White crystalline solid (A) reacts with `H_2` to form a highly associated liquid (B) and a monoatomic , colourless gas (C ). The liquid (B) is used for etching glass. Compound (A) undergoes hydrolysis slowly to form (C ), (B) and a diatomic gas (D ) whose IE is almost similar to that of (C ). (B) forms an addition compound with KF to form (E) which is electrolysed in the molten state to form a most reactive gas (F) which combines with (C ) in 2:1 ratio of produce (A).
The compound 'A' reacts with sulphur to form a compound in which hybridisation state of sulphur atom is :

Options

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

Answer

To solve the question step by step, let's break down the information provided and derive the necessary compounds and their characteristics.

### Step 1: Identify Compound A
The question states that a white crystalline solid (A) reacts with hydrogen (H₂) to form a highly associated liquid (B) and a monoatomic, colorless gas (C). Given that liquid (B) is used for etching glass, we can deduce that:

- **Liquid B** is hydrogen fluoride (HF), which is known for its ability to etch glass.

### Step 2: Determine the Reaction
The reaction of compound A with H₂ can be represented as:
\[ \text{A} + \text{H}_2 \rightarrow \text{B} + \text{C} \]
Since we identified B as HF, we can write:
\[ \text{A} + \text{H}_2 \rightarrow \text{HF} + \text{C} \]

### Step 3: Identify Monoatomic Gas C
From the reaction, we can infer that the monoatomic gas (C) produced is xenon (Xe), which is colorless and monoatomic.

### Step 4: Identify Compound A
Given that A reacts with H₂ to form HF and Xe, we can conclude that:
- **Compound A** is xenon difluoride (XeF₂).

### Step 5: Hydrolysis of Compound A
The question mentions that compound A undergoes hydrolysis to form C, B, and a diatomic gas (D). The hydrolysis of XeF₂ can be represented as:
\[ \text{XeF}_2 + 2 \text{H}_2\text{O} \rightarrow \text{Xe} + 2 \text{HF} + \text{O}_2 \]
Here, the diatomic gas (D) is oxygen (O₂).

### Step 6: Identify Addition Compound E
The liquid B (HF) forms an addition compound with KF, which is:
\[ \text{HF} + \text{KF} \rightarrow \text{KHF}_2 \]
Thus, compound E is potassium bifluoride (KHF₂).

### Step 7: Electrolysis of Compound E
When KHF₂ is electrolyzed in the molten state, it produces:
\[ \text{KHF}_2 \rightarrow \text{F}_2 + \text{K} \]
The most reactive gas (F) produced is fluorine (F₂).

### Step 8: Reaction of F with C to produce A
The question states that F combines with C in a 2:1 ratio to produce A:
\[ \text{F}_2 + \text{Xe} \rightarrow \text{XeF}_2 \]
This confirms that A is indeed XeF₂.

### Step 9: Reaction of A with Sulfur
Finally, the question asks about the hybridization state of the sulfur atom when compound A reacts with sulfur (S₈):
\[ \text{XeF}_2 + \text{S}_8 \rightarrow \text{SF}_6 + \text{Xe} \]
The product is sulfur hexafluoride (SF₆).

### Step 10: Determine Hybridization of Sulfur in SF₆
The hybridization of sulfur in SF₆, which has an octahedral geometry, is:
- **Hybridization**: sp³d²

### Final Answer
The hybridization state of the sulfur atom in the compound formed when A reacts with sulfur is **sp³d²**.

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