Chemistry worksheet for practicing Lewis dot symbols and structures of elements and molecules.

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⁻
- Leftover: 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 for S)

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

---

#### 2. NOBr (Nitryl bromide)

- N: 5
- O: 6
- Br: 7
- Total = 5 + 6 + 7 = 18 electrons
- Central atom: N (least electronegative, but O and Br are both more electronegative; typically N is central)
- Try structure: O–N–Br
- Bonding: likely double bond between N and O, single bond between N and Br
- N=O (double bond: 4 e⁻), N–Br (single: 2 e⁻) → 6 e⁻ used
- Remaining: 18 - 6 = 12 e⁻ → lone pairs
- O (double bonded): 2 lone pairs (4 e⁻)
- Br: 3 lone pairs (6 e⁻)
- N: only 8 e⁻ so far (4 from double bond, 2 from single bond) → needs one lone pair (2 e⁻)
- But 4 (O) + 6 (Br) + 2 (N) = 12 → perfect!

But check formal charges:
- N: 5 - (0 + 4) = +1 (bonded to 4 e⁻ in bonds: 2 from single, 4 from double → 6/2 = 3 bonds → valence = 5, assigned = 3 → FC = +2? Wait — let’s recalculate properly.

Better approach: assign electrons.

Final structure:
```
..
:O=N-Br:
.. ..
```
But better to show:
- Double bond between N and O
- Single bond between N and Br
- Lone pairs:
- O: 2 lone pairs (4 e⁻)
- Br: 3 lone pairs (6 e⁻)
- N: 1 lone pair (2 e⁻)

Total electrons:
- Bonds: 4 (N=O) + 2 (N–Br) = 6 e⁻
- Lone pairs: O(4) + Br(6) + N(2) = 12 → total 18 ✔️

Formal charges:
- O: 6 - (4 + 2) = 0
- N: 5 - (2 + 3) = 0? Wait:
- N has 2 lone e⁻ (1 pair), and 3 bonds (double bond counts as 2, single as 1 → total 3 bonds) → bonding electrons = 6 → half = 3
- Formal charge = 5 - (2 + 3) = 0
- Br: 7 - (6 + 1) = 0

So neutral molecule. Good.

✔ Structure:
```
..
:O=N-Br:
.. ..
```
With lone pairs on O (2), Br (3), and one lone pair on N.

---

#### 3. N₂O (Nitrous oxide)

- N: 5 × 2 = 10
- O: 6
- Total = 16 electrons
- Linear molecule: possible structures: N≡N–O or N=N=O or N–N≡O
- Most stable: N≡N–O with resonance

Try: N≡N–O
- N≡N: 6 e⁻
- N–O: 2 e⁻ → total bonding: 8 e⁻
- Remaining: 16 - 8 = 8 e⁻ → lone pairs
- Terminal N: 1 lone pair (2 e⁻)
- O: 3 lone pairs (6 e⁻)
- Middle N: no lone pairs (already 8 e⁻: 3 bonds = 6 e⁻, plus 2 from triple bond → wait, triple bond gives 6 e⁻, single bond gives 2 → total 8 e⁻)

But terminal N has 3 bonds → 6 e⁻ from bonds → needs 2 more → 1 lone pair → good.

O has 1 bond → needs 3 lone pairs → 6 e⁻ → good.

But formal charges:
- Terminal N (left): 5 - (2 + 3) = 0
- Middle N: 5 - (0 + 4) = +1 (has 4 bonds: triple bond = 3, single bond = 1 → 4 bonds → 8 bonding e⁻ → half = 4)
- O: 6 - (6 + 1) = -1

So charge: +1 on middle N, -1 on O → net zero.

Alternative: N=N=O with formal charges better?

Actually, best structure is N≡N⁺–O⁻ or ⁻N=N⁺=O

Resonance structures:
1. `:N≡N⁺–O⁻:`
2. `⁻N=N⁺=O:`

But most common is :N≡N⁺–O⁻: with lone pairs.

✔ Best Lewis structure:
```
.. ..
:N≡N⁺–O⁻:
.. ..
```
- Left N: 1 lone pair, triple bond to center N
- Center N: triple bond to left N, single bond to O, no lone pairs → formal charge +1
- O: 3 lone pairs, single bond → formal charge -1

Total electrons:
- Triple bond: 6 e⁻
- Single bond: 2 e⁻
- Lone pairs: N (2) + O (6) = 8 → total 16 ✔️

---

#### 4. SiF₃⁻ (Trifluorosilicate ion)

- 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⁻
- Each F gets 3 lone pairs → 3 × 6 = 18 e⁻
- Left: 20 - 18 = 2 e⁻ → 1 lone pair on Si
- Si has 3 bonds (6 e⁻) + 1 lone pair (2 e⁻) = 8 e⁻ → octet complete
- Negative charge on Si

✔ Structure:
```
..
:F:
/
:F--Si--F:
\
..
```
With 3 lone pairs on each F, 1 lone pair on Si, and -1 charge.

---

#### 5. BrF₃ (Bromine trifluoride)

- Br: 7
- F: 7 × 3 = 21
- Total = 28 e⁻
- Br central
- 3 single bonds → 6 e⁻ used
- Remaining: 22 e⁻ → lone pairs
- Each F: 3 lone pairs → 3 × 6 = 18 e⁻
- Left: 22 - 18 = 4 e⁻ → 2 lone pairs on Br
- Br has 3 bonds + 2 lone pairs → 10 e⁻ → expanded octet (allowed)

✔ Structure:
```
..
:F:
/
:F--Br--F:
\
..
```
With 2 lone pairs on Br, 3 lone pairs on each F.

---

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

- S: 6
- Cl: 7 × 3 = 21
- Subtract 1 e⁻ for +1 charge → total = 6 + 21 - 1 = 26 electrons
- S central
- 3 single bonds → 6 e⁻ used
- Remaining: 20 e⁻
- Each Cl: 3 lone pairs → 3 × 6 = 18 e⁻
- Left: 20 - 18 = 2 e⁻ → 1 lone pair on S
- S has 3 bonds (6 e⁻) + 1 lone pair (2 e⁻) = 8 e⁻ → octet complete
- Positive charge on S

✔ Structure:
```
..
:Cl:
/
:Cl--S--Cl:
\
..
```
With one lone pair on S, 3 lone pairs on each Cl, +1 charge.

---

#### 7. CO₃²⁻ (Carbonate ion)

- C: 4
- O: 6 × 3 = 18
- Add 2 e⁻ for -2 charge → total = 4 + 18 + 2 = 24 electrons
- C central
- Three O atoms attached
- Resonance: one double bond, two single bonds with formal charges

Structure:
- One C=O double bond
- Two C–O⁻ single bonds
- Each single-bonded O has 3 lone pairs (6 e⁻), double-bonded O has 2 lone pairs (4 e⁻)
- C has no lone pairs

Electron count:
- Double bond: 4 e⁻
- Two single bonds: 4 e⁻
- Lone pairs:
- Double-bonded O: 4 e⁻
- Two single-bonded O: 2 × 6 = 12 e⁻
→ total = 4 + 4 + 4 + 12 = 24 ✔️

Formal charges:
- C: 4 - (0 + 4) = 0
- Double-bonded O: 6 - (4 + 2) = 0
- Single-bonded O: 6 - (6 + 1) = -1 → two such O → total -2

✔ Resonance structure:
```
..
:O:
||
:O--C--O⁻:
| |
.. ..
```
With resonance: double bond rotates among three O atoms.

---

#### 8. NH₄⁺ (Ammonium ion)

- N: 5
- H: 1 × 4 = 4
- Subtract 1 e⁻ for +1 charge → total = 5 + 4 - 1 = 8 electrons
- N central
- Four single bonds to H → 4 × 2 = 8 e⁻ → all electrons used
- No lone pairs on N
- Formal charge: N: 5 - (0 + 4) = +1

✔ 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 attached
- Two S=O double bonds, two S–O⁻ single bonds
- Each single-bonded O has 3 lone pairs (6 e⁻), double-bonded O has 2 lone pairs (4 e⁻)
- S has no lone pairs

Electron count:
- 2 double bonds: 2 × 4 = 8 e⁻
- 2 single bonds: 2 × 2 = 4 e⁻
- Lone pairs:
- 2 double-bonded O: 2 × 4 = 8 e⁻
- 2 single-bonded O: 2 × 6 = 12 e⁻
→ total = 8 + 4 + 8 + 12 = 32 ✔️

Formal charges:
- S: 6 - (0 + 6) = 0
- Double-bonded O: 6 - (4 + 2) = 0
- Single-bonded O: 6 - (6 + 1) = -1 → two → total -2

✔ Resonance structure:
```
O
║
O--S--O⁻
║
O⁻
```
But actually, double bonds can resonate among any two O atoms.

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✔ Final Summary Table

| Molecule | Lewis Structure Description |
|--------|----------------------------|
| SF₃⁻ | S central, bonded to 3 F, 2 lone pairs on S, 3 lone pairs on each F, -1 charge |
| NOBr | N central, double bond to O, single bond to Br, lone pairs on O (2), Br (3), N (1) |
| N₂O | Resonance: `:N≡N⁺–O⁻:` or `⁻N=N⁺=O` |
| SiF₃⁻ | Si central, 3 F, 1 lone pair on Si, 3 lone pairs on each F, -1 charge |
| BrF₃ | Br central, 3 F, 2 lone pairs on Br, 3 lone pairs on each F |
| SCl₃⁺ | S central, 3 Cl, 1 lone pair on S, 3 lone pairs on each Cl, +1 charge |
| CO₃²⁻ | C central, 3 O, one double bond, two single bonds with -1 formal charge on each, resonance |
| NH₄⁺ | N central, 4 H, no lone pairs, +1 charge |
| SO₄²⁻ | S central, 4 O, two double bonds, two single bonds with -1 charge, resonance |

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Would you like me to provide a visual ASCII version of these structures for clarity?