Draw Lewis structures for molecules and ions, apply the octet rule, identify resonance structures, and calculate formal charges.
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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: CO₂O−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)
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.
2Problem 2medium
❓ Question:
Draw the Lewis structure for sulfate ion (SO₄²⁻). (a) Calculate the formal charge on each atom. (b) Does this structure obey the octet rule? (c) Why might resonance structures be important for this ion?
💡 Show Solution
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.
3Problem 3medium
❓ Question:
Draw the Lewis structure for the carbonate ion (CO₃²⁻). Include all resonance structures and calculate the formal charge on each atom in one resonance structure.
💡 Show Solution
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):
O−C−O∣O
Step 3: Count electrons in bonds
3 single bonds × 2 = 6 electrons used
Remaining: 24 - 6 = 18 electrons
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.
4Problem 4hard
❓ Question:
For the cyanate ion (OCN⁻), three possible Lewis structures can be drawn with different atom arrangements: [O-C≡N]⁻, [O=C=N]⁻, and [O≡C-N]⁻. Use formal charge to determine which structure is most likely to contribute to the actual bonding.
Best structure: Structure 1 or 2 (tie based on formal charge magnitudes)
However, Structure 1 is slightly preferred because:
The negative charge is on oxygen (more electronegative than nitrogen)
In Structure 2, the less electronegative N bears the negative charge
Answer: [O-C≡N]⁻ is the best Lewis structure
5Problem 5hard
❓ Question:
Draw the Lewis structure for sulfur hexafluoride (SF₆). Calculate the formal charge on the central sulfur atom and explain how sulfur can accommodate more than 8 electrons.
💡 Show Solution
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:
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:
:F:∣:F:−S−: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:FC=V−L−2B
For S:
V=6 (sulfur has 6 valence electrons)
L=0 (no lone pairs on S)
B=12 (six single bonds = 12 bonding electrons)
FCS=6−0−2
For each F:
V=7
L=6 (3 lone pairs)
B=2 (1 single bond)
FCF=7−6−2
Sum of formal charges: 0 + 6(0) = 0 ✓ (neutral molecule)
Step 7: Explain expanded octet
Why can sulfur have 12 electrons?
Period 3 element: S is in period 3, which has access to d orbitals (3d)
Electron configuration: S can use 3s, 3p, and 3d orbitals for bonding
Hybridization: S uses sp³d² hybridization to accommodate 6 bonds
1 s orbital
3 p orbitals
2 d orbitals
= 6 hybrid orbitals → 6 bonds
Geometry: Octahedral shape (90° angles)
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.
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