The intermediate product (X) formed in the following reaction is
`B_(2)H_(6)+6NH_(3)rarr 3K overset("heat")rarr 2B_(3)N_(3)H_(6)+12H_(2)`

Options

A. `[BH(NH_(3))_(3)]^(+)[BH_(4)]^(-)`
B. `[BH_(2)(NH_(3))_(4)]^(+)[BH_(4)]^(-)`
C. `[BH(NH_(3))_(4)]^(+)[BH_(4)]^(-)`
D. `[BH_(2)(NH_(3))_(2)]^(+)[BH_(4)]^(-)`

Answer

To determine the intermediate product (X) formed in the reaction sequence given, we need to analyze the reaction step by step.

### Step 1: Identify the Reactants
The reactants in the reaction are:
- **B2H6** (Diborane)
- **NH3** (Ammonia)

### Step 2: Understand the Reaction
The reaction can be summarized as:
\[ B_2H_6 + 6NH_3 \rightarrow 3K \overset{\text{heat}}{\rightarrow} 2B_3N_3H_6 + 12H_2 \]

Here, we are looking for the intermediate product (X) formed before the final products (B3N3H6 and H2) are produced.

### Step 3: Formulate the Intermediate Product
When diborane (B2H6) reacts with ammonia (NH3), it forms a complex. The expected intermediate product can be represented as:
\[ B_2H_6 \cdot 2NH_3 \]

This can also be expressed in a more detailed way:
- The diborane can be viewed as consisting of two boron atoms and six hydrogen atoms.
- The ammonia molecules coordinate with the boron atoms, forming a complex.

### Step 4: Write the Structure of the Intermediate
The structure of the intermediate product (X) can be represented as:
\[ B_2H_6 \cdot 2NH_3 \]
or in a more ionic form:
\[ BH_2 \cdot NH_3 \] (where the boron atoms are bonded to the ammonia)

### Step 5: Conclusion
Thus, the intermediate product (X) formed in the reaction is:
\[ X = B_2H_6 \cdot 2NH_3 \]

### Final Answer
The intermediate product (X) is \( B_2H_6 \cdot 2NH_3 \).

---