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ChemWiki: The Dynamic Chemistry Hypertext > Core > Organic Chemistry > Organic Chemistry With a Biological Emphasis > Chapter 1: Introduction to organic structure and bonding I > Section 1.5: Valence bond theory: \(sp\), \(sp^2\), and \(sp^3\) hybrid orbitals

Section 1.5: Valence bond theory: \(sp\), \(sp^2\), and \(sp^3\) hybrid orbitals

Note - this section has been updated - see the updated version here


As we have talked about using Lewis structures to depict the bonding in organic compounds, we have been very vague in our language about the actual nature of the chemical bonds themselves. We know that a covalent bond involves the ‘sharing’ of a pair of electrons between two atoms - but how does this happen, and how does it lead to the formation of a bond holding the two atoms together?

In this section, we will use a model called valence bond theory to describe bonding in organic 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.

Valence bond theory is adequate for describing many aspects of organic structure.  In some cases, however, chemists need to use a different model, called molecular orbital  (MO) theory, to  talk about covalent bonds in which electrons are shared not just between two atoms, but between several, or even over an entire molcule. The MO model will be introduced in chapter 2, when these particular chemical phenomena are first discussed. For now, we will concentrate on the more straightforward valence bond theory.

1.5A: Formation of sigma bonds: the \(H_2\) molecule

The simplest case to consider is the hydrogen molecule, H2. When we say that the two electrons from each of the hydrogen atoms are shared to form a covalent bond between the two atoms, what we mean in valence bond theory terms is that the two spherical 1s orbitals overlap, allowing the two electrons to form a pair within the two overlapping orbitals. 


These two electrons are now attracted to the positive charge of both of the hydrogen nuclei, with the result that they serve as a sort of ‘chemical glue’ holding the two nuclei together. 

How far apart are the two nuclei? That is a very important issue to consider. If they are too far apart, their respective 1s orbitals cannot overlap, and thus no covalent bond can form - they are still just two separate hydrogen atoms. As they move closer and closer together, orbital overlap begins to occur, and a bond begins to form. This lowers the potential energy of the system, as new, attractive positive-negative electrostatic interactions become possible between the nucleus of one atom and the electron of the second.

But something else is happening at the same time: as the atoms get closer, the repulsive positive-positive interaction between the two nuclei also begins to increase. 


At first this repulsion is more than offset by the attraction between nuclei and electrons, but at a certain point, as the nuclei get even closer, the repulsive forces begin to overcome the attractive forces, and the potential energy of the system rises quickly. When the two nuclei are ‘too close’, we have a very unstable, high-energy situation. There is a defined optimal distance between the nuclei in which the potential energy is at a minimum, meaning that the combined attractive and repulsive forces add up to the greatest overall attractive force. This optimal internuclear distance is the bond length.  For the H2 molecule, this distance is 74 x 10-12 meters, or 0.74 Å (Å means angstrom, or 10-10 meters). Likewise, the difference in potential energy between the lowest state (at the optimal internuclear distance) and the state where the two atoms are completely separated is called the  bond energy.  For the hydrogen molecule, this energy is equal to about 104 kcal/mol.

Every covalent bond in a given molecule has a characteristic length and strength. In general, carbon-carbon single bonds are about 1.5 Å long (Å means angstrom, or 10-10 meters) while carbon-carbon double bonds are about 1.3 Å, carbon-oxygen double bonds are about 1.2 Å, and carbon-hydrogen bonds are in the range of 1.0 – 1.1 Å. Most covalent bonds in organic molecules range in strength from just under 100 kcal/mole (for a carbon-hydrogen bond in ethane, for example) up to nearly 200 kcal/mole. You can refer to tables in reference books such as the CRC Handbook of Chemistry and Physics for extensive lists of bond lengths, strengths, and many other data for specific organic compounds.

It is not  accurate, however, to picture covalent bonds as rigid sticks of unchanging length - rather, it is better to picture them as springs which have a defined length when relaxed, but which can be compressed, extended, and bent. This ‘springy’ picture of covalent bonds will become very important in chapter 4, when we study the analytical technique known as infrared (IR) spectroscopy. 

One more characteristic of the covalent bond in H2 is important to consider at this point. The two overlapping 1s orbitals can be visualized as two spherical balloons being squashed together. This means that the bond is cylindrically symmetrical: if we were to take a cross-sectional plane of the bond, it would form a circle. This type of bond is referred to as a σ (sigma) bond.

We can take a similar approach in using valence orbital theory to describe the covalent bond in hydrofluoric acid, HF. In the electron configuration of an isolated fluorine atom, the 2sand two of the three 2porbitals are filled with two electrons each. The third 2pz orbital, however, is occupied only by a single electron, and thus is available for bonding. In the HF molecule, a covalent bond is formed from overlap of the 1s orbital in hydrogen with the half-full 2pz orbital in fluorine. 


This bond, like that in H2, has symmetrical symmetry (picture a spherical balloon being pressed directly into one of the ends of an elongated balloon), and thus is classified as a sigma bond. The covalent bond in molecular fluorine, F2, is treated as the overlap of two half-filled 2pz orbitals, one from each fluorine atom. This also is a sigma bond.


1.5B: Hybrid orbitals - \(sp^3\) hybridization and tetrahedral bonding

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


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.


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.


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).


While previously we drew a Lewis structure of methane in two dimensions using lines to denote each covalent bond, we can now draw a more accurate structure in three dimensions, showing the tetrahedral bonding geometry. To do this on a two-dimensional page, though, we need to introduce a new drawing convention: the solid / dashed wedge system. In this convention, a solid wedge simply represents a bond that is meant to be pictured emerging from the plane of the page. A dashed wedge represents a bond that is meant to be pictured pointing into, or behind, the plane of the page. Normal  lines imply bonds that lie in the plane of the page. 


This system takes a little bit of getting used to, but with practice your eye will learn to immediately ‘see’ the third dimension being depicted.

Example 1.17

Imagine that you could distinguish between the four hydrogens in a methane molecule, and labeled them Ha through Hd. In the images below, the exact same methane molecule is rotated and flipped in various positions. Draw the missing hydrogen atom labels. (It will be much easier to do this if you make a model.)


Example 1.18

Describe, with a picture and with words, the bonding in chloroform, CHCl3.


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.


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. 


In chapter 3 we will learn more about the implications of rotational freedom in sigma bonds, when we discuss the ‘conformation’ 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. 


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.


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 sbonds 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.


Some experimental evidence, however, suggests that the bonding orbitals on the oxygen are actually unhybridized 2p orbitals rather than sp3 hybrids. Although this would seem to imply that the H-O-H bond angle should be 90˚ (remember that p orbitals are oriented perpendicular to one another), it appears that electrostatic repulsion has the effect of distorting this p-orbital angle to 104.5˚. Both the hybrid orbital and the nonhybrid orbital models present reasonable explanations for the observed bonding arrangement in water, so we will not concern ourselves any further with the distinction.

Example 1.19

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


1.5C: Formation of \(\pi\) bonds - \(sp^2\) 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.


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.


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).


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


spacefilling image of bonding in ethene

The pi bond does not have symmetrical symmetry. 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. This argument extends to larger alkene groups: in each case, the six atoms of the group form a single plane.


Conversely, sbonds such as the carbon-carbon single bond in ethane (CH3CH3) exhibit free rotation, and can assume many different conformations, or shapes - this is one of the main subjects of Chapter 3.

Example 1.20

Circle the six atoms in the molecule below that are ‘locked’ into the same plane.


Example 1.21
What kinds of orbitals are overlapping in bonds a-d indicated below?




Example 1.22

What is wrong with the way the following structure is drawn?




A similar picture can be drawn for the bonding in carbonyl groups, such as formaldehyde. In this molecule, the carbon is sp2-hybridized, and we will assume that the oxygen atom is also sp2 hybridized. The carbon has three sigma bonds: two are formed by overlap between two of its sp2 orbitals with the 1sorbital from each of the hydrogens, and the third sigma bond is formed by overlap between the remaining carbon sp2 orbital and an sp2 orbital on the oxygen.  The two lone pairs on oxygen occupy its other two sp2 orbitals.


The pi bond is formed by side-by-side overlap of the unhybridized 2pz orbitals on the carbon and the oxygen. Just like in alkenes, the 2pz orbitals that form the pi bond are perpendicular to the plane formed by the sigma bonds. 

Example 1.23

Describe and draw the bonding picture for the imine group shown below. Use the drawing of formaldehyde above as your guide.




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. 


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.




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). 


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. 


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 1.24
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."




1.5D: The valence bonding picture in carbocations, carbanions, and carbon free radicals

Although they are almost always high-energy, unstable species, carbocations, carbanions, and carbon free radical species play important - if short-lived – roles in many organic reactions.  If a carbon atom bears a positive formal charge, or has an unpaired electron, it is generally considered to be sp2 hybridized, meaning that its three sigma bonds are arranged with trigonal planar geometry. That just leaves the perpendicular, unhybridized 2pz orbital to account for.  If the carbon bears a positive charge, the 2pz orbital is considered to be empty. In the case of a radical, the 2pz orbital is occupied by a single unpaired electron.


The picture for a negatively-charged carbon - a carbanion - depends on the broader molecular context, specifically the presence of conjugated pi bonds (section 2.1B).  In the simple case of a methyl anion, however, however, the bonding is pictured as sp3, with one of the four orbitals holding the lone pair of electrons.

The orbital and bonding arrangements for carbocation, radicals, and carbanions will be very important to consider when we begin to consider the idea of resonance (section 2.2).

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