Draw the structures of `BCl_(3).NH_(3) and AlCl_(3)` (dimer).

Answer

To draw the structures of \( BCl_3 \cdot NH_3 \) and the dimer of \( AlCl_3 \), we will follow these steps:

### Step 1: Draw the Structure of \( BCl_3 \)

1. **Identify the central atom**: The central atom in \( BCl_3 \) is boron (B).
2. **Count the valence electrons**: Boron has 3 valence electrons, and each chlorine (Cl) has 7 valence electrons. Since there are 3 chlorine atoms, the total number of valence electrons is:
\[ 3 (B) + 3 \times 7 (Cl) = 3 + 21 = 24 \text{ valence electrons} \]
3. **Form bonds**: Boron forms three single covalent bonds with the three chlorine atoms. Each bond uses 2 electrons, thus using 6 electrons in total.
4. **Distribute remaining electrons**: After forming the bonds, there are \( 24 - 6 = 18 \) electrons left. These electrons are used to complete the octets of the chlorine atoms.
5. **Final structure**: The structure of \( BCl_3 \) is trigonal planar with boron at the center and the three chlorine atoms at the corners.

### Step 2: Draw the Structure of \( NH_3 \)

1. **Identify the central atom**: The central atom in \( NH_3 \) is nitrogen (N).
2. **Count the valence electrons**: Nitrogen has 5 valence electrons, and each hydrogen (H) has 1 valence electron. Since there are 3 hydrogen atoms, the total number of valence electrons is:
\[ 5 (N) + 3 \times 1 (H) = 5 + 3 = 8 \text{ valence electrons} \]
3. **Form bonds**: Nitrogen forms three single covalent bonds with the three hydrogen atoms. Each bond uses 2 electrons, thus using 6 electrons in total.
4. **Distribute remaining electrons**: After forming the bonds, there are \( 8 - 6 = 2 \) electrons left, which are placed as a lone pair on nitrogen.
5. **Final structure**: The structure of \( NH_3 \) is pyramidal due to the lone pair on nitrogen.

### Step 3: Combine \( BCl_3 \) and \( NH_3 \)

1. **Dative bond formation**: The lone pair on nitrogen in \( NH_3 \) can be donated to the electron-deficient boron in \( BCl_3 \) to form a dative bond.
2. **Final structure**: The combined structure \( BCl_3 \cdot NH_3 \) shows boron at the center with three chlorine atoms and one nitrogen atom bonded through a dative bond.

### Step 4: Draw the Dimer of \( AlCl_3 \)

1. **Identify the dimerization**: Two \( AlCl_3 \) molecules combine to form \( Al_2Cl_6 \).
2. **Dative bond formation**: Each aluminum atom is electron-deficient and can accept a lone pair from the chlorine atoms.
3. **Structure**: The two aluminum atoms are connected by two dative bonds from two chlorine atoms, resulting in a structure where each aluminum is surrounded by three chlorine atoms.
4. **Final structure**: The structure of \( Al_2Cl_6 \) shows two aluminum atoms at the center, with four chlorine atoms forming dative bonds and two chlorine atoms completing the octets.

### Final Structures

- **For \( BCl_3 \cdot NH_3 \)**:
```
Cl
|
Cl-B-NH3
|
Cl
```

- **For \( AlCl_3 \) dimer (\( Al_2Cl_6 \))**:
```
Cl
|
Cl-Al-Cl
|
Al-Cl
|
Cl
```