• 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

$H - N - H$ $|$ $H$

Step 3: Count electrons used in bonds

3 single bonds × 2 electrons = 6 electrons used Remaining: 8 - 6 = 2 electrons

Step 4: Complete octets of outer atoms

H atoms follow duet rule (2 electrons each) - already satisfied ✓

Step 5: Place remaining electrons on central atom

Place 2 remaining electrons on N as lone pair:

$H - N - H$ $|$
$H$

(N has one lone pair on top, not shown with dots here)

Step 6: Verify octets

• Each H: 2 electrons (1 bond) ✓ (duet satisfied)
• N: 8 electrons (3 bonds + 1 lone pair) ✓ (octet satisfied)

Lewis Structure:

$..$ $H - N - H$ $|$ $H$

Or more clearly: H-\overset{..{N}-H} $|$ $H$

Step 7: Calculate formal charges

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

For nitrogen (N):

• $V = 5$ (nitrogen has 5 valence electrons)
• $L = 2$ (one lone pair = 2 electrons)
• $B = 6$ (three single bonds = 6 bonding electrons)

$\text{FC}_N = 5 - 2 - \frac{6}{2} = 5 - 2 - 3 = 0$

For each hydrogen (H):

• $V = 1$ (hydrogen has 1 valence electron)
• $L = 0$ (no lone pairs)
• $B = 2$ (one single bond = 2 bonding electrons)

$\text{FC}_H = 1 - 0 - \frac{2}{2} = 1 - 0 - 1 = 0$

Step 8: Verify sum of formal charges

Sum = 0 (N) + 0 (H) + 0 (H) + 0 (H) = 0 ✓

Matches molecular charge (neutral) ✓

Answer:

Lewis structure shows N bonded to three H atoms with one lone pair on N.

Formal charges: All atoms have FC = 0

This structure is optimal because all formal charges are zero.