Write the resonance structures for $SO_3$, $NO_2$ and $NO_3^-$.

Write the resonance structures for SO₃, NO₂ and NO₃^-.

Exercises · 4.13

SO_3

Uttar Pradesh UpmspTextbookSubjective· 2mImportance★★★★★est

15% · 17/112 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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 …

Start your 14-day free trial to unlock the full solution →

Resonance structures are multiple Lewis structures that differ only in the placement of π bonds and lone pairs, representing the same molecule. For $SO_3$ , $NO_2$ , and $NO_3^-$ , the key is to satisfy octets and minimize formal charges — each has 3, 2, and 3 equivalent resonance forms respectively.

Why Resonance Structures Matter

Resonance is not about the molecule flipping between forms — it’s a way to describe delocalized electrons when a single Lewis structure fails to capture the true bonding. The actual molecule is a hybrid of all resonance contributors, with bond lengths and charges averaged out. For exam purposes, you need to draw all valid structures that obey the octet rule (or expanded octet where possible) and minimize formal charges.

Let’s tackle each species step by step.

1. $SO_3$ (Sulfur Trioxide)

Step 1: Count valence electrons.

Sulfur has 6, each oxygen has 6 — total = $6 + 3 \times 6 = 24$ electrons.

Step 2: Draw the skeleton.

Sulfur is central, three oxygens around it. Connect each with a single bond (uses 6 electrons). Remaining: $24 - 6 = 18$ electrons.

Step 3: Complete octets on oxygens.

Each oxygen needs 6 more electrons (3 lone pairs). That uses all 18 remaining electrons. Now sulfur has only 6 electrons (from three single bonds) — it’s short of an octet.

Step 4: Form double bonds to satisfy sulfur’s octet.

Sulfur can expand its octet (it’s in period 3). Move one lone pair from an oxygen to form a S=O double bond. This gives sulfur 8 electrons. But you can do this for any one of the three oxygens — and you can even do it for two or all three oxygens, creating multiple resonance forms.

Tip

The most stable resonance structures for $SO_3$ have zero formal charge on all atoms. That happens when sulfur forms double bonds with all three oxygens — but that gives sulfur 12 electrons (expanded octet). This is perfectly valid for period 3 elements.

The three equivalent resonance structures:

Each has one S=O double bond and two S–O single bonds, with the double bond “rotating” among the three oxygens. Formal charges:

Double-bonded oxygen: 0
Single-bonded oxygens: −1 each
Sulfur: +1

These three forms are equivalent in energy and contribute equally to the hybrid.

Watch out

A common mistake is to draw only one structure and stop. For $SO_3$ , you must show all three — the double bond can be on any oxygen. Also, don’t forget that the “all double bonds” structure (with S=O three times) is a valid contributor too, though it gives sulfur a formal charge of 0 and each oxygen 0 — but it’s less important because it uses an expanded octet heavily. The three single-double bond forms are the standard answer.

2. $NO_2$ (Nitrogen Dioxide)

Step 1: Count valence electrons.

Nitrogen has 5, each oxygen has 6 — total = $5 + 2 \times 6 = 17$ electrons. Odd number → radical species.

Step 2: Draw the skeleton.

N in center, O–N–O. Connect with single bonds (uses 4 electrons). Remaining: $17 - 4 = 13$ electrons.

Step 3: Complete octets on oxygens first.

Each oxygen needs 6 more electrons (3 lone pairs) — that uses 12 electrons. Remaining: 1 electron. Place this unpaired electron on nitrogen.

Now check octets: Each oxygen has 8. Nitrogen has only 5 (from two single bonds + one lone electron) — it’s short by 3.

Step 4: Form a double bond to reduce nitrogen’s deficit.

Move one lone pair from an oxygen to form N=O. This gives nitrogen 7 electrons (still one short of octet, but that’s okay — it’s a radical). You can do this with either oxygen, giving two resonance structures.

The two resonance structures:

Structure A: N=O (left), N–O (right), unpaired electron on N.
Structure B: N–O (left), N=O (right), unpaired electron on N.

Formal charges:

In both, the double-bonded oxygen: 0
Single-bonded oxygen: −1
Nitrogen: +1 — by the standard formula $FC = V - (N_{nonbonding} + \frac{1}{2}N_{bonding})$ : N has $V = 5$ , one nonbonding (unpaired) electron, and 6 bonding electrons (one single + one double bond), so $FC = 5 - (1 + 3) = +1$ .

Important

$NO_2$ is a radical — its 17 valence electrons leave one electron unpaired, sitting on N. When applying the formal-charge formula, count that unpaired electron as one nonbonding electron on N. Always verify the total: the formal charges must sum to the species' overall charge.

Checking every atom: double-bonded O: $6 - (4 + 4/2) = 0$ ; single-bonded O: $6 - (6 + 2/2) = -1$ ; N: $+1$ . Sum $= +1 + 0 - 1 = 0$ — matching the neutral molecule. …

Start your 14-day free trial to unlock the full solution →

←Previous question · Exercises4.12 — H₃PO₃ can be represented by structures 1 and 2 shown below. Can these …→Next question · Exercises4.14 — Use Lewis symbols to show electron transfer between the following atom…

Q.Write the resonance structures for SO3SO_3SO3​, NO2NO_2NO2​ and NO3−NO_3^-NO3−​.

Why Resonance Structures Matter

1. SO3SO_3SO3​ (Sulfur Trioxide)

2. NO2NO_2NO2​ (Nitrogen Dioxide)

🎓Unlock everything free for 14 days

Q.Write the resonance structures for $SO_3$ , $NO_2$ and $NO_3^-$ .

1. $SO_3$ (Sulfur Trioxide)

2. $NO_2$ (Nitrogen Dioxide)

Unlock everything free for 14 days