`B_2 H_6 + NH_3 overset(120^@ C)to X` . Where 'X' is

Options

A. `[BH_2 (NH_3)_2]^+ [BH_4]^-`
B. `[BH_2(NH_3)_2]^+ [BH_3]^-`
C. `(BH_4)^+ [BH_2 (NH_3)_2]^-`
D. `(BH_3)^+ [BH_3(NH_3)_2 ]^-`

Answer

To solve the question regarding the reaction of \( B_2H_6 \) (Diborane) with ammonia (\( NH_3 \)) at 120°C, we can follow these steps:

### Step 1: Identify the Reactants
The reactants in this reaction are \( B_2H_6 \) (Diborane) and \( NH_3 \) (Ammonia).

### Step 2: Understand the Reaction Conditions
The reaction occurs at a temperature of 120°C, which is significant as it indicates that the reaction will proceed to form more complex products.

### Step 3: Determine the Initial Reaction Products
When \( B_2H_6 \) reacts with \( NH_3 \), it forms a cation and an anion:
- The cation formed is \( [BH_2NH_3]^+ \) (a complex with ammonia).
- The anion formed is \( [BH_4]^- \) (tetrahydroborate ion).

### Step 4: Write the Balanced Reaction
The balanced reaction can be represented as:
\[ 3 B_2H_6 + 6 NH_3 \rightarrow 3 [BH_2NH_3]^+ + 3 [BH_4]^- \]

### Step 5: Further Heating of the Products
Upon further heating, the products can react to form a more complex compound:
\[ 3 [BH_2NH_3]^+ + 3 [BH_4]^- \rightarrow B_3N_3H_6 + 3 H_2 \]
This compound \( B_3N_3H_6 \) is known as Borazole.

### Step 6: Conclusion
Thus, the final product \( X \) formed from the reaction of \( B_2H_6 \) with \( NH_3 \) at 120°C is \( B_3N_3H_6 \) (Borazole).

### Final Answer:
The product \( X \) is \( B_3N_3H_6 \).

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