Drawing Lewis Structures Worksheet Study Resources - Free Printable

Let’s go step by step to check each Lewis structure in the worksheet and see if they are correct or not.

We’ll look at each group one by one: Simple Molecules, Polyatomic Ions, Multiple Bonds.

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Group 1: Simple Molecules

These are neutral molecules with single bonds only (except where noted).

1. CH₄ (Methane)
- Carbon has 4 valence electrons → needs 4 bonds.
- Each H has 1 valence electron → needs 1 bond.
- Structure shows C bonded to 4 H atoms → ✔ Correct.

2. NH₃ (Ammonia)
- N has 5 valence electrons → forms 3 bonds + 1 lone pair.
- Each H forms 1 bond.
- Structure shows N with 3 H’s and 1 lone pair → ✔ Correct.

3. H₂O (Water)
- O has 6 valence electrons → forms 2 bonds + 2 lone pairs.
- Each H forms 1 bond.
- Structure shows O with 2 H’s and 2 lone pairs → ✔ Correct.

4. HF (Hydrogen Fluoride)
- F has 7 valence electrons → forms 1 bond + 3 lone pairs.
- H forms 1 bond.
- Structure shows H–F with 3 lone pairs on F → ✔ Correct.

5. NCl₃ (Nitrogen Trichloride)
- N has 5 valence electrons → forms 3 bonds + 1 lone pair.
- Each Cl has 7 valence electrons → forms 1 bond + 3 lone pairs.
- Structure shows N bonded to 3 Cl’s, each Cl has 3 lone pairs, N has 1 lone pair → ✔ Correct.

✔ All Group 1 structures are correct.

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Group 2: Polyatomic Ions

These have charges — we must account for extra or missing electrons.

1. PO₄³⁻ (Phosphate ion)
- P has 5 valence e⁻, each O has 6 → total = 5 + 4×6 = 29, plus 3 for charge → 32 e⁻ total.
- In structure: P bonded to 4 O’s. One O is double-bonded? Wait — actually, in the drawing, it looks like all are single bonds, but that would give P only 8 electrons (ok), but formal charges?
- If all single bonds: P has +1 formal charge, each O has -1 → total charge -4 → too much.
- But here, one O is double-bonded → then P has 0 formal charge, three O’s have -1, one O has 0 → total charge -3 → ✔ Correct.
- Also, octets satisfied → ✔

2. ClO⁻ (Hypochlorite ion)
- Cl has 7, O has 6 → total 13 + 1 for negative charge = 14 e⁻.
- Structure: Cl–O single bond. Cl has 3 lone pairs (6e), O has 3 lone pairs (6e) → total 2 (bond) + 6 + 6 = 14 → ✔
- Formal charge: Cl = 7 - 6 - 1 = 0; O = 6 - 6 - 1 = -1 → matches ion charge → ✔

3. SO₃²⁻ (Sulfite ion)
- S has 6, each O has 6 → 6 + 18 = 24 + 2 for charge = 26 e⁻.
- Structure: S bonded to 3 O’s. One double bond? Actually, drawn as S with one double bond to O, two single bonds to O⁻, and one lone pair on S.
- Electrons: double bond = 4e, two single bonds = 4e, lone pair on S = 2e, each single-bonded O has 3 lone pairs (6e each → 12e), double-bonded O has 2 lone pairs (4e) → total = 4+4+2+12+4 = 26 → ✔
- Formal charges: S = 6 - 2 - 3 = +1? Wait — let’s recalculate:
- S: valence 6, nonbonding 2, bonding 6 (3 bonds × 2) → formal charge = 6 - 2 - 3 = +1
- Double-bonded O: 6 - 4 - 2 = 0
- Each single-bonded O: 6 - 6 - 1 = -1 → total charge = +1 + 0 -1 -1 = -1 → but should be -2! ✘ Problem!

Wait — this might be wrong. Let me think again.

Actually, sulfite ion SO₃²⁻ usually has resonance with one double bond and two single bonds, but sulfur can expand octet. However, in many textbooks, it’s drawn with S having one lone pair, one double bond, two single bonds to O⁻ → total charge -2.

But formal charge calculation:

- S: 6 valence - 2 nonbonding - 3 bonds (since double bond counts as 2, but in formal charge, we count number of bonds, not electrons) → wait, formal charge formula: FC = V - N - B/2

Better way:

Formal Charge = Valence electrons - Nonbonding electrons - ½ Bonding electrons

For S in SO₃²⁻ (as drawn):

- Valence = 6
- Nonbonding = 2 (lone pair)
- Bonding electrons: one double bond (4e) + two single bonds (4e) = 8e → so ½ × 8 = 4
- FC = 6 - 2 - 4 = 0

Double-bonded O: valence 6, nonbonding 4, bonding 4 → FC = 6 - 4 - 2 = 0

Each single-bonded O: valence 6, nonbonding 6, bonding 2 → FC = 6 - 6 - 1 = -1

Total charge: 0 + 0 + (-1) + (-1) = -2 → ✔ Correct!

I made a mistake earlier — yes, it’s correct.

4. NO₂⁻ (Nitrite ion)
- N has 5, each O has 6 → 5 + 12 = 17 + 1 for charge = 18 e⁻.
- Structure: N bonded to two O’s. One double bond? Drawn as N with one double bond to O, one single bond to O⁻, and one lone pair on N.
- Electrons: double bond 4e, single bond 2e, lone pair on N 2e, double-bonded O has 2 lone pairs (4e), single-bonded O has 3 lone pairs (6e) → total = 4+2+2+4+6 = 18 → ✔
- Formal charges:
- N: 5 - 2 - 3 = 0? Wait: bonding electrons = 6 (double + single = 3 bonds → 6e) → ½×6=3 → FC=5-2-3=0
- Double-bonded O: 6-4-2=0
- Single-bonded O: 6-6-1=-1 → total charge -1 → ✔ Correct.

✔ All Group 2 structures are correct.

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Group 3: Multiple Bonds

1. HCN (Hydrogen Cyanide)
- H-C≡N
- H: 1 bond → ok
- C: 4 bonds (1 to H, 3 to N) → ok
- N: 3 bonds + 1 lone pair → 8 electrons → ok
- Total valence: H=1, C=4, N=5 → 10 e⁻
- In structure: H-C single (2e), C≡N triple (6e), lone pair on N (2e) → total 10e → ✔ Correct.

2. C≡N⁻ (Cyanide ion)
- Should be [C≡N]⁻
- Valence: C=4, N=5 → 9 +1 =10 e⁻
- Structure: C≡N with lone pair on C? Or on N?
- In drawing: it shows :C≡N: with negative sign — probably meaning lone pair on C.
- But standard is [:C≡N:]⁻ with negative on carbon? Actually, nitrogen is more electronegative, so negative charge should be on carbon? No — in cyanide, the negative charge is on carbon because carbon is less electronegative and holds the extra electron better in this case? Wait, no — actually, in Lewis structure, we put the negative charge on the atom that can best stabilize it. Nitrogen is more electronegative, so it should hold the negative charge? But in reality, cyanide ion is written as [C≡N]⁻ with the charge distributed, but conventionally, we draw the triple bond and put the lone pair and negative charge on carbon.

Let’s calculate:

If C≡N, and we add one electron → total 10 e⁻.

Triple bond uses 6e. Remaining 4e → two lone pairs.

If we put both lone pairs on N: N would have 2 (from bond) + 4 = 6 electrons? No — in triple bond, N shares 6 electrons, plus if it has one lone pair (2e), total 8e. Then C has only 6e from triple bond → needs more.

Standard Lewis structure for CN⁻ is: [:C≡N:]⁻ with the negative charge on carbon, and carbon has one lone pair, nitrogen has one lone pair.

Valence electrons: C=4, N=5, +1 charge =10.

Structure: C≡N triple bond (6e), lone pair on C (2e), lone pair on N (2e) → total 10e.

Formal charges:

- C: valence 4, nonbonding 2, bonding 6 → FC = 4 - 2 - 3 = -1
- N: valence 5, nonbonding 2, bonding 6 → FC = 5 - 2 - 3 = 0
→ Total charge -1 → ✔ Correct.

In the worksheet, it’s drawn as :C≡N: with negative sign — assuming the colon represents lone pairs, and it's implied that C has the negative charge → ✔ Correct.

3. CO₂ (Carbon Dioxide)
- O=C=O
- C has 4 valence, each O has 6 → total 16 e⁻
- Structure: two double bonds → each double bond 4e → 8e in bonds.
- Each O has two lone pairs → 4e per O → 8e → total 16e → ✔
- Octets: C has 8e (two double bonds), each O has 8e → ✔ Correct.

4. COCl₂ (Phosgene)
- Structure shown: O=C with two Cl’s attached to C? And lone pairs.
- Valence: C=4, O=6, Cl=7 each → 4+6+14=24 e⁻
- Structure: C double bond to O, single bonds to two Cl’s.
- Bonds: C=O (4e), two C-Cl (4e) → 8e in bonds.
- O has two lone pairs (4e)
- Each Cl has three lone pairs (6e each → 12e)
- Total: 8 + 4 + 12 = 24e → ✔
- Octets: C has 8e (double + two singles), O has 8e, each Cl has 8e → ✔ Correct.

✔ All Group 3 structures are correct.

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So after checking every single structure carefully...

All Lewis structures in the worksheet are correctly drawn!

Final Answer: All Lewis structures in the worksheet are correct.