Imagine you're trying to draw a photograph of a friend who is laughing. A single still frame captures one expression, but it misses the movement, the energy, the in-between of the laugh. A single Lewis structure does the same thing to certain molecules — it freezes them into one arrangement of electrons, but the real molecule is more like a short video clip, with electrons moving smoothly between positions.
Take ozone, O3. If you try to draw a Lewis structure, you get a dilemma. You can put the double bond on the left:
O=O−O
Or on the right:
O−O=O
Both satisfy the octet rule. Both have the same atoms. But which one is correct? Neither, alone. The real ozone molecule has two identical O−O bonds — each is halfway between a single and a double bond. No single Lewis picture can show that.
The Solution: Resonance Structures
Resonance structures are a set of two or more Lewis structures that collectively describe the actual electronic structure of a molecule where a single Lewis structure is inadequate. They are connected by a double-headed arrow (↔) to show they are not different molecules, but different ways of drawing the same molecule.
Important
Resonance structures are not real, separate molecules that flip back and forth. They are imaginary "snapshots" that we average together to get the true structure. The real molecule is a resonance hybrid — a blend of all contributing structures.
The Rules (Precise Statement)
Same atomic positions. Only electrons (pi bonds and lone pairs) move; atoms never move.
Same total number of electrons. You are redistributing, not adding or removing.
Valid Lewis structures. Each resonance form must obey the octet rule (for second-period elements) and have correct formal charges.
Curved arrows show electron movement. An arrow from a lone pair or a pi bond points to where those electrons go next.
How to Draw Them: The Curved Arrow Method
Take the nitrate ion, NO3−. Start with one valid Lewis structure:
O∣∣O−N=O−
Now, push electrons:
Take the lone pair on the top oxygen (the one with the negative charge) and push it down to form a double bond with nitrogen.
Simultaneously, push the existing double bond on the right up to become a lone pair on that oxygen.
You get a second structure:
O=N−O∣O−−
Repeat the process from this new structure, and you get a third. All three are resonance structures of NO3−.
Tip
A quick way to spot resonance: look for a pi bond next to an atom with a lone pair (or a pi bond next to a positive charge). That's the classic "conjugated system" that allows electrons to delocalize.
The Hybrid: What the Molecule Actually Looks Like
The resonance hybrid is not an average of the bond lengths — it is the actual molecule. In NO3−, all three N−O bonds are identical, with a bond order of 131 (one and one-third). The negative charge is spread equally over all three oxygens, not stuck on one.
You represent the hybrid by drawing dashed lines for partial bonds and placing the charge in a circle (or using fractional charges) to show delocalization.
Linearity here follows the classic electron-count rule: 16-valence-electron triatomics with no lone pair on the central atom (BeCl₂, CS₂) are linear; NO₂ (17 electrons, odd electron on N) is bent, and NCO⁺ (14 valence electrons) does not adopt the linear 16-electron geometry. The answer is (i) and (iv).
Species by species
BeCl2 — beryllium contributes two bond pairs and keeps no lone pair; the two Be–Cl bonds spread to 180∘. Linear ✓
NCO+ — valence electrons: 5+4+6−1=14. The familiar linear species of this family (CO₂, NCO⁻, N₂O) all have 16 valence electrons; removing two electrons from cyanate changes the electronic structure so that the 16-electron linear picture no longer applies. Not grouped with the linear pair. ✗ …
Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.
KCET 2026Set D31 markMCQ
Q.With respect to resonance structures of CO32− ion, which of the following statements are correct?
(a) All C-O bonds in CO32− are equivalent
(b) There are three resonance structures possible for CO32− ion
(c) The position of carbon and oxygen should change in every resonance structure
(d) The formal charge on carbon atom is -2
(A) a, b and c
(B) a and b only
(C) b and d only
(D) a, b and d
›Reveal solutionSolution
Resonance in CO32− makes all three C-O bonds equivalent through three contributing structures, but resonance only redistributes electrons — never the positions of the nuclei — and the formal charge on carbon is zero.
Step 1 — (a) All C-O bonds equivalent — TRUE
Because CO32− is a resonance hybrid of structures where the C=O double bond is located on each of the three oxygens in turn, the actual molecule has all three C–O bonds with equal, intermediate bond order (and hence equal bond length), confirmed experimentally.
Step 2 — (b) Three resonance structures possible — TRUE
Each resonance contributor has one C=O double bond and two C-O(-) single bonds; since the double bond can be drawn to any one of the three chemically identical oxygens, there are exactly three equivalent resonance structures.
Step 3 — (c) Position of carbon and oxygen changes in every resonance structure — FALSE …
Q.The percentage of s-character in the hybrid orbitals of nitrogen in NO2+, NO2− and NH4+ respectively are :
(A) 25%,50%,33.3%
(B) 33.3%,50%,25%
(C) 33.3%,25%,50%
(D) 50%,33.3%,25%
›Reveal solutionSolution
The s-character in hybrid orbitals depends on the steric number (SN) around nitrogen. For NO2+ (SN=2, sp), s-character = 50%; for NO2− (SN=3, sp²), s-character = 33.3%; for NH4+ (SN=4, sp³), s-character = 25%. The correct option is (D).
The key idea is that the percentage of s-character in a hybrid orbital is directly determined by the type of hybridisation. For an spⁿ hybrid, the s-character is n+11×100%. So the problem reduces to finding the hybridisation of nitrogen in each species, which in turn depends on the number of sigma bonds and lone pairs around it — the steric number.
Let’s work through each ion step by step.
NO2+ (nitronium ion)
Draw the Lewis structure: Nitrogen is the central atom, bonded to two oxygen atoms. There are no lone pairs on nitrogen because the positive charge indicates a deficiency of electrons. The steric number (number of sigma bonds + lone pairs) is 2.
Hybridisation: sp (two hybrid orbitals).
s-character: 1+11×100%=50%.
NO2− (nitrite ion)
Nitrogen is central, bonded to two oxygen atoms. There is one lone pair on nitrogen (the negative charge adds an electron). Steric number = 2 sigma bonds + 1 lone pair = 3.
Hybridisation: sp² (three hybrid orbitals).
s-character: 2+11×100%=33.3%.
NH4+ (ammonium ion)
Nitrogen is bonded to four hydrogen atoms via sigma bonds. No lone pairs (the positive charge removes one electron from the lone pair of NH₃). Steric number = 4. …
Q.Resonance effect is not observed in
(A) CH2=CH−CH=CH2
(B) CH2=CH−Cl
(C) CH2=CH−C≡N
(D) CH2=CH−CH2−NH2
›Reveal solutionSolution
Resonance requires a conjugated system of alternating single and multiple bonds. In option (D), the −NH2 group is separated from the double bond by a −CH2− group, breaking conjugation — so resonance is not observed. The correct option is (D).
The key idea behind resonance is that electrons can be delocalised over a system of atoms only when there is a continuous chain of overlapping p-orbitals. This happens when double bonds, triple bonds, or lone pairs are separated by exactly one single bond — that is, they are conjugated. If a saturated carbon (like −CH2−) sits between them, the p-orbital overlap is broken, and resonance cannot occur.
Let’s examine each option.
Option (A): CH2=CH−CH=CH2
This is 1,3-butadiene. The two double bonds are separated by a single bond, so the four carbon atoms form a conjugated system. The π electrons are delocalised across the entire chain. Resonance is clearly observed — in fact, this is a textbook example.
Option (B): CH2=CH−Cl
Here, the chlorine atom has lone pairs. The double bond and the lone pairs on Cl are separated by a single bond (the C−Cl bond). The p-orbital on Cl can overlap with the π system of the double bond, allowing delocalisation. This is why vinyl chloride shows resonance — the C−Cl bond has partial double-bond character.
Option (C): CH2=CH−C≡N
The triple bond in the cyano group is conjugated with the double bond through the single bond between them. The π electrons of the C≡N and the C=C can delocalise. Resonance is observed — for example, the structure can be written with a negative charge on nitrogen and a positive charge on the adjacent carbon.