10.3: Orbital Overlap in Multiple Bonds

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Learning Objectives

To explain double and triple bonds in terms of orbital overlap

So far in our valence bond orbital descriptions we have not dealt with polyatomic systems with multiple bonds. To do so, we can use an approach in which we describe \(\sigma\) bonding using localized electron-pair bonds formed by hybrid atomic orbitals, and \(\pi\) bonding using unhybridized np atomic orbitals.

Multiple Bonding

We begin our discussion by considering the bonding in ethylene (C₂H₄). Experimentally, we know that the H–C–H and H–C–C angles in ethylene are approximately 120°. This angle suggests that the carbon atoms are sp² hybridized, which means that a singly occupied sp² orbital on one carbon overlaps with a singly occupied s orbital on each H and a singly occupied sp² lobe on the other C. The sp² hybridization can be represented as follows:

With this hybridization, each carbon forms a trigonal planar set of three \(\sigma\) bonds: two C–H (sp² + s) and one C–C (sp² + sp²) (part (a) in Figure \(\PageIndex{1}\)). After hybridization, however, each carbon still has one unhybridized 2p_z orbital that is perpendicular to the hybridized lobes and contains a single electron (part (b) in Figure \(\PageIndex{1}\)). These two singly occupied 2p_z orbitals can overlap in a side-to-side fashion to form a \(\pi\) bond. The orbitals overlap both above and below the plane of the molecule but form just one bonding orbital space. The C–C \(\pi\) bond plus the hybridized C–C \(\sigma\) bond together form a double bond.

Figure \(\PageIndex{1}\): Bonding in Ethylene. (a) The \(\sigma\)-bonded framework is formed by the overlap of two sets of singly occupied carbon sp² hybrid orbitals and four singly occupied hydrogen 1s orbitals to form electron-pair bonds. This uses 10 of the 12 valence electrons to form a total of five \(\sigma\) bonds (four C–H bonds and one C–C bond). (b) One singly occupied unhybridized 2p_z orbital remains on each carbon atom to form a carbon–carbon \(\pi\) bond. (Note: by convention, in planar molecules the axis perpendicular to the molecular plane is the z-axis.)

Triple bonds, as in acetylene (C₂H₂), can also be explained using a combination of hybrid atomic orbitals and unhybridized np orbitals. The four atoms of acetylene are collinear, which suggests that each carbon is sp hybridized. If one sp lobe on each carbon atom is used to form a C–C \(\sigma\) bond and one is used to form the C–H \(\sigma\) bond, then each carbon will still have two unhybridized 2p orbitals (a 2p_x,y pair), each with one electron (part (a) in Figure \(\PageIndex{3}\)).

The two 2p orbitals on each carbon can align with the corresponding 2p orbitals on the adjacent carbon to simultaneously form a pair of \(\pi\) bonds (part (b) in Figure \(\PageIndex{2}\)). Because each of the unhybridized 2p orbitals has a single electron, four electrons are available for \(\pi\) bonding, which is enough to occupy only the bonding molecular orbitals. Acetylene must therefore have a carbon–carbon triple bond, which consists of a C–C \(\sigma\) bond and two mutually perpendicular \(\pi\) bonds. Acetylene does in fact have a shorter carbon–carbon bond (120.3 pm) and a higher bond energy (965 kJ/mol) than ethane and ethylene, as we would expect for a triple bond.

Figure \(\PageIndex{3}\): Bonding in Acetylene (a) In the formation of the \(\sigma\)-bonded framework, two sets of singly occupied carbon sp hybrid orbitals and two singly occupied hydrogen 1s orbitals overlap. (b) In the formation of two carbon–carbon \(\pi\) bonds in acetylene, two singly occupied unhybridized 2p_x,y orbitals on each carbon atom overlap. With one \(\sigma\) bond plus two \(\pi\) bonds, the carbon–carbon bond order in acetylene is 3.

Example \(\PageIndex{1}\)

Describe the bonding in HCN using a combination of hybrid atomic orbitals and unhybridized p orbitals. The HCN molecule is linear.

Given: chemical compound and molecular geometry

Asked for: bonding description using hybrid atomic orbitals and p orbitals

Strategy:

From the geometry given, predict the hybridization in HCN. Use the hybrid orbitals to form the \(\sigma\)-bonded framework.
Use any remaining unhybridized p orbitals to form \(\pi\) bonds.

Solution:

\(\sigma\)-bonding framework: an unhybridized s orbital on hydrogen overlaps with an sp hybrid orbital on carbon, sp hybrid orbital on carbon overlaps with an sp hybrid orbital on nitrogen
\(\pi\) bonding: unhybridized p atomic orbitals on carbon and nitrogen overlap to form a \(\pi\) bond.

Exercise \(\PageIndex{1}\)

Describe the bonding in formaldehyde (H₂C=O), a trigonal planar molecule, using a combination of hybrid atomic orbitals and molecular orbitals.

Answer

\(\sigma\)-bonding framework: Carbon and oxygen are sp² hybridized. Two sp² hybrid orbitals on oxygen have lone pairs, two sp² hybrid orbitals on carbon form C–H bonds, and one sp² hybrid orbital on C and O forms a C–O \(\sigma\) bond.
\(\pi\) bonding: Unhybridized, singly occupied 2p atomic orbitals on carbon and oxygen overlap to form a \(\pi\) bond.

Orbital Overlap and Resonance Structures

Resonance structures can be used to describe the bonding in molecules such as ozone (O₃) and the nitrite ion (NO₂⁻). Ozone can be represented by either of these Lewis electron structures:

Although the VSEPR model correctly predicts that both species are bent, it gives no information about their bond orders.

Experimental evidence indicates that ozone has a bond angle of 117.5°. Because this angle is close to 120°, it is likely that the central oxygen atom in ozone is trigonal planar and sp² hybridized. If we assume that the terminal oxygen atoms are also sp² hybridized, then we obtain the \(\sigma\)-bonded framework shown in Figure \(\PageIndex{4}\). Two of the three sp² lobes on the central O are used to form O–O sigma bonds, and the third has a lone pair of electrons. Each terminal oxygen atom has two lone pairs of electrons that are also in sp² lobes. In addition, each oxygen atom has one unhybridized 2p orbital perpendicular to the molecular plane. The \(\sigma\) bonds and lone pairs account for a total of 14 electrons (five lone pairs and two \(\sigma\) bonds, each containing 2 electrons). Each oxygen atom in ozone has 6 valence electrons, so O₃ has a total of 18 valence electrons. Subtracting 14 electrons from the total gives us 4 electrons that must occupy the three unhybridized 2p orbitals.

Figure \(\PageIndex{4}\): Bonding in Ozone. (a) In the formation of the \(\sigma\)-bonded framework, three sets of oxygen sp² hybrid orbitals overlap to give two O–O \(\sigma\) bonds and five lone pairs, two on each terminal O and one on the central O. The \(\sigma\) bonds and lone pairs account for 14 of the 18 valence electrons of O₃. (b) One unhybridized 2p_z orbital remains on each oxygen atom that is available for \(\pi\) bonding. The unhybridized 2p_z orbital on each terminal O atom has a single electron, whereas the unhybridized 2p_z orbital on the central O atom has 2 electrons.

Example \(\PageIndex{2}\)

Describe the bonding in the nitrite ion in terms of a combination of hybrid atomic orbitals and unhybridized p orbitals. Lewis dot structures and the VSEPR model predict that the NO₂⁻ ion is bent.

Given: chemical species and molecular geometry

Strategy:

From the structure, predict the type of atomic orbital hybridization in the ion the central atom.
Use hybrid orbitals to form the \(\sigma\)-bonding framework.
Use hybrid orbitals to form the \(\pi\) bonding.

Solution:

A The lone pair of electrons on nitrogen and a bent structure suggest that the bonding in NO₂⁻ is similar to the bonding in ozone. The bent structure implies that the nitrogen is sp² hybridized.

B If we assume that the oxygen atoms are sp² hybridized as well, then we can use two sp² hybrid orbitals on each oxygen and one sp² hybrid orbital on nitrogen to accommodate the five lone pairs of electrons. Two sp² hybrid orbitals on nitrogen form \(\sigma\) bonds with the remaining sp² hybrid orbital on each oxygen. The \(\sigma\) bonds and lone pairs account for 14 electrons. We are left with three unhybridized 2p orbitals, one on each atom, perpendicular to the plane of the molecule, and 4 electrons. Just as with ozone, these three 2p orbitals can overlap to form pi bonds.

Exercise\(\PageIndex{2}\)

Describe the bonding in the formate ion (HCO₂⁻), in terms of a combination of hybrid atomic orbitals and molecular orbitals.

Answer: Like nitrite, formate is a planar polyatomic ion with 18 valence electrons. The \(\sigma\) bonding framework can be described in terms of sp² hybridized carbon and oxygen, which account for 14 electrons. The three unhybridized 2p orbitals (on C and both O atoms) overlap to form \(\pi\) bonds. The overall C–O bond order is therefore \(frac{3}{2}\)

Summary

Polyatomic systems with multiple bonds can be described using hybrid atomic orbitals for \(\sigma\) bonding and unhybridized p orbitals to describe \(\pi\) bonding.