Lewis diagrams — AP Chemistry Study Guide

For: AP Chemistry candidates sitting AP Chemistry.

Covers: Counting valence electrons, total electron calculation, single/double/triple bonds, lone pairs, formal charge, octet rule exceptions, and drawing Lewis diagrams for covalent and polyatomic ionic compounds.

You should already know: Valence electron configuration for main group elements. Basic definitions of ionic and covalent bonding. Net charge calculation for ionic species.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Chemistry style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Lewis diagrams?

Lewis diagrams (also commonly called Lewis dot structures or Lewis structures) are two-dimensional schematic representations of covalent molecules and polyatomic ionic compounds that show bonding connectivity between atoms and the arrangement of all valence electrons in the species. Per the AP Chemistry Course and Exam Description (CED), this topic is part of Unit 2: Molecular and Ionic Compound Structure and Properties, which contributes 7–9% of the total AP exam score. Lewis diagram questions appear in both multiple-choice (MCQ) and free-response (FRQ) sections, and the skill is also used as a prerequisite for almost all other Unit 2 questions. Standard notation uses element symbols to represent the atomic nucleus and core electrons, single lines for single covalent bonds (2 shared electrons), double lines for double bonds (4 shared electrons), triple lines for triple bonds (6 shared electrons), and dots around element symbols to represent non-bonding lone pairs of electrons. Lewis diagrams are the starting point for all predictions of molecular shape, polarity, reactivity, and bond strength, making this skill one of the most foundational in AP Chemistry.

2. Step-by-Step Lewis Diagram Construction

The standard algorithm for drawing correct Lewis diagrams applies to most neutral covalent compounds and charged polyatomic ions, and follows 5 core rules. First, determine atomic connectivity: the least electronegative atom is almost always the central atom, while hydrogen and halogens are almost always terminal (they can only form one covalent bond). Second, calculate the total number of valence electrons available: add the number of valence electrons for each atom (equal to the group number for main group elements: 1 for H, 3 for group 13, 6 for group 16, etc.). For anions, add one electron per unit of negative charge; for cations, subtract one electron per unit of positive charge. Third, draw one single bond between each connected pair of atoms, which uses 2 electrons per bond. Subtract the total number of bonding electrons from the total available to get the number of remaining non-bonding electrons. Fourth, distribute remaining electrons as lone pairs starting with terminal atoms, to satisfy the octet guideline (8 valence electrons per atom, 2 for H) before assigning any leftover lone pairs to the central atom. Fifth, if the central atom still lacks a full octet, convert terminal lone pairs into multiple bonds (double or triple bonds) to complete the octet.

Worked Example

Draw the Lewis diagram for the hypochlorite ion, ${\text{ClO}}^{-}$.

1. Connectivity: There are only two atoms, so they are bonded directly to each other. The ion charge is -1, so we will add brackets and charge as a final step.
2. Total valence electrons: Cl (group 17) has 7 valence electrons, O (group 16) has 6, add 1 for the -1 charge: Total $=7+6+1=14$ valence electrons.
3. Bonding electrons: One single bond uses 2 electrons, so remaining non-bonding electrons $=14-2=12$.
4. Distribute lone pairs: Each atom needs 6 additional electrons to reach an octet (they already have 2 from the single bond). $6+6=12$, which uses all remaining electrons.
5. Check octets: Both atoms have 8 valence electrons, so no multiple bonds are needed. Add brackets to get the final structure: $[:\underset{..}{\mathrm{Cl}}-\underset{..}{\stackrel{..}{O}}:{]}^{-}$.

Exam tip: When adjusting total valence electrons for charge, explicitly write "add electrons for negative charge, subtract for positive charge" next to your work. It is easy to flip this rule under exam pressure, and an incorrect electron count guarantees an incorrect structure.

3. Formal Charge and Preferred Lewis Structures

Many molecules and ions can be drawn with multiple valid Lewis structures that all satisfy the octet guideline. To identify the most stable (preferred) structure, we use formal charge, a hypothetical charge assigned to each atom that assumes equal sharing of bonding electrons. The formula for formal charge ($FC$) is: $$FC=V-N-\frac{B}{2}$$ Where $V$ = number of valence electrons in the neutral free atom, $N$ = number of non-bonding (lone pair / unpaired) electrons on the atom in the structure, and $B$ = total number of bonding electrons shared by the atom. Three rules determine the preferred structure: (1) the most stable structure has the fewest atoms with non-zero formal charge, (2) any negative formal charge should be located on the most electronegative atom, (3) adjacent atoms should not have formal charges of the same sign.

Worked Example

Three possible Lewis structures for the thiocyanate ion ${\text{SCN}}^{-}$ are given below. Calculate the formal charge for each atom and identify the preferred structure.

1. Structure 1: $:\underset{..}{S}-\text{C}\equiv \text{N}:$

2. Structure 2: $:\underset{..}{S}=\text{C}=\underset{..}{N}:$

3. Structure 3: $:\text{S}\equiv \text{C}-\underset{...}{\stackrel{...}{N}}:$

4. Formal charge for Structure 1: $F{C}_S=6-6-(2/2)=-1$; $F{C}_C=4-0-(8/2)=0$; $F{C}_N=5-2-(6/2)=0$. Sum $=-1$, matching the ion charge.

5. Formal charge for Structure 2: $F{C}_S=6-4-(4/2)=0$; $F{C}_C=4-0-(8/2)=0$; $F{C}_N=5-2-(4/2)=-1$. Sum $=-1$, matching the ion charge.

6. Formal charge for Structure 3: $F{C}_S=6-2-(6/2)=+1$; $F{C}_C=4-0-(8/2)=0$; $F{C}_N=5-6-(2/2)=-2$. Sum $=-1$, matching the ion charge.

7. Compare structures: Structure 3 has two non-zero formal charges with a magnitude of 2, so it is eliminated. Between Structures 1 and 2, both have one non-zero formal charge. Nitrogen is more electronegative than sulfur, so negative formal charge on nitrogen is preferred. The preferred structure is Structure 2.

Exam tip: Always check that the sum of all formal charges equals the net charge of the species. If your sum does not match, you made an arithmetic error in calculation, so correct it before selecting the preferred structure.

4. Octet Rule Exceptions

The octet rule is a general guideline derived from the stability of full valence s and p orbitals, not a physical law. Three common classes of exceptions, all regularly tested on the AP exam, are: (1) Electron deficient species: Central atoms have fewer than 8 valence electrons, almost always occurring for group 13 elements (boron, aluminum) that only have 3 valence electrons to share. (2) Odd-electron species (free radicals): The total number of valence electrons is odd, so at least one atom will have an unpaired electron and only 7 valence electrons. Common examples include $\text{NO}$, ${\text{NO}}_2$, and ${\text{ClO}}_2$. (3) Expanded octet (hypervalent) species: Central atoms have more than 8 valence electrons. This is only possible for central atoms in period 3 or lower, because these elements have empty d-orbitals in their valence shell that can accommodate extra electrons. Period 2 elements can never have expanded octets.

Worked Example

Draw the Lewis diagram for boron trifluoride, ${\text{BF}}_3$, and confirm why it is an octet exception.

1. Connectivity: Boron is the least electronegative atom, so it is the central atom with three terminal fluorine atoms bonded to it.
2. Total valence electrons: B (group 13) has 3 valence electrons, each F (group 17) has 7, so total $=3+(3\times 7)=24$ valence electrons.
3. Bonding electrons: Three single bonds use $3\times 2=6$ electrons, so remaining non-bonding electrons $=24-6=18$.
4. Distribute lone pairs: Each terminal F needs 6 more electrons to complete its octet, so $3\times 6=18$ electrons, which uses all remaining electrons.
5. Check octets: All three F atoms have a full octet, but the central B atom only has 3 bonds = 6 valence electrons. There are no remaining lone pairs to form a double bond, so B is electron deficient, making ${\text{BF}}_3$ an exception to the octet rule.

Exam tip: Any AP exam multiple-choice option that shows an expanded octet on a period 2 central atom is automatically incorrect. Period 2 elements do not have accessible d-orbitals to accommodate extra electrons, so this is never allowed.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Adding electrons to the total valence count for a cation (e.g., adding 1 electron to ${\text{NH}}_4^{+}$ instead of subtracting 1). Why: Students confuse anion and cation charge adjustment, memorizing "add for charge" without checking the sign. Correct move: Always write the rule explicitly next to your calculation: "add electrons for negative charge, subtract electrons for positive charge" before proceeding.
• Wrong move: Placing the most electronegative atom as the central atom (e.g., putting O central in ${\text{H}}_2\text{CO}$ instead of C). Why: Students reverse the connectivity rule, assuming more electronegative atoms attract more electrons so they belong in the center. Correct move: Memorize the mnemonic "least electronegative is central, H and halogens are always terminal" and confirm connectivity before counting electrons.
• Wrong move: Drawing an expanded octet for a period 2 central atom (e.g., 10 valence electrons on N in ${\text{NO}}_3^{-}$). Why: Students add extra electrons to get more favorable formal charges, forgetting the orbital restriction. Correct move: Always check the period of the central atom first; if it is period 2, cap the valence electron count at 8.
• Wrong move: Forgetting to enclose polyatomic ions in square brackets and write the net charge outside the brackets. Why: Students focus on getting the electron arrangement right and skip notation requirements. Correct move: Add brackets and charge as the final step of drawing any Lewis diagram for a charged species.
• Wrong move: Counting bonding electrons twice when checking per-atom octet completion (e.g., counting 4 electrons for one bond between two atoms, for a total of 4 for one atom's octet). Why: Students confuse total molecule electron count with per-atom octet count. Correct move: For per-atom octet checks, count all bonding electrons shared by the atom (each bond contributes 2 electrons to the atom's count).
• Wrong move: Choosing a structure with negative formal charge on a less electronegative atom when two structures have the same number of non-zero formal charges. Why: Students stop after counting non-zero formal charges and forget the second rule. Correct move: If two structures have the same number of non-zero formal charges, always confirm the negative charge is on the most electronegative atom.

6. Practice Questions (AP Chemistry Style)

Question 1 (Multiple Choice)

What is the correct formal charge on the sulfur atom in the sulfate ion ${\text{SO}}_4^{2-}$, for the Lewis structure with one double bond between S and oxygen, and three single bonds between S and oxygen? A) -2 B) 0 C) +1 D) +2

Worked Solution: Use the formal charge formula $FC=V-N-B/2$. Sulfur is a group 16 element, so $V=6$. In the described structure, sulfur has no lone pairs, so $N=0$. The total bonding electrons are 4 from the double bond plus 2 from each of the three single bonds, for a total of $4+3(2)=10$, so $B=10$. Plugging in gives $FC=6-0-10/2=6-5=+1$. The correct answer is C.

Question 2 (Free Response)

The nitric oxide molecule $\text{NO}$ is an important biological signaling molecule. (a) Draw the complete Lewis diagram for $\text{NO}$. (b) Explain why $\text{NO}$ is an exception to the octet rule. (c) Two possible structures are proposed: Structure 1 (double bond, unpaired electron on N), Structure 2 (double bond, unpaired electron on O). Calculate formal charge for both and identify the more stable structure.

Worked Solution: (a) Total valence electrons = 5 (N) + 6 (O) = 11. The correct Lewis diagram has a double bond between N and O, one lone pair (2 electrons) and one unpaired electron on N, and two lone pairs (4 electrons) on O, giving N 6 total valence electrons and O 8 total valence electrons. (b) $\text{NO}$ has 11 total valence electrons, an odd number. A full octet for all atoms requires 16 total valence electrons (8 per atom), an even number, so it is impossible for all atoms to have a full octet. This makes $\text{NO}$ an exception. (c) Structure 1: $F{C}_N=5-3-4/2=0$; $F{C}_O=6-4-4/2=0$. Structure 2: $F{C}_N=5-2-4/2=+1$; $F{C}_O=6-5-4/2=-1$. Structure 1 has all formal charges equal to zero, so it is the more stable structure.

Question 3 (Application / Real-World Style)

Stratospheric ozone ${\text{O}}_3$ absorbs harmful ultraviolet radiation from the sun before it reaches Earth's surface. Two resonance structures can be drawn for neutral ${\text{O}}_3$. Draw both Lewis resonance structures, calculate the formal charge on the central oxygen atom in both structures, and explain what the formal charges indicate about the relative stability of ${\text{O}}_3$ compared to diatomic ${\text{O}}_2$ (which has formal charge 0 on both atoms).

Worked Solution: Total valence electrons for ${\text{O}}_3=3\times 6=18$. Both resonance structures have a central oxygen atom, one single bond to a terminal oxygen, and one double bond to the other terminal oxygen, with one lone pair on the central oxygen. The double-headed resonance arrow separates the two structures that swap the position of the single and double bonds. For the central oxygen: $V=6$, $N=2$ (one lone pair), $B=2+4=6$ bonding electrons. $FC=6-2-6/2=+1$, which is the same for both resonance structures. ${\text{O}}_3$ has non-zero formal charges (+1 on central O, -1 on single-bonded terminal O) while ${\text{O}}_2$ has all formal charges equal to zero. This means ozone is less stable and more reactive than diatomic oxygen, consistent with its role as a reactive UV absorber in the stratosphere.

7. Quick Reference Cheatsheet

8. What's Next

Lewis diagrams are the non-negotiable foundation for all remaining topics in Unit 2. Next, you will use Lewis diagrams to identify resonance structures, predict molecular geometry via VSEPR theory, and calculate bond polarity — all heavily tested skills on the AP exam. Without correctly drawing and interpreting Lewis diagrams, you cannot accurately predict molecular shape, polarity, or intermolecular forces, which are core topics across Units 2, 3, and 8. Lewis structures also underpin formal charge reasoning in FRQ questions and reaction mechanism predictions in organic chemistry (Unit 9). Follow-up topics to study next: VSEPR and molecular geometry Resonance structures Bond polarity and dipole moments Intermolecular forces