`B_(2)H_(6)` reacts with `NH_(3)` to form

Options

A. `B_(3)N_(3)H_(6)`
B. `BN`
C. `[B_(2)H_(5)]^(Theta)NH_(4)^(o+)`
D. `All of these

Answer

To solve the question regarding the reaction of \( B_2H_6 \) with \( NH_3 \), we will follow these steps:

### Step 1: Write the Reactants
Identify the reactants involved in the reaction. Here, we have:
- Diborane: \( B_2H_6 \)
- Ammonia: \( NH_3 \)

### Step 2: Identify the Products
When \( B_2H_6 \) reacts with \( NH_3 \), it forms borazine, which is \( B_3N_3H_6 \), and hydrogen gas \( H_2 \).

### Step 3: Write the Chemical Equation
The unbalanced reaction can be written as:
\[ B_2H_6 + NH_3 \rightarrow B_3N_3H_6 + H_2 \]

### Step 4: Balance the Chemical Equation
To balance the equation, we need to ensure that the number of atoms of each element is the same on both sides.

1. Start with boron (B):
- Left side: 2 (from \( B_2H_6 \))
- Right side: 3 (from \( B_3N_3H_6 \))
- We need 3 boron atoms, so we can put a coefficient of 3 in front of \( B_2H_6 \):
\[ 3B_2H_6 + NH_3 \rightarrow B_3N_3H_6 + H_2 \]

2. Now balance nitrogen (N):
- Left side: 1 (from \( NH_3 \))
- Right side: 3 (from \( B_3N_3H_6 \))
- We need 3 nitrogen atoms, so we can put a coefficient of 3 in front of \( NH_3 \):
\[ 3B_2H_6 + 3NH_3 \rightarrow B_3N_3H_6 + H_2 \]

3. Finally, balance hydrogen (H):
- Left side: \( 3 \times 6 + 3 \times 1 = 18 + 3 = 21 \)
- Right side: \( 6 (from B_3N_3H_6) + 2 (from H_2) = 6 + 2 = 8 \)
- To balance hydrogen, we need to adjust the coefficient in front of \( H_2 \) to ensure both sides have the same number of hydrogen atoms.

After balancing, the final balanced equation is:
\[ 2B_2H_6 + 6NH_3 \rightarrow 2B_3N_3H_6 + 6H_2 \]

### Step 5: Identify the Correct Product
From the balanced equation, we can see that the primary product formed is \( B_3N_3H_6 \) (borazine), along with hydrogen gas \( H_2 \).

### Conclusion
Thus, the correct answer to the question is that \( B_2H_6 \) reacts with \( NH_3 \) to form \( B_3N_3H_6 \).

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