AP Chemistry Unit 2 Notes: Building Molecules from Lewis Structures to 3D Geometry

Lewis Diagrams

What a Lewis diagram is (and what it is not)

A Lewis diagram (Lewis structure) is a 2D accounting system for valence electrons in a molecule or polyatomic ion. It shows:

• Bonding electrons as lines (single, double, triple bonds)
• Nonbonding electrons as lone pairs (pairs of dots)
• Sometimes formal charges on atoms (especially when needed to evaluate competing structures)

Lewis diagrams matter because they are the starting point for predicting:

• Typical bonding patterns (single vs multiple bonds)
• Where lone pairs are located
• Molecular geometry and polarity (once combined with VSEPR)
• Whether resonance is likely (and therefore whether bond lengths/strengths are “averaged”)

A key limitation: a Lewis diagram is not a photo of the molecule. Real molecules are 3D, electron density can be delocalized, and bonds are not always perfectly “single” or “double” in the way a single drawing suggests.

The core idea: the octet guideline as an electron bookkeeping tool

For many main-group atoms (especially in Period 2: C, N, O, F), stable structures often correspond to having 8 electrons around the atom (an octet) when you count shared electrons in bonds plus lone pairs. Hydrogen is the big exception—it seeks a duet (2 electrons).

It helps to treat the octet idea as a strong guideline rather than an unbreakable law. Later in this section you’ll see systematic exceptions (incomplete octets, odd-electron species, and expanded octets).

How to draw Lewis diagrams (step-by-step method)

A reliable process keeps you from “guessing” bonds.

Step 1: Count total valence electrons

Add the valence electrons of each atom (using the periodic table group number for main-group elements). Then adjust for charge:

• Add 1 electron for each negative charge
• Subtract 1 electron for each positive charge

Example counting: For NO_3^-, nitrogen has 5 valence electrons, each oxygen has 6, so total is 5 + 3(6) + 1 = 24.

Step 2: Choose a reasonable skeletal structure

In many AP problems, the least electronegative atom (other than H) is central.

• H is almost never central (it forms one bond)
• Halogens (F, Cl, Br, I) are usually terminal (one bond in many stable molecules)
• C is often central when present

Connect atoms with single bonds to make the skeleton. Each single bond represents 2 electrons.

Step 3: Complete octets on terminal atoms first

Place lone pairs on terminal atoms (typically surrounding atoms) to give them octets (or duet for H).

Step 4: Place remaining electrons on the central atom

If electrons remain after terminal octets are satisfied, put them as lone pairs on the central atom.

Step 5: If the central atom lacks an octet, form multiple bonds

If the central atom has fewer than 8 electrons, convert lone pairs from adjacent atoms into bonding pairs (make double/triple bonds) until the central atom reaches an octet.

This step is where many students go wrong: they either add electrons that don’t exist, or they force multiple bonds without checking the total electron count.

Exceptions you are expected to recognize

These exceptions come up frequently enough that you should be comfortable identifying them.

Incomplete octets (electron-deficient centers)

Some central atoms are stable with fewer than 8 electrons:

• Be often forms 2 bonds and has 4 electrons around it (e.g., BeCl_2)
• B and Al often form 3 bonds and have 6 electrons around them (e.g., BF_3)

On AP-style questions, you typically do not “force” these atoms to complete an octet by inventing extra electrons.

Odd-electron species (radicals)

Molecules or ions with an odd number of valence electrons cannot give every atom a perfect octet (e.g., NO, ClO_2). You’ll draw one unpaired electron. These are less common but fair game.

Expanded octets (Period 3 and beyond)

Atoms in Period 3 or lower (like P, S, Cl, Br, I, Xe) can sometimes have more than 8 electrons around them in common Lewis depictions (e.g., SF_6, PCl_5, SO_4^{2-}). In AP Chemistry, you should be able to draw and interpret these structures, especially for common oxyanions and sulfur/phosphorus halides.

A common misconception is “expanded octets are always wrong.” In many AP problems, allowing an expanded octet helps reduce formal charges and better matches observed stability.

Worked example: Drawing CO_2

1) Valence electrons: C has 4, each O has 6.
Total = 4 + 2(6) = 16.

2) Skeleton: O–C–O (C central).

3) Single bonds use 4 electrons; remaining 12.

4) Give each O an octet: each O needs 6 more electrons (3 lone pairs). That uses all 12 remaining.

5) Now C only has 4 electrons (two single bonds). Convert one lone pair from each O into a bond to C, making two double bonds.

Final: O=C=O, each atom has an octet, and electron count matches 16.

Worked example: Drawing NH_4^+

1) Valence electrons: N has 5, H has 1 each.
Total = 5 + 4(1) - 1 = 8.

2) Skeleton: N central with four single bonds to H.

3) Four bonds use 8 electrons; none left for lone pairs.

N has an octet (8 electrons in four bonding pairs), each H has a duet.

Exam Focus

• Typical question patterns:
  • “Draw the Lewis structure for X and identify lone pairs and multiple bonds.”
  • “How many valence electrons are in X? Show your count.”
  • “Which atom is central / which arrangement is most reasonable?”
• Common mistakes:
  • Forgetting to adjust electron count for ionic charge (especially negative charges).
  • Giving hydrogen an octet or placing H as a central atom.
  • Changing the number of electrons when forming double bonds (multiple bonds redistribute electrons; they don’t create new ones).

Resonance and Formal Charge

Why one Lewis structure is sometimes not enough

Some molecules and ions cannot be accurately represented by a single Lewis structure because the electrons that form the bond framework are delocalized—spread over more than two atoms. Resonance is the practice of drawing multiple valid Lewis structures (called resonance forms) to represent one real molecule.

Resonance matters because it changes how you interpret bonds:

• The real structure is a resonance hybrid, not a rapid flip between drawings.
• Bonds may have fractional bond order (for example, “between a single and double bond”).
• Equivalent resonance forms lead to equal bond lengths in the real molecule.

A very common misconception is that resonance forms are different molecules in equilibrium. They are not. They are different electron-bookkeeping drawings for the same arrangement of atoms.

What can change in resonance (and what cannot)

When you draw resonance forms:

• Atom positions stay the same (the skeletal structure does not change).
• Only electrons move: lone pairs, double bonds, and formal charges can shift.

If moving electrons forces you to rearrange atoms, you’re not drawing resonance—you’re drawing a different compound.

Formal charge: a tool for evaluating Lewis structures

Formal charge assigns a “bookkeeping charge” to each atom based on the assumption that bonding electrons are shared equally.

You calculate it using:

FC = V - (N + B/2)

Where:

• FC is the formal charge on the atom
• V is the number of valence electrons for the neutral atom
• N is the number of nonbonding (lone-pair) electrons on that atom in the structure
• B is the number of bonding electrons shared in bonds to that atom (count 2 per bond line)

Formal charge matters because it helps you choose the most plausible Lewis structure among several that satisfy electron counting.

How to use formal charge to choose “best” structures

In AP Chemistry, you typically prefer structures that:

1) Minimize the magnitude of formal charges (avoid large positive/negative charges when possible)
2) Place negative formal charge on more electronegative atoms (like O, F, Cl) and positive charge on less electronegative atoms
3) Have minimal charge separation (fewer atoms carrying nonzero charge)

These are not arbitrary rules—they reflect that putting electron density on electronegative atoms and avoiding unnecessary charge separation generally corresponds to lower-energy electron distributions.

Resonance in action: NO_3^- (nitrate)

Step 1: Draw a valid Lewis structure

Total valence electrons: 24 (as counted earlier).
Skeleton: N central bonded to three O atoms.

If you draw three single bonds and complete octets on O, the central N ends up with only 6 electrons. To fix that, you form one double bond between N and one O.

You can place the double bond to any of the three oxygen atoms, giving three equivalent resonance forms.

Step 2: Use formal charges to interpret

In each resonance form:

• N typically has a formal charge of +1
• Two singly bonded O atoms typically have -1 each
• The double-bonded O has 0

Because there are three equivalent ways to place the double bond, the real nitrate ion has three equal N–O bonds, each with a bond order of about 1.33 (you can think of it as the “extra” bond being shared among the three oxygens).

A common error is to conclude that nitrate has one short double bond and two long single bonds. Resonance tells you the bonds are equivalent.

Worked example: Formal charge in CO

Carbon monoxide is a good formal-charge exercise because the best Lewis structure includes a triple bond.

Total valence electrons: C has 4, O has 6, total 10.
A structure that satisfies octets is a triple bond between C and O with one lone pair on each atom.

Compute formal charges:

For C: V = 4. Nonbonding electrons N = 2 (one lone pair). Bonding electrons in a triple bond B = 6.

FC_C = 4 - (2 + 6/2) = 4 - (2 + 3) = -1

For O: V = 6. N = 2. B = 6.

FC_O = 6 - (2 + 6/2) = 6 - (2 + 3) = +1

So this Lewis structure places -1 on carbon and +1 on oxygen. That might feel “backward” because oxygen is more electronegative, but the structure is still commonly used because it satisfies octets and matches key features of CO bonding. The takeaway is that formal charge is a guide, not an absolute guarantee of everything about real electron density.

Resonance vs. “just multiple bonds”: spotting when resonance is needed

You should suspect resonance when:

• You can draw multiple valid Lewis structures by moving a double bond and lone pair(s) around without changing the skeleton.
• The molecule/ion has a pattern like “central atom bonded to multiple equivalent atoms,” especially oxyanions (e.g., nitrate, carbonate, sulfate).

Notation reference (common in AP questions)

(If you draw resonance, make sure you use the correct resonance arrow—AP questions often expect you to distinguish resonance from equilibrium.)

Exam Focus

• Typical question patterns:
  • “Draw all resonance structures for X and identify the resonance hybrid implication (equal bonds).”
  • “Calculate the formal charge on each atom and choose the best Lewis structure.”
  • “Which structure is more stable and why (formal charge + electronegativity)?”
• Common mistakes:
  • Moving atoms instead of electrons when drawing resonance forms.
  • Choosing a structure with larger formal charges when a lower-formal-charge option exists.
  • Interpreting resonance as the molecule rapidly alternating between structures rather than being a hybrid.

VSEPR and Bond Hybridization

From Lewis diagrams to 3D shape: why geometry matters

Once you have a Lewis structure, you know the connectivity (who is bonded to whom) and where lone pairs are. But molecules exist in 3D, and their geometry controls many properties you care about in AP Chemistry:

• Polarity (whether bond dipoles cancel)
• Intermolecular forces (shape affects how molecules pack and attract)
• Reactivity and how molecules approach each other

The key bridge from a 2D Lewis diagram to 3D shape is VSEPR theory.

VSEPR: electron domains repel

VSEPR (Valence Shell Electron Pair Repulsion) theory says that regions of electron density around a central atom arrange themselves to minimize repulsion. In VSEPR language, each electron domain counts as one region of electron density:

• One single bond = 1 domain
• One double bond = 1 domain
• One triple bond = 1 domain
• One lone pair = 1 domain

That “multiple bonds count as one domain” rule is crucial. Students often incorrectly count a double bond as two domains.

Electron geometry vs. molecular geometry

You will usually determine shape in two stages:

1) Electron geometry: the arrangement of all electron domains (bonding + lone pairs)
2) Molecular geometry: the arrangement of atoms only (ignore lone pairs when naming the molecular shape)

Lone pairs still matter because they push bonding pairs closer together, changing bond angles.

Predicting shape (the standard AP approach)

Step 1: Count electron domains around the central atom

Use the Lewis structure to count bonds (each bond counts once) plus lone pairs on the central atom.

Step 2: Determine electron geometry

Common electron geometries:

• 2 domains: linear
• 3 domains: trigonal planar
• 4 domains: tetrahedral
• 5 domains: trigonal bipyramidal
• 6 domains: octahedral

Step 3: Determine molecular geometry by “hiding” lone pairs

Examples (common AP set):

• Tetrahedral electron geometry:
  • 0 lone pairs: tetrahedral
  • 1 lone pair: trigonal pyramidal
  • 2 lone pairs: bent
• Trigonal planar electron geometry:
  • 0 lone pairs: trigonal planar
  • 1 lone pair: bent
• Trigonal bipyramidal electron geometry:
  • 0 lone pairs: trigonal bipyramidal
  • 1 lone pair: seesaw
  • 2 lone pairs: T-shaped
  • 3 lone pairs: linear
• Octahedral electron geometry:
  • 0 lone pairs: octahedral
  • 1 lone pair: square pyramidal
  • 2 lone pairs: square planar

Lone pairs repel more strongly (so angles shrink)

VSEPR repulsion strength is often taught qualitatively as:

• lone pair–lone pair repulsion is strongest
• lone pair–bonding pair is intermediate
• bonding pair–bonding pair is weakest

So compared to ideal angles, lone pairs usually compress bond angles.

Example: In a perfect tetrahedral arrangement, bond angles are about 109.5 degrees. In NH_3 (one lone pair), the H–N–H angle is smaller (about 107 degrees). In H_2O (two lone pairs), the H–O–H angle is even smaller (about 104.5 degrees). You don’t need to memorize every number, but you should understand the trend and be able to compare.

Worked example: Shape and angles for SO_2

1) Lewis structure: sulfur is central, bonded to two oxygens; there is resonance (often drawn with one double bond and one single bond in each form). Importantly for VSEPR, each S–O bond counts as one domain, regardless of bond order.

2) Count domains on S: two bonding domains + one lone pair = 3 domains.

3) Electron geometry: trigonal planar.

4) Molecular geometry: bent (because one domain is a lone pair).

5) Angle: less than 120 degrees due to lone pair compression.

Trigonal bipyramidal: why “axial vs equatorial” matters

In a trigonal bipyramidal electron geometry (5 domains), positions are not all equivalent:

• Equatorial positions (3 of them) are 120 degrees apart
• Axial positions (2 of them) are 180 degrees apart from each other and 90 degrees to equatorial

Lone pairs prefer equatorial positions because that minimizes strong 90-degree interactions. This is why, for example, a seesaw shape places the lone pair equatorial.

Polarity connection (often paired with VSEPR questions)

Geometry helps you decide whether individual bond dipoles cancel.

• A molecule can have polar bonds but be nonpolar overall if the geometry is symmetric (e.g., CO_2 linear, dipoles cancel).
• Less symmetric shapes (bent, trigonal pyramidal, seesaw) often produce a net dipole.

AP questions frequently ask you to justify polarity using both electronegativity (bond polarity) and shape (dipole cancellation).

Bond hybridization: a model linking electron domains to orbitals

Hybridization is a model describing how atomic orbitals on a central atom mix to form a set of equivalent hybrid orbitals oriented in space for bonding. In AP Chemistry, hybridization is primarily used as a pattern that matches electron geometry.

Hybridization matters because it connects:

• The number of electron domains (VSEPR)
• The geometry
• The types of orbitals used for sigma bonding (and where pi bonds come from)

A helpful way to think about it: VSEPR tells you where electron density wants to point; hybridization is a way to describe the orbital “tools” that point in those directions.

The key mapping: electron domains to hybridization

For the central atom, the hybridization is determined by the number of electron domains (often called the steric number in some courses):

Important: Resonance and multiple bonds usually do not change the domain count. A double bond still counts as one domain, so it typically does not change hybridization relative to a single bond.

Sigma and pi bonds (how multiple bonds fit in)

A sigma bond is formed by end-to-end overlap along the internuclear axis. A pi bond is formed by side-to-side overlap of unhybridized p orbitals.

• Single bond: 1 sigma
• Double bond: 1 sigma + 1 pi
• Triple bond: 1 sigma + 2 pi

Hybridization describes the orbitals used to form the sigma framework. The pi bonds usually come from leftover unhybridized p orbitals.

Worked example: Hybridization in C_2H_4 (ethylene)

Each carbon in ethylene is bonded to:

• two H atoms (two single bonds)
• one other C atom via a double bond (counts as one domain)

So each carbon has 3 electron domains total.

• Electron geometry around each carbon: trigonal planar
• Hybridization: sp2

In an sp2 model, each carbon uses three sp2 hybrid orbitals to make three sigma bonds (two C–H and one C–C sigma). The remaining unhybridized p orbital on each carbon overlaps side-by-side to form the pi bond.

This explains why the molecule is planar: the pi bond requires parallel p orbitals, which locks the geometry.

Worked example: Hybridization and shape in NH_3

Nitrogen in NH_3 has:

• 3 N–H bonds (3 domains)
• 1 lone pair (1 domain)

Total = 4 domains.

• Electron geometry: tetrahedral
• Molecular geometry: trigonal pyramidal
• Hybridization on N: sp3

A typical misconception is that “trigonal pyramidal means sp3 is wrong.” Hybridization follows electron geometry (domains), not the molecular geometry name.

A note on expanded octets and hybridization

For molecules like SF_6 (6 domains) or PCl_5 (5 domains), AP Chemistry often uses the sp3d2 and sp3d hybridization labels to match VSEPR geometries. You should be able to apply the domain-counting mapping above when asked.

(At more advanced levels, the bonding in some hypervalent molecules can be described with different models, but AP questions typically expect the VSEPR + hybridization mapping approach.)

Exam Focus

• Typical question patterns:
  • “Given a Lewis structure, predict electron geometry, molecular geometry, and approximate bond angles.”
  • “Determine whether the molecule is polar; justify using geometry and bond polarity.”
  • “Identify the hybridization of the central atom (or multiple atoms) based on electron domains.”
• Common mistakes:
  • Counting a double bond as two electron domains in VSEPR (it is one domain).
  • Mixing up electron geometry and molecular geometry (forgetting to account for lone pairs when naming the molecular shape).
  • Assigning hybridization from the molecular geometry name rather than from the total number of electron domains.