Resonance Structures

In ozone O₃ you can place the double bond on the left oxygen or the right one. Both drawings obey every rule, so neither can be preferred. Experiment shows the two O-O bonds are identical in length, intermediate between a single and a double bond, so neither drawing alone is the truth.

In one ozone contributor, use a central O with one lone pair, a double-bonded terminal O with two lone pairs, and a single-bonded terminal O with three lone pairs. Bonds use six electrons and the six lone pairs use twelve, totaling eighteen. The formal charges are +1 on the central O, −1 on the singly bonded terminal O and 0 on the other. In the other contributor the terminal roles exchange without changing the O–O–O skeleton.

Resonance contributors are alternative Lewis drawings used to describe a single species with delocalized electrons. The resonance arrow ↔ connects drawings; an equilibrium arrow describes different interconverting chemical species. Ozone is not spending half its time with a left double bond and half with a right double bond. Its equivalent O–O bonds have the same intermediate character at a given molecular geometry.

“One structure” does not mean the nuclei never vibrate or rotate. Resonance is a statement about the electronic description, not a denial of ordinary molecular motion. Nor do the contributors exist as separable molecules that could be collected in separate flasks.

For equivalent contributors in the introductory Lewis-resonance model, average the bond orders at a fixed pair of atoms with equal weights. In nitrate, that N–O pair is double once and single twice across three equivalent contributors, giving (2 + 1 + 1)/3 = 4/3. Unequal contributors cannot automatically be averaged equally; a weighted model would require defensible weights. These fractional Lewis bond orders summarize delocalization and are not unique quantitative measurements of electron density.

For a fixed N–O pair in nitrate, follow that exact pair through the three equivalent contributors: its orders are 2, 1 and 1, so the equal-weight Lewis average is 4/3. Do not divide by the number of atoms in the whole ion or average unrelated bonds. This fractional number summarizes a model of delocalization; it is not a literal fractional bond line switching on and off. Unequal contributors, such as those for N₂O, cannot automatically be assigned equal weights.

When the contributors are not equivalent, the ones with lower formal charges and with negative formal charge on the more electronegative atom contribute more to the hybrid. That idea is developed further in the formal charge lessons.

Delocalization often stabilizes a species relative to an otherwise comparable localized description, but it does not make every resonance-stabilized substance unreactive. Ozone is reactive, and benzene undergoes appropriate reactions. Chemical reactivity also depends on reagents, pathways and activation barriers. Use resonance to explain equivalent bonds and charge distribution without turning it into a universal reactivity prediction.

1. Keep the atom connectivity unchanged; do not move an H from one atom to another.
2. Move existing lone pairs and π pairs, preserving the total electron count.
3. Preserve the overall charge and check required octets in each contributor.
4. Recalculate formal charges after moving electrons.

For nitrite, a lone pair on the single-bonded O becomes an N=O π pair while the original π pair becomes a lone pair on the other O. Moving only the first pair would leave N with too many electrons. The two drawings are equivalent because the O atoms occupy equivalent positions. By contrast, changing which atom is bonded to H creates a different connectivity and is not resonance. All contributors must be individually valid Lewis bookkeeping drawings, even though none alone shows the full electron distribution.