18, 2012 #1
    I was solving an exercise my teacher asked us, to determinate the hibridization of the molecules. He give the molecule and I have to tell which hibridization it is. I want to know if I can do this to every molecule, I mean, are all molecules in the world hybridized? If so, why do we have to learn the geometry of s, p, d and f orbitals if all that works is the VSPER geometry for hybridized orbitals? I tried to see if the hybridized theory works and what I found out?

    Water hydrogen angle is 104.5 degrees (~109.5)
    NH3 angle is 107 degrees (~109.5)

    Although, when I compare the other molecules in further periods, it seems that the hybridization decreeases.

    H2S is 92.1 degrees
    H2Se is 91 degrees
    H2Te is 90 degrees

    PH3 93.5 degrees
    AsH3 91.8 degrees

    Why does it occur? Why are small molecules hybridized and bigger aren't? How can I know when a molecule will hibridizate?
  2. jcsd
  3. Dec 19, 2012 #2


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    Hybridization is not a property of the molecule but only a possible description of bonding in these molecules. As a matter of fact, in none of the examples you were giving is it necessary to invoke hybridization. Even H20 and NH3 can as well be described using the p orbitals only for bonding. The use of hybrids in these latter cases is often motivated using the maximal overlapp criterion to reproduce the experimental bond angles. However this criterion has only a very weak basis, and, as you correctly noticed, VSEPR arguments are completely sufficient to explain the bonding geometries.
    The main problem is that chemistry teachers are not required to take classes in quantum chemistry and are thus perpetuating long outdated concepts of bonding.

    This having been said, there are some reasons why hybrids are really not an even an option in the higher main group elements: In P, As, S, Se etc. the s and p orbitals are of vastly different size, while in O and N the sizes are quite similar. So mixing them up in a hybrid does not increase the overlap in the case of the heavier elements.
  4. Dec 26, 2012 #3
    Molecules don’t get much larger than covalent crystals.

    The carbon atoms in both diamond crystals and graphite crystals hybridize. The covalent bonds that hold the atoms are not unperturbed atomic orbitals. They are hybridized atomic orbitals. This is why they have the angles that they do.

    Here is a link.
    “Within diamond, one s-orbital and three p-orbitals undergo a SP3 hybridization.

    Within graphite, one s-orbital and two p-orbitals undergo a SP2 hybridization.

    These hybridized atomic orbitals form chemical bonds to form diamond, graphite, and buckminsterfullerenes.”

    In fact, most organic compounds have tetrahedral bonds coming from the carbon atoms. This implies that they are undergoing the precise same hybridization as diamond crystals. So each carbon atom in a protein molecule or carbohydrate molecule is hybridized. Carbon in organic molecules usually forms covalent bonds with either SP3 hybridization or SP2 hubridization.

    “Examples involving organic molecules:”

    I don’t know any example where carbon forms covalent bonds with unperturbed atomic orbitals. If you can find such examples, then please post a reference. Molecules with chains of carbon atoms can be quite large.

    So molecule size does not significantly affect hybridization.
  5. Dec 27, 2012 #4


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    That's not what OP had on mind. Question was not about hydbridization of orbitals as a function of the size of a carbon molecule, but about comparing compounds like NH3 (small molecule, with HNH angles almost exactly equal to those predicted by hybrydization) and analogous, but much larger ones (like AsH3) where HAsH angles are close to 90°, even if we could expect them (by analogy) to be similar to those in ammonia.
  6. Dec 27, 2012 #5
    In the examples that you gave, the important interaction that "causes" the hybridization is the electrostatic repulsion between the hydrogen atoms. All the compounds are two hydrogen atoms covalently bonded to a column VI (periodic table) atom. The unperturbed atomic orbitals of the column VI atoms have one pair of electrons in an s-orbital, one pair of electrons in a p orbital, and unpaired electrons in two more p orbitals. The probability distribution of the p orbitals are orthogonal to each other.

    The hydrogen bonds to the VI element through the unpaired p orbitals. So if there was no interaction between the hydrogen atoms, positions of the hydrogen atoms relative to the VI compound would be 90 degrees. An "unhybridized" VI atom would have two ligands connected at 90 degrees.

    However, there is an interaction that would force some degree of hybridization to the covalent bonds. This is the electrostatic repulsion of the hydrogen atoms. The VI column atom pulls electrons from the hydrogen, making the hydrogen atoms positive. The larger the angle between the two hydrogen atoms, the larger the distance between the two atoms and the weaker the electrostatic repulsion.

    Oxygen is the most electronegative atom of the series. The oxygen pulls on the hydrogen atoms, making them effectively positive. The positive hydrogen atoms (ions?) repel each other because like charges repel. The repulsion can be decreased only increasing the angle from 90 degrees (unhybridized p orbital) to 102 degrees (hybridized sp3).

    Tellurium is much less electronegative then oxygen. You can consider tellurium as having an effective electronegativity of zero (i.e., same electonegativity as the hydrogen atoms). Therefore, the hydrogen atoms in H2Te are uncharged. Therefore, the hydrogen atoms don't repel each other. Therefore, the electrons in the Te atom don't hybridize. Therefore, the chemical bonds are comprised of unhybridized p-orbitals. Therefore, the chemical bonds are at 90 degrees to each other.

    Figuring out when to use hybrid orbitals is rather complicated. The general rule of thumb is this.

    The hybridization occurs to relieve some stress in the molecule caused by some type of repulsion. In the cases that you presented, the stress was due to the electrostatic repulsion of hydrogen atoms (ions?). In other cases, the stress would be due to repulsion between electrons spinning in the same direction. This would be called spin-spin interactions. In still other cases, there may be some steric forcing due to one atom being forced too close to the other. However, in all cases there is some stress that has to be relieved.

    I can't give you any simple rules that are also reliable. However, the examples that you gave provided a simple example. Electrostatic repulsion was the stressor in those examples.

    Here are a few links that may help.

    http://www.foothill.edu/psme/larson/1A_assets/Chapter 9 - Hybridization.pdf
    Valence Bond Theory (Hybridization)
    “Central atoms do not use atomic (s, p, d, f) orbitals to form sigma bonds.
    Central atoms mix or hybridize their valence atomic orbitals to form new bonding orbitals
    called hybrid orbitals.

    1. Multiple covalent bonds (double and triple) form when more then one orbital from each
    atom overlap.
    2. This additional overlap occurs using UNHYBRIDIZED atomic orbitals, not hybrid
    orbitals. This overlap is called a pi (π) bond formation.

    Sigma bond (σ): the first bond formed between two atoms.”

    “The analogy between the spin and orbital angular momentum operators is well known and the commutation relations are the same. When the chemical bond has formed, the spin of the electrons is anti-parallel by Hound’s rule. The function of two interacting atoms can be compared with the intermolecular dispersion force and has a similar function. The minimum of the function describing their interaction is at the minimum energy of the induced spin-spin interaction.”
  7. Sep 21, 2013 #6
    This is a bit of an old thread, but just for anyone who comes across it, we should make it clear that "hybridization" is a model for explaining covalent bonding by simple localized orbitals which are simple superpositions of atomic orbitals. It's not that some molecules "do" it and others "don't do it", it's whether or not it's convenient to variationally solve the schrodinger equation in this way. Usually you'd solve the equations numerically for the canonical molecular orbitals that diagonalize the fock matrix and then use some localization scheme (Boys, Weinhold's natural orbitals etc) to get something like the simple "hybrid" orbitals, so basically any and every molecule both "has" and "doesn't have" hybrid orbitals. It all depends on how you want to rotate the basis.
  8. Sep 22, 2013 #7


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    You are mixing up valence bond and MO theory. Hybridization is not a concept of MO theory. Even localized molecular orbitals are delocalized over two adjacent atoms while valence bond theory formulates bonding essentially in terms of atomic orbitals (which may be hybridized).
    There are quite some dedicated programs available to do VB calculations. For the small molecules considered in this thread this is not even computationally demanding and leads usually to better results than MO theory at Hartree Fock level.
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