Drawing the resonance hybrid

Drawing the Resonance Hybrid

Resonance is a fundamental concept in understanding the stability and behavior of molecules with delocalized electrons. When a molecule has multiple valid Lewis structures, the true structure is a resonance hybrid—a blend of these resonance forms. In this section, we'll explore how to visualize and draw the resonance hybrid, focusing on key steps to represent the delocalization accurately.  We will use acetic acid as an example.

1. Identifying Resonance Forms

Begin by drawing all significant resonance structures. Each resonance form should follow the standard rules of Lewis structures, where all atoms maintain their usual bonding preferences, and formal charges are minimized where possible.

For our acetate ion example, here are the two most significant resonance structures.  This is the pattern discussed earlier, in which there is a pi bond next to an element with a full octet and lone pair electrons.

2. Analyzing Electron Delocalization

Examine the resonance forms to identify where electrons are shared among atoms. Pay particular attention to pi bonds and lone pairs on adjacent atoms, as these typically represent areas of electron delocalization.

In these two resonance structures, you can see there's a double bond (=) between the C and O atoms.  In the first structure, there's a double bond between the C and top O atom, while in the second structure there's a double bond between the C and the bottom most O.  Likewise, the top O on the left is negative, while the bottom O is negative on the resonance structure on the right.

3. Depicting the Hybrid

• Dashed Bonds: Use dashed lines to represent bonds that are partially present in the hybrid form, showing that electron density is shared across multiple atoms.  Since the left resonance structure indicates there's a double bond between the C and top O and the right resonance structure there's a single bond between the C-O, we can assume the is about 1 and a 1/2 bonds between them in the resonance hybrid.  Likewise, for the C and the bottom most O atom we would have 1 and 1/2 bonds.  How did I determine 1 and 1/2 bonds.  Well, if you add up all the bonds in all the resonance structures between the C and the uppermost C, you would have three bonds.  Dividing 3 by the number of resonance structures, 2 in this case we get 3/2 (i.e. 1 and 1/2 bonds).  This is depicted as a partial double bond with dashes, as shown below.
• Partial Charges: If any atoms consistently bear partial charges across resonance forms, indicate these charges as partial (δ+ or δ−) in the hybrid.  Since the O atom on the top of the left structure is neutral and the top oxygen on the right is -1 charge, we know the charge is somewhere between 0 and -1, so we assign a partial negative δ^- charge to this oxygen.  Similarly, the bottom most O has a partial negative charge (δ^-).
• Symmetry and Bond Lengths: Emphasize that bonds involved in resonance tend to equalize in length between single and double bond lengths, reflecting the resonance stabilization.