Multiple Bonds

In some molecules, two or more atoms can form more than one bond. Valence bond theory and VSEPR theory together can be used to explain multiple bonds. VSEPR theory treats a double or triple bond the same way as a single bond and considers only the arrangement of atoms around a central atom. Valence bond theory, on the other hand, must reconcile multiple bonds with hybrid orbitals. This can be accomplished by combining valence bond theory and VSEPR theory. For example, ethylene (C₂H₄) has a carbon-carbon double bond and two hydrogen atoms bound to each carbon atom. According to VSEPR, an ethylene molecule has two central carbon atoms. All four valence electrons per carbon atom are engaged in bonding. Two electrons are involved in carbon-carbon double bonds, and two electrons are involved in carbon-hydrogen bonds. There are three bonding structures around each carbon atom. This arrangement corresponds to a trigonal planar geometry in VSEPR theory.

This result can be used to infer the hybridization scheme. Trigonal planar electron-pair geometry requires three sp² hybrid orbitals, formed from the 2s, 2p_x, and 2p_y orbitals. Each of these hybrid orbitals requires an unpaired electron in order to be able to form three bonds. Hybridization begins with the lowest electrons in the lowest energy state. This means the carbon atom’s paired 2s electrons and the 2p electron combine to form the three sp² orbitals using the p_x and p_y orbitals. These sp² orbitals overlap end to end with the s orbitals of the hydrogen atoms and sp² orbital of the other carbon atom, forming $\sigma$ bonds.

But carbon has four valence electrons. This means each carbon atom has one more electron in an unhybridized p_z orbital. These two p_z orbitals from the carbon atoms overlap along the x-axis in the side-by-side manner of a $\pi$ bond and are observed as a density of electrons above and below the plane of the molecule. This combination of $\sigma$ and $\pi$ bonds produces the double bond. Triple bonds are modeled in a similar way. Consider the structure of ethyne (C₂H₂). Its Lewis structure is linear with triple-bonded carbon atoms. A linear geometry requires sp hybridization, which requires a minimum of two electrons, one for each sp orbital. In the case of ethyne, the two electrons in the s orbital of the carbon atom hybridize with the empty p orbital to form the two sp orbitals. The sp hybrids form $\sigma$ bonds with the hydrogen atom and the other carbon atom. The two leftover electrons in the p orbitals form pi bonds with their counterparts on the other carbon atom. Thus, the triple bond is formed by one $\sigma$ bond and two $\pi$ bonds. The two $\pi$ bonds form a sort of tube of electron density around the straight $\sigma$ bond.

The $\pi$ bonds do not affect the shape of the molecule, but they introduce rigidity. Atoms are not able to easily rotate around a double bond in the way they can rotate around a single bond, and a triple bond is more rigid still.