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Drawing resonance structures is only half the question. JEE and NEET then ask for a number: the bond order of an N–O bond in nitrate, or which ion has the shortest bond. Our post on ranking resonance contributors worked carbonate, acetate and benzene. This one gives a counting rule you can use on any ion with equivalent structures, and applies it to nitrate, ozone, the nitrogen oxyanions and the third-period oxyanions.
Where is this in the syllabus?
NCERT's Class 11 unit Chemical Bonding and Molecular Structure (2026-27 reprint) covers it in sections 4.3.4 (Bond Order) and 4.3.5 (Resonance Structures). It defines bond order, in the Lewis picture, as "the number of bonds between the two atoms in a molecule", and Exercise 4.13 asks you to write resonance structures for SO₃, NO₂ and NO₃⁻. The JEE (Main) 2026 syllabus lists "resonance" and "the concept of bond order, bond length and bond energy" in Unit 3, and the NEET (UG) 2026 syllabus has the same two items.
The same chapter has a second bond order, from molecular orbital theory, defined as half the difference between bonding and antibonding electrons. That one is for diatomic molecules. This post is about the resonance average.
What is the counting rule?
For a set of equivalent bonds from one central atom:
average bond order = (total number of bonds to those atoms, counting a double bond as 2) ÷ (number of links in the set)
Use one resonance structure to count, because every equivalent structure gives the same total. The rule only works when the contributors are equivalent; if one structure dominates, the bonds are not equal and a simple average is wrong.
Worked example 1: nitrate, NO₃⁻
Each structure has one N=O and two N–O bonds: 2 + 1 + 1 = 4 bonds over 3 links, so each N–O has bond order 4/3 ≈ 1.33.
Check the structure with formal charges (valence electrons − lone-pair electrons − half the bonding electrons). Nitrogen: 5 − 0 − 4 = +1. Each single-bonded oxygen: 6 − 6 − 1 = −1. Double-bonded oxygen: 6 − 4 − 2 = 0. Total: +1 − 1 − 1 + 0 = −1, the charge on the ion.
Worked example 2: ozone, and what the data show
Ozone has two equivalent structures, each with one O–O and one O=O: 3 bonds over 2 links, so bond order 1.5. NCERT gives the evidence: the normal O–O and O=O bond lengths are 148 pm and 121 pm, and both oxygen–oxygen bonds in O₃ measure 128 pm.
Notice that 128 pm is not the midpoint of 148 and 121 (that would be 134.5 pm). Bond length falls as bond order rises, as NCERT states, but not in a straight line. Use bond orders to rank lengths, not to calculate them.
Worked example 3: ranking the N–O bonds
NO⁺: bond order 3 (NCERT lists it as isoelectronic with N₂ and CO).
NO₂⁺: O=N=O, one structure, bond order 2.
NO₂⁻: two equivalent structures, 3 bonds over 2 links, 1.5.
NO₃⁻: 4/3.
Higher bond order means a shorter bond, so the N–O length increases in the order NO⁺ < NO₂⁺ < NO₂⁻ < NO₃⁻. The same reasoning ranks carbon–oxygen bonds: CO (3) < CO₂ (2) < HCOO⁻ (1.5) < CO₃²⁻ (4/3) for length.
What about sulphate, phosphate and perchlorate?
NCERT allows an expanded octet for elements in and beyond the third period. With those structures:
SO₄²⁻: two S=O and two S–O⁻, 6 bonds over 4 links, 1.5.
PO₄³⁻: one P=O and three P–O⁻, 5 over 4, 1.25.
ClO₄⁻: three Cl=O and one Cl–O⁻, 7 over 4, 1.75.
These numbers belong to the expanded-octet model. An all-single-bond structure with formal charges would give 1 for each, so state which picture you used. And compare bond orders only between the same pair of elements: a Cl–O bond and a P–O bond differ in length for reasons that have nothing to do with resonance.
Which slips should you watch for?
Averaging over structures that are not equivalent.
Counting a C–H or other bond outside the equivalent set, as in formate, where only the two C–O bonds share the resonance.
Mixing the molecular-orbital bond order of O₂ with the resonance average of O₃.
Forgetting that the formal charges must add up to the charge on the ion.
Treating bond length as a linear function of bond order.
If the idea of a resonance hybrid itself is shaky, the Fermi Concept Library covers every concept in physics, chemistry, biology, mathematics and computing, each with its own lesson, so you can rebuild it before returning to the counting.
A practice routine
For NO₃⁻, NO₂⁻, NO₂⁺, CO₃²⁻, HCOO⁻, O₃, SO₄²⁻, PO₄³⁻, ClO₄⁻ and ClO₃⁻, draw one structure, count bonds and divide by links.
Check every structure with formal charges.
Rank the N–O, C–O and halogen–oxygen sets by bond length.
Do NCERT Exercise 4.13, and for SO₃ work out the answer in both the octet and the expanded-octet pictures.
For the person behind the idea of resonance in chemistry, read our profile of Linus Pauling and the chemical bond. Start with nitrate tonight: one structure, one count, one division.
