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9.3: Hybridization of Atomic Orbitals

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    518979
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    Introduction

    Hybridization was introduced to explain molecular structure when the valence bond theory failed to correctly predict them. It is experimentally observed that bond angles in organic compounds are close to 109°, 120°, or 180°. According to Valence Shell Electron Pair Repulsion (VSEPR) theory, electron pairs repel each other and the bonds and lone pairs around a central atom are generally separated by the largest possible angles.

    In this section, we will use a model called valence bond theory to describe bonding in molecules. In this model, bonds are considered to form from the overlapping of two atomic orbitals on different atoms, each orbital containing a single electron. In looking at simple inorganic molecules such as H2 or HF, our present understanding of s and p atomic orbitals will suffice. In order to explain the bonding in organic molecules, however, we will need to introduce the concept of hybrid orbitals.

    Interactive website on hybrid orbitals.

    Hybrid orbitals: sp3 hybridization and tetrahedral bonding

    Now let’s turn to methane, the simplest organic molecule. Recall the valence electron configuration of the central carbon:

    image168.png

    This picture, however, is problematic. How does the carbon form four bonds if it has only two half-filled p orbitals available for bonding? A hint comes from the experimental observation that the four C-H bonds in methane are arranged with tetrahedral geometry about the central carbon, and that each bond has the same length and strength. In order to explain this observation, valence bond theory relies on a concept called orbital hybridization. In this picture, the four valence orbitals of the carbon (one 2s and three 2p orbitals) combine mathematically (remember: orbitals are described by equations) to form four equivalent hybrid orbitals, which are named sp3 orbitals because they are formed from mixing one s and three p orbitals. In the new electron configuration, each of the four valence electrons on the carbon occupies a single sp3 orbital.

    image170.png

    The sp3 hybrid orbitals, like the p orbitals of which they are partially composed, are oblong in shape, and have two lobes of opposite sign. Unlike the p orbitals, however, the two lobes are of very different size. The larger lobes of the sp3 hybrids are directed towards the four corners of a tetrahedron, meaning that the angle between any two orbitals is 109.5o.

    image172.png

    This geometric arrangement makes perfect sense if you consider that it is precisely this angle that allows the four orbitals (and the electrons in them) to be as far apart from each other as possible. This is simply a restatement of the Valence Shell Electron Pair Repulsion (VSEPR) theory that you learned in General Chemistry: electron pairs (in orbitals) will arrange themselves in such a way as to remain as far apart as possible, due to negative-negative electrostatic repulsion.

    Each C-H bond in methane, then, can be described as an overlap between a half-filled 1s orbital in a hydrogen atom and the larger lobe of one of the four half-filled sp3 hybrid orbitals in the central carbon. The length of the carbon-hydrogen bonds in methane is 1.09 Å (1.09 x 10-10 m).

    image174.png

     

    interactive 3D model of the bonding in methane

    In the ethane molecule, the bonding picture according to valence orbital theory is very similar to that of methane. Both carbons are sp3-hybridized, meaning that both have four bonds arranged with tetrahedral geometry. The carbon-carbon bond, with a bond length of 1.54 Å, is formed by overlap of one sp3 orbital from each of the carbons, while the six carbon-hydrogen bonds are formed from overlaps between the remaining sp3 orbitals on the two carbons and the 1s orbitals of hydrogen atoms. All of these are sigma bonds.

    image180.png

    Because they are formed from the end-on-end overlap of two orbitals, sigma bonds are free to rotate. This means, in the case of ethane molecule, that the two methyl (CH3) groups can be pictured as two wheels on a hub, each one able to rotate freely with respect to the other.

    image182.png

    Rotational freedom in sigma bonds leads to the multiple conformations of organic molecules.

    The sp3 bonding picture is also used to described the bonding in amines, including ammonia, the simplest amine. Just like the carbon atom in methane, the central nitrogen in ammonia is sp3-hybridized. With nitrogen, however, there are five rather than four valence electrons to account for, meaning that three of the four hybrid orbitals are half-filled and available for bonding, while the fourth is fully occupied by a (non-bonding) pair of electrons.

    image184.png

    The bonding arrangement here is also tetrahedral: the three N-H bonds of ammonia can be pictured as forming the base of a trigonal pyramid, with the fourth orbital, containing the lone pair, forming the top of the pyramid.

    image186.png

    Recall from your study of VSEPR theory in General Chemistry that the lone pair, with its slightly greater repulsive effect, ‘pushes’ the three N-H sigma bonds away from the top of the pyramid, meaning that the H-N-H bond angles are slightly less than tetrahedral, at 107.3˚ rather than 109.5˚.

    VSEPR theory also predicts, accurately, that a water molecule is ‘bent’ at an angle of approximately 104.5˚. It would seem logical, then, to describe the bonding in water as occurring through the overlap of sp3-hybrid orbitals on oxygen with 1sorbitals on the two hydrogen atoms. In this model, the two nonbonding lone pairs on oxygen would be located in sp3 orbitals.

    image188.png

    Exercise

    Draw, in the same style as the figures above, an orbital picture for the bonding in methylamine.

    Solution

    Mathematical form of sp3 hybrids1

    The four unnormalized hybrid sp3 orbitals are the following combinations of the valence s and p orbitals:

    \[h_1 = s + p_x + p_y + p_z \qquad h_2 = s - p_x - p_y + p_z \nonumber\]

    \[h_3 = s - p_x + p_y - p_z \qquad h_2 = s + p_x - p_y - p_z \nonumber\]

    Live 3-D view based on quantum computations: https://cms.gutow.uwosh.edu/gutow/Orbitals/N/sp3%20hybrid.shtml

    Formation of pi bonds - sp2 and sp hybridization

    The valence bond theory, along with the hybrid orbital concept, does a very good job of describing double-bonded compounds such as ethene. Three experimentally observable characteristics of the ethene molecule need to be accounted for by a bonding model:

    1. Ethene is a planar (flat) molecule.
    2. Bond angles in ethene are approximately 120o, and the carbon-carbon bond length is 1.34 Å, significantly shorter than the 1.54 Å single carbon-carbon bond in ethane.
    3. There is a significant barrier to rotation about the carbon-carbon double bond.

    image190.png

    Clearly, these characteristics are not consistent with an sp3 hybrid bonding picture for the two carbon atoms. Instead, the bonding in ethene is described by a model involving the participation of a different kind of hybrid orbital. Three atomic orbitals on each carbon – the 2s, 2px and 2py orbitals – combine to form three sp2 hybrids, leaving the 2pz orbital unhybridized.

    image192.png

    The three sp2 hybrids are arranged with trigonal planar geometry, pointing to the three corners of an equilateral triangle, with angles of 120°between them. The unhybridized 2pz orbital is perpendicular to this plane (in the next several figures, sp2 orbitals and the sigma bonds to which they contribute are represented by lines and wedges; only the 2pz orbitals are shown in the 'space-filling' mode).

    image194.png

    The carbon-carbon double bond in ethene consists of one sigma bond, formed by the overlap of two sp2 orbitals, and a second bond, called a π (pi) bond, which is formed by the side-by-side overlap of the two unhybridized 2pz orbitals from each carbon.

    image196.png

    interactive model

    another interactive model

    The pi bond does not have cylindrical symmetry about the bond. Because they are the result of side-by-side overlap (rather then end-to-end overlap like a sigma bond), pi bonds are not free to rotate. If rotation about this bond were to occur, it would involve disrupting the side-by-side overlap between the two 2pz orbitals that make up the pi bond. The presence of the pi bond thus ‘locks’ the six atoms of ethene into the same plane.

    Finally, the hybrid orbital concept applies well to triple-bonded groups, such as alkynes and nitriles. Consider, for example, the structure of ethyne (common name acetylene), the simplest alkyne.

    image210.png

    This molecule is linear: all four atoms lie in a straight line. The carbon-carbon triple bond is only 1.20Å long. In the hybrid orbital picture of acetylene, both carbons are sp-hybridized. In an sp-hybridized carbon, the 2s orbital combines with the 2px orbital to form two sp hybrid orbitals that are oriented at an angle of 180°with respect to each other (eg. along the x axis). The 2py and 2pz orbitals remain unhybridized, and are oriented perpendicularly along the y and z axes, respectively.

    image212.png

    image214.png

    The C-C sigma bond, then, is formed by the overlap of one sp orbital from each of the carbons, while the two C-H sigma bonds are formed by the overlap of the second sp orbital on each carbon with a 1s orbital on a hydrogen. Each carbon atom still has two half-filled 2py and 2pz orbitals, which are perpendicular both to each other and to the line formed by the sigma bonds. These two perpendicular pairs of p orbitals form two pi bonds between the carbons, resulting in a triple bond overall (one sigma bond plus two pi bonds).

    image216.png

    another view

    Mathematical form of sp and sp2 hybrid1

    sp

    The two unnormalized sp hybrids are: 

    \[h_1 = s + p_z \qquad h_2 = s - p_z\] 

    Giving two orbitals with their major lobes pointing 180˚ apart. By convention these lobes are chosen to point along the z-axis.

    Live 3-D view based on quantum computations: https://cms.gutow.uwosh.edu/gutow/Orbitals/N/sp%20hybrid.shtml

    sp2

    The three unnormalized sp2 hybrids are: 

    \[h_1 = s + \sqrt{1/2}p_y \qquad h_2 = s +\sqrt{3/2}p_x - \sqrt{1/2}p_z \qquad h_3 = s - \sqrt{3/2}p_x - \sqrt{1/2}p_z\]

    Live 3-D view based on quantum computations: https://cms.gutow.uwosh.edu/gutow/Orbitals/N/sp2%20hybrid.shtml

    Hybridization and bond lengths

    The hybrid orbital concept nicely explains another experimental observation: single bonds adjacent to double and triple bonds are progressively shorter and stronger than ‘normal’ single bonds, such as the one in a simple alkane. The carbon-carbon bond in ethane (structure A below) results from the overlap of two sp3 orbitals.

    image218.png

    In alkene B, however, the carbon-carbon single bond is the result of overlap between an sp2 orbital and an sp3 orbital, while in alkyne C the carbon-carbon single bond is the result of overlap between an sp orbital and an sp3 orbital. These are all single bonds, but the bond in molecule C is shorter and stronger than the one in B, which is in turn shorter and stronger than the one in A.

    The explanation here is relatively straightforward. An sp orbital is composed of one s orbital and one p orbital, and thus it has 50% s character and 50% p character. sp2 orbitals, by comparison, have 33% s character and 67% p character, while sp3 orbitals have 25% s character and 75% p character. Because of their spherical shape, 2s orbitals are smaller, and hold electrons closer and ‘tighter’ to the nucleus, compared to 2p orbitals. Consequently, bonds involving sp + sp3 overlap (as in alkyne C) are shorter and stronger than bonds involving sp2 + sp3 overlap (as in alkene B). Bonds involving sp3-sp3overlap (as in alkane A) are the longest and weakest of the group, because of the 75% ‘p’ character of the hybrids.

    Example

    For each of the bonds indicated by arrows a-e in the figures below, describe the bonding picture by completing this sentence:

    "The sigma bond indicated by this arrow is formed by the overlap of an ________ orbital of a _________atom and an ________orbital of a _______atom."

    image220.png

    Solution

    Resonance

    The valence bond model does not account well for electron delocalization. Just as with Lewis structures this is accounted for by making resonance hybrids. In this case the overall wavefunction is a weighted linear combination of the valence bond wavefunctions for all of the resonance hybids:

    \[\Psi_{resonance\,hybrid} = a\psi_1 + b\psi_2 + ...\nonumber\]

    An example of where this is important is in peptide bonds that link amino acids to form polypeptides and proteins.

    image072.png

    Critical to the structure of proteins is the fact that, although it is conventionally drawn as a single bond, the C-N bond in a peptide linkage has a significant barrier to rotation, almost as if it were a double bond.

    image074.png

    This, along with the observation that the bonding around the peptide nitrogen has trigonal planar geometry, strongly suggests that the nitrogen is sp2-hybridized. An important resonance contributor has a C=N double bond and a C-O single bond, with a separation of charge between the oxygen and the nitrogen.

    image076.png

    Although B is a minor contributor due to the separation of charges, it is still very relevant in terms of peptide and protein structure – our proteins would simply not fold up properly if there was free rotation about the peptide C-N bond.

    Contributors and Attributions

    1. P. Atkins, J. De Paula, Physical Chemistry for the Life Sciences, 2ed. (W.H. Freeman and Company, New York, 2011) Pp. 369 - 372.


    This page titled 9.3: Hybridization of Atomic Orbitals was last modified on Mon, 30 Jun 2025 17:14:40 GMT and is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Jonathan Gutow.

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