Compound `(X)` on reduction with `LiAlH_(4)` gives a hydride `(Y)` containing `21.72 %` hydrogen along with other products. The compound `(Y)` reacts with air explosively resulting in formation of boron trioxide. Identify `(X)` and `(Y)`.
Give balanced reactions involved in the formation of `(Y)` and its reaction with air. Give the structure of `(Y)`.

Options

A. `BCI_(3), B_(2)H_(6)`
B. `B_(2)H_(6), BCI_(3)`
C. `BF_(3),AI_(2)O_(6)`
D. `B_(2) H_(6), BF_(3)`

Explanation

To solve the problem step by step, we need to identify the compounds \(X\) and \(Y\), as well as write the balanced reactions involved in their formation and reaction with air.

### Step 1: Identify the Compound \(Y\)
The problem states that the hydride \(Y\) contains \(21.72\%\) hydrogen. To find the molecular formula of \(Y\), we can use the percentage composition to determine the molar ratio of boron to hydrogen.

1. **Calculate the molar mass of \(Y\)**:
- Let the molar mass of \(Y\) be \(M\).
- The percentage of hydrogen in \(Y\) is given as \(21.72\%\).
- Therefore, the mass of hydrogen in \(Y\) is \(0.2172 \times M\).
- The mass of boron in \(Y\) is \(M - 0.2172M = 0.7828M\).

2. **Molar mass of hydrogen and boron**:
- Molar mass of hydrogen (H) = \(1 \, \text{g/mol}\)
- Molar mass of boron (B) = \(10.81 \, \text{g/mol}\)

3. **Set up the ratio**:
\[ \frac{0.7828M}{0.2172M} = \frac{B}{H} = \frac{10.81}{1} \]
Simplifying gives:
\[ \frac{0.7828}{0.2172} \approx 3.6 \]
This suggests a ratio of approximately \(2:6\) for boron to hydrogen, leading us to conclude that \(Y\) is likely \(B_2H_6\) (diborane).

### Step 2: Identify the Compound \(X\)
The problem states that \(X\) is reduced to form \(Y\) using \(LiAlH_4\). A common precursor to diborane is boron trichloride (\(BCl_3\)).

1. **Identify \(X\)**:
- Therefore, \(X\) is \(BCl_3\).

### Step 3: Write the Balanced Reaction for the Formation of \(Y\)
The balanced reaction for the reduction of \(BCl_3\) with \(LiAlH_4\) to form \(B_2H_6\) is:

\[ 2BCl_3 + 6LiAlH_4 \rightarrow B_2H_6 + 6LiCl + 3AlCl_3 \]

### Step 4: Write the Reaction of \(Y\) with Air
When diborane (\(B_2H_6\)) reacts with oxygen in air, it produces boron trioxide (\(B_2O_3\)) and hydrogen gas.

1. **Balanced reaction**:
\[ 2B_2H_6 + 7O_2 \rightarrow 4B_2O_3 + 6H_2 \]

### Step 5: Structure of \(Y\)
The structure of diborane (\(B_2H_6\)) features unique three-center two-electron bonds, often referred to as "banana bonds." The structure can be represented as follows:

```
H H
\ /
B---B
/ \
H H
```

### Summary of Identifications and Reactions
- **Compound \(X\)**: \(BCl_3\)
- **Compound \(Y\)**: \(B_2H_6\) (diborane)
- **Formation Reaction**:
\[ 2BCl_3 + 6LiAlH_4 \rightarrow B_2H_6 + 6LiCl + 3AlCl_3 \]
- **Reaction with Air**:
\[ 2B_2H_6 + 7O_2 \rightarrow 4B_2O_3 + 6H_2 \]
- **Structure of \(Y\)**: Diborane (\(B_2H_6\)) has a unique structure with three-center two-electron bonds.