Write Lewis structure of the following compounds and show formal charge on each atom. HNO3, NO2, H2SO4

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Q. Write Lewis structure of the following compounds and show formal charge on each atom. HNO3, NO2, H2SO4

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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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The key idea for drawing stable Lewis structures for these compounds is to minimize formal charges on atoms, which often involves forming multiple bonds and, for elements in period 3 and beyond (like Sulfur), expanding the octet. Resonance structures are also important for $\text{HNO}_3$ and $\text{NO}_2$ .

Here are the Lewis structures with formal charges:

1. $\text{HNO}_3$ (Nitric Acid)

Total valence electrons: $1(\text{H}) + 5(\text{N}) + 3 \times 6(\text{O}) = 24$ electrons.
Skeletal structure: Hydrogen is bonded to an oxygen, which is bonded to nitrogen. Nitrogen is bonded to the other two oxygens.
Electron distribution: Form single bonds, then distribute lone pairs to satisfy octets. To minimize formal charges, one oxygen forms a double bond with nitrogen. This leads to resonance.
Formal charges:

\text{H}-\ddot{\text{O}}^{\text{0}}-\text{N}^{\text{0}} \quad || \\ \quad \ddot{\text{O}}^{\text{0}} \\ \quad | \\ \quad :\ddot{\text{O}}:^{-\text{1}}

(One of the terminal oxygens is double-bonded, the other is single-bonded. The structure shown is one of two resonance forms.)

2. $\text{NO}_2$ (Nitrogen Dioxide)

Total valence electrons: $5(\text{N}) + 2 \times 6(\text{O}) = 17$ electrons (an odd-electron species).
Skeletal structure: Oxygen-Nitrogen-Oxygen.
Electron distribution: Form single bonds, then distribute lone pairs. Nitrogen will have an incomplete octet and a single unpaired electron. To minimize formal charges, one oxygen forms a double bond with nitrogen. This leads to resonance.
Formal charges:

:\ddot{\text{O}}^{\text{0}}=\dot{\text{N}}^{+\text{1}}-\ddot{\text{O}}:^{-\text{1}}

(The unpaired electron is on nitrogen. This is one of two resonance forms.)

3. $\text{H}_2\text{SO}_4$ (Sulfuric Acid)

Total valence electrons: $2 \times 1(\text{H}) + 6(\text{S}) + 4 \times 6(\text{O}) = 32$ electrons. …

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Lewis structures depict valence electrons and bonding, while formal charges help evaluate the stability of these structures. For $\text{HNO}_3$ , the most stable resonance structure has formal charges of $0$ on H and on two of the O atoms, $+1$ on N, and $-1$ on the remaining singly-bonded O. For $\text{NO}_2$ , it's a radical with formal charges of $0$ on one O, $+1$ on N, and $-1$ on the other O. For $\text{H}_2\text{SO}_4$ , the most stable structure has all atoms with a formal charge of $0$ .

Understanding how atoms bond to form molecules is fundamental in chemistry. Lewis structures provide a visual representation of these bonds and the distribution of valence electrons. Formal charges, in turn, help us assess the relative stability of different possible Lewis structures for a given molecule.

What is a Lewis Structure?

A Lewis structure (also known as a Lewis dot structure or electron dot structure) is a diagram that shows the bonding between atoms of a molecule and the lone pairs of electrons that may exist in the molecule. They are based on the concept of the octet rule, which states that atoms tend to gain, lose, or share electrons to achieve a stable configuration of eight valence electrons (like noble gases), though there are exceptions.

What is Formal Charge?

Formal charge is a theoretical charge assigned to an atom in a molecule, assuming that electrons in a chemical bond are shared equally between the atoms, regardless of their relative electronegativity. It helps in determining the most plausible Lewis structure among several possibilities. A structure with formal charges closest to zero on all atoms, and negative formal charges on more electronegative atoms, is generally considered more stable.

Formal Charge (FC) on an atom = (Number of valence electrons in the free atom) - (Number of non-bonding electrons) - (1/2 * Number of bonding electrons)

General Steps to Draw Lewis Structures and Calculate Formal Charges:

Count Total Valence Electrons: Sum the valence electrons of all atoms in the molecule. For polyatomic ions, add one electron for each negative charge and subtract one for each positive charge.
Identify Central Atom: The least electronegative atom (excluding hydrogen) is usually the central atom. Hydrogen is always a terminal atom.
Draw Skeleton Structure: Connect the central atom to the terminal atoms with single bonds.
Distribute Remaining Electrons: Place lone pairs on terminal atoms first to satisfy their octets. Then, place any remaining electrons on the central atom.
Form Multiple Bonds (if needed): If the central atom does not have an octet, convert lone pairs from terminal atoms into double or triple bonds to satisfy the octet rule for the central atom.
Calculate Formal Charges: Apply the formal charge formula to each atom in the structure.
Evaluate Stability: Choose the structure that minimizes formal charges, especially placing negative formal charges on more electronegative atoms. For elements in Period 3 and beyond, the octet rule can be expanded to minimize formal charges.

Let's apply these steps to the given compounds:

1. $\text{HNO}_3$ (Nitric Acid)

Count Total Valence Electrons:
- H: 1 valence electron
- N: 5 valence electrons
- O: 6 valence electrons
- Total = $1 + 5 + (3 \times 6) = 1 + 5 + 18 = 24$ valence electrons.
Identify Central Atom: Nitrogen (N) is the least electronegative atom (excluding H), so it is the central atom. Hydrogen is bonded to an oxygen atom, which is then bonded to nitrogen.
Draw Skeleton Structure:

The typical arrangement for oxyacids is H-O-X, where X is the central atom. So, H-O-N, and the other two oxygens are bonded to N.

H - O - N - O

|

O
Distribute Remaining Electrons:
- We used 4 single bonds, which accounts for $4 \times 2 = 8$ electrons.
- Remaining electrons = $24 - 8 = 16$ .
- Distribute these 16 electrons to satisfy octets, starting with terminal oxygens.
  - The O bonded to H and N needs 4 more electrons (2 lone pairs).
  - The other two terminal O atoms each need 6 more electrons (3 lone pairs).
  - This uses $4 + 6 + 6 = 16$ electrons.
- Now, check octets: H (2), O (bonded to H, 8), two terminal O (8 each). Nitrogen has only 6 electrons (3 single bonds).
Form Multiple Bonds:
- Nitrogen needs 2 more electrons to complete its octet. We can take a lone pair from one of the terminal oxygens and form a double bond with nitrogen. This leads to resonance.
Let's consider two resonance structures:

Structure A:

One terminal oxygen forms a double bond with nitrogen.

\text{H} - \text{O} - \text{N} = \text{O} \quad \quad \quad | \quad \quad \quad \text{O}

(with lone pairs distributed)

**Structure B:**
The other terminal oxygen forms a double bond with nitrogen.

\text{H} - \text{O} - \text{N} - \text{O} \quad \quad \quad || \quad \quad \quad \text{O}

(with lone pairs distributed)

Let's draw Structure A with all electrons:

\text{H} - \ddot{\text{O}} - \text{N} = \ddot{\text{O}} \quad \quad \quad \quad | \quad \quad \quad \quad :\ddot{\text{O}}:

Calculate Formal Charges for Structure A:
- H: $1 - 0 - (1/2 \times 2) = 0$
- O (bonded to H and N): $6 - 4 - (1/2 \times 4) = 0$
- N: $5 - 0 - (1/2 \times 8) = 5 - 4 = +1$
- O (double bonded to N): $6 - 4 - (1/2 \times 4) = 0$
- O (single bonded to N): $6 - 6 - (1/2 \times 2) = -1$
The sum of formal charges is $0 + 0 + 1 + 0 + (-1) = 0$ , which matches the charge of the molecule.

The Lewis structure for $\text{HNO}_3$ showing formal charges and resonance is:

\text{H} - \ddot{\text{O}} - \underset{\text{+1}}{\text{N}} = \ddot{\text{O}} \longleftrightarrow \text{H} - \ddot{\text{O}} - \underset{\text{+1}}{\text{N}} - \ddot{\text{O}} \quad \quad \quad \quad | \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad || \quad \quad \quad \quad :\underset{\text{-1}}{\ddot{\text{O}}}: \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \ddot{\text{O}}

(Formal charges are shown above/below the atoms where they are non-zero.)

2. $\text{NO}_2$ (Nitrogen Dioxide)

Count Total Valence Electrons:
- N: 5 valence electrons
- O: 6 valence electrons
- Total = $5 + (2 \times 6) = 5 + 12 = 17$ valence electrons.
- This is an odd number of electrons, indicating a radical molecule.
Identify Central Atom: Nitrogen (N) is the central atom.
Draw Skeleton Structure:

O - N - O
Distribute Remaining Electrons:
- We used 2 single bonds, accounting for $2 \times 2 = 4$ electrons.
- Remaining electrons = $17 - 4 = 13$ .
- Distribute 13 electrons:
  - Each terminal O needs 6 electrons (3 lone pairs) to complete its octet. This uses $6 + 6 = 12$ electrons.
  - One electron remains. Place it on the central atom (N).
- Now, check octets: Each O has 8 electrons. N has $2+2+1 = 5$ electrons.
Form Multiple Bonds:
- Nitrogen needs 3 more electrons to complete an octet (or get closer to it). We can take a lone pair from one of the oxygens and form a double bond. This will also lead to resonance.
Let's consider two resonance structures:

Structure A:

One oxygen forms a double bond with nitrogen.

\ddot{\text{O}} = \text{N} - \ddot{\text{O}}:

(with lone pairs and the odd electron distributed)

Let's draw Structure A with all electrons:

:\ddot{\text{O}} = \dot{\text{N}} - \ddot{\text{O}}:

(The single dot on N represents the odd electron.)

6. Calculate Formal Charges for Structure A:

* O (double bonded to N): $6 - 4 - (1/2 \times 4) = 0$

* N: $5 - 1 - (1/2 \times 6) = 5 - 1 - 3 = +1$ (1 non-bonding electron, 6 bonding electrons)

* O (single bonded to N): $6 - 6 - (1/2 \times 2) = -1$

The sum of formal charges is $0 + 1 + (-1) = 0$, which matches the charge of the molecule.

The Lewis structure for $\text{NO}_2$ showing formal charges and resonance is:

:\ddot{\text{O}} = \underset{\text{+1}}{\dot{\text{N}}} - :\underset{\text{-1}}{\ddot{\text{O}}}: \longleftrightarrow :\underset{\text{-1}}{\ddot{\text{O}}} - \underset{\text{+1}}{\dot{\text{N}}} = \ddot{\text{O}}:

(Formal charges are shown above/below the atoms where they are non-zero.)

> [!WARNING]
> $\text{NO}_2$ is an example of a stable radical due to its odd number of valence electrons. The odd electron typically resides on the central atom.

3. $\text{H}_2\text{SO}_4$ (Sulfuric Acid)

Count Total Valence Electrons:
- H: 1 valence electron
- S: 6 valence electrons
- O: 6 valence electrons
- Total = $(2 \times 1) + 6 + (4 \times 6) = 2 + 6 + 24 = 32$ valence electrons. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

KCET 2026Set D31 markMCQ
$CO_3^{2-}$
(a) All C-O bonds in $CO_3^{2-}$ are equivalent
(b) There are three resonance structures possible for $CO_3^{2-}$ 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 $CO_3^{2-}$ 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 $CO_3^{2-}$ 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 …
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KCET 2020Set A-11 markMCQ
$\text{NO}_2^+$
›Reveal solutionSolution
The s-character in hybrid orbitals depends on the steric number (SN) around nitrogen. For $\text{NO}_2^+$ (SN=2, sp), s-character = 50%; for $\text{NO}_2^-$ (SN=3, sp²), s-character = 33.3%; for $\text{NH}_4^+$ (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 $\frac{1}{n+1} \times 100\%$ . 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.

$\text{NO}_2^+$ (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: $\frac{1}{1+1} \times 100\% = 50\%$ .

$\text{NO}_2^-$ (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: $\frac{1}{2+1} \times 100\% = 33.3\%$ .

$\text{NH}_4^+$ (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. …
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KCET 2019Set A-11 markMCQ
$\mathrm{CH_2=CH-CH=CH_2}$
›Reveal solutionSolution
Resonance requires a conjugated system of alternating single and multiple bonds. In option (D), the $-\mathrm{NH_2}$ group is separated from the double bond by a $-\mathrm{CH_2}-$ 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 $-\mathrm{CH_2}-$ ) sits between them, the p-orbital overlap is broken, and resonance cannot occur.

Let’s examine each option.

Option (A): $\mathrm{CH_2=CH-CH=CH_2}$

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 $\pi$ electrons are delocalised across the entire chain. Resonance is clearly observed — in fact, this is a textbook example.

Option (B): $\mathrm{CH_2=CH-Cl}$

Here, the chlorine atom has lone pairs. The double bond and the lone pairs on Cl are separated by a single bond (the $\mathrm{C-Cl}$ bond). The p-orbital on Cl can overlap with the $\pi$ system of the double bond, allowing delocalisation. This is why vinyl chloride shows resonance — the $\mathrm{C-Cl}$ bond has partial double-bond character.

Option (C): $\mathrm{CH_2=CH-C\equiv N}$

The triple bond in the cyano group is conjugated with the double bond through the single bond between them. The $\pi$ electrons of the $\mathrm{C\equiv N}$ and the $\mathrm{C=C}$ 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.

Option (D): $\mathrm{CH_2=CH-CH_2-NH_2}$ …
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Q.Write Lewis structure of the following compounds and show formal charge on each atom. HNO3, NO2, H2SO4

What is a Lewis Structure?

What is Formal Charge?

General Steps to Draw Lewis Structures and Calculate Formal Charges:

1. HNO3\text{HNO}_3HNO3​ (Nitric Acid)

2. NO2\text{NO}_2NO2​ (Nitrogen Dioxide)

3. H2SO4\text{H}_2\text{SO}_4H2​SO4​ (Sulfuric Acid)

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Q.Write Lewis structure of the following compounds and show formal charge on each atom.
HNO3, NO2, H2SO4

1. $\text{HNO}_3$ (Nitric Acid)

2. $\text{NO}_2$ (Nitrogen Dioxide)

3. $\text{H}_2\text{SO}_4$ (Sulfuric Acid)

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