Lewis Dot Structure worksheet for chemistry students to practice drawing covalent compound structures.

I'll solve this step by step, explaining how to draw the Lewis dot structures for each of the given covalent compounds.

## General Steps for Drawing Lewis Structures:
1. Count total valence electrons
2. Identify central atom (usually least electronegative)
3. Connect atoms with single bonds
4. Distribute remaining electrons as lone pairs
5. Check octet rule (except for exceptions)
6. Adjust if needed (double/triple bonds)

---

PF₅ (Phosphorus pentafluoride)

- P: 5 valence e⁻, F: 7 each × 5 = 35
- Total: 5 + 35 = 40 e⁻
- P is central, bonded to 5 F atoms
- Each F has 3 lone pairs, P has no lone pairs
- P expands octet (10 electrons)

```
F
|
F--P--F
|
F
|
F
```

Each F has 3 lone pairs.

---

NO (Nitric oxide)

- N: 5, O: 6 → Total: 11 e⁻ (odd number → radical)
- N and O double bonded, N has 1 unpaired electron
- O has 2 lone pairs, N has 1 lone pair

```
N=O
```
With one unpaired electron on N.

---

I₃⁻ (Triiodide ion)

- I: 7 each × 3 = 21, plus 1 for negative charge = 22 e⁻
- Central I bonded to two terminal I atoms
- Terminal I have 3 lone pairs each, central I has 2 lone pairs
- Central I has expanded octet (10 electrons)

```
I
|
I-I
|
I
```
Central I bonded to both, each terminal I has 3 lone pairs.

---

HCN (Hydrogen cyanide)

- H: 1, C: 4, N: 5 → Total: 10 e⁻
- H bonded to C, C triple bonded to N
- H: no lone pairs, C: no lone pairs, N: 1 lone pair

```
H-C≡N
```
N has one lone pair.

---

BrF₅ (Bromine pentafluoride)

- Br: 7, F: 7×5 = 35 → Total: 42 e⁻
- Br central, bonded to 5 F atoms
- Br has 1 lone pair, each F has 3 lone pairs
- Br expands octet (12 electrons)

```
F
|
F-Br-F
|
F
|
F
```

---

CO (Carbon monoxide)

- C: 4, O: 6 → Total: 10 e⁻
- Triple bond between C and O
- C has 1 lone pair, O has 1 lone pair
- Formal charges: C⁺, O⁻

```
C≡O
```
Each has one lone pair.

---

BrF₃ (Bromine trifluoride)

- Br: 7, F: 7×3 = 21 → Total: 28 e⁻
- Br central, bonded to 3 F atoms
- Br has 2 lone pairs, each F has 3 lone pairs
- Br has expanded octet (10 electrons)

```
F
|
F-Br-F
|
(lone pairs on Br)
```

---

CO₂⁻ (Carbon dioxide ion)

- C: 4, O: 6×2 = 12, plus 1 for charge = 17 e⁻
- But CO₂⁻ isn't stable; likely meant to be CO₂ or CO₃²⁻
- Assuming CO₂: C=O double bonds, each O has 2 lone pairs
- For CO₂⁻, add extra electron → one O has 3 lone pairs

```
O=C=O
```
One oxygen has three lone pairs.

---

SCl₄ (Sulfur tetrachloride)

- S: 6, Cl: 7×4 = 28 → Total: 34 e⁻
- S central, bonded to 4 Cl atoms
- S has 1 lone pair, each Cl has 3 lone pairs

```
Cl
|
Cl-S-Cl
|
Cl
```

---

AlH₃ (Aluminum hydride)

- Al: 3, H: 1×3 = 3 → Total: 6 e⁻
- Al central, bonded to 3 H atoms
- Al has only 6 electrons (electron deficient)
- Each H has 0 lone pairs

```
H
|
H-Al-H
```

---

BF₃ (Boron trifluoride)

- B: 3, F: 7×3 = 21 → Total: 24 e⁻
- B central, bonded to 3 F atoms
- B has only 6 electrons (electron deficient)
- Each F has 3 lone pairs

```
F
|
F-B-F
```

---

C₂H₄ (Ethene)

- C: 4×2 = 8, H: 1×4 = 4 → Total: 12 e⁻
- Two C atoms double bonded, each C bonded to 2 H
- No lone pairs

```
H₂C=CH₂
```

---

PH₃ (Phosphine)

- P: 5, H: 1×3 = 3 → Total: 8 e⁻
- P central, bonded to 3 H atoms
- P has 1 lone pair

```
H
|
H-P-H
|
(lone pair on P)
```

---

C₂H₆ (Ethane)

- C: 4×2 = 8, H: 1×6 = 6 → Total: 14 e⁻
- Two C atoms single bonded, each C bonded to 3 H
- No lone pairs

```
H₃C-CH₃
```

---

NH₄⁺ (Ammonium ion)

- N: 5, H: 1×4 = 4, minus 1 for positive charge = 8 e⁻
- N central, bonded to 4 H atoms
- N has no lone pairs (only 8 electrons)

```
H
|
H-N-H
|
H
```
With positive charge on N.

---

Key Concepts Used:
- Octet rule (with exceptions)
- Expanded octets for elements in period 3+
- Formal charge minimization
- Lone pair placement
- Multiple bonding
- Electron-deficient molecules

Would you like me to provide visual representations or clarify any specific structure?