Practice worksheet for drawing Lewis dot structures of various covalent compounds and ions.

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 from charge → 22 e⁻
- Central I bonded to two terminal I atoms
- Terminal I: 3 lone pairs each, central I: 2 lone pairs
- Central I has expanded octet

```
[I]⁻--I--[I]⁻
```
Central I has 2 lone pairs, terminal I have 3 each.

---

HCN (Hydrogen cyanide)

- H: 1, C: 4, N: 5 → Total: 10 e⁻
- H-C≡N, triple bond between C and N
- C has no lone pairs, N has 1 lone pair

```
H-C≡N:
```

---

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
- Both have 1 lone pair each
- Formal charges: C(-1), O(+1)

```
C≡O:
```

---

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
- T-shaped geometry

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

---

CO₂⁻ (Carbon dioxide ion)

- C: 4, O: 6×2 = 12, plus 1 → Total: 17 e⁻
- But CO₂⁻ is actually carbonate ion? Wait - likely meant CO₃²⁻?
- Assuming it's CO₂⁻: C double bonded to both O atoms
- One O has negative charge, other has formal charge

Wait - better check: CO₂⁻ isn't standard. Likely a typo.
Assuming it's CO₃²⁻ (carbonate):
- C: 4, O: 6×3 = 18, plus 2 → 20 e⁻
- Resonance structures with double bond and two single bonds

But since it says CO₂⁻, let's do that:
- C: 4, O: 6×2 = 12, plus 1 → 17 e⁻
- C double bonded to one O, single bonded to other O⁻
- O⁻ has 3 lone pairs, double-bonded O has 2 lone pairs

```
O=C-O⁻
```

---

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 incomplete octet (6 electrons)
- No lone pairs on H

```
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 incomplete octet (6 electrons)
- Each F has 3 lone pairs

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

---

C₂H₄ (Ethylene)

- C: 4×2 = 8, H: 1×4 = 4 → Total: 12 e⁻
- Double bond between two C atoms
- Each C bonded to 2 H atoms

```
H H
\ /
C=C
/ \
H H
```

---

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
```

---

C₂H₆ (Ethane)

- C: 4×2 = 8, H: 1×6 = 6 → Total: 14 e⁻
- Single bond between two C atoms
- Each C bonded to 3 H atoms

```
H H
| |
H-C-C-H
| |
H H
```

---

NH₄⁺ (Ammonium ion)

- N: 5, H: 1×4 = 4, minus 1 for charge → 8 e⁻
- N central, bonded to 4 H atoms
- No lone pairs on N (positive charge)

```
H
|
H--N⁺--H
|
H
```

---

These are the Lewis structures for all the compounds listed. The key principles applied include:
- Octet rule (with exceptions for expanded octets and incomplete octets)
- Formal charge minimization
- Correct placement of lone pairs
- Handling of ions and radicals

Would you like me to provide any specific structure in more detail or clarify any particular compound?