Free Printable Lewis Dot Structure Worksheets - Free Printable

Let's solve each molecule step by step: Lewis structure, molecular structure (electron geometry), and molecular geometry.

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🔹 1. H₂O (Water)

#### Step 1: Count valence electrons
- H: 1 × 2 = 2
- O: 6
- Total = 8 valence electrons

#### Step 2: Draw Lewis Structure
- Oxygen is the central atom.
- Two single bonds to H atoms → 2×2 = 4 electrons used.
- Remaining 4 electrons go as two lone pairs on O.

```
H
\
O:
/
H
```

(With two lone pairs on O)

#### Step 3: Electron Geometry
- 4 electron domains around O: 2 bonding pairs + 2 lone pairs → Tetrahedral electron geometry

#### Step 4: Molecular Geometry
- Only 2 atoms bonded → Bent (angular) shape

✔ Answer:
- Lewis Structure: O in center, bonded to two H atoms with two lone pairs on O.
- Molecular Structure: Tetrahedral electron geometry
- Molecular Geometry: Bent

---

🔹 2. SO₃ (Sulfur trioxide)

#### Step 1: Valence electrons
- S: 6
- O: 6 × 3 = 18
- Total = 24 valence electrons

#### Step 2: Lewis Structure
- S is central.
- Three double bonds to O atoms: S=O × 3 → uses 6 electrons per bond? Wait — each double bond is 4 electrons → 3×4 = 12 electrons.
- But total is 24 → remaining 12 electrons → 3 lone pairs on each O (each O gets 2 lone pairs = 4 electrons), so 3×4 = 12 electrons.

So:
- S has 3 double bonds → no lone pairs.
- Each O has 2 lone pairs.

But sulfur can expand octet.

Resonance structures exist – one common form is:

```
O
║
O═S═O
```

But better to draw with resonance.

Actually, the most stable form has one double bond and two dative bonds or three resonance structures with double bonds.

We usually draw three resonance structures where each O is double-bonded to S in turn.

But for simplicity, we can represent it as:

```
O
║
O—S—O⁺
│
(with resonance)
```

But standard Lewis structure is trigonal planar with three double bonds and formal charges minimized.

Best Lewis structure: S double bonded to all three O atoms. S has expanded octet (12 electrons).

Total bonds: 3 double bonds → 3×4 = 12 electrons used in bonding.

Each O has 2 lone pairs → 3×4 = 12 electrons → total 24.

✔ Lewis Structure: Central S with three double bonds to O atoms; no lone pairs on S; each O has two lone pairs.

#### Electron Geometry: 3 bonding domains → Trigonal planar

#### Molecular Geometry: Trigonal planar

✔ Answer:
- Lewis Structure: S with three double bonds to O atoms; each O has two lone pairs.
- Molecular Structure: Trigonal planar
- Molecular Geometry: Trigonal planar

---

🔹 3. HCN (Hydrogen cyanide)

#### Step 1: Valence electrons
- H: 1
- C: 4
- N: 5
- Total = 10

#### Step 2: Lewis Structure
- C is central (H only bonds to C, N bonds to C).
- H–C≡N → triple bond between C and N.

Check:
- H–C: single bond (2 e⁻)
- C≡N: triple bond (6 e⁻)
- Remaining electrons: 10 - 8 = 2 → place on N as a lone pair.

But N has 3 bonds (triple bond) + 1 lone pair = 8 electrons → OK.

C has 4 bonds → OK.

Structure:

```
H–C≡N:
```

(N has one lone pair)

✔ Lewis Structure: H–C≡N with lone pair on N.

#### Electron Geometry: Linear (only 2 electron domains: H–C and C≡N)

#### Molecular Geometry: Linear

✔ Answer:
- Lewis Structure: H–C≡N with one lone pair on N.
- Molecular Structure: Linear
- Molecular Geometry: Linear

---

🔹 4. SiCl₄ (Silicon tetrachloride)

#### Step 1: Valence electrons
- Si: 4
- Cl: 7 × 4 = 28
- Total = 32

#### Step 2: Lewis Structure
- Si is central.
- Four single bonds to Cl atoms → 4×2 = 8 electrons used.
- Each Cl needs 6 more electrons → 3 lone pairs each → 4×6 = 24 electrons.
- Total: 8 + 24 = 32 → perfect.

Structure:

```
Cl
|
Cl–Si–Cl
|
Cl
```

Each Cl has three lone pairs.

#### Electron Geometry: 4 bonding domains → Tetrahedral

#### Molecular Geometry: Tetrahedral

✔ Answer:
- Lewis Structure: Si bonded to four Cl atoms; each Cl has three lone pairs.
- Molecular Structure: Tetrahedral
- Molecular Geometry: Tetrahedral

---

🔹 5. CO₃²⁻ (Carbonate ion)

#### Step 1: Valence electrons
- C: 4
- O: 6 × 3 = 18
- Charge: +2 electrons → total = 4 + 18 + 2 = 24 electrons

#### Step 2: Lewis Structure
- C is central.
- Three O atoms attached.
- Resonance structures: One double bond and two single bonds (with negative charges on single-bonded O’s).

Structure:

```
O⁻
|
O=C
|
O⁻
```

But with resonance: The double bond rotates among the three O atoms.

Each resonance structure has:
- One C=O double bond
- Two C–O⁻ single bonds (each with negative charge)
- C has 4 bonds → OK
- Each O has 8 electrons: double-bonded O has 2 lone pairs, single-bonded O has 3 lone pairs.

Total electrons:
- Bonds: 1 double bond (4e⁻) + 2 single bonds (2×2 = 4e⁻) → 8e⁻ in bonds
- Lone pairs: 2 O have 3 lone pairs (6e⁻ each) → 12e⁻, one O has 2 lone pairs (4e⁻) → total = 16e⁻
- Total = 8 + 16 = 24 → correct.

Resonance makes all C–O bonds equivalent (partial double bond character).

#### Electron Geometry: 3 bonding domains → Trigonal planar

#### Molecular Geometry: Trigonal planar

✔ Answer:
- Lewis Structure: Central C bonded to three O atoms; one double bond, two single bonds with negative charges; resonance structures.
- Molecular Structure: Trigonal planar
- Molecular Geometry: Trigonal planar

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✔ Final Table Summary:

| Molecular Formula | Lewis Structure | Molecular Structure | Molecular Geometry |
|-------------------|------------------|----------------------|--------------------|
| H₂O | O with two single bonds to H and two lone pairs | Tetrahedral (electron) | Bent |
| SO₃ | S with three double bonds to O (no lone pairs on S) | Trigonal planar | Trigonal planar |
| HCN | H–C≡N with lone pair on N | Linear | Linear |
| SiCl₄ | Si bonded to four Cl atoms (each Cl has 3 lone pairs) | Tetrahedral | Tetrahedral |
| CO₃²⁻ | Resonance: C with one double bond and two single bonds to O⁻ | Trigonal planar | Trigonal planar |

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📌 Note: For drawing, you should sketch:
- Dots for lone pairs
- Lines for bonds
- Include formal charges if needed (especially in CO₃²⁻ and SO₃)

Let me know if you'd like a visual version of the Lewis structures!