Draw the nitrate ion with Formal Charge and you face a choice: which of the three oxygens gets the double bond? All three choices are equally good, and none is right on its own. Experiments show all three N–O bonds are the same length.
Key idea
When two or more valid Lewis structures differ only in where the electrons go (not the atoms), the real species is a single resonance hybrid, a blend of them. Its bonds are in between: the average bond order is $$\text{bond order} = \frac{\text{total bonds to the outer atoms}}{\text{number of outer atoms sharing them}}.$$
Why it works
Lewis structures can only draw electron pairs in fixed places, one pair or two pairs between atoms. Nitrate's extra bonding pair is really spread over all three N–O links at once. Each resonance structure is a partial description, and the double-headed arrow between them means "the truth is a blend", not "flips back and forth". Spreading electrons over more atoms lowers their energy, which is why resonance makes ions like nitrate and carbonate unusually stable. It also spreads charge: in every structure two oxygens carry a formal charge of $-1$ (and nitrogen $+1$), so in the blend each oxygen carries $-\tfrac{2}{3}$.
Worked example: Bond order in nitrate
What is the N–O bond order in $\mathrm{NO_3^-}$?
- Each structure has one double and two single bonds: 4 bonds in all between N and the three O atoms.
- Shared over 3 positions: $4/3 \approx 1.33$.
Answer: $\tfrac{4}{3}$
Worked example: Ozone
Ozone, $\mathrm{O_3}$, has two resonance structures. Predict its O–O bond order and estimate its bond length, given O–O (single) 148 pm and O=O (double) 121 pm.
- Each structure has one double and one single bond: 3 bonds over 2 positions, bond order $1.5$.
- The length should fall between 148 and 121 pm. Measured: both bonds are 128 pm.
Answer: Bond order $1.5$; about 128 pm, between single and double.
Worked example: Which bond is longest?
Rank the C–O bonds in $\mathrm{CO}$, $\mathrm{CO_2}$ and $\mathrm{CO_3^{2-}}$ from shortest to longest.
- Bond orders: CO 3 (triple), $\mathrm{CO_2}$ 2 (two double bonds), $\mathrm{CO_3^{2-}}$ $4/3$ (resonance).
- Higher bond order, shorter bond.
Answer: $\mathrm{CO} < \mathrm{CO_2} < \mathrm{CO_3^{2-}}$ (113, 116, 129 pm).
Common mistake
Thinking the molecule flips back and forth between its resonance structures, spending a third of its time in each. There is only one real structure, the hybrid, at every instant: all three N–O bonds are always equal. A mule is not a horse on Monday and a donkey on Tuesday; it is always a mule. Check: if the structures were real and flipping, measurements would find two different bond lengths. They find one.