Printable chemistry worksheet for practicing Lewis dot symbols and structures.

Let's solve this step-by-step. We'll begin with Part 1: Electron Dot Structures for Elements, then move to Part 2: Lewis Structures for Molecules.

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🔹 Part 1: Electron Dot Structures (Lewis Symbols)

We draw the valence electrons as dots around the element symbol. The number of valence electrons is determined by the group number in the periodic table.

| Element | Group | Valence Electrons | Lewis Dot Structure |
|----------|-------|-------------------|---------------------|
| Nitrogen (N) | 15 | 5 | :N· (with 5 dots: one on each side, and one lone pair on top) |
| Bromine (Br) | 17 | 7 | :Br: (with 7 dots: three lone pairs and one single dot) |
| Oxygen (O) | 16 | 6 | :Ö: (with 6 dots: two lone pairs and two single dots) |
| Carbon (C) | 14 | 4 | ·C· (with 4 dots: one on each side) |
| Sulfur (S) | 16 | 6 | :S: (same as oxygen, 6 dots) |
| Fluorine (F) | 17 | 7 | :F: (same as bromine, 7 dots) |

#### ✔ Drawn Lewis Symbols:
```
Nitrogen (N): ·
·N·
·

Bromine (Br): ··
:Br·
··

Oxygen (O): ··
:O·
··

Carbon (C): ·
·C·
·

Sulfur (S): ··
:S·
··

Fluorine (F): ··
:F·
··
```

> Note: Dots are placed on four sides (top, right, bottom, left), with no more than two dots per side.

---

🔹 Part 2: Lewis Structures for Molecules

Now we’ll draw full Lewis structures using bonding and lone pairs. Remember:

- Total valence electrons = sum of valence e⁻ from all atoms + electrons from negative charge – electrons removed for positive charge.
- Bonds are shared pairs (lines), lone pairs are unshared dots.
- Octet rule applies (except H: duet rule).

---

#### 1. SF₃⁻ (Sulfur trifluoride anion)

- S: 6 valence e⁻
- F: 7 × 3 = 21
- Add 1 e⁻ for negative charge → total = 6 + 21 + 1 = 28 electrons
- S is central atom (less electronegative)
- Each F forms a single bond with S → 3 bonds = 6 electrons used
- Remaining electrons: 28 - 6 = 22 → distribute as lone pairs
- Each F needs 3 lone pairs (6 e⁻) → 3 F × 6 = 18 e⁻
- Left: 22 - 18 = 4 e⁻ → 2 lone pairs on S
- S has 3 bonds (6 e⁻) + 2 lone pairs (4 e⁻) = 10 e⁻ → expanded octet allowed (S can have >8)

✔ Structure:
```
F
|
F--S--F
:
:
```
With 3 lone pairs on each F, and 2 lone pairs on S. Charge: -1 on the ion.

---

#### 2. NOBr (Nitryl bromide)

- N: 5
- O: 6
- Br: 7
- Total: 5 + 6 + 7 = 18 electrons

Central atom: N (least electronegative, except O/Br)

Try structure: N bonded to O and Br.

But N has only 2 bonds → need double bond or resonance.

Better: N=O double bond, N–Br single bond.

- N=O: 4 e⁻
- N–Br: 2 e⁻
- Total bonding: 6 e⁻ → 12 e⁻ left
- O: already has double bond → needs 2 lone pairs (4 e⁻)
- Br: single bond → needs 3 lone pairs (6 e⁻)
- So far: 4 (O) + 6 (Br) = 10 e⁻ used → 12 - 10 = 2 e⁻ left → place on N as lone pair

Now check formal charges:

- N: valence = 5, bonds = 4 (double bond counts as 2), lone pair = 1 → FC = 5 - 4 - 1 = 0
- O: 6 - 4 - 2 = 0
- Br: 7 - 1 - 6 = 0

Good! But wait — total electrons:
- N=O: 4 e⁻
- N–Br: 2 e⁻
- O lone pairs: 4 e⁻
- Br lone pairs: 6 e⁻
- N lone pair: 2 e⁻
→ 4+2+4+6+2 = 18 ✓

✔ Structure:
```
O
║
N--Br
:
```
With 2 lone pairs on O, 3 on Br, and 1 lone pair on N.

---

#### 3. N₂O (Nitrous oxide)

- N: 5 × 2 = 10
- O: 6
- Total: 16 electrons

Structure: N≡N–O or N=N=O? Try both.

Best: N≡N–O with formal charges.

Try: N≡N–O, with O having 3 lone pairs.

- N≡N: 6 e⁻
- N–O: 2 e⁻
- Total bonding: 8 e⁻ → 8 e⁻ left
- O: 3 lone pairs = 6 e⁻
- One N (terminal): triple bond → needs 1 lone pair = 2 e⁻
- Central N: 4 bonds (triple + single) → no lone pairs
- Total electrons: 8 (bonds) + 6 (O) + 2 (end N) = 16 ✓

Formal charges:
- Terminal N (≡): 5 - 3 - 2 = 0
- Central N: 5 - 4 - 0 = +1
- O: 6 - 1 - 6 = -1

So overall charge: 0 → good.

Alternative: N=N=O with resonance.

But most stable is N≡N⁺–O⁻

✔ Structure:
```
:N≡N–O:
:
```
With lone pairs:
- Terminal N: 1 lone pair
- O: 3 lone pairs
- Central N: no lone pairs
- Charge: N⁺ and O⁻

---

#### 4. SiF₃⁻ (Silicon trifluoride anion)

- Si: 4
- F: 7 × 3 = 21
- Add 1 e⁻ → total = 4 + 21 + 1 = 26 electrons

Si central. 3 single bonds to F → 6 e⁻ used

Remaining: 26 - 6 = 20 e⁻ → 3 F × 6 e⁻ (3 lone pairs) = 18 e⁻ → 2 e⁻ left → 1 lone pair on Si

Si has 3 bonds + 1 lone pair → 8 e⁻ → octet satisfied

Charge: -1 on ion

✔ Structure:
```
F
|
F--Si--F
:
```
Each F has 3 lone pairs, Si has 1 lone pair.

---

#### 5. BrF₃ (Bromine trifluoride)

- Br: 7
- F: 7 × 3 = 21
- Total: 28 e⁻

Br central. 3 single bonds → 6 e⁻ used

Remaining: 22 e⁻ → 3 F × 6 = 18 e⁻ → 4 e⁻ left → 2 lone pairs on Br

Br has 3 bonds + 2 lone pairs = 10 e⁻ → expanded octet

✔ Structure:
```
F
|
F--Br--F
:
:
```
Each F has 3 lone pairs, Br has 2 lone pairs.

---

#### 6. SCl₃⁺ (Sulfur trichloride cation)

- S: 6
- Cl: 7 × 3 = 21
- Subtract 1 e⁻ for + charge → total = 6 + 21 - 1 = 26 electrons

S central. 3 single bonds → 6 e⁻ used

Remaining: 20 e⁻ → 3 Cl × 6 = 18 e⁻ → 2 e⁻ left → 1 lone pair on S

S has 3 bonds + 1 lone pair → 8 e⁻ → octet satisfied

Charge: +1 on S

✔ Structure:
```
Cl
|
Cl--S--Cl
:
```
Each Cl has 3 lone pairs, S has 1 lone pair.

---

#### 7. CO₃²⁻ (Carbonate ion)

- C: 4
- O: 6 × 3 = 18
- Add 2 e⁻ for charge → total = 4 + 18 + 2 = 24 electrons

C central. Three O atoms.

Resonance structures: one double bond, two single bonds with formal charges.

Try: C=O (double bond), two C–O⁻ (single bonds with extra electron)

- C=O: 4 e⁻
- Two C–O: 4 e⁻
- Total bonding: 8 e⁻ → 16 e⁻ left

Lone pairs:
- Double-bonded O: 2 lone pairs (4 e⁻)
- Each single-bonded O⁻: 3 lone pairs (6 e⁻) → 2 × 6 = 12 e⁻
- Total lone pairs: 4 + 12 = 16 e⁻ → perfect

Formal charges:
- C: 4 - 4 (bonds) - 0 = 0
- Double-bonded O: 6 - 4 - 2 = 0
- Single-bonded O: 6 - 1 - 6 = -1 → two of them → total -2 → matches charge

✔ Resonance structure:
```
O⁻
|
O=C
\
O⁻
```
With lone pairs:
- Double-bonded O: 2 lone pairs
- Each single-bonded O: 3 lone pairs

(Three resonance forms; double bond rotates)

---

#### 8. NH₄⁺ (Ammonium ion)

- N: 5
- H: 1 × 4 = 4
- Subtract 1 e⁻ → total = 5 + 4 - 1 = 8 electrons

N central. Four single bonds to H → 8 e⁻ used → all electrons used

No lone pairs on N

Formal charge on N: 5 - 4 - 0 = +1 → matches ion charge

✔ Structure:
```
H
|
H--N--H
|
H
```
All single bonds, no lone pairs, +1 charge on N

---

#### 9. SO₄²⁻ (Sulfate ion)

- S: 6
- O: 6 × 4 = 24
- Add 2 e⁻ → total = 6 + 24 + 2 = 32 electrons

S central. Four O atoms.

Typical structure: two S=O double bonds, two S–O⁻ single bonds (with resonance)

- Two S=O: 8 e⁻
- Two S–O: 4 e⁻
- Total bonding: 12 e⁻ → 20 e⁻ left

Lone pairs:
- Double-bonded O: 2 lone pairs each → 2 × 4 = 8 e⁻
- Single-bonded O⁻: 3 lone pairs each → 2 × 6 = 12 e⁻
- Total lone pairs: 8 + 12 = 20 e⁻ → perfect

Formal charges:
- S: 6 - 4 (bonds) - 0 = +2? Wait — actually S has 4 bonds (2 double, 2 single) → 8 bonds → 8 e⁻ → FC = 6 - 8 - 0 = -2? No.

Wait — better: each bond counts as 1 electron for FC.

- S: 6 valence, 8 bonds → 8 bonding e⁻, 0 lone pairs → FC = 6 - 8 - 0 = -2 → not good.

Actually, sulfate uses resonance with formal charges.

Standard: S with 4 bonds (no lone pairs), two double bonds, two single bonds to O⁻.

Then:
- S: 6 - 4 (bonds) - 0 = +2
- Double-bonded O: 6 - 4 - 2 = 0
- Single-bonded O: 6 - 1 - 6 = -1 → two of them → -2
- Total charge: +2 -2 = 0 → but ion is -2 → wrong.

Ah — mistake: we need total charge -2.

Better: all four bonds are equivalent via resonance — two double bonds, two single bonds with charges.

But correct: S has expanded octet — 12 electrons.

Standard Lewis structure:

- S bonded to four O atoms
- Two S=O double bonds
- Two S–O⁻ single bonds
- S has no lone pairs
- Formal charge on S: 6 - 8 - 0 = -2 → too low

Wait — let’s recalculate properly.

In reality, sulfate has four resonance structures, each with two double bonds.

But formal charge:
- S: 6 valence
- Bonds: 8 bonding electrons → count as 4 bonds → FC = 6 - 4 - 0 = +2
- Each double-bonded O: 6 - 4 - 2 = 0
- Each single-bonded O⁻: 6 - 1 - 6 = -1 → two → -2
- Total: +2 -2 = 0 → but ion is -2 → missing.

Wait — we forgot the extra electrons.

Total valence electrons:
- S: 6
- O: 4×6 = 24
- Add 2 → 32 e⁻

Use:
- 4 single bonds: 8 e⁻
- 4 lone pairs on O: 4×6 = 24 e⁻ → total 32 e⁻

But that gives all single bonds → S would have 8 e⁻, each O has 8 e⁻ → but then S has FC = 6 - 4 - 0 = +2, each O has FC = 6 - 1 - 6 = -1 → total charge = +2 -4 = -2 → yes!

So: four single bonds, S with no lone pairs, each O has three lone pairs, and two O atoms carry -1 charge.

But this gives S only 8 e⁻ → octet, but it's possible.

However, actual structure has two double bonds to reduce formal charge.

Better: two S=O double bonds, two S–O⁻ single bonds

- Bonds: 2×4 + 2×2 = 8 + 4 = 12 e⁻
- Lone pairs:
- Double-bonded O: 2 lone pairs → 4 e⁻ each → 2×4 = 8
- Single-bonded O⁻: 3 lone pairs → 6 e⁻ each → 2×6 = 12
- Total lone pairs: 8 + 12 = 20
- Total electrons: 12 + 20 = 32 ✓

Formal charges:
- S: 6 - 8 - 0 = -2? No — bonding electrons: 8 → count as 4 bonds → FC = 6 - 4 - 0 = +2
- Double-bonded O: 6 - 4 - 2 = 0
- Single-bonded O: 6 - 1 - 6 = -1 → two → -2
- Total: +2 -2 = 0 → but ion is -2 → contradiction.

Wait — the total charge is -2, so we must have net -2.

But if we have two O⁻, that’s -2, and S at +2 → net 0 → not good.

Ah! Mistake: In sulfate, the sulfur is in +6 oxidation state, so formal charge is often +2.

But total charge is -2 → so the oxygens must contribute -4?

No — let's fix:

Correct approach:

In sulfate, the best Lewis structure has:

- S bonded to 4 O atoms
- Two S=O double bonds
- Two S–O⁻ single bonds
- But S has no lone pairs
- Each O has 8 e⁻

Electron count:
- 2× S=O: 8 e⁻
- 2× S–O: 4 e⁻
- Total bonds: 12 e⁻
- Lone pairs:
- Double-bonded O: 2 lone pairs → 4 e⁻ each → 8 e⁻
- Single-bonded O⁻: 3 lone pairs → 6 e⁻ each → 12 e⁻
- Total: 12 + 8 + 12 = 32 ✓

Formal charges:
- S: 6 valence - 4 bonds (since 8 bonding e⁻ → 4 bonds) - 0 lone pairs = +2
- Double-bonded O: 6 - 4 - 2 = 0
- Single-bonded O: 6 - 1 - 6 = -1 → two → -2
- Total charge: +2 -2 = 0 → but should be -2 → ✘

Wait — this is wrong.

Ah! The ion is SO₄²⁻, so total charge -2, so we need net -2.

But if S is +2 and two O are -1 each → +2 -2 = 0 → not -2.

So where is the error?

Answer: The sulfur has expanded octet — it has 12 electrons.

Better: Two double bonds, two single bonds, but the single-bonded oxygens have -1 charge, and sulfur has +2 formal charge → net: +2 -2 = 0 → still not -2.

Wait — we are missing something.

Let’s do it again:

Total valence electrons:
- S: 6
- 4×O: 24
- +2 electrons from charge → 32 electrons

Now, suppose:
- 4 single bonds: 8 e⁻
- Then remaining: 24 e⁻ → 6 lone pairs per O → 4×6 = 24 e⁻ → total 32

So: all single bonds, S has 8 e⁻, each O has 8 e⁻

Then formal charges:
- S: 6 - 4 - 0 = +2
- Each O: 6 - 1 - 6 = -1 → 4×(-1) = -4
- Total: +2 -4 = -2 → ✔

So this structure has S with +2 formal charge, each O with -1, total charge -2.

But this violates octet? No — S has 8 e⁻.

But this is not the most stable — real sulfate has resonance with double bonds.

But in Lewis terms, we can draw resonance structures where two O are double bonded (0 charge), two are single bonded (-1), and S is +2 → total charge: +2 -2 = 0 → ✘

No — that doesn't work.

Ah! The key: when you have a double bond, the oxygen has only two lone pairs, so it has 0 formal charge.

So:

- S: 6 valence
- Bonds: 8 bonding e⁻ → 4 bonds → FC = 6 - 4 - 0 = +2
- Double-bonded O: 6 - 4 - 2 = 0
- Single-bonded O: 6 - 1 - 6 = -1 → two → -2
- Total: +2 -2 = 0 → not -2

But we need total charge -2.

So this structure cannot exist.

Wait — I made a mistake in counting electrons.

Let’s try:

Suppose two double bonds and two single bonds.

- Bonding electrons: 2×4 (double) + 2×2 (single) = 8 + 4 = 12 e⁻
- Lone pairs:
- Double-bonded O: 2 lone pairs → 4 e⁻ each → 8 e⁻
- Single-bonded O: 3 lone pairs → 6 e⁻ each → 12 e⁻
- Total: 12 + 8 + 12 = 32 e⁻ ✓

Now formal charges:
- S: 6 - 8 - 0 = -2? No — formal charge = valence - nonbonding - ½ bonding
- Nonbonding = 0
- Bonding = 12 e⁻ → 6 bonds → ½×12 = 6
- FC = 6 - 0 - 6 = 0
- Double-bonded O: 6 - 4 - ½×4 = 6 - 4 - 2 = 0
- Single-bonded O: 6 - 6 - ½×2 = 6 - 6 - 1 = -1 → two → -2
- Total charge: 0 + 0 + (-2) = -2 → ✔

Yes! So formal charge on S is 0.

How?

- S has 12 bonding electrons → 6 bonds → ½×12 = 6 electrons counted
- Valence = 6
- Nonbonding = 0
- FC = 6 - 0 - 6 = 0

Perfect.

So the structure is:
- Two S=O double bonds
- Two S–O⁻ single bonds
- S has no lone pairs
- Each double-bonded O has 2 lone pairs
- Each single-bonded O has 3 lone pairs
- Total charge: -2

And resonance: the double bonds can rotate.

✔ Final structure:
```
O⁻
|
O=S=O
|
O⁻
```
With lone pairs:
- Double-bonded O: 2 lone pairs
- Single-bonded O: 3 lone pairs
- S: no lone pairs
- Resonance between positions

---

✔ Summary of Answers

#### Part 1: Lewis Symbols

| Element | Lewis Symbol |
|----------|--------------|
| Nitrogen | :N· (5 dots) |
| Bromine | :Br: (7 dots) |
| Oxygen | :O: (6 dots) |
| Carbon | ·C· (4 dots) |
| Sulfur | :S: (6 dots) |
| Fluorine | :F: (7 dots) |

#### Part 2: Lewis Structures

1. SF₃⁻:
```
F
|
F-S-F
:
:
```
S has 2 lone pairs, each F has 3 lone pairs, charge -1

2. NOBr:
```
O
║
N-Br
:
```
N has 1 lone pair, O has 2, Br has 3

3. N₂O:
```
:N≡N-O:
:
```
Resonance: N≡N⁺–O⁻

4. SiF₃⁻:
```
F
|
F-Si-F
:
```
Si has 1 lone pair, each F has 3 lone pairs

5. BrF₃:
```
F
|
F-Br-F
:
:
```
Br has 2 lone pairs

6. SCl₃⁺:
```
Cl
|
Cl-S-Cl
:
```
S has 1 lone pair

7. CO₃²⁻:
```
O⁻
|
O=C
\
O⁻
```
Resonance: double bond rotates

8. NH₄⁺:
```
H
|
H-N-H
|
H
```
All single bonds, no lone pairs, +1 charge

9. SO₄²⁻:
```
O⁻
|
O=S=O
|
O⁻
```
Resonance: two double bonds, two single bonds with -1 charge

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Let me know if you'd like these drawn in ASCII or visual format!