The product obtained in the reaction of diborane with excess of ammonia at low temperature is:

Options

A. `B_(2)H_(6).NH_(3)`
B. `B_(2)H_(6).2NH_(3)`
C. `(BN)_(x)`
D. Borazine

Explanation

To solve the question regarding the product obtained from the reaction of diborane with excess ammonia at low temperature, we can follow these steps:

### Step 1: Write the chemical formula for diborane
Diborane is represented by the chemical formula \( B_2H_6 \).

### Step 2: Identify the reactants
The reactants in this reaction are diborane (\( B_2H_6 \)) and ammonia (\( NH_3 \)). Since ammonia is in excess, we will consider that in our reaction.

### Step 3: Write the reaction equation
When diborane reacts with ammonia, the reaction can be represented as follows:
\[ B_2H_6 + NH_3 \rightarrow ? \]

### Step 4: Determine the products
At low temperatures, diborane reacts with ammonia to form a complex. The expected product is:
\[ B_2H_6 + 2 NH_3 \rightarrow B_2H_6(NH_3)_2 \]

### Step 5: Balance the reaction
To balance the equation, we note that there are 2 nitrogen atoms from 2 ammonia molecules and 2 boron atoms from diborane. The balanced equation is:
\[ B_2H_6 + 2 NH_3 \rightarrow B_2H_6(NH_3)_2 \]

### Step 6: Identify the final product
The product formed is \( B_2H_6(NH_3)_2 \), which indicates that diborane forms a complex with two ammonia molecules.

### Conclusion
The product obtained in the reaction of diborane with excess ammonia at low temperature is \( B_2H_6(NH_3)_2 \).

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