Free Printable Lewis Dot Structure Worksheets - Free Printable

Let's go through each compound and draw its Lewis structure, step by step, based on the rules of Lewis structures:

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🔹 Rules for Drawing Lewis Structures:

1. Count total valence electrons.
2. Place the least electronegative atom in the center (except H).
3. Connect atoms with single bonds (each bond = 2 electrons).
4. Distribute remaining electrons to satisfy octets (duet for H).
5. If needed, form double or triple bonds to satisfy octets.

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Now let’s solve each one:

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1. H₂

- Valence electrons: H has 1 → 2 × 1 = 2
- Two H atoms share a single bond.
- Structure: H–H
(Each H has 2 electrons — duet rule satisfied)

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2. HBr

- H: 1, Br: 7 → Total = 8 e⁻
- H bonded to Br with a single bond (2 e⁻)
- Remaining 6 e⁻ go on Br as lone pairs (3 pairs)
- Structure: H–Br: (with three lone pairs on Br)

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3. H₂O

- O: 6, H: 1×2 = 2 → Total = 8 e⁻
- O is central, bonded to two H atoms (two single bonds = 4 e⁻ used)
- Remaining 4 e⁻ → two lone pairs on O
- Structure:
```
H
|
:O:
|
H
```
(Two lone pairs on O)

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4. CO₂

- C: 4, O: 6×2 = 12 → Total = 16 e⁻
- C is central, bonded to two O atoms
- To satisfy octets: C=O double bonds
- Each O has two lone pairs
- Structure: O=C=O
(Each O has two lone pairs)

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5. O₃ (Ozone)

- 3 O atoms → 6×3 = 18 e⁻
- Central O bonded to two terminal O atoms
- One double bond, one single bond (resonance)
- Lone pairs: terminal O (double-bonded) has 2 lone pairs; single-bonded O has 3 lone pairs; central O has 1 lone pair
- Resonance structures:
```
O=O–O ↔ O–O=O
```
(With lone pairs adjusted accordingly)

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6. CHCl₃ (Chloroform)

- C: 4, H: 1, Cl: 7×3 = 21 → Total = 4 + 1 + 21 = 26 e⁻
- C is central, bonded to H and three Cl atoms
- Four single bonds → 8 e⁻ used
- Remaining 18 e⁻ → 3 lone pairs per Cl (3 Cl × 6 e⁻ = 18)
- Structure:
```
Cl
|
Cl–C–H
|
Cl
```
(Each Cl has 3 lone pairs)

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7. CO (Carbon monoxide)

- C: 4, O: 6 → Total = 10 e⁻
- Triple bond between C and O (6 e⁻), plus one lone pair on each
- C has 2 e⁻ (lone pair), O has 2 e⁻ (lone pair)
- Structure: C≡O with lone pairs on both ends

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8. C₂H₂ (Acetylene)

- C: 4×2 = 8, H: 1×2 = 2 → Total = 10 e⁻
- Linear molecule: H–C≡C–H
- Triple bond between C atoms, each C bonded to one H
- No lone pairs
- Structure: H–C≡C–H

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9. NCl₃

- N: 5, Cl: 7×3 = 21 → Total = 26 e⁻
- N central, bonded to 3 Cl atoms → 3 single bonds = 6 e⁻
- Remaining 20 e⁻ → 3 lone pairs on each Cl (3×6 = 18), and 1 lone pair on N
- Structure:
```
Cl
|
Cl–N–Cl
|
(lone pair on N)
```

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10. PH₃

- P: 5, H: 1×3 = 3 → Total = 8 e⁻
- P central, bonded to 3 H atoms → 3 single bonds = 6 e⁻
- Remaining 2 e⁻ → one lone pair on P
- Structure:
```
H
|
H–P–H
|
(lone pair on P)
```

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11. AlH₃

- Al: 3, H: 1×3 = 3 → Total = 6 e⁻
- Al central, bonded to 3 H atoms → 3 single bonds = 6 e⁻
- Al has only 6 electrons → electron-deficient (no octet)
- Structure:
```
H
|
H–Al–H
```
(No lone pairs; Al has incomplete octet)

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12. C₂H₆ (Ethane)

- C: 4×2 = 8, H: 1×6 = 6 → Total = 14 e⁻
- Two C atoms bonded together (single bond), each bonded to 3 H atoms
- All single bonds → no lone pairs
- Structure:
```
H H
| |
H–C–C–H
| |
H H
```

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13. IBr (Iodine bromide)

- I: 7, Br: 7 → Total = 14 e⁻
- Single bond between I and Br → 2 e⁻
- Remaining 12 e⁻ → 3 lone pairs on each atom
- Structure: I–Br
(Each has 3 lone pairs)

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14. NI₃ (Nitrogen triiodide)

- N: 5, I: 7×3 = 21 → Total = 26 e⁻
- N central, bonded to 3 I atoms → 3 single bonds = 6 e⁻
- Remaining 20 e⁻ → 3 lone pairs on each I (3×6 = 18), and 1 lone pair on N
- Structure:
```
I
|
I–N–I
|
(lone pair on N)
```

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15. CH₂F₂ (Difluoromethane)

- C: 4, H: 1×2 = 2, F: 7×2 = 14 → Total = 20 e⁻
- C central, bonded to 2 H and 2 F
- Four single bonds → 8 e⁻ used
- Remaining 12 e⁻ → 3 lone pairs on each F (2×6 = 12)
- Structure:
```
H
|
F–C–F
|
H
```
(Each F has 3 lone pairs)

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✔ Summary Table (Lewis Structure Descriptions):

| Compound | Lewis Structure Description |
|--------|----------------------------|
| H₂ | H–H |
| HBr | H–Br with 3 lone pairs on Br |
| H₂O | H–O–H with 2 lone pairs on O |
| CO₂ | O=C=O with 2 lone pairs on each O |
| O₃ | Resonance: O=O–O ↔ O–O=O (lone pairs adjusted) |
| CHCl₃ | C bonded to H and 3 Cl; each Cl has 3 lone pairs |
| CO | C≡O with lone pair on each atom |
| C₂H₂ | H–C≡C–H |
| NCl₃ | N bonded to 3 Cl; each Cl has 3 lone pairs; N has 1 lone pair |
| PH₃ | P bonded to 3 H; P has 1 lone pair |
| AlH₃ | Al bonded to 3 H; no lone pairs; electron-deficient |
| C₂H₆ | H₃C–CH₃ (all single bonds) |
| IBr | I–Br; each has 3 lone pairs |
| NI₃ | N bonded to 3 I; each I has 3 lone pairs; N has 1 lone pair |
| CH₂F₂ | C bonded to 2 H and 2 F; each F has 3 lone pairs |

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📌 Notes:

- Use dots (:) for lone pairs.
- Lines (=, –, ≡) represent bonds.
- For resonance (like O₃), draw both major structures.
- Hydrogen always forms one bond (duet rule).

Let me know if you'd like a visual version (ASCII art or diagrams) for any specific molecule!