Skip to main content
United StatesChemistrySyllabus dot point

How do we draw a Lewis diagram that correctly shows the bonding and lone pairs in a molecule or ion?

Topic 2.5 Lewis Diagrams: draw Lewis diagrams for molecules and polyatomic ions, applying the octet rule and accounting for valence electrons, multiple bonds, and common exceptions.

A focused answer to AP Chemistry Topic 2.5, covering counting valence electrons, the octet rule, single and multiple bonds, lone pairs, polyatomic ions, and common octet exceptions, with a full worked drawing procedure.

Generated by Claude Opus 4.810 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

Jump to a section
  1. What this topic is asking
  2. What a Lewis diagram shows
  3. The drawing procedure
  4. Multiple bonds and exceptions
  5. Try this

What this topic is asking

The College Board (Topic 2.5) wants you to draw Lewis diagrams for molecules and polyatomic ions: structures that show every bonding pair and lone pair of valence electrons. You apply the octet rule (most atoms want eight valence electrons around them, hydrogen wants two), use multiple bonds when needed, and recognize common exceptions. A correct Lewis diagram is the gateway to resonance, formal charge (Topic 2.6) and molecular shape (Topic 2.7).

What a Lewis diagram shows

The goal is a structure in which each atom (except hydrogen) is surrounded by eight valence electrons, the octet rule, reflecting the stability of a full outer shell. Hydrogen is the exception that wants only two (a full first shell).

The drawing procedure

For a polyatomic ion, draw the final structure in brackets with the overall charge written outside. Counting electrons correctly is the step students most often get wrong, so add for negative charge and subtract for positive before you start placing electrons.

Multiple bonds and exceptions

When there are not enough electrons to give the central atom an octet using single bonds, convert outer-atom lone pairs into double or triple bonds. Carbon dioxide (O=C=O\text{O}=\text{C}=\text{O}) and nitrogen (N≡N\text{N}\equiv\text{N}) are examples.

A few species break the octet rule, and the College Board expects you to recognize them:

  • Incomplete octets: boron and beryllium often have fewer than eight electrons (for example BF3\text{BF}_3 has six around boron).
  • Expanded octets: atoms in period 3 and beyond (such as sulfur or phosphorus) can hold more than eight electrons (for example SF6\text{SF}_6).
  • Odd-electron species (radicals): molecules with an odd total, such as NO\text{NO}, cannot give every atom a full octet.

A good Lewis diagram is the foundation for the rest of the unit. Once you have the bonding and lone pairs right, formal charge (Topic 2.6) tells you which of several possible structures is best, and the count of bonding regions and lone pairs on the central atom feeds straight into VSEPR (Topic 2.7) to predict the shape. Because so much downstream reasoning depends on it, it is worth checking your electron total twice, especially the charge adjustment for ions.

Try this

Q1. State the total number of valence electrons to use when drawing the Lewis diagram of the ammonium ion, NH4+\text{NH}_4^{+}. [1 point]

  • Cue. N (5) ++ 4 H (4×14 \times 1) −- 1 (for the ++ charge) =8= 8 electrons.

Q2. Identify the octet-rule exception shown by boron trifluoride, BF3\text{BF}_3. [1 point]

  • Cue. An incomplete octet; boron has only six valence electrons around it.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2023 (style)3 marksSection II (short FRQ). (a) Draw the Lewis diagram of the carbonate ion, CO32−\text{CO}_3^{2-}. (b) State the total number of valence electrons you used. (c) Explain how the 2−2- charge is accounted for in the electron count.
Show worked answer →

A 3-point FRQ on drawing a polyatomic ion.

(a) Lewis diagram (1 point): carbon in the center bonded to three oxygens, with one C=O double bond and two C-O single bonds; each singly bonded O carries three lone pairs and the double-bonded O carries two; the whole structure is enclosed in brackets with a 2−2- charge.
(b) Electron count (1 point): C contributes 4, each O contributes 6 (3×6=183 \times 6 = 18), and the 2−2- charge adds 2, giving 4+18+2=244 + 18 + 2 = 24 valence electrons (12 pairs).
(c) Explain (1 point): the ion has gained two electrons relative to the neutral atoms, so two extra electrons are added to the total valence count before distributing them.

Markers reward a valid octet-satisfying structure, the correct total of 24 electrons, and adding electrons for the negative charge.

AP 2021 (style)1 marksSection I (multiple choice). How many lone pairs are on the central nitrogen atom in the ammonia molecule, NH3\text{NH}_3? (A) 0 (B) 1 (C) 2 (D) 3. Justify your choice.
Show worked answer →

A 1-point conceptual MCQ. The answer is (B).

Nitrogen has 5 valence electrons. In NH3\text{NH}_3 it forms three N-H single bonds, using three of its electrons (one per bond), which leaves two electrons as one lone pair. So the central nitrogen has exactly one lone pair, which is why NH3\text{NH}_3 is trigonal pyramidal rather than trigonal planar.

Related dot points

Sources & how we know this