Draw the structure of `B_(2)H_(6)` and give its reaction with `HCl`.

Answer

To solve the question regarding the structure of \( B_2H_6 \) (Diborane) and its reaction with HCl, we will follow these steps:

### Step 1: Draw the Structure of \( B_2H_6 \)

1. **Identify the Atoms**: Diborane consists of 2 boron (B) atoms and 6 hydrogen (H) atoms.
2. **Determine the Bonding**: In \( B_2H_6 \), each boron atom forms 4 bonds. However, due to the electron deficiency of boron, \( B_2H_6 \) has a unique structure.
3. **Construct the Structure**: The structure of \( B_2H_6 \) can be visualized as follows:
- Two boron atoms are connected by two bridging hydrogen atoms.
- Each boron atom is also bonded to two terminal hydrogen atoms.

The structure can be represented as:
```
H H
\ /
B - B
/ \
H H
\
H
```
Here, the two central hydrogen atoms are bridging hydrogen atoms, while the outer hydrogen atoms are terminal.

### Step 2: Write the Reaction of \( B_2H_6 \) with HCl

1. **Identify the Reactants**: The reactants are \( B_2H_6 \) and HCl.
2. **Write the Reaction**: When diborane reacts with hydrochloric acid (HCl), it forms \( B_2H_5Cl \) and hydrogen gas (H2).

The balanced reaction can be written as:
\[ B_2H_6 + HCl \rightarrow B_2H_5Cl + H_2 \]

### Final Answer

1. The structure of \( B_2H_6 \) is represented as:
```
H H
\ /
B - B
/ \
H H
\
H
```

2. The reaction with HCl is:
\[ B_2H_6 + HCl \rightarrow B_2H_5Cl + H_2 \]