During our review of valence bond theory we alluded to the fact that discrepencies between theory and experimental fact require modification or even replacement of the theory. The inability of valence bond theory to account for the fact that dioxygen is paramagnetic prompted our discussion of molecular orbital theory . Now we'll consider another situation in which valence bond theory comes up short- compounds for which more it is possible to draw more than one valid Lewis structure. It's important at the outset to understand that we are not talking about isomers. Rather, the alternative formulations consist of an invariant atomic framework which is held together by s bonds. One structure differs from the next by the way in which p and/or non-bonding electrons are distributed over that atomic framework. We'll examine an inorganic compound, sodium carbonate, Na2CO3, first.
Figure 1 shows one of three valid Lewis structures for this simple compound. In this structure every atom has a filled valence shell, two of the oxygen atoms have a formal charge of -1, and a sodium ion is associated with each of the negatively charged oxygen atoms.
1/3 of the Lewis Structures for Sodium Carbonate
Exercise 1 Draw the other two valid Lewis structures for sodium carbonate. Be sure to include all non-bonding electron pairs.
Exercise 2 According to VSEPR theory, what is the geometry of the carbonate ion?
Exercise 3 The formula for sodium bicarbonate (baking soda) is NaHCO3. Draw two valid Lewis structures for this formula. Hint-The H is bonded to an O, not to the C.
According to valence bond theory, the carbonate ion contains two types of C-O bonds; one of them is a double bond and the other two are single bonds. In other words, valence bond theory predicts that two of the C-O bonds in sodium carbonate should be the same length, while the other should be different. Specifically, the C-O double bond should be shorter than the two C-O single bonds. You might also expect that the O=C-O bond angle would be different than the O-C-O angle. These predictions are contradicted by the experimental facts: all three C-O bonds in sodium carbonate are the same length; all three O-C-O bond angles are 120o.
This discrepency required modification of valence bond theory. That modification is what we now call resonance theory. Check out Figure 2.
The Correct Answer to Exercise 1
Here we're ignoring the sodium ions because we want to focus our attention on the structure of the carbonate ion. It should be apparent that formulations 1, 2, 3 are related to each other by a C3 rotational axis: clockwise rotation of structure 1 by 120o about an axis that passes through the carbon atom perpendicular to the page produces orientation 2. A second 120o rotation generates 3, which, in turn, yields 1 upon another 120o rotation. With a little thought it should also become apparent that the difference between structures 1 and 2 is simply the placement of two electron pairs. Figure 3 animates the conversion of structure 1 into 2 by repositioning two electron pairs.
Move 'em In, Move 'em Out
Exercise 4 Consider the intermediate structure shown in red in Figure 3. How many electrons are there around the carbon in this structure?
There are three key features about the first half of this animation:
- A non-bonding pair of electrons becomes a bonding pair.
- The resultant structure (shown in red) violates the octet rule.
- The positions of the nuclei do not change.
In order to avoid violating the octet rule, the conversion of a non-bonding pair of electrons to a bonding pair must be followed by the conversion of a bonding pair to a non-bonding pair. The net result is the transfer of the negative charge from one oxygen atom to another. Figure 4 diagrams a convention for depicting this redistribution of electron density.
Here the arrow labeled 1 is used to indicate the movement of a non-bonding pair of electrons from the oxygen atom toward the carbon. In order to avoid violating the octet rule, a bonding pair of electrons moves out onto the oxygen atom as depicted by the arrow labeled 2. The result of this sequence is that the oxygen atom that was initially negatively charged becomes neutral, while the oxygen that was initially neutral becomes negatively charged. You should review the rules for calculating the formal charge on an atom and convince yourself that the charges shown in Figure 4 are correct.
Exercise 5 What is the formal charge on the carbon atom in structures 1 and 2 in Figure 4?
Exercise 6 In the step labeled 1 in Figure 4, a non-bonded pair of electrons on the oxygen becomes a bonded pair on the carbon.
a. How many electrons does the carbon atom gain in this step? 0 1 2
b. How many electrons does the oxygen lose? 0 1 2
Exercise 7 In the step labeled 2 in Figure 4, a bonded pair of electrons on the carbon becomes a non-bonded pair on the oxygen.
a. How many electrons does the carbon atom lose in this step? 0 1 2
b. How many electrons does the oxygen gain? 0 1 2
Exercise 8 Using the convention outlined in Figure 4, show how the negative charge in structure 2 is transferred from one oxygen to another to produce structure 3.
This brings us to the central thesis of resonance theory. Whenever there are two or more valid Lewis structures for the same compound, the true structure of the molecule is best represented as a "hybrid" of the individual structures. Figure 5 presents the resonance hybrid that results from "mixing" the resonance contributors 1, 2, and 3 shown in Figure 2.
Carbonate Ion as a Resonance Hybrid
Because formulations 1, 2, and 3 are related by symmetry, each of them contributes equally to the hybrid structure of the carbonate ion. In the resonance hybrid all three C-O bonds are identical. The O-C-O bond angles are all 120o. The dashed lines indicate that the C-O bonds are not complete double bonds. They are intermediate in character between a single bond and a double bond. This is consistent with the experimental fact that the C-O bond lengths in sodium carbonate are longer than a C-O single bond but shorter than a C-O double bond when compared to reference compounds.
While the carbonate ion bears a net charge of -2, that charge is not localized on specific oxygen atoms; rather it is distributed over all three oxygen atoms as implied by the brackets surrounding the hybrid structure. Charge delocalization is a characteristic feature of resonance theory. Two pictures of the electron distribution in the carbonate ion as calculated by molecular orbital theory are shown in Figure 6. Both pictures are color coded to indicate relative electron density; red implies high electron density, blue low. Picture 2 is a transparent version of 1 in which the positions of the carbon and the three oxygen nuclei are visible.
An MO Picture of Carbonate
Now let's turn our attention to...
Nitromethane, CH3NO2, is a clear, colorless liquid that boils at 101oC. It's used as a fuel additive in race cars. Figure 7 shows a valid Lewis structure for this compound. Notice that every atom has a filled valence shell. Notice, too, that this structure depicts the positive charge as being localized on the nitrogen atom and the negative charge on one of the oxygens.
Start Your Engines
Exercise 9 According to valence bond theory, should the two N-O bonds in nitromethane have the same length? Yes No
Exercise 19 Using the arrow pushing formalism illustrated in Figure 4, show how the negative charge may be transferred from the lower oxygen atom to the upper one in the structure in Figure 6.
Exercise 11 Following the format presented in Figure 5, draw the structure of the resonance hybrid of nitromethane.
As our final example, let's consider benzene, a clear, colorless liquid that boils at 80oC. Its molecular formula is C6H6. The structure of this simple hydrocarbon puzzled chemists for many years. Elucidation of the structure is a story with many interesting chapters. We'll read part of that story later. For the moment, consider the two bond-line diagrams that organic chemists commonly use to represent benzene. They are shown in Figure 8.
Two Formulations of Benzene
Exercise 12 Using the arrow pushing formalism illustrated in Figure 4, show how the left-hand structure may be converted to the right-hand structure in Figure 8.
Both of these formulations imply that benzene contains three C-C single bonds and three C-C double bonds. All six C-C bonds in benzene have the same bond length, 1.40 pm. The average bond length of a C-C single bond is 1.48 pm, while that for a C-C double bond is 1.34 pm. Since the C-C bond length in benzene is intermediate between that of a single and double bond, it is reasonable to assume that the bond itself is intermediate between that of a single and double bond. Figure 9 offers three alternative pictures of benzene as a resonance hybrid of the two formulations given in Figure 8.
Benzene as a Resonance Hybrid
Note the similarity in the formalism of the left-hand structure to that for the carbonate ion in Figure 5.
Exercise 13 Assuming its carbon framework constitutes a perfect hexagon, how many C2 symmetry axes are there in benzene? How many C6 axes? How many symmetry planes?
As you can see from the plot of the electron density shown in Figure 10, benzene is a beautiful molecule. In pictures 1 and 2 you are looking down on the molecule. Notice how the electron density is symmetrically distributed over all 6 carbon nuclei. In picture 3 you are viewing the molecule edge-on. All the nuclei lie in the same plane, while the electron density is highest above and below that plane.
Some Pretty Pictures of Benzene
The redistribution of electron density shown in Figure 4 requires the overlap of orbitals on the oxygen atoms with an orbital on the carbon. There are strict, but easily recognizable, requirements for the orbital overlap that is involved in resonance. The three most common situations are
- a pi bond adjacent to a pi bond
- a non-bonded pair adjacent to a pi bond
- a sigma bond adjacent to a pi bond
Figure 11 illustrates each of these possibilities.
The Structural Requirements for Resonance Interactions
Figure 12 offers another perspective of the structures shown in Figure 11. Here you have to imagine that you are looking along the bond axis that joins the two atoms indicated by the colored arrows in Figure 11.
Same Thing, Different Vantage Point
Notice that in all three views, the dihedral angle between the blue orbital and the adjacent red orbital orbital is 0o. In other words, the interacting orbitals line up side-to-side as they do for an ordinary p bond. However, since the distance between the interacting orbitals is greater than a normal double bond, the orbital overlap is not as large. Finally, the orbitals involved in resonance are perpendicular to the orbitals of the s-bonded atomic framework. Orbitals that are perpendicular to each other are said to be orthogonal. The overlap between orthogonal orbitals is always zero, which means that we can treat the p system of a molecule separately from its s-bonded components.
We need to consider two more aspects of resonance theory before ending this topic. First, it is important to develop the ability to evaluate the importance of resonance contributors to the resonance hybrid. To do this, check out Evaluation of Resonance Contributors. Second, we should reiterate that resonance theory was developed to rationalize experimental data that valence bond theory could not explain satisfactorily. One area in which resonance theory has proven extremely useful is in the rationalization of chemical shifts in 1H-NMR spectra of aromatic molecules.