Draw the resonating structure of (i) Ozone molecule (ii) Nitrate ion

NCERT Exemplar · Q52

Q. Draw the resonating structure of

(i) Ozone molecule

(ii) Nitrate ion

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Concept understanding — Resonance Structures

The Problem: One Picture Isn't Enough

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, $O_3$ . 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 ( $\leftrightarrow$ ) 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, $NO_3^-$ . Start with one valid Lewis structure:

\begin{bmatrix} O \\ || \\ O - N = O \end{bmatrix}^-

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:

\begin{bmatrix} O = N - O \\ | \\ O^- \end{bmatrix}^-

Repeat the process from this new structure, and you get a third. All three are resonance structures of $NO_3^-$ .

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 $NO_3^-$ , all three $N-O$ bonds are identical, with a bond order of $1\frac{1}{3}$ (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.

Common Mistakes to Avoid …

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Resonance structures show equivalent ways to arrange electrons when a single Lewis structure cannot capture the true bonding. Ozone has two equivalent structures with alternating single/double bonds; nitrate ion has three equivalent structures with the negative charge and double bond rotating among the three oxygens.

Why resonance structures matter

Some molecules cannot be represented by a single Lewis structure. The actual molecule is a hybrid—a blend of all the resonance forms—where electrons are delocalized over several atoms. Drawing resonance structures reveals this delocalization and explains why, for instance, all three N–O bonds in nitrate are identical in length despite our drawings showing one double and two single bonds.

The key rules: resonance structures differ only in the placement of electrons (never in the position of nuclei), obey the octet rule where possible, and the actual molecule is more stable than any single contributing structure.

(i) Ozone molecule ( $\ce{O3}$ )

Count valence electrons.

Each oxygen contributes 6 valence electrons: $3 \times 6 = 18$ electrons total.
Sketch the skeleton.

The central oxygen is bonded to two terminal oxygens in a bent arrangement:

\ce{O - O - O}

Distribute electrons to satisfy octets.

Place a double bond between the central oxygen and one terminal oxygen, and a single bond to the other. Assign lone pairs so every oxygen has eight electrons:

Structure I:
```
     ··O̅ — O⁺ = O··
```
The left oxygen carries a formal charge of −1 (three lone pairs, one bond = 7 electrons vs. 6 valence), the central oxygen +1 (one lone pair, three bonds = 5 electrons vs. 6 valence), and the right oxygen 0.
Draw the equivalent resonance structure.

Swap the positions of the double and single bonds:

Structure II:
```
     ··O = O⁺ — O̅··
```
Now the double bond is on the left, the single bond on the right, with formal charges mirrored.
Represent the resonance hybrid.

The double-headed arrow $\leftrightarrow$ connects the two structures:

\begin{array}{ccc} \ce{^{-}O} & \!\!-\!\! & \ce{O^{+}} & \!\!=\!\! & \ce{O} \\ & & \updownarrow & & \\ \ce{O} & \!\!=\!\! & \ce{O^{+}} & \!\!-\!\! & \ce{^{-}O} \end{array}

In reality, both O–O bonds are equivalent with a bond order of 1.5, and the negative charge is shared equally over the two terminal oxygens.

Tip

Formal charge = (valence electrons) − (lone-pair electrons) − ½(bonding electrons). Structures with the smallest formal charges, and negative charges on more electronegative atoms, contribute most to the hybrid.

(ii) Nitrate ion ( $\ce{NO3^-}$ )

Count valence electrons.

Nitrogen contributes 5, each oxygen 6, plus 1 for the negative charge: $5 + 3(6) + 1 = 24$ electrons.
Sketch the skeleton.

Nitrogen is the central atom bonded to three oxygens in a trigonal planar arrangement:

\ce{O - N - O}

\phantom{\ce{O - }}\ce{|}

\phantom{\ce{O - }}\ce{O}

Distribute electrons.

Place a double bond to one oxygen and single bonds to the other two. Assign lone pairs to complete octets. Each singly bonded oxygen carries a formal charge of −1, the doubly bonded oxygen 0, and nitrogen +1.

Structure I:
```
       O̅
       |
   O = N — O̅
```
Draw the other two equivalent structures. …

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Q.Draw the resonating structure of

Why resonance structures matter

(i) Ozone molecule (OX3\ce{O3}OX3​)

(ii) Nitrate ion (NOX3X−\ce{NO3^-}NOX3​X−)

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(i) Ozone molecule ( $\ce{O3}$ )

(ii) Nitrate ion ( $\ce{NO3^-}$ )

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