Worksheet featuring exercises on drawing Lewis structures for various chemical compounds.

You're asked to draw Lewis structures for five molecules: BSF, HBr, C₂H₅OH (ethanol), N₂F₄, and SF₆.

Let’s go through each one step by step, explaining the process and then providing the final Lewis structure description (since I can’t draw images, I’ll describe them clearly so you can sketch them).

---

1) BSF

This is a rare or possibly hypothetical molecule. Boron (B), sulfur (S), and fluorine (F). Let’s assume it's a linear triatomic molecule: B–S–F.

#### Step 1: Count valence electrons
- B: 3
- S: 6
- F: 7
→ Total = 3 + 6 + 7 = 16 valence electrons

#### Step 2: Skeleton structure
Assume central atom is S (least electronegative after B, but B often forms electron-deficient compounds). Try B–S–F.

#### Step 3: Distribute bonds and lone pairs
- Single bond B–S: 2 e⁻
- Single bond S–F: 2 e⁻
→ Used 4 e⁻ → 12 left

Now complete octets:
- F needs 6 more electrons → 3 lone pairs (6 e⁻)
- S already has 2 bonds (4 e⁻) → needs 4 more → 2 lone pairs (4 e⁻)
- B has only 2 e⁻ from bond → no octet, but boron is okay with incomplete octet.

Total used: 4 (bonds) + 6 (F lone pairs) + 4 (S lone pairs) = 14 e⁻

Left: 2 e⁻ → put on B as a lone pair? But B usually doesn’t hold lone pairs well in stable compounds.

Alternatively, maybe double bond B=S?

Try:
- B=S double bond: 4 e⁻
- S–F single bond: 2 e⁻ → total bonding: 6 e⁻
- F: 3 lone pairs (6 e⁻)
- S: now has 3 bonds (double + single = 3 bonds → 6 e⁻) → needs 2 more → 1 lone pair (2 e⁻)
- B: double bond = 4 e⁻ → needs 4 more → 2 lone pairs (4 e⁻)

Total: 6 (bonds) + 6 (F) + 2 (S) + 4 (B) = 18 e⁻ → too many!

So original single-bond version with 16 e⁻ is best.

✔ Final Lewis Structure for BSF:

```
..
F — S — B
.. :
```

- F has 3 lone pairs (6 e⁻)
- S has 2 lone pairs (4 e⁻) and 2 single bonds → 8 e⁻ total
- B has 1 single bond (2 e⁻) and 1 lone pair (2 e⁻) → 4 e⁻ total (incomplete octet — acceptable for B)

> Note: BSF is not a common stable compound; this is a theoretical structure based on valence rules.

---

2) HBr

Hydrogen bromide — simple diatomic molecule.

#### Valence electrons:
- H: 1
- Br: 7
→ Total = 8 valence electrons

#### Structure:
Single bond H–Br uses 2 e⁻ → 6 left → all go on Br as 3 lone pairs.

✔ Lewis Structure:

```
H — Br :
..
..
```

- Br has 3 lone pairs (6 e⁻) + 1 bond (2 e⁻) = 8 e⁻ (octet)
- H has 2 e⁻ (duet) — satisfied

---

3) C₂H₅OH (Ethanol)

Molecular formula: CH₃CH₂OH

#### Valence electrons:
- C: 4 × 2 = 8
- H: 1 × 6 = 6 (5 H in ethyl + 1 in OH)
- O: 6
→ Total = 8 + 6 + 6 = 20 valence electrons

#### Skeleton:
Carbon chain: C–C, with OH group on second carbon.

Structure: H₃C–CH₂–OH

#### Bonds:
- C–C: 1 bond → 2 e⁻
- Each C–H: 5 bonds → 10 e⁻
- C–O: 1 bond → 2 e⁻
- O–H: 1 bond → 2 e⁻
→ Total bonding electrons = 2+10+2+2 = 16 e⁻

Left: 20 – 16 = 4 e⁻ → place on O as 2 lone pairs.

✔ Lewis Structure:

```
H H
| |
H — C — C — O — H
| ..
H
```

With two lone pairs (..) on oxygen.

All atoms have complete octets (except H, which has duet).

---

4) N₂F₄

Dinitrogen tetrafluoride. Two nitrogen atoms bonded together, each bonded to two fluorines.

#### Valence electrons:
- N: 5 × 2 = 10
- F: 7 × 4 = 28
→ Total = 38 valence electrons

#### Structure:
Typical structure: F₂N–NF₂, with N–N single bond.

Each N bonded to 2 F atoms and 1 N.

#### Bonds:
- N–N: 1 bond → 2 e⁻
- 4 N–F bonds → 8 e⁻
→ Total bonding = 10 e⁻

Left: 38 – 10 = 28 e⁻ → distribute as lone pairs.

Each F needs 3 lone pairs → 4 F × 6 e⁻ = 24 e⁻

Left: 28 – 24 = 4 e⁻ → put 2 lone pairs (4 e⁻) on the two N atoms (2 e⁻ each).

Each N has:
- 3 bonds (1 to N, 2 to F) → 6 e⁻
- 1 lone pair → 2 e⁻
→ Total 8 e⁻ → octet satisfied.

✔ Lewis Structure:

```
F F
| |
F — N — N — F
: :
```

Each N has one lone pair (represented by `:`), each F has three lone pairs (not shown for simplicity, but implied).

Actual structure may have slight variations (like trans or cis), but this is the basic Lewis structure.

---

5) SF₆

Sulfur hexafluoride — classic hypervalent molecule.

#### Valence electrons:
- S: 6
- F: 7 × 6 = 42
→ Total = 48 valence electrons

#### Structure:
Sulfur central atom, surrounded by 6 fluorine atoms.

Each S–F bond: 6 bonds × 2 e⁻ = 12 e⁻

Left: 48 – 12 = 36 e⁻ → assign to F atoms.

Each F needs 3 lone pairs → 6 F × 6 e⁻ = 36 e⁻

Perfect!

✔ Lewis Structure:

Imagine sulfur in center, 6 fluorines around it (octahedral geometry).

Each F has 3 lone pairs.

```
F
|
F — S — F
|
F
```

(with two more F atoms above and below the plane — 3D octahedron)

Sulfur has 12 electrons around it (6 bonds) — expanded octet, allowed for period 3+ elements.

---

## ✔ Summary of All Lewis Structures:

1. BSF: B–S–F with lone pairs on S and F; B has incomplete octet.
2. HBr: H–Br with 3 lone pairs on Br.
3. C₂H₅OH: H₃C–CH₂–OH with 2 lone pairs on O.
4. N₂F₄: F₂N–NF₂ with 1 lone pair on each N, 3 lone pairs on each F.
5. SF₆: Octahedral S with 6 F atoms, each F has 3 lone pairs.

You can now draw these using dots for lone pairs and lines for bonds. Remember to count total valence electrons and ensure octets (except for H and sometimes B).

Let me know if you want the formal charges calculated or 3D shapes explained!