sp³ Hybrid orbitals and tetrahedral bonding

Now let’s look more carefully at bonding in organic molecules, starting with methane, CH₄. Recall the valence electron configuration of a carbon atom:

This picture is problematic when it comes to describing the bonding in methane. 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 wave equations) to form four equivalent hybrid orbitals, which are called 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.

Click to see an interactive 3D model (select ‘load sp3‘ and ‘load H 1s’ to see orbitals).

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 a sigma bond formed by 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 109 pm.

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.

How does this bonding picture extend to compounds containing carbon-carbon bonds?  In ethane (CH₃CH₃), both carbons are sp³-hybridized, meaning that both have four bonds with tetrahedral geometry.  The carbon-carbon bond, with a bond length of 154 pm, 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 an axle, each one able to rotate 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 nonbonding pair (lone 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 s 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˚. The bonding in water results from overlap of two of the four sp³ hybrid orbitals on oxygen with 1s orbitals on the two hydrogen atoms. The two nonbonding electron pairs on oxygen are located in the two remaining sp³orbitals.

See a video tutorial on sp³ orbitals and sigma bonds (Note: This is the video linked to in the previous section)

sp² and sp hybrid orbitals and pi bonds

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 120^o, and the carbon-carbon bond length is 134 pm, significantly shorter than the 154 pm 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, 2p_x and 2p_y orbitals – combine to form three sp² hybrids, leaving the 2p_z 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 2p_z 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 2p_z orbitals are shown in the ‘space-filling’ mode).

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 pi bond, which is formed by the side-by-side overlap of the two unhybridized 2p_z orbitals from each carbon.

animation	spacefilling image	video tutorial	interactive 3D model (select ‘show resulting pi orbital’)

 This illustration (from University of Florida) shows the sigma and pi bonds in ethene.

Unlike a sigma bond, a pi bond does not have cylindrical symmetry. If rotation about this bond were to occur, it would involve disrupting the side-by-side overlap between the two 2p_z 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, six atoms lie in the same plane.

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 sp²hybridized. The carbon has three sigma bonds: two are formed by overlap between sp² orbitals with 1s orbitals from hydrogen atoms, 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.

interactive 3D model

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