Compound (A) on reduction with `LiAIH_4` gives a hydribe (P) containing 21.72% hydrogen along with other products.The one mole of hydride (P) and 2 moles of ammonia at higher temperature gives a compound (Q) which is known as inorganic benzene. (A) hydrolyses incompletely and forms a compound (R) and `H_3BO_2`
The hybridisation of central atom of compound (R) is :

Options

A. `sp^2`
B. `sp^3`
C. `sp`
D. `sp^3d`

Explanation

To solve the question step by step, we will analyze the information provided and deduce the necessary compounds and their properties.

### Step 1: Identify Compound (A)
We know that compound (A) on reduction with lithium aluminum hydride (LiAlH4) gives a hydride (P) containing 21.72% hydrogen.

To find the molecular formula of hydride (P), we can use the percentage of hydrogen:
- The molar mass of hydrogen (H) = 1 g/mol.
- Let the molar mass of hydride (P) be M.

Using the formula:
\[ \text{Percentage of H} = \left( \frac{\text{mass of H}}{\text{molar mass of compound}} \right) \times 100 \]
We have:
\[ 21.72 = \left( \frac{n \times 1}{M} \right) \times 100 \]
Where \( n \) is the number of hydrogen atoms in the hydride.

Assuming \( n = 6 \) (as B2H6 is a common hydride formed from boron compounds), we can calculate:
\[ 21.72 = \left( \frac{6 \times 1}{M} \right) \times 100 \implies M = \frac{600}{21.72} \approx 27.6 \text{ g/mol} \]

This suggests that compound (A) could be B2H6 (Diborane), which is formed from boron trifluoride (BF3) upon reduction.

### Step 2: Identify Compound (R)
The question states that compound (A) hydrolyzes incompletely to form compound (R) and H3BO3.

From the hydrolysis of BF3, we know that it can form boric acid (H3BO3) and other products. The incomplete hydrolysis of BF3 leads to the formation of HBF4 (fluoroboric acid) as compound (R).

### Step 3: Determine the Hybridization of the Central Atom in Compound (R)
Now we need to determine the hybridization of the central atom in compound (R), which is HBF4.

1. **Identify the central atom**: The central atom in HBF4 is Boron (B).
2. **Count the number of bonds**: Boron forms four bonds with four fluorine atoms and one hydrogen atom.
3. **Determine the hybridization**: Since boron is forming four sigma bonds, it is sp³ hybridized.

### Conclusion
The hybridization of the central atom of compound (R) (HBF4) is **sp³**.

### Final Answer
The hybridization of the central atom of compound (R) is **sp³**.

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