D11.6 Hybridization in Resonance Structures

The assignment of hybridization and molecular geometry for molecules that have two or more major resonance structures is the same as the process for a single Lewis structure. There is one caveat: the hybridization (and hence molecular geometry) assigned to one resonance structure must be the same as all other resonance structures in the set. This is because a set of resonance structures describes a single molecule.

Therefore, when assigning hybridization, consider all the major resonance structures. If an atom’s n_hyb is different in one resonance structure from another, then use the smaller n_hyb to determine hybridization of that atom.

In formamide, the delocalized π molecular orbital extends over the oxygen, carbon, and nitrogen atoms (see figure below). This is also described by the set of resonance structures, where there is a partial double-bond character between O and C as well as between C and N. An unhybridized 2p atomic orbital on the N atom must contribute to this π bond. Therefore, the N atom must have sp² hybridization (it forms three σ bonds) and a trigonal planar local geometry. If N had an sp³ hybridization, it would not be able to participate in the π bond.

When looking at the left resonance structure, you might be tempted to assign sp³ hybridization to N given its similarity to ammonia (NH₃). However, this is a resonance structure; the set of resonance structures describes a molecule that cannot be described correctly by a single Lewis structure. All atoms must remain in the same positions from one resonance structure to another in a set of resonance structures. There cannot be a N atom that is trigonal pyramidal in one resonance structure and trigonal planar in another resonance structure, because the atoms attached to the N would have to change positions.

Experimental evidence and quantum mechanics calculations show that formamide is a planar molecule, hence, the hybridization and geometry assignments are valid. This also means that the planar geometry must be more stable than the geometry involving a trigonal pyramidal sp³ N. In other words, participation in the π bonding stabilizes the sp² hybridization of N.