# Lewis diagrams

> AP Chemistry · Unit 2: Molecular and Ionic Compound Structure and Properties
> Source: https://www.owlsprep.com/study/ap-chemistry-u2-lewis-diagrams/

This guide covers all Lewis diagram concepts tested on the AP Chemistry exam, including step-by-step drawing, formal charge, octet rule exceptions, worked examples, and common exam pitfalls aligned to College Board expectations.

**Prerequisites:** Valence electron configuration for main group elements; Basic definitions of ionic and covalent bonding; Net charge calculation for ionic species

## Learning objectives

- Count total valence electrons correctly for neutral molecules and charged polyatomic ions
- Draw valid Lewis diagrams following the standard step-by-step algorithm
- Calculate formal charge for all atoms in a Lewis structure and select the most stable preferred structure
- Identify and explain the three common classes of octet rule exceptions tested on the AP exam
- Avoid common exam pitfalls when drawing and evaluating Lewis diagrams

## Lewis Diagram Basics: Step-by-Step Construction

Lewis diagrams (also called Lewis dot structures) are 2D representations of covalent molecules and polyatomic ionic compounds that show how atoms are bonded together and where all valence electrons are located. Standard notation uses element symbols for the atomic nucleus and core electrons, single lines for single bonds (2 shared electrons), double lines for double bonds (4 shared electrons), triple lines for triple bonds (6 shared electrons), and dots for non-bonding lone pairs.

**Lewis Diagram (Lewis Dot Structure)** — A schematic representation of bonding connectivity and all valence electrons in a covalent or polyatomic ionic species

*Notation:* Element symbols, lines for bonds, dots for lone pairs, $[\ ]^Q$ for charged polyatomics

*Example:* $\text{H}-\underset{..}{\overset{..}{\text{O}}}-\text{H}$ for water

1. Determine atomic connectivity: the least electronegative atom is almost always the central atom
2. Calculate total valence electrons: add valence electrons for all atoms, add 1 per negative charge, subtract 1 per positive charge
3. Draw one single bond between each connected atom pair, subtract bonding electrons from total to get remaining non-bonding electrons
4. Distribute remaining electrons as lone pairs starting with terminal atoms to satisfy the octet guideline, then assign leftover to the central atom
5. If the central atom lacks a full octet, convert terminal lone pairs into multiple bonds to complete the octet

> **mnemonic**
>
> Least electronegative is central, H and halogens are always terminal

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

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

> **Exam tip:** When adjusting for charge, explicitly write 'add electrons for negative charge, subtract for positive' next to your work to avoid flipping the rule under exam pressure.

## 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.

**Formal Charge** — A hypothetical charge assigned to each atom that assumes equal sharing of bonding electrons, used to rank the stability of alternative Lewis structures

*Notation:* $FC$

*Example:* FC = 0 for carbon in methane

The formula for formal charge is:

$$FC = V - N - \frac{B}{2}$$

Where $V$ = number of valence electrons in the neutral free atom, $N$ = number of non-bonding electrons on the atom, $B$ = 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. Calculate formal charge for each atom and identify the preferred structure. 1. $:\underset{..}{\text{S}} - \text{C} \equiv \text{N}:$, 2. $:\underset{..}{\text{S}} = \text{C} = \underset{..}{\text{N}}:$, 3. $:\text{S} \equiv \text{C} - \underset{...}{\overset{...}{\text{N}}}:$

1. Structure 1 formal charge calculation:
2. $$FC_S = 6 - 6 - (2/2) = -1; FC_C = 4 - 0 - (8/2) = 0; FC_N = 5 - 2 - (6/2) = 0$$
3. Sum of formal charges = -1, matching the ion charge.
4. Structure 2 formal charge calculation:
5. $$FC_S = 6 - 4 - (4/2) = 0; FC_C = 4 - 0 - (8/2) = 0; FC_N = 5 - 4 - (4/2) = -1$$
6. Sum of formal charges = -1, matching the ion charge.
7. Structure 3 formal charge calculation:
8. $$FC_S = 6 - 2 - (6/2) = +1; FC_C = 4 - 0 - (8/2) = 0; FC_N = 5 - 6 - (2/2) = -2$$
9. Structure 3 has two large non-zero formal charges, so it is eliminated. Nitrogen is more electronegative than sulfur, so negative formal charge on nitrogen is preferred. Structure 2 is the preferred structure.

> **Exam tip:** Always check that the sum of all formal charges equals the net charge of the species to catch arithmetic errors before selecting your answer.

## 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 are regularly tested on the AP exam:

- **Electron deficient species**: Central atoms have fewer than 8 valence electrons, almost always for group 13 elements (boron, aluminum) with only 3 valence electrons to share
- **Odd-electron species (free radicals)**: Total valence electrons is odd, so at least one atom has an unpaired electron and only 7 valence electrons. Common examples: $\text{NO}$, $\text{NO}_2$
- **Expanded octet (hypervalent) species**: Central atoms have more than 8 valence electrons, only possible for period 3 or lower central atoms (have empty d-orbitals to 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. 1. Connectivity: Boron is the least electronegative atom, so it is the central atom with three terminal fluorine atoms bonded to it.
2. 2. Total valence electrons: B (group 13) has 3, each F (group 17) has 7:
3. $$3 + (3 \times 7) = 24$$
4. 3. Three single bonds use 6 electrons, so remaining non-bonding electrons: $24 - 6 = 18$.
5. 4. Each terminal F needs 6 more electrons to complete its octet, so $3 \times 6 = 18$, which uses all remaining electrons.
6. 5. Check octets: All F have full octets, but central B only has 3 bonds = 6 valence electrons. No remaining electrons to form a multiple bond, so B is electron deficient, making $\text{BF}_3$ an octet exception.

> **Exam tip:** Any AP multiple-choice option showing an expanded octet on a period 2 central atom is automatically incorrect.

## AP-Style Concept Check

**Check your understanding**

Test your understanding with this AP-style question:

1. 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

   *Why:* Using the formula $FC = V - N - B/2$: V = 6 (group 16), N = 0 (no lone pairs), B = 10 (4 from double bond + 2×3 from single bonds). $FC = 6 - 0 - 10/2 = +1$.

## Common pitfalls

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

## Cheatsheet

| Category | Formula / Rule | Notes |
| --- | --- | --- |
| Total Valence Electrons | $\sum V_i + C$ | Add $C$ for negative charge, subtract $C$ for positive charge |
| Formal Charge | $FC = V - N - \frac{B}{2}$ | Sum of $FC$ = net charge of the species |
| Preferred Structure | 1. Minimize non-zero FC   2. Negative FC on most electronegative   3. No adjacent same-sign FC | Applies for multiple valid octet-satisfying structures |
| Connectivity | Least electronegative atom = central | H and halogens are always terminal |
| Octet Guideline | Most atoms have 8 valence e⁻, H has 2 | General guideline, not a physical law |
| Electron Deficient Exception | Central atom < 8 valence e⁻ | Almost always group 13 (B, Al) central atoms |
| Odd-Electron Exception | One atom has 7 valence e⁻ | Occurs when total valence e⁻ is odd; species called free radicals |
| Expanded Octet Exception | Central atom > 8 valence e⁻ | Only allowed for period 3+ central atoms (have d-orbitals) |
| Polyatomic Ion Notation | Enclose in $[\ ]^Q$ | Required for full FRQ credit |
| Resonance Notation | Separate structures with $\leftrightarrow$ | All equivalent structures contribute to the actual structure |

## What's next

Lewis diagrams are the non-negotiable foundation for all remaining topics in AP Chemistry Unit 2 and beyond. Correctly drawing and interpreting Lewis structures is required to identify resonance, predict molecular geometry via VSEPR theory, calculate bond polarity, and determine intermolecular forces, all of which are heavily tested in both multiple-choice and free-response sections. Lewis diagram reasoning also underpins formal charge arguments in FRQs and reaction mechanism predictions in organic chemistry (Unit 9). Mastery of this topic is essential for scoring a 5 on the AP Chemistry exam.

- [Resonance and Formal Charge](https://www.owlsprep.com/study/ap-chemistry-u2-resonance-and-formal-charge/)
- [VSEPR and Bond Hybridization](https://www.owlsprep.com/study/ap-chemistry-u2-vsepr-and-bond-hybridization/)
- [Intermolecular Forces and Properties Overview](https://www.owlsprep.com/study/ap-chemistry-u3-overview/)

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