Lewis structures (or Lewis dot diagrams) show:

• Valence electrons as dots or lines
• Bonding electrons as lines (each line = 2 electrons)
• Lone pairs (nonbonding electrons) as pairs of dots
• Arrangement of atoms in a molecule

Purpose: Visualize electron distribution and predict molecular properties

Drawing Lewis Structures

Step-by-Step Method

Step 1: Count total valence electrons

Sum valence electrons from all atoms:

• For neutral molecules: sum valence electrons
• For cations (+): subtract electrons
• For anions (-): add electrons

Example: CO₂

• C: 4 valence electrons
• O: 6 valence electrons (×2)
• Total: 4 + 12 = 16 electrons

Step 2: Determine the skeletal structure

• Central atom: Usually the least electronegative (except H)
• H is always terminal (outer atom)
• C is usually central in organic molecules
• Connect atoms with single bonds

Example: CO₂ $O - C - O$

Step 3: Count electrons used so far

Each single bond = 2 electrons

Example: CO₂

• 2 bonds × 2 electrons = 4 electrons used
• Remaining: 16 - 4 = 12 electrons

Step 4: Complete octets of outer atoms

Add lone pairs to terminal atoms (except H gets 2 electrons)

Example: CO₂ $:O - C - O:$ (add 6 electrons to each O)

Each O now has 8 electrons (2 bonding + 6 nonbonding)

• Used: 4 (bonds) + 12 (lone pairs) = 16 electrons ✓

Step 5: Place remaining electrons on central atom

Example: CO₂ No electrons remaining (all 16 used)

Step 6: Check central atom's octet

If central atom has < 8 electrons, form multiple bonds

Example: CO₂ C currently has only 4 electrons (2 bonds) → need double bonds

$O=C=O$

Now C has 8 electrons (4 bonds) ✓

Final structure:

$:O=C=O:$

Each O: 4 bonding + 4 nonbonding = 8 ✓ C: 8 bonding electrons = 8 ✓

The Octet Rule

Statement: Atoms tend to gain, lose, or share electrons to achieve 8 valence electrons (like noble gases)

Exceptions:

1. Hydrogen and Helium

• Follow duet rule (2 electrons)
• H forms 1 bond

2. Incomplete Octets (Less than 8)

• Boron (B): Often has 6 electrons (3 bonds)
  • Example: BF₃ has 6 electrons around B
• Beryllium (Be): Often has 4 electrons (2 bonds)
  • Example: BeCl₂ has 4 electrons around Be

3. Expanded Octets (More than 8)

• Elements in Period 3 and beyond can have > 8 electrons
• Use d orbitals to accommodate extra electrons
• Common in P, S, Cl, Br, I, Xe

Examples:

• PCl₅: P has 10 electrons (5 bonds)
• SF₆: S has 12 electrons (6 bonds)
• ClF₃: Cl has 10 electrons (3 bonds + 2 lone pairs)

4. Odd-Electron Species (Radicals)

• Total number of electrons is odd
• Cannot satisfy octet for all atoms
• Examples: NO (11 electrons), NO₂ (17 electrons), ClO₂

Formal Charge

Definition: The charge an atom would have if all bonding electrons were shared equally

Purpose: Determine the most stable Lewis structure

Formula:

$\text{FC} = V - (L + \frac{B}{2})$

Where:

• $V$ = number of valence electrons (from periodic table)
• $L$ = number of lone pair electrons
• $B$ = number of bonding electrons

Alternative formula:

$\text{FC} = V - L - \frac{1}{2}B$

or

$\text{FC} = V - (\text{nonbonding e}^-) - \frac{1}{2}(\text{bonding e}^-)$

Simplified thinking:

• Each lone pair electron counts fully (×1)
• Each bonding electron counts as half (×0.5)

Calculating Formal Charge

Example: CO₂

$:O=C=O:$

For left O:

• $V = 6$ (oxygen has 6 valence electrons)
• $L = 4$ (two lone pairs)
• $B = 4$ (two double bonds)
• $\text{FC} = 6 - 4 - \frac{4}{2} = 6 - 4 - 2 = 0$

For C:

• $V = 4$ (carbon has 4 valence electrons)
• $L = 0$ (no lone pairs)
• $B = 8$ (four bonds total)
• $\text{FC} = 4 - 0 - \frac{8}{2} = 4 - 0 - 4 = 0$

For right O: Same as left O, FC = 0

Sum of formal charges = 0 ✓ (matches molecular charge)

Rules for Formal Charge

1. Sum of formal charges = overall molecular charge

   • Neutral molecule: sum = 0
   • Cation: sum = positive charge
   • Anion: sum = negative charge

2. Best Lewis structure minimizes formal charges

   • Structures with formal charges of 0 are preferred
   • Minimize magnitude of formal charges

3. Negative formal charges on more electronegative atoms

   • If formal charges exist, negative FC should be on more EN atoms
   • Positive FC should be on less EN atoms

4. Adjacent formal charges of same sign are unfavorable

   • Avoid structures with +1 next to +1, or -1 next to -1

Resonance Structures

Definition: Multiple valid Lewis structures for the same molecule with same atom positions but different electron arrangements

Key points:

• Actual structure is a hybrid (average) of all resonance forms
• Electrons are delocalized (spread out)
• All resonance structures must have same arrangement of atoms
• Only electrons move, not atoms
• Use double-headed arrow (↔) between structures

Drawing Resonance Structures

Example: Ozone (O₃)

$O=O-O: \leftrightarrow :O-O=O$

Both structures:

• Same atom positions
• Different electron positions
• Both have same total charge (0)
• Both contribute to actual structure

Actual structure: Hybrid with 1.5 bonds between each O-O pair

Resonance and Stability

More resonance structures = more stable molecule

• Delocalization lowers energy
• Electrons spread over larger area
• More equivalent structures = more delocalization

Example: Benzene (C₆H₆)

Has two major resonance structures with alternating double/single bonds:

$\text{Resonance 1} \leftrightarrow \text{Resonance 2}$

Actual structure: All C-C bonds are equivalent (1.5 bond order)

Equivalent vs. Non-equivalent Resonance

Equivalent resonance structures:

• Same energy
• Contribute equally to hybrid
• Example: CO₃²⁻ has 3 equivalent structures

Non-equivalent resonance structures:

• Different energies
• Lower formal charges contribute more
• Example: Some structures of CO₂⁻ radical

Common Molecular Examples

Water (H₂O)

Valence electrons: 2(1) + 6 = 8

$H-O-H$

With lone pairs:

H-\overset{..{O}-H} (two lone pairs on O)

Ammonia (NH₃)

Valence electrons: 3(1) + 5 = 8

$H-N-H$ with one lone pair on N $|$ $H$

Carbon dioxide (CO₂)

Valence electrons: 4 + 2(6) = 16

$O=C=O$ (two double bonds)

Sulfate ion (SO₄²⁻)

Valence electrons: 6 + 4(6) + 2 = 32

Central S with 4 O atoms, multiple resonance structures

Nitrate ion (NO₃⁻)

Valence electrons: 5 + 3(6) + 1 = 24

Central N with 3 O atoms, 3 equivalent resonance structures

Expanded Octet Examples

Sulfur hexafluoride (SF₆)

Valence electrons: 6 + 6(7) = 48

• S (period 3) can expand octet
• S has 12 electrons (6 bonds)
• Each F has 8 electrons (1 bond + 3 lone pairs)

Phosphorus pentachloride (PCl₅)

Valence electrons: 5 + 5(7) = 40

• P (period 3) can expand octet
• P has 10 electrons (5 bonds)
• Each Cl has 8 electrons (1 bond + 3 lone pairs)

Xenon tetrafluoride (XeF₄)

Valence electrons: 8 + 4(7) = 36

• Xe (period 5) can expand octet
• Xe has 12 electrons (4 bonds + 2 lone pairs)
• Each F has 8 electrons (1 bond + 3 lone pairs)

Tips and Tricks

1. Central atom: Usually least electronegative (except H)
2. Check your math: Total electrons must match valence electron count
3. Symmetry: More symmetrical structures often preferred
4. Multiple bonds: Form when central atom needs more electrons
5. Resonance: Look for possibility to move lone pairs to create multiple bonds
6. Formal charge sum: Must equal molecular charge
7. Period matters: Period 3+ can expand octet; period 2 cannot

Common Mistakes to Avoid

1. Forgetting to add/subtract electrons for ions
2. Giving H more than 2 electrons
3. Not forming multiple bonds when central atom needs them
4. Incorrect formal charge calculations
5. Placing too many electrons (exceeding valence count)
6. Not considering resonance structures
7. Moving atoms (not electrons) between resonance structures

(b) Octet rule: Yes, all atoms satisfy the octet rule:

• Each O has 8 electrons
• S has 8 electrons (can expand octet to 12 using d orbitals)

(c) Resonance: SO₄²⁻ has 6 resonance structures. The double bonds can be on any 2 of the 4 oxygen atoms. Resonance delocalizes the negative charge equally over all 4 oxygens, making all S-O bonds equivalent with bond order of 1.5.

Step 3: Count electrons in bonds

3 single bonds × 2 = 6 electrons used Remaining: 24 - 6 = 18 electrons

Step 4: Complete octets of outer atoms (O)

Each O needs 6 more electrons (3 lone pairs):

Each O: 2 (bonding) + 6 (lone pairs) = 8 ✓

Used: 6 (bonds) + 18 (lone pairs) = 24 electrons ✓

Step 5: Check central atom

C currently has only 6 electrons (3 bonds) → needs 2 more

Step 6: Form double bond

Convert one lone pair from an O to make C=O double bond:

$:O:$ $|$ $:O = C - O:$

Now C has 8 electrons (4 bonds) ✓

Step 7: Draw resonance structures

The double bond can be between C and any of the three O atoms:

Resonance Structure 1: $[:O:]^- - C = O - [:O:]^-$

Resonance Structure 2: $O = C - [:O:]^- - [:O:]^-$

Resonance Structure 3: $[:O:]^- - [:O:]^- - C = O$

(Shown linearly for simplicity; actual structure is trigonal planar)

All three structures are equivalent!

Step 8: Calculate formal charges (using Structure 1)

Structure 1: Left O with single bond, C with double bond to middle O, right O with single bond

For C (double bonded to one O, single bonded to two O):

• $V = 4$
• $L = 0$ (no lone pairs)
• $B = 8$ (1 double bond + 2 single bonds = 8 bonding electrons)
• $\text{FC} = 4 - 0 - \frac{8}{2} = 4 - 0 - 4 = 0$

For double-bonded O (middle):

• $V = 6$
• $L = 4$ (2 lone pairs)
• $B = 4$ (1 double bond = 4 bonding electrons)
• $\text{FC} = 6 - 4 - \frac{4}{2} = 6 - 4 - 2 = 0$

For each single-bonded O (left and right):

• $V = 6$
• $L = 6$ (3 lone pairs)
• $B = 2$ (1 single bond = 2 bonding electrons)
• $\text{FC} = 6 - 6 - \frac{2}{2} = 6 - 6 - 1 = -1$

Step 9: Verify sum of formal charges

Sum = 0 (C) + 0 (double-bonded O) + (-1) (left O) + (-1) (right O) = -2 ✓

Matches ion charge (-2) ✓

Answer:

Three equivalent resonance structures with double bond between C and each O in turn.

Formal charges in each structure:

• C: 0
• Double-bonded O: 0
• Single-bonded O (two of them): -1 each

Actual structure: Hybrid with all three C-O bonds equivalent (bond order = 1.33)

Key insight: The negative charges are delocalized over all three oxygen atoms, so each oxygen actually bears -2/3 charge in the resonance hybrid.

Best structure: Structure 1 or 2 (tie based on formal charge magnitudes)

However, Structure 1 is slightly preferred because:

1. The negative charge is on oxygen (more electronegative than nitrogen)
2. In Structure 2, the less electronegative N bears the negative charge

Answer: [O-C≡N]⁻ is the best Lewis structure

Step 3: Connect atoms with single bonds

6 single bonds (S-F) × 2 electrons = 12 electrons used Remaining: 48 - 12 = 36 electrons

Step 4: Complete octets of outer atoms (F)

Each F needs 6 more electrons (3 lone pairs):

• 6 F atoms × 6 electrons = 36 electrons for lone pairs

Each F now has: 2 (bonding) + 6 (lone pairs) = 8 ✓

Total used: 12 (bonds) + 36 (lone pairs) = 48 ✓

Step 5: Check central atom (S)

S has 12 electrons from 6 bonds → expanded octet

This is acceptable because S is in period 3.

Lewis Structure:

$:F:$ $|$ $:F: - S - :F:$ $|$ $:F:$

(Plus 2 more F atoms in 3D, front and back)

Each F has 3 lone pairs (shown as :), S has 6 single bonds

Step 6: Calculate formal charge on sulfur

Formula: $\text{FC} = V - L - \frac{B}{2}$

For S:

• $V = 6$ (sulfur has 6 valence electrons)
• $L = 0$ (no lone pairs on S)
• $B = 12$ (six single bonds = 12 bonding electrons)

$\text{FC}_S = 6 - 0 - \frac{12}{2} = 6 - 0 - 6 = 0$

For each F:

• $V = 7$
• $L = 6$ (3 lone pairs)
• $B = 2$ (1 single bond)

$\text{FC}_F = 7 - 6 - \frac{2}{2} = 7 - 6 - 1 = 0$

Sum of formal charges: 0 + 6(0) = 0 ✓ (neutral molecule)

Step 7: Explain expanded octet

Why can sulfur have 12 electrons?

1. Period 3 element: S is in period 3, which has access to d orbitals (3d)

2. Electron configuration: S can use 3s, 3p, and 3d orbitals for bonding

3. Hybridization: S uses sp³d² hybridization to accommodate 6 bonds

   • 1 s orbital
   • 3 p orbitals
   • 2 d orbitals
   • = 6 hybrid orbitals → 6 bonds

4. Geometry: Octahedral shape (90° angles)

5. Period 2 limitation: Elements in period 2 (C, N, O, F) cannot expand octets

   • No d orbitals available in valence shell
   • Maximum 8 electrons (using 2s and 2p only)
   • Example: NF₃ exists, but NF₅ does NOT

General rule:

• Period 2: Maximum 8 electrons (octet rule strictly followed)
• Period 3+: Can expand octet using d orbitals
  • Common in P, S, Cl, Br, I, Xe
  • Examples: PCl₅ (10 e⁻), SF₆ (12 e⁻), IF₇ (14 e⁻)

Answer:

Lewis structure has S bonded to 6 F atoms, each F with 3 lone pairs.

Formal charge on S: FC = 0

Expanded octet explanation: Sulfur is in period 3 and can use d orbitals (sp³d² hybridization) to accommodate 12 electrons (6 bonds). Period 2 elements cannot do this.

Note: Recent theoretical studies suggest the role of d orbitals may be overstated, and that ionic character and polarization also contribute, but sp³d² hybridization remains the standard AP Chemistry explanation.