Worksheet 11: Electron Dot Structures (Lewis Structures) - Students draw correct Lewis structures for various chemical formulas, following the octet rule.

To solve the problem of drawing Lewis structures for the given molecules and ions, we need to follow these steps:

1. Determine the total number of valence electrons:
- Count the valence electrons for each atom in the molecule or ion.
- For ions, adjust the total by adding (for negative charges) or subtracting (for positive charges) the appropriate number of electrons.

2. Choose a central atom:
- Typically, the least electronegative atom is chosen as the central atom.
- Hydrogen (H) is never the central atom.

3. Draw a skeleton structure:
- Arrange the atoms around the central atom.
- Connect the atoms with single bonds.

4. Distribute the remaining electrons:
- Place lone pairs on the terminal atoms first to satisfy the octet rule.
- If the central atom does not have an octet, use multiple bonds (double or triple bonds) to complete its octet.

5. Check the formal charges:
- Calculate the formal charge for each atom to ensure the structure is reasonable.
- The structure with the lowest formal charges (ideally zero) is preferred.

Let's go through a few examples from the list:

Example 1: \( \text{SiH}_4 \)

- Total valence electrons: Si (4) + 4 × H (1) = 8
- Structure: Si is the central atom, bonded to 4 H atoms.
- Lewis Structure:
```
H
|
H---Si---H
|
H
```

Example 2: \( \text{C}_2\text{H}_4 \)

- Total valence electrons: 2 × C (4) + 4 × H (1) = 12
- Structure: Two C atoms are double-bonded, with each C bonded to 2 H atoms.
- Lewis Structure:
```
H
|
H---C==C---H
|
H
```

Example 3: \( \text{N}_2 \)

- Total valence electrons: 2 × N (5) = 10
- Structure: Two N atoms are triple-bonded.
- Lewis Structure:
```
:N≡N:
```

Example 4: \( \text{NH}_3 \)

- Total valence electrons: N (5) + 3 × H (1) = 8
- Structure: N is the central atom, bonded to 3 H atoms, with one lone pair on N.
- Lewis Structure:
```
H
/
H---N
\
H
```

Example 5: \( \text{HNO}_2 \)

- Total valence electrons: H (1) + N (5) + 2 × O (6) = 18
- Structure: N is the central atom, bonded to one H and two O atoms. One O has a double bond with N, and the other O has a single bond with N and a lone pair.
- Lewis Structure:
```
O
||
H---N---O
:
```

Example 6: \( \text{CH}_3\text{O} \)

- Total valence electrons: C (4) + 3 × H (1) + O (6) = 14
- Structure: C is the central atom, bonded to 3 H atoms and one O atom. The O atom has a lone pair.
- Lewis Structure:
```
O
|
H---C---H
|
H
```

Example 7: \( \text{NO}^- \)

- Total valence electrons: N (5) + O (6) + 1 (from the negative charge) = 12
- Structure: N is the central atom, double-bonded to O, with one lone pair on N and two lone pairs on O.
- Lewis Structure:
```
:N=O:
```

Example 8: \( \text{F}_2 \)

- Total valence electrons: 2 × F (7) = 14
- Structure: Two F atoms are single-bonded, with three lone pairs on each F.
- Lewis Structure:
```
:F---F:
```

Example 9: \( \text{C}_2\text{H}_2\text{NO} \)

- Total valence electrons: 2 × C (4) + 2 × H (1) + N (5) + O (6) = 18
- Structure: This is acrylonitrile (CH₂=CH-CN). The structure involves a carbon-carbon double bond and a cyano group (CN).
- Lewis Structure:
```
H
|
H---C==C---N
|
H
:
```

Final Answer:

The Lewis structures for all the given molecules and ions can be drawn following the steps above. Here is a summary of the final structures:

1. \( \text{SiH}_4 \):
```
H
|
H---Si---H
|
H
```

2. \( \text{C}_2\text{H}_4 \):
```
H
|
H---C==C---H
|
H
```

3. \( \text{H}_2\text{S} \):
```
H
/
S---H
\
H
```

4. \( \text{N}_2 \):
```
:N≡N:
```

5. \( \text{NH}_3 \):
```
H
/
H---N
\
H
```

6. \( \text{SiH}_3\text{P} \):
```
H
|
H---Si---P
|
H
```

7. \( \text{CS}_2 \):
```
S
||
C---S
||
S
```

8. \( \text{CH}_3\text{O} \):
```
O
|
H---C---H
|
H
```

9. \( \text{CH}_4\text{O} \):
```
O
|
H---C---H
/ \
H H
```

10. \( \text{NH}_3\text{S} \):
```
H
/
H---N---S
\
H
```

11. \( \text{BrCN} \):
```
Br
|
C≡N
```

12. \( \text{C}_2\text{F}_2\text{H} \):
```
F
|
H---C==C---F
|
H
```

13. \( \text{NO}^- \):
```
:N=O:
```

14. \( \text{CH}_2\text{O}_2 \):
```
O
|
H---C---O
|
H
```

15. \( \text{F}_2 \):
```
:F---F:
```

16. \( \text{C}_2\text{H}_2\text{NO} \):
```
H
|
H---C==C---N
|
H
:
```

Thus, the final answer is:

\boxed{\text{See detailed structures above}}