What are X and Y in the reaction?
`3B_(2)H_(6) + 6X to 3[BH_(2) (X)_(2)]^(+) [BH_(4)]^(-) overset("heat")(to) Y + 12 H_(2)`

Options

A. `X = NH_(3), Y = B_(3)N_(3)H_(6)`
B. `X = CO, Y = BH_(3)CO`
C. `X = NaH, Y = NaF`
D. `X = NF_(3), Y = B_(3)N_(3)`

Explanation

To determine the values of X and Y in the given reaction, we can analyze the reaction step by step.

### Step 1: Identify the reactants and products
The reaction provided is:
\[ 3B_2H_6 + 6X \rightarrow 3[BH_2(X)_2]^+ + [BH_4]^− \]
This reaction involves diborane (\(B_2H_6\)) reacting with a substance \(X\) to form a complex and a borohydride ion.

### Step 2: Determine the nature of \(X\)
In the reaction, \(X\) is acting as a Lewis base that coordinates with boron. The common Lewis base that reacts with diborane is ammonia (\(NH_3\)). Therefore, we can propose:
\[ X = NH_3 \]

### Step 3: Write the complex formation
When diborane reacts with ammonia, it forms a complex:
\[ 3B_2H_6 + 6NH_3 \rightarrow 3[BH_2(NH_3)_2]^+ + [BH_4]^− \]

### Step 4: Analyze the further heating of the products
Upon further heating, the products decompose to form a new compound \(Y\) and hydrogen gas (\(H_2\)). The product formed from the reaction of boron and nitrogen under heat is borazine, which has the formula:
\[ Y = B_3N_3H_6 \]

### Conclusion
Thus, we can summarize:
- \(X = NH_3\)
- \(Y = B_3N_3H_6\)

### Final Answer
- \(X = NH_3\)
- \(Y = B_3N_3H_6\)

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