Solution:

(a) Lewis structure and formal charges:

Central S with 4 O atoms:

• S: 6 valence electrons
• 4 O: 4 × 6 = 24 valence electrons
• Charge: -2 (add 2 electrons)
• Total: 6 + 24 + 2 = 32 valence electrons

Structure: O=S(=O)(-O⁻)(-O⁻) with double bonds to 2 oxygens, single bonds to 2 oxygens

Formal charge = V - (L + B/2) where V = valence e⁻, L = lone pair e⁻, B = bonding e⁻

• S: FC = 6 - (0 + 12/2) = 0
• O (double bonded): FC = 6 - (4 + 4/2) = 0
• O (single bonded): FC = 6 - (6 + 2/2) = -1

(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.

Solution:

Given: NH₃ (ammonia) Find: Lewis structure and formal charges

Step 1: Count total valence electrons

• N: 5 valence electrons
• H: 1 valence electron (×3)
• Total: 5 + 3 = 8 electrons

Step 2: Determine skeletal structure

N is central (less electronegative than H is irrelevant; H is always terminal):

$\ce{H - N - H}$ $\ce{ | }$ $\ce{ H }$

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:

$\ce{H - N - H}$ $\ce{ | }$
$\ce{ H }$

(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:

$\ce{ .. }$ $\ce{H - N - H}$ $\ce{ | }$ $\ce{ H }$

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

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.

Solution:

(a) Lewis structure and formal charges:

Central S with 4 O atoms:

• S: 6 valence electrons
• 4 O: 4 × 6 = 24 valence electrons
• Charge: -2 (add 2 electrons)
• Total: 6 + 24 + 2 = 32 valence electrons

Structure: O=S(=O)(-O⁻)(-O⁻) with double bonds to 2 oxygens, single bonds to 2 oxygens

Formal charge = V - (L + B/2) where V = valence e⁻, L = lone pair e⁻, B = bonding e⁻

• S: FC = 6 - (0 + 12/2) = 0
• O (double bonded): FC = 6 - (4 + 4/2) = 0
• O (single bonded): FC = 6 - (6 + 2/2) = -1

(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.

Solution:

Total valence electrons: O(6) + C(4) + N(5) + 1(charge) = 16 electrons

Structure 1: [O-C≡N]⁻

• O: FC = 6 - (6 + 2/2) = -1
• C: FC = 4 - (0 + 8/2) = 0
• N: FC = 5 - (2 + 6/2) = 0 Sum of |FC| = 1

Structure 2: [:O=C=N:]⁻

• O: FC = 6 - (4 + 4/2) = 0
• C: FC = 4 - (0 + 8/2) = 0
• N: FC = 5 - (4 + 4/2) = -1 Sum of |FC| = 1

Structure 3: [O≡C-N:]²⁻

• O: FC = 6 - (2 + 6/2) = +1
• C: FC = 4 - (0 + 8/2) = 0
• N: FC = 5 - (6 + 2/2) = -2 Sum of |FC| = 3

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

Solution:

Total valence electrons: O(6) + C(4) + N(5) + 1(charge) = 16 electrons

Structure 1: [O-C≡N]⁻

• O: FC = 6 - (6 + 2/2) = -1
• C: FC = 4 - (0 + 8/2) = 0
• N: FC = 5 - (2 + 6/2) = 0 Sum of |FC| = 1

Structure 2: [:O=C=N:]⁻

• O: FC = 6 - (4 + 4/2) = 0
• C: FC = 4 - (0 + 8/2) = 0
• N: FC = 5 - (4 + 4/2) = -1 Sum of |FC| = 1

Structure 3: [O≡C-N:]²⁻

• O: FC = 6 - (2 + 6/2) = +1
• C: FC = 4 - (0 + 8/2) = 0
• N: FC = 5 - (6 + 2/2) = -2 Sum of |FC| = 3

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

Solution:

Given: CO₃²⁻ (carbonate ion) Find: Lewis structure with resonance forms and formal charges

Step 1: Count total valence electrons

• C: 4 valence electrons
• O: 6 valence electrons (×3)
• Charge: -2 means add 2 electrons
• Total: 4 + 18 + 2 = 24 electrons

Step 2: Determine skeletal structure

C is central (less electronegative than O):

$\ce{O - C - O}$ $\ce{ | }$ $\ce{ O }$

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:

$\ce{ :O: }$ $\ce{ | }$ $\ce{:O = C - O:}$

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: $\ce{[:O:]^- - C = O - [:O:]^-}$

Resonance Structure 2: $\ce{O = C - [:O:]^- - [:O:]^-}$

Resonance Structure 3: $\ce{[:O:]^- - [:O:]^- - C = O}$

(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.

Solution:

Given: SF₆ (sulfur hexafluoride) Find: Lewis structure, formal charge on S, explanation for expanded octet

Step 1: Count total valence electrons

• S: 6 valence electrons
• F: 7 valence electrons (×6)
• Total: 6 + 42 = 48 electrons

Step 2: Determine skeletal structure

S is central (less electronegative than F):

Six F atoms arranged around S:

$\ce{F - S - F}$ (with 4 more F atoms above, below, front, back)

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:

$\ce{ :F: }$ $\ce{ | }$ $\ce{:F: - S - :F:}$ $\ce{ | }$ $\ce{ :F: }$

(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.