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.
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.
This discrepency required modification of valence bond theory. That modification is what we now call resonance theory. Check out Figure 2.
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.
There are three key features about the first half of this animation:
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 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.
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.
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.
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.
Note the similarity in the formalism of the left-hand structure to that for the carbonate ion in Figure 5.
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
Figure 11 illustrates each of these possibilities.
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.
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.
Additional exercises
