What Are Lewis Dot Structures?

A Lewis dot structure is a diagram that represents the valence electrons — the outermost shell electrons — of the atoms in a molecule. Valence electrons earn their own special designation because they are the electrons directly involved in chemical reactions, specifically in the making and breaking of chemical bonds. Every electron closer to the nucleus than the valence shell is simply classified as a core electron.

When valence electrons are shared between two atoms, we represent that sharing as a covalent bond. When electrons transfer entirely from one atom to another, the result is an ionic bond. This lesson focuses on drawing Lewis dot structures for molecular (covalent) compounds, including how to handle the octet rule exceptions and resonance structures that appear on every general chemistry exam.

The number of valence electrons for any main-group element follows directly from its group number on the periodic table:

How Do Ionic and Covalent Bonds Differ?

Consider sodium chloride (NaCl). Sodium, a Group 1 metal, has just one valence electron. We draw it as a single dot on one side of the atom — it does not matter which side. Chlorine, a halogen with seven valence electrons, gets dots on all four sides with pairs forming on three of them, leaving one unpaired electron.

When a metal meets a non-metal, the atoms organize their electrons to satisfy the octet rule — the tendency for atoms to achieve eight valence electrons in their outermost shell. Metals have low ionization energies, making it relatively easy for them to lose electrons, while non-metals have highly negative electron affinities, meaning they readily accept electrons. The solution is straightforward: sodium transfers its lone valence electron to chlorine. This gives chlorine a complete octet (and a −1 charge) while leaving sodium with a +1 charge. Opposite charges attract, and that electrostatic attraction is the essence of an ionic bond.

Covalent Bonding: Sharing Between 2 Non-Metals

Now consider two chlorine atoms, each with seven valence electrons. Both need just one more electron to complete their octets — but neither has a low enough ionization energy to simply give up an electron. That is the fundamental predicament with two non-metals: both want to gain electrons, and neither will yield. The solution? Sharing.

Each chlorine contributes one electron to a shared pair placed between them. This creates two categories of electrons:

• Bonding electrons (bonding pair) — shared between atoms
• Non-bonding electrons (lone pairs) — not shared; they remain on one atom

Each chlorine in Cl₂ has three lone pairs and one bonding pair. We commonly represent the shared pair as a line between the two atoms: Cl—Cl. Every line in a Lewis dot structure represents exactly two electrons.

Some molecules require more than one shared pair. Four shared electrons constitute a double bond (drawn as two lines), and six shared electrons make a triple bond (three lines). The 6-step rules for drawing Lewis structures will tell us when multiple bonds are needed.

The Octet Rule and Its 3 Exceptions

The octet rule states that atoms tend to gain, lose, or share electrons until they achieve eight valence electrons. This guideline correctly predicts the Lewis dot structures of the vast majority of molecules. However, three notable octet rule exceptions exist.

Exception 1: Atoms That Fall Short of the Octet

Not every atom aims for eight electrons. Helium is the only noble gas with just two valence electrons rather than eight — its first shell holds only one s orbital, maxing out at two electrons. Hydrogen, seeking to resemble its nearest noble gas (helium), is satisfied with just two electrons. Hydrogen will never follow the octet rule.

Two other elements regularly fall short as well. Beryllium has two valence electrons, shares one with each of two atoms, and stops at four electrons total — it simply runs out of electrons to share. Boron and aluminum each have three valence electrons, share one with each of three atoms, and plateau at six electrons. While boron and aluminum can occasionally achieve a full octet, they most commonly stop at six.

Exception 2: Expanded Octets

Some atoms can exceed eight electrons in their Lewis dot structure. The octet rule arises from the capacity of one s orbital and three p orbitals, which together hold a maximum of eight electrons. However, d orbitals become available starting in the third shell — there is no such thing as a 1d or 2d orbital. Elements in Period 3 or below (aluminum, silicon, phosphorus, sulfur, chlorine, and beyond) can place extra electrons into these d orbitals, exceeding the octet.

Expanded octets are not required for these elements — they simply become possible. Sulfur, for example, commonly makes two bonds (following the octet rule with two lone pairs), but it can also make four or even six bonds, pushing well beyond eight electrons.

Exception 3: Odd Numbers of Electrons

The rarest octet rule exception occurs when a molecule has an odd total number of valence electrons. The classic example is nitric oxide (NO): nitrogen contributes 5 electrons and oxygen contributes 6, totaling 11 — an odd number. Since you cannot distribute an odd number of electrons into complete pairs around both atoms, it is mathematically impossible for every atom to achieve a full octet.

How to Predict Bonds and Choose the Central Atom

The octet rule lets us predict how many covalent bonds an atom typically forms when drawing Lewis structures. Simply count how many electrons the atom needs to reach its "full" count — that number equals its typical bond count.

Identifying the Central Atom

Most molecules we draw have a central atom bonded to all the other atoms around it. The central atom needs to be the one that can form the most bonds. A reliable rule: place the least electronegative element in the center.

If a molecule contains multiple atoms of the same element (such as two different halogens), the larger, less electronegative one typically goes in the center because it can accommodate more bonds.

6 Steps to Draw Any Lewis Structure

Follow this six-step procedure to draw the Lewis dot structure for any molecule or polyatomic ion:

The best way to internalize these rules is to work through examples — so let us do exactly that.

Basic Lewis Structure Examples: CCl₄ and NF₃

CCl₄ (Carbon Tetrachloride)

Carbon has four valence electrons and is much less electronegative than chlorine, so it takes the central position. Each chlorine forms one bond to the central carbon.

The skeleton uses 4 bonds = 8 electrons. Next, fill the octets of the outside chlorines. Each chlorine needs 6 more electrons (3 lone pairs): 4 × 6 = 24. Total used: 8 + 24 = 32 — exactly what we have.

Check the central atom: carbon has 2 + 2 + 2 + 2 = 8 electrons around it. Full octet. Carbon is happy, and we are done. The Lewis structure has 8 bonding electrons and 24 non-bonding electrons distributed as lone pairs on the four chlorines.

NF₃ (Nitrogen Trifluoride)

Nitrogen is less electronegative than fluorine and goes in the center, bonded to three fluorines.

The skeleton uses 3 bonds = 6 electrons. Fill the fluorines: each needs 6 more electrons, so 3 × 6 = 18. Total so far: 24 electrons. We have 2 electrons remaining — place them on nitrogen as a lone pair.

Check the central atom: nitrogen has 2 + 2 + 2 + 2 = 8 electrons. Full octet. Nitrogen has three bonding pairs and one lone pair of non-bonding electrons. Done.

Lewis Structures with Multiple Bonds: HCN (Hydrogen Cyanide)

Carbon is the least electronegative element here (hydrogen cannot be central), so carbon goes in the middle with hydrogen on one side and nitrogen on the other.

Fill the outside atoms first. Hydrogen already has 2 electrons from its bond — it is full (hydrogen only wants 2). Nitrogen needs 6 more electrons to complete its octet: add 3 lone pairs. Electron tally: 2 (H—C bond) + 2 (C—N bond) + 6 (N lone pairs) = 10. All used up.

Check the central atom: carbon only has 4 electrons (two bonds). Not enough. We have used all our electrons, so we cannot simply add more. Instead, we must convert lone pairs from adjacent atoms into shared bonds.

Hydrogen has no lone pairs to offer, but nitrogen does. Convert one of nitrogen's lone pairs into a bonding pair between carbon and nitrogen — carbon now has 6 electrons. Still short. Convert another lone pair into a bond — carbon now has 8 electrons. Full octet.

The result is H—C≡N: a single bond to hydrogen and a triple bond between carbon and nitrogen. Both carbon and nitrogen have complete octets, and hydrogen has its two electrons. Everyone is satisfied.

What Is Formal Charge? Resonance in CO₂

Carbon goes in the center. Fill the outside oxygens first: 8 electrons each = 16 total. We have used all 16 electrons, but carbon only has 4 electrons (two single bonds). It needs four more — which means making two additional bonds.

Here is where it gets interesting: there are 3 possible ways to add those extra bonds.

1. Triple bond to the left oxygen, single bond to the right
2. Double bond to each oxygen
3. Single bond to the left oxygen, triple bond to the right

All three Lewis structures give every atom a satisfied octet. When multiple valid structures exist, we call this resonance. To determine which structure best represents reality, we use formal charge.

What Is Formal Charge?

Formal charge is not the same as oxidation state. It measures how the electron distribution in a Lewis structure compares to the atom's normal valence electron count.

Applying Formal Charge to CO₂

Structure 1 (triple bond left, single bond right): The left oxygen has 2 dots + 3 lines = 5, so 6 − 5 = +1. Carbon has 0 dots + 4 lines = 4, so 4 − 4 = 0. The right oxygen has 6 dots + 1 line = 7, so 6 − 7 = −1.

Structure 2 (double bond to each oxygen): Each oxygen has 4 dots + 2 lines = 6, so 6 − 6 = 0. Carbon has 0 dots + 4 lines = 4, so 4 − 4 = 0. All formal charges are zero.

Structure 3 mirrors Structure 1 with the bonds reversed.

Structure 2 has zero formal charges on all atoms, making it the clear winner. The correct Lewis dot structure for CO₂ is O=C=O — double bonds to both oxygens. This one shows up so frequently on exams that it is worth committing to memory.

How to Draw the Lewis Structure for N₂O (Nitrous Oxide)

This molecule is a classic exam question designed to separate the A students from the B students. It is tricky for two reasons.

Fill the outside atoms: use all 16 electrons on the terminal nitrogen and oxygen. The central nitrogen has only 4 electrons — it needs two more bonds. Three resonance structures are possible:

1. Triple bond between the two nitrogens, single bond to oxygen (N≡N—O)
2. Double bond to each outer atom (N=N=O)
3. Single bond to the left nitrogen, triple bond to oxygen (N—N≡O)

Formal Charge Analysis for N₂O

Structure 1 (N≡N—O): Left N = 0, Central N = +1, O = −1 → 2 atoms with formal charges

Structure 2 (N=N=O): Left N = −1, Central N = +1, O = 0 → 2 atoms with formal charges

Structure 3 (N—N≡O): Left N = −2, Central N = +1, O = +1 → 3 atoms with formal charges, plus a −2

Applying the rules: Rule 1 eliminates Structure 3 (three atoms with formal charges vs. two). Rule 2 also eliminates Structure 3 (a −2 is farther from zero than −1). Rule 3 distinguishes between Structures 1 and 2: both have a +1 on the central nitrogen and a −1 on one atom. In Structure 1, the −1 sits on oxygen (more electronegative). In Structure 2, the −1 sits on the terminal nitrogen (less electronegative). Oxygen should carry the negative charge.

The major resonance contributor for N₂O is therefore N≡N—O, with +1 on the central nitrogen and −1 on oxygen.

Expanded Octets in Lewis Structures: SF₄ and XeF₄

SF₄ (Sulfur Tetrafluoride)

Nothing in the formula tells you this will be an expanded octet — you have to follow the rules and see where they lead. Sulfur is in the third row of the periodic table, so it can exceed the octet rule. It is also the least electronegative atom, so it goes in the center.

Fill the four fluorines: 4 × 8 = 32 electrons. Two electrons remain — place them on sulfur as a lone pair.

Sulfur now has 2 + 2 + 2 + 2 + 2 = 10 electrons around it. That exceeds the octet — but sulfur is in Period 3 and is allowed to have an expanded octet. Formal charge check: each fluorine has 6 dots + 1 line = 7, giving 7 − 7 = 0. Sulfur has 2 dots + 4 lines = 6, giving 6 − 6 = 0. No formal charges anywhere. This is the correct Lewis structure for SF₄.

XeF₄ (Xenon Tetrafluoride)

A noble gas in a compound? It is unusual, but xenon is large enough and fluorine is electronegative enough to make it happen. This is one of the few noble gas compounds. If a noble gas — which already has a full octet — forms bonds, it is guaranteed to exceed the octet rule. Xenon goes in the center.

Fill the four fluorines: 32 electrons used. Four electrons remain — place them on xenon as two lone pairs.

Xenon now has 4 + 2 + 2 + 2 + 2 = 12 electrons around it. As a Period 5 element, xenon has plenty of d orbitals available. Formal charge check: each fluorine is 7 − 7 = 0. Xenon is 8 − 8 = 0. No formal charges. This is the correct Lewis dot structure for XeF₄.

Lewis Structures for Ions: SO₄²⁻ (Sulfate Ion)

When drawing Lewis dot structures for polyatomic ions, adjust the electron count: add electrons for anions, subtract for cations.

Sulfur is the least electronegative and goes in the center. Fill the four oxygens: 4 × 8 = 32 electrons, using them all. Sulfur has 8 electrons (four single bonds) — technically a filled octet.

But notice: sulfur normally makes two bonds, yet here it is making four. Whenever a Period 3+ element makes more bonds than usual, it is worth checking formal charges.

Improving the Lewis Structure with Formal Charge

All single bonds: Each oxygen has 6 dots + 1 line = 7, so 6 − 7 = −1. Sulfur has 0 dots + 4 lines = 4, so 6 − 4 = +2. That is five atoms with formal charges and a +2 on sulfur — not ideal.

Since sulfur can have an expanded octet, we can reduce these charges by converting a lone pair on one oxygen into a double bond. This drops that oxygen's formal charge to 0 and sulfur's to +1. Do it again with a second oxygen: sulfur drops to 0, and we now have two oxygens at 0 (double-bonded) and two at −1 (single-bonded).

Could we add a third double bond? The oxygen would drop to 0, but sulfur would become −1 — and we do not want the negative charge on the less electronegative atom. So we stop at two double bonds.

3 Equivalent Resonance Structures: NO₃⁻ (Nitrate Ion)

Nitrogen goes in the center. Fill the three oxygens: 3 × 8 = 24 electrons, using them all. Nitrogen has only 6 electrons (three single bonds) — it needs one more bond. One of the adjacent oxygens must share a lone pair to form a double bond.

But which oxygen? Unlike N₂O — where one outer atom was nitrogen and the other was oxygen — here all three outer atoms are identical oxygens. Any of the three could form the double bond, producing three possible resonance structures that are exactly equivalent in every respect.

Applying formal charge: in each resonance structure, nitrogen has +1 and two oxygens have −1 each. The formal charges are identical across all three structures — formal charge cannot distinguish between them.

Resonance Hybrids and Delocalized Electrons

Here is where things get a little counterintuitive. None of the three resonance structures for NO₃⁻ is an accurate depiction of the real molecule. The actual molecule is a kind of average of all three structures simultaneously — and we call this average the resonance hybrid.

Bond Order in Resonance Hybrids

Each bond in the NO₃⁻ resonance hybrid is somewhere between a single and a double bond: shorter than a single bond, longer than a double bond. We can quantify this with bond order: across all three resonance structures, there are a total of 4 bonds (one double + two singles = 2 + 1 + 1) distributed across 3 positions. Each bond has an order of 4/3, or approximately 1.33 — about one-third stronger than a single bond.

Delocalized Electrons in the Resonance Hybrid

To draw the resonance hybrid, we use dashed lines at each bond position (indicating partial bonds) and the partial charge symbol (δ) to show the distribution of charge. In NO₃⁻, each oxygen carries a partial negative charge of δ− ≈ −2/3 (the average of 0 in one resonance structure and −1 in the other two).

The electrons in these partial bonds are delocalized — they exist simultaneously across all three bond positions rather than being confined to a single location. This is what the dashed line in the resonance hybrid represents, and it is one of the most important concepts in general chemistry.